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3

They are popular (in the sense of: people would like to use them, some methods of dealing with nonlinear PDE are based on the theory, for example), but in some respects more difficult to handle than the finite dimensional manifolds. B-Space can be, e.g., non-reflexive, which requires additional care when you want to study the cotangent space. Also, in ...


0

Since the expanded $n$-symplex is the maximal cross-section of the $(n+1)$-cross polytope its hypervolume is given by: $$ V_n = \sqrt{n+1}\int_{x_1,\ldots,x_n\in I}\mathbb{1}_I(x_1+\ldots+x_n)\,d\mu, $$ where $I=(-1,1)$. The integral is just $2^n$ times the probability that the sum of $n$ indipendent random variables, uniformly distributed over $I$, lies in ...


0

Let F be any point on the line segment DE. Join B and F. In triangle ABF, $AB+BF > AF ..(i)$ In triangle BDF, $BD+BF > DF ..(ii)$ Subtracting (ii) from (i), $AB-BD > AD ..(iii)$ Similarly, from triangles CBF and EBF, $BC-BE > CE (iv)$ Adding (iii) and (iv), $(AB+BC) -(BD+BE) > 2AD$ Since $AD >0$, $(AB+BC) - (BD+BE) > 0$ ...


0

Assuming you want the solution to the practical (rather than the mathematical) problem, and since you don't specify the exact shape of the barrel, the easy solution would be to pour in known quantities of water and mark your dipstick after each addition.


1

It's easy to see that $$d(S,p) + d(p,p') \ge d(S,p') \ge d(S,p) - d(p,p')$$ so $d(S,p)$ is continuous (this works for any nonempty set $S$ in any metric space). As for $f(p)$: If $v = f(p) - f(p')$, $\langle p - f(p), v\rangle \ge 0$ (otherwise $t f(p) + (1-t) f(p')$ would be closer to $p$ than $f(p)$ for some $t \in (0,1)$). Similarly, $\langle p' - ...


1

Let the equation of the plane be $$ \textbf{a}^T\textbf{x} = b $$ where $\textbf{a}$ is a unit vector normal to the plane and b is the plane's distance from the origin. Transform to a coordinate system centred on a point $\textbf{r}$: $$ \textbf{x}' = \textbf{x} -\textbf{r} $$ Then in the dashed coordinate system: $$ \textbf{a}^T(\textbf{x}' + \textbf{r}) ...


0

Convert the equation of ellipse from rectangular to polar coordinates. $ (x/a)^2 + (y/b)^2 = 1 $ , $ x = r cos(\theta), y = r sin(\theta)$. $ (cos(\theta)/a )^2+ (sin(\theta)/b)^2 =1/r^2 $. $(a,b,\theta )$ are given. Find out 1/$r^2$ and $r$.


1

There are two complementary angle formed by the diagonal of a rectangle: $\alpha$ $\beta = 90^\circ - \alpha$ where $$\tan \alpha = \frac {h}{\mathcal l}\implies \alpha = \arctan\left(\frac h{\mathcal l}\right)$$ (Note: $\tan \beta = \frac{\mathcal l}{h} \implies \beta = \arctan\left(\frac {\mathcal l}{h}\right), \;\text{ and } \alpha + \beta = ...


1

The locus of the points $P$ of the plane such that $AP+PC=AB+BC$ is the ellipse $\Gamma$ through $B$ with foci in $A$ and $C$. Since such ellipse is convex, for any point $Q$ inside $\Gamma$ we have $AQ+QC < AB+BC$.


0

Draw $F$ so that $EBDF$ is a parallelogram. Then the triangles $AEB$ and $CDF$ are congruent (side-angle-side) so that $AB=CF$. Now, the required inequality is equivalent to $$ BC+CF>BD+DF.\tag{*} $$ Extend $BD$ to touch $CF$ at $G$. Then, by the triangle inequalities, $$ BC+CG>BG=BD+DG,\\ DG+GF>DF. $$ If you add the above inequalities and ...


1

Why not decompose direct connections in a vector notational way: $$ AB = AE + EB $$ Then it's simple to say $$ AB + BC = AE + EB + BE + EC \\ AD + DC = AE + ED + DE + EC $$ And now use $AB = BA$ for all $A, B$, so in the end $$ AB + BC \geq AD + BC \\ \Leftrightarrow EB \geq ED $$ which is certainly true by construction.


