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GalinKa [24]
3 years ago
8

The blueprint scale is 1 inch : 9 feet if the gym's area is 60 square inches on the blueprint, what is the actual square footage

of the gym

Mathematics
2 answers:
netineya [11]3 years ago
6 0
4860 ft^2 that’s the answer and can u help me with some of mine to pls
WITCHER [35]3 years ago
4 0

Answer:

4860 ft^2

Step-by-step explanation:

9 feet is represented by 1 inch.

● 9 => 1

Square both sides

● 81 => 1

So a 81 ft^2 square is represented by a 1 in^2 square (picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The gym's area in the blueprint is 60 in^2

That 60 1in^2 squares.

● 1 => 81

● 1×60 => 81×60

We have multiplied both sides by 60

● 60 => 4860

So the original are is 4860 ft^2

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1. A sum of money is divided between Laura and Joan in the ratio 2:3. Laura’s share is $8. Calculate Joan’s share?
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2 years ago
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The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How
DedPeter [7]

Answer:

a) Q(t) = 180e^{-0.023t}

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

Q(t) = Q(0)e^{-rt}

0.5Q(0) = Q(0)e^{-30r}

e^{-30r} = 0.5

Applying ln to both sides of the equality.

\ln{e^{-30r}} = \ln{0.5}

-30r = \ln{0.5}

r = \frac{\ln{0.5}}{-30}

r = 0.023

So

Q(t) = Q(0)e^{-0.023t}

180-mg sample, so Q(0) = 180

Q(t) = 180e^{-0.023t}

(b) How much of the sample remains after 120 years?

This is Q(120).

Q(t) = 180e^{-0.023t}

Q(120) = 180e^{-0.023*120}

Q(120) = 11.4

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

Q(t) = 180e^{-0.023t}

1 = 180e^{-0.023t}

e^{-0.023t} = \frac{1}{180}

e^{-0.023t} = 0.00556

Applying ln to both sides

\ln{e^{-0.023t}} = \ln{0.00556}

-0.023t = \ln{0.00556}

t = \frac{\ln{0.00556}}{-0.023}

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Answer:

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12(x – 17)

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12*x - 12*17

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