One half of the total cost

Foram prescritos 500mg de dipirona para uma criança com febre.Na unidade tem disponivel ampola de 1g/2ml.Quantos g vão ser admin

Valentin [98]

De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL

O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.

Fazendo a classica regra de 3, podemos chegar no volume desejado:

(atentar que 500mg = 0,5g)

g mL

1 --------- 2

0,5 --------- X

1 . X = 0,5 . 2

8 0

2 years ago

Michael has a set of 24 math problems to do for homework. Michael completed 1/4 of his homework in 3/4 of an hour. If he continu

gtnhenbr [62]

Given:

Total number of problems = 24

Michael completed $\dfrac{1}{4}$ of his homework in $\dfrac{3}{4}$ of an hour.

To find:

What fraction of his homework would he complete in one hour.

Solution:

We have,

Part of work done in $\dfrac{3}{4}$ of an hour $=\dfrac{1}{4}$

Part of work done in an hour $=\dfrac{4}{3}\times \dfrac{1}{4}$

$=\dfrac{1}{3}$

Therefore, he complete $\dfrac{1}{3}$ of his homework in 1 hour. Hence, all options are incorrect.

3 0

2 years ago

Q5 Q3.) How far, to the nearest foot, is the ship from the statue's base?

dusya [7]

Distance from base = 305 / tan(29.5deg)
= 539 ft

7 0

2 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

2 years ago

Write an equation, and express the product in standard form.

rosijanka [135]

Answer: 3 copies of 5 tenths = 3*0.5 = 1.5

Step-by-step explanation:

Ok, if we have:

"3 copies of a number B"

This is written as:

3*B

And we have that the number B is "5 tenths"

Now, a tenth is the first digit after the decimal point.

Then 5 tenths = 5*0.1 = 0.5

B = 0.5

3 copies of 5 tenths = 3*0.5 = 1.5

7 0

2 years ago