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sp2606 [1]
3 years ago
12

Section 1 - Topic 9 Parallel and Perpendicular Lines - Part 2 (please help)

Mathematics
1 answer:
RUDIKE [14]3 years ago
6 0

9514 1404 393

Answer:

  1. y = -1/7x
  2. y = 2x -7

Step-by-step explanation:

1. The slope of the given line is the x-coefficient, 7. The slope of the perpendicular line is the negative reciprocal of this: -1/7. Since the line goes through the origin, the y-intercept is zero.

  y = -1/7x

__

2. The first part of this problem is to find the solution to the given system of equations. We can substitute for x to do that:

  3(5 -y)-2y = 10

  15 -5y = 10 . . . simplify

  -3 +y = -2 . . . . divide by -2

  y = 1, x = 4 . . . add 3 to find y; subtract y from 5 to find x

__

The second part of this problem requires we note the slope of the parallel line. It is the x-coefficient, 2. The y-intercept can be found from ...

  b = y -mx = 1 -2(4) = -7

So, the desired line in y=mx+b form is ...

  y = 2x -7

_____

The first attachment shows the perpendicular lines of problem 1.

The second attachment shows the parallel lines of problem 2.

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The point slope form of a line is written as y - y1 = m(x -x1)

Where m is the slope and x1 and y1 are the points on the line.

You are told the slope is 3 and the point is (2,-1/2)

x1 is 2 and y1 is -1/2

Replace those in the equation to get:

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Simplify to get the final answer:

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Write the equation of the graph shown below in factored form. a graph that starts at the top left and continues down through the
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Answer:

  (d)  f(x) = (x − 3)^2(x − 2)(x − 1)

Step-by-step explanation:

In this context, a crossing of the axis at x=p means there is a factor of (x-p). A "touch" of the axis at x=q means there is a factor of (x -q)^2.

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2 years ago
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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Step-by-step explanation:

i don't know if correct

nakalimutan kuna yan kung pano gawin sorry

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