(a) If f(t) = x1e−kt cos(ωt) + x2e−kt sin(ωt), then f0(t) has the same form with coefficients y1,y2. (i) Find the matrix A such that y = Ax. (ii) Find the characteristic polynomial of A. (iii) What can you say about eigenvalues of A? (iv) Interpret your answer to (iii) as a calculus statement. That is, explain how your answer to (iii) could have been predicted from a basic fact of calculus. (b) If f(t) = x1 sin(t) + x2 cos(t) + x3tsin(t) + x4tcos(t), then f0(t) has the same form with coefficients y1,...,y4. Same questions (a)-(d) as previously.

341

step-by-step explanation:

all you have to do is 9x31 (which equals 279) and then 20x31 (which equals 620). then you have to subtract. 620-279=341.

answer: give more info

step-by-step explanation:

a is 8.b is 10.c is 10.5. d is -.5

step-by-step explanation:

just minus the smallest number from the biggest.

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

  region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

  region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true