What multiplies to 16 adds to -24 ?

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

a)

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

__

Using these formulas on problem (d), we get ...

d)

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

__

And for problem (f), we get ...

f)

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

Comment on problem g

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

2 years ago

Does anyone know this? I’m really stuck

S_A_V [24]

p = 21

Step-by-step explanation:

When you have a midsegment, the base is going to be two times as large as the midsegment. You can find the midsegment if you multiply it by two and set it equal to the base. I have attached a photo below that may help you.

This is the equation we will be using.

$2(p-13)=p-5$

Now let's solve! Let's start by distributing the 2 by what's in the parentheses.

$2p-26=p-5$

Subtract p from both sides.

$p-26=-5$

Add 26 to both sides.

p = 21

7 0

2 years ago

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why is using the metric system superior to using the standard system of measurement... Please someone help I need this and I tri

PilotLPTM [1.2K]

Metric is used world-wide and is used everywhere except in english places (britan and USA)

some examples why metric is nice

base 10 system
1cm=10mm=0.1m=0.01km
1cm^3=1ml
1000cm^3=1L
1000ml=1L

0 C⁰=freezing tempurature
100 C⁰=boinling tempurature

in english
12in=1foot
3ft=1yd
16oz=1lb
8oz=1cup
3tsbs=1tbs
2c=1pt
2pt=1qt
4qt=1gal
32 F⁰=freezing
2hundred something=boiling

metric is nice because everything is in multiplules of 10

and also this exerpt from a book
"In metric, one milliliter of water occupies one cubic centimeter, weighs one gram, and requires one calorie of energy to heat up by one degree centigrade which is 1 per cent of the difference between its freezing point and its boiling point. An amount of hydrogen weighing the same amount has exactly one mole of atoms in it. Whereas in the American system the answer to 'How much energy does it take to boil a room-temperature gallon of water?' is go **** yourself because you can't directly relate any of those quantities "

7 0

3 years ago

A transformation T:(x,y) (x-5,y+3) the image of A(2,-1) is

kodGreya [7K]

According to the transformation, the image of (x,y) is (x-5, y+3). Therefore, plug in the coordinates of A, or x=2, y=-1 into (x-5, y+3), and the image of A is (2-5, -1+3)=(-3, 2).

3 0

3 years ago

Rectangle ABCD is dilated by a scale factor of 2 with the origin as the center of dilation, resulting in the image A'BCD. If the

ludmilkaskok [199]

Answer:

Step-by-step explanation:

hello i am knew to this app i am trying to figure how this app works

3 0

2 years ago

Read 2 more answers