<span>Traveled Downstream a distance of 33 Mi and then came right back. If the speed of the current was 12 mph and the total trip took 3 hours and 40 minutes.
Let S = boat speed in still water then (s + 12) = downstream speed (s -12) = upstream speed
Given Time = 3 hours 40 minutes = 220 minutes = (220/60) h = (11/3) h Time = Distance/Speed
33/(s +12) + 33/(s-12) = 11/3 3{33(s-12) + 33(s +12)} = 11(s+12) (s -12) 99(s -12 + s + 12) = 11(</span> s^{2} + 12 s -12 s -144) 99(2 s) = 11(s^{2} -144) 198 s/11 = (s^{2} -144) 18 s = (s^{2} -144) (s^{2} - 18 s - 144) = 0 s^{2} - 24 s + 6 s -144 =0 s(s- 24) + 6(s -24) =0 (s -24) (s + 6) = 0 s -24 = 0, s + 6 =0 s = 24, s = -6 Answer) s = 24 mph is the average speed of the boat relative to the water.

7 0

3 years ago

I NEED HELP ASAP!!!!

abruzzese [7]

Answer is 4/7

You divide the length of C'D' over the length of CD
C'D'/CD = 8/14 = 4/7

The scale factor is smaller than 1, so C'D' is shorter than CD

8 0

4 years ago

Sqaure root of 5 (u-1)(u^5+u^4+u^3+u^2+u+1)

Anna71 [15]

GIVEN:

$5(u - 1)( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1)$

remember:

$\sqrt{u} = {u}^{ \frac{1}{2} }$

And

${u}^{n} \times {u}^{m} = {u}^{n + m}$

SOLVE:

start by multiplying the factors:

$5( ({u}^{6} + {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u ) - ( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1))$

simplify by combing like terms. Most terms subtract off, leaving:

$5( {u}^{6} - 1)$

This can be factored, but it is not a perfect square, which is really what we need to take the square root.

$5( {u}^{3} - 1)( {u}^{3} + 1)$

I'm not exactly sure what form they want the answer in...

so taking the square root:

$\sqrt{5( {u}^{6} - 1) } = {(5( {u}^{6} - 1))}^{ \frac{1}{2} }$

so my best answer is:

${5}^{ \frac{1}{2} } \times {( {u}^{6} - 1)}^{ \frac{1}{2} }$
or the more factored form:

${5}^{ \frac{1}{2} } { ({u}^{3} - 1)}^{ \frac{1}{2} } { ({u}^{3} + 1 )}^{ \frac{1}{2} }$

I'm not sure how else to solve it. Taking the square root doesn't work out super well, so I left it in the most simple form I could.

sorry for not coming to a definitive answer!

7 0

3 years ago