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galina1969 [7]
3 years ago
11

Find the slope of the line from the table:

Mathematics
1 answer:
Nostrana [21]3 years ago
7 0

Answer:

Y =  3x - 1

Step-by-step explanation:

Y = mx + b    

Y =  3   - 1

       ---

        1

Y =  3x - 1

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Step-by-step explanation:

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Reflect the point (7,0) across the y-axis
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4 years ago
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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3 years ago
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Answer:

If I read this right

3 (n -12) + 14 = 4 (7 - 5n)

First, remove the parentheses

3 n - 36 + 14 = 28 - 20 n

Combining terms

23 n = 28 - 14 + 36 = 50

n = 50 / 23 = 2 4/23 = 2.17

7 0
3 years ago
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