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iragen [17]
3 years ago
10

Linda must calculate the cost of filling's her car's 12-gallon gas tank. She calculates the difference between how much gasoline

her gas tank will hold and the number of gallons of gas, g, already in the tank. Then she multiplies the difference by the price, p, of one gallon of gas. What expression does Linda use to calculate the cost to fill her gas tank?
Mathematics
1 answer:
Nimfa-mama [501]3 years ago
5 0
I think
the answer would be: p (12-g)
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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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