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How do you evaluate #(5x ^ { - 3} y ) ( 6x ^ { 7} y ^ { - 4} )#?

Answer 1

Hazel Anglin

#(5x^(-3)y)(6x^7y^(-4))=color(blue)((30x^4)/(y^3)#

Simplify:

#(5x^(-3)y)(6x^7y^(-4))#

Simplify the parentheses.

#5x^(-3)yxx6x^7y^(-4)#

Multiply #5# and #6#.

#30x^(-3)yx^7y^(-4)#

Gather like variables.

#30x^(-3)x^(7)yy^(-4)#

Apply the product rule of exponents: #a^ma^n=a^(m+n)#. (No exponent is understood to be #1#.

#30x^(-3+7)y^(1-4)#

Simplify.

#30x^(4)y^(-3)#

Apply the negative exponent rule: #a^(-m)=1/a^m#.

#(30x^4)/(y^3)# #larr# Answer

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Answer 2

Avery Alexander

The expression simplifies to ( \frac{30x^4}{y^3} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.