Skip to content

Commit 2c207c4

Browse files
author
StephenElston
committed
Spell checking
1 parent d49cd44 commit 2c207c4

4 files changed

Lines changed: 22 additions & 22 deletions

File tree

‎Module01/01-05-Polynomials.ipynb‎

Lines changed: 4 additions & 11 deletions
Original file line numberDiff line numberDiff line change
@@ -33,7 +33,7 @@
3333
"So why do we care? Well, polynomials have some useful properties that make them easy to work with. for example, if you multiply, add, or subtract a polynomial, the result is always another polynomial.\n",
3434
"\n",
3535
"## Standard Form for Polynomials\n",
36-
"Techbnically, you can write the terms of a polynomial in any order; but the *standard form* for a polynomial is to start with the highest *degree* first and constants last. The degree of a term is the highest order (exponent) in the term, and the highest order in a polynomial determines the degree of the polynomial itself.\n",
36+
"Technically, you can write the terms of a polynomial in any order; but the *standard form* for a polynomial is to start with the highest *degree* first and constants last. The degree of a term is the highest order (exponent) in the term, and the highest order in a polynomial determines the degree of the polynomial itself.\n",
3737
"\n",
3838
"For example, consider the following expression:\n",
3939
"\\begin{equation}3x + 4xy^{2} - 3 + x^{3} \\end{equation}\n",
@@ -205,7 +205,7 @@
205205
"\n",
206206
"\\begin{equation}(x^{4} + 2)(2x^{2} + 3x - 3) \\end{equation}\n",
207207
"\n",
208-
"Let's do the first step and multiply each term in the first polynomial by each term in the second polynomial. The first term in the first polynomial is x<sup>4</sup>, and the first term in the second polynomial is 2x<sup>2</sup>, so multiplying these gives us 2x<sup>6</sup>. Then we can multiply the first term in the first polynomial (x<sup>4</sup>) by the second term in the second polynomial (3x), which gives us 3x<sup>5</sup>, and so on until we've multipled all of the terms in the first polynomial by all of the terms in the second polynomial, which results in this:\n",
208+
"Let's do the first step and multiply each term in the first polynomial by each term in the second polynomial. The first term in the first polynomial is x<sup>4</sup>, and the first term in the second polynomial is 2x<sup>2</sup>, so multiplying these gives us 2x<sup>6</sup>. Then we can multiply the first term in the first polynomial (x<sup>4</sup>) by the second term in the second polynomial (3x), which gives us 3x<sup>5</sup>, and so on until we've multiplied all of the terms in the first polynomial by all of the terms in the second polynomial, which results in this:\n",
209209
"\n",
210210
"\\begin{equation}2x^{6} + 3x^{5} - 3x^{4} + 4x^{2} + 6x - 6 \\end{equation}\n",
211211
"\n",
@@ -264,7 +264,7 @@
264264
"\n",
265265
"\\begin{equation}\\frac{4x}{2x} + \\frac{6x^{2}}{2x}\\end{equation}\n",
266266
"\n",
267-
"Then we can simplify each fraction and add the results. For the first fraction, 2x goes into 4x twice, so the fraction simplifies to 2; and for the second, 6x<sup>2</sup> is 2x mutliplied by 3x. So our answer is 2 + 3x:\n",
267+
"Then we can simplify each fraction and add the results. For the first fraction, 2x goes into 4x twice, so the fraction simplifies to 2; and for the second, 6x<sup>2</sup> is 2x multiplied by 3x. So our answer is 2 + 3x:\n",
268268
"\n",
269269
"\\begin{equation}2 + 3x\\end{equation}\n",
270270
"\n",
@@ -372,13 +372,6 @@
372372
"(x**2 + 2*x -3)/(x-2) == x + 4 + (5/(x-2))\n",
373373
" "
374374
]
375-
},
376-
{
377-
"cell_type": "code",
378-
"execution_count": null,
379-
"metadata": {},
380-
"outputs": [],
381-
"source": []
382375
}
383376
],
384377
"metadata": {
@@ -393,7 +386,7 @@
393386
"mimetype": "text/x-r-source",
394387
"name": "R",
395388
"pygments_lexer": "r",
396-
"version": "3.4.1"
389+
"version": "3.5.0"
397390
}
398391
},
399392
"nbformat": 4,

