What is the solution set for 2X - 1 > 7?

2 answers:

8 0

2x - 1 > 7 ; Start
2x > 8 ; Add 1 to each side of the inequality
x > 4 ; Divide both sides by 2

So, x have to be less than 4 in order to be in the solution set.

Svetach [21]3 years ago

8 0

2X - 1 > 7
 2X > 7 + 1
2X > 8
 X > 8/2
 X>4

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sukhopar [10]

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Step-by-step explanation:

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8 0

3 years ago

One more time!

CaHeK987 [17]

Since q(x) is inside p(x), find the x-value that results in q(x) = 1/4

$\frac{1}{4} = 5 - x^2\ \Rightarrow\ x^2 = 5 - \frac{1}{4}\ \Rightarrow\ x^2 = \frac{19}{4}\ \Rightarrow \\
x = \frac{\sqrt{19} }{2}$

so we conclude that
$q(\frac{\sqrt{19} }{2} ) = 1/4$

therefore

$p(1/4) = p\left( q\left(\frac{ \sqrt{19} }{2} \right) \right)$

plug $x=\sqrt{19}/2$ into p( q(x) ) to get answer

$p(1/4) = p\left( q\left( \frac{ \sqrt{19} }{2} \right) \right)\ \Rightarrow\ \dfrac{4 - \left( \frac{\sqrt{19} }{2}\right)^2 }{ \left( \frac{\sqrt{19} }{2}\right)^3 } \Rightarrow \\ \\ \dfrac{4 - \frac{19}{4} }{ \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{8\left(4 - \frac{19}{4}\right) }{ 8 \cdot \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{32 - 38}{19\sqrt{19}} \Rightarrow \dfrac{-6}{19\sqrt{19}} \cdot \frac{\sqrt{19}}{\sqrt{19}}\Rightarrow$

$\dfrac{-6\sqrt{19} }{19 \cdot 19} \\ \\ \Rightarrow -\dfrac{6\sqrt{19} }{361}$

$p(1/4) = -\dfrac{6\sqrt{19} }{361}$

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3 years ago

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