Your answer would have to be 91 :)
(open picture for the explanation)
G(x) = ax^2
y = ax^2
16 = a1^2
a = 16
Therefore, g(x) = 16x^2
Answer:
QS=SR ( being PS median of triangle PQR,psㅗQR)
<QPS=<SPR (Median PS bisects the angle P)
HOPE THIS WILL HELP YOU...!
This is a question of resolving the forces in question into horizontal and vertical and then finding the action of the resultant force.
The first force acts at an angle less than 90° and thus its resolved forces are positive. The second force acts at angle larger than 90° and is incident and thus its horizontal value is positive while its vertical value is negative.
Therefore;
For force of 300 N at 30°;
Horizontal value = 300*Cos 30 = 259.81 N
Vertical value = 300*Sin 30 = 150 N
For 150 N at 135°;
Horizontal value = 150*Cos (180-135) = 106.07 N
Vertical value = -150*Sin (180-135) = -106.07 N
Then,
Resultant horizontal value = 259.81+106.07 = 365.88 N
Resultant vertical value = 150-106.07 = 43.93 N
Therefore,
Resultant force, v = Sqrt (365.88^2+43.93^2) = 368.51 N
Angle of action measured from horizontal = tan ^-1(43.93/365.88) 6.85°
Then,
v = 368.51 N at 6.85° from horizontal
Its either A or C they both have the same spot when it is graphed.
