............
[5^(+2) x^(+6) y^(-8)]*[(x^(-5) y^(+3)]=
[5^(+2) x^(+6) y^(-8)]*[(x^(-5) y^(+3)]
**************************************…
.............................1
5^(+2) x^(+6)y^(+3)
**********************=
.........y^8x^5
25x^6y^3
***********=..............ask the two questions about x and about y
..y^8x^5
25x
****
y^5
5) w^3.w^(-5) OR w^3/w^(-5)
w^3*w^(-5)=
w^3
****=.................ask the two questions
w^5
...1
*****=..................ANSWER
w^2
IF............ w^3/w^(-5)
w^3/w^(-5)=
..w^3
*******=..................moved the negative exponent factor up top
w^(-5)
w^3*w^(+5)=
w^8.................ANSWER
THAT WAS FUN................THANKS
Basically, if you see a negative exponent, just move it to the other side of the fraction and change the exponent to a positive one. Add the exponent of the same coefficient when they're on the same side of the fraction. Note how anything to the power of zero is 1. For example, the first problem:
1) (8p^-5 q^0 r^6) / (16p q^2 r^10)
= 1 / (2p^6 q^2 r^4)
I'll just go ahead and solve the rest for ya.
2) (36x^10 y^-5) / (12x^-3 y^-7)
= 3x^13 y^2
3) 2^-3 + 4^-1
1/2^3 + 1/4
1/8 + 1/4
1/8 + 2/8
= 3/8
4) (5^-1 x^-3 y^4)^-2 (x^5 y^-3)^-1
(5^2 x^6 y^-8) (x^-5 y^3)
25xy^-5
= 25x / y^5
5) w^3 / w^-5
= w^8
when variable has the power 0....eg: 9^0, ans is 1. So you can ignore the ^0 no's.
when dividing like in 1 or 2, you bring the variables from the bottom to the top, and change sign
eg: 4p^-4 / 6 p^-2 ans would be 2 p^-2 / 3 ( p^-4+2..... .. p^-2 ).The fraction is just to be simplified 4/6 becomes 2/3.
when adding like in 3) negative power is cahnged to positive by inverting the term eg:
2^-3 + 4^-1 = 1/(2^3) + 1/(4^1) = 1/8 + 1/4.........you can solve this i guess.....
when multiplying, same powers are added or subtracted according to signs, w^3.w^-5 = w^-2
I ve seen double A 's solution , the last one is wrong
