Question

Find the value of the monomial 2m^2np^3 for: m=3, n=1, p=2

Find the value of the monomial 2m^2np^3 for: m=0.5, n=4, p=1

Rewrite as a monomial in standard form:
(− 2/3 x^2y)^3*(− 3/4 xy^2)^2

Answer 1

• 144
• 2
• -1/6 x⁸y⁷

Explanation:

For the first two, fill in the given values and do the arithmetic.

... 2(3²)(1)(2³) = 2×9×8 = 144

... 2(0.5²)(4)(1³) = 2×(1/4)×4 = 2

_____

The outside exponents are distributed over the inside factors according to the rule:

... (a^b)^c = a^(b·c)

Then the rule ...

... (a^b)(a^c) = a^(b+c)

applies to the product.

... (− 2/3 x²y)³·(− 3/4 xy²)² = (-2/3)³x⁶y³·(-3/4)²x²y⁴

... = (-8/27)(9/16)x⁽⁶⁺²⁾y⁽³⁺⁴⁾

... = -1/6 x⁸y⁷