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I get the following:

Let A = (4, -1), B = (6, 2), and C = (-1, 2). There exists a point X and a constant k such that for any point P, \(PA^2 + PB^2 + PC^2 = 3PX^2 + k\). Find the constant \(k\).

1.) Angle DEF is from an equilateral triangle. Since all sides are the same, then all angles are the same, and the angles of a triangle add up to 180º. So 180º/3 = 60º. So, Angle DEF = 60º.

2.) Angles ADE and AED are the same because of that triangle being isosceles. So we can find ADE easier by using the fact that line CDF is one large angle of 180º. All angles in square ABCD are 90º. And, we already know angle EDF is 60º because of it belonging to the equilateral triangle. So to get the value of angle ADE we consider the value of the angle of the line, 180º, then subtract the 90º from the square, and subtract the 60º from the equilateral triangle. 180º-90º-60º=30º. So, angle ADE is 30º. 

Since angle ADE = angle AED, AED=30º.

3.) Angle AEF = AED + DEF = 30º + 60º = 90º. 

Hence, the size of angle AEF is 90º.

-Zamarronics

Find the 3 smallest positive x-intercepts of the graph of \(y = \cos(12x) + \cos(14x)\) and list them in increasing order.

Formula: \(\boxed{\cos(x) + \cos(y) = 2\cos\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x-y}{2}\right) }\)

\(\begin{array}{|rcll|} \hline && \mathbf{\cos(12x) + \cos(14x)} \\ &=& \cos(14x) + \cos(12x) \\ &=& 2\cos\left(\dfrac{14x+12x}{2}\right)\cos\left(\dfrac{14x-12x}{2}\right) \\ &=& 2\cos\left(\dfrac{26x}{2}\right)\cos\left(\dfrac{2x}{2}\right) \\ &=& \mathbf{2\cos(13x)\cos(x)} \\ \hline \end{array}\)

 x-intercepts of the graph

\(\begin{array}{|rcll|} \hline \mathbf{2\cos(13x)\cos(x)} &=& 0 \quad | \quad : 2 \\ \cos(13x)\cos(x) &=& 0 \\ \hline \mathbf{\cos(x)} &=& \mathbf{0} \\ x &=& \pm \arccos(0) +2k\pi \qquad k\in \mathbb{Z} \\ x &=& \pm \dfrac{\pi}{2} +2k\pi \\\\ x &=& \mathbf{+} \dfrac{\pi}{2}+2k\pi \qquad k = 0 \\ x &=& + \dfrac{\pi}{2} \\ \mathbf{x} &=& \mathbf{1.57079632679}\ \text{rad} \\\\ x &=& \mathbf{-} \dfrac{\pi}{2} +2k\pi \qquad k = 1 \\ x &=& - \dfrac{\pi}{2} +2 \pi \\ \mathbf{x} &=& \mathbf{4.71238898038}\ \text{rad} \\ \hline \mathbf{\cos(13x)} &=& \mathbf{0} \\ 13x &=& \pm \arccos(0) +2k\pi \qquad k\in \mathbb{Z} \\ 13x &=& \pm \dfrac{\pi}{2} +2k\pi \\\\ 13x &=& \mathbf{+}\dfrac{\pi}{2} +2k\pi \\ x &=& \dfrac{\pi}{2*13}+\dfrac{2\pi k}{13} \qquad k = 0 \\ x &=& \dfrac{\pi}{26} \\ \mathbf{x} &=& \mathbf{0.12083048668}\ \text{rad} \\\\ x &=& \mathbf{+}\dfrac{\pi}{26}+\dfrac{2\pi k}{13} \qquad k = 1 \\ x &=& \dfrac{\pi}{26}+\dfrac{2\pi}{13} \\ x &=& \dfrac{5\pi}{26} \\ \mathbf{x} &=& \mathbf{0.60415243338}\ \text{rad} \\\\ 13x &=& \mathbf{-}\dfrac{\pi}{2} +2k\pi \\ x &=& -\dfrac{\pi}{2*13}+\dfrac{2\pi k}{13} \qquad k = 1 \\ x &=& -\dfrac{\pi}{26}+\dfrac{2\pi}{13} \\ x &=& \dfrac{3\pi}{26} \\ \mathbf{x} &=& \mathbf{0.36249146003}\ \text{rad} \\ \hline \end{array}\)

The 3 smallest positive x-intercepts of the graph:
\(\mathbf{\dfrac{\pi}{26} },\ \mathbf{\dfrac{3\pi}{26} },\ \mathbf{\dfrac{5\pi}{26} } \)

 (b): Let f be of the form f(x) = (rx+s)/(2x+t).Find an expression for f(x).

Hello daddypig!

