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Mashcka [7]
3 years ago
13

Please help me answer this question​

Mathematics
1 answer:
sukhopar [10]3 years ago
4 0

Answer:

her first mistake was on line 3.

line 1: 80 + [2(3 1/2 + 1 1/2)]

line 2: 80 + [2(5)]

line 3: 80 + 10

line 4: 90

Step-by-step explanation:

on line 3 she added 80 + 2, but she should have multiplied 2 by 5. the correct steps would be:

line 1: 80 + [2(3 1/2 + 1 1/2)]

line 2: 80 + [2(5)]

line 3: 80 + 10

line 4: 90

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Help on both please ! Im a new student at this school and didn't learn this
larisa86 [58]
For the first one, you have to convert the fractions to an improper fraction. To do that you need to multiply the bottom denominator number (3) by the whole number (1) then you need to add the numinator, so 3x1+2= 5. You have to keep the denominator so 1 2/3 is equal to 5/3. Then do the same to 2 1/5, and you get 11/5. Now you have to find a common denominator, that's basically the smallest number that both numbers can go Into, the lowest common denominator for 3 and 5 is 15. So 3x5= 15, so we have to multiply the top number by 5 which is 25. So 5/3 is equal to 25/15, then 5x3= 15, so you need to multiply 11 by 3 which is 33. So 11/5 is equal to 33/15. Then you add them. Add the numinators (25+33=58. Then you keep the denominator 15. So when u add it it's 58/15 then you need to simplify that and you get 3 13/15.

The second one you turn them into improper fractions like I told you how to before (multiply the bottom number by the whole number then add he top number, then add he same denominator.) do that for both. Then you line them right next to each other and multiply across. (I just realized that they were the same number so they are equal to 5/3 and 11/5)
Then you do 5x11 and you get 55 then do 3x5 and you get 15. 55/15 is your answer, but you need to simplify it, you need to divide 55 by 15, (not all the way just the first number) so you do 15x3 and that's 45, then you subtract that from 55, and you get 10, so then you take your denominator (15) and you answer is 3 10/15. But when you simplify it it's 3 and 2/3


Hope I helped sorry it's so long and sorry for any typos it's so long I didn't go back and check
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3 years ago
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After releasing of radioactive material into.The atnosphere from.A nuclear.Powe plant in a country in 1988, the hay in that coun
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Answer:

the question is incomplete, so I looked for a similar one:

<em>After the release of radioactive material into the atmosphere  from a nuclear power plant, the hay was contaminated by iodine  131 ( half-life, 8 days). If it is all right to feed the hay to cows  when 10% of the iodine 131 remains, how long did the farmers  need to wait to use this hay? </em>

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0.5A₀ = A₀eᵇˣ

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we eliminate A₀ from both sides

0.5 = eᵇ⁸

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since the farmers need to wait until only 10% of the iodine remains (we already calculated b, now we need to find x):

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3 years ago
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asambeis [7]

Answer:

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