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Fantom [35]
3 years ago
12

Calculate the area of area of a rectangular with a base of 10cm and a height of 6cm​

Mathematics
2 answers:
nadya68 [22]3 years ago
8 0

Answer:

<em>60cm²</em>

Step-by-step explanation:

Length = 6cm

Breadth = 10cm

Area of rectangle = l×b

=( 10 ×6)cm

= <u>60cm²</u>

Hope it helps.

DanielleElmas [232]3 years ago
7 0

Answer:

\Huge \boxed{\mathrm{60 \ cm^2 }}

Step-by-step explanation:

Area of a rectangle is length × width.

The length of the rectangle is 10 cm.

The width of the rectangle is 6 cm.

\Rightarrow 10 \cdot 6 \\\\\\ \Rightarrow 60

The area of the rectangle is 60 cm².

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If the diameter of an entire coconut is 5 inches, and the coconut meat is 1 inch thick.
Arturiano [62]

Answer:

V=33.51in^3

Step-by-step explanation:

5inch-1 in

=4inch

divide by 2 because the problem gives diameter, not radius. (radius is half of diameter)

The space inside is a sphere so it will be

V=4/3πr^3

V=4/3*pi*(2)^3

V=33.51 inch cubed

6 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
PLEASE HELP !!!!!!!!!! Jonas used linear combination to solve the system of equations shown. He did so by multiplying the first
OverLord2011 [107]
This question is worded slightly strangely but I believe I understand it. 
To eliminate x, multiply the second equation by 8, so that it becomes 4x + .8y= 96
Then you can subtract 4x on the top by 4x on the bottom to eliminate it. 
7 0
3 years ago
Evaluate: 10x when x=10<br> The answer is_______( just type in your final number answer)​
Kisachek [45]

Answer:

100

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Please help me, I CANT fail this.
Annette [7]

Answer:

B

Step-by-step explanation:

4 0
3 years ago
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