It would be C, because a straight line equals to 180 degrees, and since 3 is 86, and lines m and n are parallel, 180-86 = C.94. Hope this helps!
Answer:
H=9;8)
B=(5;4) (ball)
R=(7;0) (hit point)
B'=symetric of B axis perpendicular of x in R
B'=(7+(7-5);4)=(9;4)
Equation BR: y-4=(0-4)/(7-5)(x-5)==>y=-2x+14
Equation RB': y-4=(4-0)/(9-7)(x-9)==>y=2x-14
Is H a point of RB'? y=2x-14 : 8=? 2*9-14==>8=?4 No!
you will not make your putt
Step-by-step explanation:
ASA
Angle-Side-Angle
AAS is the same but the order in different.
SAS is a no because there are no two S.
Can't determine bah! You can determine.
PLEASE BRAINLIEST
Answer:
a. 25.98i - 15j mi/h
b. 1.71i + 4.7j mi/h
c. 27.69i -10.3j mi/h
Step-by-step explanation:
a. Identify the ship's vector
Since the ship moves through water at 30 miles per hour at an angle of 30° south of east, which is in the fourth quadrant. So, the x-component of the ship's velocity is v₁ = 30cos30° = 25.98 mi/h and the y-component of the ship's velocity is v₂ = -30sin30° = -15 mi/h
Thus the ship's velocity vector as ship moves through the water v = v₁i + v₂j = 25.98i + (-15)j = 25.98i - 15j mi/h
b. Identify the water current's vector
Also, since the water is moving at 5 miles per hour at an angle of 20° south of east, this implies that it is moving at an angle 90° - 20° = 70° east of north, which is in the first quadrant. So, the x-component of the water's velocity is v₃ = 5cos70° = 1.71 mi/h and the y-component of the water's velocity is v₄ = 5sin70° = 4.7 mi/h
Thus the ship's velocity vector v' = v₃i + v₄j = 1.71i + 4.7j mi/h
c. Identify the vector representing the ship's actual motion.
The velocity vector representing the ship's actual motion is V = velocity vector of ship as ship moves through water + velocity vector of water current.
V = v + v'
= 25.98i - 15j mi/h + 1.71i + 4.7j mi/h
= (25.98i + 1.71i + 4.7j - 15j )mi/h
= 27.68i -10.3j mi/h
The answer is -3 because - and - are together plus