Simplify the number using the imaginary unit i:

2 answers:

7 0

We have to simplify:
√ ( - 28 )
We know that √ (-1 ) = i ( the imaginary unit )
√ ( - 28 ) = √ 28 * ( - 1 ) = i √ ( 28 ) = i √ ( 4 * 7 ) = 2 i √( 7 )
But there is not such solution among the options.
Only correct answer ( using the imaginary unit ) is:
C. i (square root) 28

Effectus [21]4 years ago

7 0

Answer:

CONNEXUS: Lesson 7 Unit 5 Complex Numbers ANSWERSS!

Step-by-step explanation:

Just took the quiz and i have all the answers so here you go

1) D: 2i(square root)7

2)B:second graph (4,2)

3)B:11-2i

4)C: 20-4i

5)A:31+i/26

6)C: 1(plus or minus) i(square root)11

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Inverse laplace of L^-1 {5s/s^2 + 3s - 4}

Salsk061 [2.6K]

Answer:

$e^t+e^{-4t}$

Step-by-step explanation:

We have to simplify the original function using partial fraction, hence:

$\frac{5s}{s^2+3s-4} =\frac{5s}{(s-1)(s+4)}\\\\=\frac{A}{s-1}+\frac{B}{s+4}\\\\Therefore:\\\\\frac{A(s+4)+B(s-1)}{(s-1)(s+4)}=\frac{5s}{(s-1)(s+4)}\\\\Eliminating\ the \ denominator:\\\\A(s+4)+B(s-1)=5s\\\\substitute\ s=1:\\\\A(1+4)+B(1-1)=5(1)\\\\5A=5\\\\A=1\\\\tsubstitute\ s=-4:\\\\A(-4+4)+B(-4-1)=5(-4)\\\\-5B=-20\\\\B=4\\\\Therefore\ substituting\ A\ and\ B\ gives:\\\\\frac{5s}{s^2+3s-4}=\frac{1}{s-1}+ \frac{4}{s+4}\\\\$

$From\ Laplace\ inverse:\\\\But\ L^{-1}[\frac{1}{s-a} ]=e^{at}\\\\Hence:\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=L^{-1}[\frac{1}{s-1} ]+L^{-1}[\frac{4}{s+4} ]=e^{t}+4e^{-4t}\\\\L^{-1} [\frac{5s}{s^2+3s-4}]=e^{t}+4e^{-4t}$