If this question is asking: is the point (6,9) a solution of the equation y = 12x + 6, then NO, it's not a solution.
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69
The quadratic equation provides solutions for the generalized equation ax2 + bx + c = 0. The solution(s) are (-b +/- square root ( b2 - 4ac)) / 2a. Plugging in a=1, b=3, and c=-15 (from x2 + 3x -15 = 0) you get... (-3 +/- square root (32 - 4(1)(-15)) / 2(1) ... or ... (-3 +/- square root (9 + 60)) / 2 ... or ... (-3 +/- square root (69)) / 2 ... or ... about -3 +/- 8.3 Since 69 is positive, the square root (69) is real, at 8.3 This equation has two real roots, x=5.3 and x=-11.3 If the discriminant (b2 - 4ac) were zero, there would be one real root. If it were negative, there would be one real root and one imaginary root, i.e. a complex conjugate.
69 + 69 = 138.
Presumably this is a simultaneous equation in the form of: 3r-4s = 0 2r+5s = 23 Multiply all the terms in the top equation by 2 and in the bottom equation by 3: 6r-8s = 0 6r+15s = 69 Subtract the bottom equation from the top equation: -23s = -69 Divide both sides by -23 to find the value of s remembering that a minus number divided into a minus number is equal to a plus number: s = 3 Substitute the value of s into the original equations to find the value of r: Therefore: r = 4 and s = 3
69% = 69/100 = 0.69/1 = 0.69