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𝐑𝐢𝐠𝐡𝐭 𝐬𝐜𝐚𝐥𝐞𝐧𝐞 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞.

𝐔𝐬𝐞 𝐭𝐡𝐞 𝐋𝐚𝐰 𝐨𝐟 𝐒𝐢𝐧𝐞𝐬 𝐨𝐫 𝐭𝐫𝐢𝐠 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬 𝐭𝐨 𝐟𝐢𝐧𝐝 𝐭𝐡𝐞 𝐦𝐢𝐬𝐬𝐢𝐧𝐠 𝐬𝐢𝐝𝐞𝐬:

𝐒𝐢𝐝𝐞𝐬: 𝐚 = 𝟕 𝐛 = 𝟔.𝟓𝟕𝟖 𝐜 = 𝟐.𝟑𝟗𝟒

𝐀𝐫𝐞𝐚: 𝐓 = 𝟕.𝟖𝟕𝟒
𝐏𝐞𝐫𝐢𝐦𝐞𝐭𝐞𝐫: 𝐩 = 𝟏𝟓.𝟗𝟕𝟐
𝐒𝐞𝐦𝐢𝐩𝐞𝐫𝐢𝐦𝐞𝐭𝐞𝐫: 𝐬 = 𝟕.𝟗𝟖𝟔

𝐀𝐧𝐠𝐥𝐞 ∠ 𝐀 = α = 𝟗𝟎° = 𝟏.𝟓𝟕𝟏 𝐫𝐚𝐝
𝐀𝐧𝐠𝐥𝐞 ∠ 𝐁 = β = 𝟕𝟎° = 𝟏.𝟐𝟐𝟐 𝐫𝐚𝐝
𝐀𝐧𝐠𝐥𝐞 ∠ 𝐂 = γ = 𝟐𝟎° = 𝟎.𝟑𝟒𝟗 𝐫𝐚𝐝

That is not quite right because the cards are not replaced.

Yes it would be 4 and that is the period.

Would that be 4?

One point is (0,0), And from that point the curve goes up.

At what next point does this happen?

I mean what is the x value when is the y value next at 0 with the curve on its way up,

I think it's either 2 or 3?

How far does it go in the x direction before it starts repeating?  That is the period.

Tell me what you think it is and i. or someone else, will tell you if you are right.

𝐉𝐮𝐬𝐭 𝐦𝐮𝐥𝐭𝐢𝐩𝐥𝐲:

𝟏.𝟓 𝐱 𝟒 =𝟔 𝐦𝐢𝐥𝐞𝐬

𝐘𝐨𝐮 𝐜𝐚𝐧 𝐠𝐞𝐧𝐞𝐫𝐚𝐭𝐞 𝐭𝐡𝐞 𝐧𝐭𝐡 𝐭𝐞𝐫𝐦 𝐚𝐬 𝐟𝐨𝐥𝐥𝐨𝐰𝐬:

𝐚(𝐧) = 𝐧^𝟐 + 𝟒

Integration

Suppose that \(f'(x) = \dfrac{4}{\sqrt{1-x^2}}\) and that \(f(0) = 2\).
Find \(f\left(\dfrac{\sqrt{3}}{2}\right)\).

\(\begin{array}{|rcll|} \hline && \mathbf{\large{ \int } \dfrac{4}{\sqrt{1-x^2}} \ dx} \\\\ &=& 4\mathbf{\large{ \int } \dfrac{1}{\sqrt{1-x^2}} \ dx} \qquad \boxed{ \text{substitute: } \\ x = \sin(\theta),\ dx=\cos(\theta)d\theta } \\\\ &=& 4 \large{ \int } \dfrac{\cos(\theta)}{ \sqrt{1-\Big(\sin(\theta)\Big)^2} }\ d\theta \\\\ &=& 4 \displaystyle \int \dfrac{\cos(\theta)}{ \cos(\theta) }\ d\theta \\\\ &=& 4 \displaystyle \int d\theta \\ &=& 4 \theta + c \quad | \quad \theta = \arcsin(x) \\ &=& 4 \Big(\arcsin(x)\Big) + c \\ \hline f(x) &=& 4 \Big(\arcsin(x)\Big) + c \quad | \quad x=0 \\ f(0) &=& 4 \Big(\arcsin(0)\Big) + c \quad | \quad f(0) = 2 \\ 2 &=& 4 \Big(\arcsin(0)\Big) + c \\ 2 &=& 4 \cdot 0 + c \\ 2 &=& 0 + c \\ \mathbf{c} &=& \mathbf{2} \\ \hline \mathbf{f(x)} &=& \mathbf{4 \Big(\arcsin(x)\Big) + 2} \\\\ f\left(\dfrac{\sqrt{3}}{2}\right) &=& 4 \Big(\arcsin\left(\dfrac{\sqrt{3}}{2}\right)\Big) + 2 \\\\ f\left(\dfrac{\sqrt{3}}{2}\right) &=& 4\cdot \Big(\dfrac{\pi}{3}\Big) + 2 \\\\ f\left(\dfrac{\sqrt{3}}{2}\right) &=& \dfrac{4\pi}{3} + 2 \\\\ \mathbf{f\left(\dfrac{\sqrt{3}}{2}\right)} &=& \mathbf{6.18879020479} \\ \hline \end{array}\)

