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2 years ago

Find the perimeter of this semi-circle with diameter,

Mathematics

1 answer:

SVEN [57.7K]2 years ago

6 0

Answer:

58.1cm

Step-by-step explanation:

perimeter of a semi-circle = πd/2

d is the diameter

Given

diameter = 37cm

Perimeter of the semi circle = π(37)/2

Perimeter of the semi circle = 18.5(3.14)

Perimeter of the semi circle = 58.09 cm

Hence the perimeter of the semi-circle is 58.1cm

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Answer:

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Step-by-step explanation:We use the formule: t=S/v. So:

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kotegsom [21]

Answer:

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Explanation:

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Step-by-step explanation:

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2 years ago

How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title

kotegsom [21]

Let $t=\sqrt y$ , so that $t^2=y$ , $t^4=y^2$ , and $\mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}$ . Then

$\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}$

Now let $y=\sqrt2\tan z$ , so that $\mathrm dy=\sqrt2\sec^2z\,\mathrm dz$ . Then

$\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C$

Transform back to $y$ to get

$\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C$

and again to get back a result in terms of $t$ .

$\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C$

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3 years ago