False because you should always make sure a company has medical and detal procedures
Option A: z + 1
Option B: 6 + w
Option D: 
Solution:
Let us first define the polynomial.
A polynomial can have constants, variables, exponents and fractional coefficients.
A polynomial cannot have negative exponents, fractional exponents and never divided by a variable.
<u>To find which expressions are polynomial:</u>
Option A: z + 1
By the definition, z + 1 is a polynomial.
It is polynomial.
Option B: 6 + w
By the definition, 6 + w is a polynomial.
It is polynomial.
Option C: ![y^{2}-\sqrt[3]{y}+4](https://tex.z-dn.net/?f=y%5E%7B2%7D-%5Csqrt%5B3%5D%7By%7D%2B4)
![y^{2}-\sqrt[3]{y}+4=y^{2}-{y}^{1/3}+4](https://tex.z-dn.net/?f=y%5E%7B2%7D-%5Csqrt%5B3%5D%7By%7D%2B4%3Dy%5E%7B2%7D-%7By%7D%5E%7B1%2F3%7D%2B4)
Here, y have fractional exponent.
So, it is not a polynomial.
Option D: 
By the definition,
is a polynomial.
It is polynomial.
Hence z + 1, 6 +w and
are polynomials.
Answer:
Invalid
Step-by-step explanation:
This is invalid because you don't have to go some were to love something.
Answer:
The answer is C because the two figures are not congruent.
Since there are two halves in a whole, you can times each number by 2 to get the number of halves it has.
12*2=24
10*2=20
13*2=26
15*2=30
8*2=16
5*2=10