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lidiya [134]
2 years ago
5

working alongside the pharmacist , one of the duties a medication reconciliation technician would perform is to a ) deal with va

rious drug suppliers . b ) promote the assigned drugs in the healthcare communities such as physician clinics and pharmacies . keep needed medications on hand . d ) ensure that the medication orders align with the medication that's sent home with a patient .
Physics
1 answer:
Vanyuwa [196]2 years ago
5 0

Answer:

okkiikkkkkkkkoiiiiiiii8iiii

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A 200kg bucket of cement<br>​
ozzi

Answer:

Yes. A 200 kg bucket of cement = About 440.925 pounds of cements.

Explanation:

7 0
3 years ago
Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir
Svet_ta [14]
Let's ask this question step by step:
 Part A) 
 a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
 ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 Part (c)
 (a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
 (a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
 (a + b) b = 10 + 36
 (a + b) b = 46
 Part (d)
 comp (ba) = (a.b) / lbl
 a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
 lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
 comp (ba) = 26 / root (20)
 answer
 2k
 26
 46
 26 / root (20)
3 0
3 years ago
What type of charge is acquired when a rubber is rubbed with a fur
Rus_ich [418]

Answer:

The triboelectic charging process (a.k.a., charging by friction) results in a transfer of electrons between the two objects that are rubbed together. Rubber has a much greater attraction for electrons than animal fur.

Explanation:

HOPE IT HELP^_^

THANKS^_^

4 0
2 years ago
To avoid using the variable "a"
sdas [7]

Answer:

option c .

Explanation:

..............................

6 0
3 years ago
A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance,
Lisa [10]

Answer:

Explanation:

The two charges are q and Q - q. Let the distance between them is r

Use the formula for coulomb's law for the force between the two charges

F = \frac{Kq_{1}q_{2}}{r^{2}}

So, the force between the charges q and Q - q is given by

F = \frac{K\left ( Q-q \right )q}}{r^{2}}

For maxima and minima, differentiate the force with respect to q.

\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )

For maxima and minima, the value of dF/dq = 0

So, we get

q = Q /2

Now \frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}

the double derivate is negative, so the force is maxima when q = Q / 2 .

6 0
4 years ago
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