F(x)= 0.02x^2 + 0.08x + 30.6
so for f(0) all that means is where you see x plug 0 for x..
f(0)= 0.02(0)^2+ 0.08(0) + 30.6
f(0)= 30.6
f(4)= 0.02(4)^2 + 0.08(4) + 30.6
f(10)= 0.02(10)^2 + 0.08(10) + 30.6
attached are calculated answers
Hello,
(tg(x))'=1/cos²(x)
(arcsin(x))'=1/√(1-x²)
cos²(arcsin(x))=1-sin²(arcsin(x))=1-x²
Let x,y be two different numbers
suppose x^2=y^2
then x^2-y^2=0
which yields (x+y)(x-y)=0
so either x=y or x=-y
In any case, x and y must be the same value
also when a vairable is squared like y=x^2
we must note that there are 2 possible solutions
x=(+/-)sqrt(y)
Step-by-step explanation:
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Hope these helps. There are many other important factors as well. You can ask any time for any help in this. Would be glad to help. :)
There's no diagram, so I made my own.
But I can't prove AD = 2*AB
