Consider the following data set: 9,8,8,2,0,0 find the median
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and $\theta$ lies in $2^{\text{th}}$ quadrant.
where, $x-$ coordinate is negative, and $y-$ coordinate is positive
so it can't a.
now, cosine means, side adjacent over the hypotenuse, in Cartesian plane, that will be $x-$ coordinate over the distance from origin.
Assume the triangle , with base $2$ units and hypotenuse $\sqrt{29}$ and it's in second quadrant. (so )
now, the leftmost point on $x-$ axis is , obviously $(-2,0)$
and by Pythagoras theorem, we can find the perpendicular side, that will be $y^2=(\sqrt{29})^2-(2)^2\implies y=5$
so the coordinates of the upper vertex is $(-2,5)$, each point lying on this "ray" should have equal ratio of respective coordinates. i.e. $\frac25=\left|\frac xy\right| $
and it should lie on second quadrant, so $x<0 \, y>0$
Option d satisfies this.
Answer: b = 4
Step-by-step explanation:
<u>Given</u>
4b + 5 = 1 + 5b
<u>Subtract 4b on both sides</u>
4b + 5 - 4b = 1 + 5b - 4b
5 = 1 + b
<u>Subtract 1 on both sides</u>
5 - 1 = 1 + b - 1
Hope this helps!! :)
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