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2 years ago

What is standard form for 6.43999001 x 10^8

1 answer:

5 0

6.43999001
1st you move the decimal 8 times to your right
once you are done moving the decimal you should end up with a whole number
you answer is (643, 999, 001)

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PLS help with this immediately no links or guesses

shusha [124]

Answer:

B all real numbers greater than 0

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3 years ago

For each ordered pair, determine whether it is a solution to 2x - 5y=11.

Sidana [21]

Answer:

(3, -1)

Step-by-step explanation:

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3 years ago

Factor each completely. <br><br>10a² - 9a + 2 <br><br>

Sophie [7]

Answer:

a= $\frac{2}{5}$ or $\frac{1}{2}$

Step-by-step explanation:

Equate the equation to zero

$10a^{2}$ -9a+2=0

$10a^{2}$ -5a-4a+2=0

5a(2a-1)-2(2a-1)=0

Write out the factors

(5a-2)(2a-1)=0

Then equate each factor to zero

5a-2=0 OR 2a-1=0

5a=2 OR 2a=1

a= $\frac{2}{5}$ OR a= $\frac{1}{2}$

a= $\frac{2}{5}$ or $\frac{1}{2}$

3 0

2 years ago

These tables show equivalent ratios. The value missing in each table is replaced with X. Choose ALL tables in which 12 replaces

marishachu [46]

Answer:

Step-by-step explanation:

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2 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