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larisa [96]
4 years ago
13

One hundred grams of radium are stored in a container. The amount RR (in grams) of radium present after tt years can be modeled

by R=100e−0.00043tR=100e−0.00043t. After how many years will only 5 grams of radium be present? Round your answer to the nearest whole year.
Mathematics
1 answer:
Alona [7]4 years ago
5 0
Ending Amount = Beginning Amount * e^k*t
5 grams = 100 grams * e^-.00043 * years

We have to solve the equation for "years"
(You can find the formula here: http://www.1728.org/halflif2.htm
or you can just read the next line

time = natural log (bgng amt / endg amt) / k
time = natural log (100 / 5) / .00043
time = natural log (20) / .00043
time = 2.9957322736 / .00043
time =  <span> <span> <span> 6,967</span></span></span><span> years
</span>
We should double-check this.
First, we need the half-life
k = ln(.5) / half-life
half-life = <span>-.693147 / -.00043
half-life = 1,612 years
Now let's see how many half-lives that is:
</span><span>6,967 / 1,612 years = 4.2 half-lives
So basically, after 4 half-lives the mass should go from

100
50  1 half life
25</span><span><span>  2 half lives</span>
12.5 </span><span>3 half lives
</span> 6.25 <span>4 half lives</span>

6.25 is very close to 5 grams so we can assume we have calculated correctly.


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The sum of the roots of 2x squared + 8x -3=0 is?
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4 0
3 years ago
Let R be the region bounded by
loris [4]

a. The area of R is given by the integral

\displaystyle \int_1^2 (x + 6) - 7\sin\left(\dfrac{\pi x}2\right) \, dx + \int_2^{22/7} (x+6) - 7(x-2)^2 \, dx \approx 9.36

b. Use the shell method. Revolving R about the x-axis generates shells with height h=x+6-7\sin\left(\frac{\pi x}2\right) when 1\le x\le 2, and h=x+6-7(x-2)^2 when 2\le x\le\frac{22}7. With radius r=x, each shell of thickness \Delta x contributes a volume of 2\pi r h \Delta x, so that as the number of shells gets larger and their thickness gets smaller, the total sum of their volumes converges to the definite integral

\displaystyle 2\pi \int_1^2 x \left((x + 6) - 7\sin\left(\dfrac{\pi x}2\right)\right) \, dx + 2\pi \int_2^{22/7} x\left((x+6) - 7(x-2)^2\right) \, dx \approx 129.56

c. Use the washer method. Revolving R about the y-axis generates washers with outer radius r_{\rm out} = x+6, and inner radius r_{\rm in}=7\sin\left(\frac{\pi x}2\right) if 1\le x\le2 or r_{\rm in} = 7(x-2)^2 if 2\le x\le\frac{22}7. With thickness \Delta x, each washer has volume \pi (r_{\rm out}^2 - r_{\rm in}^2) \Delta x. As more and thinner washers get involved, the total volume converges to

\displaystyle \pi \int_1^2 (x+6)^2 - \left(7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \pi \int_2^{22/7} (x+6)^2 - \left(7(x-2)^2\right)^2 \, dx \approx 304.16<em />

d. The side length of each square cross section is s=x+6 - 7\sin\left(\frac{\pi x}2\right) when 1\le x\le2, and s=x+6-7(x-2)^2 when 2\le x\le\frac{22}7. With thickness \Delta x, each cross section contributes a volume of s^2 \Delta x. More and thinner sections lead to a total volume of

\displaystyle \int_1^2 \left(x+6-7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \int_2^{22/7} \left(x+6-7(x-2)^2\right) ^2\, dx \approx 56.70

7 0
2 years ago
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