Let f,g∈F(R) be two functions bounded in R and let h(x)=(f∘g)(x)sinx+g(x)[1+sinx+cos(2x)],∀x∈R Show that h is bounded in R. Please explain any proofs/methodologies listed in the solution so I can better understand it. A detailed solution will result in a thumbs up. Listed below is an example you can use as well as a rubric for how the solution should look. Understand this is how the class is taught and how solutions are meant to look.

The equation shows |h(x)| ≤ C+3B for all x ∈ R, so h is bounded in R by the constant M = C+3B

To show that h is bounded in R, we need to show that there exists a constant M such that

|h(x)| ≤ M for all x ∈ R.

First, note that since f and g are bounded in R, there exist constants A, B, C, and D such that

|f(x)| ≤ A and |g(x)| ≤ B for all x ∈ R,

and

|f(g(x))| ≤ C for all x ∈ R.

As a rubric example,

Now consider the term

(f∘g)(x)sinx.

Since sinx is bounded between -1 and 1, we can write

|(f∘g)(x)sinx| ≤ C |sinx| ≤ C for all x ∈ R.

Similarly, we can write

|g(x)(1+sinx+cos(2x))| ≤ B(1+1+1) = 3B for all x ∈ R.

Therefore, we have

|h(x)| ≤ C+3B for all x ∈ R,

so h is bounded in R by the constant

M = C+3B.

Thus, we have shown that h is bounded in R.

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We have |h(x)| ≤ M1 + 3M2 for all x in R. Let M = M1 + 3M2, then |h(x)| ≤ M for all x in R. Hence, h(x) is bounded in R.

To show that h(x) is bounded in R, we need to find a constant M such that |h(x)| ≤ M for all x in R.

First, note that since f and g are bounded in R, there exist constants M1 and M2 such that |f(x)| ≤ M1 and |g(x)| ≤ M2 for all x in R.

Then, we can bound each term in h(x) separately:

|f(g(x))sin(x)| ≤ M1|sin(x)| ≤ M1

|g(x)(1+sin(x)+cos(2x))| ≤ M2(1+|sin(x)|+|cos(2x)|) ≤ M2(1+1+1) = 3M2

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is -2 9/11 a rational number or irrational number​

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Which of these division problems have the same quotient as 12.3 divided by 0.4? Check all that apply.

A. 123 ÷ 4
B. 1.23 ÷ 0.4
C. 1,230 ÷ 40
D. 123 ÷ 0.04

HELPPP!!!! 50 points​

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Lucas and Mitchell were asked to add the numbers 3.54 and 12.09.


Lucas stated that he completed the addition problem by using the following method:

First, I broke the numbers apart. I broke 3.54 into 3.00+0.50+0.04 and 12.09 into 12.00+0.09.

Then, I combined 0.04 and 0.09 to get 0.13.

Next, I combined 0.13 and 0.50 to get 0.63.

Then, I combined 3.00+12.00 to get 15.00.

Finally, I added the decimal to get 15.63.

Mitchell stated that he completed the addition problem by using the following method:

First, I added the whole numbers to get 3.00+12.00=15.00.

Next, I added the numbers in the tenths place, 0.50+0.00=0.50.

Then, I added the numbers in the hundredths place, 0.04+0.09=0.013.

I then added 0.50 and 0.013 to get 0.513.

Lastly, I added the decimal to the whole number to get 15.513.

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A coordinate grid is shown with x and y axes labeled from 0 to 7 at increments of 1. A straight line joins the ordered pair 0, 1 with the ordered pair 6, 7.


a
1

b
5

c
6

d
7

A. [(-30)+(-5)]+[45÷(-5)]
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E.[23+(-19)]+[(-5)×8×(-3){
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