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23%/100% * 123 = 28.29

Point p is   1/3 of the way from AP to BP     this will nake the ratio   AP:BP  1:2     (1+2) =3

Let's find x coordinate first    from 1    to    5   is 4 units    we need   1/3      4 x 1/3 = 4/3      add this to A

1 + 4 /3 =   7/3  = x

Now for y =   1 to 6 is 5     5 x 1/3 = 5/3    add this to  A

1+5/3 = 8/3 = y

7/3, 8/3   = p = (2.3,2.7)     rounded

A WHOLE circle is 360 degrees....we need to find the fraction  60 degrees/360 degrees

Area of whole circle = pi r^2 = pi 12^2

the fraction of this whole circle (60 degrees worth):      60 /360 * pi 12^2 = 24 pi

I thinks its probs 12

john is twice as old as his friend peter.peter is 5 years older than alice. In 5 years, john will be 3 times as old as alice. how old is peter now?

J=2P

P=A+5

J+5=3(A+5)

Find P

J=2P=2(A+5) 

so

2(A+5) +5=3(A+5)

5=A+5

A=0

P=5

J=10

check

john is twice as old as Peter.    True

Peter is 5 years older than Alice    True

In 5 years john will be 15 and Alice will be 5. John will be 3 times older.   True

Peter is 5 now.

Imagine this.

You have a big solid paper cylinder.   The volume is the area of the base* height    =      \(\pi*Radius^2*height\)

Now you drill out the centre of it.  This is a smaller cylider, the hollow bit in the middle. 

   The volume of the the small drilled out cylinder is     also     \(\pi*\color{red}{r} \color{black}adius^2*height\)      I used a little r this time becaus it is a smaller radius.

Anyway I don't want the middle bit, so I throw it away.

The volume of paper is   

  \(V=\pi R^2h \quad- \quad \pi\color{red}{r} \color{black}^2h\\\)

This is the same as Rom's answer - his answer is just factorised.

[x/2]+[x/4]+[x/8]+507=x

[x/2]+[x/4]+[x/8]+507=x

[x/2],[x/4],[x/8]- INTEGER DIVISIONS!!!

I assume the "Floor Function": \(\left\lfloor\dfrac{x}{2}\right\rfloor+\left\lfloor\dfrac{x}{4}\right\rfloor+\left\lfloor\dfrac{x}{8}\right\rfloor+507=x\)

So x is an integer.

We rearrange:

\(\left\lfloor\dfrac{4x}{8}\right\rfloor+\left\lfloor\dfrac{2x}{8}\right\rfloor+\left\lfloor\dfrac{x}{8}\right\rfloor+507=x \)

We substitute: \(y= \dfrac{x}{8}\)
\(\left\lfloor 4y \right\rfloor+\left\lfloor 2y \right\rfloor+\left\lfloor y \right\rfloor+507=x\)

\(\text{The fractional part of $\mathbf{y}$ is $\alpha$ and $0 < \alpha < 1$ } \\ \text{The integer part of $\mathbf{y}$ is $n$ }\\ \text{So $\mathbf{y = n+\alpha}$} \\ \text{$\mathbf{x = \lfloor 8y \rfloor} = \lfloor 8(n+\alpha)\rfloor=8n+\lfloor 8\alpha\rfloor$}\)

\(\begin{array}{|rcll|} \hline \left\lfloor 4(n+\alpha)\right\rfloor+\left\lfloor 2(n+\alpha) \right\rfloor+\left\lfloor n+\alpha\right\rfloor+507 &=& 8n+\lfloor 8\alpha\rfloor \\\\ \mathbf{\left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507}&\mathbf{=}& \mathbf{8n+\lfloor 8\alpha\rfloor} \\ \hline \end{array}\)

We divide alpha into 8 parts: 

\(0 < \alpha < \frac{1}{8},\ \frac{1}{8} \le \alpha < \frac{2}{8},\ \frac{2}{8} \le \alpha < \frac{3}{8},\ \frac{3}{8} \le \alpha < \frac{4}{8} ,\ \frac{4}{8} \le \alpha < \frac{5}{8},\ \frac{5}{8} \le \alpha < \frac{6}{8},\ \frac{6}{8} \le \alpha < \frac{7}{8},\ \frac{7}{8} \le \alpha < \frac{8}{8} \)

