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3 years ago

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Suppose the function D(p)=1500-25p represents the demand for hot dogs, whose price is p, at a baseball game. Find the domain of

the function.

1 answer:

TiliK225 [7]3 years ago

3 0

The function is given by

$\\ \tt\hookrightarrow D(p)=1500-25p$

Price must be greater than 0

$\\ \tt\hookrightarrow 1500-25p\geqslant 0$

$\\ \tt\hookrightarrow 1500\geqslant 25p$

$\\ \tt\hookrightarrow p\leqslant 60$

Now

$\\ \tt\hookrightarrow D_f\in (-\infty,60]$

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<h2><u>Exercise 34.</u></h2>

1. Forming the inequality.

Let's make the left side be a calculation of the total earning based on the number of tickets sols. Also, let's make the right side the cost of the dance to compare and find an adecuate minimum value of x.

$4t+72\geq 400$ .

<em>We use the </em><em>"greater or equal to"</em><em> symbol because the side of x expresses the earning of the committee, and these earnings will have to be equal or greater than the cost of the dance to be able to organize the dance.</em>

2. Solving the inequality.

$4t+72\geq 400\\ \\4t\geq 400-72\\ \\4t\geq 328\\ \\t\geq\frac{328}{4} \\ \\t\geq82$

3. Express your results.

$t\leq82$ , therefore, the dance committee will have to sell at least 82 tickets to be able to cover the cost of the dance.

<h2><u>Exercise 35.</u></h2>

$2-6(-5t+1)$

<em>Apply the associative property of multiplication.</em>

<em>Rule to be applied: </em> $a(b+c)=ab+ac$ .

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A 10 ft ladder is being pulled away from a wall at a rate of 3 ft/sec. What is the rate of change in the area beneath the ladder

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The rate of change in the area beneath the ladder is 5.25 ft²/s

Step-by-step explanation:

Area of triangle is given by;

$A = \frac{1}{2}bh\\\\2A = bh\\\\take \ derivative \ of \ both \ sides \ with \ respect \ to \ "t"\\\\2\frac{dA}{dt} = h\frac{db}{dt} + b\frac{dh}{dt}\\\\divide \ through \ by \ 2\\\\\frac{dA}{dt} = (\frac{h}{2} )\frac{db}{dt} + (\frac{b}{2}) \frac{dh}{dt}$

where;

b is the base of the triangle, given as 6 ft

h is the height of the triangle, determined by applying Pythagoras theorem.

h² = 10² - 6²

h² = 100 - 36

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h = √64

h = 8 ft

Determine the rate of change of the height;

$h^2 + b^2 = 10^2\\\\h^2 + b^2 =100\\\\2h\frac{dh}{dt} + 2b\frac{db}{dt} =0\\\\h\frac{dh}{dt} + b\frac{db}{dt} =0\\\\h \frac{dh}{dt} = -b\frac{db}{dt} \\\\\frac{dh}{dt} = (\frac{-b}{h} )\frac{db}{dt}\\\\\frac{dh}{dt} =(\frac{-6}{8})(3)\\\\\frac{dh}{dt} = -\frac{9}{4} \ ft/s$

Finally, determine the rate of change of area beneath the ladder;

$\frac{dA}{dt} = (\frac{h}{2} )\frac{db}{dt} + (\frac{b}{2}) \frac{dh}{dt}\\\\\frac{dA}{dt} = (\frac{8}{2} )(3) + (\frac{6}{2}) (\frac{-9}{4})\\\\\frac{dA}{dt} = 12 - \frac{27}{4} \\\\\frac{dA}{dt} = \frac{48-27}{4}\\\\\frac{dA}{dt} = \frac{21}{4} \ ft^2/s\\\\\frac{dA}{dt} = 5.25 \ ft^2/s$

Therefore, the rate of change in the area beneath the ladder is 5.25 ft²/s

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