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185 | Let $f(x) = 2x^{2}-x-6$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $2x^{2}-x-6 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 2, b = -1, c = -6$ $ x = \dfrac{+ 1 \pm \sqrt{(-1)^{2} - 4 \cdot 2 \cdot -6}}{2 \cdot 2}$ $ x = \dfrac{1 \pm \sqrt{49}}{4}$ $ x = \dfrac{1 \pm 7}{4}$ $x =2,-\frac{3}{2}$ |
185 | Let $f(x) = -4x^{2}-5x+2$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-4x^{2}-5x+2 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -4, b = -5, c = 2$ $ x = \dfrac{+ 5 \pm \sqrt{(-5)^{2} - 4 \cdot -4 \cdot 2}}{2 \cdot -4}$ $ x = \dfrac{5 \pm \sqrt{57}}{-8}$ $ x = \dfrac{5 \pm \sqrt{57}}{-8}$ $x =\dfrac{5 \pm \sqrt{57}}{-8}$ |
185 | Let $f(x) = x^{2}-7x-10$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $x^{2}-7x-10 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 1, b = -7, c = -10$ $ x = \dfrac{+ 7 \pm \sqrt{(-7)^{2} - 4 \cdot 1 \cdot -10}}{2 \cdot 1}$ $ x = \dfrac{7 \pm \sqrt{89}}{2}$ $ x = \dfrac{7 \pm \sqrt{89}}{2}$ $x =\dfrac{7 \pm \sqrt{89}}{2}$ |
185 | Let $f(x) = 5x^{2}+7x+2$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $5x^{2}+7x+2 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 5, b = 7, c = 2$ $ x = \dfrac{-7 \pm \sqrt{7^{2} - 4 \cdot 5 \cdot 2}}{2 \cdot 5}$ $ x = \dfrac{-7 \pm \sqrt{9}}{10}$ $ x = \dfrac{-7 \pm 3}{10}$ $x =-\frac{2}{5},-1$ |
185 | Let $f(x) = 9x^{2}-9x-3$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $9x^{2}-9x-3 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 9, b = -9, c = -3$ $ x = \dfrac{+ 9 \pm \sqrt{(-9)^{2} - 4 \cdot 9 \cdot -3}}{2 \cdot 9}$ $ x = \dfrac{9 \pm \sqrt{189}}{18}$ $ x = \dfrac{9 \pm 3\sqrt{21}}{18}$ $x =\dfrac{3 \pm \sqrt{21}}{6}$ |
185 | Let $f(x) = 6x^{2}+9x-9$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $6x^{2}+9x-9 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 6, b = 9, c = -9$ $ x = \dfrac{-9 \pm \sqrt{9^{2} - 4 \cdot 6 \cdot -9}}{2 \cdot 6}$ $ x = \dfrac{-9 \pm \sqrt{297}}{12}$ $ x = \dfrac{-9 \pm 3\sqrt{33}}{12}$ $x =\dfrac{-3 \pm \sqrt{33}}{4}$ |
185 | Let $f(x) = 6x^{2}+9x-6$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $6x^{2}+9x-6 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 6, b = 9, c = -6$ $ x = \dfrac{-9 \pm \sqrt{9^{2} - 4 \cdot 6 \cdot -6}}{2 \cdot 6}$ $ x = \dfrac{-9 \pm \sqrt{225}}{12}$ $ x = \dfrac{-9 \pm 15}{12}$ $x =\frac{1}{2},-2$ |
185 | Let $f(x) = 7x^{2}+x-4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $7x^{2}+x-4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 7, b = 1, c = -4$ $ x = \dfrac{-1 \pm \sqrt{1^{2} - 4 \cdot 7 \cdot -4}}{2 \cdot 7}$ $ x = \dfrac{-1 \pm \sqrt{113}}{14}$ $ x = \dfrac{-1 \pm \sqrt{113}}{14}$ $x =\dfrac{-1 \pm \sqrt{113}}{14}$ |
185 | Let $f(x) = 8x^{2}+3x-2$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $8x^{2}+3x-2 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 8, b = 3, c = -2$ $ x = \dfrac{-3 \pm \sqrt{3^{2} - 4 \cdot 8 \cdot -2}}{2 \cdot 8}$ $ x = \dfrac{-3 \pm \sqrt{73}}{16}$ $ x = \dfrac{-3 \pm \sqrt{73}}{16}$ $x =\dfrac{-3 \pm \sqrt{73}}{16}$ |
185 | Let $f(x) = 2x^{2}-8x-4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $2x^{2}-8x-4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 2, b = -8, c = -4$ $ x = \dfrac{+ 8 \pm \sqrt{(-8)^{2} - 4 \cdot 2 \cdot -4}}{2 \cdot 2}$ $ x = \dfrac{8 \pm \sqrt{96}}{4}$ $ x = \dfrac{8 \pm 4\sqrt{6}}{4}$ $x =2 \pm \sqrt{6}$ |
185 | Let $f(x) = -5x^{2}-3x+4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-5x^{2}-3x+4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -5, b = -3, c = 4$ $ x = \dfrac{+ 3 \pm \sqrt{(-3)^{2} - 4 \cdot -5 \cdot 4}}{2 \cdot -5}$ $ x = \dfrac{3 \pm \sqrt{89}}{-10}$ $ x = \dfrac{3 \pm \sqrt{89}}{-10}$ $x =\dfrac{3 \pm \sqrt{89}}{-10}$ |
185 | Let $f(x) = 10x^{2}+4x-10$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $10x^{2}+4x-10 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 10, b = 4, c = -10$ $ x = \dfrac{-4 \pm \sqrt{4^{2} - 4 \cdot 10 \cdot -10}}{2 \cdot 10}$ $ x = \dfrac{-4 \pm \sqrt{416}}{20}$ $ x = \dfrac{-4 \pm 4\sqrt{26}}{20}$ $x =\dfrac{-1 \pm \sqrt{26}}{5}$ |
185 | Let $f(x) = -7x^{2}-3x+1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-7x^{2}-3x+1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -7, b = -3, c = 1$ $ x = \dfrac{+ 3 \pm \sqrt{(-3)^{2} - 4 \cdot -7 \cdot 1}}{2 \cdot -7}$ $ x = \dfrac{3 \pm \sqrt{37}}{-14}$ $ x = \dfrac{3 \pm \sqrt{37}}{-14}$ $x =\dfrac{3 \pm \sqrt{37}}{-14}$ |
185 | Let $f(x) = 10x^{2}-8x-3$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $10x^{2}-8x-3 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 10, b = -8, c = -3$ $ x = \dfrac{+ 8 \pm \sqrt{(-8)^{2} - 4 \cdot 10 \cdot -3}}{2 \cdot 10}$ $ x = \dfrac{8 \pm \sqrt{184}}{20}$ $ x = \dfrac{8 \pm 2\sqrt{46}}{20}$ $x =\dfrac{4 \pm \sqrt{46}}{10}$ |
185 | Let $f(x) = 3x^{2}-4x-9$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $3x^{2}-4x-9 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 3, b = -4, c = -9$ $ x = \dfrac{+ 4 \pm \sqrt{(-4)^{2} - 4 \cdot 3 \cdot -9}}{2 \cdot 3}$ $ x = \dfrac{4 \pm \sqrt{124}}{6}$ $ x = \dfrac{4 \pm 2\sqrt{31}}{6}$ $x =\dfrac{2 \pm \sqrt{31}}{3}$ |
185 | Let $f(x) = x^{2}-8x-1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $x^{2}-8x-1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 1, b = -8, c = -1$ $ x = \dfrac{+ 8 \pm \sqrt{(-8)^{2} - 4 \cdot 1 \cdot -1}}{2 \cdot 1}$ $ x = \dfrac{8 \pm \sqrt{68}}{2}$ $ x = \dfrac{8 \pm 2\sqrt{17}}{2}$ $x =4 \pm \sqrt{17}$ |
185 | Let $f(x) = 4x^{2}+6x-4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $4x^{2}+6x-4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 4, b = 6, c = -4$ $ x = \dfrac{-6 \pm \sqrt{6^{2} - 4 \cdot 4 \cdot -4}}{2 \cdot 4}$ $ x = \dfrac{-6 \pm \sqrt{100}}{8}$ $ x = \dfrac{-6 \pm 10}{8}$ $x =\frac{1}{2},-2$ |
185 | Let $f(x) = -10x^{2}+9x+1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-10x^{2}+9x+1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -10, b = 9, c = 1$ $ x = \dfrac{-9 \pm \sqrt{9^{2} - 4 \cdot -10 \cdot 1}}{2 \cdot -10}$ $ x = \dfrac{-9 \pm \sqrt{121}}{-20}$ $ x = \dfrac{-9 \pm 11}{-20}$ $x =-\frac{1}{10},1$ |
