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Serjik [45]
2 years ago
12

How to simplify (81^-0.25)^3

Mathematics
1 answer:
IRISSAK [1]2 years ago
4 0
(81^-0.25)^3 = ( 1 / (81^0.25) )^3 

<span>81^.025 is the 4th root of 81 which is 3 </span>
<span>Therefor </span>
<span>( 1 / (81^0.25) )^3 = (1/3)^3 </span>

<span>(1/3)^3 = 1/27 <-----

Hope I Helped You!!! :-)

Have A Good Day!!!</span>
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A collection of 20 coins made up of only nickels, dimes, and quarters has a total value of $3.35. If the dimes were nickels, the
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Answer:  The required number of quarters in the collection is 11.

Step-by-step explanation:  Given that a collection of 20 coins made up of only nickels, dimes and quarters has a total value of $3.35.

If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of $2.75.

We are to find the number of quarters in the collection.

Let x, y and z represents the number of nickels, dimes and quarters respectively in the collection.

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1 nickel = $ 0.05,   1 dime = $ 0.10   and    1 quarter = $0.25.

Then, according to the given information, we have

x+y+z=20\\\\\Rightarrow x=20-y-z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\0.05x+0.10y+0.25z=3.35\\\\\Rightarrow 5x+10y+25z=335\\\\\Rightarrow x+2y+5z=67~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\0.05y+0.25x+0.10z=2.75\\\\\Rightarrow 5y+25x+10z=275\\\\\Rightarrow y+2z+5x=55~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)

Substituting the value of x from equation (i) in equations (ii) and (iii), we have

(20-y-z)+2y+5z=67\\\\\Rightarrow y+4z=47\\\\\Rightarrow y=47-4z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iv)

and

y+2z+5(20-y-z)=55\\\\\Rightarrow -4y-3z=-45\\\\\Rightarrow y=\dfrac{45-3z}{4}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(v)

Comparing the values of y from equations (iv) and (v), we get

47-4z=\dfrac{45-3z}{4}\\\\\Rightarrow 188-16z=45-3z\\\\\Rightarrow 16z-3z=188-45\\\\\Rightarrow 13z=143\\\\\Rightarrow z=\dfrac{143}{13}\\\\\Rightarrow z=11.

Thus, the required number of quarters in the collection is 11.

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