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Elena L [17]
3 years ago
5

Solve the system of equations by elimination: −5x+y+5z=27 5x+5y−4z=−12 −6y−z=−16

Mathematics
1 answer:
VladimirAG [237]3 years ago
7 0

Answer:

no asnwer

Step-by-step explanation:

first step:

-5x+y+5z=27

+ 5x+5y-4z=-12

which is

6y+z=15

add -6y-z=-16

which will be 0=-1

this means that there is no answer and that the statement is false.

hope this helped!

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Mass = density x volume

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2 years ago
One number is 5 greater than another. The product of the numbers are 84. Find both the two positive and two negative sets of num
Pepsi [2]

Let x,y be the two numbers.

Given that one number is 5 greater than another.

Let x be the smaller number ans y be the greater number.

That is y=x+5. Let this be the first equation.

And also given that product of the two numbers is 84.

That is x*y = 84, let us plugin y=x+5 here.

           x*(x+5) = 84

           x^2 + 5x -84 = 0.

           x^2+12x-7x-84 = 0

          x(x+12)-7(x+12) =0

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That is x= 7 or -12.

If x=7, y= 7+5=12.

If x=-12, y= -12+5 = -7.

Hence two positive numbers corresponding to given conditions are 7,12.

And two negative numbers corresponding to given conditions are -12,-7.

3 0
3 years ago
Read 2 more answers
This answer needs to be rounded to the nearest hundredth. Example:12.34
lara [203]
Answer is 14.14
Reason
Using Pythagorean theory
a^2 + b^2 = c^2
The two legs are a and b, the longest diagonal opposite the right angle is c
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2 years ago
Please help me answer question B!
skelet666 [1.2K]

Answer:

  • The program counter holds the memory address of the next instruction to be fetched from memory
  • The memory address register holds the address of memory from which data or instructions are to be fetched
  • The memory data register holds a copy of the memory contents transferred to or from the memory at the address in the memory address register
  • The accumulator holds the result of any logic or arithmetic operation

Step-by-step explanation:

The specific contents of any of these registers at any point in time <em>depends on the architecture of the computer</em>. If we make the assumption that the only interface registers connected to memory are the memory address register (MAR) and the memory data register (MDR), then <em>all memory transfers of any kind</em> will use both of these registers.

For execution of the instructions at addresses 01 through 03, the sequence of operations may go like this.

1. (Somehow) The program counter (PC) is set to 01.

2. The contents of the PC are copied to the MAR.

3. A Memory Read operation is performed, and the contents of memory at address 01 are copied to the MDR. (Contents are the LDA #11 instruction.)

4. The MDR contents are decoded (possibly after being transferred to an instruction register), and the value 11 is placed in the Accumulator.

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6. The contents of the PC are copied to the MAR.

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8. The MDR contents are decoded and the value 05 is placed in the MAR.

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11. The PC is incremented to 03.

12. The contents of the PC are copied to the MAR.

13. A Memory Read operation is performed, and the contents of memory at address 03 are copied to the MDR. (Contents are the STO 06 instruction.)

14. The MDR contents are decoded and the value 06 is placed in the MAR.

15. The Accumulator value is placed in the MDR, and a Memory Write operation is performed. Memory address 06 now holds the value 8.

16. The PC is incremented to 04.

17. Instruction fetch and decoding continues. This program will go "off into the weeds", since there is no Halt instruction. Results are unpredictable.

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Note that decoding an instruction may result in several different data transfers and/or memory and/or arithmetic operations. All of this is usually completed before the next instruction is fetched.

In modern computers, memory contents may be fetched on the speculation that they will be used. Adjustments need to be made if the program makes a jump or if executing an instruction alters the data that was prefetched.

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