The equation 2x^2+12x=4 is written in the form of 2(x-p)^2+q=0. what is the value of q?

1) Factor out the coefficient 2:  2x^2+12x=4 => 2(x^2 + 6x) = 4

2) Complete the square of x^2 + 6x:  x^2 + 6x + 9 - 9.  Then:

                                                    2(x^2 + 6x + 9 - 9) = 4

3) Rewrite the square as the square of a binomial:

                                                    2( [x+3]^2 - 9) = 4, or

                                                     2([x+3]^2)       = 22, or

                                                     2( [x+3]^2 ) - 22 = 0

The value of q is -22.

< 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⭐

24/56=hx^7

step-by-step explanation: