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aev [14]
4 years ago
5

princess wanted to plant 28 lillies in rows. all rows contain the same number of plants. there are between 5 and 12 plants on ea

ch row. how many plants are in each row
Mathematics
1 answer:
marysya [2.9K]4 years ago
6 0
Add 5 12 and 28 and there is ur answer

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If the hot dog eating champion's goal is to eat 1 hot dog every 13 seconds, how many whole hot dogs does he expect to eat in 12
sweet [91]
He wil eat a total of 55.2 hot  dogs. there 60 seconds in a minute . he plans to eat 1 hotdog in every  13 seconds . you divide 60 by 13 . he will eat 4.6 hot dogs in  a minute . then you multiply 4.6 by 12.
6 0
3 years ago
Solve each equation for x
Andrews [41]
BY taking the square root you can find each of them as follows:

1. sqrt(144) = 12

2. sqrt(25/289) = 5/17
6 0
3 years ago
Solve 10 + 6(–9 – 4x) = 10(x – 12) + 8.
Luden [163]

Answer: x = 2

Step-by-step explanation:

<em>first, we remove the parentheses. </em>

10 - 54 - 24x = 10x - 120 + 8

<em>then, we just calculate that out.</em>

-44 - 24x = 10x - 112

<em>next, we move the terms.</em>

-24x - 10x = -112 + 44

<em>then, we collect the like terms and calculate it.</em>

-34x = -68

<em>finally, we divide both sides.</em>

x = 2

3 0
3 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

3 0
2 years ago
Find lim ?x approaches 0 f(x+?x)-f(x)/?x where f(x) = 4x-3
Whitepunk [10]

If f(x)=4x-3:

\displaystyle\lim_{\Delta x\to0}\frac{(4(x+\Delta x)-3)-(4x-3)}{\Delta x}=\lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4

If f(x)=4x^{-3}:

\displaystyle\lim_{\Delta x\to0}\frac{\frac4{(x+\Delta x)^3}-\frac4{x^3}}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac{4x^3-4(x+\Delta x)^3}{x^3(x+\Delta x)^3}}{\Delta x}

\displaystyle=\lim_{\Delta x\to0}\frac{4x^3-4(x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3)}{x^3\Delta x(x+\Delta x)^3}

\displaystyle=\lim_{\Delta x\to0}\frac{-12x^2\Delta x-12x(\Delta x)^2-4(\Delta x)^3}{x^3\Delta x(x+\Delta x)^3}=-\frac{12}{x^4}

7 0
4 years ago
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