Greater than, as four is more than zero.
3x^2 + 8x + 4 = (x+2) (3x+2) x = -2 x = -2/3 So there are no complex roots, they are real. You can test this by b^2 - 4ac if greater than 0, it is real if equal, there will be 2 identical roots. if less than 0 you get imaginary roots.
y = x^(2) + 4x + 5 Find the vertex , differentiate and equate to zero. dy/dx = 0 = 2x + 4 2x + 4 - 0 2x = -4 x = -2 To find if the vertex is at a max/min differentiate are second time. If the answer is positive(+)/Negative(-), then it is a minimum/maximum. Hence dy/dx = 2x + 4 d2y/dx2 = (+)2 Positive(+) so the parabola is at a minimum. at x = -2.
x2 - 3x - 4 = 0 so (x+1)(x-4) = 0, so x+1 = 0 and x-4 = 0, so x = -1 or 4
-4, -2, 0, and 2 are the four consecutive even integers. When you add them up they equal -4.
x=2, 3, 4...
greater than
Yes, 0 is less than 4. In mathematical terms, 0 is on the left side of 4 on the number line, indicating that it is smaller. Additionally, the inequality symbol "<=" represents "less than or equal to," meaning that 0 is also equal to 4.
No. Zero is less than 4.
0
0 <= x <= 6 gives -1 <= Y <= 14.
1-4 equal less than and greater than = -3
p is greater than or equal to 4.
a is greater than or equal to 3
25
depends, when c>0, yes. when c=0 it's equal 4*0=0 and when c<0 it is smaller, because c=4: 4*-2=-8 and -8 is smaller than -2
-5, -4, -3, -2, -1, 0 , 1, 2, ...
3 multiples of 4 that is greater than 0 is 8,12, and 16.