Answer:
titutex=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)…
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Answer:
15.8183
Step-by-step explanation:
just round it you can do that ;)
Answer:
a. Yes
b. No
Step-by-step explanation:
To find if an ordered pair is a solution to the inequality, substitute its x and y values for the x and y in the inequality and solve. If the equation is true, then it is a solution. If it is not, then it is not a solution.
1) First, substitute the x and y values of (3, 1) into the inequality. So, substitute 3 for x and 1 for y:
1 is greater than or equal to 1, so (3,1) is a solution.
2) Second, do the same with the point (1, -4). Substitute 1 for x and -4 for y:
However, -4 is not greater than or equal to -3, thus (1, -4) is not a solution.
Answer:
y = -12x + 53
Step-by-step explanation:
don't mind this hehgajhakwuzgbsldj
Answer:
y=5x-3
Step-by-step explanation:
it just is
