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There are 360 people in my school.

15 take calculus, physics, and chemistry, and

15 don't take any of them.

180 take calculus.

Twice as many students take chemistry as take physics.

75 take both calculus and chemistry, and

75 take both physics and chemistry.

Only 30 take both physics and calculus.

How many students take physics? 

\(\begin{array}{|c|c|c|rl|} \hline \text{calculus} & \text{physics} & \text{chemistry} & \\ \hline x & x & x & 15 & \text{ take calculus, physics, and chemistry} \\ \hline x & & x & 75 & \text{ take both calculus and chemistry} \\ \hline x & x & & 30 & \text{ take both physics and calculus} \\ \hline x & & & 180-15-75-30 = \mathbf{60} & \text{ take only calculus} \\ \hline & & & 180 & \text{ take calculus} \\ \hline \end{array} \)

\(\begin{array}{|c|c|c|rl|} \hline \text{chemistry} & \text{physics} & \text{calculus} & \\ \hline x & x & x & 15 & \text{ take calculus, physics, and chemistry} \\ \hline x & & x & 75 & \text{ take both calculus and chemistry} \\ \hline x & x & & 75 & \text{ take both physics and chemistry} \\ \hline x & & & v & \text{ take only chemistry} \\ \hline & & & 15+75+75+v \\ & & & =165+v & \text{ take chemistry} \\ \hline \end{array}\)

\(\begin{array}{|c|c|c|rl|} \hline \text{physics} & \text{chemistry} & \text{calculus} & \\ \hline x & x & x & 15 & \text{ take calculus, physics, and chemistry} \\ \hline x & & x & 30 & \text{ take both physics and calculus} \\ \hline x & x & & 75 & \text{ take both physics and chemistry} \\ \hline x & & & u & \text{ take only physics} \\ \hline & & & 15+30+75+u \\ & & & =120+u & \text{ take physics} \\ \hline \end{array}\)

Twice as many students take chemistry as take physics:

\(\begin{array}{|rcll|} \hline 165+v &=& 2(120+u) \\ &=& 240 +2u \\ \mathbf{v} &=& \mathbf{2u+75} \\ \hline \end{array}\)

There are 360 people in my school:

\(\begin{array}{|rcll|} \hline \overbrace{180}^{\text{take calculus}} + \overbrace{u+v+ \underbrace{75}_{\text{take both physics and chemistry} } + \underbrace{15}_{\text{don't take any of them}} }^{\text{take no calculus}} &=& 360 \\ 180 + u+v+ 75 +15 &=& 360 \\ u+v+ 90 &=& 180 \\ u+v &=& 90 \\ \mathbf{u} &=& \mathbf{90-v} \quad | \quad v=2u+75 \\ u &=& 90-(2u+75) \\ u &=& 90-2u-75 \\ 3u &=& 15 \\ \mathbf{u} &=& \mathbf{5} \\\\ v &=& 2u+75 \\ v &=& 2*5 + 75 \\ \mathbf{v} &=& \mathbf{85} \\ \hline \end{array} \)

How many students take physics?

\(\begin{array}{|rcll|} \hline && 120+u \text{ take physics} \\ &=& 120 +5 \\ &=& 125 \text{ take physics} \\ \hline \end{array} \)

A robot moved 10 meters towards north. It then turned 44 degrees to its right and moved another 10 meters.

It then turned 55 degress to its right and moved another 10 meters.

Last, it turned x degrees to its right.

The robot was facing south after the last move.

What is x

\(\begin{array}{|rcll|} \hline 44^\circ+55^\circ+x &=& 180^\circ \\ 99^\circ+x &=& 180^\circ \\ x &=& 180^\circ -99^\circ \\ \mathbf{x} &=& \mathbf{81^\circ} \\ \hline \end{array} \)

The complex numbers a and b satisfy  \(a\overline{b} = -1 + 5i\).
Find \(\overline{a} b\).

