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Converting a PDF to a Word Document
PDFs are extremely useful files but, sometimes, the need arises to edit or deliver the content in them in a Microsoft Word file format. Here’s a quick look at how this can be done.
Know the Software You Need
The PDF (portable document format) file format was created by Adobe in the early 1990s as a standardized method of distributing and viewing digital documents, regardless of what computer or device is used in sending or receiving. Viewing PDFs is simple and can be done with simple PDF readers like Adobe Acrobat Reader DC or even a web browser. Editing PDFs isn’t always as simple and, sometimes, it’s convenient or desirable to convert a PDF into a Microsoft Word file. There are two ways that this can be done: using Adobe’s Acrobat DC software or using Microsoft Word. You need to own a copy of one of these pieces of software to follow the steps in this tutorial.
Option A: Using Adobe Acrobat DC, Part 1
The first step is open Adobe Acrobat DC, then click File, and choose Open. Navigate to and select the PDF file that you intend to convert into a Word file _ alternatively, you could navigate to the file first within Windows Explorer and double-click it to launch Adobe Acrobat DC and open the file. Next, click File, and select Export To.
Option A: Using Adobe Acrobat DC, Part 2
In the Export To menu, there are several exporting options. Select Microsoft Word, and then choose one of the two available options: Word Document, or Word 97-2003 Document. Use the first option, which delivers a “.docx” file, for most modern uses or choose the second option, which supplies a “.doc,” if you have a specific need to save the PDF in Microsoft’s older Word format. Enter a name for the new file and click Save. It may take Acrobat a moment to make the conversion.
Option B: Using Microsoft Word, Part 1
Alternatively, you can use Microsoft Word 2013 or later to convert a PDF file to Word, although it’s possible that using this method may introduce minor formatting inconsistencies into your new version. This process may work better if the PDF you’re attempting to convert is mostly text.
Option B: Using Microsoft Word, Part 2
To start, open Microsoft Word, and then open the PDF file you would like to convert. In this case, be sure open Word first. Word will make a copy of your PDF file so that you still have the original, and then run the conversion process to Word format on the copy. You can now edit the PDF in Word and save it as a Word file format.
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QUADRATIC EQUATION WORD PROBLEMS WORKSHEET WITH ANSWERS
Problem 1 :
Difference between a number and its positive square root is 12. Find the number.
Problem 2 :
If the difference between a number and its reciprocal is ²⁴⁄₅ , find the number.
Problem 3 :
A piece of iron rod costs $60. If the rod was 2 meter shorter and each meter costs $1 more, the cost would remain unchanged. What is the length of the rod?
Problem 4 :
Divide 25 in two parts so that sum of their reciprocals is ⅙ .
Problem 5 :
The hypotenuse of a right angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.
Problem 6 :
The sides of an equilateral triangle are shortened by 12 units, 13 units and 14 units respectively and a right angle triangle is formed. Find the length of each side of the equilateral triangle.
1. Answer :
Let x be the required number.
Its positive square root is √x.
Given : Difference between x and √x is 12.
x - √x = 12
x - 12 = √x
(x - 12) 2 = x
x 2 - 24x + 144 = x
Subtract x from both sides.
x 2 - 25x + 144 = 0
x 2 - 9x - 16x + 144 = 0
x(x - 9) - 16(x - 9) = 0
(x - 9)(x - 16) = 0
x = 9 or x = 16
x = 9 does not satisfy the condition given in the question.
Then,
x = 16
Therefore, the required number is 16.
2. Answer :
Let y be the required number.
Then, its reciprocal is ¹⁄ y .
Given : D ifference between a number and its reciprocal is ²⁴⁄₅ .
y - ¹⁄ y = ²⁴⁄₅
Multiply both sides by 5y to get rid of the denominators y and 5.
5y(y - ¹⁄ y ) = 5y( ²⁴⁄₅ )
5y(y) - 5y( ¹⁄ y ) = 24y
5y 2 - 5 = 24y
Subtract 24y from boths sides.
5y 2 - 24y - 5 = 0
Solve by factoring.
5y 2 - 25y + y - 5 = 0
5y(y - 5) + 1(y - 5) = 0
(y - 5)(5y + 1) = 0
y - 5 = 0 or 5y + 1 = 0
y = 5 or y = ⁻¹⁄₅
Justification y - 5 = 0 or
Both the values y = 5 and y = ⁻¹⁄₅ satisfy the condition given in the question.
