perimeter=2(85)+π(12.5)=220+2×12.5π
=170+25×3.14
=170+78.5
=248.5
distance ran=3×248.5=745.5 yards.
i hope this work for you
The answer is 4x that the answer
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!!!
Answer:
y =C₁eˣ + C₂e⁻ˣ + C₃ cos x + C₄ sin x
Step-by-step explanation:
given,
y⁽⁴⁾ - y = 0
( D⁴ - 1 ) y = 0..............(1)
writing characteristics equation
m⁴ -1 = 0
(m² - 1)(m² +1) = 0
(m - 1)(m+1)(m²+1) = 0
hence, roots of the equation comes out to be
= 1, -1, ± i
so the general solution will be
y =C₁eˣ + C₂e⁻ˣ + C₃ cos x + C₄ sin x
