What is the equation of the line that is perpendicular to y=2/3x-5 and passes through (6,-1)?

2y + 3x - 16 = 0

Step-by-step explanation:

Equation of the line  is    y = 2x/3 - 5

From the equation , the slope m₁ = 2/3. therefore recall, from condition for perpendicularity, m₁m₂ = -1

The product of their gradient must be (-1).

Now since m₁ = 2/3 and m₁m₂ =-1

                          2m₂/3 = -1

Therefore , m₂ = -3/2.

Since the equation passes through the coordinate of (6,-1)

we now substitute for x and y in the equation of a line to get the y intercept (c)

  y = mx + c

  -1 = -3x/2 +c

  -1 = -3/2 x 6 + c

  -1 = -9 + c

  -1 + 9 = c

 Therefore c = 8

Now to get the equation of line that is perpendicular to y = 2x/3 - 5

 y = -3x/2 + 8

making it a linear equation,

 2y = -3x + 16

 2y +3x - 16 = 0

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)

no

step-by-step explanation: