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rachelleigh · Aug 27, 2007, 05:04 PM

How do I solve

3(4+3)to the second power-1??

Gahh.
I hate algebraa!

Skell · Aug 27, 2007, 05:12 PM

Give you a hint. Do the bit in the brackets first!!

Lenovo · Aug 27, 2007, 05:33 PM

3(3+4)to power of-1= 3(7)to power of-1= 3x7to power of -1=21 to power of -1= 21x-1=
-21
See how I did that?

galactus · Aug 27, 2007, 05:41 PM

I don't see much algebra. Just basic arithmetic.

interinfinity · Jan 25, 2008, 02:03 PM

so would that be (21^-1)^-1? I hate computer math!

PolluxCastor · Jan 25, 2008, 08:05 PM

Originally Posted by Lenovo

3(3+4)to power of-1= 3(7)to power of-1= 3x7to power of -1=21 to power of -1= 21x-1=
-21
See how I did that?

Um, no.

Originally Posted by interinfinity

so would that be (21^-1)^-1 ?? i hate computer math!

Again, no.

Originally Posted by rachelleigh

3(4+3)to the second power-1???

This could be either x^2 - 1
or ax^2 - 1.

I think the later, you try it from here.

PolluxCastor · Jan 28, 2008, 05:13 AM

3(4+3)to the second power-1
3(4+3)^2 -1
3(7)^2-1
3*49 - 1
147 - 1
146

The other possibility would be:
(3*(4+3))^2 - 1
(3*7)^2 -1
21^2 -1
441 - 1
440

Any questions?