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natka813 [3]
3 years ago
6

Help me please i need some help.

Mathematics
1 answer:
photoshop1234 [79]3 years ago
4 0

Answer:

8, -5, -4, 6

Step-by-step explanation:

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He will need 109 tiles
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ASAP can someone help me solve this problem​
sveta [45]

Answer:

Step-by-step explanation:

x° + y° = 180°

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3 0
3 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

5 0
2 years ago
Please help!! Just with 13 please. please try explain it and show work if possible, I really dont understand ​
Mashcka [7]

Answer:

0.0987

Step-by-step explanation:

8 0
3 years ago
Multiply (x + 1)^3
Shalnov [3]
<h3>Answer: x3 +2 +x+x+</h3>

Step-by-step explanation:

6 0
3 years ago
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