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IrinaVladis
3 months ago
7

A merry-go-round has a radius of 18 feet. If a passenger gets on a

Mathematics
1 answer:
Zina [12.3K]3 months ago
6 0

Answer:

The rotation angle measures 2.11 °

Step-by-step explanation:

Stated as follows:

The radius of the circular path = r = 18 feet

The distance rolled by the wheel = l = 38 feet

Let us denote the angle of rotation as Ф

Now, according to the problem:

∵ the length of an arc at the center corresponds to an angle Ф

Thus,

distance rolled by the wheel = π × radius × \frac{\Theta }{180^{\circ}}

As 180° represents π radians

And π approximates to 3.14

Thus, distance rolled by the wheel = 180 °× radius × \frac{\Theta }{180^{\circ}}

That is l = r × Ф

So, Ф = \frac{l}{r}

Consequently, Ф = \frac{38 feet}{18 feet}

Therefore, Ф = 2.11 °

Thus, the rotation angle is Ф = 2.11 °

Hence, the rotation angle is 2.11 ° Answer

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Determine the multiplicity of the roots of the function k(x) = x(x + 2)3(x + 4)2(x − 5)4. 0, -2, -4, 5
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Answer:

The polynomial expression k(x)=x(x+2)^3(x+4)^2(x-5)^4

To ascertain the multiplicity of 0, -2, -4, 5.

The multiplicity refers to how many times a root appears in a function.

First, identify the function's roots by setting it to zero.

x(x+2)^3(x+4)^2(x-5)^4=0

Thus, the roots are x=0, -2, -4, and 5

To determine the roots' multiplicities:

The factor of x will give a root of x=0 with a multiplicity of 1

likewise,  x=-2 with a multiplicity of 3

x=-4 with a multiplicity of 2

and x=5 with a multiplicity of 4.



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Explain how to multiply 28 and 36 using the set of steps you learned.
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Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the p
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Answer:

Colin has 8 sheets remaining for his third class.

Step-by-step explanation:

Let's summarize the details:

Total number of sheets of paper = x

The number of sheets used in the first class is five less than half of the total number.

We create the equation:

\text{Number of pieces of papers used for 1st class =} \dfrac{x}{2} -5...... (1)

Moreover, for the second class, the number of sheets used is two more than the amount used in the first class.

\text{Number of pieces of papers used for 2nd class =} \dfrac{x}{2} -5+2 = \dfrac{x}2 -3...... (2)

Therefore, the total sheets remaining for the third class is calculated as: Total sheets - Sheets used in the first class - Sheets used in the second class

\text{number of pieces of papers left for the third class = }x-(\dfrac{x}{2}-5)-(\dfrac{x}{2}-3)\\\Rightarrow x-\dfrac{x}2-\dfrac{x}2+5+3\\\Rightarrow x-x+5+3\\\Rightarrow 8

Consequently, the final result is:

Colin has 8 sheets remaining for his third class.

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