Jeremy and Gerardo run at a constant speeds. Jeremy can run 1 mile in 8 minutes,and Gerardo can 3 miles in 33 minutes. Jeremy started running 10 minutes after Gerardo. assuming they run the same path, when will Jeremy catch up with Gerardo? write the linear equation that represents Jeremy's constant speed.

not sure about the linear equation but Jeremy catches up in 3 min.
1. divide 33 by 3 = 11 so Gerardo runs 1 mile in 11min
2. 11-8 =3 so Jeremy catches up in 3 min

Answer Link

ap8997154 ap8997154 · Answer 2 · 2022-01-17T11:51:22+01:00

They will meet after 26.6667 mins.

Given to us

Jeremy and Gerardo run at constant speeds.
Jeremy can run 1 mile in 8 minutes, and Gerardo can 3 miles in 33 minutes.
Jeremy started running 10 minutes after Gerardo.
assuming they run the same path.

Speed of Jeremy

[tex]\rm{Speed =\dfrac{Distance}{Time}[/tex]

[tex]\rm{speed = \dfrac{1\ miles}{8\ min}= \dfrac{1}{8}\ mile/min.[/tex]

Speed of Gerardo

[tex]\rm{Speed =\dfrac{Distance}{Time}[/tex]

[tex]\rm{speed = \dfrac{3\ miles}{33\ min}= \dfrac{1}{11}\ mile/min.[/tex]

Assumption

Let assume that it takes t mins for Jeremy to cover the distance between Jeremy and Gerardo.

Time of Jeremy

[tex]\rm{Distance=speed \times Time[/tex]

[tex]{Distance=\dfrac{1}{8}\times t[/tex]

Time of Gerardo

[tex]\rm{Distance=speed \times Time[/tex]

[tex]{Distance=(\dfrac{1}{11}\times t)+(\dfrac{1}{11}\times 10)[/tex]

Equating time,

[tex]\dfrac{1}{8}\times t=(\dfrac{1}{11}\times t)+(\dfrac{1}{11}\times 10)\\\\\dfrac{t}{8} = \dfrac{t}{11}+\dfrac{10}{11}\\\\\dfrac{t}{8}=\dfrac{t+10}{11}\\\\11 \times t = 8 \times (t+10)\\11t = 8t +80\\11t-8t = 80\\3t = 80\\t = \dfrac{80}{3}\\t= 26.6667[/tex]

Hence, they will meet after 26.6667 mins.

Learn more about Speed, distance, and time:

https://brainly.com/question/15100898