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Lelechka [254]
3 years ago
15

Drag steps of the solution into order to solve for x in the equation 9(x + 3) = 108.

Mathematics
2 answers:
OlgaM077 [116]3 years ago
7 0
9(x+3)=108
9x+27=108
9x=81
x=9
Hope this helps!
Marat540 [252]3 years ago
3 0

Answer:

Given expression is,

9(x+3) = 108

In solving the equation the steps are as follows,

Step 1 : 9x + 27 = 108      ( By distributive property )

Step 2 : 9x = 81               ( Subtracting 108 on both sides )

Step 3 : x = 9                   ( Divide both sides by 9)

Hence, by the above explanation we can drag the steps of solution in correct order.

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Lisa [10]
<h3>Answer: 10</h3>

=======================================================

Explanation:

We'll start off converting each mixed number into an improper fraction.

The formula to use is A \frac{b}{c} = \frac{A*c+b}{c}

So,

A \frac{b}{c} = \frac{A*c+b}{c}\\\\2 \frac{1}{8} = \frac{2*8+1}{8}\\\\2 \frac{1}{8} = \frac{16+1}{8}\\\\2 \frac{1}{8} = \frac{17}{8}\\\\

And,

A \frac{b}{c} = \frac{A*c+b}{c}\\\\2 \frac{2}{3} = \frac{2*3+2}{3}\\\\2 \frac{2}{3} = \frac{6+2}{3}\\\\2 \frac{2}{3} = \frac{8}{3}\\\\

So the task of computing 2 \frac{1}{8} \times 2 \frac{2}{3} is exactly the same as computing \frac{17}{8} \times \frac{8}{3}

Notice how we have an 8 up top and an 8 down below. Those 8's cancel out and we're left with \frac{17}{3}. That fraction does not reduce any further.

The last step is to convert that improper fraction result to a mixed number.

\frac{17}{3} = \frac{15+2}{3}\\\\\frac{17}{3} = \frac{15}{3}+\frac{2}{3}\\\\\frac{17}{3} = 5+\frac{2}{3}\\\\\frac{17}{3} = 5 \frac{2}{3}\\\\

Or you could note that 17/3 leads to 5 remainder 2. The 5 is the whole part and the 2 forms the numerator of the fractional part 2/3.

The value is in A \frac{b}{c} form where

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Therefore, A+b+c = 5+2+3 = 10

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How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?
MrRissso [65]
\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3}  \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ =  \frac{(3x)^{3 \times  \frac{5}{3} }}{(2y^3)^{3 \times  \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}

Now, If the exponent was negative like you asked....

\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}

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