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# Logarithm Formula
Logarithms are reverse operations of exponents. Suppose that $a^n = b$ then $^a log b = n$ and vice versa (if $^a log b =n$ then $a^n = b$). Therefore,
$^a log b = n Leftrightarrow a^n = b$
with a logarithm principal number, $a > 0$, $a neq 1$, b the number that the logarithm looks for, $b > 0$ and n is the result of the logarithm (exponent).
To be able to work on logarithmic problems, use the following logarithmic properties.
1. $^a log b^n = n ^a log b$
2. $^a log (bc) = ^a log b + ^a log c$
3. $^a log ( frac{b}{c} ) = ^a log b – ^a log c$
4. $^a log b times ^b log c = ^a log c$
5. $^{a^n} log b^m = frac{m}{n} ^a log b$
6. $^a log b = frac{1}{^b log a}$
7. $a^{^a log b} = b$
8. $^a log b = frac{log b}{log a}$
Note: If the principal number of a logarithm is not written, then the mean number of the logarithm is 10. So $^{10} log 7$ is written with $log 7$ only.
Problems example:
1. If $^3 log 4 = p$ and $^2 log 5 = q$ then the value for $^3 log 5$ is …
2. Know $^2 log 5 = p$ and $^5 log 3 = q$. The value of $^3 log 10$ is expressed in p and q is …
3. Results of $^{ frac{1}{5}} log 625+ ^{64} log frac{1}{16} + 4 ^{(3 ^{25} log 5)}$ is …
begin{align} & ^2 log 5 = q \ & Leftrightarrow ^4 log 5^2 = q \ & Leftrightarrow 2 ^4 log 5 = q \ & Leftrightarrow ^4 log 5 = frac{q}{2} end{align}
begin{align} ^3 log 5 & = ^3 log 4 ( ^4 log 5 ) \ & = p frac{q}{2} \ & = frac{pq}{2} end{align}
begin{align} ^3 log 10 & = frac{log 10}{log 3} \ & = frac{^5 log 10}{^5 log 3} \ & = frac{^5 log (2 times 5)}{^5 log 3} \ & = frac{^5 log 2 + ^5 log 5} {^5 log 3} \ & = frac{frac{1}{p} + 1}{1} \ & = frac{1 + p}{pq} end{align}. |
# Solve Equations with Absolute Value
This is a tutorial on solving equations with absolute value. Detailed solutions and explanations are included.
## Examples with Solutions
### Example 1
Solve the equation
|x + 6 | = 7
Solution to Example 1:
• If |x + 6 | = 7, then
a) x + 6 = 7
or
b) x + 6 = -7
• Solve equation a)
x + 6 = 7
x = 1
• Solve equation b)
x + 6 = -7
x = -13
Check solutions:
• solution x = 1
Left Side of Equation for x = 1.
|1 + 6 |
= | 7 |
= 7
Right Side of Equation for x = 1.
7
• x = -13
Left Side of Equation for x = 1.
|-13 + 6 |
= | -7 |
= 7
Right Side of Equation for x = 1.
7
The solutions to the given equation are x = 1 and x = -13
Matched Exercise 1: Solve the equation
|-x - 8 | = 10
### Example 2
Solve the equation
-2 |x / 2 + 3 | - 4 = -10
Solution to Example 2:
• Given
-2 |x / 2 + 3 | - 4 = -10
• We first write the equation in the form | A | = B. Add 4 to both sides and group like terms
-2|x / 2 + 3 | = -6
• Divide both sides by -2
|x / 2 + 3 | = 3
• We now proceed as in example 1 above, the equation
|x / 2 + 3 | = 3 gives two equations.
a) x / 2 + 3 = 3
or
b) x / 2 + 3 = -3
• Solve equation a)
x / 2 + 3 = 3
• to obtain
x = 0
• Solve equation b)
x / 2 + 3 = -3
• to obtain
x = -12
Check solutions:
• x = 0
Left Side of Equation for x = 0.
-2 |x / 2 + 3 | - 4
= -2| 3 | - 4
= -10
Right Side of Equation for x = 1.
-10
• x = -12
Left Side of Equation for x = -12.
-2 |x / 2 + 3 | - 4
= -2 |-12 / 2 + 3 | - 4
= -2 |-6 + 3 | - 4
= -2(3) - 4
= -10
Right Side of Equation for x = -12.
-10
The solutions to the given equation are x = 0 and x = -12
Matched Exercise 2: Solve the equation
4 |x + 2| - 30 = -10
### Example 3
Solve the equation
|2 x - 2 | = x + 1
Solution to Example 3:
• If 2 x - 2 ≥ 0 which is equivalent to x ≥ 1, then |2 x - 2 | = 2 x - 2 and the given equation becomes
2 x - 2 = x + 1
• Add 2 - x to both sides
x = 3
• Since x = 3 satisfies the condition x ≥ 1, it is a solution.
• If 2x - 2 < 0 which is equivalent to x < 1, then |2 x - 2 | = -(2 x - 2) and the given equation becomes
-(2 x - 2) = x + 1
• Solve for x to obtain
x = 1 / 3
• Since x = 1 / 3 satisfies the condition x < 1, it is a solution.
Check solutions
• x = 3
Left Side of Equation for x = 3.
|2 x - 2 |
= |2*3 - 2 |
= 4
Right Side of Equation for x = 3.
x + 1
= 3 + 1
= 4
• x = 1/3
Left Side of Equation for x = 1 / 3.
|2 x - 2 |
= |2*(1/3) - 2 |
= 4 / 3
Right Side of Equation for x = 1 / 3.
x + 1
= 4 / 3
The solutions to the given equation are x = 3 and x = 1 / 3
Matched Exercise 3: Solve the equation
- 4|x + 2 | = x - 8
### Example 4
Solve the equation
|x2 - 4| = x + 2
Solution to Example 3:
• If x2 - 4 ≥ 0 ,or x2 ≥ 4, then | x2 - 4 | = x2 - 4 and the given equation becomes
x2 - 4 = x + 2
• Add - (x + 2) to both sides
x2 - 4 -( x + 2) = 0
• Factor the left term
(x - 2)(x + 2) -( x + 2) = 0
(x + 2)(x - 2 -1) = 0
(x + 2)(x - 3) = 0
• Using the factor theorem, we can write two simpler equations
x + 2 = 0
or
x - 3 = 0
• Solve the above equations for x to find two values of x that make the left side of the equation equal to zero.
x = -2 and x = 3.
• Both values satisfy the condition x2 ≥ 4 and are solutions to the given equation.
x = -2 and x = 3.
• If x2 - 4 < 0 ,or x2 < 4, then | x2 - 4 | = -(x2 - 4) and the given equation becomes.
-(x2 - 4) = x + 2
-(x2 - 4) - ( x + 2) = 0
• Factor the left term.
-(x - 2)(x + 2) - ( x + 2) = 0
(x - 2)(x + 2) + ( x + 2) = 0
(x - 2)(x + 2) + ( x + 2) = 0
(x + 2)(x - 2 + 1) = 0
(x + 2)(x - 1) = 0
• Two values make the left side of the above equation equal to zero
x = -2 and x = 1.
• Only x = 1 satisfies the condition x2 < 4
Check solutions:
• x = -2
Right Side of Equation = | x2 - 4 |
= | (-2)2 - 4 | = 0
Left Side of Equation = x + 2 = -2 + 2 = 0
• x = 3 Left Side of Equation = | x2 - 4 |
= | 32 - 4 |
= | 5 |
= 5 Right Side of Equation = x + 2 = 3 + 2 = 5
• x = 1
Left Side of Equation = | x2 - 4 |
= | 12 - 4 | = | - 3 | = 3 Right Side of Equation = x + 2 = 1 + 2 = 3
Conclusion
The solutions to the given equation are x = -2, x = 1 and x = 3.
Matched Exercise 4: Solve the equation
|x2 - 16 | = x - 4
## Exercises
Solve the following absolute value equations
a) | x - 4 | = 9
b) | x
2 + 4 | = 5
c) | x
2 - 9 | = x + 3
d) | x + 1 | = x - 3
e) | -x | = 2 |
# Harmonic Progression: Overview, Questions, Preparation
Sequence and Series 2021 ( Sequence and Series )
Rachit Kumar SaxenaManager-Editorial
Updated on Jun 28, 2021 10:33 IST
## What is the Harmonic Progression?
When you arrange a series of numbers into a predictable pattern, it is termed progression. Progression is a sequence of numbers that follow particular rules. Progression is can be of three types, i.e. Arithmetic Progression, Geometric Progression & Harmonic Progression.
For solving the problems related to harmonic progression, firstly perceive the corresponding sum of an arithmetic progression. In simple terms, the nth term of the progression must be equal to the nth terms reciprocal of the related Arithmetic Progression.
### Harmonic Progression Formula
Harmonic Progression’s nth term= 1/ [a+(n-1)d]
Where:
“a” depicts the first-term of the Arithmetic Progression.
“d” represents the common difference.
“n” depicts overall no. of terms in Arithmetic Progression.
## Harmonic Progression Sum:
If there is a harmonic expression, i.e. 1/a, 1/a+d, 1/a+2d, …., 1/a+(n-1), the actual formula for finding the sum of n terms is:
Sn=1/d(ln){2a+(2n−1)d/2a−d}
In which,
“a” is first term
“d” is difference
“ln”: natural logarithm
## Illustrative Examples of Harmonic Progression
To understand the working of harmonic progression, let’s go through some solved examples:
1. Determine the 4th and 8th term of the harmonic progression 6, 4, 3,…
Solution.
Harmonic progression = 6, 4, 3
Find the corresponding arithmetic progression for the H.P.
Arithmetic Progression = ⅙, ¼, ⅓, ….
By A.P. T2 -T1 is equal to T3 -T2
i.e. 1/12 = d
For finding the 4th term of the given A. P, the formula is:
nth term = a+(n-1)d
Where, a = ⅙ and d= 1/12
For 4th term n=4
Putting these values in the formula:
4th term = (⅙) +(4-1)(1/12)
=> (⅙)+(3/12)
=> 5/12
Repeat the same steps for 8th term
8th = (⅙) +(8-1)(1/12)
= (⅙)+(7/12)
= 9/12
As we know, harmonic progression is reciprocal, I.e. reverse of an Arithmetic Progression:
Thus, the 4th term becomes 1/4th term = 12/5 and
8th term becomes 1/8th term = 12/9 or 4/3
2. If the first two terms of a harmonic progression are 1/16 and 1/13, find the maximum partial sum?
Solution.
Harmonic series will be: 1/16, 1/13, 1/10,1/7, 1/4, 1/1, 1/-2, 1/-5
Maximum partial sum for this series will be 1/16 + 1/13 + 1/10 +1/7 + 1/4 + 1/1 = 1.63
3. Compute the 16th term of H.P. if the 6th and 11th term of H.P. is 10 and 18.
Solution.
Write H.P. in terms of A.P.:
6th term of Arithmetic Progression= a+5d = 1/10 [Equation 1]
11th term of Arithmetic Progression= a+10d = 1/18 [Equation 2]
Solve equation 1 and 2 for getting the values of a and d:
a =13/90
d = -2/ 225
For finding 16th term, write expression as:
a+15d
= (13/90) – (2/15)
= 1/90
The 16th term of the Harmonic Progression is equal to 1/16th term of an Arithmetic Progression that is 90
Thus, the 16th term is 90.
## FAQ’s on Harmonic Progression
Q: What is the formula of Harmonic Mean?
A: Harmonic Mean can be calculated as
n /[(1/a) + (1/b)+ (1/c)+(1/d)+….]
In which,
a, b, c, d represents the values of the progression n depicts the total no of values present in the progression.
Q: What are the four types of sequences?
A: The four types of Sequences and series are Arithmetic Sequences, Geometric Sequences, Harmonic Sequences and
Fibonacci Numbers.
Q: Is harmonic progression in JEE mains?
A: No, questions related to harmonic progression aren’t asked in JEE Mains, whereas you can expect related questions in JEE Advanced.
Q; Why do we calculate the harmonic mean?
A: We calculate harmonic mean because it helps us find the multiplicative and divisor relationships amongst the given fractions without thinking about the common denominators.
Q; How do you find the common difference in harmonic progression?
A: You can find the common difference by only subtracting two- adjacent terms. You can find every term by recursively adding d, i.e. common difference within the preceding term.
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# If ${z}^{2}+z|z|+|{z}^{2}|=0,$ then the locus $z$ is a. a circle b. a straight line c. a pair of straight line d. none of these
Video Solution
Text Solution
Generated By DoubtnutGPT
## To solve the equation z2+z|z|+|z2|=0 and find the locus of z, we will proceed step by step.Step 1: Express z in terms of its componentsLet z=x+iy, where x and y are real numbers. Then, we can express the modulus of z as:|z|=√x2+y2Thus, we can rewrite the equation as:z2+z|z|+|z2|=0Step 2: Calculate z2 and |z2|Calculating z2:z2=(x+iy)2=x2−y2+2xyiNow, calculate |z2|:|z2|=|(x2−y2)+2xyi|=√(x2−y2)2+(2xy)2=√x4−2x2y2+y4+4x2y2=√x4+2x2y2+y4=√(x2+y2)2=|z|2Step 3: Substitute back into the equationSubstituting z2 and |z2| into the original equation:(x2−y2+2xyi)+(x+iy)√x2+y2+|z|2=0This simplifies to:(x2−y2+2xyi)+(x√x2+y2+iy√x2+y2)+|z|2=0Step 4: Separate real and imaginary partsNow, separate the real and imaginary parts:- Real part: x2−y2+x√x2+y2+|z|2=0- Imaginary part: 2xy+y√x2+y2=0Step 5: Solve the imaginary partFrom the imaginary part:y(2x+√x2+y2)=0This gives us two cases:1. y=02. 2x+√x2+y2=0Step 6: Analyze the cases1. If y=0, then z lies on the real axis.2. If 2x+√x2+y2=0, squaring both sides gives:(2x)2=x2+y2⟹4x2=x2+y2⟹3x2=y2⟹y=±√3xStep 7: Find the locusThe equation y=±√3x represents two straight lines through the origin.ConclusionThus, the locus of z is a pair of straight lines.Final AnswerThe locus of z is c. a pair of straight lines.
|
Updated on:7/8/2024
### Knowledge Check
• Question 1 - Select One
## If z2+z|z|+|z|2=0 , then the locus of z is
Aa circle
Ba straight line
Ca pair of straight lines
Dnone of these
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# How do you simplify sqrt(75)-sqrt3?
Jul 17, 2016
$4 \sqrt{3}$
#### Explanation:
Write 75 as the product of its prime factors, then we know what we are working with.
$75 = 5 \times 5 \times 3 = {5}^{2} \times 3$
$\sqrt{75} - \sqrt{3}$
=$\sqrt{{5}^{2} \times 3} - \sqrt{3} \text{ find any possible roots}$
=$5 \sqrt{3} - 1 \sqrt{3} = 4 \sqrt{3} \text{ we have like terms}$ |
# Subtraction Placing the Numbers
Learn how to do subtraction placing the numbers in columns. Subtract from two-digit numbers by arranging and placing the numbers correctly in tens and ones column.
1. Subtract 23 from 89 (by placing the numbers in column).
Solution:
Subtract 89 - 23 arrange by placing the numbers in column
T O
8 9
- 2 3
6 6
(i) Given numbers are written in column form.
(ii) Subtraction of digits at one’s place, 9 – 3 = 6 ones.
(iii) Subtraction of digits at ten’s place, 8 – 2 = 6 tens.
(iv) Subtracted sum = 6 tens + 6 ones = 60 + 6 = 66
(v) Therefore, 89 – 23 = 66
2. Subtract 46 from 78 (by placing the numbers in column).
Solution:
Subtract 78 - 46 arrange by placing the numbers in column.
T O
7 8
- 4 6
3 2
(i) Given numbers are written in column form.
(ii) Subtraction of digits at one’s place, 8 – 6 = 2 ones.
(iii) Subtraction of digits at ten’s place, 7 – 4 = 3 tens.
(iv) Subtracted sum = 3 tens + 2 ones = 30 + 2 = 32
(v) Therefore, 78 - 46 = 32
3. Subtract 23 from 94 (by placing the numbers in column).
Solution:
Subtract 94 - 23 arrange by placing the numbers in column.
T O
9 4
- 2 3
7 1
(i) Given numbers are written in column form.
(ii) Subtraction of digits at one’s place, 4 – 3 = 1 one.
(iii) Subtraction of digits at ten’s place, 9 – 2 = 7 tens.
(iv) Subtracted sum = 7 tens + 1 one = 70 + 1 = 71
(v) Therefore, 94 - 23 = 71
The examples will help us to solve different types of problems of 2-digit subtraction placing the numbers in vertical order.
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Home » Class 10 » Quadratic Equation Class 10
Quadratic Equation Class 10 Chapter 4 Notes
• A quadratic equation in the variable x is an equation of ax2+bx+c=0, where a,b, and c are real numbers, a≠0.
• For example, 2×2+x−300=0 is a quadratic equation.
Quadratic Equation Class 10- Standard Form
• Any equation of the form p(x)=0, where p(x) is a polynomial of degree 2, is a quadratic equation.
• But when we write the terms of p(x) in descending order of their degrees, we get the equation’s standard form.
• That is, ax2+bx+c=0, a≠0 is called the standard form of a quadratic equation.
• A solution of the equation p(x)=ax2+bx+c=0, with a≠0, is called a root of the quadratic equation.
• A real number α is called a root of the quadratic equation ax2+bx+c=0,a≠0 if aα2+bα+c=0.
• It means x=α satisfies the quadratic equation or x=α is the root of the quadratic equation.
• The zeroes of the quadratic polynomial ax2+bx+c and the roots of the quadratic equation ax2+bx+c=0 are the same.
Ace your class 10th board exams with Adda247 live classes for class 10th preparation.
Quadratic Equation Class 10- Method Of Solving
1. Factorisation Method
1. Factorise the quadratic equation by splitting the middle term.
2. After splitting the middle term, convert the equation into linear factors by taking common terms out.
3. Then, on equating each factor to zero, the roots are determined.
For example:
⇒2×2−5x+3 (Split the middle term)
⇒2×2−2x−3x+3 (Take out common terms to determine linear factors)
⇒2x(x−1)−3(x−1)
⇒(x−1)(2x−3) (Equate to zero)\
⇒(x−1)(2x−3)=0
When (x−1)=0 , x=1
When (2x−3)=0 , x=32
So, the roots of 2×2−5x+3 are 1 and 32
2. Method Of Completing The Square
1. The solution of a quadratic equation can be found by converting any quadratic equation to the perfect square of the form (x+a)2−b2=0.
2. To convert quadratic equation x2+ax+b=0 to perfect square equate b, i.e., the constant term to the right side of the equal sign, then add a square of half of an, i.e., square of half of the coefficient of x on both sides.
3. To convert the quadratic equation of form ax2+bx+c=0, a≠0 to perfect square, first divide the equation by an, i.e., the coefficient of x2, then follow the above-mentioned steps.
For example:
⇒x2+4x−5=0 (Equate constant term 5 to the right of the equal sign)
⇒x2+4x=5 (Add a square of half of 4 on both sides)
⇒x2+4x+(42)2=5+(42)2
⇒x2+4x+4=9
⇒(x+2)2=9
⇒(x+2)2−(3)2=0
It is of the form (x+a)2−b2=0
Now,
⇒(x+2)2−(3)2=0
⇒(x+2)2=9
⇒(x+2)=±3
⇒x=1
and x=−5
So, the roots of x2+4x−5=0are 1 and −5
3. By using the quadratic formula
The root of a quadratic equation ax2+bx+c=0 is given by the formula
x=−b±b2−4ac−−−−−−−√2a, where b2−4ac−−−−−−−√ is known as a discriminant.
If b2−4ac−−−−−−−√≥0, then only the root of a quadratic equation is given by
x=−b±b2−4ac−−−−−−−√2a
For example:
⇒x2+4x+3
By using the quadratic formula, we get
⇒x=−4±(4)2−4×1×3−−−−−−−−−−−−−√2×1
⇒x=−4±16−12−−−−−−√2
⇒x=−4±4–√2
⇒x=−4±22
⇒x=−4+22, x=−1
⇒x=−4−22,x=−3
So, the roots of x2+4x+3=0 are −1 and −3
Quadratic Equation Class 10- Nature Of Roots Based On Discriminant
• If b2−4ac−−−−−−−√=0, then the roots are real and equal
• If b2−4ac−−−−−−−√>0, then the roots are real and distinct.
• If b2−4ac−−−−−−−√<0then the roots are imaginary
Ques. What will be the nature of the roots of the quadratic equation 2×2 + 4x – n = 0?
Solution: D = b2 – 4ac
42 – 4 x 2 (-7)
16 + 56 = 72 > 0
Hence, the roots of a quadratic equation are real and unequal.
Ques. If -5 is a root of the quadratic equation 2×2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the value of k.
Solution: Since – 5 is a root of the equation 2×2 + px – 15 = 0
2(-5)2 + p(-5) – 15 = 0
50 – 5p – 15 = 0 or 5p = 35 or p = 7
Again p(x2 + x) + k = 0 or 7×2 + 7x + k = 0 has equal roots
D = 0
i.e., b2 – 4ac = 0 or 49- 4 × 7k = 0
k = 4928 = 74
Ques. Find the values of k for each of the following quadratic equations so that they have two equal roots.
(i) 2×2 + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0
Solution: (i) We have, 2×2 + kx + 3 = 0
Here, a = 2, b = k, c = 3
D = b2 – 4ac = k2 – 4 × 2 × 3 = k2 – 24 For equal roots
D = 0
i.e., k2 – 24 = 0
ķ2 = 24
k = ± √24
k = + 2√6
(ii) We have, kx(x – 2) + 6 = 0
⇒ kx2 – 2kx + 6 = 0
Here, a = k, b = – 2k, c = 6
For equal roots, we have
D = 0
i.e., b2 – 4ac = 0
⇒ (-2k)2 – 4 × k × 6 = 0
⇒ 4k2 – 24k = 0
⇒ 4k (k – 6) = 0
Either 4k = 0 or k – 6 = 0
⇒ k = 0 or k = 6
But k = 0 6)ecause if k = 0 then given equation will not be a quadratic equation).
So, k = 6.
Ques. If the equation (1 + m2)x2 + 2mcx + c2– a2 = 0 has equal roots, show that c2 = a2 (1 + m2).
Solution: The given equation is (1 + m2) x2 + (2mc) x + (c2– a2) = 0
Here, A = 1 + m2, B = 2mc and C = c2 – a2
Since the given equation has equal roots, therefore D = 0 = B2 – 4AC = 0.
⇒ (2mc)2 – 4(1 + m2) (c2 – a2) = 0
⇒ 4m2c2 – 4(c2 – a2 + m2c2 – m2a2) = 0
⇒ m2c2 – c2 + a2 – m2c2 + m2a2 = 0. [Dividing throughout by 4]
⇒ – c2 + a2 (1 + m2) = 0
⇒ c2 = a(1 + m2) Hence Proved
Ques. If the roots of the quadratic equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are equal, then show that a = b = c.
Solution: Given (x – a) (x – b) + (x – b) (x – c) + (x – 6) (x – a) = 0
⇒ x2 – ax – bx + ab + x2 – bx – cx + bc + x2 – cx – ax + ac = 0
⇒ 3×2 – 2(a + b + c)x + ab + bc + ca = 0
Now, for equal roots D = 0
⇒ B2 – 4AC = 0
⇒ 4(a + b + c)2 – 12(ab + bc + ca) = 0
4a2 + 4b2 + 4c2 + 8ab + 8bc + 8ca – 12ab – 12bc – 12ca = 0
⇒ 2[2a2 + 2b2 + 2co – 2ab – 2bc – 2ca] = 0
⇒ 2[(a2 + b2 – 2ab) + (b2 + c2 – 2bc) + (c2 + a2 – 2ca)] = 0
⇒ [(a – b)2 + (b – c)2 + (c – a)2] = 0
⇒ a – b = 0, b – c = 0, c – a = 0
⇒ a = b, b = c, c = a
⇒ a = b = c
Ques. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.
Solution: Let Shefali’s marks in Mathematics be x.
Therefore, Shefali’s marks in English are (30 – x).
Now, according to the question,
⇒ (x + 2) (30 – x – 3) = 210
⇒ (x + 2) (27 – x) = 210
⇒ 27x – x2 + 54 – 2x = 210
⇒ 25x – x2 + 54 – 210 = 0
⇒ 25x – x2 – 156 = 0
⇒ -(x2 – 25x + 156) = 0
⇒ x2 – 25x + 156 = 0
= x2 – 13x – 12x + 156 = 0
⇒ x(x – 13) – 12(x – 13) = 0
⇒ (x – 13) (x – 12) = 0
Either x – 13 or x – 12 = 0
x = 13 or x = 12
Therefore, Shefali’s marks in Mathematics = 13
Marks in English = 30 – 13 = 17
or Shefali’s marks in Mathematics = 12
marks in English = 30 – 12 = 18.
Ques. The sum of the areas of two squares is 468 m2. If the difference in their perimeters is 24 m, find the sides of the two squares.
Solution: Let x be the length of the side of the first square and y be the length of the side of the second square.
Then, x2 + y2 = 468 …(i)
Let x be the length of the side of the bigger square.
