/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 (a) Evaluate $$ \left(1+\fra... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Evaluate $$ \left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{5}\right) . $$ (b) Evaluate $$ \sqrt{1+\frac{1}{2}} \times \sqrt{1+\frac{1}{3}} \times \sqrt{1+\frac{1}{4}} \times \sqrt{1+\frac{1}{5}} \times \sqrt{1+\frac{1}{6}} \times \sqrt{1+\frac{1}{7}} $$ (c) We write the product " \(4 \times 3 \times 2 \times 1 "\) as "4!" (and we read this as "4 factorial"). Using only pencil and paper, how quickly can you work out the number of weeks in \(10 !\) seconds? \(\Delta\)

Short Answer

Expert verified
(a) The result is 3. (b) The result is 2. (c) 10! seconds equals about 6 weeks.

Step by step solution

01

Simplify Each Parenthesis Expression (a)

First, evaluate each separate term in the expression: \(1 + \frac{1}{2} = \frac{3}{2}\),\(1 + \frac{1}{3} = \frac{4}{3}\),\(1 + \frac{1}{4} = \frac{5}{4}\),\(1 + \frac{1}{5} = \frac{6}{5}\).
02

Multiply the Simplified Fractions (a)

Now multiply the fractions obtained from Step 1: \[ \left(\frac{3}{2}\right) \times \left(\frac{4}{3}\right) \times \left(\frac{5}{4}\right) \times \left(\frac{6}{5}\right) \]
03

Cancel and Simplify the Multiplication (a)

Cancel the common factors:- \(3\) in the numerator of the first fraction and the denominator of the second,- \(4\) in the numerator of the second and the denominator of the third,- \(5\) in the numerator of the third and the denominator of the fourth.This leaves: \[ \frac{3 \times 4 \times 5 \times 6}{2 \times 3 \times 4 \times 5} = \frac{6}{2} = 3. \]
04

Evaluate Each Square Root Expression (b)

Calculate each expression inside the square root:\(\sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}\),\(\sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}}\),\(\sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}}\),\(\sqrt{1 + \frac{1}{5}} = \sqrt{\frac{6}{5}}\),\(\sqrt{1 + \frac{1}{6}} = \sqrt{\frac{7}{6}}\),\(\sqrt{1 + \frac{1}{7}} = \sqrt{\frac{8}{7}}\).
05

Multiply the Square Root Expressions (b)

Multiply the square root expressions obtained:\[ \sqrt{\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \frac{6}{5} \times \frac{7}{6} \times \frac{8}{7}}. \]
06

Cancel and Simplify the Multiplication (b)

Cancel the overlapping terms:- \(3\), \(4\), \(5\), \(6\), and \(7\) cancels similarly with both numerator and denominator.This simplifies to:\(\sqrt{\frac{8}{2}}\ = \sqrt{4} = 2.\)
07

Understanding Factorial and Time Units (c)

Understand that a factorial \(n!\) represents the product of all positive integers up to \(n\). Recall that 10! means the product of all numbers from 10 down to 1. This is purely numeric calculation unrelated to time units but can be calculated for curiosity or exercises only.
08

Calculate 10! as a practice (c)

Compute \(10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\). A quick mental estimation or look-up for illustration purposes returns 3,628,800.
09

Convert Time Numerically to Weeks (c)

