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3 years ago

Write the equation of the line yhat is perpendicular to 5x+8y=16 and passes through (-5,7)

Mathematics

1 answer:

Viefleur [7K]3 years ago

5 0

Answer:

$y-7=\frac{8}{5}(x+5)$

$y=\frac{8}{5} x +15$

Step-by-step explanation:

To write the equation, convert the line to slope intercept form to find the slope.

5x+8y=16

8y=16-5x

y=-5/8 x +2

The slope of this line is -5/8. The slope of the line perpendicular to this line will be the negative reciprocal or 8/5.

Write the equation using the slope m=8/5 and the point slope form.

$y-y_1=m(x-x_1)\\y-7=\frac{8}{5}(x--5)\\y-7=\frac{8}{5}(x+5)$

This is the equation of the line. You can convert it to slope intercept form:

$y-7=\frac{8}{5}(x+5)\\y-7 = \frac{8}{5}x + 8\\y=\frac{8}{5}x +15$

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3 years ago

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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:

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This implies that:

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Also;

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

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

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

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

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

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

Thus, x = 2 is an irregular singular point.

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