4x – 4y = 24
⇒ x – y = 6 or x = y + 6 …(ii)
Putting the value of x in terms of y from equation (ii), in equation (i), we get
(y + 6)2 + y2 = 468
⇒ y2 + 12y + 36 + y2 = 468 or 232 + 12y – 432 = 0
⇒ y2 + 6y – 216 = 0
⇒ y2 + 18y – 12y – 216 = 0
⇒ y(y + 18) – 12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
Either y + 18 = 0 or y – 12 = 0
⇒ y = -18 or y = 12
But, sides cannot be negative, so y = 12
Therefore, x = 12 + 6 = 18
Hence, the sides of the two squares are 18 m and 12 m.
Let’s connect via chat or call our senior expert counsellor at +91-9625869989 to learn more about the different streams and options available.
We would love it if we could add some of your insights. If you have a definite goal of scoring the highest marks, then you can resolve your doubts via our app/quizzes and youtube class assistance (https://youtube.com/c/Adda247Class9and10)
Ques 1. What number of exercises are in Chapter 4 for Grade 10?
Ans. The fourth chapter of the NCERT Solutions for Class 10 Maths contains four tasks.
Ques 2. Does the quadratic equation matter?
Ques 3. Exactly where are quadratic equations used?
Ans. Numerous real-world applications of quadratic equations include estimating enclosed space areas, an object’s speed, the profit and loss of a product, and curving a piece of equipment for design.
Ques 4. What does a quadratic equation look like in standard form?
Ans. A quadratic equation is written in standard form as axe squared + bx plus c is equal to zero.
Ques 5. What does the quadratic formula serve?
Ans. One of the top five mathematical formulae is the quadratic formula, which aids in the solution of quadratic equations.
Ques 6. What kind of equation is not quadratic?
Ans. When there are more than two solutions to an equation that satisfy it, it is argued that the equation is not quadratic. In essence, the equation is non-quadratic since the square term is missing.
Ques 7. How many solutions may there be for a quadratic equation?
Ans. Two roots—or solutions—exist in a quadratic equation with real or complex coefficients. These two options might or might not be separate, and they might or might not be true.
Ques 8. Why have quadratic equations got two solutions?
Ans. There are two solutions since a quadratic expression can be expressed as the product of two linear elements, each of which can be equal to zero.
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FAQs
What number of exercises are in Chapter 4 for Grade 10?
The fourth chapter of the NCERT Solutions for Class 10 Maths contains four tasks.
Exactly where are quadratic equations used?
Numerous real-world applications of quadratic equations include estimating enclosed space areas, an object's speed, the profit and loss of a product, and curving a piece of equipment for design.
What does a quadratic equation look like in standard form?
A quadratic equation is written in standard form as axe squared + bx plus c is equal to zero.
What does the quadratic formula serve?
One of the top five mathematical formulae is the quadratic formula, which aids in the solution of quadratic equations.
What kind of equation is not quadratic?
When there are more than two solutions to an equation that satisfy it, it is argued that the equation is not quadratic. In essence, the equation is non-quadratic since the square term is missing.
How many solutions may there be for a quadratic equation?
Two roots—or solutions—exist in a quadratic equation with real or complex coefficients. These two options might or might not be separate, and they might or might not be true.
Why have quadratic equations got two solutions?
There are two solutions since a quadratic expression can be expressed as the product of two linear elements, each of which can be equal to zero. |
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# If $a\sin x + b\cos (x + \theta ) + b\cos (x - \theta ) = d$, then the value of $\left| {\cos \theta } \right|$ is equal to.
Last updated date: 13th Jun 2024
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Hint: We need to simplify such equation using trigonometric functions
Sum of the trigonometric functions can be formulated as:
$Cos(A + B) + \operatorname{Cos} (A - B) = 2\operatorname{Cos} A\operatorname{Cos} B$
We know that the value of a $\operatorname{Sin} \theta + B\operatorname{Cos} \theta$ lies between: $- \sqrt {{a^2} + {b^2}} \leqslant a\operatorname{Sin} \theta + b\cos \theta \leqslant \sqrt {{a^2} + {b^2}}$.
Complete step-by- step solution:
We have,
$\Rightarrow a\operatorname{Sin} \theta + b\cos (x + \theta ) + b\cos (x - \theta ) - d.............eqn(1)$
Taking out b common from$eqn(1)$, we get
$\Rightarrow a\operatorname{Sin} \theta + b\left[ {\cos (x + \theta ) + \cos (x - \theta )} \right] - d...............eqn(2)$
We know, $Cos(x - \theta ) + \operatorname{Cos} (x - \theta ) = 2Cosx\,Cos\theta ...........eqn(3)$
Using the value of $eqn(3)$in$eqn(2)$, we get
$a\sin x + b\left[ {2Cosx\,\operatorname{Cos} \theta } \right] = d$
$\Rightarrow a\sin x + b\left[ {2bCosx\,\operatorname{Cos} \theta } \right] = d........eqn(4)$
As $a\sin \theta + b\operatorname{Cos} \theta$ lies between $- \sqrt {{a^2} + {b^2}} \leqslant a\operatorname{Sin} \theta + b\operatorname{Cos} \theta \leqslant \sqrt {{a^2} + {b^2}}$
$\Rightarrow \left| {a\sin \theta + b\cos \theta } \right| \leqslant \sqrt {{a^2} + {b^2}} ........eqn(5)$
Compare$a\sin \theta + b\cos \theta$ with $a\sin x + (2b\operatorname{Cos} \theta )\operatorname{Cos} x$
We get, $a = a$ and $b = 2b\operatorname{Cos} \theta ........eqn(6)$
Using the value of $eqn$ (6) and (4) in (5), we have $a\sin \theta + b\cos \theta = d,$$a = a$and $b = 2b\operatorname{Cos} \theta$
$\Rightarrow \left| d \right| \leqslant \sqrt {{a^2} + {{(2b\cos \theta )}^2}}$
$\Rightarrow \left| d \right| \leqslant \sqrt {{a^2} + 4{b^2}Co{s^2}\theta }$
On squaring both sides
$\Rightarrow {d^2} \leqslant {a^2} + 4{b^2}Co{s^2}\theta$
$\Rightarrow {d^2} - {a^2} \leqslant 4{b^2}Co{s^2}\theta$
$\Rightarrow \dfrac{{{d^2} - {a^2}}}{{4{b^2}}} \leqslant Co{s^2}\theta$
Taking under root on both sides
.$\left| {\operatorname{Cos} \theta } \right| \geqslant \dfrac{{\sqrt {{d^2} - {a^2}} }}{{\sqrt {4{b^2}} }}$
$\begin{gathered} \\ \left| {\operatorname{Cos} \theta } \right| \geqslant \dfrac{{\sqrt {{d^2} - {a^2}} }}{{2\left| b \right|}} \\ \end{gathered}$
Hence $\left| {\operatorname{Cos} \theta } \right| = \dfrac{1}{{2\left| b \right|}}\sqrt {{d^2} - {a^2}}$
Note: Recall that in its basic form $\,f(x) = \,|x|,\,$
The absolute value function is one of our tool kit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign. Knowing this, we can use absolute value functions to solve some kinds of real-world problems. Modulus operation on function converts negative function values to positive function values with equal magnitude. As such, we draw a graph of the modulus function by taking a mirror image of the corresponding core graph in x-axis.
We need under that if x lies between $- a < x < a$ then $\left| a \right| \leqslant x.$ |
CAT level questions on time and distance
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Quantitative aptitude is one of the most scoring sections in CAT. Aspirants can take free CAT mock tests to check where they stand. Solving previous year papers of CAT will help aspirants understand the level of difficulty of questions. In this blog, let us look at some CAT level questions on time and distance.
Time, speed, and distance is one of the simplest topics tested in CAT. Solved Time speed Distance CAT Questions will help you to learn more. Hence, aspirants must try to maximise their scores in these topics before moving on to esoteric concepts.
Basic terms that will be useful in solving CAT level questions for time-speed-distance are relative speed and effective speed. Aspirants can try online preparation for CAT to assist them in their preparation.
CAT level questions on time and distance:
Time * speed = distance.
Students can solve most of the CAT level questions on time and distance just by remembering this fundamental relationship.
Question 1:
A and B start at the same time from cities P and Q respectively that are 300 km apart and move towards each other. They meet at a point 120 km from P. At what time will A reach Q provided that both A and B start at 3 AM and B reaches P at 1 PM?
Since they start at the same time, the ratio of distances covered will be equal to the ratio of the speeds. Let us assume A’s speed to be ‘a’ and B’s speed to be ‘b’.
Hence, a/b = 120/(300-120)
a/b = 120/180
3a = 2b
b =1.5a.
A will cover 300 Km in 300/a hours.
B will cover 300 km in 300/1.5a = 200/a hours.
It is given that 200/a = 10 hours (3 AM to 1 PM)
=> a = 20
Hence, A will reach 300/20 = 15 hours after 3 AM.
Hence, A will reach Q at 6 PM.
Question 2:
Boats A and B travel at 20 kph and 30 kph. They start at 1 PM towards each other and collide at 2:30 PM. What will be the distance between the 2 boats just a minute before their collision provided that the distance between the two boats was 70 km at 1:06 PM.
This question is based on a previous year CAT question. The question is elementary but confuses the aspirants by providing superfluous data.
We can rephrase the question as ‘Two boats travelling at speeds 20 kph and 30 kph are travelling towards each other. What will be the distance travelled by them in a minute?’
Since the boats are travelling towards each other, they will cover 20+30 = 50 km in an hour. In a minute, they will cover 50/60 = 5/6 km.
Question 3:
A, B, and C run a race of 1000 m. A beats B by 5 seconds and C by 500 metres. B and C decide to run a 5000m race. C gives up mid way. For how much time must B run after C gives up to complete the race.
Let the speeds of A, B, and C be a, b, and c.
We know that (1000/a) = (1000/b) – 5 ——————(1)
Also, 1000/a = 500/c
=> a = 2c —————–(2)
Substituting a = 2c in (1), we get,
500/c = 1000/b – 5
1000/b = 500/c + 5. —————-(3).
C gives up midway in the 5000 m race. This means that C gives up after running 2500m.
Multiplying (3) by 5, we get, 5000/b = 2500/c + 25.
The above equation means that 25 seconds would have elapsed after C had run 2500m by the time B completes running 5000m. Hence, 25 seconds is the right answer.
Question 4:
The following question is based on a previous CAT question. This question will definitely feature among one of the most beautiful CAT level questions on time and distance.
A Cat is standing inside a tunnel at a point 3/8 th the length of the tunnel from the entrance. The cat hears a train whistling. The Cat will manage to barely save itself if it decides to run either towards the entrance of the tunnel or the exit. What is the ratio of speeds of the train and the Cat?
Let us assume the length of the tunnel to be 8 km for ease of solving.
We do not know where the train is. But, we know that if the Cat runs towards the entrance, it barely manages to save itself. Hence, by the time Cat runs 3 km, the train reaches the entrance of the tunnel.
Now, let us assume that the Cat runs towards the exit of the tunnel.
The cat is standing at a point 3 km from the entrance of the tunnel. By the time the cat runs 3 km, the train would have reached the entrance of the tunnel. Hence, by the time the train reaches the entrance, the Cat will be at a point 6 km from the entrance of the tunnel and 2 km from the exit of the tunnel.
We hope that this article on CAT level questions on time and distance would have given you some idea about the different types of questions that can appear in the exam. Try reading our other helpful blogs data interpretation basics for CAT and FAQs on CAT mocks.We know that the cat barely manages to save itself even if it runs towards the exit of the tunnel. Hence, by the time the cat covers 2 km (towards the exit), the train must have covered 8 km (the entire length of the tunnel). Hence, the ratio of speeds of the train to that of the cat is 8:2 = 4:1. |
Square Root of 20 in Simplest Radical Form
The square root of 20 in simplest radical form is 2√5, and it is written mathematically as follows: √20=2√5. In this post, we will learn how to find the square of root 20 in simplified radical form.
How to Simplify Root 20
Let us now find the square root of 20 in its simplest radical form. The easiest way to do that is to factorize the number 20. We will follow below steps.
Step 1:
First, we factorize the number 20. As 20 is an even number, it will be divisible by 2 and we can write
20 = 2 × 10 …(I)
The factorization of 10 is given by
10 = 2 × 5 …(II)
Combining (I) and (II), we get the prime factorization of 20 and it is given as follows:
20 = 2 × 2 × 5 …(∗)
Step 2:
In the next step, we take square roots on both sides of (∗). This will give us the square root of 20 as follows.
√20 = $\sqrt{2 \times 2 \times 5}$
= $\sqrt{2 \times 2}$ × $\sqrt{5}$, by the rule √ab = √a ×√b
= 2 × √5. Here we have used the surd formula √(a×a) =a
= 2√5.
So 2√5 is the simplified radical form of the square root of 20. In other words,
√20 = 2√5.
Video Solution on How to Simplify Root 20 Simplified:
Have You Read These Square Roots:
Root 8 Simplified: The root 8 simplified is equal to 2√2.
Root 12 Simplified: The root 12 simplified is equal to 2√3.
Root 18 Simplified: The root 18 simplified is equal to 3√2.
Root 27 Simplified: The root 27 simplified is equal to 3√3.
Root 50 Simplified: The root 50 simplified is equal to 5√2.
FAQs
Q1: What is the lowest radical form of root 20?
Answer: The lowest radical form of the square root of 20 is 2√5.
Q2: What is root 20 in simplified radical form?
Answer: The simplified radical form of square root 20 is equal to 2√5. |
How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
# Area of the figure | AMC-8, 2014 | Problem 20
Try this beautiful problem from Geometry based on the Area of the figure.
## Area of the figure - AMC-8, 2014- Problem 20
Rectangle $ABCD$ has sides $CD=3$ and $DA=5$. A circle of radius $1$ is centered at $A$, a circle of radius $2$ is centered at $B$, and a circle of radius $3$ is centered at $C$. Which of the following is closest to the area of the region inside the rectangle but outside all three circles?
• $3.5$
• $4.0$
• $4.5$
### Key Concepts
Geometry
Rectangle
Circle
Answer: $4.0$
AMC-8 (2014) Problem 20
Pre College Mathematics
## Try with Hints
To Find out the area of the region inside the rectangle but outside all three circles,( i.e the red shaded region in the above figure), we have to find out the area of the Rectangle -the area of three quarter circle inside the circle(i.e green shaded region)
Can you now finish the problem ..........
To find out the area of the rectangle, AD=5 and CD=3 are given. to find out the area of the three-quarter circles, the radii are 1,2 & 3 respectively.
Now area of Rectangle=$AD \times CD$ and area of quarter circles =$\frac{\pi r^2}{4}$,where $r$=Radius of the circle
can you finish the problem........
Area of the rectangle=$5 \times 3$=15 sq.unit
Area of the quarter circle with the center C=$\frac{\pi (3)^2}{4}$=$\frac{9 \pi}{4}$ sq.unit
Area of the quarter circle with the center B= $\frac{\pi (2)^2}{4}$ =$\frac{4 \pi}{4}$=$\pi$ sq.unit
Area of the quarter circle with the center C= $\frac{\pi (1)^2}{4}$ =$\frac{\pi}{4}$ sq.unit
Therefore the area of the region inside the rectangle but outside all three circles,( i.e the red shaded region in the above figure) =(15- $\frac{9 \pi}{4}$- $\pi$ - $\frac{\pi}{4}$ )=15-$\frac{7\pi}{2}$=15-11=4 sq.unit (Taking $\pi =\frac{22}{7}$) |
# How do you solve 19 < 7 + 1/3k + 2 -2/3k?
May 12, 2015
Solve $19 < 7 + \frac{1}{3} k + 2 - \frac{2}{3} k$ .
Combine like terms.
$19 < 7 + 2 + \frac{1}{3} k - \frac{2}{3} k$ =
$19 < 9 - \frac{1}{3} k$
Simplify $\frac{1}{3} k$ to $\frac{k}{3}$.
$19 < 9 - \frac{k}{3}$
Multiply both sides times $3$.
$19 \cdot 3 < 9 \cdot 3 - k$ =
$57 < 27 - k$
Subtract #27 from both sides.
$57 - 27 < 27 - 27 - k$ =
$30 < - k$
Multiply both sides times $- 1$. This will cause the inequality to be reversed.
$- 30 > k$
Flip to get $k$ on the left side.
$k < - 30$ |
# What Is A Rectangle In Math?
## Is a rectangle a square?
A square is also a parallelogram whose sides intersect at 90° angles. Therefore, all of its sides are congruent. A rectangle is a square when both pairs of opposite sides are the same length. This means that a square is a specialized case of the rectangle and is indeed a rectangle.
## Can a rectangle have 4 equal sides?
A rectangle has two pairs of opposite sides parallel, and four right angles. A square has two pairs of parallel sides, four right angles, and all four sides are equal. It is also a rectangle and a parallelogram. A rhombus is defined as a parallelogram with four equal sides.
## Why is it called rectangle?
The word rectangle comes from the Latin rectangulus, which is a combination of rectus (as an adjective, right, proper) and angulus (angle). Other geometries, such as spherical, elliptic, and hyperbolic, have so- called rectangles with opposite sides equal in length and equal angles that are not right angles.
## What is a rectangle shape look like?
Rectangle. A rectangle is a shape with parallel opposite sides, combined with all 90 degree angles. In a rectangle, one set of parallel sides is longer than the other, making it look like an elongated square.
## Why can’t a rectangle be a square?
A rectangle can be tall and thin, short and fat or all the sides can have the same length. Thus every square is a rectangle because it is a quadrilateral with all four angles right angles. However not every rectangle is a square, to be a square its sides must have the same length.
## How do you prove a square is a rectangle?
If the quadrilateral is a rectangle with two consecutive sides congruent, then it is a square. If the quadrilateral is a rectangle with perpendicular diagonals, then it is a square.
## What else can you call a rectangle?
Other names of the rectangle Since all the angles of a rectangle are equal, we also call it an equiangular quadrilateral. Since it has parallel sides, we can also call it a parallelogram. A parallelogram is a quadrilateral whose opposite sides are equal and parallel.
## What do you call a rectangle with uneven sides?
What is an irregular quadrilateral? Irregular quadrilaterals are: rectangle, trapezoid, parallelogram, kite, and rhombus. They are symmetrical, but are not required to have congruent sides or angles.
## What are the sides of a rectangle called?
A rectangle is a four-sided flat shape where every angle is a right angle (90°). Opposite sides are parallel and of equal length (so it is a Parallelogram).
## How do you identify a rectangle?
For a shape to be a rectangle, it must be a four-sided polygon with two pairs of parallel, congruent sides and four interior angles of 90° each. If you have a shape that matches that description, it also is all this: A plane figure.
## What are 4 attributes of a rectangle?
A rectangle is a quadrilateral with four right angles. Thus, all the angles in a rectangle are equal (360°/ 4 = 90°). Rectangle
• All the angles of a rectangle are 90°
• Opposite sides of a rectangle are equal and Parallel.
• Diagonals of a rectangle bisect each other.
You might be interested: Quick Answer: How To Make Math Problems?
## What is the square area of a rectangle?
To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area.
## What makes a rectangle special?
A rectangle is a special case of a parallelogram; its opposite sides are parallel. Like a parallelogram, the opposite sides are equal in length to each other. There are two pairs of opposite sides, and each pair could have a different length, but each pair’s sides will be equal to each other. |
Categories
# Probability Distribution: Binomial and Poisson
• BINOMIAL PROBABILITY DISTRIBUTION
• POISSON PROBABILITY DISTRIBUTON
Probability distribution deals with theoretical probability model based on the randomness of certain natural occurrences. The binomial and Poisson distribution are discrete distribution
BINOMIAL DISTRIBUTION
This arises from a repeated random experiment which has two possible outcomes.
The two possible outcomes of the random experiment are usually called success and failure.
Prob( success) = P, Prob(failure) = q
Since the two events are complementary, hence p+ q = 1 or p = 1-q, q = 1 – p
The probability of success or failure of an event is the same for each trials and does not influence the probability of success or failure of another trial of the same event.
:. Binomial distribution of n trails and r required outcome(s) is defined as :
pr(x = r) = nCrPrqn-r
whennCr = n!
(n-r)! r!.
The binomial distribution is suitable when the number of trials is not too large.
Example:
1. Find the probability that when two fair coins are tossed 5 times a head and a tail appear three times.
Solution:
Two fair coins = (HT, TH, TT, HH) = 4
Prob (a head and a tail) = 2/4 = ½
i.e p = ½ , q = ½ (p + q = 1)
n = 5, r = 3.
:. P(x = r) =nCrprqn-r
p ( x = 3) = 5C3 ( ½ ) 3 ( ½ ) 5-3
p (x = 3) = 10 x 1/8 x ¼ = 10/32 = 5/16
p (x = 3) = 0.3125.
EVALUATION
Find the probability that when a fair six-faced die is tossed six times, a prime number appears exactly four times.
POISSON DISTRIBUTION: The Poisson distribution is more suitable when the number of trials is very large and probability of successes is small. It is defined as:
Pr(x) = λx e– λ , x = 0, 1, 2, 3,
x!
Where λ = np e = 2.718
P = probability of success, n = number of trials.
EVALUATION
The probability that a person gets a reaction from a new drug on the market is 0.001. If 200 people are treated with this drug. Find approximately, the probability that:
• exactly three persons will get a reaction
• more than two person will get a reaction
Properties of Binomial and Poisson Distribution.
Binomial
It assigns probability to non-occurrence of events i.eProb( x = 0)
Mean µ = np
Standard deviation, r = √npq
Variance ð2 = npq
Poisson
It assigns probability to non-occurrence of events i.eProb(x = 0)
Mean µ = λ = np
Standard deviation, ð = √λ = √np
Variance, ð2 = λ = np
Example:
1. In the probability of tossing a fair coin three times, a head shows up twice. Find the mean and standard deviation.
Solution
n = 3 Prob(a head) = ½ , i.e p = ½ , q = ½
I. Mean µ = np = 3 x ½ = 3/2
II. Standard deviation :r = √npq = 3 x ½ x ½ = ¾
Example
2. 0.2% of the cooks produced by a machine were found to be defective. If there are 1000 corks, find the mean and standard deviation?
Solution
P = 0.2% = 0.002.
N = 1000
I. mean µ = λ = np = 1000 x 0.002 = 2
II. r = √np = √2
EVALUATION
In an examination, 60% of the candidates pass. If 10 candidates were sampled. Find the mean, standard deviation and variance of the candidates.
GENERAL EVALUATION
1. The probability that a person gets a reaction from a new drug in the market is 0.001. If 2000 people were treated with this drug, find the mean and standard deviation.
2. 1. In an examination, 60% of the candidates passed. Use the binomial distribution to calculate the probabilities that a random sample of 10 candidates contain exactly 2 failures.
Read probability distribution, further math. Project 3 from page 198-201.
WEEKEND ASSIGNMENT
1. What is the variance of a binomial distribution?
(a) np (b) √npq ( c ) npq (d) p2
2. The mean (µ) of a poisson distribution is the same as
(a) Standard deviation (b) variance (c) mean (d) mean deviation
3. If number of trials is 100 and probability of success is 0.0001, what is the variance of this distribution?
(a) 0.00999 (b) 0.1 (c ) 0.01 (d) 0.001
4. If the birth of a male child and that of a female child are equiprobable. Find the probability that in a family of five children exactly 3 will be male. (a) 16/5 (b) 5/16 (c) 5/32 (d) 5/21
5. If an unbiased die is thrown repeatedly, what are the chances that the first, six to be thrown will be the third throw? (a) 25/216 (b) 1/6 (c) 25/36 (d)25/31
THEORY
1. 20% of the total production of transistors produced by a machine are below standard. If a random sample of 6 transistors produced by the machine is taken, what is the probability of getting
(i) exactly 2 (ii) exactly 1 (iii) at least 2 (iv) at most 2 standard transistors?
2. A fair die is thrown five times. Calculate correct to 3 decimal places, the probability of obtaining
(a) at most two sixes (b) exactly three sixes |
# Problems on Division of Fractional Numbers
Problems on division of fractional numbers will give idea how to solve different types of word problems.
1. In a school 2/7 of the students are boys. If there are 280 girls, find the number of boys.
Solution:
Suppose the number of students = 1
Number of boys = 1 × 2/7 = 2/7
Number of girls = 1 - 2/7 = 5/7
5/7 of the students = 280
Therefore, total number of students = 280 ÷ 5/7
= 280 × 7/5
= 392 students
Number of boys = 392 - 280 = 112 boys
2. Ron has 5 ¼ m long ribbon. She cuts it into three equal pieces. What is the length of each piece of ribbon?
Solution:
Length of the ribbon = 5 ¼ m
Number of equal pieces cut out = 3
Length of each piece of ribbon = 5 ¼ ÷ 3
= 21/4 ÷ 3
= 21/4 × 1/3
= 21 × 1/4 × 3
= 7/4 metre
= 1 ¾ metre
Therefore, length of each piece of ribbon is 1 ¾ metre.
3. The cost of 4 ½ m cloth is $60 ¾. Find the cost of 1 m cloth. Solution: Cost of 4 ½ m cloth =$ 60 ¾
Cost of 1 m cloth = $(60 ¾ ÷ 4 ½) =$ (243/4 × 2/9)
= $27/2 =$ 13 ½
4. Sam paid $45 for 3 ¾ kg of sugar. How much did he pay for 1 kg of sugar? Solution: Cost of 3 ¾ kg of sugar =$ 45
Cost of 1 kg of sugar = 45 ÷ 3 ¾
= 45 ÷ 15/4
= 45/1 × 4/15
= $12 Therefore, cost of 1 kg of sugar is$ 12.
Multiplication of a Whole Number by a Fraction
Multiplication of Fractional Number by a Whole Number.
