PLEASE HELP. Quadratic Functions - Pre-Calculus 11.

Mathematics

1 answer:

ohaa [14]2 years ago

5 0

Ans: The new price from each donut results in most revenue.

Work in the picture:

You might be interested in

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

Find the volume of this cylinder give your answer to 1dp 20cm 9cm

vlada-n [284]

Answer:

this is a required answer.