Recent content by soroban

Bert's speed is twice that of Ermie. Hence, Bert's distance (for a particular time) is twice that of Ernie. When they first meet, their total distance is the perimeter, P = 12x meters. Bert's distance is \tfrac{2}{3}P. Ernie's distance is \tfrac{1}{3}P Therefore . . .

The grid is misleading. The earrings are drawn without replacement, There are 5\cdot4 = 20 outcomes, not 25. \begin{array}{cccc}W_1W_2 & W_1W_3 & W_1B_1 & W_1B_2 \\ W_2W_1 & W_2W_3 & W_2B_1 & W_2B_2 \\ W_3W_1 & W_3W_2 & W_3B_1 & W_3B_2 \\ B_1W_1 & B_1W_2 & B_1W_3 & B_1B_2 \\ B_2W_1 & B_2W_2 &...

Since two pairs of angles are congruent, the third pair is also congruent. . . That is: \angle BAC = \angle EDF. We have: \angle ACB = \angle DFE.\;AC = DF,\;\angle BAC = \angle EDF. The triangles are congruent by ASA. .

An equivalent statement: B \cap A'

I think I've found the typo . . . \begin{array}{cccc} \text{We have:} & \dfrac{csc x}{1+\csc x} - \dfrac{csc x}{1 - \csc x} &=&50\\ \\ & \dfrac{\csc x(1-\csc x) - \csc x)1+\csc x)}{(1+\csc x)(1 - \csc x)} &=& 50\\ \\ & \dfrac{-2\csc^2x}{1-\csc^2x} &=& 50\\ \\ & \dfrac{\frac{-2}{\sin^2x}}{1...

By Venn diagrams? . Truth tables? . Other?

Has anyone else seen this approach to Partial Fractions? \frac{1}{x(x+1)^2}\;=\;\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2} Multiply through by the LCD: .1 \;=\;A(x+1)^2+Bx(x+1)+CxNow select values of x: . . \text{Let }x = -1:\;1 \:=\:A(0) + B(0) + C(-1) \quad\Rightarrow\quad \boxed{C = -1}...

\begin{array}{cc}\log_{2x-5}125 \:=\: \log_28 \\ \log_{2x-5}125 \:=\:3 \\ (2x-5)^3 \:=\:125 \\ 2x-5 \:=\: 5 \\ 2x \:=\: 10 \\ x \:=\:5 \end{array} - - Updated - - -

I've never seen logarithms written like that . . . \begin{array}{ccc}\text{We have:} & \log_{3x-2}100 \:=\:\log_24 \\ & \log_{3x-2}100 \:=\:2 \\ & (3x-2)^2 \:=\:100 \\ & 3x-2 \:=\:10 \\ & 3x\:=\:12 \\ & x \:=\:4 \end{array}

z = \tan\frac{x}{2} \quad\Rightarrow\quad x = 2\arctan z \quad\Rightarrow\quad dx = \frac{2\,dz}{1+z^2} \quad\Rightarrow\quad \cos x = \frac{1-z^2}{1+z^2} \displaystyle \int\sec x\,dx \;=\;\int \frac{1+z^2}{1-z^2}\,\frac{2\,dz}{1+z^2} \;=\; \int\frac{2\,dz}{1-z^2} \;=\;\int\left(\frac{1}{1-z}...

\begin{array}{ccc} \tan^2\theta - \sin^2\theta &=& \dfrac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta \\ \\ & = & \sin^2\theta\left(\dfrac{1}{\cos^2\theta} - 1\right) \\ \\ & = & \sin^2\theta\left(\dfrac{1-\cos^2\theta}{\cos^2\theta}\right) \\ \\ & = & \sin^2\theta...

RHS \;\;=\;\;\frac{1}{\sin^2\theta} - \frac{1}{\sin\theta}\frac{\cos\theta}{\sin\theta} \;\;=\;\;\frac{1-\cos\theta}{\sin^2\theta} \;\;=\;\;\frac{1-\cos\theta}{1-\cos^2\theta} . . . . . =\;\;\frac{1-\cos\theta}{(1-\cos\theta)(1+\cos\theta)} \;\;=\;\;\frac{1}{1+\cos\theta} \;\;=\;\; LHS

\text{We have: }\;f(x) \;=\; \dfrac{\frac{1}{x} + \frac{1}{7}}{x+7} \;=\;\dfrac{7+x}{7x(x+7)} \;=\; \frac{1}{7x} \text{Then: }\;f(-7) \;=\;-\frac{1}{7(-7)} \text{Therefore: }\:C = -\frac{1}{49}

\text{We have: }\;I \;=\; 9\int x^8\ln x\,dx \begin{array}{ccccccccc} u&=& \ln x && dv &=& x^8dx \\ du &=& \frac{dx}{x} && v &=& \frac{1}{9}x^9 \end{array} I \;=\; 9\left[\tfrac{1}{9}x^9\ln x - \tfrac{1}{9}\int x^8dx\right] \;=\; 9\left[\tfrac{1}{9}x^9\ln x - \tfrac{1}{81}x^9\right] + C I...

\text{Take logs: }\; \ln y \;=\;\ln\left[\frac{x(x^2+1)^{\frac{1}{2}}}{(x+1)^{\frac{2}{3}}}\right] \;=\;\ln x + \ln(x^2+1)^{\frac{1}{2}} - \ln(x+1)^{\frac{2}{3}} . .. .. .. . . . \ln y \;=\;\ln x + \tfrac{1}{2}\ln(x^2+1) - \tfrac{2}{3}\ln(x+1) . . \frac{y'}{y} \;=\;\frac{1}{x} +...