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Lana71 [14]
3 years ago
8

F(x) = 2x + 3and g(x) = 0 - 2 find (fxg)(-3)

Mathematics
1 answer:
Helga [31]3 years ago
3 0

Step-by-step explanation:

<u>f</u><u>(</u><u>X)</u><u>-</u><u>2</u>

<u>2</u><u>(</u><u>-</u><u>2</u><u>)</u><u>+</u><u>3</u>

<u>-</u><u>1</u><u> </u><u>answer</u>

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A derangement of rooks. Eight mutually antagonistic rooks are placed randomly on a standard chessboard with all arrangements equ
salantis [7]

Answer:

The solution is attached below:

Step-by-step explanation:

4 0
4 years ago
What is the approximate side length of each square
alexandr1967 [171]

Answer: 9\ units

Step-by-step explanation:

You need to use the following formula for calculate the area of a square:

A=s^2

Where "s" is the lenght of any side of the square.

You know that the area of a mural is 81\ units^2. Then, you can substitute this area into the formula and solve for "s", in order to find the side lenght of the mural.

Therefore, this is:

81\ units^2=s^2\\\\s=\sqrt{(81\ units^2)}\\\\s=9\ units

8 0
3 years ago
The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in
Marina86 [1]

Answer:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3

We can express this formula like this:

\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

If we operate and we take out the 1/4 as a factor we got this:

\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}

We can cancel n^2 and we got

\lim_{n\to\infty} \frac{(n+1)^2}{n^2}

We can reorder the terms like this:

\lim_{n\to\infty} (\frac{n+1}{n})^2

We can do some algebra and we got:

\lim_{n\to\infty} (1+\frac{1}{n})^2

We can solve the square and we got:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

3 0
3 years ago
Simplify. 14+{−2+3[1+3(−6−2)]} show how to solve as well with answer
frez [133]
It’s -57 thank me later
3 0
4 years ago
Read 2 more answers
Am I doing these correctly?
Masteriza [31]

Answer:

Yes

Keep it up

7 0
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