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3 years ago

6

PLEASE HELP WILL GIVE BRAINSLIST

1 answer:

Zina [86]3 years ago

6 0

Answer: 3. 55° 2. 225° 1. 125°

Step-by-step explanation: don't want brainliest

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The shirt cost $113 less than the coat. The shirt cost $50. How much is the coat?

olga nikolaevna [1]

Shirt: $50
Coat: $113

$50+ $113= $163

The coat costs $163, hope this helps!

7 0

3 years ago

Substitutions <br> y=x+4<br> y=3x+0

cluponka [151]

Answer:

x = 2

Step-by-step explanation:

since both equal y, you can set them equal to each other

x + 4 = 3x + 0

subtract x on both sides

4 = 2x

divide by 2 on both sides

2 = x

♡♡ hope this helped ♡♡

4 0

3 years ago

Read 2 more answers

Help me plzzz will give brainlest!!!!

Leno4ka [110]

Answer:b,they are not proportional

Step-by-step explanation:

8 0

3 years ago

Integers are natural numbers.

Stells [14]

Answer: False

Step-by-step explanation:

Hope this helps

5 0

4 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