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A motorboat took 3 hours to make a downstream trip with a 4 mph
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A motorboat took 3 hours to make a downstream trip with a 4 mph current. The return trip against the same current took 5 hours. Find the speed of the boat in still water. Can you solve this? (Pic unrelated)
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Rewrite as sums or differences of logarithms -> log base d of (x^5*y^6*z)=
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Log base 5 of (5x-6)=4 (solve for x)
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16mph
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Log x = -6 (solve for x)
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6^7x=36 (solve for x)
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25^x*5^3x=25 (solve for x)
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Any help on these problems would be amazing, this algebra class makes me feel like an imbecile
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5 minutes left on this test
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>>7679944
Trust your feelings anon.

Make a picture of the problem.
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>>7679927
>A motorboat
>with a 4 mph current.
Cannot be solved.
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>>7679927
distance is 60. velocity is 16. this is not college level stuff. this is retard level stuff.
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>>7680118
that wasnt very nice of me... its a little tricky actually. just takes some algebra. 2 d=v*t equations. eliminate d, solve for v with v= (vboat+/-vcurrent)
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d = (v+5)*t
d = (v-5)*t
Those two equations are equal to eachother so just do [eqn] (v+5)*(3)=(v-5)*(5) [/eqn] Simplify this to [eqn] 3v+15 = 5v-25 [/eqn] Now just solve for v
[eqn] v = 5 [/eqn] Congrats you have now solved a grade 9 level math problem!
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>>7679927
>still water
0
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>>7679929
[eqn] 4log_dx+6log_dy+log_dz [/eqn]
>>7679931
[eqn]5^4=5x-6[/eqn]
[eqn]5^3+6/5=x[/eqn]
>>7679934
x=10^-6
>>7679937
[eqn] (log_636)/7 = x [/eqn]
>>7679939
[eqn] (log_525)/5=x [/eqn]

>>7680146
I put the current as 5 instead of 4 so fix that.
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>>7680149
>not reading the question
It says the speed of the boat not the speed of the water retard.
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>>7680146
lmao i also suck at algebra and solved this wrong
>>
In still water a 60mi trip takes 3:45.
What's the name of the vessel.
Thread replies: 19
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