Given O A ‾ ⊥ O C ‾ OA ⊥ OC start overline, O, A, end overline, \perp, start overline, O, C, end overline m ∠ B O C = 6 x − 6 ∘ m∠BOC=6x−6 ∘ m, angle, B, O, C, equals, 6, x, minus, 6, degrees m ∠ A O B = 5 x + 8 ∘

x = 8

m∠BOC = 37°

m∠AOB = 53°

Step-by-step explanation:

If OA is perpendicular to OC, this means that <AOC = 90°

Given

m∠BOC=6x−6

m ∠AOB = 5 x + 8

The expression is true

m∠BOC+m∠AOB= m∠AOC

6x-6+5x+8 = 90

Find x:

11x + 2 = 90

11x = 90-2

11x = 88

x = 88/11

x =8

Get m∠BOC:

m∠BOC= 6x-6

m∠BOC= 6(8)-11

m∠BOC = 48-11

m∠BOC= 37°

Get m∠AOB;

m∠AOB = 90-m∠BOC

m∠AOB =90-37

m∠AOB = 53°

b

step-by-step explanation:

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.