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I don't get these questions. please help me with them, thanks :)

 

first question

 

 

second question:

 Mar 22, 2019
edited by Guest  Mar 22, 2019
 #1
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1150 + 150n     n from 1 to 12

n = 12    an = 1150+1800=2950

 

s= n (a1+ an)/2

12(1150+2950)/2=24600

 Mar 22, 2019
 #4
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(1300, 1450, 1600, 1750, 1900, 2050, 2200, 2350, 2500, 2650, 2800, 2950) =$25,500

Guest Mar 22, 2019
 #2
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Well the resturant makes 2950 in 12 months here is the math\

 

Well the resturant starts with $1,300

 

so each month it earns $150 

 

so 

150*11=1650

 

Then 

1,650        =                          2,950 a year I should say that is how much it makes in 12 months 

1,300        =

 

 

Hope This helps (;

 Mar 22, 2019
 #3
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Sum =N/2[2*F + (N - 1)*D, where N=12, F=1,300, D=150
Sum =12/2[2*1,300 + (12 - 1)*150
Sum =6[2,600 + (11*150)]
Sum =6[2,600 + 1,650]
Sum =6*[4,250]
Sum=$25,500 - Restaurant's profit in 12-month period.

 Mar 22, 2019
 #5
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How is that

Nickolas  Mar 22, 2019
 #6
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Because they get a 150 profit every year you took it as if th profit started in febuary..

HiylinLink  Mar 22, 2019
 #7
avatar+1013 
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So I am wrong then???

Nickolas  Mar 22, 2019
 #8
avatar+476 
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Slightly....

HiylinLink  Mar 23, 2019
 #9
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What is the correct answer then, HiyinLink?

Guest Mar 23, 2019
 #10
avatar+1013 
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Well two people agreed on mine it is not right to say mine is wrong without prof 2 people said mine was right 2

Nickolas  Mar 23, 2019
 #11
avatar+476 
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Well I have no Idea which guest is talking but it was both guest users the equation Idea I though was an excellent way to put it but unfortuanatly 6th graders aren't familar with full algrbra concepts yet...

HiylinLink  Mar 23, 2019
 #12
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I'm the one who posted the question that you are all replying to :) So I'm the one asking HiyinLink what the correct answer is.

Guest Mar 23, 2019
edited by Guest  Mar 23, 2019
 #13
avatar+476 
-1

Cool well good XD yup I liked the equation it was an excellent Idea we can just simplify it for him if he's not familiar...

HiylinLink  Mar 23, 2019
 #14
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The subject this question is from is arithmetic series & applications, and i'm in algebra 2.

Guest Mar 23, 2019
edited by Guest  Mar 23, 2019
 #15
avatar+1013 
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I see you speaking

\

Nickolas  Mar 23, 2019
 #16
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Just simplify my answer I am sorry about the un accracy well just simplthy my answer XD goodnight

Nickolas  Mar 23, 2019
 #17
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Can anybody help me with the second question?

Guest Mar 23, 2019
 #18
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-2

um someone did

Nickolas  Mar 23, 2019
 #19
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Who did lol, HiyinLink or the guest that first posted?

Guest Mar 23, 2019
 #20
avatar+476 
-1

I am in pre-Algebra so eight grade...

HiylinLink  Mar 23, 2019
 #21
avatar+476 
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I might be able to help you tommorrow I GTG though if not someone else that may get to it before me...

 Mar 23, 2019
 #22
avatar+476 
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But  trick fo the final 1 thre is an increase of three evry row so going down sven you can do it like this you just addd the new values everytime you go down..

 Mar 23, 2019
 #23
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Ok, syt, but the answer should be 168? If I do 15+18+21+24+27+30+33 I get 168.

Guest Mar 23, 2019
 #24
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Second question :

 

We have the following arithmetic sequence

15, 18, 21

 

The 7th row has     15 + 3(n - 1)   =  15 + 3(7 - 1)   =  15 + 3(6)  = 33 logs

 

So.....the total number of logs is

 

( 15 + 33) * 7 / 2 =

 

48 * 7  / 2  =

 

168 logs in total

 

 

cool cool cool

 Mar 23, 2019

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