Ajet flew from montreal to san francisco, a distance of 4000km. on the return trip the speed was increased by 10km/h. if the total trip took 13h, what was the speed from montreal to san francisco?

answer:

you can put this solution on your website!

this is an rtd problem. we can create a table based on the given information

jet . . . rate . . . . . time . . . . . . . . distance

m-> s . . r . . . . . . 4000/r . . . . . .. . . 4000

s-> m . . r + 300 . . . .4000/(r + 300) . . . .. .4000

totals . . . . . . . . . . .13

so, from the table, we get

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}

i think the amount would be $140.62