0.5 / 100 * X = 3
X = 3 * 100 / 0.5 = 300 / 0.5 = 600
A.) False
b.) True
c.) True
d.) True
well, we know the ceiling is 6+2/3 high, and Eduardo has 4+1/2 yards only, how much more does he need, well, is simply their difference, let's firstly convert the mixed fractions to improper fractions and then subtract.
![\stackrel{mixed}{6\frac{2}{3}}\implies \cfrac{6\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{20}{3}} ~\hfill \stackrel{mixed}{4\frac{1}{2}}\implies \cfrac{4\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{9}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{20}{3}-\cfrac{9}{2}\implies \stackrel{using ~~\stackrel{LCD}{6}}{\cfrac{(2\cdot 20)-(3\cdot 9)}{6}}\implies \cfrac{40-27}{6}\implies \cfrac{13}{6}\implies\blacktriangleright 2\frac{1}{6} \blacktriangleleft](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B6%5Cfrac%7B2%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B6%5Ccdot%203%2B2%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B20%7D%7B3%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B9%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B20%7D%7B3%7D-%5Ccfrac%7B9%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Busing%20~~%5Cstackrel%7BLCD%7D%7B6%7D%7D%7B%5Ccfrac%7B%282%5Ccdot%2020%29-%283%5Ccdot%209%29%7D%7B6%7D%7D%5Cimplies%20%5Ccfrac%7B40-27%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B13%7D%7B6%7D%5Cimplies%5Cblacktriangleright%202%5Cfrac%7B1%7D%7B6%7D%20%5Cblacktriangleleft)
The correct answer is A: 3. This is because...
3 * 4 = 12
12 * 5 = 60
Therefore, the answer is 60.
First aquarium dimensions:
Length = 6 m.
Width = 4 m and
Height = 2 meter.
Second aquarium dimensions:
Length = 8 m.
Width = 9 m and
Height = 3 meter.
We know formula for volume of a cuboidal box = Length*Width*Height.
Plugging values of length, width and height of first aquarium in formula of volume. We get
V1 = 6*4*2 = 48 m^3.
Plugging values of length, width and height of second aquarium in formula of volume. We get
V2 = 8*9*3 = 216 m^3.
In order to find the total cubic meters of space do the sea turtles have in their habitat, we need to add both volumes.
Therefore, Total voulme of both aquarium = V1 +V2 = 48+216 = 264 m^3.
Therefore, total 264 m^3 cubic meters of space the sea turtles have in their habitat.
