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solong [7]
2 years ago
9

Ryan wrote a total of 8 pages over 2 hours. How many hours will Ryan have to spend writing

Mathematics
1 answer:
Reika [66]2 years ago
6 0

Answer:

12 Hours

Step-by-step explanation:

So it says he writes 8 pages in 2 hours. Which would mean he takes an hour to write 4 pages. You divide 48 pages by 4 hours and you get 12.

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Gerard ran out of tile for his patio.The width of the remaining area 2 2/9 feet.The length of the remaning area is 7 feet.How mu
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Length X Width =
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2/9 x 7 = 44/9
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He has 18 8/9 feet left to tile.
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5) 5 ft<br> A) 38 ft<br> C) 18 ft<br> B) 15<br> D) 16 ft
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the answer is a

Step-by-step explanation:

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the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
2 years ago
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