Answer:
The description for the given question is described in the explanation section below.
Explanation:
I would like to reinforce in advanced or complex concepts such as documents as well as channels, internet programming, multi-threading, after that last lesson.
- I am interested in learning web development to develop applications or software. I would also like to explore those concepts by using open source tools.
- Course concepts will have to develop models for handling.
- No there is no subject matter or definition you provide further clarity for.
- I'm interested in studying java as well as web development in comparison to C++ so I can use it in my contract work.
A write blocker is any tool that permits read-only access to data storage devices without compromising the integrity of the data. A write blocker, when used properly, can guarantee the protection of the data chain of custody. NIST‘s general write blocking requirements hold that:
<span>The tool shall not allow a protected drive to be changed.The tool shall not prevent obtaining any information from or about any drive.<span>The tool shall not prevent any operations to a drive that is not protected.</span></span>
Answer:
hello your question is incomplete attached is the complete question and solution
answer : The solution is attached below
Explanation:
Below is a program named Derivations.java that creates an array with the deviations from average of another array.
Are you trying to change the shape of an item within your program by entering the values into text boxes that become variables?
What language are you using?
Answer:
A = 120
B = 40
C = 70
Solution:
As per the question:
Manufacturer forced to make 10 more type C clamps than the total of A and b:
10 + A + B = C (1)
Also, 3 times as many type B as type A clamps are:
A = 3B (2)
The total no. of clamps produced per day:
A + B + C = 330 (3)
The no. of each type manufactured per day:
Now, from eqn (1), and (3):
A + B + 10 + A + B = 330
2A + 2B = 320
A + B = 160 (4)
Now, from eqn (2) and (4):
3B + B = 160
B = 40
Since, A = 3B
A = 
A = 120
Put the values of A and C in eqn (3):
120 + 40 + C = 330
C = 70