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guapka [62]
3 years ago
7

An island has some portion submerged in water. Point A on the island is at negative 7 over 3 feet from the surface of the sea. P

oint B on the island is at negative 5 over 2 feet from the surface of the sea. Because negative 7 over 3 feet > negative 5 over 2 feet, what is true about the distance of the two points from the surface of the sea?
Mathematics
2 answers:
Tasya [4]3 years ago
8 0

Answer:

1 / 6 feet

Step-by-step explanation:

sashaice [31]3 years ago
3 0

Answer:

1/6feet

Step-by-step explanation:

Given

Point A on the island = -7/3 feet from the sea surface

Point B on the island = -5/2 from the sea surface

The distance between the two points will be Point A - Point B

Distance between the two points = -7/3 - (-5/2)

= -7/3+5/2

Find the LCM

= {2(-7)+3(5)}/6

= -14+15/6

= 1/6

Hence the distance of the two points from the surface of the sea is 1/6feet

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-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.

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