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help solve the following?ab^2c+5 divided by x^0y where a=3 b=-2 c=10 x=-7 y=5

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2 Answers

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

 

Hi Dianne,

I get 25. I have spotted a few errors.

"3 times -2 squared times 10 plus 5 = -115"

This is:

3 • (-2)2 • 10 + 5

= 3 • 4 • 10 + 5

= 12 • 10 + 5

= 120 + 5

= 125

For the divisor (denominator), you say "-7 to the 0 power is -1 times 5 = -5"

Keep in mind that the problem was originally x0y with x = -7 and y = 5.

(-7)0 • 5

= 1 • 5

= 5

The zero exponent applied to the x so it applies to the whole -7. Any number raised to the zero power is 1 (except 00, which is undefined).


Putting it all together you have

125 ÷ 5

= 25

I hope that helps. Good luck.

 

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