Answered: |(5e" + 5z* - 6 sin z)dz =- 5e" + z5 –… | bartleby

Algebra & Trigonometry with Analytic Geometry

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter2: Equations And Inequalities

Section2.1: Equations

Problem 76E

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|(5e" + 5z* – 6 sin z)dz =
5e" + z5 – 6 cos z+c
5e" + 5 – 6 sin z+e
5e - 2° + 6 cos z+e
5e° + 25 + 6 sin z+c
5e + z +6 cosz+e
out of
question
on 10
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Mag question
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Show that (cos x + isin x)n= cos nx + isin nx whenever n is a positive integer. (Here i is the square root of −1)
1. Express as a sum or difference a.) sin 7t sin t b.) cos 6u cos (-4u) c.) cos 2t sin 6t d.) 2 sin 5θ cos 3θ e.) 2 sin 7θ sin 5θ f.) 3 cos x sin 2x g.) 5 cos 4u cos 5u
3. The following is an incorrect solution to evaluating the integral ofdx/(1 + 4x^2)^3/2.Set x = tan θ. Then the integral becomes:the integral of dx/(1 + 4x^2)^3/2 = integral of dθ^(1 + 4 tan2 θ)^3/2 = intgral of dθ/ (sec^2 θ)^3/2 =integral of dθ/sec^3 θ =intgral of cos^3θ dθ = integral of (1 − sin^2 θ) cos θ dθ =integral of (1 − u^2) du = u − u^3/3+ C = sin θ − sin^3 θ/3 + C*Use u = sin θ, du = cos θ dθ. (a) There are MULTIPLE errors in the above work. Identify them all. (b) Provide a correct solution to evaluating this integral.
Suppose that the position of one particle at time t is given by x1 = 3 sin(t), y1 = 2 cos(t), 0 ≤ t ≤ 2? and the position of a second particle is given by x2 = −3 + cos(t), y2 = 1 + sin(t), 0 ≤ t ≤ 2?. (a) Graph the paths of both particles. How many points of intersection are there? __ points of intersection (b) Are any of these points of intersection collision points? That is, are the particles ever at the same place at the same time? If so, find the collision points. (Enter your answers as a comma-separated list of ordered pairs of the form (x, y). If there are no collision points, enter DNE.) (x, y) =__ (c) Describe what happens if the path of the second particle is given by x2 = 3 + cos(t), y2 = 1 + sin(t), 0 ≤ t ≤ 2?. The circle is centered at (x, y) = __ There are _ intersection point(s), and there are _ collision point(s).
a) solve over 0<x<360° for 4sin^2 x - 3sinx -1 = 0 cos x + sec x = 0 b) include restrictions for cos x + sec x
Let k and n be positive integers. Prove each of the following:a) π−πsin2(nx) dx = π−πcos2(nx) dx = π.b) π−πsin(kx) cos(nx) dx = 0.c) π−πsin(kx) sin(nx) dx = π−πcos(mx) cos(nx) dx = 0 ifm = n.
Prove using the IVT cosx = cos-1x has a solution in (0, 1).

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