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3 years ago

If f(x) = 3x - 9 and g(x) = x², what is (gºf)(5)?

Mathematics

1 answer:

Xelga [282]3 years ago

3 0

Answer:

Step-by-step explanation:

Here, f(x) is the input to g(x), forming a composite function.

To evaluate (gºf)(5),

1) find the value of f(5). It is f(5) = 6.

2) Use this result as the input to g(x): g(6) = (6)^2 = 36

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Jennifer and her friends were bored one day and were playing with a tank of water whose dimensions were 48 centimeters long 30 c

Diano4ka-milaya [45]

Draw an illustration describing the problem as shown in the attached picture. The black box drawn represents the tank of water with dimensions labeled 30cm×48cm×36 cm. Jennifer and her friends threw heavy solid objects represented by the blue box. It has dimensions of 12cm×12m×20 cm. These boxes occupy the bottom of the tank up to a height of 28 cm drawn by the blue lines. Now, I think the question will be how many of those heavy solid objects will fill up the bottom of the tank. The other question would be, how much is the volume of water displaced by the heavy solid objects.

First, let's compute the volume of each solid object labeled as v.
v = 12×12×20
v = 2,880 cm³

Next, let's solve the volume to be filled up by these solid objects, labeled as V.
V = 30×48×28
V = 40,320 cm³

This is the volume of water displaced. When Jennifer and her friends drop the solid objects, each volume of object dropped corresponds to the same volume of water that overflows from the tank.

Next, we solve how many solid objects does it take to occupy 40,320 cm³.
Number of blocks = 40,320 cm³/2,880 cm³
Number of blocks = 14

Therefore, it would take 14 heavy solid objects to displace 40,320 cm³ of water.

3 0

4 years ago

Consider the following number line:

JulijaS [17]

Answer:

The pink A goes to the Green C on the number line. the green C goes to the pink A on the number line. The Blue B goes to the Blue B in the number line.

Step-by-step explanation:

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7 0

3 years ago

Read 2 more answers

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

Baxter is thinking about buying a car. The table below shows the projected value of two different cars for three years.

saw5 [17]

Car 1: It is an exponential function that is given as P = 18,500 (0.9595)ⁿ and the price after 10 years is $12,235.5.

Car 2: It is a linear function that is given as 2000x + 3y = 55500 and the price after 10 years is $11,833.33.

And Yes, there is a significant difference.

<h3>What is a function?</h3>

A function is a statement, rule, or law that establishes the connection between two variables. In mathematics, functions are everywhere and are necessary for constructing physical connections.

Baxter is thinking about buying a car. The table below shows the projected value of two different cars for three years.

Car 1 (value in dollars)

Year 1: 18,500

Year 2: 17,390

Year 3: 16,346.60

Car 2 (value in dollars)

Year 1: 18,500

Year 2: 17,500

Year 3: 16,500

The exponential function describes car 1.

Then the function will be

$\rm P = 18500\times (0.95995)^n$

The linear function describes car 2.

$\rm y \ - \ 18500 = \dfrac{-2000}{3}(x - 0)\\\\\\3y - 55500 = -2000x\\\\\\2000x +3y = 55500$

Then the value of the car 1 after 10 years will be

$\rm P = 18500\times (0.95995)^{10}\\\\\\P = \$ \ 12,235.5$

Then the value of the car 2 after 10 years will be

$\rm 2000 \times 10 +3y = 55500\\\\y = \$ \ 11,833.33$

Yes, there is a significant difference.

More about the function link is given below.

brainly.com/question/5245372

#SPJ1

5 0

2 years ago

Which statement could be used to prove to provide that the function f(x)= -4x+6 grows by equal differences over any x-interval o

Ivahew [28]

Find the Greatest Common Factorthe largest number that divides evenly into 4x4x and -6−6?
It is 22.
the highest degree of xx that divides evenly into 4x4x and -6−6?
It is 1, since xx is not in every term.Multiplying the results above,
The GCF is 22.

2(24x−26)


−2(2x−3)

3 0

3 years ago

If f(x) = 3x - 9 and g(x) = x², what is (gºf)(5)?​

If f(x) = 3x - 9 and g(x) = x², what is (gºf)(5)?