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zubka84 [21]
3 years ago
7

Solve and check 2/3y =6 A.1/9 B.1/4 C.9 D.4

Mathematics
2 answers:
kotegsom [21]3 years ago
8 0
\frac{2}{3}y=6\ \ \ \ |divide\ both\ sides\ by\ \frac{2}{3}\ /multiply\ by\ \frac{3}{2}/\\\\y=6\cdot\frac{3}{2}\\\\y=\frac{6}{1}\cdot\frac{3}{2}\\\\y=\frac{\not6^3\cdot3}{1\cdot\not2_1}\\\\y=\frac{3\cdot3}{1\cdot1}\\\\y=\frac{9}{1}\\\\\boxed{y=9}\leftarrow answer\ \boxed{C}

check:\\\\/put\ the\ value\ of\ y=9\ to\ the\ equation\ \frac{2}{3}y=6/\\\\\frac{2}{3}\cdot9=6\\\\\frac{2}{3}\cdot\frac{9}{1}=6\\\\\frac{2}{\not3_1}\cdot\frac{\not9^3}{1}=6\\\\\frac{2}{1}\cdot\frac{3}{1}=6\\\\2\cdot3=6\\\\\boxed{6=6}\leftarrow TRUE


Archy [21]3 years ago
5 0

\frac{2}{3} *y=6  \\  \\ y=6: \frac{2}{3} =6* \frac{3}{2} = \frac{18}{2} =9



So the answer is C. 9

2/3* 9=(2*9)/3=18/3= 6 <em>(true)</em>

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Step-by-step explanation:

1. Let's review the information provided to us for solving the questions:

Power capacity of the wind farms = 2,200 Megawatts or 2.2 Gigawatts

2. Let's resolve the questions A and B:

Part A

Assuming wind farms typically generate 25​% of their​ capacity, how much​ energy, in​ kilowatt-hours, can the​ region's wind farms generate in one​ year?

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Given that the average household in the region uses about​ 10,000 kilowatt-hours of energy each​ year, how many households can be powered by these wind​ farms?

For calculating the amount of households we divide the total amount of energy the wind farms can generate (4,818'000,000 kilowatt-hours) and we divide it by the average household consumption (10,000 kilowatt-hours)

<u>Amount of households =  4,818'000,000/10,000 = 481,800</u>

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