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A model rocket is launched upward with an initial velocity of 230 feet per second. The height, in feet, of the rockett seconds after the launch is given by h - -16+2300. How many seconds after the launch
will the rocket be 330 feet above the ground? Round to the nearest tenth of a second. (Enter your answers as a comma-separated list.)
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step-by-step explanation:

where is the questions

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet