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klasskru [66]
3 years ago
15

Which statements can be used to describe the original functions f(x) and g(x)? Select three options. (See picture attached) PLEA

SE ANSWER QUICKLY I AM TIMED!!!!

Mathematics
1 answer:
NeX [460]3 years ago
6 0

Answer:

THE THIRD ONE

Step-by-step explanation:

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Let y=f(x)=(3x+7)/(x-2) <br> find a rule for f^-1
Murrr4er [49]

Answer:

f^{-1}(x)=\frac{2x+7}{x-3}.

Step-by-step explanation:

The given function is f(x)=\frac{3x+7}{x-2},\:x\ne2.

We want to find a rule for f^{-1}(x).

Let y=\frac{3x+7}{x-2}.

Interchange x and y.

x=\frac{3y+7}{y-2}.

Solve for y.

x(y-2)=3y+7.

Expand

xy-2x=3y+7.

Group y-terms

xy-3y=7+2x.

Factor

(x-3)y=7+2x.

y=\frac{2x+7}{x-3}.

\therefore f^{-1}(x)=\frac{2x+7}{x-3},\:x\ne3.

8 0
3 years ago
HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP DUE SOON
Iteru [2.4K]

Answer:

ez

Step-by-step explanation:

For a all you have to do is multiply by 3 so the answer would be -15

For b all you have to do is subtract 12 then divide by -1 so the answer would be -0.5

for c all you have to do is divide by -10 so the answer would be -0.1/-10 or 0.1/10

7 0
3 years ago
Express cos I as a fraction in simplest terms.​
Serga [27]

Answer:

Solution given

Cos I=adjacent/hypotenuse=16/34

<u>8</u><u>/</u><u>1</u><u>7</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>required</u><u> </u><u>answer</u><u>.</u>

4 0
3 years ago
How to do multiplying radical expressing
8_murik_8 [283]
The answer is
8 \sqrt{6}
. I've demonstrated the 'split and combine' method in the image attached.
This principal is based on the fact that you can multiply and divide radicals as you would with regular integers, but you remember that the answers remain as radicals unless the number you're dealing with is a square (e.g. 9, 16, 25).

6 0
3 years ago
Solve the equation for x, where x is a real number (5 points): <br> -2x^2 + 5x - 10 = 3
KatRina [158]
Subtract 3 from both sides so that the equation becomes -2x^2 + 5x - 13 = 0.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
                                      x = [ -b ± √(b^2 - 4ac) ] / (2a)
                                      x = [ -5 ± √((5)^2 - 4(-2)(-13)) ] / ( 2(-2) )
                                      x = [-5 ± √(25 - (104) ) ] / ( -4 )
                                      x = [-5 ± √(-79) ] / ( -4)
Since √-79 is nonreal, the answer to this question is that there are no real solutions.
3 0
3 years ago
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