Answer:
Step-by-step explanation:
The formula for the length of a vector/line in your case.
Answer:
Step-by-step explanation:
9) PQR Is an isosceles triangle
=> ∠PRQ = (180° - x)/2
PRS is an isosceles right triangle
=> ∠PRS = 45°
Have: ∠PRS + ∠PRQ = 115°
=>
=> 180 - x = (115 - 45).2 = 140
<=> x = 180 - 140 = 40
10) ABD is an isosceles right triangle => ∠ABD = 45°
BCD is an equilateral triangle => ∠CBD = 60°
have: x = ∠ABD + ∠CBD = 45° + 60° = 105°
11) have: x = y (2)
PQT is an isosceles triangle => ∠PQT = 180 - 70.2 = 40
QTS is an isosceles triangle => ∠TQS = 180 -2x
QRS is an isosceles triangle => ∠RSQ = y
have: 40 + 180 - 2x + y = 180 => 2x - y = 40 (1)
(1)(2) =>
=> x + y = 80
12) EFJ Is an equilateral triangle => ∠FJE = 60
∠FJE is the outer angle of the triangle FHJ but FHJ is an isosceles triangle
=> 60 = 2.∠JHF => ∠JHF = 30°
∠JHF is the outer angle of the triangle FHG
=> 30° = 2x
<=> x = 15°
Answer:
-127
Step-by-step explanation:
s = 101x-41+50
80 = 101x - 4+ 50
-101x
-181 = x - 4 +50
-181 = x -54
+ 54
-127 = x
I hope this helps