185 | Let $f(x) = -9x^{2}+2x+1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-9x^{2}+2x+1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -9, b = 2, c = 1$ $ x = \dfrac{-2 \pm \sqrt{2^{2} - 4 \cdot -9 \cdot 1}}{2 \cdot -9}$ $ x = \dfrac{-2 \pm \sqrt{40}}{-18}$ $ x = \dfrac{-2 \pm 2\sqrt{10}}{-18}$ $x =\dfrac{-1 \pm \sqrt{10}}{-9}$ |
185 | Let $f(x) = -6x^{2}-3x+1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-6x^{2}-3x+1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -6, b = -3, c = 1$ $ x = \dfrac{+ 3 \pm \sqrt{(-3)^{2} - 4 \cdot -6 \cdot 1}}{2 \cdot -6}$ $ x = \dfrac{3 \pm \sqrt{33}}{-12}$ $ x = \dfrac{3 \pm \sqrt{33}}{-12}$ $x =\dfrac{3 \pm \sqrt{33}}{-12}$ |
185 | Let $f(x) = 3x^{2}+3x-4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $3x^{2}+3x-4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 3, b = 3, c = -4$ $ x = \dfrac{-3 \pm \sqrt{3^{2} - 4 \cdot 3 \cdot -4}}{2 \cdot 3}$ $ x = \dfrac{-3 \pm \sqrt{57}}{6}$ $ x = \dfrac{-3 \pm \sqrt{57}}{6}$ $x =\dfrac{-3 \pm \sqrt{57}}{6}$ |
185 | Let $f(x) = -x^{2}+2x+8$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-x^{2}+2x+8 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -1, b = 2, c = 8$ $ x = \dfrac{-2 \pm \sqrt{2^{2} - 4 \cdot -1 \cdot 8}}{2 \cdot -1}$ $ x = \dfrac{-2 \pm \sqrt{36}}{-2}$ $ x = \dfrac{-2 \pm 6}{-2}$ $x =-2,4$ |
185 | Let $f(x) = -9x^{2}+3x+8$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-9x^{2}+3x+8 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -9, b = 3, c = 8$ $ x = \dfrac{-3 \pm \sqrt{3^{2} - 4 \cdot -9 \cdot 8}}{2 \cdot -9}$ $ x = \dfrac{-3 \pm \sqrt{297}}{-18}$ $ x = \dfrac{-3 \pm 3\sqrt{33}}{-18}$ $x =\dfrac{-1 \pm \sqrt{33}}{-6}$ |
185 | Let $f(x) = -10x^{2}-5x+7$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-10x^{2}-5x+7 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -10, b = -5, c = 7$ $ x = \dfrac{+ 5 \pm \sqrt{(-5)^{2} - 4 \cdot -10 \cdot 7}}{2 \cdot -10}$ $ x = \dfrac{5 \pm \sqrt{305}}{-20}$ $ x = \dfrac{5 \pm \sqrt{305}}{-20}$ $x =\dfrac{5 \pm \sqrt{305}}{-20}$ |
185 | Let $f(x) = -7x^{2}+9x+8$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-7x^{2}+9x+8 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -7, b = 9, c = 8$ $ x = \dfrac{-9 \pm \sqrt{9^{2} - 4 \cdot -7 \cdot 8}}{2 \cdot -7}$ $ x = \dfrac{-9 \pm \sqrt{305}}{-14}$ $ x = \dfrac{-9 \pm \sqrt{305}}{-14}$ $x =\dfrac{-9 \pm \sqrt{305}}{-14}$ |
185 | Let $f(x) = -8x^{2}+9x+7$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-8x^{2}+9x+7 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -8, b = 9, c = 7$ $ x = \dfrac{-9 \pm \sqrt{9^{2} - 4 \cdot -8 \cdot 7}}{2 \cdot -8}$ $ x = \dfrac{-9 \pm \sqrt{305}}{-16}$ $ x = \dfrac{-9 \pm \sqrt{305}}{-16}$ $x =\dfrac{-9 \pm \sqrt{305}}{-16}$ |
185 | Let $f(x) = -3x^{2}+3x+7$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-3x^{2}+3x+7 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -3, b = 3, c = 7$ $ x = \dfrac{-3 \pm \sqrt{3^{2} - 4 \cdot -3 \cdot 7}}{2 \cdot -3}$ $ x = \dfrac{-3 \pm \sqrt{93}}{-6}$ $ x = \dfrac{-3 \pm \sqrt{93}}{-6}$ $x =\dfrac{-3 \pm \sqrt{93}}{-6}$ |
185 | Let $f(x) = 5x^{2}-10x+4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $5x^{2}-10x+4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 5, b = -10, c = 4$ $ x = \dfrac{+ 10 \pm \sqrt{(-10)^{2} - 4 \cdot 5 \cdot 4}}{2 \cdot 5}$ $ x = \dfrac{10 \pm \sqrt{20}}{10}$ $ x = \dfrac{10 \pm 2\sqrt{5}}{10}$ $x =\dfrac{5 \pm \sqrt{5}}{5}$ |
185 | Let $f(x) = x^{2}+9x+8$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $x^{2}+9x+8 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 1, b = 9, c = 8$ $ x = \dfrac{-9 \pm \sqrt{9^{2} - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$ $ x = \dfrac{-9 \pm \sqrt{49}}{2}$ $ x = \dfrac{-9 \pm 7}{2}$ $x =-1,-8$ |
185 | Let $f(x) = -4x^{2}+6x+4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-4x^{2}+6x+4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -4, b = 6, c = 4$ $ x = \dfrac{-6 \pm \sqrt{6^{2} - 4 \cdot -4 \cdot 4}}{2 \cdot -4}$ $ x = \dfrac{-6 \pm \sqrt{100}}{-8}$ $ x = \dfrac{-6 \pm 10}{-8}$ $x =-\frac{1}{2},2$ |
185 | Let $f(x) = -4x^{2}+3x+1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-4x^{2}+3x+1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -4, b = 3, c = 1$ $ x = \dfrac{-3 \pm \sqrt{3^{2} - 4 \cdot -4 \cdot 1}}{2 \cdot -4}$ $ x = \dfrac{-3 \pm \sqrt{25}}{-8}$ $ x = \dfrac{-3 \pm 5}{-8}$ $x =-\frac{1}{4},1$ |
185 | Let $f(x) = 2x^{2}+2x-6$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $2x^{2}+2x-6 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 2, b = 2, c = -6$ $ x = \dfrac{-2 \pm \sqrt{2^{2} - 4 \cdot 2 \cdot -6}}{2 \cdot 2}$ $ x = \dfrac{-2 \pm \sqrt{52}}{4}$ $ x = \dfrac{-2 \pm 2\sqrt{13}}{4}$ $x =\dfrac{-1 \pm \sqrt{13}}{2}$ |
185 | Let $f(x) = -7x^{2}+8x+2$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-7x^{2}+8x+2 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -7, b = 8, c = 2$ $ x = \dfrac{-8 \pm \sqrt{8^{2} - 4 \cdot -7 \cdot 2}}{2 \cdot -7}$ $ x = \dfrac{-8 \pm \sqrt{120}}{-14}$ $ x = \dfrac{-8 \pm 2\sqrt{30}}{-14}$ $x =\dfrac{-4 \pm \sqrt{30}}{-7}$ |
185 | Let $f(x) = -10x^{2}-10x+7$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-10x^{2}-10x+7 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -10, b = -10, c = 7$ $ x = \dfrac{+ 10 \pm \sqrt{(-10)^{2} - 4 \cdot -10 \cdot 7}}{2 \cdot -10}$ $ x = \dfrac{10 \pm \sqrt{380}}{-20}$ $ x = \dfrac{10 \pm 2\sqrt{95}}{-20}$ $x =\dfrac{5 \pm \sqrt{95}}{-10}$ |
185 | Let $f(x) = 2x^{2}-6x-2$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $2x^{2}-6x-2 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 2, b = -6, c = -2$ $ x = \dfrac{+ 6 \pm \sqrt{(-6)^{2} - 4 \cdot 2 \cdot -2}}{2 \cdot 2}$ $ x = \dfrac{6 \pm \sqrt{52}}{4}$ $ x = \dfrac{6 \pm 2\sqrt{13}}{4}$ $x =\dfrac{3 \pm \sqrt{13}}{2}$ |
185 | Let $f(x) = -8x^{2}+9x+3$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-8x^{2}+9x+3 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -8, b = 9, c = 3$ $ x = \dfrac{-9 \pm \sqrt{9^{2} - 4 \cdot -8 \cdot 3}}{2 \cdot -8}$ $ x = \dfrac{-9 \pm \sqrt{177}}{-16}$ $ x = \dfrac{-9 \pm \sqrt{177}}{-16}$ $x =\dfrac{-9 \pm \sqrt{177}}{-16}$ |