\(\begin{array}{|r|r|rcl|} \hline \text{part} & \text{domain} & \\ \hline 1. & 0 < \alpha < \frac{1}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +2n+n+507 &=& 8n+0 \\ && n &=& 507 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 507 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4056} \\ \hline 2. & \frac{1}{8} \le \alpha < \frac{2}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +2n+n+507 &=& 8n+1 \\ && n &=& 506 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 506+1 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4049} \\ \hline 3. & \frac{2}{8} \le \alpha < \frac{3}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +1+2n+n+507 &=& 8n+2 \\ && n &=& 506 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 506+2 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4050} \\ \hline 4. & \frac{3}{8} \le \alpha < \frac{4}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +1+2n+n+507 &=& 8n+3 \\ && n &=& 505 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 505+3 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4043} \\ \hline 5. & \frac{4}{8} \le \alpha < \frac{5}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +2+2n+1+n+507 &=& 8n+4 \\ && n &=& 506 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 506+4 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4052} \\ \hline 6. & \frac{5}{8} \le \alpha < \frac{6}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +2+2n+1+n+507 &=& 8n+5 \\ && n &=& 505 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 505+5 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4045} \\ \hline 7. & \frac{6}{8} \le \alpha < \frac{7}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +3+2n+1+n+507 &=& 8n+6 \\ && n &=& 505 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 505+6 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4046} \\ \hline 8. & \frac{7}{8} \le \alpha < \frac{8}{8} & \left\lfloor 4n+4\alpha\right\rfloor+\left\lfloor 2n+2\alpha \right\rfloor+\left\lfloor n+\alpha \right\rfloor+507& =& 8n+\lfloor 8\alpha\rfloor \\ && 4n +3+2n+1+n+507 &=& 8n+7 \\ && n &=& 504 \\ && x &=& 8n + \lfloor 8\alpha\rfloor \\ && x &=& 8\cdot 504+7 \\ && \mathbf{x} & \mathbf{=} & \mathbf{4039} \\ \hline \end{array}\)

Graph the system of equation on graph paper.

\(\text{he'll need the radius of the roll }R\\ \text{the radius of the inner cardboard tube }r\\ \text{and the height of the roll }h\\ V = \pi h (R^2 - r^2)\)

(2|7)

Yes, thx. I just from Russia and i didn't know how it translete right on Eng))) I just gave you the literal translation))

I took the AMC 10 A. 

Tip: chill when you can't figure something out. Also, if you are sure you don't know how to solve any of the problems left, then double check your answers for the problems you did do.

Good luck!

Thank you!

Image is question pls help me

Thanks, CPhill

Ok lets take a look - I'll start from the beginning.

\(3 r^2 -16r-7 = 5\\ \text{you need it to equal 0 so take five from BOTH sides}\\ 3 r^2 -16r-7-5 = 5-5\\ \color{red}{3 r^2 -16r-12 = 0}\\ \text{Now you have it in general from for a quadratic equation}\\ ar^2+br+c=0\\ \color{red}{a=3\qquad b=-16 \qquad c=-12}\\~\\ \text{Now use the quadratic formula to find the answer}\\ r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \)

Now substitute the numbers into the equation and simplify to find your answers. :)

[that was meant to be general FORM not general from ]

When you have a QUADRATIC EQUATION of the form     a x^2 +  b x + c = 0    (or in this case  a r^2 + b r + c = 0)    

   you can use the QUADRATIC FORMULA to find the values of x (or  r in this case)

Melody showed you the FORMULA....Now do you know what   a   b   and c  are?

   Put those values in the FORMULA and solve it.....    OK ?  

BINGO !    

is it the second one?

Im having a hard time with math.

The solution to the system of equations, is the common point on the graphs.....the point where they intersect......you can see that can't ya?

What do you think the coordintes of that point of intersection are?????