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In the solution for 

https://web2.0calc.com/questions/question-about-conics-pls-help#r3, what do you do to get to \(-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}=1\)

The sum of the ages of Sam and Tomas is 27 years.

Seven years ago, Tomas was three years more than one- fourth as old as Sam.

Find there present age.

x = sam age
y = thomas age

\(\begin{array}{|rclcrcl|} \hline x+y &=& 27 \quad \text{or} \quad \mathbf{y = 27-x} \\\\ \mathbf{y-7} &=& \mathbf{\dfrac{1}{4}(x-7) + 3} \\ y &=& \dfrac{1}{4}(x-7) + 3 +7 \\\\ y &=& \dfrac{1}{4}(x-7) + 10\quad | \quad y = 27-x \\\\ 27-x &=& \dfrac{1}{4}(x-7) + 10 \\\\ -x &=& \dfrac{1}{4}(x-7) + 10-27 \\\\ -x &=& \dfrac{1}{4}(x-7) -17 \\\\ x + \dfrac{1}{4}(x-7) &=& 17 \quad | \quad *4 \\\\ 4x + x-7 &=& 4*17 \\\\ 5x &=& 4*17 +7 \\\\ 5x &=& 75 \\\\ \mathbf{x} &=& \mathbf{15} \\ \hline \\ \mathbf{y} &=& \mathbf{27-x} \\ y &=& 27-15 \\ \mathbf{y} &=& \mathbf{12} \\ \hline \end{array}\)

sam is 15 and thomas is 12

Balance of his account statement on Oct.31 =$2,865.45

$2,865.45 x 29.99% =$859.35 - interest charge on his balance for 1 year.

$859.35 / 12 =$71.61 - interest charge on his balance for 1 month.

$2,865.45 + $71.61 =$2,937.06 - Balance of account statement on Nov. 30, principal + interest.

$2,937.06 x 5% =$146.85 - Minimum monthly pmt. that he must make.

$2,937.06 - $146.85 =$2,790.21 - Balance of his acct. statement on Nov.30th after the minimum monthly payment.

Heureka, what do you do to get to \(-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}=1\)

\(\begin{array}{|l|} \hline \mathbf{y} = \mathbf{\dfrac{b^2}{y_t}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}x}\quad \boxed{\text{compare with tangent line }\\ y=k+1x} \\ \hline \end{array} \\ \begin{array}{rclcrl} y&=&\dfrac{b^2}{y_t}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}x &\color{darkgrey}\text{compare}& y&=&k+1*x \\\\ y&=& {\color{red}\dfrac{b^2}{y_t}}{\color{green}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}}x &\color{darkgrey}\text{compare}& y&=&{\color{red}k}{\color{green}+1}*x \\ &&&\begin{array}{|c|} \hline {\color{red}\dfrac{b^2}{y_t}} = {\color{red}k} \\\\ {\color{green}-\dfrac{b^2}{a^2}\dfrac{x_t}{y_t}} = {\color{green}+1}\\ \hline \end{array} \end{array}\\\)

1324  1432  3124  3412  4132  4312  Total =  6 such numbers.

Dragan, could you please put your answer with mine and Heueka's on the original thread and then explain how you came by it please?

Was it really that easy?

Triangle ABC with vertices A(−2, 0), B(1, 4) and C(−3, 2) is reflected over the y-axis to form triangle A'B'C'. What is the length of a segment drawn from C to C'?

Thanks Heureka, 

here is another coordinate geometry approach.

I just found the midpoint of the horizontal line x=-2 and drew a vertical axis of symmetry through it.

then the midpoint of the vertical points y=-0.5 and drew a horizontal axis of symmetry through it.

Where these axes of symmetry meet is the centre.  (-2,-0.5)

Now just use the distance formula between the centre and any point on the circumference to find the radius.

Then do what you need to do to find the diameter.

Diameter = sqrt ( 8² + 1² )   

BC is tangent to both a circle with center at A and a circle with center at D.
The area of the circle with center at A is 225p and the area of the circle with center at D is 36p.
If BC=16, find the distance between the centers of the two circles.