185 | Let $f(x) = 8x^{2}-4x-9$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $8x^{2}-4x-9 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 8, b = -4, c = -9$ $ x = \dfrac{+ 4 \pm \sqrt{(-4)^{2} - 4 \cdot 8 \cdot -9}}{2 \cdot 8}$ $ x = \dfrac{4 \pm \sqrt{304}}{16}$ $ x = \dfrac{4 \pm 4\sqrt{19}}{16}$ $x =\dfrac{1 \pm \sqrt{19}}{4}$ |
185 | Let $f(x) = -2x^{2}+7x+10$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-2x^{2}+7x+10 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -2, b = 7, c = 10$ $ x = \dfrac{-7 \pm \sqrt{7^{2} - 4 \cdot -2 \cdot 10}}{2 \cdot -2}$ $ x = \dfrac{-7 \pm \sqrt{129}}{-4}$ $ x = \dfrac{-7 \pm \sqrt{129}}{-4}$ $x =\dfrac{-7 \pm \sqrt{129}}{-4}$ |
185 | Let $f(x) = -x^{2}-x+1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-x^{2}-x+1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -1, b = -1, c = 1$ $ x = \dfrac{+ 1 \pm \sqrt{(-1)^{2} - 4 \cdot -1 \cdot 1}}{2 \cdot -1}$ $ x = \dfrac{1 \pm \sqrt{5}}{-2}$ $ x = \dfrac{1 \pm \sqrt{5}}{-2}$ $x =\dfrac{1 \pm \sqrt{5}}{-2}$ |
185 | Let $f(x) = 2x^{2}-6x-5$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )? | The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $2x^{2}-6x-5 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 2, b = -6, c = -5$ $ x = \dfrac{+ 6 \pm \sqrt{(-6)^{2} - 4 \cdot 2 \cdot -5}}{2 \cdot 2}$ $ x = \dfrac{6 \pm \sqrt{76}}{4}$ $ x = \dfrac{6 \pm 2\sqrt{19}}{4}$ $x =\dfrac{3 \pm \sqrt{19}}{2}$ |
dividing-by-8 | ${72} \div {8} = {?}$ | If we split ${72}$ circles into $8$ equal rows, how many circles are in each row? ${8}$ ${\color{#29ABCA}{1}}$ ${\color{#29ABCA}{2}}$ ${\color{#29ABCA}{3}}$ ${\color{#29ABCA}{4}}$ ${\color{#29ABCA}{5}}$ ${\color{#29ABCA}{6}}$ ${\color{#29ABCA}{7}}$ ${\color{#29ABCA}{8}}$ ${\color{#29ABCA}{9}}$ ${7}$ ${\color{#29ABCA}{10}}$ ${\color{#29ABCA}{11}}$ ${\color{#29ABCA}{12}}$ ${\color{#29ABCA}{13}}$ ${\color{#29ABCA}{14}}$ ${\color{#29ABCA}{15}}$ ${\color{#29ABCA}{16}}$ ${\color{#29ABCA}{17}}$ ${\color{#29ABCA}{18}}$ ${6}$ ${\color{#29ABCA}{19}}$ ${\color{#29ABCA}{20}}$ ${\color{#29ABCA}{21}}$ ${\color{#29ABCA}{22}}$ ${\color{#29ABCA}{23}}$ ${\color{#29ABCA}{24}}$ ${\color{#29ABCA}{25}}$ ${\color{#29ABCA}{26}}$ ${\color{#29ABCA}{27}}$ ${5}$ ${\color{#29ABCA}{28}}$ ${\color{#29ABCA}{29}}$ ${\color{#29ABCA}{30}}$ ${\color{#29ABCA}{31}}$ ${\color{#29ABCA}{32}}$ ${\color{#29ABCA}{33}}$ ${\color{#29ABCA}{34}}$ ${\color{#29ABCA}{35}}$ ${\color{#29ABCA}{36}}$ ${4}$ ${\color{#29ABCA}{37}}$ ${\color{#29ABCA}{38}}$ ${\color{#29ABCA}{39}}$ ${\color{#29ABCA}{40}}$ ${\color{#29ABCA}{41}}$ ${\color{#29ABCA}{42}}$ ${\color{#29ABCA}{43}}$ ${\color{#29ABCA}{44}}$ ${\color{#29ABCA}{45}}$ ${3}$ ${\color{#29ABCA}{46}}$ ${\color{#29ABCA}{47}}$ ${\color{#29ABCA}{48}}$ ${\color{#29ABCA}{49}}$ ${\color{#29ABCA}{50}}$ ${\color{#29ABCA}{51}}$ ${\color{#29ABCA}{52}}$ ${\color{#29ABCA}{53}}$ ${\color{#29ABCA}{54}}$ ${2}$ ${\color{#29ABCA}{55}}$ ${\color{#29ABCA}{56}}$ ${\color{#29ABCA}{57}}$ ${\color{#29ABCA}{58}}$ ${\color{#29ABCA}{59}}$ ${\color{#29ABCA}{60}}$ ${\color{#29ABCA}{61}}$ ${\color{#29ABCA}{62}}$ ${\color{#29ABCA}{63}}$ ${1}$ ${\color{#29ABCA}{64}}$ ${\color{#29ABCA}{65}}$ ${\color{#29ABCA}{66}}$ ${\color{#29ABCA}{67}}$ ${\color{#29ABCA}{68}}$ ${\color{#29ABCA}{69}}$ ${\color{#29ABCA}{70}}$ ${\color{#29ABCA}{71}}$ ${\color{#29ABCA}{72}}$ ${72} \div {8} = {9}$ |
dividing-by-8 | ${32} \div {8} = {?}$ | If we split ${32}$ circles into $8$ equal rows, how many circles are in each row? ${8}$ ${\color{#29ABCA}{1}}$ ${\color{#29ABCA}{2}}$ ${\color{#29ABCA}{3}}$ ${\color{#29ABCA}{4}}$ ${7}$ ${\color{#29ABCA}{5}}$ ${\color{#29ABCA}{6}}$ ${\color{#29ABCA}{7}}$ ${\color{#29ABCA}{8}}$ ${6}$ ${\color{#29ABCA}{9}}$ ${\color{#29ABCA}{10}}$ ${\color{#29ABCA}{11}}$ ${\color{#29ABCA}{12}}$ ${5}$ ${\color{#29ABCA}{13}}$ ${\color{#29ABCA}{14}}$ ${\color{#29ABCA}{15}}$ ${\color{#29ABCA}{16}}$ ${4}$ ${\color{#29ABCA}{17}}$ ${\color{#29ABCA}{18}}$ ${\color{#29ABCA}{19}}$ ${\color{#29ABCA}{20}}$ ${3}$ ${\color{#29ABCA}{21}}$ ${\color{#29ABCA}{22}}$ ${\color{#29ABCA}{23}}$ ${\color{#29ABCA}{24}}$ ${2}$ ${\color{#29ABCA}{25}}$ ${\color{#29ABCA}{26}}$ ${\color{#29ABCA}{27}}$ ${\color{#29ABCA}{28}}$ ${1}$ ${\color{#29ABCA}{29}}$ ${\color{#29ABCA}{30}}$ ${\color{#29ABCA}{31}}$ ${\color{#29ABCA}{32}}$ ${32} \div {8} = {4}$ |
dividing-by-8 | ${8} \div {8} = {?}$ | If we split ${8}$ circles into $8$ equal rows, how many circles are in each row? ${8}$ ${\color{#29ABCA}{1}}$ ${7}$ ${\color{#29ABCA}{2}}$ ${6}$ ${\color{#29ABCA}{3}}$ ${5}$ ${\color{#29ABCA}{4}}$ ${4}$ ${\color{#29ABCA}{5}}$ ${3}$ ${\color{#29ABCA}{6}}$ ${2}$ ${\color{#29ABCA}{7}}$ ${1}$ ${\color{#29ABCA}{8}}$ ${8} \div {8} = {1}$ |
dividing-by-8 | ${40} \div {8} = {?}$ | If we split ${40}$ circles into $8$ equal rows, how many circles are in each row? ${8}$ ${\color{#29ABCA}{1}}$ ${\color{#29ABCA}{2}}$ ${\color{#29ABCA}{3}}$ ${\color{#29ABCA}{4}}$ ${\color{#29ABCA}{5}}$ ${7}$ ${\color{#29ABCA}{6}}$ ${\color{#29ABCA}{7}}$ ${\color{#29ABCA}{8}}$ ${\color{#29ABCA}{9}}$ ${\color{#29ABCA}{10}}$ ${6}$ ${\color{#29ABCA}{11}}$ ${\color{#29ABCA}{12}}$ ${\color{#29ABCA}{13}}$ ${\color{#29ABCA}{14}}$ ${\color{#29ABCA}{15}}$ ${5}$ ${\color{#29ABCA}{16}}$ ${\color{#29ABCA}{17}}$ ${\color{#29ABCA}{18}}$ ${\color{#29ABCA}{19}}$ ${\color{#29ABCA}{20}}$ ${4}$ ${\color{#29ABCA}{21}}$ ${\color{#29ABCA}{22}}$ ${\color{#29ABCA}{23}}$ ${\color{#29ABCA}{24}}$ ${\color{#29ABCA}{25}}$ ${3}$ ${\color{#29ABCA}{26}}$ ${\color{#29ABCA}{27}}$ ${\color{#29ABCA}{28}}$ ${\color{#29ABCA}{29}}$ ${\color{#29ABCA}{30}}$ ${2}$ ${\color{#29ABCA}{31}}$ ${\color{#29ABCA}{32}}$ ${\color{#29ABCA}{33}}$ ${\color{#29ABCA}{34}}$ ${\color{#29ABCA}{35}}$ ${1}$ ${\color{#29ABCA}{36}}$ ${\color{#29ABCA}{37}}$ ${\color{#29ABCA}{38}}$ ${\color{#29ABCA}{39}}$ ${\color{#29ABCA}{40}}$ ${40} \div {8} = {5}$ |
