X=cos\alp,\alp∈[0:;π] \displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\L eftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+
1−x
2


∣=
2

(2x
2
−1)\Leftright∣cos\alp+sin\alp∣=
2

(2cos
2
\alp−1)
\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|= \N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N
2

cos(\alp−
4
π

)∣=N
2

cos(2\alp)\Right\alp∈[0;
4
π

]∪[
4
3π

;π]
1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;

PLEASE SIMPLIFY

what?

Step-by-step explanation:

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18