3x + 2 = 0 3x = -2 x = -2/3
The number of diagonals for a shape with x sides is (x²-3x)/2 The number of sides is x You want to know if x is ever less than (x²-3x)/2 or else (x²-3x)/2 - x > 0 (x² - 3x) / 2 - x > 0 (x² - 3x)/2 - (2x)/x > 0 x² - 3x - 2x > 0 x² - 5x > 0 x(x-5)>0 This only true if x<0 or x>5, and since a polygon cannot have less than 2 sides, x<0 is meaningless. This means that if x<5 then the number of diagonals is less than the number of sides, equal for x=5, and more for x>5. ■
x3+3x2+3x+2 divided by x+2 equals x2+x+1
3x + 4y - 4 = 0 y = -2 3x + 4(-2) - 4 = 0 3x - 8 - 4 = 0 3x = 12 x = 4 (4,-2)
I think you mean to solve:(-2 cos23x) - (3 sin 3x) = 0cos2 x + sin2 x = 1⇒ cos2x = 1 - sin2 x⇒ -2 cos2 3x - 3 sin 3x = -2(1 - sin2 3x) - 3 sin 3x = 0⇒ 2 sin2 3x - 3 sin 3x - 2 = 0⇒ (2 sin 3x + 1)(sin 3x - 2) = 0⇒ sin 3x = -1/2 or 2sin 3x = 2 is impossible as the range of sine is -1 ≤ sine ≤ 1Thus:sin 3x = -1/2⇒ 3x = 2nπ - π/6 or 2nπ - 5π/6⇒ x = 2/3nπ -π/18 or 2/3nπ -5π/18
(3x2 - 6x)/3x = 3x(x-2)/3x = x-2, for x<>0
If 2 divided by 3x it is equal to 2/3x
3x + 2 = 0 3x = -2 x = -2/3
The number of diagonals for a shape with x sides is (x²-3x)/2 The number of sides is x You want to know if x is ever less than (x²-3x)/2 or else (x²-3x)/2 - x > 0 (x² - 3x) / 2 - x > 0 (x² - 3x)/2 - (2x)/x > 0 x² - 3x - 2x > 0 x² - 5x > 0 x(x-5)>0 This only true if x<0 or x>5, and since a polygon cannot have less than 2 sides, x<0 is meaningless. This means that if x<5 then the number of diagonals is less than the number of sides, equal for x=5, and more for x>5. ■
(x - 3x - 6) / 6 = (3x + 2) / 2 ∴2(3x + 2) = 6(3x + 2) ∴6x + 4 = 18x = 12 ∴12x = -8 ∴x = -2/3
x^2 + 3x + 2/(-3x) - x^2 - 2=3x-2+2/(-3x)=3x-2-2/(3x)
1.5
y = -1x/2 3x + 6y = 0 -3x -3x __________ 6y = 0 - 3x --- -------- 6 6 y= -3x/6 y= -1x/2
2
3x - 3x -1 + 2x -1 = 0 0 - 1 + 2x - 1 = 0 2x - 2 = 0 2x = 2 x = 1
9x2 + 18x - 16 = 0 ⇒ (3x - 2)(3x + 8) = 0 ⇒ 3x - 2 = 0 → x = 2/3 or 3x + 8 = 0 → x = -22/3
x3+3x2+3x+2 divided by x+2 equals x2+x+1