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pishuonlain [190]
1 year ago
6

Many eighteenth- and nineteenth-century Romantic

Mathematics
1 answer:
Zigmanuir [339]1 year ago
5 0

Many eighteenth- and nineteenth-century Romantic poets <u>were believers in rebellion against social conventions, express strong emotion</u>, and the power of imagination. (Option E: believed in rebellion against social conventions, the expression of strong emotions)

<h3>What is sentence correction?</h3>

Sentence correction or sentence improvement is a type of grammatical practice where a sentence is given with a word or a phrase that requires grammatical changes or improvement.

Now,

  • The lack of a parallel structure in this statement is the problem; in order to affirm a parallel structure, the three items provided in the sentence should be phrased similarly.
  • Expressing strong emotions and "believers in revolt" are not synonymous in this context. The proper order and thus the sentence correction is "rebellion," "expression," and "power.", which is visible in Option E.

Hence, Option E is the most appropriate option, i.e., Many eighteenth- and nineteenth-century Romantic poets <u>were believers in rebellion against social conventions, express strong emotion</u>, and the power of imagination. (Option E: believed in rebellion against social conventions, the expression of strong emotions).

To learn more about sentence correction, refer to the link: brainly.com/question/14632568

#SPJ4

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Noah has a summer tree-trimming business. Based on experience, Noah knows that his profit, P, in dollars, can be modelled by = −
crimeas [40]

Solving quadratic equations, it is found that he needs to charge:

1. He needs to charge $40 to break even.

2. He needs to charge $30 for a profit of $600.

<h3>What is a quadratic function?</h3>

A quadratic function is given according to the following rule:

y = ax^2 + bx + c

The solutions are:

  • x_1 = \frac{-b + \sqrt{\Delta}}{2a}
  • x_2 = \frac{-b - \sqrt{\Delta}}{2a}

In which:

\Delta = b^2 - 4ac

The profit equation in this problem is:

P(x) = -3x² + 150x - 1200.

He breaks even when P(x) = 0, hence:

-3x² + 150x - 1200 = 0.

The coefficients are a = -3, b = 150, c = -1200, hence:

  • \Delta = 150^2 - 4(-3)(-1200) = 8100
  • x_1 = \frac{-150 + \sqrt{8100}}{-6}
  • x_2 = \frac{-150 - \sqrt{8100}}{-6} = 40

He needs to charge $40 to break even.

For a profit of $600, we have that P(x) = 600, hence:

-3x² + 150x - 1200 = 600.

-3x² + 150x - 1800 = 0.

The coefficients are a = -3, b = 150, c = -1800, hence:

  • \Delta = 150^2 - 4(-3)(-1800) = 900
  • x_1 = \frac{-150 + \sqrt{900}}{-6}
  • x_2 = \frac{-150 - \sqrt{900}}{-6} = 30

He needs to charge $30 for a profit of $600.

More can be learned about quadratic equations at brainly.com/question/24737967

#SPJ1

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1 year ago
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