All categories

3 years ago

Pleas hurry i need this fast whats the other?

Mathematics

2 answers:

sergejj [24]3 years ago

7 0

The answer is supplementary because they add to 180(a straight line)

dexar [7]3 years ago

5 0

Answer:

Same side interior (consecutive)

Step-by-step explanation:

Ignore line "m" completely, it is not relevant to this problem.

Now, we look at lines "L" and "N", we can see that angles 2 nd 5 are in-between the lines, therefore they are interior angles.

Since they are both on top of the line, they are on the same side. They are same side interior angles.

You might be interested in

I bet you 10 points you will not answer this I think its B i need assurance

Flauer [41]

Answer:

(a) seems to be the closest to being correct of these five statements.

Step-by-step explanation:

Let's go through the list of possible descriptors:

a) There are no outliers. This seems to be the response most likely to be correct.

b) The distribution is not skewed left. It's skewed right.

c) The center is not 44. From what I see, the center is 48.

d) This distribution is not bimodal; it does not have two peaks.

d) The spread is not 38 to 67; it's 29 to 67.

5 0

3 years ago

Plane A leaves eastaboga traveling east at a speed twice that of plane be which is flying west. how fast is each plane traveling

ivanzaharov [21]

Your doing great keep up the good work!!

5 0

3 years ago

Plz help me quickly I’ll give brainlist

Svetlanka [38]

Answer:

$5\frac{5}{6}$

Step-by-step explanation:

3 1/2 x 1 2/3 = 35/6

6 x 5 + 5 = 35/6

4 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.

6 0

2 years ago

How to factor a quadratic

Romashka [77]

Answer:

To factor a quadratic polynomial use the equation

Step-by-step explanation:

F(x) = k{x2 - (sum of zeroes)x + (product of zeroes)}

3 0

3 years ago