Answer:
A=16+j12…'B=6+j10.4
Explanation:
add the following vector given in
Answer: N has to be lesser than or equal to 1666.
Explanation:
Cost of parts N in FPGA = $15N
Cost of parts N in gate array = $3N + $20000
Cost of parts N in standard cell = $1N + $100000
So,
15N < 3N + 20000 lets say this is equation 1
(cost of FPGA lesser than that of gate array)
Also. 15N < 1N + 100000 lets say this is equation 2
(cost of FPGA lesser than that of standardcell)
Now
From equation 1
12N < 20000
N < 1666.67
From equation 2
14N < 100000
N < 7142.85
AT the same time, Both conditions must hold true
So N <= 1666 (Since N has to be an integer)
N has to be lesser than or equal to 1666.
Answer: precision
Explanation: Because accuracy is where you keep on getting it right but precision is where you get closer and closer
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:
Answer:
Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point
