Help with integer exponents, please!!!?

Question

Trying to complete yet another Algebra homework assignment, and I am stuck on 4 problems. Would someone mind helping me solve the problems, and also explain why you did what you did? THANK YOU for your help! "Answers" has been a blessing when it comes to my Algebra homework!!! :o)

1. p^-5q^-4 / 9r^-3

2. (p^-4q / r^-3)^-3

3. (m^8n^-4)^2 / m^-2n^5

4. (wz^-5 / w^-3z)^-2

THANK YOU AGAIN TO ALL YOU MATH WHIZZES!!! :o)

Answer 1

1. p^-5q^-4/qr-3 = r^3/9p^5q^4

2. (p^-4q/r^-3)^-3 = (r^3q/p^4)^-3 = (P^4/r^3q)^3 = p^12/r^9q^3

3. (m^8n^-4)^2/m^-2n^5 = m^16n^-8/m^-2n^5 = m^16m^2/n^5n^8 = m^8/n^13

4. (wz^-5/w^-3z)^{-2} = (ww^3/zz^5)^{-2} = (w^4/z^6)^{-2} = z^6/w^4)^2 = z^12/w^8

Answer 2

hi, it relatively is incredibly shopper-friendly. There are 2 themes first one is base and 2nd one is exponent. think of you have a form 3^2. Then 3 is the backside and a couple of is the exponent. on the behalf of there 2. There are some properties- a million. If the bases are comparable and are in multiplication, then skill/exponent are greater. as an party- a^3 *a^2=a^5 2. If there is skill on skill, then powers get greater ideal. as an party- (a^3 )^2= a^6. 3. If the bases are diverse, powers are comparable, then you certainly ought to freshen up it relatively. 4. If the skill is 0 then the top result would be one continuously in spite of the cost of base is taken.

Answer 3

1.) p^-5q^-4/qr-3 = r^3/9p^5q4
2.) (p^-4q/r^-3)^-3 = (r^3q/p^4)^-3 = (P^4/r^3q)3
= p^12/r^9q^3
3.) (m^8n^-4)^2/m^-2n^5 = m^16n^-8/m^-2n^5
= m^16m^2/n^5n^8 = m^8/n^13
4.) (wz^-5/w^-3z)^-2 = (ww^3/zz^5)^-2 = (w^4/z^6)^-2
= z^6/w^4)^2 = z^12/w^8