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Phoenix [80]
3 years ago
6

Solve the enquality 2x+1>11​

Mathematics
2 answers:
lord [1]3 years ago
8 0
X>5
Since 2x>11-1 is 2x>10 divide by 2
And we get x>5
saveliy_v [14]3 years ago
8 0

Step-by-step explanation:

2x+1>11 <em>1</em><em> </em>is positive integers so it have to be cancelled by negative integer <em>1</em>

<em>2</em><em>x</em><em>></em><em>1</em><em>1</em><em>-</em><em>1</em><em> </em>

<em>2</em><em>x</em><em>></em><em>1</em><em>0</em><em> </em><em>two </em><em>will </em><em>be </em><em>cancelled</em><em> by</em><em> </em><em>two </em><em>to </em><em>get </em><em>x</em>

<em>then </em><em>we </em><em>divide </em><em>2</em><em> </em><em>from </em><em>1</em><em>0</em>

<em>x</em><em>></em><em>5</em><em> </em><em>this </em><em>means </em><em>x </em><em>is </em><em>greater </em><em>than </em><em>5</em><em> </em>

<em>{</em><em>x=</em><em>6</em><em>,</em><em>7</em><em>,</em><em>8</em><em>,</em><em>.</em><em>.</em><em>.</em><em>}</em>

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
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Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

4.

(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

6.

2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

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