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Misha Larkins
2 months ago
12

The sharks are fed three times a day. During the morning feeding , 2/15 tons of fish is fed. During the afternoon feeding, the w

eight of fish fed will be 1/15 ton more than the fish fed during the morning. If the total weight of fish fed in a day is 1/2 ton, how much is fed during the feeding at night?
Mathematics
1 answer:
zzz [12.3K]2 months ago
7 0

Answer:

1/6 ton of fish is provided during the night feeding.

Step-by-step explanation:

The fish consumed in the morning is \dfrac{2}{15} tons, and the amount of fish served in the afternoon is \dfrac{1}{15} greater than in the morning. This means

\dfrac{2}{15} +\dfrac{1}{15}

Let’s represent the quantity of fish given at night as x, and if the total fish fed throughout the day equals \dfrac{1}{2}, we have

\dfrac{2}{15}+(\dfrac{2}{15} +\dfrac{1}{15})+x=\dfrac{1}{2}

By summing the numerators on the left side, we derive:

\dfrac{5}{15} +x=\dfrac{1}{2}

and subtracting \dfrac{5}{15} from both sides allows us to isolate x:

x=\dfrac{1}{2} -\dfrac{5}{15}

Since \dfrac{5}{15} =\dfrac{1}{3}, we derive

x=\dfrac{1}{2}-\dfrac{1}{3}

The common denominator for the fractions is 6; hence, we write the equation in the form

x=\dfrac{3}{6}-\dfrac{2}{6}

and simplifying the numerators results in:

\boxed{x=\dfrac{1}{6} }

which indicates the amount of tons fed during the night feeding.

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Henri has $24,000 invested in stocks and bonds. The amount in stocks is $6,000 more than three times the amount in bonds. Call t
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Answer:

The solution to this problem indicates that Hernry invested $18,000 in stocks and $6,000 in bonds.

Step-by-step explanation:

First, you need to multiply 6 by 3. The outcome is 18, meaning it is three times higher than the bond amount. Next, subtract $24,000 from $18,000, which gives you $6,000, representing the bond investment. Hence, it is three times less than the stocks as the question describes.

4 0
2 months ago
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A high school drama teacher organizes a musical production. He wants to record the number of students involved in each part of t
PIT_PIT [12445]

I can’t stay awake; I need sleep and have 4 papers to finish for my portfolio. I’m unable to assist, but I wish you the best

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5 0
2 months ago
Herbert plans to use the earnings from his lemonade stand according to the table above, for the first month of operations. If he
Svet_ta [12734]

The table is included below

Based on the table, we can observe that the cost of lemons contributes to 35% of the expenses.

Let x represent the expenses; hence, the lemon cost is 70.

The lemon cost is 35% of x.

So, 70 = 35% of x (where 35% = \frac{35}{100} = 0.35).

Thus, 70 = 0.35 * x.

Solving that gives us x = \frac{70}{0.35}

x = 200, meaning the expense is $200.

Given that the profit margin is 15% = \frac{15}{100} = 0.15,

we calculate profit as the product of the profit percentage and expenses.

Profit equals 0.15 * 200 = 30.

Therefore, the profit amounts to $30


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1 month ago
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Question: Sara started her homework at 5:30 p.M. And finished it at 9:15 p.M. How much time did she take to finish her homework?
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Explicación detallada:

530 p.m. = 17:50 en 24 horas/tipo militar

915 p.m. = 21:15 en 24 horas/tipo militar

por lo tanto..

24 horas/tipo militar. 21:15 - 17:50 = 3.65 horas

en nuestro tiempo. 3 horas 45 minutos

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2 months ago
Given: <br> AB tangent at D, AD = OD = 4<br> Find: Area of the shaded region
lawyer [12517]
<span>With AB tangent at D, where AD = OD = 4
, triangle OAD forms a right angle with sides measuring 4 and 4.
So, the area of triangle OAD is computed as 1/2 * 4 * 4 = 8

The angles AOD and DAO equal 45 degrees.
Therefore, the area of circular sector OCD is calculated as the area of circle O multiplied by 45/360
= pi * 4 * 4 * 45/360
= 2pi

The shaded area ACD results from the triangle OAD minus the area of circular sector OCD
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</span>
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1 month ago
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