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Rasek [7]
4 years ago
15

97.57349 rounded to four decimal places is

Mathematics
2 answers:
jeyben [28]4 years ago
8 0
97.5725 hope this helps!!! :)
Umnica [9.8K]4 years ago
7 0

Answer:

97.5735

Step-by-step explanation:

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Step-by-step explanation:

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6 0
2 years ago
-3|9x-7|=2 Find the answer for x
labwork [276]
<h2>Solving Equations with Absolute Expressions</h2><h3>Answer:</h3>

<u>No Solutions</u>

<h3>Step-by-step explanation:</h3>

Given:

-3|9x -7| = 2

Rewriting the given equation:

-3|9x -7| = 2 \\ |9x -7| = -\frac{2}{3}

We have to realize that the right side of the equation, |9x -7|, will always be positive no matter what real values of x (because we're taking the absolute value of the expression) and we are equating it to a <em>negative</em> constant number, -\frac{2}{3}\\. Something that is always positive will never be negative so there's no value for x that satisfies the solution.

\rule{6.5cm}{0.5pt}

<em>You</em><em> </em><em>may</em><em> </em><em>not</em><em> </em><em>read</em><em> </em><em>the</em><em> </em><em>following</em><em> passage</em><em> </em><em>that</em><em> </em><em>I</em><em> </em><em>have</em><em> </em><em>written.</em>

\rule{6.5cm}{0.5pt}

Solving by positive of the expression:

9x -7 = -\frac{2}{3} \\ 9x = -\frac{2}{3} +7 \\ 9x = -\frac{2}{3} +\frac{21}{3} \\ 9x = \frac{19}{3} \\ 9x \times \frac{1}{9} = \frac{19}{3} \times \frac{1}{9} \\ x = \frac{19}{27}

Solving by the negative of the expression:

-(9x -7)= -\frac{2}{3} \\ 9x -7 = \frac{2}{3} \\  9x = \frac{2}{3} +7 \\ 9x = \frac{2}{3} +\frac{21}{3} \\ 9x = \frac{23}{3} \\ 9x \times \frac{1}{9} = \frac{23}{3} \times \frac{1}{9} \\ x = \frac{23}{27}

Checking: x = \frac{19}{27}\\

-3|9(\frac{19}{27}) -7| \stackrel{?}{=} 2 \\ -3|\frac{19}{3} -7| \stackrel{?}{=} 2 \\ -3|\frac{19}{3} -\frac{21}{3}| \stackrel{?}{=} 2 \\ -3|-\frac{2}{3}| \stackrel{?}{=} \\ -3(\frac{2}{3}) \stackrel{?}{=} 2 \\ -2 \stackrel{?}{=} 2 \\ -2 \neq 2

x = \frac{19}{27}\\ is an extraneous solution.

Checking: x = \frac{23}{27}\\

-3|9(\frac{23}{27}) -7| \stackrel{?}{=} 2 \\ -3|\frac{23}{3} -7| \stackrel{?}{=} 2 \\ -3|\frac{23}{3} -\frac{21}{3}| \stackrel{?}{=} 2 \\ -3|\frac{2}{3}| \stackrel{?}{=} \\ -3(\frac{2}{3}) \stackrel{?}{=} 2 \\ -2 \stackrel{?}{=} 2 \\ -2 \neq 2

x = \frac{23}{27}\\ is an extraneous solution.

8 0
3 years ago
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