Is a 12 ounce bottle of generic mouthwash for 1.39 and a 24 ounce mouth wash for 2.79 which one is the best buy

In converting 10 inches to centimeters, what unit (omit the number) would you place in the numerator of your ratio?

belka [17]

Answer:

centimeters

Step-by-step explanation:

The numerator of the conversion factor always has the "to" units. The denominator has the "from" units. That way, when you multiply, the "from" units cancel:

(xx from) · (yy to)/(zz from) = xx·yy/zz · (from/from) · to = xx·yy/zz · to

Here, you want to convert to centimeters, so centimeters will be the units in the numerator.

10 in · (2.54 cm)/(1 in) = 25.4 cm

_____

The conversion factor is always "1". That is, the numerator and denominator are always equal in value. Here, 2.54 cm = 1 in, so (2.54 cm)/(1 in) = 1. You can multiply by 1 anytime you like. For units conversion, it only has the effect of changing the units.

5 0

3 years ago

The price of a coat increased from $78 to $92. What is the price increase percentage?

svetlana [45]

Answer:

18%

Step-by-step explanation:

Percent change can be found using the following formula:

$\frac{original-new}{original}*100$

Original represents the initial cost where new represents the new cost. We multiply the answer by 100 to get the percentage:

$\frac{92-78}{78}*100$ = 18%

6 0

3 years ago

If

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

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