See the attached figure.
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AB = 10 , FD = 3
∵ D is the midpoint of AB, and F is the mid point of CB
∴ FD // AC , FD = 0.5 AC
∵ Δ ABC is a right triangle at C
∴ FD ⊥ BC
∴ BD = 0.5 AB = 5
∴ in Δ FDB ⇒⇒ BF² = BD² - FD² = 5² - 3² = 16
∴ BF = √16 = 4
∵ F is the mid point of CB
∴ CF = BF = 4 , and CB = 2 BF = 2*4 = 8
∵ D is the midpoint of AB, and E is the mid point of AC
∴ DE // CB , and DE = 0.5 CB = 0.5 * 8 = 4
∴ T<span>he length of line ED is 4
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Use the formula 0 = b^2-4ac
0 = -b^2-4(2)(-9)
0 = -b^2+72
b^2 = 72
b = square root of 72
b = 2 real numbers
Answer:
I believe it is c
Step-by-step explanation:
Correct me if I'm wrong but since the two lines look the same length I believe it is c
Answer:
1.25+√3 or 2.9821 to the nearest ten thousandth.
Step-by-step explanation:
sin 60 = √3/2, tan 45 = 1 and cos^2 60 = (1/2)^2 = 1/4
So we get 2 *√3/2 + 1 + 1/4
= √3 + 1.25