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
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
i don't have a graphing calculator. that's why i posted the question so someone could help.
ok, do you know how to do augmented matrices?
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.
Alright. Do you know what they are?
know what ?
What an augented matrice is?
no
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
Your matrix would look like 4 2 1 450 16 4 1 143 100 10 1 869
Now you have to solve and such to transform you r matrix into the one above
you have to get the ones in the order of 2,1 3,1 3,2 2,3 1,3 1,2
then what?
correction of the order up above, it should be 1 0 0 A 0 1 0 B 0 0 1 C
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.
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
then?
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
there's no easier way to find the answer?
only if you have a graphind calculator
now try solving for the third zero
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.
I have never heard of gaussian before
you should gett the second zero by multiplying r1 x -4 and then adding it to r2.
'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.
would you be willing to help me with the 5 questions?
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 = ....
well it's not so much of advice, i have no idea how to do it.
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.
i can't. it only shows one question at a time
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.
425?
I believe so.
that was correct. so do you think you could help with the others
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.
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.
Does it say you have to use a quadratic like before, or is this linear or something? It effects how you go about it.
quadratic. they all are.
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