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Alchen [17]
1 year ago
14

In sentence 7, the phrase by all accounts is best

Mathematics
1 answer:
hram777 [196]1 year ago
5 0

<u>In sentence 7, the </u><u>phrase</u><u> by all accounts is best replaced by </u><u>however.</u>

Correct option is A

  • The sentence that precedes 7 states something that is in contrast with it.
  • Therefore 'by all accounts' in sentence should be changed to 'however' as it is generally used to present something that is different from or contrasts with the previous statement.
  • Thus A is the best answer. The other options are incorrect because they do not convey the contrast that is required in the given context.

What is a phrase in grammar?

  • In syntax and grammar, a phrase is a group of words which act together as a grammatical unit.
  • For instance, the English expression "the very happy squirrel" is a noun phrase which contains the adjective phrase "very happy".
  • Phrases can consist of a single word or a complete sentence.

Learn more about Phrases

brainly.com/question/15806900

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Sheri surveyed eighty students and then generated three random samples of twenty-five student responses. The means of the sample
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2 years ago
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Artemon [7]

Answer:

x=9 and z=83

Step-by-step explanation:

First you subtract 97 from 180 because that gives you z because 180 is a flat angle and 83 plus 97 gives 180. For x the opposite angle is also 97 so you set the equation 5x + 52 = 180 and you solve for x to get the x value. Hope this helps :)

6 0
3 years ago
Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop
xxMikexx [17]

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

 f(x)=log (3 x) +5 x²

f'(x)=\frac{1}{3x}+10 x

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

  x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248

x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012

x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057

x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052

So, root of the equation =0.205 (Approx)

Approximate relative error

                =\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41

 Approximate relative error in terms of Percentage

   =0.41 × 100

   = 41 %

7 0
2 years ago
If the radius of the cylinder is doubled does the volume of the cylinder double?
Anastaziya [24]
The answer to your question is yes
6 0
3 years ago
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