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alekssr [168]
2 years ago
15

Find the area of the shaded portion in the equilateral triangle with sides 6. Show all work for full credit. (Hint: Assume that

the central point of each arc is its corresponding vertex.) 

Mathematics
2 answers:
Vadim26 [7]2 years ago
8 0
So... if you notice the picture below

each circle, has their central angle at the vertex of the triangle
that simply means, 3 circles with a radius of 3, overlapping the triangle

now, the area in the middle, the shaded one, will be, the whole area of the triangle MINUS those 3 circle sectors

hmmmm each sector has 60°, that means, all three of them will then be 60+60+60 or 180°, so the area of those three sectors, can be combined into a 180° sector, well, hell, 180° is really half a circle

so.... the area of those three sectors of 60° each, all three combined, is the same area of half a circle with a radius of 3

so    \bf \textit{area of an equilateral triangle}\\\\
A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\textit{length of one side}\\\\
-----------------------------\\\\
\textit{area of a circle}\\\\
A=\pi r^2\qquad r=radius\\\\
\textit{area of half a circle}\\\\
A=\cfrac{\pi r^2}{2}\\\\
-----------------------------\\\\

\bf \textit{now, let us use the side of 6, and radius of 3}
\\\\\\

\begin{array}{clclll}
\cfrac{6^2\sqrt{3}}{4}&-&\cfrac{\pi 3^2}{2}\\
\uparrow &&\uparrow \\
triangle's&&semi-circle's
\end{array}\impliedby \textit{area of shaded area}\\\\
-----------------------------\\\\
\boxed{\cfrac{36\sqrt{3}}{4}-\cfrac{9\pi }{2}}

you can, add the fractions if you want, or leave them like that, or get their difference by using their decimal format

Mamont248 [21]2 years ago
7 0

Answer:

9 sqrt 3 - 4.5 pi

Step-by-step explanation:

So, first off, you need to find the area of the whole triangle. Since this is an equilateral triangle, the formula for it would be:  

A = 1/4 * s^2 * sqrt 3 ----> A = 1/4 * 6^2 * sqrt 3 ----> A = 9 sqrt 3. So the area of the whole triangle is 9 sqrt 3.

Next, imagine that there are three circles that are part of the sectors (I am using the same picture as jdoe0001 attached). So to find the area of each sector, you would do 60o (o = degree sign) divided by 360o --> 60/360. You should get 1/6. Then you would multiply this by the area of the circle, which you find by doing the following: A = pi * r^2 ----> A = 9 pi. So when you would multiply 1/6 by the area of the circle, you would get 1.5 pi. Since this is just one sector, you need to multiply 1.5 pi by 3, because there are 3 sectors. You should get 4.5 pi.

Finally, to get the area of the shaded portion, you would simply do this: the area of the whole triangle minus the area of the sectors ----> <u>9 sqrt 3 - 4.5 pi</u>. This is the exact answer. The approximate answer would be 1.45.

<em>Hope this helps! :)</em>

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Answer:

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Step 2:

Rule for piece 1: y=-x

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Step-by-step explanation:

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6 0
2 years ago
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In choice situations of this type, subjects often exhibit the "center stage effect," which is a tendency to choose the item in t
victus00 [196]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability that he or she would choose the pair of socks in the center position is   p =\frac{1}{5}

The correct answer choice is

X has a binomial distribution with parameters n=100 and p=1/5  

b

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The standard deviation is \sigma=4

c

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d

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

Using the R the probability Pe = 0.0003

The probabilities P \approx Pe

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

                  p =\frac{1}{5}

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The mean of this distribution is mathematical represented as

           \mu = np

substituting the value

         \mu = 100 * 0.2

             \mu = 20

The standard deviation is mathematically represented as

         \sigma = \sqrt{np (1-p)}

substituting the value

           = \sqrt{100 * 0,2 (1-0.2)}

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By standardizing the normal approximation we have that

              P(X \ge 34) \approx P(Z \ge z)

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               z = \frac{x-\mu}{\sigma }

Substituting values

             z = \frac{34-20}{4}

               =3.5

So  using the z table the P(Z \ge 3.5) is  0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small  So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

 

6 0
2 years ago
If y is the circumcenter is angle STU find each measure
snow_tiger [21]

Answer:

Measures are SV=9 units., SY=14 units, YW=\sqrt75units , YW=\sqrt27units

Step-by-step explanation:

Given Y is the circumcenter of ΔSTU. we have to find the measures SV, SY, YW and YX.

As Circumcenter is equidistant from the vertices of triangle and also The circumcenter is the point at which the three perpendicular bisectors of the sides of the triangle meet.

Hence, VY, YW and YX are the perpendicular bisectors on the sides ST, TU and SU.

Given ST=18 units.

As VY is perpendicular bisector implies SV=9 units.

Also in triangle VTY

YT^{2}=VY^{2}+VT^{2}

⇒ 14^{2}=VY^{2}+9^{2}

⇒ VY^{2}=115

As vertices of triangle are equidistant from the circumcenter

⇒ SY=YT=UY=14 units

Hence, SY is 14 units

In ΔUWY, UY^{2}=YW^{2}+UW^{2}

⇒ 14^{2}=YW^2+11^{2}

⇒ YW^2=196-121=75 ⇒ YW=\sqrt75units

In ΔYXU, UY^{2}=YX^{2}+XU^{2}

⇒ 14^{2}=YX^2+13^{2}

⇒ YX^2=196-169=27 ⇒ YW=\sqrt27units

Hence, measures are SV=9 units., SY=14 units, YW=\sqrt75units , YW=\sqrt27units





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Then
Number of cats drawn by Caleb in the first instance = (18 * 6) cats
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Again Caleb draws 5 more dogs on each poster with dogs.
Then,
The number of Dogs drwn in the second instance = (5 * 4) dogs
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So, the total number of dogs drawn by Caleb = ( 56 + 20) dogs
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