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2 years ago

HELP PLEASE I'll mark you brainliest

Mathematics

2 answers:

kogti [31]2 years ago

8 0

The answer is the first one . A

insens350 [35]2 years ago

6 0

The correct answer is A

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Aidan has 20 ft of fence with which to build a rectangular dog run. What is the maximum area he can enclose? Enter your answer i

irina1246 [14]

<h2>Maximum area is 25 m²</h2>

Explanation:

Let L be the length and W be the width.

Aidan has 20 ft of fence with which to build a rectangular dog run.

Fencing = 2L + 2W = 20 ft

L + W = 10

W = 10 - L

We need to find what is the largest area that can be enclosed.

Area = Length x Width

A = LW

A = L x (10-L) = 10 L - L²

For maximum area differential is zero

So we have

dA = 0

10 - 2 L = 0

L = 5 m

W = 10 - 5 = 5 m

Area = 5 x 5 = 25 m²

Maximum area is 25 m²

7 0

2 years ago

4+5(-2x+3)+3x=61<br> if possible could you also show me how to get to the answer?

Ket [755]

Answer:

I need this answer too !!!

4 0

2 years ago

which of the following correctly describes the end behavior of the polynomial function, f(x)=2x^4-3x^2+2x

inna [77]

Answer:

Both ends go up.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Answer Question number 5 <br><br>Plzzzz

Softa [21]

Step-by-step explanation:

Consider x⅓ =a

then the following expression can be written as

a²+a-2

a²+2a-a-2

a(a+2)-1(a+2)

(a+2)(a-1)

a= -2 or, a= 1

Putting the value of a

x⅓ = -2 and, x⅓ =1

6 0

3 years ago

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A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$

5 0

3 years ago