# Khan/khan-exercises

Fix malformed and un-wrapped text in exercises to keep graphie blocks…

… from entering in to translations.
jeresig committed Jan 23, 2013
1 parent dcca838 commit c04bbc9d993881434f6aa4b7dd9845d884a8ca42
 @@ -59,13 +59,13 @@
- Given
+
Given
\qquad m ANGLE_ONE = COEF_1x + CONST_1, and
\qquad m ANGLE_TWO = COEF_2x + CONST_2,
, and
\qquad m LARGE_ANGLE = largeAngle,
-

+

init({
 @@ -104,11 +104,11 @@

- Remember that the measure of the angles in a triangle sum to 180°. + Remember that the measure of the angles in a triangle sum to 180°. Solve for \color{orange}{\angle{BCA}} by subtracting the measures of angles \color{purple}{\angle{BAC}} and \color{green}{\angle{ABC}} from 180°. We find that - \color{orange}{\angle{BCA}} = Tri_X°. + \color{orange}{\angle{BCA}} = Tri_X°. // label angle BAC arc([2.5, 3], .75, 180, 220, { stroke: "orange" }); @@ -117,10 +117,9 @@

- Solve for - \color{blue}{\angle{DAF}} by using the fact - that it is a corresponding angle to + Solve for \color{blue}{\angle{DAF}} by using the fact + that it is a corresponding angle to \color{orange}{\angle{BCA}}. LABEL.remove(); LABEL = label([-3.3, 0], @@ -129,20 +128,20 @@ - \color{blue}{\angle{CAE}} by using the fact - that it is an alternate interior angle to + Solve for \color{blue}{\angle{CAE}} by using the fact + that it is an alternate interior angle to \color{orange}{\angle{BCA}}. LABEL.remove(); LABEL = label([1, 0], "\\color{blue}{∠CAE} = " + Tri_X + "°", "above", { color: "blue"}); - \color{orange}{\angle{BCA}}. That means those + That means those angles are equal because they are both created by the same set of parallel lines \overline{BC} and \overline{DE}, and transversal line - \overline{CF}. + \overline{CF}.

@@ -249,9 +248,9 @@ parallel lines bisected by a single line.

- First solve for \color{orange}{\angle{AGH}}. We know that + First solve for \color{orange}{\angle{AGH}}. We know that \color{orange}{\angle{AGH}} = \color{purple}{\angle{EGB}} - because opposite angles are equal. + because opposite angles are equal. arc([1,2], .88, 180, 225, {stroke:"orange"}); label([0,2], "\\color{orange}{" + X + "°}", "below left"); @@ -268,11 +267,11 @@ parallel lines bisected by a single line.

- First solve for \color{orange}{\angle{AGH}}. We know that + First solve for \color{orange}{\angle{AGH}}. We know that \color{orange}{\angle{AGH}} = 180° - \color{purple}{\angle{BGH}} , - because angles along a line plane add up to 180°. + because angles along a line plane add up to 180°. arc([1,2], .88, 180, 225, {stroke:"orange"}); label([0,2], "\\color{orange}{" + X + "°}", "below left"); @@ -289,23 +288,22 @@ formed by parallel lines, and a single bisecting lines.

- Therefore, - \angle{GHD} = X° + Therefore, \angle{GHD} = X°. LABEL.remove(); label([-2, -2], "\\color{blue}{\\angle{GHD}}=" + X + "°", "above right"); - \angle{CHF} = X° + Therefore, \angle{CHF} = X°. LABEL.remove(); label([-4, -2.5], "\\color{blue}{\\angle{CHF}}=" + X + "°", "below left"); - . +

@@ -395,11 +393,11 @@

- + \color{purple}{\angle{BHD}} = 180° - \color{green}{\angle{BDC}} - \color{orange}{\angle{DBE}} = Tri1_Z°. - This is because the interior angles of a triangle add up to 180°. + This is because the interior angles of a triangle add up to 180°. // label angle BHD arc([3.2, 1.3], .75, 118, 220, { stroke: "purple" }); @@ -544,8 +542,8 @@

- \color{purple}{\angle{DIJ}} = 180° - \color{orange}{\angle{AIC}}. - This is because angles along a line total 180°. + \color{purple}{\angle{DIJ}} = 180° - \color{orange}{\angle{AIC}}. + This is because angles along a line total 180°. // label angle JID arc([0, -1.2], .75, 143, 220, { stroke: "purple" }); @@ -554,8 +552,8 @@

- \color{purple}{\angle{HGC}} = 180° - \color{orange}{\angle{FGH}}. - This is because angles along a line or flat plane total 180°. + \color{purple}{\angle{HGC}} = 180° - \color{orange}{\angle{FGH}}. + This is because angles along a line or flat plane total 180°. // label angle HGC arc([1.8, 5], 1, 280, 0, { stroke: "purple" }); @@ -565,9 +563,9 @@

- \color{teal}{\angle{DJI}} = 180° - \color{green}{\angle{BDC}} - + \color{teal}{\angle{DJI}} = 180° - \color{green}{\angle{BDC}} - \color{purple}{\angle{DIJ}}. - We know this because the sum of angles inside a triangle add up to 180°. + We know this because the sum of angles inside a triangle add up to 180°. // label angle JID arc([-3.2, 1.3], .75, 260, 320, { stroke: "teal" }); @@ -576,9 +574,9 @@

