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4 years ago

A sequence is defined to be An = n^2 + 2. What is the 5th term of this sequence?

Mathematics

1 answer:

STatiana [176]4 years ago

4 0

Answer:

Step-by-step explanation:

To find the fifth term substitute n = 5 into the rule, that is

$A_{5}$ = 5² + 2 = 25 + 2 = 27

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Solve for the equation x. 2x + 22 = 4(x + 3)

Phoenix [80]

Answer:

Step-by-step explanation:

4(x+3) is 4x+12

2x+22=4x+12

-2x from both sides

get 22=2x+12

-12 from both sides

get 10=2x

divide by 2 to both sides

get 5

7 0

3 years ago

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

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3 years ago

Use the graph below to identify the following key features.

o-na [289]

Answer:

(-4,-4)min, x-intercepts (-2,0) (-6,0), y-intercept (0,2)

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3 years ago

When the Glacier Zone Ice Rink opened, Betsy bought 18 ice-skating lessons during their grand-opening sale. She paid 387 in all.

snow_tiger [21]

Answer:

$21.50