\(\begin{array}{|rcll|} \hline a\overline{b} &=& -1 + 5i \\ \overline{a\overline{b}} &=& \overline{-1 + 5i} \quad | \quad \overline{a\overline{b}} = \overline{a}\overline{\overline{b}} \\ \overline{a}\overline{\overline{b}} &=& \overline{-1 + 5i} \quad | \quad \overline{a}\overline{\overline{b}} = \overline{a} b \\ \overline{a} b &=& \overline{-1 + 5i} \quad | \quad \overline{-1 + 5i} = -1 - 5i \\ \mathbf{\overline{a} b} &=& \mathbf{ -1 - 5i } \\ \hline \end{array}\)

Thank you for your reply.

When the measures of the angles of a triangle are placed in order,
the difference between the middle angle and smallest angle is equal
to the difference between the middle angle and largest angle.
If one of the angles of the triangle has measure 23, then
what is the measure in degrees of the largest angle of the triangle?

\(\text{Let $\alpha+\beta+\gamma=180^\circ$ with $\alpha < \beta < \gamma$ } \)

\(\begin{array}{|rcll|} \hline \gamma-\beta &=& \beta-\alpha \\ \mathbf{\gamma} &=& \mathbf{2\beta-\alpha} \quad | \quad \alpha=180^\circ-\beta-\gamma \\ \gamma &=& 2\beta-(180^\circ-\beta-\gamma) \\ \gamma &=& 2\beta-180^\circ+\beta+\gamma \quad | \quad -\gamma \\ 0 &=& 2\beta-180^\circ+\beta \\ 180^\circ &=& 3\beta \\ \beta &=& \dfrac{180^\circ}{3} \\ \mathbf{\beta} &=& \mathbf{60^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \gamma &=& 2\beta-\alpha \quad | \quad \alpha = 23^\circ,\ \beta=60^\circ \\ \gamma &=& 2\cdot 60^\circ-23^\circ \\ \gamma &=& 120^\circ-23^\circ \\ \mathbf{\gamma} &=& \mathbf{97^\circ} \\ \hline \end{array}\)

The measure in degrees of the largest angle of the triangle is \(\mathbf{97^\circ}\)

help!

\(\text{Let $f(n)$ be the base-10 logarithm of the sum of the elements of the $n$th row in Pascal's triangle.$\\$Express $\frac{f(n)}{\log_{10} 2}$ in terms of $n$. Recall that Pascal's triangle begins}\)

\(\begin{array}{|rcll|} \hline f(n) &=& \log_{10} 2^n \\ \mathbf{f(n)} &=& \mathbf{n\log_{10} 2} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dfrac{f(n)}{\log_{10} 2} &=& \dfrac{n\log_{10} 2}{\log_{10} 2} \\\\ \mathbf{\dfrac{f(n)}{\log_{10} 2} } &=& \mathbf{n} \\ \hline \end{array}\)

Two circles are internally tangent at a point T and have radii of 1 and 3.
The maximum possible area for a triangle with one vertex at T,
another vertex on the small circle,
and the third on the large circle can be expressed in the form a*sqrt(b)/c,
where a,b, and c are positive integers,
b is not divisible by the square of any prime,
and a and c are relatively prime.
Find a+b+c.

\(\text{Let $P_1$ point on small circle } \\ \text{Let $P_2$ point on large circle } \\ \text{Let $A$ area of the triangle $T-P_1-P_2$ }\)

