Which is 0.54 converted to a simplified fraction?

Suppose y = 2x + 1 , where x and y are functions of t. (a) if dx/dt = 3, find dy/dt when x = 4. dy dt = (b) if dy/dt = 2, find d

qwelly [4]

Answer:

Step-by-step explanation:

If y = 2x + 1, then dy/dt = 2(dx/dt).

If y = 2x + 1, then y = 2(40) + 1 when 40 is substituted for x. y = 81.

(a) if dx/dt = 3, find dy/dt when x = 4:

Replacing dx/dt with 3 in dy/dt = 2(dx/dt) yields dy/dt = 2(3) = 6.

(b) if dy/dt = 2, find dx/dt when x = 40:

Replacing dy/dt with 2 in dy/dt = 2(dx/dt) results in 2 = 2(dx/dt), so dx/dt must be 1.

3 0

2 years ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago

Does 3,4, and 5 make a right triangle or not a right triangle?

S_A_V [24]

Answer:

I believe it does

Step-by-step explanation:

sorry if i am wrong

6 0

3 years ago

Read 2 more answers

Please help meeeee. I literally learned this last year but my dum mb self forgot hot to do it

Aleks [24]

It would be 30 because 10^4 is 10000 and 10000 times 12 is 120000 and 10^3 is 1000 and 1000 times 4 is 4000 so 120000 divided by 4000 is 30

8 0

3 years ago

PLZ HELP IWILL GIVE BRAINLIESTTT

liq [111]

Answer:

Choice B

A = b x h

Step-by-step explanation:

In other words the area of a parallelogram is the base times height

8 0

2 years ago