CAN SOMEONE PLEASE HELP ME?
I think you can do part A yourself.... go for it. (But you got to explain yourself, not just yes or no).
no... because ,,, the x valuse are the same for more than one ordered pair...?
is that right, @SolomonZelman
@SolomonZelman ...? plz help me...
yes correct, so part A is no !
yes i was right
but what about the ressst
Part B. \(\normalsize\color{blue}{ f(x)=-x+6 }\) when you say f(4) you just plug in 4 insteaf of x, into `-x+6`, can you do that for me?
instead of x.
-2 = -x...?
so it becomes x = 2, right...?
Yes, part B is also correct, \(\normalsize\color{black}{f(4)=-2 }\) !!!
yaay! but wut bout c
For Part C it looks as though, as you just plug in p instead of x, because x is the distance from the hoop and p is also (the new) distance from the hoop.
O_O i have no idea what that means
Looks like it should be, \(\normalsize\color{black}{f(x)=-p+6 }\), but it wouldn't make sense. I am kind of stuck on this part. Sorry-:(
me too lolz but thx anyway
Did whatever I can, I have some very significant 'holes' in my knowledge in math.... you welcome though.
lol
I NEEED HELP ON PART C
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