Is the difference between two rational numbers always rational number

Mathematics

2 answers:

vladimir1956 [14]3 years ago

7 0

Not always. It depends on the teacher and the school. Depends on the rules and how your being tought.

Alenkinab [10]3 years ago

6 0

Hello,

An integer is a rational number (ex 5=5/1=10/2...)
The sum of 2 rationals is a rational.
Substract a rational is adding its opposite.
The difference of 2 rationals is always a rational (substraction is a intern law)

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Blake wants to break apart the 7 in 5 × 7. Complete the equation to show two different ways Blake could break apart the 7. Choos

Aleks [24]

5+2=7, so the first blank is 2.
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Ad libitum [116K]

125 because it’s symmetrical between both

3 0

2 years ago

Read 2 more answers

Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .