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kobusy [5.1K]
3 years ago
6

WHAT ARE COMMON FACTORS

Mathematics
1 answer:
AlekseyPX3 years ago
7 0
"Common" means "same for both". If two numbers have a factor that's the same, then it's a common factor of them. Example: 2 is a common factor of 6 and 10. It's NOT the only factor of either number, but it's a factor that they share ... a 'common' one.
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Help me! I'm the most stupidest person ever!
vovangra [49]
Alright here's the breakdown:
your trying to find how many miles per day 
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6 0
3 years ago
Read 2 more answers
Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe
ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for n\geq 78

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: \sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4

For n=1: \sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4

From this point we will assume that n\geq 2

As we can see, \sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4 and (4n)^4=256n^4. Then,

\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4

Now, we will use the formula for the sum of the first 4th powers:

\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}

Therefore:

\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0

and, because n \geq 0,

465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1

Observe that, because n \geq 2 and is an integer,

n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5

In concusion, the statement is true if and only if n is a non negative integer such that n\leq 77

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  (4n)^4- \sum^{n}_{i=0} (2i)^4 for 77 and 78 you will obtain:

(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064

(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992

7 0
3 years ago
Of the 20 plants in Melissa's
vagabundo [1.1K]

Answer:

35 % are herbs

Step-by-step explanation:

7    = 35

20  = 100

6 0
3 years ago
PLEASE HELP I WILL MARK BRAINLIEST
krok68 [10]

Answer:

Option 2

Step-by-step explanation:

Because it shows that it is going up in a half not a whole.

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2 years ago
Drag the tiles to the correct locations. Each tile can be used more than once, but not all tiles will be used. One or more locat
Korolek [52]

Answer:

.. .. ..

o = N - F :

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Under the N just leave empty

Step-by-step explanation:

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2 years ago
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