‎Module01/01-06-Factorization.ipynb‎

Lines changed: 6 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -127,7 +127,7 @@
127127
"\n",
128128
"So what's the greatest common factor of these two expressions?\n",
129129
"\n",
130-
"It helps to break the expressions into their consitituent components. Let's deal with the coefficients first; we have 15 and 9. The highest value that divides evenly into both of these is **3** (3 x 5 = 15 and 3 x 3 = 9).\n",
130+
"It helps to break the expressions into their constituent components. Let's deal with the coefficients first; we have 15 and 9. The highest value that divides evenly into both of these is **3** (3 x 5 = 15 and 3 x 3 = 9).\n",
131131
"\n",
132132
"Now let's look at the ***x*** terms; we have x<sup>2</sup> and x. The highest value that divides evenly into both is these is **x** (*x* goes into *x* once and into *x*<sup>2</sup> *x* times).\n",
133133
"\n",
@@ -141,7 +141,7 @@
141141
"- The *largest* numeric factor of the numeric coefficients in the polynomial expressions (in this case 3)\n",
142142
"- The *smallest* exponential of each variable (in this case, x and y, which technically are x<sup>1</sup> and y<sup>1</sup>.\n",
143143
"\n",
144-
"You can check your answer by dividing the original expressions by the GCF to find the coefficent expressions for the GCF (in other words, how many times the GCF divides into the original expression). The result, when multiplied by the GCF will always produce the original expression. So in this case, we need to perform the following divisions:\n",
144+
"You can check your answer by dividing the original expressions by the GCF to find the coefficient expressions for the GCF (in other words, how many times the GCF divides into the original expression). The result, when multiplied by the GCF will always produce the original expression. So in this case, we need to perform the following divisions:\n",
145145
"\n",
146146
"\\begin{equation}\\frac{15x^{2}y}{3xy}\\;\\;\\;\\;\\;\\;\\;\\;\\frac{9xy^{3}}{3xy}\\end{equation}\n",
147147
"\n",
@@ -184,7 +184,7 @@
184184
"\n",
185185
"\\begin{equation}6x + 15y \\end{equation}\n",
186186
"\n",
187-
"To factor this, we need to find an expression that divides equally into both of these expressions. In this case, we can use **3** to factor the coefficents, because 3 &bull; 2x = 6x and 3&bull; 5y = 15y, so we can write our original expression as:\n",
187+
"To factor this, we need to find an expression that divides equally into both of these expressions. In this case, we can use **3** to factor the coefficients, because 3 &bull; 2x = 6x and 3&bull; 5y = 15y, so we can write our original expression as:\n",
188188
"\n",
189189
"\\begin{equation}6x + 15y = 3(2x) + 3(5y) \\end{equation}\n",
190190
"\n",
@@ -360,7 +360,7 @@
360360
"metadata": {},
361361
"source": [
362362
"### Perfect Squares\n",
363-
"A *perfect square* is a number multiplied by itself, for example 3 multipled by 3 is 9, so 9 is a perfect square.\n",
363+
"A *perfect square* is a number multiplied by itself, for example 3 multiplied by 3 is 9, so 9 is a perfect square.\n",
364364
"\n",
365365
"When working with equations, the ability to factor between polynomial expressions and binomial perfect square expressions can be a useful tool. For example, consider this expression:\n",
366366
"\n",
@@ -383,7 +383,7 @@
383383
"\n",
384384
"When you combine the two 5x terms we get back to our original expression of x<sup>2</sup> + 10x + 25.\n",
385385
"\n",
386-
"Now we have an expression multipled by itself; in other words, a perfect square. We can therefore rewrite this as:\n",
386+
"Now we have an expression multiplied by itself; in other words, a perfect square. We can therefore rewrite this as:\n",
387387
"\n",
388388
"\\begin{equation}(x + 5)^{2}\\end{equation}\n",
389389
"\n",
@@ -440,7 +440,7 @@
440440
"mimetype": "text/x-r-source",
441441
"name": "R",
442442
"pygments_lexer": "r",
443-
"version": "3.4.1"
443+
"version": "3.5.0"
444444
}
445445
},
446446
"nbformat": 4,

