Ask question

Ask question

All categories

3 years ago

6

Cody bagged the plastic bottles after a recycling drive. He placed 15 bottles in the first bag, 90 bottles in the second bag, 36

0 bottles in the third bag, and 720 bottles in the fourth bag. What kind of sequence is this?

1 answer:

stepan [7]3 years ago

3 0

Answer: repeating sequence

Step-by-step explanation:

You might be interested in

What is 4 divide blank equal 4<br> ?

soldi70 [24.7K]

1 is the answer to nthis/...

3 0

3 years ago

Read 2 more answers

A company pays a bonus to four employees A, B, C, and D. A gets four times as much as B. B gets 50% of the amount paid to C. C a

Tomtit [17]

Answer:

A = 800, B = 200, C = 400 Andy D = 400

Step-by-step explanation:

5 0

3 years ago

How to put 40,023,032 in expanded form

Valentin [98]

40,023,032 = 40,000,000 + 20,000 + 3,000 + 30 + 2

5 0

3 years ago

1/5 divided by 2 = ___

Vlad1618 [11]

Answer:

0.1

Step-by-step explanation:

According to my calculator lol

6 0

3 years ago

Read 2 more answers

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

4 years ago