If an object is thrown at 19.6 m/s from a 58.8m tall platform the equation for the objects height h meters at time t seconds after launch is h(t)= -4.9 t^2 +19.6t+58.8 when does the object strike the ground

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

7/2

(decimal: 3.5)

welcome ; )

step-by-step explanation:

i'm sorry mate, but i am not seeing any figures. i am unable to solve this for you.