Hey there :)
Now lets name each house
A B C D E F
500ft <-> 500ft <-> 500ft <-> 500ft <-> 500ft <-> 2000ft
Let's say the bus stops at A
= 0 ( from A ) + 500 ( from B ) + 500 + 500 ( from C ) + 500 + 500 + 500 ( from D ) + 500 + 500 + 500 + 500 ( from E ) + 500 + 500 + 500 + 500 + 2000 ( from F )
= 0 + 500 + 1000 + 1500 + 2000 + 4000
= 9000 ft
at B
= 500 ( from A ) + 0 ( from B ) + 500 ( from C ) + 500 + 500 ( from D ) + 500 + 500 + 500 ( from E ) + 500 + 500 + 500 + 2000 ( from F )
= 500 + 0 + 500 + 1000 + 1500 + 3500
= 7000 ft
at C
= 500 ( from A ) + 500 + 500 ( from B ) + 0 ( from C ) + 500 ( from D ) + 500 + 500 ( from E ) + 500 + 500 + 2000 ( from F )
= 500 + 1000 + 0 + 500 + 1000 + 3000
= 6000 ft
Do the same for D , E and F
at D
= 6000 ft
at E
= 7000 ft
at F
= 15000 ft
You will find that the bus should stop at either C or D to make the sum of distances from every house to the stop as small as possible.
Answer:
B) 5y ( x-2 )
Step-by-step explanation:
= 5xy -10y
= 5y ( x-2 )
The answer is 5 .
3 3/4 / 3/4 = 5
Hope this helps
The first graph is what i would think
-- Find how much 'y' changes from the first point to the second one.
-- Find how much 'x' changes from the first point to the second one.
-- The slope of the line going from the first point to the second one is
(change in 'y') / (change in 'x') .
