16 to 18 cashews per serving
If you would like to simplify (z/1) / (1/7), you can do this using the following steps:
<span>(z/1) / (1/7) = z / (1/7) = z * (7/1) = z * 7 = 7z
</span>
The correct result would be d. 7z.
Composite is a number that is NOT prime. So we want to find an odd perfect square that IS prime to be a counterexample.
√36 = Not odd
√49 = 7 Factors only 1 * 7 so it is prime
√81 = 9 Factors 1, 3, 9 composite
√225 = 15 Factors 1,3, 5,9,15, 25, 45, 75, 225 composite
Counterexample for 2∧n - 1 Is prime
2^6 - 1 = 64 - 1 = 63 NOT PRIME Factors are: 1, 3, 7, 9,21, 63
2^5 - 1 = 32 - 1 = 31 Prime
2^3 - 1 = 8 - 1 = 7 Prime
2^2 - 1 = 4 - 1 = 3 Prime
Answer:
Therefore the auxiliary solution is
![y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t]](https://tex.z-dn.net/?f=y%3DC_1e%5E%7Bt%7D%2Be%5E%7B-t%7D%5BC_2sin%20%5C%20t%2BC_3%20cos%20%5C%20t%5D)
Step-by-step explanation:
To find auxiliary equation we have to put 
in the given differential equation.
The degree of the differential equation is 3.
Therefore the number of root of the differential equation is 3.
Let 
, 
and 
be three roots of the auxiliary equation.
- If three roots are real and equal.
Then 
- If three roots are real and distinct.
Then 
- If two roots imaginary and one root real ,
Then 
Now, ,
Given differential equation is
The auxiliary equation is
Here 
Therefore ,
![y=C_1e^{1.t}+e^{-1.t}[C_2sin \ (1.t)+C_3 cos \ (1.t)]](https://tex.z-dn.net/?f=y%3DC_1e%5E%7B1.t%7D%2Be%5E%7B-1.t%7D%5BC_2sin%20%5C%20%281.t%29%2BC_3%20cos%20%5C%20%281.t%29%5D)
![\Rightarrow y=C_1e^{t}+e^{-t}[C_2sin \ t+C_3 cos \ t]](https://tex.z-dn.net/?f=%5CRightarrow%20y%3DC_1e%5E%7Bt%7D%2Be%5E%7B-t%7D%5BC_2sin%20%5C%20t%2BC_3%20cos%20%5C%20t%5D)
this is the answer I came up with hopes this helps
