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3 years ago

True or false the edges of a polyhedron are the line segments bordering each face?

2 answers:

8 0

True, the edges of a polyhedron are the line segments bordering each face. When two faces come together in a 3D shape a line is formed. This line is called an edge.

san4es73 [151]3 years ago

3 0

<h2>Answer:</h2>

The given statement " the edges of a polyhedron are the line segments bordering each face" is true.

Step-by-step explanation:

A polyhedron is a 3 D solid figure possessing straight edges and flat sides. The flat sides are called the faces and the corners are called vertices. The edges of a polyhedron are the lines at folds, where these faces meet each other. The two faces are joined, it joins in a line.

Therefore, the given statement is true.

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Which of the following is the quotient of the rational expressions shown below? 2x-3/5x^2 divided by x-5/3x-1

Kamila [148]

The expression in fraction form will be:

$\frac{ \frac{2x-3}{5 x^{2} } }{ \frac{x-5}{3x-1} }$

The denominator can be multiplied to the numerator by taking its reciprocal as shown below:

$\frac{2x-3}{5 x^{2} } * \frac{3x-1}{x-5} \\ \\ 
= \frac{(2x-3)(3x-1)}{5 x^{2}(x-5) }$

There are no common factors, so the expression can not be simplified any further.

6 0

3 years ago

Read 2 more answers

I need help plzzzzzzz :))

SVEN [57.7K]

Answer:

C. Both Functions Have a y-intercept of -2

Step-by-step explanation:

In Function F it shows us that it has a y-intercept of -2, and in the equation of Function G it shows us that you would subtract 2 out of what the initial value would be:

(3 x 0 = 0)

0 - 2 would be -2 so it would have a y-intercept of -2.

4 0

2 years ago

If X is a r.v. such that E(X^n)=n! Find the m.g.f. of X,Mx(t). Also find the ch.f. of X,and from this deduce the distribution of

astraxan [27]

$M_X(t)=\mathbb E(e^{Xt})$
$M_X(t)=\mathbb E\left(1+Xt+\dfrac{t^2}{2!}X^2+\dfrac{t^3}{3!}X^3+\cdots\right)$
$M_X(t)=\mathbb E(1)+t\mathbb E(X)+\dfrac{t^2}{2!}\mathbb E(X^2)+\dfrac{t^3}{3!}\mathbb E(X^3)+\cdots$
$M_X(t)=1+t+t^2+t^3+\cdots$
$M_X(t)=\displaystyle\sum_{k\ge0}t^k=\frac1{1-t}$

provided that $|t|$ .

Similarly,

$\varphi_X(t)=\mathbb E(e^{iXt})$
$\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots$
$\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=\dfrac{1+it}{1+t^2}=\dfrac1{1-it}$

You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{e^{itx}\varphi_X(-t)-e^{-itx}\varphi_X(t)}{it}\,\mathrm dt$

The integral can be rewritten as

$\displaystyle\int_0^\infty\frac{2i\sin(tx)-2it\cos(tx)}{it(1+t^2)}\,\mathrm dt$

so that

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt$

There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to $x$ , which gives

$\displaystyle\mathcal L_s\left\{\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt\right\}=\int_0^\infty\frac{1-s}{(1+t^2)(s^2+t^2)}\,\mathrm dt$
$=\displaystyle\frac{\pi(1-s)}{2s(1+s)}$

and taking the inverse transform returns

$F_X(x)=\dfrac12+\dfrac1\pi\left(\dfrac\pi2-\pi e^{-x}\right)=1-e^{-x}$

which describes an exponential distribution with parameter $\lambda=1$ .

6 0

3 years ago

A diameter of a circle has endpoints P(-10,-2) and Q(4,6).

Lesechka [4]

Check the picture below.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ P(\stackrel{x_1}{-10}~,~\stackrel{y_1}{-2})\qquad Q(\stackrel{x_2}{4}~,~\stackrel{y_2}{6}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{4-10}{2}~~,~~\cfrac{6-2}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{4}{2} \right)\implies \stackrel{\textit{center}}{(-3~,~2)} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{\textit{center}}{(\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})}\qquad Q(\stackrel{x_2}{4}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[4-(-3)]^2+[6-2]^2}\implies r=\sqrt{(4+3)^2+(6-2)^2} \\\\\\ r=\sqrt{49+16}\implies r=\sqrt{65} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{2}{ k})\qquad \qquad radius=\stackrel{\sqrt{65}}{ r} \\[2em] [x-(-3)]^2+[y-2]^2=(\sqrt{65})^2\implies (x+3)^2+(y-2)^2=65$

5 0

3 years ago

Sum of 13 and 3 times a number is 31. What is the number?

Volgvan

Answer:

13 + 3x= 31

Step-by-step explanation:

6 0

2 years ago