One number = x
Another number = y
x = y + 5
x + y = 37
(y + 5) + y = 37
2y + 5 = 37
2y = 32
y = 16
x = (16) + 5
x = 21
The numbers are 16 and 21.
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
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Hello :
<span>f(x)=x²+4x-5
</span><span>The axis of symmetry for a function in the form f(x)=x^2+4x-5 is x=-2 :
</span>f(x) = (x+2)² + b
f(x) x²+4x+4+b= x² +4x-5
4+b= -5
b = -9
the vertex is : (2 , -9)
I think the answer would be 81.60
Answer:
35 because i had this problem before and the answer was 35 cuz yeah