1. Which of the following equations has a solution of 6?

Mathematics

1 answer:

matrenka [14]3 years ago

4 0

Answer:

i dkn

Step-by-step explanation:

so hi 8X equals 56x equals blank check does 8 equal 56 this is correct

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Consider the NASCAR race, the Indianapolis 500. Its name refers to the total race distance of 500 miles. There are 44 cars are r

lukranit [14]

Answer: $5289.35

Step-by-step explanation:

Since the cars have an average mileage of 6.78 miles per gallon and the race is 500 miles, the number of gallons each car will take will be:

= 500/6.78

= 73.75 gallons

Since they're 44 cars, the total gallons of fuel will be:

= 44 × 73.75

= 3245 gallons

Since the fuel cost is $1.63 per gallon, the total fuel cost, will be:

= $1.63 × 3245

= $5289.35

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3 years ago

While he hikes in the Gold Country of northern California, Sid dreams about the adventurers that walked the same trails years ag

olga2289 [7]

I did. Undestand the probelm

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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: