Thanks, Lag....
Here is the graph...https://www.desmos.com/calculator/tarsovokoa
The solutions are as Lag said....!!!
No prob !!!
Note, josh....if we drew a horizontal line at y = 1, we would divide the figure up into two triangles at the top and a trapezoid at the bottom
The top left triangle has a base of 3 and a height of 2, so its area = (1/2)base * height = (1/2)(3)(2) = 3
The top right triangle has a base of 5 and a height of 1....so the area = (1/2)(5)(1) = 5/2 = 2.5
The trapezoid formed has a height of 4 and two parallel bases of 8 and 11
So...its area = (1/2)(height)(sum of the bases) = (1/2)(4)(8 + 11) = 2(19) = 38
So....the total area = 3 + 2.5 + 38 = 43.5 [ units^2 ]
No pic, Nickolas .....
1. See the graph here, PF : https://www.desmos.com/calculator/xshyczl312
The solutions occur at x = -3 and x = 1
2. 2x^3 + 3x
If we can put -x into the function and come back to what we started with...it is even
So..... 2(-x)^3 + 3(-x) = -2x^3 - 3x = f(-x) ......not what we started with...so..not even
If f(-x) = -f (x) ......then it is odd
-f(x) = - [ 2x^3 + 3x ] = -2x^3 - 3x = f(-x)....so...it's odd
3. (f + g) x means f(x) + g(x) = [ x^2 - 36 ] + [ x^3 + 2x^2 -10 ] = x^3 + 3x^2 - 46
Nice, Melody.....
Nice, heureka !!!!
First one
The roots are x = - 3 and x = 5
So....the equation of the parabola is y = - ( x - 5) (x + 3)
y = -x^2 + 2x - 15
Because of symmetry....the x coordinate of the vertex is x = [ -3 + 5] / 2 = 2/2 = 1
So....to find the y coordinate we have
y = -(1)^2 + 2(1) + 15
y = -1 + 2 + 15
y = 16
So...the vertex is (1, 16)
Here's a graph : https://www.desmos.com/calculator/5szp8sqydj
Nice, Rom !!!!
A = ( 1, 0, -1) B = (2, 3, 1) C = (0, 1, 1)
vector AB = ( 2 - 1, 3 - 0 , 1 - -1 ) = ( 1, 3, 2)
vector AC = ( 0 - 1, 1 - 0, 1 - -1) = ( -1, 1, 2)
vector BC = ( 0 - 2, 1 - 3, 1 - 1) = ( -2, -2, 0 )
