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The size of a certain population of wild horses varies because of certain ecological pressures.
In 1999, the number of horses was 100.
In 2002, the population was down to 60,
but in 2005, the population was back up to 100.

Assume that 100 was the maximum number of horses and 60 was the minimum.
Write an equation that models the size of the population of horses,
in terms of what year it is,
starting with 1999.

\(\text{Let $x$ is the year}\)

\(\begin{array}{|rcll|} \hline f(x) &=& 20\sin\left(\frac{\pi}{3}(x-1999) + \frac{\pi}{2} \right ) + 80 \quad & | \quad \pi \mathrel{\hat{=}} 180^{\circ} \\\\ &=& 20\sin\left(\frac{180^{\circ}}{3}(x-1999) + \frac{180^{\circ}}{2} \right ) + 80 \\ \mathbf{f(x)} & \mathbf{=} & \mathbf{ 20\sin( 60^{\circ}(x-1999) + 90^{\circ} ) + 80 } \\ \hline \end{array}\)

we can reduce the the problem to the following

\((\text{length of the blue line})^2 =2^2 + (\text{length of the red line }-1)^2\)

\((\text{length of blue line})^2 = (t-2)^2 + (t-2)^2 = 2(t-2)^2\\ (\text{length of red line}-1)^2 =(\sqrt{ (t-3)^2+t^2}-1)^2 \\ \text{a modest amount of algebra gets us }\\ 6+2t = 2\sqrt{2 t^2-6 t+9}\\ 3+t=\sqrt{2 t^2-6 t+9}\\ 9+6t+t^2 = 2t^2 - 6t+9\\ 12t=t^2\\ t=12 \)

2 ways

1) click tan and hold it, the button will change to atan

2) click on the button labeled "2nd", the buttons will all change including the tan button

\(\text{the hole is created by }\dfrac{x-4}{x-4}\\ \text{the vertical asymptote is created by }\dfrac{1}{x-2}\\ \text{as we end up with a single power of }x \text{ in the denominator }\\ \text{the horizontal asymptote is created by }2x^2+x \\ \text{combining all this we have}\\ f(x) = \dfrac{(x-4)(2x^2+x)}{(x-4)(x-2)} = \dfrac{x(x-4)(2x+1)}{(x-4)(x-2)}\)

\(f(x) = \dfrac{2x^3-7x^2-4x}{x^2-6x+8}\)

\(\text{veranda }=320 cm\times 640cm\\ \text{we want to find the greatest common factor of }320 \text{ and }640\\ 320=5\times 2^6\\ 640=5 \times 2^7\\ \text{so the LCM of the two is }5\times 2^6 = 320\\ \text{we use square tiles }320cm \times 320cm = 3.2m \times 3.2m\\ \text{we'll need 2 tiles}\)

Enter 470/27 into the calculator.
You should get 17.41 yards/minute.

If using sig fig rules, you should get 17 yards/minute.

If you cut the fabric once, you end up with 2 parts.
If you cut the fabric two times, you end up with 3 parts. 

If you want sections that are 2 yards long, you need to end up with
20yards/2yards = 10 parts.
 

If you cut the fabric nine times, you end up with 10 parts of 2 yards each. 

\(\text{I'm not seeing anything clever}\\ \begin{array}{lr} z_1=1+i &|z_1-i| = 1\not < \dfrac 1 2\\ z_2 = \dfrac{2}{2}+i = 1+i &|z_2-i|=1 \not < \dfrac 1 2 \\ z_3 = \dfrac{2}{3}+i &|z_3-i| = \dfrac 2 3 \not < \dfrac 1 2\\ z_4 = \dfrac{13}{36}+i &|z_4-i| = \dfrac{13}{36} < \dfrac 1 2 \end{array}\\ \text{the least value of }n \text{ such that }|z_n-i|<\dfrac 1 2 \text{ is }n=4\)

Thank you so much!!! It helped a ton!

what you do with these is multiply top and bottom by the complex conjugate of the denominator

\(\dfrac{7+9i}{1+3i}=\dfrac{7+9i}{1+3i} \dfrac{1-3i}{1-3i} = \\ \dfrac{(7+27)+(9-21)i}{10} = \dfrac{34-12i}{10} = \\ \dfrac{17}{5} -\dfrac 6 5 i\)

well...

the answer is t=12

The algebra involved is formidable w/o software so there is probably a very clever geometric solution.

Stay tuned.

Another means:

x+y = 12    so x = -12-y

x*y=-45     Sub x=12-y in to this

(-12-y)y =-45

-y^2-12y +45 = 0 

y^2+12y-45=0            now use quadratic formula to find x = -15 and 3

3.2 = 32/10 = 16/5  as an improper fraction

The corners of the square are

 \(\left(\pm \dfrac s 2, \pm \dfrac s 2\right) so \\ \dfrac{s^2}{12}+\dfrac{s^2}{24}=1\\ 3s^2 = 24 \\ s^2 = 8 =\text{ area of the square}\)

This is just: \(3\frac{2}{10}=\boxed{3{\frac{1}{5}}}\) .

\((x+y)=-12\\ xy = -45\\ y=-12-x\\ x y = x(-12-x)=-45\\ x(12+x) = 45\\ x^2 + 12x -45 = 0\\ (x+15)(x-3) = 0\\ x=-15 \text{ or }x=3 \\ y=3 \text{ or }y=-15\\ \text{so we see that the two numbers are }-3 \text{ and }15\)

\(\text{Assume the corner is }(0,0)\\ \text{The equation of the table edge is }\\ (x+r)^2 + (y+r)^2 = r^2\\ \text{we are given that the point }(-18,-25) \text{ is on the table edge}\\ (-18+r)^2 + (-25+r)^2 = r^2\\ r^2 - 86r + 949 = 0\\ (r-73)(r-13) = 0\\ r=73 \text{ or }r=13\)

Both these answers are correct.  If the radius is 73 in then the mark is to the right of the center of the circle.  If the radius is 13 in then the mark is to the left of it.

Here is the graph:

I come up with r=13

Since the circle sits agains both walls (the x and y axes), the center  h,k is where   x = y

  the distance from the center to the point is the radius length and is also equal to the distance from the y axis to the center ...namely x    equate these

sqrt [(x-25)^2 + (x-18)^2] =  x    square both sides

(x-25)^2 + (x-18)^2 = x^2

x^2 -50x+625 + x^2 -36x+324 = x^2

x^2-86x+949 = 0     

    Quadratic formula finds x = 73 or 13      73 is extraneous (too big) so x = 13

h,k then = 13,13   and radius = 13

(x-13)^2 + (y-13)^2 = 13^2          r= 13  

Good point from Rom:

Both these answers are correct.  If the radius is 73 in then the mark is to the right of the center of the circle.  If the radius is 13 in then the mark is to the left of it.       Although that would be a big table with ~ 12 foot diameter! But possible....

So the Mode is the number(s) that repeat the most frequently. For example 1, 2, 4, 4, 8, the mode in this set of numbers would be 4 because there are two of them. Does that clear anything up?

Asset A costs 85   returns 100 in a year for an increase $ 15    15/85= 17.65%  ROR

          B           95                                                                  5       5/95 =  5.26% ROR