Does anyone know the equation to this graph

y = -x - 2

Step-by-step explanation:

You can start with the equation of a line in the slope-intercept form.

y = mx + b

m = slope

b = y-intercept

Look at the graph to find the y-intercept. At what point on the y-axis does the graph intersect the y-axis? -2. The y-intercept is -2. The value of b is -2.

Now we hjave

y = mx + (2)

or simply

y = mx - 2

Now we need to find the slope.

The slope is the ratio of the rise (difference in y) over the run (difference in x).

Look at the point (6, -7) that is shown in your graph. Go up 1 unit vertically. That is a rise of 1. Now go left 1 unit horizontally. That is a run of -1.

slope = m = rise/run = 1/(-1) = 1

The slope is 1. Now we replace m with -1.

y = mx - 2

y = -1x - 2

y = x - 2

y = -x - 2

step-by-step explanation:

the first 5 right here the next 5 i will upload as separate question

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