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konstantin123 [22]
3 years ago
7

Reduce -56/-70 to lowest terms •-2/3 •-4/5 •4/5 •8/10

Mathematics
2 answers:
Kaylis [27]3 years ago
4 0
4/5 is the right answer.....
d1i1m1o1n [39]3 years ago
3 0

Answer:

option c, 4/5

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A village experienced 2% population growth, compounded continuously, each year for 10 years. At the end of the 10 years, the pop
pantera1 [17]

Answer:

The initial population at the beginning of the 10 years was 129.

Step-by-step explanation:

The population of the village may be modeled by the following function.

P(t) = P_{0}e^{rt}

In which P is the population after t hours, P_{0} is the initial population and r is the growth rate, in decimal.

In this problem, we have that:

P(10) = 158, r = 0.02.

So

158 = P_{0}e^{0.02*10}

P_{0} = 158*e^{-0.2}

P_{0} = 129

The initial population at the beginning of the 10 years was 129.

6 0
3 years ago
Read 2 more answers
Which answer choice shows the number of hours in 51/2 days
liq [111]
Hey there,

The answer would be 132 hours..There is 24hours in a day so you multiply 24•5=120 then a half if a day is 12 hours, so you then add 12 to 120 which you get 132..
4 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
3 years ago
The number of games in the first round of a chess tournament is equal to 2×2×2×2×2×2. Write the number of games using an exponen
Naily [24]

Answer:

2^6

Step-by-step explanation:

2 is being multiplied 6 times meaning it would be 2 to the 6th power.

8 0
3 years ago
two rectangular properties share a common side. Lot is 33 feet wide and 42 feet long. The combined area of the lots is 1,848 squ
klio [65]

First, find the area of lot A.

To find the area, multiply the length by the width.

33*42=1,386

Now, subtract the area of lot a from the total area. This will give us the area of lot b.

1,848-1,386=462

The area of lot b is 462 square feet.

Finally, divide the area of lot b (462) by its length to find the width. Since they both share a side, we know that it’s length is 42 feet.

462/42=11

Lot b is 11 feet wide.

Hope this helps!

4 0
3 years ago
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