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Maslowich
4 years ago
15

Evaluate sigma from (n-1 to 14) 3n+2

Mathematics
2 answers:
Harman [31]4 years ago
8 0
The given series 3n + 2 represents an Arithmetic sequence. 

Consider the first few terms of the sequence.

For n = 1, the term is 5
For n = 2,  the term is 8
For n = 3, the term is 11

Notice that the difference between the terms is constant. Hence it is an arithmetic sequence. We are to find the sum of first 14 terms of the sequence.

The formula for the sum of AP is:

S_{n}=  \frac{n}{2}(2a_{1}+(n-1)*d)

n = number of terms = 14
a1 = first term = 5
d = common difference = 3

Using the values, we get:

S_{14}= \frac{14}{2}(2*5+(13)*3)   \\  \\ 
S_{14}= 343

So, the answer to this question is 343
Vanyuwa [196]4 years ago
6 0
Arithmetic sequence
You can use the formula
Sum 14 terms  = (14/2) [ first term + last term]
 = 7 (5 + 44) = 7 *49 = 343 answer
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3 years ago
Factor by grouping:<br> a^2 + 2a + 6a + 12=
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Answer:

<em>Answer is </em><em>given below with explanations</em><em>. </em>

Step-by-step explanation:

{a}^{2} + 2 a + 6a + 12 = 0 \\   a(a + 2) + 6(a + 2) = 0 \\ (a + 6)(a + 2) = 0 \\ a =  - 6 \:  \:  \:  \:  \: (or) \:  \: a =  - 2

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6 0
3 years ago
Describe the dimensions of two different prisms that each have a volume of 2400 cubic centimeters using work
kirza4 [7]

Answer:

Step-by-step explanation:

Alright, lets get started.

We have given the volume of a prism that is 2400 cubic centimeters.

we have not given which type of prism it is, so we are taking it as a rectangular prism.

The formula of volume of rectangular prism is :

Volume = l*w*h

Where l is length of base

w is width of base

h is height of prism

So,  2400=l*w*h

So, if we factor 2400 in two different manners, we could have two different dimensions of this prism having same volume that is 2400.

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Hence we have two dimensions of prism, 20  20  6  and another one is 40  30  2.    :  Answer

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3 0
4 years ago
How do I solve this?
KengaRu [80]

Answer with Step-by-step explanation:

The given differential euation is

\frac{dy}{dx}=(y-5)(y+5)\\\\\frac{dy}{(y-5)(y+5)}=dx\\\\(\frac{A}{y-5}+\frac{B}{y+5})dy=dx\\\\\frac{1}{100}\cdot (\frac{10}{y-5}-\frac{10}{y+5})dy=dx\\\\\frac{1}{100}\cdot \int (\frac{10}{y-5}-\frac{10}{y+5})dy=\int dx\\\\10[ln(y-5)-ln(y+5)]=100x+10c\\\\ln(\frac{y-5}{y+5})=10x+c\\\\\frac{y-5}{y+5}=ke^{10x}

where

'k' is constant of integration whose value is obtained by the given condition that y(2)=0\\

\frac{0-5}{0+5}=ke^{20}\\\\k=\frac{-1}{e^{20}}\\\\\therefore k=-e^{-20}

Thus the solution of the differential becomes

 \frac{y-5}{y+5}=e^{10x-20}

5 0
3 years ago
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