To ease your problem, consider "L" as you x-axis
Then the coordinate become:
A(- 4 , 3) and B(1 , 2) [you notice that just the y's changed]
This is a reflection problem.
Reflect point B across the river line "L" to get B', symmetric of B about L.
The coordinates of B'(1 , -1) [remember L is our new x-axis]
JOIN A to B' . AB' intersect L, say in H
We have to find the shortest way such that AH + HB = shortest.
But HB = HB' (symmetry about L) , then I can write instead of
AH + HB →→ AH + HB'. This is the shortest since the shortest distance between 2 points is the straight line and H is the point requiered
we are given
differential equation as
we are given
Firstly, we will find y' , y'' and y'''
those are first , second and third derivative
First derivative is
Second derivative is
Third derivative is
now, we can plug these values into differential equation
and we get
now, we can factor out common terms
we can move that term on right side
now, we can factor out
now, we can set them equal
so, we will get
...............Answer
Step-by-step explanation:
open the bracket
6={(1/5×4y) + 1/5×10}
6=4/5y+2
6-2=4/5y
4=4/5y
divide both sides by 4/5
y=1/5
Answer:
A
Step-by-step explanation:
ghfgkjhydrk;l'klhh
