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2 years ago

Why is upendrakishore roy chouwdhury remembered

Mathematics

1 answer:

soldi70 [24.7K]2 years ago

4 0

Answer:

Upendrakishore Roy Chowdhury (Bengali: উপেন্দ্রকিশোর রায়চৌধুরী; 12 May 1863 – 20 December 1915) was a Bengali writer and painter. ... He was the first person who introduced the colour printing in bangal. He started the first colour childrens magazine Sandesh in 1913.

Born: Kamadaranjan Ray; 12 May 1863; Moshua, Kishoreganj District in Bengal Pre...

Children: Shukhalata Rao Sukumar Ray Punyalata Chakrabarty Subinoy Ray Shanti...

Known for: Writer, painter

Nationality: Bengali

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The variables y and x have a proportional relationship, and y = 9 when x = 2.

Aloiza [94]

Answer: $y=13.5$

Step-by-step explanation:

1. By definition, if $y=9$ when $x=2$ , then:

$\frac{y}{x}=\frac{9}{2}$

2. Keeping this on mind, to find the value of $y$ when $x=3$ , you must apply the following proccedure:

$\frac{y}{3}=\frac{9}{2}$

3. Now, you must solve for $y$ , as following:

$y=3(\frac{9}{2})\\y=\frac{27}{2}\\y=13.5$

4. Then, the result is:

$y=13.5$

4 0

4 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.

5 0

3 years ago

IF YOU SOLVE THIS I WILL LOVE YOU FOREVER!!!!!!!!!!!!!!!!!!!!!!

ryzh [129]

The third option is how you find the average

7 0

4 years ago

Is 11.5 equal to 11.05

azamat

Answer:

false

Step-by-step explanation: