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3 years ago

F(x) = 2x g(x) = x + 5 h(X) = 3x - 7 What is the correct set up for h(f(x))?? *

Mathematics

2 answers:

Usimov [2.4K]3 years ago

7 0

Answer:

Step-by-step explanation:

-Dominant- [34]3 years ago

3 0

B is correct (3x-7)+5 i hope this helps and u have a great day:)

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Use the general slicing method to find the volume of the following solids. The solid whose base is the region bounded by the cur

Y_Kistochka [10]

Answer:

The volume is $V=\frac{64}{15}$

Step-by-step explanation:

The General Slicing Method is given by

<em>Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on [a, b]. The volume of the solid is</em>

$V=\int\limits^b_a {A(x)} \, dx$

Because a typical cross section perpendicular to the x-axis is a square disk (according with the graph below), the area of a cross section is

The key observation is that the width is the distance between the upper bounding curve $y = 2 - x^2$ and the lower bounding curve $y = x^2$

The width of each square is given by

$w=(2-x^2)-x^2=2-2x^2$

This means that the area of the square cross section at the point x is

$A(x)=(2-2x^2)^2$

The intersection points of the two bounding curves satisfy $2 - x^2=x^2$ , which has solutions x = ±1.

$2-x^2=x^2\\-2x^2=-2\\\frac{-2x^2}{-2}=\frac{-2}{-2}\\x^2=1\\\\x=\sqrt{1},\:x=-\sqrt{1}$

Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is

$V=\int\limits^{1}_{-1} {(2-2x^2)^2} \, dx\\\\V=\int _{-1}^14-8x^2+4x^4dx\\\\V=\int _{-1}^14dx-\int _{-1}^18x^2dx+\int _{-1}^14x^4dx\\\\V=\left[4x\right]^1_{-1}-8\left[\frac{x^3}{3}\right]^1_{-1}+4\left[\frac{x^5}{5}\right]^1_{-1}\\\\V=8-\frac{16}{3}+\frac{8}{5}\\\\V=\frac{64}{15}$

5 0

3 years ago

What is the area of the shaded part?

iVinArrow [24]

Answer:

Step-by-step explanation:

(8x-3)^2-x^2

64x^2-48x+9-x^2

63x^2-48x+9

8 0

2 years ago

A skydiver is falling at about 176 feet per second. How many feet per minute?

bixtya [17]

176 feet per sec × 6o secs in a minute = 10560

3 0

3 years ago

What does the solution to an equation with one solution look like?

mihalych1998 [28]

I believe it's b. Hope this helps

3 0

3 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

Oksana_A [137]

Answer:

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

${n \choose 5 } = \frac{n!}{5!(n-5)!}$

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 ${n \choose 4} .$

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is $4 * {n \choose 4} .$

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is ${n \choose 3} .$

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have ${3 \choose 2 } = 3$ possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of $6 * {n \choose 3}$

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of ${n \choose 2}$ possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.

The total amount of possibilities for this case is

$4 * {n \choose 2}$

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

I hope that works for you!

4 0

3 years ago