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masha68 [24]
4 years ago
10

A circle with a radius of one unit is inscribed in an equilateral triangle with an area of 4√3 square units. Determine the exact

area of the shaded region.
4√3 - 2pi square units
4√3 + 2pi square units
4√3 - pi square units
4√3 + pi square units

Mathematics
1 answer:
chubhunter [2.5K]4 years ago
7 0

Answer:

4√3 -π square units

Step-by-step explanation:

The triangle area is given. The area of the circle is πr², where r=1, so is π square units.

If the shaded area is the difference, then its area is ...

... 4√3 - π

_____

<em>Comment on the problem statement</em>

Normally, when a figure is "inscribed", the size of the inscribed figure is limited by the enclosing figure. Here, the circle is not "inscribed" in the triangle, as the triangle is larger than would be required for an inscribed circle of radius 1.

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I have to show ALL work
cluponka [151]

Answer:

Answer:C = 8

Answer:C = 8Step-by-step explanation:

-2(4c+16)^1/4+2=-2

Move the constant to the right and change the sign

-2(4c+16)^1/4= -2-2

transform expression

- 2  \sqrt[4]{4c - 16}  =  - 4

Divide each side by -2

\sqrt[4]{4c - 16 }  = 2

raise both side of the equation by power of 4

4c-16=16

move constant to the right and change the sign

4c=16+16

4c=32

THEN DIVIDE BOTH SIDES BY 4

C = 8

5 0
3 years ago
If DE=1.5m-9 and EF=4.5m-42 what is the value of DF (equilateral triangle)
11111nata11111 [884]

Answer:

DF = 7.5

Step-by-step explanation:

Since the 3 sides are congruent, then

EF = DE , that is

4.5m - 42 = 1.5m - 9 ( subtract 1.5m from both sides )

3m - 42 = - 9 ( add 42 to both sides )

3m = 33 ( divide both sides by 3 )

m = 11

Thus

DE = 1.5m - 9 = 1.5(11) - 9 = 16.5 - 9 = 7.5

Then

DF = 7.5

3 0
3 years ago
This is AP Statistics Question. Can you help me with this?
LenKa [72]
The confidence interval is given by

\mu_s\pm Z_{\alpha_0}\sigma_s

where \mu_s is the sample mean and \sigma_s is the standard error of the mean. In turn, the standard error of the mean is \sigma_s=\dfrac\sigma{\sqrt n} where n is sample size.

We have

\sigma_s=\dfrac6{\sqrt{25}}=1.2

The endpoints of the confidence interval correspond to the finite endpoints of the rejection region. That is,

\begin{cases}\mu_s-1.2Z_{\alpha_0}=28\\\mu_s+1.2Z_{\alpha_0}=32\end{cases}

for which we can solve for Z_{\alpha_0}. We get

Z_{\alpha_0}=\dfrac53\approx1.67

which is the critical value for a confidence level of \alpha_0\approx90.44\%.
6 0
3 years ago
HELPP ASaPP I’ll mark you as brainlister
nika2105 [10]

Answer:

BC ≈ 8.9 units

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos36° = \frac{adjacent}{hypotenuse} = \frac{BC}{AB} = \frac{BC}{11} ( multiply both sides by 11 )

11 × cos36° = BC , then

BC ≈ 8.9 ( to the nearest tenth )

5 0
3 years ago
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For the function y= -2x+5sin(pi/12(x-2))what is the maximum value?
Lena [83]

Answer:

239

Step-by-step explanation:


8 0
3 years ago
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