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Inessa [10]
2 years ago
5

Solve the equation pleasse someone please help will give brainliest pleeeasse ​

Mathematics
1 answer:
NISA [10]2 years ago
7 0

Answer:

45 it is 45 45 um yea I ben did thiss

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The Gordon family plans to buy a TV. One TV has a purchase price of $330 and an estimated yearly operating cost of $14. The othe
garri49 [273]

Answer:

Step-by-step explanation:

They should purchase the $369 one with the $9 plan.

330 + (14x8) =442

369+ (9x8) = 441

Hope this helps!

6 0
3 years ago
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Micah’s bill for dinner was $35.50. He left the server an 18% tip. What is the total amount that Micah paid for dinner including
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Answer:

$41.89

Step-by-step explanation:

Can I get brainliest please?

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2 years ago
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Kelvin claims that a pol yhedron can have 30 edges , 18 faces and 12 vertices . Do you agree with him ? Explain your answer .
AURORKA [14]
Number of faces + (Number of vertices - Number of edges) = 2
So plugging in the numbers it would then be 
18+(12-30)=2
18-18=2
0=2
Kelvin would be wrong and I disagree with him. (Because ...*plugin work cited above*)
5 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\\\\&#10;x_1+(x_2'+1)+(x_3'+1)+...+(x_{n-1}'+1)+x_n+x_{n+1}=m\\\\&#10;x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-(n-1)\\\\&#10;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!\\\\&#10;\boxed{\boxed{\dfrac{m!(m+1)!}{(m-n+1)!}}}

4 0
3 years ago
Need help please!
Readme [11.4K]

Answer:

4) v=kx^2/y^3

2= k(4^2)/(3^3)

2 =k16/27

k = 2(27/16) = 27/8

v = (27/8)(3^2)/(2^3 = (27/8)(9/8) = 243/64

v = 243/64 when x=3 and y=2

Step-by-step explanation:

here

5 0
2 years ago
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