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hodyreva [135]
3 years ago
11

Find the equation of the line in standard form containing a point P(-3.3) and slope m= -4/3​

Mathematics
1 answer:
bixtya [17]3 years ago
5 0

Answer:

4x + 3y = -3.

Step-by-step explanation:

First substitute into the point-slope form:

y - y1 = m(x - x1):

y - 3 = (-4/3)(x - -3)

y - 3 = (-4/3)(x + 3)

y - 3 = -4/3 x - 4

y = -4/3 x - 4 + 3

y = -4/3x - 1   Multiply through be 3:

3y = -4x - 3

4x + 3y = -3  (answer).

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Lori is three times as old as Michelle. In five years Michelle's age will be one half of Lori's age. How old are Lori and Michel
AURORKA [14]

Answer:

Lori's age is 15 years old and Michelle's age is 5 years old now

Step-by-step explanation:

Let

x-----> Lori's age

y ----> Michelle's age

we know that

x=3y-----> equation A

(y+5)=\frac{1}{2}(x+5) ----> equation B

substitute equation A in equation B and solve for y

(y+5)=\frac{1}{2}(3y+5)

2y+10=(3y+5)

3y-2y=10-5

y=5

Find the value of x

x=3(5)=15

therefore

Lori's age is 15 years old and Michelle's age is 5 years old now

7 0
3 years ago
Mike earned $11.76 per hour for working 23.5 hours last week. how much money did mike earn last week?
densk [106]
Just multiply.

11.76 * 23.5= 276.36

So Mike earned $276.36 last week
6 0
3 years ago
Ac=4 bx=y+4 what is the value of y
Kitty [74]
Ac=4 bx=y+4 doesn't make sense.  Perhaps you meant <span>Ac-4bx=y+4.  If this is not correct, ensure that you have copied down the original problem correctly.

</span>To solve  Ac-4bx=y+4  for y, subtract 4 from both sides of this equation.  You'll get:

Ac-4bx-4 = y+4 - 4, or y = ac-4bx - 4.
6 0
3 years ago
Write 34% as a fraction in simplest form
max2010maxim [7]

Answer:

The answer is that the simplest fraction form of 34% is 17/50

Step-by-step explanation:

Both the numerator and denominator of the fraction 34/100 can only be divided by 2, to satisfy creating its simplest form.

34% = 17/50

Have an awesome day! :D

7 0
2 years ago
Read 2 more answers
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.

5. The PC is incremented to 02.

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

7. A Memory Read operation is performed, and the contents of memory at address 02 are copied to the MDR. (Contents are the SUB 05 instruction.)

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

9. A Memory Read operation is performed and the contents of memory at address 05 are copied to the MDR. (Contents are the value 3.)

10. The Accumulator contents are replaced by the difference of the previous contents (11) and the value in the MDR (3). The accumulator now holds the value 11 -3 = 8.

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.

_____

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.

4 0
3 years ago
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