Answer:
As you noted, ∠PRS=80°. Take the triangle △TRS: we know two angles out of three, so that ∠TSR=45°. Now take the isosceles triangle △PTS: the two angles adjacent to the base PT are equal to 35°, so the third angle ∠PST=110° and then ∠PSR=65°. This is one of the two angles adjacent to the base of an isosceles trapezoid: knowing it, you can easily complete the solution.
The distance is increasing at a rate that is the speed of the plane multiplied by the cosine of the angle between its flight path and the direct line to the radar station. That cosine is 4/√(3²+4²) = 4/5, so the distance is increasing at
440 mi/h × 4/5 = 352 mi/h
Answer:
wheres the problem?
Step-by-step explanation:
18+0.07x18= 18+1.26 = 19.26
20-19.26= 0.74
So yes she does
Answer:
x=-1
Step-by-step explanation:
3x+7=7x+8-3x
3x+7=4x+8
-4x -4x
-x+7=8
-7 -7
<u>-x=1</u>
-1
x=-1
