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Rainbow [258]
3 years ago
6

The side length of a cube is √7 cm.

Mathematics
1 answer:
Genrish500 [490]3 years ago
6 0
<span>The answer is The total area of the top and bottom squares. s = √7 cm is irrational value and we need rational value. Let's check all choices: A. The volume. V = (s)^3. If s = √7, V = (√7)^3 = 7√7 - irrational value. B. The perimeter of the front square. P = 4 * s = 4 * √7 = 4√7 - irrational value. C. The diagonal of the back square. d = s√2 = √7 * √2 = √(7*2) = √14 - irrational value. D. The total area of the top and bottom squares. A = 2 * s^2 = 2 * (√7)^2 = 2 * 7 = 14 - rational value.</span>
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Savatey [412]

Answer:

$482.625 ≈ $482.63

Step-by-step explanation:

\frac{y}{450}: \frac{7.25}{100}

y · 100 = 450 · 7.25

100y = 3262.5

100y ÷ 100 = 3262.5 ÷ 100

y = 32.625

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5 0
3 years ago
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<img src="https://tex.z-dn.net/?f=%20%28%7B%20%7Bx%7D%5E%7B2%7D%20%20-%204%7D%29%5E%7B5%7D%20%28%20%7B4x%20-%205%7D%29%5E%7B4%7D
Makovka662 [10]

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)

=2(x^2-4)^4(4x-5)^3(28x^2-57)

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