(b)

\(\color{blue}f(x)=\frac{1}{x-3}-4\\ f(x)=\frac{1}{x-3}-\frac{4(x-3)}{x-3}\\ f(x)=\frac{1-(4x-12)}{x-3} \)

              \(2x+t=x-3\\ t=-x-3\)

\(\large f(x)=\frac{-4x+13}{2x+(-x-3)}\)

r = - 4

s = 13

t = (- x - 3)

  !

Yes, the solve function seems to be disabled at present.  Don't know why.  With luck it will be up and running again soon!

(a): Let f be of the form f(x) = (ax+b)/(x+c).Find an expression for f(x).

Hello daddypig!

(a)

\(\color{blue}f(x)=\frac{1}{x-3}-4\\ f(x)=\frac{1}{x-3}-\frac{4(x-3)}{x-3}\\ f(x)=\frac{1-(4x-12)}{x-3}\)

\( f(x)=\frac{-4x+13}{x-3}\)   

\(\large f(x)=\frac{-4x+13}{x-3}\)  

a = - 4

b = 13

c = - 3

  !

This is \(\sum_{i=1}^34\times(\frac{1}{2})^{i-1}=4\times 1+4\times \frac{1}{2}+4\times \frac{1}{4}\)

You should be able to finish from here.

A mathematician works for \(t\) hours per day and solves \(p\) problems per hour, where \(t\) and \(p\) are positive integers and \(1 < p < 20\) .
One day, the mathematician drinks some coffee and discovers that he can now solve \(3p+7\) problems per hour.
In fact, he only works for \(t-4\) hours that day, but he still solves twice as many problems as he would in a normal day.
How many problems does he solve the day he drinks coffee?

Formula: \((3p+7)(t-4) = 2tp\)

\(\begin{array}{|rcll|} \hline (3p+7)(t-4) &=& 2tp \\ 3pt-12p+7t-28 &=& 2tp \\ pt-12p+7t-28 &=& 0 \\ t(p+7)-12p-28 &=& 0 \\ t(p+7)&=& 12p+28 \\ t(p+7)&=& 4(3p+7) \\ t &=& 4\left(\dfrac{3p+7}{p+7}\right) \\ t &=& 4\left(\dfrac{3p+21-14}{p+7}\right) \\ t &=& 4\left(\dfrac{3(p+7)-14}{p+7}\right) \\ \mathbf{t} &=& \mathbf{4\left(3-\dfrac{14}{p+7}\right)} \\ \hline \end{array} \)

\(t\) and \(p\) are positive integers

\(\begin{array}{|rcll|} \hline \dfrac{14}{p+7} &=& 0 \qquad t > 0 \checkmark \\ \dfrac{14}{p+7} &=& 1 \qquad t > 0 \checkmark \\ \dfrac{14}{p+7} &=& 2 \qquad t > 0 \checkmark \\ \dfrac{14}{p+7} &=& 3 \qquad t = 0 \\ \dfrac{14}{p+7} &>& 3 \qquad t < 0 \\ \hline \end{array} \)

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

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{14}{p+7}} &=& \mathbf{0} \\ 14 &=& 0(p+7) \\ 14 &=& 0 \qquad | \text{ does not yield any value of p.} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{14}{p+7}} &=& \mathbf{1} \\ 14 &=& 1(p+7) \\ 14 &=& p+7 \\ \mathbf{p} &=& \mathbf{7}\ \checkmark \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{14}{p+7}} &=& \mathbf{2} \\ 14 &=& 2(p+7) \\ 14 &=& 2p+14 \\ 0 &=& 2p \\ p &=& 0\qquad 1 < p < 20,\ \text{ no solution, p > 1!} \\ \hline \end{array}\)

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

\(\begin{array}{|rcll|} \hline t &=& 4\left(\dfrac{3p+7}{p+7}\right) \quad | \quad \mathbf{p=7} \\ t &=& 4\left(\dfrac{3*7+7}{7+7}\right) \\ t &=& 4\left(\dfrac{4*7}{2*7}\right) \\ t &=& 4\left(\dfrac{4 }{2 }\right) \\ t &=& 4*2 \\ \mathbf{t} &=& \mathbf{8} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 2tp &=& 2*8*7 \\ \mathbf{2tp} &=& \mathbf{112} \\ \hline \end{array}\)

The mathematician solved 112 problems the day he drank coffee.

A mathematician works for \(t\) hours per day and solves \(p\) problems per hour, where \(t\) and \(p\) are positive integers and \(1< p < 20\) .
One day, the mathematician drinks some coffee and discovers that he can now solve \(3p+7\) problems per hour.
In fact, he only works for \(t-4\) hours that day, but he still solves twice as many problems as he would in a normal day.
How many problems does he solve the day he drinks coffee?

Your graph is correct and your logic seems sound.

