Simplify. consider all the cases.
example:
x−5=

x−5, if x> 5
0, if x=5
5−x, if x< 5

2x+7

a) if x > , then 2x + 7 =
b)if x = , then 2x + 7 =
c) if x < , then 2x + 7 =
(x−4)+7

if x > , then (x − 4) + 7 =
if x = , then (x − 4) + 7 =
if x < , then (x − 4) + 7 =

7−(x−7)

if x > , then 7 − (x − 7) =
if x = , then 7 − (x − 7) =
if x < , then 7 − (x − 7) =

7−(12−x)

if x > , then 7 − (12 − x)=
if x = , then 7 − (12 − x =
if x < , then 7 − (12 − x)=

< 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⭐

sadsdda

step-by-step explanation:

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180

its 60

4 x 12 = 48

3 x 4 = 12

48 + 12 = 60

yeeeet