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salantis [7]
4 years ago
15

with vertices D(2, 5), E(6, 4), and F(3, 3), is reflected across the line A student determined one of the vertices on the image

to be (2, –5). Evaluate the student’s answer
Mathematics
1 answer:
Nezavi [6.7K]4 years ago
6 0
The vertices of the image reflected over the line y = x are:
D´( 5 , 2 ),  E´( 4, 6 ) and F´( 3, 3 ).
So ( 2, -5 ) is incorrect and it could be incorrectly reflected across over the x-axis.

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Let u=x^2-4 and v=4x-5. By the product rule,

\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}

By the power rule, we have (u^5)'=5u^4 and (v^4)'=4v^3, but u,v are functions of x, so we also need to apply the chain rule:

\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}

\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}

and we have

\dfrac{\mathrm du}{\mathrm dx}=2x

\dfrac{\mathrm dv}{\mathrm dx}=4

So we end up with

\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3

Replace u,v to get everything in terms of x:

\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3

We can simplify this by factoring:

10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)

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Gala2k [10]
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Read 2 more answers
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Answer:

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474 = 2(s) + 2(s + 13)

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474 = 2s + 2s + 26

Combine like terms

474 = 4s + 26

Subtract 26 from both sides

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Divide both sides by 4

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4 years ago
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