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iragen [17]
3 years ago
13

(w+14)/(4w+6)=(3)/(4) show work plz

Mathematics
1 answer:
sergey [27]3 years ago
7 0

Answer:

w = -1.44 and w = -14.06

Step-by-step explanation:

That "(3) / (4)" in  (w+14)/(4w+6)=(3)/(4) looks fishy.  Assuming that you really meant 3/4, then we have   (w+14)/(4w+6)=3/4.

Mult. both sides by 4 to remove the fraction:  4(w+14)/(4w+6) = 3.

Multiply out (w+14)/(4w+6):  4w^2 + 6w + 56w + 84.  Combine the w terms, obtaining 4w^2 + 62w + 84.  Then the original equation becomes:

4w^2 + 62w + 84 = 3, or  4w^2 + 62w + 81 = 0.

This is a quadratic equation, with a = 4, b = 62 and c = 81.  The two roots (values of

w) are as follows:

        -62 plus or minus √(62^2-4(4)(81)

   ----------------------------------------------------------

                               8

         -62 plus or minus √(2548)           -62 plus or minus 50.48

w = -------------------------------------------   =  ---------------------------------------

                            8                                                  8

Simplifying, we get w = -1.44 and w = -14.06

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son4ous [18]

Answer:

y=\frac{5}{4}x+\frac{15}{4}

Step-by-step explanation:

x | y

-------

1     5

5   10

9    15

13   20

This one is linear because as x goes up by the same number so does y. So the ratio of difference of y to difference of x is the same per pair of points.

So a linear equation in slope-intercept form is y=mx+b where m is the slope and b is the y-intercept.

To find the slope, I'm going to line up the points vertically and subtract, then put 2nd difference over 1st difference. Like so,

( 1    ,   5)

-( 5   ,  10)

----------------

-4          -5

So the slope is 5/4 which makes sense since the y's are going up by 5 each time and the x's are going up by 4 each time.

So we have m=5/4. Let's plug that into our y=mx+b.

y=5/4 x+b

To find b, we need to use y=5/4 x+b along with one of the given points.

Choose; it doesn't matter.  I like (1,5) I guess.

y=5/4 x +b with (1,5)

5=5/4 (1)+b

5=5/4    +b

Subtract 5/4 on both sides:

5-5/4=b

20/4-5/4=b  (Found a common denominator)

15/4=b

The y-intercept is 15/4 so b=15/4.

So the equation for the line in slope-intercept form is y=5/4 x +15/4.

y=\frac{5}{4}x+\frac{15}{4}

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<em></em>

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

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\large\begin{array}{l} \textsf{From the picture, we get}\\\\ \mathsf{tan\,\theta=\dfrac{2}{3}}\\\\ \mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{2}{3}}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\qquad\mathsf{(i)} \end{array}


\large\begin{array}{l} \textsf{Square both sides of \mathsf{(i)} above:}\\\\ \mathsf{(3\,sin\,\theta)^2=(2\,cos\,\theta)^2}\\\\ \mathsf{9\,sin^2\,\theta=4\,cos^2\,\theta}\qquad\quad\textsf{(but }\mathsf{cos^2\theta=1-sin^2\,\theta}\textsf{)}\\\\ \mathsf{9\,sin^2\,\theta=4\cdot (1-sin^2\,\theta)}\\\\ \mathsf{9\,sin^2\,\theta=4-4\,sin^2\,\theta}\\\\ \mathsf{9\,sin^2\,\theta+4\,sin^2\,\theta=4} \end{array}

\large\begin{array}{l} \mathsf{13\,sin^2\,\theta=4}\\\\ \mathsf{sin^2\,\theta=\dfrac{4}{13}}\\\\ \mathsf{sin\,\theta=\sqrt{\dfrac{4}{13}}}\\\\ \textsf{(we must take the positive square root, because }\theta \textsf{ is an}\\\textsf{acute angle, so its sine is positive)}\\\\ \mathsf{sin\,\theta=\dfrac{2}{\sqrt{13}}} \end{array}

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\large\begin{array}{l} \textsf{From (i), we find the value of }\mathsf{cos\,\theta:}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{2}\,sin\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{\sqrt{13}}}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\sqrt{13}}}\\\\ \end{array}

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\large\begin{array}{l} \textsf{Since sine and cosecant functions are reciprocal, we have}\\\\ \mathsf{sin\,2\theta\cdot csc\,2\theta=1}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{sin\,2\theta}\qquad\quad\textsf{(but }}\mathsf{sin\,2\theta=2\,sin\,\theta\,cos\,\theta}\textsf{)}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\,sin\,\theta\,cos\,\theta}}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\cdot \frac{2}{\sqrt{13}}\cdot \frac{3}{\sqrt{13}}}} \end{array}

\large\begin{array}{l} \mathsf{csc\,2\theta=\dfrac{~~~~1~~~~}{\frac{2\cdot 2\cdot 3}{(\sqrt{13})^2}}}\\\\ \mathsf{csc\,2\theta=\dfrac{~~1~~}{\frac{12}{13}}}\\\\ \boxed{\begin{array}{c}\mathsf{csc\,2\theta=\dfrac{13}{12}} \end{array}}\qquad\checkmark \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2150237


\large\textsf{I hope it helps.}


Tags: <em>trigonometry trig function cosecant csc double angle identity geometry</em>

</span>
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