<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>att</em><em>ached</em><em> </em><em>picture</em>
<em>hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em>
Answers:
x = 4
EF = 14
CF = 7
EC = 7
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Work Shown:
C is the midpoint of segment EF. This means that EC = CF. In other words, the two pieces are congruent.
Use substitution and solve for x
EC = CF
5x-13 = 3x-5
5x-13+13 = 3x-5+13
5x = 3x+8
5x-3x = 3x+8-3x
2x = 8
2x/2 = 8/2
x = 4
Now that we know that x = 4, we can use this to find EC and CF
Let's compute EC
EC = 5x - 13
EC = 5*x - 13
EC = 5*4 - 13 ... replace x with 4
EC = 20 - 13
EC = 7
Let's compute CF
CF = 3x - 5
CF = 3*x - 5
CF = 3*4 - 5 ... replace x with 4
CF = 12 - 5
CF = 7
As expected, EC = CF (both are 7 units long).
By the segment addition postulate, we can say EC+CF = EF
EC+CF = EF
EF = EC+CF
EF = 7+7
EF = 14
Answer:
sinΘ = 
Step-by-step explanation:
using the identity
sin²x + cos²x = 1 ( subtract cos²x from both sides )
sin²x = 1 - cos²x ( take square root of both sides )
sinx = ± 
given
cosΘ = -
, then
sinΘ = ± 
= ± 
= ± 
= ± 
since Θ is in quadrant II where sinΘ > 0 , then
sinΘ = 
Y= 3 - 3x/2
Y+6= - 3/2 (x-6)
3x+2y-6=0
Y intercept 3
X intercept 2