How many years will it take for a person in the United States eat 1,120 pounds of turkey (14 pounds)

1 answer:

3 0

If I understand you, a person eats about 14 pounds of turkey a year. So we will divide all that turkey (1120 pounds) by 14 to find out how many years.

1,120 ÷ 14 = 80 So it will take 80 years. (Since babies can't eat that much turkey a year it will actually take more than 80 years - or the person can eat more than 14 pounds in some years to make up for their early years when they ate less) I know it's a math question, but I also like to be realistic about it. Who measures the amount of turkey they eat yearly anyway? :-)

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I'll give you brainliest! Just need help with 5-10! (Least common factors)

umka2103 [35]

Answer:

Step-by-step explanation:

$12,\:30\\\\\mathrm{Prime\:factorization\:of\:}12:\quad 2\times\:2\times\:3\\\mathrm{Prime\:factorization\:of\:}30:\quad 2\times\:3\times\:5\\\\=2\times\:2\times\:3\times\:5\\\\=60$

$6,\:10\\\\\mathrm{Prime\:factorization\:of\:}6:\quad 2\times\:3\\\mathrm{Prime\:factorization\:of\:}10:\quad 2\times \:5\\\\=2\times \:3\times \:5\\\\=30$

$16,\:24\\\\\mathrm{Prime\:factorization\:of\:}16:\quad 2\times \:2\times\:2\times\:2\\\mathrm{Prime\:factorization\:of\:}24:\quad 2\times\:2\times\:2\times\:3\\\\=2\times\:2\times\:2\times\:2\times\:3\\\\=48$

$14,21\\\\\mathrm{Prime\:factorization\:of\:}14:\quad 2\times\:7\\\mathrm{Prime\:factorization\:of\:}21:\quad 3\times\:7\\\\=2\times\:7\times\:3\\\\=42$

$9,15\\\\\mathrm{Prime\:factorization\:of\:}9:\quad 3\times\:3\\\mathrm{Prime\:factorization\:of\:}15:\quad 3\times\:5\\\\=3\times\:3\times\:5\\\\=45$

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$5,\:11\\\\\mathrm{Prime\:factorization\:of\:}5:\quad 5\\\mathrm{Prime\:factorization\:of\:}11:\quad 11\\\\=5\times \:11\\\\=55$

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3 years ago

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lesantik [10]

To calculate distance between two points we use the distance formula sqrt((x2−x1)^2+(y2−y1)^2).

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Now we square 2 and 7 and add them together to get 4 + 49 = 53.

The last thing we do to find the distance is take the square root of 53. 53 is not a perfect square and is also a prime number so our answer in simplest form is still sqrt53.<span />

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pogonyaev

The Least Common Denomiator Is 12
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4 years ago

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We know that there are (4^9)^5 ⋅ 4^0 at the library, so we just need to simplify this to get the answer.
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