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myrzilka [38]
4 years ago
14

The graphs below have the same shape. What is the equation for the red graph?

Mathematics
2 answers:
Serhud [2]4 years ago
8 0
Your answer would be B because even though it's the same shape the red shape is 2 times bigger.......your exponent 4 stays 4
spayn [35]4 years ago
3 0

<u>Answer-</u>

\boxed{\boxed{g(x)=6-x^4}}

<u>Solution-</u>

Here,

f(x)=3-x^4

As it is given that the graphs have the same shape, so it is neither stretched nor shrank i.e nothing is multiplied with x or y.

So, f(x) must have only translated, in order to get g(x).

As the g(x) is above f(x), so it must have been translated up.

The vertex of f(x) is at (0, 3) and vertex of g(x) is at (0, 6).

Hence, it is translated 3 units upward.

So,

g(x)=f(x)+3

=3-x^4+3

=6-x^4

Therefore, the equation of g(x) is g(x)=6-x^4

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3 years ago
Need help on these, thank you!
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Answer:

5x+2 , 2x^2+5x-2,\frac{x^2+5x}{2-x^2}

Step-by-step explanation:

We are given f(x) and g(x)

1. (f+g)(x)

(f+g)(x) = f(x) + g(x)

           = x^2+5x+2-x^2

           = 5x+2

Domain : All real numbers as it there exists a value of (f+g)(x) f every x .

2. (f-g)(x)

(f-g)(x) = f(x)-g(x)

          = x^2+5x-2+x^2

          =2x^2+5x-2

Domain : All real numbers as it there exists a value of (f-g)(x) f every x .

Part 3 .

(\frac{f}{g})(x)\\(\frac{f}{g})(x) = \frac{f(x)}{g(x)}\\=\frac{x^2+5x}{2-x^2}

Domain : In this case we see that the function is not defined for values of x for which the denominator becomes 0 or less than zero . Hence only those values of x are defined for which

2-x^2>0

or 2>x^2

   Hence taking square roots on both sides and solving inequality we get.

-\sqrt{2}

8 0
4 years ago
Jack sold 14 adult tickets and 20 student tickets. the adult tickets cost $ 10 and the student tickets cost $ 5. how much money
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3 years ago
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Answer:

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Step-by-step explanation:

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