Ask question

Ask question

All categories

Elena-2011 [213]

3 years ago

13

HELP PLEASEEEE!!!

2 answers:

beks73 [17]3 years ago

8 0

Answer:

0.4

Step-by-step explanation:

I took the test

polet [3.4K]3 years ago

4 0

Answer:this is a little late but i think its 0.4

Step-by-step explanation:

You might be interested in

Tomtit [17]

Teh answer is 4 cm^2

5 0

4 years ago

Read 2 more answers

a carnival charges $8.50 per ticket for adults and $4.50 per ticket for children. which expression represents the total price a

ra1l [238]

The answer is D <span>8.5x + 4.5y</span>

3 0

3 years ago

19. Heather invests $4,900 in an account with a 3.5% interest rate, making no other deposits or withdrawals. What will Heather’s

denis-greek [22]

Answer:

$\$5,828.28$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=5\ years\\ P=\$4,900\\ r=0.035\\n=2$

substitute in the formula above

$A=\$4,900(1+\frac{0.035}{2})^{2*5}$

$A=\$4,900(1.0175)^{10}=\$5,828.28$

5 0

3 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

4 years ago

Please help, and if you could show work too that would be amazing!! :)

luda_lava [24]

Step-by-step explanation:

you first need to know what is mean .mean is when you add all the numbers up then divide them by 2

3 0

3 years ago