In trapezoid ABCD, \(\overline{AB}\parallel \overline{CD}\), and \(AB < CD\).
The base \(\overline{CD}\) has a length of \(8\).
The legs \(\overline{AD}\) and \(\overline{BC}\) have lengths of \(7\), and the diagonal \overline{BD} has a length of \(9\).
Find the area of the trapezoid.

\(\text{Let $\overline{CE}=\dfrac{8-\color{red}x}{2}$} \\ \text{Let $\overline{DE}=8-\dfrac{8-\color{red}x}{2}$} \\ \text{Let $\overline{BE}=h$} \\ \text{Let $\overline{AB}=\color{red}x$} \\ \text{Let area of the trapezoid $=A $ } \)

Pythagorean Theorem:

\(\begin{array}{|rcll|} \hline \mathbf{\left(\dfrac{8-x}{2}\right)^2 + h^2} &=& \mathbf{7^2} \\\\ h^2 &=& 7^2 - \left(\dfrac{8-x}{2}\right)^2 \\\\ \mathbf{h^2} &=& \mathbf{7^2 - \dfrac{\Big(8-x\Big)^2}{4}} \qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\Big(8-\left(\dfrac{8-x}{2}\right)\Big)^2 + h^2} &=& \mathbf{9^2} \\\\ \left(\dfrac{16-(8-x)}{2}\right)^2 + h^2 &=& 9^2 \\\\ \left(\dfrac{8+x}{2}\right)^2 + h^2 &=& 9^2 \\\\ \dfrac{\Big(8+x\Big)^2}{4} + h^2 &=& 9^2 \\\\ \mathbf{h^2} &=& \mathbf{9^2 - \dfrac{\Big(8+x\Big)^2}{4}} \qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)=(2): & h^2 = 7^2 - \dfrac{\Big(8-x\Big)^2}{4} &=& 9^2 - \dfrac{\Big(8+x\Big)^2}{4} \\\\ & 7^2 - \dfrac{\Big(8-x\Big)^2}{4} &=& 9^2 - \dfrac{\Big(8+x\Big)^2}{4} \\\\ & \dfrac{\Big(8+x\Big)^2}{4} - \dfrac{\Big(8-x\Big)^2}{4} &=& 9^2-7^2 \\\\ & \dfrac{\Big(8+x\Big)^2-\Big(8-x\Big)^2}{4} &=& 32 \quad | \quad \cdot 4 \\\\ & \Big(8+x\Big)^2-\Big(8-x\Big)^2 &=& 32\cdot 4 \\\\ & 64+16x+x^2-(64-16x+x^2) &=& 32\cdot 4 \\ & 64+16x+x^2-64+16x-x^2 &=& 32\cdot 4 \\ & 16x +16x &=& 32\cdot 4 \\ & 32x &=& 32\cdot 4 \quad | \quad :32 \\ & \mathbf{ x } &=& \mathbf{4} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{h^2} &=& \mathbf{7^2 - \dfrac{\Big(8-x\Big)^2}{4}} \quad | \quad \mathbf{x=4} \\\\ & h^2 &=& 7^2 - \dfrac{\Big(8-4\Big)^2}{4} \\\\ & h^2 &=& 7^2 - \dfrac{4^2}{4} \\\\ & h^2 &=& 7^2 - 4 \\ & h^2 &=& 45 \\ & h &=& \sqrt{45} \\ & \mathbf{h} &=& \mathbf{3\sqrt{5}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{A} &=& \mathbf{\left(\dfrac{8+x}{2}\right)\cdot h} \quad | \quad \mathbf{x=4},\ \mathbf{h=3\sqrt{5}} \\\\ A &=& \left(\dfrac{8+4}{2}\right)\cdot 3\sqrt{5} \\\\ A &=& \left(\dfrac{12}{2}\right)\cdot 3\sqrt{5} \\\\ A &=& 6\cdot 3\sqrt{5} \\\\ \mathbf{A} &=& \mathbf{18\sqrt{5}} \\\\ \mathbf{A} &=& \mathbf{40.2492235950} \\ \hline \end{array} \)

Find the area of the trapezoid.

