Graph the first six terms of a sequence where a1 = −10 and d = 3. (2 points)
2 answers:
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The series 7 + 16 + 25 +34 +43 +52 + 61 is an illusration of arithmetic series
The sigma notation of the series is:
The series is given as:
7 + 16 + 25 +34 +43 +52 + 61
The above series is an arithmetic series, with the following parameters
- First term, a = 7
- Common difference, d = 9
- Number of terms, n = 7
Start by calculating the nth term using:
a(n) = a + (n - 1) * d
This gives
a(n) = 7 + (n - 1) * 9
Evaluate the product
a(n) = 7 - 9 + 9n
Evaluate the difference
a(n) = 9n - 2
So, the sigma notation is:
Read more about arithmetic series at:
Answer:
These type of questions are super tricky b/c you have to remember all the different versions of the identities, and then they put the question in some odd form, I feel like this should land math professors in jail , for dishonesty , b/c it's really a form of "how tricky can I make a question and still have a way to solve it" anyway,
Step-by-step explanation:
a)
next the question asks 1-cos 2A and this is total abuse of notation. the way this should be written is 1- cos( 2A) so we know that the A is part of the cosine functions input... btw.. in any computer program, it would never ever let you get away with that top form of the expression. :/ anyway... I keep ranting.. huh... sorry :P
1-cos(2A) is an odd form of the identity 1/2(1-cos(2A) = (A) the 1/2 is missing but we can add that pretty easy, we just have to remember to take it out too. I usually forget to do that. and my professor marks me off completely, totally wrong, but I just miss one small thing :/ anyway....
our 1-cos(2A) needs the 1/2 added to it. or if we move that 1/2 to the other side it looks like 2*(A) = 1-cos(2A) and this is that "odd" from of the identity that I was talking about.
next let's deal with sin(2A) it has an identity of 2 sin(A)cos(A) which is really nice for us b/c it will cancel out the 2 in then numerator for us, nice !
now our fraction looks like [2* (A)] / 2 sin(A)cos(A)
so cancel out one of the sines
2*sin(A) / 2 cos(A)
cancel the 2s
Sin(A) / Cos(A) = Tan(A)
nice it worked out :P
b)
by the above that we just worked out, then
Tan(15) = Sin(15) / Cos(15)
I had to look up what sin of 15 is b/c it's not one of those special angles but it does have an exact form of
Sin(15) = (√3 - 1) / 2√2
Cos(A) = (√3 + 1) / 2√2
you can use rule of Cos(A-B) = Cos(A)Cos(B)+Sin(A)Sin(B) to get the above and a similar rule for Sin(A-B)
back to our problem, the 2√2 will cancel out
then we have
Tan(15) = (√3 - 1) /(√3 + 1)
in the form that is above that's exact, the roots could be approximated but i'll just leave that in the form that is exact. Most math professors like that form.