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siniylev [52]
4 years ago
8

The only swimming pool at the El Cheapo Motel is outdoors. It is 5.0 m wide and 12.0 m long. If the weekly evaporation is 2.35 i

n., how many gallons of water must be added to the pool if it does not rain?
Mathematics
1 answer:
Alika [10]4 years ago
3 0

Answer:

946.10 gallons per week

Step-by-step explanation:

1 cm = 0.393701 inch

Width = 5.0m = 196.85 inch

Length = 12.0 m = 472.44 inch

The volume evaporated weekly is given by:

V = L*W*2.35 = 196.85*472.33*2.35\\V=218,550.12\ in^3

Converting to gallons:

1\ gal = 231\ in^3\\V=\frac{218,550.12}{231}\\V= 946.10\ gal

946.10 gallons of water must be added to the pool each week.

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Answer:

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Step-by-step explanation:

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3 years ago
Marcus looked up the path that he had taken on his run this morning on the map shown below.
AURORKA [14]

Answer: 6 miles.


Step-by-step explanation:

1. To solve this problem you only need to add the distances shown in the map attached, which shown the path Marcus had taken. Then, you have:

Total_{distance}=1in+3inches+2inches+2.3inches\\Total_{distance}=8.3inches

2. Now, you must convert the result obtained from inches to miles. According to the problem 4inches=3miles, therefore, you have:

Total_{distance}=\frac{8,3inches*3miles}{4inches}\\Total_{distance}=6.22miles

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6 0
4 years ago
Read 2 more answers
Given f(x) = 52x, evaluate f-1), f(o), and K2).<br><br> 01/25, 1, 625
lbvjy [14]

Answer:

f(-1) = \frac{5}{2}

f(0) = 5

f(2) = 20

Step-by-step explanation:

Given

f(x) = 5(2^x)

Required

Determine f(-1); f(0) and f(2)

Solving f(-1)

In this case, we simply take x as;

x = -1

Substitute -1 for x in f(x) = 5(2^x)

f(-1) = 5(2^{-1})

Apply law of indices

f(-1) = 5 * \frac{1}{2}

f(-1) = \frac{5}{2}

Solving f(0)

In this case, we simply take x as;

x = 0

Substitute 0 for x in f(x) = 5(2^x)

f(0) = 5(2^0)

f(0) = 5(1)

f(0) = 5

Solving f(2)

In this case, we simply take x as;

x = 2

Substitute 2 for x in f(x) = 5(2^x)

f(2) = 5(2^2)

f(2) = 5(4)

f(2) = 20

4 0
4 years ago
At the start of a year, company XYZ's stock is $40 per share. At the end, the company's stock is $60 per share. What was the sto
e-lub [12.9K]

We have been given that at the start of a year, company XYZ's stock is $40 per share. At the end, the company's stock is $60 per share. We are asked to find the percent increase.

We will use percent increase formula to solve our given problem.

\text{Percent increase}=\frac{\text{Final}-\text{Initial}}{\text{Initial}}\times 100\%

\text{Percent increase}=\frac{60-40}{40}\times 100\%

\text{Percent increase}=\frac{20}{40}\times 100\%

\text{Percent increase}=0.5\times 100\%

\text{Percent increase}=50\%

Therefore, the stock price's rate of return was 50%.

4 0
4 years ago
Marcia claims that the GCF for (2x2 + 4xy + 8xy4) is 8x2y4.
miskamm [114]
She is not correct.
The GCF for 2x^2 + 4xy + 8xy^4
is 2x
The first term does not have a 'y' in it so you cannot have a 'y' in the GCF.
The second and third terms both only have 1 'x', so you can only take out 1 'x', not 2.
The highest constant that all three terms have in common is a 2.
8 0
4 years ago
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