heureka

For the following system of equations, what is the value of c?
\(a+b+c+d = 88 \\ a+b+c+e = 84 \\ a+b+d+e = 82 \\ a+c+d+e = 78 \\ b+c+d+e = 72 \)

\(\begin{array}{|lrcll|} \hline (1) & a+b+c+d = 88 \\ (2) & a+b+c+e = 84 \\ (3) & a+b+d+e = 82 \\ (4) & a+c+d+e = 78 \\ (5) & b+c+d+e = 72 \\ \hline (1)+(2)+(3)+(4)+(5): & 4a+4b+4c+4d+4e &=& 88 + 84+82+78+72 \\ & 4(a+b+c+d+e) &=& 404 \quad | \quad : 4 \\ (6) & \mathbf{ a+b+c+d+e } &=& \mathbf{101} \\ \hline (6)-(3): & (a+b+c+d+e)-(a+b+d+e) &=& 101 - 82 \\ & \mathbf{c} &=& \mathbf{19} \\ \hline \end{array}\)

Suppose a,b,c are positive reals such that

ab=2a + 2b,

ac= 3a + 3c,

bc = 4b + 4c

Find a+b+c

\(\begin{array}{|lrcll|} \hline (1) & 2a + 2b &=& ab \\ & 2a-ab &=& -2b \\ & a(2-b) &=& -2b \\ & a &=& \dfrac{-2b}{2-b} \\ &\mathbf{ a } &=& \mathbf{ \dfrac{2b}{b-2} } \\ \hline (3) & 4b + 4c &=& bc \\ & 4c-bc &=& -4b \\ & c(4-b) &=& -4b \\ & c &=& \dfrac{-4b}{4-b} \\ &\mathbf{ c } &=& \mathbf{ \dfrac{4b}{b-4} } \\ \hline (2) & 3a + 3c &=& ac \\ & 3(a+c) &=& ac \\ & 3 \left(\dfrac{2b}{b-2}+\dfrac{4b}{b-4} \right) &=& \left(\dfrac{2b}{b-2}\right) \left(\dfrac{4b}{b-4}\right) \\ & 3 \left(\dfrac{2b(b-4)+4b(b-2)}{(b-2)(b-4)} \right) &=& \dfrac{8b^2}{(b-2)(b-4)} \\ & 3 \Big(2b(b-4)+4b(b-2)\Big) &=& 8b^2 \\ & 6b(b-4)+12b(b-2) &=& 8b^2 \quad | \quad : 2 \\ & 3b(b-4)+6b(b-2) &=& 4b^2 \\ & 3b^2-12b+6b^2-12b &=& 4b^2 \\ & 5b^2-24b &=& 0 \\ & b(5b-24) &=& 0 \\\\ & b &=& 0 \quad | \quad \text{no solution, while b is a positive real } \\\\ & 5b-24 &=& 0 \\ & \mathbf{b} &=& \mathbf{\dfrac{24}{5}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{a} &=& \mathbf{ \dfrac{2b}{b-2} } \\\\ a &=& \dfrac{2\times\dfrac{24}{5}}{\dfrac{24}{5}-2} \\\\ a &=& \dfrac{48}{5\times\dfrac{(24-10)}{5}} \\\\ a &=& \dfrac{48}{14} \\\\ \mathbf{a} &=& \mathbf{\dfrac{24}{7}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{c} &=& \mathbf{ \dfrac{4b}{b-4} } \\\\ c &=& \dfrac{4\times\dfrac{24}{5}}{\dfrac{24}{5}-4} \\\\ c &=& \dfrac{96}{5\times\dfrac{(24-20)}{5}} \\\\ c &=& \dfrac{96}{4} \\\\ \mathbf{c} &=& \mathbf{24} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline a+b+c &=& \dfrac{24}{7} + \dfrac{24}{5} + 24 \\ &=& 24\times \left( \dfrac{1}{7} + \dfrac{1}{5} + 1 \right) \\ &=& 24\times \left( \dfrac{47}{35} \right) \\ \mathbf{a+b+c} &=& \mathbf{\dfrac{1128}{35}} \\ \hline \end{array} \)

Find all integers X for which there exists an integer Y such that

1/X + 1/Y = 1/7
(In other words, find all ordered pairs of integers (X, Y) that satisfy this equation, then enter just the X's from these pairs.)

