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Mathematics

2 answers:

nekit [7.7K]3 years ago

6 0

Answer:

$c. \: {w}^{64}$

Sorry if i get it wrong

vovangra [49]3 years ago

6 0

C. I believe, if I am wrong I’m sorry.

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Divide f(x) by d(x), and write a summary statement in the form indicated.

lukranit [14]

( x^4 - 4 x³ + 2 x² - 4 x + 1 ) : ( x² + 1 ) = x² - 4 x + 1
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------------------------
- 4 x³ + x² - 4 x
4 x³ + 4 x
-----------------------
x² + 1
- x² - 1
----------------
R ( x ) = 0
( x^4 - 4 x³ + 2 x² - 4 x + 1) = ( x² + 1 ) ( x² - 4 x + 1 )

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3 years ago

At a movie theater, 360 people can see a particular movie over a period of 3 hours. At this

Agata [3.3K]

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Step-by-step explanation:cause i said so

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3 years ago

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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 $n \geq 0$ ,

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