He subtracted the 29fron the 30
Answer:
B
Step-by-step explanation:
Answer:
See proof below
Step-by-step explanation:
An equivalence relation R satisfies
- Reflexivity: for all x on the underlying set in which R is defined, (x,x)∈R, or xRx.
- Symmetry: For all x,y, if xRy then yRx.
- Transitivity: For all x,y,z, If xRy and yRz then xRz.
Let's check these properties: Let x,y,z be bit strings of length three or more
The first 3 bits of x are, of course, the same 3 bits of x, hence xRx.
If xRy, then then the 1st, 2nd and 3rd bits of x are the 1st, 2nd and 3rd bits of y respectively. Then y agrees with x on its first third bits (by symmetry of equality), hence yRx.
If xRy and yRz, x agrees with y on its first 3 bits and y agrees with z in its first 3 bits. Therefore x agrees with z in its first 3 bits (by transitivity of equality), hence xRz.
Answer:
The simplified form of the expression is ![\sqrt[3]{2x}-6\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7Bx%7D)
Step-by-step explanation:
Given : Expression ![7\sqrt[3]{2x}-3\sqrt[3]{16x}-3\sqrt[3]{8x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D-3%5Csqrt%5B3%5D%7B16x%7D-3%5Csqrt%5B3%5D%7B8x%7D)
To Simplified : The expression
Solution :
Step 1 - Write the expression
Step 2- Simplify the roots and re-write as
and 
![7\sqrt[3]{2x}-3\times2\sqrt[3]{2x}-3\times2\sqrt[3]{x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D-3%5Ctimes2%5Csqrt%5B3%5D%7B2x%7D-3%5Ctimes2%5Csqrt%5B3%5D%7Bx%7D)
Step 3- Solve the multiplication
![7\sqrt[3]{2x}-6\sqrt[3]{2x}-6\sqrt[3]{x}](https://tex.z-dn.net/?f=7%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7B2x%7D-6%5Csqrt%5B3%5D%7Bx%7D)
Step 4- Taking ![\sqrt[3]{2x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D)
common from first two terms
![\sqrt[3]{2x}(7-6)-6\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2x%7D%287-6%29-6%5Csqrt%5B3%5D%7Bx%7D)
Therefore, The simplified form of the expression is
