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3 years ago

How is a recursive sequence different from an arithmetic or geometric sequence?

1 answer:

4 0

An arithmetic or geometric is a recursive sequence, because you can write it in terms of the previous term adding or multiplying some constant respectively. A recursive sequence however, is not always arithmetic or geometric.

You might be interested in

Write 1,270 in expanded form

Vikentia [17]

Answer:

1,000 + 200 + 70

Step-by-step explanation:

Expanded form is a way of writing numbers to see the math value of individual digits.

so for example:

5,000 + 300 + 20 + 3

is 5,323.

4 0

3 years ago

Two turtles race. The speed of the first is 4 1/ 2 inches per minute and the speed of the other is 5 inches per minute How far a

RSB [31]

Answer:

(a) 1.5 inches

(b) 2.5 inches

Step-by-step explanation:

speed of first turtle = 4 1/2 inches per minute = 9/2 inches per minute

speed of the second turtle = 5inches per minute

distance = speed x time

(a) Distance traveled by first turtle in 3 minutes

s = 9/2 x 3 = 13.5 inches

distance traveled by second turtle in 3 minutes

s' = 5 x 3 = 15 inches

So, the gap is

s'- s = 15 - 13.5 = 1.5 inches

(b) Distance traveled by first turtle in 5 minutes

s = 9/2 x 5 = 22.5 inches

distance traveled by second turtle in 5 minutes

s' = 5 x 5 = 25 inches

So, the gap is

s'- s = 25 - 22.5 = 2.5 inches

s = 9/2 x 10 = 45 inches

distance traveled by second turtle in 10 minutes

s' = 5 x 10 = 50 inches

So, the gap is

s'- s = 50 - 45 = 5 inches

6 0

3 years ago

If lim x-> infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b

MAXImum [283]

We have

$\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}$

$\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0$

The rational term vanishes as x gets arbitrarily large, so we can ignore that term, leaving us with

$\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0$

and this happens if a = 1 and b = -1.

To confirm, we have

$\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0$

as required.

3 0

3 years ago

Q3. A student plans a method to prepare pure crystals of copper sulfate.

andreev551 [17]

Answer:

1) The student didn't add copper sulfate but instead he added calcium carbonate.

2) The heated solution should be left to get bigger crystals and should not be instantly allowed to evaporate.

3) He should use copper carbonate not calcium carbonate.

4) Sulfuric acid was added instead of hydrochloric acid.

5) Add solid excessively so that all of the acid is retained.

6) Filter the solution to remove unreacted solid.

7) Heat gently, so the crystals will start to appear.

3 0

3 years ago

Striped fNbric:

DIA [1.3K]

Answer:

sssk

Step-by-step explanation:

sksks

4 0

3 years ago