Tuesday

March 3, 2015

March 3, 2015

Posted by **Anonymous** on Saturday, September 14, 2013 at 6:36pm.

-3x^2+8=-4

- PreCal -
**Steve**, Saturday, September 14, 2013 at 6:46pm-3x^2 = -12

x^2 = 4

x = ±2

- PreCal -
**Anonymous**, Saturday, September 14, 2013 at 6:49pmthanks and what about this one ?

Simplify the expression

(3x^-1y^2)^3(5x^2y)0

I'm getting confused because of the negative exponent

- PreCal -
**Steve**, Saturday, September 14, 2013 at 6:56pmanything^0 = 1, so we just have

(3x^-1y^2)^3 = 27 x^-3 y^6 = 27y^6/x^3

negative exponents indicate a swap between numerator and denominator

x^-3 = 1/x^3

1/x^-3 = x^3

recall that x^a/x^b = x^(a-b)

**Answer this Question**

**Related Questions**

precal - Solve for y: (y-(10/y))^2+6(y-(10/y))-27=0

precal - solve for x: (x^2-2x)^2 -11(x^2-2x)+24

PreCal - Solve The Inequality 0 < -9x/x-x^2

precal - solve by subsituting 5x+4y=7 x-3y=9

precal 2 - solve the equation 4cos^2x-1=0

precal - How do you solve: sin[arccos(-2/7)] ?

PreCal - Solve Using Log Properties: 1/4 = 8^3x+5

Precal - Solve the exponential equation algebraically 5^(-t/2) = 0.20

PreCal - I'm suppose to solve log(3)-27=x I get 3^x=-27 I can't go any father ...

precal 2 - use the law of cosines to solve the triangle a=55, c=72, b=25