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3 years ago

Helpppp!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! pleaseeeeeee

Mathematics

2 answers:

Umnica [9.8K]3 years ago

5 0

All I can give you is all of them are x12

Galina-37 [17]3 years ago

4 0

Answer:

Step-by-step explanation:

y= money earned , x is the number of hours

k is the constant=y/x=12

b- if she worked 18 hours:

y=12(18)

<h2>y=216 dollars</h2>

c- if she earned 330 dollars:

330=12x

x=330/12

<h2>x=27.5 hours</h2>

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(2x - 1) ft<br> (x + 4) ft<br> 7 ft

anygoal [31]

What is the question?

7 0

3 years ago

Solve for x.<br> (17x - 23)<br> (&r - 4)*<br> -(3x + 17)

Finger [1]

Answer:

17rx2−23rx−71x+75

Step-by-step explanation:

(17x−23)(xr−4)−(3x+17)

=(17x−23)(xr−4)+−1(3x+17)

=(17x−23)(xr−4)+−1(3x)+(−1)(17)

=(17x−23)(xr−4)+−3x+−17

=(17x)(xr)+(17x)(−4)+(−23)(xr)+(−23)(−4)+−3x+−17

=17rx2+−68x+−23rx+92+−3x+−17

=(17rx2)+(−23rx)+(−68x+−3x)+(92+−17)

=17rx2+−23rx+−71x+75

6 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

Which of the following are irrational numbers?<br><br> 3.14<br> √ 5<br> 0.3333<br> √ -4<br> 1/9

Galina-37 [17]

Answer:

√ -4

Step-by-step explanation:

all square roots of negative numbers are irrational

6 0

3 years ago

HELP PLSASE: Find the solution of the system of equations.

Mamont248 [21]

Answer:

Step-by-step explanation:

−

Explanation:

−

Subtract the bottom equation from the top one so the

and

−

cancel each other out.

−

3 0

3 years ago