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Sidana [21]
3 years ago
12

You are in charge of purchases at the student-run used-book supply program at your college, and you must decide how many introdu

ctory calculus, history, and marketing texts should be purchased from students for resale. Due to budget limitations, you cannot purchase more than 700 of these textbooks each semester. There are also shelf-space limitations: Calculus texts occupy 2 units of shelf space each, history books 1 unit each, and marketing texts 4 units each, and you can spare at most 1,200 units of shelf space for the texts. If the used book program makes a profit of $10 on each calculus text, $4 on each history text, and $8 on each marketing text, how many of each type of text should you purchase to maximize profit? HINT [See Example 3.]
calculus text(s) =

history text(s) =

marketing text(s) =

What is the maximum profit the program can make in a semester?
Mathematics
1 answer:
mariarad [96]3 years ago
7 0

Answer:

  • Calculus texts: 600
  • History texts: 0
  • Marketing texts: 0

Step-by-step explanation:

Each Calculus text returns $10/2 = $5 per unit of shelf space. For History and Marketing texts, the respective numbers are $4/1 = $4 per unit, and $8/4 = $2 per unit. Using 1200 units of shelf space for 600 Calculus texts returns ...

  $5/unit × 1200 units = $6000 . . . profit

Any other use of units of shelf space will reduce profit.

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5 x 100 = ? for easy points
zimovet [89]

Answer:

500! Just multiply 5 x 1 then add the remaining 2 zeros :)

Step-by-step explanation:

Thank you!

8 0
3 years ago
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A surveyor leave her base camp and drive 42 km on a bearing of 032, she then drive 28km on a bearing of 154.How far is she then
aleksklad [387]

Answer:

Step-by-step explanation:

The scenario is represented in the attached photo. Triangle ABC is formed. AB represents her distance from her base camp. We would determine BC by applying the law of Cosines which is expressed as

a² = b² + c² - 2abCosA

Where a,b and c are the length of each side of the triangle and B is the angle corresponding to b. It becomes

AB² = AC² + BC² - 2(AC × BC)CosC

AB² = 42² + 28² - 2(42 × 28)Cos58

AB² = 1764 + 784 - 2(1176Cos58)

AB² = 2548 - 1246.37 = 1301.63

AB = √1301.63

AB = 36.08 km

To find the bearing, we would determine angle B by applying sine rule

AB/SinC = AC/SinB

36.08/Sin58 = 42/SinB

Cross multiplying, it becomes

36.08SinB = 42Sin58

SinB = 42Sin58/36.08 = 0.987

B = Sin^-1(0.987)

B = 81°

Therefore, her bearing from the base camp is

360 - 81 = 279°

8 0
3 years ago
We use the properties of equality to rewrite equations and solve equations for a variable. These properties allow us to maintain
Katyanochek1 [597]
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3 years ago
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3 years ago
The circumference of a circle is 65?. In terms of pi, what is the area of the circle?
iVinArrow [24]

Answer:

1056.25π square units

Step-by-step explanation:

A few formulas an definitions which will help us:

(1) \pi=\frac{c}{d}, where c is the circumference of a circle and d is its diameter

(2) A=\pi r^2, where A is the area of a circle with radius r. To put it in terms of d, remember that a circle's diameter is simply twice its radius, or mathematically, (3) d=2r \rightarrow r=\frac{d}{2}.

We can rearrange equation (1) to put d in terms of π and c, giving us (4) d = \frac{c}{\pi}, and we can make a few substitutions in (2) using (3) and (4) to get use the area in terms of the circumference and π:

A=\pi r^2\\=\pi\left(\frac{d}{2}\right)^2\\=\pi\left(\frac{d^2}{4}\right)\\=\pi\left(\frac{(c/\pi)^2}{4}\right)\\=\pi\left(\frac{c^2/\pi^2}{4}\right)\\=\pi\left(\frac{c^2}{4\pi^2}\right)\\\\=\frac{\pi c^2}{4\pi^2}\\ =\frac{c^2}{4\pi}

We can now substitute c for our circumference, 65, to get our answer in terms of π:

A=\dfrac{65^2}{4\pi}=\dfrac{4225}{4\pi}=1056.25\pi

8 0
3 years ago
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