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Nutka1998 [239]
3 years ago
7

Lucas wants to expand his vocabulary to better understand the sentence below. The insidious burglar moved through the house unhe

ard and unseen. Which clue best helps define insidious? burglar, unheard and unseen, moved through, house
Computers and Technology
2 answers:
adell [148]3 years ago
7 0

Answer:

B is correct because the top one said so

Explanation: :)

olga nikolaevna [1]3 years ago
3 0

Answer:

B unheard & unseen

Explanation:

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What are the arguments for writing efficient programs even though hardware is relatively inexpensive?
Ainat [17]

Answer: Even though the hardware is inexpensive the writing of program is not efficient through this method as proper development of program is necessary for the clear execution due to factors like:-

  • The facility of writing program even the cost of hardware is less but it is not a free facility.
  • It also has a slower processing for the execution of the program
  • The construction of the efficient program is necessary for the compilation and execution of it rather than poorly constructed program is worthless and inefficient in working.

7 0
3 years ago
There will be 10 numbers stored contiguously in the computer at location x 7000 . Write a complete LC-3 program, starting at loc
Artist 52 [7]

Answer:

The LC-3 (Little Computer 3) is an ISA definition for a 16-bit computer. Its architecture includes physical memory mapped I/O via a keyboard and display; TRAPs to the operating system for handling service calls; conditional branches on N, Z, and P condition codes; a subroutine call/return mechanism; a minimal set of operation instructions (ADD, AND, and NOT); and various addressing modes for loads and stores (direct, indirect, Base+offset, PC-relative, and an immediate mode for loading effective addresses). Programs written in LC-3 assembler execute out of a 65536 word memory space. All references to memory, from loading instructions to loading and storing register values, pass through the get Mem Adr() function. The hardware/software function of Project 5 is to translate virtual addresses to physical addresses in a restricted memory space. The following is the default, pass-through, MMU code for all memory references by the LC-3 simulator.

unsigned short int get Mem Adr(int va, int rwFlg)

{

unsigned short int pa;

// Warning: Use of system calls that can cause context switches may result in address translation failure

// You should only need to use gittid() once which has already been called for you below. No other syscalls

// are necessary.

TCB* tcb = get TCB();

int task RPT = tcb [gettid()].RPT;

pa = va;

// turn off virtual addressing for system RAM

if (va < 0x3000) return &memory[va];

return &memory[pa];

} // end get MemAdr

Simple OS, Tasks, and the LC-3 Simulator

We introduce into our simple-os a new task that is an lc3 Task. An lc3 Task is a running LC-3 simulator that executes an LC-3 program loaded into the LC-3 memory. The memory for the LC-3 simulator, however, is a single global array. This single global array for memory means that alllc3 Tasks created by the shell use the same memory for their programs. As all LC-3 programs start at address 0x3000 in LC-3, each task overwrites another tasks LC-3 program when the scheduler swaps task. The LC-3 simulator (lc3 Task) invokes the SWAP command every several LC-3 instruction cycles. This swap invocation means the scheduler is going to be swapping LC-3 tasks before the tasks actually complete execution so over writing another LC-3 task's memory in the LC-3 simulator is not a good thing.

You are going to implement virtual memory for the LC-3 simulator so up to 32 LC-3 tasks can be active in the LC-3 simulator memory without corrupting each others data. To implement the virtual memory, we have routed all accesses to LC-3 memory through a get Mem Adr function that is the MMU for the LC-3 simulator. In essence, we now have a single LC-3 simulator with a single unified global memory array yet we provide multi-tasking in the simulator for up to 32 LC-3 programs running in their own private address space using virtual memory.

We are implementing a two level page table for the virtual memory in this programming task. A two level table relies on referring to two page tables both indexed by separate page numbers to complete an address translation from a virtual to a physical address. The first table is referred to as the root page table or RPT for short. The root page table is a fixed static table that always resides in memory. There is exactly one RPT per LC-3 task. Always.

The memory layout for the LC=3 simulator including the system (kernel) area that is always resident and non-paged (i.e., no virtual address translation).

The two figures try to illustrate the situation. The lower figure below demonstrates the use of the two level page table. The RPT resident in non-virtual memory is first referenced to get the address of the second level user page table or (UPT) for short. The right figure in purple and green illustrates the memory layout more precisely. Anything below the address 0x3000 is considered non-virtual. The address space is not paged. The memory in the region 0x2400 through 0x3000 is reserved for the RPTs for up to thirty-two LC-3 tasks. These tables are again always present in memory and are not paged. Accessing any RPT does not require any type of address translation.

The addresses that reside above 0x3000 require an address translation. The memory area is in the virtual address space of the program. This virtual address space means that a UPT belonging to any given task is accessed using a virtual address. You must use the RPT in the system memory to keep track of the correct physical address for the UPT location. Once you have the physical address of the UPT you can complete the address translation by finding the data frame and combining it with the page offset to arrive at your final absolute physical address.

A Two-level page table for virtual memory management.

x7000 123F x7000 0042

x7001 6534 x7001 6534

x7002 300F x7002 300F

x7003 4005 after the program is run, memory x7003 4005

x7004 3F19

7 0
3 years ago
Read 2 more answers
// This pseudocode segment is intended to compute and display
Finger [1]

The pseudocode to calculate the average of the test scores until the user enters a negative input serves as a prototype of the actual program

<h3>The errors in the pseudocode</h3>

The errors in the pseudocode include:

  • Inclusion of unusable segments
  • Incorrect variables
  • Incorrect loops

<h3>The correct pseudocode</h3>

The correct pseudocode where all errors are corrected and the unusable segments are removed is as follows:

start

Declarations

     num test1

     num test2

     num test3

     num average

output "Enter score for test 1 or a negative number to quit"

input test1

while test1 >= 0

     output "Enter score for test 2"

     input test2

     output "Enter score for test 3"

     input test3

     average = (test1 + test2 + test3) / 3

     output "Average is ", average

     output "Enter score for test 1 or a negative number to quit"

     input test1

endwhile

output "End of program"

stop

Read more about pseudocodes at:

brainly.com/question/11623795

3 0
2 years ago
PLS CAN ANYONE HELP ME
Nesterboy [21]

The exercise contains 15 questions. The solution is provided for each question.

Each question contains the necessary skills you need to learn.

I have added tips and required learning resources for each question, which helps you solve the exercise. When you complete each question, you get more familiar with a control structure, loops, string, and list.

Use Online Code Editor to solve exercise questions.

Also, try to solve the basic Python Quiz for beginners

Exercise 1: Given two integer numbers return their product. If the product is greater than 1000, then return their sum

Reference article for help:

Accept user input in Python

Calculate an Average in Python

Given 1:

number1 = 20

number2 = 30

Expected Output:

The result is 600

Given 2:

number1 = 40

number2 = 30

Expected Output:

The result is 70

Explanation:

4 0
2 years ago
Formatting changes only the appearance of data ___it does not affect the data itself.
Vladimir [108]

Answer:

The answer to this question is "true".

Explanation:

In this question, the answer is true because The use of formatting provides changes in the data presentation, not in the data. It is also known as layout.

The layout or formatting is important for two reasons that can be given as:  

1) It makes your data like a formatted data.

2) It helps to make your data more readable.

8 0
3 years ago
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