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yulyashka [42]
4 years ago
13

Find the Laplace transformation of each of the following functions. In each case, specify the values of s for which the integral

converges. 2et a. b. 3e 5t-3 C. 2/-e 3 cos 5 10sin 6t 6sin 2t -5 cos 2 (P+1) (sin-cost) d. e. f. g. h. i.
Mathematics
1 answer:
MAVERICK [17]4 years ago
7 0

Answer:

a. \frac {2} {s-1} converges to s> 1.

b. \frac{3}{e^3 \left(s-5 \right)} converges to s> 5.

c. - \frac {2}{s + 3} converges to s> - 3.

d. \frac {s}{s^2 + 25} converges to s> 0.

e. \frac {10} {s^2 + 1} converges even s> 0.

f. \frac {12}{s^2 + 4} converges to s> 0.

g. -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4} converges to s> 0.

h. \frac {1} {s ^ 2 + 4} converges to s> 0.

Step-by-step explanation:

a. L \left\{2e^t \right\} = 2L \left\{e^t \right\} = 2 \cdot \frac {1} {s-1} = \frac {2} {s-1} converges to s> 1.

b. L \left\{3e^{5t-3} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = \frac{3}{e^3 \left(s-5 \right)} converges to s> 5.

c. L \left\{-2e^{-3t} \right\} = -2L \left\{e^{-3t} \right\} = - \frac {2}{s + 3} converges to s> - 3.

d. L \left\{\cos\left (5t \right)\right\} = \frac {s}{s^2 + 25} converges to s> 0.

e. L \left\{10 \sin\left(t\right)\right\} = 10L\left\{\sin\left(t\right)\right\} = \frac {10} {s^2 + 1} converges even s> 0.

f. L \left\{6\sin \left(2t \right) \right\} = 6L\left\{\sin\left (2t\right)\right\} = \frac {12}{s^2 + 4} converges to s> 0.

g. L \left\{-5\cos\left(2t + 1\right) \right\} = -5L\left\{\cos\left(2t + 1 \right)\right\} = -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4} converges to s> 0.

h. L\left\{\sin \left(t\right)\cos \left(t\right)\right\} = L\left\{\sin\left(2t\right)\frac{1}{2}\right\} =\frac{1}{2}\cdot \frac{2}{s^2+4} = \frac {1} {s ^ 2 + 4} converges to s> 0.

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Hamdi lives 452 miles from her
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Answer:

It's 317 miles.

Step-by-step explanation:

You just take the amount of miles it is from her grandparents house and the amount of miles to her cousins house and then you subtract them.

 452-135=317 miles apart.

7 0
3 years ago
Rhonda is drawing a 3-inch square inside an isosceles triangle with two side lengths measuring 7 inches and 14 inches. Which of
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7

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5 0
3 years ago
If two of these shapes are randomly chosen, one after the other without replacement, what is the probability that the first will
aleksandrvk [35]

Answer:

P(T\ n\ S) = \frac{1}{14}

Step-by-step explanation:

*Missing Part of the Question*

4 stars

5 triangles

3 circles

3 squares

Required

Determine the probability of triangle being first then square being second

Total = 4 + 5 + 3 + 3

Total = 15

Represent the triangle with T and square with S

So, we're solving for P(T n S)

P(T\ n\ S) = P(T) * P(S)

Solving for P(T)

P(T) = \frac{n(T)}{Total}

P(T) = \frac{5}{15}

Solving for P(S)

The question implies a probability without replacement;

Hence Total has now been reduced by 1

Total = 14

P(S) = \frac{n(S)}{Total}

P(S) = \frac{3}{14}

Recall that

P(T\ n\ S) = P(T) * P(S)

P(T\ n\ S) = \frac{5}{15} * \frac{3}{14}

P(T\ n\ S) = \frac{15}{15 * 14}

P(T\ n\ S) = \frac{1}{14}

Hence, the required probability is

P(T\ n\ S) = \frac{1}{14}

3 0
3 years ago
Please help me to answer these questions!! Thank you in advance.
oksano4ka [1.4K]
1. 
5u - 2u = 3u (diff. between longest and shortest)
3u = 87cm
1u = 87cm ÷ 3 = 29cm
5u + 4u + 2u=11u
11u = 11 x 29cm = 319cm

Ans: 319cm

2.
9u = 1350
1u = 1350 ÷ 9 = 150
6u + 5u = 11u
11u = 11 x 150 = 1650

Ans: 1650


8 0
4 years ago
Can 15x-10 be factored
Anna [14]
Yes because removing the common factor is important. There are a few steps, then you can factor by regrouping. Hope this helps.
6 0
4 years ago
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