Therefore, the required number is 5 or ⁻¹⁄₅ .
3. Answer :
Let x be the length of the given rod. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). Cost of one meter of the given rod is
= ⁶⁰⁄ₓ
Cost of one meter of the rod which is 2 meter shorter is
= ⁶⁰⁄₍ₓ ₋ ₂₎
Given : If the rod was 2 meter shorter and each meter costs $1 more.
That is, 60/(x-2) is $1 more than 60/x.
⁶⁰⁄₍ₓ ₋ ₂₎ - ⁶⁰⁄ₓ = 1
Multiply both sides by x(x - 2) to get rid of the denominators (x - 2) an x.
x(x - 2) [⁶⁰⁄₍ₓ ₋ ₂₎ - ⁶⁰⁄ₓ] = x(x - 2)
x(x - 2) [⁶⁰⁄₍ₓ ₋ ₂₎] - x(x - 2) [⁶⁰⁄ₓ] = x(x - 2)
60x - 60(x - 2) = x 2 - 2x
60x - 60x + 120 = x 2 - 2x
120 = x 2 - 2x
0 = x 2 - 2x - 120
x 2 - 2x - 120 = 0
x 2 - 12x + 10x - 120 = 0
x(x - 12) + 10(x - 12) = 0
(x - 12)(x + 10) = 0
x - 12 = 0 or x + 10 = 0
x = 12 or x = -10
Because length can not be a negative number, we can ignore x = -10.
Therefore, the length of the given rod is 12 m.
4. Answer :
Let x be one of the parts of 25.
Then, the other part is (25 - x). Given : Sum of the reciprocals of the parts is 1/6.
¹⁄ₓ + ¹⁄₍₂₅ ₋ ₓ₎ = ⅙
Multiply both sides by 6x(25 - x) to get rid of the denominators x, (25 - x) and 6.
6x(25 - x) [ ¹⁄ₓ + ¹⁄₍₂₅ ₋ ₓ₎ ] = 6x(25 - x) [⅙]
6x(25 - x) [ ¹⁄ₓ ] - 6x(25 - x) [ ¹⁄₍₂₅ ₋ ₓ₎ ] = x(25 - x)
6(25 - x) - 6x = 25x - x 2
150 - 6x + 6x = 25x - x 2
150 = 25x - x 2
x 2 - 2 5x + 150 = 0
x 2 - 1 5x - 10x + 150 = 0
x(x - 15) - 10(x - 15) = 0
(x - 15)(x - 10) = 0
x - 15 = 0 or x - 10 = 0
x = 15 or x = 10
When x = 15,
25 - x = 25 - 15
When x = 10,
25 - x = 25 - 10
Therefore, the two parts of 25 are 10 and 15.
5. Answer :
Let x and (x + 4) be the lengths of other two sides. Using Pythagorean theorem, we have
(x + 4) 2 + x 2 = 20 2
x 2 + 8x + 16 + x 2 = 400
2 x 2 + 8x + 16 = 400
Subtract 400 from both sides.
2 x 2 + 8x - 384 = 0
Divide both sides by 2.
x 2 + 4x - 192 = 0
x 2 - 12x + 16x - 192 = 0
x(x - 12) + 16(x - 12) = 0
(x - 12)(x + 16) = 0
x - 12 = 0 or x + 16 = 0
x = 12 or x = -16
x = -16 can not be accepted. Because length can not be negative.
x + 4 = 12 + 4
Therefore, the other two sides of the triangle are 12 cm and 16 cm.
6. Answer :
Let x be the length of each side of the equilateral triangle. Then, the sides of the right angle triangle are
(x - 12), (x - 13) and (x - 14)
In the above three sides, the side represented by (x - 12) is hypotenuse (Because that is the longest side). Using Pythagorean theorem, we have
(x - 12) 2 = (x - 13) 2 + (x - 14) 2
x 2 - 24x + 144 = x 2 - 26x + 169 + x 2 - 28x + 196
x 2 - 24x + 144 = 2x 2 - 54x + 365
0 = x 2 - 34x + 221
x 2 - 34x + 221 = 0
x 2 - 13x - 17x + 221 = 0
x(x - 13) - 17(x - 13) = 0
(x - 13)(x - 17) = 0
x - 13 = 0 or x - 17 = 0
x = 13 or x = 17
x = 13 can not be accepted.
Because, if x = 13, the side represented by (x - 14) is negative. Therefore, the length of each side of the equilateral triangle is 17 units.
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