A model rocket is launched with an initial velocity of 250 ft per second. The height h, in feet, of the rocket t seconds after t
1 answer:
9514 1404 393
Answer:
{2.96, 12.66}
Step-by-step explanation:
A graph shows the rocket will be 600 feet up after 2.96 seconds, and again at 12.66 seconds.
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You can solve the equation h = 600 to find the times.
-16t^2 +250t = 600
-16(t^2 -15.625t) = 600
-16(t^2 -15.625t +7.8125²) = 600 -16(7.8125²)
-16(t -7.8125)² = -376.5625
t -7.8125 = √(376.5625/16) ≈ ±4.8513
t = 7.8125 ± 4.8513 ≈ {2.9612, 12.6638}
The rocket will be 600 ft above the ground 2.96 and 12.66 seconds after launch.
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Answer:
Step-by-step explanation:
To start calculating, we first need to make some proof.
Firstly, since AB = AC, we know that ΔABC is isosceles, which means that ∠ABC = ∠ACB.
Now, looking only to ΔBDE and ΔCDF, we can see that they are similar, because the two of its angles are congruent:
∠BED=∠CFD
∠DBE=∠DCF
To make it easier to visualize which are the corresponding vertexes, we can draw them like this:
And we need to remember that BC is 24, so:
BD+CD=24
Since the triangles are similar, their corresponding sides have constant ratio, which we can calculate from the corresponding sides DE and CF:
This ratio is the same for the other corresponding sides, so we can apply that for BD and CD:
Thus, the measure of CF is approximately 13, alternative D.
