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ella [17]
4 years ago
7

Find the area of each polygon. 4 ft 14 ft 10ft 18ft

Mathematics
1 answer:
Anna [14]4 years ago
5 0

Answer: 112 square feet

Step-by-step explanation:

(4*10)+(18*4)= 40+72=112

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The function f(x), shown in the graph, represents an exponential growth function. Compare the average rate of change of
svetoff [14.1K]

Answer:0

Step-by-step explanation:

6 0
3 years ago
Help me please. <br> F(x) = 4 - 3x <br> f(10) =
tigry1 [53]

Answer:

f(10) = - 26

Step-by-step explanation:

To evaluate f(10), substitute x = 10 into f(x), that is

f(10) = 4 - 3(10) = 4 - 30 = - 26

8 0
3 years ago
ASAP HELP<br> 10.Simplify the following expression: 2n+4-0.3n-3
Oksana_A [137]
1.7n +1
I think the is the answer
I hope this helps :)
7 0
3 years ago
Pls help find the amount at the end of 6 hours
marshall27 [118]

Answer:

A. 7,348

Step-by-step explanation:

P = le^kt

intitial population = 500

time = 4 hrs

end population = 3,000

So we have all these variables and we need to solve for what the end population will be if we change the time to 6 hours. First, we need to find the rate of the growth(k) so we can plug it back in. The given formula shows a exponencial growth formula. (A = Pe^rt) A is end amount, P is start amount, e is a constant that you can probably find on your graphing calculator, r is the rate, and t is time.

A = Pe^rt

3,000 = 500e^r4

now we can solve for r

divide both sides by 500

6 = e^r4

now because the variable is in the exponent, we have to use a log

log_{e}(6) = 4r

ln(6) = 4r

we can plug the log into a calculator to get

1.79 = 4r

divide both sides by 4

r = .448

now lets plug it back in

A = 500e^(.448)(6 hrs)

A = 7351.12

This is closest to answer A. 7,348

4 0
3 years ago
Suppose a tank contains 400 gallons of salt water. If pure water flows into the tank at the rate of 7 gallons per minute and the
Strike441 [17]

Answer:

Step-by-step explanation:

This is a differential equation problem most easily solved with an exponential decay equation of the form

y=Ce^{kt}. We know that the initial amount of salt in the tank is 28 pounds, so

C = 28. Now we just need to find k.

The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is \frac{dy}{dt}. Thus, the change in the concentration of salt is found in

\frac{dy}{dt}= inflow of salt - outflow of salt

Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:

3(\frac{y}{400})

Therefore,

\frac{dy}{dt}=0-3(\frac{y}{400}) or just

\frac{dy}{dt}=-\frac{3y}{400} and in terms of time,

-\frac{3t}{400}

Thus, our equation is

y=28e^{-\frac{3t}{400} and filling in 16 for the number of minutes in t:

y = 24.834 pounds of salt

6 0
3 years ago
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