$\bf y=(2x-3)^5(2-x^4)^3\\\\
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\cfrac{dy}{dx}=\stackrel{\textit{product rule}}{[5(2x-3)^4\cdot 2(2-x^4)3]~~+~~[(2x-3)^5[3(2-x^4)^2(-4x^3)]]}
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\cfrac{dy}{dx}=10(2x-3)^4(2-x^4)^3~~-~~12x^3(2x-3)^5(2-x^4)^2
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\cfrac{dy}{dx}=\stackrel{\textit{common factor}}{2(2x-3)^4(2-x^4)^2}~[5(2-x^4)-6x^3(2x-3)]
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\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2~[10-5x^4-12x^4+18x^3]
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\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2(10-17x^4+18x^3)$

5 0

3 years ago

3. The vertices of ABC are A(2, -1), B(0, -3), and C(4, -2). Find the vertices of the triangle after a translation 2 units right

kow [346]

Answer:

Step-by-step explanation:

5 0

3 years ago

The solution to 3x2 – 12x 24 = 0 is

zzz [600]

$3x^2 - 12x + 24 = 0 \\ x^2 - 4x + 8 = 0 \\ x= \frac{-b \pm \sqrt{ b^{2} -4ac} }{2a} 
$ ; where a = 1, b = -4 and c = 8
$x= \frac{-(-4) \pm \sqrt{ (-4)^{2} -4 \times 1 \times 8} }{2 \times 1} \\ = \frac{4 \pm \sqrt{ 16 -32} }{2} \\ =\frac{4 \pm \sqrt{ -16} }{2} \\ =\frac{4 \pm 4i }{2} \\ =2+2i \ or \ 2-2i$

7 0

3 years ago

Read 2 more answers

*High School problem not Middle School

devlian [24]

Should be c ? can you send me a picture please ? and it’s dice .

4 0

3 years ago