All categories

3 years ago

PLZ HELP ASAP AREA OF A SECTOR

Mathematics

1 answer:

xxTIMURxx [149]3 years ago

5 0

The relationship of arcs is:
S '/ S = ((1/9) * pi * 3) / (2 * pi * 3)
Rewriting we have:
S '/ S = ((1/9)) / (2)
S '/ S = 1/18
Therefore, the area of the shaded region is:
A '= (S' / S) * A
Where A: area of the complete circle:
A '= (1/18) * pi * r ^ 2
A '= (1/18) * pi * (3) ^ 2
A '= (1/18) * pi * 9
A '= (1/2) * pi
Answer:
The area of the shaded region is:
A '= (1/2) * pi

You might be interested in

Mademuasel [1]

I hope this helps you

take both of sides paranthesis square

(2a+7b)^2=(11)^2

4a^2+2.4a.7b+49b^2=121

4a^2+49b^2+56.2=121

4a^2+49b^2=9

5 0

3 years ago

Read 2 more answers

Δ HFM- Δ PST What is the perimeter of triangle PST? Enter your answer in the box.

fgiga [73]

Step-by-step explanation:

$\triangle HFM \sim \triangle PST... (Given) \\\\\therefore \frac{HF}{PS} = \frac{FM}{ST} = \frac{HM}{PT}...(csst) \\ \\ \therefore \frac{30}{PS} = \frac{35}{ST} = \frac{45}{9}\\ \\ \therefore \frac{30}{PS} = \frac{35}{ST} = \frac{5}{1} \\ \\ \therefore \frac{30}{PS} = \frac{5}{1} \\ \\ \therefore PS = \frac{30}{5} \\ \\\boxed{ \therefore PS = 6 }\\ \\ \because \: \frac{35}{ST} = \frac{5}{1} \\ \\ \therefore \: ST = \frac{35}{5} \\ \\ \boxed{\therefore \: ST = 7} \\ \\ perimeter \: of \: \triangle PST \\ = 9 + 7 + 6 \\ = 22 \: units$

4 0

3 years ago

Square ABCD and square EFGH share a common center on a coordinate plane. is parallel to diagonal . Across how many lines of refl

Oduvanchick [21]

Square ABCD and square EFGH will reflect onto themselves across 8 lines of reflection

7 0

3 years ago

Read 2 more answers

The line graph with gradient 2 and y-intercept(0,5)

Varvara68 [4.7K]

Answer:

The line graph is:

y=2x+5

4 0

3 years ago

Point A is at (6,-6) and point C is at(-6,-2). find the coordinates of B on AC so that AB= 3/4AC.

loris [4]

Answer:

B(-6, 0)

Step-by-step explanation:

You want to find B such that ...

(B -A) = (3/4)(C -A) . . . . the required distance relation

4(B -A) = 3(C -A) . . . . . . multiply by 4

4B = 3C +A . . . . . . . . . . add 4A, simplify

Now, we can solve for B and substitute the given coordinates:

B = (3C +A)/4 = (3(-6, -2) +(-6, 6))/4 = (-24, 0)/4 = (-6, 0)

The coordinates of point B are (-6, 0).

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

6 0

3 years ago