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
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Answer:
f(x) × g(x) = 15x^6 - 6x^4
Step-by-step explanation:
hello :
f(x) = 5x^3 -2x and g(x) =3x^3.
f(x) × g(x) = 3x^3(5x^3 -2x) = 15x^6 - 6x^4
Your answer is: A. 4/2
Hope this helps and happy holidays!!
28 + vector b = -12
vector b = -40
vector b has magnitude 40 units in the negative y direction.