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Marta_Voda [28]
3 years ago
5

Colin and amil Each built boxes with a volume of 8 ft.³ Collins box is a cube almils box is a rectangular prism the length of am

ils box is twice the length of Collins box all of the side measures are whole numbers what could be the length width and height of amils box
Mathematics
1 answer:
SOVA2 [1]3 years ago
3 0

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

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   Amount after 15 years translates to, if rate of interest is 4.83 %

{\text{Using the formula}} P(1+\frac{R}{100})^n\\\\ 128,000(1 +\frac{4.83}{100})^{15}\\\\ =128,000 \times (1.0483)^{15}\\\\=128,000 \times 2.0290\\\\=259713.0404

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