For this case we have the following polynomials:
3x2
x2y + 3xy2 + 1
We have then:
For 3x2:
Classification: polynomial of one variable:
Degree: 2
For x2y + 3xy2 + 1:
Classification: polynomial of two variables
Degree: 2 + 1 = 3
Answer:
The polynomial 3x2 is of one variable with a degree of 2.
The polynomial x2y + 3xy2 + 1 is of two variables a with a degree of 3.
y < - 8 or y > 4
inequalities of the form | x | > a always have solutions of the form
x < - a or x > a
we have to solve
y + 2 < - 6 or y + 2 > 6
y + 2 < - 6 ( subtract 2 from both sides )
y < - 8
or
y + 2 > 6 ( subtract 2 from both sides )
y > 4
these can be combined using interval notation
y ∈ (- ∞, - 8 ) ∪ (4, ∞ )
As a check
substitute chosen values of x from each interval
y = - 10 : | - 10 + 2 | = | - 8 | = 8 > 6 this is true
y = 12 : | 12 + 2 | = | 14 | = 14 > 6 which is also true
Answer:
M=5
Step-by-step explanation:
2 ( 2m ) + 4 ( 4m + 2 ) = 108
step 1. do distributive property
4m + 16m + 8 = 108
step 2. add 4m and 16m and subtract 8 from 108
20m = 100
step 3. divide 100 by 20
m = 5
