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Bumek [7]
3 years ago
12

Solve F(x) for the given domain. Include all of your work in your final answer. Submit your solution.

Mathematics
2 answers:
amm18123 years ago
5 0
Hello : 
<span>F(x) = x 2 + 2 the domain is : R
F(x - h) = (x - h )² +2 = x² - 2xh +h² +2</span>
Anna71 [15]3 years ago
4 0

Answer:

As per the given statement:

F(x) = x^2+2     for all x belongs to R              ......[1]

Then, find F(x-h);

Substitute x-h in place of x in [1] we get;

F(x-h) = (x-h)^2+2

Using identity rule:

(a-b)^2 = a^2 +b^2 -2ab

Apply this rule we have;

F(x-h) = x^2+h^2-2xh+2

Therefore, final answer is, F(x-h) = x^2+h^2-2xh+2

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Galina-37 [17]

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7

Step-by-step explanation:


3 0
3 years ago
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Find the zeros of the function f (x) = 4x^2-16
s2008m [1.1K]

Answer:

x=-2

x=2

Step-by-step explanation:

We can think of this equation like a statement

"4 times what squared minus sixteen is equal to zero"

First, we need to rearrange the equation:

4x^2=16

Now using our common knowledge, we know that 4 times 4 is equal to 16

now we have another statement:

"x times x is going to be equal to 4"

*Remember that the two numbers have to be the same since x is squared*

Again, we can use common knowledge and say that 2x2 is equal to 4

This means that x=2

Now when you square a variable, the variable can have to values, a positive, and a negative value

So in this case, the other value for x would be x=-2 since -2 squared is also equal to 4

So we can come to a conclusion that

x=2

AND

x=-2

4 0
2 years ago
50 x 2 x 34 x5 who has insta if so what is it and btw you dont have to answer this
nignag [31]
Are you a guy or a girl
6 0
3 years ago
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Chapter Name :-
iragen [17]

The value for A^{20226397596} is        \left[\begin{array}{cc}\omega^{20226397596}&0\\0&\omega^{20226397596}\end{array}\right].

<h3>Procedure - Determination of the power of a matrix</h3>

Let be A a <em>diagonal</em> matrix. By linear algebra we know that the n-th power of <em>diagonal</em> matrix of the form \left[\begin{array}{cc}\omega&0\\0&\omega\end{array}\right] is equal to:

A^{n} = \left[\begin{array}{cc}\omega^{n}&0\\0&\omega^{n}\end{array}\right], \forall\, n \ge 1, n\in \mathbb{N} \forall \,\omega \in \mathbb{R} (1)

Hence, we have the following result for A^{20226397596}:

A^{20226397596} = \left[\begin{array}{cc}\omega^{20226397596}&0\\0&\omega^{20226397596}\end{array}\right]

The value for A^{20226397596} is        \left[\begin{array}{cc}\omega^{20226397596}&0\\0&\omega^{20226397596}\end{array}\right]. \blacksquare

To learn more on matrices, we kindly invite to check this verified question: brainly.com/question/4470545

4 0
2 years ago
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Solve this equation 3p-6&gt;21
Rus_ich [418]
3p - 6 > 21

First, our goal is to get the variable (p) by itself on one side of the problem. We can start out by adding 6 to each side.
3p \textgreater 21 + 6

Second, we can now add 21 + 6 which can easily be done by either counting on your fingers, in your head, or using a calculator. 21 + 6 = 27.
3p \textgreater 27

Third, our next step is to divide each side by 3. After doing this, the variable is now standing alone. 
p \textgreater  \frac{27}{3}

Fourth, our last step is to simplify the fraction. To do so, what multiples into 3 to get 27? The answer is 9. (9 times 3 = 27).
p \textgreater 9

Answer: \fbox {p \textgreater 9}
6 0
3 years ago
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