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2 years ago

The vertices of a triangle are located at X(-2,-1), Y(2,5) and Z(4,3). What is the perimeter of this triangle?

1 answer:

3 0

If you plot the points you will find you have to use the distance formula. Pick out 2 pairs label them (x1,y1) (x2,y2) then plug it into the distance formula. The repeat for the other sides. So you should get (2,5) (4,3) plug into distance formula what is get it distance between those two points, then (4,3) (-2,-1) plug into distance formula to get an answer, lastly (-2,-1) (2,5) plug in. Now you have three answers add all together and here is your perimeter of a triangle.

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ludmilkaskok [199]

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Step-by-step explanation:

Given that:

The differential equation; $(x^2-4)^2y'' + (x + 2)y' + 7y = 0$

The above equation can be better expressed as:

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The pattern of the normalized differential equation can be represented as:

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This implies that:

$p(x) = \dfrac{(x+2)}{(x^2-4)^2} \$

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$q(x) = \dfrac{7}{(x^2-4)^2}$

$q(x) = \dfrac{7}{(x+2)^2(x-2)^2}$

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

$\lim \limits_{x \to-2} (x+ 2) p(x) = \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}$

$\implies \dfrac{1}{16}$

$\lim \limits_{x \to-2} (x+ 2)^2 q(x) = \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{7}{(x-2)^2}$

$\implies \dfrac{7}{16}$

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

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Let the number of apples be x and that of pears be y, then:
0.64x + 0.45y = 5.26 . . . (1)
0.32x + 0.39y = 3.62 . . . (2)

(2) x 2 => 0.64x + 0.78y = 7.24 . . . (3)
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