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steposvetlana [31]
3 years ago
12

A farmer had twice as many chickens as ducks on his farm. After he sold 166 chickens and ducks, he had half as many chickens as

ducks left. How many chickens did the farmer have at first
Mathematics
1 answer:
mamaluj [8]3 years ago
7 0

Answer:

The farmer have at first <u>202</u> chickens.

Step-by-step explanation:

Given:

A farmer had twice as many chickens as ducks on his farm. After he sold 166 chickens and ducks, he had half as many chickens as ducks left.

Now, to find the chickens farmer have at first.

Let the chickens be x.

And, the ducks be y.

<em>As, the farmer had twice as many chickens as ducks on his farm.</em>

So, x=2y    ......(1)

<em>As, given the farmer after selling 166 chickens and ducks, he had half as many chickens as ducks left.</em>

According to question:

x-166=(y-29)\times \frac{1}{2}

x-166=\frac{y-29}{2}

Substituting the value of x from equation (1) we get:

2y-166=\frac{y-29}{2}

<em>By cross multiplying we get:</em>

4y-332=y-29

<em>Adding both sides 332 we get:</em>

4y=y+303

<em>Subtracting both sides by </em>y<em> we get:</em>

3y=303

<em>Dividing both sides by 3 we get:</em>

y=101.

Now, to get the number of chickens substituting the value of y in equation (1):

x=2y\\\\x=2\times 101\\\\x=202.

Therefore, the farmer have at first 202 chickens.

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Answer: Word problem: Mya goes to a carnival and buys food for her and her friends.  She buys 5 funnel cakes that were $20 each. She buys 10 buckets of popcorn that were 8 dollars each.

Regular problem: $20+$15+$20+$100+$25=180

Step-by-step explanation: I didn't know what you needed. I'm sorry if that's not it, but hope it helps.

7 0
2 years ago
Makato and his friends rolled two fair number cubes 500 times and recorded the sum of the numbers shown on the cubes each time.
denpristay [2]

Answer:

2:12

Step-by-step explanation:

6 and 4 = 10

5 and 5 = 10

8 0
3 years ago
1-What is the sum of the series? ​∑j=152j​ Enter your answer in the box.
tangare [24]

Answer:

Please see the Step-by-step explanation for the answers

Step-by-step explanation:

1)

∑\left \ {{5} \atop {j=1}} \right. 2j

The sum of series from j=1 to j=5 is:

∑ = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)

  =  2 + 4 + 6 + 8 + 10

∑ = 30

2)

This question is not given clearly so i assume the following series that will give you an idea how to solve this:

∑\left \ {{4} \atop {k=1}} \right. 2k²

The sum of series from k=1 to j=4 is:

∑ = 2(1)² + 2(2)² + 2(3)² + 2(4)²

  = 2(1) + 2(4) + 2(9) + 2(16)

  =  2 + 8 + 18 + 32

∑ = 60

∑\left \ {{4} \atop {k=1}} \right. (2k)²

∑ = (2*1)² + (2*2)² + (2*3)² + (2*4)²

  = (2)² + (4)² + (6)² + (8)²

  = 4 + 16 + 36 + 64

∑ = 120

∑\left \ {{4} \atop {k=1}} \right. (2k)²- 4

∑ = (2*1)²-4 + (2*2)²-4 + (2*3)²-4 + (2*4)²-4

  = (2)²-4 + (4)²-4 + (6)²-4 + (8)²-4

  = (4-4) + (16-4) + (36-4) + (64-4)

  = 0 + 12 + 32 + 60

∑ = 104

∑\left \ {{4} \atop {k=1}} \right. 2k²- 4

∑ = 2(1)²-4 + 2(2)²-4 + 2(3)²-4 + 2(4)²-4

  = 2(1)-4 + 2(4)-4 + 2(9)-4 + 2(16)-4

  = (2-4) + (8-4) + (18-4) + (32-4)

  = -2 + 4 + 14 + 28

∑ = 44

3)

∑\left \ {{6} \atop {k=3}} \right. (2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)  

  = (6-10) + (8-10) + (10-10) + (12-10)

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∑ = -4

4)

1+1/2+1/4+1/8+1/16+1/32+1/64

This is a geometric sequence where first term is 1 and the common ratio is 1/2 So

a = 1

This can be derived as

1/2/1 = 1/2 * 1 = 1/2

1/4/1/2 = 1/4 * 2/1 = 1/2

1/8/1/4 = 1/8 * 4/1  = 1/2

1/16/1/8 = 1/16 * 8/1  = 1/2

1/32/1/16 = 1/32 * 16/1  = 1/2

1/64/1/32 = 1/64 * 32/1  = 1/2

Hence the common ratio is r = 1/2

So n-th term is:

ar^{n-1} = 1(\frac{1}{2})^{n-1}

So the answer that represents the series in sigma notation is:

∑\left \ {{7} \atop {j=1}} \right. (\frac{1}{2})^{j-1}

5)

−3+(−1)+1+3+5

This is an arithmetic sequence where the first term is -3 and the common difference is 2. So  

a = 1

This can be derived as

-1 - (-3) = -1 + 3 = 2

1 - (-1) = 1 + 1 = 2

3 - 1 = 2

5 - 3 = 2

Hence the common difference d = 2

The nth term is:

a + (n - 1) d

= -3 + (n−1)2

= -3 + 2(n−1)

= -3 + 2n - 2

= 2n - 5

So the answer that represents the series in sigma notation is:

∑\left \ {{5} \atop {j=1}} \right. (2j−5)

6 0
3 years ago
Evaluate. 4-6 | 3-5 | /3
riadik2000 [5.3K]

Answer:

0

Step-by-step explanation:

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8 0
3 years ago
A jewelry maker has 24 jade beads and 30 teak beads.What is the greatest number of necklaces she can make? How many beads of eac
gulaghasi [49]

Jewelery maker has

24 jade beads

30 teak Beads

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So we need to find the highest common factor for both 24 and 30

The factors for both numbers are as follows

24 -1,2,3,4,6,8,12,24

30 - 1,2,3,5,6,10,15,30

The highest common factor for both is 6

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Number of jade beads - 24/6 = 4 jade beads

Number of teak beads - 30/6 = 5 teak beads

So each necklace will have 4 jade beads and 5 teak beads

6 0
3 years ago
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