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kicyunya [14]
3 years ago
10

What is this number in standard form?

Mathematics
1 answer:
sdas [7]3 years ago
4 0

Answer:

the answer is 71.092

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0.46 meters

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just devide 46 by 100 (because it's1%), the metric system is quite easy to use

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Ben had 10 football jerseys. After his birthday, he increased his collection by 4 jerseys. Caleb, bens best friend had 6 jersey
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3 years ago
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}

t = 225.8

225.8 years.

8 0
3 years ago
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Brrunno [24]

Answer:

1-11  2-4   3-14  4-9  5-8  6-12  7-15

Step-by-step explanation:plz mark brainliest

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