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lions [1.4K]
3 years ago
7

I need help with this please

Mathematics
2 answers:
Vadim26 [7]3 years ago
7 0

Answer:

Choice B

Step-by-step explanation:

The reason why < 3 is equal to < 6 is because they are alternate interior angles.

The first choice are supplementary angles they add up to 180 so they can't be equal.

The third one is false because we don't actually know the measurements of the angles but they both need to be 45° to add up to 90 since they are equal to each other.

The last one is also incorrect because they are vertical angles just like the third choice and they aren't on a line so they can't add up to 180.

expeople1 [14]3 years ago
4 0

Answer:

b

Step-by-step explanation:

there's a property for Alternate Interior Angles

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\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt

Now the integral is

\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt

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\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt

Decompose the integrand into partial fractions:

\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}

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\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt

which are all standard integrals. You end up with

\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt
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To try to get the terms to match up with the available answers, let's add and subtract \ln\left|1+\tan\dfrac x2\right| to get

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