I am having problems solving this problem:    Using only basic geometry is easy to go here:    And propose 4 equations: $$ x+y+70=180$$ $$x+w+40=180 $$ $$u+y+50=180 $$ $$u+w+20=180 $$ And it doesn't make sense to me because, they are linear dependent, we can reduce it to 3 equations that depends (for example) of $u$.Giving us: $$y=130-u$$ $$w=160-u $$ $$x=u-20 $$  But for the construction of the problem, it don't understand how can this problem depends of a variable, and don't know how to solve it. Please help with this.
2724
The following is a geometry puzzle from a math school book. Even though it has been a long time since I finished school, I remember this puzzle quite well, and I don't have a nice solution to it.  So here is the puzzle:    The triangle $ABC$ is known to be isosceles, that is, $AC=BC$. The labelled angles are known to be $\alpha=\gamma=20°$, $\beta=30°$. The task is to find the angle labelled "?".  The only solution that I know of is to use the sine formula and cosine formula several times. From this one can obtain a numerical solution. Moreover this number can be algebraically shown to be correct (all sines and cosines are contained in the real subfield of the 36th cyclotomic field). So in this sense I solved the problem, but the solution is kind of a brute force attack (for example, some of the polynomials that show up in the computation have coefficients > 1000000). Since the puzzle originates from a book that deals only with elemetary geometry (and not even trigonometry if I remember correctly) there has to be a more elegant solution.
99525
So my suggestion was:  Assume: Let $S = \{u_1, u_2, ....u_r\}$ be a set of vectors in $\mathbb{R}^n$ . If $r &gt; n$ then the vectors $u_1, u_2 \ldots u_r$ must be linearly dependent.  If one writes the linear system corresponding to $c_1u_1 + c_2u_2 + ... + c_ru_r$ , one will have a homogeneous system of $n$ equations in $r$ unknowns. We know that such a system has infinitely many solutions. Thus in $\mathbb{R}^n$ , a set which is linearly independent cannot contain more than $n$ vectors.
62448
Good Night ... Here I am editing the question and explaining the purpose of the same ... Mathematics course, and I love this science, I am very curious, and these days, studying a book of calculus found in (book from the library of the University) a sheet full of challenges, to share with a colleague, he gave me this challenge, which he also failed to respond ... I tried, I tried, I used the formulas of quadratic equation, and I've been trying since Friday (here in Brazil is Sunday at 22:33), and could not, and I ask your help, and please believe me, I struggled, tried to make substitutions, tried both, and I'm confused ... I must find the values of $ a, b $ and $ c $?  Consider a parabola whose roots, other are $x'$ and $x''$, and such that $x'=-x''$ where $x'&gt; 0$. Let $$A = (x', 0)$$ $$B = (x'', 0)$$ and $V$ the vertex of the parabola. Suppose that the equation of this parabola is $$y = ax^2 +bx+ c$$. Calculate the value of the discriminant $$\Delta=b^2-4ac$$, so that the triangle is equilateral AVB.
19786
How do I solve these? $$|-2w-4|&lt;12$$  and  $$(-1/3)|6-5x|+2&lt;1$$  for the first problem i got $-8&lt;w&lt;4$ is that right?
79310
The problem is what are the integer solutions to  $$7x^2 - 40xy + 7y^2 = (|(x - y)| + 2)^3$$  I only got $x - y = 4 (\mod 13)$. Also, how to solve it?
57729
I know the formula $x^2 + 2xy + y^2 = (x + y)^2$ by heart and in order to understand it better I am trying to solve the formula myself; however, I don't understand how step 4 is derived from step 3.  $$x^2+2xy+y^2\qquad1$$  $$x^2+xy+xy+y^2\qquad2$$  $$x(x+y) + y(x+y) \qquad3$$  $$(x+y)(x+y) \qquad 4$$  How does one get from step 3 to step 4? I don't know how to do that process.
78303
$$u = (r,1,1), \ v = (1,s,2s), \ w = (1,2,2)$$  I know to assume the set will be linearly dependent, equate them to the $0$ vector and find the values not equal to what $r$ and $s$ is. But I can't seem to reduce the matrix to echelon form.  I've gotten to the point of:  $$     \begin{pmatrix}     r &amp; 1 &amp; 1 \\     1 &amp; s &amp; 2 \\     1 &amp; 2s &amp; 2 \\     \end{pmatrix} $$  $R_2 = R_2 - R_3$ $$     \begin{pmatrix}     r &amp; 1 &amp; 1 \\     1 &amp; -s &amp; 0 \\     1 &amp; 2s &amp; 2 \\     \end{pmatrix} $$ $R_3 = R_3 +2R_2$  $$     \begin{pmatrix}     r &amp; 1&amp; 1 \\     0 &amp; -s &amp; 0 \\     1 &amp; 0 &amp; 2 \\     \end{pmatrix} $$  EDIT: Thank you @LordSharktheUnknown  $R_3 = r \times R_3 - R_1$  $$     \begin{pmatrix}     r &amp; 1 &amp; 1 \\     0 &amp; -s &amp; 0 \\     0 &amp; -1 &amp; 2r-1 \\     \end{pmatrix} $$  And I'm stuck. I know $s ≠ 0$ for it to be linearly independent but I have no idea how to reduce the rows any further to get to $r$ by itself.  Any help is appreciated.  Thank you.
