Find the possible value or values of n in the quadratic equation 2n2 – 7n + 6 = 0. a. n = −2, n = −32 b. n = 32, n = 2 c. n = 6, n = 0 d. n = 1, n = 3

B. n = 3/2, n = 2.

Step-by-step explanation:

2n^2 – 7n + 6 = 0

Use the ac method:-

2 * 6 = 12 , We need 2 numbers whose product is 12 and whose sum = -7.

They are -3 and - 4, so we write:-

2n^2 - 4n - 3n + 6 = 0

2n(n - 2) - 3(n - 2) = 0

(2n - 3)(n - 2) = 0

n = 3/2, 2.

don't know sorry bru

step-by-step explanation:

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