For second order des, the roots of the characteristic equation may be real or complex. if the roots are real, the complementary solution is the weighted sum of real exponentials. use c1 and c2 for the weights, where c1 is associated with the root with smaller magnitude. if the roots are complex, the complementary solution is the weighted sum of complex conjugate exponentials, which can be written as a constant times a decaying exponential times a cosine with phase. use c1 for the constant and phi for the phase. (note: some equations in the text give the constant multiplying the decaying exponential as 2c1. this was done for the derivation. the constant for this problem should be c1 alone.)

all numerical angles(phases) should be given in radian angles (not degrees). given the differential equation y′′+10y′+61y=6u(t)

write the functional form of the complementary solution, yc(t)
yc(t)=(c1e^(-5t)cos(6t+phi))u(t) < - solved correct

find the particular solution, yp(t).
yp(t)=(6/61)u(t) < -solved, correct

find the total solution, y(t)y(t) for the initial conditions y(0)=9 and y′(0)=4
y(t)=

all numerical angles(phases) should be given in radian angles (not degrees).

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation:

the terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. a recursive formula for a geometric sequence with common ratio r is given by an=ran−1 a n = r a n − 1 for n≥2 n ≥ 2 .

step-by-step explanation:

$2416.6

step-by-step explanation:

per hour = $ 25

total work time = 35

commission = 15%

total she has to make = $ 4,500

let she sold x dollar products

so her commission will be 15 % of x

now

total = weekly basic pay + commission

4500= (25*35) + 15 % of x

4500=875 + * x

4500-875 =   * x

3625= * x

multiplying wwith 100 on both sides gives

36250 = 15 x

now

x =  

x= $2416.6