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Vikentia [17]
3 years ago
13

Raise 5 to the 3rd power, then find the quotient of the result and s

Mathematics
1 answer:
Art [367]3 years ago
8 0

Answer:

125/s

Step-by-step explanation:

5^3 = 5 x 5 x 5 = 25 x 5 = 125

Find the quotient of the result and s

125 ÷ s

= 125/s

Hope this helps :)

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Write the equation of the line passing through points (2,-2) and (1,0) in slope intercept form.
katrin2010 [14]

Answer:

y=-2x+2

Step-by-step explanation:

hope it helps!

8 0
2 years ago
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If a wheelchair-marathon racer moving at 13.1 miles per hour expends energy at a rate of 645 calories per hour, how much energy
Anna71 [15]

Answer:

1290 calories

Step-by-step explanation:

Given: The racer moves at 13.1 miles per hour expends energy at a rate of 645 calories per hour.

To find: Energy in calories, required to complete a marathon race 26.2 miles at this pace.

Solution: We have,

The racer moves at 13.1 miles per hour.

The racer expends energy at a rate of 645 calories per hour.

So, energy expended while moving 13.1 miles =645 calories.

Now, energy expended while moving 1 mile =\frac{645}{13.1} calories.

So, energy expended while moving 26.2 miles =\frac{645}{13.1}\times 26.2=645\times 2=1290 calories.

Hence, 1290 calories of energy is required to complete a marathon race 26.2 miles at this pace.

6 0
3 years ago
Is 1/4 of 8 equal to 2
Svet_ta [14]
Yes, 1/4 of 8 is equal to two.
1/4x8=2
Hope this helps! :)
5 0
3 years ago
Read 2 more answers
Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species
kotegsom [21]

Answer:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

Step-by-step explanation:

The logistic equation is the following one:

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that P(0) = 80, K = 2000.

The number of fish tripled in the first year. This means that P(1) = 3P(0) = 3(80) = 240.

Using the equation for P(1), that is, P(t) when t = 1, we find the value of r.

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}

280*(2000 + 80(e^{r} - 1)) = 160000e^{r}

280*(2000 + 80e^{r} - 80) = 160000e^{r}

280*(1920 + 80e^{r}) = 160000e^{r}

537600 + 22400e^{r} = 160000e^{r}

137600e^{r} = 537600

e^{r} = \frac{537600}{137600}

e^{r} = 3.91

Applying ln to both sides.

\ln{e^{r}} = \ln{3.91}

r = 1.36

This means that the expression for the size of the population after t years is:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

4 0
3 years ago
What is the answer for this problem if f(x)= 3x+10x and g(x)=5x-3, find (f+g)(x).
hammer [34]
(f+g)(x) =f(x)+g(x)=3x+10x+5x-3=\boxed{18x-3}\ D.
6 0
3 years ago
Read 2 more answers
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