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3 years ago

A line goes through the points (5, -1) and (7, -5). What is a possible equation to a new line that is perpendicular?

1 answer:

7 0

Ok, so you are given an equation in standard form, 5x+4y=2, and a point, (7, 5). You are being asked to write the equation for a line that is parallel to the equation in standard from, and that includes the point (7, 5).

First, let's start by finding the slope of your new line. We know that it needs to have the same slope as 5x+4y=2, because parallel lines have the same slopes. To do that, we need to put the equation into slope-intercept form (y=mx+b), which means we need to isolate the "y."

5x+4y=2

-5x -5x (subtract 5x from both sides to move it to the right side of your equation)

4y = 2 - 5x

/4 /4 /4 (divide all the terms by 4 to get "y" by itself)

y = (1/2) - (5/4)x ... I suggest leaving your slope as a fraction.

Now, we know that our slope, m , is going to be -(5/4).

Next, we are going to use our slope, -(5/4), and point, (7, 5), to find the b-value (y-intercept) of your new line. Let's plug in what we know:

y = 5

m = -(5/4)

x = 7

y=mx+b

5=(-5/4)(7) + b -> I plugged in what we knew for y, m, and x.

5 = -(35/4) + b -> I multiplied the numbers in the numerator (5 x 7) to get 35/4

20/4 = -(35/4) + b -> I converted 5 into a faction with a denominator of 4 by multiplying by (4/4)

+(35/4) +(35/4) -> I add (35/4) to both sides to isolate b

55/4 = b ... or b = 13.74

Answer:

m = -(5/4)

b = 13.75

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Find the point (x,y) on the line y = x that is equidistant from the points (-6,2) and (10,-9)

Nezavi [6.7K]

Answer:

(14.1, 14.1)

Step-by-step explanation:

If the point (x,y) is on the line y=x, then it has coordinates (x,x).

Find the distances from this point to points (-6,2) and (10,-9):

distance from (x,x) to (-6,2): $\sqrt{(x-(-6))^2+(x-2)^2}=\sqrt{(x+6)^2+(x-2)^2}$
distance from point (x,x) to (10,-9): $\sqrt{(x-10)^2+(x-(-9))^2}=\sqrt{(x-10)^2+(x+9)^2}$

Equate these distances:

$\sqrt{(x+6)^2+(x-2)^2}=\sqrt{(x-10)^2+(x+9)^2}$

Square this equation:

$(x+6)^2+(x-2)^2=(x-10)^2+(x+9)^2\\ \\x^2+12x+36+x^2-4x+4=x^2-20x+100+x^2+18x+81\\ \\8x+40=-2x+181\\ \\8x+2x=181-40\\ \\10x=141\\ \\x=14.1$

So, the coordinates are (14.1, 14.1)

6 0

4 years ago

What is 1/5 expressed as a percent?

juin [17]

1/5 is expressed 20%

6 0

3 years ago

State the vertical asymptote of the rational function. f(x) =((x-9)(x+7))/(x^2-4)

Lubov Fominskaja [6]

$D:x^2-4\not=0\\
D:x^2\not=4\\
D:x\not=-2 \wedge x\not =2\\\\
\displaystyle
\lim_{x\to-2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\
\dfrac{(-2-9)(-2+7)}{(-2^-)^2-4}=\dfrac{-11\cdot5}{4^+-4}=\dfrac{-55}{0^+}=-55\cdot\infty=-\infty\\
\dfrac{(-2-9)(-2+7)}{(-2^+)^2-4}=\dfrac{-11\cdot5}{4^--4}=\dfrac{-55}{0^-}=-55\cdot(-\infty)=\infty
$

$\displaystyle
\lim_{x\to2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\
\dfrac{(2-9)(2+7)}{(2^-)^2-4}=\dfrac{-7\cdot9}{4^--4}=\dfrac{-63}{0^-}=-63\cdot(-\infty)=\infty\\
\dfrac{(2-9)(2+7)}{(2^+)^2-4}=\dfrac{-7\cdot9}{4^+-4}=\dfrac{-63}{0^+}=-63\cdot\infty=-\infty\\$

So, the vertical asymptotes are $x=\pm 2$

5 0

4 years ago

Read 2 more answers

What fractions are equivalent or equal to 1/2

Savatey [412]

Answer:

some examples are 2/4, 3/6, 4/8, 5/10, 6/12, 7/14 and so on.

Step-by-step explanation:

any fraction with the numerator (number on the top) half of the denominator (number on the bottom) would apply to this.

4 0

3 years ago

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Differentiating Functions of Other Bases In Exercise, find the derivative of the function.

aniked [119]

Answer:

$\dfrac{dy}{dx} = \frac{2x-3}{\ln 16(x^2 - 3x)}$

Step-by-step explanation:

We are given the following in the question:

$y = \ln {16}(x^2 - 3x)$

We have to find the derivative of the given expression.

$y = \ln 16(x^2 - 3x)\\\text{Using the log propert}\\\\\log_a b = \dfrac{\log b}{\log a}\\\\dfrac{d(x^n)}{dx} = nx^{n-1}\\\\\dfrac{d(\log x)}{dx} = \dfrac{1}{x}\\\\\text{\bold{Differentiating we get}}\\\\\displaystyle\frac{dy}{dx} = \frac{d(\ln 16(x^2-3x))}{dx}\\\\= \frac{1}{16(x^2-3x)})\frac{d(x^2-3x)}{dx}\\\\=\frac{1}{\log 16(x^2-3x)}(2x - 3)\\\\= \frac{2x-3}{\ln 16(x^2 - 3x)}$

$\dfrac{dy}{dx} = \frac{2x-3}{\ln 16(x^2 - 3x)}$

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4 years ago