What is the equation of a line perpendicular to y=2x-6 that passes through (5,4)?

what are the

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation:

1) the property is   ixi = a ⇔x = a   o x = -a

4(x-5)+3= 15⇒4x-20+3 = 15⇒4x-17= 15⇒4x = 15+17⇒4x = 32⇒x =32/4= 8

4(x-5) +3= -15⇒4x-20 +3 = -15⇒4x-17= -15⇒4x =-15+17⇒4x = 2⇒x =2/4= 1/2

the values are x =8 and x = 0,5

11) obtuse

13)right

15) acute