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Gwar [14]
3 years ago
12

Why in division there is sometimes remaindres?

Mathematics
2 answers:
Ksenya-84 [330]3 years ago
7 0
In division problems, if the division results in a decimal or fraction, (not a whole number), than it has a remainder.  If the division results in a whole number, then there is no remainder, or a remainder of zero. 

Examples:

6/3=2

6/4=1.5
MAXImum [283]3 years ago
3 0
The reason is because  the number you are trying to divide into parts is not a factor of the number you are trying to divide into parts. so there fore there are left overs. Example:  28/3 ( 28 divided by 3).  3 only goes into 28 9 time (because 9*3 is 27) and then your answer would be 9 REMAINDER of 1 because the 3 cannot equally divide into 28. Hope this helps!

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Super easy plz answer all for 20 points just adding
sergejj [24]
7/8+9/10=71/40=1 31/40
1/4+2/3=11/12
1/2+1/4 is different from 1/2+1/3 because 
1/2+1/4= 2/4+1/4=  5/1 =1  1/4
1/2+1/3=3/6+2/6=5/6
6 0
3 years ago
Plz.. Help me.. True or false?
zimovet [89]

Answer:

\boxed{\mathrm{False}}

Step-by-step explanation:

(p-q)^2

(p-q)(p-q)

Use <em>FOIL</em> method.

p^2-pq-pq +q^2

p^2-2pq +q^2

6 0
3 years ago
store one is selling three notebooks for $5.13 store to is selling 5 notebooks for $8.40 store 3 7/8 notebooks for $14 in which
sdas [7]

Answer:

store 2

Step-by-step explanation:

This is because store one sells notebook for 5.13. Store 2 sells one notebook for 1.68, and the last stores sells one notebook for 2 or 1.75. Meaning that store 2 sells it for cheap.

5 0
3 years ago
Please help me!
Sindrei [870]
Well, let's check if your inequality is true. We have to add the 3 scores up and divide them by 3 to know the average.

73 + 81 + 86 = 240
240 / 3 = 80

80%

That's true, now let's check 101.

73 + 81 + 101 = 255
255 / 3 = 85

85%

That's true.

Your inequality is correct.
5 0
3 years ago
8x + y = -39<br> 2x - 5y = 27<br> Solve the system by elimination
Ahat [919]

Answer:

(-4, -7)

Step-by-step explanation:

21y=-147

y=-7

2x-5x(-7)=27

x=-4

5 0
3 years ago
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