Answer:
$2025
Step-by-step explanation:
Hope that helps!
Answer:
124
Step-by-step explanation:
because yes
Since f(x) is (strictly) increasing, we know that it is one-to-one and has an inverse f^(-1)(x). Then we can apply the inverse function theorem. Suppose f(a) = b and a = f^(-1)(b). By definition of inverse function, we have
f^(-1)(f(x)) = x
Differentiating with the chain rule gives
(f^(-1))'(f(x)) f'(x) = 1
so that
(f^(-1))'(f(x)) = 1/f'(x)
Let x = a; then
(f^(-1))'(f(a)) = 1/f'(a)
(f^(-1))'(b) = 1/f'(a)
In particular, we take a = 2 and b = 7; then
(f^(-1))'(7) = 1/f'(2) = 1/5
Answer:
six is 3 pointers 4 is the 2 pointers
Step-by-step explanation:
(1-cos^2(x)) csc^2(x)=1
one of the trigonometry rules is sin^2(x) + cos^2(x) = 1 if you rearrange this you realize that sin^2= 1-cos^2(x)
we also know that csc^2(x)= 1/sin^2(x) so now you can rewrite your equation as:
sin^2(x) x 1/sin^2(x) = 1
sin^2(x)/sin^2(x) =1
the LHS (left hand side) can cancel down to 1 because the numerator and denominator are the same
so then 1=1 Therefore LHS=RHS
Hope this helps
