Melody

In Latex I just use       ^\circ

\(18^\circ\)

Whatever works for you is good.

I am sure some people will take note of what you have suggested and learn from you.  

I actually have only heard of the 1st one and I have never used any of them.

If I could remember them they would be very useful.

Please do not give answers only.

We do not want to overtly encourage cheating and how is that supposed to help someone learn?

It is better to give partial answers.

How did you come by those values Geno?

Ok,

It is a learning curve.

I actually did not think you were intentionally rude. 

I have seen how you have behaved here and I think you are a valuable addition to our forum.;

Next time you want to repost please do not add the full question again.

Just add a link to the original question,

say you still would still very much appreciate an answer

 and ask people to answer on the original thread.

Once the original post is answered you can put a note on both to say thanks to the answerer.

Why have you deleted your question Helpmeeeee?

I can see it has not been answered but even so this greatly discouraged as it is not polite.

Can you please put it back.

If you have the answer you could add that in as an answering post.

\(48∗4^𝑥+27=𝑎+𝑎∗4^𝑥+2\\ 48∗4^𝑥-𝑎∗4^𝑥=𝑎+2-27\\ 4^x(48-a)=𝑎-25\\ 2^{2x}(48-a)=𝑎-25\\ 2^{2x}=\frac{𝑎-25}{48-a}\\ log_2{2^{2x}}=log_2\frac{𝑎-25}{48-a}\\ 2x=log_2\left(\frac{𝑎-25}{48-a}\right)\\ x=\frac{1}{2}log_2\left(\frac{𝑎-25}{48-a}\right)\\\)

BUT you cannot find the log of a negative number so 

\(\frac{𝑎-25}{48-a}>0 \qquad and \quad 48-a\ne0\\ \frac{𝑎-25}{48-a}>0 \qquad and \quad a\ne48\\ 25

\(x=\frac{1}{2}log_2\left(\frac{𝑎-25}{48-a}\right)\qquad where\;\;25

Here is the graph

What is the standard form of the quadratic function that has a vertex at (3,-2) and goes through the point (4,0)?

Here is the easy way:

the axis of symmetry is x=3 with vertex at y=-2

one root is at 4   

3+1=4

3-1=2

so the other root is at 2

\(y=k(x-2)(x-4)\\ sub\;in\; (3,-2)\\ -2=k*1*-1\\ k=2\\ y=2(x-2)(x-4)\\ y=2(x^2-6x+8)\\ y=2x^2-12x+16\)

Or if you prefer the harder way.  then...

\((x-h)^2=4a(y-k)\\ (x-3)^2=4a(y+2)\\ sub\;\; in\;\; (4,0)\\ (4-3)^2=4a(0+2)\\ 1=8a\\ a=0.125\\~\\ (x-3)^2=4*0.125(y+2)\\ x^2-6x+9=0.5y+1\\ x^2-6x+8=0.5y\\ y=2x^2-12x+16 \)

Coding:

y=k(x-2)(x-4)\\
sub\;in\; (3,-2)\\
-2=k*1*-1\\
k=2\\
y=2(x-2)(x-4)\\
y=2(x^2-6x+8)\\
y=2x^2-12x+16

(x-h)^2=4a(y-k)\\
(x-3)^2=4a(y+2)\\
sub\;\; in\;\; (4,0)\\
(4-3)^2=4a(0+2)\\
1=8a\\
a=0.125\\~\\
(x-3)^2=4*0.125(y+2)\\

x^2-6x+9=0.5y+1\\

x^2-6x+8=0.5y\\

y=2x^2-12x+16

What have you done to solve this?

Which quadrants can 3x be in?

If you let    \(\theta=3x\)    Can you draw a right-angled triangle where     \(\csc(\theta)=\frac{2\sqrt3}{3}\)

answer what you can and then get back to me with your responses.

Please no one else come in with a more full answer

     unless guest responds with more than just a 'nothing' answer.