112.5
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2/5 is closer to 0.
9x^4 -18x^2 + 5=0 Let y be x^2 9y^2-18y+5=0 (3y-5)(3y-1)=0 3y-5=0 or 3y-1=0 3x^2-5=0 3x^2-1=0 x^2=±√5/3 x=±1/√3
Of the points: (5, 3) (3, 0) (5, 2) and (0, 2) which means (x, y) thus the y values are 3, 0, 2 and 2 * * * * * True, but that does not answer the question that was asked. Only one of these points is on the y-axis and that is (0, 2).
3b2 + b - 10 = 0 This quadratic equation can be factored. (3b -5)(b +2) = 0 The equation can be solved when either (3b - 5) = 0, or (b + 2) = 0. If 3b - 5 = 0 then 3b = 5 : b = 5/3 If b + 2 = 0 then b = -2 The roots or solutions to this equation are b = 5/3 and b = -2
x = -1/2 or -5 2x2 + 11x + 5 = 0 ⇒ (2x + 1)(x + 5) = 0 ⇒ (2x + 1) = 0 → x = -1/2 or (x + 5) = 0 → x = -5