Answer:
$12
Step-by-step explanation:
(-8)/(2y-8)=(5/(y+4))-7y+(8/(y^2-16))
(-4)/(y-4)=(5/(y+4))-7y+(8/(y+4)(y-4))
((-4)(y+4))/((y+4)(y-4))=((5(y-4))/(y+4)(y-4))-(7y(y+4)(y-4))/(y+4)(y-4))+(8/(y+4)(y-4))
(-4(y+4))=(5(y-4))-(7y(y+4)(y-4))+8
-4y-16=5y-20-(7y(y^2-16))+8
-4y-16=5y-20-7y^3+112y+8
-4y-16=117y-7y^3-12
-4=(121-7y^2)(y)
None of these choices would be equal to -4
.21 as a fraction would be 21/100
Answer:
The minimum number of different tanks needed to safely house all the fish is:
Step-by-step explanation:
To identify the minimum number of different tanks, we're gonna concentrate in a fish species, in this case can be the A: as you see in the table, the A species can live with all the fish excepting the F and G, by their side, the F and G can't live together , by this reason, this three species must live in a different tank, in the next form:
- Tank 1: <em>A</em>
- Tank 2: <em>F</em>
- Tank 3: <em>G</em>
Now the B species, it can live with A, F and G, but for this example we can put in the tank 1 (the tank of the A species). The C especies can live with A, F and G, but how we have A and B together, we're gonna put the C especies in the tank 3 (the tank of the G especies). The D species can live with A and G, we're gonna put in the tank 1 because can live with B species too. The E species can live with A and F, we're gonna put in the tank 2 (the tank of the F species) because the E species can't live with D that is in the in the tank 1. Al last, the H species just can live with A, E, F, and H species, by this reason, the only tank that can be put is the tank 2. In this form, the order is the next:
- Tank 1: <em>A, B, D</em>.
- Tank 2: <em>F, E, H</em>.
- Tank 3: <em>G, C</em>.
And t<u>he owner of the pet store must buy three different tanks to display these tropical fish</u>.
