A rocket is launched with an initial upward velocity of 320 feet per second from an initial height of 15 feet. The function h (t) = -1672 + 320t + 15 models the height of the rocket, in feet, t seconds after it was
launched. For how many seconds is the height of the rocket greater than or equal to 1.039 feet?

The height of the rocket is greater than or equal to 1.039 feet during first 19.936 seconds.

Step-by-step explanation:

The height of the rocket launched from the ground is modelled after the following expression:

(1)

Where:

- Time, measured in seconds.

- Height, measured in feet.

After a careful reading to the statement, we translate all relevant information into the following mathematical inequation:

(2)

After some algebra, we get the equivalent inequation:

Which means that the following conditions must be observed to satisfy the inequation above:

The height of the rocket is greater than or equal to 1.039 feet during first 19.936 seconds.

step-by-step explanation:

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g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)