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2 years ago

7

the perimeter of a rectangular swimming pool is 86m and the area is 450 what equation represents the perimeter of the swimming p

ool

1 answer:

zlopas [31]2 years ago

7 0

Answer:

Step-by-step explanation:b

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How do you convert 6.5 pounds to a fraction

fomenos

6.5 lbs/100lbs but what is it out of? 6.5 lbs out of what? 6.5/10 6.5/10000000lbs???

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3 years ago

AD=<br> DC=<br> Angle a=<br> Angle b=<br> Angle c=

Nadya [2.5K]

9514 1404 393

Answer:

AD= 8

DC= 15

Angle a= 112°

Angle b= 68°

Angle c= 112°

Step-by-step explanation:

We have to assume the figure is a parallelogram.

Opposite sides of a parallelogram are congruent, so ...

BC = AD = 8

AB = DC = 15

Opposite angles are congruent, and adjacent angles are supplementary.

180° - ∠D = Angle a = Angle c = 112°

∠D = Angle b = 68°

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2 years ago

A triangle has three angles 1x, 11x and 8x. as described below. Find the value of x

Rom4ik [11]

Answer:

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2 years ago

Estimate 7.411×47hhhh

Vinil7 [7]

answer is 348.31 h^4

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3 years ago

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

2 years ago