It is both 1 and 2. Both repeat the same amount of times
-Make variables
-Write equation given using variables
-Plugin the equation for L with 2w to solve for W.
-plugin W value to solve for L
Answer:
Follows are the solution to the given point:
Step-by-step explanation:
In point a:
¬∃y∃xP (x, y)
∀x∀y(>P(x,y))
In point b:
¬∀x∃yP (x, y)
∃x∀y ¬P(x,y)
In point c:
¬∃y(Q(y) ∧ ∀x¬R(x, y))
∀y(> Q(y) V ∀ ¬ (¬R(x,y)))
∀y(¬Q(Y)) V ∃xR(x,y) )
In point d:
¬∃y(∃xR(x, y) ∨ ∀xS(x, y))
∀y(∀x>R(x,y))
∃x>s(x,y))
In point e:
¬∃y(∀x∃zT (x, y, z) ∨ ∃x∀zU (x, y, z))
∀y(∃x ∀z)>T(x,y,z)
∀x ∃z> V (x,y,z))
The entire race is 1 full race.
One full race is 1.
He ran 3/8 of 1, so he ran 3/8.
He needs to run the rest of the race.
The rest is unknown, so we call it x.
When you add 3/8 to the unknown, you get the full race.
The equation is
x + 3/8 = 1
Change 1 to a denominator of 8.
x + 3/8 = 8/8
Subtract 3/8 from both sides.
x = 5/8
Answer: He still needs to run 5/8 of the race.
the possibilities are probably b, using the elimination method