The equation for line t can be written as y=7/6x-10. line u, which is parallel to line t, includes the point (2,5). which is the equation of line u?

Hi, there--

I am not sure if my answer is going to help you but I am also sure it will, frankly my answer is probaly not what you want but it took some time to write this so I hope this helps!

The Problem:

Consider the line y=7x-6

Find the equation of the line that is parallel to this line and passes through the point (6,-5).

Find the equation of the line that is perpendicular to this line and passes through the point (6,-5).

A Solution:

When two lines are parallel, they have the same slope, but the y-intercepts are different.

{1} parallel line

Your original equation is in slope-intercept form: (y=mx+b). The slope m is the coefficient of the

x term; it's  7.

We know that the parallel line has a slope of 7 and that it passes through the point (6,-5).

We'll substitute these known values into the slope-intercept form and solve for the

y-intercept b.

y = mx + b

-5 = (7)(6) + b

-5 = 42 + b

b = -47

The equation of the parallel line is y = 7x - 47.

{2} perpendicular line

If a line has a slope m, then the line perpendicular to it has the slope -1/m.

Since the original equation has a slope of 7, the perpendicular line has a slope of -1/7.

From here the procedure is quite similar to finding the parallel line's equation. Use the

slope-intercept form to find the y-intercept.

y = mx + b

-5 = (-1/7)(6) + b

-5 = (-6/7) + b

b = -5 + 6/7

b = -29/7

The equation for the perpendicular line is y = (-1/7)x - 29/7.

I hope this helps you!!!

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

ver más en -

step-by-step explanation:

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~