asinus

What are the coordinates of the points where the graphs  of f(x)=x^3 + x^2 - 3x + 5 and g(x) = x^3 + 2x^2 intersect?

Hello ABJelly!

$x^3 + x^2 - 3x + 5 = x^3 + 2x^2\\ x^2+3x-5=0\\ x_{f,g}=-1.5\pm\sqrt{2.25+5}\\ x_{f,g}=\frac{1}{2}(3\pm\sqrt{29})\\ x_{f,g}\in \{ -4.19258,\ 1.19258\}$

$y_g\in \{-38.540659,4.540659\}\\ \color{blue}So\ ( -4.19258,-38.540659),(1.19258,4.540659)$

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$\frac{1}{1 \cdot 4} + \frac{1}{7 \cdot 10} + \frac{1}{37 + 40}\\ =\frac{1}{1 \cdot 2^2} + \frac{1}{7 \cdot 2\cdot 5} + \frac{1}{7\cdot 11}\\ =\frac{5\cdot7\cdot 11+2\cdot 11+2^2\cdot 5}{2^2\cdot 5\cdot7\cdot 11}\\ =\frac{427}{2^2\cdot 5\cdot7\cdot 11}\\ =\frac{7\cdot 61}{2^2\cdot 5\cdot7\cdot 11}\\ \color{blue}=\frac{61}{2^2\cdot 5\cdot 11}$

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$1 + \dfrac{3}{6} + \dfrac{15}{6^2} + \dfrac{27}{6^3}\\ =\dfrac{6^3+3\cdot 6^2+15\cdot 6+27}{6^3}\\ =\dfrac{441}{6^3}\\ =\dfrac{3^2\cdot 7^2}{2^3\cdot 3^3}\\ \color{blue}=\dfrac{7^2}{2^3\cdot3}$

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Find the measure of angle MON, in degrees.

$\alpha ,\beta ,\gamma\ be\ \angle \ of\ \triangle ABC\\\ \angle ACO\ is\ \angle 1\\ \angle BCO\ is \ \angle 2$

$\angle 1=\alpha -45°\\ \angle 2=\beta-45°\\ \angle AOB=90°\\ \gamma =\frac{1}{2}\cdot 90°=45°\ (Peripheral\ angle\ on\ the\ perimeter)\\ \gamma=\angle 1+ \angle 2=45°\\ \alpha +\beta=180°-\gamma=180°-45°=135°$

$\angle\ MON=(180°-90°-\angle 1)+(180°-90°-\angle 2)\\ \angle\ MON=180°-(\angle 1 +\angle 2)=180°-45°\\ \color{blue}\angle\ MON=135°$

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What is the value of the first fraction he wrote?

Let the desired denominator be x.

$\frac{x+2}{x}\cdot \frac{x+3}{x+1}\cdot \frac{x+4}{x+2}=10\\ \frac{(x+3)(x+4)}{x(x+1)}=10\\ \color{blue}x=1$

$\color{blue}The\ first\ fraction\ he\ wrote\ \frac{3}{1}\ .$

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The radius of the circle is r.

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

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$A(a,a^2)\\ B(b,b^2)\\ \dfrac{b^2-a^2}{b-a}=2\\ \dfrac{(b+a)(b-a)}{b-a}=2$

$\color{blue}b+a=2\\ $

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${\color{blue}\overline{CM}}=\overline{AB}=\sqrt{4^2+4^2}=\sqrt{32}\color{blue}=5.65685$

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Find the radius of the circle.

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

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

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Find the sum of the coordinates of the fourth vertex.

A(-3,-1)

C(5,-2)

D(0,0)

$m_{da}=\frac{y_d-y_a}{x_d-x_a}=\frac{0-(-1)}{0-(-3)}=\frac{1}{3}=m_{cb}\\ m_{dc=}\frac{y_d-y_c}{x_d-x_c}=\frac{0-(-2)}{0-5}=-\frac{2}{5}=m_{ab}$

$f_{ab}(x)=m_{ab}(x-x_a)+y_a=-\frac{2}{5}(x-(-3))+(-1)=-\frac{2}{5}(x+3)-1\\ \color{blue}f_{ab}(x)=-\frac{2}{5}x-\frac{11}{5}\\ f_{cb}(x)=m_{cb}(x-x_c)+y_c=\frac{1}{3}(x-5)+(-2)=\frac{1}{3}x-\frac{11}{3}\\ \color{blue}f_{cb}(x)=\frac{1}{3}x-\frac{11}{3}$

$-\frac{2}{5}x-\frac{11}{5}=\frac{1}{3}x-\frac{11}{3}\\ (\frac{1}{3}+\frac{2}{5})x=\frac{11}{3}-\frac{11}{5}\\ x=\frac{\frac{11}{3}-\frac{11}{5}}{\frac{1}{3}+\frac{2}{5}}\\ \color{blue}x_b=2$

$f_{cb}(x)=y_b=\frac{1}{3}x-\frac{11}{3}=\frac{1}{3}\cdot 2-\frac{11}{3}\\ \color{blue}y_b=-3$

-3 + 2 = -1

The sum of the coordinates of the fourth vertex is -1.

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Sorry, I wrote down the coordinates of my point A incorrectly.