1: the one that looks like an x
2:regular pentagon
3: 4
4: 1
5: 1 line of symmetry
We have to find the" ratio of the area of sector ABC to the area of sector DBE".
Now,
the general formula for the area of sector is
Area of sector= 1/2 r²θ
where r is the radius and θ is the central angle in radian.
180°= π rad
1° = π/180 rad
For sector ABC, area= 1/2 (2r)²(β°)
= 1/2 *4r²*(π/180 β)
= 2r²(π/180 β)
For sector DBE, area= 1/2 (r)²(3β°)
= 1/2 *r²*3(π/180 β)
= 3/2 r²(π/180 β)
Now ratio,
Area of sector ABC/Area of sector DBE =
= 4/3
If the legs of a right triangle are a and b and the hypotnuse is c then
a²+b²=c²
so we are given
the legs are x and √7
and the hypotnuse is √19
so
x²+(√7)²=(√19)²
x²+7=19
minus 7 both sides
x²=12
sqrt both sides
x=√12
x=√(4*3)
x=(√4)(√3)
x=2√3
I believe the answer to be 0.0016
Answer:
Substitute
Step-by-step explanation:
Example
y=3
11+y=2x
The first thing you should do is put 3 in place of y in the second equation.
11+3=2x
14/2=2x/2
x=7