Answer:
R V
1 9π in³
2 36π in³
3 81π in³
Step-by-step explanation:
For, r = 1
V = π(1)²9 = 9π in³
For, r = 2
V = π(2)²9 = 4*9π in³ = 36π in³
For r = 3
V = π(3)²9 = 9*9π in³ = 81π in
R V
1 9π in³
2 36π in³
3 81π in³
<span>2x^2 +3x -4 + 8 - 3x -5x^2 +2
answer is
</span>-3x^2 +6
hope that helps
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.
If they’re all weighted equally, then your average grade will be 79.9%
Answer:
1
Step-by-step explanation:
||-4|- |-3||
|4-3|
|1|
