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pentagon [3]
3 years ago
13

6n-3>-18 please show work

Mathematics
2 answers:
maxonik [38]3 years ago
6 0
First off, add 3 to both sides of the equation.
6n - 3 + 3 > -18 + 3
Then, your equation should look like this: 6n > -15
Next, divide both sides by 6.
6n/6 > -15/6
Your answer is n > -5/2
kolezko [41]3 years ago
4 0
Your answer : n > -\frac{5}{2}


How I FOUND my answer ↓


Rearrange⇒ 6 × n- 3 - (-18) > 0 

Take out the like terms ⇒ 6n + 15 = 3 × (2n + 5)

Divide both sides by  3 and 2 

n + (\frac{5}{2} ) > 0

Then subtract both sides by -\frac{5}{2} ⇒ n -\frac{5}{2} 

Lastly, 6.000 x + 15.000 > 0


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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+

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∣=

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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

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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)…

1

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6 0
4 years ago
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