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Marina86 [1]
3 years ago
10

Please help me please help me if you don't know gently move your hand

Mathematics
1 answer:
Zinaida [17]3 years ago
8 0

Answer:

Diameter: 4km

Radius: 2km (divide diameter by two)

Circumference Formula: c = dπ or c = 2rπ

Circumference Value = 4π or 2(2π)

(the actual answer is 12.56, just do the math)

Step-by-step explanation:

You might be interested in
Question 1 of 10
Sonbull [250]

The statement regarding the expansion of (x + y)" that is correct is D. The coefficients of yn - 1 and yn – 1 both equal 1.

<h3>How to expand?</h3>

It should be noted that for any positive integer, n, the expression of (x + y)^n will be C(n,0)x^n + C+n, 1)x^n-1 .... C(n, n)y^n.

Here, the statement regarding the expansion of (x + y)" that is correct is that the coefficients of yn - 1 and yn – 1 is both equal to 1.

Learn more about expansion on:

brainly.com/question/29114

#SPJ1

8 0
2 years ago
How do I complete this equation
Nesterboy [21]

Answer:

130°

Step-by-step explanation:

All the interior angles should add up to 540 degrees because it's a pentagon

∠T ≅∠S, so ∠T is also 115°

∠P and ∠R are both 90°

Add everything up and solve for ∠Q:

115 + 115 + 90 + 90 + m∠Q = 540

410 + m∠Q = 540

m∠Q = 130°

8 0
2 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
Suppose an airplane climbs 15 feet up for every 40 feet it moves forward. What is the slope of this airplanes ascent?
Anastasy [175]

the rise over run is 15/40 so the slope of the airplanes ascent is 0.375 feet per foot is move forward

5 0
3 years ago
I WILL MARK BRAINLIST!!
Soloha48 [4]

Answer:

For (2x^2 + 8x - 8), the group is CLOSED as it a polynomial.

Step-by-step explanation:

Here, the given polynomials are:

P(x)=(5x^2+2x-6)\\Q(x)=(3x^2-6x+2)

Now, a group G is said to be <u>CLOSED UNDER SUBTRACTION</u> if,

a and b are he elements of G ⇒ (a-b) is ALSO AN ELEMENT of G

Now, here:

R(x)  = P(x) - Q(x) = (5x^2+2x-6) - (3x^2-6x+2) = (2x^2 + 8x - 8)

or, R(x)  = (2x^2 + 8x - 8)

Also, R(x) is a POLYNOMIAL.

So, R(x) is an Element of group G.

So, the set G of polynomials.

Hence, for the polynomial  (2x^2 + 8x - 8), it will be a polynomial the group is CLOSED.

3 0
3 years ago
Read 2 more answers
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