asinus

Find the area of the strip.

Let a be the distance of the base 10 touched by the strip.

Let alpha be the slope angle of the strip.

\(tan(\alpha)=\frac{8}{10-a}\\ sin(\alpha )=\frac{1}{a}\\ \alpha =atan(\frac{8}{10-a})=asin(\frac{1}{a})\\ atan(\frac{8}{10-a})-asin(\frac{1}{a})=0\)

\(\color{green}WolframAlpha\\ \color{green}a=\frac{8\sqrt{163}}{63}-\frac{10}{63}\\\)

\(\color{blue}a=1.4625\\ A=8\cdot 1.4625\\ \color{blue}A=11.7\)

\(\color{blue}The\ area\ of\ the\ strip\ is\ 11.7\ .\)

\(A\in \mathbb R\ |0< A< \infty\)

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What is the radius of the circle?

\(\dfrac{83^\circ \cdot \pi r^2}{360^\circ}=60 \pi\\ r=\sqrt{\frac{60\pi \cdot 360}{83\pi}}=16.132\\ \color{blue}The\ radius\ of\ the\ circle\ is\ 16.132.\)

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The length of AP is 4, and the circle has a radius of 5.  Find the length AB.

\(f_{circle}(x)=\sqrt{5^2-x^2}\\ \color{blue}A\ (0,5)\\ \color{blue}P\ (4,5)\\ f_{PB}(x)=-\sqrt{4^2-(x-4)^2}+5=-\sqrt{16-x^2+8x-16}+5\\ f_{PB}(x)=-\sqrt{8x-x^2}+5 \)

\(\sqrt{25-x^2}=-\sqrt{8x-x^2}+5\\ \sqrt{5-x}\sqrt{x+5}=5-\sqrt{8-x}\sqrt{x}\\ (WolframAlpha)\\ x_{B}\in \{0,\frac{200}{41}\}\\ y_b=\sqrt{25-4.\overline{87804}\ ^2}=1.0976\\ B\ (\frac{200}{41},\sqrt{20.\overline{12195}}) \\ \color{blue}B\ (4.\overline{87804},1.0976)\)

 \(\overline{AB}= \sqrt{(y_A-y_B)^2+(x_B-x_A)^2}=\sqrt{(5-1.0976)^2+(4.\overline{87804}-0)^2}\\ \color{blue}\overline{AB}=6.2469\)

\(\color{blue}The\ length\ \overline{AB}\ is\ 6.2469.\)  

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\(The\ length\ of\ \overline{AB}\ is\ 4,\ find\ the\ length\ AB.\\ What's\ this\ nonsense?\)

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 How many miles did I drive?

\(s=40mph\cdot t=55mph\cdot (t-\frac{25}{60}h)\\ 40mph\cdot t=55mph\cdot t-55mph\cdot \frac{25}{60}h\\ (55-40)mph\cdot t=55mph\cdot \frac{25}{60}h\\ t=\dfrac{55\cdot \frac{25}{60}h}{15}\\ t=1.52\overline 7h\)

\(s=40mph\cdot1.52\overline 7h\\ s=61\frac{1}{9}\ miles\\ \color{blue}61\frac{1}{9}\ miles\ dit\ i\ drive.\\ \)

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 Find the area of triangle AMN.

\(\overline{AB} = c=5, \overline{AC} = b=5\ and\ \overline{BC}=a = 8\)

\( A = s ( s − a ) ( s − b ) ( s − c )\ Heron's\ formula\\ s=\dfrac{a+b+c}{2}=\dfrac{8+5+5}{2}=9\\ A_{ABC} = 9 ( 9 − 8 ) ( 9 − 5) ( 9 − 5 )\\ A_{ABC}=144\\ Any\ line\ through\ the\ center\ of\ gravity\ of\ a\ triangle\ divides\ it\ into\ equal\ parts.\\A_{AMN=}\frac{1}{2}\cdot A_{ABC}=\frac{1}{2}\cdot 144\\ \color{blue}A_{AMN}=72\)

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 PQ = 8, QR = 5, and PR = 10.

\(f_{PR}(x)=0\\ 5^2=8^2+10^2-2\cdot 8\cdot 10\cdot cos\ P\\ 160\cdot cos\ \ P=64+100-25\\ cos\ P=\frac{64+100-25}{160}\\ P=29.686° \)

\(f_{PX}(x)=tan\ (\frac{29.686}{2})\cdot x\\ f_{PX}(x)=0.265x\\ sin\ R:sin\ P=8:5\\ sin\ \ R= \dfrac{8\cdot sin\ P }{5}=\dfrac{8\cdot sin\ 29.686° }{5}\\ R=52.4105°\\ f_{RQ}(x)=tan\ (-R)\cdot (x-10)\\ f_{RQ}(x)=tan\ (-52,4105°)\cdot (x-10)\\ f_{RQ}(x)=-1,3\cdot (x-10)\\\)

\(f_{PX}(x)=f_{PQ}(x)\\ 0.265x=-1.3(x-10)\\ 0.265x=-1.3x+13\\ x=\dfrac{13}{1.565}\\ \color{blue}x_X=8.3067\\ \color{blue}y_X=2.2013\)

\(f_{XY}(x)=m(x-x_X)+y_X\\ m=tan\ (90°-R)=tan\ (90°-52.4105°)\\ m=0.7698\\ f_{XY}(x)=0.7698(x-8.3067)+2.2013\\ f_{XY}(x)=0.7698x-6.3946+2.2013=0\\ \color{blue}x_Y=5.4472\\ \color{blue}y_Y=0\)

\(L_{XY}=\sqrt{y_X^2+(x_X-x_Y)^2}=\sqrt{2.2013^2+(8.3067-5.4472)^2}\\ \color{blue}L_{XY}=3.6087\)

\(\color{blue}The\ length\ of\ \overline{XY}\ is\ 3.6807.\)

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The question can be solved by specifying PR=10 instead of PR=1.

