Which expression is correctly developed to use the Product Rule of Exponents?(1 point)

A.10^(8) * 10^(8)
B.32^(7)
C.(5^(2))^(9)
D.6^(2) * 7^(3)

screw u bot y u use them just post the answers bro

To simplify the expression 100^(6)⋅100^(7), you can use the Product Rule of Exponents, which states that a^(m)⋅a^(n) = a^(m+n).

Therefore, we can simplify the expression as:

100^(6)⋅100^(7) = 100^(6+7) = 100^(13)

So the correct expression that uses the Product Rule of Exponents to simplify the expression 100^(6)⋅100^(7) is 100^(13).

Bot?

To correctly use the Product Rule of Exponents, you need to multiply two or more exponential terms with the same base. The rule states that when multiplying exponential terms with the same base, you keep the base and add the exponents.

Let's examine each expression to see which one follows this rule:

A. 10^(8) * 10^(8)

In this expression, both terms have the same base of 10. Applying the product rule, we keep the base 10 and add the exponents 8 and 8. Therefore, the result is 10^(8 + 8) = 10^(16).

B. 32^(7)

In this expression, there is only one exponential term, so the product rule does not apply. It is not correctly developed to use the product rule of exponents.

C. (5^(2))^(9)

In this expression, we have an expression raised to a power, but there is no multiplication happening. The product rule is not applicable here.

D. 6^(2) * 7^(3)

In this expression, we have two exponential terms with different bases. The product rule requires the same base. Therefore, it is not correctly developed to use the product rule of exponents.

Based on the analysis, the expression that is correctly developed to use the Product Rule of Exponents is option A. 10^(8) * 10^(8).

Which of the following correctly used the Product Rule of Exponents to simplify the expression 100^(6)⋅100^(7)?(1 point)

Find the numerical equivalent of 102 ⋅103(1 point)

Responses

100,000
100,000

10
10

1,100
1,100

105

The expression that is correctly developed to use the Product Rule of Exponents is D. 6^(2) * 7^(3).