A young equestrian loved to ride her horse all day long, dreaming of the day she would ride in the Kentucky

5 answers
Question:

A young equestrian loved to ride her horse all day long, dreaming of the day she would ride in the Kentucky Derby. One day in early April, she left home at 10:00 am and rode her stallion at a nice, steady 10 miles per hour away from her house. At noon a call came in from the Derby saying she was in! Her father immediately jumped on his scooter and rides at a breezy 15 miles per hour to try and track her down and tell her the good news. On what scale would you measure the number of minutes the father must ride his scooter before catching up to his daughter? A. Hundreds B. Thousands C. Ten thousands D. Tens

Answers

A. Determine how far from home the daughter is when her father

leaves.

Ten thousands

Step-by-step explanation:

Poor family could afford a horse and a scooter but not a cellphone. Kidding aside, The girl had a two-hour head start, and was riding her horse at 10 mph. Her father was riding his scooter at 15 mph. As they will meet up, their distances would be the same. Looking at the formula for distance: distance = speed*time, we can equate the distance for the two people as they would be the same. This is shown below:

Let: x = number of hours

distance of girl = distance of dad
10(x + 2) = 15x
x = 4

Therefore, they will meet after 4 hours after 12:00pm, or at 4:00 pm.

The first step is:
- determine how far from the house is young equestrian at noon. 
2 x 10 = 20 miles
15 t = 20 + 10 t
15 t - 10 t = 20
5 t = 20
t = 20 : 5
t = 4 hours
at 4:00 PM. 

The answer fam is  A. Hundreds

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