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3 years ago

Help 10 points for it

Mathematics

2 answers:

Brums [2.3K]3 years ago

8 0

Answer:

it as actually 5 soo.

Step-by-step explanation:

polet [3.4K]3 years ago

3 0

Answer: (2,1) duuuuuh

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Equivalent fractions to 1/3

pochemuha

= 2/6 = 3/9 = 12/36 = 24/72

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3 years ago

Use the net to find the surface area of the regular pyramid.

Svetradugi [14.3K]

Answer:

7 or $7^{2}$

Step-by-step explanation:

Each triangle has the surface area of 1.5, because 1 $*$ 3 = 3 / 2 = 1.5. You divide by two because it is half a rectangle instead of a full one. You multiply by four which would get you 1.5 $*$ 4 = 6. The square in the middle has the surface area of 1 because 1 $*$ 1 = 1. So you would get 7.

6 0

2 years ago

The slope of a line is undefined, and the x-intercept is -7. what is the equation of the line?

Andreyy89

The equation of line with undefined slope and x - intercept -7 is x = -7

Solution:

Given that the slope of line is undefined and the x intercept is -7

To find: Equation of line

If a line has undefined slope, then it is a vertical line

A vertical line has undefined slope because all points on the line have the same x-coordinate

The equation of a vertical line always takes the form:

x = b

where, "b" is the x - intercept

Given x - intercept is -7

Thus the equation of line is:

$x = -7$

Therefore, the equation of line is found

7 0

3 years ago

7 1/3 ÷ 4 Please help

oksian1 [2.3K]

Answer: 1.83

Step-by-step explanation:

6 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

3 years ago