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3 years ago

How many mEq of KCI will the patient receive in 24 hours?

Mathematics

1 answer:

Brilliant_brown [7]3 years ago

8 0

10% = 10g/100ml. Ergo, 15ml/100ml (10g) = 1500mg of KCL
.9% NACL = 9g/1000mL, Ergo = 9000mg NaCl
Now we have to figure out how many mEqs of each salt we have.
MW of NaCl = 58.5
MW of KCl = 74.5
So, in moles, we have:
1500 KCl / 74.5 = 20.13 mmol KCl
9000 NaCl / 58.5 = 153.85mmol NaCl
Because the valence of both ions is one, as an example, it would be both 20.13 mEq of Cl- and K+ from 20.13 mmoles of KCL.
So, just add the two together, 20.13 + 153.85 = 174-ish.

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Plz g0ive me solution

Serga [27]

Answer:

separate the x from the numbers it will make the equation easier

8 0

3 years ago

PLEASE HELP ASAP!

scoray [572]

Answer:

AM = 900

CI = 500

I have to be honest, this is the most ridiculously worded question of this type that I have ever seen... ask you teacher, why it was so important to force you to solve it using only one variable?

Step-by-step explanation:

(AM) + (1400-AM) = 1400

2(AM) + 3(1400 - AM) = 3300

2AM + 4200 -3AM = 3300

-AM = -900

AM = 900

CI = 500

4 0

3 years ago

A cosine function is a reflection of its parent function over the x-axis. The amplitude of the function is 11, the vertical shif

omeli [17]

We have been given that a cosine function is a reflection of its parent function over the x-axis. The amplitude of the function is 11, the vertical shift is 9 units down, and the period of the function is $\frac{7\pi}{12}$ . The graph of the function does not show a phase shift. We are asked to write the equation of our function.

We know that general form a cosine function is $y=A\cos(b(x-c))-d$ , where,

A = Amplitude,

$\frac{2\pi}{b}$ = Period,

c = Horizontal shift,

d = Vertical shift.

The equation of parent cosine function is $y=\cos(x)$ . Since function is reflected about x-axis, so our function will be $y=-\cos(x)$ .

Let us find the value of b.

$\frac{2\pi}{b}=\frac{7\pi}{12}$

$7\pi\cdot b=24\pi$

$\frac{7\pi\cdot b}{7\pi}=\frac{24\pi}{7\pi}$

$b=\frac{24}{7}$

Upon substituting our given values in general cosine function, we will get:

$f(x)=-11\cos(\frac{24}{7}x)-9$

Therefore, our required function would be $f(x)=-11\cos(\frac{24}{7}x)-9$ .

7 0

3 years ago

Evaluate 10 m +\frac {n^2}410m+ 4 n 2 10, m, plus, start fraction, n, start superscript, 2, end superscript, divided by, 4, en

DerKrebs [107]

Evaluate the expression $10m + \frac{n^2}{4}$ where m=5 and n=4

We replace the value of m and n to evaluate

m=5 and n=4

$10m + \frac{n^2}{4}$

$10(5) + \frac{4^2}{4}$

$50 + \frac{16}{4}$

50 + 4 = 54

The value of the given expression is 54 when m=5 and n= 4

5 0

3 years ago

Analyzing a Graph In Exercise, analyze and sketch the graph of the function. Lable any relative extrema, points of inflection, a

Sladkaya [172]

Answer:

$y=(\ln{x})^2$

point of extremity: (1,0)

vertical asymptote: along the y-axis (x = 0)

point of inflection: (e,1)

Solution:

Although all these points can be directly observed from the graph below, but these are the analytical solutions if you're curious!

1) Extreme point can be found by differentiating 'y' once and equating to zero. solving for x:

$\dfrac{dy}{dx}=\dfrac{dy}{dx}((\ln{x})^2)$

$\dfrac{dy}{dx}=2\ln{x}\left(\dfrac{1}{x}\right)$

substitute dy/dx = 0, and solve for x

$0=2\ln{x}\left(\dfrac{1}{x}\right)$

$0=2\ln{x}$

$x=e^0$

$x=1$

use this value of x back in y, to find the y-coordinate of the extreme point

$y=(\ln{1})^2$

$y=0$

The extreme point = (1,0)

2) Differentiate y twice to find the inflection point.

$\dfrac{dy}{dx}=2\ln{x}\left(\dfrac{1}{x}\right)$

$\dfrac{d^2y}{dx^2}=2\ln{x}\left(-\dfrac{1}{x^2}\right)+\left(\dfrac{1}{x}\right)\left(\dfrac{2}{x}\right)$

$\dfrac{d^2y}{dx^2}=\dfrac{2}{x^2}\left(-\ln{x}+1}\right)$

substitute d2y/dx2 = 0, and solve for x

$0=\dfrac{2}{x^2}\left(-\ln{x}+1}\right)$

$0=-\ln{x}+1$

$\ln{x}=1$

$x = e$

use this value of x back in y, to find the y-coordinate of the inflection point

$y=(\ln{e})^2$

$y=1$

The extreme point = (e,1)

6 0

4 years ago