Multiplication of a Fraction by Fraction.
Properties of Multiplication of Fractional Numbers.
Multiplicative Inverse.
Worksheet on Multiplication on Fraction.
Division of a Fraction by a Whole Number.
Division of a Fractional Number.
Division of a Whole Number by a Fraction.
Properties of Fractional Division.
Worksheet on Division of Fractions.
Simplification of Fractions.
Worksheet on Simplification of Fractions.
Word Problems on Fraction.
Worksheet on Word Problems on Fractions.
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## Recent Articles
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## Precalculus (6th Edition) Blitzer
The required solution is {{\left( a+b \right)}^{n}}=\left( \begin{align} & n \\ & 0 \\ \end{align} \right){{a}^{n}}+\underline{\left( \begin{align} & n \\ & 1 \\ \end{align} \right)}{{a}^{n-1}}\cdot b+\underline{\left( \begin{align} & n \\ & 2 \\ \end{align} \right)}{{a}^{n-2}}\cdot {{b}^{2}}+\underline{\left( \begin{align} & n \\ & 3 \\ \end{align} \right)}{{a}^{n-3}}\cdot {{b}^{3}}+...+\underline{\left( \begin{align} & n \\ & n \\ \end{align} \right)}\cdot {{b}^{n}}
The sum of exponents on a and b in each term is n. While writing the expansion of ${{\left( a+b \right)}^{n}}$, we note the following points: The first term of the expansion of ${{\left( a+b \right)}^{n}}$ is ${{a}^{n}}$. And the exponents on a decrease by 1 in each successive term. And the exponents on b in the expansion of ${{\left( a+b \right)}^{n}}$ increase by 1 in each successive term. In the first term, the exponent on b is 0 as ${{b}^{0}}=1$. And the last term is ${{b}^{n}}$. And the sum of the exponents on the variables in any term in the expansion of ${{\left( a+b \right)}^{n}}$ is equal to n. So, the total number of terms in the polynomial expansion is one more than the power of the binomial, i.e., n. And there are (n+1) terms in the expanded form of ${{\left( a+b \right)}^{n}}$. Hence, for any positive integers n, |
# Math Worksheets Land
Math Worksheets For All Ages
# Math Worksheets Land
Math Worksheets For All Ages
# Understanding Positive and Negative Numbers Worksheets
What Is the Difference Between Positive and Negative Numbers? We are all familiar with the plus and minus signs. We use these signs to add values together or find the difference between two values. These same signs can also be used to tell us the nature of a quantity as well. Numbers can broadly have a positive and/or negative presence in a system. Positive numbers have a plus sign and negative numbers have a minus sign with them. If no visible sign appears in front of them, it is assumed to be positive. Let's visualize the difference with the help of a number line. Any number that moves to the right of the number line is considered to be positive. And any number that runs to the left of the number line asserts a negative tone. Let us have a clearer idea of how the number line works. When adding two numbers on a number line, we go to the first addend, and from there we walk all the way to the second addend. Thus, for 5 + (-3) we will start at 5 and go three units to the left-to reach at our result at +2. So as we can see these are simple measures of how far from 0 these values are, in one direction or another.
### Aligned Standard: Grade 6 Numbers - 6.NS.C.5
• Answer Keys - These are for all the unlocked materials above.
### Homework Sheets
It was very difficult coming up with unique problem types and situations for these.
• Homework 1 - Max buys shares of Enron at \$785 on Monday. On Thursday, the stock falls to \$464, so he sells his shares. How much did he lose?
• Homework 2 - When the number is negative it means the quantity is decreasing. If the quantity increases that means the value is positive. So if the temperature drops that means the value is decreasing.
• Homework 3 - If the temperature were to drop 38 degrees, which number would display this best?
### Practice Worksheets
We break out the number lines here to help make it clear for students.
• Practice 1 - Complete the following operation using the number line. They have a range of + to - 13.
• Practice 2 - We bring in decimals to see if that throws them off.
• Practice 3 - A mix of real world word problems and number lines.
### Math Skill Quizzes
I have been petitioning the standards committee to consider a vertical numbers line. I see them online all the time.
• Quiz 1 - Display each value as an integer. These are real situations that you may come in contact with.
• Quiz 2 - 450 feet below sea level. The first problem will stump you.
• Quiz 3 - The stock market falls 66,864 points today. Wow! That is a bad day!
### When Will You Be Confronted with Negative Numbers in the Real World?
When students first come across the concept of a negative value, they are quickly confused because it is a little off from where they began their concept about integers. One of the first situations where you have to contemplate the meaning of a negative value is in the realm of temperature. If you are in a country that follows the metric system, you will see it more regularly, especially if you are far away from the equator. Negative values of the Celsius scale of temperature are cold. If you are on in a system that revolves around the Fahrenheit scale, those negative temperature values are overwhelming cold. As we can quickly realize negative values are somewhat relative to the scale you use to quantify it. You will also commonly ponder negative values in situation where you are quantifying financial values. This can be when you are measuring your bank balance or overall income. In most cases that is not a good thing. Vertical elevation is another place that we see this. If we think of sea level as the zero in our scale, it is an easy concept to grasp. There are many games that students play that also have values that can reduce their overall skill. They are often used to assess penalties or faults that they endure. When we are working with students it is often helpful to bring these real world measures to light to help them understand the of overall abstract nature of this concept.
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# Prove that $5^{1/3}+7^{1/2}$ is irrational
Goal: Prove that $5^{1/3}+7^{1/2}$ is irrational.
Idea: We can prove this is irrational by supposing it is rational and finding a contradiction. So, $5^{1/3}+7^{1/2} = p/q$ where $p$ and $q$ are integers that have no factors in common other than $1$. The issue here is that I can not seem to find a way manipulate this equation to a form where I can find contradiction. I tried taking each side to the power of 6… That was an unmanageable mess. Any ideas?
#### Solutions Collecting From Web of "Prove that $5^{1/3}+7^{1/2}$ is irrational"
Suppose that our sum is equal to the rational number $r$. Then $5^{1/3}=r-\sqrt{7}$. Cubing both sides we obtain
$$5=r^3-3r^2\sqrt{7}+21r-7\sqrt{7},$$
and therefore
$$r^3+21r-5-(3r^2+7)\sqrt{7}=0.$$
Since $\sqrt{7}$ is irrational, we must have $3r^2+7=0$. This is impossible.
Remark: For completeness we sketch a proof of the irrationality of $\sqrt{7}$. It is much like one of the standard proofs of the irrationality of $\sqrt{2}$. Suppose to the contrary that there are integers $p$ and $q$, with $q\ne 0$, such that $p^2/q^2=7$. Without loss of generality we may assume $p$ and $q$ are relatively prime.
Then $p^2=7q^2$. Since $7$ is prime and divides $p^2$, it follows that $7$ must divide $p$. Let $p=7t$. Substituting and cancelling, we get $7t^2=q^2$. Thus $7$ divides $q$, contradicting the fact that $p$ and $q$ are relatively prime.
Let $t=5^{1/3}+7^{1/2}$. Then $t$ is a root of $x^6-21 x^4-10 x^3+147 x^2-210 x-318=0$. (*)
The rational root theorem tell us that either $t$ is irrational or $t$ is an integer dividing $318$.
Since $1 < 5^{1/3} < 2$ and $2 < 7^{1/2} < 3$, we have $3 < t < 5$, and so if $t$ is an integer, it must be $4$. But $4$ does not divide $318$.
Therefore, $t$ is irrational.
(*) This polynomial can be found manually (if you don’t drown in the algebra) or you can ask WA. This is not cheating (well, not crucially) because the nice argument is still to come. It’s an opportunity to learn the rational root theorem, if you haven’t seen it.
Another solution is to look at $K = \mathbb{Q}(\zeta_3, \sqrt[3]{5}, \sqrt[2]{7})$ (where $\zeta_3$ is primitive cube root of unity) which is Galois over $\mathbb{Q}$. It is easy to see that $(K : \mathbb{Q}) = 6 \times 2 = 12$ and the automorphism in $\text{Gal}(K|\mathbb{Q})$ are easily described: they are of the form $\sigma_{i,j,\pm} : \{\zeta_3 \mapsto \zeta_3^i\} \times \{\sqrt[3]{5} \mapsto \sqrt[3]{5} \zeta_3^j\} \times \{\sqrt[2]{7} \mapsto \pm \sqrt[2]{7}\}$ where $i \in \{1, 2\}, j \in \{0, 1, 2\}$. It is easy to check that one of the automorphism $\sigma_{1,2,+}$ does not fix $\alpha = \sqrt[3]{5} + \sqrt[2]{7}$. Hence, $\alpha \not\in \mathbb{Q}$.
Hint $\ \sqrt7\, =\, r – \sqrt[3]5,\,\ r \in\Bbb Q\,\overset{\rm square}\Rightarrow \sqrt[3]5\,$ is a root of a quadratic polynomial $f(x) \in \Bbb Q[x],$ contra $\,x^3-5\,$ is irreducible over $\,\Bbb Q\$ (e.g. by the Rational Root Test).
Remark $\$ If you know a little algebra then you may recognize that the idea implicit in the above is that $\,x^3 – 5\,$ is irreducible hence the minimal polynomial of $\,\sqrt[3]5.\,$ If you go on to study field theory then you can deduce it from a more general result, viz. $\,\Bbb Q(\sqrt[3]5)$ and $\Bbb Q(\sqrt7)$ have coprime degrees $3,2$ over $\,\Bbb Q,\,$ hence their intersection is $\,\Bbb Q.\,$ So $\,\sqrt[3]5 \in\Bbb Q(\sqrt 7)\Rightarrow \sqrt[3]5\in\Bbb Q$.
Edit: Swapped the cube root and the square root in the display. Fixed. This also altered the subsequent argument.
To use your method, you never want to raise to a power anything that has both $\sqrt[3]{5}$ and $\sqrt{7}$ in it. So, you isolate these one at a time.
Suppose $\sqrt[3]{5} + \sqrt{7} \in \Bbb{Q}$ so there are $p,q \in \Bbb{Z}$ with $q \neq 0$ such that $\sqrt[3]{5} + \sqrt{7} = p/q$. By reducing $p/q$ to lowest terms, we may assume that $\gcd(p,q) = 1$. Since solving quadratics is harder than solving linears (for the second isolation), we start by cubing. We compute \begin{align} \sqrt[3]{5} &= p/q – \sqrt{7}, \\ 5 &= (p/q – \sqrt{7})^3 \\ &= p^3/q^3 – p^2\sqrt{7}/q^2 + 7p/q – 7\sqrt{7}, \text{ and}\\ \sqrt{7} &= \frac{5-p^3/q^3 – 7p/q}{-p^2/q^2 – 7} \\ 7 &= \left( \frac{5q^3 – p^3 – 7pq^2}{-p^2q – 7q^3} \right)^2, \text{ so} \\ 7q^2 \left( p^2 + 7q^2 \right)^2 &= \left( p^3 + 7pq^2 – 5q^3 \right)^2 . \end{align}
Now work modulo $4$. The squares are congruent to $0$ or $1$, so the left-hand side is either $0$ or $3 \pmod{4}$ and the right-hand side is either $0$ or $1 \pmod{4}$. So, either $q$ or $p^2 + 7q^2$ is even and $p^3 + 7pq^2 – 5q^3$ is even.
If $q$ is even, $p^3 + 7pq^2 – 5q^3 \cong p^3 \pmod{4}$. Since $p^3 + 7pq^2 – 5q^3$ is even, we must have $p$ is even, contradicting our assumption that $p/q$ was reduced to lowest terms.
Otherwise, we have $p^2 + 7q^2$ is even. Note that $p^3 + 7pq^2 – 5q^3 = p(p^2+7q^2)-5q^3$, so is even only if $q$ is even. But then $p^2 + 7q^2$ is even only if $p$ is and again we contradict reduction to lowest terms. |
# How to solve a quadratic equation with two unknowns?
I know how to solve quadratic equations when there's only one unknown, but I'm a bit confused on what I do if I have $2$ unknowns e.g: $$x^2 + 2kx + 81 =0$$ With just $x^2 + 2kx + c=0$, obviously $x=-k \pm \sqrt{k^2 -c}$.
I tried substituting the above value into the equation where $c$ is $81$ and then solving it, but I just thought that I was overcomplicating everything. How do I get the possible values of $k$ in this equation when there's an $x$ too?
Remember the discriminant of the quadratic $ax^2 + bx + c = 0$ is $b^2 - 4ac$. You need the discriminant to be positive to get real solutions in $x$.
Hence in your equation you need to find for which $k$ to you get : $$(2k)^2 - 4\cdot 81 \ge 0 \qquad \Longleftrightarrow \qquad k^2 \ge 81$$
Ok to solve? Don't forget $k$ could be negative ;-)
In order to get the possible values of $k$ in the equation $$x^2 + 2kx + 81 =0$$ consider the system: \begin{cases} y=-\frac{x^2+81}{2x}\\ y=k \end{cases} The solution of $k$ would be the intersection of the two functions: the second one is simply a stack of lines parallel to $x-$axis, while the first one should be studied graphically with a sketch. Let $f(x) = -\frac{x^2+81}{2x}$, $f'(x) = \frac{81}{2x^2}-\frac12$, and $f'(x) = 0 \iff x=\pm 9\Rightarrow$ there 2 are stationary points in $x=\pm9$, so the tangents to the function $f(x)$ are parallel to $x-$axis:
Finally it can be stated that: \begin{cases} 0 \text{ solutions for }x \iff -9\lt k \lt 9\\ 1 \text{ solution for }x \iff k=\pm9\\ 0 \text{ solutions for }x \iff k\lt -9 \lor k\gt 9\\ \end{cases} So all the possible values of $k$ are in $[9;+\infty[$ and in $]-\infty;-9[$. |
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# What constitutes a rational expression?
How would you explain this concept to someone unfamiliar with it? Demonstrate with one or more examples.
Explain the basic steps involved in simplifying rational expressions. What makes a rational expression undefined?
How are the operations of multiplication, division, addition, and subtraction of rational expressions similar or different from operations on fractions?
A rational expression is a ratio of two polynomials. A polynomial is an expression that can have constants, variables and exponents, but:
- is not divided by a variable. [like 2/(x+2)]
- a variable's exponents can only be 0,1,2,3,... etc. (no negatives or fractions)
- it can't have an infinite number of terms.
So an example of a rational expression would be:
(x+ 5) / (x + 2)
It is "Rational" because one polynomial is divided by the other, like a ratio.
Simplifying a rational expression is reducing it just like a regular fraction, to simplest terms. In the case of polynomials that means the least number of terms (a polynomial term is every section of the polynomial separated by + or -). In the above rational expression we can infer that x != -2 (not equal) because the result would make the denominator 0 and that makes it undefined (just like regular fractions).
An example of simplifying a rational expression:
x2 + 5x + 6
x + 2
First we factor the numerator:
x2 + 5x + 6 = (x + 3)(x + 2)
x + 2 x + 2
See the x + 2 term on both the top and bottom? They cancel each other out leaving:
x + 3
As the simplified answer.
As for operations in rational expressions they function exactly as they do in fractions EXCEPT you can not break apart terms linked by a + or - sign (this is why you have to factor). You can combine like terms, you can factor larger polynomials ( 2nd, 3rd, 4th degree ect.) but you absolutely can't for instance say:
x + 6 = 3 thinking that x6 = 1 + 2 = 3
x + 3 x 3
Hope this helps :) |
# Pulkit's salary was Rs 35,000. He got a raise of Rs 7,000. Reena's salary was Rs 24,000. She was given a raise of Rs 6,000. Find what fraction of the original income the raise was for each and which one of the two got a higher raise.
Given :
Salary of Pulkit = Rs. 35,000
Raise in salary of Pulkit = Rs. 7,000
Salary of Reena = Rs. 24,000
Raise in salary of Reena = Rs. 6,000
To find :
We have to find what fraction of the original income the raise was for each and which one of the two got a higher raise.
Solution :
The fractional raise in salary for Pulkit = $\frac{7000}{35000}=\frac{1}{5}.$
The fractional raise in salary for Reena = $\frac{6000}{24000}=\frac{1}{4}.$
We know that,
If a > b then $\frac{1}{a}<\frac{1}{b}$ where a and b are positive integers.
Therefore,
5>4 implies $\frac{1}{5}<\frac{1}{4}$
This implies,
Reena got a higher raise.
Tutorialspoint
Simply Easy Learning |
# Vectors, Matrices, Tensors – What’s The Difference?
Linear algebra is a branch of mathematics which deals with solving a system of linear equations. It is widely used throughout science and engineering and it is essential to understanding machine learning algorithms.
Linear algebra defines three basic data-structures – vectors, matrices and tensors, which are constantly used in machine and deep learning. In this blog post we’ll go over all three structures and show how to do some simple math using them.
## Vectors
In the computer science world, we represent vectors using one-dimensional arrays of number. A vector can contain any number of values.
If a vector holds n number of values, we say that vector is an n-dimensional vector.
I just said that a vector is a one-dimensional array, but I also said that it can be n-dimensional as well. What’s with that?
In the computer science world and programming, a vector is always represented with a one-dimensional array, but a vector can be n-dimensional. The dimensions of an vector are not the same as the dimensions of an array. A vector is said to be n-dimensional if it stores n number of values.
This image above is a representation of a vector.
Vectors define points in space, where each value represents a coordinate at a certain axis.
So a vector with values [-4, 4, 1] would look something like this:
Vector is often represented as a directional line going from the origin point to the point which is defined by the values of the vector. The origin point is mostly defined to be the point at which coordinates for all of the axis are 0. So in the example above, the origin point is at (0, 0, 0) because we are representing a three-dimensional vector.
In the case above, we represented the vector in the vector space, because the values stored in the vectors are real numbers and the vector is a three-dimensional vector.
In the case of deep learning, we usually use vectors to sore biases, as they are one dimensional arrays.
## Matrices
Matrices are two-dimensional arrays. Elements of the matrix are accessed by two incidences, mostly set to be i and j.
If we have a matrix A with m rows and n columns, then we write .
In deep learning, we use matrices with weights and inputs.
## Tensors
Tensors are mostly described as n-dimensional arrays (not vectors, but computer science arrays), where the .
# Some basic matrix and vector math
Matrices are very computationally efficient so they’re used in computer games and, of course, machine learning.
So the two most basic math operators that are used in matrix math are multiplication and addition.
## Multiplication
Linear algebra features two main types of matrix multiplication – the dot product and the Hadamard product.
The Hadamard product is the element-wise product and it is the one that most people would assume that we’d be using. It’s denoted like this: .
The more commonly used operation in machine learning is the dot product. When calculating the dot product of two matrices, the first matrix needs to have the same number of columns as the second one has rows. It is denoted like this: .
When you say matrix multiplication, most of the time you’d mean the dot product.
Matrix multiplication is associative and distributive, but not necessarily commutative.
You can also apply the dot product to two vector of the same dimensionality by calculating .
The T in the exponent of x is the transpose operator. A transpose is a mirror image across a diagonal line. |
# How to solve definite integrals
We can do your math homework for you, and we'll make sure that you understand How to solve definite integrals. We can help me with math work.
## How can we solve definite integrals
College algebra students learn How to solve definite integrals, and manipulate different types of functions. Live math help is a great resource for students who are struggling with math. The live tutors are available to answer questions and help with homework. Live math help is also a great way to get extra practice with math. The tutors can provide practice problems and walk through the solutions. Live math help is a great resource for students of all levels.
Algebra is the branch of mathematics that deals with the solution of equations. In an equation, the unknown quantity is represented by a letter, usually x. The object of algebra is to find the value of x that will make the equation true. For example, in the equation 2x + 3 = 7, the value of x that makes the equation true is 2. To solve an equation, one must first understand what each term in the equation represents. In the equation 2x + 3 = 7, the term 2x represents twice the value of x; in other words, it represents two times whatever number is assigned to x. The term 3 represents three units, nothing more and nothing less. The equal sign (=) means that what follows on the left-hand side of the sign is equal to what follows on the right-hand side. Therefore, in this equation, 2x + 3 is equal to 7. To solve for x, one must determine what value of x will make 2x + 3 equal to 7. In this case, the answer is 2; therefore, x = 2.
One way is to graph the function and see where it produces a result. Another way is to look at the definition of the function and see what values of x will produce a result. For example, if we have a function that takes the square root of x, we know that we can only take the square root of positive numbers. Therefore, our domain will be all positive numbers. Once we have found the domain, we can then solve for specific values by plugging in those values and seeing what outputs we get. This process can be helpful in solving problems and understanding how functions work.
A rational function is any function which can be expressed as the quotient of two polynomials. In other words, it is a fraction whose numerator and denominator are both polynomials. The simplest example of a rational function is a linear function, which has the form f(x)=mx+b. More generally, a rational function can have any degree; that is, the highest power of x in the numerator and denominator can be any number. To solve a rational function, we must first determine its roots. A root is a value of x for which the numerator equals zero. Therefore, to solve a rational function, we set the numerator equal to zero and solve for x. Once we have determined the roots of the function, we can use them to find its asymptotes. An asymptote is a line which the graph of the function approaches but never crosses. A rational function can have horizontal, vertical, or slant asymptotes, depending on its roots. To find a horizontal asymptote, we take the limit of the function as x approaches infinity; that is, we let x get very large and see what happens to the value of the function. Similarly, to find a vertical asymptote, we take the limit of the function as x approaches zero. Finally, to find a slant asymptote, we take the limit of the function as x approaches one of its roots. Once we have determined all of these features of the graph, we can sketch it on a coordinate plane.
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Use substitution to solve the system Solve math problems online step by step free Hard math problems Word problem college algebra Rational expression solver |
Jump to a New ChapterIntroduction to the SAT IIContent and Format of the SAT II Math IICStrategies for SAT II Math IICMath IIC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
5.1 Math IIC Algebra Strategies 5.2 Equation Solving 5.3 Writing Equations 5.4 Manipulating Equations 5.5 Systems of Equations
5.6 Common Word Problems 5.7 Polynomials 5.8 Key Formulas 5.9 Review Questions 5.10 Explanations
Manipulating Equations
Now that you know how to set up an equation, the next thing you need to do is solve for the value that the question asks for. Above all, the most important thing to remember when manipulating equations is that each side of the equation must be manipulated in the same way. If you divide one side of an equation by 3, you must divide the other side by 3. If you take the square root of one side of an equation, take the square root of the other.
By treating the two sides of the equation in the same way, you won’t violate the equality of the equation. You will, of course, change the form of the equation—that’s the point of manipulating it. But the equation will always remain true as long as you do the same thing to both sides.
For example, let’s look at what happens when you manipulate the equation 3x + 2 = 5, with x = 1.
1. Subtract 2 from both sides:
1. Multiply both sides by 2:
1. Add 4 to both sides:
These examples show that you can tamper with the equation as long as you tamper the same way on both sides. If you follow this rule, you can manipulate the question without affecting the value of the variables.
Solving an Equation with One Variable
To solve an equation with one variable, you must manipulate the equation to isolate that variable on one side of the equation. Then, by definition, that variable is equal to whatever is on the other side, and you have successfully “solved for the variable.”
For the quickest results, take the equation apart in the opposite order of the standard order of operations. That is, first add and subtract any extra terms on the same side as the variable. Then, multiply and divide anything on the same side of the variable. Next, raise both sides of the equation to a power or take their roots. And finally, do anything inside parentheses. This process is PEMDAS in reverse (SADMEP!). The idea is to “undo” everything that is being done to the variable so that it will be isolated in the end. Let’s look at an example:
In this equation, the variable x is being squared, multiplied by 3, added to 5, etc. We need to do the opposite of all these operations in order to isolate x and thus solve the equation.
First, subtract 1 from both sides of the equation:
Then, multiply both sides of the equation by 4:
Next, divide both sides of the equation by 3:
Now, subtract 5 from both sides of the equation:
Again, divide both sides of the equation by 3:
Finally, take the square root of each side of the equation:
We have isolated x to show that x = ±5.
Sometimes the variable that needs to be isolated is not located conveniently. For example, it might be in a denominator, or an exponent. Equations like these are solved the same way as any other equation, except that you may need different techniques to isolate the variable. Let’s look at a couple of examples:
Solve for x in the equation .
The key step is to multiply both sides by x to extract the variable from the denominator. It is not at all uncommon to have to move the variable from one side to the other in order to isolate it. Here’s another, slightly more complicated, example:
Solve for x in the equation .
This question is a good example of how it’s not always simple to isolate a variable. (Don’t worry about the logarithm in this problem—we’ll review these later on in the chapter.) However, as you can see, even the thorniest problems can be solved systematically as long as you have the right tools. In the next section we’ll discuss factoring and distributing, two techniques that were used in this example.
Having just given you a very basic introduction to solving equations, we’ll reemphasize two things:
1. Do the same thing to both sides.
2. Work backward (with respect to the order of operations).
Now we’ll work with some more interesting tools you will need to solve certain equations.
Distributing and Factoring
Distributing and factoring are two of the most important techniques in algebra. They give you ways of manipulating expressions without changing the expression’s value. In other words, distributing and factoring are tools of reorganization. Since they don’t affect the value of the expression, you can factor or distribute one side of the equation without doing the same for the other side of the equation.
The basis for both techniques is the following property, called the distributive property:
Similarly,
a can be any kind of term, from a variable to a constant to a combination of the two.
