Tuesday

December 1, 2015
Posted by **tim** on Sunday, April 10, 2011 at 10:36pm.

- math -
**Jai**, Sunday, April 10, 2011 at 11:17pmfor (1),

a^2 * b^(-3)

note that when a term is raised to a negative exponent, we get its reciprocal and raise it by the positive exponent,, for example,

2^(-4) = 1/(2^4) ; thus we can rewrite

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

therefore,

a^2 * (1/(b^3))

a^2 / b^3

for (2),

(5/3)^-3

we do the same thing,, get the reciprocal, make the exponent positive and simplify:

(3/5)^3

27 / 125

for (3),

(a^2b^0c)^0

note that any number (except zero) raised to zero is equal to 1,, thus assuming a, b and c are not equal to 0,

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

for (4),

do you mean x^(-2) * y^(3) ? if so,

it's the same as #1:

(1/(x^2)) * y^3

y^3 / x^2

for (5),

3x^-2

it's also the same procedure as #1:

3*(1 / (x^2))

3/(x^2)

for (6),

(n^2) / (n^(-3))

same procedure as #1,,

(n^2) / (1/(n^3))

simplifying,

(n^2)*(n^3)

note that when multiplying terms raised to exponents but have the same base, we can just add the exponents,, for example

2^3 * 2^5 = 2^(3+5) = 2^8

note that the base here is 2.

therefore,

(n^2)*(n^3) = n^(2+3) = n^5

hope this helps~ :)

- math -
**tim**, Monday, April 11, 2011 at 2:43pmthanks! i really appreciate this :)