Alan

$$$$\frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}$$\\
Multiply top and bottom of 1st term by $\sqrt2$\\
$$\frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}=\frac{\sqrt2c}{\sqrt2\sqrt{2x}}+\frac{y}{2\sqrt{x}}$$\\
The denominator of the first term can be rewritten as $\sqrt2\sqrt{2x}=2\sqrt{x}$ so
$$ \frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}=\frac{\sqrt2c}{2\sqrt{x}}+\frac{y}{2\sqrt{x}}$$\\
Both terms now have a common denominator so:
$$\frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}=\frac{\sqrt2c+y}{2\sqrt{x}}$$

Let h be the cost of a highlighter.  We are told that:

6*h + 6*3.48 = 23.06

Subtract 6*3.48 from both sides:

6*h = 23.06 - 6*3.48

Divide both sides by 6:

h = (23.06 - 6*3.48)/6

$${\mathtt{h}} = {\frac{\left({\mathtt{23.06}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{3.48}}\right)}{{\mathtt{6}}}} \Rightarrow {\mathtt{h}} = {\mathtt{0.363\: \!333\: \!333\: \!333\: \!333\: \!3}}$$ dollars

or, multiplying by 100 to get cents:

h = 36 cents to the nearest cent.

I think this should be written as (cot²B - cos²B)/(csc²B - 1) 

$$$$\frac{\cot^2B-\cos^2B}{\csc^2B-1}=\frac{\frac{\cos^2B}{\sin^2B}-\cos^2B}{\frac{1}{sin^2B}-1}$$\\
$$=\frac{\cos^2B-\sin^2B\cos^2B}{1-\sin^2B}$$\\
$$=\frac{\cos^2B(1-\sin^2B)}{1-\sin^2B}$$\\
$$=\cos^2B$$

$$$$\sin^4{A}-\cos^4{A}=(\sin^2{A}-\cos^2{A})(\sin^2{A}+\cos^2{A})$$\\
\\
$$\sin^2{A}+\cos^2{A}=1$$\\
$$\sin^2{A}-\cos^2{A}=1-\cos^2{A}-\cos^2{A}=1-2\cos^2{A}$$\\
\\
So, using the last two lines in the first line we have:\\
$$\sin^4{A}-\cos^4{A}=1-2\cos^2{A}$$

This is just a single point at x-coordinate 6 and y-coordinate 4.

5n² = -42n

One solution is n = 0

Another is found when n is not 0. In that case you can divide both sides by n to get

5n = -42

Divide both sides by 5

n = -42/5

or n = -8.4

Enter 1 + 3/4 + 5 + 4/9 in the calculator.

Type alt 0176 on the numeric keypad (keeping your finger on the alt key while you type the numbers).

If tan(A)=0.333 then A = tan^-1(0.333)  (on a calculator tan^-1 is usually represented by atan) so:

$${\mathtt{A}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.333}}\right)} \Rightarrow {\mathtt{A}} = {\mathtt{18.417\: \!758\: \!370\: \!539^{\circ}}}$$

A ≈18.4° This is the result in first quadrant.

Because tan also has the same value in the 3rd quadrant the solution could also be A = 180°+18.4° = 198.4°

Not quite rosala.

If 5=6/x, multiply both sides by x to get 5x = 6.

Now divide both sides by 5 to get x = 6/5 (or x = 1.2)

Do it one step at a time.