Use the work shown below to write the equation for a line that has a slope of Negative two-fifths and passes through (15, –2). y = mx + b

1. Substitute the known values: negative 2 = negative two-fifths (15) + b. 2. Solve for b: negative 2 = negative 6 + b. b = 4.
What is the equation of the line in slope-intercept form?

Negative 2 = negative two-fifths x + 4

y = negative two-fifths (15) + 4

y = negative two-fifths x minus 4

y = negative two-fifths x + 4

y=negative two fifths x plus 4

Step-by-step explanation:

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step-by-step explanation:

< 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

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