- anonymous

the company discovered that it costs $45 to produce 2 calculators, $143 to produce 4 calculators and $869 to produce 10 calculators. find the cost of producing 7 calculators. use the quadratic function to solve

- schrodinger

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- anonymous

Put the numbers into \[y =ax ^{2}+bx+c\] form. So your points would be (2, 450) (4, 143) and (10, 869). Plug the x in for the x in the equation and y in for y in the equation and then enter it into your matrix (3x4) in the graphing calculator, and it will give you 3 numbers. Put those in for a b and c and you have your equation to find the y for 7

- anonymous

i don't have a graphing calculator. that's why i posted the question so someone could help.

- anonymous

ok, do you know how to do augmented matrices?

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## More answers

- anonymous

i don't know how to do it at all. i have 5 questions on the assignment. and don't know how to do it. the online tutuorial doesn't explain how to do it very well or enough for me to figure it out.

- anonymous

Alright. Do you know what they are?

- anonymous

know what ?

- anonymous

What an augented matrice is?

- anonymous

no

- anonymous

It is a table kind of that gives you the a,b, c of the quadratic equation. You have to get it to be
101a
010b
101c

- anonymous

Your matrix would look like
4 2 1 450
16 4 1 143
100 10 1 869

- anonymous

Now you have to solve and such to transform you r matrix into the one above

- anonymous

you have to get the ones in the order of 2,1 3,1 3,2 2,3 1,3 1,2

- anonymous

then what?

- anonymous

correction of the order up above, it should be
1 0 0 A
0 1 0 B
0 0 1 C

- anonymous

to get the zeros, you have to do stuff like Row 1 x -25 ( multiply each number in row 1 by -25) and then row 1 + row 3=Row 3 so the first number in row 3 cancels out to 0. and then you can divide row 1 by -25 to get the original row again. But make sure when you add rows you add all of the numbers to the corresponding numbers.

- anonymous

You keep doing this to each place where you need a zero but make sure you don't lose any zeros making new zeros otherwise you'll get a continuous loop of math

- anonymous

then?

- anonymous

you do this to get all the zeros. so next you would do row 1 x -5 and then add it to row 3 to cancel out the 10 and make it 0

- anonymous

there's no easier way to find the answer?

- anonymous

only if you have a graphind calculator

- anonymous

now try solving for the third zero

- anonymous

Unless I'm missing something, can't the initial equations:
45 = (2^2)a+2b+c
143 = (4^2)a + 4b + c
869 = (10^2)a + 10b + c
...be solved in about 5 lines with normal Gaussian elimination? I may, of course, have misunderstood the set-up of this, so feel free to correct me.

- anonymous

I have never heard of gaussian before

- anonymous

you should gett the second zero by multiplying r1 x -4 and then adding it to r2.

- anonymous

'Gaussian elimination' is just a fancy way of saying use two equations to eliminate c, then the others to eliminate a or b, and then you have one. put that back in... Matrices are good in theory but almost always terribly laborious in practice if it's not a computer doing it.
To start:
45 = (2^2)a+2b+c (*)
143 = (4^2)a + 4b + c (**)
869 = (10^2)a + 10b + c (***)
(**) - (*) => 98 = 12 a + 2b
(***) - (**) => 726 = 84a + 6b
Repeat for these two and you have it done.

- anonymous

would you be willing to help me with the 5 questions?

- anonymous

Are they all a similar thing? Because if so seeing one should help you to do the others, but I could offer you more advice if you need it I guess. As there are 4 more I may just finish this one off for you:
Carrying on:
98 = 12 a + 2b (1)
726 = 84a + 6b (2)
(2) = 3 x (1)
=> 432 = 48 a => a = 9.
It follows that b = ... and c = ....

- anonymous

well it's not so much of advice, i have no idea how to do it.

- anonymous

OK post one more and I'll see what I can do to help. But I'm going to assume whatever you have to do, matrices is not it.

- anonymous

i can't. it only shows one question at a time

- anonymous

OK - do you need to give the answer to get the next one?
------
Well if you solve the equations as above, you should find a = 9, b = -5 c = 19
So the quadratic is "Cost = 9x^2 - 5x + 19", where x is the number of calculators. See if you can work out the cost of 7.

- anonymous

425?

- anonymous

I believe so.

- anonymous

that was correct. so do you think you could help with the others

- anonymous

Yes I probably can help for a bit if you post the next one, but I will try a little more this time to help you get the answer instead of giving it, so then you should be able to do the rest.

- anonymous

After a second, the rock is 58 feet in the air; after two seconds , it is 112 feet in the air. find the height, in feet, of the rock after 10 seconds.

- anonymous

Does it say you have to use a quadratic like before, or is this linear or something? It effects how you go about it.

- anonymous

quadratic. they all are.

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