Raul and Esteban just started working at their uncle's farm on the weekends. Their first task was to count the ostriches and llamas. When
they reported to their uncle, Raul said, "I counted 47 heads."
Esteban added, "I counted 122 legs."
"How many are ostriches? How many are llamas?" asked their uncle.
"It's getting dark and I promised your mother I'd get you home for dinner.
There's no time to count again. You'll have to figure out how many
ostriches and how many llamas there are from that information when you
get home. Can you give me a call after dinner and let me know your
answer?"
How did Raul and Esteban figure out how many ostriches and how many
llamas there were?

because they use the linear equation

Step-by-step explanation:

first, let's turn ostrich into the variable "o" and ilamas into the variable "i"

this is a linear euation problem, so we need at least 2 equation in order to solve the problem

this is head

i + o = 47     why? , beacause both ostrich and ilamas have only  1 head each

this is legs

4i + 2o = 122   why? because ilamas have 4 legs, and ostrich have 2 legs

let's find the "point" that would solve 2 of these equation

i + o = 47 let's multiply this top equation by -2 (so the o can be cancel out)

:

-2i - 2o = -94

4i + 2o = 122

2i  = 28

i = 14

there are 14 ilamas

let's subtitute "14" into this equation

i + o = 47

14 + 0 = 47

then, subtract 14 from both sides

14+ 0 = 47

-14         -14

0 = 33

there are 33 ostrich

still not sure? let's subtitute into this equation to see if it's true or not

4i + 2o = 122

4( 14) +2(33) = 122

56 + 66 = 122

and yeah, we get the same resulte!

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}

6/20

step-by-step explanation: