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

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

(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.

5 0

3 years ago

What Should i write down basically i just need the answers and let me copy and paste

Anastaziya [24]

Answer:

✔️Ratio of weight to mass in simplest form: 49/5 or 49 : 5

✔️x = 110

Step-by-step explanation:

Since the ratio of weight to mass is constant, therefore we can use any pair of values from the table to find the ratio that represents the table. Let's pick the first row:

(20, 196)

Ratio of weight to mass = 196 : 20

= 196/20

= 49/5

Or 49 : 5

✔️Let's use this ratio/proportion to find x in the table

Ratio of weight to mass = 49:5

Thus:

1078/x = 49/5

Cross multiply

x*49 = 1078*5

49x = 5,390

x = 5,390/49

x = 110

5 0

3 years ago

JUST HELP ME ITS DUE IN5 MINUTES:"( ASAP <3

Citrus2011 [14]

Answer:

A. 25%

25% is equivalent to 1/4 and when you multiply 75 by 4 you get 300. Which means 75 is equivalent to 1/4 and 1/4 is equivalent to 25%. So the answer is 25%. I'm so sorry if this explanation is bad

5 0

3 years ago