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3 years ago

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The diagram shows two congruent equilateral triangles whose overlap is a hexagon. The areas of the smaller triangles, which are

also equilateral, are 1 , 1 , 9 , 9 , 16 and 16 , as shown. What is the area of the inner hexagon?

2 answers:

seraphim [82]3 years ago

6 0

Answer:

Step-by-step explanation:

sweet [91]3 years ago

5 0

Answer:

38

Step-by-step explanation:

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Help!!!! Please just number 11 worth 90 points!!!!!

timurjin [86]

Answer:

6.75 hrs are missing

Step-by-step explanation:

First, change all fractions into decimals:

8 = 8

7 1/4 = 7 + 0.25 = 7.25

8 1/2 = 8 + 0.50 = 8.50

Next, add all numbers gotten together:

8 + 7.25 + 8.50 = 23.75

Subtract the number gotten from the total

30.5 - 23.75 = 6.75

~

3 0

3 years ago

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3x-9-2x+4. answer????

Firlakuza [10]

The answer is x-5 First collect like term so it will be x-9+4 and then subtract 9-4= 5 so now it will be x-5

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3 years ago

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .

The area (in square inches) of the circle that the hexagon did not cover would be:

$\begin{aligned}&\text{Area of circle} - \text{Area of hexagon} \\ &= 100\pi - 150\sqrt{3}\end{aligned}$ .

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3 years ago

Use the Angle Addition Postulate to find the measure of each angle.

Simora [160]

For number 15 and 16, you just have to find the absolute difference between the two points along the calibration of the protractor.

15. ∠BXC = |B - C| = |140° - 110°| = 30°
16. ∠BXE = |B - E| = |140° - 30°| = 110°

For numbers 20 and 21, apply the Angle Addition Postulate. This is when you add the individual interior angles to equate to the total angle.

20. ∠PQS = ∠PQR + ∠RQS
112° = 72°+ 10x°
x = 4

21. ∠KLM = ∠KLN + ∠NLM
135° = 47°+ 16y°
y = 5.5

5 0

3 years ago

Solve for x. Round to the nearest tenth if necessary (will give brainliest PLEASE help!!!)

Lyrx [107]

Answer:

I think it's 30.1

Step-by-step explanation:

Since the hypothesis is always the longest side of the triangle

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3 years ago

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