- \color{teal}{\angle{CHG}} = 180° - \color{green}{\angle{ACD}} - + \color{teal}{\angle{CHG}} = 180° - \color{green}{\angle{ACD}} - \color{purple}{\angle{HGC}}. - We know this, because the sum of angles inside a triangle add up to 180°. + We know this, because the sum of angles inside a triangle add up to 180°. // label angle CHG arc([3.2, 1.3], .75, 38, 120, { stroke: "teal" }); @@ -598,24 +596,23 @@

- Therefore, - \angle{AJF} = Tri2_X° + Therefore, \angle{AJF} = Tri2_X°. LABEL.remove(); label([-3.7, 2.5], "\\color{blue}{\\angle{AJF}}=" + Tri2_X + "°", "above"); - \angle{IHE} = Tri2_X° + Therefore, \angle{IHE} = Tri2_X°. LABEL.remove(); label([4.0, -0.3], "\\color{blue}{\\angle{IHE}}=" + Tri2_X + "°", "below left"); - . +

 @@ -29,10 +29,10 @@ randFromArray([ "student", "teacher", "person" ]) - TOTAL plural( RESPONDENT, TOTAL ) were asked about their favorite SUBJECT. +

TOTAL plural( RESPONDENT, TOTAL ) were asked about their favorite SUBJECT. toSentence( shuffle( \$.map( DATA, function( num, i ) { return "<code>" + num + "</code> " + plural( RESPONDENT, num ) + " said " + CATEGORIES[ i ]; - }) ) ). + }) ) ).

Create a bar chart showing everyone's favorite plural( SUBJECT ):

 @@ -138,7 +138,7 @@ graph.pointB.toFront(); - Click and drag the points to move the lines. +

Click and drag the points to move the lines.

x = X

 @@ -139,8 +139,8 @@
- We know that \blue{SINGLE_PCT \%} of the time, he'll make - his first shot. +

We know that \blue{SINGLE_PCT \%} of the time, he'll make + his first shot.

init({ range: [[-1, 11], [-1, 1]] @@ -155,11 +155,10 @@ line([0,0], [PROB*10,0]); line([PROB*10,-0.2], [PROB*10,0.2]);
-
- Then SINGLE_PCT \% of +

Then SINGLE_PCT \% of the time he makes his first shot, he will also make his second shot, and ((1-PROB)*100).toFixed(0) \% of the time he makes his - first shot, he will miss his second shot. + first shot, he will miss his second shot.

init({ range: [[-1, 11], [-1, 1]] @@ -191,9 +190,9 @@ PROB \cdot PROB = (PROB*PROB).toFixed(3).

- We can repeat this process again to get the probability of making three free throws in a row. We simply take +

We can repeat this process again to get the probability of making three free throws in a row. We simply take SINGLE_PCT\% of probability that he makes two in a row, which we know from the previous step is - (PROB*PROB).toFixed(3) \approx \orange{TWO_PCT\%}. + (PROB*PROB).toFixed(3) \approx \orange{TWO_PCT\%}.

init({ range: [[-1, 11], [-1, 1]] @@ -210,10 +209,9 @@ line([0,0], [PROB*PROB*10,0]); line([PROB*PROB*10,-0.2], [PROB*PROB*10,0.2]);
-
- SINGLE_PCT\% of \orange{TWO_PCT\%} is +

SINGLE_PCT\% of \orange{TWO_PCT\%} is PROB \cdot (PROB*PROB).toFixed(3) = Math.pow(PROB,3).toFixed(3), or - about \green{THREE_PCT\%}: + about \green{THREE_PCT\%}:

init({ range: [[-1, 11], [-1, 1]] @@ -252,9 +250,9 @@
- We know that \blue{SINGLE_PCT \%} of the time, he'll miss +

We know that \blue{SINGLE_PCT \%} of the time, he'll miss his first shot - (100 \% - (PR*100).toFixed(0) \% = SINGLE_PCT \%). + (100 \% - (PR*100).toFixed(0) \% = SINGLE_PCT \%).

init({ range: [[-1, 11], [-1, 1]] @@ -269,11 +267,10 @@ line([0,0], [PROB*10,0]); line([PROB*10,-0.2], [PROB*10,0.2]);
-
- Then SINGLE_PCT \% of +

Then SINGLE_PCT \% of the time he misses his first shot, he will also miss his second shot, and ((1-PROB)*100).toFixed(0) \% of the time he misses his - first shot, he will make his second shot. + first shot, he will make his second shot.

init({ range: [[-1, 11], [-1, 1]] @@ -305,9 +302,9 @@ PROB \cdot PROB = (PROB*PROB).toFixed(3).

- We can repeat this process again to get the probability of missing three free throws in a row. We simply take +

We can repeat this process again to get the probability of missing three free throws in a row. We simply take SINGLE_PCT\% of probability that he misses two in a row, which we know from the previous step is - (PROB*PROB).toFixed(3) \approx \orange{TWO_PCT\%}. + (PROB*PROB).toFixed(3) \approx \orange{TWO_PCT\%}.

init({ range: [[-1, 11], [-1, 1]] @@ -324,10 +321,9 @@ line([0,0], [PROB*PROB*10,0]); line([PROB*PROB*10,-0.2], [PROB*PROB*10,0.2]);
-
- SINGLE_PCT\% of \orange{TWO_PCT\%} is +

SINGLE_PCT\% of \orange{TWO_PCT\%} is PROB \cdot (PROB*PROB).toFixed(3) = Math.pow(PROB,3).toFixed(3), or - about \green{THREE_PCT\%}: + about \green{THREE_PCT\%}:

init({ range: [[-1, 11], [-1, 1]]