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stealth61 [152]
3 years ago
12

45.5 meters in 13 second

Mathematics
2 answers:
expeople1 [14]3 years ago
5 0
45.5 meters in 13 seconds : so all u do is divide 45.5 by 13 
 
and 45.5 divided by 13 = 3.5

hope this helps!
joja [24]3 years ago
3 0

Answer:

3.5 meters/second

Step-by-step explanation:

45.5 meters in 13 second

We will divide 45.5 by 13 to get unit rate.

Speed = \frac{Distance}{time}

           =   \frac{45.5}{13}

           = 3.5 meters/second

3.5 meters per second.

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3 years ago
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The point(3,7) is reflected in the y axis what are the coordinates now
lapo4ka [179]
When a point P(a, b) is reflected about the y-axis, the coordinates of the reflected point are P'(-a, b).

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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

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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}

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

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3 years ago
Solve for h.<br><br> –2 − 10h = –11h − 20
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。☆✼★ ━━━━━━━━━━━━━━  ☾  

-2 - 10h = -11h - 20

+ 2

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+ 11h

h = -18

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