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3 years ago

Can someone please help me with the square root question?

Mathematics

1 answer:

vovangra [49]3 years ago

7 0

Answer:

a ≈ 9.8 yd³

Step-by-step explanation:

Given: V ≈ 941 yd³

Using the formula: V = a³

Solving for a: A = $V^{\frac{1}{3} }$ = $941^{\frac{1}{3} }$ ≈ 9.79933

Round to the nearest tenth: a ≈ 9.8

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Step-by-step explanation:

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Find a parametric representation for the part of the cylinder y2 + z2 = 49 that lies between the planes x = 0 and x = 1. x = u y

sweet-ann [11.9K]

Answer:

The equation for z for the parametric representation is $z = 7 \sin (v)$ and the interval for u is $0\le u\le 1$ .

Step-by-step explanation:

You have the full question but due lack of spacing it looks incomplete, thus the full question with spacing is:

Find a parametric representation for the part of the cylinder $y^2+z^2 = 49$ , that lies between the places x = 0 and x = 1.

$x=u\\ y= 7 \cos(v)\\z=? \\ 0\le v\le 2\pi \\ ?\le u\le ?$

Thus the goal of the exercise is to complete the parameterization and find the equation for z and complete the interval for u

Interval for u

Since x goes from 0 to 1, and if x = u, we can write the interval as

$0\le u\le 1$

Equation for z.

Replacing the given equation for the parameterization $y = 7 \cos(v)$ on the given equation for the cylinder give us

$(7 \cos(v))^2 +z^2 = 49 \\ 49 \cos^2 (v)+z^2 = 49$

Solving for z, by moving $49 \cos^2 (v)$ to the other side

$z^2 = 49-49 \cos^2 (v)$

Factoring

$z^2 = 49(1- \cos^2 (v))$

So then we can apply Pythagorean Theorem:

$\sin^2(v)+\cos^2(v) =1$

And solving for sine from the theorem.

$\sin^2(v) = 1-\cos^2(v)$

Thus replacing on the exercise we get

$z^2 = 49\sin^2 (v)$

So we can take the square root of both sides and we get

$z = 7 \sin (v)$

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3 years ago