Question :-
A boat goes downstream and covers the distance between two ports in 5 hours and
returns upstream in 7 hours. If the speed of the water flowing in the stream is 2 kmph,
find the speed of the boat in still water. Also, find the distance between the ports.
Answers
Solution :-
let the speed of the boat in still water = x kmph
speed of the stream = 2 kmph.
upstream
Speed of the boat = ( x - 2 ) kmph
Time = 7 hours
Speed = Distance/Time
⟹ x - 2 = Distance/7
⟹ 7 ( x - 2 ) = Distance
Downstream
Speed of the boat = ( x + 2 ) kmph
Time = 5 hours
Speed = Distance/time
⟹ x + 2 = Distance/5
⟹ 5 ( x + 2 ) = Distance
Distance covered upstream = Distance covered downstream
⟹ 7 ( x - 2 ) = 5 ( x + 2 )
⟹ 7x - 14 = 5x + 10
⟹ 7x - 5x = 10 + 14
⟹ 2x = 24
⟹ x = 12
∴ Speed of the boat in still water = 12 kmph.
Distance between the ports = 7 ( x - 2 )
= 7 ( 12 - 2 )
= 70 km.
Answer:
Account A: Decreasing at 8 % per year
Account B: Decreasing at 10.00 % per year
The amount f(x), in dollars, in account A after x years is represented by the function below:
f(x) = 10,125(1.83)x
Account B shows the greater percentage
change
Step-by-step explanation:
Part A: Percent change from exponential
formula
f(x) = 9628(0.92)*
The general formula for an exponential
function is
y = ab^x, where
b = the base of the exponential function.
if b < 1, we have an exponential decay
function.
f(x) decreases as x increases.
Account A is decreasing each year.
We can rewrite the formula for an
exponential decay function as:
y= a(1 – b)”, where
1- b = the decay factor
b = the percent change in decimal
form
If we compare the two formulas, we find
0.92 = 1- b
b = 1 - 0.92 = 0.08 = 8 %
The account is decreasing at an annual rate of 8%. The account is decreasing at an annual rate of 10.00%.
Account B recorded a greater percentage change in the amount of money over the previous year.