Please help asap 25 pts
2 answers:
First
Better than solving it, I'll put a graph at the bottom of this answer. First of all it is an absolute value graph.
Remark
All members of that family are shaped like a v with both arms symmetrical about the vertex.
Second
You use the same rule for determining the vertex with an absolute value graph that you would use for a quadratic graph.
if the graph is of the form
y= abs(x + a) Then if a > 0 then the graph moves to the left (contrary to what you might think)
If a < 0 then the graph moves right again contrary to what you might think.
If a = 0 then the graph is on the same line as 0,0 depending on what else is with the absolute value formula.
Answer D <<<<<
Better than solving it, I'll put a graph at the bottom of this answer. First of all it is an absolute value graph.
Remark
All members of that family are shaped like a v with both arms symmetrical about the vertex.
Second
You use the same rule for determining the vertex with an absolute value graph that you would use for a quadratic graph.
if the graph is of the form
y= abs(x + a) Then if a > 0 then the graph moves to the left (contrary to what you might think)
If a < 0 then the graph moves right again contrary to what you might think.
If a = 0 then the graph is on the same line as 0,0 depending on what else is with the absolute value formula.
Answer D <<<<<
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<h2>Answer with explanation:</h2>
We are asked to prove by the method of mathematical induction that:
where n is a positive integer.
- Let us take n=1
then we have:
Hence, the result is true for n=1.
- Let us assume that the result is true for n=k
i.e.
- Now, we have to prove the result for n=k+1
i.e.
<u>To prove:</u>
Let us take n=k+1
Hence, we have:
( Since, the result was true for n=k )
Hence, we have:
Also, we know that:
(
Since, for n=k+1 being a positive integer we have:
)
Hence, we have finally,
Hence, the result holds true for n=k+1
Hence, we may infer that the result is true for all n belonging to positive integer.
i.e.
where n is a positive integer.