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3 years ago

Please help me out, I will give brainliest, this is due at 4 PM, please don't take too long?

Mathematics

2 answers:

Leya [2.2K]3 years ago

3 0

Answer:

Tampa

Step-by-step explanation:

Because they have the most significant change out of all of them.

Makovka662 [10]3 years ago

3 0

Answer:

Tampa

Step-by-step explanation:

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Water is pumped out of a holding tank at a rate of 6-6e^-0.13t liters/minute, where t is in minutes since the pump is started. I

astraxan [27]

Procedure:

1) Integrate the function, from t =0 to t = 60 minutues to obtain the number of liters pumped out in the entire interval, and

2) Substract the result from the initial content of the tank (1000 liters).

Hands on:

Integral of (6 - 6e^-0.13t) dt ]from t =0 to t = 60 min =

= 6t + 6 e^-0.13t / 0.13 = 6t + 46.1538 e^-0.13t ] from t =0 to t = 60 min =

6*60 + 46.1538 e^(-0.13*60) - 0 - 46.1538 = 360 + 0.01891 - 46.1538 = 313.865 liters

2) 1000 liters - 313.865 liters = 613.135 liters

Answer: 613.135 liters

3 0

3 years ago

What expression is equivalent to (10x)^-3

irakobra [83]

Answer:

1/ (1000x^3)

Step-by-step explanation:

yes

6 0

3 years ago

Read 2 more answers

Subtract the mixed numbers

Lisa [10]

Answer: 4 1/2 - 1 11/12 = 2 1/3 but in decimal form 2.3

8 0

2 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

3 years ago

What equals to 2/3 the answers denominator is 9

GuDViN [60]

The answer would be 6/9

explanation:

2/3 ?/9

if you multiply 3 and 3, it gives you 9, which is the denominator. then you also have to multiply the numerator by the same thing you multiply the denominator with. therefore, 2x3 would give you 6, which makes your answer:

6/9

8 0

3 years ago