∫¹₀ min (1, n/y)dy = ∫ⁿ₀ (1, n/y)dn + ∫¹n min (1, n/y) dy
Hope this helps
Answer:

Step-by-step explanation:
To find the distance between a point (m, n ) and the line
Ax + By + C = 0
d = 
Here (m, n) = (6, 2) and rearranging the line
6x - y + 3 = 0 ← in general form
with A = 6, B = - 1 and C = 3 , then
d = 
= 
=
Rationalise the denominator by multiplying numerator/ denominator by 
=
×
=
← cancel 37 on numerator/ denominator
= 
Step-by-step explanation:
This can be modeled by the equation f(x)=7+5x
Because if f(1)=7+5=12, f(2)=7+10=17, and f(x)=7+15=22
If x=50 then f(50)=7+5(50)
f(50)=$257
It would be 4!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
You know 1 is not a root because the sum of the coeffcients of the equation is 14, not zero.
It is fairly easy to try 3 by synthetic division (see attachment), which tells you that 3 is a root and the remaining quadratic factor is x²-3x-5. The quadratic formula tells you the roots of that factor are
... x = (-b±√(b²-4ac))/(2a) = (3±√29)/2
The appropriate choices are
... C. (3-√29)/2
... D. (3+√29)/2
... F. 3
_____
The quadratic formula tells you the solution to
... ax²+bx+c=0
is x = (-b±√(b²-4ac))/(2a)
We have a=1, b=-3, c=-5.