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polet [3.4K]
3 years ago
8

Evaluate the following expression using the values given: Find -a^2-3b^3+c^2+2b^3-c^2 if a=3, b=2, and c=-3.

Mathematics
1 answer:
erastovalidia [21]3 years ago
7 0
Ok so substitute the values. a=3 b=2 and c=-3
-3^(2)-3(2)^(3)+(-3)^(2)+2(2)^(3)-(-3)^(2)=-9-24+9+16-9=-33+25-9=-42+25=-17
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The planning department of abstract office supplies has been asked to determine what is the company should introduce a new compu
BabaBlast [244]

Answer:

<em>The company needs to sell 40 desks to break even</em>

Step-by-step explanation:

<u>Application of Equations</u>

There is virtually no limit to the possible situations where equations can help to find the solution of specific problems related to areas like economy, where one could need to establish some important indicators about the business.

B. The fixed cost for Abstract Office Supplies to sell a new computer desk is $14,000. Each desk will cost $150 to produce. The cost function to produce X desks is

C(x)=150x+14,000

A. The revenue for each desk is estimated at $500, for X desks will be

R(x)=500x

C. The company will break even when the cost and the revenue are the same. We'll find how many desks need to be sold for that to happen. We equate

C(x)=R(x)

Or equivalently

150x+14,000=500x

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500x-150x=14,000

350x=14,000

Solving for x

x=14,000/350= 40

The company needs to sell 40 desks to break even

4 0
3 years ago
Which of the following does not belong to the solution set of -3x + 7 &lt; 11?
erik [133]
-3x+7<11

-3x<4

x>-4/3

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With your update:

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7 0
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. 3 Przeczytaj zadanie.
sattari [20]

Answer:

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3 0
2 years ago
Please help me I'm stuck on this question.
valentinak56 [21]
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Step2247 [10]

Answer:

p=10\sqrt[3]{54}x^2

Step-by-step explanation:

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Since, two opposite sides of any parallelogram are always equal

Let's assume first side =a

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so, we get

a=2\sqrt[3]{54}x^2

b=3\sqrt[3]{54}x^2

now, we can find perimeter

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so, we get

p=2(2\sqrt[3]{54}x^2)+2(3\sqrt[3]{54}x^2)

we can simplify it

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p=10\sqrt[3]{54}x^2................Answer


6 0
3 years ago
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