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2 years ago

REwaRDiNG BRAINLY TO CORRECT

Mathematics

2 answers:

vfiekz [6]2 years ago

6 0

Answer:

54 millimeters

Step-by-step explanation:

The median length would be in the middle of the values in order, so you could order the numbers 50,50,52,53,54,55,55,55,57. You would then find the middle value which would be the 5th value in a list of 9. In this case, it would be 54 millimeters.

il63 [147K]2 years ago

6 0

Answer:

Step-by-step explanation:

Put the numbers in order: 50,50,52,53,54,55,55,55,57 and the middle one is your median in this case 54 is in the middle so ye

Good luck!

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The least common multiple is 96

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Three less than a numbers divided by 4 is five

mariarad [96]

Answer:

x/4 (-3) = 5

Step-by-step explanation:

Simply translate into numbers. :D

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3 years ago

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Find the measure of the angle<br> I-ready<br> Please help

Anon25 [30]

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Step-by-step explanation:

if you see that line in between 90, 100 that is a five and it is past 90 making it 95

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%

xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

$\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }$

Let a = 1 and b the cosine product, and write them as

$\dfrac{a - b}{x^2}$

with

$b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}$

Now we use the identity

$a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$

to rationalize the numerator. This gives

$\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}$

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

$\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}$

For the remaining limit, use the Taylor expansion for cos(x) :

$\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)$

where $\mathcal{O}(x^4)$ essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2$

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)$

$\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)$

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

$\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52$

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0

2 years ago

A class of 32 students is organised in 33 teams every team consists of 3 students and there are no identical teams . show that t

Tresset [83]

Answer:

Step-by-step explanation:

Let's start by making up as many teams as we can with the 32 student. Given that each team is different, we can make 10 teams of 3 each. (we still have 23 more teams to make).

The last two people make a team of only 2. No matter which student from the 30 other students is picked, the team of two and the one the student is coming from will have one student in common. Though there are more borrowings that take place (many more), the results remain as stated. At least 2 teams will have 1 person in common.

The method is called the pigeon hole method.

6 0

3 years ago