Welcome to the forum.

This doesn't seem to be a linear differential equation, but just an equation.

ae^x cannot be zero so you can cancel it out. What's left simplifies quickly to a solution.

Bob

]]>https://www.youtube.com/watch?v=FTAyNkwrPys

]]>The principle is rather easy: with a base font size, create a sequence of font sizes give a chosen ratio:

```
scale
ratio factor description
----- ------ -----------
16:15 1.067 minor second => growing a font size nice and slow
9:8 1.125 major second
6:5 1.2 minor third
5:4 1.25 major third
4:3 1.333 perfect fourth => still appealing, but already growing too fast for comfort
√2:1 1.414 aug. fourth / dim. fifth
3:2 1.5 perfect fifth
8:5 1.6 minor sixth
1.618:1 1.618 golden ratio (ϕ)/Fibonacci
5:3 1.667 major sixth
16:9 1.778 minor seventh
15:8 1.875 major seventh
2:1 2 octave
5:2 2.5 major tenth
8:3 2.667 major eleventh
e 2.718 Euler's number (2.718281828..)
3:1 3 major twelfth
4:1 4 double octave => massive effect as each next font size is 4 times larger
```

Current math used:

**My question is: what mathematical way can I use to slow down the growth of a font size given some ratio.Something like: the higher the 'scale factor', the larger the 'slow down factor'.**

As this is for CSS, the only usable math operators are: + - / *

This is the current (pseudo) code (working as expected, btw):

```
size0 = basesize * a11y-scale
size1 = size0 * scalefactor
size2 = size1 * scalefactor
size3 = size2 * scalefactor
etc...
size-1 = size0 / scalefactor
size-2 = size1 / scalefactor
size-3 = size2 / scalefactor
etc...
```

Combined with a generic *document scale factor* I will be using this to create a tutorial on how to make a website more accessible (The a11y Project) to the **Vision Impaired**.

So, much obliged in advance!

]]>Welcome to the forum.

If you have worked out h (well done!) then you know the full equation of both circles, and hence the centre of C2.

This means you can work out the equation of AB, and so you can substitute for y in each circle equation and so work out the coordinates of A and separately, B. This will also give you the coordinates of the second point on AB where it crosses C2, lets call it point D.

Because AB gives the line that is a diameter of each circle, the tangents must go through A and D, so you have all you need to form the equations of these new circles, centre and a point on the circumference.

I haven't tried the question yet. If you are still having trouble with any part, post again and I'll try to help some more.

Bob

]]>Welcome to the forum.

Compuhigh has asked us not to publish their copyright worksheets so I have removed the questions you posted. They have better lawyers than we do so it's best not to upset them.

We will certainly help you with any topic in general but we won't do your homework for you.

For questions like these it is essential to include a diagram too.

Looking forward to helping once these matters are sorted out.

Bob

]]>I suddenly like this idea of making a larger thing smaller

I will start to practice from arithmetic when i get some free time .

And i will come here and ask if i have more questions

Thanks a lot .

]]>Latin grammar sounds interesting, I have not had the occasion to encounter it until now. Yes, I do suspect a typographical error.

As for ‘commonly encountered in the Workplace’, I can only guess. Either these Latin proverbs WERE common 95 years ago or the publisher overestimated the number of Latin proverbs the common office-goer is aware of and uses frequently. If I remember rightly, the list included over 120 proverbs, both Latin and of languages other than English such as French and German.

The idea for the signature sounds promising. I’ll try thinking of something along those lines.

]]>Welcome to the forum.

Are you sure you've posted the problem correctly? As it stands, I can't see how the latter statement is true. We have:

and sosince obviously But then the problem statement suggests that we should then have whereas the polynomial doesn't have any real roots. (You can show this by differentiating it once to find the stationary point, then again to show that it's a minimum -- then showing that at this minimum, is strictly positive, i.e. doesn't ever cut theAttempt to continuously hone my skills in mathematics, programming, and more broadly problem solving.]]>

Hello,

See:

https://i.postimg.cc/rs1py38L/mathclarificationopaque.png

thank you a lot

]]>**Welcome to the forum!**