X+2y=10. x+y=6 SOLVE for the value of x and y.

1 answer:

7 0

<span>X+2y=10 ----Eq.1
</span><span>x+y=6---------Eq.2
from Eq2.
x=6-y
Subtituting the value of X in the first equation
6-y+2y=10
6+y=10
y=4
Plug y=4 in equation 2.
</span><span>x+4=6
</span>X= 2

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Find g^-1(x) <br><br> A<br> B<br> C<br> D

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<h2>Answer:</h2>

$a. \ g^{-1}=-1 \pm\sqrt{\frac{3}{x}+1}$

<h2>Step-by-step explanation:</h2>

A function has an inverse function if and only if passes the Horizontal Line Test for Inverse Functions. This test tells us that a function $f$ has an inverse function if and only if there is no any horizontal line that intersects the graph of $f$ at more than one point. So the function is called one-to-one. The graph of $g$ is shown below. As you can see, this function does not pass the Horizontal Line Test, therefore the inverse is not a function. However, let's find $g^-{1}(x)$ :

$g(x)=\frac{3}{x^2+2x} \\ \\ Substitute \ g(x) \ by \ y \\ \\y=\frac{3}{x^2+2x} \\ \\ Interchange \ x \ and \ y: \\ \\ x=\frac{3}{y^2+2y} \\ \\ Solve \ for \ y: \\ \\y^2+2y=\frac{3}{x} \\ \\ Completing \ square \\ \\y^2+2y\mathbf{+ 1}\mathbf{-1}=\frac{3}{x} \\ \\(y+1)^2=\frac{3}{x}+1 \\ \\y+1=\pm\sqrt{\frac{3}{x}+1} \\ \\ y=-1 \pm\sqrt{\frac{3}{x}+1}$