For question 9 - It is stating that AE is equal to 38. So the length of the entire line is 38. We then need to look at which figures are actually useful to us, which in this case is AC which is 10. Therefore the answer is as simple as 38-10, which gives us an answer of 28 for the other half of the line, notated as CE.
<u><em>Answer:</em></u>
Radius of the ball is approximately 6.5 cm to the nearest tenth
<u><em>Explanation:</em></u>
The ball has the shape of a sphere
<u>Surface area of a sphere can be calculated using the following rule:</u>
Surface area of sphere = 4πr² square units
<u>In the given problem, we have:</u>
Surface area of the ball = 531 cm²
<u>Substitute with the area in the above equation and solve for the radius as follows:</u>
which is approximately 6.5 cm to the nearest tenth
Hope this helps :)
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.