dividing-by-8 | ${64} \div {8} = {?}$ | If we split ${64}$ circles into $8$ equal rows, how many circles are in each row? ${8}$ ${\color{#29ABCA}{1}}$ ${\color{#29ABCA}{2}}$ ${\color{#29ABCA}{3}}$ ${\color{#29ABCA}{4}}$ ${\color{#29ABCA}{5}}$ ${\color{#29ABCA}{6}}$ ${\color{#29ABCA}{7}}$ ${\color{#29ABCA}{8}}$ ${7}$ ${\color{#29ABCA}{9}}$ ${\color{#29ABCA}{10}}$ ${\color{#29ABCA}{11}}$ ${\color{#29ABCA}{12}}$ ${\color{#29ABCA}{13}}$ ${\color{#29ABCA}{14}}$ ${\color{#29ABCA}{15}}$ ${\color{#29ABCA}{16}}$ ${6}$ ${\color{#29ABCA}{17}}$ ${\color{#29ABCA}{18}}$ ${\color{#29ABCA}{19}}$ ${\color{#29ABCA}{20}}$ ${\color{#29ABCA}{21}}$ ${\color{#29ABCA}{22}}$ ${\color{#29ABCA}{23}}$ ${\color{#29ABCA}{24}}$ ${5}$ ${\color{#29ABCA}{25}}$ ${\color{#29ABCA}{26}}$ ${\color{#29ABCA}{27}}$ ${\color{#29ABCA}{28}}$ ${\color{#29ABCA}{29}}$ ${\color{#29ABCA}{30}}$ ${\color{#29ABCA}{31}}$ ${\color{#29ABCA}{32}}$ ${4}$ ${\color{#29ABCA}{33}}$ ${\color{#29ABCA}{34}}$ ${\color{#29ABCA}{35}}$ ${\color{#29ABCA}{36}}$ ${\color{#29ABCA}{37}}$ ${\color{#29ABCA}{38}}$ ${\color{#29ABCA}{39}}$ ${\color{#29ABCA}{40}}$ ${3}$ ${\color{#29ABCA}{41}}$ ${\color{#29ABCA}{42}}$ ${\color{#29ABCA}{43}}$ ${\color{#29ABCA}{44}}$ ${\color{#29ABCA}{45}}$ ${\color{#29ABCA}{46}}$ ${\color{#29ABCA}{47}}$ ${\color{#29ABCA}{48}}$ ${2}$ ${\color{#29ABCA}{49}}$ ${\color{#29ABCA}{50}}$ ${\color{#29ABCA}{51}}$ ${\color{#29ABCA}{52}}$ ${\color{#29ABCA}{53}}$ ${\color{#29ABCA}{54}}$ ${\color{#29ABCA}{55}}$ ${\color{#29ABCA}{56}}$ ${1}$ ${\color{#29ABCA}{57}}$ ${\color{#29ABCA}{58}}$ ${\color{#29ABCA}{59}}$ ${\color{#29ABCA}{60}}$ ${\color{#29ABCA}{61}}$ ${\color{#29ABCA}{62}}$ ${\color{#29ABCA}{63}}$ ${\color{#29ABCA}{64}}$ ${64} \div {8} = {8}$ |
dividing-by-8 | ${56} \div {8} = {?}$ | If we split ${56}$ circles into $8$ equal rows, how many circles are in each row? ${8}$ ${\color{#29ABCA}{1}}$ ${\color{#29ABCA}{2}}$ ${\color{#29ABCA}{3}}$ ${\color{#29ABCA}{4}}$ ${\color{#29ABCA}{5}}$ ${\color{#29ABCA}{6}}$ ${\color{#29ABCA}{7}}$ ${7}$ ${\color{#29ABCA}{8}}$ ${\color{#29ABCA}{9}}$ ${\color{#29ABCA}{10}}$ ${\color{#29ABCA}{11}}$ ${\color{#29ABCA}{12}}$ ${\color{#29ABCA}{13}}$ ${\color{#29ABCA}{14}}$ ${6}$ ${\color{#29ABCA}{15}}$ ${\color{#29ABCA}{16}}$ ${\color{#29ABCA}{17}}$ ${\color{#29ABCA}{18}}$ ${\color{#29ABCA}{19}}$ ${\color{#29ABCA}{20}}$ ${\color{#29ABCA}{21}}$ ${5}$ ${\color{#29ABCA}{22}}$ ${\color{#29ABCA}{23}}$ ${\color{#29ABCA}{24}}$ ${\color{#29ABCA}{25}}$ ${\color{#29ABCA}{26}}$ ${\color{#29ABCA}{27}}$ ${\color{#29ABCA}{28}}$ ${4}$ ${\color{#29ABCA}{29}}$ ${\color{#29ABCA}{30}}$ ${\color{#29ABCA}{31}}$ ${\color{#29ABCA}{32}}$ ${\color{#29ABCA}{33}}$ ${\color{#29ABCA}{34}}$ ${\color{#29ABCA}{35}}$ ${3}$ ${\color{#29ABCA}{36}}$ ${\color{#29ABCA}{37}}$ ${\color{#29ABCA}{38}}$ ${\color{#29ABCA}{39}}$ ${\color{#29ABCA}{40}}$ ${\color{#29ABCA}{41}}$ ${\color{#29ABCA}{42}}$ ${2}$ ${\color{#29ABCA}{43}}$ ${\color{#29ABCA}{44}}$ ${\color{#29ABCA}{45}}$ ${\color{#29ABCA}{46}}$ ${\color{#29ABCA}{47}}$ ${\color{#29ABCA}{48}}$ ${\color{#29ABCA}{49}}$ ${1}$ ${\color{#29ABCA}{50}}$ ${\color{#29ABCA}{51}}$ ${\color{#29ABCA}{52}}$ ${\color{#29ABCA}{53}}$ ${\color{#29ABCA}{54}}$ ${\color{#29ABCA}{55}}$ ${\color{#29ABCA}{56}}$ ${56} \div {8} = {7}$ |
dividing-by-8 | ${24} \div {8} = {?}$ | If we split ${24}$ circles into $8$ equal rows, how many circles are in each row? ${8}$ ${\color{#29ABCA}{1}}$ ${\color{#29ABCA}{2}}$ ${\color{#29ABCA}{3}}$ ${7}$ ${\color{#29ABCA}{4}}$ ${\color{#29ABCA}{5}}$ ${\color{#29ABCA}{6}}$ ${6}$ ${\color{#29ABCA}{7}}$ ${\color{#29ABCA}{8}}$ ${\color{#29ABCA}{9}}$ ${5}$ ${\color{#29ABCA}{10}}$ ${\color{#29ABCA}{11}}$ ${\color{#29ABCA}{12}}$ ${4}$ ${\color{#29ABCA}{13}}$ ${\color{#29ABCA}{14}}$ ${\color{#29ABCA}{15}}$ ${3}$ ${\color{#29ABCA}{16}}$ ${\color{#29ABCA}{17}}$ ${\color{#29ABCA}{18}}$ ${2}$ ${\color{#29ABCA}{19}}$ ${\color{#29ABCA}{20}}$ ${\color{#29ABCA}{21}}$ ${1}$ ${\color{#29ABCA}{22}}$ ${\color{#29ABCA}{23}}$ ${\color{#29ABCA}{24}}$ ${24} \div {8} = {3}$ |
dividing-by-8 | ${16} \div {8} = {?}$ | If we split ${16}$ circles into $8$ equal rows, how many circles are in each row? ${8}$ ${\color{#29ABCA}{1}}$ ${\color{#29ABCA}{2}}$ ${7}$ ${\color{#29ABCA}{3}}$ ${\color{#29ABCA}{4}}$ ${6}$ ${\color{#29ABCA}{5}}$ ${\color{#29ABCA}{6}}$ ${5}$ ${\color{#29ABCA}{7}}$ ${\color{#29ABCA}{8}}$ ${4}$ ${\color{#29ABCA}{9}}$ ${\color{#29ABCA}{10}}$ ${3}$ ${\color{#29ABCA}{11}}$ ${\color{#29ABCA}{12}}$ ${2}$ ${\color{#29ABCA}{13}}$ ${\color{#29ABCA}{14}}$ ${1}$ ${\color{#29ABCA}{15}}$ ${\color{#29ABCA}{16}}$ ${16} \div {8} = {2}$ |
dividing-by-8 | ${80} \div {8} = {?}$ | If we split ${80}$ circles into $8$ equal rows, how many circles are in each row? ${8}$ ${\color{#29ABCA}{1}}$ ${\color{#29ABCA}{2}}$ ${\color{#29ABCA}{3}}$ ${\color{#29ABCA}{4}}$ ${\color{#29ABCA}{5}}$ ${\color{#29ABCA}{6}}$ ${\color{#29ABCA}{7}}$ ${\color{#29ABCA}{8}}$ ${\color{#29ABCA}{9}}$ ${\color{#29ABCA}{10}}$ ${7}$ ${\color{#29ABCA}{11}}$ ${\color{#29ABCA}{12}}$ ${\color{#29ABCA}{13}}$ ${\color{#29ABCA}{14}}$ ${\color{#29ABCA}{15}}$ ${\color{#29ABCA}{16}}$ ${\color{#29ABCA}{17}}$ ${\color{#29ABCA}{18}}$ ${\color{#29ABCA}{19}}$ ${\color{#29ABCA}{20}}$ ${6}$ ${\color{#29ABCA}{21}}$ ${\color{#29ABCA}{22}}$ ${\color{#29ABCA}{23}}$ ${\color{#29ABCA}{24}}$ ${\color{#29ABCA}{25}}$ ${\color{#29ABCA}{26}}$ ${\color{#29ABCA}{27}}$ ${\color{#29ABCA}{28}}$ ${\color{#29ABCA}{29}}$ ${\color{#29ABCA}{30}}$ ${5}$ ${\color{#29ABCA}{31}}$ ${\color{#29ABCA}{32}}$ ${\color{#29ABCA}{33}}$ ${\color{#29ABCA}{34}}$ ${\color{#29ABCA}{35}}$ ${\color{#29ABCA}{36}}$ ${\color{#29ABCA}{37}}$ ${\color{#29ABCA}{38}}$ ${\color{#29ABCA}{39}}$ ${\color{#29ABCA}{40}}$ ${4}$ ${\color{#29ABCA}{41}}$ ${\color{#29ABCA}{42}}$ ${\color{#29ABCA}{43}}$ ${\color{#29ABCA}{44}}$ ${\color{#29ABCA}{45}}$ ${\color{#29ABCA}{46}}$ ${\color{#29ABCA}{47}}$ ${\color{#29ABCA}{48}}$ ${\color{#29ABCA}{49}}$ ${\color{#29ABCA}{50}}$ ${3}$ ${\color{#29ABCA}{51}}$ ${\color{#29ABCA}{52}}$ ${\color{#29ABCA}{53}}$ ${\color{#29ABCA}{54}}$ ${\color{#29ABCA}{55}}$ ${\color{#29ABCA}{56}}$ ${\color{#29ABCA}{57}}$ ${\color{#29ABCA}{58}}$ ${\color{#29ABCA}{59}}$ ${\color{#29ABCA}{60}}$ ${2}$ ${\color{#29ABCA}{61}}$ ${\color{#29ABCA}{62}}$ ${\color{#29ABCA}{63}}$ ${\color{#29ABCA}{64}}$ ${\color{#29ABCA}{65}}$ ${\color{#29ABCA}{66}}$ ${\color{#29ABCA}{67}}$ ${\color{#29ABCA}{68}}$ ${\color{#29ABCA}{69}}$ ${\color{#29ABCA}{70}}$ ${1}$ ${\color{#29ABCA}{71}}$ ${\color{#29ABCA}{72}}$ ${\color{#29ABCA}{73}}$ ${\color{#29ABCA}{74}}$ ${\color{#29ABCA}{75}}$ ${\color{#29ABCA}{76}}$ ${\color{#29ABCA}{77}}$ ${\color{#29ABCA}{78}}$ ${\color{#29ABCA}{79}}$ ${\color{#29ABCA}{80}}$ ${80} \div {8} = {10}$ |
