Let 'x' be the distance from THE far bank where 700 is the distance to the NEAR bank
boat one has travelled 700 (rate = 700/unit time) boat two has travelled x rate = x / unit time
boat one then travels x + 400 more and boat two travels 700 + (700+x -400) more when they meet
The time is the same rate x time = distance distance/rate = time equate the distances divided by the respective rates
(700 + x + 400)/700 = ( x + 700 + (700+x-400) )/x
1100x + x^2 = 1400x + 700000
x^2-300x -700000 = 0 quadratic formula yields x = 1000
One boat travels 700 the other 1000 whe they first meet.....width of river = 700+ 1000 = 1700 m
Answer:
Interest rate ≈ 3.48%
Step-by-step explanation:
The answer is 1680. To get this you do 8*7*6*5 because there are 8 options for the first character, and then you can repeat that for the second one with 7, 6, and 5
Answer:The value of the bulldozer after 3 years is $121950
Step-by-step explanation:
We would apply the straight line depreciation method. In this method, the value of the asset(bulldozer) is reduced linearly over its useful life until it reaches its salvage value. The formula is expressed as
Annual depreciation expense =
(Cost of the asset - salvage value)/useful life of the asset.
From the given information,
Useful life = 23 years
Salvage value of the bulldozer = $14950
Cost of the new bulldozer is $138000
Therefore
Annual depreciation = (138000 - 14950)/ 23 = $5350
The value of the bulldozer at any point would be V. Therefore
5350 = (138000 - V)/ t
5350t = 138000 - V
V = 138000 - 5350t
The value of the bulldozer after 3 years would be
V = 138000 - 5350×3 = $121950