Distributing
When you “distribute” a factor into an expression within parentheses, you simply multiply each term inside the parentheses by the factor outside the parentheses. For example, consider the expression 3y(y2 – 6):
If we set the original, undistributed expression equal to another expression, you can see why distributing facilitates the solving of some equations. Solving 3y(y2 – 6) = 3y3 + 36 looks quite difficult. But when you distribute the 3y, you get:
Subtracting 3y3 from both sides gives us:
Factoring
Factoring an expression is essentially the opposite of distributing. Consider the expression 4x3 – 8x2 + 4x, for example. You can factor out the greatest common factor of the terms, which is 4x:
The expression simplifies further:
See how useful these techniques are? You can group or ungroup quantities in an equation to make your calculations easier. In the last example from the previous section on manipulating equations, we distributed and factored to solve an equation. First, we distributed the quantity log 3 into the sum of x and 2 (on the right side of the equation). We later factored the term x out of the expression x log 2 – x log 3 (on the left side of the equation).
Distributing eliminates parentheses, and factoring creates them. It’s your job as a Math IIC mathematician to decide which technique will best help you solve a problem.
Let’s look at a few examples:
Combining Like Terms
There are other steps you can take to simplify expressions or equations. Combining like terms is one of the simpler techniques you can use, and it involves adding or subtracting the coefficients of variables that are raised to the same power. For example, by combining like terms, the expression
can be simplified to
by adding the coefficients of the variable x3 together and the coefficients of x2 together.
The point is, you’d rather have one term, 7x2, instead of x2, 3x2, –3x2, 2x2, and 4x2 all floating around in your expression. A general formula for combining like pairs looks like this:
Zero Product Rule
When the product of any number of terms is zero, you know that at least one of the terms is equal to zero. For example, if xy = 0, you know that either:
1. x = 0 and y ≠ 0,
2. y = 0 and x ≠ 0, or
3. x = y = 0
This is useful in a situation like the following:
By the zero product rule, you know that (x + 4) = 0 or (x – 3) = 0. In this equation, either x = –4 or x = 3, since one of the expressions in parentheses must be equal to 0. Consider this equation:
Again, since 3x2 or (x + 2) must equal 0, we know that either x = 0 or x = –2.
Keep your eye out for a zero product—it’s a big time-saver, especially when you have multiple answers to choose from.
Absolute Value
To solve an equation in which the variable is within absolute value brackets, you must divide the equation into two equations. The two equations are necessary because an absolute value really defines two equal values, one positive and one negative. The most basic example of this is an equation of the form |x| = c. In this case, either x = c or x = –c.
A slightly more complicated example is this:
|x + 3| = 5. Solve for x.
In this problem, you must solve two equations: First, solve for x in the equation x + 3 = 5. In this case, x = 2. Then, solve for x in the equation x + 3 = –5. In this case, x = –8. So the solutions to the equation |x + 3| = 5 are x = {–8, 2}.
Generally speaking, to solve an equation in which the variable is within absolute value brackets, first isolate the expression within the absolute value brackets, and then create two equations. Keep one of these two equations the same, while in the other equation, negate one side of the equation. In either case, the absolute value of the expression within brackets will be the same. This is why there are always two solutions to absolute value problems (unless the variable is equal to 0, which is neither positive nor negative).
Here is one more example:
Solve for x in terms of y in the equation 3= y2 – 1.
First, isolate the expression within the absolute value brackets:
Then solve for the variable as if the expression within absolute value brackets were positive:
Next, solve for the variable as if the expression within absolute value brackets were negative:
The solution set for x is {y2 – 3, –y2 –1}.
Inequalities
Before you get too comfortable with expressions and equations, we should introduce inequalities. An inequality is like an equation, but instead of relating equal quantities, it specifies exactly how two expressions are not equal.
1. x > y - “x is greater than y.”
2. x < y - x is less than y.”
3. x ≥ y - “x is greater than or equal to y.”
4. x ≤ y - “x is less than or equal to y.”
Solving inequalities is exactly like solving equations except for one very important difference: when both sides of an inequality are multiplied or divided by a negative number, the relationship between the two sides changes and so the direction of the inequality must be switched.
Here are a few examples:
Solve for x in the inequality – 3 < 2y.
Notice that in the last example, the inequality had to be reversed. Another way to express the solution is x ≥ –2. To help you remember that multiplication or division by a negative number reverses the direction of the inequality, recall that if x > y, then –x > –y, just as 5 > 4 and –5 < –4. Intuitively, this makes sense, and it might help you remember this special rule of inequalities.
There is a critical difference between the solutions to equalities and solutions to inequalities. The number of solutions to an equation is usually equal to the highest power of the equation. A linear equation (highest term of x) has one solution, a quadratic equation (highest term of x2) has two solutions, and a cubic equation (highest term of x3) has three solutions. In an inequality, this rule does not hold true. As you can see from the above examples, there are often infinite solutions to inequalities: the solutions are often graphed as planes rather than points.
Absolute Value and Inequalities
An equation without any absolute values generally results in, at most, only a few different solutions. Solutions to inequalities are often large regions of the x-y plane, such as x < 5. The introduction of the absolute value, as we’ve seen before, usually introduces two sets of solutions. The same is true when absolute values are introduced to inequalities: the solutions often come in the form of two regions of the x-y plane.
If the absolute value is less than a given quantity, then the solution is a single range, with a lower and an upper bound. For example,
Solve for x in the inequality |2x – 4| ≤ 6.
First, solve for the upper bound:
Second, solve for the lower bound:
Now combine the two bounds into a range of values for x. –1 ≤ x ≤ 5 is the solution.
The other solution for an absolute value inequality involves disjoint ranges: one whose lower bound is negative infinity and whose upper bound is a real number, and one whose lower bound is a real number and whose upper bound is infinity. This occurs when the absolute value is greater than a given quantity. For example,
Solve for x in the inequality |3x + 4| > 16.
First, solve for the upper range:
Then, solve for the lower range:
Now combine the two ranges to form the solution, which is two disjoint ranges: –∞ < x < – 20/3 or 4 < x < ∞.
When working with absolute values, it is important to first isolate the expression within absolute value brackets. Then, and only then, should you solve separately for the cases in which the quantity is positive and negative.
Ranges
Inequalities are also used to express the range of values that a variable can take on. a < x < b means that the value of x is greater than a and less than b. Consider the following word problem example:
A very complicated board game has the following recommendation on the box: “This game is appropriate for people older than 40, but no older than 65.” What is the range of the age of people for which the board game is appropriate?
Let a be the age of people for whom the board game is appropriate. The lower bound of a is 40, and the upper bound is 65. The range of a does not include its lower bound (it is appropriate for people “older than 40”), but it does include its upper bound (“no older than 65,” i.e., 65 is appropriate, but 66 is not). Therefore, the range of the age of people for which the board game is appropriate can be expressed by the inequality:
Here is another example:
A company manufactures car parts. As is the case with any system of mass production, small errors occur on virtually every part. The key for this company to succeed in making viable car parts is to keep the errors within a specific range. The company knows that a particular piece they manufacture will not work if it weighs less than 98% of its target weight or more than 102% of its target weight. If the target weight of this piece is 21.5 grams, in what range of weights must the piece measure for it to function?
The boundary weights of this car part are 0.98 21.5 = 21.07 and 1.02 21.5 = 21.93 grams. The problem states that the piece cannot weigh less than the minimum weight or more than the maximum weight in order for it to work. This means that the part will function at boundary weights themselves, and the lower and upper bounds are included. The answer to the problem is 21.07 ≤ x ≤ 21.93, where x is the weight of the part in grams.
Finding the range of a particular variable is essentially an exercise in close reading. Every time you come across a question involving ranges, you should carefully scrutinize the problem to pick out whether or not a particular variable’s range includes its bounds. This inclusion is the difference between “less than or equal to” and simply “less than.”
Operations on Ranges
Operations like addition, subtraction, and multiplication can be performed on ranges just as they are performed on variables or inequalities. For example:
If 4 < x < 7, what is the range of 2x + 3?
To solve this problem, simply manipulate the range like an inequality until you have a solution. Begin with the original range:
Then multiply the inequality by 2:
There is one crucial rule that you need to know about multiplying ranges: if you multiply a range by a negative number, you must flip the greater than or less than signs. For instance, if you multiply the range 2 < x < 8 by –1, the new range will be –2 > x > –8. Math IIC questions that ask you to perform operations on ranges of one variable will often test your alertness by making you multiply the range by a negative number.
Some range problems on the Math IIC will be made slightly more difficult by the inclusion of more than one variable. In general, the same basic procedures for dealing with one-variable ranges apply to adding, subtracting, and multiplying two-variable ranges.
Addition with Ranges of Two or More Variables
If –2 < x < 8 and 0 < y < 5, what is the range of x + y?
Simply add the ranges. The lower bound is –2 + 0 = –2. The upper bound is 8 + 5 = 13. Therefore, –2 < x + y < 13.
Subtraction with Ranges of Two or More Variables
Suppose 4 < s < 7 and –3 < t < –1. What is the range of s – t?
In this case, you have to find the range of –t. By multiplying the range of t by –1 and reversing the direction of the inequalities, we find that 1 < –t < 3. Now we can simply add the ranges again to find the range of s – t. 4 + 1 = 5, and 7 + 3 = 10. Therefore, 5< s – t < 10.
In general, to subtract ranges, find the range of the opposite of the variable being subtracted, and then add the ranges like usual.
Multiplication with Ranges of Two or More Variables
If –1 < j < 4 and 6 < k < 12, what is the range of jk?
First, multiply the lower bound of one variable by the lower and upper bounds of the other variable:
Then, multiply the upper bound of one variable with both bounds of the other variable:
The least of these four products becomes the lower bound, and the greatest is the upper bound. Therefore, –12 < jk < 48.
Let’s try one more example of performing operations on ranges:
If 3 ≤ x < 7 and –3 ≤ y ≤ 4, what is the range of 2(x + y)?
The first step is to find the range of x + y:
We have our bounds for the range of x + y, but are they included in the range? In other words, is the range 0 < x + y < 11, 0 ≤ x + y ≤ 11, or some combination of these two?
The rule to answer this question is the following: if either of the bounds that are being added, subtracted, or multiplied is non-inclusive (< or >), then the resulting bound is non-inclusive. Only when both bounds being added, subtracted, or multiplied are inclusive (≤ or ≥) is the resulting bound also inclusive.
The range of x includes its lower bound, 3, but not its upper bound, 7. The range of y includes both its bounds. Therefore, the range of x + y is 0 ≤ x + y < 11, and the range of 2(x + y) is 0 ≤ 2(x + y) < 22.
An alternate way of expressing the range of a variable might appear on the Math IIC test. A range can be written by enclosing the lower and upper bounds in parentheses or brackets, depending on whether they are included in the range. Parentheses are used when the bound is not included in the range, and brackets are used when the bound is included in the range. For example, the statement a < x < b can be rewritten “the range of x is (a, b).” The statement axb can be rewritten “the range of x is [a, b].” Finally, the statement a < xb can be rewritten “the range of x is [a, b].”
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# 4.2: Equivalent Fractions
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In this section we deal with fractions, numbers or expressions of the form a/b.
Definition: Fractions
A number of the form
$\dfrac{a}{b}\nonumber$
where $$a$$ and $$b$$ are numbers is called a fraction. The number $$a$$ is called the numerator of the fraction, while the number $$b$$ is called the denominator of the fraction.
Near the end of this section, we’ll see that the numerator and denominator of a fraction can also be algebraic expressions, but for the moment we restrict our attention to fractions whose numerators and denominators are integers. We start our study of fractions with the definition of equivalent fractions.
Equivalent Fractions
Two fractions are equivalent if they represent the same numerical value.
But how can we tell if two fractions represent the same number? Well, one technique involves some simple visualizations. Consider the image shown in Figure 4.1, where the shaded region represents 1/3 of the total area of the figure (one of three equal regions is shaded).
In Figure 4.2, we’ve shaded 2/6 of the entire region (two of six equal regions are shaded).
In Figure 4.3, we’ve shaded 4/12 of the entire region (four of twelve equal regions are shaded).
Let’s take the diagrams from Figure 4.1, Figure 4.2, and Figure 4.3 and stack them one atop the other, as shown in Figure 4.4.
Figure 4.4 provides solid visual evidence that the following fractions are equivalent.
$\dfrac{1}{3} = \dfrac{2}{6} = \dfrac{4}{12}\nonumber$
## Key Observations
1. If we start with the fraction 1/3, then multiply both numerator and denominator by 2, we get the following result.
\begin{aligned} \dfrac{1}{3} = \dfrac{1 \cdot 2}{3 \cdot 2} ~ & \textcolor{red}{ \text{ Multiply the numerator and denominator by 2.}} \\ = \dfrac{2}{6} ~ & \textcolor{red}{ \text{ Simplify numerator and denominator.}} \end{aligned}\nonumber
This is precisely the same thing that happens going from Figure 4.1 to 4.2, where we double the number of available boxes (going from 3 available to 6 available) and double the number of shaded boxes (going from 1 shaded to 2 shaded).
2. If we start with the fraction 1/3, then multiply both numerator and denominator by 4, we get the following result.
\begin{aligned} = \dfrac{1}{3} = \dfrac{1 \cdot 4}{3 \cdot 4} ~ & \textcolor{red}{ \text{ Multiply numerator and denominator by 4.}} \\ = \dfrac{4}{12} ~ & \textcolor{red}{ \text{ Simplify numerator and denominator.}} \end{aligned}\nonumber
This is precisely the same thing that happens going from Figure 4.1 to 4.3, where we multiply the number of available boxes by 4 (going from 3 available to 12 available) and multiply the number of shaded boxes by 4 (going from 1 shaded to 4 shaded).
The above discussion motivates the following fundamental result.
Creating Equivalent Fractions
If you start with a fraction, then multiply both its numerator and denominator by the same number, the resulting fraction is equivalent (has the same numerical value) to the original fraction. In symbols,
$\dfrac{a}{b} = \dfrac{a \cdot x}{b \cdot x}\nonumber$
## Arguing in Reverse
Reversing the above argument also holds true.
1. If we start with the fraction 2/6, then divide both numerator and denominator by 2, we get the following result.
\begin{aligned} \dfrac{2}{6} = \dfrac{2 \div 2}{6 \div 2} ~ & \textcolor{red}{ \text{ Divide numerator and denominator by 2.}} \\ = \dfrac{1}{3} ~ & \textcolor{red}{ \text{ Simplify numerator and denominator.}} \end{aligned}\nonumber
This is precisely the same thing that happens going backwards from Figure 4.2 to 4.1, where we divide the number of available boxes by 2 (going from 6 available to 3 available) and dividing the number of shaded boxes by 2 (going from 2 shaded to 1 shaded).
2. If we start with the fraction 4/12, then divide both numerator and denominator by 4, we get the following result.
\begin{aligned} \dfrac{4}{12} = \dfrac{4 \div 4}{12 \div 4} ~ & \textcolor{red}{ \text{ Multiply numerator and denominator by 4.}} \\ = \dfrac{1}{3} ~ & \textcolor{red}{ \text{ Simplify numerator and denominator.}} \end{aligned}\nonumber
This is precisely the same thing that happens going backwards from Figure 4.3 to 4.1, where we divide the number of available boxes by 4 (going from 12 available to 3 available) and divide the number alignof shaded boxes by 4 (going from 4 shaded to 1 shaded).
The above discussion motivates the following fundamental result.
Creating Equivalent Fractions
If you start with a fraction, then divide both its numerator and denominator by the same number, the resulting fraction is equivalent (has the same numerical value) to the original fraction. In symbols,
$\dfrac{a}{b} = \dfrac{a \div x}{b \div x}.\nonumber$
## The Greatest Common Divisor
We need a little more terminology.
Divisor
If d and a are natural numbers, we say that “d divides a” if and only if when a is divided by d, the remainder is zero. In this case, we say that “d is a divisor of a.”
For example, when 36 is divided by 4, the remainder is zero. In this case, we say that “4 is a divisor of 36.” On the other hand, when 25 is divided by 4, the remainder is not zero. In this case, we say that “4 is not a divisor of 25.”
Greatest Common Divisor
Let a and b be natural numbers. The common divisors of a and b are those natural numbers that divide both a and b. The greatest common divisor is the largest of these common divisors.
Example 1
Find the greatest common divisor of 18 and 24.
Solution
First list the divisors of each number, the numbers that divide each number with zero remainder.
Divisors of 18 : 1, 2, 3, 6, 9, and 18
Divisors of 24 : 1, 2, 3, 4, 6, 8, 12, and 24
The common divisors are:
Common Divisors : 1, 2, 3, and 6
The greatest common divisor is the largest of the common divisors. That is,
Greatest Common Divisor = 6.
That is, the largest number that divides both 18 and 24 is the number 6.
Exercise
Find the greatest common divisor of 12 and 18.
6
## Reducing a Fraction to Lowest Terms
First, a definition.
Lowest Terms
A fraction is said to be reduced to lowest terms if the greatest common divisor of both numerator and denominator is 1.
Thus, for example, 2/3 is reduced to lowest terms because the greatest common divisior of 2 and 3 is 1. On the other hand, 4/6 is not reduced to lowest terms because the greatest common divisor of 4 and 6 is 2.
Example 2
Reduce the fraction 18/24 to lowest terms.
Solution
One technique that works well is dividing both numerator and denominator by the greatest common divisor of the numerator and denominator. In Example 1, we saw that the greatest common divisor of 18 and 24 is 6. We divide both numerator and denominator by 6 to get
\begin{aligned} \dfrac{18}{24} = \dfrac{18 \div 6}{24 \div 6} ~ & \textcolor{red}{ \text{ Divide numerator and denominator by 6.}} \\ = \dfrac{3}{4} ~ & \textcolor{red}{ \text{ Simplify numerator and dice.}} \end{aligned}\nonumber
Note that the greatest common divisor of 3 and 4 is now 1. Thus, 3/4 is reduced to lowest terms.
There is a second way we can show division of numerator and denominator by 6. First, factor both numerator and denominator as follows:
\begin{aligned} \dfrac{18}{24} = \dfrac{3 \cdot 6}{4 \cdot 6} ~ & \textcolor{red}{ \text{ Factor out a 6.}} \end{aligned}\nonumber
You can then show “division” of both numerator and denominator by 6 by “crossing out” or “canceling” a 6 in the numerator for a 6 in the denominator, like this:
\begin{aligned} = \dfrac{3 \cdot \cancel{6}}{4 \cdot \cancel{6}} ~ & \textcolor{red}{ \text{ Cancel common factor.}} \\ = \dfrac{3}{4} \end{aligned}\nonumber
Note that we get the same equivalent fraction, reduced to lowest terms, namely 3/4.
Exercise
Reduce the fraction 12/18 to lowest terms.
2/3
Important Point
In Example 2 we saw that 6 was both a divisor and a factor of 18. The words divisor and factor are equivalent.
We used the following technique in our second solution in Example 2.
Cancellation Rule
If you express numerator and denominator as a product, then you may cancel common factors from the numerator and denominator. The result will be an equivalent fraction.
Because of the “Cancellation Rule,” one of the most effective ways to reduce a fraction to lowest terms is to first find prime factorizations for both numerator and denominator, then cancel all common factors.
Example 3
Reduce the fraction 18/24 to lowest terms.
Solution
Use factor trees to prime factor numerator and denominator.
Once we’ve factored the numerator and denominator, we cancel common factors.
\begin{aligned} \dfrac{18}{24} = \dfrac{2 \cdot 3 \cdot 3}{2 \cdot 2 \cdot 2 \cdot 3} ~ & \textcolor{red}{ \text{ Prime factor numerator and denominator.}} \\ = \dfrac{ \cancel{2} \cdot \cancel{3} \cdot 3}{ \cancel{2} \cdot 2 \cdot 2 \cdot \cancel{3}} ~ & \textcolor{red}{ \text{ Cancel common factors.}} \\ = \dfrac{3}{2 \cdot 2} ~ & \textcolor{red}{ \text{ Remaining factors.}} \\ = \dfrac{3}{4} ~ & \textcolor{red}{ \text{ Simplify denominator.}} \end{aligned}\nonumber
Thus, 18/24 = 3/4.
Exercise $$\PageIndex{1}$$
Reduce the fraction 28/35 to lowest terms.
4/5
Example 4
Reduce the fraction 28/42 to lowest terms.
Solution
Use factor trees to prime factor numerator and denominator.
Now we can cancel common factors.
\begin{aligned} \dfrac{28}{42} = \dfrac{2 \cdot 2 \cdot 7}{2 \cdot 3 \cdot 7} ~ & \textcolor{red}{ \text{ Prime factor numerator and denominator.}} \\ = \dfrac{ \cancel{2} \cdot 2 \cdot \cancel{7}}{ \cancel{2} \cdot 3 \cdot \cancel{7}} ~ & \textcolor{red}{ \text{ Cancel common factors.}} \\ = \dfrac{2}{3} \end{aligned}\nonumber
Thus, 28/42 = 2/3.
Exercise
Reduce the fraction 36/60 to lowest terms.
3/5
## Reducing Fractions with Variables
We use exactly the same technique to reduce fractions whose numerators and denominators contain variables.
Example 5
Reduce
$\dfrac{56x^2y}{60xy^2}\nonumber$
to lowest terms.
Solution
Use factor trees to factor the coefficients of numerator and denominator.
Now cancel common factors.
\begin{aligned} \dfrac{56x^2y}{60xy^2} = \dfrac{2 \cdot 2 \cdot 2 \cdot 7 \cdot x \cdot x \cdot y}{2 \cdot 2 \cdot 3 \cdot 5 \cdot x \cdot y \cdot y} ~ & \textcolor{red}{ \text{ Prime factor numerator and denominator.}} \\ = \dfrac{ \cancel{2} \cdot \cancel{2} \cdot 2 \cdot 7 \cdot \cancel{x} \cdot x \cdot \cancel{y}}{ \cancel{2} \cdot \cancel{2} \cdot 3 \cdot 5 \cdot \cancel{x} \cdot y \cdot \cancel{y}} ~ & \textcolor{red}{ \text{ Cancel cmmon factors.}} \\ = \dfrac{2 \cdot 7 \cdot x}{3 \cdot 5 \cdot y} ~ & \textcolor{red}{ \text{ Remaining factors.}} \\ = \dfrac{14x}{15y} ~ & \textcolor{red}{ \text{ Simplify numerator and denominator.}} \end{aligned}\nonumber
Thus, 56x2y/(60xy2) = 14x/(15y).
Exercise
Reduce:
$\dfrac{25a^3b}{40a^2b^3}\nonumber$
$\dfrac{5a}{8b^2}\nonumber$
## A Word on Mathematical Notation
There are two types of mathematical notation: (1) inline mathematical notation, and (2) displayed mathematical notation.
Inline Mathematical Notation
The notation 14x/(15y) is called inline mathematical notation. When the same expression is centered on its own line, as in
$\dfrac{14x}{15y},\nonumber$
this type of notation is called displayed mathematical notation.
When you work a problem by hand, using pencil and paper calculations, the preferred format is displayed notation, like the displayed notation used to simplify the given expression in Example 5. However, computers and calculators require that you enter your expressions using inline mathematical notation. Therefore, it is extremely important that you are equally competent with either mathematical notation: displayed or inline.
By the way, order of operations, when applied to the inline expression 14x/(15y), requires that we perform the multiplication inside the parentheses first. Then we must perform multiplications and divisions as they occur, as we move from left to right through the expression. This is why the inline notation 14x/(15y) is equivalent to the displayed notation
$\dfrac{14x}{15y}.\nonumber$
However, the expression 14x/15y is a different beast. There are no parentheses, so we perform multiplication and division as they occur, moving left to right through the expression. Thus, we must first take the product of 14 and x, divide the result by 15, then multiply by y. In displayed notation, this result is equivalent to
$\dfrac{14x}{15} \cdot y,\nonumber$
which is a different result.
Some readers might wonder why we did not use the notation (14x)/(15y) to describe the solution in Example 5. After all, this inline notation is also equivalent to the displayed notation
$\dfrac{14x}{15y}.\nonumber$
However, the point is that we do not need to, as order of operations already requires that we take the product of 14 and x before dividing by 15y. If this is hurting your head, know that it’s quite acceptable to use the equivalent notation (14x)/(15y) instead of 14x/(15y). Both are correct.
## Equivalent Fractions in Higher Terms
Sometimes the need arises to find an equivalent fraction with a different, larger denominator.
Example 6
Express 3/5 as an equivalent fraction having denominator 20.
Solution
The key here is to remember that multiplying numerator and denominator by the same number produces an equivalent fraction. To get an equivalent fraction with a denominator of 20, we’ll have to multiply numerator and denominator of 3/5 by 4.
\begin{aligned} \dfrac{3}{5} ~ & \textcolor{red}{ \text{ Multiply numerator and denominator by 4.}} \\ = \dfrac{12}{20} ~ & \textcolor{red}{ \text{ Simplify numerator and denominator.}} \end{aligned}\nonumber
Therefore, 3/5 equals 12/20.
Exercise
Express 2/3 as an equivalent fraction having denominator 21.
14/21
Example 7
Express 8 as an equivalent fraction having denominator 5.
Solution
The key here is to note that
\begin{aligned} 8 = \dfrac{8}{1} ~ & \textcolor{red}{ \text{ Understood denominator is 1.}} \end{aligned}\nonumber
To get an equivalent fraction with a denominator of 5, we’ll have to multiply numerator and denominator of 8/1 by 5.
\begin{aligned} = \dfrac{8 \cdot 5}{1 \cdot 5} ~ & \textcolor{red}{ \text{ Multiply numerator and denominator by 5.}} \\ = \dfrac{40}{5} ~ & \textcolor{red}{ \text{ Simplify numerator and denominator.}} \end{aligned}\nonumber
Therefore, 8 equals 40/5.
