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The cubic is y = -x^3 - 36x.

Idk-

Hi Melody and asinus,

Thank you for you help!

I have plotted in f and g onto Desmos but I am unsure what to do now. How should I continue?

4 sqrt 9 for what?

Because it asks for side lengths

3/4 is the constant term....it has no variable (m,n x, y   etc )  in it

The laws of logarithms that you need are as follows:

ln(a*b) = ln(a) + ln(b)    (and ln(a/b) = ln(a) - ln(b))

ln(aⁿ)  = n*ln(a)

ln(e) = 1  

See if you can take it from here.

Arc AB = 180 - 2 * 35 = 110 degrees

Rubbish Pangolin,

Guest gave you some good advice.

Draw it and count!

You have made an error asinus, can you find it?

Here is another way, (though Heron's formula is probably easier.

Let X be the point of intersection of the circle and the tangent

Let r be the radius

Now you have triangles BCX  and  BAX

The angles at X are right angles so you can use pythagoras's theorum to find XC  and  XA  in terms of r

added they equal 12

solve

(it is a bit tricky)

Let f(x) and g(x) be functions with domain $(0,\infty)$ . Suppose $f(x)=x^2$   and the tangent line to f(x) at x=a is perpendicular to the tangent line to g(x) at x=a for all positive real numbers a. Find all possible functions g(x).

Sei f (x) und g (x) Funktionen mit Domäne $(0,\infty)$. $f(x)= x^2$ und die Tangentenlinie zu f (x) sei bei x = a  für alle positiven reellen Zahlen a senkrecht zur Tangentenlinie zu g (x). Finde alle möglichen Funktionen g (x).

Hello yeliah!

$f(x)=x^2\\ f'(x)=2x$

$g'(x)=-\frac{1}{f'(x)}\\ g'(x)=-\frac{1}{2x}$

$g(x)=-\frac{1}{2}\cdot \int x^{-1}\cdot dx$

$g(x)=-\frac{1}{2}\cdot ln |x|+C$

$\{possible\ g(x)\}\in\{(-\frac{1}{2}\cdot ln |\mathbb Q|+\mathbb Q)\}$  Is this the correct answer?     

[Added by Melody:  No, this isn't right.  Look at my next post for other points]

The tangent line to f(x) at  x=a $\large ?$ is perpendicular to the tangent line to g(x) at x=a $\large ?$ for all positive real numbers a. $\large ?$ Find all possible functions g(x).

I do not understand this text. Is a the same as | x |? Why?

  !

Let f(x) and g(x) be functions with domain $(0,\infty)$. Suppose $f(x)=x^2$  and the tangent line to f(x) at x=a is perpendicular to the tangent line to g(x) at x=a for all positive real numbers a. Find all possible functions g(x).

ok you already have said that the GRADIENT OF THE TANGENT TO F(X) is

f'(x)=2x

So the gradient of the tangent to g(x) at any given point must be

   $g'(x)= \frac{-1}{2x}$

So you have to integrate that to get the value of   g(x)

Can you do that?

Do you completely understand ?

When you get an answer plot f and g on desmos.   Check you understand what happens.

Is it not  4sqrt(9)?? that would work like a charm

the sides would be  5,5,6  and  6,6,6

Hint:  Use Heron's formula to calculate the area, then use this formula  (h = 2*A / a) to find the radius.

The altitude from point B to line segment AC is the radius of a circle.

Please think your answer out before you post it. Five is impossible.

thank you so much!

I understand now!

Triangle ABC, inscribed in a circle, has AB = 15 and BC = 35. A tangent to the circle is drawn at B, and a line through A parallel to this tangent intersects BC at D. Find DC. 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Hello, Guest! I think that BC should be 25 instead of 35. If BC = 25 then DC = 16

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let angle ABC = 90º

Angle BAD = arctan(15 / 35)

BD = tan∠BAD * AB = 6.428571429

DC = BC - BD = 28.57142857

Start by dividing top and bottom by 2x

and then simplifying of course.

then show us what you get.

You would probably be best off to repost the second one

I'll answer the first one.

If   f'''  = 0 

then

f'' =k  \

k can be any6 constant but 6 is as good as another

f' = 6x+t     

t can be any constant but -8 is as good as any other

f' = 6x-8

f=   3x^2 -8x + q

q can be any constant but 1 is as good as any other

f=   3x^2 -8x + 1

So there is one.

In fact any quadratic f function would work

sorry, should I repost?

Please just ask one question per post.