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Aleks [24]
3 years ago
12

Find the sum of the summation of 3 i minus 15, from i equals 2 to 7.

Mathematics
2 answers:
Mrac [35]3 years ago
3 0
The summation symbol means we use the variable i in the equation from the given range. In this case, we are given with the equation 3 i - 15 and i ranges from 2 to 7. The summation using simply the calculator is equal to -9. The answer to this problem is -9.
Shkiper50 [21]3 years ago
3 0

Answer:

\sum _{i=2}^73i-15=-9

Step-by-step explanation:

Given : \sum _{i=2}^73i-15

We have to evaluate the given summation.

Consider \sum _{i=2}^73i-15

using, we have,

\sum _{k=m}^n\:=\:\sum _{k=1}^n\:-\:\sum _{k=1}^{m-1}

=\sum _{i=1}^73i-15-\sum _{i=1}^13i-15      ..............(1)

Consider, \sum _{i=1}^73i-15

Apply sum rule,

\quad \sum a_n+b_n=\sum a_n+\sum b_n , we have,

=\sum _{i=1}^73i-\sum _{i=1}^715

Now, first consider, \sum _{i=1}^73i

Apply constant multiplication rule,\quad \sum c\cdot a_n=c\cdot \sum a_n we have,

=3\cdot \sum \:_{i=1}^7i

=3\cdot \:28\\\\=84

Also, \sum _{i=1}^715=105

Thus, \sum _{i=1}^73i-15=84-105=-21

Similarly, \sum _{i=1}^13i-15=-12

Substitute in (1) , we have,

Thus, \sum _{i=2}^73i-15=-21-(-12)=-21+12=-9

Thus, \sum _{i=2}^73i-15=-9

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Rewrite in simplest terms: 9(m-9)-4(10m+2)9(m−9)−4(10m+2)
trasher [3.6K]

The required simplified expression for 9(m - 9) - 4(10m + 2) is  -31m - 90.

Given that,
To rewrite in simplest terms: 9(m-9)-4(10m+2).


<h3>What is simplification?</h3>

The process in mathematics to operate and interpret the function to make the function simple or more understandable is called simplifying and the process is called simplification.

Question asked to simplify the term 9(m-9)-4(10m+2).

= 9(m - 9) - 4(10m + 2)
= 9m - 81 - 40m - 8
=  -31m - 90

Thus, the required simplified expression for 9(m - 9) - 4(10m + 2) is  -31m - 90.

Learn more about simplification here: brainly.com/question/12501526

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5(x+2)-7=-32 solve variable. include steps, also check your answer!
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Answer:

Step-by-step explanation:

5x+10-7=-32

5x+3=-32

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Check:

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