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Dmitry_Shevchenko [17]
3 years ago
9

Farmer Doug has some pigs and chickens. One day he counted 24 legs and 7 heads in the barnyard. How many Of each animal did Farm

er Doug count?
Mathematics
2 answers:
boyakko [2]3 years ago
8 0
5 pigs and 2 chickens
Ostrovityanka [42]3 years ago
7 0

Answer: Farmer Doug counts 2 chickens and 5 pigs.

Step-by-step explanation:

Let x be the number of chickens and y be the number of pigs.

Since a chicken has 2 legs and a pig has 4 legs

According to the question , we have two equations :-

x+y=7.......................(1)\\\\2x+4y=24.........................(2)

Divide 2 on both sides in equation (2), we get

x+2y=12...............(3)

Now, Subtract equation (1) from (3), we get

y=5

Put y=5 in (1) , we get

x+5=7\\\\\Rightarrow\ x=7-5=2

Hence, Farmer Doug counts 2 chickens and 5 pigs.

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Make sure it's fully reduced.<br><br> 2 - x = 5/8
postnew [5]

Isolate the x. Note the equal sign. What you do to one side, you do to the other.

First, subtract 2 from both sides

2 (-2) - x = (5/8) - 2

-x = 5/8 - 2

Make sure they have the same denominator. Remember that what you multiply to the denominator, you multiply the numerator:

-x = 5/8 - 16/8

-x  = - 11/8

Isolate the x. Divide -1 from both sides

(-x)/-1 = (-11/8)/-1

x = 11/8

11/8 is your answer for x

<em>~Rise Above the Ordinary</em>

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4 years ago
The data represent the birth weights, in kilograms, for 10 lambs. 1.1, 1.1, 1.1, 1.1, 1.3, 1.5, 1.5, 1.9, 1.9, 1.9 which histogr
ch4aika [34]
I think is the 1.1and the 1.9 data is the answer.
5 0
3 years ago
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Translate the inequality and solve. Three times a number is greater than six. 1 point​
Inessa05 [86]

Answer:

3x > 6

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Step-by-step explanation:

3 0
3 years ago
In how many ways can you place 50 identical balls in six distinct urns such that each urn contains an odd number of balls
mezya [45]

Answer:

80730

Step-by-step explanation:

The answer is c(27,5)=27*26*25*24*23/5!. Here is why.

Let us solve a simpler problem. In how many ways 25 identical balls can be placed in six distinct urns such that no urn is empty?

25 balls in a row make 24 places for five dividers, so the answer is c(24,5).

2. Next problem. In how many ways 50 identical balls can be placed in six distinct urns such that no urn is empty and each urn contains even number of balls.

We group 50 balls in 25 pairs, so the answer is the same as in the previous problem which is

c(48/2,5)=c(24,5).

Let us denote the content of each urn as

u(1),u(2),…u(6). Note that each u(i) is even. Denote u(i) as u(i)=2k(i) where k is a natural number. We have

2k(1)+2k(2)+2k(3)+2k(4)+2k(5)+2k(6)=50.

3. Now let us proceed to the original problem where odd number of balls in each urn is required.

We have

2k(1)-1+2k(2)-1+2k(3)-1+2k(4)-1+2k(5)-1+2k(6)-1=50 or

2k(1)+2k(2)+2k(3)+2k(4)+2k(5)+2k(6)=56.

So the original problem reduces to previous problem where even number of balls in each urn is required but the total number of balls is 56 instead of 50.

Its answer is c(54/2,5)=c(27,5).

8 0
3 years ago
What is the sign of s^67/t^9 when s &lt; 0 and t &gt; 0?
Sphinxa [80]

Answer:

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Step-by-step explanation:

The sign of a product or quotient cannot be determined by the positive number (t), so we can ignore it. The sign of the expression will be negative if and only if there are an odd number of negative factors.

Here, there are 67 (an odd number) negative factors, so the expression will be negative.

4 0
1 year ago
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