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3 years ago

What is the solution to the equation x2 + 2x = 12?

Mathematics

2 answers:

Tems11 [23]3 years ago

7 0

Is the x squared?? The pic I uploaded is the answer if it is x squared.

Shtirlitz [24]3 years ago

3 0

Step-by-step explanation:

The solution would be 6

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Given the circle with the equation (x - 3)2 + y2 = 49, determine the location of each point with respect to the graph of the cir

Bas_tet [7]

To find out if a point is inside, on, or outside a circle, we need to substitute the ordered pair into the equation of the circle:
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.

If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.

For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2

A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]

B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]

C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]

5 0

3 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.

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

The sum of two consecutive integers is 75. Find the number

Luba_88 [7]

Divide 75 by 2:

75 / 2 = 37.5

Now you need the whole number below 37.5 and the whole number above 37.5:

37 + 38 = 75

The numbers are 37 and 38

4 0

3 years ago

Find the value of each variable in the parallelogram.

kenny6666 [7]

Answer:

n = 12 , m = 5

Step-by-step explanation:

since it is said it is a parallelogram so opposite sides are equal

n = 12 (being opposite sides of parallelogram)

m + 1 = 6 (being opposite sides of parallelogram)

m = 6 - 1

m = 5

Hope it will help :)

7 0

3 years ago

Simplify (5x^-6)^3 leave your answer as a fraction in index form

Elan Coil [88]

Answer: 125/x^18

Step-by-step explanation: this should be the answer

5 0

3 years ago

Read 2 more answers