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Evaluate the infinite geometric series ​\(0.4 + 0.12 + 0.036 + 0.0108 + \dotsb\)

\(a_1=0.4\\ q=\dfrac{0.12}{0.40}=0.3\\ \dfrac{lim\ s_n}{n\to \infty}= \dfrac{a_1}{1-q}=\dfrac{0.4}{1-0.3}\\ \color{blue}\dfrac{lim\ s_n}{n\to \infty}=s= \dfrac{4}{7}=0.\overline{571428}\)

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How many miles did I drive?

Ich fuhr mit einer Geschwindigkeit von 40 Meilen pro Stunde zum Strand. Wenn ich stattdessen mit einer Geschwindigkeit von 55 Meilen pro Stunde gefahren wäre, wäre ich 25 Minuten früher angekommen. Wie viele Meilen bin ich gefahren?

\(40\ mph\cdot t=55\ mph(t-25\ min)\\ 40t=55t-1375\ min\\ 15t=1375\ min\\ t=91\frac{2}{3}\ min\)

\(s=v\cdot t=40\ mph\cdot 91\frac{2}{3}\ min\cdot \frac{h}{60\ min}\\ \color{blue}s=61.111\ miles\)

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\(Find\ the\ measure\ of\ \angle MON\ in\ degrees.\)

\(\angle CAO+\angle CBO=45°\ |\ peripheral\ \angle \ to\ 90°\ (\angle AOB)\\ \angle MON=90°-\angle CAO+90°-\angle CBO\\ =180°-(\angle CAO+\angle CBO)=180°-45°\\ \color{blue}\angle MON=135° \)

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Find AE.

\(\triangle DEF\ is\ equilateral. \\ DA = 3,\ AF = 2,\ EF=5\\ \angle AFE=60°\)

Pythagorean trigonometric theorem: \(a²=b²+c²-2bc\cdot cos\ \alpha \)

\(\overline{AE}\ ^2=\overline{AF}\ ^2+\overline{EF}\ ^2-cos\ (AFE)\\\overline{AE}\ ^2=2^2+5^2-2\cdot 5\cdot cos\ 60^°\\ \overline{AE}= \sqrt{29-2\cdot 5\cdot0.5}\\ \color{blue }\overline{AE}= \sqrt{24}=4.899\)

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Find the area of triangle ABC.

\(Point\ X\ is\ on\ side\ BC.\\ AX=13,\ BX=10,\ CX=10\\ \triangle XAB\ symmetrical\ \triangle XAC \\ \color{blue}A_{ABC}= \dfrac{1}{2}\cdot (10+10)\cdot 13=130\)

What is BX?

Angle B is 90°. Angle BXA is 90°.

So BX is the height of the Thales circle with radius 5.

\(\color{blue}\overline {BX}=5\)

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Find the length of the shortest altitude in this triangle.

\(a_a:a_b:a_c=\frac{1}{a}:\frac{1}{b}:\frac{1}{c}\\ a_a:a_b:a_c=\frac{1}{10}:\frac{1}{12}:\frac{1}{15}\\\)

\({\color{blue}The\ shortest\ altitude}\ in\ this\ triangle\ \color{blue}is\ a_c.\\ \overline{AC}=12\ ,\ \overline{BC}=10\ ,\ \overline{AB}=15\)

            C

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             \(a_c\)                      Triangle ABC

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A     x    .     15-x    B

\( a_c\ ^2=12^2-x^2=10^2-(15-x)^2\\ 144-x^2=100-(225-30x+x^2)\\ 144-100+225=30x\\ x=8.9\overline 6\)

\(a_c\ ^2=12^2-8.9\overline 6\ ^2\\ a_c=\sqrt{12^2-8.9\overline 6\ ^2}\\ \color{blue}a_c=7.97489\)

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Write the equation of this line in the form Ax + By = C

\(m=\dfrac{y_p-y_q}{x_p-x_q}=\dfrac{2-7}{-8-4}=\dfrac{5}{12}\\ y=m(x-x_q)+y_q\\ y=\frac{5}{12}(x-4)-7\\ y=\frac{5}{12}x-\frac{5}{3}-7\\ -\frac{5}{12}x+y=-\frac{5}{3}-7\\ \dfrac{-5x+12y}{12}=\dfrac{-20-84}{12}\\ \dfrac{5x-12y}{12}=\dfrac{20+84}{12}\)

\(Ax+By=C\\ \color{blue}5x-12y=104\)

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A(-2,-4), B(-3,12), C(4,y) find y.

\(m= \dfrac{y_B-y_A}{x_B-x_A}=\dfrac{12-(-4)}{-3-(-2)}=-16\\ y=m(x-x_A)+y_A=-16(x-(-2))+(-4)=-16x-32-4\\ y=-16x-36\\ y_C=-16x_C-36=-16\cdot 4-36\\ \color{blue}y_C=-100\)

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 \(Compute\ a^3b + ab^3\)

\(2x^2 - 8x + 7 = x^2 - 4x + 13\\ x^2-4x-6=0\\ x=2\pm\sqrt{4+6}\\ x\in \{2-\sqrt{10}\ ,\ 2+\sqrt{10}\}=\{a,b\}\)

\(a^3b + ab^3\\ =(2-\sqrt{10})^3(2+\sqrt{10})+(2-\sqrt{10})(2+\sqrt{10})^3\\ \color{blue}=-168\)