asinus

Find WX.

\(h_{yz}=\sqrt{r^2-2^2}\\ h_{uv}=\sqrt{r^2-3^2}\\ h_{yz}-h_{uv}=2\\ \sqrt{r^2-2^2}-\sqrt{r^2-3^2}=2\\ r=\frac{\sqrt{145}}{4}=3.0104\ (WolframAlpha)\)

\(h_{yz}=2.25\\ h_{uv}=0.2500\\ h_{wx}=(2.25+0.250)/2=1.25\\ \overline{WX}=2 \sqrt{r^2-h_{wx}^2}=2\sqrt{3.0104^2-1.25^2}\\ \color{blue}\overline{WX}=5.4772\)

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May 18 2024

https://web2.0calc.com/questions/geometry_18743#r1

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Find a + b.

\({\color{blue}m_{AB}=2}=\frac{y_b-y_a}{x_b-x_a}=\frac{b^2-a^2}{b-a}=\frac{(b+a)(b-a)}{b-a}\color{blue}=b+a\\ \color{blue}a+b=2\)

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\(\color{black }Compute\ the\ slope\ of\ line\ segment\ \overline{PQ}.\)

\(Let\ \overline{ OQ}=\overline{OP} =5\\ f_c(x)=+\sqrt{5^2-x^2}\\ f_{oq}(x)=5x\\ f_{op}(x)=4x\)

\(\sqrt{25-x^2}=5x\\ 25-x^2=25x^2\\ x_q= \sqrt{\frac{25}{26}}=0.9806\\ y_q=\sqrt{\frac{25}{26}}\cdot 5=4.903\\ \sqrt{25-x^2}=4x\\ x_p=\sqrt{\frac{25}{17}}=1.2127\\ y_p=\sqrt{\frac{25}{17}}\cdot 4=4.8507\)

\(m_{PQ}=\frac{y_q-y_p}{x_q-x_p}=\frac{4.9029-4.8507}{0.9806-1.2127}\\ \color{blue }m_{PQ}=-0.2249\\ \color{blue }The\ slope\ of\ line\ segment\ \overline{PQ}\ is\ -0.2249.\)

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The two circles below?

\(\overline{PQ}=\sqrt{(R+r)^2-(R-r)^2}=\sqrt{R^2+2Rr+r^2-R^2+2Rr-r^2}\\ \color{blue}\overline{PQ}=2 \sqrt{Rr}\)

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\(What\ is\ the\ coefficient\ of\ x^2\ ? \\ .\\ (x^3 + x^2 + x + 1)(x^4 + 3x^3 + 8x + 7x + 1)\\ =1 + 16 x + {\color{blue}16 x^2} + 19 x^3 + 19 x^4 + 4 x^5 + 4 x^6 + x^7\)

\(\color{blue }The\ coefficient\ of\ x^2\ is\ 16.\)

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\(Find\ \angle YXZ, in\ degrees.\)

If angle XWZ = 60°, angle WXY = 180° - 60° = 120° because XY and WZ are parallels.

Angle ZXW must be less than 120°.

The question is not solvable.

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Fill in the blanks.

\( x\in\{-5,5\}\\ (x+5)(x-5)=0\\ (x+5)(x-5)=\color{blue}x^2-25\)

x² - 25 = x² + ___x + __

x² - 25 = x² + 0x + (-25)

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Find the radius of the circle.

\((1+r)^2=(\dfrac{\overline{AC}}{2}-r)^2+1^2\\ (1+r)^2=(\dfrac{4}{2}-r)^2+1^2\\ 1+2r+r^2=4-4r+r^2+1\\ 6r=4\\ \color{blue}r=\dfrac{2}{3}\)

The radius of the circle is \(\color{blue}\frac{2}{3}\ .\)

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Find PM.

\(f_Q(x)=\sqrt{5^2-x^2}\\ f_R(x)=\sqrt{8^2-(x-11)^2}\\ 25-x_P^2=64-x_P^2+22x_P-121\\ x_P=\dfrac{121+25-64}{22}\\ x_P=3.\overline{72}\\ y_P=3.333\)

\(P(3.\overline{72},\ 3.333)\\ M(5.5,\ 0)\\ \overline{PM}=\sqrt{(x_M-x_P)^2+y_P^2}=\sqrt{(5.5-3.727)^2+3.333^2}\\ \color{blue}\overline{PM}=3.775\)

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What is UX?

\(\color{blue}S(0;0)\\\color{blue} T(13,0)\\ \color{blue}M(6.5, 0)\\f_S(x)=\pm \sqrt{14^2-x^2}\\ f_T(x)=\pm \sqrt{4^2-(x-13)^2}\\ x^2-14^2=(x-13)^2-4^2\\ x^2-14^2=x^2-26x+13^2-4^2\\ 26x=169-16+196\\ x_U=\dfrac{349}{26}\\ x_U=13.4\overline{230769}\\ y_U=3.9776\)

\(\color{blue}U(13.423,\ 3.978)\\ m_{TU}=\dfrac{y_U}{x_U-x_T}=\dfrac{3.9776}{13.42308-13}=9.4015\\ m_{NS}=-\dfrac{1}{m_{TU}}=-0.10637\)

\({\color{blue}f_{SN}(x)=}m_{NS}(x-x_S)+y_S=\color{blue}-0.10637x\)

\(m_{UM}=\dfrac{y_U}{x_U-x_M}=\dfrac{3.9776}{13.423-6.5}\\ m_{UM}=0.5387\)

\(f_{UM}(x)=m_{UM}(x-x_M)+y_M\\ f_{UM}(x)=0.5387x-0.5387\cdot 6.5\\ f_{UM}(x)=0.5387x-3.502\)

\(f_{UM}(x)=f_{SN}(x)\\ 0.5387x-3.5020=-0.10637x\\ x_X=\dfrac{3.5020}{0.5387+0.10637}\\ x_X=5.4285\\y_X=-0.57743\\ \color{blue}X(5.4285, -0.57743)\)

\(\overline{UX}=\sqrt{(y_U-y_X)^2+(x_U-x_X)^2}\\ \overline{UX}=\sqrt{(3.9776-(-0.57743))^2+(13.4231-5.4285)^2}\\ \color{blue}\overline{UX}=9.2012\)

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Find the coordinates of the center of the circle.

A(22,15)

B(-25,0)

C(22,-29)

\(M_{AB}(-1.5,7.5)\\ M_{AC}(22,-7)\\ f_{AB}=-\frac{1}{\frac{15}{47}}(x-(-1.5))+7.5\\ f_{AB}=-\frac{47}{15}(x+1.5)+7.5\\ f_{AB}=-\frac{47}{15}x-4.7+7.4\\ f_{AB}=-\frac{47}{15}x+2.7\)

\(f_{AC}=-7\\ f_{AB}=f_{AC}\\ -\frac{47}{15}x+2.7=-7\\ x=\frac{15(7+2.7)}{47}\\ x=3.0957\\ y=-7\\\)

The coordinates of the center of the circle are (3.0957, -7).

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\((1+r)^2=(\overline{AB}-r)^2+1^2\\(1+r)^2=(2-r)^2+1\\ 1+2r+r^2=4-4r+r^2+1\\ 6r=4\\\color{blue}r=\frac{2}{3}\)

\(\color{blue}The\ radius\ of\ the\ circle\ is\ \frac{2}{3}.\)

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