1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Elina [12.6K]
3 years ago
6

Is the following sequence a geometric sequence? 4, 12, 36, 108

Mathematics
1 answer:
Lyrx [107]3 years ago
5 0

Answer:

yes it is

Step-by-step explanation:

it is since there is a common ratio between each term

You might be interested in
A) Evaluate the limit using the appropriate properties of limits. (If an answer does not exist, enter DNE.)
Gelneren [198K]

For purely rational functions, the general strategy is to compare the degrees of the numerator and denominator.

A)

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \boxed{\frac27}

because both numerator and denominator have the same degree (2), so their end behaviors are similar enough that the ratio of their coefficients determine the limit at infinity.

More precisely, we can divide through the expression uniformly by <em>x</em> ²,

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \lim_{x\to\infty} \frac{2-\dfrac5{x^2}}{7+\dfrac1x-\dfrac3{x^2}}

Then each remaining rational term converges to 0 as <em>x</em> gets arbitrarily large, leaving 2 in the numerator and 7 in the denominator.

B) By the same reasoning,

\displaystyle \lim_{x\to\infty} \frac{5x-3}{2x+1} = \boxed{\frac52}

C) This time, the degree of the denominator exceeds the degree of the numerator, so it grows faster than <em>x</em> - 1. Dividing a number by a larger number makes for a smaller number. This means the limit will be 0:

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \boxed{0}

More precisely,

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \lim_{x\to-\infty}\frac{\dfrac1x-\dfrac1{x^2}}{1+\dfrac8{x^2}} = \dfrac01 = 0

D) Looks like this limit should read

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2}

which is just another case of (A) and (B); the limit would be

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = -1

That is,

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = \lim_{t\to\infty}\frac{\dfrac1{t^{3/2}}+1}{\dfrac3t-1} = \dfrac1{-1} = -1

However, in case you meant something else, such as

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t+t^2}}{3t-t^2}

then the limit would be different:

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t^2}\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{t\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{\sqrt{\dfrac1t+1}}{3-t} = 0

since the degree of the denominator is larger.

One important detail glossed over here is that

\sqrt{t^2} = |t|

for all real <em>t</em>. But since <em>t</em> is approaching *positive* infinity, we have <em>t</em> > 0, for which |<em>t</em> | = <em>t</em>.

E) Similar to (D) - bear in mind this has the same ambiguity I mentioned above, but in this case the limit's value is unaffected -

\displaystyle \lim_{x\to\infty} \frac{x^4}{\sqrt{x^8+9}} = \lim_{x\to\infty}\frac{x^4}{\sqrt{x^8}\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac{x^4}{x^4\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac1{\sqrt{1+\dfrac9{x^8}}} = \boxed{1}

Again,

\sqrt{x^8} = |x^4|

but <em>x</em> ⁴ is non-negative for real <em>x</em>.

F) Also somewhat ambiguous:

\displaystyle \lim_{x\to\infty}\frac{\sqrt{x+5x^2}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{x\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{\dfrac1x+5}}{3-\dfrac1x} = \dfrac{\sqrt5}3

or

\displaystyle \lim_{x\to\infty}\frac{\sqrt{x}+5x^2}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{\sqrt x}+5x}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{x^{3/2}}+5}{3-\dfrac1x} = \frac53\lim_{x\to\infty}x = \infty

G) For a regular polynomial (unless you left out a denominator), the leading term determines the end behavior. In other words, for large <em>x</em>, <em>x</em> ⁴ is much larger than <em>x</em> ², so effectively

\displaystyle \lim_{x\to\infty}(x^4-2x) = \lim_{x\to\infty}x^4 = \boxed{\infty}

6 0
3 years ago
Can someone help me understand how to name polynomials??
Serhud [2]

A polynomial can have  

constants

variables

exponents

that can be combined using addition, subtraction, multiplication and division ...

... except ...

... not division by a variable

8 0
3 years ago
If x equals -6, then find the matching value for y.<br> 2x+2y=-2
BabaBlast [244]

2(-6) + 2y = -2

-12 + 2y =-2

add 12 to each side

2y =10

10/2 = 5

y=5


8 0
3 years ago
Please help me I really need this answer.
Sindrei [870]
Reduce planning cycle times, optimize operational plans and forecast accurately. Have Interactive Content, Dynamic Reporting, Guided Experience, and Automatic Recommendations. Hopes this helps like you helped me :D
8 0
4 years ago
Your hands are slow,
aivan3 [116]

Answer:

uh

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Other questions:
  • What's the value of (1/8)^-1/3
    10·2 answers
  • (Show work please &amp; explain the steps ❤️)
    9·2 answers
  • Is 7cups less than,greater than or equal to 3 1/4 pint
    5·1 answer
  • Joeys grade has a goal to collect 4,000 cans for the school canned food drive . There are 486 students in joeys grade. If each s
    14·1 answer
  • Find 3 consecutive numbers that equal 45
    14·2 answers
  • What is the function rule for the line?
    13·1 answer
  • Please solve this!<br><br> Simplify:<br><br> 7x/2x
    6·1 answer
  • A rectangle has a diagonal of 277 centimeters and a length of 252cm. How long is the width?
    8·2 answers
  • Please help solve for RN
    11·1 answer
  • Randy jog 3 1/2 miles in 3/4 hours write a common ratio as miles to hours
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!