Answer:

Find the midsegment of the triangle which is parallel to CA.

Tip
- A midsegment of a triangle is a segment connecting the midpoints of two sides of a triangle.
- This segment has two special properties. It is always parallel to the third side, and the length of the midsegment is half the length of the third side.
- If two segments are congruent, then they have the same length or measure.In other words, congruent sides of a triangle have the same length.

We have to find the segment which is parallel to CA.
From the given data,
The segment EG is the midsegment of the triangle
ABC.
So we have,
A midsegment of a triangle is a segment connecting the midpoints of two sides of a triangle. This segment has two special properties. It is always parallel to the third side.

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Well first you have to simplify the denominators with x, by multiplying the denominator on the left times the top and bottom of the middle, and vice versa to get 10x/(4x^2-4x)-9(2x-2)/(4x^2-4x)=-1/4 and then you combine the fractions on the left to get 2(9-4x)/(4x^2-4x)=-1/4 and then you cross multiply the fractions to get 8(9-4x)=-4x^2+4x and then simplify to get 72-32x=-4x^2+4x and then 4x^2-36x+72=0 which then we can turn into 4(x-6)(x-3)=0 so x is 6 and 3
Answer is
<span>A. 3 pages are edited every 5 min</span>
Answer:
d
Step-by-step explanation:
because in 2 they make 600 and 600 times 3 is 1,800 then do 2 times 3 witch is 6 so 6 hours after 10:30 is 3:30. I hope that makes sense <3
Answer:
Option (1)
Step-by-step explanation:
By the property of alternate segment theorem,
"Angle formed between the tangent and the chord in a circle measures the half of the measure of the intercepted arc"
m(∠EFG) =
× m(minor arc FG)
minor arc FG = 2m(∠EFG)
= 2(76°)
= 152°
Therefore, Option (1) will be the correct option.