SA=201.7cm^2
I used the equation
SA=pi^1/3(6*V)^2/3 in case you have any more questions like this
1=51
2=34
3= 95 (i think)
4=38
A. 1/5 fish
We know that 2/5 of Mike's fish are clownfish. Therefore, 3/5 are not clownish as 5/5 – 2/5 = 3/5. Also, if you look at the model, there are five pieces. If we assume that model represents the whole of Mike's fish and you take away two pieces, you are left with 3/5. So we know that the remaining fish is 3/5
Next, we know that of these 3/5 fish, 1/3 is damselfish, so we need to find 1/3 of 3/5. To do this, we must multiply 1/3 by 3/5 as "of" means multiply in Math.
So: 1/3 • 3/5 = 1 • 3/3 • 5 = 3/15 3 ÷ 3 = 1 and 15 ÷ 3 = 5 3/15 = 1/5
1/5 of Mike's fish are damsel fish
B. 2/5 fish
Now we know that 1/5 of Mike's fish is damselfish, and 2/5 is clownfish. To find the fraction of his fish that are neither, therefore, we must subtract their sum from the whole.
First, we add 1/5 and 2/5 together. Adding the numerators, 1 and 2, we get 1/5 + 2/5 = 3/5
Next, we subtract: 5/5 – 3/5 = 2/5, so 2/5 of his fish are neither clownfish or damselfish
And if you look at the model again, you can see that if you cross out 1 piece for the damselfish, and 2 pieces for the clownfish, you are left with 2/5
