XY bisects ∠AXB. If m∠AXB = -18x° and m∠AXY = (3x2 - 12)°. Find m∠BXY. Please help, I'm blanking!!!

m<BXY = 36degrees

Step-by-step explanation:

If XY bisects ∠AXB, then;

<AXY + <BXY = <AXB

Given

m∠AXY = (3x^2 - 12)°

m∠AXB = -18x°

Required

Find m∠BXY.

From the formula above;

Find m∠BXY = <AXB - <AXY

m<BXY = -18x - (3x^2-12)

m<BXY = -18x - 3x^2 + 12

m<BXY =  -3x^2 -18x + 12

Also <AXY = m<BXY

3x^2 - 12 =   -3x^2 -18x + 12

6x^2 + 18x -24 = 0

x^2+3x-4 = 0

Factorize

x = -3±√9+16/2

x = -3±5/2

x = -3+5/2 and -3-5/2

x = 2/2 and -8/2

x = 1 and -4

Substitute x = 1 into m<BXY

m<BXY =  -3x^2 -18x + 12

m<BXY =  -3(1)^2 -18(1) + 12

m<BXY =  -3 -18+ 12

m<BXY =  -9

when x= -4

m<BXY =  -3(-4)^2 -18(-4) + 12

m<BXY =  -3(16) +72+ 12

m<BXY =  -48+84

m<BXY = 36degrees

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

bull

step-by-step explanation: