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erik [133]
3 years ago
15

How do I simplify 1/-5z^-5?

Mathematics
1 answer:
belka [17]3 years ago
3 0
\bf \cfrac{1}{-5z^{-5}}\\\\
-------------------------\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\\ \quad \\
%  negative exponential denominator
a^{{ n}} \implies \cfrac{1}{a^{- n}}
\qquad \qquad 
\cfrac{1}{a^{- n}}\implies \cfrac{1}{\frac{1}{a^{ n}}}\implies a^{{ n}} \\\\
-------------------------\\\\
thus


\bf -\cfrac{1}{5}\cdot \cfrac{1}{z^{-5}}\implies -\cfrac{1}{5}\cdot \cfrac{1}{\frac{1}{z^5}}\implies -\cfrac{1}{5}\cdot \cfrac{\frac{1}{1}}{\frac{1}{z^5}}\implies -
\cfrac{1}{5}\cdot \cfrac{1}{1}\cdot \cfrac{z^5}{1}
\\\\\\
-\cfrac{z^5}{5}
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3 years ago
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pav-90 [236]
STEP 1:
determine equations needed

x= # of $10 tickets
y= # of $14 tickets

QUANTITY EQUATION
x + y= 800

COST EQUATION
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STEP 2:
multiply quantity equation by -14

-14(x + y)= -14(800)
-14x - 14y= -11,200


STEP 3:
add new quantity equation in step 2 to cost equation of step 1 to solve for x using elimination

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divide both sides by -4

x= 540 ten dollar tickets


STEP 4:
substitute x value in step 3 into either original equation

x + y= 800
540 + y= 800
subtract 540 from both sides

y= 260 fourteen dollar tickets


ANSWER: There were 540 $10 tickets and 260 $14 tickets sold.

Hope this helps! :)
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