Reflection Lines

Mathematics

1 answer:

mel-nik [20]3 years ago

3 0

Answer:

Midpoint

Step-by-step explanation:

Equidistant from the pre = image AA would be the middle

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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

Find the ratio of the area inside the square but outside the circle to the area of the square in the picture

Eduardwww [97]

Answer:

0.2146

Step-by-step explanation:

From the picture:

The radius of the circle = r. This means that the area of the circle = πr²

Also For the square, the length of the square = 2r, Therefore the area of the square = Length × length = 2r × 2r = 4r²

The area inside the square but outside the circle = Area of square - Area of circle = 4r² - πr² = r²(4 - π) = 0.8584r²

The ratio of the area inside the square but outside the circle to the area of the square = r²(4 - π) / 4r² = (4 - π) / 4 = 1 - π/4 = 0.2146