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>
19/8 is the improper fraction of 2 3/8.
3 5/6
I think, I haven’t done this in a long time.
Answer: x = -1 and y = -2
Step-by-step explanation:
So if you know y = x-1 you can plug that in where y is in the first equation
-4x + 3(x-1) = -2
Distribute the 3:
-4x +3x -3 = -2
Combine like terms:
-x -3 = -2
Add 3 to both sides:
-x = 1
Multiply by -1:
X = -1
Then you plug that into the other equation to get y=(-1)-1
Which is -2
