asinus

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

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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