What is the equation of a line that passes through the point (8,1) and is perpendicular to the line whose equation is y= -2/3 x + 5? a) 2x+3y= 15 b) y= -2/3x + 11 c) y= 3/2x+ 11 d) y= 3/2x - 11 i am thinking its b but am not sure. i don't wanna retake this so i wanna be 100%.

The equation of this line would be y = 3/2x - 11, which is the answer D.

Step-by-step explanation:

In order to find this equation we must first find the slope of the original line. The original slope (the coefficient of x) is -2/3, which means the new slope will also be 3/2 because perpendicular lines have opposite and reciprocal slopes. Now, we can use this slope along with the point in point-slope form to find the equation of the line.

y - y1 = m(x - x1)

y - 1 = 3/2(x - 8)

y - 1 = 3/2x - 12

y = 3/2x - 11

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