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3 years ago

Factor. m2 + 12m+36

Mathematics

2 answers:

Art [367]3 years ago

8 0

<h3>Answer: Choice A which is (m+6)^2</h3>

Explanation:

Think of two numbers that multiply to 36 and add to 12. Those two numbers are 6 and 6

6+6 = 12

6*6 = 36

So we can factor it into (m+6)(m+6). Use the FOIL rule to expand it out to get m^2+12m+36 back again.

The expression (m+6)(m+6) is the same as (m+6)^2

asambeis [7]3 years ago

7 0

Answer:

$(m-6)^{2}$

<em>Plug into the calculator. </em>

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Read 2 more answers

How is the series 5+11+17…+251 represented in summation notation?

NISA [10]

Answer:

$\displaystyle \large{\sum_{n=1}^{42}(6n-1)$

Step-by-step explanation:

Given:

Series 5+11+17+...+251

To find:

Summation notation of the given series

Summation Notation:

$\displaystyle \large{\sum_{k=1}^n a_k}$

Where n is the number of terms and $\displaystyle \large{a_k}$ is general term.

First, determine what kind of series it is, there are two main series that everyone should know:

Arithmetic Series

A series that has common difference.

Geometric Series

A series that has common ratio.

If you notice and keep subtracting the next term with previous term:

11-5 = 6
17-11 = 6

Two common difference, we can in fact say that the series is arithmetic one. Since we know the type of series, we have to find the number of terms.

Now that brings us to arithmetic sequence, we know that first term is 5 and last term is 251, we’ll be finding both general term and number of term using arithmetic sequence:

<u>Arithmetic Sequence</u>

$\displaystyle \large{a_n=a_1+(n-1)d}$

Where $\displaystyle \large{a_n}$ is the nth term, $\displaystyle \large{a_1}$ is the first term and $\displaystyle \large{d}$ is the common difference:

So for our general term:

$\displaystyle \large{a_n=5+(n-1)6}\\\displaystyle \large{a_n=5+6n-6}\\\displaystyle \large{a_n=6n-1}$

And for number of terms, substitute $\displaystyle \large{a_n}$ = 251 and solve for n:

$\displaystyle \large{251=6n-1}\\\displaystyle \large{252=6n}\\\displaystyle \large{n=42}$

Now we can convert the series to summation notation as given the formula above, substitute as we get:

$\displaystyle \large{\sum_{n=1}^{42}(6n-1)$

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2 years ago

Factor. m2 + 12m+36​

Factor. m2 + 12m+36