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3 years ago

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In the figures below, the cube-shaped box is 6 inches wide and the rectangular box is 10 inches long, 4 inches wide, and 4 inche

s high. How much greater is the volume of the cube-shaped box than the rectangular box?

1 answer:

11111nata11111 [884]3 years ago

4 0

The difference in volume= $v^{3} -V*L*H$
= $6^{3}-(10*4*4)$
=216-160
= 56 cubic inches

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<u><em>Step-by-step explanation:</em></u>

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<u><em>{ 11 , 10 , 17 , 18 }</em></u>

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It is important to know that the sum of two rational numbers is rational. Similarly, the sum between a rational and an irrational is irrational, but not always. Similarly, the sum of two irrational numbers is sometimes irrational, not always.

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At last, the product of two irrational numbers is sometimes irrational.

The following image shows the diagram

As you can observe, rational and irrational numbers don't have common elements, so they don't intersect.

Hence, the true statements are

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1 year ago

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

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Answer:

Step-by-step explanation:

From the given information:

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A)

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

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

B)

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

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C)

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

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

D)

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

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3 years ago