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Aleonysh [2.5K]
3 years ago
5

Select the correct answer. The radius of a nitrogen atom is 5. 6 × 10-11 meters, and the radius of a beryllium atom is 1. 12 × 1

0-10 meters. Which atom has a larger radius, and by how many times is it larger than the other? A. The radius of nitrogen is 20 times larger than the radius of beryllium. B. The radius of beryllium is 40 times larger than the radius of nitrogen. C. The radius of nitrogen is 4 times larger than the radius of beryllium. D. The radius of beryllium is 2 times larger than the radius of nitrogen.
Mathematics
1 answer:
Grace [21]3 years ago
7 0

A negative exponent means the number of the exponent divide by the number. The the radius of beryllium atom is larger than the radius of a nitrogen atom by two times.

Given information-

The radius of a nitrogen atom is 5.6\times10^{-11} meters.

The radius of a beryllium atom is 1.12 \times10^{-10} meters

<h3>Negative exponents</h3>

A negative exponent means the number of the exponent divide by the number.

As the radius of the nitrogen atom is 5.6\times10^{-11} meters. Convert it in the power of negative exponent of 10.

Let the radius of the nitrogen is r mm. Thus,

r=5.6\times 10^{-11}\\&#10;r=5.6\times 10^{-10-1}\\&#10;r=5.6\times 10^{-10}\times10^{-1}\\&#10;r=\dfrac{5.6\times 10^{-10}}{10} \\&#10;r=0.56\times 10^{-10}

Now the base of the exponents of both the atoms are same. As the radius of the nitrogen atom is 0.56\times 10^{-10} meters and the radius of the beryllium atom is 1.12 \times10^{-10}. Thus the radius of beryllium atom is larger.

To calculate how many times, the radius of beryllium atom should be divided by the radius of a nitrogen atom. Thus,

=\dfrac{1.12\times10^{-10}}{0.56\times10^{-10}} \\&#10;=2

Thus the the radius of beryllium atom is larger than the radius of a nitrogen atom by two times.

Learn more about the negative exponents here;

https://brainly.in/question/1221452

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A company manufactures and sells x television sets per month. The monthly cost and​ price-demand equations are ​C(x)equals72 com
solmaris [256]

Answer:

Part (A)

  • 1. Maximum revenue: $450,000

Part (B)

  • 2. Maximum protit: $192,500
  • 3. Production level: 2,300 television sets
  • 4. Price: $185 per television set

Part (C)

  • 5. Number of sets: 2,260 television sets.
  • 6. Maximum profit: $183,800
  • 7. Price: $187 per television set.

Explanation:

<u>0. Write the monthly cost and​ price-demand equations correctly:</u>

Cost:

      C(x)=72,000+70x

Price-demand:

     

      p(x)=300-\dfrac{x}{20}

Domain:

        0\leq x\leq 6000

<em>1. Part (A) Find the maximum revenue</em>

Revenue = price × quantity

Revenue = R(x)

           R(x)=\bigg(300-\dfrac{x}{20}\bigg)\cdot x

Simplify

      R(x)=300x-\dfrac{x^2}{20}

A local maximum (or minimum) is reached when the first derivative, R'(x), equals 0.

         R'(x)=300-\dfrac{x}{10}

Solve for R'(x)=0

      300-\dfrac{x}{10}=0

       3000-x=0\\\\x=3000

Is this a maximum or a minimum? Since the coefficient of the quadratic term of R(x) is negative, it is a parabola that opens downward, meaning that its vertex is a maximum.

Hence, the maximum revenue is obtained when the production level is 3,000 units.

And it is calculated by subsituting x = 3,000 in the equation for R(x):

  • R(3,000) = 300(3,000) - (3000)² / 20 = $450,000

Hence, the maximum revenue is $450,000

<em>2. Part ​(B) Find the maximum​ profit, the production level that will realize the maximum​ profit, and the price the company should charge for each television set. </em>

i) Profit(x) = Revenue(x) - Cost(x)

  • Profit (x) = R(x) - C(x)

       Profit(x)=300x-\dfrac{x^2}{20}-\big(72,000+70x\big)

       Profit(x)=230x-\dfrac{x^2}{20}-72,000\\\\\\Profit(x)=-\dfrac{x^2}{20}+230x-72,000

ii) Find the first derivative and equal to 0 (it will be a maximum because the quadratic function is a parabola that opens downward)

  • Profit' (x) = -x/10 + 230
  • -x/10 + 230 = 0
  • -x + 2,300 = 0
  • x = 2,300

Thus, the production level that will realize the maximum profit is 2,300 units.

iii) Find the maximum profit.

You must substitute x = 2,300 into the equation for the profit:

  • Profit(2,300) = - (2,300)²/20 + 230(2,300) - 72,000 = 192,500

Hence, the maximum profit is $192,500

iv) Find the price the company should charge for each television set:

Use the price-demand equation:

  • p(x) = 300 - x/20
  • p(2,300) = 300 - 2,300 / 20
  • p(2,300) = 185

Therefore, the company should charge a price os $185 for every television set.

<em>3. ​Part (C) If the government decides to tax the company ​$4 for each set it​ produces, how many sets should the company manufacture each month to maximize its​ profit? What is the maximum​ profit? What should the company charge for each​ set?</em>

i) Now you must subtract the $4  tax for each television set, this is 4x from the profit equation.

The new profit equation will be:

  • Profit(x) = -x² / 20 + 230x - 4x - 72,000

  • Profit(x) = -x² / 20 + 226x - 72,000

ii) Find the first derivative and make it equal to 0:

  • Profit'(x) = -x/10 + 226 = 0
  • -x/10 + 226 = 0
  • -x + 2,260 = 0
  • x = 2,260

Then, the new maximum profit is reached when the production level is 2,260 units.

iii) Find the maximum profit by substituting x = 2,260 into the profit equation:

  • Profit (2,260) = -(2,260)² / 20 + 226(2,260) - 72,000
  • Profit (2,260) = 183,800

Hence, the maximum profit, if the government decides to tax the company $4 for each set it produces would be $183,800

iv) Find the price the company should charge for each set.

Substitute the number of units, 2,260, into the equation for the price:

  • p(2,260) = 300 - 2,260/20
  • p(2,260) = 187.

That is, the company should charge $187 per television set.

7 0
3 years ago
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