Question

how do i do this

Answer 1

You need to simplify the numerator:

\([59(1 + h) - 0.83(1 + h)^2] - 58.17\)

Answer 2

First, square the binonmial \((1 + h)^2\).
If you don;t remember how top square a binomial, just write is as a product of two binomials an use FOIL.

\((1 + h)^2 = (1 + h)(1 + h) \)

Answer 3

\[h ^{2}+2h+1\]

Answer 4

so is the answer for the two blanks .83

Answer 5

Great.
Now distribute the 59 through the first set of parentheses and the -0.83 through the second set of parentheses.

Answer 6

Distribute each number in red by the set of parentheses that follows it.

\([\color{red}{59}(1 + h) \color{red}{- 0.83}(h^2 + 2h + 1)] - 58.17\)

Answer 7

i got the answer thanks for the help

Answer 8

why is this wrong

Answer 9

sorry wrong picture

Answer 10

what you need help with?

Answer 11

\(\dfrac{[59(1 + h) - 0.83(h^2 + 2h + 1)] - 58.17}{h} =\)

\(\dfrac{59 + 59h - 0.83h^2 -1.66 h -0.83 - 58.17}{h} =\)

\(\dfrac{57.34h - 0.83h^2 }{h} =\)

\(57.34 - 0.83h\)

Answer 12

estimating the speed when t=1

Answer 13

I figured out that equation mathstudent55 I don't understand how to estimate the speed when t=1

Answer 14

You are not asked for the average speed at t = 1.
You are being asked the average speed between t = 1 and t = 1.1, then t = 1
and t = 1.01, then t = 1 and t = 1.001.

Answer 15

so would the speed at t=1 be 57.34

Answer 16

What you calculated in the first part is how to find the average speed between t = 1 and t = 1 + h, h being an interval.
For t = 1, h = 0.
You are correct, at t = 0, v = 57.34

Answer 17

Then when you want the average speed from t = 1 to t = 1.1, use h = 1.1 - 1 = 0.1 in the formula
v = 57.34 - 0.83h

Answer 18

ok thanks

Answer 19

yw