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olganol [36]
3 years ago
8

The volume of a rectangular prism is (x3 – 3x2 + 5x – 3), and the area of its base is (x2 – 2). If the volume of a rectangular p

rism is the product of its base area and height, what is the height of the prism?
Mathematics
2 answers:
lesya692 [45]3 years ago
8 0

Answer:

The answer is A

Step-by-step explanation:

The way that you get this answer is by divided x2 to x3 and you get x and then you multiply x to x2-2 and you get x3-2x and the you subtract that from x3-3x2+5x-3 and then you repeat that process and the final answer is x-3 + 7x-9/x2-2

sweet-ann [11.9K]3 years ago
3 0

Answer:

height  = (x^3 – 3x^2 + 5x – 3) / (x^2 – 2)

Step-by-step explanation:

We know that the volume of a rectangular prism is the product of its base area and height

Volume_prism = area_base *  height

height =  Volume_prism /  area_base

Volume_prism  = (x^3 – 3x^2 + 5x – 3)

area_base = (x^2 – 2)

height =  Volume_prism /  area_base

height  = (x^3 – 3x^2 + 5x – 3) / (x^2 – 2)

Please see attached image

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Svet_ta [14]

Answer:

a. $1.03

b. $0.93

c. 0.98

d. 2 workers

Step-by-step explanation:

a. Given that:

  • 1 computer : 1 worker : inventory 150 items per hour
  • 1 computer : 2 workers : inventory 200 items per hour
  • 1 computer : 3 workers : inventory 220 items per hour
  • 1 computer : 4+ workers : fewer than 235 items per hour
  • Cost: $100 per computer ; $25 per worker

The fixed production factor in the warehouse is the computer used:

-One computer used, but the number of users is varied to inventory a specified number of items.

-The variable production factor is the number of workers assigned per one computer.

#The cost of inventorying a single item by one worker is:

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_1=\frac{125+30}{150}\\\\\\=1.03

Hence, the cost of inventorying a single item is $1.03

b. Using the information provided above, the cost of inventorying a single item when two workers are assigned is :

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_2=\frac{125+2\times30}{200}\\\\\\=0.925

Hence, the cost of inventorying a single item is $0.93

c.Using the information provided above, the cost of inventorying a single item when three workers are assigned is :

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_3=\frac{125+3\times30}{220}\\\\\\=0.98

Hence, the cost of inventorying a single item is $0.98

d. To determine the most cost-effective job assignment, we calculate the cost of 4+ workers.

Take any number less than 235(say 234) as the inventory units:

Cost=\frac{C_{pc}+Wage}{Items} \ , C_{pc}=\$125\\\\Cost_4=\frac{125+4\times30}{234}\\\\\\=1.05

From our calculations, it's clear that two workers per computer costs the least amount($0.93) per unit item. Hence, it is best to assign two workers per computer.

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3 years ago
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Bogdan [553]

Answer:

\boxed{\bold{z=-60}}

Step-by-step explanation:

Switch Sides

\bold{2+\frac{z}{-6}=12}

Subtract 2 From Both Sides

\bold{2+\frac{z}{-6}-2=12-2}

Simplify

\bold{\frac{z}{-6}=10}

Multiply Both Sides By -6

\bold{\frac{z\left(-6\right)}{-6}=10\left(-6\right)}

Simplify

\bold{z=-60}

3 0
3 years ago
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the answer is A because it shows.
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