To Complete: the statement

Given:

Figure is shown below.

Nann
2021-02-11
Answered

To Complete: the statement

Given:

Figure is shown below.

You can still ask an expert for help

faldduE

Answered 2021-02-12
Author has **109** answers

Calculation:

In$\mathrm{\u25b3}RWZ{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\mathrm{\u25b3}ZWS$ .

$\mathrm{\angle}WRZ\stackrel{\sim}{=}\mathrm{\angle}WZS\therefore$ (Given)

$\mathrm{\angle}W\stackrel{\sim}{=}\mathrm{\angle}W\therefore$ (Common)

$WZ\stackrel{\sim}{=}WZ\therefore$ (Common)

By AAS similarity,$\mathrm{\u25b3}RWZ\sim \mathrm{\u25b3}ZWS$ .

Therefore,

$\frac{RW}{ZW}=\frac{ZR}{SZ}=\frac{WZ}{WS}$

Therefore, the answer is ZW, SZ and WS.

In

By AAS similarity,

Therefore,

Therefore, the answer is ZW, SZ and WS.

asked 2021-05-17

Find the point on the line

asked 2022-07-21

Is there an easier way to solve a "Find the locus" problem?

Note: I am not concerned with the accuracy of my solution so you don't need to redo any of my calculations.

Original question: Suppose ABC is an equilateral triangular lamina of side length unity, resting in two-dimensions. If A and B were constrained to move on the x- and y-axis respectively, then what is the locus of the centre of the triangle?

My solution: Let the vertices of $\mathrm{\Delta}ABC$ be $A=({x}_{0},0)$, $B=(0,{y}_{0})$ and $C=(x,y)$. Let the centre of the triangle be $D=(X,Y)$. We wish to find the locus of D under the constraints:

$\begin{array}{rl}{x}_{0}^{2}+{y}_{0}^{2}& =0\\ (x-{x}_{0}{)}^{2}+{y}^{2}& =0\\ {x}^{2}+(y-{y}_{0}{)}^{2}& =0\end{array}$

Since D is the centre of the triangle, we know

$X=\frac{x+{x}_{0}}{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Y=\frac{y+{y}_{0}}{3}$

Parametrizing: Let $\theta $ be the angle that the edge AB makes with the x-axis. We use (theta) as our parameter of choice to write, $\begin{array}{rl}{x}_{0}& =\mathrm{cos}\theta \\ {y}_{0}& =\mathrm{sin}\theta \end{array}$

for $\theta \in [0,2\pi )$

This implies, $(x-\mathrm{cos}\theta {)}^{2}+{y}^{2}=0$

We can parametrize this with another parameter $\varphi \in [0,2\pi )$ as

$\begin{array}{rl}x& =\mathrm{cos}\theta +\mathrm{cos}\varphi \\ y& =\mathrm{sin}\varphi \end{array}$

Plugging these back into third equation gives

$(\mathrm{cos}\theta +\mathrm{cos}\varphi {)}^{2}+(\mathrm{sin}\varphi -\mathrm{sin}\theta {)}^{2}=1$

After lots of algebraic manipulations...

$\begin{array}{rlrl}X& =\frac{1}{2}\mathrm{cos}\theta \pm \frac{1}{2\sqrt{3}}\mathrm{sin}\theta & Y& =\frac{1}{2}\mathrm{sin}\theta \pm \frac{1}{2\sqrt{3}}\mathrm{cos}\theta \end{array}$

Note: I am not concerned with the accuracy of my solution so you don't need to redo any of my calculations.

Original question: Suppose ABC is an equilateral triangular lamina of side length unity, resting in two-dimensions. If A and B were constrained to move on the x- and y-axis respectively, then what is the locus of the centre of the triangle?

My solution: Let the vertices of $\mathrm{\Delta}ABC$ be $A=({x}_{0},0)$, $B=(0,{y}_{0})$ and $C=(x,y)$. Let the centre of the triangle be $D=(X,Y)$. We wish to find the locus of D under the constraints:

$\begin{array}{rl}{x}_{0}^{2}+{y}_{0}^{2}& =0\\ (x-{x}_{0}{)}^{2}+{y}^{2}& =0\\ {x}^{2}+(y-{y}_{0}{)}^{2}& =0\end{array}$

Since D is the centre of the triangle, we know

$X=\frac{x+{x}_{0}}{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Y=\frac{y+{y}_{0}}{3}$

Parametrizing: Let $\theta $ be the angle that the edge AB makes with the x-axis. We use (theta) as our parameter of choice to write, $\begin{array}{rl}{x}_{0}& =\mathrm{cos}\theta \\ {y}_{0}& =\mathrm{sin}\theta \end{array}$

for $\theta \in [0,2\pi )$

This implies, $(x-\mathrm{cos}\theta {)}^{2}+{y}^{2}=0$

We can parametrize this with another parameter $\varphi \in [0,2\pi )$ as

$\begin{array}{rl}x& =\mathrm{cos}\theta +\mathrm{cos}\varphi \\ y& =\mathrm{sin}\varphi \end{array}$

Plugging these back into third equation gives

$(\mathrm{cos}\theta +\mathrm{cos}\varphi {)}^{2}+(\mathrm{sin}\varphi -\mathrm{sin}\theta {)}^{2}=1$

After lots of algebraic manipulations...

$\begin{array}{rlrl}X& =\frac{1}{2}\mathrm{cos}\theta \pm \frac{1}{2\sqrt{3}}\mathrm{sin}\theta & Y& =\frac{1}{2}\mathrm{sin}\theta \pm \frac{1}{2\sqrt{3}}\mathrm{cos}\theta \end{array}$

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For a triangle ABC, prove that angle C is a right angle if and only if the median of side c is c/2

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For the following statement, either prove that they are true or provide a counterexample:

Let a, b, m,$n\in Z$ such that m, n > 1 and $n\mid m$ . If $a\equiv b\left(\text{mod}m\right)$ , then

$a\equiv b\left(\text{mod}n\right)$

Let a, b, m,

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Finding the volume of revolution using the method of shells

I'm trying to find the volume of the solid generated by revolving the region bounded by $y={x}^{2}$ and $y=6x+7$ about x-axis using the shell method. I applied the method and I got 15864/5 multiplied by $\pi $ but it's not correct.

Details: I integrated ${\int}_{1}^{49}y(\sqrt{y}-\frac{y-7}{6})dy$

I'm trying to find the volume of the solid generated by revolving the region bounded by $y={x}^{2}$ and $y=6x+7$ about x-axis using the shell method. I applied the method and I got 15864/5 multiplied by $\pi $ but it's not correct.

Details: I integrated ${\int}_{1}^{49}y(\sqrt{y}-\frac{y-7}{6})dy$

asked 2021-02-16

Consider the line represented by the equation

asked 2022-07-17

Mathematically prove that a Beta prior distribution is conjugate to a Geometric likelihood function

I have to prove with a simple example and a plot how prior beta distribution is conjugate to the geometric likelihood function. I know the basic definition as

'In Bayesian probability theory, a class of distribution of prior distribution θ is said to be the conjugate to a class of likelihood function $f(x|\theta )$ if the resulting posterior distribution is of the same class as of $f(\theta )$.'

But I don't know how to prove it mathematically.

I have to prove with a simple example and a plot how prior beta distribution is conjugate to the geometric likelihood function. I know the basic definition as

'In Bayesian probability theory, a class of distribution of prior distribution θ is said to be the conjugate to a class of likelihood function $f(x|\theta )$ if the resulting posterior distribution is of the same class as of $f(\theta )$.'

But I don't know how to prove it mathematically.