0

Use B as centre, radius = BD, draw an arc cutting BC at P. Then, BP = BD. Therefore BC = BP + PC > BD Similar result for BA > BE The required result follows. Note that the said given is not really necessary.


0

Try Graphing Calculator 3D. It's both powerful and easy to use. It plots beautiful math graphs in 3D and the interactivity is super fast. http://www.runiter.com/graphing-calculator/


2

The slope measure form 0 to 1 is basically an angle (lets call it $\alpha$), the other one is the slope, lets call it $m$. The relation is $$ m = \tan\left( \alpha \cdot \frac{\pi}{2} \right)$$ if you are calculating with radians or $$ m = \tan\left( \alpha \cdot 90° \right)$$ if you are calculating in degrees.


0

Assume that the lines are the perpendicular bisectors of AB and AC (your question was unclear on this point). Let A be the point $(a,b)$. Reflect it in the two lines to get the points $(-b,-a)$ and $(\frac{40}{41}a+\frac{9}{41}b,\frac{9}{41}a-\frac{40}{41}b)$ as B and C. They must be collinear with $(g,f)$. So using the usual determinant condition we get ...


1

In geometry, a "figure" is a set of points in the plane. So, two figures are equal if they have the same points. In other words, two equal figures are exactly equal: the same figure. Congruent figures have the same shape and size (informally) but possibly different points. No diagram is needed for this explanation. Part of the confusion between "equal" ...


0

Let's have a rectangle with sides $5$m and $4$m. How many squares with a side of length $1$m (unit squares) can you put in the rectangle? The answer is $20$, which can be seen if you drew straight lines that would divide the rectangle into squares: \begin{array}{|c|c|c|c|} \hline \hphantom{i}& \hphantom{i} & \hphantom{i} &\hphantom{i} & ...


0

Using Gimp (or a similar graphics program), assigning coordinates is fairly easy. Gimp gives me (95,263) for the bottom point, and (211,78) for the top one. Then, referring back to the previous answer (the one by vladimirm), we get $$ \Delta x = 116\quad \Delta y = 185\quad \text{angle} = \arctan \frac{185}{116} \approx 1.0107407\text{ radians} \approx ...


0

Hint: The slope of the line joining the two points is $tan \theta$, and it is given by $\frac{y_2-y_1}{x_2-x_1}$ where $(x_1,y_1)$ and $(x_2,y_2)$ are the coordinates of the start and end points of the line segment. Can you find $\theta$ from here?


0

Assign some coordinates to your points $A$ and $B$, and let the red line be parralel to the $x$ axis, then $\Delta x = B_x - A_x$, $\Delta y = B_y - A_y$, Angle $= \arctan\left(\frac{\Delta y}{\Delta x}\right) $


-2

4/3 * 1/2*12*9 = 72 area of triangle of medians = 1/2*12*9=54


1

Please refer to the diagram shown below::--


0

Let's relabel the four points in your diagram as $\mathbf{P}_1 = \mathbf{A}$, $\mathbf{P}_2 = \mathbf{B}$, $\mathbf{P}_3 = \mathbf{C}$, $\mathbf{P}_4 = \mathbf{D}$, to avoid fiddling with subscripts. So, according to the Catmull-Rom construction, we are dealing with a cubic segment that starts at the point $\mathbf{B}$, with a start derivative of ...


0

Parabola is a special case of an ellipse or hyperbola, one focus is at infinity. How do you define a "simple" function? If the distances d(P,F1)= u , d(P,F2)= v , then apart from those you mentioned ( u+v, u-v, u*v, u/v ) there can be many functions that can be prescribed to remain constant to generate loci, it is up to your combinatorial imagination. u^2 ...


2

The containing square If we imagine rotating the triangle by an angle $\theta$ and measuring the width (minimum $x$ coordinate to maximum $x$ coordinate, as could be measured with a vernier caliper), we get a width function $w(\theta)$ which can be expressed in terms of the side lengths $s_1,s_2,s_3$ and side headings $\phi_1, \phi_2, \phi_3$ as ...


2

I do not think using squares is a good idea. If you use instead the ratio $r/R$ of the inradius to the circumradius, you get a very clean formula: $$ 4\sin(A/2)\sin(B/2)\sin(C/2), $$ where $A, B, C$ are the angles of the triangle. See for instance here.


0

for P satisfies the condition: $$(x-a)^2+(y-b)^2+(x-b)^2+(y+a)^2=(a-b)^2+(a+b)^2$$ Solve: $$(x^2+a^2-2ax)+(y^2+b^2-2bx)+(x^2+b^2-2bx)+(y^2+a^2+2ay)=2a^2+2b^2$$ $$- a x - 2 b x + x^2 + a y + y^2 = 0$$ $$\left(x-\frac{a+2b}2\right)^2+(y+a/2)^2=\left(\frac{\sqrt{2a^2+4b^2+4ab}}{2}\right)^2$$ The last equation is the locus of point P which apparently is a ...


0

Given known coordinates for $A,$ $B,$ and $C,$ you can compute the lengths $\overline{AC}$ and $\overline{BC}$ by the Pythagorean Theorem. Given known coordinates for $P$ and $Q,$ and given the lengths $\overline{PR}$ and $\overline{QR},$ apply the Pythagorean Theorem to the unknown coordinates of $R$ and the known coordinates of $P$ and $Q,$ and set the ...


1

$$\begin{align} \triangle O_1RP \sim \triangle O_2SP &\qquad\implies\qquad \frac{r}{|O_1P|}=\frac{s}{|O_2P|} \quad=: k\\[6pt] \triangle O_1PQ \sim \triangle O_1BA &\qquad\implies\qquad \frac{|AB|}{|PQ|}=\frac{r}{|O_1P|} = k \quad\to\quad |AB| = k\;|PQ|\\[6pt] \triangle O_2PQ \sim \triangle O_2DC &\qquad\implies\qquad ...


1

I'd iterate over $u$ and $v$ instead, since the point which actually has easy rational coordinates is not exactly at a pixel position. double d; for (int u = 0; u <= 100; ++u) { for (int v = 0; (d = u*u+v*v+1) <= 100; ++v) { markBlack(2*u/d, 2*v/d); // arguments are double in [0,1] } } or, if you want to stay in integer arithmetic: int d; ...


1

I take back what I said in the comment above. It's not entirely impossible to answer your question. Here is an attempt. $12800 \text{ cm}^2$ is about $1984$ square inches. A suitable shape for an area of $1948$ square inches would be 46 inches wide by 43 inches tall. I'm pretty sure a 46 inch width will fit in a Jeep, because most US cars are designed with ...


2

Since $O_2 Y\perp O_1 Y$ and $O_1 Z\perp O_2 Z$, $O_1 Y Z O_2$ is a cyclic quadrilateral. This gives that $JO_1 Y$ and $JO_2 Z$ are similar triangles, so: $$\frac{JO_1}{O_1 A}=\frac{JO_1}{O_1 Y}=\frac{JO_2}{O_2 Z}=\frac{JO_2}{O_2 C}\tag{1}$$ but $(1)$ gives that $AC$ is parallel to $O_1 O_2$ due to Thales' theorem. Obviously the same holds for $BD$, and both ...


0

Assuming you have Cartesian coordinates $(x,y,z)$ for each point, to reduce this to a two-dimensional problem, you can simply choose a coordinate axis that is not parallel to the plane in which the polygon lies, and remove that coordinate from all points. For example, if the plane is not parallel to the $z$ axis, consider just the coordinates $(x,y)$ at each ...


0

Here's a quick, although not efficient, solution: Find the median of the polygon points; call it $P$. For each of the polygon points $Q_i$, define $r_i = Q_i - P.$ Compute $$ n = r_1 \times r_2\\ u = r_1\\ w = n \times v $$ Express the $r_i$ in terms of $u$ and $w$; fior the unknown point $Q$, express it that way too: $$ x_i = r_i \cdot u \\ y_i = r_i ...


2

The C.M. itself is the mass weighted average of the position vector. For object with uniform density, it is the same as the volume weighted average. What you need to do is figure out is the Volume and C.M. of individual pieces. For the top cylinder and middle box, the C.M. are at their symmetry center. For the bottom cone, it is at $\frac34$ of its height. ...


3

Lets assume density is equal to 1. This doesn't affect the location of the center of mass. Then $$ M_c = \text{Mass of cylinder} = \pi \times \text{radius}^2 \times \text{height} = \pi \times 144 \times 50 = 22619.4671 $$ $$ M_b = \text{Mass of box} = \text{height} \times \text{width} \times \text{depth} = 70\times 56\times 70 = 274400 $$ $$ M_k = ...


