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
Given:
Area(A)= 63m^2
Length (L)= 2W+5
Width (W)=?
A=LxW
63=(2W+5)(W)
63=2W^2 + 5W
0=2W^2+5W-63
0=(2W-9)(W+7)
2W-9=0 then W=4.5 and W+7=0 then W=-7
Can only use 4.5 since it is positive and distance is positive.
W= 4.5 m
L=2W+5=2 (4.5)+5=9+5=14m
Answer:y=4x+2
Step-by-step explanation:m= rise and b=run rise /run when u look at the graph start at the pount to a count how many u rise to get to 2 then cout how many u run to get to 2 again so that would be y=4x+2.
Archeologists are the ones that study archeology.
archeology is a study, an archeologist is a description of a person . they are alike becausethey are alike because they consist of architecture,biofacts, artifacts, etc
