All categories

3 years ago

Who can solve this I'm lost

Mathematics

1 answer:

pentagon [3]3 years ago

5 0

I can't solve it but you can enter the equation into www.quickmath.com and it will give you the answer. Hope it helps.

You might be interested in

I don't know how to do this. Can someone help me please?

Svetach [21]

1. y =220- x
150=220-70

5 0

3 years ago

What is the value of X

skelet666 [1.2K]

X is 3 your are welcome

8 0

2 years ago

Read 2 more answers

Can someone help me out

Rufina [12.5K]

Answer:

Step-by-step explanation:

6 0

3 years ago

2. Peter drew two rays, AC and AP with A as a common endpoint. Which of the following statements

Ksivusya [100]

Complete question is;

Peter drew two rays, AC and AP with A as a common endpoint. Which of the following statements

might describe Peter's drawing?

I. AC and AP are parallel.

II. PAC is an angle

III. AC and AP are perpendicular

A. I and II

B. II and III

C. I and 111

D. I, II, and III

Answer:

Option B: II & III

Step-by-step explanation:

We are told that Peter drew two rays which are AC and AP.

We are told that A is a common endpoint.

If A is a common endpoint, it means the 2 rays interaction point at A is at an angle.

The angle could also be 90° which means it's possible that the rays AC and AP are perpendicular.

Thus, the correct statements that describe his drawing are: II & III

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