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emmainna [20.7K]
3 years ago
14

Explain why the system of equations - 2x + y=6 and y-2x = 4 has no solution

Mathematics
1 answer:
Zolol [24]3 years ago
7 0
Hi, so basically when you solve the system, you will end up with 6=4 which is clearly not true so the system has no solution. i’ve put the working in the image.

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Find the Laplace transformation of each of the following functions. In each case, specify the values of s for which the integral
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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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Read 2 more answers
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