Answer:
(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
t t t
t f f
f f t
f f f
next step is getting the opposite
(p∧q)'
<em>f</em>
<em> t</em>
<em> t</em>
<em> t</em>
Then we get p' V q', V is true (t) when the first or the second is true.
p' V q'
f <em>f</em> f
f <em>t</em> t
t <em>t</em> f
t <em>t</em> t
Let's compare them, ≡ is true if the first is equal to the second one.
(p∧q)' ≡ (p' V q')
<em>f f </em>
<em> t t</em>
<em> t t</em>
<em> t t</em>
Both are true, so
(p ∧ q)’ ≡ p’ ∨ q’
Answer:
your answer will be <em><u>B. HL Theorem </u></em>
Step-by-step explanation:
hope it helps you...
(x-5)(x+3)=0
x^2+3x-5x-15=0
(x^2+3x) (-5x-15)
x(x+3) -5(x+3)
x-5=0 x+3=0
x=5 OR -3
Answer:
5unit
Step-by-step explanation: since area of rectangle = length*width
Here length is (x+2) and width is(2x+3)
(x+2)(2x+3)=91
On further solving
2x^2+75x-85=0
2x^2-10x+17x-85=0
2x(x-5)+17(x-5)=0
(2x+17)(x-5)=0
x=5unit
