Tom jugs three fifth mile on Monday and to 6 mile on Tuesday how much farther did Tom jog on Monday, then on Tuesday.

First thing to do is to get all the animals in terms of cows, since that is our main unknown. All the other animals' numbers are based on cows. So let's call cows "x". Then sheep is 3 times the number of cows, so sheep is 3x. The number of goats is 45 more than the number of cows, so goats is x + 45. Let's say, for the sake of ease, that the number of animals is "a".

Here's what we have so far:

# of animals: a

# of cows: x

# of sheep: 3x

# of goats: x + 45

We are told that 2/5 of the animals are goats. In an equation, that is:

2/5 (animals) = goats, where the word "of" means to multiply, and the word "are" is the equals sign. Filling in from our key above:

$\frac{2}{5}a=x+45$

That says that 2/5 of the animals are goats when goats is x + 45.

If 2/5 of the animals are goats, then 3/5 of the animals have to be cows and sheep. In an algebraic equation:

$\frac{3}{5}a=x + 3x$ .

Let's go back to the first equation and get rid of the fraction by multiplying everything by 5. When we do that we get

2a = 5(x + 45) so

2a = 5x + 225

Now let's get rid of the fraction in the second equation by multiplying everything by 5. When we do that we get

3a = 5(x + 3x) or

3a = 5(4x) so

3a = 20x. Solve that equation for a.

$a=\frac{20}{3}x$

Now that we have a in terms of x, we can sub that value in for a in the first equation to solve for x, the number of cows:

$2(\frac{20}{3}x)=5x+225$ and, simplifying:

$\frac{40}{3}x=5x+225$ and

$\frac{40}{3}x-5x=225$ . Get rid of the fraction here by multiplying everything by 3:

40x - 15x = 675 and

25x = 675 so

x = 27

That means that there are 27 cows.

Sheep is 3 times that, so there are 81 sheep.

Goats is 45 more than that, so there are 72 goats.

The total number of animals then is 27 + 81 + 72 which is 180. Let's check that to see if the number of goats, 72, is 2/5 of 180.

$\frac{2}{5}(180) = 72$

So we got it! Yay! That was hard!!

7 0

3 years ago

Steve invests in a circus production. The cost includes an overhead of $84,000, plus production costs of $2000 per performance.

Dima020 [189]

Answer:

14

Step-by-step explanation:

Given that Steve invests in a circus production. The cost includes an overhead of $84,000, plus production costs of $2000 per performance. A sold-out performance brings in $8000.

Sales per performance =8000

Production cost per performance =2000

Contribution per performance = $8000-2000=6000$