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3 years ago

What is the radius and diameter of a circle 4.2 cm

Mathematics

1 answer:

Tema [17]3 years ago

3 0

R=2.1
Diameter would be half of that

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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:

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

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

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

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

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

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

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

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

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2 years ago

A car driving cross-country traveled 180 miles in 4 hours.At this rate of speed,how many miles farther will the car travel in th

kompoz [17]

180/4 * 3 = 45 * 3 = 135

6 0

3 years ago

Suppose that a box is being towed up an inclined plane as shown in the figure. Find the force w needed to make the component of

Lina20 [59]

You need to find, first, the horizontal component of the inclined force (call it Wx), using an angle of 15 degrees (which is the angle between the inclined plane and the horizontal).

=> cos(15°) = Wx / 2.5 lb => Wx = 2.5 lb * cos(15°).

Second, you can find the inclined force , W, needed to produce the horizontal component Wx.

cos(33°) = Wx / W => W = Wx / cos (33°) = 2.5 lb*cos(15°) / cos(33°) = 2.88 lb

Answer: 2.88 lb

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