$$\small{
\begin{array}{rrrrrrr}
1.& 85g =& & 4 \times 15g &+ 1 \times 25g& \\
2.& 85g =& 1 \times 5g &+ 2 \times 15g & & + 1 \times 50g\\
3.& 85g =& 1 \times 5g &+ 2 \times 15g & + 2 \times 25g \\
4.& 85g =& 2 \times 5g & & + 1 \times 25g & + 1 \times 50g\\
5.& 85g =& 2 \times 5g & & + 3 \times 25g \\
6.& 85g =& 2 \times 5g &+ 5 \times 15g \\
7.& 85g =& 3 \times 5g &+ 3 \times 15g & + 1 \times 25g \\
8.& 85g =& 4 \times 5g &+ 1 \times 15g & & + 1 \times 50g\\
9.& 85g =& 4 \times 5g &+ 1 \times 15g & + 2 \times 25g \\
10.& 85g =& 5 \times 5g &+ 4 \times 15g \\
11.& 85g =& 6 \times 5g &+ 2 \times 15g & + 1 \times 25g \\
12.& 85g =& 7 \times 5g & & & + 1 \times 50g \\
13.& 85g =& 7 \times 5g & & + 2 \times 25g \\
14.& 85g =& 8 \times 5g &+ 3 \times 15g \\
15.& 85g =& 9 \times 5g &+ 1 \times 15g & + 1 \times 25g \\
16.& 85g =& 11 \times 5g &+ 2 \times 15g \\
17.& 85g =& 12 \times 5g & & + 1 \times 25g \\
18.& 85g =& 14 \times 5g &+ 1 \times 15g \\
19.& 85g =& 17 \times 5g \\
\end{array}
}$$

$$ln (6.2)= log_e (6.2)$$

or in words it is  log base e of 6.2

$${ln}{\left({\mathtt{6.2}}\right)} = {\mathtt{1.824\: \!549\: \!292\: \!051\: \!045\: \!9}}$$

---------------------------------------------------------

e is a special number (with special properties) just like pi is.   Both these numbers are irrational numbers.

$${\mathtt{e}} = {\mathtt{2.718\: \!281\: \!828\: \!459\: \!045\: \!2}}$$

That is the approx value of e.  It cannot be written exactly as a decimal or as a fraction because it is irrational 

copy it straight into the home page calc

-32.9^2-4(.35)(-925)

you may need to add the times signs and the brackets around the power  

-32.9^(2)-4*(.35)*(-925)   press =

arc sine is the same as inverse sine so use    asin

The brackets are not balanced   

anyway you can just type it straight into the calc on the home page.

Only use round brackets

[1,000/(800]1/20 - 1

becomes

(1000/800)*1/20 - 1

--------------------------------------

actually now I am looking at it it looks simple

(1000/800)*1/20 - 1

=5/4)*1/20 - 1

1/4*1/4-1

1/16-1

-15/16

See the question was: Height of a right triangle without area :

$$\\(1) \quad Area = \dfrac{h_c*c}{2} \\\\
(2) \quad Area = \dfrac{a*b}{2}\\\\
\dfrac{h_c*c}{2} = \dfrac{a*b}{2} \\\\
h_c*c = a*b \\\\
h_c= \dfrac{a*b}{c} \\\\$$

log to base e is also known as natural logarithm, ln

log(e) = 1

where log denotes log to base e

😾 😾

Use the home page calculator :)

Type

atan(0.75)    and press equals 

arc tan is the same as inverse tan

Inverse sine is the same as arc sine 

So on the web2 calc you use asin

$$\\x=\sqrt6*20 \\
x=20\sqrt6$$

you cannot simplify this any more.    

Well that is an interesting little observation.  Thank you 

$$\frac{3^{-1}}{4}=\frac{1*3^{-1}}{4}=\frac{1}{4*3^{+1}}=\frac{1}{4*3}=\frac{1}{12}$$

METHOD:

Any number or letter  that is raised to a negative indice is swapped to the other side of the equal sign and the neg is changed to a positive.

$$\\\sqrt{128}=\sqrt{4*32}\\
=\sqrt{4*16*2}\\
=\sqrt{4}\sqrt{16}\sqrt{2}\\
=2*4*\sqrt{2}\\
=8\sqrt{2}$$

Whats the pattern for 1 -3 9 ? 81 ? -243. ? ?

$$\begin{array}{lrcl}
n& & & \\
1& 1 & = & 3^0 \\
2& -3 &=& (-3)^1 \\
3& 9 &=& 3^2\\
4& -27 &=& (-3)^3 \\
5& 81 &=& 3^4\\
6&-243 &=& (-3)^5\\
7&729 &=& 3^6\\
8&-2187 &=& (-3)^7\\
\dots
\end{array}
\\
\small{\text{
The formula is: $(-3)^{n-1}$
}}$$

It might be that the numerator and denominator each represent magnitude and phase angle (electrical engineers do this a lot).  Usually this would be written as: magnitude∠phase 

It means

 $$magnitude\times e^{i\times phase}$$

To do the division you divide the magnitudes and subtract the angle in the denominator from that in the numerator:

(14.73/26.08)∠(-37.66°-122.47°)  =  0.5648∠-160.13°

However, the poster of the question would need to say if this is what was intended.

. 

Assuming you know the length of all three sides you could make use of Heron's formula.  If the sides are of length a, b and c, and b is the base then 

$$\frac{1}{2}\times b\times h=\sqrt{s(s-a)(s-b)(s-c)}$$

where h is the height and s is the semi-perimeter (=(a+b+c)/2).  This can be rearranged to find h.

.

Another alternative is to note that, since the triangle is equilateral, all the angles measure 60 degrees

So, using  A  = (1/2)(s^2)[ sin60 ]   and letting s = 6, we have

A = (1/2)[(6)^2][√3/2]= (36/4)√3  = 9√3 sq units

Drop a perpendicular down from the apex and it will cat the base in half.

Now you have 2 right angled triangles  each has the hypotenuse of 6 and the short side of 3.  

The other side is the height.

Use pythagoras's theorum to get the height,   (or you could use trigonometry)

Now use the area of a triangle formula  A = 1/2 *b*h

------------------

OR

you could use Heron's formula.

I will let you google that one  

Really Heureka??  

 If I had more time I would try and work out how that could be demonstrated.

I did try a little but I had no success.  

I don't remember seeing anything like it before.   

Can you show me a proof of this?

Yes that is correct - thanks anon  

Take a look at this

6(x+1) = 6*(x+1)

3xy=3*x*y

If there is an operator missing it will be an invisable  multiply sign   

The intersection is where all thes graphs are true at the same time.