All categories

3 years ago

How to work out this (64/27)1/3

Mathematics

1 answer:

Paladinen [302]3 years ago

6 0

You need to answer what is in the parentheses first, so 64 divided by 27 is about 2.37.

Multiply 2.37 by 1/3, which is .018.

Good luck and hope this helps! :)

You might be interested in

5.1.2 Exam: Semester 2 Exam

Nikolay [14]

Answer:

See attached picture to view the graph

Step-by-step explanation:

Start by analyzing how this average idea works:

If only one member goes to the trip, it will cost him/her $1000+$200 = $1200.

If two members go to the trip, then they will share the cost as per the following: ($1000+ $200 + $200 = $1400) which they will be dividing into two people, thus costing each of them $700.

Notice that the general function that represents such average will be given by: $f(x) = \frac{1000+200x}{x}$

Plot such function in the two dimensional plane, and you will get the asymptotic behavior shown in the attached image.

8 0

2 years ago

Which is equivalent to cos^-1(0) answer in radian

ruslelena [56]

Answer:

Step-by-step explanation:

cos ^-1(0)=(2n+1) π/2

where n is an integer.

if 0 ≤α ≤2 π

n=0,1

cos ^-1(0)=π/2,3π/2

8 0

3 years ago

Whih of the following is equivalent to (81m^6)^1/2

Setler [38]

Can you list the answer choices?

$\sqrt{81m^6}$ would be equivalent

4 0

3 years ago

Read 2 more answers

<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

Which is the range of the function f(x) =(9)x?

Annette [7]

f(x) =(9)x = 9x

It's linear function. Domain and range are all real numbers

Answer

Range is all real numbers

5 0

3 years ago

Read 2 more answers