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2 years ago

What is the multiplicative rate of change of the function?

Mathematics

1 answer:

BartSMP [9]2 years ago

8 0

Answer:

We know that the multiplicative rate of change of a function is the number by which each next term of an exponential function is increasing or decreasing. We can find multiplicative rate of change by dividing any term of the function by its previous term.

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What is the 7-intercept of 2x-4y=12

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Answer:

-3

Step-by-step explanation:

2x - 4y = 12

4y = 2x - 12

y = ½x - 3

Y intercept is -3

X intercept is 6

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3 years ago

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2 years ago

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2 years ago

Using Laplace transforms, solve x" + 4x' + 6x = 1- e^t with the following initial conditions: x(0) = x'(0) = 1.

professor190 [17]

Answer:

The solution to the differential equation is

$X(s)=\cfrac 1{6} -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)$

Step-by-step explanation:

Applying Laplace Transform will help us solve differential equations in Algebraic ways to find particular solutions, thus after applying Laplace transform and evaluating at the initial conditions we need to solve and apply Inverse Laplace transform to find the final answer.

Applying Laplace Transform

We can start applying Laplace at the given ODE

$x''(t)+4x'(t)+6x(t)=1-e^t$

So we will get

$s^2 X(s)-sx(0)-x'(0)+4(sX(s)-x(0))+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Applying initial conditions and solving for X(s).

If we apply the initial conditions we get

$s^2 X(s)-s-1+4(sX(s)-1)+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Simplifying

$s^2 X(s)-s-1+4sX(s)-4+6X(s)=\cfrac 1s -\cfrac1{s-1}$

$s^2 X(s)-s-5+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Moving all terms that do not have X(s) to the other side

$s^2 X(s)+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}+s+5$

Factoring X(s) and moving the rest to the other side.

$X(s)(s^2 +4s+6)=\cfrac 1s -\cfrac1{s-1}+s+5$

$X(s)=\cfrac 1{s(s^2 +4s+6)} -\cfrac1{(s-1)(s^2 +4s+6)}+\cfrac {s+5}{s^2 +4s+6}$

Partial fraction decomposition method.

In order to apply Inverse Laplace Transform, we need to separate the fractions into the simplest form, so we can apply partial fraction decomposition to the first 2 fractions. For the first one we have

$\cfrac 1{s(s^2 +4s+6)}=\cfrac As + \cfrac {Bs+C}{s^2+4s+6}$

So if we multiply both sides by the entire denominator we get

$1=A(s^2+4s+6) + (Bs+C)s$

At this point we can find the value of A fast if we plug s = 0, so we get

$1=A(6)+0$

So the value of A is

$A = \cfrac 16$

We can replace that on the previous equation and multiply all terms by 6

$1=\cfrac 16(s^2+4s+6) + (Bs+C)s$

$6=s^2+4s+6 + 6Bs^2+6Cs$

We can simplify a bit

$-s^2-4s= 6Bs^2+6Cs$

And by comparing coefficients we can tell the values of B and C

$-1= 6B\\B=-1/6\\-4=6C\\C=-4/6$

So the separated fraction will be

$\cfrac 1{s(s^2 +4s+6)}=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6}$

We can repeat the process for the second fraction.

$\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac A{s-1} + \cfrac {Bs+C}{s^2+4s+6}$

Multiplying by the entire denominator give us

$1=A(s^2+4s+6) + (Bs+C)(s-1)$

We can plug the value of s = 1 to find A fast.

$1=A(11) + 0$

So we get

$A = \cfrac1{11}$

We can replace that on the previous equation and multiply all terms by 11

$1=\cfrac 1{11}(s^2+4s+6) + (Bs+C)(s-1)$

$11=s^2+4s+6 + 11Bs^2+11Cs-11Bs-11C$

Simplifying

$-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C$

And by comparing coefficients we can tell the values of B and C.

$-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C\\-1=11B\\B=-\cfrac{1}{11}\\5=-11C\\C=-\cfrac{5}{11}$

So the separated fraction will be

$\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac {1/11}{s-1} + \cfrac {-s/11-5/11}{s^2+4s+6}$

So far replacing both expanded fractions on the solution

$X(s)=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6} -\cfrac {1/11}{s-1} -\cfrac {-s/11-5/11}{s^2+4s+6}+\cfrac {s+5}{s^2 +4s+6}$

We can combine the fractions with the same denominator

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {-s/6-4/6+s/11+5/11+s+5}{s^2 +4s+6}$

Simplifying give us

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{s^2 +4s+6}$

Completing the square

One last step before applying the Inverse Laplace transform is to factor the denominators using completing the square procedure for this case, so we will have

$s^2+4s+6 = s^2 +4s+4-4+6$

We are adding half of the middle term but squared, so the first 3 terms become the perfect square, that is

$=(s+2)^2+2$

So we get

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{(s+2)^2 +(\sqrt 2)^2}$

Notice that the denominator has (s+2) inside a square we need to match that on the numerator so we can add and subtract 2

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2-2)/66+316 /66}{(s+2)^2 +(\sqrt 2)^2}\\X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66+194 /66}{(s+2)^2 +(\sqrt 2)^2}$

Lastly we can split the fraction one more

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66}{(s+2)^2 +(\sqrt 2)^2}+\cfrac {194 /66}{(s+2)^2 +(\sqrt 2)^2}$

Applying Inverse Laplace Transform.

Since all terms are ready we can apply Inverse Laplace transform directly to each term and we will get

$\boxed{X(s)=\cfrac 1{6} -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)}$

6 0

3 years ago

Factor

AfilCa [17]

Answer:

The answer is 6ab³c(3ab + 4c)

Step-by-step explanation:

You have to collect the like-term :

18a²b⁴c + 24ab³c²

= 6(3a²b⁴c + 4ab³c²)

= 6a(3ab⁴c + 4b³c²)

= 6ab³(3abc + 4c²)

= 6ab³c(3ab + 4c)

6 0

2 years ago