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3 years ago

9

Help quick!! Pleaseee!

1 answer:

Elina [12.6K]3 years ago

4 0

Answer:

ahaha would love to help but can't see the photo :/

Step-by-step explanation:

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If you have 12 customers in a lawn care business, what is the minimum number of times you need to mow per customer each month to

Trava [24]

I believe the answer would be D. Because, 10*12= 120 and 400/120= 3.33
Since you are talking about money, you would round up.

Meaning that the answer should be D.

7 0

4 years ago

Given the Arithmetic sequence A1,A2,A3,A4 53, 62, 71, 80 What is the value of A38?

Murrr4er [49]

Answer:

$A_{38} = 350$

Step-by-step explanation:

The 5th term of the arithmetic sequence is 53. We can write the equation:

$a + 4d = 53...(1)$

The 6th term of the arithmetic sequence is 62. We can write the equation:

$a + 5d = 62...(2)$

Subtract the first equation from the second one to get:

$5d - 4d = 62 - 53$

$d = 9$

The first term is

$a + 4(9) = 53$

$a + 36 = 53$

$a = 53 - 36$

$a = 17$

The 38th term of the sequence is given by:

$A_{38} = a + 37d$

$A_{38} = 17+ 37(9)$

$A_{38} = 350$

3 0

3 years ago

Read 2 more answers

What is the value of x in the equation 8 + 4 = 2(x - 1)?

kramer

Answer:

Step-by-step explanation:

8 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

What does reflects most nearly mean

Gala2k [10]

Either to mirror, or to copy.

5 0

3 years ago

Read 2 more answers