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Alexxandr [17]
3 years ago
9

What is the value of the expression below? 3.5 x 103

Mathematics
2 answers:
alekssr [168]3 years ago
7 0

Answer:

the box would be 3

Step-by-step explanation:

so if you were to multiply 0.583 x 1000 you would get 583.

1000 has 3 zeros so you can write 1000 as 10^3 (10x10x10 =1000)

if you get a similar question u could just divide 583 by 0.583 which would be 1000 which you should know is 10^3

grigory [225]3 years ago
3 0

Answer:

360.5

Step-by-step explanation:

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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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3 years ago
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ycow [4]

Answer:

7x - 5 = 11

Step-by-step explanation:

To solve for x, substitute y = -5 into the equation 7x + y = 11.

This becomes 7x + (-5) = 11 or 7x - 5 = 11.

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