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We have two equations: x+y=55, x=9+y, so 9+y+y=55, 2y=46, y=23.

x=55-23=32, so 23 and 32.

because it won't let me upload it.

It's on the internet

Thanks!

We have: $56x+569=54x+811$ . So, $2x=242$, and $x=121$ . 

The total cost is $56(121)+569=7345.$

If the party had $\boxed{121}$ attendees, the cost at each venue would be $\boxed{\$7345}$

Yeah, we need a picture for this.

We have 5a=6p and a=15+p

So, 5(15+p)=6p, 75+5p=6p, p=75 cents.

Now, 5a=6(75)

5a=450 cents

So, 450+450=900 cents or 9 dollars. 

No.                  

how to solve 9x^10 - 10x^9 +1 ⋮ (x-1)^2

$\mathbf{(9x^{10} - 10x^9 + 1) : (x^2 - 2x + 1) = 9x^8 + 8x^7 + 7x^6 + 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1}$

The example in your first paragraph is not consistent with the words! Assuming the example should be 6138 goes to 1386, then:

Prior to the last * the number was 324

The square root of this is 18

Prior to the first * the number was 81.

The original number was therefore the square root of this, namely, 9.

Yes it is different.  It is a computer language convention, from the days when all computer languages were text only, with a limited numbers of characters on a line.  It is often seen as the upper case E as well.

Thanks Alan. Is this "e" different from the maths constant "e constant / Euler's number"?

Hmm. If I use LaTeX here in the answer some of it displays properly in the question.  If I delete the LaTeX here the question looks a mess again!!

So:  Use the LaTeX button and copy and paste your questions into the resulting box, having removed the $ signs. e.g.

 If $z = \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }$ find $\lfloor z \rfloor$

"I did a calculation and got the answer 1.349e-10. Is this the same as 1.349*10^-10? How do I get this value in plain numbers? "

1.349e-10. Is this the same as 1.349*10^-10?  Yes.

Plain number:   Move the decimal point 10 places to the left;  0.000 000 000 134 9

All the people who post here do so with no payment. 

Personally I do it because I have a love of mathematics and also a love of teaching. 

I can do both here. I don't get renumeration, but very occasionally I get greatly rewarded by the people I have helped. 

The best rewards, the ones that mean the most emotionally, are the positive impacts that we have on another's life, and it is shown to us physically or verbally, not financially. 

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Of course most people need to be paid for what they do and I am not criticizing that. But it is not the paycheck that determines the degree of job satisfaction that a person gets.

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I will add that I get to interact with other like minded people. 

There is a great community of very generous people here.

It is an great honour that I can be here with them. 

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Thanks for recognising the dedicated people who are here :)

In the sequence 1,2,2,4,8,32,256...

each term (starting from the third term) is the product of the two terms before it. For example, the seventh term is 256  , which is the product of the fifth term (8) and the sixth term (32).

This sequence can be continued forever, though the numbers very quickly grow enormous! (For example, the 14'th term is close to some estimates of the number of particles in the observable universe.)

What is the last digit of the  term of the sequence

$\begin{array}{|lcll|} \hline a_1 &=& 1 \\ a_2 &=& 2\\ a_3 = a_2\cdot a_1 &=& a_2^1 \\ a_4 = a_3\cdot a_2 = a_2^1\cdot a_2 &=& a_2^2 \\ a_5 = a_4\cdot a_3 = a_2^2\cdot a_2^1 &=& a_2^3 \\ a_6 = a_5\cdot a_4 = a_2^3\cdot a_2^3 &=& a_2^5 \\ a_7 = a_6\cdot a_5 = a_2^5\cdot a_2^3 &=& a_2^8 \\ a_8 = a_7\cdot a_6 = a_2^8\cdot a_2^5 &=& a_2^{13}\\ a_9 = a_8\cdot a_7 = a_2^{13}\cdot a_2^8 &=& a_2^{21} \\ \ldots \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \text{Let } \mathcal{F} \text{ are the Fibonacci number }\\\\ \mathcal{F}_1 &=& 1\\ \mathcal{F}_2 &=& 1\\ \mathcal{F}_3 &=& 2\\ \mathcal{F}_4 &=& 3\\ \mathcal{F}_5 &=& 4\\ \mathcal{F}_6 &=& 5\\ \mathcal{F}_7 &=& 13\\ \mathcal{F}_8 &=& 21\\ \mathcal{F}_9 &=& 34\\ \mathcal{F}_{10} &=& 55\\ \mathcal{F}_{11} &=& 89\\ \mathcal{F}_{12} &=& 144\\ \mathcal{F}_{13} &=& 233\\ \mathcal{F}_{14} &=& 377\\ \ldots \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a_n &=& a_2^{\mathcal{F}_{n-1}} \quad & | \quad a_2 = 2 \\ a_n &=& 2^{\mathcal{F}_{n-1}} \\ \hline \end{array}$

If n = 14:

