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svlad2 [7]
3 years ago
10

Which formula can be used to describe the sequence below? 27,9,3,1,1/3

Mathematics
2 answers:
Gnesinka [82]3 years ago
8 0

Answer:

D. an=3an-1-1/3;a1=27

Step-by-step explanation:

givi [52]3 years ago
5 0
This is a geometric sequence with an initial term of 27 and a common ratio of 1/3

This just means that each term is 1/3 the term preceding it.

Any geometric sequence can be expressed as:

a(n)=ar^(n-1), a=initial term, r=common ratio, n=term number in this case:

a(n)=27(1/3)^(n-1)
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The sum of 3 odd consecutive integers are 93. find the 3 integers
Solnce55 [7]
Let, Your Integers = x, x+2, x+4
Now, x + x+2 + x+4 = 93
3x + 6 = 93
3x = 93 - 6
x = 87/3
x = 29
Then, x+2 = 31, & x+4 = 33

In short, Your Integers would be 29, 31, 33

Hope this helps!
4 0
3 years ago
Is square root 72 rational or irrational explain
Shalnov [3]

Answer:

irrational

Step-by-step explanation:

it is not a perfect square.

square root of 72= 8.48528137424...

7 0
4 years ago
PLEASE HELP ME ASAP I REALLY NEED HELP!!!
padilas [110]

1) \overline{UP}

2) \overline{UP} \cong \overline{UP}

3) \overline{UR}

4) \overline{PR}

5) \triangle OUP \cong \triangle RUP by SSS

6) CPCTC

7 0
2 years ago
5 left! Help please! So close to being done! a. 75. b. 66 c. 122 d. 98. thank you
user100 [1]

Answer:

98

Step-by-step explanation:

180- 122 = 58

58 + 24= 82

180- 82= 98

5 0
3 years ago
Read 2 more answers
For what value of constant c is the function k(x) continuous at x = 0 if k =
nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x)  is continuous, we take the limit of the parameter as follows:

\mathbf{ \lim_{x \to 0^-} k(x) =  \lim_{x \to 0^+} k(x) =  0 }

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= c }

Provided that:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }

Using l'Hospital's rule:

\mathbf{\implies  \lim_{x \to 0} \ \  \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}=  \lim_{x \to 0}   sec \ x  \ tan \ x = 0}

Therefore:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}=0 }

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

5 0
2 years ago
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