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Aug 28 2013 07:27pm
Been doing homework and studying literally all day, going brain dead and can't think.

Quote
Find the value of c such that the point P ( a, b ) lies on the graph of the function f.

f(x) = x√(16 − x²)+ c; P(3, 6)


Thanks much, might hand out a couple fg if quick answer.

This post was edited by KitsuneYosh on Aug 28 2013 07:28pm
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Aug 28 2013 07:36pm
y=x√(16 − x²)+ c

substitute in y=6, x=3, solve for c.
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Aug 28 2013 07:39pm
Quote (Exx @ Aug 28 2013 08:36pm)
y=x√(16 − x²)+ c

substitute in y=6, x=3, solve for c.


Jfc I knew I was gonna be pissed when I saw how simple it was.

Also you followin me?
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Aug 28 2013 07:43pm
Quote (KitsuneYosh @ 29 Aug 2013 01:27)
Been doing homework and studying literally all day, going brain dead and can't think.
Thanks much, might hand out a couple fg if quick answer.


so x is 3 and f(x)=6 [assuming that is what P(3,6) is meant to say]
now the right side is 3.sqrt(16-9)+c =3.sqrt(7)+c
which means:

6-3.sqrt(7) = c = -1.937....
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Aug 28 2013 07:51pm
Quote (brmv @ Aug 28 2013 08:43pm)
so x is 3 and f(x)=6 [assuming that is what P(3,6) is meant to say]
now the right side is 3.sqrt(16-9)+c =3.sqrt(7)+c
which means:

6-3.sqrt(7) = c = -1.937....


The problem I'm having here is this is an online assignment and I have to type the exact answers into boxes.

What the hell am I supposed to enter for a repeating decimal?
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Aug 28 2013 08:07pm
Quote (KitsuneYosh @ Aug 28 2013 06:51pm)
The problem I'm having here is this is an online assignment and I have to type the exact answers into boxes.

What the hell am I supposed to enter for a repeating decimal?


I'm going to go out on a limb and suggest there might be a way to enter the radical. Most of those kind of sites have a set of instructions somewhere (look on the main page) on how to enter stuff like fractions and radicals. BTW I hate these sites for this exact reason. :mad:

might need to write it as 6 - 3*(7)^(1/2)
or maybe 6 - 3*\sqrt{7}
or maybe there is a input/formatting bar to get symbols you need

This post was edited by Azrad on Aug 28 2013 08:12pm
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Aug 29 2013 02:22am
f(x) = x√(16 − x²)+ c; P(3, 6)

f(x) = x√(16 - x²) + c
y = x√(16 - x²) + c
- c = x√(16 - x²) - y
c = - x√(16 - x²) + y
c = - 3√(16 - 3²) + 6
c = 3√7 + 6

that's as close as you can get i believe
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Aug 29 2013 02:57am
Quote (Penguins0690 @ 29 Aug 2013 08:22)
f(x) = x√(16 − x²)+ c; P(3, 6)
f(x) = x√(16 - x²) + c
y = x√(16 - x²) + c
- c = x√(16 - x²) - y
c = - x√(16 - x²) + y
c = - 3√(16 - 3²) + 6
c = 3√7 + 6

that's as close as you can get i believe


1. you ever read other posts?
2. you always drop the "-" in the end? :lol:
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