2 years ago

Answer question below answer question below

Mathematics

1 answer:

podryga [215]2 years ago

5 0

Answer:

r = sqrt(11)

Step-by-step explanation:

Area of sector, A = (theta) r^2/ 2 = 11 cm^2

Perimeter of sector, p= r theta+2r

Length of arc AB, a= r theta

We are given that p = 2a, so equate

r theta + 2r = 2(r theta)

2r = r( theta )

theta = 2 (radians)

Solve for r from area

Area = (theta) r^2 /2 = 11

substitute theta = 2

2 r^2 /2 =11

r^2 = 11

r = sqrt(11)

You might be interested in

What are the excluded values? b = 0 and -10 b = 2 and 5 b = 3 and 7 none of the above

Dominik [7]

b= 2 and 5

is your answer to the above question.

8 0

2 years ago

The triangular entrance to a tent is 2/3 as wide as it is tall. The area of the entrance is 12ft^2. Find the height and the base

Reika [66]

X+ $\frac{2}{3} x$ =12
5x=36
x=7.2
So, the triangular entrance to the tent is 7.2 feet tall and 4.8 feet wide

4 0

3 years ago

Will give brainliest if answered *correctly** picture attached

lakkis [162]

Answer:

A, C, E

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the y and x intercept 6x-8y=48

Crank

6x-8y=48

to find the x intercept set y=0

6x - 8*0 = 48

6x = 48

divide by 6 on each side

x =48/6

x=8

(8,0)

to find the y intercept set x=0

6(0) - 8y = 48

0 - 8y = 48

-8y = 48

divide by -8 on each side

y = -6

(0,-8)

8 0

3 years ago

The base of a pyramid has n sides <br> write an expression for the number of faces of the pyramid

Keith_Richards [23]

N=5. its a pyramid. simple.

6 0

3 years ago

Read 2 more answers