Use a truth table to verify the first de morgan law (p ∧ q)’ ≡ p’ ∨ q’.

(p ∧ q)’ ≡ p’ ∨ q’

Step-by-step explanation:

First, p and q have just four (4) possibilities, p∧q is true (t) when p and q are both t.

p∧    q

ttt

tff

fft

fff

next step is getting the opposite

(p∧q)'

    f

    t

    t

    t

Then we get p' V q', V is true (t) when the first or the second is true.

p'V  q'

f  f  f

f  t  t

t  t  f

t  t  t

Let's compare them, ≡ is true if the first is equal to the second one.

(p∧q)'   ≡     (p' V q')

    f              f

    t              t

    t              t

    t              t

Both are true, so

(p ∧ q)’ ≡ p’ ∨ q’

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

the ratio of balloons to tables is 5: 2 and the constant variation is 5.

the equation is y=5x