Given vectors \(\mathbf{v}\) and \(\mathbf{w}\) such that \(\|\mathbf{v}\| = 3,\ \|\mathbf{w}\| = 7,\) and \(\mathbf{v} \cdot \mathbf{w} = 10,\) then find \(\|\operatorname{proj}_{\mathbf{w}} \mathbf{v}\|.\)

\(\begin{array}{|rcll|} \hline \|\operatorname{proj}_{\mathbf{w}} \mathbf{v}\| &=& \left\| \dfrac{\mathbf{v} \cdot \mathbf{w}}{\left\|\mathbf{w}\right\|^2}\mathbf{w} \right\| \\\\ &=& \left\| \dfrac{10}{7^2}\mathbf{w} \right\| \\\\ &=& \dfrac{10}{7^2} \| \mathbf{w} \| \\\\ &=& \dfrac{10}{7^2} \cdot 7 \\\\ \|\operatorname{proj}_{\mathbf{w}} \mathbf{v}\| &=& \mathbf{\dfrac{10}{7}} \\ \hline \end{array}\)

Thank you, CPhill !

1.

\(\lim \limits_{x \to 0}\left(\dfrac{\dfrac{1}{\sqrt{1+x}}-1}{x}\right) \)

\(\begin{array}{|rcll|} \hline && \mathbf{\lim \limits_{x \to 0}\left(\dfrac{\dfrac{1}{\sqrt{1+x}}-1}{x}\right) } \\\\ &=&\lim \limits_{x \to 0}\left(\dfrac{ \left(1+x \right)^{-\frac{1}{2}} -1}{x}\right) \quad | \quad \text{L'Hospital's rule} \\\\ &=&\lim \limits_{x \to 0} \left( \dfrac{ \dfrac{ d\ \left( \left(1+x \right)^{-\frac{1}{2}} -1 \right) } { dx } } { \dfrac{d\ (x)}{dx} } \right) \\\\ &=&\lim \limits_{x \to 0} \left( \dfrac{ -\frac{1}{2} \left(1+x \right)^{-\frac{1}{2}-1} } {1} \right) \quad | \quad x\to 0 \\\\ &=& \mathbf{-\dfrac{1}{2}} \\ \hline \end{array}\)

Find the projection of \(\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) onto the plane \(3x - y + 4z = 0\).

\(\text{Let $\vec{P}= \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$}\)

1.

Normal vector \(\mathbf{\vec{n}}\)

\(\begin{array}{|rcll|} \hline \vec{n} &=& \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \quad | \quad \text{plane: }{\color{red}3}x {\color{red}-1}y + {\color{red}4}z = 0 \\ \hline \end{array}\) 

2. \(\mathbf{\vec{P}_{Proj.}}\)

\(\begin{array}{|rcll|} \hline \mathbf{\vec{P}_{Proj.}} &=& \mathbf{\vec{P}+\lambda\vec{n}} \\\\ \vec{P}_{Proj.} &=& \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}+\lambda\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \\\\ x_{Proj.} &=& 1+3\lambda \\ y_{Proj.} &=& 2-\lambda \\ z_{Proj.} &=& 3+4\lambda \\ \hline \end{array} \)

3. \(\mathbf{\lambda}\)

\(\begin{array}{|rcll|} \hline 3x_{Proj.} - y_{Proj.} + 4z_{Proj.} &=& 0 \\ 3(1+3\lambda) - (2-\lambda) + 4(3+4\lambda) &=& 0 \\ 3+9\lambda-2+\lambda+12+16\lambda &=& 0 \\ 26\lambda &=& -13 \\\\ \lambda &=& -\dfrac{13}{26} \\\\ \mathbf{\lambda} &=& -\mathbf{\dfrac{1}{2}} \\ \hline \end{array}\)

4.
\(\mathbf{\vec{P}_{Proj.}}\)

