anonymous
  • anonymous
The ratio of the surface areas of two cubes is 49/81 What is the ratio of their volumes?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
thats fairly easy
anonymous
  • anonymous
surface area of a cube is 6a^2 , where a is the side length
anonymous
  • anonymous
so let one cube have side length x, the other have side length y

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anonymous
  • anonymous
so 6x^2 / 6y^2 = 49 / 81
anonymous
  • anonymous
now we want the ratio of volume volume of a cube is a^3
anonymous
  • anonymous
therefore the ratio of volumes is (x^3 / y^3 )
anonymous
  • anonymous
from the first equation we know that x^2 / y^2 = 49/81 so (x/y) = +- 7/9 ( by taking sqrts ) , but we are dealing with lengths ( which must be positive, so take the + case
anonymous
  • anonymous
so the ratio of volumes = (x^3/y^3) = (x^2 / y^2 ) times (x/y) = 49/81 x 7/9
anonymous
  • anonymous
= 0.471 as a decimal to 3 decimal places
anonymous
  • anonymous
i'm alittle confuse
anonymous
  • anonymous
with what ?
dumbcow
  • dumbcow
\[SA \rightarrow \frac{6a ^{2}}{6b ^{2}} = \frac{a ^{2}}{b ^{2}}=\frac{49}{81}\] \[\sqrt{\frac{a ^{2}}{b^{2}}} = \sqrt{\frac{49}{81}}\] \[\frac{a}{b} = \frac{7}{9}\] \[V \rightarrow \frac{a^{3}}{b^{3}} = \frac{7^{3}}{9^{3}}\]
anonymous
  • anonymous
surface area is area, so it is two dimensional, while volume is three dimensional. (square units vs cubed units) you have a ratio of two squares, 49 and 81 being 7^2 and 9^2. therefore the ratio of the lengths of the sides must be 7 to 9 and the ratio of the volumes must be 7^3 to 9^3. dumbcow(nice handle) said it nicely.
anonymous
  • anonymous
so the ratio is 7:9
anonymous
  • anonymous
no :|

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