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Elanso [62]
1 year ago
14

The price p (in dollars) and the demand X for a particular clock radio are related by the equation X = 3000 -10p. Find and answe

r A - D

Mathematics
1 answer:
alukav5142 [94]1 year ago
6 0

The price p based on the equation given is p = 3000 - 0.1x.

<h3>How to express the price?</h3>

The equation given in the question is x = 3000 - 10p. The price(p) will be:

x = 3000 - 10p.

Make p the subject of the formula

x + 10p = 3000

10p = 3000 - x

p = (3000 - x)/10

p = 3000 - 0.1x

The revenue will be price multiplied by quantity. This will be:

= (3000 - 0.1x) × x

= 3000x - 0.1x²

The marginal revenue will be calculated after differentiating. This will be:

= 3000x - 0.1x²

= 3000 - 0.2x

The demand which is the quantity based on the information will be:

3000 - 0.2x = 0

0.2x = 3000

x = 1500

Learn more about demand on:

brainly.com/question/1245771

#SPJ1

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The complete version of question:

<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26. What is the solution of this problem.</em>

Answer:

5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}

Step-by-step explanation:

As the description of the statement is:

'<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26'.</em>

<em />

As

  • <em>Five times the sum of a number and 27  </em>is written as: 5(x + 27)
  • <em>greater than or equal </em>is written as:  \geq
  • <em>six times the sum of that number and 26'  </em>is written as: 6(x + 26)

so lets combine the whole statement:

<em />5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)<em />

solving

<em />5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)<em />

<em />5x+135\ge \:6x+156<em />

<em />5x+135-135\ge \:6x+156-135<em />

<em />5x\ge \:6x+21<em />

<em />5x-6x\ge \:6x+21-6x<em />

<em />-x\ge \:21<em />

<em />\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}<em />

<em />\left(-x\right)\left(-1\right)\le \:21\left(-1\right)<em />

<em />x\le \:-21<em />

Therefore,

5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}

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Step-by-step explanation:

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