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brilliants [131]
3 years ago
10

I don’t know what to do on here it’s really hard and I need help for the answer

Mathematics
1 answer:
Alla [95]3 years ago
5 0
Ok so our original fraction is:
\frac{36 {x}^{12} }{6 {x}^{9} }
To simplify this fraction, look for instances where the values on the top and bottom can be reduced:

For example, 36 over 6 is the same as 6 over 1, so we can simplify the fraction so it is:
\frac{6 {x}^{12} }{ {x}^{9} }
We can also eliminate the denominator by dividing the nominator by x^9 so:
\frac{6 {x}^{12} }{ {x}^{9} }  \div  \frac{ {x}^{9} }{ {x}^{9} }
= 6 {x}^{3}
And that is the simplified answer of the fraction

Hope this helped
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Answer:

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Step-by-step explanation:

Answer(s) is above. Hope it helps!

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Convert 2 1/3 into improper fraction: *<br> 7/3<br> O 7/6<br> O 6/3<br> O 3/6
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Step-by-step explanation:

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public class person { 4. private double salary; 5. public person() { 6. salary = 1000.0; 7. } what type of constructor is illust
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brainly.com/question/29744767

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1 year ago
A health food company sells packs of assorted nuts. The company incurs a monthly fixed cost of $1,400 for ingredients and raw ma
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3 years ago
Read 2 more answers
What is the solution of x^2+x-6/x-7&lt;0 PLEASE HELP ME :( (it’s not the 2nd one i already tried it)
blsea [12.9K]

Answer:

the first one  \:x\le \:-3\quad \mathrm{or}\quad \:2\le \:x

Step-by-step explanation:

Factor x^2+x-6 :

Break the expression into groups

\left(x^2-2x\right)+\left(3x-6\right)

\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2-2x\mathrm{:\quad}

x\left(x-2\right)

\mathrm{Factor\:out\:}3\mathrm{\:from\:}3x-6\mathrm{:\quad}

3\left(x-2\right)

x\left(x-2\right)+3\left(x-2\right)

\mathrm{Factor\:out\:common\:term\:}x-2

\left(x-2\right)\left(x+3\right)

\frac{\left(x-2\right)\left(x+3\right)}{x-7}\\

\frac{\left(x-2\right)\left(x+3\right)}{x-7}\le \:0

\mathrm{Find\:the\:signs\:of\:the\:factors\:of\:}\frac{\left(x-2\right)\left(x+3\right)}{x-7}

\mathrm{Identify\:the\:intervals\:that\:satisfy\:the\:required\:condition:}\:\le \:\:0

x

Merge Overlapping Intervals

x\le \:-3\quad \mathrm{or}\quad \:2\le \:x

6 0
3 years ago
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