It is: 122-4*(9*-2) = 216
9x2...I guess that means 9x squared ? The problem: 9x(squared)+12x+4 Here's the answer: (3x+2)(3x+2) or even more simplified: (3x+2) squared
12x + 31 = 2x + 1710x = -14x = - 1.4
12x + 17 = 7 Subtract 17 from both sides: 12x = -10 Divide both sides by 12: x = -10/12 = -5/6
Do you mean y2=x2-12x+17 Well you take the square root from both sides to get y by itself.
9x2 + 12x + 4 = 9x2 + 6x + 6x + 4 = 3x(3x + 2) + 2(3x + 2) = (3x + 2)(3x + 2) = (3x + 2)2
It is: 122-4*(9*-2) = 216
9x2...I guess that means 9x squared ? The problem: 9x(squared)+12x+4 Here's the answer: (3x+2)(3x+2) or even more simplified: (3x+2) squared
-9x 2 -12x+5 =
12x + 31 = 2x + 1710x = -14x = - 1.4
12x-3=2x+17 10x-3=17 10x=20 x=2
12x + 17 = 7 Subtract 17 from both sides: 12x = -10 Divide both sides by 12: x = -10/12 = -5/6
There's a chance this was notated incorrectly. It looks like it wants to be some combination of 3x and 2. 9x2 + 12x + 4 = (3x + 2)(3x + 2) or (3x + 2)2 -9x2 - 12x - 4 = -(3x + 2)2 9x2 - 12x + 4 = (3x - 2)2 -9x2 + 12x - 4 = -(3x - 2)2
Do you mean y2=x2-12x+17 Well you take the square root from both sides to get y by itself.
Given, 3x2 - 4x = -2 Then, 9x2 - 12x = -6; 9x2 - 12x + 4 = -2 {completing the square} ; 3x - 2 = ±i√2 {sq rt of both sides} ; and 3x = 2 ± i√2. Therefore, x = ⅓(2 ± i√2).
(3x)2/(12x)6(3x)2=32x2=9x2(12x)6=126x6=2985984x6(3x)2/(12x)6=9x2/2985984x69/2985984=1/331776.x2/x6=x2 * x-6=x-4=1/x4.9x2/2985984x6=1/331776 * 1/x4=1/331776x4
If you mean: 9x squared-12x+4 when x = 3 Then it is: (9*3*3)-(12*3)+4 = 49