a) The formula S=16t^2 is used to approximate the distance S, in feet, that an object falls freely from rest in t seconds. The height of a building is 1164 feet. How many seconds would it take for an object to fall from the top?
b) Keely's den is 6 feet longer than it is wide, if the den's area is 135 square feet, what are the dimensions of the room? What are the dimensions (in feet) of the room?
c) Two cars leave town, one driving north and the other east. They are 26 miles apart when one of them is 14 miles farther from the town than the other. At that time, how far were they each from the town?
Type the distance driven by the two cars (in miles)
d) A Rocket is launched from the top of a 70 foot cliff with an initial velocity of 100 feet per second. The height, h, of the rocket after t seconds is given by the equation h=-16t^2+100t+70. How long after the rocket is launched will it be 20 feet from the ground?
t ≈
b) Keely's den is 6 feet longer than it is wide, if the den's area is 135 square feet, what are the dimensions of the room? What are the dimensions (in feet) of the room?
c) Two cars leave town, one driving north and the other east. They are 26 miles apart when one of them is 14 miles farther from the town than the other. At that time, how far were they each from the town?
Type the distance driven by the two cars (in miles)
d) A Rocket is launched from the top of a 70 foot cliff with an initial velocity of 100 feet per second. The height, h, of the rocket after t seconds is given by the equation h=-16t^2+100t+70. How long after the rocket is launched will it be 20 feet from the ground?
t ≈
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a.
1164=16t^2
72.75=t^2
t=8.53sec.
b.
let : wide be x then length will be x+6
x(x+6) = 135
x^2+6x - 135=0
(x-9)(x+15)=0
x=9ft => wide
9+6=15ft=> length
c.
let: go north be x then east will be x+14
x^2 + (x+14)^2 = 26^2
x^2+x^2+28x-480=676
2x^2+28x-480=0
2(x^2+14x-240)=0
2(x-10)(x+24)=0
x=10 miles =>north
10+14=24 miles => east
d.
20==-16t^2+100t+70.
-16t^2+100t+70-20=0
-16t^2+100t+50=0
by Quadratic Formula we can get t =6.72 sec.
1164=16t^2
72.75=t^2
t=8.53sec.
b.
let : wide be x then length will be x+6
x(x+6) = 135
x^2+6x - 135=0
(x-9)(x+15)=0
x=9ft => wide
9+6=15ft=> length
c.
let: go north be x then east will be x+14
x^2 + (x+14)^2 = 26^2
x^2+x^2+28x-480=676
2x^2+28x-480=0
2(x^2+14x-240)=0
2(x-10)(x+24)=0
x=10 miles =>north
10+14=24 miles => east
d.
20==-16t^2+100t+70.
-16t^2+100t+70-20=0
-16t^2+100t+50=0
by Quadratic Formula we can get t =6.72 sec.
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a. your equation is s=16t^2+1164, because 1164 is the height you're dropping it from.
now just use the quadratic formula (-b+/-(√b^2-4ac)/2a) . the number with t^2 is a (16), the number multiplied by t (0 for this one) is b, and the number by itself (1164) is c. just plug in the numbers into your equation and find the bigger number out of the two solutions.
b. equation is x(x+6)=135
multiply x by everything in parenthesis and make equal to 0:
x^2+6x-135=0 and use the quadratic formula again.
now just use the quadratic formula (-b+/-(√b^2-4ac)/2a) . the number with t^2 is a (16), the number multiplied by t (0 for this one) is b, and the number by itself (1164) is c. just plug in the numbers into your equation and find the bigger number out of the two solutions.
b. equation is x(x+6)=135
multiply x by everything in parenthesis and make equal to 0:
x^2+6x-135=0 and use the quadratic formula again.