<span>Partial products are different in regrouping in terms of how numbers are clustered from a set equation as a whole delivering it individual but naturally to all the numbers involved in the set. </span>
Regrouping is just like the commutative or associative property of numbers.
<span>Associative property of addition is used when you want to group addends. This is mainly used to cluster set of numbers or in this case, addends. How do you use the associative property when you break apart addends? Simple you group them using the open and closed parentheses or brackets. Take for an example 1 + 1 + 2 = 4. Using the associative property you can have either (1 + 1) + 2 = 4 or 1 + (1 + 2) = 4 clustered into place.
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Answer:
2801 boxes
Step-by-step explanation:
You have 1 red box
You have 7 blue boxes, so far 8 boxes
There are 49 total black boxes (7*7), making 57 total boxes
There are 49*7 = 343 yellow boxes, making 400 boxes
There are 343*7 = 2401 gold boxes, making for a total of 2801 boxes.
Answer:
(2b - 5) + b + (b + 80) = 983
Step-by-step explanation:
Given that,
Total score of three teams = 983
Since the teams' scores are given in reference to team B's score, let 'b' be the score of team B.
So,
The scores of Team A = (2 * b) - 5
The scores of Team B = b
The scores of Team C = (b + 80)
Thus,
The equation for determining the total points of Team B would be:
(2b - 5) + b + (b + 80) = 983
On solving,
(2b - 5) + b + (b + 80) = 983
⇒ 2b + b + b = 983 + 5 - 80
⇒ 4b = 908
⇒ b = 908 ÷ 4
⇒ b = 227
Team B's score = 227
Team A's score = (2 * 227 - 5)
= 449
Team C's score = 227 + 80
= 307
Total ⇒ 449 + 227 + 307 = 983
Hence proved.
Step-by-step explanation:
Maths
Bookmark
if 2
2008
−2
2007
−2
2006
+2
2005
=k ⋅2
2005
then the value of k is equal to
Answer
Correct option is
B
3
2
2008
−2
2007
−2
2006
+2
2005
=k⋅2
2005
⇒[2
2008
−2
2006
]−[2
2007
−2
2005
]=k⋅2
2005
⇒2
2006
(2
2
−1)−2
2005
(2
2
−1)=k⋅2
2005
⇒3(2
2006
−2
2005
)=k⋅2
2005
⇒3[2
2005
(2−1)]=k⋅2
2005
⇒3(2
2005
)=k⋅2
2005
On comparing, we get
k=3
Hence, Option B is correct.
