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Aleksandr-060686 [28]
2 years ago
7

The function f(x)=−4x+1 has a horizontal asymptote at?

Mathematics
1 answer:
a_sh-v [17]2 years ago
3 0

The function has a horizontal asymptote at y = 1

<h3>How to determine the horizontal asymptote?</h3>

The function is given as:

f(x) = -4/x + 1

Set the radicand to 0.

So, we have:

y = 0 + 1

Evaluate

y = 1

Hence, the function has a horizontal asymptote at y = 1

Read more about horizontal asymptote at:

brainly.com/question/4084552

#SPJ1

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Solve differential equation:<br><br> y'''+4y''-16y'-64y=0 y(0)=0, y'(0)=26, y''(0)=-16
Ipatiy [6.2K]

Answer:  The required solution of the given differential equation is

y(x)=3e^{4x}-3e^{-4x}+2xe^{-4x}.

Step-by-step explanation:  We are given to solve the following differential equation :

y^{\prime\prime\prime}+4y^{\prime\prime}-16y^\prime-64y=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\\y(0)=0,~y^\prime(0)=26,~y^{\prim\prime}(0)=-16.

Let, y=e^{mx} be an auxiliary solution of equation (i).

Then, y^\prime=me^{mx},~~y^{\prime\prime}=m^2e^{mx},~~y^{\prime\prime\prime}=m^3e^{mx}.

Substituting these values in equation (i), we get

m^3e^{mx}+4m^2e^{mx}-16me^{mx}-64e^{mx}=0\\\\\Rightarrow (m^3+4m^2-16m-64)e^{mx}=0\\\\\Rightarrow m^3+4m^2-16m-64=0,~~~~~~~~~[\textup{since }e^{mx}\neq 0]\\\\\Rightarrow m^2(m-4)+8m(m-4)+16(m-4)=0\\\\\Rightarrow (m-4)(m^2+8m+16)=0\\\\\Rightarrow (m-4)(m+4)^2=0\\\\\Rightarrow m-4=0,~~(m+4)^2=0\\\\\Rightarrow m=4,~m=-4,~-4.

So, the general solution is given by

y(x)=Ae^{4x}+Be^{-4x}+Cxe^{-4x}.

Then, we have

y^\prime=4Ae^{4x}-4Be^{-4x}-4Cxe^{-4x}+Ce^{-4x},\\\\y^{\prime\prime}=16Ae^{4x}+16Be^{4x}+16Cxe^{-4x}-4Ce^{-4x}-4Ce^{-4x}\\\\\Rightarrow y^{\prime\prime}=16Ae^{4x}+16Be^{4x}+16Cxe^{-4x}-8Ce^{-4x}.

With the conditions given, we get

y(0)=A+B+C\times 0\\\\\Rightarrow A+B=0\\\\\Rightarrow A=-B~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

y^\prime(0)=4A-4B+C\\\\\Rightarrow 4A-4B+C=26\\\\\Rightarrow 4(A+A)+C=26~~~~~~~~~~~~~~~~[\textup{using equation (i)}]\\\\\Rightarrow C=26-8A~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)

and

y^{\prime\prime}(0)=16A+16B-8C\\\\\Rightarrow 16A-16A-8C=-16~~~~~~~~~~~~[\textup{using equation (ii)}]\\\\\Rightarrow -8C=-16\\\\\Rightarrow C=2.

From equation (iii), we get

C=26-8A\\\\\Rightarrow 2=26-8A\\\\\Rightarrow 8A=24\\\\\Rightarrow A=3.

From equation (ii), we get

B=-3.

Therefore, the required solution of the given differential equation is

y(x)=3e^{4x}-3e^{-4x}+2xe^{-4x}.

4 0
3 years ago
Several friends each had 2/5 of a bag of peanuts left over from the baseball game. They realize that they could have bought 2 fe
pishuonlain [190]
..................................
6 0
3 years ago
HELP PLEASE! Find the standard deviation for Data set 1 and 2. round to the nearest hundred. DATA SET ONE: 8.2, 11.6 8.7, 10.6,
Elan Coil [88]

Answer:

0.92 and 1.25 respectively

Step-by-step explanation:

0.92 and 1.25 respectively

first the mean of each value which is 9.7 and 9.8 respectively

standard deviation = square root of the mean deviation of each value

then deduct the mean from each element

and square each value and add them together. not you should square each deviation value before you add them

at last divide the result by the number of frequency and find it's square root

3 0
3 years ago
20.) Which inequality represents the solution on the graph?*<br><br> 
VMariaS [17]

Answer:

Option 2

Step-by-step explanation:

You can see that the graph of the inequality is x <= -3, so you want to find why inequality matches that. Solve each one:

1

.3x \geq  -9\\x \geq -3

doesn't match

2.

-7x\geq 21\\x\leq -3

Does match, so this is the answer.

7 0
2 years ago
What is the lcd of (3x)/(x + 3), 3/(x ^ 2 - 9)
Rudik [331]

Answer:

Simplify the denominator.

Tap for more steps...

x

(

x

+

3

)

(

x

−

3

)

⋅

3

x

x

2

−

5

x

+

6

Factor

x

2

−

5

x

+

6

using the AC method.

Tap for more steps...

x

(

x

+

3

)

(

x

−

3

)

⋅

3

x

(

x

−

3

)

(

x

−

2

)

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

(

x

+

3

)

(

x

−

3

)

,

(

x

−

3

)

(

x

−

2

)

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number

1

is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of

1

,

1

is the result of multiplying all prime factors the greatest number of times they occur in either number.

1

The factor for

x

+

3

is

x

+

3

itself.

(

x

+

3

)

=

x

+

3

(

x

+

3

)

occurs

1

time.

The factor for

x

−

3

is

x

−

3

itself.

(

x

−

3

)

=

x

−

3

(

x

−

3

)

occurs

1

time.

The factor for

x

−

2

is

x

−

2

itself.

(

x

−

2

)

=

x

−

2

(

x

−

2

)

occurs

1

time.

The LCM of

x

+

3

,

x

−

3

,

x

−

3

,

x

−

2

is the result of multiplying all factors the greatest number of times they occur in either term.

(

x

+

3

)

(

x

−

3

)

(

x

−

2

)

Step-by-step explanation:

there does that help

6 0
3 years ago
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