The answer is D (y+10)(y-6)
because
<h3><u>Answer</u> :</h3>
See the attachment for better understanding.
<u>G</u><u>i</u><u>v</u><u>e</u><u>n</u><u> </u>: Length of string (AC) = 300ft
<u>T</u><u>o</u><u> </u><u>F</u><u>i</u><u>n</u><u>d</u><u> </u>: Height of the kite from ground (AB)
Hope it helps !
Answer:
try and angle it more upwards so that it goes more through the second dot from the top and those 2 dots towards the bottom of the graph above the line.
Step-by-step explanation:
Answer:
The second time when Luiza reaches a height of 1.2 m = 2 08 s
Step-by-step explanation:
Complete Question
Luiza is jumping on a trampoline. Ht models her distance above the ground (in m) t seconds after she starts jumping. Here, the angle is entered in radians.
H(t) = -0.6 cos (2pi/2.5)t + 1.5.
What is the second time when Luiza reaches a height of 1.2 m? Round your final answer to the nearest hundredth of a second.
Solution
Luiza is jumping on trampolines and her height above the levelled ground at any time, t, is given as
H(t) = -0.6cos(2π/2.5)t + 1.5
What is t when H = 1.2 m
1.2 = -0.6cos(2π/2.5)t + 1.5
0.6cos(2π/2.5)t = 1.2 - 1.5 = -0.3
Cos (2π/2.5)t = (0.3/0.6) = 0.5
Note that in radians,
Cos (π/3) = 0.5
This is the first time, the second time that cos θ = 0.5 is in the fourth quadrant,
Cos (5π/3) = 0.5
So,
Cos (2π/2.5)t = Cos (5π/3)
(2π/2.5)t = (5π/3)
(2/2.5) × t = (5/3)
t = (5/3) × (2.5/2) = 2.0833333 = 2.08 s to the neareast hundredth of a second.
Hope this Helps!!!
Your answer is at the top good luck!