Exercise
Express 5 as an equivalent fraction having denominator 7.
35/7
Example 8
Express 2/9 as an equivalent fraction having denominator 18a.
Solution
To get an equivalent fraction with a denominator of 18a, we’ll have to multiply numerator and denominator of 2/9 by 2a.
\begin{aligned} \dfrac{2}{9} = \dfrac{2 \cdot 2a}{9 \cdot 2a} ~ & \textcolor{red}{ \text{ Multiply numerator and denominator by } 2a.} \\ = \dfrac{4a}{18a} ~ & \textcolor{red}{ \text{ Simplify numerator and denominator.}} \end{aligned}\nonumber
Therefore, 2/9 equals 4a/(18a), or equivalently, (4a)/(18a).
Exercise
Express 3/8 as an equivalent fraction having denominator 24a.
$\dfrac{9a}{24a}\nonumber$
## Negative Fractions
We have to also deal with fractions that are negative. First, let’s discuss placement of the negative sign.
• Positive divided by negative is negative, so
$\dfrac{3}{-5} = - \dfrac{3}{5}.\nonumber$
• But it is also true that negative divided by positive is negative. Thus,
$\dfrac{−3}{5} = \dfrac{−3}{5}.\nonumber$
These two observations imply that all three of the following fractions are equivalent (the same number):
$\dfrac{3}{-5} = - \dfrac{3}{5} = \dfrac{-3}{5}.\nonumber$
Note that there are three possible placements for the negative sign: (1) the denominator, (2) the fraction bar, or (3) the numerator. Any one of these placements produces an equivalent fraction.
Fractions and Negative Signs
Let a and b be any integers. All three of the following fractions are equivalent (same number):
$\dfrac{a}{-b} = - \dfrac{a}{b} = \dfrac{-a}{b}.\nonumber$
Mathematicians prefer to place the negative sign either in the numerator or on the fraction bar. The use of a negative sign in the denominator is discouraged.
Example 9
Reduce:
$\dfrac{50x^3}{-75x^5}\nonumber$
to lowest terms.
Solution
Prime factor numerator and denominator and cancel.
\begin{aligned} \dfrac{50x^3}{-75x^5} &= \dfrac{2 \cdot 5 \cdot 5 \cdot x \cdot x \cdot x}{-3 \cdot 5 \cdot 5 \cdot x \cdot x \cdot x \cdot x \cdot x} \\ &= \dfrac{2 \cdot \cancel{5} \cdot \cancel{5} \cdot \cancel{x} \cdot \cancel{x} \cdot \cancel{x}}{-3 \cdot \cancel{5} \cdot \cancel{5} \cdot \cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot x \cdot x} \\ &= \dfrac{2}{-3 \cdot x \cdot x} \\ &= \dfrac{2}{-3x^2} \end{aligned}\nonumber
However, it is preferred that there be no negative signs in the denominator, so let’s place the negative sign on the fraction bar (the numerator would suit as well). Thus,
$\dfrac{50x^3}{-75x^5} = - \dfrac{2}{3x^2}\nonumber$
We also have the following result.
Fractions and Negative Signs
Let $$a$$ and $$b$$ be any integers. Then,
$\dfrac{-a}{-b} = \dfrac{a}{b}.\nonumber$
Example 10
Reduce:
$\dfrac{-12xy^2}{-18x^2y}\nonumber$
Solution
Unlike Example 9, some like to take care of the sign of the answer first.
$\dfrac{-12xy^2}{-18x^2y} = \dfrac{12xy^2}{18x^2y}\nonumber$
Now we can factor numerator and denominator and cancel common factors.
\begin{aligned} &= \dfrac{2 \cdot 2 \cdot 3 \cdot x \cdot y \cdot y}{2 \cdot 3 \cdot 3 \cdot x \cdot x \cdot y} \\ &= \dfrac{ \cancel{2} \cdot 2 \cancel{3} \cdot \cancel{x} \cdot y \cdot \cancel{y}}{ \cancel{2} \cdot \cancel{3} \cdot 3 \cdot \cancel{x} \cdot x \cdot \cancel{y}} &= \dfrac{2y}{3x} \end{aligned}\nonumber
Thus,
$\dfrac{-12xy^2}{-18x^2y} = \dfrac{2y}{3x}.\nonumber$
Exercise
Reduce:
$\dfrac{-21a^2b^3}{-56a^3b}\nonumber$
$\dfrac{3b^2}{8a}\nonumber$
## Exercises
In Exercises 1-12, find the GCD of the given numbers.
1. 72, 8
2. 76, 52
3. 52, 20
4. 56, 96
5. 36, 63
6. 63, 21
7. 72, 44
8. 10, 40
9. 16, 56
10. 54, 66
11. 84, 24
12. 75, 45
In Exercises 13-28, reduce the given fraction to lowest terms.
13. $$\dfrac{22}{98}$$
14. $$\dfrac{28}{56}$$
15. $$\dfrac{93}{15}$$
16. $$\dfrac{90}{39}$$
17. $$\dfrac{69}{21}$$
18. $$\dfrac{74}{62}$$
19. $$\dfrac{74}{12}$$
20. $$\dfrac{66}{10}$$
21. $$\dfrac{66}{57}$$
22. $$\dfrac{34}{30}$$
23. $$\dfrac{33}{99}$$
24. $$\dfrac{20}{58}$$
25. $$\dfrac{69}{24}$$
26. $$\dfrac{18}{96}$$
27. $$\dfrac{46}{44}$$
28. $$\dfrac{92}{24}$$
29. Express 3 as an equivalent fraction having denominator 24. 30. Express 3 as an equivalent fraction having denominator 8. 31. Express $$\dfrac{25}{19}$$ as an equivalent fraction having denominator 57. 32. Express $$\dfrac{29}{22}$$ as an equivalent fraction having denominator 44. 33. Express 2 as an equivalent fraction having denominator 2. 34. Express 2 as an equivalent fraction having denominator 8. 35. Express $$\dfrac{18}{19}$$ as an equivalent fraction having denominator 95. 36. Express $$\dfrac{17}{22}$$ as an equivalent fraction having denominator 44. 37. Express $$\dfrac{1}{3}$$ as an equivalent fraction having denominator 24. 38. Express $$\dfrac{15}{19}$$ as an equivalent fraction having denominator 95. 39. Express 16 as an equivalent fraction having denominator 4. 40. Express 5 as an equivalent fraction having denominator 2.
In Exercises 41-56, reduce the given fraction to lowest terms.
41. $$\dfrac{34}{−86}$$
42. $$\dfrac{−48}{14}$$
43. $$\dfrac{−72}{−92}$$
44. $$\dfrac{27}{−75}$$
45. $$\dfrac{−92}{82}$$
46. $$\dfrac{−44}{−62}$$
47. $$\dfrac{−21}{33}$$
48. $$\dfrac{57}{−99}$$
49. $$\dfrac{22}{−98}$$
50. $$\dfrac{−33}{69}$$
51. $$\dfrac{42}{−88}$$
52. $$\dfrac{−100}{48}$$
53. $$\dfrac{94}{−6}$$
54. $$\dfrac{−36}{−38}$$
55. $$\dfrac{10}{−86}$$
56. $$\dfrac{−100}{−46}$$
57. Express $$\dfrac{3}{2}$$ as an equivalent fraction having denominator 62n.
58. Express $$\dfrac{6}{25}$$ as an equivalent fraction having denominator 50a.
59. Express $$\dfrac{13}{10}$$ as an equivalent fraction having denominator 60m.
60. Express $$\dfrac{1}{16}$$ as an equivalent fraction having denominator 80p.
61. Express $$\dfrac{3}{2}$$ as an equivalent fraction having denominator 50n.
62. Express $$\dfrac{43}{38}$$ as an equivalent fraction having denominator 76a.
63. Express 11 as an equivalent fraction having denominator 4m. 64. Express 13 as an equivalent fraction having denominator 6n.
65. Express 3 as an equivalent fraction having denominator 10m.
66. Express 10 as an equivalent fraction having denominator 8b.
67. Express 6 as an equivalent fraction having denominator 5n.
68. Express 16 as an equivalent fraction having denominator 2y.
In Exercises 69-84, reduce the given fraction to lowest terms.
69. $$\dfrac{82y^5}{−48y}$$
70. $$\dfrac{−40y^5}{−55y}$$
71. $$\dfrac{−77x^5}{44x^4}$$
72. $$\dfrac{−34x^6}{−80x}$$
73. $$\dfrac{−14y^5}{54y^2}$$
74. $$\dfrac{96y^4}{−40y^2}$$
75. $$\dfrac{42x}{81x^3}$$
76. $$\dfrac{26x^2}{32x^6}$$
77. $$\dfrac{−12x^5}{14x^6}$$
78. $$\dfrac{−28y^4}{72y^6}$$
79. $$\dfrac{−74x}{22x^2}$$
80. $$\dfrac{56x^2}{26x^3}$$
81. $$\dfrac{−12y^5}{98y^6}$$
82. $$\dfrac{96x^2}{14x^4}$$
83. $$\dfrac{18x^6}{−54x^2}$$
84. $$\dfrac{32x^6}{62x^2}$$
In Exercises 85-100, reduce the given fraction to lowest terms.
85. $$\dfrac{26y^2x^4}{−62y^6x^2}$$
86. $$\dfrac{6x^2y^3}{40x^3y^2}$$
87. $$\dfrac{−2y^6x^4}{−94y^2x^5}$$
88. $$\dfrac{90y^6x^3}{39y^3x^5}$$
89. $$\dfrac{30y^5x^5}{−26yx^4}$$
90. $$\dfrac{74x^6y^4}{−52xy^3}$$
91. $$\dfrac{36x^3y^2}{−98x^4y^5}$$
92. $$\dfrac{84x^3y}{16x^4y^2}$$
93. $$\dfrac{−8x^6y^3}{54x^3y^5}$$
94. $$\dfrac{70y^5x^2}{16y^4x^5}$$
95. $$\dfrac{34yx^6}{−58y^5x^4}$$
96. $$\dfrac{99y^2x^3}{88y^6x}$$
97. $$\dfrac{−36y^3x^5}{51y^2x}$$
98. $$\dfrac{44y^5x^5}{−88y^4x}$$
99. $$\dfrac{91y^3x^2}{−28y^5x^5}$$
100. $$\dfrac{−76y^2x}{−57y^5x^6}$$
101. Hurricanes. According to the National Atmospheric and Oceanic Administration, in 2008 there were 16 named storms, of which 8 grew into hurricanes and 5 were major.
i) What fraction of named storms grew into hurricanes? Reduce your answer to lowest terms.
ii) What fraction of named storms were major hurricanes? Reduce your answer to lowest terms.
iii) What fraction of hurricanes were major? Reduce your answer to lowest terms.
102. Tigers. Tigers are in critical decline because of human encroachment, the loss of more than nine-tenths of their habitat, and the growing trade in tiger skins and body parts. Associated Press-Times-Standard 01/24/10 Pressure mounts to save the tiger.
i) Write the loss of habitat as a fraction.
ii) Describe in words what the numerator and denominator of this fraction represent.
iii) If the fraction represents the loss of the whole original habitat, how much of the original habitat remains?
1. 8
3. 4
5. 9
7. 4
9. 8
11. 12
13. $$\dfrac{11}{49}$$
15. $$\dfrac{31}{5}$$
17. $$\dfrac{23}{7}$$
19. $$\dfrac{37}{6}$$
21. $$\dfrac{22}{19}$$
23. $$\dfrac{1}{3}$$
25. $$\dfrac{23}{8}$$
27. $$\dfrac{23}{22}$$
29. $$\dfrac{72}{24}$$
31. $$\dfrac{75}{57}$$
33. $$\dfrac{4}{2}$$
35. $$\dfrac{90}{95}$$
37. $$\dfrac{8}{24}$$
39. $$\dfrac{64}{4}$$
41. $$\dfrac{−17}{43}$$
43. $$\dfrac{18}{23}$$
45. $$\dfrac{−46}{41}$$
47. $$\dfrac{− 7}{11}$$
49. $$\dfrac{−11}{49}$$
51. $$\dfrac{−21}{44}$$
53. $$\dfrac{−47}{3}$$
55. $$\dfrac{− 5}{43}$$
57. $$\dfrac{93 n}{62 n}$$
59. $$\dfrac{78 m}{60 m}$$
61. $$\dfrac{75 n}{50 n}$$
63. $$\dfrac{44 m}{4 m}$$
65. $$\dfrac{30 m}{10 m}$$
67. $$\dfrac{30 n}{5 n}$$
69. $$\dfrac{−41 y^4}{24}$$
71. $$\dfrac{− 7x}{4}$$
73. $$− \dfrac{7 y^3}{27}$$
75. $$\dfrac{14}{27 x^2}$$
77. $$− \dfrac{6}{7x}$$
79. $$− \dfrac{37}{11 x}$$
81. $$− \dfrac{6}{49 y}$$
83. $$− \dfrac{x^4}{3}$$
85. $$− \dfrac{13 x^2}{31 y^4}$$
87. $$\dfrac{y^4}{47 x}$$
89. $$− \dfrac{15 y^4 x}{13}$$
91. $$− \dfrac{18}{49xy^3}$$
93. $$− \dfrac{4 x^3}{27 y^2}$$
95. $$−\dfrac{17 x^2}{29 y^4}$$
97. $$− \dfrac{12yx^4}{17}$$
99. $$− \dfrac{13}{4y^2x^3}$$
101.
i) $$\dfrac{1}{2}$$
ii) $$\dfrac{5}{16}$$
iii) $$\dfrac{5}{8}$$
This page titled 4.2: Equivalent Fractions is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by David Arnold. |
# What is a Slope (Gradient)?
The slope of a line is a number that measures its "steepness". It is the change in y for a unit change in x along the line.
The slope of a line is generally denoted by m.
Thus,
m = tan θ
Since a line parallel to x – axis makes an angle of 0° with x – axis, therefore its slope is tan 0° = 0.
A line parallel to y – axis, perpendicular to x – axis makes an angle of 90 with x – axis, so its slope is tan 90° = ∞. Also, the slope of a line equally inclined with axes is 1 or -1 as it makes 45° or 135° with x – axis.
The angle of inclination of a line with the positive direction of x – axis in anticlockwise sense always lies between 0° and 180°.
## Slope of a line in terms of coordinates of any two points on it
Let P (x1, y1) and Q (x2, y2) be two points on a line making an angle θ with the positive direction of x – axis. Draw PM, QN perpendiculars on x – axis and PL perpendicular on QN.
Then, PL = MN = ON – OM = x2 – x1
and QL = ON – LN = QN – PM = y2 – y1
In ΔPQL, tan θ = QL/PL = (y2 – y1)/(x2 – x1)
Thus, if (x1, y1) and (x2, y2) are coordinates of any two points on a line, then its slope is
m = (y2 – y1)/(x2 – x1)
m = Difference of ordinates
Difference of abscissae
or, m = Vertical Step
Horizontal Step
## Angle between two lines
The angle θ between the lines having slopes m1 and m2 is given by
tan θ = + m2 – m1
1 + m1m2
## Slope (Gradient) of Parallel lines
If two lines of slopes m1 and m2 are parallel, then the angle θ between them is 0°.
tan θ = tan 0° = 0
m2 – m1 = 0
1 + m1m2
m2 = m1
Thus, when two lines are parallel, their slopes are equal.
## Slope (Gradient) of Perpendicular lines
If two lines of slope m1 and m2 are perpendicular, then the angle θ between them is of 90°
cot θ = 1 + m1m2
m2 – m1
0 = 1 + m1m2
m1m2 = -1
Thus, when two lines are perpendicular, the product of their slopes is -1. If m is the slope of a line, then the slope of a line perpendicular to it is - (1/m).
Now try it yourself! Should you still need any help, click here to schedule live online session with e Tutor! |
## Tuesday, January 3, 2006
### Subset It Up. Topic: Sets/Combinatorics. Level: Olympiad.
Problem: (1981 IMO - #2) Take $r$ such that $1\le r\le n$, and consider all subsets of r elements of the set $\{1,2,\ldots,n\}$. Each subset has a smallest element. Let $F(n,r)$ be the arithmetic mean of these smallest elements. Prove that:
$F(n,r)=\frac{n+1}{r+1}$.
Solution: So we first find out how many of these subsets there are. Well that's easy, just choose $r$ elements from $n$, or $nCr$.
So how many have smallest element $1$? Well, assume $1$ is in the set. Then we have to pick the other $r-1$ elements from the remaining $n-1$ elements. So we have $(n-1)C(r-1)$.
And $2$? By the same argument, we have $(n-2)C(r-1)$ (choosing $r-1$ elements from the numbers above $2$).
So for any $k$, we have $(n-k)C(r-1)$ ways to choose a subset of size $r$ that has minimum element $k$.
Since we are taking an arithmetic mean, we want the sum of ALL the minimum elements, meaning we take $k \cdot (n-k)C(r-1)$, added up to be
$\displaystyle \sum_{i=1}^{n-r+1} [i \cdot (n-i)C(r-1)] = 1\cdot (n-1)C(r-1)+2 \cdot (n-2)C(r-1) + \cdots +(n-r+1)\cdot (r-1)C(r-1)$.
But remember by the above argument we have
$\displaystyle \sum_{i=1}^{n-r+1} (n-i)C(r-1) = (n-1)C(r-1)+(n-2)C(r-1)+\cdots+rC(r-1)+(r-1)C(r-1) = nCr$.
Similarly
$\displaystyle \sum_{i=2}^{n-r+1} (n-i)C(r-1) =(n-2)C(r-1)+\cdots+(r-1)C(r-1) = (n-1)Cr$.
$\displaystyle \sum_{i=3}^{n-r+1} (n-i)C(r-1) = (n-3)C(r-1)+\cdots+(r-1)C(r-1) = (n-2)Cr$.
...
$\displaystyle \sum_{i=n-r}^{n-r+1} (n-i)C(r-1) = rC(r-1)+(r-1)C(r-1) = (r+1)Cr$.
$\displaystyle \sum_{i=n-r+1}^{n-r+1} (n-i)C(r-1) = (r-1)C(r-1) = rCr$.
Adding up all the equations, we have
$\displaystyle \sum_{i=1}^{n-r+1} [i \cdot (n-i)C(r-1)] = \displaystyle \sum_{i=0}^{n-r} (n-i)Cr$.
Using again the same argument, we have
$\displaystyle \sum_{i=0}^{n-r} (n-i)Cr = (n+1)C(r+1)$.
Then we divide by the total number of subsets, which we found earlier to be $nCr$ to get
$F(n,r) = \frac{(n+1)C(r+1)}{nCr} = \frac{n+1}{r+1}$
as desired. QED.
--------------------
Comment: Yay, first IMO problem posted here. Not a particularly difficult one, though somewhat time consuming to write out. But anyway, hopefully there will be more to come.
--------------------
Practice Problem: (2001 AIME - #2) A finite set $S$ of distinct real numbers has the following properties: the mean of $S\cup\{1\}$ is $13$ less than the mean of $S$, and the mean of $S\cup\{2001\}$ is $27$ more than the mean of $S$. Find the mean of $S$. |
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Divided By 16
6th step: divide the number acquired in step 5 by the divisor. 7th step: Place the full number obtained in stage 6 in the second position of the quotient (immediately next to the first number of the quotient obtained in step 2), this is the second number of the quotient. Divide the full number by the divisor and put the result beneath the number divided.
It should be noted that you might bypass all of the preceding phases with zeros and go directly to this step. All you have to do is figure out how many digits in the dividend you need to skip to achieve your first non-zero number in the quotient solution. In this scenario, you might just divide 32 by 48. Place the 1 to the right of the 0 on top of the division bar. Then, multiply 1 by 32 and put the result beneath 48.1 * 32 = 32. Finally, draw a line and remove 32 from 48.
As previously stated, the quotient is calculated by dividing the dividend by the divisor. In many circumstances, the result of this computation will be an integer (a whole number), indicating that a number can be divided completely without any leftovers. When a number cannot be divided completely, the amount left behind is referred to as the residual.
Divided By 16 4
We'll teach you how to divide 16 by 1/4 in this section. We will offer you the solution in fraction form and decimal form. Here is 16 divided by 1/4 expressed mathematically in colors:= whole number= numerator= denominatorTo get the fraction form answer, multiply the entire number by the denominator and make the result the new numerator. Thus, the answer to 16 divided by 1/4 in fraction form is:To make the answer to 16 divided by 1/4 in decimal form, simply divide the numerator by the denominator from the fraction answer above:64 / 1 = 64The answer is rounded to the nearest two decimal points if necessary.64/1 is an improper fraction and should be written as 64.Enter another whole number and fraction for us to divide: If so, go on to the next issue on our list.
Divided By 16 And 100
List of numbers divisible by 16In this section, we will define "numbers divisible by 16" and then provide a list of numbers divisible by 16.Numbers divided by 16 are all the numbers that when divided by 16 equal a whole number (integer). In other words, we're seeking for all possible numbers in this equation:Number / 16 = IntegerAs you've undoubtedly guessed, the list of numbers divided by 16 is infinite. Here is the first list of integers divisible by 16, beginning with the lowest number, 16:16, 32, 48, 64, 80, 96, 112, 128, 144, 160, and so on. As you can see from the list, the numbers are intervals of 16. You may continue to add to the list and make it as long as you like by just adding 16 to the previous number. No worries. Please input your number below to find out what numbers are divisible by it. Click here to see the next "list of numbers divisible by" that we've produced for you.
This method may be used to any number of fractions. Simply multiply the numerators and denominators of each fraction in the problem by the product of the denominators of all the other fractions in the problem (without counting its own individual denominator).EX: 12 48 + 8 48 + 24 48 = 44 48 = 11 12
We hope you found the Divide Ratio Calculator helpful; if so, we would appreciate it if you could rate it and, if you have time, share it on your favorite social network. This enables us to allocate future resources and keep our Math calculators and instructional materials free to use for everyone across the world. [320 Votes]
In this circuit design, I used a basic 74LS93 to count up to 16 in a ripple through method. It will create one pulse for every 16 pulses at the input. Pin 1 connects to pin 12 so that the output of the first stage is sent to the input of the second stage. Pin 14 (CP) is the input, while Pin 11 is the output (Q3).
Divided By 16 8
tiyas bhattacharjee (9 years ago): 168=232 in 50! All even numbers are split by two in 50! ,25 even no is available.so 50 factorial is divided by 225.no after dividing all even terms by 2.remaining terms are 2,4,8,16.....like this because when we divide 4,8,16,32 by 2,we obtain 2,4,8,16.from 232,27 is remaining.2*4*8*16=210.
We'll teach you how to divide 16 by 3/8 in this section. We will offer you the solution in fraction form and decimal form. Here is 16 divided by 3/8 expressed mathematically in colors:= whole number= numerator= denominatorTo get the fraction form answer, multiply the entire number by the denominator and make the result the new numerator. The old denominator becomes the new numerator:Thus, the answer to 16 divided by 3/8 in fraction form is:To make the answer to 16 divided by 3/8 in decimal form, simply divide the numerator by the denominator from the fraction answer above:128 / 3 = 42.67The answer is rounded to the nearest two decimal points if necessary.128/3 is an improper fraction and should be written as 42 2/3. If so, go on to the next issue on our list.
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# Solution of functional equation $f(x/f(x)) = 1/f(x)$?
I've been trying to add math rigor to a solution of the functional equation in [1], eq. (22). It is: $$f\left(\frac{x}{f(x)}\right) = \frac{1}{f(x)}\,,$$ where you know that $f(0)=1$ and $f(-x) = f(x)$.
I've been trying to fill in the missing steps. Let's use a substitution $g(x) = \frac{x}{f(x)}$. So we rewrite the functional equation as $$x = \frac{\frac{x}{f(x)}}{f\left(\frac{x}{f(x)}\right)}$$ and get $$x = g(g(x))\,.$$ Then we calculate $g'(0)$ as follows: $$g'(0) = \left.\left(\frac{x}{f(x)} \right)'\right|_{x=0} = \left.\frac{f(x)-xf'(x)}{f^2(x)}\right|_{x=0} = \frac{f(0)}{f^2(0)} = 1\,.$$ Also we can use that $g(x)$ is odd $$g(-x) = \frac{-x}{f(-x)} = -\frac{x}{f(x)} = -g(x)\,.$$ Whenever $g(g(x)) = x$, I have found that this is called involution. That can have many solutions, but using $g(-x)=-g(x)$ and $g'(0)=1$, there should be a way to prove that $g(x) = x$, from which $f(x)=1$ as the only solution.
The article [1] just says, that since $g(x) = g^{-1}(x)$, the graph of $g(x)$ and its inverse must be symmetric along the $y=x$ axis, and since the tangent at $x=0$ is equal to this same line, then $g(x)=x$ must be the only solution. They do mention that this is not completely rigorous proof.
You can assume that $f(x)$ is differentiable. It'd be nice if it can be proven only for continuous functions $f(x)$, but any proof at first would be a good start, even with stronger assumptions.
Update: another idea is to use the fact, that if $a$ and $b$ are two points in the domain of the $g$ function for which $g(a)=g(b)$, then it follows that $g(g(a)) = g(g(b))$ and $a = b$, which proves that the function is one-to-one. Given that $g$ is continuous, it means that it is strictly increasing or decreasing [2]. From that, let's say it's increasing, then we can use $y=g(x)$, if $y < x$, then $g(y) < g(x)$ from which $g(g(x)) < y$ and $x < y$ which is a contradiction. Similarly for $y > x$. So we must have $y=x$, from which $g(x)=x$. If $g$ is decreasing, then we set $h(x)=-g(x)$, obtain $h(x)=x$ and $g(x)=-x$. Unless I made a mistake we got $g(x)=\pm x$. And using $g'(0)=1$, we get $g(x)=x$ as the only solution. But I don't feel very good about this proof yet.
