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2 years ago

PLSPLSPLSPLSPLSPLSPLSPLS

Mathematics

1 answer:

Oksi-84 [34.3K]2 years ago

6 0

Answer:

reason

1. symmetrical property

sides can be interchanged and its still true

statement

1. if triangle kjm is congruent to klm

then klm is congruent to klm

reason

2. reflexive property

triangle is congruent to itself

statement

2. triangle kjm is congruent to triangle kjm

reason

3. transitive property

if the 1st 2 triangles are congruent to each other and congruent to a third

then the first triangle is congruent to the third triangle

statement

3. if kjm is congruent to klm and

klm is congruent to abc then

kjm is congruent to abc

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A) Use the definition of Laplace transform to find L{f }. (Do the integrals.) For what values of s is L{f } defined?f(t) = (2t+1

kiruha [24]

For the given function f(t) = (2t + 1) using definition of Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

As given in the question,

Given function is equal to :

f(t) = 2t + 1

Simplify the given function using definition of Laplace transform we have,

L(f(t))s = $\int\limits^\infty_0 {f(t)e^{-st} } \, dt$

= $\int\limits^\infty_0[2t +1] e^{-st} dt$

= $2\int\limits^\infty_0 te^{-st} + \int\limits^\infty_0e^{-st} dt$

= 2 L(t) + L(1)

L(1) = $\int\limits^\infty_0e^{-st} dt$

= (-1/s) ( 0 -1 )

= 1/s , ( s > 0)

2L ( t ) = $2\int\limits^\infty_0 te^{-st}$

= $2[t\int\limits^\infty_0 e^{-st} - \int\limits^\infty_0 ({(d/dt)(t) \int\limits^\infty_0e^{-st} \, dt )dt]$

= 2/ s²

Now ,

L(f(t))s = 2 L(t) + L(1)

= 2/ s² + 1/s

Therefore, the solution of the given function using Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

Learn more about Laplace transform here

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Check all the ordered pairs that satisfy the equation below. Y=6x

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Answer:

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Answer the question with explanation;

PSYCHO15rus [73]

Answer:

The statement in the question is wrong. The series actually diverges.

Step-by-step explanation:

We compute

$\lim_{n\to\infty}\frac{n^2}{(n+1)^2}=\lim_{n\to\infty}\left(\frac{n^2}{n^2+2n+1}\cdot\frac{1/n^2}{1/n^2}\right)=\lim_{n\to\infty}\frac1{1+2/n+1/n^2}=\frac1{1+0+0}=1\ne0$

Therefore, by the series divergence test, the series $\sum_{n=1}^\infty\frac{n^2}{(n+1)^2}$ diverges.

EDIT: To VectorFundament120, if $(x_n)_{n\in\mathbb N}$ is a sequence, both $\lim x_n$ and $\lim_{n\to\infty}x_n$ are common notation for its limit. The former is not wrong but I have switched to the latter if that helps.