If r(x) = 3x - 1 and s(x) = 2x + 1, which expression is equivalent to (r/s)(6)?

Mathematics

1 answer:

zavuch27 [327]4 years ago

4 0

Answer:

17/13

Step-by-step explanation:

r(6) = 3(6) - 1 = 17 and s(6) = 2(6) + 1 = 13.

Then (r/s)(6) = 17/13.

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Does the following infinite series converge or diverge? 1/3+2/9+4/27+8/81+...

PilotLPTM [1.2K]

Note that A and D are ludicrous choices, so you can throw them away outright. (Any divergent series cannot have a sum, and any convergent series must have a sum.)

The sum is certainly convergent because it can be written as a geometric sum with common ratio between terms that is less than 1 in absolute value.

$S=\dfrac13+\dfrac29+\dfrac4{27}+\dfrac8{81}+\cdots$
$S=\dfrac13\left(1+\dfrac23+\dfrac{2^2}{3^2}+\dfrac{2^3}{3^3}+\cdots\right)$

We can then find the exact value of the sum:

$\dfrac23S=\dfrac13\left(\dfrac23+\dfrac{2^2}{3^2}+\dfrac{2^3}{3^3}+\dfrac{2^4}{3^4}+\cdots\right)$

$\impliesS-\dfrac23S=\dfrac13$
$\implies\dfrac13S=\dfrac13$
$\implies S=1$

So the answer is B.

8 0

3 years ago

I really need help on this question can someone help me I will give Brainliest for correct answer!!!

lawyer [7]

Answer:

The rectangle with 5 and 4

The triangle with 10 and 4

Brainliest pls :)

3 0

3 years ago

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

FromTheMoon [43]

Answer:

The Taylor series is $\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}$ .

The radius of convergence is $R=3$ .

Step-by-step explanation:

<em>The Taylor expansion.</em>

Recall that as we want the Taylor series centered at $a=3$ its expression is given in powers of $(x-3)$ . With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of $\ln(1+x)$ .