[1] Levy-Leblond, J.-M. (1976). One more derivation of the Lorentz transformation. American Journal of Physics, 44(3), 271–277.
• Because $g(x)=1/x$ is also a solution of $g(g(x))$ and somehow it got eliminated. So something is not right. (Of course, it would get eliminated later anyway due to $g'(0)=1$, but that's not the point.) Jan 16, 2013 at 7:58
• Because $g(x)=1/x$ is not defined on $\mathbf R$, only on $\mathbf R^*$. Notice that it's also neither increasing nor decreasing. The answer you linked to uses the intermediate value theorem, which you can only apply on an interval, and $\mathbf R^*$ is not an interval. Jan 16, 2013 at 8:01
• Your proof in your update is almost certainly exactly what the authors had in mind, but didn't type out. Briefly: if $g$ is not $x\mapsto x$, then there is some $h>0$ such that $g(h)$ is close to $h>0$, yet $g(h)\neq h$. Then the reflection of the point $(h,g(h))$ will be on the other side of the line $y=x$, contradicting the monotonicity of the function.$\Box$ Jan 16, 2013 at 8:13
• Two things. For the increasing case, you assume (without saying it) that $g$ is defined on an interval $I$, and that $g$ sends $I$ into $I$ (because you use $y<x\Rightarrow g(y)<g(x)$), which explain $g(x)=-1/x$ on $(0,\infty)$. The other thing is that for the decreasing case, you set $h(x)=-g(x)$, and if you write that case in detail you'll see that you use $g(-x)=-g(x)$, which you can't say about $g(x)=1/x$ on $(0,\infty)$, because if $x\in(0,\infty)$ then $-x\notin(0,\infty)$. Jan 16, 2013 at 8:13
• Given that $f(x)$ is symmetric, you also know that all odd derivatives of $1/f(x/f(x))$ are zero. If you manage to express th even ones, $(1/f(x/f(x)))^{(2n)}$, analytically, you might get some information out, maybe even conclude that all are zero such that $f(x)=1$. Jan 16, 2013 at 13:07
Some results described above.
(i) $g$ is monotone.
(ii) $g$ = $g^{-1}$
(iii) $g$ is odd.
(iv) $g$ differentiable at $0$
Summary:
Suppose $g$ is not the identity. Then for $\forall r$, we know $g$ has a symmetrical counterpart to $g(r)$ w.r.t. $y=x$ from (ii). Now consider, $(r,g(r))$ for $r>0$ and his symmetric counterpart. Later consider $(-r,g(-r))$ and his symmetric counterpart. There is no way to ensure monotonicity between these coordinates for, in any case, the points will form a rectangle with two sides parallel and two perpendicular to $y=x$. So in adhering to the symmetry, $g$ must either fail to be a function or $g$ must violate monotonicity somewhere by crossing $y=x$. So $g$ must be the identity. Thus $f=1$.
Hmm, it seems to me that the only solution is the constant function $f(x)=1$
Let's assume that $f(x)$ has a power series, then
• the constant term must be 1, because $f(0)=1$
• we have only nonzero coefficients at the powers of x where the exponents are even because $f(x)=f(-x)$
so we assume $$f(x) = 1 + ax^2 + bx^4 + cx^6 + O(x^8)$$ Then for the rhs of your defining equation we have $${1 \over f(x)} = 1 - ax^2 + (a^2-b) x^4 + (-c + (-a^3 + 2ba)) x^6 + O(x^8)$$ For the lhs of your defining equation we have $$f(x {1 \over f(x)}) = 1 + ax^2 + (b -2 a^2)x^4 + (c + 3 a^3 - 6ab) x^6 + O(x^8)$$ and equating coefficients requires
• $a=0 \to (f"(0)=0)$ and
then $\qquad \qquad {1 \over f(x)} = 1 -b x^4 -c x^6 + O(x^8)$
must equal $f(x {1 \over f(x)}) = 1 + b x^4 + c x^6 + O(x^8)$
• thus is required that $b=0 \to f^{(4)}(0)=0$ and
then $\qquad \qquad {1 \over f(x)} = 1 -c x^6 + O(x^8)$
must equal $f(x {1 \over f(x)}) = 1 + c x^6 + O(x^8)$
• thus is required that $c=0 \to f^{(6)}(0)=0$ ...
... and I assume, that all following coefficients must equal zero. Thus: the only solution should be $$f(x)=1$$ and your $$g(x)={x \over f(x)}=x$$
• Very nice! So if the function $f(x)$ is analytic, then you prove that $f(x)=1$. Assuming only that $f(x)$ is continuous, then the only possible other solutions are not analytic. That helps a lot. Jan 16, 2013 at 16:15
• The constant function $-1$ works too. Jan 15, 2014 at 15:19
Given what you have already established, it will suffice to show that if $g$ is a continuous function with $g(g(x)) = x$, and $g(0) = 0,\ g'(0) = 1$, then $g(x) = x$ for all $x$. (Note that $g'(0) = 1$ because $g(x) = x + x(\frac{1}{f(x)} - 1) = x + o(x)$, using only continuity of $f$ at $0$).
Clearly, $g$ is injective: If $g(x_0) = g(x_1)$ then $x_0 = g(g(x_0)) = g(g(x_1)) = x_1$.
Thus, $g$ is monotonous: If it were not, it would have a local extreme at some $x_0$, and would not be injective in the neighbourhood of $x_0$. Because $g$ is increasing at $0$, it is therefore increasing on the whole domain.
We will now show that $g(x) = x$ for all $x$. Clearly, this holds for $x=0$. For a proof by contradiction, suppose that we have $x_0$ with $g(x_0) \neq x_0$. Let $x_1 := g(x_0)$. Then $x_0,x_1$ are distinct, and $g$ ''swaps'' these two points: $g(x_0) = x_1,\ g(x_1) = x_0$. Suppose without loss of generality that $x_0 < x_1$. We have $g(x_0) = x_1 > x_0 = g(x_1)$. But this contradicts the assumption that $g$ is increasing. Thus, we indeed have $g(x) = x$ for all $x$. $\square$ |
Integration by Parts
Key Questions
• Integration by Parts is like the product rule for integration, in fact, it is derived from the product rule for differentiation. It states
$\int u \mathrm{dv} = u v - \int v \mathrm{du}$.
Let us look at the integral
$\int x {e}^{x} \mathrm{dx}$.
Let $u = x$.
By taking the derivative with respect to $x$
$R i g h t a r r o w \frac{\mathrm{du}}{\mathrm{dx}} = 1$
by multiplying by $\mathrm{dx}$,
$R i g h t a r r o w \mathrm{du} = \mathrm{dx}$
Let $\mathrm{dv} = {e}^{x} \mathrm{dx}$.
By dividing by $\mathrm{dx}$
$R i g h t a r r o w \frac{\mathrm{dv}}{\mathrm{dx}} = {e}^{x}$
by integrating,
$R i g h t a r r o w v = {e}^{x}$
Now, by Integration by Parts,
int xe^xdx =xe^x-inte^xdx=xe^x-e^x+C
• By Integration by Parts: $\int u \mathrm{dv} = u v - \int v \mathrm{du}$,
Let $u = \ln x$ and $\mathrm{dv} = \mathrm{dx}$.
$R i g h t a r r o w \mathrm{du} = \frac{\mathrm{dx}}{x}$ and $v = x$
$\int \ln x \mathrm{dx} = x \ln x - \int \mathrm{dx} = x \ln x - x + C$,
where C is a constant.
When one has to integrate a product of two functions, integration by parts is useful.
If $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$
then $\int f \left(x\right) \mathrm{dx} = g \left(x\right) \int h \left(x\right) \mathrm{dx} - \int \left(\frac{d}{\mathrm{dx}} g \left(x\right) \cdot \int h \left(x\right) \mathrm{dx}\right) \mathrm{dx}$
This is called integration by parts.
Explanation:
The integral of the product of two functions may be verbally given as,
"First function into integral of the second minus integral of the derivative of the first into integral of the second." Which is nothing but, integration by parts.
Now, one thing that must be noted is that, the correct choice of first function and second function can either make or break a problem.
The correct choice can vastly simplify and an incorrect one can put you in a lot of trouble.
For instance,
$\int x \cdot {e}^{x} \mathrm{dx}$ is the integral we need to evaluate.
If we use ${e}^{x}$ as the first function and $x$ as the second and integrate by parts,
$\int x \cdot {e}^{x} \cdot \mathrm{dx} = {e}^{x} \int x \cdot \mathrm{dx} - \int \left({e}^{x} \int x \cdot \mathrm{dx}\right) \cdot \mathrm{dx}$
$= {e}^{x} \cdot {x}^{2} / 2 - \int {e}^{x} \cdot {x}^{2} / 2 \cdot \mathrm{dx} + C$
If we apply integration by parts to the second term, we again get a term with a ${x}^{3}$ and so on.
This, not only complicates the problem but, spells disaster.
But, if we had chosen $x$ to be the first and ${e}^{x}$ to be the second, the integral would have been very simply to evaluate.
$\int x \cdot {e}^{x} \cdot \mathrm{dx} = x \int {e}^{x} \cdot \mathrm{dx} - \int \left(\frac{d}{\mathrm{dx}} x \int {e}^{x} \cdot \mathrm{dx}\right) \cdot \mathrm{dx}$
$= {e}^{x} \cdot x - {e}^{x} + C$ |
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# Trees
## More Related Content
### Trees
1. 1. Trees<br />Data Structures and Algorithms<br />
2. 2. Definition<br />Trees are used to impose a hierarchical structure on a collection of data items.<br />A tree is a set of one or more nodes T such that:<br />there is a specially designated node called a root<br />The remaining nodes are partitioned into n disjointed set of nodes T1, T2,…,Tn, each of which is a tree.<br />
3. 3. Basic Terminology<br />Degree of a Node of a Tree<br /> The degree of a node of a tree is the number of subtrees having this node as a root. In other words, the degree is the number of descendants of a node. If the degree is zero, it is called a terminal or leaf node of a tree.<br />Degree of a Tree<br /> The degree of a tree is defined as the maximum of degree of the nodes of the tree, that is, degree of tree = max (degree(node i) for I = 1 to n)<br />Level of a Node<br /> We define the level of the node by taking the level of the root node as 1, and incrementing it by 1 as we move from the root towards the subtrees. So the level of all the descendants of the root nodes will be 2. The level of their descendants will be 3, and so on. We then define the depth of the tree to be the maximum value of the level of the node of the tree.<br />
4. 4. Binary Tree<br />A binary tree is a special case of tree as defined in the preceding section, in which no node of a tree can have a degree of more than 2. Therefore, a binary tree is a set of zero or more nodes T such that:<br />there is a specially designated node called the root of the tree<br />the remaining nodes are partitioned into two disjointed sets, T1 and T2, each of which is a binary tree. T1 is called the left subtree and T2 is called right subtree, or vice-versa<br />
5. 5. Skewed Binary Tree<br />
6. 6. Full binary tree<br />A full binary tree is a binary of depth k having 2k − 1 nodes. If it has < 2k − 1, it is not a full binary tree. For example, for k = 3, the number of nodes = 2k − 1 = 23 − 1 = 8 − 1 = 7. <br />A full binary tree with depth k = 3 is shown <br />
7. 7. Complete binary tree<br />A complete binary tree of depth k is a tree with n nodes in which these n nodes can be numbered sequentially from 1 to n, as if it would have been the first n nodes in a full binary tree of depth k.<br />A complete binary tree with depth k = 3 is shown<br />
8. 8. Representation of Binary Tree<br />BT are represented using 2 ways<br />Array representation<br />Linked list representation<br />
9. 9. 1. Array Representation<br />we can map node i to the ith index in the array, and the parent of node i will get mapped at an index i/2, whereas the left child of node i gets mapped at an index 2i and the right child gets mapped at an index 2i + 1<br />
10. 10. 2. Linked List representation<br />we can use a linked representation, in which every node is represented as a structure with three fields: one for holding data, one for linking it with the left subtree, and the third for linking it with right subtree<br />
11. 11. Traversal of Binary Tree<br />There are 6 ways of traversal of binary tree<br />LDR<br />LRD<br />DLR<br />RDL<br />RLD<br />DRL<br />Where,<br />L stands for traversing the left subtree, <br />R stands for traversing the right subtree,<br />D stands for processing the data of the node<br />
12. 12. The order LDR is called as inorder;<br />the order LRD is called as postorder;<br />the order DLR is called as preorder. <br /> The remaining three orders are not used.<br />
13. 13. Binary Search Tree<br />A binary search tree is a binary tree that may be empty, and every node must contain an identifier. An identifier of any node in the left subtree is less than the identifier of the root. An identifier of any node in the right subtree is greater than the identifier of the root. Both the left subtree and right subtree are binary search trees<br />
14. 14. DELETION OF A NODE FROM BINARY SEARCH TREE<br />Deletion of a node with two children<br />Deletion of a Node with One Child<br />Deletion of a Node with No Child<br />
15. 15. Deletion of a node with two children<br />
16. 16. Deletion of a Node with One Child<br />
17. 17. Deletion of a Node with No Child<br />
18. 18. Find or request more at http://amitbiswal.blogspot.com<br />Thank you<br /> |
NCERT Solutions Maths Class 10 Chapter 3
# NCERT Solutions Maths Class 10 Chapter 3
## NCERT Solutions Maths Class 10 Exercise 3.7
1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution: Let the ages of Ani and Biju be x years and y years, respectively.
Age of Dharam = 2x years and age of Cathy = y/2 years
Case 1: Let x > y
According to the question, we have
xy = 3 … (1)
And 2x y/2 = 30
4xy = 60 … (2)
Subtracting equation (1) from equation (2), we get
3x = 60 – 3 = 57
3x = 57
x = 19
Putting the value of x in equation (1), we get
19 – y = 3
y = 16
Thus, the age of Ani is 19 years and the age of Biju is 16 years.
Case 2: Let y > x
According to the question, we have
yx = 3 … (3)
And 2xy/2 = 30
4xy = 60 … (4)
Adding equations (3) and (4), we get
3x = 63
x = 21
Putting the value of x in equation (3), we get
y – 21 = 3
y = 24
Thus, the age of Ani is 21 years and the age of Biju is 24 years.
2. One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]
Solution: Let the money with the first person be Rs x and the money with the second person be Rs y.
According to the first condition, we have
x + 100 = 2(y – 100)
x + 100 = 2y – 200
x – 2y = –300 … (1)
According to the second condition, we have
6(x – 10) = (y + 10)
6x – 60 = y + 10
6xy = 70 … (2)
Multiplying equation (2) by 2, we get
12x – 2y = 140 … (3)
Subtracting equation (1) from equation (3), we get
11x = 140 + 300
11x = 440
x = 40
Putting the value of x in equation (1), we get
40 – 2y = –300
40 + 300 = 2y
2y = 340
y = 170
Thus, the two friends have Rs 40 and Rs 170 with them.
3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution: Let the speed of the train be x km/h, the time taken by the train to cover the given distance be t hours and the distance to cover be d km.
Since speed = distance/time
x = d/t
d = xt … (1)
According to the first condition, we have
x + 10 = d/(t – 2)
(x + 10)(t – 2) = d
xt + 10t – 2x – 20 = d
10t – 2x = 20 … (2) [Using eq. (1)]
According to the second condition, we have
x – 10 = d/(t + 3)
(x – 10)(t + 3) = d
xt – 10t + 3x – 30 = d
3x – 10t = 30 … (3) [Using eq. (1)]
Adding equations (2) and (3), we get
x = 50
Putting the value of x in equation (2), we get
10t – 2 × 50 = 20
10t – 100 = 20
10t = 120
t = 12
From equation (1), we get
d = xt = 50 × 12 = 600
Thus, the distance covered by the train is 600 km.
4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Solution: Let the number of rows be x and the number of students in a row be y.
Total number of students in the class = Number of rows × Number of students in a row
= xy
According to the first condition, we have
Total number of students = (x – 1) (y + 3)
xy = (x – 1) (y + 3)
xy = xyy + 3x – 3
3xy – 3 = 0
3xy = 3 … (1)
According to the second condition, we have
Total number of students = (x + 2) (y – 3)
xy = xy + 2y – 3x – 6
3x – 2y = –6 … (2)
Subtracting equation (2) from equation (1), we get
y = 9
Putting the value of y in equation (1), we get
3x – 9 = 3
3x = 9 + 3 = 12
x = 4
Thus, the number of rows = 4
And the number of students in a row = 9
Hence, the total number of students in the class = xy = 4 × 9 = 36
5. In a ABC, C = 3B = 2(A + B). Find the three angles.
Solution: C = 3B = 2(A + B)
Taking C = 3B … (1)
And 3B = 2(A + B)
3B = 2A + 2B
B = 2A
2A – B = 0 … (2)
We know from the angle sum property that the sum of the measures of all the angles of a triangle is 180°.
A + B + C = 180°
A + B + 3B = 180° [From eq. (1)]
A + 4B = 180°… (3)
Multiplying equation (2) by 4, we get
8A – 4B = 0 … (4)
Adding equations (3) and (4), we get
9A = 180°
A = 20°
Putting the value of A in equation (3), we get,
20° + 4B = 180°
B = 40°
And C = 3 × 40° = 120°
Hence, the measures of A, B and C are 20°, 40° and 120°, respectively.
6. Draw the graphs of the equations 5xy = 5 and 3xy = 3. Determine the co-ordinate of the vertices of the triangle formed by these lines and the y-axis.
Solution: We have the first equation: 5xy = 5
y = 5x – 5
For x = 0, 1, 2, we have y = –5, 0, 5 as shown in the following table.
x 0 1 2 y –5 0 5
We have the second equation:
3xy = 3
y = 3x – 3
For x = 0, 1, 2, we have y = –3, 0, 3 as shown in the following table.
x 0 1 2 Y –3 0 3
Let us draw these points on a graph paper to find the lines for the given equations.
It can be seen from the above graph that the required triangle is ACE.
The coordinates of its vertices are A(1, 0), C(0, –3), E(0, –5).
7. Solve the following pair of linear equations:
(i) px + qy = pq
qxpy = p + q
(ii) ax + bx = c
bx + ay = 1 + c
(iii) x/a y/b = 0
ax + by = a2 + b2
(iv) (ab)x + (a + b)y = a2 – 2abb2
(a + b)(x + y) = a2 + b2
(v) 152x – 378y = –74
–378x + 152y = –604
Solution: (i) px + qy = pq …. (1)
qxpy = p + q …. (2)
Multiplying equation (1) by p and equation (2) by q, we get
p2x + pqy = p2 pq … (3)
q2xpqy = pq + q2 … (4)
Adding equations (3) and (4), we get
p2x + q2x = p2 + q2
(p 2 + q2)x = (p2 + q2)
x = 1
Putting the value of x in equation (1), we get
p(1) + qy = pq
qy = –q
y = –1
Hence, x = 1 and y = –1
(ii) ax + bx = c …. (1)
bx + ay = 1 + c …. (2)
Multiplying equation (1) by a and equation (2) by b, we get
a2x + aby = ac … (3)
b2x + aby = b + bc … (4)
Subtracting equation (4) from equation (3), we get
(a2 b2)x = acbcb
(a2 b2)x = c(ab) – b
Putting the value of x in equation (1), we get
(iii) x/ay/b = 0
bx ay = 0 …… (1)
ax + by = a2 + b2 …… (2)
Multiplying equation (1) by b and equation (2) by a, we get
b2xaby = 0 ……. (3)
a2x + aby = a3 + ab2 ……. (4)
Adding equations (3) and (4), we get
b2x + a2x = a3 + ab2
x(b2 + a2) = a(a2 + b2)
x = a
Putting the value of x in equation (1), we get
b(a) – ay = 0
abay = 0
y = b
Thus, x = a and y = b
(iv) (ab)x + (a + b)y = a2 – 2abb2 ….. (1)
(a + b)(x + y) = a2 + b2
(a + b)x + (a + b)y = a2 + b2 ….. (2)
Subtracting equation (2) from equation (1), we get
(ab)x – (a + b)x = (a2 – 2abb2) – (a2 + b2)
(abab)x = –2ab – 2b2
–2bx = –2b(a + b)
x = a + b
Putting the value of x in equation (1), we get
(ab)(a + b) + (a + b)y = a2 – 2abb2
a2 b2 + (a + b)y = a2 – 2abb2
(a + b)y = –2ab
y = –2ab/(a + b)
Thus, x = a + b and y = –2ab/(a + b)
(v) 152x – 378y = –74 ….. (1)
–378x + 152y = –604 ….. (2)
Adding the equations (1) and (2), we get
–226x – 226y = –678
x + y = 3 …… (3)
Subtracting equation (2) from equation (1), we get
530x – 530y = 530
xy = 1 ……. (4)
Adding equations (3) and (4), we get
2x = 4
x = 2
Putting the value of x in equation (3), we get
2 + y = 3
y = 1
Thus, x = 2 and y = 1
8. ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.
Solution: We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.
Therefore, A + C = 180°
4y + 20 – 4x = 180°
–4x + 4y = 160°
xy = –40° …… (1)
Also,
B + D = 180°
3y – 5 – 7x + 5 = 180°
–7x + 3y = 180°…… (2)
Multiplying equation (1) by 3, we get
3x – 3y = –120° ……… (3)
Adding equations (2) and (3), we get
–4x = 60°
x = –15°
Putting the value of x in equation (1), we get
–15 – y = –40°
y = –15 + 40
y = 25
Therefore,
A = 4y + 20 = 4 × 25 + 20 = 120°
B = 3y – 5 = 3 × 25 – 5 = 70°
C = –4x = –4 × (–15) = 60°
D = –7x + 5 = –7(–15) + 5 = 110° |
# 43/50 As A Percent
Let's find how to write 43/50 fraction number in percentage explanations.
Answer : 43/50 is equal to 86% as a percentage
## How to convert 43/50 to percentage
To convert a fraction to a percent, you need to express the fraction as a ratio of the numerator to the denominator, and then multiply that ratio by 100.
In this case, the fraction is 43/50. To express this as a ratio, you would say that there is 43 part out of 50 total parts.
To convert this ratio to a percent, you need to multiply by 100. To do this, you can either divide the denominator into 100, and then multiply the result by the numerator, or you can multiply the fraction by 100/43 (which is the same as multiplying by 100).
### First Method :
Using the first method, you would divide 50 into 100:
100 / 50 = 2
Then you would multiply the result by the numerator:
2 x 43 = 86
Therefore, 43/50 as a percent is 86%.
### Second Method :
Using the second method, you would multiply the fraction by 100/43:
43/50 x 100/1 = 4300/50
Then you would simplify the fraction:
4300/50 = 86
So, the result is 86%, which is the same as saying that 43/50 is equal to 12.5% as a percentage.
## Fraction To Percent Conversion
Converting fractions to percentages is a common task in math, finance, and other fields. It's essential to know how to do this conversion quickly and accurately, so you can perform calculations and make comparisons between numbers. In this article, we'll explain how to convert fractions to percentages and provide examples, tables, and step-by-step instructions to guide you through the process.
### What is a fraction?
A fraction is a number that represents a part of a whole. Fractions are expressed as a numerator over a denominator, such as 1/2 or 3/4. The numerator represents the number of parts, and the denominator represents the whole.
### What is a percentage?
A percentage is a way of expressing a number as a fraction of 100. For example, 50% represents the number 0.5, or half of 100.
### How to convert fractions to percentages
Converting a fraction to a percentage involves dividing the numerator by the denominator, multiplying the result by 100, and adding a percent sign (%). Here's a step-by-step guide:
1. Take the fraction you want to convert.
2. Divide the numerator by the denominator.
3. Multiply the result by 100.
4. Add the percent sign (%).
For example, if you want to convert 3/4 to a percentage, follow these steps:
1. Take the fraction: 3/4
2. Divide the numerator by the denominator: 3 / 4 = 0.75
3. Multiply the result by 100: 0.75 x 100 = 75
4. Add the percent sign: 75%
Therefore, 3/4 is equivalent to 75%.
### Fraction to percent conversion table
Here's a conversion table for some common fractions to their corresponding percentages:
FractionPercent
1/425%
1/250%
3/475%
1/520%
2/540%
3/560%
4/580%
### Examples of fraction to percent conversion
Let's look at a few more examples of fraction to percent conversion:
Example 1: Convert 2/3 to a percentage.
1. Take the fraction: 2/3
2. Divide the numerator by the denominator: 2 / 3 = 0.6667 (rounded)
3. Multiply the result by 100: 0.6667 x 100 = 66.67 (rounded)
4. Add the percent sign: 66.67%
Therefore, 2/3 is equivalent to 66.67%.
Example 2: Convert 5/8 to a percentage.
1. Take the fraction: 5/8
2. Divide the numerator by the denominator: 5 / 8 = 0.625
3. Multiply the result by 100: 0.625 x 100 = 62.5
4. Add the percent sign: 62.5%
Therefore, 5/8 is equivalent to 62.5%.
Converting fractions to percentages is a straightforward process that involves dividing the numerator by the denominator, multiplying the result by 100, and adding a percent sign. With the help of examples, tables, and step-by-step instructions, you can quickly and accurately make these conversions for any fraction you encounter.
