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Schach [20]
1 year ago
7

ANSWER FIRST=BRAINIEST

Mathematics
2 answers:
Nataly [62]1 year ago
5 0

Answer:

d

Step-by-step explanation:

-4x; 2 x -4 =-8, 1 x -4= -4

Svetllana [295]1 year ago
3 0

Answer:

D

Step-by-step explanation:

Examine each individual column of the table. How do you get from y to x? Divide by -4.

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Will someone explain what i did wrong here?
tensa zangetsu [6.8K]

The actual inverse function is:

f^{-1}(x) = x^2 + 3

And the domain is [0, ∞).

<h3>Where is the mistake?</h3>

Remember that for a given function f(x) with a domain D and a range R.

For the inverse function, f⁻¹(x) the domain is R and the range is D.

Here, for the given function the domain is x ≥ 3 and the range is [0, ∞).

Then for the inverse function, which is:

f^{-1}(x) = x^2 + 3

(to check this, you must have that):

f^{-1}(f(x)) = x\\\\f(f^{-1}(x)) = x

The domain will be [0, ∞) and the range x ≥ 3

If you want to learn more about inverse functions:

brainly.com/question/14391067

#SPJ1

7 0
2 years ago
Help me with this please.
Aneli [31]

6 is the correct answer.

3 0
3 years ago
What is 2.4166666666666 as a fraction
larisa86 [58]
Suppose that 2,4166666 is just 2,416. So the fraction will be 2 in front of a fraction 416 (in nominator) and 1000 ( in denominator) I hooope you understood
5 0
3 years ago
I need you to answer with a, b, c, d
solong [7]

To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:

\begin{gathered} ax^2+bx+c=0 \\  \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}

For the given function:

f(x)=2x^2-10x-3x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}x=\frac{10\pm\sqrt[]{100+24}}{4}\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\  \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\  \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\  \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\  \end{gathered}\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\  \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\  \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}

Then, the zeros of the given quadratic function are:

\begin{gathered} x=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \\  \\ x_{}=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}

Answer: Third option

8 0
1 year ago
How do you work out 5/6y-2=8
aliya0001 [1]
5/6y-2=8
       +2  +2
Add the reciprocal to cancel out the twos andd 8+2=10
so you have left 5/6y=10 and you'll take the reciprocal of 5 divided by 6 times five and cancels the 5/ out and also multiply 10*5 and you're left with 6y=50 then you just divide by six and there you go you have y=12

hope this helped

3 0
3 years ago
Read 2 more answers
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