# 0.10 Writing mathml  (Page 3/3)

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Step 5 : Looking at the equation, it is easy to note that we need to have two fraction term with nominator and denominator. Thus, copy the block to display vertical term i.e. copy and paste “mfrac” code lines 9 to12 and modify according to the requirement(new lines 17 to 20). Copy and paste code lines 13 to 16 and modify as required (new lines 21 to 24). Finally copy and paste code lines 9 to 12 again as shown(new lines 25 to 28).

1: <m:math display="block"> 2: <m:mi> x </m:mi> 3: <m:mo> = </m:mo> 4: <m:mo> [ </m:mo> 5: <m:mn> 3200 </m:mn> 6: <m:mo> ( </m:mo> 7: <m:mn> 1 </m:mn> 8: <m:mo> - </m:mo> 9: <m:mfrac> 10: <m:mn> 5 </m:mn> 11: <m:mn> 100 </m:mn> 12: </m:mfrac> 13: <m:mo> ) </m:mo> 14: <m:mo> ( </m:mo> 15: <m:mn> 1 </m:mn> 16: <m:mo> + </m:mo> 17: 9: <m:mfrac> 18: 10: <m:mn> 10 </m:mn> 19: 11: <m:mn> 100 </m:mn> 20: 12: </m:mfrac> 21: 13: <m:mo> ) </m:mo> 22: 14: <m:mo> ( </m:mo> 23: 15: <m:mn> 1 </m:mn> 24: 16: <m:mo> + </m:mo> 25: 9: <m:mfrac> 26: 10: <m:mn> 5 </m:mn> 27: 11: <m:mn> 100 </m:mn> 28: 12: </m:mfrac> </m:math>

What you have coded till now : $x=\left[3200\left(1-\frac{5}{100}\right)\left(1+\frac{10}{100}\right)\left(1+\frac{5}{100}$

What is to be coded ultimately : $x=\left[32000\left(1-\genfrac{}{}{0.2ex}{}{5}{100}\right)\left(1+\genfrac{}{}{0.2ex}{}{10}{100}\right)\left(1+\genfrac{}{}{0.2ex}{}{\frac{25}{2}}{100}\right)\right]$

Step 6 : Note that last “mfrac” display requires another “mfrac” implementation. Its numerator itself is a “mfrac” display. Thus, we would require to change the line 26 placed near bottom with a block of “mfrac” codes that shall display “25/2”. For this we replace code line 26 with code block consisting of code lines 9 to 12. In order to make this block as the numerator of the parent “mfrac” element, we need to enclose this block with “mrow” element so that “mfrac” element considers the block as one argument. Change the “mo’ content (lines 10 and 11 at the bottom) to reflect the ratio as 25/2. Finally, add a closing parenthesis and a bracket at the end (lines 29 and 30) as shown.

13: <m:mo> ) </m:mo> 14: <m:mo> ( </m:mo> 15: <m:mn> 1 </m:mn> 16: <m:mo> + </m:mo> 17: 9: <m:mfrac> 18: 10: <m:mn> 10 </m:mn> 19: 11: <m:mn> 100 </m:mn> 20: 12: </m:mfrac> 21: 13: <m:mo> ) </m:mo> 22: 14: <m:mo> ( </m:mo> 23: 15: <m:mn> 1 </m:mn> 24: 16: <m:mo> + </m:mo> 25: 9: <m:mfrac> ---- <m:mrow> 9: <m:mfrac> 10: <m:mn> 25 </m:mn> 11: <m:mn> 2 </m:mn> 12: </m:mfrac> </m:mrow> ----- 27: 11: <m:mn> 100 </m:mn> 28: 12: </m:mfrac> 29: <m:mo> ) </m:mo> 30: <m:mo> ] </m:mo> </m:math>

Step 7 : Save the file as test.xml. The code (after renumbering for reference purpose) at this stage looks like :

13: <m:mo> ) </m:mo> 14: <m:mo> ( </m:mo> 15: <m:mn> 1 </m:mn> 16: <m:mo> + </m:mo> 17: <m:mfrac> 18: <m:mn> 10 </m:mn> 19: <m:mn> 100 </m:mn> 20: </m:mfrac> 21: <m:mo> ) </m:mo> 22: <m:mo> ( </m:mo> 23: <m:mn> 1 </m:mn> 24: <m:mo> + </m:mo> 25: <m:mfrac> 26: <m:mrow> 27: <m:mfrac> 28: <m:mn> 25 </m:mn> 29: <m:mn> 2 </m:mn> 30: </m:mfrac> 31: </m:mrow> 32: <m:mn> 100 </m:mn> 33: </m:mfrac> 34: <m:mo> ) </m:mo> 35: <m:mo> ] </m:mo> </m:math>

What you have coded till now : $x=\left[3200\left(1-\frac{5}{100}\right)\left(1+\frac{10}{100}\right)\left(1+\frac{5}{100}\right)\right]$

What is to be coded ultimately : $x=\left[32000\left(1-\genfrac{}{}{0.2ex}{}{5}{100}\right)\left(1+\genfrac{}{}{0.2ex}{}{10}{100}\right)\left(1+\genfrac{}{}{0.2ex}{}{\frac{25}{2}}{100}\right)\right]$

Step 8 : Comparing what we have achieved so far and what is the expected, we need few tweaking here and there. First we see that the brackets have not grown to the vertical height of the terms, composing the equation. In order to do this, we club the part of equation in the brackets within “mrow” tags. Thus, we insert “mrow” tag before line 4 and at the end of the code after line 35. Also, the bars of the three central "mfrac" elements have to be distinguished from the nested one. We use "linethickness" attribute on "mfrac" element and set the same to "medium". The final code and display are shown in the example.

## Indices element : mroot

<m:math display="block"> <m:mi> x </m:mi> <m:mo> = </m:mo> <m:mrow> <m:mo> [ </m:mo> <m:mn> 32000 </m:mn> <m:mo> ( </m:mo> <m:mn> 1 </m:mn> <m:mo> - </m:mo> <m:mfrac linethickness="2"> <m:mn> 5 </m:mn> <m:mn> 100 </m:mn> </m:mfrac> <m:mo> ) </m:mo> <m:mo> ( </m:mo> <m:mn> 1 </m:mn> <m:mo> + </m:mo> <m:mfrac linethickness="2"> <m:mn> 10 </m:mn> <m:mn> 100 </m:mn> </m:mfrac> <m:mo> ) </m:mo> <m:mo> ( </m:mo> <m:mn> 1 </m:mn> <m:mo> + </m:mo> <m:mfrac linethickness="2"> <m:mrow> <m:mfrac> <m:mn> 25 </m:mn> <m:mn> 2 </m:mn> </m:mfrac> </m:mrow> <m:mn> 100 </m:mn> </m:mfrac> <m:mo> ) </m:mo> <m:mo> ] </m:mo> </m:mrow> </m:math>

Save the file after editing as “test.xml”. The display looks like :

$x=\left[32000\left(1-\genfrac{}{}{0.2ex}{}{5}{100}\right)\left(1+\genfrac{}{}{0.2ex}{}{10}{100}\right)\left(1+\genfrac{}{}{0.2ex}{}{\frac{25}{2}}{100}\right)\right]$

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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Sherica
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Sherica
right! what he said ⤴⤴⤴
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