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Harlamova29_29 [7]
3 years ago
11

for each month of a year Selena saving an extra $100 from her paycheck by the end of the year she had saved 1800 write and solve

an equation to determine how much she took typically safe from her paycheck
Mathematics
2 answers:
maxonik [38]3 years ago
8 0
100+1800 would be good
kipiarov [429]3 years ago
5 0

Answer: It took 18 months to Selena to save $1800 from her paycheck.

Step-by-step explanation:

Hi, to answer this question we have to write an equation.

The amount of money she saves per month (100) multiplied by the number of months (m) must be equal to $1800.

m 100 = 1800

Solving for m

m = 1800/100

m = 18 months

It took 18 months to Selena to save $1800 from her paycheck.

Feel free to ask for more if needed or if you did not understand something.

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Bond [772]

Answer:

Step-by-step explanation:

608 square inches

5 0
3 years ago
Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea
Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

d) 0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}

And if we simplify we got this:

p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we simplify we got:

p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And if we set equal to 0 we got:

0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

7 0
3 years ago
HELP PLEASE
Jet001 [13]

Answer:

c $11,751

Step-by-step explanation:

The function that models Alicia's car is

f(x) = 21000(0.93)^{x}

where x represents the number of years since she purchased the car.

We want to find the value of Alicia's car after 8 years.

We substitute x=8 into the formula to get:

f(8) = 21000(0.93)^{8}

f(8) = 21000(0.5596)

f(8) = 11751.218

We round to the nearest dollar to get:

f(8) =11751

The correct answer is C.

5 0
3 years ago
Can Someone plz help me with this question? <br> I will mark brainliest
Gala2k [10]
The answer to your question is

No
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2 years ago
The graph below represents the rate at which Melanie listens to songs. What is the equation that represents the relationship bet
alexira [117]

Answer:

4x

Step-by-step explanation:

So what you wanna do is look everywhere a point is on a line perfectly. Then see how much it takes per point that's on a line perfectly

8 0
2 years ago
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