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.
Answer:
15 = x
Step-by-step explanation:
Separate the vectors into their <em>x</em>- and <em>y</em>-components. Let <em>u</em> be the vector on the right and <em>v</em> the vector on the left, so that
<em>u</em> = 4 cos(45°) <em>x</em> + 4 sin(45°) <em>y</em>
<em>v</em> = 2 cos(135°) <em>x</em> + 2 sin(135°) <em>y</em>
where <em>x</em> and <em>y</em> denote the unit vectors in the <em>x</em> and <em>y</em> directions.
Then the sum is
<em>u</em> + <em>v</em> = (4 cos(45°) + 2 cos(135°)) <em>x</em> + (4 sin(45°) + 2 sin(135°)) <em>y</em>
and its magnitude is
||<em>u</em> + <em>v</em>|| = √((4 cos(45°) + 2 cos(135°))² + (4 sin(45°) + 2 sin(135°))²)
… = √(16 cos²(45°) + 16 cos(45°) cos(135°) + 4 cos²(135°) + 16 sin²(45°) + 16 sin(45°) sin(135°) + 4 sin²(135°))
… = √(16 (cos²(45°) + sin²(45°)) + 16 (cos(45°) cos(135°) + sin(45°) sin(135°)) + 4 (cos²(135°) + sin²(135°)))
… = √(16 + 16 cos(135° - 45°) + 4)
… = √(20 + 16 cos(90°))
… = √20 = 2√5
D would be the right answer
A = 6
tn = a + (n - 1)d
t4 = 6 + 3d = 12
3d = 12 - 6 = 6
d = 6/3 = 2
f(n + 1) = f(n) + 2
