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3 years ago

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We will find the solution to the following lhcc recurrence: an=8an−1−16an−2 for n≥2 with initial conditions a0=4,a1=7. The first

step as usual is to find the characteristic equation by trying a solution of the "geometric" format an=rnan=rn. (We assume also r≠0). In this case we get: rn=8r^n−1−16r^n−2. Since we are assuming r≠0r≠0 we can divide by the smallest power of r, i.e., rn−2 to get the characteristic equation:
r^2=8r−16. (Notice since our lhcc recurrence was degree 2, the characteristic equation is degree 2.)

This characteristic equation has a single root rr. (We say the root has multiplicity 2). Find r.
r=?

1 answer:

bagirrra123 [75]3 years ago

5 0

Answer:

idk try to figure it pouy

Step-by-step explanation:

You might be interested in

How many solution does 3=5

swat32

Answer:

None.

Step-by-step explanation:

The equation 3=5 is false, therefore, there are no solutions.

3≠5

3 0

3 years ago

The larger of two numbers is 12 more than the smaller number. if the sum of the two numbers is 74 find the two numbers.

STatiana [176]

Hi there!

Let th' two number be ' a ' and ' b ' [ a > b ]

According to th' question :-

→ The largest of two numbers is 12 more than the smaller number.

a = 12 + b ---(i)

→ Sum of the two numbers is 74.

a + b = 74

b = 74 - a ---(ii)

Substitute th' value of ' b in Eqn. :-

a = 12 + (74 - a)

a = 86 - a

a + a = 86

2a = 86

a = $\dfrac {86}{2}$ = 43

Now,
Substitute th' value of ' a ' in Eqn. (ii)

b = 74 - b

b = 74 - 43

b = 31

Hence,
The required answer is :-

• The smaller number is 31
• The largest number is 43

~ Hope it helps!

7 0

3 years ago

Read 2 more answers

4/5 - 1/6 in simplest form

sladkih [1.3K]

5 and 6 both go into 30

4/5= 24/30
1/6= 5/30

24/30-5/30= 19/30

4 0

4 years ago

Read 2 more answers

14, 16, and 20 using elimination method showing work. Thanks so much

Nady [450]

14) x=0, y=3, z=-2

Solution Set (0,3,-2)

16) x=1, y=1 and z=1

Solution set = (1,1,1)

20) x = -263/31, y=164/31 ,z=122/31

Solution set (-263/31, 164/31 ,122/31)

Step-by-step explanation:

14)

$x-y+2z=-7\\y+z=1\\x=2y+3z$

Rearranging and solving:

$x-y+2z=-7\,\,\,eq(1)\\y+z=1\,\,\,eq(2)\\x-2y-3z=0\,\,\,eq(3)$

Eliminate y:

Adding eq(1) and eq(2)

$x-y+2z=-7\,\,\,eq(1)\\ 0x+y+z=1\,\,\,eq(2)\\-------\\x+3z=-6\,\,\,eq(4)$

Multiply eq(2) with 2 and add with eq(3)

$0x+2y+2z=2\,\,\,eq(2)\\\\x-2y-3z=0\,\,\,eq(3)\\--------\\x-z=2\,\,\,eq(5)$

Eliminate x:

Subtract eq(4) and eq(5)

$x+3z=-6\,\,\,eq(4)\\x-z=2\,\,\,eq(5)\\-\,\,\,+\,\,\,\,\,\,-\\---------\\4z=-8\\z= -2$

So, value of z = -2

Now putting value of z in eq(2)

$y+z=1\\y+(-2)=1\\y-2=1\\y=1+2\\y=3$

So, value of y = 3

Now, putting value of z and y in eq(1)

$x-y+2z=-7\\x-(3)+2(-2)=-7\\x-3-4=-7\\x-7=-7\\x=-7+7\\x=0$

So, value of x = 0

So, x=0, y=3, z=-2

S.S(0,3,-2)

16)

$3x-y+z=3\\\x+y+2z=4\\x+2y+z=4$

Let:

$3x-y+z=3\,\,\,eq(1)\\x+y+2z=4\,\,\,eq(2)\\x+2y+z=4\,\,\,eq(3)$

Eliminating y:

Adding eq(1) and (2)

$3x-y+z=3\,\,\,eq(1)\\x+y+2z=4\,\,\,eq(2)\\---------\\4x+3z=7\,\,\,eq(4)$

Multiply eq(1) by 2 and add with eq(3)

$6x-2y+2z=6\,\,\,eq(1)\\x+2y+z=4\,\,\,eq(3)\\---------\\7x+3z=10\,\,\,eq(5)$

Now eliminating z in eq(4) and eq(5) to find value of x

Subtracting eq(4) and eq(5)

$4x+3z=7\,\,\,eq(4)\\7x+3z=10\,\,\,eq(5)\\-\,\,\,-\,\,\,\,\,\,\,\,\,\,-\\-----------\\-3x=-3\\x=-3/-3\\x=1$

So, value of x = 1

Putting value of x in eq(4) to find value of x:

$4x+3z=7\\4(1)+3z=7\\4+3z=7\\3z=7-4\\z=3/3\\z=1$

So, value of z = 1

Putting value of x and z in eq(2) to find value of y:

$x+y+2z=4\\1+y+2(1)=4\\1+y+2=4\\y+3=4\\y=4-3\\y=1$

So, x=1, y=1 and z=1

Solution set = (1,1,1)

20)

$x+4y-5z=-7\\3x+2y+2z=-7\\2x+y+5z=8$

Let:

$x+4y-5z=-7\,\,\,eq(1)\\3x+2y+2z=-7\,\,\,eq(2)\\2x+y+5z=8\,\,\,eq(3)$

Solving:

Eliminating z :

Adding eq(1) and eq(3)

$x+4y-5z=-7\,\,\,eq(1)\\2x+y+5z=8\,\,\,eq(3)\\---------\\3x+5y=1\,\,\,eq(4)$

Multiply eq(1) with 2 and eq(2) with 5 and add:

$2x+8y-10z=-14\,\,\,eq(1)\\15x+10y+10z=-35\,\,\,eq(2)\\----------\\17x+18y=-49\,\,\,eq(5)$

Eliminate y:

Multiply eq(4) with 18 and eq(5) with 5 and subtract:

$54x+90y=18\\85x+90y=-245\\-\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\\-------\\-31x=158\\x=-\frac{263}{31}$

So, value of x = -263/31

Putting value of x in eq(4)

$3x+5y=1\\3(-\frac{263}{31})+5y=1\\-\frac{789}{31}+5y=1 \\5y=1+\frac{789}{31}\\5y=\frac{820}{31}\\y=\frac{820}{31*5}\\y=\frac{164}{31}$

Now putting x = -263/31 and y=164/31 in eq(1) and finding z:

We get z=122/31

So, x = -263/31, y=164/31 ,z=122/31

Solution set (-263/31, 164/31 ,122/31)

Keywords: Solving system of Equations

Learn more about Solving system of Equations at:

#learnwithBrainly

4 0

3 years ago

Discrete MathA coin is flipped four times. For each of the events described below, express the event as a set in roster notation

Vladimir79 [104]

Answer:

a) 25%

b) 50%

c) 87.5%

Step-by-step explanation:

There are $2^4 = 16$ number of outcome in total.

a)First and last flips come up heads. There are 4 outcomes here. Probability of the event is 4/16 = 1/4, or 25%.

- HHHH

- HHTH

- HTHH

- HTTH

b) at least 2 Consecutive flips that is heads. There are 8 outcomes. Probability of event is 8/16 = 1/2, or 50%

- HHHH

- HHHT

- HHTH

- HHTT

- THHH

- THHT

- HTHH

- TTHH

c) At least 2 consecutive flips that are the same. There are 14 outcomes. Probability of 14/16 = 7/8, or 87.5%

- HHHH

- HHHT

- HHTH

- HHTT

- THHH

- THHT

- HTHH

- TTHH

- TTTT

- TTTH

- TTHT

- HTTT

- HTTH

- THTT

6 0

3 years ago