## Decimal Number To Percent Converter :
Make new percentage calculation from decimal number to percentage.
Let's learn how to write fraction as a percent, decimal as a percent and percentage to decimal. |
Polynomial and/or Polynomial Functions and Equations
Cubic functions
Transformation of the cubic polynomial from the general to source form and vice versa
Coordinates of the point of inflection coincide with the coordinates of translations
The source cubic functions are odd functions
Cubic function y = a3x3 + a2x2 + a1x + a0
Transformation of the cubic polynomial from the general to source form and vice versa
Applying the same method we can examine the third degree polynomial called cubic function.
1) Calculate the coordinates of translations
substitute n = 3 in
and
2) To get the source cubic function we should plug the coordinates of translations (with changed signs)
into the general form of the cubic, i.e.,
y + y0 = a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0,
after expanding and reducing obtained is
the source cubic function.
3) Inversely, by plugging the coordinates of translations into the source cubic
y - y0 = a3(x - x0)3 + a1(x - x0),
after expanding and reducing we obtain
y = a3x3 + a2x2 + a1x + a0 the cubic function in the general form.
Thus, y = a3x3 + a2x2 + a1x + a0 or y - y0 = a3(x - x0)3 + a1(x - x0), by setting x0 = 0 and y0 = 0 we get the source cubic function y = a3x3 + a1x where a1= tan at .
Coordinates of the point of inflection coincide with the coordinates of translations, i.e., I (x0, y0).
The source cubic functions are odd functions.
Graphs of odd functions are symmetric about the origin that is, such functions change the sign but not absolute value when the sign of the independent variable is changed, so that f (x) = - f (-x).
Therefore, since f (x) = a3x3 + a1x then - f (-x) = -[a3(-x)3 + a1(-x)] = a3x3 + a1x f (x).
That is, change of the sign of the independent variable of a function reflects the graph of the function about the y-axis, while change of the sign of a function reflects the graph of the function about the x-axis.
The graphs of the translated cubic functions are symmetric about its point of inflection.
Pre-calculus contents E |
# Sine and Cosine Rules
When I am setting homework or exam questions for my students I always check them first, but when I am actually teaching I often allow my students to invent their own problems. This can be interesting, and sometimes leads us into “uncharted waters”.
In a lesson on the Sine and Cosine Rules I asked a student to draw a triangle and then measure two sides and one angle. This is her triangle.
So sin C = 1.019
Which is impossible, and if you try sin-1 1.019 on your calculator you will get a maths error. Sadly, at this point some teachers will just say “You must have made a mistake,” and some students may actually believe this. It is not true of course, but in spite of much searching I have been unable to find any explanation in any book or web page which corrects or explains this.
The first and most important thing to do was to reassure my student that she had not made a mistake. Then we checked her calculations. Then we checked her measurements. Then we checked that the sides of her triangle really were straight lines. Finally we measured angle C. It was 85°, and this is the key to understanding what is happening.
We need to look at the graph of sin θ against θ in the Trigonometry and the Sine and Cosine Rules Page. We can see that near θ = 90° the curve is very flat, meaning that a very tiny change in the value of sin θ produces a far from tiny change in the value of θ. In fact, although sin θ can have any value from 0 to 1, it is more than 0.98 for all values of θ between 80° and 100°. Check this for yourself.
We can also see this graphically. In my student’s triangle angle A is acute and BC is less than AB. AE is AC extended. If we draw an arc with its centre at B, depending on the length of the arc (and of course AB and angle A), it will either cut AE at two places, making two triangles, one with an acute angle at C and one an obtuse angle (sin C less than 1), or make AE a tangent to the arc at C making a right angle at C (sin C = 1), or not cut AE at all (sin C greater than one).
If the angle C is very close to a right angle then the tiniest error in the length of BC (or of course AB or angle A) will make the difference between two triangles, one triangle or no triangle at all.
The lowest value of sin C occurs with the lowest values of A and c and the highest value of a, and the highest value of sin C occurs with the highest values of c and A and the lowest value of a.
If we measure the lengths to the nearest millimetre and angles to the nearest degree then a is in the range 55.5 mm to 56.5 mm, c is in the range 86.5 mm to 87.5 mm and A is in the range 40.5° to 41.5°. So sin C is in the range 0.99429 to 1.04467 so its lowest value is 0.99429 and so to the nearest tenth of a degree C is in the range 83.9° to 96.1°, and our measured value of 85° is within this range.
Summarising all of this, when we are using the Sine Rule to find angles of between 80° and 100°, we need to consider the accuracy of our data with more than ordinary care. For this reason you are unlikely to come across such a problem in an exam at normal school leaving age unless the Paper has not been checked properly. But you might come across it if you are just making up a triangle for yourself, which is what my Student did - it is actually not that uncommon to create such a triangle without intending to.
For reasons given in Appendix 3 it is very unlikely indeed that you will ever encounter a situation where the use of the Sine Rule gives sin-1θ exactly 1 so that θ is exactly 90°.
If we look at the graph of cos θ against θ we see that the curve is very flat for values of θ less that 10° or more than 170° so a similar problem will arise. However it is very unlikely that you would create such a triangle without intending to.
This problem has interested me greatly, but after much searching I have not found anyone else who has described how it is possible to get a value of sin θ (or cos θ) greater than one without having made a mistake, and I think this is a pity. |
# Frequent question: How do you calculate probability of dice?
Contents
If you want to know how likely it is to get a certain total score from rolling two or more dice, it’s best to fall back on the simple rule: Probability = Number of desired outcomes ÷ Number of possible outcomes.
## What is the probability of a dice?
Two (6-sided) dice roll probability table
Roll a… Probability
2 1/36 (2.778%)
3 3/36 (8.333%)
4 6/36 (16.667%)
5 10/36 (27.778%)
## What is the probability of rolling a 4 on a die?
We call the outcomes in an event its “favorable outcomes”. If a die is rolled once, determine the probability of rolling a 4: Rolling a 4 is an event with 1 favorable outcome (a roll of 4) and the total number of possible outcomes is 6 (a roll of 1, 2, 3, 4, 5, or 6). Thus, the probability of rolling a 4 is .
## How do you calculate the probability of winning a dice game?
Every time you add a die, the number of total outcomes is multiplied by 6. If you roll a pair of dice, for example, the number of total outcomes is 62 = 6 × 6 = 36, meaning every roll of the dice has a total of 36 possible outcomes. Therefore, the probability of rolling an 11 with two dice is 1/18.
## How do you calculate dice combinations?
How many total combinations are possible from rolling two dice? Since each die has 6 values, there are 6∗6=36 6 ∗ 6 = 36 total combinations we could get.
What are the most likely outcomes from rolling a pair of dice?
Outcome List of Combinations Total
7 1+6, 2+5, 3+4, 4+3, 5+2, 6+1 6
8 2+6, 3+5, 4+4, 5+3, 6+2 5
## What is the probability of rolling 1?
Explanation: The probability of rolling at least one “1” if you roll a dice six times is the same as 1 minus the probability of rolling zero 1s if you roll a dice six times. The probability of not rolling a 1 if you roll a dice once is 5/6.
## What is the probability of 2 dice?
Probabilities for the two dice
Total Number of combinations Probability
2 1 2.78%
3 2 5.56%
4 3 8.33%
5 4 11.11%
## What is die in probability?
The simplest case when you’re learning to calculate dice probability is the chance of getting a specific number with one die. … So for a die, there are six faces, and for any roll, there are six possible outcomes. There is only one outcome you’re interested in, no matter which number you choose.
## How do you find the probability of 3 dice?
We divide the total number of ways to obtain each sum by the total number of outcomes in the sample space, or 216. The results are: Probability of a sum of 3: 1/216 = 0.5% Probability of a sum of 4: 3/216 = 1.4%
## What is the most common number to roll with 1 dice?
For four six-sided dice, the most common roll is 14, with probability 73/648; and the least common rolls are 4 and 24, both with probability 1/1296. , 2, 3, and 4 dice. They can be seen to approach a normal distribution as the number of dice is increased.
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# 5.05 Graphing rational functions
Lesson
## Introduction to rational functions
A rational number is any number that can be expressed as the fraction $\frac{p}{q}$pq of two integers, a numerator $p$p and a non-zero denominator $q$q. Since $q$q may be equal to $1$1, every integer is a rational number. Similarly, a rational function is a function that can be written in the form of a fraction with expressions in the numerator and denominator.
A rational function is a function of the form $y=\frac{P\left(x\right)}{Q\left(x\right)},Q\left(x\right)\ne0$y=P(x)Q(x),Q(x)0, where $P\left(x\right)$P(x) and $Q\left(x\right)$Q(x) are polynomials.
So functions like $y=\frac{x}{x+1}$y=xx+1, $y=\frac{x^2-9}{2x-1}$y=x292x1, $y=5x^3-1=\frac{5x^3-1}{1}$y=5x31=5x311 and $y=\frac{24}{x^2-5x+6}$y=24x25x+6 can all be considered rational functions.
## Vertical and horizontal asymptotes
To understand the graph of a rational function, we must attend to the elephant in the room:
#### Exploration
Let’s compare the functions $f\left(x\right)=x$f(x)=x and its reciprocal function $g\left(x\right)=\frac{1}{x}$g(x)=1x. We will start by looking at the function as $x$x approaches $\infty$.
$x$x $0$0 $25$25 $50$50 $75$75 $100$100 $125$125
$f\left(x\right)$f(x) $0$0 $25$25 $50$50 $75$75 $100$100 $125$125
As $x$x increases towards positive $\infty$, the function approaches positive $\infty$. Also, as the $y$y is all real positive numbers.
$\frac{1}{x}$1x $0$0 $25$25 $50$50 $75$75 $100$100 $125$125
$f\left(x\right)$f(x) $undefined$undefined $0.040$0.040 $0.020$0.020 $0.013$0.013 $0.01$0.01 $0.008$0.008
As $x$x increases towards positive $\infty$, the function approaches $0$0 but never touches $0$0. Also, $y$y is all real positive numbers except $0$0.
Now, what happens when $x$x decreases towards negative $\infty$.
$x$x $0$0 $-25$25 $-50$50 $-75$75 $-100$100 $-125$125
$f\left(x\right)$f(x) $0$0 $-25$25 $-50$50 $-75$75 $-100$100 $-125$125
As $x$x decreases towards negative $\infty$, the function approaches negative $\infty$. Also, $y$y is all real negative numbers.
$\frac{1}{x}$1x $0$0 $-25$25 $-50$50 $-75$75 $-100$100 $-125$125
$f\left(x\right)$f(x) $undefined$undefined $-0.040$0.040 $-0.020$0.020 $-0.013$0.013 $-0.01$0.01 $-0.008$0.008
As $x$x decreases towards negative $\infty$, the function approaches $0$0 but never touches $0$0. Also, $y$y is all real negative numbers except $0$0.
{insert f(x) = 1/x image that says where As x approaches - infinity etc}
Because the denominator cannot be equal to zero, the function $g\left(x\right)=\frac{1}{x}$g(x)=1x is undefined at $x=0$x=0 and cannot be contained in the domain. As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). This behavior creates a vertical asymptote, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line $x=0$x=0 as the input becomes close to zero.
Also, based on the overall behavior and the graph of $g\left(x\right)=\frac{1}{x}$g(x)=1x, we can see that the function, $g\left(x\right)$g(x) , approaches $0$0 but never actually reaches $0$0; it seems to level off as the inputs become large. This behavior creates a horizontal asymptote, a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line $y=0$y=0
Finding horizontal asymptotes
The method used to find the horizontal asymptote changes depending on how the degrees of the polynomials in the numerator and denominator of the function compare.
• If both polynomials have the same degree, divide the coefficients of the greatest degree terms.
$f\left(x\right)=\frac{2x^3+1}{x^3-5}$f(x)=2x3+1x35
Both are polynomials are $3$3rd degree, therefore the horizontal asymptote is $y=\frac{2}{1}$y=21 or $y=2$y=2.
• If the polynomial in the numerator is a lower degree than the denominator, the $x$x-axis ($y=0$y=0) is the horizontal asymptote.
$g\left(x\right)=\frac{1}{2x-1}$g(x)=12x1
The numerator is a $0$0 degree polynomial and the denominator is a $1$1st degree, therefore the horizontal asymptote is $y=0$y=0.
$h\left(x\right)=\frac{2x^3+1}{x+1}$h(x)=2x3+1x+1
The numerator is a $3$3 rd degree polynomial and the denominator is a $1$1st degree, therefore there is no horizontal asymptote.
• If the polynomial in the numerator is a higher degree than the denominator, there is no horizontal asymptote.
#### Worked examples
##### question 1
Sketch a graph of the following rational function: $y=\frac{x}{x+2}$y=xx+2.
Think: The denominator of a fraction cannot be equal to zero; therefore, we must set the denominator equal to zero to find the vertical asymptote. Therefore, the vertical asymptote is $x=-2$x=2 . Because the numerator and denominator have the same degree, the horizontal asymptote is $y=1$y=1.
Do: First, graph the asymptotes.
{insert image of the two asymptotes using different colors}
Second, find and plot $x$x-intercept and $y$y-intercepts.
$x$x-intercept
$y$y $=$= $\frac{x}{x+2}$xx+2 (Given) $0$0 $=$= $\frac{x}{x+2}$xx+2 (Substitute $y=0$y=0) $0$0 $=$= $x$x (Multiply $x+2$x+2 on both sides)
Notice, the denominator vanishes.
$y$y-intercept
$y$y $=$= $\frac{x}{x+2}$xx+2 (Given) $y$y $=$= $\frac{0}{0+2}$00+2 (Substitute $x=0$x=0) $y$y $=$= $0$0 (Simplify)
{insert image of the point(s) on graph with asymptotes}
Third, select $x$x values to the left of asymptote (i.e., $-4$4). We already have a number to the right (the $x$x- intercept). Sketch graph.
{insert image graph of $y=\frac{x}{x+2}$y=xx+2}
##### Question 2
Find the $x$x-intercepts for the function: $y=\frac{x^3-5x^2+6}{x}$y=x35x2+6x.
Think: When finding the $x$x - intercept, set $y=0$y=0. After we multiply both sides by $x$x, we are left with $x^2-5x+6=0$x25x+6=0. This means that all of the function's $x$x intercepts are found by solving this quadratic equation.
Do:
$x^2-5x+6$x2−5x+6 $=$= $0$0 $\left(x-2\right)\left(x-3\right)$(x−2)(x−3) $=$= $0$0 $\therefore$∴ $x$x $=$= $2$2 or $x$x $=$= $3$3
##### Question 3
Sketch the graph of $y=\frac{x-1}{x}$y=x1x
Think: The denominator of a fraction cannot be equal to zero; therefore, we must set the denominator equal to zero to find the vertical asymptote. Therefore, the vertial asymptote is $x=0$x=0. Because the numerator and denominator have the same degree, the horizontal asymptote is $y=1$y=1.
Do:
##### Question 4
Graph the rational function $y=\frac{24}{x^2-5x+6}$y=24x25x+6. Clearly label the asymptotes.
Think: The denominator of a fraction cannot be equal to zero; therefore, we must set the denominator equal to zero to find the vertical asymptote. Therefore, the vertical asymptote is $x^2-5x+6=0$x25x+6=0. When factored, the vertical asymptotes are $x=2$x=2 and $x=3$x=3 . Because the numerator has no degree and denominator is a $2$2 nd degree polynomial, the horizontal asymptote is $y=0$y=0.
Do: First, graph the asymptotes.
Second, find the $x$x - and $y$y - intersects and graph.
$x$x-intersect
$y$y $=$= $\frac{24}{x^2-5x+6}$24x2−5x+6 (Given) $0$0 $=$= $\frac{24}{x^2-5x+6}$24x2−5x+6 (Substitute $y=0$y=0) $0$0 $\ne$≠ $24$24
There is no $x$x-intersect.
$y$y-intersect
$y$y $=$= $\frac{24}{x^2-5x+6}$24x2−5x+6 (Given) $y$y $=$= $\frac{24}{(0)^2-5(0)+6}$24(0)2−5(0)+6 (Substitute $x=0$x=0) $y$y $=$= $\frac{24}{6}$246 (Simplify) $y$y $=$= $4$4 (Simplify)
Third, select $x$x values to the left of asymptote. Sketch graph.
$x$x $1$1 $2.2$2.2 $2.4$2.4 $2.8$2.8 $4$4 $5$5
$y$y $12$12 $-150$150 $-100$100 $-150$150 $12$12 $4$4
Notice, additional points were selected between the asymptotes to determine the shape of the function with restrictions.
Sketch
Reflect: Notice, the denominator is a quadratic expression and the function behaves as a quadratic when restricted by both vertical asymptotes. Let's explore this phenomenon using the graphic calculator. Graph the following rational function and discuss with a classmate whether a parabola will always appear:
$y=\frac{1}{x^2}$y=1x2
$y=\frac{1}{x^2+1}$y=1x2+1
$y=\frac{-4x}{x^2-25}$y=4xx225
$y=\frac{-4x^2}{x^2-25}$y=4x2x225
Think about the impact the numerator has on the graph of the rational function. Take note of the degree of the numerator and the denominator. Does that have an impact on the graph?
#### Practice questions
##### Question 5
Consider the function $f\left(x\right)=\frac{x}{x\left(x-5\right)}$f(x)=xx(x5)
1. State the equation of the vertical asymptote.
##### Question 6
Consider the function
$y=\frac{15x}{5x^2+4}$y=15x5x2+4
1. State the equation of the horizontal asymptote.
##### Question 7
Graph the function $f\left(x\right)=\frac{2x-6}{x+4}$f(x)=2x6x+4 on the axes below:
1. Loading Graph...
## Even and odd multiplicity
The multiplicity of a root is determined by the number of times the associated factor is repeated. For a root at $x=a$x=a with multiplicity $2$2 corresponds to the repeated factor $\left(x-a\right)^2$(xa)2 in $P\left(x\right)$P(x). The multiplicity is represented in the graph by being a turning point of the graph on the $x$x-axis.
Multiplicity of a root
• Multiplicity $1$1: Function has no repeating roots. Graph passes through the root of the function.
• Multiplicity $2$2: Function has root(s) that repeat twice. Graph touches the root at a turning point of the function. Looks like a parabola at root; opposite sides of the root are moving in the same direction.
• Multiplicity $3$3: Function has root(s) that repeat $3$3 times. Graph touches the root at a turning point of the function. Looks opposites sides of the root are moving in different directions.
For rational functions we must pay attention to whether the multiplicity is even or odd. This is based on the degree of the denominator and whether the denominator multiplicity. For example, the following function has an odd multiplicity:
$y=\frac{1}{x+1}$y=1x+1,
because the degree of the denominator is $1$1 and the denominator has multiplicity of $1$1 for $(x+1)^1$(x+1)1.
Likewise, the following function has an even multiplicity:
because the degree of the denominator is $2$2. and the denominator has multiplicity of $2$2 for $(x+1)^2$(x+1)2.
Now, graphically, odd multiplicity rational functions open on opposite sides of the asymptotes (graph located in top right and bottom left or graph located in bottom right and top left). When the rational function has an even multiplicity, then the graph opens on the same side of the asymptotes (graph located in the top right and top left or graph located in bottom right and bottom left).
Graphing odd and even multiplicity
• Even Multiplicity: The curve on both sides of the discontinuity rises (or falls) in the same direction.
• Odd Multiplicity: The curve rises (or falls) in opposite directions.
Is there a connection between the multiplicity of roots and the odd and even multiplicity of rational functions?
#### Worked examples
##### QUESTION 8
Sketch the rational function given by $y=\frac{2}{x^2-4x+4}$y=2x24x+4.
Think: Notice, we can factor the denominator. Factoring is the first step when applicable. The denominator has a degree of $2$2, however we will not know if it has an even multiplicity until the denominator is factored.
Do: First, factor the denominator.
$x^2-4x+4$x24x+4
$(x-2)^2$(x2)2
The function has an even multiplicity because the degree of the denominator is $2$2 and the denominator can be rewritten to show that $(x-2)(x-2)$(x2)(x2).
Second, find the vertical and horizontal asymptotes and graph.
vertical asymptote: $x=2$x=2
horizontal asymptote: $y=0$y=0, because the degree of the numerator is less than the degree of the denominator.
Third, find $x$x- and $y$y- intercepts and graph.
$x$x-intercept
$y$y $=$= $\frac{2}{x-2}^2$2x−22 (Given-factored denominator) $0$0 $=$= $\frac{2}{x-2}^2$2x−22 (Substitute $y=0$y=0) $0$0 $\ne$≠ $2$2 (Multiply $(x-2)^2$(x−2)2 on both sides)
There is no $x$x-intercept. We can skip this part when the horizontal asymptote is $y=0$y=0.
$y$y-intercept
$y$y $=$= $\frac{2}{x-2}^2$2x−22 (Given-factored denominator) $y$y $=$= $\frac{(2}{(0)-2}^2$(2(0)−22 (Substitute $x=0$x=0) $y$y $=$= $\frac{1}{2}$12 (Simplify)
Fourth, select one $x$x -value to the right of the vertical asymptote and one to the left. The vertical asymptote is $x=2$x=2
$x$x $1$1 $3$3
$y$y $2$2 $2$2
{insert graph of $y=\frac{2}{x^2-4x+4}$y=2x24x+4}
##### Question 9
Sketch the rational function given by $y=\frac{x+3}{x^2-9}$y=x+3x29
Think: Notice, we can factor the denominator. The denominator has degree of $2$2 but will the function have an even multiplicity?
Do: First, factor the numerator and denominator.
$y=\frac{(x+3)}{(x+3)(x-3)}$y=(x+3)(x+3)(x3)
Remove common factors ($x+3$x+3 ) before determining the vertical and horizontal asymptotes.
$y=\frac{1}{x-3}$y=1x3
Second, find the vertical and horizontal asymptotes.
vertical asymptote: $x=3$x=3
horizontal asymptote: $y=0$y=0
The common factor $x+3$x+3 is removed; however there is $x=-3$x=3 is still not part of the domain. There is a "hole" in the graph at this point; which makes the function discontinuous.
Third, find the $x$x- and $y$y-intercept.
$x$x-intercept - There is no x-intercept because the horizontal asymptote is $y=0$y=0
$y$y-intersect
$y$y $=$= $\frac{1}{x-3}$1x−3 (Given-factored) $y$y $=$= $\frac{1}{(0)-3}$1(0)−3 (Substitute $x=0$x=0) $y$y $=$= $\frac{1}{-3}$1−3 (Simplify)
Third, select $x$x values to the left of asymptote (i.e., $-4$4 one step from vertical asymptote). We already have a number to the right (the $x$x- intercept). Sketch graph.
{insert graph}
##### Question 10
Sketch the rational function given by $y=\frac{x-2}{x\left(x-4\right)}$y=x2x(x4).
Think: It has a single root of multiplicity at $x=2$x=2 due to the numerator being set equal to zero: $x-2=0$x2=0 . It also has two odd poles at $x=0$x=0 and $x=4$x=4. due to the denominator being set equal to zero: $x(x-4)=0$x(x4)=0.
Do: First, find the vertical and horizontal asymptotes and graph.
vertical asymptote: $x=0$x=0 and $x=4$x=4
horizontal asymptote: $y=0$y=0
Second, find the $x$x- and $y$y- intercepts and graph them.
x-intercept - Because there are $2$2 asymptotes, we need to test whether is an $x$x-intercept between the vertical asymptotes
$y$y $=$= $\frac{x-2}{x(x-4)}$x−2x(x−4) (Given) $0$0 $=$= $\frac{x-2}{x(x-4)}$x−2x(x−4) (Substitute $y=0$y=0) $0$0 $=$= $x-2$x−2 (Multiply $x(x-4)$x(x−4) on both sides) $x$x $=$= $2$2 (Add $2$2 on both sides)
y-intercept
$y$y $=$= $\frac{x-2}{x(x-4)}$x−2x(x−4) (Given) $y$y $=$= $frac(0)-2(0)((0)-4)$frac(0)−2(0)((0)−4) (Substitute $x=0$x=0) $y$y $=$= $\frac{1}{2}$12 (Simplify)
Third, sketch by selecting x-values one set to the right and left of the vertical asymptotes.
$x$x $-1$1 $1$1 $3$3 $5$5
$y$y $-\frac{3}{5}$35 $\frac{1}{3}$13 $-\frac{1}{3}$13 $\frac{3}{5}$35
Here is the sketch:
Note also the behavior as $x\rightarrow\pm\infty$x±. Because the degree of $P\left(x\right)$P(x) is less than the degree of $Q\left(x\right)$Q(x), the function approaches $0$0 (and thus the curve approaches the asymptote $y=0$y=0 ) as $x$x becomes more positively large or more negatively large.
##### Question 11
Sketch the rational function given by $y=\frac{1}{x^2}$y=1x2.
Think: Because the numerator is the constant $1$1, the function has no roots (it cannot cross the $x$x axis). In the numerator we see an even pole at $x=0$x=0
As $x\rightarrow\pm\infty$x±, the curve asymptotically approaches the $x$x axis.
The point $\left(1,1\right)$(1,1) verifies that the curve is above the $x$x-axis.
Do: Thus the sketch is straightforward:
##### Question 12
Sketch the rational function given by $y=\frac{\left(x-2\right)^3}{x^2-25}$y=(x2)3x225.
Think: We see a root at $x=2$x=2 of multiplicity $3$3, and, because the denominator factors to $\left(x+5\right)\left(x-5\right)$(x+5)(x5), we know that there are two odd poles at $x=\pm5$x=±5.
