You and your friends are having a race, but since you are faster than he is, you decide to give him a head start. He starts 10 m In front if the starting line, while you start on the starting line. You run at 6 meters/ second, while your friend runs at 5.5 meters/ second. After how many seconds will you and your friend be same distance from the starting line?

the number of seconds that you and your friend would be have the similar distance from the starting line is 20 seconds

Step-by-step explanation:

The computation of the number of seconds that you and your friend would be have the similar distance from the starting line is shown below:

As we know that

Speed = Distance ÷ Time

SO,

Time = Distance ÷ speed

= 10 meters ÷ difference in speed

= 10 meters ÷ (6 meters per second - 5.5 meters per second)

= 10 meters ÷ 0.5 meters per second

= 20 seconds

hence, the number of seconds that you and your friend would be have the similar distance from the starting line is 20 seconds

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}

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5