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Leona [35]
3 years ago
12

The water in a river moves south at 10 mi/hr. A motorboat travels due east at a speed of 20 mi/hr relative to the shore. Determi

ne the speed and direction of the boat relative to the moving water.
Mathematics
1 answer:
weeeeeb [17]3 years ago
4 0

Answer:22.36 mi/hr

Step-by-step explanation:

Given

water in a river moves south at (V_w)10 mi/hr

motorboat travels due to east at a speed of (V_b)20 mi/hr

Velocity of boat relative to the moving water V_{bw}

V_{bw}=V_b-V_w

V_{bw}=20i-(-10j)=20i+10j

Magnitude of V_{bw}=\sqrt{20^2+10^2}=\sqrt{500}

V_{bw}=10\sqrt{5} \approx 22.36 mi/hr

tan\theta =\frac{10}{20}

\theta =26.56^{\circ} with the x axis.

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Step-by-step explanation:

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Write the computer program for;urgent​
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ANSWER:

#include <stdio.h>

#define PAYRATE 10 //basic pay rate per hour

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{

overHours = hours - 40;

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Answer:

Step-by-step explanation:

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Answer:

Our two intersection points are:

\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)

Step-by-step explanation:

We want to find where the two graphs given by the equations:

\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1

Intersect.

When they intersect, their <em>x-</em> and <em>y-</em>values are equivalent. So, we can solve one equation for <em>y</em> and substitute it into the other and solve for <em>x</em>.

Since the linear equation is easier to solve, solve it for <em>y: </em>

<em />\displaystyle y = -\frac{3}{4} x + \frac{1}{4}<em />

<em />

Substitute this into the first equation:

\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16

Simplify:

\displaystyle (x+1)^2 + \left(-\frac{3}{4} x  + \frac{9}{4}\right)^2 = 16

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\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16

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