Melody

I wanted to do this answer exact but it is getting all too difficult and I don't know if what I have done is correct anyway. 

$$\\ED=1.5h=1.5*112\approx 168\\
KD=\sqrt3*h = \sqrt3*112\approx 194\\
\triangle AKD \equiv \triangle HKD \;\;\; $Three equal side test$\\\\
so\;\; KDA=30^0\\\\
But $Now consider triangle EKD$\\
EK^2=168^2+194^2-2*168*194*cos(78.2)$$

$${\sqrt{{{\mathtt{168}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{194}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{168}}{\mathtt{\,\times\,}}{\mathtt{194}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{78.2}}^\circ\right)}}} = {\mathtt{229.194\: \!522\: \!964\: \!077\: \!551\: \!3}}$$

$$\\EK \approx 229.2\\\\
Let \;\;\alpha $ be the angle of elevation sort, that is, $ tan\alpha = \frac{HK}{EK}\\\\
tan\alpha = \frac{h}{229.2}\\\\
tan\alpha = \frac{112}{229.2}\\\\
\alpha=tan^{-1}\frac{112}{229.2}\\\\$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{112}}}{{\mathtt{229.2}}}}\right)} = {\mathtt{26.042\: \!733\: \!855\: \!708^{\circ}}}$$

So if i have not made any stupid mistakes (which I probably did)  then  the minimum angle of elevation is approx 26 degrees and it is from the point E

a)

AK=h     (right isosceles triangle AKH)

$$\\tan30=\frac{HK}{KD}\\\\
\frac{1}{\sqrt3}=\frac{h}{KD}\\\\
KD=\sqrt3\;h\\\\\\
AH^2+KD^2=AD^2\\
h^2+(\sqrt3\;h)^2=AD^2\\
AD^2=h^2+3h^2\\
AD^2=4h^2\\
AD=2h$$

b)

It takes 20 min to walk AD

15min to walk DE

and 15min to walk AE

so triangle ADE is isoceles and the ratio of the sides is 4:3:3

AD is 2h  so   DE=AE= (2h/4)*3 = 1.5h

Usine cosine rule 

$$\\cos(E)=\frac{3^2+3^2-4^2}{2*3*3}\\\\
cos(E)=\frac{9+9-16}{18}\\\\
cos(E)=\frac{1}{9}\\$$

$$\\$Area of \triangle ADE = $0.5*AE*ED*sinE$$

$$\\6272\sqrt5=0.5*1.5h*1.5h*sin(cos^{-1}\;\frac{1}{9}\;)\\\\
6272\sqrt5=0.5*1.5h*1.5h*\frac{\sqrt{80}}{9}\\\\
6272\sqrt5=\frac{1}{2}*\frac{3h}{2}*\frac{3h}{2}*\frac{4\sqrt{5}}
{9}\\\\
6272\sqrt5=\frac{9h^2}{8}*\frac{4\sqrt{5}}{9}\\\\
6272=\frac{h^2}{2}\\\\
h^2=12544\\\\
h=112$$

I simultaneously worked the problem and wrote the code so there could easily be error.

h did turn out to be a 'nice' number, so maybe it is right :)))

Feel free to ask questions 

I just realized that it is not finished yet.

I think the minumum angle of elevation TO H   is from E because that point is furthest away

I need to find distance EK

$$\\KD=\sqrt3*h\\
AE=ED=1.5h$$

This is not just falling out and I don't have more time now.  I shall return if another doesn't answer first. 

with a car dealership price of $19,192, excluding sales tax the sales tax rate is 8.375%

a. what growth factor is associated with sales tax rate?  1+0.08375 = 1.08375   (or  108.375%)

b.total price of the car is $?      $19,192 * 1.08375 = and you can finish it :)

-20-12 = -32

Does that mean that you are double the trouble?    LOL

Looks good to me Chris :)

Desmos is for graphing not for solving.  So there is no mystery about that :))

Yes that is true :))

Thanks Chris,

I often forget how clever our own site calculator is. 

Thank you Mr Massow for making it for us.  

y=(-6/y)²-1

$$\\y=\left(\frac{-6}{y}\right)^2-1\\\\
y=\frac{36}{y^2}-1\\\\
y^3=36-y^2\qquad y\ne 0\\\\
y^3+y^2-36=0\\\\$$

now I am going to use reaminder theorum to look for a factor

If y=3 then  27+9-36=0  great (y-3) is a factor and 3 is a root

$$\\(y^3+y^2-36)\div (y-3)=y^2+4y+12 \qquad $I used polynomial division to get this$\\\\
so\\
y^3+y^2-36=(y-3)(y^2+4y+12)\\\\\\
$Now find roots of $y^2+4y+12\\\\
\triangle = 16-48<0\qquad $therefore there are no real roots to this$\\\\
so\\
$The only real solution is $y=3$$

check