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3 years ago

Beth is writing out the steps using the "Shortest Route Algorithm". She just finished writing out all

Mathematics

1 answer:

JulijaS [17]3 years ago

5 0

Answer:

ACBD; 7

Step-by-step Explanation:

The Route Algorigtm finds the short route that is efficient, in order to find the solution following pathways.

From the pictures from the complete question,

STEP #1, there exist, three routes, which are:

AC,

and AD.

*AB is not viable because of the negative distance, so it can be removed .

* the AC distance is 3,and AD distance is 8. Therefore, AC can be selected and circled.is because of AC efficiency is high compare to AD.

*from selected AC, the possible routes are ACB; 3 and ACE; 6. which are route shown in #2

* ACB is circled, because it's more shorter than ACE

*The possible routes are ACBE; 8 and ACBD; 7. which are route shown in #3 . Therefore, route ACBD is circled because it's more shorter.

From the analysis of the algorithm, the route ACBD route is the shortest and the most efficient

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dentify all of the following solutions of square root of x minus 8 end root plus 8 equals x. (1 point)

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Answer:gh

Step-by-step explanation:

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3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

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3 years ago

Hey I hope this helps many people but how do you do this problem <br><br> 5x=0.4

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3 years ago

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