)consider the initial value problem y′−5y=15t+3et, y(0)=y0 y′−5y=15t+3et, y(0)=y0 (a) solve the initial value problem. (enter y0 for y0y0). y(t)y(t)= equation editorequation editor (b) determine the value of y0y0 that separates solutions that grow positively as t→[infinity]t→[infinity] from those that grow negatively.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

5+2x=a

asdasdsaadsaddsasds

168

step-by-step explanation:

first off, we know that he drove 3.5 hours at a rate of 48 mph.

knowing that, we want to see how many miles he drove all-together.

in this case, we can just multiply 48 by 3.5.

48 * 3.5 = 168