How to simplify the following algebra equations.?

1)
10^-4
= 1/10^4
= 1/(10 x 10 x 10 x 10)
= 1/10000 (0.0001)

2)
a^-3b^2
= (1/a^3)(b^2)
= b^2/a^3

3)
(3/5)^-3
= 1/(3/5)^3
= 1/(3^3/5^3)
= 1/(27/125)
= 1(125/27)
= 125/27

4)
(a^2b^0c)^0
= 1

5)
x^-2y^3
= (1/x^2)(y^3)
= y^3/x^2

6)
3x^-2
= 3(1/x^2)
= 3/x^2

7)
n^-2/n^3
= n^(-2 - 3)
= n^-5
= 1/n^5

whilst multiplying radicals (sq. roots), you are able to multiply the radicands (radicand= the style below the sq. root image), and then take the sq. root of that. whilst taking the sq. root of a fraction, you are able to take the sq. root of the numerator and the denominator one after the other. (ex: ?(7 / one hundred) = ?7 / ?one hundred = ?7 / 10) whilst elevating an intensive to a power, you are able to rewrite the unconventional as a fractional power ( sq. root is comparable to the a million/2 power, cubed root is comparable to the a million/third power, the nth root is comparable to the a million/n power, in maximum circumstances.) So, once you enhance a power to a power, you multiply the powers, such as you're able to do with entire style powers. (ex: ?forty 5 ^4 = (forty 5^(a million/2))^4 = forty 5^(4/2) = forty 5^2 = 2025) finally, (i think of) you are able to only upload radical expressions that share a undemanding base. (think of of the unconventional it relatively is related to the completed style as an x, so which you would be able to only upload issues that are related to x's. Ex: 2?2 + 3?2 = 5?2, yet 2?2 + 3?3 won't be able of be simplified.)

1) 10^ -4 =1/10^4=0.0001| 2) a^-3/b^2= b^2/a^3 | 3)( 3/5 )^-3=5/3 ^3=125/27| 4) 1 5)y^3/x^2 6) 3/x^2 7) n^-5=1/n^5

1. 10^-4
= 1/10^4
= 1/10,000
= 0.0001

2. a^-3 b^2
= 1/a^3 * b^2
= b^2/a^3

3. (3/5)^-3
= 1/(3/5)^3
= (5/3)^3
= 125/27
or = 4.63(approx.)

4. (a^2 b^0 c)^0
= 1
[Since, any constant or variable with power 0 = 1]

5. x^-2 y^3
= 1/x^2 * y^3
= y^3/x^2

6. 3x^-2
= 1/3x^2

7. n^-2 / n^3
= 1/n^2 * 1/n^3
= 1/n^(2+3)
= 1/n^5

1. 10^-4=1/10^4=0.0001
2. a^-3b^2=b^2/a^3
3. (3/5)^-3=(5/3)^3=125/27=4.629
4. (a^2b^0c)^0=1
5.,6.,7., are like above ones.

hi
1/(10)4 ........b^2/a^3..........(5/3)^3......1..............y^3/x^2
3/x^2............1/(n^2*n^3)=1/n^5