i need help on algebra 1 hw?

Question

a fishing boat traveled 3hours against a 6km/hr current. the return trip took only 2 hours. find the speed of the boat in still water.

please help if u can and can u explain how u got it thanks

Answer 1

Let the sped of the boat in still water be x kmph ,
Therefore,the resultant speed against the current would be x-6 kmph and in favour of the current would be x+6 kmphlet the distance travelled be d km.
By the problem
d/(x-6)=3
or d=3(x-6)
=>d=3x-18.....(1)
from the second condition we get
d/(x+6)=2
=> d=2(x+6) [by cross-multiplication]
=>d=2x+12....(2)
From the first eqn. we get d=3x-18 and from the second one we get d=2x+12
Therefore, 3x-18=2x+12
=>3x-2x=12+18
=>x=30
Therefore,the speed of the boat in still water is 30 kmph

Answer 2

You'll use the distance formula "distance = rate x time" or d = rt.
Since the distance 'there' and 'back' are the same, we'll set the two distances equal to each other, or say that the rate times the time there is equal to the rate times the time back.

The boat is traveling x km/hr. The current will make you go slower on the way there, so your rate is x - 6. The rate coming back will be faster because of the current, so the rate back is x + 6. The time there is 3 and the time back is 2.

d(there) = (x - 6)(3)
d(back) = (x + 6)(2)

(x - 6)(3) = (x + 6)(2)
3x - 18 = 2x + 12 (distributive property)
-2x -2x

x - 18 = 12
+ 18 + 18

x = 30 km/hr.

This means that you were going 24 km/hr there for 3 hours
d = 24 x 3 = 72 km

and 36 km/hr back for 2 hours: d = 36 x 2 = 72 km

Answer 3

x= the speed of the boat in still water
the equation is (i think...)

3x-6(3) = 2x

each side represents the distant that the boat travels.
on the left, it was traveling for 3 hours at (x) miles per hour. but you had to subtract the distance that the wind took away from it. on the left side, you have the same distance, but it only traveled for two hours and had no wind.

this would be the finished equation
3x-6(3) = 2x
3x-18=2x
x = 18 (you have to bring the 2x over and the -18 over)

so the boat travels 18 km/hr in still water

Answer 4

You solve it by finding the 'point of intersection'. Basically, set up two equasions using the Distance = (Speed)(Time):

Let 'x' represent speed of boat in still water
Let 'y' represent the distance the boat travelled
With current: y = (x + 6)(2)
Against current: y = (x - 6)(3)

They both equal 'y', therefore they equal eachother, so:

(x + 6)(2) = (x - 6)(3)

After that, just solve for 'x' and there ya have it. That should work, if I remember my math correctly (hopefully ^^). Good luck!

Answer 5

on the 1st question you additionally could make x as buddy one and a million.5x the different, provided that he's pay a million.5 circumstances what the different guy or woman (x) will pay. So, your equation could look like: x + a million.5x = $840 Now, basically resolve from there the 2nd is comparable. The length is two circumstances the dimensions of the width. So, 2w is the dimensions and w is the width. provided that perimeter is each and all the standards extra up, then your equation could look like 2w +2w+w+w = 40 two. Now, on your innovations think of of each W as having a one in front of it. so, 2w+2w+1W+1W = 40 two or 6w=40 two - now divide the two factors via 6 and you get w=7 ok? try something on your individual. optimistically the above supply you the assumption.

Answer 6

oh..... i see how you got the speed, but in still water.... thats a little hard... lets see, you find the average, and thats, oh, well good luck.