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3 years ago

5

Larry, Mary, and Terry each had a full glass of juice. Larry drink three force of his. Mary drink 3/8 of hers.Terry drink 7/10 o

f his. Who drink less than half of their juice?

2 answers:

kati45 [8]3 years ago

3 0

Mary, She drank 3/8 and half would be 4/8. So 3/8 is less than 4/8. Please mark brainiest.

Dmitry [639]3 years ago

3 0

Mary because 4 is half of 8 and she drank 3/8

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What is the exact solution to the equation 2^2x=5^x−1 ?

Ann [662]

2^2x=5^x−1

Take the log pf both sides:

ln(2^2x) = ln(5^x-1)

Expand the logs by pulling the exponents out:

2xln(2) = (x-1)ln(5)

Simpligy the right side:

2xln(2) = ln(5)x - ln(5)

Now solve for x:

Subtract ln(5)x from both sides:

2xln(2) - ln(5)x = -ln(5)

Factor x out of 2xln(2)-ln(5)x

x(2ln(2) - ln(5)) = -ln(5)

Divide both sides by (2ln(2) - ln(5))

X = - ln(5) / (2ln(2) - ln(5))

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3 years ago

Read 2 more answers

30 points ! <br><br> Please answer the question in the picture

vaieri [72.5K]

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Step-by-step explanation:

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2 years ago

Plz help will give brainliest

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3 years ago

a baseball team played 35 games and won 4/7 of them. How many games were won? How many games were lost?

nikklg [1K]

Won: 4/7 x 35 = 20
lost: 3/7 x 35 = 15

6 0

3 years ago

Suppose <img src="https://tex.z-dn.net/?f=m" id="TexFormula1" title="m" alt="m" align="absmiddle" class="latex-formula"> men and

ollegr [7]

Firstly, we'll fix the postions where the $n$ women will be. We have $n!$ forms to do that. So, we'll obtain a row like:

$\underbrace{\underline{~~~}}_{x_2}W_2 \underbrace{\underline{~~~}}_{x_3}W_3 \underbrace{\underline{~~~}}_{x_4}... \underbrace{\underline{~~~}}_{x_n}W_n \underbrace{\underline{~~~}}_{x_{n+1}}$

The n+1 spaces represented by the underline positions will receive the men of the row. Then,

$x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m~~~(i)$

Since there is no women sitting together, we must write that $x_2,x_3,...,x_{n-1},x_n\ge1$ . It guarantees that there is at least one man between two consecutive women. We'll do some substitutions:

$\begin{cases}x_2=x_2'+1\\x_3=x_3'+1\\...\\x_{n-1}=x_{n-1}'+1\\x_n=x_n'+1\end{cases}$

The equation (i) can be rewritten as:

$x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m\\\\
x_1+(x_2'+1)+(x_3'+1)+...+(x_{n-1}'+1)+x_n+x_{n+1}=m\\\\
x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-(n-1)\\\\
x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-n+1~~~(ii)$

We obtained a linear problem of non-negative integer solutions in (ii). The number of solutions to this type of problem are known: $\dfrac{[(n)+(m-n+1)]!}{(n)!(m-n+1)!}=\dfrac{(m+1)!}{n!(m-n+1)!}$

[I can write the proof if you want]

Now, we just have to calculate the number of forms to permute the men that are dispposed in the row: $m!$

Multiplying all results:

$n!\times\dfrac{(m+1)!}{n!(m-n+1)!}\times m!\\\\
\boxed{\boxed{\dfrac{m!(m+1)!}{(m-n+1)!}}}$

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3 years ago