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3 years ago

What is the sum of the geometric sequence 1,-6,36... if there are 6 terms

1 answer:

7 0

$a_1=1;\ a_2=-6;\ a_3=36;\ ...\\\\r=a_2:a_1\to r=-6:1=-6\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\a_1=1;\ r=-6;\ n=6\\\\subtitute\\\\S_6=\dfrac{1(1-(-6)^6)}{1-(-6)}=\dfrac{1-46656}{1+6}=\dfrac{46655}{7}=6665$

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Find x! I have no idea what it is

mash [69]

102+28=130 180-130=50, answer is 50°

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2 years ago

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What is the solution to this equation?

pychu [463]

Answer:

A. x = 3

Step-by-step explanation:

5x+15 + 2x = 24 +4x

7x+15=24+4x

7x+15-15=24+4x-15

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7x-4x=4x+9-4x

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3x/3=9/3

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4 0

1 year ago

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#12<br> The distance between the points (3, 3) and (7, 3) is

Marina CMI [18]

Solution:

We have been asked to find the distance between the points (3, 3) and (7, 3).

we can find the distance between the two points using the distance formula.

The distance formula is given as

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \text{where } (x_1, y_1) \ and \ (x_2, y_2) \ are \ two \ points\\$

Now substitute the given values we get

$D=\sqrt{(7-3)^2+(3-3)^2}\\ \\ \ D=\sqrt{4^2+0}=4\\$

Hence the distance between the given points is 4

5 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.

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2 years ago

May someone please help me!!! Please don’t answer wrong this is really important for a test :( and also PLEASE don’t answer with

mars1129 [50]

Answer:

7^8

Step-by-step explanation:

when multiplying exponents with the same base, you add the exponents together to get your answer. you can even check it with a calculator.

6 0

2 years ago

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