‎Module01/01-07-Quadratic Equations.ipynb‎

Lines changed: 4 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -224,7 +224,7 @@
224224
"\n",
225225
"\\begin{equation}2(x - 1) = 0\\end{equation}\n",
226226
"\n",
227-
"If we distrbute this, we get:\n",
227+
"If we distribute this, we get:\n",
228228
"\n",
229229
"\\begin{equation}2x - 2 = 0\\end{equation}\n",
230230
"\n",
@@ -322,7 +322,7 @@
322322
"\n",
323323
"\\begin{equation}y = 3x^{2} - 12\\end{equation}\n",
324324
"\n",
325-
"Note that this is in the standard quadratic form, but there is no *b* term; in other words, there's no term that contains a coeffecient for ***x*** to the first power. This type of equation can be easily solved using the square root method. Let's restate it so we're solving for ***x*** when ***y*** is 0:\n",
325+
"Note that this is in the standard quadratic form, but there is no *b* term; in other words, there's no term that contains a coefficient for ***x*** to the first power. This type of equation can be easily solved using the square root method. Let's restate it so we're solving for ***x*** when ***y*** is 0:\n",
326326
"\n",
327327
"\\begin{equation}3x^{2} - 12 = 0\\end{equation}\n",
328328
"\n",
@@ -334,7 +334,7 @@
334334
"\n",
335335
"\\begin{equation}x^{2} = 4\\end{equation}\n",
336336
"\n",
337-
"No we can isolate ***x*** by taking the square root of both sides. However, there's an additional consideration because this is a quadratic equation. The ***x*** variable can have two possibe values, so we must calculate the *principle* and *negative* square roots of the expression on the right:\n",
337+
"No we can isolate ***x*** by taking the square root of both sides. However, there's an additional consideration because this is a quadratic equation. The ***x*** variable can have two possible values, so we must calculate the *principle* and *negative* square roots of the expression on the right:\n",
338338
"\n",
339339
"\\begin{equation}x = \\pm\\sqrt{4}\\end{equation}\n",
340340
"\n",
@@ -411,7 +411,7 @@
411411
"## Solving Quadratics Using the Completing the Square Method\n",
412412
"In quadratic equations where there is a *b* term; that is, a term containing **x** to the first power, it is impossible to directly calculate the square root. However, with some algebraic manipulation, you can take advantage of the ability to factor a polynomial expression in the form *a<sup>2</sup> + 2ab + b<sup>2</sup>* as a binomial *perfect square* expression in the form *(a + b)<sup>2</sup>*.\n",
413413
"\n",
414-
"At first this might seem like some sort of mathematical sleight of hand, but follow through the steps carefull and you'll see that there's nothing up my sleeve!\n",
414+
"At first this might seem like some sort of mathematical sleight of hand, but follow through the steps careful and you'll see that there's nothing up my sleeve!\n",
415415
"\n",
416416
"The underlying basis of this approach is that a trinomial expression like this:\n",
417417
"\n",

‎Module01/01-08-Functions.ipynb‎

Lines changed: 8 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -86,7 +86,7 @@
8686
"cell_type": "markdown",
8787
"metadata": {},
8888
"source": [
89-
"Of course, the value returned by a function depends on the input; and you can graph this with the iput (let's call it ***x***) on one axis and the output (***f(x)***) on the other."
89+
"Of course, the value returned by a function depends on the input; and you can graph this with the input (let's call it ***x***) on one axis and the output (***f(x)***) on the other."
9090
]
9191
},
9292
{
@@ -464,6 +464,13 @@
464464
"source": [
465465
"Note that the ***p(x)*** values in the plot drop exponentially for ***x*** values that are negative, and then rise exponentially for positive ***x*** values; but the minimum value returned by the function (for an *x* value of 0) is **1**."
466466
]
467+
},
468+
{
469+
"cell_type": "code",
470+
"execution_count": null,
471+
"metadata": {},
472+
"outputs": [],
473+
"source": []
467474
}
468475
],
469476
"metadata": {

0 commit comments

Comments
 (0)