The intervals included in solution are:
[tex]\frac{1}{x} + \frac{1}{x}-10\ge \frac{2}{24}\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:0<x\le \frac{24}{121}\:\\ \\\:\mathrm{Decimal:}&\:0<x\le \:0.19834\dots \\\\ \:\mathrm{Interval\:Notation:}&\:(0,\:\frac{24}{121}]\end{bmatrix}[/tex]
Solution:
Given that,
A boat tour guide expects his tour to travel at a rate of x mph on the first leg of the trip
On the return route, the boat travels against the current, decreasing the boat's rate by 10 mph
The group needs to travel an average of at least 24 mph
Given inequality is:
[tex]\frac{1}{x} + \frac{1}{x} - 10\geq \frac{2}{24}[/tex]
We have to solve the inequality
[tex]\frac{1}{x} + \frac{1}{x} - 10\geq \frac{2}{24}\\\\\frac{2}{x} - 10\geq \frac{2}{24}[/tex]
[tex]\mathrm{Subtract\:}\frac{2}{24}\mathrm{\:from\:both\:sides}\\\\\frac{2}{x}-10-\frac{2}{24}\ge \frac{2}{24}-\frac{2}{24}\\\\Simplify\\\\\frac{2}{x}-10-\frac{2}{24}\ge \:0[/tex]
[tex]\frac{2}{x}-\frac{10}{1}-\frac{2}{24} \geq 0\\\\\frac{ 2 \times 24}{x \times 24} -\frac{10 \times 24}{1 \times 24} - \frac{2 \times x }{24 \times x}\geq 0\\\\\frac{48}{24x}-\frac{240x}{24x}-\frac{2x}{24x}\geq 0\\\\Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions\\\\\frac{48-240x-2x}{24x}\geq 0\\\\Add\:similar\:elements\\\\\frac{48-242x}{24x}\ge \:0[/tex]
[tex]\mathrm{Multiply\:both\:sides\:by\:}24\\\\\frac{24\left(48-242x\right)}{24x}\ge \:0\cdot \:24\\\\Simplify\\\\\frac{48-242x}{x}\ge \:0\\\\Factor\ common\ terms\\\\\frac{-2\left(121x-24\right)}{x}\ge \:0\\\\\mathrm{Multiply\:both\:sides\:by\:}-1\mathrm{\:\left(reverse\:the\:inequality\right)}[/tex]
When we multiply or divide both sides by negative number, then we must flip the inequality sign
[tex]\frac{\left(-2\left(121x-24\right)\right)\left(-1\right)}{x}\le \:0\cdot \left(-1\right)\\\\\frac{2\left(121x-24\right)}{x}\le \:0\\\\\mathrm{Divide\:both\:sides\:by\:}2\\\\\frac{\frac{2\left(121x-24\right)}{x}}{2}\le \frac{0}{2}\\\\Simplify\\\\\frac{121x-24}{x}\le \:0[/tex]
[tex]\mathrm{Find\:the\:signs\:of\:the\:factors\:of\:}\frac{121x-24}{x}\\[/tex]
This is attached as figure below
From the attached table,
[tex]\mathrm{Identify\:the\:intervals\:that\:satisfy\:the\:required\:condition:}\:\le \:\:0\\\\0<x<\frac{24}{121}\quad \mathrm{or}\quad \:x=\frac{24}{121}\\\\Merge\:Overlapping\:Intervals\\\\0<x\le \frac{24}{121}[/tex]
Therefore, solution set is given as:
[tex]\frac{1}{x} + \frac{1}{x}-10\ge \frac{2}{24}\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:0<x\le \frac{24}{121}\:\\ \\\:\mathrm{Decimal:}&\:0<x\le \:0.19834\dots \\\\ \:\mathrm{Interval\:Notation:}&\:(0,\:\frac{24}{121}]\end{bmatrix}[/tex]
Answer:
The correct answer is
Step-by-step explanation:
1st one (0,4] and (10,30]
and the 2nd question is (10,30] only