dividing-by-8 | ${48} \div {8} = {?}$ | If we split ${48}$ circles into $8$ equal rows, how many circles are in each row? ${8}$ ${\color{#29ABCA}{1}}$ ${\color{#29ABCA}{2}}$ ${\color{#29ABCA}{3}}$ ${\color{#29ABCA}{4}}$ ${\color{#29ABCA}{5}}$ ${\color{#29ABCA}{6}}$ ${7}$ ${\color{#29ABCA}{7}}$ ${\color{#29ABCA}{8}}$ ${\color{#29ABCA}{9}}$ ${\color{#29ABCA}{10}}$ ${\color{#29ABCA}{11}}$ ${\color{#29ABCA}{12}}$ ${6}$ ${\color{#29ABCA}{13}}$ ${\color{#29ABCA}{14}}$ ${\color{#29ABCA}{15}}$ ${\color{#29ABCA}{16}}$ ${\color{#29ABCA}{17}}$ ${\color{#29ABCA}{18}}$ ${5}$ ${\color{#29ABCA}{19}}$ ${\color{#29ABCA}{20}}$ ${\color{#29ABCA}{21}}$ ${\color{#29ABCA}{22}}$ ${\color{#29ABCA}{23}}$ ${\color{#29ABCA}{24}}$ ${4}$ ${\color{#29ABCA}{25}}$ ${\color{#29ABCA}{26}}$ ${\color{#29ABCA}{27}}$ ${\color{#29ABCA}{28}}$ ${\color{#29ABCA}{29}}$ ${\color{#29ABCA}{30}}$ ${3}$ ${\color{#29ABCA}{31}}$ ${\color{#29ABCA}{32}}$ ${\color{#29ABCA}{33}}$ ${\color{#29ABCA}{34}}$ ${\color{#29ABCA}{35}}$ ${\color{#29ABCA}{36}}$ ${2}$ ${\color{#29ABCA}{37}}$ ${\color{#29ABCA}{38}}$ ${\color{#29ABCA}{39}}$ ${\color{#29ABCA}{40}}$ ${\color{#29ABCA}{41}}$ ${\color{#29ABCA}{42}}$ ${1}$ ${\color{#29ABCA}{43}}$ ${\color{#29ABCA}{44}}$ ${\color{#29ABCA}{45}}$ ${\color{#29ABCA}{46}}$ ${\color{#29ABCA}{47}}$ ${\color{#29ABCA}{48}}$ ${48} \div {8} ={6}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $4.9\cdot 10^{5} - 5.8 \cdot 10^{4} = $ | $\phantom{=}{ 4.9\cdot 10^{5}} - 5.8 \cdot 10^{4}$ $={49\cdot 10^{4}} - 5.8 \cdot 10^{4}$ $= (49-5.8)\cdot10^{4}$ $=43.2\cdot10^{4}$ $=4.32\cdot10^{5}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $5.4 \cdot 10^5 + 6.7 \cdot 10^4 = $ | $\phantom{=} {5.4 \cdot 10^5} + 6.7 \cdot 10^4$ $={54 \cdot 10^{4}} + 6.7\cdot10^{4} $ $= (54+6.7)\cdot10^{4}$ $=60.7\cdot10^{4}$ $= 6.07\cdot10^{5}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $8.2 \cdot 10^{-5} + 0.0004 = $ | $\phantom{=} 8.2 \cdot 10^{-5} + {0.0004}$ $=8.2 \cdot 10^{-5} +{4\cdot10^{-4}}$ $=8.2 \cdot 10^{-5} + {40\cdot10^{-5}} $ $= (8.2+40)\cdot10^{-5}$ $=48.2\cdot10^{-5}$ $=4.82\cdot10^{-4}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $2.3 \cdot 10^8 + 4.7 \cdot 10^7 = $ | $\phantom{=} {2.3 \cdot 10^8} + 4.7 \cdot 10^7$ $={23 \cdot 10^{7}} + 4.7\cdot10^{7} $ $= (23+4.7)\cdot10^{7}$ $=27.7\cdot10^{7}$ $= 2.77\cdot10^{8}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $6.4 \cdot 10^5 + 36{,}000 = $ | $\phantom{=}{6.4 \cdot 10^5} + 36{,}000$ $={6.4 \cdot 10^{5}} +3.6\cdot10^{4} $ $={64 \cdot 10^{4}} + 3.6\cdot10^{4} $ $= (64+3.6)\cdot10^{4}$ $=67.6\cdot10^{4}$ $=6.76\cdot10^5$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $2.2 \cdot 10^{6} - 4.3 \cdot 10^{5} = $ | $\phantom{=}{2.2 \cdot 10^{6}} - 4.3 \cdot 10^{5}$ $={22 \cdot 10^{5}} - 4.3 \cdot 10^{5}$ $= (22-4.3)\cdot10^{5}$ $=17.7\cdot10^{5}$ $=1.77\cdot10^{6}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $7.4 \cdot 10^{-8} - 6.7 \cdot 10^{-9} = $ | $\phantom{=}{7.4 \cdot 10^{-8}} - 6.7 \cdot 10^{-9}$ $={74 \cdot 10^{-9}} - 6.7 \cdot 10^{-9}$ $= (74-6.7)\cdot10^{-9}$ $=67.3\cdot10^{-9}$ $=6.73\cdot10^{-8}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $5.8 \cdot 10^{1} - 7.4 \cdot 10^{0} = $ | $\phantom{=}{5.8 \cdot 10^{1}} - 7.4 \cdot 10^{0}$ $={58 \cdot 10^{0}} - 7.4 \cdot 10^{0}$ $= (58-7.4)\cdot10^{0}$ $= 50.6\cdot10^{0}$ $=5.06\cdot10^{1}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $7.9 \cdot 10^7 + 6.5 \cdot 10^6 = $ | $\phantom{=} {7.9 \cdot 10^7} + 6.5 \cdot 10^6$ $={79 \cdot 10^{6}} + 6.5\cdot10^{6} $ $= (79+6.5)\cdot10^{6}$ $=85.5\cdot10^{6}$ $= 8.55\cdot10^{7}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $8.8 \cdot 10^7 + 3.1 \cdot 10^8 = $ | $\phantom{=} 8.8 \cdot 10^7 + {3.1 \cdot 10^8}$ $=8.8 \cdot 10^{7} + {31\cdot10^{7} }$ $= (8.8+31)\cdot10^{7}$ $=39.8\cdot10^{7}$ $=3.98\cdot10^8$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $4.9 \cdot 10^4 - 8200 = $ | $\phantom{=} {4.9 \cdot 10^4} - 8200$ $={4.9 \cdot 10^4} - 8.2 \cdot10^3 $ $={49 \cdot 10^3} -8.2 \cdot10^3 $ $= (49-8.2)\cdot10^{3}$ $= 40.8\cdot10^{3}$ $=4.08\cdot10^4$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $9.3\cdot 10^{7} - 3.4 \cdot 10^{6} = $ | $\phantom{=} {9.3\cdot 10^{7}} - 3.4 \cdot 10^{6} $ $={93\cdot 10^{6}} - 3.4 \cdot 10^{6}$ $= (93-3.4)\cdot10^{6}$ $=89.6 \cdot10^6$ $=8.96\cdot10^{7}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $1.8 \cdot 10^{-2} - 3.9 \cdot 10^{-3} = $ | $\phantom{=}{1.8 \cdot 10^{-2}} - 3.9 \cdot 10^{-3}$ $={18 \cdot 10^{-3}} - 3.9 \cdot 10^{-3}$ $= (18-3.9)\cdot10^{-3}$ $=14.1\cdot10^{-3}$ $=1.41\cdot10^{-2}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $4.1 \cdot 10^{-2} - 2.6 \cdot 10^{-3} = $ | $\phantom{=}{4.1 \cdot 10^{-2}} - 2.6 \cdot 10^{-3}$ $={41 \cdot 10^{-3}} -2.6\cdot10^{-3}$ $= (41-2.6)\cdot10^{-3}$ $=38.4\cdot10^{-3}$ $=3.84\cdot10^{-2}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $3.3 \cdot 10^{5} + 44{,}000 = $ | $\phantom{=} {3.3 \cdot 10^{5}} + 44{,}000$ $={3.3 \cdot 10^{5}} +4.4\cdot10^{4} $ $={33 \cdot 10^{4}} + 4.4\cdot10^{4} $ $= (33+4.4)\cdot10^{4}$ $=37.4\cdot10^{4}$ $=3.74\cdot10^{5}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $2.8\cdot10^{-3} -0.00065 = $ | $\phantom{=}{ 2.8\cdot10^{-3}} -0.00065 $ $={2.8 \cdot 10^{-3}} - 6.5\cdot10^{-4} $ $={28 \cdot 10^{-4}} - 6.5\cdot10^{-4} $ $= (28-6.5)\cdot10^{-4}$ $=21.5\cdot10^{-4}$ $= 2.15\cdot10^{-3}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $1.4\cdot 10^{9} - 8.6 \cdot 10^{8} = $ | $\phantom{=}{ 1.4\cdot 10^{9}} - 8.6 \cdot 10^{8}$ $={14\cdot 10^{8}} - 8.6 \cdot 10^{8} $ $= (14-8.6)\cdot10^{8}$ $= 5.4\cdot10^{8}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $0.00045 - 2.5\cdot10^{-5} = $ | $\phantom{=} {0.00045} - 2.5\cdot10^{-5} $ $={4.5 \cdot 10^{-4}} - 2.5 \cdot10^{-5} $ $={45 \cdot 10^{-5}} - 2.5 \cdot10^{-5} $ $= (45-2.5)\cdot10^{-5}$ $= 42.5\cdot10^{-5}$ $=4.25\cdot10^{-4}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $4.9 \cdot 10^{-5} + 0.0005 = $ | $\phantom{=}4.9 \cdot 10^{-5} + {0.0005}$ $=4.9 \cdot 10^{-5} +{5\cdot10^{-4}} $ $=4.9 \cdot 10^{-5} + {50\cdot10^{-5}} $ $= (4.9+50)\cdot10^{-5}$ $=54.9\cdot10^{-5}$ $=5.49\cdot10^{-4}$ |
