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tatiyna
3 years ago
13

The equations x + 5 y = 10, 3 x minus y = 1, x minus 5 y = 10, and 3 x + y = 1 are shown on the graph below. On a coordinate pla

ne, there are 4 lines. Green line goes through (0, negative 1) and (1, 2). Blue line goes through (0, 1) and (1, negative 2). Pink line goes through (0, 2), and (2, 1.5). Orange line goes through (negative 2, negative 2.5) and (2, negative 1.5). Which is the approximate solution for the system of equations x + 5 y = 10 and 3 x + y = 1? (–0.3, 2.1) (–0.3, –2.1) (0.9, –1.8) (0.9, 1.8)
Mathematics
2 answers:
nadezda [96]3 years ago
5 0

Answer:

A: (–0.3, 2.1)

sveticcg [70]3 years ago
5 0

Answer:a

Step-by-step explanation:

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ALSO WORTH 50 POINTS Which graph matches the inequality: 2(4x - 5) < 14x + 8?
ss7ja [257]

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The correct answer is: C

Step-by-step explanation:

2(4x - 5) < 14x + 8

First multiply 2 by 4x and -5

8x - 10 < 14x + 8

Now find out what x is

Move 14x to the left

8x - 14x - 10 = 8

Then move 10 to the right and change it's sign to plus

8x - 14x = 8 + 10

Now combine like terms

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Now divide 18 by 6 to get x

x = 18/-6

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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

  • If β < 1, then the series converges
  • 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

- 1 - 2 < x < 1 - 2

- 3 < x < - 1

 

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.

Learn more about radius of convergence here:

brainly.com/question/14394994

#SPJ1

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1 year ago
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