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ipn [44]
2 years ago
5

The quadratic function f has a vertex at (3,4) and opens upward. The quadratic function g is shown below as g(x)=2(x-4)2+3. Whic

h statement is true?  A. The maximum value of f is greater than the maximum value of g. B. The minimum value of f is greater than the minimum value of g. C. The maximum value of g is greater than the maximum value of f. D. The minimum value of g is greater than the minimum value of f.
Mathematics
1 answer:
oksano4ka [1.4K]2 years ago
7 0

The vertex coordinates of g(x) is (4,3). Indeed, the function is given as g(x)=2(x-4)^{2}+3 and when we set x = 4, then g(x) = 3 because the function is given in a closed form. In g(x) and f(x) the values 3 and 4 respectively are the minimum values. The functions are going upward. The value of f(x) is above the value of g(x). The correct answer is B

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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 trav
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<em><u>The intervals included in solution are:</u></em>

\frac{1}{x} + \frac{1}{x}-10\ge \frac{2}{24}\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:0

<em><u>Solution:</u></em>

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

<em><u>Given inequality is:</u></em>

\frac{1}{x} + \frac{1}{x} - 10\geq \frac{2}{24}

<em><u>We have to solve the inequality</u></em>

\frac{1}{x} + \frac{1}{x} - 10\geq \frac{2}{24}\\\\\frac{2}{x}  - 10\geq \frac{2}{24}

\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

\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

\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)}

When we multiply or divide both sides by negative number, then we must flip the inequality sign

\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

\mathrm{Find\:the\:signs\:of\:the\:factors\:of\:}\frac{121x-24}{x}\\

This is attached as figure below

From the attached table,

\mathrm{Identify\:the\:intervals\:that\:satisfy\:the\:required\:condition:}\:\le \:\:0\\\\0

<em><u>Therefore, solution set is given as</u></em>:

\frac{1}{x} + \frac{1}{x}-10\ge \frac{2}{24}\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:0

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Domain:

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y>0
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