Knowing there are 604,800 seconds in a week by multiplying seconds in a minute (60) \(\times\) minutes in an hour (60) \(\times\) hours in a day (24) \(\times\) days in a week (7). Divide \(10!\) by 604,800 for curiosity or fun, noting calculators help with exactness.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorials
Imagine you have a row of ten books and you want to know how many different ways you can arrange them. Factorials are a way of answering this kind of question. The factorial of a number, represented as "n!", is the product of all positive integers up to that number. For example, "4!" is equal to 4 multiplied by 3, multiplied by 2, and then by 1, which gives us 24. It's a shortcut for saying "these are all the ways to arrange four items." Computing these factorials is particularly handy in various mathematical problems and also in statistics, where it helps determine combinations and permutations.
  • "0!" is defined as 1, which is a standard convention for consistency in equations.
  • The larger the number, the faster the factorial grows. For example, "10!" equals 3,628,800.
Factorials teach us about multiplicative patterns and are a crucial part of understanding probability, algebra, and calculus. Though they don't directly relate to time, imagining the factorial in different contexts can deepen your grasp of the concept.
Square Roots
Understanding square roots is like finding out what number, when squared, gives you a specific number. For example, the square root of 9 is 3, because 3 squared (\(3^2\)) equals 9.In mathematical notation, the square root of a number is represented as \(\sqrt{\ }\). It's like asking, "What number times itself equals this number?"Square roots often appear in problems involving areas, physics principles like acceleration, and geometry. They transform seemingly complex calculations into manageable ones. For instance:
  • \(\sqrt{16} = 4\), because 4 \(\times\) 4 = 16.
  • \(\sqrt{1+\frac{1}{2}} = \sqrt{\frac{3}{2}}\), simplifying further calculations.
This simplicity demonstrates their usefulness in mathematics, as breaking a number into equal factors often uncovers more straightforward paths to a solution.
Fraction Multiplication
When you multiply fractions, you multiply straight across, multiplying the numerators together and the denominators together. If you have two fractions, \(\frac{a}{b}\) and \(\frac{c}{d}\), their product is \(\frac{a \times c}{b \times d}\).This process looks simple but can add layers of complexity if not approached methodically. Here’s a bright side, though! Simplification often happens within the multiplication process itself. For example:
  • \(\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \frac{6}{5} = \frac{6}{2} = 3\).
  • Notice how overlapping factors, like 3, 4, and 5, can cancel out neatly, making computations easier!
Fraction multiplication is highly valued in areas like probability and algebra, where relationships between whole parts are frequently analyzed. By making fractional multiplication familiar, it opens a wider understanding of proportional reasoning itself.
Simplification
Simplification in mathematics is all about making expressions as straightforward as possible. This process helps us to understand problems more quickly and find solutions with less effort.Think of simplification as cleaning up a room to find a lost item—the fewer distractions, the easier it is. In mathematical terms, this might mean:
  • Reducing fractions to their simplest form.
  • Cancelling common factors in a product.
  • Reorganizing equations so that relations are clear.
Simplifying an expression can turn a complex-looking situation into something more accessible. For example, with fractions, simplifying through cancellation at steps like\(\left(\frac{3}{2}\right) \times \left(\frac{4}{3}\right) \)...removes unnecessary clutter.The ultimate aim of simplification is clarity. It reduces the components of a problem, thereby sharpening the focus, helping to solve problems efficiently.

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Most popular questions from this chapter

Let \(A B C\) be a triangle. We use the standard labelling convention, whereby the side \(B C\) opposite \(A\) has length \(a\), the side \(C A\) opposite \(B\) has length \(b,\) and the side \(A B\) opposite \(C\) has length \(c\). Prove that, if \(c^{2}=a^{2}+b^{2},\) then \(\angle B C A\) is a right angle.

Reveals the triple (3,4,5) as the first instance \((m=1)\) of a one-parameter infinite family of triples, which continues $$ (5,12,13)(m=2),(7,24,25)(m=3),(9,40,41)(m=4), \ldots $$ whose general term is $$ (2 m+1,2 m(m+1), 2 m(m+1)+1) $$ The triple (3,4,5) is also the first member of a quite different "one- parameter infinite family" of triples, which continues $$ (6,8,10),(9,12,15), \ldots $$ Here the triples are scaled-up versions of the first triple (3,4,5) . In general, common factors simply get in the way: If \(a^{2}+b^{2}=c^{2}\) and \(H C F(a, b)=s,\) then \(s^{2}\) divides \(a^{2}+b^{2},\) and \(a^{2}+b^{2}=c^{2} ;\) so \(s\) divides \(c .\) And if \(a^{2}+b^{2}=c^{2}\) and \(H C F(b, c)=s,\) then \(s^{2}\) divides \(c^{2}-b^{2}=a^{2},\) so \(s\) divides \(a .\) Hence a typical Pythagorean triple has the form \((s a, s b, s c)\) for some scale factor \(s,\) where \((a, b, c)\) is a triple of integers, no two of which have a common factor: any such triple is said to be primitive (that is, basic - like prime numbers). Every Pythagorean triple is an integer multiple of some primitive Pythagorean triple. The next problem invites you to find a simple formula for all primitive Pythagorean triples.

