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2 years ago

Help me please please.

Mathematics

1 answer:

Romashka [77]2 years ago

8 0

9514 1404 393

Answer:

$6.56

Step-by-step explanation:

The value is found by multiplying:

32% of $20.51 = 0.32 × $20.51 = $6.5632 ≈ $6.56

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Bad White [126]

first polygon

ext. angle=180°-120°

=60°

$ext \: ang = \frac{360}{n}$

n=360°/60°

n=6

second polygon

n=2(6)=12

ext. ang= 360°/n = 360°/12° = 30°

int. ang = 180°-30°= 150°

answer is C

6 0

3 years ago

1. Name of the missing segment.<br> 2.Length of the missing segment.

allochka39001 [22]

Answer:

Name: FE

Length: 13

Step-by-step explanation:

The total length is 34, so you would do 34-21, which equals 13.

6 0

2 years ago

Read 2 more answers

The length of a rectangular poster is 2 more inches than three times its width. The area of the poster is 33 square inches. Solv

laiz [17]

Answer:

Step-by-step explanation:

Area of the rectangle = 33 inches sq

Area of a rectangle (formula) = l*b

let the width be x

length = 2 inches + 3x = 2+3x

Area of the rectangle = l*b = 2+3x + x

33 inches sq = 2+3x + x

33 in sq = 2+4x

33/2 = 4x

16.5/4 = x

x = 4.125

width= <u>4.125 inches</u>

length = 2+3x = 2+3(4.125) = <u>14.375</u>

<u></u>

(unless your poster is 32 sq inches and you made a mistake, just replace 33 with 32 and solve it)

7 0

1 year 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:

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