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4 years ago

A square deck has an area of 81 square feet? How long is each side of the deck?

Mathematics

2 answers:

ss7ja [257]4 years ago

4 0

All 4 sides of a square are of the same length.

The formula for the area of a square is s*s = area where s is the length of one side of the square.

s^2 = 81

s = 9 or -9

-9 does not apply here but 9 does.

Each side of the square deck is 9 feet long.

Veseljchak [2.6K]4 years ago

3 0

Formula for area of a square A=s^2 81=s^2 square root of 81 is 9 all the sides are the same 9 ft hope this helps

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3 years ago

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The best estimate for this correlation would be B) 0.9.

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Read more on Brainly.com - brainly.com/question/11162318#readmore

Step-by-step explanation:

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4 years ago

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What is the area of this figure? :)

aliya0001 [1]

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3 years ago

g A window is being built and the bottom is a rectangle and the top is a semi-circle. If there is 12 meters of framing materials

photoshop1234 [79]

Answer:

Semicircle of radius of 1.6803 meters

Rectangle of dimensions 3.3606m x 1.6803m

Step-by-step explanation:

Let the radius of the semicircle on the top=r

Let the height of the rectangle =h

Since the semicircle is on top of the window, the width of the rectangular portion =Diameter of the Semicircle =2r

The Perimeter of the Window

=Length of the three sides on the rectangular portion + circumference of the semicircle

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The area of the window is what we want to maximize.

Area of the Window=Area of Rectangle+Area of Semicircle

$=2hr+\frac{\pi r^2}{2}$

We are trying to Maximize A subject to $2h+2r+\pi r=12$

$2h+2r+\pi r=12\\h=6-r-\frac{\pi r}{2}$

The first and second derivatives are,

Area, A(r) $=2r(6-r-\frac{\pi r}{2})+\frac{\pi r^2}{2}}=12r-2r^2-\frac{\pi r^2}{2}$

Taking the first and second derivatives

$A'\left( r \right) = 12 - r\left( {4 + \pi } \right)\\A''\left( r \right) = - 4 - \pi$

From the two derivatives above, we see that the only critical point of r

$A'\left( r \right) = 12 - r\left( {4 + \pi } \right)=0$

$r = \frac{{12}}{{4 + \pi }} = 1.6803$

Since the second derivative is a negative constant, the maximum area must occur at this point.

$h=6-1.6803-\frac{\pi X1.6803}{2}=1.6803$

So, for the maximum area the semicircle on top must have a radius of 1.6803 meters and the rectangle must have the dimensions 3.3606m x 1.6803m ( Recall, The other dimension of the window = 2r)

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