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aleksklad [387]
3 years ago
6

4/3 b + 2 equals 4 - 1/3 be​

Mathematics
2 answers:
igomit [66]3 years ago
8 0

For this case we have the following equation:

\frac {4} {3} b + 2 = 4- \frac {1} {3} b

Subtracting 2 from both sides we have:

\frac {4} {3} b = 4- \frac {1} {3} b-2

We add similar terms from the right side of the equation:

\frac {4} {3} b = 2- \frac {1} {3} b

Adding\frac {1} {3} b to both sides:

\frac {4} {3} b + \frac {1} {3} b = 2\\\frac {4 + 1} {3} b = 2\\\frac {5} {3} b = 2

Multiplying by 3 on both sides:

5b = 3 * 2\\5b = 6

Dividing between 5 on both sides:

b = \frac {6} {5}

Answer:

b = \frac {6} {5} = 1.2

DIA [1.3K]3 years ago
3 0

I do not understand what you are asking but I am going to assume that you are trying to simplify the equation.

First, write the equation.

(4/3)b + 2 = 4 - (1/3)b

Add (1/3)b to each side.

(5/3)b + 2 = 4

Subtract 2 on each side.

(5/3)b = 2

Divide by (5/3) on each side

b = 1.2

b = 1.2 is your final answer.

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Read 2 more answers
(a) Suppose anxn has finite radius of convergence R and an ≥ 0 for all n. Show that if the series converges at R, then it also c
valina [46]

Answer:

a) See the proof below.

b) \sum \frac{(-x)^n}{n}

Step-by-step explanation:

Part a

For this case we assume that we have the following series \sum a)n x^n and this series has a finite radius of convergence R and we assume that a_n \geq 0 for all n, this information is given by the problem.

We assume that the series converges at the point x= R since w eknwo that converges, and since converges we can conclude that:

\sum a)n R^n < \infty

For this case we need to show that converges also for x=-R

So we need to proof that \sum a_n (-R)^n < \infty

We can do some algebra and we can rewrite the following expression like this:

\sum a_n (-R)^n = \sum (-1)^n a)n R^n and we see that the last series is alternating.

Since we know that \sum a_n x^n converges then the sequence {a_n R^n} must be positive and we need to have lim_{n\to \infty} a^n R^n = 0

And then by the alternating series test we can conclude that \sum a_n (-R)^n also converges. And then we conclude that the power series a_n x^n converges for x=-R ,and that complete the proof.

Part b

For this case we need to provide a series whose interval of convergence is exactly (-1,1]

And the best function for this \frac{(-x)^n}{n}

Because the series \sum \frac{(-x)^n}{n} converges to -ln(1+x) when |x| using the root test.

But by the properties of the natural log the series diverges at x=-1 because \sum \frac{1}{n} =\infty and for x=1 we know that converges since \sum \frac{-1}{n} is an alternating series that converges because the expression tends to 0.

6 0
3 years ago
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