$Use\ (a+b)^2=a^2+2ab+b^2\qquad(*)\\\\y=x^2+6x+10=x^2+2(x)(3)+10=\underbrace{x^2+2(x)(3)+3^2}_{(*)}-3^2+10\\\\=(x+3)^2-9+10=(x+3)^2+1\\\\Answer:\ \boxed{y=(x+3)^2+1}$

3 0

3 years ago

The owner of a new pet store wishes to display tropical fish in display tanks. The above table shows the species that cannot liv

Pepsi [2]

Answer:

The minimum number of different tanks needed to safely house all the fish is:

3 tanks.

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: A
Tank 2: F
Tank 3: G

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: