In january, bangkok is hotter than cape town while cape town is hotter than moscow. which inequality correctly compares the temperatures of the three cities? a. bangkok's temperature < cape town's temperature < moscow's temperature b. bangkok's temperature < moscow's temperature < cape town's temperature c. bangkok's temperature > moscow's temperature > cape town's temperature d. bangkok's temperature > cape town's temperature > moscow's temperature

The correct option is D.

Explanation:

It is given that in January, Bangkok is hotter than Cape Town it means the temperature of Bangkok is greater than the temperature of Cape Town. It can be written in the inequality form as shown below,

Bangkok's temperature > Cape Town's temperature

It is the conclusion (1).

It is also given that the Cape Town is hotter than Moscow it means the temperature of Cape Town is greater than Moscow. It can be written in the inequality form as shown below,

Cape Town's temperature > Moscow's temperature

It is the conclusion (2).

When we combine conclusion (1) and (2). We get

Bangkok's temperature > Cape Town's temperature > Moscow's temperature

This statement is shown in option D, therefore the correct option is D.

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}

figure c

step-by-step explanation:

hope this .