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3 years ago

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write an equation of the line perpendicular to the given line that contains P in point slope form? P (4,8) y=4x-7

2 answers:

astraxan [27]3 years ago

8 0

Answer:

y - 8 = -1/4(x - 4)

Step-by-step explanation:

Perpendicular lines have negative-reciprocal slopes.

The line contains point P(4, 8)

<u>Point slope form:</u>

y - y₁ = m(x - x₁)

<u>Given line:</u>

y = 4x - 7

The slope of the perpendicular line is m = -1/4.

<u>The line in point-slope form is:</u>

y - 8 = -1/4(x - 4)

gregori [183]3 years ago

8 0

Answer:

y - 8 = -1/4(x - 4)

Step-by-step explanation:

If two lines are perpendicular to each other, they have opposite slopes.

The first line is y = 4x - 7. Its slope is 4. A line perpendicular to this one will have a slope of -1/4, the reciprocal (opposite) of the given equation.

Plug this value (-1/4) into your standard point-slope equation of y - y₁ = m(x - x₁)

y - y₁ = -1/4(x - x₁)

To find x₁ and y₁, we want to plug in a value that we know is on this line: in this case, it is P (4, 8). Plug in the x and y values into the x₁ and y₁ of the standard equation.

y - 8 = -1/4(x - 4)

This equation is perpendicular to your given equation (y = 4x - 7) and contains point P (4, 8)

Hope this helps!

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Graph this rational equation. Identify the points of discontinuity, holes, vertical asymptotes, x-intercepts, and horizontal asy

irakobra [83]

Step-by-step explanation:

We have given,

A rational function : f(x) = $\frac{x-2}{x-4}$

W need to find :

Point of discontinuity : - At x = 4, f(x) tends to reach infinity, So we get discontinuity point at x =4.

For no values of x, we get indetermined form (i.e $\frac{0}{0}$ ), Hence there is no holes

Vertical Asymptotes:

Plug y=f(x) = ∞ in f(x) to get vertical asymptote {We can us writing ∞ = $\frac{1}{0}$ }

i.e ∞ = $\frac{x-2}{x-4}$

or $\frac{1}{0}=\frac{x-2}{x-4}$

or x-4 =0

or x=4, Hence at x = 4, f(x) has a vertical asymptote

X -intercept :

Plug f(x)=0 , to get x intercept.

i.e 0 = $\frac{x-2}{x-4}$

or x - 2 =0

or x = 2

Hence at x=2, f(x) has an x intercept

Horizontal asymptote:

Plug x = ∞ in f(x) to get horizontal asymptote.

i.e f(x) = $\frac{x-2}{x-4}$ = $\frac{x(1-\frac{2}{x} )}{x(1-\frac{4}{x} )}$

or f(x) = $\frac{(1-\frac{2}{∞} )}{(1-\frac{4}{∞} )}$

or f(x) = 1 = y

hence at y =f(x) = 1, we get horizontal asymptote

4 0

3 years ago

Can someone help me please?

lana [24]

Step-by-step explanation:

#12)

1 5/8= 8x1 = 8 + 5 = 13

13/8 = 13÷8 = 1.625 and it not a repeating decimal

#13)

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3 years ago

Find the slope of the line passing through (3, 7) and (6, 11). Note: Use the y-y/x-x formula!

lana [24]

Answer:

Slope of the line $m = \frac{4}{3}$

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given points are ( 3,7) and (6,11)

Slope of the line

$m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

$m = \frac{11-7}{6-3}$

$m = \frac{4}{3}$

Slope of the line $m = \frac{4}{3}$

6 0

3 years ago

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jek_recluse [69]

Answer:

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Step-by-step explanation:

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cross multiplying

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erastovalidia [21]

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2 years ago

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