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OverLord2011 [107]
3 years ago
15

Write the first five terms of the sequence defined by the recursive formula an=2*(an-1)+3 with a1=-2

Mathematics
1 answer:
Tasya [4]3 years ago
5 0

Answer:

<u>The first five terms of the sequence are 5, 7, 9, 11 and 13.</u>

Step-by-step explanation:

The first five terms of the sequence are:

a₁ = 2 * (2 - 1) + 3 = 2 * 1 + 3 = 2 + 3 = 5

a₂ = 2 * (3 - 1) + 3 = 2 * 2 + 3 = 4 + 3 = 7

a₃ = 2 * (4 - 1) + 3 = 2 * 3 + 3 = 6 + 3 = 9

a₄ = 2 * (5 - 1) + 3 = 2 * 4 + 3 = 8 + 3 = 11

a₅ = 2 * (6 - 1) + 3 = 2 * 5 + 3 = 10 + 3 = 13

<u>The first five terms of the sequence are 5, 7, 9, 11 and 13.</u>

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Alinara [238K]

Answer:

The circulation of the field f(x) over curve C is Zero

Step-by-step explanation:

The function f(x)=(x^{2},4x,z^{2}) and curve C is ellipse of equation

16x^{2} + 4y^{2} = 3

Theory: Stokes Theorem is given by:

I= \int \int\limits {{Curl f\cdot \hat{N }} \, dx

Where, Curl f(x) = \left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right]

Also, f(x) = (F1,F2,F3)

\hat{N} = grad(g(x))

Using Stokes Theorem,

Surface is given by g(x) = 16x^{2} + 4y^{2} - 3

Therefore, tex]\hat{N} = grad(g(x))[/tex]

\hat{N} = grad(16x^{2} + 4y^{2} - 3)

\hat{N} = (32x,8y,0)

Now,  f(x)=(x^{2},4x,z^{2})

Curl f(x) = \left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\F1&F2&F3\end{array}\right]

Curl f(x) = \left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\frac{∂}{∂x} &\frac{∂}{∂y} &\frac{∂}{∂z} \\x^{2}&4x&z^{2}\end{array}\right]

Curl f(x) = (0,0,4)

Putting all values in Stokes Theorem,

I= \int \int\limits {Curl f\cdot \hat{N} } \, dx

I= \int \int\limits {(0,0,4)\cdot(32x,8y,0)} \, dx

I= \int \int\limits {(0,0,4)\cdot(32x,8y,0)} \, dx

I=0

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3 0
3 years ago
A group consists of 6 men and 5 women. Five people are selected to attend a conference. In how many ways can 5 people be selecte
cluponka [151]

Answer:

462

6

1.2%

Step-by-step explanation:

Since the questions are combinations, we must apply the combination formula, which is as follows:

n C r = (n!) / (r! (n-r)!)

Because there are 6 men and 5 women, there are a total of 11 people.

Thus:

n = 11

In the first question in how many ways can 5 people be selected from this group of 11, r = 5.

Replacing in the formula:

11 C 5 = (11!) / (5! * (11-5)!)

11 C 5 = (11!) / (5! * 6!)

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In the second question, in how many ways can 5 men be selected from the 6 men, here n = 6 and r = 5, replacing we are left with:

6 C 5 = (6!) / (5! * (6-5)!)

6 C 5 = (6!) / (5! * 1!)

6 C 5 = 6

In the last question of what is the probability that the selected group is all men, we have that it is the combination of the two previous questions. Since the total would be part A it would be the total of the combinations of choosing 5 of 11 people and part B of the 6 men that there are the combinations of choosing 5.

Divide the two values ​​from parts A and B to get ...

(result from part B) / (result from part A) = (# of ways to pick 5 men) / (# of ways to pick 5 people)

(result from part B) / (result from part A) = 6/462

(result from part B) / (result from part A) = 0.012

In other words, the probability is 1.2%

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3 years ago
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Answer:

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Answer:

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Step-by-step explanation:

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Kisachek [45]

Answer:

Step-by-step explanation:

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