31.6 An older proof of Theorem 31.3 goes as follows, which we outline for c = 0. Assume x > 0, let M be as in the proof of Theorem 31.3, and let F(t) = f(t) + n −1 k=1 (x − t)k k! f(k) (t) + M · (x − t)n n! for t in [0, x]. Show F is differentiable on [0, x] and F (t) = (x − t)n−1 (n − 1)! [f(n) (t) − M].

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.

answer:

3 bags of apples and 2 bags of oranges.

step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

what do u need to know

step-by-step explanation: