Use linear approximation, i. e. the tangent line, to approximate 1/0.201 as follows: let f (x) = 1/x and find the equation of the tangent line to f(x) at a "nice" point near 0.201 . then use this to approximate 1/0.201.

f(0.201) ≈ 4.975  

Step-by-step explanation:

1. Calculate the position of the point and the slope of the tangent

When x = 0.2

So, the point is at (0.2, 5) and the slope of the tangent is -25.

2. Calculate the equation of the tangent line

y = mx+ b

Substitute the coordinates of the point

5 = -25×0.2 + b

5 = -5 + b

b = 10

The equation of the tangent line is

y = -25x + 10

3. Calculate y when x = 0.201

y = -25×0.201 + 10 = -5.025 + 10 = 4.975

4. How good is the approximation?

The actual value of f(0.201) is 4.975 124 …

the right system of equations to describe the situation would

be on the form:

x1 = 8000 + y1*t

and

x2 = 8000 + y2*t

where x1 and x2 represents the total money of imogene and her

friend respectively at the end of t years.

now for the value of amount earned, y1 and y2:

y1=8000*0.08

y2=2000*√(t-2)

lol.

it's ok we all do this sometimes.

-aparri