The correct response is ;
<span>multiply the term number by 2, subtract 1 from the result, multiply by 2 times the term number, and divide the result by 2.
if term number is x,
the number in the pattern for xth term is ;
multiply term number by 2 - 2x
subtract 1 - 2x - 1
multiply by 2 times the term number - (2x-1) * 2x
divide by 2 - (</span>(2x-1) * 2x)/2
so first term x=1
((2*1-1) * 2*1)/2
2/2 = 1
second term x = 2
((2*2-1) * 2*2)/2
(3*4)/2 = 6
third term x = 3
((2*3-1) * 2*3)/2
5*6/2 = 15
fourth term x= 4
((2*4-1) * 2*4)/2
56/2 = 28
Therefore the above response shows the correct pattern
Answer and Step-by-step explanation:
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1. Let's write out the equation to subtract.
7t - 2u - 3v - (t - 3v)
Distribute the negative to the t and -3v.
7t - 2u - 3v - t + 3v (The negatives cancel out)
Now simplify by combining like terms.
6t - 2u
This is the answer because the 3v and -3v cancel out.
2. I don't really understand what this is saying. Is there answer choices for this? But what I think its saying is that the lift has a constant of 2.
3. To find out the amount of terms, we would simplify the equation.
2x + 3y - 5x + yz - x
-4x + 3y + yx
Here, we can see that we have 3 terms in this expression.
-4x is the first term, +3y is the second term, and +yx is the third term.
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#teamtrees #WAP (Water And Plant)
Answer:
its 8, so insert to forms of 8!!! so a and c!
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.
It doesn’t bother me that much but it’s understandable
