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Dmitry_Shevchenko [17]
3 years ago
5

If cos =5\13 and 90°≤ø≤180,evaluate cotø,secø,cosø​

Mathematics
1 answer:
kotegsom [21]3 years ago
6 0

Answer:  \bold{\cot\theta=-\dfrac{5}{12}\qquad \sec\theta = \dfrac{13}{5}\qquad \cos\theta = \dfrac{5}{13}}

<u>Step-by-step explanation:</u>

90° ≤ θ ≤ 180° means that it is in Quadrant II →   x is + ,  y is -

\cos \theta = \dfrac{5}{13}\quad \rightarrow\quad x = 5, \ r = 13\\\\\\\text{Use Pythagorean Theorem to find y}:\\x^2+y^2=r^2\quad \rightarrow \quad 5^2+y^2=13^2\quad \rightarrow \quad y = -12\\\\\\\cot\theta=\dfrac{x}{y}\quad =\dfrac{5}{-12}\quad =\large\boxed{-\dfrac{5}{12}}\\\\\\\sec\theta=\dfrac{r}{x}\quad = \dfrac{13}{5}\quad = \large\boxed{\dfrac{13}{5}}\\\\\\\cos\theta =\dfrac{5}{13}\quad \text{(Given)}\\

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It takes 3 hours to paddle a kayak 12 miles downstream and 4 hours for the return trip upstream. Find the rate of the kayak in s
Furkat [3]

The rate of the kayak in still water is 3.5 mph

<h3><u>Solution:</u></h3>

Given that It takes 3 hours to paddle a kayak 12 miles downstream

Distance covered in downstream = 12 miles

Time taken to cover downstream = 3 hours

Also given that it takes 4 hours for the return trip upstream

Distance covered in upstream = 12 miles

Time taken to cover upstream = 4 hours

<em><u>Formula to remember:</u></em>

If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then: Speed downstream = (u + v) km/hr and Speed upstream = (u - v) km/hr

Let the speed of Kayak in still water = x mph

And The speed of current = y mph

For downstream:

Speed downstream = x + y

We know that speed = \frac{distance}{time}

\frac{distance}{time} = x + y

x + y = \frac{12}{3}

x + y = 4  ------ eqn 1

<em><u>For upstream:</u></em>

Speed upstream = x - y

\frac{12}{4} = x - y

x - y = 3 -------- eqn 2

Now let us eqn 1 and eqn 2

Add eqn 1 and eqn 2

x + y + x - y = 4 + 3

2x = 7

x = 3.5

speed of Kayak in still water = x mph  = 3.5 mph

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Solve linear equations 1/2x+y=-6 and y=3/5x+5
ser-zykov [4K]
1/2x+y=-6   y=3/5x+5

substitute
1/2x+3/5x+5=-6

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Which statement best reflects the solution(s) of the equation?
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x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation \frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1} and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0

Now, factoring the term:

x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2

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Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, \frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}  the denominator becomes zero i.e

\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2  in the equation,

\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}

\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

  • brainly.com/question/1626495
  • brainly.com/question/2959656
  • brainly.com/question/2456302

#learnwithBrainly

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