\(\begin{array}{|rcll|} \hline T &=& \begin{pmatrix} 0\\ 0 \end{pmatrix} \\ \hline P_1 = \begin{pmatrix} x_1\\ y_1 \end{pmatrix} &=& \begin{pmatrix} r_1\cos(\varphi_1)+r_1 \\ r_1\sin(\varphi_1) \end{pmatrix} \\\\ &=& \begin{pmatrix} 1*\cos(\varphi_1)+1 \\ 1*\sin(\varphi_1) \end{pmatrix} \\\\ \begin{pmatrix} x_1\\ y_1 \end{pmatrix} &=& \begin{pmatrix} \cos(\varphi_1)+1 \\ \sin(\varphi_1) \end{pmatrix} \\ \hline P_2 = \begin{pmatrix} x_2\\ y_2 \end{pmatrix} &=& \begin{pmatrix} r_2\cos(\varphi_2)+r_2 \\ r_2\sin(\varphi_2) \end{pmatrix} \\\\ &=& \begin{pmatrix} 3\cos(\varphi_2)+3 \\ 3\sin(\varphi_2) \end{pmatrix} \\\\ \begin{pmatrix} x_2\\ y_2 \end{pmatrix} &=& \begin{pmatrix} 3(\cos(\varphi_2)+1) \\ 3\sin(\varphi_2) \end{pmatrix} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 2A &=& |P_1\times P_2| \\ 2A &=& x_1y_2-y_1x_2 \\ &=& (\cos(\varphi_1)+1)3\sin(\varphi_2)-\sin(\varphi_1)3(\cos(\varphi_2)+1) \\ 2A&=& 3\Big( (\cos(\varphi_1)+1)\sin(\varphi_2)-\sin(\varphi_1)(\cos(\varphi_2)+1) \Big) \\\\ \mathbf{A} = f(\varphi_1,\varphi_2) &=& \mathbf{\dfrac{3}{2}\Big( (\cos(\varphi_1)+1)\sin(\varphi_2)-\sin(\varphi_1)(\cos(\varphi_2)+1) \Big)} \\ \hline \end{array}\)

maximum:

\(\begin{array}{|rcll|} \hline f_{\varphi_1} &=& \dfrac{\partial f(\varphi_1,\varphi_2)}{\partial \varphi_1}=0 \\ &=& \mathbf{\dfrac{3}{2}\Big( -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)(1+\cos(\varphi_2)) \Big)} \quad & (1) \\ \hline f_{\varphi_2} &=& \dfrac{\partial f(\varphi_1,\varphi_2)}{\partial \varphi_2}=0 \\ &=& \mathbf{\dfrac{3}{2}\Big( \cos(\varphi_2)(1+\cos(\varphi_1))-\sin(\varphi_1)(-\sin(\varphi_2)) \Big)} \quad & (2) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{3}{2}\Big( -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)(1+\cos(\varphi_2)) \Big) &=& 0 \quad | \quad : \dfrac{3}{2} \\ & -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)(1+\cos(\varphi_2)) &=& 0 \\ & -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)-\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\\\ (2) & \dfrac{3}{2}\Big( \sin(\varphi_1)\sin(\varphi_2)+\cos(\varphi_2)(1+\cos(\varphi_1)) \Big) &=& 0 \quad | \quad : \dfrac{3}{2} \\ & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)(1+\cos(\varphi_1)) &=& 0 \\ & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)+\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1) & -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)-\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\ (2) & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)+\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\ \hline (1)+(2): & -\cos(\varphi_1)+\cos(\varphi_2) &=& 0 \\ & \cos(\varphi_2) &=& \cos(\varphi_1) \\ & \varphi_2 &=& \pm \arccos(\cos\varphi_1) \\\\ & \varphi_2 &=& \varphi_1 \Rightarrow \text{minimum} \\\\ & \varphi_2 &=& -\varphi_1 \Rightarrow \text{maximum} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (2) & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)+\cos(\varphi_1)\cos(\varphi_2) &=& 0 \quad | \quad \varphi_2 = -\varphi_1 \\ & \sin(\varphi_1)\sin(-\varphi_1) +\cos(-\varphi_1)+\cos(\varphi_1)\cos(-\varphi_1) &=& 0 \\ & -\sin(\varphi_1)\sin(\varphi_1) +\cos(\varphi_1)+\cos(\varphi_1)\cos(\varphi_1) &=& 0 \\ & \cos^2(\varphi_1)-\sin^2(\varphi_1)+\cos(\varphi_1)&=& 0 \quad | \quad \sin^2(\varphi_1) = 1-\cos^2(\varphi_1)\\ & \cos^2(\varphi_1)-\left(1-\cos^2(\varphi_1)\right)+\cos(\varphi_1)&=& 0 \\ & \cos^2(\varphi_1)-1+\cos^2(\varphi_1)+\cos(\varphi_1)&=& 0 \\ & 2\cos^2(\varphi_1)+\cos(\varphi_1)-1 &=& 0 \\\\ & \cos(\varphi_1) &=& \dfrac{-1\pm \sqrt{1-4\cdot 2\cdot(-1)}}{2\cdot 2} \\ & \cos(\varphi_1) &=& \dfrac{-1\pm \sqrt{9}}{4} \\ & \cos(\varphi_1) &=& \dfrac{-1\pm 3}{4} \\\\ &\cos(\varphi_1) &=&\dfrac{-1+ 3}{4} &\cos(\varphi_1) &=&\dfrac{1}{2} \\ & \varphi_1 &=& \arccos(\dfrac{1}{2}) \\ & \mathbf{\varphi_1} &=& \mathbf{60^\circ} \Rightarrow \text{ maximum} \\\\ &\cos(\varphi_1) &=&\dfrac{-1- 3}{4} &\cos(\varphi_1) &=& -1 \\ & \varphi_1 &=& \arccos(-1 ) \\ & \varphi_1 &=& 180^\circ \Rightarrow \text{ minimum} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \varphi_2 &=& -\varphi_1 \quad | \quad \mathbf{\varphi_1 = 60^\circ } \\ \mathbf{\varphi_2} &=& \mathbf{- 60^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{A} &=& \mathbf{\dfrac{3}{2}\Big( (\cos(\varphi_1)+1)\sin(\varphi_2)-\sin(\varphi_1)(\cos(\varphi_2)+1) \Big)} \quad | \quad \varphi_1 = 60^\circ,\ \varphi_2 = -60^\circ \\ &=&\dfrac{3}{2}\Big( (\cos(60^\circ)+1)\sin(-60^\circ)-\sin(60^\circ)(\cos(-60^\circ)+1) \Big) \\ &=&\dfrac{3}{2}\Big( (\cos(60^\circ)+1)\left(-\sin(60^\circ)\right) -\sin(60^\circ)(\cos(60^\circ)+1) \Big)\quad | \quad \cos(60^\circ) = \dfrac{1}{2},\ \sin(60^\circ)=\dfrac{\sqrt{3}}{2} \\ &=&\dfrac{3}{2}\Big( (\dfrac{1}{2}+1)\left(-\dfrac{\sqrt{3}}{2}\right) -\dfrac{\sqrt{3}}{2}(\dfrac{1}{2}+1) \Big) \\\\ &=&\dfrac{3}{2}\Big( -(\dfrac{1}{2}+1) \dfrac{\sqrt{3}}{2} -\dfrac{\sqrt{3}}{2}(\dfrac{1}{2}+1) \Big) \\\\ &=&\dfrac{3}{2}\Big( -2(\dfrac{1}{2}+1) \dfrac{\sqrt{3}}{2} \Big) \\\\ &=&\dfrac{3}{2}\Big( - (\dfrac{1}{2}+1) \sqrt{3} \Big) \\\\ &=&-\dfrac{3}{2}\Big(\dfrac{3}{2} \sqrt{3} \Big) \\\\ &=&-\dfrac{9\sqrt{3}}{4} \\\\ \mathbf{|A|} &=& \mathbf{\dfrac{9\sqrt{3}}{4}} \\\\ && \boxed{a=9,\ b=3,\ c=4 } \\\\ a+b+c &=& 9+3+4 \\ \mathbf{a+b+c} &=& \mathbf{16} \\ \hline \end{array}\)

Thank you, Roxettna !

