Answer:
a. ∀ x≤0 ∧ ∀ y ∈ IR + ∪ 0
b. ∀ x≤0 ∧ ∀ y ∈ IR -
c. ∀ x≤0 ∧ ∀ y -∞ → +∞
Step-by-step explanation:
P=4l ⇒ 28=4l ⇔ l=28/4 =7
A=l.l=7*7=49[tex]u^{2}[/tex]
As can be seen in the graph so that x is negative, it must be fulfilled that its values are located in the second and third quadrants and their values must be less than zero and also the values of y can be all positive and negative real numbers
Possible coordinates:
a. ∀ x≤0 ∧ ∀ y ∈ IR + ∪ 0
for all x less iqual than zero and all positive real y including zero
b. ∀ x≤0 ∧ ∀ y ∈ IR -
for all x less iqual than zero and all y negative reals
c. ∀ x≤0 ∧ ∀ y -∞ → +∞
for all x less equal than zero and all y from minus infinite to plus infinite
