Answer:
a) Эx ∈ R ⊇ x³ = 2
b) ∀x ∈ R, x² ≥ 0
c) Эx ∈ R ⊇ x³ = x
d) ∀x ∈ R, x ≤ x²
Step-by-step explanation:
Given the data in the question;
let us first go through some symbols and their possible meanings;
Э ⇒ there exists
∀ ⇒ for all
∈ ⇒ belongs to or set membership or element of the set
⊇ ⇒ such that
now;
a) There is a number whose cube is equal to 2
let x represent the number; so
Эx ∈ R ⊇ x³ = 2
b) The square of every number is at least 0
x² ≥ 0, ∀x ∈ R
∀x ∈ R, x² ≥ 0
c) There is a number that is equal to its square
Эx ∈ R ⊇ x³ = x
d) Every number is less than or equal to its square.
x ≤ x², ∀x ∈ R
∀x ∈ R, x ≤ x²
Hello from MrBillDoesMath!
Answer: (1/7) * ( 4 +\- sqrt(5) i)
where i = sqrt(-1)
Discussion:
The solutions of the quadratic equation ax^2 + bx + c = 0 are given by
x = ( -b +\- sqrt(b^2 - 4ac) )/2a.
The equation 7 x^2 + 3 = 8x can be rewritten as
7x^2 - 8x + 3 = 0.
Using a = 7, b = -8 and c = 3 in the quadratic formula gives:
x = (8 +\- sqrt ( (-8)^2 - 4*7*3) ) / (2*7)
= ( 8 +\- sqrt( 64 - 84)) /(2*7)
= ( 8 +\- sqrt( -20) ) / (2*7)
= ( 8 +\- sqrt( -20) ) / 14
= 8/14 +\- sqrt(5 *4 * -1) /14
= 4/7 +\- 2 sqrt(5) *i /14
As 2/14 = 1/7 in the second term
= 4/7 +\- sqrt(5) *i /7
Factor 1/7 from each term.
= (1/7) * ( 4 +\- sqrt(5) i)
Thank you,
MrB
Answer:
x = 52
Step-by-step explanation:
To answer this question, we need to set up a system of linear equations:
x = 4y - 8
(1/2)x + y = 11 (Note: Please use ( ) around fractions such as this 1/2, to
eliminate possible ambiguity)
We want to find the x-coordinate, so should start by elilminating y from this system of equations.
Do this by multiplying the 2nd equation by -4. This results in
-2x - 4y = -44
Combine this result with x = 4y - 8:
x = 4y - 8
-2x - 4y = -44
----------------------
-x = -52 (Notice how the "y" terms cancel each other out.)
Dividing both sides by -1 produces the desired answer: x = 52
3.100 I think if I am correct
