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ale4655 [162]
3 years ago
10

A circle with radius \pink{6}6start color #ff00af, 6, end color #ff00af has a sector with a central angle of \purple{48^\circ}48

∘ start color #9d38bd, 48, degrees, end color #9d38bd. What is the area of the sector? Either enter an exact answer in terms of \piπpi or use 3.143.143, point, 14 for \piπpi and enter your answer as a decimal rounded to the nearest hundredth.
Mathematics
2 answers:
alexandr1967 [171]3 years ago
7 0

Answer:

Therefore,

The area of the sector is 15.09 unit².

Step-by-step explanation:

Given:

Circle with,

radius = r = 6 unit

central angle = θ = 48°

pi = 3.143

To Find:

Area of sector = ?

Solution:

If  'θ' is in degree the area of sector is given as

\textrm{Area of Sector}=\dfrac{\theta}{360}\times \pi r^{2}

Substituting the values we get

\textrm{Area of Sector}=\dfrac{48}{360}\times 3.143\times 6^{2}

\textrm{Area of Sector}=15.0864=15.09\ unit^{2} rounded to nearest hundredth

Therefore,

The area of the sector is 15.09 unit².

matrenka [14]3 years ago
3 0

Answer:

24/5 pi

Step-by-step explanation:

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Based on the sample results of the population the percentage of population that  prefers reading a hard copy i fht e digital copy is 4, the hard copy is 21 and n equas 25 is 84%

Step-by-step explanation:

From the above question  we know that .

<u>Total number of population, n = 25</u>

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3 years ago
Find the absolute value of 16
myrzilka [38]

Answer:

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Please help this is important for me.
Anit [1.1K]
<h3>Answer: point Q is located at (-1, 1)</h3>

=================================================

Explanation:

Check out the diagram below.

Plot R(-3,7) and T(3,-11) on the same xy grid.

Draw a vertical line through R and a horizontal line through T. A right triangle forms. At the intersection point is point S(-3,-11)

----------------------

Now measure the distance from point S to point T. You can count out the spaces or subtract the x coordinates and use absolute value.

|x1-x2| = |-3-3| = |-6| = 6

From point S to point T is 6 units.

We want to subdivide this horizontal length in the ratio 1:2

What this means is that we want to plot a point U somewhere such that SU:UT = 1:2

In other words,

SU = x

UT = 2x

SU+UT = ST

x+2x = 6

3x = 6

x = 6/3

x = 2

So we must move 2 spaces to the right from point S to get to U(-1,-11)

Going from point U(-1,-11) to T(3,-11) is 4 spaces

We have SU:UT = 2:4 = 1:2 to help confirm we have the correct location for point U

From point U, we then move straight up to the line segment RT

We'll land on Q(-1,1)

----------------------

Another way to find the y coordinate of point Q is to subdivide the segment RS into the ratio 1:2 similar to how we divided ST up

Segment RS is 18 units long since we go from y = 7 to y = -11 when going from R to S.

If V was on segment RS such that

RV:VS = 1:2

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then

RV+VS = RS

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3y = 18

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RV = y = 6

VS = 2y = 2*6 = 12

So you'll move 6 units down from y = 7 to land on y = 1 (when going from the y coordinate of R to the y coordinate of Q)

3 0
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