thelizzybeth

\(\frac{5(x+3)}{(x+3)(x-3)}=\frac{5}{x-3}\)

\(\frac{y-4}{4-y}=-1\)

\(\frac{5}{x-3}*-1=\frac{5}{3-x}\) which is answer choice \(\fbox a\)

Instead of trying to figure out the value of x, we can use clever squaring.

Square the first equation.

\((x+\frac{1}{x})^2= 3^2\)

\( x^2+1+1+\frac{1}{x^2}=9\)

\(x^2+\frac{1}{x^2}=7\)

Square this equation again.

\((x^2+\frac{1}{x^2})^2=7^2\)

\(x^4+1+1+\frac{1}{x^4}=49\)

\(x^4+\frac{1}{x^4}=47\)

f(-2.2) = -2.2 - y = integer

We're trying to get rid of the fractional part 0.2. Since y > 0, y = 0.8

-2.2 + 0.8 = -1.4

First divide both sides by 2.

\(ny^2=n^2y-48\)

Then move all the variables to one side.

\(n^2y-ny^2=48\)

\(ny(n-y)=48\)

I factorized 48 and used trial and error to find

\(ny=12\)

\(n-y=4\)

Solve and find \(n=6, y=2\)

Your questions have already been answered in this thread:

Your questions have already been answered in this thread:

I'm not sure how you went from \(y(2x-6)=rx+s\)  to  \(2xy-4x=6y+s\). 

You expanded this part wrong, so let's start from this point.

\(y(2x-6)=rx+s\)

\(2xy-6y=rx+s\)

\(2xy-rx=6y+s\)

\(x(2y-r)=6y+s\)

\(x=\frac{6y+s}{2y-r}\)

Using the horizontal asymptote of y = -4, we find that r = -8. 

Now use the point (1, 0) to find s. 

At 9:40, the minute hand is at 4 and the hour hand is between 9 and 10. 

There is \(30 ^\circ*5=150^\circ\)between the minute hand and 9 o'clock. 

The hour hand is \(\frac{20}{60}*30^\circ=10^\circ\) past 9 o'clock.

So our answer is \(150^\circ+10^\circ=160^\circ\)

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The inverse function is the function that "undoes".

We can find the inverse function by setting f(x) as y and solving for x in terms of y.

\(y(2x-1)=x\)

\(2xy-y=x\)

\(y = 2xy-x\)

\( y=x(2y-1) \)

\(x=\frac{y}{2y-1}\)

So the inverse function is \(\frac{x}{2x-1}\)