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Alchen [17]
3 years ago
8

The sum of ages of Nira before 10 years and after 10 years is 60.find her present age.

Mathematics
2 answers:
mojhsa [17]3 years ago
4 0

Answer:

<h3>Nira's present age= 30</h3>

Step-by-step explanation:

her \: age \: before \: 10years \:  = x - 10 \\ her \: age \: after \: 10years \: = x + 10 \\ therefore.(x - 10) + (x + 10) = 60 \\ x - 10 + x + 10 = 60 \\ 2x = 60 \\ x = 60 \div 2  \\ x = 30 \\ hope \: it \: helps....

vodomira [7]3 years ago
3 0

Answer:

I think Nira would be 30 years old

Step-by-step explanation:

So 10 years before and after her present age added together equals 60. So if she was 30, it would be 20+40 which =60.

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3 years ago
Describe the results of a similarity transformation
chubhunter [2.5K]

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Step-by-step explanation:

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3 years ago
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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)

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\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2

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\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.

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\tan \theta = \frac{\sqrt{1 - x^{2}}}{x}

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To learn more on unit circles: brainly.com/question/12100731

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