A projectile is launched at a height of h feet above the ground at an angle of with the horizontal. The initial velocity is Vo feet per second, and the path of the projectile is modeled by the parametric
equations
x= (Vo cos theta ). and y=h+(Vo sin theta )– 16t^2

A= 109. Vo = 70 feet per second
• Determine the maximum height (to one decimal place).
feet
• Give the time the object reaches this height (to one decimal place).
seconds
• Give the approximate horizontal distance the objects is at this point (to one decimal place).
feet
• Determine the time the object will hit the ground after it is launched
one decimal place).
seconds

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )

a reflection across the x axis and a 90 degree rotation left and right.

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.

step-by-step explanation:

gym b charges $1.05 less per hour than gym a.