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3 years ago

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Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding re

ctangular equation by eliminating the parameter.
x = 3t - 5 y = 5t + 1

1 answer:

raketka [301]3 years ago

8 0

Answer:

(See explanation below for further details).

Step-by-step explanation:

Let be a parametric curve represented by $x = 3\cdot t - 5$ and $y = 5\cdot t + 1$ , where $t$ is the parametric variable.

The curve is represented graphically with the help of a graphing tool, whose outcome is included in the image attached below. The corresponding rectangular equation is found by eliminating t of each equation.

$t = \frac{x+5}{3}$ and $t = \frac{y-1}{5}$

$\frac{x+5}{3} = \frac{y-1}{5}$

$5\cdot (x+5) = 3\cdot (y-1)$

$5\cdot x +25 = 3\cdot y - 3$

$5\cdot x -3\cdot y = -28$

The parametric equations represents a linear function (first-order polynomial).

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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

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

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4 years ago

What is the missing constant term in the perfect square that starts with x^2-12x

Ipatiy [6.2K]

Essentially, we are trying to find the missing constant term of $(x - a)^2$ (remember that we are subtracting $a$ due to the negative sign in front of the second term). Let's expand this to see what we can work with:

$(x - a)^2$

Set up

$x^2 - ax - ax + a^2 = x^2 - 2ax + a^2$

FOIL and simplify

Now, we know the second term is $12x$ , so let's set the second term in the polynomial we just found equal to $12x$ :

$2ax = 12x$

Set up

$2a = 12$

Divide both sides of the equation by $x$

$a = 6$

Divide both sides of the equation by 2

We have found $a = 6$ . We know the missing constant term is $a^2$ , according to the polynomial we found earlier. Thus, the missing term is:

$a^2 \Rightarrow 6^2 = 36$

The missing constant term is 36.

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3 years ago

Simplify<br> 6(-5n + 7 thank you

Studentka2010 [4]

Answer:

-30n + 42

Step-by-step explanation:

6( -5n + 7)

-30n + 42

Hope this helps!

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3 years ago

The sum of two numbers is 27 and their product is 50. Find the numbers.?<br><br>

san4es73 [151]

Answer:

a + b = 27

=> a = 27 - b

ab = 50

So, (27 - b) x b = 50

27b - b^2 = 50

b^2 - 27b + 50 = 0

Solving, b = 25, 2

Hence,

Case 1: b = 25, a = 2

Case 2, b = 2, a = 25

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Answer:

I'm not sure the answer but I can tell you how to solve.

Step-by-step explanation:

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3 years ago