The length of a rectangle has a measure of x and the width of a rectangle has a measure of (x-2). Find x if the area is 35.

Mathematics

1 answer:

Oliga [24]3 years ago

4 0

Answer:

length = 7

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Step-by-step explanation:

A = Lw

35 = x (x -2)

35 = x2 - 2x

x2 -2x -35 = 0

(x-7)(x+5) = 0

x-7=0

x=7 (we won't use the other factor since it would give us a negative length

so length = 7

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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

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