adding-and-subtracting-in-scientific-notation | Express your answer in scientific notation. $3600 - 6.3\cdot10^{2} = $ | $\phantom{=} {3600} - 6.3\cdot10^{2}$ $={3.6 \cdot 10^{3}} - 6.3 \cdot10^{2} $ $={36 \cdot 10^{2}} - 6.3 \cdot10^{2}$ $= (36-6.3)\cdot10^{2}$ $= 29.7\cdot10^{2}$ $= 2.97\cdot10^{3}$ |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} 2xy & x^2 & 0 \\ \\ 0 & -1 & 1 \\ \\ 1 & 0 & 1 \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $\left( 1, \dfrac{1}{4}, 2 \right)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} 2xy & x^2 & 0 \\ \\ 0 & -1 & 1 \\ \\ 1 & 0 & 1 \end{bmatrix} \right) \\ \\ &= 2xy(-1 - 0) - x^2(0 - 1) + 0(0 + 1) \\ \\ &= x^2 - 2xy \end{aligned}$ If we evaluate $|J(f)|$ at $\left( 1, \dfrac{1}{4}, 2 \right)$, we get $\dfrac{1}{2}$. Because the Jacobian determinant here has an absolute value less than $1$ but not $0$, we can conclude that $f$ will finitely contract the space around $\left( 1, \dfrac{1}{4}, 2 \right)$. To recap, the Jacobian determinant of $f$ is $x^2 - 2xy$, and $f$ will finitely contract the space around the point $\left( 1, \dfrac{1}{4}, 2 \right)$. |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} 8x & 8y \\ \\ 3y & 3x \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $(2, 3)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} 8x & 8y \\ \\ 3y & 3x \end{bmatrix} \right) \\ \\ &=24x^2 - 24y^2 \\ \\ &= 24(x^2 - y^2) \end{aligned}$ If we evaluate $|J(f)|$ at $(2, 3)$, we get $-120$. Because the Jacobian determinant here has an absolute value greater than $1$, we can conclude that $f$ will finitely expand the space around $(2, 3)$. To recap, the Jacobian determinant of $f$ is $24(x^2 - y^2)$, and $f$ will finitely expand the space around the point $(2, 3)$. |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} \dfrac{1}{x} & 0 \\ \\ 1 & 1 \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $\left( \dfrac{1}{9}, \dfrac{1}{3} \right)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} \dfrac{1}{x} & 0 \\ \\ 1 & 1 \end{bmatrix} \right) \\ \\ &= \dfrac{1}{x} \end{aligned}$ If we evaluate $|J(f)|$ at $\left( \dfrac{1}{9}, \dfrac{1}{3} \right)$, we get $9$. Because the Jacobian determinant here has an absolute value greater than $1$, we can conclude that $f$ will finitely expand the space around $\left( \dfrac{1}{9}, \dfrac{1}{3} \right)$. To recap, the Jacobian determinant of $f$ is $\dfrac{1}{x}$, and $f$ will finitely expand the space around the point $\left( \dfrac{1}{9}, \dfrac{1}{3} \right)$. |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} 3x^2 - 4 & 0 \\ \\ 0 & 3y^2 - 4 \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $(1, -1)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} 3x^2 - 4 & 0 \\ \\ 0 & 3y^2 - 4 \end{bmatrix} \right) \\ \\ &= (3x^2 - 4)(3y^2 - 4) \end{aligned}$ If we evaluate $|J(f)|$ at $(1, -1)$, we get $1$. Because the Jacobian determinant here has an absolute value equal to $1$, we can conclude that $f$ will leave the space around $(1, -1)$ the same, not expanding nor contracting it. To recap, the Jacobian determinant of $f$ is $(3x^2 - 4)(3y^2 - 4)$, and $f$ will leave the space around the point $(1, -1)$ the same, not expanding nor contracting it. |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} \cos(\theta) & -r\sin(\theta) \\ \\ \sin(\theta) & r\cos(\theta) \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $r = \dfrac{1}{2}, \theta = \dfrac{\pi}{2}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} \cos(\theta) & -r\sin(\theta) \\ \\ \sin(\theta) & r\cos(\theta) \end{bmatrix} \right) \\ \\ &= r \cos^2(\theta) + r \sin^2(\theta) \\ \\ &= r (\cos^2(\theta) + \sin^2(\theta)) \\ \\ &= r \end{aligned}$ If we evaluate $|J(f)|$ at $r = \dfrac{1}{2}, \theta = \dfrac{\pi}{2}$, we get $\dfrac{1}{2}$. Because the Jacobian determinant here has an absolute value less than $1$ but not $0$, we can conclude that $f$ will finitely contract the space around $r = \dfrac{1}{2}, \theta = \dfrac{\pi}{2}$. To recap, the Jacobian determinant of $f$ is $r$, and $f$ will finitely contract the space around the point $r = \dfrac{1}{2}, \theta = \dfrac{\pi}{2}$. |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} y & x - 1 \\ \\ y - 1 & x \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $(2, -1)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} y & x - 1 \\ \\ y - 1 & x \end{bmatrix} \right) \\ \\ &= xy - (x - 1)(y - 1) \\ \\ &= xy - (xy - x - y + 1) \\ \\ &= x + y - 1 \end{aligned}$ If we evaluate $|J(f)|$ at $(2, -1)$, we get $0$. Because the Jacobian determinant here is equal to $0$, we can conclude that $f$ will infinitely contract the space around $(2, -1)$. To recap, the Jacobian determinant of $f$ is $x + y - 1$, and $f$ will infinitely contract the space around the point $(2, -1)$. |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} -\sin(x) & -\cos(y) \\ \\ \cos(x) & \sin(y) \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $(\pi, \pi)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} -\sin(x) & -\cos(y) \\ \\ \cos(x) & \sin(y) \end{bmatrix} \right) \\ \\ &= -\sin(x)\sin(y) + \cos(x)\cos(y) \end{aligned}$ If we evaluate $|J(f)|$ at $(\pi, \pi)$, we get $1$. Because the Jacobian determinant here has an absolute value equal to $1$, we can conclude that $f$ will leave the space around $(\pi, \pi)$ the same, not expanding nor contracting it. To recap, the Jacobian determinant of $f$ is $-\sin(x)\sin(y) + \cos(x)\cos(y)$, and $f$ will leave the space around the point $(\pi, \pi)$ the same, not expanding nor contracting it. |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} 1 & 0 & 0 \\ \\ y & x & 0 \\ \\ yz & xz & xy \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $(2, 0, 3)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} 1 & 0 & 0 \\ \\ y & x & 0 \\ \\ yz & xz & xy \end{bmatrix} \right) \\ \\ &= -1 (x \cdot xy - 0 \cdot xz) \\ \\ &- 0 (y \cdot xy - 0 \cdot yz) \\ \\ &+ (-1)(y \cdot xy - x \cdot yz) \\ \\ &= x^2y \end{aligned}$ If we evaluate $|J(f)|$ at $(2, 0, 3)$, we get $0$. Because the Jacobian determinant here is equal to $0$, we can conclude that $f$ will infinitely contract the space around $(2, 0, 3)$. To recap, the Jacobian determinant of $f$ is $x^2y$, and $f$ will infinitely contract the space around the point $(2, 0, 3)$. |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} \sin(y) & x\cos(y) \\ \\ -y\sin(x) & \cos(x) \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $\left( \dfrac{\pi}{2}, \dfrac{-\pi}{2} \right)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} \sin(y) & x\cos(y) \\ \\ -y\sin(x) & \cos(x) \end{bmatrix} \right) \\ \\ &= \cos(x)\sin(y) + xy\cos(y)\sin(x) \end{aligned}$ If we evaluate $|J(f)|$ at $\left( \dfrac{\pi}{2}, \dfrac{-\pi}{2} \right)$, we get $0$. Because the Jacobian determinant here is equal to $0$, we can conclude that $f$ will infinitely contract the space around $\left( \dfrac{\pi}{2}, \dfrac{-\pi}{2} \right)$. To recap, the Jacobian determinant of $f$ is $\cos(x)\sin(y) + xy\cos(y)\sin(x)$, and $f$ will infinitely contract the space around the point $\left( \dfrac{\pi}{2}, \dfrac{-\pi}{2} \right)$. |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \\ \\ \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $(1, 0)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \\ \\ \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \end{bmatrix} \right) \\ \\ &= \left( \dfrac{\sqrt{3}}{2} \right)^2 + \left(\dfrac{1}{2} \right)^2 \\ \\ &= 1 \end{aligned}$ If we evaluate $|J(f)|$ at $(1, 0)$, we get $1$. Because the Jacobian determinant here has an absolute value equal to $1$, we can conclude that $f$ will leave the space around $(1, 0)$ the same, not expanding nor contracting it. To recap, the Jacobian determinant of $f$ is $1$, and $f$ will leave the space around the point $(1, 0)$ the same, not expanding nor contracting it. |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} -1 & 0 & -1 \\ \\ 3 & 0 & 0 \\ \\ 0 & \cos(y) & 1 \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $(0, \pi, 0)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} -1 & 0 & -1 \\ \\ 3 & 0 & 0 \\ \\ 0 & \cos(y) & 1 \end{bmatrix} \right) \\ \\ &= -1 (0 \cdot 1 - 0 \cdot \cos(y)) \\ \\ &- 0 (3 \cdot 1 - 0 \cdot 0) \\ \\ &+ (-1)(3 \cdot \cos(y) - 0 \cdot 0) \\ \\ &= -3\cos(y) \end{aligned}$ If we evaluate $|J(f)|$ at $(0, \pi, 0)$, we get $3$. Because the Jacobian determinant here has an absolute value greater than $1$, we can conclude that $f$ will finitely expand the space around $(0, \pi, 0)$. To recap, the Jacobian determinant of $f$ is $-3 \cos(y)$, and $f$ will finitely expand the space around the point $(0, \pi, 0)$. |
jacobian-determinant | Let $f$ be a transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} 0 & -z\sin(y) & \cos(y) \\ \\ \cos(z) & 0 & -x\sin(z) \\ \\ -y\sin(x) & \cos(x) & 0 \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $\left( \dfrac{\pi}{3}, 0, \dfrac{\pi}{3} \right)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely | The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} 0 & -z\sin(y) & \cos(y) \\ \\ \cos(z) & 0 & -x\sin(z) \\ \\ -y\sin(x) & \cos(x) & 0 \end{bmatrix} \right) \\ \\ &= z\sin(y)(0-xy\sin(z)\sin(x)) + \cos(y)(\cos(z)\cos(x)-0) \\ \\ &= -xyz\sin(x)\sin(y)\sin(z) + \cos(x)\cos(y)\cos(z) \end{aligned}$ If we evaluate $|J(f)|$ at $\left( \dfrac{\pi}{3}, 0, \dfrac{\pi}{3} \right)$, we get $\dfrac{1}{4}$. Because the Jacobian determinant here has an absolute value less than $1$ but not $0$, we can conclude that $f$ will finitely contract the space around $\left( \dfrac{\pi}{3}, 0, \dfrac{\pi}{3} \right)$. To recap, the Jacobian determinant of $f$ is $-xyz\sin(x)\sin(y)\sin(z) + \cos(x)\cos(y)\cos(z)$ and $f$ will finitely contract the space around the point $\left( \dfrac{\pi}{3}, 0, \dfrac{\pi}{3} \right)$. |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+4x-21 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 + 4x = 21$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $4$, half of it would be $2$, and squaring it gives us ${4}$. $x^2 + 4x { + 4} = 21 { + 4}$ We can now rewrite the left side of the equation as a squared term. $( x + 2 )^2 = 25$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-10x-96$ $f(x)=(x+$ | We want to complete $x^2{-10}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-10}}{2}\right)^2={25}$ to it: $x^2{-10}x+{25}=(x-5)^2$ In order to keep the expression equivalent, we add and subtract ${25}$, not forgetting the expression's constant term, $-96$ : $\begin{aligned} f(x)&=x^2-10x-96 \\\\ &=x^2-10x+{25}-96-{25} \\\\ &=(x-5)^2-96-25 \\\\ &=(x-5)^2-121 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 5)^2 - 121$ This is equivalent to $f(x)=(x+{-5})^2-121$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-4x = 0$ $(x + $ | We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-4$, half of it would be $-2$, and squaring it gives us ${4}$. $x^2 - 4x { + 4} = 0 { + 4}$ We can now rewrite the left side of the equation as a squared term. $( x - 2 )^2 = 4$ This is equivalent to $(x+{-2})^2=4$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-4x-32 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 - 4x = 32$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-4$, half of it would be $-2$, and squaring it gives us ${4}$. $x^2 - 4x { + 4} = 32 { + 4}$ We can now rewrite the left side of the equation as a squared term. $( x - 2 )^2 = 36$ This is equivalent to $(x+{-2})^2=36$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+8x+4$ $f(x)=(x+$ | We want to complete $x^2{+8}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+8}}{2}\right)^2={16}$ to it: $x^2{+8}x+{16}=(x+4)^2$ In order to keep the expression equivalent, we add and subtract ${16}$, not forgetting the expression's constant term, $4$ : $\begin{aligned} f(x)&=x^2+8x+4 \\\\ &=x^2+8x+{16}+4-{16} \\\\ &=(x+4)^2+4-16 \\\\ &=(x+4)^2-12 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 4)^2 - 12$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-14x+40 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 - 14x = -40$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-14$, half of it would be $-7$, and squaring it gives us ${49}$. $x^2 - 14x { + 49} = -40 { + 49}$ We can now rewrite the left side of the equation as a squared term. $( x - 7 )^2 = 9$ This is equivalent to $(x+{-7})^2=9$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+4x+41$ $f(x)=(x+$ | We want to complete $x^2{+4}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+4}}{2}\right)^2={4}$ to it: $x^2{+4}x+{4}=(x+2)^2$ In order to keep the expression equivalent, we add and subtract ${4}$, not forgetting the expression's constant term, $41$ : $\begin{aligned} f(x)&=x^2+4x+41 \\\\ &=x^2+4x+{4}+41-{4} \\\\ &=(x+2)^2+41-4 \\\\ &=(x+2)^2+37 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 2)^2 + 37$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+14x+49 = 0$ $(x + $ | The left side of the equation is already a perfect square trinomial. The coefficient of our $x$ term is $14$, half of it is $7$, and squaring it gives us ${49}$, our constant term. Thus, we can rewrite the left side of the equation as a squared term. $( x + 7 )^2 = 0$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-2x+17$ $f(x)=(x+$ | We want to complete $x^2{-2}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-2}}{2}\right)^2={1}$ to it: $x^2{-2}x+{1}=(x-1)^2$ In order to keep the expression equivalent, we add and subtract ${1}$, not forgetting the expression's constant term, $17$ : $\begin{aligned} f(x)&=x^2-2x+17 \\\\ &=x^2-2x+{1}+17-{1} \\\\ &=(x-1)^2+17-1 \\\\ &=(x-1)^2+16 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 1)^2 + 16$ This is equivalent to $f(x)=(x+{-1})^2+16$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+12x+7$ $f(x)=(x+$ | We want to complete $x^2{+12}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+12}}{2}\right)^2={36}$ to it: $x^2{+12}x+{36}=(x+6)^2$ In order to keep the expression equivalent, we add and subtract ${36}$, not forgetting the expression's constant term, $7$ : $\begin{aligned} f(x)&=x^2+12x+7 \\\\ &=x^2+12x+{36}+7-{36} \\\\ &=(x+6)^2+7-36 \\\\ &=(x+6)^2-29 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 6)^2 - 29$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-8x+7 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 - 8x = -7$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-8$, half of it would be $-4$, and squaring it gives us ${16}$. $x^2 - 8x { + 16} = -7 { + 16}$ We can now rewrite the left side of the equation as a squared term. $( x - 4 )^2 = 9$ This is equivalent to $(x+{-4})^2=9$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-14x+63$ $f(x)=(x+$ | We want to complete $x^2{-14}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-14}}{2}\right)^2={49}$ to it: $x^2{-14}x+{49}=(x-7)^2$ In order to keep the expression equivalent, we add and subtract ${49}$, not forgetting the expression's constant term, $63$ : $\begin{aligned} f(x)&=x^2-14x+63 \\\\ &=x^2-14x+{49}+63-{49} \\\\ &=(x-7)^2+63-49 \\\\ &=(x-7)^2+14 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 7)^2 + 14$ This is equivalent to $f(x)=(x+{-7})^2+14$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+2x-35 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 + 2x = 35$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $2$, half of it would be $1$, and squaring it gives us ${1}$. $x^2 + 2x { + 1} = 35 { + 1}$ We can now rewrite the left side of the equation as a squared term. $( x + 1 )^2 = 36$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-14x+49 = 0$ $(x + $ | The left side of the equation is already a perfect square trinomial. The coefficient of our $x$ term is $-14$, half of it is $-7$, and squaring it gives us ${49}$, our constant term. Thus, we can rewrite the left side of the equation as a squared term. $( x - 7 )^2 = 0$ This is equivalent to $(x+{-7})^2=0$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+12x+36 = 0$ $(x + $ | The left side of the equation is already a perfect square trinomial. The coefficient of our $x$ term is $12$, half of it is $6$, and squaring it gives us ${36}$, our constant term. Thus, we can rewrite the left side of the equation as a squared term. $( x + 6 )^2 = 0$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-8x-2$ $f(x)=(x+$ | We want to complete $x^2{-8}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-8}}{2}\right)^2={16}$ to it: $x^2{-8}x+{16}=(x-4)^2$ In order to keep the expression equivalent, we add and subtract ${16}$, not forgetting the expression's constant term, $-2$ : $\begin{aligned} f(x)&=x^2-8x-2 \\\\ &=x^2-8x+{16}-2-{16} \\\\ &=(x-4)^2-2-16 \\\\ &=(x-4)^2-18 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 4)^2 - 18$ This is equivalent to $f(x)=(x+{-4})^2-18$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+6x-78$ $f(x)=(x+$ | We want to complete $x^2{+6}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+6}}{2}\right)^2={9}$ to it: $x^2{+6}x+{9}=(x+3)^2$ In order to keep the expression equivalent, we add and subtract ${9}$, not forgetting the expression's constant term, $-78$ : $\begin{aligned} f(x)&=x^2+6x-78 \\\\ &=x^2+6x+{9}-78-{9} \\\\ &=(x+3)^2-78-9 \\\\ &=(x+3)^2-87 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 3)^2 - 87$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-6x-60$ $f(x)=(x+$ | We want to complete $x^2{-6}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-6}}{2}\right)^2={9}$ to it: $x^2{-6}x+{9}=(x-3)^2$ In order to keep the expression equivalent, we add and subtract ${9}$, not forgetting the expression's constant term, $-60$ : $\begin{aligned} f(x)&=x^2-6x-60 \\\\ &=x^2-6x+{9}-60-{9} \\\\ &=(x-3)^2-60-9 \\\\ &=(x-3)^2-69 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 3)^2 - 69$ This is equivalent to $f(x)=(x+{-3})^2-69$ |
Subsets and Splits