\(\begin{array}{|rcll|} \hline \vec{P}_{Proj.} &=& \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}+\lambda\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \quad | \quad \lambda = -\dfrac{1}{2} \\\\ &=& \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} -\dfrac{1}{2}\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \\\\ &=& \begin{pmatrix} 1-\dfrac{3}{2} \\\\ 2+\dfrac{1}{2} \\\\ 3-\dfrac{4}{2} \end{pmatrix} \\\\ \mathbf{\vec{P}_{Proj.}}&=& \begin{pmatrix} \mathbf{-\dfrac{1}{2}} \\\\ \mathbf{\dfrac{5}{2}} \\\\ \mathbf{1} \end{pmatrix} \\ \hline \end{array}\)

What is the number of degrees in the smaller angle formed by the hour and minute hands of a clock at 5:44?

\(\text{Let the angle formed by the minute hand and hour hand in degrees $\mathbf{ \Delta \alpha }$ } \\ \text{Let the time in hours $\mathbf{ t }$ }\)

The formula between the two values \(\Delta \alpha\) and \(t\) is:
\(\boxed{~ \Delta \alpha = 330 \cdot t } \)

\(\begin{array}{|rcll|} \hline \Delta \alpha &=& 330 \cdot t \quad | \quad t= 5:44 = 5+\dfrac{44}{60}= 5.7\overline{3}\ h \\ &=& 330 \cdot 5.7\overline{3} \\ &=& 1892 \\ &=& 1892 -5\cdot 360 \\ &=& 1892 - 1800 \\ \mathbf{\Delta \alpha} &=& \mathbf{ 92^\circ } \\ \hline \end{array}\)

Let a, b, c be real numbers such that \(a + b + c = 1\).

Find the minimum value of \(2a^2 + 3b^2 + 6c^2\).

Schwarz Inequality:

\(\begin{array}{|rcll|} \hline \mathbf{(x_1^2+x_2^2+\ldots + x_n^2)(y_1^2+y_2^2+\ldots + y_n^2) } &\geq& \mathbf{ \left(x_1y_1+x_2y_2+\ldots x_ny_n \right)^2 } \\\\ \left( \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6} \right)(2a^2+3b^2+6c^2 ) &\geq& \left( \dfrac{1}{\sqrt{2}} \sqrt{2}a+ \dfrac{1}{\sqrt{3}} \sqrt{3}b+ \dfrac{1}{\sqrt{6}} \sqrt{6}c \right)^2 \\\\ \left( \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6} \right)(2a^2+3b^2+6c^2 ) &\geq& \left( a+b+c \right)^2 \\\\ \left( \underbrace{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}}_{=1} \right)(2a^2+3b^2+6c^2 ) &\geq& \left(\underbrace{ a+b+c}_{=1} \right)^2 \\ 2a^2+3b^2+6c^2 &\geq& 1^2 \\ \mathbf{2a^2+3b^2+6c^2} &\geq& \mathbf{1} \\ \hline \end{array}\)

The minimum value of \(2a^2 + 3b^2 + 6c^2 \)is \(\mathbf{1}\)

Suppose with with \(A, B, C \in \mathbb{R}\).

What is \(A\)?
\(\dfrac{1}{x^3-x^2-21x+45}=\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x - 3)^2}\)

\(x^3-x^2-21x+45=(x+5)(x-3)^2\)

\(\begin{array}{|lrcll|} \hline & \mathbf{\dfrac{1}{x^3-x^2-21x+45} } &=&\mathbf{\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x - 3)^2} } \\\\ & \dfrac{1}{(x+5)(x-3)^2} &=&\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x-3)^2} \quad | \quad \times (x+5)(x-3)^2 \\\\ & 1 &=& A(x-3)^2 + B(x+5)(x-3) + C(x+5) \\ \hline x=-5: & 1 &=& A(-5-3)^2 + B(-5+5)(-5-3) + C(-5+5) \\ & 1 &=& A(-8)^2 + B(0)(-8) + C(0) \\ & 1 &=& 64A \\ & \mathbf{A} &=& \mathbf{ \dfrac{1}{64} } \\ \hline \end{array}\)