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3 years ago

What is the difference between -4=7 vs 5=5

Mathematics

1 answer:

KiRa [710]3 years ago

3 0

Answer:

{-4 = 7, 5 = 5} = {False, True}

Step-by-step explanation:

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Recall that the Fibonacci Sequence is defined by the recurrence relation, a0 = a1 = 1 and for n ≥ 2, an = an−1 + an−2 . a. Show

e-lub [12.9K]

Answer:

Step-by-step explanation:

From the given information:

$a_n = a_{n-1} + a_{n-2}; \ \ \ n \ge 2 \\ \\ a_o = 1 \\ \\ a_1 =1 \ \ \ \ \ since \ \ a_o = a_1 = 1$

$a_n - a_{n-1} - a_{n-2} = 0 \\ \\ \implies \sum \limits ^{\infty}_{n=2}(a_n -a_{n-1}-a_{n-2} ) x^n = 0 \\ \\ \implies \sum \limits ^{\infty}_{n=2} a_nx^n - \sum \limits ^{\infty}_{n=2} a_{n-1}x^n - \sum \limits ^{\infty}_{n=2}a_{n-2} x^n = 0 \\ \\ \implies (a(x) -a_o-a_1x) - (x(a(x) -a_o)) -x^2a(x) = 0 \\ \\ \implies a(x) (1 -x-x^2) -a_o-a_1x+a_ox = 0 \\ \\ \implies a(x)(1-x-x^2)-1-x+x=0 \\ \\ \implies a(x) (1-x-x^2) = 1$

$\mathbf{Generating \ Function: a(x) = \dfrac{1}{1-x-x^2}=f(x)}$

$If \ \ 1 -x-x^2 = (1 - \alpha x) ( 1- \beta x) \\ \\ \implies 1 -x - ^2 = 1 + \alpha \beta x^2 - ( \alpha + \beta )x \\ \\ \text{It implies that:} \\ \\ \alpha \beta = -1 \\ \\ \alpha + \beta = 1 \\ \\ \implies \alpha = ( 1-\beta) \\ \\ ( 1- \beta) \beta = -1 \\ \\ \implies \beta - \beta^2 = -1 \implies \beta - \beta^2 -1 = 0\\ \\ \beta = \dfrac{-(-1) \pm \sqrt{(-1)^2 -4(1)(-1)}}{2(1)}$

$\beta = \dfrac{1\pm \sqrt{5}}{2} \\ \\ \beta = \dfrac{1 + \sqrt{5}}{2} \ \ and \ \ \alpha = \dfrac{1 - \sqrt{5}}{2}$

$\dfrac{1}{1-x-x^2}= \dfrac{A}{1-\alpha x}+ \dfrac{\beta}{1-\beta x} \\ \\ = \dfrac{A(1-\beta x) + B(1-\alpha x)}{(1-\alpha x) (1 - \beta x)} \\ \\ = \dfrac{(A+B)-(A\beta+B\alpha)x}{(1-\alpha x) (1-\beta x)}$

$\text{It means:} \\ \\ A+B=1 \\ \\ B = (1-A) \\ \\ A\beta+ B \alpha =0 \\ \\ A\beta ( 1 -A) \alpha = 0 \\ \\ A( \beta - \alpha ) = -\alpha \\ \\ A = \dfrac{\alpha}{\alpha - \beta } \\ \\ \\ \\ B = 1 - \dfrac{\alpha }{\alpha - \beta} \implies \dfrac{\alpha - \beta - \alpha }{\alpha - \beta } \\ \\ =\dfrac{-\beta }{\alpha - \beta} \\ \\ \mathbf{B = \dfrac{\beta }{\beta - \alpha }}$

$\text{The formula for} a_n: \\ \\ a(x) = \dfrac{\alpha }{\alpha - \beta }\sum \limits ^{\infty}_{n=0} \alpha ^n x^n - \dfrac{\beta}{\beta - \alpha }\sum \limits ^{\infty}_{n=0} \beta x^n \\ \\ \implies \sum \limits ^{\infty}_{n =0} \dfrac{\alpha ^{n+1}- \beta ^{n+1}}{\alpha - \beta}x^n \\ \\ a_n = \dfrac{\alpha ^{n+1}- \beta ^{n+1}}{\alpha - \beta } \\ \\ \\ a_n = \dfrac{1}{\sqrt{5}} \Big (\Big( \dfrac{\sqrt{5}+1}{2}\Big)^{n+1}- \Big ( \dfrac{1-\sqrt{5}}{2}\Big) ^{n+1}\Big)$

8 0

2 years ago

Write the equation of the line that is perpendicular to the graph of 2x+y =5, and whose y-intercept is 4.

Stells [14]

Answer and Step-by-step explanation:(MAKE BRAINLIEST PLEASE!!)

Let the quantity of understudies going to the game be s and the quantity of non-understudies going to the game be n. The all out number of individuals going to the game is 500. Address the present circumstance in a condition as s + n = 500.

Every understudy pays $5 for confirmation. Along these lines, s understudies pay 5s dollars in confirmations. Each non-understudy pays $10 for affirmation. Along these lines, n non-understudies pay 10n dollars in affirmations. Thus, the all out affirmations gathered is 5s + 10n.

The aggregate sum gathered was $4,000. So the subsequent condition is 5s + 10n = 4,000.

6 0

3 years ago

Marcus hikes at a rate of 2 miles per hour if he hikes for 6 1/3 hours how many will he hike

Firdavs [7]

2 = 1 hour
x = 6 1/3

2 * 6 1/3 = 12 2/3

3 0

3 years ago

Read 2 more answers

given the equation 5y-3x=15 how would you determine another equation which would have: no solution, one solution, and infinitely

victus00 [196]

First, I would convert the equation to slope intercept form.

That would be: y = (3/5)x + 3 , now it will be easier to create the needed equations.

For no solution, keep the slope the same and change the y-intercept.
You could have: y = (3/5)x + 9

For one solution, all we have to do is change the slope.
You could have: y = (4/5)x + 3

For an infinite number of solutions, just multiply the entire equation by any number. Let's use the number 2.
2y = (6/5)x + 6

3 0

3 years ago

Which statements must be true about the image of ΔMNP after a reflection across Line E G? Select three options. The image will b

masya89 [10]

Answer:

The image will be congruent to ΔMNP.

Line EG will be perpendicular to the line segments connecting the corresponding vertices.

The line segments connecting corresponding vertices will all be parallel to each other.

Step-by-step explanation:

The reflected image will be the mirror image of ΔMNP, so it will be congruent. The lines connecting each vertex to the corresponding vertex will be parallel, and all three lines are perpendicular to EG.

4 0

2 years ago