Formula area of the circle: \(A=\pi r^2\)

\(\begin{array}{|rcll|} \hline A=\pi r^2 \\ \hline 225\pi &=& \pi R^2 \\ 225 &=& R^2 \\ \mathbf{15} &=& \mathbf{R} \\\\ 36\pi &=& \pi r^2 \\ 36 &=& r^2 \\ \mathbf{6} &=& \mathbf{r} \\ \hline \end{array} \)

Pythagoras' theorem:

\(\begin{array}{|rcll|} \hline AD^2 &=& 16^2+(R+r)^2 \\ AD^2 &=& 16^2+(15+6)^2 \\ AD^2 &=& 16^2+21^2 \\ AD^2 &=& 256+441 \\ AD^2 &=& 697 \\ AD &=& 26.4007575649 \\ \mathbf{AD} &\approx& \mathbf{26.4} \\ \hline \end{array}\)

The distance between the centers of the two circles is \(\approx \mathbf{26.4}\)

Let
\(f(x)=(x^2+6x+9)^{50}-4x+3\), and let \(r_1,r_2,\ldots,r_{100}\) be the roots of \(f(x)\).

Compute \((r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}\).

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x^2+6x+9)^{50}-4x+3} \\ &=& \left((x+3)^2\right)^{50}-4x+3 \\ \mathbf{f(x)} &=& \mathbf{\left( x+3 \right)^{100}-4x+3} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & f(r_1) =0 &=& \left( r_1+3 \right)^{100}-4r_1+3 \\ & f(r_2) =0 &=& \left( r_2+3 \right)^{100}-4r_2+3 \\ & \ldots \\ & f(r_{100}) =0 &=& \left( r_{100}+3 \right)^{100}-4r_{100}+3 \\ & \hline \\ \text{sum} & 0 &=& (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} \\ & && -4(r_1+r_2+\ldots + r_{100}) + 3\cdot 100 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ \hline \end{array}\)

\(\mathbf{\text{vieta:}}\)
For any polynomial equation
\(0=x^n+a_{n-1}·x^{n-1}+...+a_2·x^2+a_1·x^1+a_0\)
with the solutions \(r_1\dots r_n\), the relatively simple formulas for \(a_0\) and \(a_{n-1}\) are:
\(a_0=(-1)^n \prod\limits_{k=1}^{n} r_k \\ a_{n-1}= -\sum \limits_{k=1}^{n} r_k\)

\(\begin{array}{|rcll|} \hline && \left( x+3 \right)^{100} \\ \\ &=& x^{100} + \binom{100}{1}x^{99}\cdot 3 + \ldots \\ \\ &=& x^{100} + \underbrace{300}_{ \underbrace{=a_{n-1}}_{= -(r_1+r_2+\ldots + r_{100})}} x^{99} + \ldots \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ && \boxed{r_1+r_2+\ldots + r_{100} = -300} \\ (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(-300) - 300 \\ \mathbf{(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}} &=& \mathbf{-1500} \\ \hline \end{array}\)

Given that the point (9,7) is on the graph of y=f(x), there is one point that must be on the graph of 2y = (f(2x)/2) + 2. What is the sum of coordinates of that point?

You can sum the coordinates.

Four red candies and three green candies can be combined to make many different “flavors.”
Flavors are different if the percent red is different,
so “3 red / 0 green” is the same flavor as “2 red / 0 green,”
and likewise “4 red / 2 green” is the same flavor as “2 red / 1 green.”
If a flavor is to be made using some or all of the seven candies,
how many different flavors are possible?

