Precalculus: check my answers . 1. write the domain of the function given in the graph in set builder notation. (the curved graph belongs to this.) a) {x|x > = 10} b) {x|x > = 0} c) {x|x > 10} is the answer c? d) {x|x > 0} 2. write the range of the function given in the graph in interval notation. (the two line segments belongs to this.) a) [-4,8] b) (-4,3)u(3,8] c) (-4,8] d) [-4,3)u(3,8) i know it is either b or d, but i am not sure if it is b, or d.

Part a) Write the domain of the function given in the graph in set builder notation

we know that

the domain for the function given in the graph are the values of x

---------> interval [10.∞)

therefore

the answer part a) is the option

A) {x|x >= 10}

Part b) Write the range of the function given in the graph in interval notation.

the range of the function are

the first segment has a range (-4,3]

the second segment has a range (3,8]

The function has a range

(-4,3] U (3,8]= (-4,8]

the answer is the option

C) (-4,8]

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}