Write an expression that has 5 terms and that is equivalent to 5x +y.

Mathematics

1 answer:

Aleonysh [2.5K]2 years ago

4 0

Answer:

x+x+x+y+2x

Step-by-step explanation:

There are several ways you can solve this problem. I just split up the 5x into different terms.

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Find the distance of a(1,8) and point c (-8,-6) then find the midpoint of ac

mr Goodwill [35]

$\text{MIdpoint =} \bigg(\dfrac{1 + (-8)}{2}, \dfrac{8+(-6)}{2} \bigg) = \bigg(\dfrac{-7}{2}\ , \ \dfrac{2}{2} \bigg) = \bigg(-3.5 \ , \ 1 \bigg)$

Answer: The midpoint (-3.5, 1)

5 0

3 years ago

SOMEONE HELP ME PLEASE

Aloiza [94]

Answer:

it so easy that correct

Step-by-step explanation:

$500.00×$8.5

4 0

3 years ago

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

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