Thanks  MPS101

In the lab, Aldo has two solutions that contain alcohol and is mixing them with each other. Solution A is 7% alcohol and Solution B is 2% alcohol. He uses 300 milliliters of Solution A. How many milliliters of Solution B does he use, if the resulting mixture is a 5% alcohol solution?

James’s four-year-old brother is trying to arrange black and yellow Legos in rows.
He has 56 yellow Legos and 120 black Legos.
He wants to arrange them in rows so that there are either only yellow or only black Legos in each row,
and so that each row has the same number of Legos.
What is the least number of rows he can make?

\(\begin{array}{|rcll|} \hline gcd(56,120) &=& 8 \\ \dfrac{56}{8} &=& 7 \\ \dfrac{120}{8} &=& 15 \\ \hline \end{array}\)

The least number of rows he can have is 22;

7 rows of yellow and 15 rows of black.

Find the distance between the two points. (1,2) (7,6). 

Suppose that a and b are positive integers such that a-b=6 and  \(\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9 \) . Find the smallest possible value of b.

\(a-b=6\\ a=6+b\)

\(\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9\\ \text{gcd}\left(\frac{(a+b)(a^2-ab+b^2)}{a+b}, ab\right) = 9\\ \text{gcd}\left((a^2-ab+b^2), ab\right) = 9\\ \text{gcd}\left((a^2-2ab+b^2+ab), ab\right) = 9\\ \text{gcd}\left(((a-b)^2+ab), ab\right) = 9\\ \text{gcd}\left((6^2+ab), ab\right) = 9\\ \text{gcd}\left((36+ab), ab\right) = 9\\ \text{gcd}\left(9(4+\frac{ab}{9}), ab\right) = 9\\ \text{gcd}\left((4+\frac{ab}{9}), \frac{ab}{9}\right) = 1\\\)

So ab is a multiple of 9 AND ab is not even  AND   \(\frac{ab}{9}\)  is not even

If ab=9 then gcd is correct, but there are not values of a and b exactly 6 units apart

If ab=27 then gcd is correct, and 9*3=27,   9-3=6

So the smallest values of a and b are   9 and 3 respectfully. 

The smallest value of b is 3.

LaTex:

\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9\\
\text{gcd}\left(\frac{(a+b)(a^2-ab+b^2)}{a+b}, ab\right) = 9\\
\text{gcd}\left((a^2-ab+b^2), ab\right) = 9\\
\text{gcd}\left((a^2-2ab+b^2+ab), ab\right) = 9\\
\text{gcd}\left(((a-b)^2+ab), ab\right) = 9\\
\text{gcd}\left((6^2+ab), ab\right) = 9\\
\text{gcd}\left((36+ab), ab\right) = 9\\
\text{gcd}\left(9(4+\frac{ab}{9}), ab\right) = 9\\
\text{gcd}\left((4+\frac{ab}{9}), \frac{ab}{9}\right) = 1\\

The coordinates are (48/3, 12).

The area of triangle DMT is 15.

X = (3,-1) and k = 14.

I got that angle NZM + angle NPM is 135 degrees.

The number of ways is c(12,5) - c(7,2)*c(5,2) = 582.

(a) f(x) = (3x + 2)/(x + 7)

(b) f(x) = (4x + 1)/(2x - 3)

Cross-multiplication.

\(6(3x-6)=3(7x-3)\)

Divide both sides by 3

\(2(3x-6)=(7x-3)\)

Expand left bracket

\(6x-12=7x-3\)

Add 12

\(6x=7x+9\)

subtract 6x

\(x+9=0\)

Subtract 9

\(x=-9\)

Melody, CPhill, geno, anyone... could you verify this graph? Thanks.

So first we start off by making a diagram as shown above.

then, we split the quadrilateral into 2 pieces: triangle ABC and triangle ACD.

Triangle ABC is a right triangle, and use the Pythagorean Theorem to find the lengths. Then, you can find out the length of the base and height of the triangle ABC. You can find the area from there.

Next, find the length of AD and the length of C brought down. You can find the area from there.

After you're done with those, add them up.

You're done!

I know that you can do this from here. Good luck! Also, if you have any questions, don't be hesitant to ask!

-MPS101

v = <-9, 2> is transformed using the transformation matrix A = [-4, 1, 1, 2] What is the resulting vector? Show all your work.

We have the vector <-9, 2>, so we just apply the matrix A to this vector. We get:

A<-9,2> = [-4, 1, 1, 2]<-9,2> = <-4*-9 + 1 * 2, 1 * -9 + 2 * 2 > = <36 + 2, -9+4> = <38,-5>

Sorry for the formatting, I am new around typsetting languages for mathematics like LaTeX so I'm still learning how to do things(like matrices :/) 

  area = 2   x   1/2 b h   = b h

I did    BUT there are   TWO triangles in this method of solution    2  x 1/2  = 1  

I just used the wrong base   should be   2.3511   not  4.70228      Thanx for correction !