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Basile [38]
2 years ago
11

DESCRIBE THE GENERAL STRUCTURE OF THE OCEAN FLOOR! (Not math)

Mathematics
2 answers:
pychu [463]2 years ago
5 0
Ummmmm mini volcanoes (I don’t know the exact name) fish lol I am trying to tell u as much I can remember
Fynjy0 [20]2 years ago
5 0

Answer:

A lot of sand :D

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Find the average of 2m,20cm and 95cm give your answer in cm​
stepan [7]

Answer:

The average is:

105 cm

Step-by-step explanation:

1 m = 100cm

2m = 2*100 = 200cm

then:

(200 + 20 + 95)/3 = 315/3 = 105cm

3 0
3 years ago
Jack has a rectangular sticker that is 6 centimeters long and 4 centimeters wide.
Solnce55 [7]
Its c because 6*4 = 24
Thats how you find the area
8 0
3 years ago
Given a+b=7 and a–b=3, find:<br><br> 2^a*2^b
seropon [69]

Answer:

128

Step-by-step explanation:

5+2=7

5-2=3

2^5=32

2^2=4

32*4=128

Hope this helps:)

7 0
3 years ago
SOLVE FOR BRAINLIEST PLEASE
grandymaker [24]

<u>Answer</u>:

x = 20

<u>Explanation</u>:

all the interior angles in the triangle is 180°

<u>therefore</u>:

4x - 17° + 71° + 46° = 180°

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4x = 180° - 100°

4x = 80°

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3 0
2 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
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