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4 years ago

Quadrilateral abcd is inscribed in a circle with m∠a = (x)degrees, m∠b = (2x - 10)degrees, and m∠c = (3x)degree. what is m∠d?

1 answer:

6 0

Hello,

When you have a polygon inscribed in a circle, the opposite angles are supplementary.

So we can start by saying that A + C = 180. Or x + 3x = 180. This gives us a value of 45 for x.

If we input that in for angle B, we get: 2(25) - 10 = 80.

If angle B = 80, then angle D must be 100, because the angles are supplementary.

Good luck,
MrEQ

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The total surface area is 96 cm cube.

The radius is calculated as,

$\begin{gathered} r=\frac{d}{2} \\ r=\frac{2x}{2} \\ r=x\text{ cm} \end{gathered}$

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$\begin{gathered} S=2\pi rh+2\pi(r)^2 \\ 96=2\pi xh+2\pi(x^2) \\ h=\frac{96-2\pi(x^2)}{2\pi x} \end{gathered}$

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$\begin{gathered} V=\pi(r)^2h \\ =\pi(x^2)\frac{96-2\pi(x^2)}{2\pi x} \\ =\frac{x(96-2\pi(x^2)}{2} \end{gathered}$

Now, differentiate with respect to x,

$\begin{gathered} \frac{dV}{dx}^{}=\frac{d}{dx}(\frac{x(96-2\pi(x^2)}{2}) \\ =\frac{d}{dx}\mleft(x\mleft(-\pi x^2+48\mright)\mright) \\ =\frac{d}{dx}\mleft(x\mright)\mleft(-\pi x^2+48\mright)+\frac{d}{dx}\mleft(-\pi x^2+48\mright)x \\ =1\cdot\mleft(-\pi x^2+48\mright)+\mleft(-2\pi x\mright)x \\ =84-3\pi(x^2)\ldots\ldots\ldots\ldots\text{.}(1) \end{gathered}$

Now,

$\begin{gathered} \frac{dV}{dx}=0 \\ 84-3\pi(x^2)=0 \\ x^2=\frac{16}{\pi} \\ x=\sqrt[]{\frac{16}{\pi}} \end{gathered}$

Now, differentiate (1) with respect to x again,

$\begin{gathered} \frac{d^2V}{dx^2}=\frac{d}{dx}(84-3\pi(x^2)) \\ =-6\pi x \\ At\text{ x=}\sqrt[]{\frac{16}{\pi}} \\ \frac{d^2V}{dx^2}=-6\pi\sqrt[]{\frac{16}{\pi}}$

Since, the double derivative is negative.

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So, the volume becomes,

$\begin{gathered} V=\pi(x^2)h \\ V=\pi(\sqrt[]{\frac{16}{\pi}})^2h \\ V=\frac{16h}{\pi} \end{gathered}$

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1 year ago

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goldfiish [28.3K]

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