What is the equation of a line that passes through the point (6, 1) and is perpendicular to the line whose equation is y=−2x−6 ? enter your answer in the box.

y = 1/2x - 2

Look at Desmos Math pic i sent.

y =  +   3

Step-by-step explanation:

To find the equation that passes through the points (6,1) and is perpendicular to the line whose equation is y = -2x - 6, we are going to follow the steps below;

First, we determine the slope of the equation:y = -2x - 6, only then can we find the slope of our perpendicular equation.

y = -2x - 6

m = -2

The slope of the above line is   -2

So, the slope of the perpendicular line to y = -2x - 6 will have a slope equals to the negative reciprocal , that is;     =   -1

      The slope(m) of our perpendicular equation is using the above formula.

Haven gotten our  slope, next is for us to find our intercept

To get the intercept, we will use this standard equation;

y = mx + c

where m =slope(our new slope=1/2)   c=intercept   x and y are the two points through which the line passes through. That is; x=6 and y=1. So we are going to plug in all this variable into the standard equation;

y = mx + c

1  =  (6)  +   c

1  =  3  +  c          (six will divide two to give us three)

To get the value of c, subtract 3 from both-side of the equation

1 - 3  =  3+ c -3

-2  =   c

c = -2

Therefore, our new intercept is 3

So we can now plug in our new slope and intercept into y=mx+c

y =  +   3

We can re-arrange it;

- + y  =  3

the figures below are made out of circles, semicircles, quarter … find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations). remember that we'll also have to cut whatever areas we get in half, since we're only dealing

step-by-step explanation:

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17