Gcf is the grouping of all the factors common to both groups
lcm is a group of all the factors that has all the elements of both groups
factor
20x^2y^3=2*2*5*x*x*y*y*y
60x^3y^2=2*2*3*5*x*x*x*y*y
40xy^2=2*2*2*5*x*y*y
the GCF=2*2*5*x*y*y=20xy^2
the LCM=2*2*2*3*5*x*x*x*y*y*y=120x^3y^3
Taking

and differentiating both sides with respect to

yields
![\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B3x%5E2%2By%5E2%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B7%5Cbigg%5D%5Cimplies%206x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D0)
Solving for the first derivative, we have

Differentiating again gives
![\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B6x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B0%5Cbigg%5D%5Cimplies%206%2B2%5Cleft%28%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29%5E2%2B2y%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D0)
Solving for the second derivative, we have

Now, when

and

, we have

> 0
First, note that x is undefined at 5. / x ≠ 5
Second, replace the inequality sign with an equal sign so that we can solve it like a normal equation. / Your problem should look like:

= 0
Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
Forth, add 5 to both sides. / Your problem should look like: 3x = 5
Fifth, divide both sides by 3. / Your problem should look like: x =
Sixth, from the values of x above, we have these 3 intervals to test:
x <


< x < 5
x > 5
Seventh, pick a test point for each interval.
1. For the interval x <

:
Let's pick x - 0. Then,

> 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.
2. For the interval

< x < 5:
Let's pick x = 2. Then,

> 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.
3. For the interval x > 5:
Let's pick x = 6. Then,

> 0
After simplifying, we get 13 > 0, which is ture.
Keep this interval.
Eighth, therefore, x <

and x > 5
Answer: x <

and x > 5
Step-by-step explanation:
120÷24
0 24⟌120
0 24⟌120 0
0 24⟌120 -0 1
00 24⟌120 -0 12
0 24⟌120 -0 12
00 24⟌120 -0 12 - 0 12
00 24⟌120 -0 12 - 0 120
Shortest is 3/4 longest is 1 1/2. 1 1/2 -3/4=3/4