Answer:
1) V
2) V
3) V
4) V
5) V
Step-by-step explanation:
From there, you simply need algebra and a calculator that works in radians.
Take the inverse cos of both sides to get 2x = arccos(-1)
Then divide both sides by 2 to get x = arccos(-1) / 2
Put that into a calculator and you get π/2. But because your bounds are 0 to 2π, you have to add π your solution to get the solution on the other side of the unit circle, which would be (3π/2).
Now that you have the x value, put (π/2) and (3π/2) into f(x) to get the y coordinate.
f(π/2) = 2(π/2) + sin(2(π/2) = π, which means this solution is just (π/2, π)
f(3π/2 = 2(3π/2) + sin(2(3π/2) = 3π, which means this solution is (3π/2, 3π)
Answer:
C
Expand the bracets then simplfy and then factorise
(a) By the fundamental theorem of calculus,
<em>v(t)</em> = <em>v(0)</em> + ∫₀ᵗ <em>a(u)</em> d<em>u</em>
The particle starts at rest, so <em>v(0)</em> = 0. Computing the integral gives
<em>v(t)</em> = [2/3 <em>u</em> ³ + 2<em>u</em> ²]₀ᵗ = 2/3 <em>t</em> ³ + 2<em>t</em> ²
(b) Use the FTC again, but this time you want the distance, which means you need to integrate the <u>speed</u> of the particle, i.e. the absolute value of <em>v(t)</em>. Fortunately, for <em>t</em> ≥ 0, we have <em>v(t)</em> ≥ 0 and |<em>v(t)</em> | = <em>v(t)</em>, so speed is governed by the same function. Taking the starting point to be the origin, after 8 seconds the particle travels a distance of
∫₀⁸ <em>v(u)</em> d<em>u</em> = ∫₀⁸ (2/3 <em>u</em> ³ + 2<em>u</em> ²) d<em>u</em> = [1/6 <em>u</em> ⁴ + 2/3 <em>u</em> ³]₀⁸ = 1024
Answer: Y=1/8
Slope m=0, b =1/8
Step-by-step explanation:
