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Sorry, I didn't realize. Thanks for helping :)

In return, I'll do one of the problems up there.

3- g(3)=3^2-7(3)+1

=9-21+1

=-11

y = x^2 + k

First, we illustrate the problem:

I - the incenter (center of inscribed circle)

Then, draw lines like this:

Notice that the triangle has been split into 3 smaller ones, each with a height of the radius, and with a base of the sides of the large triangle. Which means that we can say:

\(a\triangle ABC=a\triangle AIB+a\triangle BIC+a\triangle CIA\)

\(a\triangle ABC={5r\over 2}+{6r\over 2}+{7r\over 2}\)

\(a\triangle ABC =9r\)

Now, we also know that \(a\triangle ABC=\sqrt{s(s-AB)(s-AC)(s-BC)}\) wherein \(s={AB+AC+BC\over2}\). This is called Heron's Formula. Substituting the values, we get:

\(\sqrt{9*4*3*2}=9r\)

\(6\sqrt{6}=9r\)

\(r={2\sqrt{6}\over3}\)

The square root of 49 is an integer, and a rational number.

Thanks Heureka. That is great !!

Hi Melody

"Find the coefficient of x^3 y^3 z^2 in the expansion of (3x+5y+7z)^8".

Set \(a = 3x\)

Set \(b = 5y\)

Set \(c = 7z\)

Set \(n = 8\)

Set \(i = 3\)

Set \(j = 3\)

Set \(k = 2\)

Trinomial Coefficient of  \(x^3 y^3 z^2\):
\(\begin{array}{|rcll|} \hline \dbinom{8}{3,3,2}(3x)^3\cdot(5y)^3\cdot (7z)^2 &=& \dfrac{8!}{3!3!2!} (3x)^3\cdot(5y)^3\cdot (7z)^2 \\\\ &=& \dfrac{8!}{3!3!2!} \cdot 3^35^37^2 x^3y^3z^2 \quad & | \quad \dfrac{8!}{3!3!2!} = 560 \\\\ &=& 560\cdot 3^35^37^2 x^3y^3z^2 \\\\ &=& 560\cdot 27 \cdot 125 \cdot 49 x^3y^3z^2 \\\\ &\mathbf{=}& \mathbf{ 92610000 x^3y^3z^2 } \\ \hline \end{array}\)

\(\text{The coefficient of $x^3 y^3 z^2$ in the expansion of $(3x+5y+7z)^8$ is $\mathbf{92~ 610~ 000}$ } \)

What is the radius of the circle inscribed in triangle ABC if AB = 5, AC=6, BC=7? 

Hello friends !

The radius of the circle is r .

\(a=7\\ b=6\\ c=5 \)

\(c^2=a^2+b^2-2ab\cdot cos\gamma\\ \gamma=arccos\frac{a^2+b^2-c^2}{2ab}=arccos\frac{7^2+6^2-5^2}{2\cdot 7\cdot 6}=arccos 0.7142857\\ \gamma=44.4153°\)

\(b^2=a^2+c^2-2ac\ cos\beta\\ \beta=arccos\frac{a^2+c^2-b^2}{2ac}=arccos\frac{7^2+5^2-6^2}{2\cdot 7\cdot 5}=arccos\ 0.54285714\\ \beta=57.12165°\)

\(a=s+t\)   (two sections on the side a,

                     on both sides of the contact with the circle)

\(tan\frac{\gamma}{2}=\frac{r}{s}\\ s=\frac{r}{tan \frac{\gamma}{2}}\)             \(tan \frac{\beta}{2}=\frac{r}{t}\\ t= \frac{r}{tan \frac{\beta}{2}}\)

\(a=\frac{r}{tan \frac{\gamma}{2}}+\frac{r}{tan \frac{\beta}{2}}=7\)

\(\frac{r}{tan \frac{44.4153°}{2}}+\frac{r}{tan \frac{57.12165°}{2}}=7\)

\(\frac{r}{0.408248}+\frac{r}{0.544331}=7\)

\(0.544331r+0.408248r=7\cdot 0.408248\cdot 0.544331\\ r=\frac{7\ \cdot\ 0.408248\ \cdot\ 0.544331}{0.544331+0.408248}\)

\(r=1.63299\)

\(The\ radius\ of\ the\ circle\ inscribed\ in\ triangle\ ABC\ is\ 1.63299.\)

  !

Here's a way of looking at the problem that doesn't require calculus:

.

How many positive four-digit integers n have the property that the three-digit number obtained by removing

the leftmost digit is one-ninth of n?

