Melody

Try watching this

$$y=2log_3(x-4)+5$$

You can BUILD this graph, 

you need to have some idea of what  $$y=log_3x$$     looks like.

You should know the basic shape of a log function and know that it will go through the point (1,0)

also since it is  the same as  $$x=3^y$$    other points will be (3,1),  (9,2), (1/3,-1)  etc

Once you have drawn   $$y=log_3x$$    you can start manipulating it.

To get   $$y=log_3(x-4)$$     you just move the whole graph 4 places to the right  (positive direction)

then

To get   $$y=2log_3(x-4)$$     all the y values are doubled so the range is doubled.

then

to get    $$y=2log_3(x-4)+5$$    just move the whole graph UP 5 places.

I have done these steps here

Great job Chris,

This venn diagram was a little tricky :)

Yep,

1/4 of 36 = 9

idk Eloise, Maybe one home gets all 10 elephants  ://     LOL

Thanks Alan,

X=Y^2-(10Y)+40

This is a sidways parabola. I know because it is the y that is squared.

It opens out in the positive direction. I know that because the coefficient of y^2 is 1 which is POSITIVE.

Now if I can get it into the form

$$(y-k)^2=4a(x-h)$$ 

then the vertex will be   (h,k)

$$\\X=Y^2-(10Y)+40\\\\
X-40=Y^2-(10Y)\\\\
Y^2-(10Y)=X-40\\\\
Y^2-(10Y)+25=X-40+25\qquad $I have completed the square$\\\\
(Y-5)^2=X-15\\\\
$Vertex is (15,5)$$$

This the same answer as shown in Alan's graph.   :)

Thanks Civonamzuk,

It should  be noted that you have assumed that the unknown side is the hypotenuse.

This may not be the case :)

Good thinking Chris !

It would have been nice if you had given us the jist of your logic anon :)

6*y^10-(3*y^6)+((8*y^4)/3y^2

You brackets are not quite right but I think that this is what you mean :)

$$\\6*y^{10}-(3*y^6)+((8*y^4)/(3y^2))\\\\
6y^{10}-3y^6+\frac{8y^4}{3y^2}\\\\
6y^{10}-3y^6+\frac{8y^2}{3}\\\\$$

This cannot be simplified any further.

Thanks Chris. :)