please help find error in simplification w/ powers, abs vals

[Otis_BF]

I hope I'm posting this into the right category. I'm not sure how best to write out this problem so.. i'll do my best

(Note: I'm using the "*" to mean "multiply", "^" to mean "to the power of", and "cbrt" to mean "the cube root of".)

Evaluate the following expression:

. . . .(1/2)|3 - 10| - (2/3)*(9/2)
. . .-----------------------------------
. . .-10^0 + cbrt[1/27] - 3^(-2)

Here is how I do it. It seems right to me, but my answer is always wrong, so....

On top:

. . .|3 - 10| = |-7| = 7

. . .(2/3)*(9/2)

The 2's cancel each other, 3 goes into 9 three times and itself once, so up top I figure I should have

. . .(1/2)*7 - 3

Down below:

. . .-10^0 = 1

. . .cbrt[1/27] = 1/3

. . .3^-2 = 1 / (3^2) = 1/9

Putting it together, I should have:

. . . .(1/2)*7 - 3
. . .----------------
. . .1 + 1/3 - 1/9

. . . . . . . . .(7/2) - 3
. . .= -------------------------
. . . . .(9/9) + (3/9) - (1/9)

. . . . ..1/2
. . .= ------
. . . . .11/9

. . .= (1/2)*(9/11)

. . .= 9/22

My final answer is 9/22, but the book says the answer is -9/14. I've no idea what I'm doing wrong. Can anyone help me find my error?

Thank you for your time!

- Otis

 

[sgtpepper]

hmm I didn't find a mistake either; I'm not too sure what exactly you did wrong - I worked it and got the same answer.

 

[wjm11]

You just dropped a minus sign in the denominator on the -10^0 = -1. Everything else is correct.

 

[pka]

You wrote \(\displaystyle - 10^0.\) The exponent is applied to the 10 and –10.
Therefore, here is what happens \(\displaystyle - 10^0 = - \left( {10^0 } \right) = - 1.\)
Now the answer in the text is correct.

 

[Otis_BF]

You just dropped a minus sign in the denominator on the -10^0 = -1. Everything else is correct.

Thank you very much, that gives me the right answer.

I was under the impression that "any nonzero integer^0 = 1"
and in my notes(which must be wrong, i've found mistakes in them before) i have

"x^0 = 1 eg: -3^0 = 1, 3^0 = 1"

but after seeing your replies can I assume that I must have missed brackets when my teacher was writing on the board...

-3^0 = -1, (-3)^0 = 1

I'm assuming that's how it should look? thanks a lot! i really appreciate your help!

 

[pka]

I was under the impression that "any nonzero integer^0 = 1

Actually that statement is correct!
But –3 to the zero power is written \(\displaystyle ( - 3 )^0.\)

 

[Otis_BF]

I was under the impression that "any nonzero integer^0 = 1

Actually that statement is correct!
But –3 to the zero power is written \(\displaystyle ( - 3 )^0.\)

I understand now, thank you!