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3 years ago

A swimming pool at the local park is 25 ft long ,20 ft wide,and 5 ft deep how much water can the pool hold

2 answers:

7 0

So you have to remember a formula, which is V=lwh

So, what you do is V=25×20×5, which is equal to 2500 ft³
So your answer is 2500ft³

Mama L [17]3 years ago

7 0

Solving this problem, you would need to follow the volume formula (v=lwh)

L= 25 ft.
W= 20 ft.
H= 5 ft.

V=lwh
V= (25)•(20)•(5)
V= 2,500 ft

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someone please explain to me how to find the area PLEASE! Find the area and perimeter of quadrilateral ABCD below. Explain your

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Answer:

Part A) The area of the figure is $24\ units^{2}$

Part B) The perimeter of the figure is $20\ units$

Step-by-step explanation:

step 1

Find the area of the figure

we know that

The area of the figure is equal to the area of triangle ABD plus the area of triangle BCD

The area of triangle is equal to

$A=\frac{1}{2}bh$

<u>Area of triangle ABD</u>

Observing the graph

$b=BD=(-2+8)=6\ units$

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substitute

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<u>Area of triangle BCD</u>

Observing the graph

$b=BD=(-2+8)=6\ units$

$h=(5-1)=4\ units$

substitute

$A=\frac{1}{2}(6)(4)=12\ units^{2}$

The area of the figure is

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step 2

Find the perimeter of the figure

we know that

The perimeter of the figure is equal to

$P=AB+BC+CD+AD$

we have

$A(-5,1),B(-8,5),C(-5,9),D(-2,5)$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$d=\sqrt{(5-1)^{2}+(-8+5)^{2}}=5\ units$

Find the distance BC

$d=\sqrt{(9-5)^{2}+(-5+8)^{2}}=5\ units$

Find the distance CD

$d=\sqrt{(5-9)^{2}+(-2+5)^{2}}=5\ units$

Find the distance AD

$d=\sqrt{(5-1)^{2}+(-2+5)^{2}}=5\ units$

substitute the values

$P=5+5+5+5=20\ units$

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3 years ago

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