Tamara J. answered 11/04/12
Math Tutoring - Algebra and Calculus (all levels)
You are trying to solve for the following: (ab2c+5)/(x0y)
when given the following: a= 3, b= -2, c= 10, x= -7, and y= 5
This is what you did:
(ab2c+5)/(x0y) = ((3*-22*10)+5)/(-70*5)
= ((3*-4*10)+5)/(-1*5)
= (-120+5)/(-5)
= -115/-5
= 23
You made 2 mistakes. The first is in the numerator, where you squared 2 and then you took its negative rather than squaring -2. Since b=-2, then b2=(-2)2=(-2)*(-2)=4. If the problem had been -b2, then -b2=-(-2)2=-(-2)*(-2)=-(4)=-4. The second mistake is in the denominator, where you raised 7 to the zero power and then took its negative rather than raising -7 to the zero power. Since x=-7, then x0=(-7)0=1. This isn't the same as -70, which implies that you are only raising 7 to the zero power (i.e., -70=-(7)0=-(1)=-1).
So,
(ab2c+5)/(x0y) = ((3*(-2)2*10)+5)/((-7)0*5)
= ((3*4*10)+5)/(1*5)
= (120+5)/(5)
= 125/5
= 25