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vfiekz [6]
3 years ago
9

Simplify this math problem show Your work

Mathematics
1 answer:
loris [4]3 years ago
5 0

9514 1404 393

Answer:

  (p -9q)/(4p² +12pq)

Step-by-step explanation:

The least common denominator will be the product of the denominators.

  \dfrac{-3}{4p}+\dfrac{1}{p+3q}=\dfrac{-3(p+3q)+1(4p)}{(4p)(p+3q)}=\boxed{\dfrac{p-9q}{4p^2+12pq}}

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Weights of golden retriever dogs are normally distributed. Samples of weights of golden retriever​ dogs, each of size nequals​15
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Answer:

B.

Step-by-step explanation:

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3 years ago
The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope
lesya692 [45]

(1) D

L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt

Integrate by parts, taking

u = t \implies \mathrm du=\mathrm dt

\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}

Then

L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}

L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}

(2) A

L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}

L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}

7 0
3 years ago
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