I hope this helps, this is what I got
x/4 +18 ( x over 4 plus 18) :)
x=2 is only solution while x=1 is extraneous solution
Option C is correct.
Step-by-step explanation:
We need to solve the equation and find values of x.
Solving:
Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)
Multiply the entire equation with x(x-1)
Now, factoring the term:
The values of x are x=1 and x=2
Checking for extraneous roots:
Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.
If we put x=1 in the equation, the denominator becomes zero i.e
which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.
If we put x=2 in the equation,
So, x=2 is only solution while x=1 is extraneous solution
Option C is correct.
Keywords: Solving Equations and checking extraneous solution
Learn more about Solving Equations and checking extraneous solution at:
#learnwithBrainly
Answer:
a) f(16) = 42
b) f(16) = 54
c) f(16) = 162
d) f(16) = 30
Step-by-step explanation:
a) y = mx + b ∧ m = (f(8) - f(4))/(8-4) ⇒ m = (18 - 6)/(8 - 4) = 3
b = y - mx = 6 - 3(4) = 6 - 12 = - 6
f(16) = 3(16) - 6 = 42
b) y = kxⁿ ∧ f(4) = 6 = k4ⁿ ∧ f(8) = 18 = k8ⁿ ⇒ 18/6 = (k8ⁿ)/(k4ⁿ) ⇒ 3 = 2ⁿ
n = ㏑(3) / ㏑(2) ⇒ k = y/xⁿ ⇒ k = 6/4ⁿ = 2/3
f(16) = 2/3 × 16ⁿ = 54
c) y = aeᵇˣ ∧ f(4) = 6 = aeᵇ⁴ ∧ f(8) = 18 = aeᵇ⁸ ⇒ 18/6 = (aeᵇ⁸)/(aeᵇ⁴) ⇒ 3 = e⁴ᵇ
b = ㏑(3/4) ∧ a = y / eᵇˣ ⇒ a = 6 / e⁴ᵇ = 2
f(16) = 2eᵇ¹⁶ = 162
d) y = a㏑(bx) ∧ f(4) = 6 = a㏑(b4) ∧ f(8) = 18 = a㏑(b8)
⇒ 18 - 6 = a㏑(b8) - a㏑(b4) ⇒ 12 = a㏑(8b/4b) ⇒ a = 12 / ㏑(2)
f(4) = 6 = a㏑(4b) ⇒ b = (√2)/4
f(16) = a㏑(b16) = 30
Answer:
duster f vying FE65820 crystal cyst