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nignag [31]
3 years ago
14

A lab technician is ordered to take a sample of your blood. As she inserts the needle, she says, "My, you have tough skin!" What

would be an equivalent translation of this statement?
Engineering
1 answer:
Anvisha [2.4K]3 years ago
7 0

Answer:

Your stratified squamous epithelium is difficult to penetrate!

Explanation:

The epithelium is the tissue formed by one or several layers of cells attached to each other that cover all the free surfaces of the organism, and constitute the internal lining of the cavities, organs, hollows, ducts of the body and skin and also form the Mucous and glands. The epithelia also forms the parenchyma of many organs, such as the liver.

Flat or squamous epithelia: Formed by flat cells, with much less height than width and a flattened nucleus. It is one of the most external being part of the epidermis and generating some inconveniences when penetrating with needles to perform blood extractions when you have certain characteristics of hardness.

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Write the following decorators and apply them to a single function (applying multiple decorators to a single function): 1. The f
natita [175]

Answer:

Complete question is:

write the following decorators and apply them to a single function (applying multiple decorators to a single function):

1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <strong> and </strong> to the argument of the decorator. The return value of the wrapper should look like: return “<strong>” + func() + “</strong>”

2. The decorator will return the wrapper per usual.

3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <em> and </em> to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “<em>” + func() + “</em>.

4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.

5. Apply both decorators (by @ operator to greetings()).

6. Invoke the greetings() function and capture the result.

Code :

def strong_decorator(func):

def func_wrapper(name):

return "<strong>{0}</strong>".format(func(name))

return func_wrapper

def em_decorator(func):

def func_wrapper(name):

return "<em>{0}</em>".format(func(name))

return func_wrapper

@strong_decorator

@em_decorator

def Greetings(name):

return "{0}".format(name)

print(Greetings("Hello"))

Explanation:

5 0
4 years ago
A) For Well A, provide a cross-section sketch that shows (i) ground elevation, (ii) casing height, (iii) depth to
Ad libitum [116K]
Don’t go on that file will give a virus! Sorry just looking out and I don’t know how to comment!
7 0
3 years ago
What is the force in kN of work done is 1.2 ms moves through 120m​
Semmy [17]

Answer:

\frac{1.2}{120}

0.01

5 0
2 years ago
Define extensive and intensive properties of thermodynamic system.
Flura [38]

Answer:

An intense property is a physical attribute of a system that is independent of the size of the system or the quantity of material it contains. An extensive property of a system, on the other hand, is dependent on the size of the system or the amount of material in it.

Explanation:

7 0
3 years ago
Read 2 more answers
6.28 A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 4 ft from the right
dimulka [17.4K]

Answer:

Assume Base free flow speed (BFFS) = 70 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, fLW = 6.6 mph

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, fLC = 0.8 mph

Interchanges/Ramps = 9/ 6 miles = 1.5 /mile

Reduction in speed corresponding to Interchanges/ramps, fID = 5 mph

No. of lanes = 3

Reduction in speed corresponding to number of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 70 – 6.6 – 0.8 – 3 – 5 = 54.6 mph

Peak Flow, V = 2000 veh/hr

Peak 15-min flow = 600 veh

Peak-hour factor = 2000/ (4*600) = 0.83

Trucks and Buses = 12 %

RVs = 6 %

Rolling Terrain

fHV = 1/ (1 + 0.12 (2.5-1) + 0.06 (2.0-1)) = 1/1.24 = 0.806

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 2000/ (0.83*3*0.806*1.0) = 996.54 ~ 997 veh/hr/ln

Vp < (3400 – 30 FFS)

S = FFS

S = 54.6 mph

Density = Vp/S = (997) / (54.6) = 18.26 veh/mi/ln

7 0
3 years ago
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