6^7/6^5 in index formhelp me
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Answer and Step-by-step explanation:
Alot of these can be solved with the table up above the problems.
<h2>A, B, and C.</h2><h2></h2>
According to the table, every time a value has an exponent of 0, it would equal to 1. Since all A B and C values follow the 0 exponent, their value is 1.
<h2>D.</h2><h2 />
According to the table, when we have a negative exponent, the number gets turned into a fraction of
Since our value is 9^-2, do as followed :
Solve what is 9^2 :
This question is different, since we have to go backwards, we have to switch the negative exponent to a positive exponent and rid the fraction.
Since we have one negative exponent, follow the fraction rule but instead of having one as the numerator, have 5a^3.
Since -3 is in parenthesis, we can rid the negative along with the parenthesis :
Follow the negative exponent rule :
Follow the negative exponent rule yet again :
The sum of the first four terms of the sequence is 22.
In this question,
The formula of sum of linear sequence is
The sum of the first ten terms of a linear sequence is 145
⇒
⇒ 145 = 5 (2a+9d)
⇒
⇒ 29 = 2a + 9d ------- (1)
The sum of the next ten term is 445, so the sum of first twenty terms is
⇒ 145 + 445
⇒
⇒ 590 = 10 (2a + 19d)
⇒
⇒ 59 = 2a + 19d -------- (2)
Now subtract (2) from (1),
⇒ 30 = 10d
⇒ d =
⇒ d = 3
Substitute d in (1), we get
⇒ 29 = 2a + 9(3)
⇒ 29 = 2a + 27
⇒ 29 - 27 = 2a
⇒ 2 = 2a
⇒ a =
⇒ a = 1
Thus, sum of first four terms is
⇒
⇒
⇒ S₄ = 2(2+9)
⇒ S₄ = 2(11)
⇒ S₄ = 22.
Hence we can conclude that the sum of the first four terms of the sequence is 22.
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