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Basile [38]
3 years ago
14

What is the greatest common factor of 4xy2 and 20x2y4? 4xy 4xy2 24xy 24xy2

Mathematics
2 answers:
icang [17]3 years ago
6 0

Answer:

the answer is B

Step-by-step explanation:

alexira [117]3 years ago
4 0

Answer:

B: 4xy2

Step-by-step explanation:

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.01 meter or 1/100 meter =<br><br> A. decimeter<br> B. centimeter<br> C. millimeter
tamaranim1 [39]

There are

100 Centemeters in a meter, and you would need 100 (1/100)'s to make 1.  The answer is centimeters.

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Simplify the expression<br><br>9x⁰y⁸<br>____<br>z⁸
Olenka [21]

Anything raised to exponent 0 equals 1

So

x^{0}=1

So we get

\frac{9 (1) y^8}{z^8} = \frac{9y^8}{z^8}

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Bogdan [553]

Answer: What is the answer you need?

Step-by-step explanation:

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3 years ago
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If k is an integer, what is <br> k + 1<br> /2<br> !<br> ? Why?
Amanda [17]

If k\in\mathbb{Z} than k+1\in\mathbb{Z} however any \mathbb{Z} divided by 2 is in \mathbb{Q}.

But it is not so simple.

If k is odd then k+1 is even but when even number is divided by 2, you get an odd number which is in \mathbb{Z}.

However if k is even then k+1 is odd and when we divide that by 2, you get a number like for example \frac{5}{2} which is not in \mathbb{Z} but rather inside \mathbb{Q}. There are fortunately no irrational numbers.

There is also a problem with zero, zero is neither odd nor even but still an integer. If your k happens to be -1, k+1=0, and zero divided by 2 is still just 0 which is in \mathbb{Z}.

So to sum up,

If k is odd and not -1, then \frac{k+1}{2}\in\mathbb{Z}.

Or to put it in math \forall k\in\mathrm{odd}-\{-1\}\cup\{0\}\implies\frac{k+1}{2}\in\mathbb{Z}.

If k is even or -1, then \frac{k+1}{2}\in\mathbb{Q}.

Or to put it in math \forall k\in\mathrm{even}\cup \{-1\}\implies\frac{k+1}{2}\in\mathbb{Q}.

<u>Notes</u>:

\mathbb{Z} is a set of integers.

\mathbb{Q} is a set of rational numbers.

\mathbb{Z}\subset\mathbb{Q} means integers are a subset of rational numbers, that is, every integer is a also a fraction, however not every fraction is an integer.

Hope this helps :)

8 0
3 years ago
I been stuck on dis for a long time
melamori03 [73]

Answer:

Direct

Step-by-step explanation:

I had this same question on my exam

3 0
3 years ago
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