Letf(x) = x + |2x − 5|.find all solutions to the equationf(x) = 10.(enter your answers as a comma-separated list. if an answer does not exist, enter dne.)x = (b) letg(x) = 4x − 6 + |x + 3|.find all values of a which satisfy the equationg(a) = 3a + 9.(enter your answers as a comma-separated list. if an answer does not exist, enter dne.)a =

Part 1 : Given equation,

f(x) = x + |2x − 5|,

We have,

f(x) = 10,

x + |2x - 5| = 10

|2x - 5| = 10 - x

2x - 5 = ±(10-x)

Case 1 : 2x - 5 = 10 - x

2x = 10 - x + 5

2x + x = 15

3x = 15

⇒ x = 5

Case 2 : 2x - 5 = -10 + x

2x = -10 + x + 5

2x - x = -5

⇒ x = -5

Thus, the value of x is 5 or -5

Part 2 : Given equation,

g(x) = 4x − 6 + |x + 3|,

Put x = a,

g(a) = 4a - 6 + |a + 3|

We have, g(a) = 3a + 9

4a - 6 + |a+3|=3a + 9

4a + |a + 3 | = 3a + 9 + 6

|a + 3| = 3a - 4a + 15

|a + 3| = -a + 15

⇒ a + 3 = ±(-a + 15)

Case 1 : a + 3 = -a + 15

a = -a + 15 - 3

a + a = 12

2a = 12

⇒ a = 6

Case 2 : a + 3 = a - 15

a = a - 15 - 3

a - a = -18

0 = -18 ( False )

Thus, the value of a is 6.

< 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⭐

8 2/3 is the answer

step-by-step explanation: just use a calculator