Solve for the length of segment AB using an equation based on the phythagorean theorem:

\(\begin{array}{|rcll|} \hline AB^2 &=& AC^2+BC^2 \quad | \quad AB = z+2,\ AC = z,\ BC=6 \\ \left(z+2\right)^2 &=& z^2+6^2 \\ \not{z^2}+4z+4 &=& \not{z^2}+ 36\\ 4z+4 &=& 36 \quad | \quad -4 \\ 4z &=& 32 \quad | \quad :4 \\\\ z &=& \dfrac{32}{4} \\\\ \mathbf{ z } &=& \mathbf{8} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline AB &=& z+2 \quad | \quad z=8 \\ AB &=& 8+2 \\ \mathbf{ AB } &=& \mathbf{10} \\ \hline \end{array}\)

check:

\(\begin{array}{|rcll|} \hline (z+2)^2 &=& z^2+6^2 \\ (8+2)^2 &=& 8^2+6^2\\ 10^2 &=& 8^2 +6^2 \\ 100 &=& 64 + 36 \\ 100 &=& 100\ \checkmark \\ \hline \end{array}\)

Find the distance from the point \((1,2,3)\) to the line described by \(\begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}\).

\(\text{Let point $\vec{p} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} $} \\ \text{Let line $\vec{x} = \begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} $} \\ \text{Let direction vector $\vec{r} = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} $} \)

\(\begin{array}{|rcll|} \hline \left( \vec{x}-\vec{p} \right) \cdot \vec{r} &=& 0 \quad | \quad (\vec{x}-\vec{p}) \perp \vec{r} \\\\ \Bigg(\begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}-\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \Bigg) \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ \Bigg(\begin{pmatrix} 6-1 \\ 7-2 \\ 7-3 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \Bigg) \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ \Bigg(\begin{pmatrix} 5 \\ 5 \\ 4 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \Bigg) \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ \begin{pmatrix} 5+3t \\ 5+2t \\ 4-2t \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} &=& 0 \\\\ 3(5+3t)+2(5+2t)-2(4-2t)&=& 0 \\ 15+9t+10+4t-8+4t &=& 0 \\ 17+17t &=& 0 \\ 17t &=& -17 \quad | \quad : 17 \\ \mathbf{ t } &=& \mathbf{ -1 } \\ \hline \end{array}\)

\(\text{foot (of a perpendicular):}\)

\(\begin{array}{|rcll|} \hline \vec{x}_{f} &=& \begin{pmatrix} 6 \\ 7 \\ 7 \end{pmatrix} + (-1) \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \quad | \quad t=-1 \\\\ &=& \begin{pmatrix} 6-3 \\ 7-2 \\ 7+2 \end{pmatrix} \\\\ \mathbf{ \vec{x}_{f} } &=& \begin{pmatrix} \mathbf{3} \\ \mathbf{5} \\ \mathbf{9} \end{pmatrix} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \vec{PL} &=& \vec{x}_{f} - \vec{p} \\\\ &=& \begin{pmatrix} 3\\ 5 \\ 9\end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \\\\ &=& \begin{pmatrix} 3-1\\ 5-2 \\ 9-3\end{pmatrix} \\\\ \mathbf{\vec{PL}} &=& \begin{pmatrix} \mathbf{2}\\ \mathbf{3} \\ \mathbf{6}\end{pmatrix} \\ \hline \end{array}\)

\(\text{distance} = |\vec{PL}|\)

\(\begin{array}{|rcll|} \hline |\vec{PL}| &=& \sqrt{2^2+3^2+6^2} \\ &=& \sqrt{4+9+36} \\ &=& \sqrt{49} \\ \mathbf{|\vec{PL}|} &=& \mathbf{7} \\ \hline \end{array}\)

The distance from the point \((1,2,3)\) to the line is 7.