All categories

3 years ago

Find the area of the shaded regions. ANSWER IN PI FORM AND DO NOT I SAID DO NOT WRITE EXPLANATION

Mathematics

2 answers:

7nadin3 [17]3 years ago

8 0

Answer: 18π

okokok gg

Anvisha [2.4K]3 years ago

6 0

Step-by-step explanation:

Here angle is given in degree.We have convert it into radian.

${1}^{\circ} =( { \frac{\pi}{180} } )^{c} \\ \therefore \: {80}^{\circ} = ( \frac{80\pi}{180} ) ^{c} = {( \frac{4\pi}{9} })^{c} \: = \theta ^{c}$

radius r = 9 cm

Area of green shaded regions = A

$\sf \: A = \frac{1}{2} { {r}^{2} }{ { \theta}^{ c} } \\ = \frac{1}{2} \times {9}^{2} \times \frac{4\pi}{9} \\ = 18\pi \: {cm}^{2}$

You might be interested in

A rectangle has an area of 104 cm ^2 and a width that measures 8 cm. Find the perimeter of the rectangle.

Dmitry_Shevchenko [17]

Perimeter is 42
~~~~~~~~~~~~~~~~~~~~~~~~~

7 0

3 years ago

Asking again! Please solve! For both! Brainliest and lots of points!

Dennis_Churaev [7]

Answer:

1 D (10, 120)

2 C (20, 40)

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Event

Nataly_w [17]

What are the following questions because they aren’t there

4 0

3 years ago

Please help Will mark Brainliest

Alchen [17]

Answer:

I'm pretty sure it's the last one !

Step-by-step explanation:

Good luck on your quiz!!

Hope you have a good day

3 0

3 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

3 years ago