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Romashka-Z-Leto [24]
3 years ago
8

Can someone help me solve #1 and #2 by working backward? Here is a picture:

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
3 0
1. 4(n+3)=16, n+3=4, n=1
2. m/4+3=24, m/4=21, m=84
1. three added to a number n implies n+3, 4 times that sum is 4 (n+3), to solve divide both sides by 4 then subtract three from both sides.
2. a number divided by 4 means n/4 adding it to three n/4+3, to solve subtract 3 from both sides then multiply both sides by 4
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How do you find a vector of length 10 in the direction of vequals=left angle 3 comma negative 2 right angle3,−2​?
Kazeer [188]
First we look for the angle of the vector, which will be given by:
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4 0
2 years ago
Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15. N an 1 4 2 −12 3 36 t
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By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is                    \sum_{n=4}^{15} 5(-2)^{n-1}$$  

<h3>What is sequence ?</h3>

Sequence is collection of  numbers with some pattern .

Given sequence

a_{1}=5\\\\a_{2}=-10\\\\\\a_{3}=20

We can see that

\frac{a_1}{a_2}=\frac{-10}{5}=-2\\

and

\frac{a_2}{a_3}=\frac{20}{-10}=-2\\

Hence we can say that given sequence is Geometric progression whose first term is 5 and common ratio is -2

Now n^{th}  term of this Geometric progression can be written as

T_{n}= 5\times(-2)^{n-1}

So summation of 15 terms can be written as

\sum_{n=4}^{15} T_{n}\\\\$\\$\sum_{n=4}^{15} 5(-2)^{n-1}$$

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is                    \sum_{n=4}^{15} 5(-2)^{n-1}$$  

To learn more about Geometric progression visit : brainly.com/question/14320920

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x=a(m-n)
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