Is 5.43 greater or less than 5.432?

What is 33<img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D" id="TexFormula1" title="\frac{1}{3}" alt="\frac{1}{3}" align=

aleksklad [387]

Answer:

33.3 Repeating

Step-by-step explanation:

1. You already know that 33 is the whole number so you have 33.

2. 1/3 is converted to .3333333 so you put 33 before the decimal point

3. You get 33.3333333

Hope this helps :)

8 0

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

PLEASE HELP!!!!! WILL GIVE BRAILIEST!!!

scoundrel [369]

SinB = 8/16
sinB = 0.5
m<B = 30
answer is A. 30

6 0

3 years ago

Fred is trying to earn $35 to buy a basketball. He has saved $11.25. He earns $2.25 per hour doing yard work for his neighbor, a

rjkz [21]

$2.25y + 7.25z + 11.25 \geq 35 \\ y = 2 \\ z = 3 \\ 2.25(2) + 7.25(3) + 11.25 \geq 35 \\ 4.50 + 21.75 + 11.25 \geq 35 \\ 37.5 \geq 35$
The answer will be the third one: Yes, because the total will be $37.50.

7 0

3 years ago

A club consisting of 6 juniors and 8 seniors is to be formed from a group of 13 juniors and 16 seniors. How many different clubs

Vilka [71]

Answer: 22,084,920 different clubs

Step-by-step explanation:

The club must have 6 juniors and 8 seniors

We have a total of 13 juniors and 16 seniors.

Now, we know that the possible combinations of N objects into a group of K is equal to:

$C = \frac{N!}{(N-K)!*K!}$