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2 years ago

For problem 8, find the area of the shaded region. The polygon is a regular polygon. please show work.

Mathematics

1 answer:

MissTica2 years ago

4 0

Answer:

34.7791 [unit²].

Step-by-step explanation:

1) area of the circle is:

πr²;

2) area of the regular polygon is:

6*r²√3*0.25=1.5√3r²;

3) the required area:

πr²-1.5√3r²=r²(π-1.5*√3)=64*(3.1415-1.5*1.7321)=34.7791

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Please help me..thank you

Sergio039 [100]

Answer:

7π cm² ≈ 22.0 cm²

Step-by-step explanation:

The shaded area is the difference between the area of Circle B and the areas of Circle C and Circle O. That is, those white areas are subtracted from the area of Circle B to get the shaded area.

The formula for the area of a circle is

A = πr²

Using this, we find ...

Circle B area = π(3 cm)² = 9π cm²

Circle C area = Circle O area = π(1 cm)² = π cm²

Then the shaded area is ...

Shaded Area = Circle B area - Circle C area - Circle O area

= (9π cm²) - (π cm²) - (π cm²)

= 7π cm² ≈ 22 cm²

5 0

3 years ago

If angle 0 is in quadrant I, what is the value of

Brilliant_brown [7]

Answer:

$the \: answer \: is \: c \: = - \frac{ \sqrt{14} }{5}$

Step-by-step explanation:

$\sin^{2}0 + ( \frac{ \sqrt{11} }{5} ) ^{2} = 1( \sin^{2}0 + \cos^{2} 0 = 1)$

${ \sin }^{2}0 + \frac{11}{25} = 1 \\ { \sin }^{2}0 = \frac{14}{25}$

$\sin0_{1} = - \frac{ \sqrt{14}}{5} while \sin0_{2} = \frac{ \sqrt{14} }{5}$

Therfore theta is QI
Therfore sin theta >0
$therefore \: \sin0 = \frac{ \sqrt{14} }{5}$

4 0

2 years ago

Segment PH is graphed on the coordinate grid where P is (-31,8) and H is (29,-10). Point E is the midpoint of segment PH. If poi

maks197457 [2]

Answer:

The coordinates of point A are (-11 , 2)

Step-by-step explanation:

- The coordinates of point P are (-31 , 8)

- The coordinates of point H are (29 , -10)

- Point E is the mid-point of segment PH

- The coordinates of the the mid-point are:

→ $x=\frac{x_{1}+x_{2}}{2}$ and $y=\frac{y_{1}+y_{2}}{2}$

∵ E is the mid-point of segment PH

∴ The x-coordinate of point E is $x=\frac{-31+29}{2}=-1$

∴ The x-coordinate of point E is -1

∴ The y-coordinate of point E is $y=\frac{8+-10}{2}=-1$

∴ The y-coordinate of point E is -1

∴ The coordinates of point E are (-1 , -1)

- Point A is graphed $\frac{2}{3}$ of the way along the segment from P to E

- That mean the distances from points P to A is 2 parts and from P to E

is 3 parts, then the distance from A to E is "3 - 2" = 1 part

- So point A divides the line segment PE at ratio 2 : 1 from P

- The coordinates of the division point are:

→ $x=\frac{x_{1}m_{2}+x_{2}m_{1}}{m_{1}+m_{2}}$ and $y=\frac{y_{1}m_{2}+y_{2}m_{1}}{m_{1}+m_{2}}$

∵ Point A divides PE at ratio 2 : 1

∵ The coordinates of point P are (-31 , 8)

∵ The coordinates of point E are (-1 , -1)

∴ The x-coordinate of point A is $x=\frac{(-31)(1)+(-1)(2)}{2+1}= -11$

∴ The x-coordinate of point A is -11

∴ The y-coordinate of point A is $x=\frac{(8)(1)+(-1)(2)}{2+1}=2$

∴ The y-coordinate of point A is 2

∴ The coordinates of point A are (-11 , 2)

* The coordinates of point A are (-11 , 2)

3 0

3 years ago

Assume y=x²+3x+6. what is the avg rate of change of y with respect to x as x changes from -4 to -2

AURORKA [14]

The average rate of change is 3

5 0

3 years ago

Elena went to a sporting goods store that was having a sale. She bought a tennis ball and 3 cans of tenis balls The tennis racke

Kazeer [188]

Answer: She will pay $52.68, including taxes

Step-by-step explanation: One can of tennis ball is $4 and there are no discount on it. Elena bought 3 cans, so:

can = 3 x 4 = 12

One tennis racket is $43 but it has a discount of 15%:

discount = 0.15 x 43 = 6.45

total racket = 43 - 6.45 = 36.55

Elena will have to pay:

pay = 12 + 36.55 = 48.55

She will also have to pay taxes, which are 8.5% over the price she paid:

pay + tax = 48.55 x 0.085 = 4.13

total to pay = 48.55 + 4.13 = 52.68

Buying 3 can of tennins ball and one racket, Elena paid $52.68, including taxes.

4 0

3 years ago

For problem 8, find the area of the shaded region. The polygon is a regular polygon. please show work.​

For problem 8, find the area of the shaded region. The polygon is a regular polygon. please show work.