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Hey old school mate! I know you
You look familiar
lol that was a prank to get you started with a bit of funXD
lol you look like a guy in my class
In question like this you need to start organizing the known values and keys that can lead to finding the final answer.
First you know that this is a quadratic equation we all know to be old school nostalgia
Ok since now that you know this is a quadratic function you find the coordinates to find in with other coordinates in the form of quadratic equation which has parabola and vertex just like your teacher's lesson
you need to assume that this is a quadratic equation and that all of the coordinates described conform with the shape of that parabola old school
You can either use your calculator to do it, just assign it to quadratic equation and enter the coordinates given. Or you can draw it on a cartesian plane to get the approximate amount.
Or you can go about calculating it by hand which would take a bit of trick ;) Let me know how it goes.
math stands for mental abuse to humans;) and you are the participant;)
I have no clue how to do what you described above
Ok. Which part? I don't understand.
any of it
let c(x) = ax^2 + bx + c general form of a quadratic function from the data 45 = a(2)^2 + 2b + c 143 = a(4)^2+ 4b + c 869 = a(10^2 + 10b + c from this we can find values of a , b and c then plug in x = 7
so the system of equations is 4a + 2b + c = 45 16a + 4b + c = 143 100a + 10b + c = 869
looks a bit scary but they can be solved
if you subtract the first one from the second then 2nd from the third you will eliminate the c and get 2 equations in a and b
12a + 2b = 98 84a + 6b = 726
now multiply first equation by -3 and add -36a - 6b = -294 84a + 6b = 726 adding:- 48a = 432 a = 432/48 = 9 a = 9
I've found a now you can find b and c gotta go