Answer:
the slope is -3/2 and intercept is 0,3
Let
F1------------------------- > airplane speed <span>150 mi / h East
F2-------------------------- > </span>wind speed <span>30 mi / h North
calculate the resultant force R
F1x=150 F1y=0 F2x= 0 F2y=30
Rx=F1x+F2x------------- >150+0 ---------------- >Rx=150
</span>Ry=F1y+F2y------------- >0+30 ---------------- >Rx=30
║R║=√150² 30²=√23400=152.97----- >153 mi/h
tan(theta)=Ry/Rx=30/150=0.20
arctan (0.20)=11.31 degrees-----------11.3 degrees
the answer is
actual speed 153 mi/h and direction 11.3 degrees North-East (I Quadrant)
Answer:
-3
Step-by-step explanation:
First - change the equation to y=mx+b format
3y/3 = 4x/3 - 17/3
y = 4/3x - 17/3
Second - then (2,y) plug x = 2
Third - solve
y = 4/3(2) - 17/3
y = 8/3 - 17/3 = -9/3 = -3
