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:
C) 10 feet
Step-by-step explanation:
h² = 8² + 6² = 64 + 36 = 100
h = √100 = 10
Answer:
$6
Step-by-step explanation:
divide 18 into three
18÷3=6
Answer:
$86.81
Step-by-step explanation:
Using the given formula, we want to compute A for ...
P = 4750
r = 0.2279
n = 365 . . . . . assuming "exact" interest
t = 1 or 30
For 1 day late:
A = 4750(1 +0.2279/365)^(365·(1/365)) = 4752.97
For 30 days late:
A = 4750(1 +0.2279/365)^(365·(30/365)) = 4839.78
The difference in these payment amounts is ...
$4839.78 -4752.97 = $86.81
You would save $86.81 in interest charges by paying only 1 day late.
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<em>Comment on the question</em>
It would be a poor choice of credit card to use one that compounds interest daily. Most do so on a monthly basis.
The answer is that y is equal to zero
