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lora16 [44]
2 years ago
7

Marnie had two pages of math homework and four pages of ELA homework. If each page had 14 questions on it, how many questions di

d she have to complete total?
Mathematics
1 answer:
ahrayia [7]2 years ago
4 0

Answer:

84

Step-by-step explanation:

First, there are 6 pages of work. then multiply 6 by the number of questions (14)

6x14=84

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Answer:b

Step-by-step explanation:

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3 years ago
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Tommy Thomas’s tankard holds 480ml when it is one quarter empty. How much does it hold when it is one quarter full?
agasfer [191]

Answer:

Tommy Thomas's tankard holds 160ml when it is one-quarter full.

Step-by-step explanation:

When Tommy's tankard holds 480ml when it's one-quarter empty. There are three other quarters in the tankard. So, you would divide 480 by 3 to see how much is in each of the other 3 quarters. The answer comes down to 160ml per quarter, which equals one-quarter full.

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2 years ago
There are 7800 grams of lodine that have a half-life of 8 days. How much lodine will remain after 40 days?
anygoal [31]

Given that we know the initial mass of Iodine and its half-life, we want to see how much will remain after 40 days.

After 40 days, 244.17 grams of Iodine will remain.

<h3 /><h3>The half-life of materials and how to use it:</h3>

The half-life of a material is the time it takes for that amount of material to reduce to its half.

We can model the amount of Iodine as:

A(t) = A*e^{k*t}

  • Where A is the initial amount, in this case, 7800g.
  • k is a constant that depends on the half-life.
  • t is the time in days.

Replacing what we know, we get:

A(t) = 7800g*e^{k*t}

Now we use the fact that the half-life is 8 days, this means that:

e^{k*8} = 1/2

ln(e^{k*8}) = ln(1/2)

k*8 = ln(1/2)

k = ln(1/2)/8 = -0.0866

Then the function is:

A(t) =  7800g*e^{-0.0866*t}

So now we just need to evaluate this in t = 40.

A(40) = 7800g*e^{-0.0866*40} = 244.17g

So, after 40 days, 244.17 grams of Iodine will remain.

If you want to learn more about half-life and decays, you can read:

brainly.com/question/11152793

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Rotation 180° counterclockwise about the origin

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