Supposing that f, g: [a, b]→r are continuous functions differentiable on (a, b)(a, b) and h(t)=(f(b)−f(a))g(t)−(g(b)−g(a))f(t )h(t)=(f(b)−f(a))g(t)−(g(b)−g(a))f( t), there is some t∈(a, b)t∈(a, b) with h′(t)=0h′(t)=0

See proof below

Step-by-step explanation:

This is known as the generalized mean value theorem. To prove it, the idea is to apply the mean value theorem to an adequate function.

Define the function . Because f and g are differentiable on (a,b) and continuous on [a,b], s is also differentiable on (a,b) and continuous on [a,b]. Therefore, applying the mean value theorem to s on [a,b], there exist some t∈(a,b) such that (b-a)s'(t)=s(b)-s(a).

From the definition of s, we see that s(b)=s(a)=0 then s(b)-s(a)=0. Furthermore, using the definions of s and h, (b-a)s'(t)=-(f(b)−f(a))g('t)+(g(b)−g(a))f'(t)=-h'(t). From this, we conclude that -h'(t)=0, then h'(t)=0 and t is the required point.

the answer is 96 degrees

step-by-step explanation:

you have to realize that a line equals 180 degrees

sas (side angle side)

aa (angle angle)

sss(side side side)

sas find two sides with an angle in between

aa is finding two angles and the third will add up to 180

sss find all sides

step-by-step explanation: