Given:
t1 = 3.6 h
t2 = 4.5 h
x = speed of boat
y = speed of water
Required:
a) Expression of distance traveled with moving water with 3.6h
Expression of distance traveled with moving water with 4.5h
b) Solve for y
c) Percent of boat's speed is the water current
Solution:
Working formula: distance = velocity*time
a) For travelling downstream, we get the equation
d = (x +y)*3.6
For travelling upstream, we get the equation
d = (x-y)*4.5
b) Setting the distance as equal for travelling upstream or downstream, we arrive at the equation of
(x+y)*3.6 = (x-y)*4.5
3.6x + 3.6y = 4.5x - 4.5y
8.1y =0.9x
y = x/9
c) percentage = 1/9*100% = 11.1%
<em>ANSWERS: a) d = (x+y)*36; d = (x-y)*4.5
</em> <em>b) y = x/9
</em> <em>c) 11.1%</em>
Answer:
x=−5
Step-by-step explanation:
Steps:
Step 1 to 4 : Simplify
Steps 5: Calculating the Least Common Multiple
Steps 6: Calculating Multipliers
The correct answer for this question is x=−5
Answer: x=−5
<em><u>Hope this helps.</u></em>
Answer:
Dear User,
Answer to your query is provided below
The time will be 8am on Wednesday when flight arrive in Tokyo.
Step-by-step explanation:
Here, You can see that when 11 a.m. in Johannesburg, it is 6 p.m. the same day in Tokyo, Japan ; this means 7Hrs. gap.
Further, A flight leaves Johannesburg at 6 a.m. on Tuesday for Tokyo takes 18 hours in the air, with an additional 1 hour stop-over on land.
So, 6a.m. + 18hr.+1hr.+7hr. = 8am on Wednesday in Tokyo (Arrival)
Answer:
19 would be the naswer
Step-by-step explanation:
Answer: The formula is x6
Step-by-step explanation: