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2 years ago

13

On a coordinate plane, triangle T V W has points (negative 4, 8), (0, 4), and (4, 4).

1 answer:

Vanyuwa [196]2 years ago

6 0

The coordinates of the endpoints of line segments T'V' are; T'(-1, 2) and V'(0, 1).

<h3>What are the coordinates of the endpoints of the segment T'V'?</h3>

It follows from the task content that the transformation involved in the formation of the image from the pre-image is dilation by a scale factor of 1/4.

On this note, given that the coordinates of T and V from the task content are; (-4, 8) and (0,4), it follows that the coordinates of the endpoints as required are; T'(-1, 2) and V'(0, 1).

Read more on dilations;

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Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

ludmilkaskok [199]

Answer:

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:

$y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0$

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

y'' + p(x)y' + q(x) y = 0

This implies that:

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

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

$p(x) = \dfrac{1}{(x+2)(x-2)^2}$

Also;

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

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

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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3 years ago

8. The expression (x+5)(x+8)+(x+5)(2x - 3) can be written as the product of (x + 5) and

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Answer:

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Step-by-step explanation:

(x - 5)(x + 8) + (x + 5)(2x - 3) ← factor out (x + 5) from each term

= (x + 5)(x + 8 + 2x - 3)

= (x + 5)(3x + 5)

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EleoNora [17]

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Step-by-step explanation:

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