A ball is thrown straight up from the top of a building 124 ft tall with an initial velocity of 64 ft per second. The height s(t) (in feet) of the ball from the ground, at time t (in seconds), is given by s(t) = 124 + 64t − 16t2. Find the maximum height attained by the ball.

246 ft is the maximum height

Step-by-step explanation:

The height h given above is a quadratic function. The graph of h as a function of time t gives a parabolic shape and the maximum height h occur at the vertex of the parabola. For a quadratic function of the form h = a t² + bt + c, the vertex is located at t = - b / 2a. Hence for h given above the vertex in the question s(t) = 124 + 64t − 16t², is at t

t = -64/2(-16) = 64/32 = 2 seconds

Thus, 2 seconds after the object was thrown, it reaches its highest point (maximum value of h) which is given by

h = -16(2)² + 64 (2) + 124 = 246eet

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

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step-by-step explanation: