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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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Which shows one way to determine the factors of 12x3 – 2x2 + 18x – 3 by grouping?
AleksAgata [21]

we are given

12x^3-2x^2+18x-3

Firstly, we will split terms

2x^2*6x-1*2x^2+3*6x-1*3

We can group first two and last two terms

(2x^2*6x-1*2x^2)+(3*6x-1*3)

we can see that 2x^2 is common in first two terms

and 3 is common in last two terms

so, we can factor factor out 2x^2 from first group and 3 from second group

we get

=2x^2(6x-1)+3(6x-1)

we can see that each terms are multiple of 6x-1

so, we can factor out 6x-1

so, we get

12x^3-2x^2+18x-3 =(2x^2+3)(6x-1)............Answer


4 0
2 years ago
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Joanna set a goal to drink more water daily. The number of ounces of water she drank each of the last seven days is shown below.
butalik [34]
(a) Data with the eight day's measurement.
Raw data:      [60,58,64,64,68,50,57,82], 
Sorted data:  [50,57,58,60,64,64,68,82]
Sample size = 8 (even)
mean            = 62.875
median         = (60+64)/2 = 62
1st quartile   = (57+58)/2 = 57.5
3rd quartile  = (64+68)/2 =  66
IQR = 66 - 57.5 = 8.5

(b) Data without the eight day's measurement.
Raw data:      [60,58,64,64,68,50,57]
Sorted data:  [50,57,58,60,64,64,68]
Sample size = 7 (odd)
mean            = 60.143
median         = 60
1st quartile   = 57
3rd quartile = 64
IQR = 64 -57 = 7

Answers:
1. The average is the same with or without the 8th day's data.  FALSE
2. The median is the same with or without the 8th day's data.  FALSE
3. The IQR decreases when the 8th day is included.                  FALSE
4. The IQR increases when the 8th day is included.                   TRUE
5. The median is higher when the 8th day is included.              TRUE

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2 years ago
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The histogram shows a city’s daily high temperatures recorded for four weeks. A graph shows temperature (degrees Fahrenheit) the
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Answer:

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Step-by-step explanation:

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One of your peers claims that boys do better in math classes than girls. Together you run two independent simple random samples
JulijaS [17]

Answer:

Step-by-step explanation:

Hello!

To test if boys are better in math classes than girls two random samples were taken:

Sample 1

X₁: score of a boy in calculus

n₁= 15

X[bar]₁= 82.3%

S₁= 5.6%

Sample 2

X₂: Score in the calculus of a girl

n₂= 12

X[bar]₂= 81.2%

S₂= 6.7%

To estimate per CI the difference between the mean percentage that boys obtained in calculus and the mean percentage that girls obtained in calculus, you need that both variables of interest come from normal populations.

To be able to use a pooled variance t-test you have to also assume that the population variances, although unknown, are equal.

Then you can calculate the interval as:

[(X[bar]_1-X[bar_2) ± t_{n_1+n_2-2;1-\alpha /2} * Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }]

Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} } = \sqrt{\frac{14*(5.6^2)+11*(6.7^2)}{15+12-2} }= 6.108= 6.11

t_{n_1+n_2-2;1-\alpha /2}= t_{15+12-2;1-0.05}= t_{25;0.95}= 1.708

[(82.3-81.2) ± 1.708* (6.11*\sqrt{\frac{1}{15}+\frac{1}{12}  }]

[-2.94; 5.14]

Using a 90% confidence level you'd expect the interval [-2.94; 5.14] to contain the true value of the difference between the average percentage obtained in calculus by boys and the average percentage obtained in calculus by girls.

I hope this helps!

3 0
2 years ago
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