Can you please tell me what’s wrong in this plot? I can’t see what.

Mathematics

1 answer:

Readme [11.4K]3 years ago

7 0

Answer:

The Upper Quartile would not be 59. It would be 60.5.

Step-by-step explanation:

The Upper Quartile would not be 59. It would be 60.5.

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Evaluate the integral using the indicated trigonometric substitution. (use c for the constant of integration.) x^3 / sqrt x^2 +

slava [35]

$\displaystyle\int\frac{x^3}{\sqrt{x^2+49}}\,\mathrm dx$

Taking $x=7\tan\theta$ gives $\mathrm dx=7\sec^2\theta\,\mathrm d\theta$ , so that the integral becomes

$\displaystyle\int\frac{(7\tan\theta)^3}{\sqrt{(7\tan\theta)^2+49}}(7\sec^2\theta)\,\mathrm d\theta$
$=\displaystyle7^4\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{49\tan^2\theta+49}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\tan^2\theta+1}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{|\sec\theta|}\,\mathrm d\theta$

When $\sec\theta>0$ , we have

$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sec\theta}\,\mathrm d\theta$
$=\displaystyle7^3\int\tan^3\theta\sec^2\theta\,\mathrm d\theta$

and from here we can substitute $u=\tan\theta$ to proceed from here.

Quick note: When we set $x=7\tan\theta$ , we are implicitly enforcing $-\dfrac\pi2$ just so that the substitution can be undone later via $\theta=\tan^{-1}\dfrac x7$ . But note that over this domain, we automatically guarantee that $\sec\theta>0$ , so the absolute value bars can be dropped immediately.

6 0

3 years ago

Helppp 1-7 ANSWERS help me get them ASAP

NNADVOKAT [17]

Answer:

I can't see the graph.....

Step-by-step explanation:

5 0

3 years ago

What are both of these?

allochka39001 [22]

Pshhhh she said math negatives I’m pretty sure.