 Ask Remember Me? vermonstervt Posts: 2, Reputation: 1 New Member #1 Jul 31, 2005, 05:49 AM
evaluate formulas
evaluate:
x^2+y^3 if x=2 y=-3
4+-27=-23
--------------------
pm^2-z^3 if p=1 m=-4 z=-2
(1)(-4^2)-(-2)^3=(1)(16)-(-8)=16+8=24
-------------------------------------
-x^2-y^3 if x=-3 y=-2
-3^2-2^3
-9-8=-17
-------------------
a^2-b^2a if a=-2 b=3
-2^2-3^2
-4-9-2=-13-2=-15
If not correct please explain step by step what I did wrong so I can get the hang of this thanks :) wzartv Posts: 402, Reputation: 21 Full Member #2 Jul 31, 2005, 08:14 AM
Hello and welcome,

Number 1 looks to be correct.

Number 2, however, I noticed an error.
(-4^2) comes out to -16, not 16.
That is the only error in that problem. Be sure to look over the problem and see where you made that mistake. Try working it out again and the answer is -8.

Number 3, I noticed an error.
The variables in the problem (x and y) have a negative sign in front of them. Also, the numbers that are replacing the variables are negative. That's a double negative, therefore making them a positive number. Look at the problem, see what I mean, and the answer comes out to 17, not -17.

Number 4, could you please explain the problem. Is there an "a" after -b^2, are there parenthesis present, was it a typo? You have written a^2-b^2a - is that the correct problem?

All right, I double checked my answers and I am quite sure that they are correct. If anyone sees any errors, please point them out. Let me know about the fourth problem. CroCivic91 Posts: 729, Reputation: 23 Senior Member #3 Jul 31, 2005, 12:43 PM Originally Posted by wzartv
Number 2, however, I noticed an error.
(-4^2) comes out to -16, not 16.
That is the only error in that problem. Be sure to look over the problem and see where you made that mistake. Try working it out again and the answer is -8.
Let me just add to this...
(-4^2) = -(4^2) = -16
but...
((-4)^2) = 16
Just watch out how the braces are placed. Originally Posted by wzartv
Number 3, I noticed an error.
The variables in the problem (x and y) have a negative sign in front of them. Also, the numbers that are replacing the variables are negative. That's a double negative, therefore making them a positive number. Look at the problem, see what I mean, and the answer comes out to 17, not -17.
I have to dissagree here.

Let's look at the braces again...
-x^2-y^3 if x=-3 y=-2
that means...
-(x^2)-(y^3) = - ((-3)^2) - ((-2)^3) = -9 - (-8) = -9 + 8 = -1

You have to be careful with the braces.

a^2-b^2a if a=-2 b=3
Let's imagine that you misstyped "-b^2a" and that it should be "-b^2".
a^2 - (b^2) = (-2)^2 - (3^2) = 4 - 9 = -5
Let's imagine that you didn't misstype that...
You have multiple options... does that mean
- (b^(2a))... (case 1)
or does it mean
- (b^2)*a... (case 2)

Case 1:
-(b^(2a)) = - (3^(2*(-2))) = - (3^(-4)) = - ( 1 / (3^4) ) = - ( 1 / 81 )

Case 2:
- (b^2)*a = - (3^2) * (-2) = (-9) * (-2) = 9 * 2 = 18

So the whole result in case 1 would be "4 - ( 1 / 82 )" and in case 2 "4 + 18 = 22" wzartv Posts: 402, Reputation: 21 Full Member #4 Jul 31, 2005, 12:50 PM
All right thanks for the posts... I guess I have to go to school again... hahaha wzartv Posts: 402, Reputation: 21 Full Member #5 Jul 31, 2005, 12:51 PM
Good point on number three... didn't think of that CroCivic91 Posts: 729, Reputation: 23 Senior Member #6 Aug 1, 2005, 02:53 AM
It's all right, everybody makes mistakes :) wzartv Posts: 402, Reputation: 21 Full Member #7 Aug 1, 2005, 07:48 AM
vermonstervt - I left you a response to your e-mail... the subject will come up as "Re: Thanks"

Check it out.

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