3) x² - 121 = 0
x² = 121
x' = +√121
x' = 11
_______________
x'' = - √121
x'' = -11
Solution ⇒ S{-11 ; 11 } or (x-11)(x+11)
4) 4x² + 144 = 0
4x² = -144
x² = -144 / 4
x² = -36
x = √-36
No solution ⇒ S = ∅
5) z²+10z+21 = 0
Δ = 10² - 4(1)(21)
Δ = 100 - 84
Δ = 16
x' = (-10+4) / 2 = -6/2 = -3
x'' = (-10-4) / 2 = -14/2 = -7
Solution ⇒ S{ -7 ; -3} or (x+3)(x+7)
Given:
The equation 
goes through the coordinates (6,5) and (-2,-5).
To find:
The values of A and B.
Solution:
We have,
It goes through the coordinates (6,5) and (-2,-5). It means these points will satisfy the equation.
...(i)
...(ii)
Adding (i) and (ii), we get
Divide both sides by 4.
Put A=5 in (i), we get
Divide both sides by 5.
Therefore, the values of A and B are 5 and -4 respectively.
Answer:
64/3 cc or 64/3 cm³
Step-by-step explanation:
The formula for the volume of a triangular pyramid is
V = (1/3)(area of base)(height)
Here we have a square prism (actually, a cube), whose square base is 4 cm by 4 cm (4 cm is the cube root of 64 cc). The height of this cube is also 4 cm.
The volume of a triangular pyramid of base area (4 cm)² and height 4 cm is
V = (1/3)(base area)(height)
= (1/3)(16 cm²)(4 cm) = 64/3 cc
The distance between them is 9.
