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I am Lyosha [343]
3 years ago
8

What is the 16th term in the sequence? 2, 6, 18, 54, 162, . . .

Mathematics
2 answers:
Gala2k [10]3 years ago
8 0
It was a hard one but the answer is 258,280,326.
hope it really helped.
Sladkaya [172]3 years ago
5 0
You could multiplied every thing by 3
Each time you got a product
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What is the asymptote for the graph of this logarithmic function?<br><br> f(x) = log3(x – 1)
Pepsi [2]

Answer:

Vertical Asymptote:

x=1

Horizontal asymptote:

it does not exist

Step-by-step explanation:

we are given

f(x)=log_3(x-1)

Vertical asymptote:

we know that vertical asymptotes are values of x where f(x) becomes +inf or -inf

we know that any log becomes -inf when value inside log is zero

so, we can set value inside log to zero

and then we can solve for x

x-1=0

we get

x=1

Horizontal asymptote:

we know that

horizontal asymptote is a value of y when x is +inf or -inf

For finding horizontal asymptote , we find lim x-->inf or -inf

\lim_{x \to \infty}  f(x)= \lim_{x \to \infty}log_3(x-1)

\lim_{x \to \infty}  f(x)=log_3(\infty-1)

\lim_{x \to \infty}  f(x)=undefined

so, it does not exist

7 0
4 years ago
Read 2 more answers
What’s the answer for this…<br><br><br> Solve for p<br><br> 2p + 3 &gt; 2 (p – 3)
andreyandreev [35.5K]

Answer:

2p+3>2(p−3)

Use the distributive property to multiply 2 by p−3.

2p+3>2p−6

Subtract 2p from both sides.

2p+3−2p>−6

Combine 2p and −2p to get 0.

-3>6

This is true for any p.

p∈R

Step-by-step explanation:

Hope this helps! :)

3 0
2 years ago
Read 2 more answers
When solving an absolute value inequality, the fact of whether it is a "greater than/greater than or equal to" as opposed to "le
Alex73 [517]
It is true hopes this helps you
7 0
3 years ago
CAN SOMEONE PLEASE HELP ME WITH THIS STATISTICS PROBLEM I DONT UNDERSTAND IT AT ALLL!!!! The mean of a normal probability distri
tankabanditka [31]

Answer:

A.)

Lower Value = 295

Upper Value = 485

B.)

Lower Value = 200

Upper Value = 580

C.)

Lower Value = 105

Upper Value = 675

Step-by-step explanation:

Given that mean (M) = 390

Standard deviation (σ) = 95

Lower Value = mean value - number of standard deviations specified

Upper Value = mean value + number of standard deviations specified

a. Μ ± 1σ of the observations lie between what two values?

Lower Value = 390 - 1(95) = 295

Upper Value = 390 + 1(95) = 485

b. Μ ± 2σ of the observations lie between what two values?

Lower Value = 390 - 2(95) = 200

Upper Value = 390 + 2(95) = 580

c. Μ ± 3σ of the observations lie between what two values?

Lower Value = 390 - 3(95) = 105

Upper Value = 390 + 3(95) = 675

4 0
3 years ago
Write an equivalent expression for 1/ y-1/4 using positive exponents
emmainna [20.7K]

\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{1}{~~y^{-\frac{1}{4}}~~}\implies y^{\frac{1}{4}}

7 0
3 years ago
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