Can anyone help me out? thanks
2 answers:
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Hello!
To solve this, first perform the opposite operation for the last operation (on the left side) on both sides. The last operation of the left side is squaring. Therefore, square root both sides.
Please note that you must include ±. This is because the square root of 64 can be either positive or negative, as a square of either a positive or negative number is positive.
Now, add 9 to both sides.
There are 2 solutions from here. One comes from adding 8, and the other subtracting 8. Therefore, the two solutions are:
y = 9 + 8 = 17
y = 9 - 8 = 1
Therefore, your two solutions are 17 and 1.
Hope this helps!
In an arithmetic sequence, consecutive terms have a fixed distance d between them. If a₁ is the first term, then
2nd term = a₂ = a₁ + d
3rd term = a₃ = a₂ + d = a₁ + 2d
4th term = a₄ = a₃ + d = a₁ + 3d
and so on, up to
nth term =
so that every term in the sequence can be expressed in terms of a₁ and d.
6. It's kind of hard to tell, but it looks like you're given a₁₃ = -53 and a₃₅ = -163.
We have
a₁₃ = a₁ + 12d = -53
a₃₅ = a₁ + 34d = -163
Solve for a₁ and d. Eliminating a₁ and solving for d gives
(a₁ + 12d) - (a₁ + 34d) = -53 - (-163)
-22d = 110
d = -5
and solving for a₁, we get
a₁ + 12•(-5) = -53
a₁ - 60 = -53
a₁ = 7
Then the nth term is recursively given by
and explicitly by
7. We do the same thing here. Use the known terms to find a₁ and d :
a₁₉ = a₁ + 18d = 15
a₃₈ = a₁ + 37d = 72
⇒ (a₁ + 18d) - (a₁ + 37d) = 15 - 72
⇒ -19d = -57
⇒ d = 3
⇒ a₁ + 18•3 = 15
⇒ a₁ = -39
Then the nth term is recursively obtained by
and explicitly by
8. I won't both reproducing the info I included in my answer to your other question about geometric sequences.
We're given that the 1st term is 3 and the 2nd term is 12, so the ratio is r = 12/3 = 4.
Then the next three terms in the sequence are
192 • 4 = 768
768 • 4 = 3072
3072 • 4 = 12,288
The recursive rule with a₁ = 3 and r = 4 is
and the explicit rule would be