I'm taking the liberty of editing your function <span>v = e5xey: It should be
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<span>v = e^5x^ey, with " ^ " indicating exponentiation.
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Did you mean e^(5x) or (e^5)x? I'll assume it's e^(5x).
The partial of v = e^(5x)e^y with respect to x is e^(5x)(5)*e^y, or 25x*e^y.
The partial of v = e^(5x)e^y with respect to y is e^(5x)e^y.
P(H)=1/2, P(1)=1/6
P(H and 1)=(1/2)(1/6)=1/12 (≈8.33%)
Step-by-step explanation:
Explanation:
Because
(2,6) is in the function, we can plug
6 as y
and
2 as x.
From that we get
6=a(2−1)2+k
6=a+k
Same for the next point:
12=a(3−1)2+k
12=22a+k
12=4a+k
Now we have this system of equations to solve:
a+k=6
4a+k=12
Solving it leads to the conclusion that
a=2,k=4 .