Question

Drag each expression to the correct location on the table.

Answer 1

The solution to the exponential expression is as follows:

• $\mathbf{\dfrac{(2\times 3^{-2})^3(5\times 3^2)^2}{(3^{-2})(5\times 2)^2} = 2}$
• $\mathbf{=(3^3)(4^0)^2(3\times 2)^{-3}(2^2)=\dfrac{1}{2}}$
• $\mathbf{\dfrac{(3^7\times 4^7)(2 \times 5)^{-3}(5^2)}{12^7\times 5^{-1}\times 2^{-4}} = 2}$
• $\mathbf{\dfrac{(2\times 3)^{-1}\times 2^0}{(2\times 3)^{-1}}= 1}$

What are exponential expressions?

Exponential expressions are convenient ways to express the mathematical power of a number in short forms.

Simplifying the following expressions given:

1.

$\mathbf{\dfrac{(2\times 3^{-2})^3(5\times 3^2)^2}{(3^{-2})(5\times 2)^2} }$

Applying the exponent rule: $\mathbf{(a\times b)^n = a^n b^n}$

$\mathbf{\implies \dfrac{(2\times 3^{-2})^3\times 5^2\times 81}{3^{-2}\times 5^2\times2^2} }$

cancel out the common factor, we have:

$\mathbf{\implies \dfrac{(2\times 3^{-2})^3\times 81}{3^{-2}\times2^2} }$

$\mathbf{\implies \dfrac{(2\times 3^{-2})^3\times 3^4}{3^{-2}\times2^2} }$

$\mathbf{\implies \dfrac{(2\times 3^{-2})^3\times 3^6}{2^2} }$

$\mathbf{\implies \dfrac{2^3 \times \dfrac{1}{729} \times 3^6}{2^2} }$

$\mathbf{\implies \dfrac{2^3 \times \dfrac{1}{729} \times729}{2^2} }$

$\mathbf{\implies \dfrac{2^3 }{2^2} = 2}$

2.

$\mathbf{=(3^3)(4^0)^2(3\times 2)^{-3}(2^2)}$

= $\mathbf{\dfrac{1}{2}}$

3.

$\mathbf{\dfrac{(3^7\times 4^7)(2 \times 5)^{-3}(5^2)}{12^7\times 5^{-1}\times 2^{-4}} = 2}$

4.

$\mathbf{\dfrac{(2\times 3)^{-1}\times 2^0}{(2\times 3)^{-1}}= 1}$

Learn more about the exponential expressions here:

brainly.com/question/12940982

#SPJ1