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3 years ago

6

When a figure is rotated, does every point move the same distance? Explain. *

2 answers:

fenix001 [56]3 years ago

8 0

Answer:

Yes. Since the angle of rotation is the same, the distance that each point moves is the same.

Step-by-step explanation:

Anna007 [38]3 years ago

5 0

Answer Yes. Since the angle of rotation is the same, the distance that each point moves is the same i think

Step-by-step explanation:

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In the figure below, two secants are drawn to a circle from exterior point U. Suppose that UW=45, UZ=30, and UX=22.5. Find UY

Alex777 [14]

UY = 25

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3 0

2 years ago

14m + 51 - 6s + 75 - 81 + 115 - 95 - 41 + 10p - 3p + 51<br> what is the coefficient of m?

densk [106]

Answer:

$14is \: coefficient \: of \: m$

4 0

2 years ago

What is the answer to this question?

MrRa [10]

Answer:

20 i think

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

One of the interior angles of a regular polygon is 144. fine the number of sides of the polygon

lora16 [44]

Answer:

10

Step-by-step explanation:

In any polygon

exterior angle + interior angle = 180° , thus

exterior angle + 144 = 180 ( subtract 144 from both sides )

exterior angle = 36°

The sum of the exterior angles = 360°

Thus number of sides n is calculated as

n = 360 ÷ 36 = 10

3 0

2 years ago

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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

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

3 0

3 years ago