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3 years ago

If f(x)=4x^2 and g(x)=x+1, find (fog)(x)

Mathematics

1 answer:

iren [92.7K]3 years ago

5 0

Answer: The value of $(f_\circ g)(x)$ is $4(x+1)^2$ .

Step-by-step explanation:

Given: $f(x) = 4x^2 \text { and } g(x) = x+1$

To find: $(f_\circ g)(x)$

As we know it is composition function which means that g(x) function is in f(x) function.

So we have

$(f_\circ g) (x) = f[g(x)]$

$\Rightarrow( f_\circ g)(x)= f(g(x)) = 4(g(x))^2$

Now substitute the value of g(x) we get

$(f_\circ g)(x)= 4(x+1)^2$

Hence, the value of $(f_\circ g)(x)$ is $4(x+1)^2$ .

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Ana has won a lottery. She was offered two options to receive the award: She can either take it in five installments of $60,000

AleksandrR [38]

Answer:

Results are below.

Step-by-step explanation:

Giving the following information:

She can either take it in five installments of $60,000 annually, starting from now; or she can take a lump-sum of $255,000 now.

First, we determine the value of the 5 installments using a 5% annual compounded rate.

We calculate the future value, and then the present value:

FV= {A*[(1+i)^n-1]}/i

A= annual payments

FV= {60,000*[(1.05^5) - 1]} / 0.05

FV= $331.537.88

PV= FV/(1+i)^n

PV= $259,768.60

At an annual rate of 5% compounded annually, she should choose the five installments instead of the $255,000.

Now, if the annual rate is 6% continuously compounded.

First, we need to calculate the effective interest rate:

r= e^i - 1

r= effective inerest rate

r= e^0.06 - 1

r= 0.0618

FV= {60,000*[(1.0618^5) - 1]} / 0.0618

FV= 339,443.23

PV= 339,443.23/1.0618^5

PV= $251,509.01

At an annual rate of 6% compounded continuously, she should choose the $255,000.

5 0

2 years ago

Drag the tiles to list the sides of △MNO from shortest to longest.

sweet [91]

The smaller the angle subtended by a side, the smaller the length of the

side.

The correct responses are;

Question 1: The list of sides from shortest to longest are;

MO/Shortest MO/Medium and MO/Longest

a) Friday

b) 70 minutes

c) 40%

d) Yes, the sum of the mean number of minutes spent on aerobic training and the mean number of minutes spent on strength training is equal to the mean total number of minutes spent training.

From the given diagram, we have, the measure of the third angle, ∠O, is

found as follows;

∠O = 180° - 54° - 61° = 65°

Therefore, ∠O = The largest angle

We get;

The longest side is opposite the largest angle, which gives;

The shortest side is the side opposite ∠N (54°)= $\frac{}{MO}$

The next shortest side is the side opposite ∠M(61°) = $\frac{}{NO}$

The longest side is the side opposite ∠O(65°) = $\frac{}{MN}$

a) The time spent training on Tuesday = 60 + 10 = 70 minutes

The time spent training on Thursday = 50 + 30 = 80 minutes

The time spent training on Friday = 45 + 40 = 85 minutes

Therefore, the day the athlete spent the longest total amount of time training is on Friday

b) The time spent training on Monday = 10 + 20 = 30 minutes

The time spent training on Wednesday = 20 + 15 = 35 minutes

Therefore, we get;

30, 35, 70, 80, and 85

The median total number of minutes the athlete spent training each day = 70 minutes

c) The time spent strength training = 20 + 10 + 15 + 30 + 45 = 120

The total number of minutes the athlete spent training = 70 + 80 + 85 + 30 + 35 = 300

$The$ $percentage$ $spent$ $on$ $strength$ $training$ = $\frac{120}{300}$ × $100 =$ $\frac{40}$ %

d) The mean number of minutes spent on strength training is found as follows;

$Mean_{strength} =\frac{120}{5} =24$

The mean number of minutes spent on aerobic training is found as follows;

$Mean_{aerobic} =\frac{10+60+20+50+40}{5} =36$

$Mean_{strength} +Mean_{aerobic} =24+36=60$

The mean total number of minutes spent training, $Mean_{total}$ = $\frac{300}{5}$ = 60

Therefore;

$Mean_{strength}+Mean_{aerobic} = Mean_{total} \\$

Learn more here:

brainly.com/question/2962546

4 0

2 years ago

Which month had the biggest difference between the high and low temperatures?A.January

salantis [7]

Answer:
C. October

Explanation:
Because all the other month only have 2-3 units between their high and low October has four units between its high and low

6 0

2 years ago

A dairy company gets milk from two dairies and then blends the milk to get the desired amount of butterfat. Milk from dairy I co

zubka84 [21]

Answer:

a) i The company should buy 40 gallons from dairy I and 60 gallons from dairy

ii) What is the maximum amount of butterfat? The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b.The excess capacity of dairy I is 10 gallons, and for dairy II it is 30 gallons.

Step-by-step explanation:

a. How much milk from each supplier should the company buy to get at most 100 gallons of milk with the maximum amount of butterfat?

From the question, we are told that:

Milk from dairy I costs $2.40 per gallon, Milk from dairy II costs $0.80 per gallon.

Let's represent:

Number of gallons of Milk from dairy I = x

Number of gallons of Milk from dairy II = y

At most $144 is available for purchasing milk.

$2.40 × x + $0.80 × y = 144

2.40x + 0.80y = 144........ Equation 1

x + y = 100....... Equation 2

x = 100 - y

2.40(100 - y) + 0.80y = 144

240 - 2.4y + 0.80y = 144

-1.60y = 144 - 240

-1.6y = -96

y = -96/-1.6

y = 60

From Equation 2

x + y = 100....... Equation 2

x + 60 = 100

x = 100 - 60

x = 40

Therefore, since number of gallons of Milk from dairy I = x and number of gallons of Milk from dairy II = y

The company should buy 40 gallons from dairy I and 60 gallons from dairy

II. What is the maximum amount of butterfat?

From the question

Dairy I can supply at most 50 gallons averaging 3.9% butterfat,

50 gallons = 3.9% butterfat

40 gallons =

Cross Multiply

= 40 × 3.9/50

= 3.12%

Dairy II can supply at most 90 gallons averaging 2.9% butterfat.

90 gallons of milk = 2.9% butter fat

60 gallons of milk =

Cross Multiply

= 60 × 2.9%/90

=1.9333333333%

≈ 1.93%

The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b. The solution from part a leaves both dairy I and dairy II with excess capacity. Calculate the amount of additional milk each dairy could produce.

Excess capacity of Diary I =

50 gallons - 40 gallons = 10 gallons

Excess capacity of Diary II =

90 gallons - 60 gallons = 30 gallons

Therefore, the excess capacity of dairy I is 10 gallons, and for dairy II it is 30 gallons.

3 0

3 years ago

What is the quotient to the problem

Schach [20]

Answer: A quotient is the answer to a division problem. The divisor is the number of parts you divide the dividend by. The dividend is the number you are dividing .

Step-by-step explanation:

7 0

3 years ago