(x0-y0)(x0 + hy0 + k)=f(x)
2006-11-26 19:15:30 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
It seems that no matter what you make X, Y, K, and H to be the answer is going to be 0=0 because when you multiply times 0 the answer is always 0.
The first component (x0-y0) is going to be 0 no matter what X and Y are. Then for the second component (x0 + hy0 + k) it really doesnt matter what any of those are either because the whole thing is still going to be multiplied by the first component, 0.
The only letter you can give a definative number to is F which would equal 0.
So for fun lets make X=1, Y=2, H=3, K=4, and F=0. Then you get:
(1x0 - 2x0)(1x0 + 3x2x0 + 4)=0x1
(0 - 0)(0 + 0 + 4)=0
0x4=0
0=0
2006-11-26 19:24:17 · answer #1 · answered by supermonkey081 2 · 1⤊ 1⤋
the equation (x0-y0)(x0 + hy0 + k)=f(x)
the answer is just zero because the term (x0 - y0)
is equal to 1-1 = 0
any number multiplied by zero is always equal to zero.
i'm sure with my answer!
2006-11-27 05:43:16 · answer #2 · answered by jdash01 3 · 0⤊ 1⤋
At the time of day that I am looking at your question, I'd rather try and answer the questions about Trains leaving New York traveling 35 mpg, and one leaving San Antonio going 37 mph..... Its been a while for math for me, but anything with a 0, multiplied out, would make even the *k* with a 0 amount, so f(x)=1? ....
2006-11-27 03:22:50 · answer #3 · answered by dontblamemeivoted 3 · 0⤊ 1⤋
this is a generalised equation that approximates derivatives, there is no specific solution as such.
the zeros at the end of the variables are infact subscripts
2006-11-27 04:31:11 · answer #4 · answered by yasiru89 6 · 1⤊ 0⤋
ok i can solve your problem. look what is your problem first
you explain perfectly
2006-11-27 03:31:49 · answer #5 · answered by ajay b 1 · 0⤊ 1⤋
Sure
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2006-11-27 03:16:56 · answer #6 · answered by COOKIE 5 · 0⤊ 1⤋