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October 19th, 2017, 01:21 PM   #1
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A small business owner has determined that the cost to produce 12 tables in a month

hello people I was able to find the 875 but I am not sure where the 2500 comes from..thanks for the help!

c(n)=875n+2500
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October 19th, 2017, 01:37 PM   #2
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you have two ordered pairs ... (number of tables, cost)

$(12,13000)$ and $(22,21750)$

$C-13000 = \left(\dfrac{21750-13000}{22-12}\right) (N - 12)$

simplify ...
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October 19th, 2017, 02:07 PM   #3
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Quote:
Originally Posted by skeeter View Post
you have two ordered pairs ... (number of tables, cost)

$(12,13000)$ and $(22,21750)$

$C-13000 = \left(\dfrac{21750-13000}{22-12}\right) (N - 12)$

simplify ...
okay thanks I would have never guessed to set it up that way so thanks! just wondering for where you but (N-12) that could have also been (N-22) or is that a no
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October 23rd, 2017, 11:51 AM   #4
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Frankly, the question, as given
"Find a formula linear function that models the number of tables made in a month, C, if the number of tables made is n"
makes no sense!

This is apparently a translation into English of a problem that was not initially in English so I can forgive the awkward "formula linear function" but taking this literally, the answer would be "C(n)= n" but you and everyone who responded here has correctly interpreted this to say "models the cost of the tables made in a month, C, if the number of tables is n".

You want the equation of the line that passes through (x, y)= (12, 13000) and (x, y)= (22, 21750). You should know that any linear function can be written in the form y= ax+ b. With x= 12 and y= 13000 that is 13000= 12a+ b. With x= 22 and y= 21750, 21750= 22a+ b. Subtracting the first equation from the second, 8750= 10a, a= 8750/10= 875. Putting that value of a into 13000= 12a+ b gives 13000= 12(875)+ b, 13000= 10500+ b, b= 13000- 10750= 2500.
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October 25th, 2017, 11:55 AM   #5
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Quote:
Originally Posted by GIjoefan1976 View Post
okay thanks I would have never guessed to set it up that way so thanks! just wondering for where you but (N-12) that could have also been (N-22) or is that a no
that is a no.


$(x_1,y_1) = (12,13000)$, $(x_2,y_2) = (22,21750)$


$m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{21750-13000}{22-12} = 875$

or ...

$m = \dfrac{y_1-y_2}{x_1-x_2} = \dfrac{13000-21750}{12-22} = 875$


point-slope form of a linear equation ...

$y - y_1 = m(x - x_1)$

$y - 13000 = m(x - 12)$

or ...

$y - y_2 = m(x - x_2)$

$y - 21750 = m(x - 22)$


... if you use $y_1$, you need to use $x_1$; same idea with $y_2$ and $x_2$. Either way, you'll wind up with the same final equation.
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