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3 years ago

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Mike and Dena are painting a fence. if Mike has done 1/3 of the job and Dena has done 2/7 of the job, what fraction of the job i

s left to be finished?

1 answer:

ludmilkaskok [199]3 years ago

3 0

Answer:

8/21

Step-by-step explanation:

The total job is 1

Mike has done 1/3 and Dena has done 2/7 of the job We still need to do x of the job to complete it

1/3 + 2/7+ x = 1

We need to get a common denominator 3*7 = 21

Multiply each side by 21

21(1/3 + 2/7+ x) = 1*21

21(1/3) + 21*(2/7)+ 21x = 21

7+ 3*2 +21x =21

7+6+21x = 21

13+21x = 21

Subtract 13 from each side

13-13+21x=21-13

21x = 8

Divide each side by 21

21x/21 = 8/21

x = 8/21

We need to complete 8/21 of the job

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Lamaj rides his bike over a piece of gum and continues riding his bike at a constant rate. At time = 1.25 seconds, the gum is at a maximum height above the ground and 1 second later the gum is on the ground again.

a. If the diameter of the wheel is 68 cm, write an equation that models the height of the gum in centimeters above the ground at any time, t, in seconds.

b. What is the height of the gum when Lamaj gets to the end of the block at t = 15.6 seconds?

c. When are the first and second times the gum reaches a height of 12 cm?

Answer:

Step-by-step explanation:

a)

We are being told that:

Lamaj rides his bike over a piece of gum and continues riding his bike at a constant rate. This keeps the wheel of his bike in Simple Harmonic Motion and the Trigonometric equation that models the height of the gum in centimeters above the ground at any time, t, in seconds. can be written as:

$\mathbf {y = 34cos (\pi (t-1.25))+34}$

where;

y = is the height of the gum at a given time (t) seconds

34 = amplitude of the motion

the amplitude of the motion was obtained by finding the middle between the highest and lowest point on the cosine graph.

$\mathbf{ \pi}$ = the period of the graph

1.25 = maximum vertical height stretched by 1.25 m to the horizontal

b) From the equation derived above;

if we replace t with 1.56 seconds ; we can determine the height of the gum when Lamaj gets to the end of the block .

So;

$\mathbf {y = 34cos (\pi (15.6-1.25))+34}$

$\mathbf {y = 34cos (\pi (14.35))+34}$

$\mathbf {y = 34cos (45.08)+34}$

$\mathbf{y = 58.01}$

Thus, the gum is at 58.01 cm from the ground at t = 15.6 seconds.

c)

When are the first and second times the gum reaches a height of 12 cm

This indicates the position of y; so y = 12 cm

From the same equation from (a); we have :

$\mathbf {y = 34 cos(\pi (t-1.25))+34}$

$\mathbf{12 = 34 cos ( \pi(t-1.25))+34}$

$\dfrac {12-34}{34} = cos (\pi(t-1.25))$

$\dfrac {-22}{34} = cos(\pi(t-1.25))$

$2.27 = (\pi (t-1.25)$

t = 2.72 seconds

Similarly, replacing cosine in the above equation with sine; we have:

$\mathbf {y = 34 sin (\pi (t-1.25))+34}$

$\mathbf{12 = 34 sin ( \pi(t-1.25))+34}$

$\dfrac {12-34}{34} = sin (\pi(t-1.25))$

$\dfrac {-22}{34} = sin (\pi(t-1.25))$

$-0.703 = (\pi(t-1.25))$

t = 2.527 seconds

Hence, the gum will reach 12 cm first at 2.527 sec and second time at 2.72 sec.

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