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lapo4ka [179]
3 years ago
15

a boat on the Potomac River travels the same distance downstream in 2/3 hour as it does going upstream in 1 hour. If the speed o

f the current is 3 mph, find the speed of the boat in still water
Mathematics
1 answer:
lord [1]3 years ago
7 0
\bf \begin{array}{ccccllll}
&distance&rate&time(hrs)\\
&-----&-----&-----\\
upstream&d&r-3&\frac{2}{3}
\\\\
downstream&d&r+3&1
\end{array}\\\\
-----------------------------\\\\
thus\qquad 
\begin{cases}
d=(r-3)\frac{2}{3}
\\\\
d=(r+3)1
\end{cases}\qquad d=d\qquad thus
\\\\\\
(r-3)\cfrac{2}{3}=(r+3)1

notice, the distance is the same, upstream or downstream, thus is "d" for both

solve for "r"
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-Dominant- [34]
We know that

 in the triangle TQS
<span>applying the Pythagorean theorem
QS</span>²=TS²+TQ²---------> TQ²=QS²-TS²--------> TQ²=18²-9x²-----> equation 1

in the triangle TRS
TS²=TR²+RS²--------------> TR²=TS²-RS²-------> TR²=9x²-144----> equation 2

in the triangle QTR
TQ²=TR²+36-----------> equation 3


<span>I substitute 1 and 2 in 3
</span>18²-9x²=9x²-144+36--------> 18x²-432=0------> x²=24-------> x=√24
x=2√6
TS=3*x------> 3*2√6-----> 6√6
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the answer is
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3 years ago
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9. Find (f.g)(a) if f(a) = . and g(a) = 2². Fully simplify your answer.
Aleks04 [339]

D is your answer ....

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F(x) = -4x2 + 4x – 4<br> Find f(3)
ValentinkaMS [17]

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F(3) = -28

I hope this helps! :)

5 0
2 years ago
You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then
gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is  d\theta_1  =(14.36p)^o

Step-by-step explanation:

From the question we are told that

    The angle of elevation  is  \theta_1  =  15 ^o =  \frac{\pi}{12}

     The height of the tree is  h

      The distance from the base is  D

h is mathematically represented as

            h  = D tan \theta       Note : this evaluated using SOHCAHTOA i,e

                                               tan\theta  =  \frac{h}{D}

Generally for small angles the series approximation of  tan \theta \  is

          tan \theta  =  \theta  + \frac{\theta ^3 }{3}

So given that \theta =  15 \ which \ is \ small

       h = D (\theta + \frac{\theta^3}{3} )

       dh = D (1 + \theta^2) d\theta

=>        \frac{dh}{h} =  \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta

Now from the question the relative error of height should be at  most

        \pm  p%

=>    \frac{dh}{h} =   \pm p

=>    \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta  = \pm p

=>      d\theta  =  \pm  \frac{\theta +  \frac{\theta^3}{3} }{1+ \theta ^2} *    \ p

 So  for   \theta_1

            d\theta_1  =  \pm  \frac{\theta_1 +  \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} *    \ p

substituting values  

          d [\frac{\pi}{12} ]  =  \pm  \frac{[\frac{\pi}{12} ] +  \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} *    \ p

 =>       d\theta_1  = 0.25 p

Converting to degree

           d\theta_1  = (0.25* 57.29) p

            d\theta_1  =(14.36p)^o

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