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3 years ago

Write an equation for a line perpendicular to y=2x+5 and passing through the point (-6,5)

Mathematics

2 answers:

ivanzaharov [21]3 years ago

5 0

Answer:

5 would be on your Y axis (go up two numbers from 5, go to the right 1 and draw a dot then connect your lines

Step-by-step explanation:

AleksandrR [38]3 years ago

4 0

Answer:

$y=-\frac{1}{2}x+2$

Step-by-step explanation:

So we want to find the equation of a line perpendicular to y=2x+5 and passes through the point (-6,5).

First, let's determine the slope of our equation. Remember that the slopes of perpendicular lines are negative reciprocals. In other words:

$m_1\cdot m_2=-1$

To find our slope, let's substitute 2 (the slope of y=2x+5) for m₁ and solve for m₂. So:

$2\cdot m_2=-1$

Divide both sides by 2:

$m_2=-\frac{1}{2}$

Therefore, the slope of our new line is -1/2.

Now, we can use the point-slope form to find the equation of our line. The point-slope form is:

$y-y_1=m(x-x_1)$

Where m is the slope and (x₁, y₂) is a point.

So, let's substitute -1/2 for m and (-6,5) for (x₁, y₂), respectively. So:

$y-5=-\frac{1}{2}(x-(-6))$

Simplify:

$y-5=-\frac{1}{2}(x+6)$

Distribute:

$y-5=-\frac{1}{2}x-3$

Add 5 to both sides:

$y=-\frac{1}{2}x+2$

So, our equation is:

$y=-\frac{1}{2}x+2$

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A copy machine makes 24 copies per minute . how long does it take to make 84 copies

malfutka [58]

Simply divide 84 by 24

$\frac{84}{24} = \frac{21}{6} = 3 \frac{3}{6} = 3 \frac{1}{2}$

$3 \frac{1}{2} \: minutes$

3 0

4 years ago

Read 2 more answers

BRAINLIESTTT ASAP!<br><br> please answer

Pie

When you add the two angles together, they should equal 90 degrees (according to the square marker). So, all you have to do is add the two angles together, make them equal to 90, and solve for x.
2x+8+5x-2=90
7x+6=90
7x=84
x=12

Now, we substitute x and solve for the angles.
2x+8
2(12)+8
32 degrees

5x-2
5(12)-2
58 degrees

To double check our answer, we can add the two angles together and see if we get 90.
32+58=90 yes

Our answer is correct

3 0

3 years ago

¿Para cuál(es) valor(es) de p las rectas de ecuación x − 1/p = 2 − y/p y x − 1/1 − p = y − 2/2 son perpendiculares? A) Solo para

Akimi4 [234]

Respuesta:

C) Solo para el -1

Explicación paso a paso:

Para resolver este problema, debemos de determinar la pendiente en cada una de las ecuaciones provistas:

$\frac{x-2}{p}=\frac{2-y}{p}$

$\frac{x-1}{1-p}=\frac{y-2}{2}$

ahora bien, necesitamos conocer el valor de la pendiente de una de las dos ecuaciones. Tomemos la primera ecuación y resolvámosla para y:

$\frac{x-2}{p}=\frac{2-y}{p}$

Multiplicamos ambos lados para p y obtenemos:

x-1=2-y

volteamos la ecuación y nos da:

2-y=x-1

pasamos el 2 a restar al otro lado y nos da:

-y=x-1-2

-y=x-3

y dividimos ambos lados de la ecuación dentro de -1

y=-x+3

esta ecuación ya tiene la forma pendiente intercepto:

y=mx+b

donde m es nuestra pendiente:

$m_{1}=-1$

Esta es la pendiente de una de las dos ecuaciones, para que la segunda ecuación sea perpendicular a la primera, su pendiente debe de ser el recíproco negativo de la pendiente de la primera ecuación, entonces la pendiente de la segunda ecuación debe ser:

$m_{2}=-\frac{1}{m_{1}}$