The actual inverse function is:
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:
(to check this, you must have that):
The domain will be [0, ∞) and the range x ≥ 3
If you want to learn more about inverse functions:
brainly.com/question/14391067
#SPJ1
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
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}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20ax%5E2%2Bbx%2Bc%3D0%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5Cend%7Bgathered%7D)
For the given function:
![x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%28-10%29%5Cpm%5Csqrt%5B%5D%7B%28-10%29%5E2-4%282%29%28-3%29%7D%7D%7B2%282%29%7D)
![x=\frac{10\pm\sqrt[]{100+24}}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B100%2B24%7D%7D%7B4%7D)
![\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}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B124%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B2%5Ccdot2%5Ccdot31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B2%5E2%5Ccdot31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B10%5Cpm2%5Csqrt%5B%5D%7B31%7D%7D%7B4%7D%20%5C%5C%20%20%5Cend%7Bgathered%7D)
![\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\ \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_1%3D%5Cfrac%7B10%7D%7B4%7D%2B%5Cfrac%7B2%5Csqrt%5B%5D%7B31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x_1%3D%5Cfrac%7B5%7D%7B2%7D%2B%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5Cend%7Bgathered%7D)
![\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\ \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_2%3D%5Cfrac%7B10%7D%7B4%7D-%5Cfrac%7B2%5Csqrt%5B%5D%7B31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x_2%3D%5Cfrac%7B5%7D%7B2%7D-%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5Cend%7Bgathered%7D)
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}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B5%7D%7B2%7D%2B%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5C%5C%20%20%5C%5C%20x_%7B%7D%3D%5Cfrac%7B5%7D%7B2%7D-%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Answer: Third option
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
