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Anton [14]
4 years ago
8

Si el cumpleaños de Anne es el 13 de enero, ¿cuántos días faltan para Nochevieja si el año es bisiesto? Utiliza la resta y su pr

ueba para com-probar que es correcta tu solución
Mathematics
1 answer:
boyakko [2]4 years ago
8 0

Answer:

353 dias

Step-by-step explanation:

31 dias de enero menos 13 dias que ya paso por su cumple seria 18 dias restantes del dia de enero. Mas los otros dias del año seria 353

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The indefinite integral can be found in more than one way. First use the substitution method to find the indefinite integral. Th
Fantom [35]

Answer:

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C

Step-by-step explanation:

To find:

∫6x^5(x^6-2)\,dx

Solution:

Method of substitution:

Let x^6-2=t

Differentiate both sides with respect to t

6x^5\,dx=dt

[use (x^n)'=nx^{n-1}]

So,

∫6x^5(x^6-2)\,dx = ∫ t\,dt = \frac{t^2}{2}+C_1 where C_1 is a variable.

(Use ∫t^n\,dt=\frac{t^{n+1} }{n+1} )

Put t=x^6-2

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C_1

Use (a-b)^2=a^2+b^2-2ab

So,

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C

where C=2+C_1

Without using substitution:

∫6x^5(x^6-2)\,dx = ∫6x^{11}-12x^5\,dx = \frac{6x^{12} }{12}-\frac{12x^6}{6}+C=\frac{x^{12} }{2}-2x^6+C

So, same answer is obtained in both the cases.

7 0
3 years ago
If sin(x) = 0 and cos(x) = 1, what is cot(x)?
Nata [24]

Answer:

undefined

Step-by-step explanation:

Sin(x) = 0 where x = 0

Cos(x) = 1 where x = 0

Cot(0) is undefined

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