Simplify 9 - {x - (7 + x)}.

Which numbers below belong to the solution set inequality?

andrew-mc [135]

the inequality goes from -30 and all number up but not including -30 because it has open circle. so mark all number greater than -30

3 0

4 years ago

Y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′′(0) = 0

Snowcat [4.5K]

Answer:

y(t) = 3u₂(t) [ $e^{-2t+4} - e^{-5t + 10)}$ ] - 4u₅(t) [ $e^{-2t+10)} - e^{-5t + 25)}$ ]

Step-by-step explanation:

To find - y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′(0) = 0

Formula used -

L{δ(t − c)} = $e^{-cs}$

L{f''(t) = s²F(s) - sf(0) - f'(0)

L{f'(t) = sF(s) - f(0)

Solution -

By Applying Laplace transform, we get

L{y″+ 5y′ + 6y} = L{3δ(t − 2) − 4δ(t −5)}

⇒L{y''} + 5L{y'} + 6L{y} = 3L{δ(t − 2)} − 4L{δ(t −5)}

⇒s²Y(s) - sy(0) - y'(0) + 5[sY(s) - y(0)] + 6Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒s²Y(s) - 0 - 0 + 5[sY(s) - 0] + 6Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒s²Y(s) + 5sY(s) + 6Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[s² + 5s + 6] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[s² + 3s + 2s + 6] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[s(s + 3) + 2(s + 3)] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒[(s + 2)(s + 3)] Y(s) = 3 $e^{-2s}$ - 4 $e^{-5s}$

⇒Y(s) = $\frac{3e^{-2s} }{(s + 2)(s + 3)} - \frac{4e^{-5s} }{(s + 2)(s + 3)}$

Now,

Let

$\frac{1}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3} \\\frac{1}{(s+2)(s+3)} = \frac{A(s + 3) + B(s+2)}{(s+2)(s+3)}\\1 = As + 3A + Bs + 2B\\1 = (A+B)s + (3A + 2B)$

By Comparing, we get

A + B = 0 and 3A + 2B = 1

⇒A = -B

and

3(-B) + 2B = 1

⇒-B = 1

⇒B = -1

So,

A = 1

∴ we get

$\frac{1}{(s+2)(s+3)} = \frac{1}{s+2} + \frac{-1}{s+3}$

So,

Y(s) = $3e^{-2s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}] - 4e^{-5s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}]$

⇒Y(s) = $3e^{-2s} \frac{1}{(s + 2)} - 3e^{-2s} \frac{1}{(s + 3)} - 4e^{-5s}\frac{1}{(s + 2)} + 4e^{-5s}\frac{1}{(s + 3)}$

By applying inverse Laplace , we get

y(t) = 3u₂(t) [ $e^{-2(t-2)} - e^{-5(t - 2)}$ ] - 4u₅(t) [ $e^{-2(t-5)} - e^{-5(t - 5)}$ ]

⇒y(t) = 3u₂(t) [ $e^{-2t+4} - e^{-5t + 10)}$ ] - 4u₅(t) [ $e^{-2t+10)} - e^{-5t + 25)}$ ]

It is the required solution.

3 0

3 years ago

7q + 2p = 29 & 2p - q =5

faltersainse [42]

Eliminate p terms
times 2nd equation by -1
q-2p=-5
add to 2nd equation

q-2p=-5
<u>7q+2p=29 +</u>
8q+0p=24

8q=24
divide both sides by 8
q=3
sub back

2p-q=5
2p-3=5
add 3
2p=8
divide by 2
p=4

p=4
q=3

7 0

4 years ago

Read 2 more answers

A safety regulation states that the height hh of a guardrail should be 106 centimeters with an absolute deviation of no more tha

iren [92.7K]

Let us say that h is the height of the guardrail. Therefore the inequality equation that we can generate from this scenario is:

| h – 106 | = ± 7

There are two ways to solve this, either the equation is positive or negative.

When the equation is positive, therefore:

| h – 106 | = 7

h = 7 + 106 = 113 cm

When the equation is negative, therefore:

| h – 106 | = - 7

h = -7 + 106 = 99 cm

So the height must be 99 cm to 113 cm

6 0

4 years ago

A sprinkler is set to water the section of lawn represented by the shaded region in the circle below. which measurement is close

Elden [556K]

Answer:

A = 64.141

Step-by-step explanation:

150/360 = 0.41667

41.667% of the circle is shaded so when we find the area we will multiply ir by 0.41667 to find only the shaded area

A=πr^2 x 0.41667

A = π(7)^2 x 0.41667

A = 64.142

7 0

3 years ago