Answer:I think the y=17 I can’t remember what was right on I-ready
Step-by-step explanation:
Answer:
4' x 2' x 1'
Step-by-step explanation:
Collins' cube has a volume of that is the length of any side, x, cubed: Vol = x^3. Since his box has 8^3, we can say that x = 2. <u>[2^3 = 8]</u>
Amil's box has one side that is 2x. That side would be 2*2 = 4 feet. His volume is also 8 ft^3. Amil's box also has a volume of 8 ft^3.
His box dimensions are therefore: (4)(X)(Y) = 8 ft^3 , where X and Y are whole-number dimensions for the other 2 dimensions of his box.
(4)(X)(Y) = 8 ft^3
X*Y = 2
The only combination of whole numbers for which this this would work is 1 and 2.
Amil's box is 4' x 2' x 1' or 8 ft^3
Answer:
They are compatible
Step-by-step explanation:
The first thing is to say that an "ace" and that it is a "coarse"
"ace" is card number 1. Group A
"coarse" is a type of the deck, found from number 1 to card 13. Group B
Thus:
Calculate A U B:
1 to 13 + 1 of the other types of cards in the deck.
At intersection B:
1 of "coarse"
Therefore, if group A is compatible with group B
