Well based on this information, I don't think that this is a valid inference. There isn't enough data to go off of in order to answer this question.
So the perimeter(P) of a rectangle would be:
P= 2L+2W
L being the length and W being the width.
The problem says the length is 4cm more than the width, so L= 4+W.
So if we substitute L with 4+W, we get:
P= 2(4+W) + 2W
Use the Distributive Property
P= 8+2W+2W
Combine like terms
P=8+4W
Since we're given the perimeter, we could replace P with 52. So:
52=8+4W
Subtract 8 to both sides
44=4W
Divide 4 to both sides
11=W
Therefore, the width is 11cm
And since the length is 4cm more than the width, we could add 4cm to 11cm to find that the length is 15cm
Thus, the dimensions of the rectangle are 15cm by 11cm
12
2 multiplied by 6 is 12 and so 1 multiplied by something is 12
Hope this helps.
It will expand. Or it will grow.
Answer:
33 newspapers per 1 pile.
Step-by-step explanation:
33+33+33=99
