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3 years ago

14

The principal of East High School wants to buy a new cover for the sand pit used in the long-jump competition. He measured the s

and pit and found that the length is 29.2 feet and the width is 9.8 feet. What will the area of the new cover be?

1 answer:

Ganezh [65]3 years ago

3 0

Answer: area is 286.16 ft²

Step-by-step explanation:

Length of the sand pit is 29.2 feet and the width is 9.8 feet.

L= 29.2 ft

W = 9.8 ft

To find out the area, all we have to do is multiply the lenght by the width:

And do not forget that to calculate the area, all units must be the same and for the area, it is always the unit squared, in this case, feet square or simply (ft²).

A = l*w

A = 29.2 * 9.8 = 286.16 ft²

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Hi there!

Since we have 2 variables in the given equation, and you only want to solve for one, the other will end up in your answer.

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Step 1) Add 8y to both sides.

$6x - 8y + 8y = 12 + 8y 6x = 8y + 12$

Step 2) Divide both sides by 6.

$\frac{6x}{6} = \frac{8y + 12}{6}$

$x = \frac{4}{3} y + 2$

Final Answer -

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if you can get 6 games online for $33.60 that would be $5.60 per game. If you buy from the toy store it would cost $17.36 more than online

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Express the number 220 as the sum of four numbers that form a geometric progression such that the third term is greater than the

siniylev [52]

Answer:

Step-by-step explanation:

The first four terms of geometric series is:

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Since, we have given information that the third number is greater than 44 that means:

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Now, using:

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Here, a=r,b=1

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The sum of first four terms is:

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$a(1+r+r^2+r^3)=220$

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Divide equation (2) by (1) we get:

$\frac{r^2+1}{r-1}=\frac{220}{44}$

$r^2+1=5r-5$

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$\Rightarrow r^2-3r-2r+6=0$

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$a(4)=a+44$

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$a(3)^2=a+44$

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From CASE2: $\frac{11}{2},\frac{11\cdot 3}{2},\frac{11\cdot 3^2}{2},\frac{11\cdot 3^3}{2}$

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Answer:

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