Atoy rocket is launched straight up into the air with an initial velocity of 60 ft/s from a table 3 ft above the ground. if acceleration due to gravity is –16 ft/s2, approximately how many seconds after the launch will the toy rocket reach the ground?

7,55 seg

Step-by-step explanation:

Initial Velocity = 60 ft/s

High = 3ft

Acceleration = -16 ft/s2

According to the next formula

H = Vi(t) - 1/2 gt2

We got a cuadratic formula which roots are

t1 = -0,04 seg and t2 = 7,55 seg

the time not negative so t= 7,55 seg

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3