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AnnZ [28]
2 years ago
14

Solve for m: 2 (m + 7) = -4- m Your answer

Mathematics
2 answers:
Leokris [45]2 years ago
8 0

Answer:

m = -6

Step-by-step explanation:

2 (m + 7) = -4 - m

2m + 14 = -4 - m

3m + 14 = -4

3m = -18

m = -6

Hopefully this helps!

Brainliest please?

frutty [35]2 years ago
8 0

Answer:

m = -6

Step-by-step explanation:

To solve this equation for m, we multiply out the left side to remove the parentheses:

2m + 14 = -4 - m

Next, we combine like terms:  2m and -m are like terms, as are 14 and -4:

3m = -18

Dividing both sides by 3 yields m = -6

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Step-by-step explanation:

13+x>11

Subtract 13 from each side

13-13+x>11-13

x>-2

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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

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