\(\begin{array}{|c|c|c|c|c|} \hline & {\color{green}0} & {\color{green}1} &{\color{green}2} &{\color{green}3} \\ \hline {\color{red}0} & - & 0+1=1 & 0+2=2 & 0+3=3 \\ & & \frac{{\color{red}0}}{1}: \frac{{\color{green}1}}{1} &\frac{{\color{red}0}}{2}: \frac{{\color{green}2}}{2}&\frac{{\color{red}0}}{3}: \frac{{\color{green}3}}{3}\\ & & {\color{red}0}: {\color{green}1} & {\color{red}0}: {\color{green}1}&{\color{red}0}: {\color{green}1}\\ \hline {\color{red}1} & 1+0=1 & 1+1=2 & 1+2=3 & 1+3=4 \\ & \frac{{\color{red}1}}{1}: \frac{{\color{green}0}}{1} & \frac{{\color{red}1}}{2}: \frac{{\color{green}1}}{2} &\frac{{\color{red}1}}{3}: \frac{{\color{green}2}}{3}&\frac{{\color{red}1}}{4}: \frac{{\color{green}3}}{4}\\ & {\color{red}1}: {\color{green}0} & \\ \hline {\color{red}2} & 2+0=2 & 2+1=3 & 2+2=4 & 2+3=5 \\ & \frac{{\color{red}2}}{2}: \frac{{\color{green}0}}{2} & \frac{{\color{red}2}}{3}: \frac{{\color{green}1}}{3} &\frac{{\color{red}2}}{4}: \frac{{\color{green}2}}{4}&\frac{{\color{red}2}}{5}: \frac{{\color{green}3}}{5}\\ & {\color{red}1}: {\color{green}0} & &\frac{{\color{red}1}}{2}: \frac{{\color{green}1}}{2} \\ \hline {\color{red}3} & 3+0=3 & 3+1=4 & 3+2=5 & 3+3=6 \\ & \frac{{\color{red}3}}{3}: \frac{{\color{green}0}}{3} & \frac{{\color{red}3}}{4}: \frac{{\color{green}1}}{4} &\frac{{\color{red}3}}{5}: \frac{{\color{green}2}}{5}&\frac{{\color{red}3}}{6}: \frac{{\color{green}3}}{6}\\ & {\color{red}1}: {\color{green}0} & & &\frac{{\color{red}1}}{2}: \frac{{\color{green}1}}{2}\\ \hline {\color{red}4} & 4+0=4 & 4+1=5 & 4+2=6 & 4+3=7 \\ & \frac{{\color{red}4}}{4}: \frac{{\color{green}0}}{4} & \frac{{\color{red}4}}{5}: \frac{{\color{green}1}}{5} &\frac{{\color{red}4}}{6}: \frac{{\color{green}2}}{6}&\frac{{\color{red}4}}{7}: \frac{{\color{green}3}}{7}\\ & {\color{red}1}: {\color{green}0} & &\frac{{\color{red}2}}{3}: \frac{{\color{green}1}}{3} \\ \hline \end{array}\)

The percent red

\(\begin{array}{|c|c|c|c|c|} \hline & {\color{green}0} & {\color{green}1} &{\color{green}2} &{\color{green}3} \\ \hline {\color{red}0} & & {\color{red}0} & {\color{red}0}&{\color{red}0}\\ \hline {\color{red}1} & {\color{red}1} & \frac{{\color{red}1}}{2} &\frac{{\color{red}1}}{3}&\frac{{\color{red}1}}{4} \\ \hline {\color{red}2} & {\color{red}1} & \frac{{\color{red}2}}{3} &\frac{{\color{red}1}}{2}&\frac{{\color{red}2}}{5} \\ \hline {\color{red}3} & {\color{red}1} & \frac{{\color{red}3}}{4} &\frac{{\color{red}3}}{5} &\frac{{\color{red}1}}{2} \\ \hline {\color{red}4} & {\color{red}1} & \frac{{\color{red}4}}{5}&\frac{{\color{red}2}}{3}&\frac{{\color{red}4}}{7} \\ \hline \end{array}\)

The different flavors are \(\{ {\color{red}0},\ \frac{{\color{red}1}}{4},\ \frac{{\color{red}1}}{3},\ \frac{{\color{red}2}}{5},\ \frac{{\color{red}1}}{2},\ \frac{{\color{red}3}}{5},\ \frac{{\color{red}4}}{7},\ \frac{{\color{red}2}}{3},\ \frac{{\color{red}3}}{4},\ \frac{{\color{red}4}}{5},\ {\color{red}1} \} \)

The graph below should help:

The digits of a four-digit positive integer add up to 14.

The sum of the two middle digits is nine, and the thousands digit minus the units digit is one.

If the integer is divisible by 11,

what is the integer?

