What are the values of x and y that make the equations true?

thats easy

Step-by-step explanation:

Substitutex=cos\alp,\alp∈[0:;π]

\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |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;  

4

π

​  

]

\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−  

4

π

​  

)=cos(2\alp)…

2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[  

4

3π

​  

;π]

\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−  

4

π

​  

)=cos(2\alp)

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb

c

-by-step explanation:

it would be c