Ask question

Ask question

All categories

2 years ago

13

D = {f, g, k}

1 answer:

atroni [7]2 years ago

7 0

Answer:

Step-by-step explanation:

D∩L={f}

D∪L={a,f,g,j,k}

You might be interested in

How do you model the square root of 16

kotegsom [21]

$\sqrt{16} =\sqrt{4*4}=\sqrt{4}*\sqrt{4}=2*2=4$

4 0

2 years ago

Read 2 more answers

Leo has already taken 1 page of notes on his own, and he will take 1 page during each hour

sattari [20]

Answer:

49 Hours

Step-by-step explanation:

If it takes him 1 hour to finish 1 page of notes and he already finished 1, he would need to do a total of 49 hours to get 49 more pages done.

Hope this helps!

8 0

2 years ago

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

3 0

3 years ago

A. When solving the system of linear equations 6x+y = 1 and 7x+2y = -3 with substitution, it is easiest to first solve the equat

nalin [4]

Answer:

6x + y = 1

y = 1 - 6x

7x + 2(1 - 6x) = -3 substitute 1 - 6x in for y

7x + 2 - 12x = -3 combine like terms

2 - 5x = -3 subtract 2 from both sides

-5x = -5 divide by -5

x = 1

3 0

2 years ago

6 3/4 % tax, in dollars, on a price of $14.29

seraphim [82]

The awnser to this would be (6.75)(14.29)

4 0

3 years ago