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Veronika [31]
2 years ago
7

Circle A has a radius of 9.0 cm. The shortest distance between B and C on the circle is 8.5 cm. What is the approximate area of

the shaded portion of circle A?
30.0 cm²

38.25 cm²

56.5 cm²

254.5 cm²

Mathematics
1 answer:
love history [14]2 years ago
5 0

Answer:

38.25 cm²

Step-by-step explanation:

We use the formula for the length of an arc to find the central angle of the sector of the circle.

Then we use the formula for the area of a sector of a circle to find the area.

Length of arc of circle of radius r:

s = \dfrac{n}{360^\circ}2 \pi r

s = arc length

n = measure of the central angle of the sector

s = \dfrac{n}{360^\circ}2 \pi r

8.5~cm = \dfrac{n}{360^\circ}2 \pi \times 9.0~cm

n = 54.1^\circ

Area of sector of circle of radius r:

A = \dfrac{n}{360^\circ} \pi r^2

A = area of sector of circle

n = measure of the central angle of the sector

A = \dfrac{54.1^\circ}{360^\circ} \pi (9.0~cm)^2

A= 38.25~cm^2

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3 years ago
Sumi rides at a constant speed of 10 ft/s. Write an equation that expresses the distance d that sumi has traveled as a function
Shtirlitz [24]

Answer:

d = 10s

Step-by-step explanation:

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Given

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3 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
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Sindrei [870]
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provide a counterexample (specific values of a,b, etc. which make the statement false) for each of the following statements. Ass
ivolga24 [154]

Answer:

1. a=3, b = 2, c = 1.

2. a=2, b=3, c=4, d=6.

3. a = 9, b = 3.

Step-by-step explanation:

1.  3 | 21   is a counterexample  because 3 does not divide into  2 or 1.

2. 23 | 46 is a counterexample because 2 does not divide into 3.

3. 9 | 3^2 is a counterexample because 9 does not divide into 3.

8 0
2 years ago
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