asinus

 If BD = 3 and CD = 5, what is AC?

I

$mx=\sqrt{8^2-(x-6)^2}$

II

$\frac{m}{2}\cdot x=\sqrt{8^2-(x-6)^2}\\ mx=2\cdot \sqrt{8^2-(x-6)^2}$

III

$\sqrt{8^2-(x-6)^2}=2\cdot \sqrt{8^2-(x-6)^2}\\ 64-x^2+12x-36=4\cdot(64-x^2+12x-36)\\ 64-x^2+12x-36=256-4x^2+48x-144\\ 3x^2+60x-84=0\\ x^2 +20x-28=0\\ x_{1,2}=-10\pm \sqrt{100+28}\\ x_C=1.3137\\ C\ (1.3137,6.4837)$

$A\ (0,0)\\ \overline {AC}=\sqrt{1.3137^2+6.4837^2}$

$\color{blue} If\ BD = 3\ and\ CD = 5, is\ AC=6.6154$

will be continued

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What is the probability that BD < 4?

$\color{blue}P=\dfrac{A_{DBC}}{A_{ABD}}$

$f_{AC}(x)=\sqrt{5^2-x^2}\\ f_{BC}(x)=\sqrt{3^2-(x-7)^2}\\ 25-x^2=9-x^2+14x-49\\ 14x=65\\ x_C=4.6429 \\ y_C=1.8558\\ C(4.6429,1.8558)\\ A(0,0)\\ B(7.0)$

$f_{AD}=\frac{1.8558}{4.6429}x\\ f_{BD}=\sqrt{4^2-(x-7)^2}\\ \frac{1.8558}{4.6429}x=4^2-(x-7)^2\\ \frac{1.8558}{4.6429}x=16-x^2+14x-49$

$x^2-13.6003x+33=0\\ x=6.8001\pm \sqrt{46.2420-33}\\ x_D=3.1661\\ y_D=1.2635\\ D(3.1661,1.2635)$

$A_{ABC}=\frac{7}{2}\cdot 1.8558\\ A_{ABC}=6.4953\\ A_{ABD}=\frac{7}{2}\cdot 1.2635\\ \color{blue}A_{ABD}=4.4224\\ A_{DBC}=A_{ABC}-A_{ABD}=6.4953-4.4224\\ \color{blue}A_{DBC}=2,0729$

$P=\dfrac{A_{DBC}}{A_{ABD}}=\frac{2.0729}{4.4224}:\frac{2.0729}{2.0729}=1:2.1334$

The probability that BD < 4 is  $\color{blue }P=1:2.1334$

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What is distance from A to B along the (minor) arc of a great circle?

$s=2\\ r=2\\ b=2\cdot \frac{\alpha}{2}\cdot \frac{2\pi r}{360^{\circ}} \\sin\frac{\alpha}{2}=\frac{1}{2}\\ b=2\cdot arcsin \frac{1}{2}\cdot\frac{2\pi \cdot 2}{360^{\circ}}\\ \color{blue}b=2.0944$

$1.11111 + 0.11111 + 0.01111 + 0.00111 +0.00011 + 0.00001\color{blue} = 1.23456$

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What is b?

$b=\sqrt{(12-7)^2+(5+2)^2}\cdot cos(30^{\circ }+arctan\frac{5}{5+2})+5\\ \color{blue}b=8.562$

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Where is the picture?

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The equation of the reflection of the line y = 2x + 1 across the line y = 5x - 3.

Intersection:

$f_1(x)=f_2(x)\\ 2x+1=5x-3\\ 3x=4\\ x_{1,2}=\frac{4}{3}\\ y_{1,2}=\frac{11}{3}=3\frac{2}{3}\\ P_{1,2}(1\frac{1}{3},3\frac{2}{3})$

gradient

$m_{1,2}=tan\ [\frac{1}{2}(\arctan 2+\arctan 5)]\\ m_{1,2}=2.9145$

Point direction equation

$y=m(x-x_P)+y_P \\ y=2.914536(x-1.\overline {33})+3.\overline{66}\\ y=2.914536x-3.886053+3.666667$

The equation of the reflection is

$\color{blue}f_{1,2}(x)= y=2.914536x-0.219381$

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If M is the midpoint of BC, then find AM^2.

$sin\ ( 22.2^{\circ})=\frac{\overline {BM}}{4}\\ \overline {BM}=4\ sin\ ( 22.5^{\circ})\\ \overline {AM}^2 =4^2-\overline {BM}^2=16- 16\ sin^2\ (22.5^{\circ })=16\ (1-sin^2\ (22.5^{\circ }))\\ \color{blue}\overline {AM}^2=13.657$

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What does that mean?

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 If CE = 3 and DF = 3, then what is BD?

Im Dreieck ABC liegen die Punkte D und F auf {AB} und E auf {AC}, sodass {DE}\parallel {BC} und {EF} parallel {CD} sind. Wenn CE = 3 und DF = 3, was ist dann BD?

Let AF = AE = c and BD = x. 

Then, according to the first ray theorem:

$\dfrac{s+3}{s}=\dfrac{s+3+x}{s+3}\\ s^2+6x+9=s^2+3s+sx\\ sx=3s+9\\ x=3+\dfrac{9}{s}$

To solve the question, it is necessary to specify an additional value, for example AE.

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