Use the virtual models to solve 5/8 x 3

There are some goats, cows, and sheep on a farm. 2/5 of the animals were goats. There are 3 times as many sheep as cows. If ther

svetlana [45]

Answer:

180

Step-by-step explanation:

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

Match the quadratic equation

marta [7]

The last choice is the correct one

4 0

3 years ago

Which ordered pair is a solution of the inequality? 3y-6<12x

BartSMP [9]

The answer is 3. (4, -2)

7 0

3 years ago

Suppose that

natka813 [3]

Answer:

I don’t know

Step-by-step explanation:

6 0

3 years ago

Suppose that a factory manufactures only tables and chairs and that the profit on one chair is $15 and on the table is $20. Each

Leya [2.2K]

Answer:

The data we have is:

Profit per chair = $15

Profit per table = $20

Chair needs: 1 large pice, 2 small pieces.

Table needs: 2 large pieces, 2 small pieces.

You have 6 large pieces and 8 small pieces, and you want to maximize the profit.

Let's define:

T = # of tables

C = # of chairs.

Total profit:

P = C*$15 + M*$20

The number of chairs that you can make is given by:

0 ≤ C ≤ 4

If you make 4 chairs, the profit is:

P = 4*$15 = $60

And there will be 2 large pieces left.

Now for the tables.

0 ≤ T ≤ 3

If you make 4 tables, the profit will be:

P = 3*$20 = $60 (same as before)

And there will be two small pieces left.

Then we want to use a medium number of tables and chairs, we can not use the maximum for any of those, the first thing we should try is:

we ideally would want to use all the pieces that we have.

8 small, 6 large:

Then we can make:

2 tables: we use 4 large pieces and 4 small pieces.

T = 2

And we have left 2 large pieces and 4 small pieces.

The leftover is enough to make two chairs.

C = 2.

In this case, where we used all our materials, the profit will be:

P = 2*$15 + 2*$20 = $30 + $40 = $70

The profit is maximized when we make 2 tables and two chairs.

5 0

3 years ago