Answer:
x ∈ (-∞,-3)∪(0.5,∞)
Step-by-step explanation:
2x^2+6x-x-3>0
2x(x+3)-1(x+3)>0
(2x-1)(x+3)>0
x ∈ (-∞,-3)∪(0.5,∞)
Answer:
Please check the explanation below.
Step-by-step explanation:
Given the equations
![6x+18y+6z=24,\:-x-3y-2z=-4,\:2x+6y+2z=8\:\:](https://tex.z-dn.net/?f=6x%2B18y%2B6z%3D24%2C%5C%3A-x-3y-2z%3D-4%2C%5C%3A2x%2B6y%2B2z%3D8%5C%3A%5C%3A)
Steps to solve
![\begin{bmatrix}6x+18y+6z=24\\ -x-3y-2z=-4\\ 2x+6y+2z=8\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D6x%2B18y%2B6z%3D24%5C%5C%20-x-3y-2z%3D-4%5C%5C%202x%2B6y%2B2z%3D8%5Cend%7Bbmatrix%7D)
![\mathrm{Subsititute\:}x=4-3y-z](https://tex.z-dn.net/?f=%5Cmathrm%7BSubsititute%5C%3A%7Dx%3D4-3y-z)
![\begin{bmatrix}-\left(4-3y-z\right)-3y-2z=-4\\ 2\left(4-3y-z\right)+6y+2z=8\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D-%5Cleft%284-3y-z%5Cright%29-3y-2z%3D-4%5C%5C%202%5Cleft%284-3y-z%5Cright%29%2B6y%2B2z%3D8%5Cend%7Bbmatrix%7D)
As
![-\left(4-3y-z\right)-3y-2z=-4](https://tex.z-dn.net/?f=-%5Cleft%284-3y-z%5Cright%29-3y-2z%3D-4)
![-z-4=-4](https://tex.z-dn.net/?f=-z-4%3D-4)
and
![2\left(4-3y-z\right)+6y+2z=8](https://tex.z-dn.net/?f=2%5Cleft%284-3y-z%5Cright%29%2B6y%2B2z%3D8)
![8=8](https://tex.z-dn.net/?f=8%3D8)
so
![\begin{bmatrix}-z-4=-4\\ 8=8\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D-z-4%3D-4%5C%5C%208%3D8%5Cend%7Bbmatrix%7D)
Isolating z for ![-z-4=-4](https://tex.z-dn.net/?f=-z-4%3D-4)
![-z-4=-4](https://tex.z-dn.net/?f=-z-4%3D-4)
![-z=0](https://tex.z-dn.net/?f=-z%3D0)
![\frac{-z}{-1}=\frac{0}{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B-z%7D%7B-1%7D%3D%5Cfrac%7B0%7D%7B-1%7D)
![z=0](https://tex.z-dn.net/?f=z%3D0)
![\mathrm{For\:}x=4-3y-z](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3A%7Dx%3D4-3y-z)
![\mathrm{Expressing\:}x\mathrm{\:in\:terms\:of\:}y](https://tex.z-dn.net/?f=%5Cmathrm%7BExpressing%5C%3A%7Dx%5Cmathrm%7B%5C%3Ain%5C%3Aterms%5C%3Aof%5C%3A%7Dy)
![x=4-3y-0](https://tex.z-dn.net/?f=x%3D4-3y-0)
![x=-3y+4](https://tex.z-dn.net/?f=x%3D-3y%2B4)
Therefore,
![\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}](https://tex.z-dn.net/?f=%5Cmathrm%7BThe%5C%3Asolutions%5C%3Ato%5C%3Athe%5C%3Asystem%5C%3Aof%5C%3Aequations%5C%3Aare%3A%7D)
![x=-3y+4,\:z=0](https://tex.z-dn.net/?f=x%3D-3y%2B4%2C%5C%3Az%3D0)
Answer: this is how you do it
Step-by-step explanation:
For example if you have a function f(x) = x - 1, then x = 1 is a zero of this function because using it as x gives 1 - 1 = 0. The Riemann Zeta function has some zeros that are easy to find which are of little interest but there are some other ones that are harder to find which is why the are called non-trivial.
Answer:P ERRA
Step-by-step explanation:
JAJJAA
8) This is <u>inverse </u>variation because you can multiply both sides by 3 to get y = 6/x. In this case, it matches up with y = k/x and k = 6.
============================================
10) This is <u>not </u>direct variation, nor inverse variation. We cannot write it in either y = kx or y = k/x form
============================================
12) This is <u>inverse</u> variation. Rewrite 5/(2x) as (5/2)/x or 2.5/x; note how 5/2 = 2.5, so in this case k = 2.5
============================================
14) This is <u>direct</u> variation. Multiply both sides by 4, and do a bit of rearranging and you should end up with y = (20/3)x, so we see it matches up with y = kx and k = 20/3.
============================================
16) Cross multiply to get xy = 16, then divide both sides by x to end up with y = 16/x. This equation is in the form y = k/x with k = 16, so it is actually an inverse variation equation.