asinus

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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$If\ BD = 8,\ BC = 3,\ CE = 1,\ AC = 19\ and\ AB = 13,\ then\ what\ is\ DE?$

$f_{AB}(x)=\sqrt{13^2-x^2}\\ f_{BC}(x)=\sqrt{3^2-(x-19)^2}\\ 169-x^2=9-x^2+38x-361\\ 38x=169-9+361=521\\ x_B=13.7105\\ y_B=\color{red }\sqrt{-18.9...}$

The triangle ABC, with sides 3, 19, 13 is not possible.

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$Compute\ \frac{\sqrt{3} - 4 \sqrt{5}}{(\sqrt{3})^2 + (\sqrt{2})^2}\\ \frac{\sqrt{3} - 4 \sqrt{5}}{(\sqrt{3})^2 + (\sqrt{2})^2}\\ =\frac{\sqrt{3} - 4 \sqrt{5}}{(\sqrt{3})^2 +\ (\sqrt{2})^2}\cdot \frac{(\sqrt{3})^2 -\ (\sqrt{2})^2}{(\sqrt{3})^2 -\ (\sqrt{2})^2}\\ =\frac{3\sqrt{3}-2\sqrt{3}-12\sqrt{5}+8\sqrt{5}}{9-4}\\ \color{blue}=\frac{\sqrt{3}-4\sqrt{5}}{5}$

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$This\ gives:\\ (2 \cdot \frac{x - 5}{x - 3} )\ {\color{red}>}\ (\frac{2x - 5}{x + 2} + 20)$

$x\in \mathbb R$(-2 < x < -1.5292)

and

$x\in \mathbb R$(2.7792 < x < 3)

Sorry. It didn't display correctly in answer 1#.

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Find all real x for

        I                II

$\color{blue}2 \cdot \frac{x - 5}{x - 3} > \frac{2x - 5}{x + 2} + 20\\ .\\ f(x)=2 \cdot \frac{x - 5}{x - 3} - \frac{2x - 5}{x + 2} -20>0\\ 2(x-5)(x+2)-(2x-5)(x-3)-20(x-3)(x+2)>0\\ (2x-10)(x+2)-(2x^2-6x-5x+15)-20(x^2+2x-3x-6)>0$

$2x^2+4x-10x-20-2x^2+11x-15-20x^2-40x+60x+120>0\\ f(x)=-20x^2+25x+85>0$

$x_{1,2} = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x_{1,2} = {-25 \pm \sqrt{25^2+4\cdot 20\cdot 85} \over -40}\\ x_0\in \{-1.5292,2.7792\}\\$

$The\ extreme\ points\ of\ the\ hyperbolas\ I\ and\ II\ are\ x_I=3\ and\ x_{II}=-2.$

$In\ the\ calculated\ range\ is\\ (2 \cdot \frac{x - 5}{x - 3} ){\color{red}\ <}\ (\frac{2x - 5}{x + 2} + 20)\\ x\in \mathbb R\ (-1.52921099245< x< 2.77921099245)\\$

$The\ extreme\ points\ of\ the\ hyperbolas\ I\ and\ II\ are\ x_I=3\ and\ x_{II}=-2.\\ \color{blue }This\ gives:\\ (2 \cdot \frac{x - 5}{x - 3} )\ {\color{red}>}\ (\frac{2x - 5}{x + 2} + 20)\\ \color{blue}x\in \mathbb R(-2 <-1.5292)\\ and\\ \color{blue}x=\mathbb R(2.7792 <3)$

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The domain of the function f(x) = sqrt(5 - 8x + x^2).

$x^2-8x+5=0\\ x\in\{(4-\sqrt{16-5}),(4+\sqrt{16-5})\}\\ x\in\{(4-\sqrt{11}),(4+\sqrt{11})\}\\ f(x)=\sqrt{(x-4+\sqrt{11})(x-4-\sqrt{11})}=0\\ x_0\in \{(4-\sqrt{11}\ ),(4+\sqrt{11})\}$

$D1_{f(x)}=\mathbb R(-inf \le x<[4-\sqrt{11}])\\ D2_{f(x)}=\mathbb R([4+\sqrt{11}] \le x< inf)$

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Express \log_{15} 72 in terms of P and Q.

$P = \log_8 3\\ 8^{log_8\ 3}=3\\ 8^P=3\\ \color{blue}8=3^{\frac{1}{P}}$

$Q = \log_3 5\\ \color{blue}3=5^{\frac{1}{Q}}$

$x=\log_{15} 72\\ \color{blue}15=72^{\frac{1}{x}}$

will be continued

I can't go any further. Sorry!

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Find the radius of the semicircle.

Let the side length of the squares be s.

Case A) Squares next to each other.

Then the radius of the semicircle is $\color{blue}r=s\sqrt{2}.$

Case B) Squares on top of each other.

Then the radius of the semicircle is:

$r^2=(2s)^2+(\frac{s}{2})^2\\ r=\sqrt{\frac{4\cdot4s^2+s^2}{4}}\\ \color{blue}r=\frac{s}{2}\sqrt{5}$

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What is a scale factor -1 ?

As far as I know, a scaling factor is a real number > 0.

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$x^3 - 4x^2 - 7x + 12=0\\ The\ equation\ is\ solved\ with\ \color{blue}WolframAlpha.\\ \color{blue}r=-2.0912\\ \color{blue}s=1.1648\\ \color{blue}t=4.9265$

$\dfrac{rs}{t^2} + \dfrac{rt}{s^2} + \dfrac{st}{r^2}\\ =\dfrac{-2.0912\cdot 1.1648}{4.9265^2} + \dfrac{-2.0912\cdot 4.9265}{1.1648^2} + \dfrac{1.1648\cdot 4.9265}{(-2.0912)^2}\\ \color{blue}=-6.38147970131$

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$What\ is\ the\ equation\ of\ the\ line\ of\ symmetry\ of\ the\ parabola\ y = f(x)?$

$f(x) = 2x^2 - 13x + 20 - 5x^2 + 19x + 7\\ f(x)= -3x^2+6x+27\\ x^2-2x-9=0\\ x_{1,2}=1\pm\sqrt{1^2+9}\\ x\in \{(1-\sqrt{10}),(1+\sqrt{10})\}\\ \dfrac{1-\sqrt{10}+1+\sqrt{10}}{2}=1$

${\color{blue}The\ equation\ of\ the\ line\ of\ symmetry}\ of\ the\ parabola\ y=f(x)\ \color{blue}is\ x=1$

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