Answer:
The first set is a set of linear equations.
The way to figure this out is pretty easy. If you want to see it visually, go search up desmos graphing calculator and put in these equations.
A linear equation is a function that has a constant slope, meaning that the rate it increases or decreases will never change. The first one is a set of linear equations because it is 2 equations with constant slopes, meaning that the slopes will never change no matter what y and x are.
The second set is not, because while the first equation is linear, the second is an inequality. While it is a straight line, it doesn't count as a linear equation.
The third set, both equations have exponents on the x, which means that the slope will change depending on x. This means that both of these are not linear equations.
The only set that is a linear set is the one that has only linear equations.
Answer:
y = 1/2
Step-by-step explanation:
<u>Step 1: Solve for y</u>
2y + 3 = 4y + 2
2y + 3 - 2 - 2y = 4y + 2 - 2 - 2y
1 / 2 = 2y / 2
1/2 = y
Answer: y = 1/2
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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