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3 years ago

Determine the correct scientific notation form of the number. 327,000,000,000

1 answer:

7 0

So this "327,000,000,000" needs to be converted into scientific notation, in which the answer that I will get would be 3.27 10^11, in calculation (as in calculator), you will get 3.27E+11.

Hope this helped!

Nate

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What’s the answer to 8-10?

vovangra [49]

Answer:

-2

Step-by-step explanation:

When you have equations with smaller number subtracting bigger numbers, it means that you are always going to end up with a negative answer. So if you do 10-8 it would just be 2, but if you flip it , it will be a negative so the the answer would be

8-10= -2

Hope that helped or it probaly just made you more confused lol

4 0

3 years ago

Please help me with these questions please I need help? I'm confused with these questions can somebody explain me please.

san4es73 [151]

Rationals are whole numbers no negative numbers or decimals or fractions and Intergers are only negativd numbers. Absolute Value is a negative number changed into a whole number. Hope this helps.

5 0

3 years ago

A railway engine is of mass 72tonnes and 11m long. An exact scale model is made of it and is 44cm long. find the mass of the mod

34kurt

Answer:

it is correct

Step-by-step explanation:

2000 N

The resistance force acting is 1N per ton.

Tension in the coupling between engine and the wagon is 2000N.

4 0

2 years ago

Read 2 more answers

PLEASE HELP I really need help please heplp

Darya [45]

Answer:

im not sure but i think its A

Step-by-step explanation:

6 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.

5 0

3 years ago