Urgent

when they are both racing on hoverboards, victoria is 3 times as fast as her brother max. when she is on foot, she is 3 times slower than max on his hoverboard. they took off on hoverboards at the same time, but after 12 minutes, victoria’s hoverboard broke and she immediately started to run. if the race was a tie, how long, in minutes, did it last from start to finish?

48 minutes

Step-by-step explanation:

We will use the equation d=rt for this situation, where d is distance, r is the rate of speed and t is the time.

Let r be Max's rate of speed and t be the total time.

With both on hoverboards, Victoria is 3 times as fast as Max; this would be represented as 3r.  She travels at this speed for 12 minutes; this gives us the expression 3r(12).

For the remainder of the race, Victoria is 3 times slower than Max.  This is represented as 1/3r.  We do not know the amount of time she travels this way; we will represent this as 1/3r(t-12), since t is the total time and she has already spent 12 minutes on the hoverboard.

Together the distance on hoverboard and the distance on foot can be represented by d=3r(12)+1/3r(t-12).

This is the same distance that Max travels.  Max's distance is represented using the equation d=rt.  Setting them equal, we have

3r(12)+1/3r(t-12) = rt

Simplifying, we have

36r+1/3r(t)-1/3r(12) = rt

36r + 1/3rt - 12/3r = rt

36r + 1/3rt - 4r = rt

Combining like terms,

32r + 1/3rt = rt

Subtract 1/3rt from each side:

32r + 1/3rt - 1/3rt = rt - 1/3rt

32r = 2/3rt

Divide both sides by r:

32r/r = (2/3rt)/r

32 = 2/3t

Divide both sides by 2/3:

32 ÷ 2/3 = 2/3t ÷ 2/3

32 ÷ 2/3 = t

32/1 × 3/2 = t

96/2 = t

48 = t

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}

a, b, c, d are correct.

step-by-step explanation:

we have g(x) = 3 and are required to find which option is a translation of g(x).

now, 'translation' is a transformation which means to shift the function in any direction and the general form of translation is g(x) + k, where k is any real number.

so, according to the options we see that all are correct as they satisfy the general form of translation.

also, it can be seen from the graph below that the function g(x) = 3 is shifted from its position in all the four options.