Answer:
The diagram of the ER and depreciation is in the attachments.
Answer:
Follows are the solution to these question:
Explanation:
A simple gullible algorithm is present. Let h mark the house to the left. Then we put a base station about 4 kilometers to the right. Now delete and repeat all the houses protected by this base station. In this algorithm, they can simply be seen to position baselines at b1, . . , bk as well as other algorithms (which may be an optimum algorithm) at
and so on. (from left to right)
That's why
.
Answer:
19.9 pF
Explanation:
Given that:
Series connection :
11pF and 21pF
C1 = 11pF ; C2 = 21pF
Cseries = (C1*C2)/ C1 + C2
Cseries = (11 * 21) / (11 + 21)
Cseries = 7.21875 pF
C1 = 22pF ; C2 = 30pF
Cseries = (C1*C2)/ C1 + C2
Cseries = (22 * 30) / (22 + 30)
Cseries = 12.6923 pF
Equivalent capacitance is in parallel, thus,
7.21875pF + 12.6923 pF = 19.91105 pF
= 19.9 pF
Answer:
Explanation:
Required
Which returns smallest integer greater than or equal to 7.3
i.e.

When executed, the result of each instruction is:
-- This returns the smallest integer greater than 7.3
--- This returns the smallest integer less than 7.3
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<em> --- there is no such thing as larger() in python</em>
--- This rounds 7.3 to the nearest integer
From the above result,
8 is the smallest integer greater than or equal to 7.3
i.e.

Hence:
is correct