In trapezoid \(PQRS\), \(\overline{PQ} \parallel \overline{RS}\).
Let \(X\) be the intersection of diagonals \(\overline{PR}\) and \(\overline{QS}\).
The area of triangle \(PQX\) is \(20\), and the area of triangle \(RSX\) is \(45\).
Find the area of trapezoid \(PQRS\).

\(\begin{array}{|rcll|} \hline \text{Let Area }A_1 &=& [PQX] \\ A_1 &=& 20 \\\\ \text{Let Area }A_2 &=& [RSX] \\ A_2 &=& 45 \\\\ \text{Let Area }A_3 &=& [PXS] \\\\ \text{Let Area }A_4 &=& [QXR] \\\\ \text{Let Area }A_5 &=& [PSR] \\ &=& A_2+A_3 \\\\ \text{Let Area }A_6 &=& [QSR] \\ A_6 &=& A_2 +A_4 \\ \hline \mathbf{A_5} &=& \mathbf{A_6} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{A_5} &=& \mathbf{A_6} \\ A_2+A_3 &=& A_2 +A_4 \\ \mathbf{A_3 }&=& \mathbf{A_4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline A_3 = \dfrac{SX*h_1}{2} && A_1 = \dfrac{QX*h_1}{2} \\ \mathbf{\dfrac{A_3}{A_1} }&=& \mathbf{\dfrac{SX}{QX}} \\\\ A_2 = \dfrac{SX*h_2}{2} && A_4 = \dfrac{QX*h_2}{2} \\ \mathbf{\dfrac{A_2}{A_4} }&=& \mathbf{\dfrac{SX}{QX}} \\\\ \dfrac{SX}{QX}=\mathbf{\dfrac{A_3}{A_1} }&=&\mathbf{\dfrac{A_2}{A_4} } \quad | \quad A_4 = A_3 \\\\ \dfrac{A_3}{A_1} &=& \dfrac{A_2}{A_3} \\\\ A_3^2 &=& A_1A_2 \\ A_3^2 &=& 20*45 \\ A_3^2 &=& 900 \\ \mathbf{A_3} &=& \mathbf{30} \\\\ A_4 &=& A_3 \\ \mathbf{A_4} &=& \mathbf{30} \\ \hline \end{array}\)

\(\text{The area of trapezoid $PQRS = A_1+A_2+A_3+A_4 = 20+45+30+30=\mathbf{125}$ }\)

Yes well this time you forgot the question.

Just as well your head is well screwed on.  

Simplify C(n,k)/C(n,k-1)

\(\begin{array}{|rcll|} \hline \dfrac{C(n,k)} {C(n,k-1)} &=& \dfrac{ \dbinom{n}{k} }{ \dbinom{n}{k - 1} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k - 1}} &=& \dfrac{n!}{(k-1)!(n-k+1)!} \quad &| \quad (k-1)!k=k!~ \text{or}~(k-1)!= \dfrac{k!}{k} \\\\ &=& k *\dfrac{n!}{k!(n-k+1)!} \quad &| \quad (n-k)!(n-k+1)=(n-k+1)! \\\\ &=& \dfrac{k}{n-k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{k}{n-k+1} \dbinom{n}{k} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{ \dbinom{n}{k} }{ \dbinom{n}{k - 1} } &=& \dfrac{ \dbinom{n}{k} }{ \dfrac{k}{n-k+1} \dbinom{n}{k} } \\\\ \dfrac{ \dbinom{n}{k} }{ \dbinom{n}{k - 1} } &=& \dfrac{1}{ \dfrac{k}{n-k+1} } \\\\ \mathbf{\dfrac{ \dbinom{n}{k} }{ \dbinom{n}{k - 1} } } &=& \mathbf{\dfrac{n-k+1}{k}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{C(n,k)} {C(n,k-1)}} &=& \mathbf{\dfrac{n-k+1}{k}} \\ \hline \end{array}\)