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3 years ago

Can i get the answer please

Mathematics

1 answer:

Sladkaya [172]3 years ago

8 0

No, because 40 x 10 is not 4000, but is is 400. It is 100x greater not 10x

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Simplify the following expression as much as you can use exponential properties. (6^-2)(3^-3)(3*6)^4

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Simplifying the expression $(6^{-2})(3^{-3})(3*6)^4$ we get $\mathbf{108}$

Step-by-step explanation:

We need to simplify the expression $(6^{-2})(3^{-3})(3*6)^4$

Solving:

$(6^{-2})(3^{-3})(3*6)^4$

Applying exponent rule: $a^{-m}=\frac{1}{a^m}$

$=\frac{1}{(6^{2})}\frac{1}{(3^{3})}(18)^4\\=\frac{(18)^4}{6^{2}\:.\:3^{3}} \\$

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Replacing terms with factors

$=\frac{(2\times3^2)^4}{(2\times 3)^{2}\:.\:3^{3}} \\=\frac{(2)^4\times(3^2)^4}{(2)^2\times (3)^{2}\:.\:3^{3}} \\$

Using exponent rule: $(a^m)^n=a^{m\times n}$

$=\frac{(2)^4\times(3)^8}{(2)^2\times (3)^{2}\:.\:3^{3}} \\=\frac{2^4\times 3^8}{2^2\times 3^{2}\:.\:3^{3}}$

Using exponent rule: $a^m.a^n=a^{m+n}$

$=\frac{2^4\times 3^8}{2^2\times 3^{2+3}}\\=\frac{2^4\times 3^8}{2^2\times 3^{5}}$

Now using exponent rule: $\frac{a^m}{a^n}=a^{m-n}$

$=2^{4-2}\times 3^{8-5}\\=2^{2}\times 3^{3}\\=4\times 27\\=108$

So, simplifying the expression $(6^{-2})(3^{-3})(3*6)^4$ we get $\mathbf{108}$

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