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Mandarinka [93]
4 years ago
9

Divide. (5a^3x/3x^5)/(a^2/9x) Simplify your answer as much as possible!

Mathematics
1 answer:
VLD [36.1K]4 years ago
5 0

Answer:

The simplified expression is  \frac{15a}{x^3}

Step-by-step explanation:

The given expression is

\frac{\frac{5a^3x}{3x^5} }{\frac{a^2}{9x}}

We change the middle bar to a normal division sign to obtain,

=\frac{5a^3x}{3x^5}\div \frac{a^2}{9x}


We multiply by the reciprocal of the second fraction to obtain,

=\frac{5a^3x}{3x^5}\times \frac{9x}{a^2}


We cancel out common factors to get,

=\frac{5a}{x^3}\times \frac{3}{1}

This simplifies to,

=\frac{15a}{x^3}



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Which of the following shows the polynomial below written in descending<br> order?
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Answer:

A. 4x¹² + 9x⁷ + 3x³ -x

Step-by-step explanation:

Hi!

<em>==================================================================</em>

To write a polynomial in descending order, we write the terms with higher degrees, or exponents, first.

3x³ + 9x⁷ -x + 4x¹²

4x¹² has the highest degree, so it is written as the first term.

⇒4x¹²

9x⁷ has the next highest degree, so it is written next.

⇒4x¹² + 9x⁷

3x³ has the next highest degree, so it is written next.

⇒4x¹² + 9x⁷ + 3x³

-x has the lowest degree, so it is written last.

⇒4x¹² + 9x⁷ + 3x³ -x

<u>4x¹² + 9x⁷ + 3x³ -x</u>

<em>==================================================================</em>

<em>Hope I Helped, Feel free to ask any questions to clarify :)</em>

<em>Have a great day!</em>

<em>        -Aadi x</em>

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3 years ago
In an arithmetic sequence the sum of the first six odd-numbered terms (a1,a3,a5,a7,a9 and a11) is 60. find the sum of the first
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Read 2 more answers
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