Answer:
The area of the associated sector is
Step-by-step explanation:
step 1
Find the radius of the circle
we know that
The circumference of a circle is equal to

we have

substitute and solve for r


step 2
Find the area of the circle
we know that
The area of the circle is equal to

we have

substitute

step 3
Find the area of the associated sector
we know that
subtends the complete circle of area 
so
by proportion
Find the area of a sector with a central angle of 

Answer:
8
Step-by-step explanation:
10+8-3+5-12=
Answer:
Step-by-step explanation:
The given system of equations is expressed as
3x + y = 9 - - - - - - - - - - - - - - -1
3x = 9 - y - - - - - - - - - - - - - -2
To apply the method of elimination, we would rearrange equation 2 so that it would take the form of equation 1. Therefore, we would add y to the left hand side and the right hand side of the equation, it becomes
3x + y = 9 - - - - - - - - - - - - - - - - -3
Subtracting equation 3 from equation 1, it becomes
0 = 0
The equations have infinitely many solutions because if we input any values of x and y that satisfies the first equation, those values will also satisfy the second equation.
Answer:
it's d
Step-by-step explanation:
because the answer would be the same on both sides
Answer:
Part 1) 
Part 2) 
Step-by-step explanation:
Part 1) Find the area
we know that
The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle
so

we have that the base and the height of triangle is equal to the radius of the circle

substitute


simplify
Factor 36

Part 2) Find the perimeter
The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle
<em>The circumference of a quarter of circle is equal to</em>

substitute the given values


The hypotenuse of right triangle is equal to (applying the Pythagorean Theorem)


simplify

Find the perimeter

simplify
Factor 6
