Yes they are equivalent because their ratio is 9
81/9= 9
63/7=9
We can re-create the quadratic equation by
(x -3) * (x -2)
x^2 -5x + 6 = 0
Answer:
Step-by-step explanation:
Reduction to normal from using lambda-reduction:
The given lambda - calculus terms is, (λf. λx. f (f x)) (λy. Y * 3) 2
For the term, (λy. Y * 3) 2, we can substitute the value to the function.
Therefore, applying beta- reduction on "(λy. Y * 3) 2" will return 2*3= 6
So the term becomes,(λf. λx. f (f x)) 6
The first term, (λf. λx. f (f x)) takes a function and an argument, and substitute the argument in the function.
Here it is given that it is possible to substitute the resulting multiplication in the result.
Therefore by applying next level beta - reduction, the term becomes f(f(f(6)) (f x)) which is in normal form.
Answer:
-20c^3
Step-by-step explanation:
((−5c)(d−4))((2cd2)2)
=
−20c3d4
/d4
=-20c^3
Answer:
See below.
Step-by-step explanation:
Given :-
x² + bx + c = 0 if x₁ + x₂ = -b and x₁x₂ = c
Solving :-
a) x₁ = 1/2 and x₂ = -3/4
=> x² + -(1/2 + -3/4)x + 1/2(-3/4)
=> x² + -(-1/4)x + (-3/8)
=> x² + 1/4x - 3/8
=> 8x² + 2x - 3 = 0
b) x₁ = 1 + √5 and x₂ = 1 - √5
=> x² + -(1 + √5 + 1 - √5)x + (1 + √5)(1 - √5)
=> x² - 2x + 1 - 5
=> x² - 2x - 4 = 0
