An object is launched from ground level with an initial velocity of 120 meters per second. For how long is the object at or above 500 meters (rounded to the nearest second)?

The equation that models the path of the object is y = -4.9t2 + 120t.

What expression is equivalent to 1/2(2+3x-6)

Can someone me out here number 4 plz

Consider this equation. |y + 6| = 2 what can be concluded of the equation? check all that apply. there will be one solution. there will be two solutions. the solution to –(y + 6) = 2 will be also be a solution to the given absolute value equation. the solution(s) will be the number(s) on the number line 2 units away from –6. the value of y must be positive since the variable is inside absolute value signs.