# Thread: need help with two problems

1. ## need help with two problems

12x - 2y -4 x equals 8 and y equals -6 Is it 80 or 104?

is (2/3) -4 (that's a negative to the fourth power equal to -3/2

2. Originally Posted by duckpluckie
12x - 2y -4 x equals 8 and y equals -6 Is it 80 or 104?
12*8 - 2*(-6) - 4 = 96 - (-12) - 4 = 96 + 12 - 4 = 104

-Dan

3. Originally Posted by duckpluckie
is (2/3) -4 (that's a negative to the fourth power equal to -3/2
Is it (2/3)^{-4} ie (2/3) to the "-4" power?

The "-" on the 4 in the exponent means "flip the fraction over" so

(2/3)^{-4} = (3/2)^4 = (3^4)/(2^4) = 81/16

-Dan

4. Originally Posted by duckpluckie
12x - 2y -4 x equals 8 and y equals -6 Is it 80 or 104?

is (2/3) -4 (that's a negative to the fourth power equal to -3/2

i don't understand your problem... are you saying y=-6
12x -[(2)*(-6)]-4x = 8
12 x -4x = 20
8x=20
x=2.5

or are you saying x=8 and y=-6
(12*8) - (2*-6)-(4*8)
= 96 +12 - 32
=76 ??????

dan

5. Originally Posted by topsquark
12*8 - 2*(-6) - 4 = 96 - (-12) - 4 = 96 + 12 - 4 = 104

-Dan

you did much better at understanding that then me.....

~dan

6. ## frustrated and flummoxed

I don't have the correct math symbols - what I am trying to do is find out if
-3/2 is equivalent to (2/3) with a - tiny 4 - (a negative 4th power) Sorry I can't do better - I am a behavioral psych who play with primates and counts on her fingers and toes

7. Originally Posted by duckpluckie
I don't have the correct math symbols - what I am trying to do is find out if
-3/2 is equivalent to (2/3) with a - tiny 4 - (a negative 4th power) Sorry I can't do better - I am a behavioral psych who play with primates and counts on her fingers and toes
First, please post responses as a reply in the original thread. It makes for easier reading. Hopefully one of the mods will move it there.

Apparently my guess in the other thread is correct. So as I showed there:
(2/3)^{-4} = 81/16 is NOT equal to -3/2.

Don't worry about not knowing the symbols. We'll figure it out sooner or later.

-Dan

8. Originally Posted by topsquark
First, please post responses as a reply in the original thread. It makes for easier reading. Hopefully one of the mods will move it there.

RonL