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3 years ago

2+363892917382827363

Mathematics

2 answers:

Zinaida [17]3 years ago

8 0

Answer:

so u type 2 then the + sign on the calculator to 363892917382827363 to get 363892917382827365

Step-by-step explanation:

pentagon [3]3 years ago

3 0

Answer:

363892917382827365

Step-by-step explanation:

thats how one adds

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The sum of the first three terms of geometric sequence is 14. If the first term is 2, find the possible values of the sum of the

Irina18 [472]

Geometric series are in the form of

$a +a*r +a*r^2+...$

Where a is the first term and r is the common ratio .

And it is given that

$2 +2*r +2*r^2=14$

$2r^2+2r-12=0$

$r^2 +r-6=0$

$(r+3)(r-2)=0$

r=-3,2

So the first five terms are

$2+2(-3)+2(-3)^2+2(-3)^3+2(-3)^4 or 2+2(2)+2(2)^2+2(2)^3+2(2)^4$

= 2-6+18-54+162 or 2+4+8+16+32

= 122 or 62

4 0

3 years ago

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What is the measure of angle A? A. 110 B. 70 C. 250 D. 55

zhuklara [117]

B. 70 is the correct answer

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3 years ago

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The population of a rabbit colony triples each month.

djverab [1.8K]

Answer:

3645 rabbits after 6 months.

Step-by-step explanation:

The population of a rabbit colony triples each month.

This means that the population of the colony after t months is given by:

$P(t) = P(0)(3)^t$

In which P(0) is the initial population.

The population started with 5 rabbits.

This means that $P(0) = 5$

$P(t) = 5(3)^t$

How many rabbits are there after 6 months?

This is P(6). So

$P(6) = 5(3)^6 = 3645$

3645 rabbits after 6 months.

3 0

3 years ago

Find the length of the diagonal of a rectangular football pitch with sides 46.6 m and 78.9 m.

ICE Princess25 [194]

Answer:

d≈91.63m

Step-by-step explanation:

d=w2+l2=78.92+46.62≈91.63389m

7 0

3 years ago

Set up the integral that represents the arc length of the curve f(x) = ln(x) + 5 on [1, 3], and then use Simpson's Rule with n =

marta [7]

Answer:

The integral for the arc of length is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

By using Simpon’s rule we get: 1.5355453

And using technology we get: 2.3020

The approximation is about 33% smaller than the exact result.

Explanation:

The formula for the length of arc of the function f(x) in the interval [a,b] is:

$\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx$

We need the derivative of the function:

$f'(x)=\frac{1}{x}$

And we need it squared:

$[f'(x)]^2=\frac{1}{x^2}$

Then the integral is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

Now, the Simposn’s rule with n=4 is:

$\displaystyle\int_a^b g(x)}dx\approx\frac{\Delta x}{3}\left( g(a)+4g(a+\Delta x)+2g(a+2\Delta x) +4g(a+3\Delta x)+g(b) \right)$

In this problem:

$a=1,b=3,n=4, \displaystyle\Delta x=\frac{b-a}{n}=\frac{2}{4}=\frac{1}{2},g(x)= \sqrt{1+\frac{1}{x^2}}$

So, the Simposn’s rule formula becomes:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\\\approx \frac{\frac{1}{3}}{3}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{1}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(1+\frac{2}{2}\right)^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{3}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$

Then simplifying a bit:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx \approx \frac{1}{9}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(\frac{3}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(2\right)^2}} +4\sqrt{1+\frac{1}{\left(\frac{5}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$

Then we just do those computations and we finally get the approximation via Simposn's rule:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\approx 1.5355453$

While when we do the integral by using technology we get: 2.3020.

The approximation with Simpon’s rule is close but about 33% smaller:

$\displaystyle\frac{2.3020-1.5355453}{2.3020}\cdot100\%\approx 33\%$

8 0

3 years ago