Finding first six terms of a sequence
2 answers:
To find the first six terms of a sequence, you first of all need to have a starting number. You will also need to have a pattern depending on the kind of sequence
For example, if our first number is 1 and our pattern is that we raise each of the natural numbers to the power of 2, we can now derive our sequence. To get the sequence, you've gotta carry out that pattern on each umber to be used. Since we are looking for the first six terms, the operation will only be carried out on 1,2,3,4,5 and 6. In the end this will give us;
and 
=
Please mark me as brainliest.
For example, if our first number is 1 and our pattern is that we raise each of the natural numbers to the power of 2, we can now derive our sequence. To get the sequence, you've gotta carry out that pattern on each umber to be used. Since we are looking for the first six terms, the operation will only be carried out on 1,2,3,4,5 and 6. In the end this will give us;
Please mark me as brainliest.
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Answer:
a) true
Step-by-step explanation:
The bottom-line numbers from synthetic division <em>alternate signs</em>, indicating that -2 is a lower bound. The given statement is true.
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If the signs were all positive, it would indicate the proposed zero is an <em>upper bound</em>.
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A graph shows all real zeros are greater than -2.
Answer:
Follows are the solution to the given point:
Step-by-step explanation:
In point a:
¬∃y∃xP (x, y)
∀x∀y(>P(x,y))
In point b:
¬∀x∃yP (x, y)
∃x∀y ¬P(x,y)
In point c:
¬∃y(Q(y) ∧ ∀x¬R(x, y)) ∀y(> Q(y) V ∀ ¬ (¬R(x,y)))
∀y(¬Q(Y)) V ∃xR(x,y) )
In point d:
¬∃y(∃xR(x, y) ∨ ∀xS(x, y))
∀y(∀x>R(x,y)) ∃x>s(x,y))
In point e:
¬∃y(∀x∃zT (x, y, z) ∨ ∃x∀zU (x, y, z))
∀y(∃x ∀z)>T(x,y,z) ∀x ∃z> V (x,y,z))