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.
<span>(14x^2+21x) = 7x(2x + 3)
A= L * W
but L = 7x so W = </span>2x + 3
<span>
answer: </span><span>width is (2x +3) feet</span><span>
</span>
Answer:
The origin is (0, 0) so let's say we have (0, 5) we find that this point is 5 units away from the origin. We find that by subtracting the origin's y coordinate from our (0, 5) point's y coordinate. This method can be used for the x-axis too.
