16.5 degrees below the starting temp, whatever that is. If its 0, then -16.5 degrees.

The first equation would be (.5)5-11=-8.5, because the metal has been cooling for 5 hours.

The device that 'aids' in the cooling would be -5-3=-8, because it is a separate variable that cools the metal, so the amount the device cools is independent of the natural cooling amount, and the equation is independent of the natural cooling equation.
You then add -8.5 and -8, because the device has lowered 8.5 degrees and 8 degrees. This equals -16.5 degrees, or a decrease of 16.5 degrees.

7 0

3 years ago

The table shows the steps for solving the given inequality for x.

slavikrds [6]

Given inequality -3(4-6x) < x+5.

We have -3 in front of Parenthesis.

That represents multiplication of -3 and Parenthesis.

The multiplication of Parenthesis could be done by applying distributive property.

On distributing, we get

-12+18x < x+5

x is added on right side of the inequality. The reverse operation of addition is subtraction. So we need to subtract x from both sides, we get

-12+18x-x < x-x+5

-12+17x < 5

Now, we need to get rid -12 from left side.

So, we need to apply addition property of equality, we need to add 12 on both sides, we get

-12+12+18x < x+5+12

17x < 17

We need to get rid 17 from left side. So we need to apply division property of equality.

On dividing both sides by 17, we get

17x/17 < 17/17

x<1.

8 0

2 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

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

Solving for F(s) on the last equation, $F(s)=\frac{s}{s^2+25}$ , then the Laplace transform we were searching is $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

3 0

2 years ago