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Assoli18 [71]
3 years ago
6

What is the pattern sequence of 13, 39, 117, 351, 1053?

Mathematics
2 answers:
Solnce55 [7]3 years ago
6 0

 the pattern is that you take each number and multiply it by 3


Oksi-84 [34.3K]3 years ago
3 0
This is a geometric sequence where the common ratio is 3.
13*3=39
39*3=117
117*3=351
351*3=1053
The next term of the sequence is the previous term multiplied by 3. The rule is a_{n} =13* 3^{n-1}
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Expression is equivalent to 8(12+30)
Fittoniya [83]

Answer:

336

Step-by-step explanation:

8 (12 + 30)

Follow PEMDAS  order of operations.

Calculate within parentheses. (12 + 30)

12 + 30

12 + 30 = 42

= 42

=8 * 42

Multiply and divide left to right

8 *42

8 * 42 = 336

= 336

7 0
3 years ago
Connect the Central Limit Theorem to the Law of Large numbers. Provide an example of your own in your explanation.
slava [35]
Sorry I don’t know if you have any other problems then I will be willing to help
5 0
3 years ago
in the interval 0° ≤ x ≤360°, find the values of x for which tan x =1.3562 Give your answers to the nearest degree
belka [17]

Answer:

The answers are 54 degrees and 234 degrees

Step-by-step explanation:

mathematically we have the tan positive on only two axes

these are the first (0-90) degrees and the 3rd quadrant (180-270)

We start by find the arc tan of the angle value

Thus;

x = arc tan (1.3562)

x = 54 degrees

On the third quadrant, we have it that;

180 + 54 = 234 degrees

4 0
3 years ago
Which expression is equivalent to the given expression?<br> (3y – 4) (2y + 7) + 1ly – 9
PilotLPTM [1.2K]

Answer:

6y²+40y--27

Step-by-step explanation:

(3y – 4) (2y + 7) + 1ly – 9

6y² + 21y -- 8y --28 + 11y --9

6y²+40y--27

8 0
3 years ago
Find the local max and min values of f(x)=x^2/x-1 using both first and second derivative tests
uranmaximum [27]
Hmm, the 2nd derivitve is good for finding concavity

let's find the max and min points
that is where the first derivitive is equal to 0
remember the difference quotient

so
f'(x)=(x^2-2x)/(x^2-2x+1)
find where it equals 0
set numerator equal to 0
0=x^2-2x
0=x(x-2)
0=x
0=x-2
2=x

so at 0 and 2 are the min and max
find if the signs go from negative to positive (min) or from positive to negative (max) at those points

f'(-1)>0
f'(1.5)<0
f'(3)>0

so at x=0, the sign go from positive to negative (local maximum)
at x=2, the sign go from negative to positive (local minimum)


we can take the 2nd derivitive to see the inflection points
f''(x)=2/((x-1)^3)
where does it equal 0?
it doesn't
so no inflection point
but, we can test it at x=0 and x=2
at x=0, we get f''(0)<0 so it is concave down. that means that x=0 being a max makes sense
at x=2, we get f''(2)>0 so it is concave up. that means that x=2 being a max make sense




local max is at x=0 (the point (0,0))
local min is at x=2 (the point (2,4))
6 0
4 years ago
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