To check for continuity at the edges of each piece, you need to consider the limit as 
approaches the edges. For example,
has two pieces, 
and 
, both of which are continuous by themselves on the provided intervals. In order for 
to be continuous everywhere, we need to have
By definition of , we have 
, and the limits are
The limits match, so is continuous.
For the others: Each of the individual pieces of 
are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.
1 7/8 is the answer
Please mark brainiest
Answer:
Step-by-step explanation:
Using the given formula, we put n = 5
h = 8 / 16
h = 1 / 2 inches
Answer:
2x+17
Step-by-step explanation:
-4y+8-3x+4y+9+5x
Combine like terms
-3x +5x -4y +4y +8+9
2x +0 +17
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.