(a) Use Pythagoras' Theorem in a square \(A B C D\) of side 1 to show that the diagonal \(A C\) has length \(\sqrt{2}\). Use this to work out in your head the exact values of \(\sin 45^{\circ}, \cos 45^{\circ}, \tan 45^{\circ}\) (b) In an equilateral triangle \(\triangle A B C\) with sides of length \(2,\) join \(A\) to the midpoint \(M\) of the base \(B C\). Apply Pythagoras' Theorem to find \(A M\). Hence work out in your head the exact values of \(\sin 30^{\circ}, \cos 30^{\circ}, \tan 30^{\circ},\) \(\sin 60^{\circ}, \cos 60^{\circ}, \tan 60^{\circ}\) (c)(i) On the unit circle with centre at the origin \(O:(0,0),\) mark the point \(P\) so that \(P\) lies in the first quadrant, and so that \(O P\) makes an angle \(\theta\) with the positive \(x\) -axis (measured anticlockwise from the positive \(x\) -axis). Explain why \(P\) has coordinates \((\cos \theta, \sin \theta)\). (ii) Extend the definitions of \(\cos \theta\) and \(\sin \theta\) to apply to angles beyond the first quadrant, so that for any point \(P\) on the unit circle, where \(O P\) makes an angle \(\theta\) measured anticlockwise from the positive \(x\) -axis, the coordinates of \(P\) are \((\cos \theta, \sin \theta) .\) Check that the resulting functions sin and cos satisfy: \(*\) sin and cos are both positive in the first quadrant, \(*\) sin is positive and cos is negative in the second quadrant, \(*\) sin and cos are both negative in the third quadrant, and \(*\) sin is negative and \(\cos\) is positive in the fourth quadrant. (iii) Use (a), (b) to calculate the exact values of \(\cos 315^{\circ}, \sin 225^{\circ}, \tan 210^{\circ}\), \(\cos 120^{\circ}, \sin 960^{\circ}, \tan \left(-135^{\circ}\right)\) (d) Given a circle of radius \(1,\) work out the exact area of a regular \(n\) -gon inscribed in the circle: (i) when \(n=3\) (ii) when \(n=4\) (iii) when \(n=6\) (iv) when \(n=8\) (v) when \(n=12\). (e) Given a circle of radius 1 , work out the area of a regular \(n\) -gon circumscribed around the circle: (i) when \(n=3\) (ii) when \(n=4\) (iii) when \(n=6\) (iv) when \(n=8\) (v) when \(n=12\). Knowing the exact values of \(\sin , \cos\) and tan for the special angles \(0^{\circ}\), \(30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}\) is like knowing one's tables. In particular, it allows one to evaluate trigonometric functions mentally for related angles in all four quadrants (using the CAST mnemonic \(-\mathrm{C}\) being in the \(\mathrm{SE}\) of the unit circle, \(A\) in the NE quadrant, \(S\) in the NW quadrant, and \(T\) in the SW quadrant - to remind us which functions are positive in each quadrant). These special angles arise over and over again in connection with equilateral triangles, squares, regular hexagons, regular octagons, regular dodecagons, etc., where one can use what one knows to calculate exactly in geometry.

Problem 39 (Shadows) Can one use the Sun's rays to produce a plane shadow of a cube: (i) in the form of an equilateral triangle? (ii) in the form of a square? (iii) in the form of a pentagon? (iv) in the form of a regular hexagon? (v) in the form of a polygon with more than six sides?

(a) In the "24 game" you are given four numbers. Your job is to use each number once, and to combine the four numbers using any three of the four basic arithmetical operations - using the same operation more than once if you wish, and as many brackets as you like (but never concatenating different numbers, such as "3" and "4" to make " \(34 "\) ). If the given numbers are \(3,3,4,4,\) then one immediately sees \(3 \times 4+3 \times 4=24 .\) With 3,3,5 , 5 it may take a little longer, but is still fairly straightforward. However, you may find it more challenging to make 24 in this way: (i) using the four numbers 3,3,6,6 (ii) using the four numbers 3,3,7,7 (iii) using the four numbers 3,3,8,8 . (b) Suppose we restrict the numbers to be used each time to "four \(4 \mathrm{~s}\) " \((4,4,4,4),\) and change the goal from "make \(24 ",\) to "make each answer from \(0-10\) using exactly four \(4 \mathrm{~s} "\). (i) Which of the numbers \(0-10\) cannot be made? (ii) What if one is allowed to use squaring and square roots as well as the four basic operations? What is the first inaccessible integer? Calculating by turning the handle deterministically (as with addition, or multiplication, or multiplying out brackets, or differentiating) is a valuable skill. But such direct procedures are usually only the beginning. Using mathematics and solving problems generally depend on the corresponding inverse procedures \(-\) where a certain amount of juggling and insight is needed in order to work backwards (as with subtraction, or division, or factorisation, or integration). For example, in applications of calculus, the main challenge is to solve differential equations (an inverse problem) rather than to differentiate known functions. Problem 14 captures the spirit of this idea in the simplest possible context of arithmetic: the required answer is given, and we have to find how (or whether) that answer can be generated. We will meet more interesting examples of this kind throughout the rest of the collection.

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