\(\begin{array}{|lrcll|} \hline (1) & n_4 &=& 10^3a + n_3 \\ \\ \hline \\ (2) & n_3 &=& \dfrac{n_4}{9} \\\\ & 9n_3 &=& n_4 \\\\ & 9n_3 &=& 10^3a + n_3 \\\\ & 8n_3 &=& 10^3a \quad & | \quad : 8 \\\\ & \mathbf{n_3} & \mathbf{=} & \mathbf{125a} \\ \hline \end{array} \)

\(\begin{array}{|r|c|l|l|} \hline & n_3=125a & \\ a & n_3\lt 1000 & n_4 &\\ \hline 1 & 125 & 1125 & \frac{1125}{9} = 125 \\ 2 & 250 & 2250 & \frac{2250}{9} = 250 \\ 3 & 375 & 3375 & \frac{3375}{9} = 375 \\ 4 & 500 & 4500 & \frac{4500}{9} = 500 \\ 5 & 625 & 5625 & \frac{5625}{9} = 625 \\ 6 & 750 & 6750 & \frac{6750}{9} = 750 \\ 7 & 875 & 7875 & \frac{7875}{9} = 875 \\ \hline \end{array}\)

Find the smallest positive  \(N\) such that

\(N\equiv6(mod 12) \\ N\equiv6(mod 18) \\ N\equiv6(mod 24) \\ N\equiv6(mod 30 ) \\ N\equiv6(mod 60) \)

\(\begin{array}{|rcll|} \hline & N &\equiv& 6 \pmod {12} \\ & N &\equiv& 6 \pmod {18} \\ & N &\equiv& 6 \pmod {24} \\ & N &\equiv& 6 \pmod {30} \\ & N &\equiv& 6 \pmod {60} \\\\ \Rightarrow & N &\equiv& 6 \pmod{\text{lcm}(12,18,24,30,60)} \\ & N &\equiv& 6 \pmod{360} \\ & \mathbf{N} & \mathbf{=} & \mathbf{6+360m,\ \quad m \in Z} \\ \hline \end{array} \)

\(\text{The smallest positive $ N = 6,\ $ if $m = 0$ }\)

When N is divided by 10, the remainder is a. When N is divided by 13, the remainder is b.
What is N modulo 130, in terms of a and b?
(Your answer should be in the form ra+sb, where r and s are replaced by nonnegative integers less than 130.)

\(\begin{array}{|rclcl|} \hline N &\equiv& a \pmod{10} \quad &\text{or}& \quad N= a+10n,\ n \in Z \\ N &\equiv& b \pmod{13} \quad &\text{or}& \quad N= a+13m,\ m \in Z \\ \\ \hline \\ N = a+10n &=& b+13m \\ 10n &=& 13m+b-a \\\\ \mathbf{n} & \mathbf{=}& \mathbf{\dfrac{13m+b-a}{10}} \\\\ n &=& \dfrac{10m+3m+b-a}{10} \\\\ n &=& \dfrac{10m}{10} +\dfrac{3m+b-a}{10} \\\\ n &=& m +\underbrace{\dfrac{3m+b-a}{10}}_{=c} \\\\ c &=& \dfrac{3m+b-a}{10} \\\\ 10c &=& 3m+b-a \\\\ 3m &=& 10c+a-b \\\\ m &=& \dfrac{10c+a-b}{3} \\\\ \mathbf{m} & \mathbf{=}& \mathbf{\dfrac{10c+a-b}{3}} \\\\ m &=& \dfrac{9c+c+a-b}{3} \\\\ m &=& \dfrac{9c}{3} + \dfrac{c+a-b}{3} \\\\ m &=& 3c + \underbrace{\dfrac{c+a-b}{3}}_{=d} \\\\ d &=& \dfrac{c+a-b}{3} \\\\ 3d &=& c+a-b \\\\ \mathbf{c} & \mathbf{=}& \mathbf{3d+b-a} \\ \hline \end{array} \)

\(\mathbf{m=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{m} & \mathbf{=}& \mathbf{\dfrac{10c+a-b}{3}} \quad & | \quad \mathbf{c} & \mathbf{=}& \mathbf{3d+b-a} \\\\ m &=& \dfrac{10(3d+b-a)+a-b}{3} \\\\ m &=& \dfrac{30d+10b-10a+a-b}{3} \\\\ m &=& \dfrac{30d+9b-9a}{3} \\\\ \mathbf{m} & \mathbf{=}& \mathbf{10d+3b-3a} \\ \hline \end{array}\)

\(\mathbf{N=\ ?}\)

\(\begin{array}{|rcll|} \hline N &=& b+13m \quad & | \quad \mathbf{m} & \mathbf{=}& \mathbf{10d+3b-3a} \\ N &=& b+13(10d+3b-3a) \\ N &=& 130d -39a + 40b \\ \hline \end{array}\)

\(\mathbf{N \pmod{130} =\ ?}\)

\(\begin{array}{|rcll|} \hline N &=& 130d -39a + 40b \quad & | \quad \pmod{130} \\ N &\equiv& 0 -39a + 40b \pmod{130} \\ N &\equiv& -39a + 130a + 40b \pmod{130} \\ \mathbf{N} & \mathbf{\equiv} & \mathbf{91a + 40b \pmod{130}} \\ \hline \end{array}\)

\(N \pmod{130} = 91a + 40b\)

7x^2014  * 2x^2012  *  9x^2007   *  8x^2009

___________________________________  =

4x^2013 * 28x^2011 * x^2008  * 6x^2010

7x  * 2x * 9  * 8

________________  =

4 * 28 * x * 6x

1008x^2

_______   =

672x^2

   3

____

   2

Just add all the powers of x on the top, and add all the powers of x on the bottom, you should get the number of 8042. Then subtract 8042 - 8042 =0. Then multiply the coefficients on the top: 7 x 2 x 9 x 8=1008. Then multiply the coefficients on the bottom: 4 x 28 x 6 =672. Then divide the top total by the bottom total and you have: 1008 / 672 = 3/2

It can be shown that the area will be  maximized when each side of the rectangle is equal to the perimeter / 4....in essence the rectangle will be a square

So...each side will  =  44/ 4  =  11 cm

And the maximum area is   11^2  =  121 cm^2

the factors of 2010 are 67, 5, 3, 2

67 is a prime number so the only other option is 67 x 30.

but there is no N such that satisfies this with integers because of 67 being prime.

not sure how to help you visualy, but do this

put a dot at (3,-4) then have it curve up and to the left like half a parabla.

the asymitope would be at y=2 so the curve would approch y=2

Did you actually read any of what we wrote?

My answer came in one minute after yours. I was working on it before you said you could do it.

You had not said that you already understood otherwise I would not have wasted my time.

You have done this too many times RP.  

Do not ask questions until you have given youself lots of time to think about it first. !!

If you do not need to ask then do not ask.

If anyone puts their time into giving you an answer then ALWAYS give feedback and say thank you.

\({x(x+1) \over 4x^{2}+6x+3}   \)

You cannot divide by 0 so the denominator cannot be 0

\(x \ne {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x \ne {-6 \pm \sqrt{36-48} \over 8}\\ x \ne {-6 \pm \sqrt{12} \;i\over 8}\\ \)

This is a complex number so there are no restriction on x in the real number system.

There answer has restricted values but I do not think that this is what the question is asking.

Here is the graph if the expression was put equal to y.

You can see that the range is quite restricted.   The domain is not restricted.

Thank you very much, this solution is very clear!

- Daisy

This similarity relationship is derived from the fact that each triangle has one right angle and acute angle in common:

\(\because \angle BQC= \angle BPH= \angle HQA = \angle CPA\\ \because \angle QBC = \angle QBC\ \&\ \angle PAC = \angle PAC \ \& \ \angle BHP=\angle AHQ\)

\(\therefore \triangle BQC\sim \triangle BPH\sim \triangle HQA\sim \triangle CPA\)

From the similarity of corresponding triangles:

\(\frac{HP}{BP}=\frac{HQ}{QA}=\frac{PC}{AP}=\frac{QC}{BQ}\) results in:  \(BP\cdot PC-AQ\cdot QC =HP\cdot AP-HQ\cdot BQ\\ =HP(HP+HA)-HQ(HQ+HB)=HP^2-HQ^2=\boxed{21}.\) 

Arithimetic sequence {\(a_n\)}  {\(b_n \)} have integer terms,  \(a_1\)=\(b_1\)=\(1\)<\(a_2\)<=\(b_2\) and\(a_n*b_n=2010\) for some \(n\). What is the largest possible value of \(n\)?

Sorry, everytime i type the same thing, it cuts off

Lol thanks guys but I said I already understand ^-^

This is not a question

Arithmetic sequence {a_n} {b_n} have integer terms, a_1=b_1=1