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3 years ago

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a) Give the definition of sequence of real numbers (Xn). Then give 2 example. b) Give the definition of convergent sequence. The

n give 2 example. c) Give the definition of Cauchy sequence. Then give an example 0. d) Show that lim n+1 72

1 answer:

jolli1 [7]3 years ago

6 0

Answer:

a) Definition of sequence of real numbers: A sequence of real number is the function from set of natural number to the set of real numbers. i.e.

f: <em>N</em> → <em>R</em>

Example: $S_{n}=\frac{1}{n}$

$S_{n}=\frac{n}{n+1}$

b) Definition of convergent sequence: A sequence is said to be convergent if for very large value of n, function will give the finite value.

Example: $S_{n}=\frac{1}{n}$

$S_{n}=\frac{n}{n+1}$

c) Definition of Cauchy sequence: A sequence is said to be Cauchy Sequence if terms of sequence get arbitrary close to one another.

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What is the value of z, rounded to the nearest tenth? Use the law of sines to find the answer. 2. 7 units 3. 2 units 4. 5 units

worty [1.4K]

The law of sine is the nothing but the relationship between the sides of the triangle to the angle of the triangle (oblique triangle).

The value of the $z$ is 3.2 (rounded to the nearest tenth). The option 2 is the correct option.

<h3>What is law of sine?</h3>

The law of sine is the nothing but the relationship between the sides of the triangle to the angle of the triangle (oblique triangle).

It can be given as,

$\dfrac{\sin A}{a} =\dfrac{\sin B}{b} =\dfrac{\sin C}{c}$

Here $A,B,C$ are the angle of the triangle and $a,b,c$ are the sides of that triangle.

Given information-

The triangle for the given problem is attached below.

In the triangle the base of the triangle is 2.6 units long.

The sine law for the given triangle can be written as,

$\dfrac{\sin X}{x} =\dfrac{\sin Y}{y} =\dfrac{\sin Z}{z}$

As the value of $y$ side is known and the value of $z$ has to be find. Thus use

$\dfrac{\sin Y}{y} =\dfrac{\sin Z}{z}$

Put the values,

$\dfrac{\sin 51}{2.6} =\dfrac{\sin 76}{z}$

Solve it for the $z$ ,

$z=\dfrac{\sin 76\times 2.6}{\sin 51}\\z=3.2462$

Hence the value of the $z$ is 3.2 (rounded to the nearest tenth). The option 2 is the correct option.

Learn more about the sine law here;

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2 years ago

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Step-by-step explanation:

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3 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

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Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
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e) 2 + 3·(-1)^n
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g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

a)

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

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The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

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For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

__

Using these formulas on problem (d), we get ...

d)

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

__

And for problem (f), we get ...

f)

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

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3 years ago

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AlladinOne [14]

Answer:

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Step-by-step explanation:

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tester [92]

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