26390
Is there an error in this question?  Let u = $(1,0,2)$, v = $(2,1,0)$ and w = $(0,2,1)$. Compute  ||u + v - w.||v = v  This is exactly what is in the question (even the dot after w which I don't know what it means).  Thanks
64533
I'm trying to find the minimal distance between the surfaces described by $z=x^2+y^2$ and $x+y-2z=8$. I would imagine there are several approaches including the use of Lagrange multipliers. I attempted to find the spot when their normal vectors are parallel ( since i believe if these vectors are not parallel there is a direction that will bring the distance lower). So if $U(x,y,z)=x^2+y^2$ and $V(x,y,z)=x+y-2z$ are functions describing these surfaces then their normal vectors are $(2x,2y,-1)$ and $(1,1-2)$ respectively. Then I want their cross product to be zero; this is at $(1/4,1/4,z)$. But when i solve for $z$ and I cannot simultaneously satisfy both equations. What gives?! Am I along the right line of thinking?
10869
No matter how I do it, I always end up with $x = 0, 90, 270$ and $360$. All of those except $270$ is right, but I can't quite figure out how to get the $270$ degrees out of the answer. I've tried using trig identities, I've tried squaring both sides, but I always end up with $$2\sin x\cos x$$ which then leads me to $x = 0, 90, 270, 360$. But $$\sin (270) + \cos (270) = -1$$ so I'm doing something wrong.
30704
I have a "proof" that has an error in it and my goal is to figure out what this error is. The proof:  If $x = y$, then  $$ \begin{eqnarray} x^2 &amp;=&amp; xy \nonumber \\ x^2 - y^2 &amp;=&amp; xy - y^2 \nonumber \\ (x + y)(x - y) &amp;=&amp; y(x-y) \nonumber \\ x + y &amp;=&amp; y \nonumber \\ 2y &amp;=&amp; y \nonumber \\ 2 &amp;=&amp; 1 \end{eqnarray} $$    My best guess is that the error starts with the line $2y = y$. If we accept that $x + y = y$ is true, then   $$ \begin{eqnarray} x + y &amp;=&amp; y \\ x &amp;=&amp; y - y \\ x &amp;=&amp; y = 0 \end{eqnarray} $$  Did I find the error? If not, am I close?
112240
Find all ($x,y,z$) such that $$ 27^{3x^2 + 2y}+27^{3y^2 + 2z}+27^{3z^2 + 2x}=1$$  I am a high school student and would appreciate it if anyone could solve it using high school mathematics. here is what I have tried till now: $$ 3x^2 +2y &lt; 0 $$ (this can also be said for the other 3 equation. it also means that x,y,z&lt;0) $$ -2y&gt;3x^2 $$ $$ 4y^2&gt;9x^2 $$ ( -2y and 3x^2 are both positive) $$ 3y^2&gt;27/4 x^2 $$ $$ -2z&gt;3y^2&gt;27/4 x^4 $$ $$ 4z^2 &gt; 729/16 x^8 $$ $$ 3z^2&gt; 2187/64 x^8 $$ $$ -128/2187&lt;x^7 $$ $$ -2/3 &lt; x $$ $$ -2/3&lt; x,y,z &lt; 0 $$ i tried this but as you can see didn't do much good
115007
No integer bigger than 70 can be expressed as a sum of 3 integers smaller than 30  My attempt: $U_1 = {x∈Z∶x&gt;70 }$ $U_2 = {y∈Z∶y&lt;30 }$ $∄x ∈ U_1 , ∀y ∈ U_2  , x = y_1+y_2+y_3$  I understand that there are many ways to write a sentence in mathematical notations, just wondering whether there is an easier way to write this? Thanks guys :)
122205
I think I have proved that that a cannot be greater than b (ie a &amp; b must be equal) then using this fact I got a quadratic in terms of a and c, trying out some values I got that (3,3,4) is a solution, but how do I prove this is the only solution or if this is not the only solution then what are the other solutions and how do I find them?
47252
I found this problem in a book and while I understand the given solution, I don't quite understand how somebody would arrive at this solution unless they did out the casework and happened to make the right guesses to solve the problem. Any guidance on solving this from scratch would be greatly appreciated!
58186
Help me understand problems of this type a bit more intuitively.   The solution $C(3+11−1,11)$ seems simple enough, but I got stuck thinking about how many integers you are choosing from within $x_1$, $x_2$, and $x_3$ and I was trying to get an $n$ - as in $C(n+r-1,r)$ - this way. Also, I though the $r$ was 3 since you have $x_1$, $x_2$, $x_3$ choices. Basically, I went wrong at every step. Help!
31570
Can someone help me with the following question?  The sum of two numbers is $18$ and their product is $45$. Find the numbers.  I know that the answer is $15$ and $3$. But how do I find that answer algebraically? It tried doing it as $x+y=18$ and $xy=45$ and then $x= 18-y$ and I substituted in $xy=45$. I got it as $(18-y)y =45$, and I don't know what to do next.