# can someone double check this for me?

• Jul 18th 2010, 12:08 PM
soyeahiknow
can someone double check this for me?
Hi, can someone double check this for me?

Also sorry about the badly written problem. What programs do you all use? I tried word but i couldn't import it over to here.

Attachment 18229

Also, when you do the power, you do that keep that sign first and then incorporate either the - or + sign next right?

So lets say x=-2, for (x^5)-(X^3)=
you do -2*-2*-2*-2*-2 which is -32 and then you do -2*-2*-2 which is -8. Then the last step is to subtract -32- (-8) but since there is two neg. signs, -8 becomes a positive 8 right? so it is -32+8

Thanks
• Jul 18th 2010, 12:11 PM
mr fantastic
Quote:

Originally Posted by soyeahiknow
Hi, can someone double check this for me?

Also sorry about the badly written problem. What programs do you all use? I tried word but i couldn't import it over to here.

Attachment 18229

Also, when you do the power, you do that keep that sign first and then incorporate either the - or + sign next right?

So lets say x=-2, for (x^5)-(X^3)=
you do -2*-2*-2*-2*-2 which is -32 and then you do -2*-2*-2 which is -8. Then the last step is to subtract -32- (-8) but since there is two neg. signs, -8 becomes a positive 8 right? so it is -32+8

Thanks

Check it here: Wolfram|Alpha&mdash;Computational Knowledge Engine
• Jul 18th 2010, 12:16 PM
soyeahiknow
I did that but I got -53/3.

I know it is 0-(64/6)-(16/4)+(-8/3), but I thought I just didn't worry about the zero and just have (64/6)-(16/4)+(-8/3). Am I wrong?
• Jul 18th 2010, 12:20 PM
undefined
Quote:

Originally Posted by soyeahiknow
Also sorry about the badly written problem. What programs do you all use? I tried word but i couldn't import it over to here.

LaTeX help subforum

Quote:

Originally Posted by soyeahiknow
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You wrote a plus sign where it should have been minus sign, right before the 8/3.

Quote:

Originally Posted by soyeahiknow
Also, when you do the power, you do that keep that sign first and then incorporate either the - or + sign next right?

I don't follow..

Quote:

Originally Posted by soyeahiknow
So lets say x=-2, for (x^5)-(X^3)=
you do -2*-2*-2*-2*-2 which is -32 and then you do -2*-2*-2 which is -8. Then the last step is to subtract -32- (-8) but since there is two neg. signs, -8 becomes a positive 8 right? so it is -32+8

That is correct. If you have a function g(x) = x^5 - x^3 then g(-2) = -24.

Note that you know in advance that (-2)^5 is negative because the exponent is odd. So it's just -2^5. (Note the lack of parentheses. Meaning take 2^5 and then negate it.)

Quote:

Originally Posted by soyeahiknow
I did that but I got -53/3.

I know it is 0-(64/6)-(16/4)+(-8/3), but I thought I just didn't worry about the zero and just have (64/6)-(16/4)+(-8/3). Am I wrong?

You seem to have typed the wrong thing into Wolfram Alpha. The answer is -4. See here.
• Jul 18th 2010, 12:23 PM
soyeahiknow
yeah sorry, I messed up that sign.

So the full problem written out is 0- {(64/6)-(16/4)+(-8/3)}, but I thought I just didn't worry about the zero and just have (64/6)-(16/4)+(-8/3). Am I wrong?
• Jul 18th 2010, 12:30 PM
undefined
Quote:

Originally Posted by soyeahiknow
yeah sorry, I messed up that sign.

So the full problem written out is 0- {(64/6)-(16/4)+(-8/3)}, but I thought I just didn't worry about the zero and just have (64/6)-(16/4)+(-8/3). Am I wrong?

$0- \left(\left(\dfrac{64}{6}\right)-\left(\dfrac{16}{4}\right)+\left(-\,\dfrac{8}{3}\right)\right) = -4$

So this is right, just make sure you get -4 as final answer.

Edit: I thought you were just evaluating the thing inside parentheses first, and then would remember to negate it afterwards. It looks like you were making the mistake pointed out in Archie Meade's post below mine though.
• Jul 18th 2010, 01:55 PM
Quote:

Originally Posted by soyeahiknow
yeah sorry, I messed up that sign.

So the full problem written out is 0- {(64/6)-(16/4)+(-8/3)}, but I thought I just didn't worry about the zero and just have (64/6)-(16/4)+(-8/3). Am I wrong?

You wouldn't have to worry about the zero if you were adding it to the three values.
Unfortunately, you are subtracting all of them from zero.

$0-\left(\frac{64}{6}-4-\frac{8}{3}\right)$

is very different to

$0-\frac{64}{6}-4-\frac{8}{3}$

which has a redundant zero.

$0-\left(\frac{64}{6}-4-\frac{8}{3}\right)=-\frac{64}{6}-(-4)-\left(-\frac{8}{3}\right)$