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mixer [17]
3 years ago
10

I kinda need a lot of help

Mathematics
1 answer:
anzhelika [568]3 years ago
8 0

Answer:600

Step-by-step explanation: 6 times 10 times 10 equals to 600 because 6 times 10 is 60 and 60 times 10 is 600

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What expression is not equivalent to 1/343
GREYUIT [131]
1 / 343
This is just equal to 
1 / 7^3 = 7^-3

A. 7^-1 * 7^-2 = 7^(-1 + -2) = 7^-3
B. 7^7 * 7^-10 = 7^( 7+ -10) = 7^-3
C. 7^-2 * 7^-5 = 7^(-2 + -5) = 7^-7
D. 7^-5 * 7^2 = 7^(-5 + 2) = 7^-3

So the correct answer is letter C. 7^-2 * 7^-5
7 0
3 years ago
What is the answer to this table?
rjkz [21]

Answer:     -3

Step-by-step explanation:

7 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
Please help woth this
jok3333 [9.3K]

Answer:

.6

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
At a local Dunkin’ Donuts, a survey showed that out of 1,300 customers eating lunch, 520 ordered coffee with their meal. What pe
k0ka [10]

First subtract the number that did, to find the number that didn't.

1300 - 520 = 780 customers did not order coffee.

Now divide the number who didn't order coffee by total number of customers:

780 / 1300 = 0.6

Multiply by 100 to get the percent:

0.6 x 100 = 60% did not order coffee.

5 0
3 years ago
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