asinus

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

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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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\(f(x) = 2x^2 - 13x + 20 - 5x^2 + 19x + 7.\\ f(x)=-3x^2+3x+27\\ x^2-x-9=0\\ x_0\in \{\frac{1}{2}\pm \sqrt{(\frac{1}{2})^2+9}\}\\ x_0\in \{\frac{1}{2}+\sqrt{9.25},\frac{1}{2}-\sqrt{9.25}\}\\ \color{blue}x_0\in \{3.541, -2.541\}\)

The x-coordinate(s) of all point(s) where the parabola y = f(x) intersects the line y = 0

are \(\color{blue}x_0\in \{3.541, -2.541\}.\)

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Find the length of the shortest altitude in this triangle.

general circle equation

\((x – x_P)² + (y – y_P)² = r²\)

\(y^2=10^2-x^2\\ y^2=9^2-(15-x)^2\\ y^2=81-225+30x-x^2\\ 100-x^2=-144+30x-x^2\\30x=244\\ x_C=8.1\overline{33}\\ \color{blue}y_C=h_C=5.818\)

\(y^2=15^2-x^2\\ y^2=10^2-(9-x)^2\\ y^2=100-x^2+18x-81\\ 225-x^2=19+18x-x^2\\ 18x=206\\ x_A=11.\overline{44}\\ y_A=h_A=9.697\)

\(y^2=9^2-x^2\\ y^2=15^2-(10-x)^2\\ y^2=225-100+20x-x^2\\ 81-x^2=125+20x-x^2\\20x=-44\\ x_B=-2.2\\ y_B=h_B=8.726\)

\(\color{blue}The\ length\ of\ the\ shortest\ altitude\ in\ this\ triangle\ is\ h_C=5.818\)

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\(An\ acute\ angle\ is\ an\ angle\ whose\ sides\ are\ within\ >\ 0^{\circ}\ and\ <\ 90^{\circ}\ \\ or\ whose\ circular\ arc\ are\ within\ >\ 0\ and\ <\ \dfrac{\pi}{2}.\\ \cos \theta = \frac{3 \sqrt{5}}{7}\\ sin \theta=\sqrt{1-cos^2}=\sqrt{1-\dfrac{9\cdot 5}{49}}=\dfrac{1}{7}\sqrt{49-45}=\dfrac{2}{7}\\ \color{blue}sin \theta=\dfrac{2}{7}\)

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If I cut the wire in half and pass $175$ volts through it, how many milliamps of current will I measure?

\(R_{wire}=\dfrac{U}{I}=\dfrac{500V}{25mA}\ \ (Ohm's\ law)\\ R_{\frac{1}{2}wire}=\dfrac{1}{2}\cdot \dfrac{500V}{25mA}\\ \dfrac {I\cdot R}{U}=1\ \to \ \dfrac{A\cdot \Omega}{V}=1\\ R_{2( \frac{1}{2}wire)}=\dfrac{1}{2}\cdot \dfrac{1}{2} \cdot \dfrac{500V}{25mA}\cdot\dfrac{1000mA}{A} \cdot \dfrac{A\cdot \Omega}{V}=5000\Omega\)

\(I=\dfrac{U}{R_{2( \frac{1}{2}wire)}} =\dfrac{175V}{5000\Omega}\cdot \dfrac{1000mA}{A}\cdot \dfrac{A\cdot \Omega }{V}=\color{ blue}35mA \)

35 milliamps of current will I measure.

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How many miles is the distance from Penthaven to Jackson?

\(v_1=\dfrac{s}{t_1}=\dfrac{s}{108min}\\ t_2=\dfrac{s}{v_1+\dfrac{1mil}{2min}}+\dfrac{s}{v_1-\dfrac{1mil}{2min}}\\ 225min=\dfrac{s}{\dfrac{s}{108min}+\dfrac{1mil}{2min}}+\dfrac{s}{\dfrac{s}{108min}-\dfrac{1mil}{2min}}\\ 225min=\dfrac{s}{\dfrac{s+54mil}{108min}}+\dfrac{s}{\dfrac{s-54mil}{108min}}\)

\(225min=\dfrac{s\cdot 108min}{s+54mil}+\dfrac{s\cdot 108min}{s-54mil}\\ 225min(s^2-54^2mil^2)=s\cdot 108min(s-54mil+s+54mil)\\ s^2-54^2mil^2=\dfrac{216s^2}{225}\\0.04s^2=54^2mil^2\\ 0.2s=54mil\\ \color{blue}s=270\ mil\)

The distance from Penthaven to Jackson is 270 miles.

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Find XY .     What is this nonsense?    

\(\color{blue}\overline {XY}=8 \)

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Find the area of triangle BXA.

\(A_{BXA}=\frac{\overline {AB}\cdot h}{2}\\ \overline {AB}=\sqrt{24^2-12^2}\\ \angle BXC=(180-60-45)^{\circ}=75^{\circ}\\ \overline{BX}=\frac{12\cdot sin 60^\circ}{sin\ \angle BXC}=\frac{12\cdot sin60^{\circ}}{sin75^\circ}\)

\(h=\frac{\overline {BX}\cdot sin45^\circ}{sin90^{\circ}} =\frac{12\cdot sin60^{\circ}}{sin75^\circ}\cdot \frac{sin45^\circ}{sin90^\circ}\\ A_{BXA}=\frac{\overline {AB}\cdot h}{2}\\ A_{BXA}=\frac{1}{2}\cdot \sqrt{24^2-12^2}\cdot \frac{12\cdot sin60^{\circ}}{sin75^\circ} \cdot \frac{sin45^\circ}{sin90^\circ}\\ \color{blue}A_{BXA}=79.061 \)

The area of triangle BXA is 79.061 .

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What will the value of the car be, to the nearest cent, after 15 years?

\(100\%-8.25\%=91.75\%=0.9175\)

\({\color{blue}v_{(after\ 15\ years)}}=25000\ USD\cdot 0.9175^{15}\color{blue}=\underline6871.18\ USD\)

Thanks T.Phill

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How many milliamps of current will I measure?

\(R=\frac{U}{I}=\frac{500V}{0.025A}\\ R=20k\Omega\)

The parallel-connected wire halves offer a quarter of the resistance of the original wire.

\(I=\frac{U}{R}=\frac{175V}{5k\Omega}=0,035A\\ \color{blue}I=35mA\)

We measure 35mA.

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\(f(x)=5x^2 - 2x + 8 - 2x^2 + 6x + 2\\ f(x)=3x^2+4x+10=0\\ \)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ b^2-4ac=16-4\cdot 3\cdot 10=-104\\ (b^2-4ac)<0\)

The discriminant is less than zero. The function has no zero, no solution.

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