1

From $CB=\lambda DA$ and $ED=\lambda AC$ we get $$CB+ED = \lambda (DA+AC) = \lambda DC = \lambda v.$$ In fact $\lambda=\varphi$ is the golden ration, see Wikipedia. So we have $$\lambda=\frac1\lambda-1.$$ So your result is not incorrect. In fact, if we combine your computation together with mine, we get the proof that $$\lambda=\frac1\lambda-1.$$


4

Use this facts: COM of Cylinder and Cube are at body centre. COM of Cone at height $h/4$ from base. Now do weighted AM of coordinates of COM three figures with volume as their weights. Let's do Calculation. Simple symmetry says COM will be along the line perpendicular to ground through the point from the cone touching ground. So let's do single ...


4

I believe, you are looking in a wrong place. Geometry and Topology are related, but different fields of mathematics. Same with Analysis, which you are trying to put under the same umbrella. You might eventually find a definition which is broad enough to cover all three areas, but then it will cover so much in mathematics that it becomes useless. Here are ...


0

Here's my construction. You may refer to this image to follow along. From top to bottom, call the input points $A$, $B$, $C$. $D$, the fourth point, is easy: draw a line through $C$ parallel to $AB$, and the intersection of that line with the front right edge is $D$. ...it occurs to me that this doesn't always happen correctly, but in this case it ...


0

An $N$-dimensional rectangular cuboid, or orthotope, or hyperrectangle...


2

HINT: Note that this is the continuous image of $[1,2]\subseteq\Bbb R$ which is compact into $\Bbb R^2$, where the map is $x\mapsto\langle x,0\rangle$.


2

It is compact. In $\Bbb R^2$, $K$ is compact if and only if $K$ is closed and bounded. This set is just the line segment from $(1,0)$ to $(2,0)$ in the plane.


1

As stated by metacompactness in the comments, the triangles $QAT$ and $SAP$ are similar since they have three congruent angles, so we have: $$\frac{PA}{AT}=\frac{PS}{QT}=\frac{QR}{QT}=\frac{5}{3}.$$ An interesting technique for solving such problems is to consider that affine maps preserve the ratios between the lengths of segments on the same line, so it is ...


0

You just have to look at the signs of those numbers. You know that $(x,y)$ is in the first quadrant, if $x > 0$ and $y > 0$; in the second quadrant, if $x < 0$ and $y > 0$; in the third quadrant, if $x < 0$ and $y < 0$; in the fourth quadrant, if $x > 0$ and $y < 0$. So, for example, (1,4) is in the first quadrant because both $1$ ...


1

You only have to look at the angle formed between the point and the $x$-axis. If you have $P = (r, \theta)$ in polar coordinates, then: $P$ is in the first quadrant $\iff 0 \leq \theta < \pi/2$; $P$ is in the second quadrant $\iff \pi/2 \leq \theta < \pi$; $P$ is in the third quadrant $\iff \pi \leq \theta < 3\pi/2$; $P$ is in the fourth quadrant ...


1

It looks to me as if in this case, the intersection will be a hexagon. The plane will, of course, intersect the cube in OTHER points than just these three. But you can get a pretty good sense of things by drawing the triangle that contains the three points; the plane is the unique plane containing that triangle. Let's label the top four edges A, B, C, D, ...


0

A different color is the usual way. You plot everything multiplied by $i$ in a different color than the real part. Try plotting Euler's identity in mathematica; Plot[E^(I x), {x, -6 Pi, 6 Pi}] which is $$y=\cos(x)+i \sin(x)$$ to get a better understanding of what I'm saying. To your original question, the roots of a polynomial with a negative discriminant ...


0

Hint: You need one station for each suburb exactly on the circumference of the city. This is the number of ways to express $R^2$ as a sum of two squares. A little searching will find this. It depends on the factorization of $R^2$


0

Intuitively you can think about the quotient space of such a good CW pair $(X,Y)$, as collapsing all cells in $Y$ to a point. So if you collapse $Y$ you get all cells of $X$ which were not in $Y$ but glued in maybe a very different way than before. So the dimension (i.e. the highest dimension where there exists a cell) of the quotient space decreases at ...



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