A carpenter uses a hand saw to cut a piece of wood in half. The length of the saw blade is 40 cm, while the wood he is cutting is 8 cm across. Each time he extends the saw out with his arm, there is 5 cm of blade between the handle and the wood. When he pulls the saw toward his self with his arm, the saw handle is 25 centimetres from the wood. Each time he extends his arm out and pulls it back in, 1 second has passed. He starts with his arm extended out. Write an equation that models how far the tip of the saw is from the wood in terms of time.

The equation that models how far from the tip of the saw is from the wood in terms of time is x(t) = 10×cos(2×π×t) + 17

Step-by-step explanation:

The given parameters are;

Length of the saw blade = 40 cm

Thickness of the wood across = 8 cm

Length of blade between wood saw handle with hand extended = 5 cm

Length of the tip from the wood at the above time = 40 - 8 - 5 = 27 cm

Length of blade between wood saw handle with saw pulled inwards = 25 cm

Length of the tip from the wood at the above time = 40 - 8 - 25 = 7 cm

Time for one complete cycle = 1 second

We note that the basic equation for oscillatory motion is of the form;

x(t) = A·cos(ωt) + d

Where:

A = Amplitude of the motion = (27 - 7)/2 = 10 cm

ω = Angular frequency = 2·π/T

ωt = Motion's phase

t = Time of the motion

d = The middle location = 27 - 10 = 17 cm

T = The time to complete a cycle = 1 s

Therefore;

ω = 2·π

Given that he stars with his arm extended out, we have;

27 = 10×cos(2×π×0) + 17

Therefore, the equation that models how far from the tip of the saw is from the wood in terms of time is x(t) = 10×cos(2×π×t) + 17.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}