Hi
.... then it is located on the perpendicular bisector.
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
To get the answer we have to tranform into y=ax+b form
-2x+4y=68 /+2x
4y=2x+68 /:4
y=
- its the result
13x+4y=284 /-13x
4y=-13x+284 /:4
y=
y=-
- its the result
Solve for A
ab - ac = 2
First, you need to factorize
a(b - c) = 2
a= 2/(b -c)
25% of 40 is 40 times 25 divided by 100 which is 10 (number of children in the club)
