helppp

Where I'm at- please help asap
http://tinypic.com/view.php?pic=50qmn6&s=8

e^x = 4
x = ln(4) = 1.39

5^(y+9) = 8
y = log5(8) - 9 = log 8 / log 5 - 9 = -7.71

r(x) = x- 4
s(x) = 3x + 6

(r*s)(x) = r(x) * s(x) = (x-4)(3x+6) = 3x^2 - 12x + 6x - 24 = 3x^2 - 6x - 24
(r + s)(x) = x - 4 + 3x + 6 = 2x +2
(r - s)(1) = 1 - 4 - 3 - 6 = -12

sqrt(50xw) * w * x^2
OR sqrt(50x^5 * w^3)

Not sure what they want exactly

I feel like you are setting yourself up for some serious trouble down the road. If this is a placement test, you are gonna get placed somewhere you don't belong and will fail in a spectacular fashion. Just my 2 cents.

This post was edited by russian on Jan 21 2015 12:39pm

First system has infinitely many solutions, because the second equation is just the first one * -1. They satisfy y = 1 - x/3

Second system has no solution.

nvm

This post was edited by Freebs on Jan 21 2015 01:29pm