How can you Factor:ab^2-ab

Mathematics

1 answer:

NeTakaya2 years ago

3 0

Answer:

ab(b−1)

Step-by-step explanation:

Factor ab out of ab ^2 − ab

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Solve for x:a(a²+b²)x²+b²x-a

m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}$

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}$

$\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}$

5 0

3 years ago

Polygon ABCD is dilated by a scale factor of 2 with the center of dilation at the origin to create A'B'C'D'. If the endpoints of

Sonja [21]

The distance formula is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Take point A(0, -7) as (x₁, x₂) and point B(8, 8) as (y₁, y₂)

Distance AB = $\sqrt{(8-0)^2+(8--7)^2}$
Distance AB = $\sqrt{8^2+15^2}$
Distance AB = $\sqrt{289}$
Distance AB = 17

The line AB is dilated to A'B' by scale factor of 2, it means the length of the line is multiplied by 2

Distance A'B' = 2 × 17 = 34 units

5 0

3 years ago

0.09 is 10 times as much as ?

ANEK [815]

To find out the answer we divide 0.09 by 10: $0.09$ ÷10=0.009. Therefore, 0.09 is 10 times as much as 0.009.