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Cerrena [4.2K]
2 years ago
13

5^2•3^-1•5^-3/3^4•5^-1•3^-3

Mathematics
1 answer:
puteri [66]2 years ago
3 0
Assuming you want the expression to be simplified. 

We begin with the following: 

5^{2} * 3^{-1} * \frac{5^-3}{3^{4}} * 5^{-1} * 3^{-3}

Simplify the first part, 5^{2}. That is 25. Now we have this: 

25*3^{-1.5} * \frac{-3}{3^{4}} * 5^{-1} * 3^{-3} 

Next, simplify 3^{-1}, which is 1/3, and get this: 

25* 1/3 * \frac{-3}{3^{4}} * 5^{-1} * 3^{-3} 

The next part is \frac{5^-3}{3^{4}}. Simplify the denominator, 3^{4}, which is 81. Simplify the numerator, which is 1/125. Then divide 1/125 by 81, which we will keep as a fraction for simplicity's sake, but simplify it to \frac{1}{10125}. Now we have: 

25* 1/3 * \frac{1}{10125} * 5^{-1} * 3^{-3}

Now simplify 5^{-1}, which is 0.2, or 1/5. Now we have: 

25* 1/3 * \frac{1}{10125} * 0.2 * 3^{-3} 

Finally, simplify 3^{-3}. That is 1/27. We have: 

25* 1/3 * \frac{1}{10125} * 0.2 * 1/27

Lastly, multiply them all together! Now we are done, with the product of: 

\frac{1}{6075}

That certainly did take a while to type in all the LaTex, so I really hope that helped!

Note- if anything isn't working with the LaTex, just tell me and I'll fix it! (:
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What is the value of i^n if the remainder of n/4 is 2?
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n=2x

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3 years ago
There are premium tickets costing $99 each and rest are regualr tickets costing $25 each. A pile of 120 tickets have a total val
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Answer:

There are 38 premium tickets and 82 regular tickets in a pile.

Step-by-step explanation:

Given,

Total number of tickets = 120

Total amount = $5812

Solution,

Let the number of premium tickets be x.

And the number of regular tickets be y.

Total number of tickets is the sum of total number of premium tickets and  total number of regular tickets.

So the equation can be written as;

x+y=120\ \ \ \ \ equation\ 1

Again,total amount  is the sum of total number of premium tickets multiplied by cost of each premium ticket and  total number of regular tickets multiplied by cost of each regular ticket.

So the equation can be written as;

99x+25y=5812\ \ \ \ equation\ 2

Now We will multiply equation 1 by 25 we get;

x+y=120\\25(x+y)=120\times25\\25x+25y= 3000 \ \ \ \ equation\ 3

Now Subtracting equation 3 from equation 2 we get;

(99x+25y)-(25x+25y)=5812-3000\\\\99x+25y-25x-25y=2812\\\\74x=2812\\\\x=\frac{2812}{74}=38

We will now substitute the value of x in equation 1 we get;

x+y=120\\38+y=120\\y=120-38\\y=82

Hence There are 38 premium tickets and 82 regular tickets in a pile.

3 0
3 years ago
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