If an object is projected upward with an initial velocity of 129 ft per sec, its height h after t seconds is h = -16^2+123t. Fin
2 answers:
Answer: 236ft.
Step-by-step explanation: To solve the given problem, we need to substitute the given time (4 seconds) in the equation to calculate the height of the object with an initial velocity of 129ft per sec. The substitution is shown as follow:
h=-256+492
h=236ft.
The height of the object after 4 seconds is 236ft.
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Answer:
Maximum (250,125) Answer
Step-by-step explanation:
This is more of a graphing problem than it is anything else.
Begin by graphing all 4 given equations.
When you do that, mark the intersection points of at least 2 lines. In this case it is exactly 2 lines for each intersecting point.
6x + 4y <= 2000 and 2x + 4y <= 1000 intersect at (250,125)
x=>0 and y=>0 intersect at (0,0)
6x + 4y <=2000 and x => 0 intersect at 333.333
2x + 4y <=1000 and y>=0 intersect at 0,250.
Any other intersection points fall outside the range of the givens. The shaded part we are interested in is sort of a very dark green/blue. It is the interior of the quadrilateral determined by the 4 vertices that are marked.
Now all you have to do is determine the maximum point using P=15x + 12y
For 0,0 P = 15*0 + 12,0 = 0
For 0,250 P = 15*0 + 12*250 = 3000
For 333.3333,0 P = 15*333.3333 + 12*0 = 5000 rounded.
For 250,125 P = 15*250 + 12*125 = 5250 Which is the maximum
Answer (250,125) produces the maximum value Answer
