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emmainna [20.7K]
3 years ago
6

A private ship carries two types of cargo: large crates and extra-large crates. Each type of crate weighs a specific amount.

Mathematics
1 answer:
Arisa [49]3 years ago
8 0

Answer:

G. 32

Step-by-step explanation:

32 x 335 = 10720

16,100 - 10720 = 5380/190 = 28.3157894737

275 x 32 = 8800

18,800 - 8800 = 10,000/400 = 25 (This is the closer).

30 x 335 = 10,050

16,100 - 10,050 = 6050/190  = 31.8421052632

30 x 275 = 8250

18,800 - 8250 = 10550/400 = 26.375 (Same reason why it does't work)

25 x 335 = 8375

16,100 - 8375 = 7725/190 = 40.6578947368

25 x 275 = 6875

18,800 - 6875 = 11925/400 = 29.8125

24 x 335 = 8040

16,100 - 8040 = 8060/190 = 42.4210526316

24 x 275 = 6600

18,800 - 6600 = 12200/400 = 30.5

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Answer: 6

Step-by-step explanation:

Given Data:

Number of people = 6

Amount of Gold to be shared = 40lbs

Subtract 2/3 from your answer.

Therefore

If six people are to share 40lbs of gold evenly

= 40lbs / 6

Divide through by 2

= 20lbs / 3

= 20lbs / 3 - 2/3

= 18/3

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Each person would get approximately 6lbs of gold

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by multiplying all numbers seperatly by each other and adding those

Step-by-step explanation:

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Answer:

ab and ac are congruent with each other

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Find the radius of convergence, then determine the interval of convergence
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The radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series. This can be obtained by using ratio test.  

<h3>Find the radius of convergence R and the interval of convergence:</h3>

Ratio test is the test that is used to find the convergence of the given power series.  

First aₙ is noted and then aₙ₊₁ is noted.

For  ∑ aₙ,  aₙ and aₙ₊₁ is noted.

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = β

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  • If β > 1, then the series diverges
  • If β = 1, then the series inconclusive

Here a_{k} = \frac{(x+2)^{k}}{\sqrt{k} }  and  a_{k+1} = \frac{(x+2)^{k+1}}{\sqrt{k+1} }

   

Now limit is taken,

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }/\frac{(x+2)^{k} }{\sqrt{k} }|

= \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }\frac{\sqrt{k} }{(x+2)^{k}}|

= \lim_{n \to \infty} |{(x+2) } }{\sqrt{\frac{k}{k+1} } }}|

= |{x+2 }|\lim_{n \to \infty}}{\sqrt{\frac{k}{k+1} } }}

= |{x+2 }| < 1

- 1 < {x+2 } < 1

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We get that,

interval of convergence = (-3, -1)

radius of convergence R = 1

Hence the radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series.

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