The distance from Humood's house to the school is 500m
<h3>How to determine the distance from Humood's house to the school?</h3>
The given parameters are:
Humood's house is (-5,7)
The school is (3,1)
The distance between both points is calculated using
Substitute the known values in the above equation
Evaluate
d = 10
Each unit in the graph is 50m.
So, we have
Distance =10 * 50m
Evaluate the product
Distance = 500m
Hence, the distance from Humood's house to the school is 500m
Read more about distance at
brainly.com/question/1872885
#SPJ1
Answer:
3789
Step-by-step explanation:
Answer:
x intercept is (3,0)
y intercept is (0,6)
Step-by-step explanation:
Answer:
Solve the system of equations by graphing, elimination or
subod
Step-by-step explanation:
Using proportions, it is found that the ordered pair that represents the location of the layover is (3.75, 1.5).
<h3>What is a proportion?</h3>
A proportion is a fraction of a total amount, and the measures are related using a rule of three.
The flight distance is partitioned in a ratio of 3:1, hence, for both the x and y coordinates, we have that:
L - Nv = 3/4(Ny - Nv)
For the x-coordinate, we have that:
x = 3.75
For the y-coordinate, we have that:
y = 1.5.
The ordered pair that represents the location of the layover is (3.75, 1.5).
More can be learned about proportions at brainly.com/question/24372153
#SPJ1
