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Anuta_ua [19.1K]
2 years ago
13

Which of the following is most likely the next step in the series?

Mathematics
2 answers:
saw5 [17]2 years ago
5 0

Answer:

C, because each time, the number of purple dots, or vertices increases by one..

2, 3, 4, ?..

the next would be five, which is C.

Step-by-step explanation:

lisabon 2012 [21]2 years ago
3 0

Answer:

here

Step-by-step explanation:

each shape has one more vertici (the purple dots) than the previous one, so then the pattern would be 2, 3, 4, .....? so the next one would be 5 which is C

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Applying the product rule gives

\displaystyle \frac{d}{dx}\int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \frac{d}{dx}\int_1^{\ln(x)}\frac{dt}{(1+t)^2}

Use the fundamental theorem of calculus to compute the remaining derivatives.

\displaystyle \frac{4x^3}{1+x^4} \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \frac{1}{x(1+\ln(x))^2}\int_1^{x^2}\frac{2t}{1+t^2}\,dt

The remaining integrals are

\displaystyle \int_1^{\ln(x)}\frac{dt}{(1+t)^2} = -\frac1{1+t}\bigg|_1^{\ln(x)} = \frac12-\frac1{1+\ln(x)}

\displaystyle \int_1^{x^2}\frac{2t}{1+t^2}\,dt=\int_1^{x^2}\frac{d(1+t^2)}{1+t^2}=\ln|1+t^2|\bigg|_1^{x^2}=\ln(1+x^4)-\ln(2) = \ln\left(\frac{1+x^4}2\right)

and so the overall derivative is

\displaystyle \frac{4x^3}{1+x^4} \left(\frac12-\frac1{1+\ln(x)}\right) + \frac{1}{x(1+\ln(x))^2} \ln\left(\frac{1+x^4}2\right)

which could be simplified further.

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<span>C) rational and real

===================
</span>
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3 years ago
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