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Maurinko [17]
3 years ago
7

I need work shown. I can't figure these out

Mathematics
1 answer:
tatuchka [14]3 years ago
6 0
9). The river is 80 meters wide. That is, the distance from shore to shore as measured straight across the river is 80 meters. The time to cross this 80 meters wide river can be determined by rearranging and substituting into the average speed equation.



Time = distance/(ave.speed)

The distance of the 80 m can be substituted into the numerator; but what about the denominator. What value should be used for average speed?

time = (80 m)/(4 m/s) = 20 s it requires 20 s for the boat to travel across the river. During this 20 s the boat also drifts downstream.

What distance downstream? 

distance = ave.speed * time = (3 m/s) * (20 s)
    distance = 60 m  The boat is carried 60 meters downstream during 20 seconds it takes to cross the river.


Resultant Velocity can be performed using the Pythagorean Theorem.




10).  



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<h3><u>Answer</u> :</h3>

We know that,

\dag\bf\:sin^2A=\dfrac{1-cos2A}{2}

\dag\bf\:sin2A=2sinA\:cosA

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

<u>Now, Let's solve</u> !

\leadsto\:\bf\dfrac{sin^2A-sin^2B}{sinA\:cosA-sinB\:cosB}

\leadsto\:\sf\dfrac{\frac{1-cos2A}{2}-\frac{1-cos2B}{2}}{\frac{2sinA\:cosA}{2}-\frac{2sinB\:cosB}{2}}

\leadsto\:\sf\dfrac{1-cos2A-1+cos2B}{sin2A-sin2B}

\leadsto\:\sf\dfrac{2sin\frac{2A+2B}{2}\:sin\frac{2A-2B}{2}}{2sin\frac{2A-2B}{2}\:cos\frac{2A+2B}{2}}

\leadsto\:\sf\dfrac{sin(A+B)}{cos(A+B)}

\leadsto\:\bf{tan(A+B)}

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3 years ago
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