$\begin{array}{|rcll|} \hline a_{14} &=& 2^{\mathcal{F}_{14-1}} \\ a_{14} &=& 2^{\mathcal{F}_{13}} \quad & | \quad \mathcal{F}_{13} = 233 \\ a_{14} &=& 2^{233} \\\\ && \text{ The last digit of the term } a_{14} \\ && 2^{233} \pmod{10} \\ &\equiv & 2^{13\cdot 17+12} \pmod{10}\\ &\equiv & \left(2^{13}\right)^{17} 2^{12} \pmod{10} \quad & | \quad 2^{13} \equiv 2 \pmod{10} \\ &\equiv & 2^{17}\cdot 2^{12} \pmod{10} \\ &\equiv & 2^{29} \pmod{10} \\ &\equiv & 2^{13\cdot 2+3} \pmod{10} \\ &\equiv & \left(2^{13}\right)^{2} 2^{3} \pmod{10} \quad & | \quad 2^{13} \equiv 2 \pmod{10} \\ &\equiv & 2^2\cdot 2^3 \pmod{10} \\ &\equiv & 2^{5} \pmod{10} \\ &\equiv & 32 \pmod{10} \\ &\mathbf{\equiv} & \mathbf{ 2 \pmod{10}} \\ \hline \end{array}$

$\text{ The last digit of the term $a_{14}$ is $\mathbf{2}$}$

A+B=88

A-B=44

THEN A/ B= ?

$\begin{array}{|rcll|} \hline A-B &=& 44 \\\\ A+B &=& 88 \\ &=& 2\cdot 44 \quad & | \quad A-B = 44 \\ &=& 2\cdot (A-B) \\ A+B &=& 2\cdot (A-B) \\ A+B &=& 2A-2B \\ B+2B &=& 2A-A \\ 3B &=& A \\ A &=& 3B \\ \mathbf{\dfrac{A}{B}} &\mathbf{=}& \mathbf{3} \\ \hline \end{array}$

What is the area, in square units, of triangle ABC in the figure shown if points A, B, C and D are coplanar,
angle D is a right angle, AC = 13, AB = 15 and DC = 5?

1.

$\begin{array}{|rcll|} \hline DA^2+DC^2 &=& AC^2 \quad & | \quad DC=5,~ AC = 13 \\ DA^2+5^2&=& 13^2 \\ DA^2 &=& 13^2-5^2 \\ DA^2 &=& 144 \\ DA^2 &=& 12^2 \\ \mathbf{DA} &\mathbf{=}& \mathbf{12} \\ \hline \end{array}$

2.

$\begin{array}{|rcll|} \hline BD^2+DA^2 &=& AB^2 \quad & | \quad DA=12,~ AB = 15 \\ BD^2+12^2&=& 15^2 \\ BD^2 &=& 15^2-12^2 \\ BD^2 &=& 81 \\ BD^2 &=& 9^2 \\ \mathbf{BD} &\mathbf{=}& \mathbf{9} \\ \hline \end{array}$

3.

$\begin{array}{|rcll|} \hline BC &=& BD-DC \quad & | \quad BD=9,~ DC = 5 \\ BC &=& 9-5 \\ \mathbf{BC} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}$

4.

$\begin{array}{|rcll|} \hline A &=& \dfrac{BC\cdot DA}{2} \quad & | \quad BC=4,~ DA = 12 \\ A &=& \dfrac{4\cdot 12}{2} \\ A &=& 2\cdot 12 \\ \mathbf{A} &\mathbf{=}& \mathbf{24} \\ \hline \end{array}$

The area of triangle ABC is 24

$f\left(\dfrac 5 6\right)^3 \left(\dfrac 6 5\right)^7 = \dfrac 5 9\\ f\left(\dfrac 6 5\right)^4 = \dfrac 5 9\\ f = \dfrac 5 9 \left(\dfrac 5 6\right)^4 = \dfrac{3125}{11664}$

Use Pythagorean Theorem to find DA. So 13^2-5^2=144, and the square root of 144 is 12. So DA must be 12.

Notice that AB has the length of 15 and DA has the length of twelve. Since some right triangles have a ratio of 3:4:5, side DB must have the length of 9

Since DC+BC is 9, and DC is 5, BC is 4.

So now we know the three sides of ABC, which is 13, 15, and 4. We can use Heron's formula to figure out the area.

Heron's formula is:

A=sqrt((S)(S-A)(S-B)(S-C))

S=(a+b+c)/2

We plug it sides into the 2nd equation, S=(13+15+4)/2, and we get S=16.

Then plug your answers into the first equation. A=sqrt((16)(16-15)(16-13)(16-4)), and we get A=24.

So the area of ABC is 24

Don't be sorry about asking so many questions.  Just DON"T ask so many qeustions.

You should be using the forum to learn from, not to cheat from.

I am not here to dou your homework for you.  The other anserers should not be either.

Where it the pic now. I mean is it on your computer in a file like maybe a jpg file, is it on the internet, or is it on paper?

the assignment is equivalent to a 10 digit base 3 number.

thus there are 3¹⁰ = 59049 different grade assignments

$\text{the smallest number we'll get is by using just the prime number }2 \text{ as a factor}\\ 2^6 = 64<100,~2^7 = 128 > 100\\ \text{so a whole number }<100 \text{ will have at most }6 \text{ non-unique prime factors}$