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36+28√25×-292(2+9-3)
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Answer:
If [math]f(2,8,5) = 24, f(3,6,9) = 36, f(4,7,8) = 28, f(5,5,8) = 15[/math] what is the value of [math]f(6,7,7)[/math] ?
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There are infinitely many solutions to this problem.
If you want f(6,7,7) to be x, you can always find a function f for which f(6,7,7) = x and also works for the other examples.
Lets say I want f(6,7,7) to be 24, then f(x, y, z) = 0.5 x^4 - 4.5 x^3 + 3x^2 + 50x - 60. Note that I did not take y and z into account. I can find a formula for every other solution you can come up with.
How I determined that formula:
The first argument of f is unique in all the examples, so I will drop the two other arguments and call it g. So g(2)=24, g(3) = 36, g(4) = 28, g(5) = 15 and g(6) = 24.
Our formula can have the following form:
g(x) = T2(x) + T3(x) + T4(x) + T5(x) + T6(x)
For which:
T2(2) = g(2) = 24, T3(2) = 0, T4(2) = 0, …
T2(3) = 0, T3(3) = g(3) = 36, T4(3) = 0, …
…
A solution for T2, T3, … could be:
T2(x) = (x-3)(x-4)(x-5)(x-6)C2
T3(x) = (x-2)(x-4)(x-5)(x-6)C3
But we still have to determine C2, C3, C4, … Lets do this for T2 by filling in x:
T2(2) = (2 - 3)(2 - 4)(2 - 5)(2 - 6)C2 = 24
-1*-2*-3*-4*C2 = 24
C2 = 24/24 = 1
All C values:
C2 = 1
C3 = -6
C4 = 7
C5 = -2.5
C6 = 1 (I was ‘lucky’ when I pick 24 as a solution)
So f(x, y, z) = g(x) =
(x-3)(x-4)(x-5)(x-6) -
6(x-2)(x-4)(x-5)(x-6) +
7(x-2)(x-3)(x-5)(x-6) -
2.5(x-2)(x-3)(x-4)(x-6)
Answer:
327004
this is the right answer
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