Considering a general quadratic equation y=ax^2 + bx + c, the x coordinate of the vertex is found from the formula x= -b/2a and the y coordinate is found from putting that x coordinate back into the original quadratic equation which in this case I am assuming is y= -2x^2 + 16x -15.
So, the x coordinate of the vertex is x=-16/(2*-2) = 4
To find the y coordinate we plug 4 back into y= -2x^2 + 16x -15 so we have y= -2 * 4^2 + 16*4 - 15. Following the order of operations we get y= -2 *16 + 64 - 15= 17
Therefore the vertex is at (4, 17).
plot 4,-5 2,2 and 7,-5 on a graph than connect the vertex
X2 - 16X + 15 = 0 (X - 1)(X - 15) = 0 So X -1 = 0 or X - 15 = 0 ie X = 1 or X = 15
2x2 + 15 = 11x ? Can be only for some values of x.2x2 + 15 = 11x subtract 11x to both sides2x2 - 11x + 15 = 0(2x - 5)(x - 3) = 02x - 5 = 0 or x - 3 = 0x = 5/2 or x = 3
speed = distance/time speed = 45/15 = 3 m/s
Area = 0.5*a*b*sin(C) so 112.5 = 0.5*15*15*sin(c) 1 = sin(C) so C = pi/2 radians (or 90 degrees).
1-16x speed = -15
plot 4,-5 2,2 and 7,-5 on a graph than connect the vertex
15x2+16x-15 = (3x+5)(5x-3) when factored
If: 16x-15-6x = 13 Then: 10x = 28 And: x = 2.8.
What is the solution set of 2x2 + x = 15
The given expressions can be simplified to: 16x-10
You need to define a domain: the set of values that x can take.Then, for each of a selection of value of x, calculate the value of y = 2x2-7x-15. Plot the point (x, y) on a coordinate plane. Finally, join the plotted points using a smooth curve.
If you mean: 2x2+11x+15 then it is (2x+5)(x+3) when factored
(x - 15)(x - 1)
2x2+13x+15 = (2x+3)(x+5)
2x2 + 11x + 15 = (x + 3)(2x + 5).
X2 - 16X + 15 = 0 (X - 1)(X - 15) = 0 So X -1 = 0 or X - 15 = 0 ie X = 1 or X = 15