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vekshin1
2 years ago
9

PLS HELP AND DONT GET IT WRONG!

Mathematics
2 answers:
ivolga24 [154]2 years ago
8 0

\frac{5(u^{-3})^2}{2u^8} =\frac{5u^{-6} }{2u^8} = \frac{5}{2} u^{-6-8} =\frac{5}{2} u^{-14} =\frac{5}{2u^{14} }

Hope that helps!

Alex2 years ago
3 0

Answer:

\frac{5}{2u14}

Sry for answering again my answer was reported.

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contradictions!!!! i dont get them help me with 5 of them pleaseee i will mark you brainlest answer and everything
vodomira [7]
Contradictions usually mean they go against established rules. Squaring a negative number is one example if it is in math. But just think of anything that goes against what is known.  
3 0
3 years ago
Solve 6^x=1,296<br><br><br> X=__________a0
Troyanec [42]

Answer:

x=4

Step-by-step explanation:

6^x = 1296

Take the log base 6 on each side

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Rewrite 1296 as 6^4

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5 0
3 years ago
Read 2 more answers
X² – x² – 3x + 3<br> all rational zeros to a polynomial
Mrrafil [7]

Answer:

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4 0
4 years ago
A. 4x^5+3x+2<br> B. 4x^2+3x+2x^5<br> C. 4x^5+3+2/x^5<br> D. 18x^5+12x+6
aalyn [17]
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3 0
3 years ago
The identification code on a bank card consists of 1 digit followed by 2 letters. The code must meet the following conditions: T
jasenka [17]

Answer:

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The second selection is a letter from the set of all the letters (27) minus the set of the vowels (5)

So here we have 27 - 5 = 22 options

The third selection is also a letter from the previous set, but because each letter can be used only one time, and in the previous selection we already selected one of the letters, in this selection we have a letter less than in the previous selection.

Here we have 22 - 1 = 21 options.

The total number of combinations (of possible codes) is equal to the product of the number of options for each selection:

C = 5*22*21 = 2310.

There are 2310 different possible codes

8 0
3 years ago
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