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Anni [7]
3 years ago
15

A circular pool if filling with water. Assuming the water level will be 4' deep and the diameter is 20' , what is the volume of

the water needed to fill the pool?
Mathematics
1 answer:
Yuliya22 [10]3 years ago
8 0

35,569.92 L of water is required to fill the pool

Step-by-step explanation:

  • Step 1: Find the area of the circular pool when radius = 20/2 = 10 ft

⇒ Area = πr² = 3.14 × 10² = 314 ft²

  • Step 2: Find the volume of the pool

Volume = Area × Depth

             = 314 × 4 = 1256 ft³

  • Step 3: Convert the volume into liters.

Volume of water = 1256 × 28.32 = 35,569.92 L (∵1 ft³ = 28.32 L)

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Jaden and cadence share a reward of $140 in a ratio of 2/5. what fraction of the total reward does Jaden get? how do you know?
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5 0
3 years ago
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If 5 + 20 times 2 Superscript 2 minus 3 x Baseline = 10 times 2 Superscript negative 2 x Baseline + 5, what is the value of x? –
sp2606 [1]

Answer:

<h3>Option D) 3 is correct</h3><h3>Therefore the value of x is 3</h3>

Step-by-step explanation:

Given equation is 5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5

<h3>To find the value of x :</h3>

First solving the given equation we have,

5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5

5+20\times (2)^{2-3x}-5=10\times (2)^{-2x}+5-5

20\times (2)^{2-3x}=10\times (2)^{-2x}

20\times (2)^2.(2)^{-3x}=10\times (2)^{-2x}

20\times 4.(2)^{-3x}=10\times (2)^{-2x}

80(2)^{-3x}=10\times (2)^{-2x}

\frac{80}{10}(2)^{-3x}=\frac{10\times (2)^{-2x}}{10}

8(2)^{-3x}=(2)^{-2x}

\frac{(2)^{-3x}}{(2)^{-2x}}=\frac{1}{8}

(2)^{-3x}.(2)^{2x}=\frac{1}{2^3} ( by using the property \frac{1}{a^{-m}}=a^m )

2^{-3x+2x}=\frac{1}{2^3} ( by using the property a^m.a^n=a^{m+n} )

2^{-x}=\frac{1}{2^3}

\frac{1}{2^x}=\frac{1}{2^3} ( by using the property a^{-m}=\frac{1}{a^m} )

Since bases are same so powers are same

Therefore we can equate the powers we get x=3

<h3>Therefore the value of x is 3</h3><h3>Option D) 3 is correct</h3>
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