\(\begin{array}{|lrcll|} \hline \text{A four-digit positive integer}:& abcd \\ \hline \text{The digits add up to $14$}: & a+b+c+d=14 \\ \text{The sum of the two middle digits is nine}: & b+c=9 \\ \text{The thousands digit minus the units digit is one}: & a-d=1 \\ \text{Divisibility by Eleven: if the difference of the alternating sum is $0$}: & a+c-(b+d)=0 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a+b+c+d &=& 14 \quad | \quad b+c=9 \\ a+9+d &=& 14 \quad | \quad a-d=1\quad \text{or} \quad a=1+d \\ 1+d+9+d &=& 14 \\ 2d+10&=& 14 \\ 2d &=& 4 \\ \mathbf{d} &=& \mathbf{2} \\ \hline a &=& 1+d \\ a &=& 1+2 \\ \mathbf{a} &=& \mathbf{3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a+c-(b+d) &=& 0 \\ a+c-b-d &=& 0 \\ a-d+c-b &=& 0 \quad | \quad a-d = 1 \\ 1+c-b &=& 0 \quad | \quad b+c=9\quad \text{or} \quad c=9-b \\ 1+9-b-b &=& 0 \\ 2b &=& 10 \\ \mathbf{b} &=& \mathbf{5} \\ \hline c &=& 9-b \\ c &=& 9-5 \\ \mathbf{c} &=& \mathbf{4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{The four-digit positive integer $abcd$ is}:& \mathbf{3542} \\ & 3542 : 11 = 322 \\ \hline \end{array} \)

In regular pentagon PQRST, X is the midpoint of segment ST. What is the measure of angle XQS?

∠XQS = 540° - ( ∠XQP + ∠QPT + ∠PTS + ∠TSR + ∠SRQ + ∠RQS )  

Please give an actual answer

In the diagram below, chords \(\overline{AB}\) and \(\overline{CD}\) are perpendicular, and meet at X.
If AX = 3, BX = 4, CX = 6, and DX = 2, then find the diameter of the circle.

\(\text{Let the center of the circle $C(x_c,\ y_c)$ }\)

\(\begin{array}{|rcll|} \hline 4+y_c &=& 3-y_c \\ 2y_c &=& -1 \\ \mathbf{y_c} &=& \mathbf{-\dfrac{1}{2}} \\ \hline \end{array} \begin{array}{|rcll|} \hline 6+x_c &=& 2-x_c \\ 2x_c &=& -4 \\ \mathbf{x_c} &=& \mathbf{-2} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & (x-x_c)^2 + (y-y_c)^2 &=& r^2 \\ P(2,0): & (2-x_c)^2 + (0-y_c)^2 &=& r^2 \quad | \quad x_c = -2,\ y_c =-\dfrac{1}{2} \\ & (2+2)^2 + \left(0+\dfrac{1}{2} \right)^2 &=& r^2 \\ & 4^2 + \dfrac{1}{4} &=& r^2 \\ & 16 + \dfrac{1}{4} &=& r^2 \quad | \quad *4 \\ & 64+1 &=& 4r^2 \\ & 65 &=& 4r^2 \quad | \quad \text{sqrt both sides} \\ & \sqrt{65} &=& 2r \\ \hline \end{array}\)

The diameter of the circle is \(\mathbf{\sqrt{65}}\)

Find constants  and  such that
\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1}\)

I will assume you got the answer correct.....now you can see the 'outliers'     20 and 229   and remove them and do the calculations without them...

   outliers are probably mis-charged  (I guess) ...or perhaps the 229 is a multi-room suite    ....who knows ?

ok! ill try!
 

Mean (Average)82.125

Median75

Range209Mode75, appeared 4 times

Sorted Data Set: 20, 60, 65, 65, 70, 70, 75, 75, 75, 75, 80, 85, 85, 90, 95, 229

thank you! 

x = sqrt(26) cos 140^o  = ..........

 y= sqrt (26) sin 140^o = ..........

3x^4-7x^3+2x^2-bx+1      put in '1' for 'x' to find  f(1)    and it has to = 1

3 (1⁴) - 7(1³) + 2 (1²) - b(1) + 1   = 1         I think you should be able to take it from here !  

*** edited ****     Thanx, Alan !