# math

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• math -

for (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 -

thanks! i really appreciate this :)