A few days ago
babygurl4lifey

Math experts please help (algebra 2)…?

Need help with these math problems please try ur best to help thanx:

1. -y(to the 5 power) z ( to the 7 power)

over(devided)

y(squared) z(to the 5 power)

2.(a(cubed)b(cubed)) (ab)(squared by a negative 2, outside of the perenthesis)

3. 3a(5a(squared)b) (6ab(cubed)

4.2c(cubed) d(3c(squared)d(to the fifth power) )

over (devided)

30c(to the fourth power) d (squared)

5. (6x(squared) y(to the fourth power) )

over (devided) >Cubed

(3x(to the fourth power) y (cubed))

6. 30a-2b(to the negative 6 power)

over (devided)

60a-(to the 6 power) b (to the negative 8 power)

Top 2 Answers
A few days ago
fivestring46

Favorite Answer

When working with variables and exponents in division, take the smaller exponent either up or down and change it to a negative and subtract.

1) -y^5z^7/y^2z^5 so take the y^2 to the top (it becomes y^-2) and subtract the exponents 5-2=3 so you have y^3 now take the z^5 to the top (it becomes z^-5) and subtract the exponents 7-5=2 and you have z^2

You end up with y^3z^2 / 1 which equals y^3z^2 ANSWER

2) To get rid of the exponent outside the (), multiply it times the exponents of everything inside. It really means: (ab)(ab).

So: a^3b^3(ab)^2 =

a^3b^3(a^2b^2) = multiply and we get:

a^5b^5 ANSWER

See if you can do the rest on your own.

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4 years ago
Anonymous
question a million. uncomplicated ingredient is 11a^3 question 2. you will discover components of acceptable triangles making use of the pythagorean theorem. this is a^2 + b^2 = c^2. So with B and C given, we are in a position to remedy for A. 2^2 is 4 and sqrt 13 squared is merely 13 a^2 + 4 = 13 a^2 = 9 a = 3 wish that helps.
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