2 meters

For each of the following sequences, • Give a formula for the nth term in the sequence, • Give a recursive definition for the se

Andrew [12]

Answer:

(a) $n^{th} = n$

$f(1) = 1$

$f(n) = f(n-1) + 1$

(b) $n^{th} = 2^{n-1}$

$f(1) = 1$

$f(n) = f(n-1) * 2$

(c) $n^{th} = n!$

$f(1) = 1$

$f(n) = f(n-1) * n$

Step-by-step explanation:

(a) This is a sequence of consecutive number

$n^{th} = n$

$f(1) = 1$

$f(n) = f(n-1) + 1$

(b) This is a sequence of 2 to the power of n - 1. The next number is twice time of this number

$n^{th} = 2^{n-1}$

$f(1) = 1$

$f(n) = f(n-1) * 2$

(c) This is factorial sequence. Where the next number is this number multiplied by $n^{th}$

$n^{th} = n!$

$f(1) = 1$

$f(n) = f(n-1) * n$

5 0

3 years ago

How many solutions does the system of equations have?

sertanlavr [38]

Y = 4 x + 3
y = 4 x + 3/2
no solution :) hope this helps

7 0

3 years ago

Read 2 more answers

Y=109x y=79x+2506 can some one solve both equations and a graph would be helpful plz

Amiraneli [1.4K]

Answer:

x ≈ 83.533

y = 9105.13

Step-by-step explanation:

Step 1: Substitution

109x = 79x + 2506

30x = 2506

x = 83.533

Step 2: Plug x in

y = 109(83.533)

y = 9105.13

Graphically:

5 0

4 years ago

Can you help me find each measure?

madam [21]

Answer:

Step-by-step explanation:

1. the measure for HJ you can already see, start from h and draw a line to J. The measure shows 63 degrees.

2. start at F and draw a line to G first. the measure shows 65 degrees. Now continue the line from where you stopped at G, to H. There's no measure but you can see it is in a semicircle. A semicircle is 180 degrees andthe other two angles are 63 and 65.

So...

180=63+65+GH

subtract to get GH

180-63-65= 52

so the measure GH is 52 degrees. The full measure you are trying to find is FGH thought. So ad the 65 and the 52.

FGH =117 degrees.

3. The meaure is CDE. CD as you can see is a right angles, so 90 degrees. But there is no measure for DE. If you look to the angle vertical from DE which is BA. It measures 40 degrees. DE and BA are vertical so they are congruent. If DE equals 40 and CD equals 90, put them together and you get 130.

CDE= 130 degrees

4. Next measure is BCD. We already know CD is a 90 degree angle but BC is blank. You can see the measure BCD is in a semi circle. A semicircle equals 180, CD equals 90, and BA equals 40

so...

180= 90+40+BC

so subtract

180-90-40= 50

BC =50

so add BC=50 and CD=90. SO, BCD is 140 degrees.

5.The angle LMN is next. MN is 30 but LM is blank. LMN is in a semi circle.

A semicircle is 180 degrees.

so...

180=105+30+LM

subtract

180-105-30=45

LM = 45

Add LM=45 and MN=30

LMN is 75 degrees.

6.The last angle is LNP

If you look at it MP is a semi circle, so 180 degrees. And LM is 45 from our last question. so 180 +45=225

LNP=225

hope this helps

4 0

3 years ago

Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions

monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

$y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2$

Now suppose $y$ has a power series expansion

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Then the ODE can be written as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

$a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}$

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

4 0

3 years ago