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Ahat [919]
3 years ago
11

What’s the answer ?

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0

Answer:

It is 9x^3.

Step-by-step explanation:

36x^4 and 45x^3.

GCF of 36 and 45 = 9

GCF of x^4 and x^3 = x^3

GCF of 36x^4 and 45x^2 = 9x^3.

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Simplify. Evaluate the numerical bases. 3^2 b^2 c^-4 *3^-1 b^-5 c^7
photoshop1234 [79]
Hope I can be of assistance!

3^2b^2c^{-4}\cdot \:3^{-1}b^{-5}c^7 \ \textgreater \  \mathrm{Apply\:exponent\:rule}:\ \:a^b\cdot \:a^c=a^{b+c}
3^{-1}\cdot \:3^2=\:3^{2-1}=\:3^1=\:3 \ \textgreater \  3b^{-5}b^2c^{-4}c^7

\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \  b^{-5}b^2=\:b^{2-5}=\:b^{-3}
3b^{-3}c^{-4}c^7

\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \  c^{-4}c^7=\:c^{-4+7}=\:c^3
3b^{-3}c^3

\mathrm{Apply\:exponent\:rule}: \:a^{-b}=\frac{1}{a^b} \ \textgreater \  b^{-3}=\frac{1}{b^3} \ \textgreater \  3\cdot \frac{1}{b^3}c^3

\mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \  \frac{1\cdot \:3c^3}{b^3}

Finally
\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \  \frac{3c^3}{b^3}

Hope this helps!
8 0
3 years ago
Please explain your answer as well. THX!!
Stells [14]

Answer:

option b

6 C 2 + 6

Step-by-step explanation:

Given in the question that,

number of pink beans = 87

number of purple beans = 74

number of white beans = 35

number of black beans = 70

number of orange beans = 25

number of green beans = 47

<h3>Step 1</h3>

if two beans selected are different

6 C 2

(6 colours of beans)

<h3>Step 2</h3>

if two beans selected are same

6 C 1 = 6

(6 colours of beans)

<h3>Step 3 </h3>

total sample space will be

6 C 2 + 6

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3 years ago
Order the following decimals from greatest to least (separated by
Feliz [49]

Answer:

0.14, 0.5, 0.71, 0.8, 0.83

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2 years ago
The local newspaper has letters to the editor from 30 people. If this number represents 8% of all of the​ newspaper's readers, h
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Step-by-step explanation:

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read the answer through and collect all the information you can

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