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polet [3.4K]
3 years ago
10

Mandy baby sat for 3 1/4 hours on Friday night and 4 5/6 hours on Saturday. How many hours did she babysit this weekend?

Mathematics
1 answer:
torisob [31]3 years ago
3 0

8 1/3 hours is the answer ^ ^

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Sam has a rectangular field whose length and breadth are in the ratio 8:7. He wants to fence the field with bamboo to make it an
USPshnik [31]

Answer:

Length = 200 m

Breadth = 175 m

Step-by-step explanation:

Perimeter of the rectangular field = Total cost ÷ rate per metre

                                                        = 12000 ÷ 16

                                                        = 750 m

Length : breadth = 8 : 7

Length = 8x

Breadth = 7x

Perimeter = 750

2*(length + breadth) = 750

2*(8x + 7x) = 750

2* 15x = 750

  30x = 750

      x = 750/30

x = 25

Length = 8x = 8*25 = 200 m

Breadth = 7*25 = 175 m

8 0
2 years ago
Blank percent of 50 shirts
Dmitrij [34]
The answer is 10 because 5 10 15 20 25 30 35 40 45 50
8 0
3 years ago
PLEASE HELP ME PLEASE HELP
aliina [53]
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140/56 = 2.5 hours

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7 0
4 years ago
Read 2 more answers
Lamaj is rides his bike over a piece of gum and continues riding his bike at a constant rate time = 1.25 seconds the game is at
Hitman42 [59]

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.

7 0
3 years ago
A doctor's office has a computer file showing the heights of 100 male patients of one of the doctors of the practice. the height
kumpel [21]
A) Yes. we don't have enough information so this can not be determined.
B) No, it would just be an outlier.
7 0
3 years ago
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