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Westkost [7]
3 years ago
6

What is 9.09 repeating as a fraction?

Mathematics
1 answer:
Alborosie3 years ago
3 0
9 1/11
or 100/11
they are both the same, it just depends on how the class wants you to write it
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Alice says that all the factors of 8 are even.
Rom4ik [11]
1 is not even yet it is a factor of 8. Alice is wrong.
6 0
2 years ago
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
How do you solve -1/12 - (-11/12)
valentinak56 [21]

-1/12 - (-11/12) would change into -1/12 + 11/12 because a double negative (bold print) would become a positive. The answer to the new problem would be 10/12 or 5/6 (depending on if it says to simplify).

7 0
3 years ago
Plllzzz im new and i neeed help
ella [17]

Answer:

4082

Step-by-step explanation:

Given

The composite object

Required

The volume

The object is a mix of a cone and a hemisphere

Such that:

<u>Cone</u>

r = 10cm ---- radius (r = 20/2)

h = 19cm

<u>Hemisphere</u>

r=10cm

The volume of the cone is:

V_1 = \frac{1}{3}\pi r^2h

V_1 = \frac{1}{3}\pi * 10^2 * 19

V_1 = \frac{1900}{3}\pi

The volume of the hemisphere is:

V_2 = \frac{2}{3}\pi r^3

V_2 = \frac{2}{3}\pi 10^3

V_2 = \frac{2000}{3}\pi

So, the volume of the object is:

V = V_1 + V_2

V = \frac{1900}{3}\pi + \frac{2000}{3}\pi

V = \frac{3900}{3}\pi

V = 1300\pi

V = 1300 * 3.14

V = 4082

8 0
3 years ago
Freeeee p0ints!! :) <br> have a good day
Tema [17]

Answer:

thx i ned the points lol

Step-by-step explanation:

bErS

8 0
2 years ago
Read 2 more answers
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