Answer: x=9
Step-by-step explanation:
The bottom triangle that has a 40° is an isosceles triangle, therefore the other angle at the bottom right is also 40°. This leaves the top angle to be 100°.
180=40+40+x
180=80+x
x=100
Now, to find ∠2, you can tell it is a supplementary angle. Therefore, the 2 angles add up to 180°
180-100=∠2
80°=∠2
The problem states that ∠2 is 9x-1. We know that ∠2 is 80°. We can solve for x.
80=9x-1
81=9x
x=9
Step-by-step explanation:
x2 - 4x - 12
Finding factors of 12 then
x2 - 6x + 2x - 12
x( x - 6) + 2 (x - 6)
So the factors are (x+2) (x-6)
Given : Angle < CEB is bisected by EF.
< CEF = 7x +31.
< FEB = 10x-3.
We need to find the values of x and measure of < FEB, < CEF and < CEB.
Solution: Angle < CEB is bisected into two angles < FEB and < CEF.
Therefore, < FEB = < CEF.
Substituting the values of < FEB and < CEF, we get
10x -3 = 7x +31
Adding 3 on both sides, we get
10x -3+3 = 7x +31+3.
10x = 7x + 34
Subtracting 7x from both sides, we get
10x-7x = 7x-7x +34.
3x = 34.
Dividing both sides by 3, we get
x= 11.33.
Plugging value of x=11.33 in < CEF = 7x +31.
We get
< CEF = 7(11.33) +31 = 79.33+31 = 110.33.
< FEB = < CEF = 110.33 approximately
< CEB = < FEB + < CEF = 110.33 +110.33 = 220.66 approximately
Answer:
2,568.39ft2
Step-by-step explanation:
We need to find the angle (rounded to the nearest whole degree) of a circle that would represent the percent of students taking mathematics.
Since a complete turn around the circle has 360º, we have the following proportion:
<em>Number of students Angle</em>
734 360º
442 x
Then, cross multiplying the above values, we obtain:
<em>Answer</em>: 217º
