1.) a 2600 lb. car travelling downhill has a grade resistance of -130 lbs. find the angle of the grade to the nearest tenth of a degree. 2.) a car travelling on a 2.7 degree uphill grade has a grade resistance of 120 lbs. determine the weight of the car to the nearest hundred poulds.

Step-by-step explanation:

Using the formula for Grade Resistance/Force in this case...

--> R/F: the Grade Resistance

--> W: Weight of the Vehicle

--> ∅: Angle of the Hill

Because the Grade Resistance in problem 1) is (-)130lbs... we know that the car is traveling DOWNhill. (If it were simply 130lbs, the car would be traveling up the hill)

to do this, we must take the INVERSE of sin.... (because we cannot simply divide both sides by 'sin' by itself... we must instead...

---> sin(-0.05) = ?

∅ ≅ -2.866 ..... (you can then divide 866 by 60' if necessary, to find degree and minute answer)...

∅ ≅ -2° 14'

Hope this will help you and others with the 2nd question as well! ^_^

DONT GIVE UP! You got this '_^

86

step-by-step explanation:

find the area of the triangle then multiply by 2 then add the area of the parollelegram multiplied by 2