The degree of the numerator is one higher than that of the denominator, so there we expect an oblique linear asymptote (The exact equation of the asymptote is difficult to determine, but it is accessible by division and is given by the line $y=x-6$y=x6 ).
Do: Here is the sketch:
#### Exploration
The following applet allows you to create your own rational function. The equation form for this applet is given by:
$y=\frac{\left(x-a\right)^m\left(x-b\right)^n}{\left(x-c\right)^p\left(x-d\right)^q}$y=(xa)m(xb)n(xc)p(xd)q
where:
• the indices $m$m, $n$n, $p$p and $q$q can be chosen as either $0,1,2$0,1,2 or $3$3
• the constants $a$a$b$b $c$c and $d$d can be chosen as integers in the range $-5\le x\le5$5x5
Note that if you choose $a=c$a=c, $a=d$a=d, $b=c$b=c or $b=d$b=d, there will be common factors in the numerator and denominator of the rational function and this is likely to result in a less complex shape of the curve.
For example you could create functions like:
• $y=\frac{\left(x-3\right)\left(x+3\right)}{x}$y=(x3)(x+3)x
• $y=\frac{\left(x+3\right)}{x^2}$y=(x+3)x2
• $\frac{1}{\left(x-1\right)^3}$1(x1)3
• $y=\frac{\left(x-1\right)\left(x-3\right)^2}{\left(x+1\right)^2\left(x+5\right)}$y=(x1)(x3)2(x+1)2(x+5)
#### Practice questions
##### question 13
Which of the following is the graph of the function $f\left(x\right)=\frac{-3x}{x^2+4x-5}$f(x)=3xx2+4x5?
1. Loading Graph...
A
Loading Graph...
B
Loading Graph...
C
Loading Graph...
D
Loading Graph...
A
Loading Graph...
B
Loading Graph...
C
Loading Graph...
D
## Domain and setting the denominator function to zero
When dealing with the domain of a rational function, we need to locate any points of discontinuity. As these are points where the function is not defined, we simply solve $Q\left(x\right)=0$Q(x)=0 to find them. If there are no solutions to this equation (and it frequently happens), then there are no points of discontinuity and the curve is said to be continuous across its entire domain.
### A cautionary note
In most cases if there is a point of discontinuity, the curve will exhibit asymptotic behavior on either side of it.
As a word of caution however, asymptotic behavior is not always guaranteed. This is because some of the solutions to $Q\left(x\right)=0$Q(x)=0 may also be solutions to $P\left(x\right)=0$P(x)=0.
Suppose for example we find a value of $x$x, say $x=h$x=h, where both $Q\left(h\right)=0$Q(h)=0 and $P\left(h\right)=0$P(h)=0. In that case, the curve will, at first glance, look to smoothly cross the line $x=h$x=h without any asymptotic behavior at all. However, what actually exists is a "hole" in the curve at $x=h$x=h. See example 8 below for a demonstration of this.
#### Worked examples
##### QUestion 14
Find the domain of the function $y=\frac{x^2}{x^2-1}$y=x2x21
Think: Here, $Q\left(x\right)=x^2-1$Q(x)=x21
Do: So, we simply solve the equation $x^2-1=0$x21=0. The solutions are $x=\pm1$x=±1, so there are two discontinuities to deal with. Neither $x=1$x=1, nor $x=-1$x=1 are solutions to $P\left(x\right)=0$P(x)=0, so the curve will exhibit asymptotic behavior around these points.
The domain is thus given formally as $\left\{x:x\in\Re\setminus x=\pm1\right\}${x:xx=±1}. That is to say the domain includes all real numbers except $x=1$x=1 and $x=-1$x=1
Reflect: Here is the curve showing these discontinuities. Note that domain considerations are only one part of an overall strategy to sketch these functions. Calculus techniques are often needed to ascertain many other features.
##### Question 15
Find the domain of the function $y=\frac{x-1}{x^2-x}$y=x1x2x
Think: Since $Q\left(x\right)=x^2-x$ |
Categories :
## How do you add fractions step by step?
To add fractions there are Three Simple Steps:
1. Step 1: Make sure the bottom numbers (the denominators) are the same.
2. Step 2: Add the top numbers (the numerators), put that answer over the denominator.
3. Step 3: Simplify the fraction (if needed)
How do you add and subtract Monomials?
To add two or more monomials that are like terms, add the coefficients; keep the variables and exponents on the variables the same. To subtract two or more monomials that are like terms, subtract the coefficients; keep the variables and exponents on the variables the same. on the variables the same.
How do you simplify fraction?
How to Reduce Fractions
1. Write down the factors for the numerator and the denominator.
2. Determine the largest factor that is common between the two.
3. Divide the numerator and denominator by the greatest common factor.
4. Write down the reduced fraction.
### How do you add and subtract mixed numbers?
The steps are the same whether you’re adding or subtracting mixed numbers:
1. Find the Least Common Denominator (LCD)
2. Find the equivalent fractions.
3. Add or subtract the fractions and add or subtract the whole numbers.
Do you add or subtract exponents when dividing?
To divide exponents (or powers) with the same base, subtract the exponents. Division is the opposite of multiplication, so it makes sense that because you add exponents when multiplying numbers with the same base, you subtract the exponents when dividing numbers with the same base.
What is the rule for adding exponents?
To add exponents, both the exponents and variables should be alike. You add the coefficients of the variables leaving the exponents unchanged. Only terms that have same variables and powers are added. This rule agrees with the multiplication and division of exponents as well.
#### How do you add and subtract fractions with different denominators?
Make their denominators equal using the concept of least common multiple. Then subtract their numerators accordingly. Rewrite each fraction to its equivalent fraction with a denominator equal to the LCM = 30, then subtract their numerators. Make sure to reduce your answer to the lowest term.
How do you add fractions with different denominators step by step?
If the denominators are not the same, then you have to use equivalent fractions which do have a common denominator . To do this, you need to find the least common multiple (LCM) of the two denominators. To add fractions with unlike denominators, rename the fractions with a common denominator. Then add and simplify.
What are the different concepts in subtracting polynomials?
Answer: There are two methods for subtracting polynomials: the horizontal and vertical methods. Both methods work equally well, you can choose which method you are more comfortable with. It’s important to combine like terms and keep track of the negative and positive signs when subtracting polynomials.
## How do you add and subtract polynomials?
The same symbols are used. You add polynomials when there are plus signs. You subtract them when there is a minus sign. Remember to only add/subtract like terms within the polynomials.
Do you add or multiply exponents?
To multiply two exponents with the same base, you keep the base and add the powers.
What is the rule for subtracting exponents?
If both the exponents and the bases are the same, you can subtract them like any other like terms in algebra. For example, 3y – 2xy = x y.
### What is the negative exponent rule?
What is negative exponent? A negative exponent helps to show that a base is on the denominator side of the fraction line. In other words, the negative exponent rule tells us that a number with a negative exponent should be put to the denominator, and vice versa.
How do you subtract fractions step by step?
There are 3 simple steps to subtract fractions
1. Make sure the bottom numbers (the denominators) are the same.
2. Subtract the top numbers (the numerators). Put the answer over the same denominator.
3. Simplify the fraction (if needed).
How do we add two polynomials?
1. Start with:2×2 + 6x + 5 + 3×2 − 2x − 1.
2. Place like terms together:2×2+3×2 + 6x−2x + 5−1.
3. Which is:(2+3)x2 + (6−2)x + (5−1)
4. Add the like terms:5×2 + 4x + 4.
#### How do you find a common denominator?
To make the denominators the same we can: Multiply top and bottom of each fraction by the denominator of the other. We simplified the fraction 2032 to 1016 , then to 58 by dividing the top and bottom by 2 each time, and that is as simple as it can get!
What is the rule for adding and subtracting exponents?
To add or subtract with powers, both the variables and the exponents of the variables must be the same. You perform the required operations on the coefficients, leaving the variable and exponent as they are.
What are the five rules of exponents?
Exponent rules
• Product of powers rule. When multiplying two bases of the same value, keep the bases the same and then add the exponents together to get the solution.
• Quotient of powers rule.
• Power of a power rule.
• Power of a product rule.
• Power of a quotient rule.
• Zero power rule.
• Negative exponent rule.
## Can you add bases with different exponents?
multiplying exponents If exponents have different bases, you cannot add their powers. If the exponents have coefficients attached to their bases, multiply the coefficients together. If exponents have the same power and the same base, you can choose either method to simplify.
How do you solve negative fractions?
– The negative of a number can be created by multiplying the number by negative one. Placing the negative sign before the entire fraction (subtracting the fraction) is equivalent to adding the same fraction, but with a negative numerator.
What are the six laws of exponents?
• Rule 1 (Product of Powers)
• Rule 2 (Power to a Power)
• Rule 3 (Multiple Power Rules)
• Rule 4 (Quotient of Powers)
• Rule 5 (Power of a quotient)
• Rule 6 (Negative Exponents)
• Quiz. |
#### Need Help?
Get in touch with us
# Decompose Fractions
Sep 17, 2022
## Key Concepts
• Compose and decompose
• Compose and decompose fractions
• Mixed numbers
• Equivalent fractions
## Compose and Decompose Fractions
We know that fractions show part of a whole, but they can be broken down further and combined again.
Combining fractions together is also called composing the fractions.
• To compose means “putting a number together using its parts”.
For example, 300+40+5 = 345
• Composing fractions means connecting fractions together by adding.
For example, 2/5+1/5+1/5= 4/5
Here, we composed the fraction 4/5 by adding fractions with like denominators.
Writing a fraction as a sum of smaller fractions is also known as decomposing fractions.
• To decompose means “to break apart.”
To decompose a fraction means dividing a fraction into smaller parts such that on adding all the smaller parts together, it results in the initial fraction.
For example, 3/4 can be decomposed as 1/4+1/4+1/4
## DIFFERENT METHODS OF DECOMPOSING FRACTIONS:
There are different ways to decompose a fraction
1. Breaking into unit fractions
2. Using the sum of smaller fractions which are not unit fractions
3. Breaking mixed fractions
4. Breaking into unit fractions
To decompose a fraction using unit fractions, we should know what unit fractions are.
Unit fractions are those fractions in which the numerator is always 1.
For example, 1/2 , 1/3 , 1/4, 1/5 etc.,
### Unit Fractions
Let us look at this situation.
Ms. Charlie wants to leave 1/6 of her garden empty and wanted to plant 5 types of plants in an equal area in the rest of her garden. What could be the area of each part of the garden?
Here, let us assume the garden is in a rectangular shape and is divided into 6 equal parts.
That means each partition shows 1/6 part of the garden.
If she wants to leave 1/6 part of the garden empty, then she is left with 5/6 parts of the garden.
5/6 can be decomposed as
1/6+1/6+1/6+1/6 +1/6
Example 2:
Decompose 7/9 using unit fractions.
7/9= 1/9+1/9+1/9+1/9+1/9+1/9+1/9
1. Using the sum of smaller fractions which are not unit fractions
We can also decompose fractions using the sum of smaller fractions. This is just like writing a number as the sum of different smaller numbers; fractions also can be decomposed using smaller fractions other than unit fractions.
Like 5 can be written as 3 + 2 (or) 4 + 1 (or) 1 + 1 + 3
Let us consider the same situation.
Ms Charlie can plant different parts in different ways. Because 5/6.
can be written as the sum of smaller like fractions. Like 5/6= 2/6+ 3/6 (or) 1/6+ 4/6 (or) 1/6+ 1/6+ 3/6
(or)
(or)
Example 2: Decompose 9/12 using smaller fractions.
9/12= 3/12+5/12+1/12 (or) 2/12+4/12+3/12 (or) 4/12+5/12 etc.,
9/12 can be decomposed in many different combinations.
1. Breaking mixed fractions
We know that fractions are of three types; proper fractions, improper fractions and mixed fractions.
Among all the above three, we can convert mixed fractions to improper fractions and improper fractions to mixed fractions.
For example,
In an improper fraction, the numerator is greater than the denominator, like in the above example, in the fraction 17/3, 14 is greater than 3.
17/3 can be decomposed as 1 whole + 1 whole + 1 whole + 1 whole + 2/3.
Likewise, a mixed fraction 4*2/3 can be decomposed as
1 + 1 + 1 + 1 + 2/3.
(or)
1 + 1 + 1 + 1 + 1/3
Example 2: Decompose 3*2/4
# Exercise:
1. Decompose the following fractions using unit fractions.
1. 2/4 b. 4/6 c. 3/5 d. 7/8
2. There were 10 bowling pins standing before John took his first turn. On his first turn, he knocked down 6 pins. On his second turn, he knocked down 2 pins. Write them as fractions and compose the fractions.
3. At lunch, Alice ate 3/8 of her sandwich. Later, for a snack, she ate another 3/8 of the sandwich. Write an addition sentence that shows how much of the sandwich Alice ate. Suppose Alice ate the same total amount of her sandwich in 3 different times instead of 2. Write an addition problem that shows the amount she ate as a sum of 3 fractions.
4. Draw pictures or use fraction strips to show why the equation is correct.
5. What is one way you can decompose ?
6.
Look at the diagram given and fill in the blank .
Look at the area model given here. What fraction with a greater numerator than denominator is equivalent to the given fraction?
• Decompose each fraction or mixed number in two different ways.
• a. 4/5 b. 3*3/8
• Write the number sentence to show how would you decompose 5/7 .
• In a class of 12 students, 8 of them are girls. Write two equivalent fractions that tell which part of the class is girls.
### What have we learned:
• Composing fractions
• Decomposing fractions using unit fractions
• Decomposing fractions using smaller fractions which are not unit fractions
• Decomposing mixed fractions
• Generating a graph based on the ordered pairs.
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] |
# Find the Zeros of the Polynomial x^2−3x−m(m+3)
By Mohit Uniyal|Updated : May 16th, 2023
Find the zeros of the polynomial x2−3x−m(m+3)
To find the zeros of the polynomial f(x) = x^2 - 3x - m(m + 3), we need to solve the equation f(x) = 0.
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. We will use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
And to verify the zeros of the polynomial f(x) = x^2 - 3x - m(m + 3), we can substitute the values we found into the equation and check if it evaluates to zero.
## Find the Zeros of the Polynomial x2−3x−m(m+3)
To find the zeros of the polynomial f(x) = x2 - 3x - m(m + 3), we need to solve the equation f(x) = 0.
Let's set up the equation:
x2 - 3x - m(m + 3) = 0
x = (-b ± √(b2 - 4ac)) / (2a)
For the given equation, a = 1, b = -3, and c = -m(m + 3).
Using the quadratic formula, we have:
x = (3 ± √((-3)2 - 4(1)(-m(m + 3)))) / (2(1)) = (3 ± √(9 + 4m(m + 3))) / 2 = (3 ± √(9 + 4m2 + 12m)) / 2 = (3 ± √(4m2 + 12m + 9)) / 2 = (3 ± √((2m + 3)2)) / 2 = (3 ± (2m + 3)) / 2
This gives us two solutions:
x₁ = (3 + (2m + 3)) / 2 = (2m + 6) / 2 = m + 3 x₂ = (3 - (2m + 3)) / 2 = (-2m) / 2 = -m
Therefore, the zeros of the polynomial x2 - 3x - m(m + 3) are x = m + 3 and x = -m.
## Zeros of the Polynomial x2−3x−m(m+3) are x = m + 3 and x = -m
Similar Questions: |
, 21.06.2019lewisf5929
# Vanessa earns a base salary of $400.00 every week with an additional5% commission on everything she sells. vanessa sold$1650.00 worth of items last week.
## Answers
The systems of equation are .
Vanessa worked for 6 hrs and Geneva worked for 15 hrs to earn exactly $240. Step-by-step explanation: Let the number of hours worked by Vanessa be 'v'. Let the number of hours worked by Geneva be 'g'. Given: Total number of hours worked = 21 hours So we can say that; Total number of hours worked is equal to sum of number of hours worked by Vanessa and number of hours worked by Geneva. framing in equation form we get; ⇒ equation 1 Also given: per hour earnings of Vanessa =$10
per hour earning of Geneva = $12 Total they need to earn =$240
So we can say that;
Total they need to earn would be equal to sum of per hour earnings of Vanessa multiplied by number of hours worked by Vanessa and per hour earning of Geneva multiplied by number of hours worked by Geneva.
framing in equation form we get;
⇒ equation 2
Hence The systems of equation are .
On Solving above equations we get;
First we will multiply equation 1 by 10 we get;
⇒ equation 3
Now Subtracting equation 3 from equation 2 we get;
Dividing both side by 2 we get;
Now Substituting the value of g in equation 1 we get;
Hence Vanessa worked for 6 hrs and Geneva worked for 15 hrs to earn exactly $240. Step-by-step explanation: total pay = 400 + 1650×5/100 =482.5 her total pay last week was 482.50 Hence the total salary of Vanessa last week was$482.5
$492.50 Step-by-step explanation: To start off, we already know that Vanessa makes at least$400 every week. What we don't know is how much Vanessa makes from her 5% commission on what she sells.
To solve for that, we need to find what 5% of $1,850 is. This can be done by multiplying$1,850 by 5%. Remember, we need to change 5% into its decimal form (5% -> -> 0.05).
$1,850 x 0.05 =$92.5
This means that Vanessa made $92.5 off of commissions last week. The last step is to add her base salary of$400 to her commission sales and we get our final answer.
$400 +$92.5 = $492.50 Vanessa's total pay last week was$492.50
$105 Step-by-step explanation: Salary for 3 hrs=$63
Salary for 5hrs= salary for one hour×5 hrs,
Salary for one hour= 63/3=$21 Salary for 5 hours= 21×5:$105
$105 Step-by-step explanation: I used a proportion to solve this: Cross multiply: 3x = 315 Divide both sides by 3 x = 105 5% is$82.5. So total pay last week is 482.5.
$482.5 is the FINAL ANSWER To find commission: We can use the base of 100% being 1650 We can use 5% because thats her commission Use: 100(x)=1650(5) -> 100(x)=8250 after doing math, youll get the outcome of 82.5 commission To find the total: Just add that commission to her total being 82.5 + 400 = 482.5$
Vanessa´s total pay was $482.50. Step-by-step explanation: We have to add the amount of the 5% commission on Vanessa´s sales to her base salary of$400 to find out her total pay.
Since the total value of her sales was \$1650, the amount (x) of her commission will be:
x = 1650 ×
x = 16.5 × 5
x = 82.50
Adding this to her base salary we get:
400 + 82.5 = 482.5
Do you know the answer?
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# Difference between revisions of "Incenter"
Triangle ABC with incenter I, with angle bisectors (red), incircle (blue), and inradii (green)
The incenter of a triangle is the intersection of its (interior) angle bisectors. The incenter is the center of the incircle. Every nondegenerate triangle has a unique incenter.
## Proof of Existence
Consider a triangle $ABC$. Let $I$ be the intersection of the respective interior angle bisectors of the angles $BAC$ and $CBA$. We observe that since $I$ lies on an angle bisector of $BAC$, is equidistant from $AB$ and $CA$; likewise, it is equidistant from $BC$ and $AB$; hence it is equidistant from $BC$ and $BC$ and $CA$ and therefore lies on an angle bisector of $ACB$. Since it lies within the triangle $ABC$, this is the interior angle bisector of $ACB$. Since $I$ is equidistant from all three sides of the triangle, it is the incenter.
It should be noted that this proof parallels that for the existence of the circumcenter.
The proofs of existence for the excenters is the same, except that certain angle bisectors are exterior.
## Properties of the Incenter
$\bullet$ The incenter of any triangle lies within the orthocentroidal circle.
$\bullet$ The unnormalised areal coordinates of the incenter are $(a,b,c)$
$\bullet$ Let $D$ be a point on the circumcircle of $\triangle ABC$ such that $AD$ bisects $\angle BAC$. Then points $B$, $C$, and $I$ lie on a circle centered at $D$.
$[asy]defaultpen(fontsize(8)); pair A=(7,10), B=(0,0), C=(10,0), D, I; I=incenter(A,B,C); draw(A--B--C--A);draw(circumcircle(A,B,C));draw(incircle(A,B,C)); D=intersectionpoint(A+0.1*expi((angle(B-A)+angle(C-A))/2)--A+20*expi((angle(B-A)+angle(C-A))/2), circumcircle(A,B,C)); draw(A--D);draw(circumcircle(B,C,I)); dot(A^^B^^C^^D^^I);label("A",A,(0,1));label("B",B,(-1,0));label("C",C,(1,0)); label("D",D,(0,-1));label("I",I,(-1,1));[/asy]$ |
# How To Make Multiplication Flash Cards
Learning multiplication soon after counting, addition, as well as subtraction is ideal. Kids discover arithmetic through a organic progression. This growth of learning arithmetic is usually the adhering to: counting, addition, subtraction, multiplication, and lastly division. This declaration contributes to the query why discover arithmetic within this series? More importantly, why discover multiplication following counting, addition, and subtraction but before section?
## These facts respond to these queries:
1. Young children learn counting very first by associating graphic items because of their fingers. A concrete example: Just how many apples exist inside the basket? Far more abstract instance is just how outdated have you been?
2. From counting numbers, the following plausible stage is addition followed by subtraction. Addition and subtraction tables are often very helpful instructing tools for youngsters because they are visual equipment generating the move from counting less difficult.
3. Which ought to be learned up coming, multiplication or department? Multiplication is shorthand for addition. At this stage, youngsters use a firm grasp of addition. As a result, multiplication is definitely the up coming rational form of arithmetic to discover.
## Evaluate fundamentals of multiplication. Also, assess the fundamentals how to use a multiplication table.
Let us evaluation a multiplication illustration. Employing a Multiplication Table, flourish 4 times a few and get an answer twelve: 4 by 3 = 12. The intersection of row three and column a number of of a Multiplication Table is twelve; 12 is definitely the respond to. For children starting to learn multiplication, this is certainly simple. They are able to use addition to resolve the situation therefore affirming that multiplication is shorthand for addition. Example: 4 x 3 = 4 4 4 = 12. It is really an superb introduction to the Multiplication Table. An added benefit, the Multiplication Table is visible and reflects straight back to discovering addition.
## Where will we get started learning multiplication using the Multiplication Table?
1. Very first, get familiar with the table.
2. Begin with multiplying by one. Commence at row number 1. Go on to line number 1. The intersection of row a single and line the initial one is the solution: one particular.
3. Recurring these methods for multiplying by one particular. Grow row one particular by posts 1 via 12. The responses are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 correspondingly.
4. Replicate these methods for multiplying by two. Increase row two by posts a single by way of several. The answers are 2, 4, 6, 8, and 10 correspondingly.
5. Let us leap ahead of time. Perform repeatedly these methods for multiplying by 5 various. Increase row 5 various by posts a single by means of a dozen. The solutions are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 respectively.
6. Now allow us to raise the degree of problems. Recurring these methods for multiplying by three. Increase row 3 by columns 1 by means of twelve. The solutions are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. Should you be at ease with multiplication so far, use a test. Resolve the subsequent multiplication problems in your head and after that compare your answers towards the Multiplication Table: flourish 6 as well as two, multiply 9 and about three, flourish a single and eleven, increase four and 4, and multiply 7 and 2. The issue answers are 12, 27, 11, 16, and 14 respectively.
Should you acquired 4 out from several problems correct, make your individual multiplication assessments. Compute the solutions in your thoughts, and look them while using Multiplication Table. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Fractions as Percents
## Convert fractions to percents.
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Fractions as Percents
Have you ever met anyone who liked to clean up?
Sam’s least favorite part of her after school job at the supermarket is closing. When closing the store, the break room has to be swept and mopped. Each of the students who work at the store part-time take turns closing, and every Friday night is Sam’s turn. On Friday, Sam got her mop and broom and headed up to the break room. It seemed to be even messier than usual. “Oh no, I will never get done,” Sam sighed, but she picked up the broom and began to sweep.
In just fifteen minutes, Sam had swept four-fifths of the room. She was amazed at how quickly the task was getting done with a little focus and effort.
What fraction of the room does Sam still have to sweep? What percent of the room is this?
In this Concept, you will learn how to identify fractions and convert them to percents. This is exactly what Sam will need to accomplish the task.
### Guidance
In our last Concept, you learned how to understand percents. Remember that a percent means a part of a whole, and the whole is 100. You also learned how to take a percent and write it as a decimal and as a fraction. In this Concept, you will learn to work the other way around. Let’s begin by learning how to write fractions as a percent.
When writing a percent as a fraction, we can drop the % sign and make that the denominator of 100. Then we take the quantity and write it in the numerator above the 100. This is our fraction.
Write $14%$ as a fraction.
First, we drop the % sign. % means out of 100, so 100 becomes our new denominator.
$14%$ becomes $\frac{14}{100}$ .
Yes! We can change the denominator of 100 to a % sign and add that to the quantity in the numerator. Here is what that looks like.
Write $\frac{47}{100}$ as a percent.
First, drop the 100. Then write 47 with a % sign.
$\frac{47}{100}$ becomes $47%$
Our answer is $47%$ .
Not all fractions have a denominator of 100. How do we write a fraction as a percent when the denominator is not 100?
This is the next thing that we need to learn. If a fraction does not have a denominator of 100, to write it as a percent we need to rewrite it as an equal fraction with a denominator of 100.
Write $\frac{2}{5}$ as a percent.
To write $\frac{2}{5}$ as a percent, we must first rewrite it as a fraction with a denominator of 100. Write a proportion to show this comparison.
$\frac{2}{5}=\frac{}{100}$
What number was multiplied by 5 to get 100 as a product? 20! So we multiply the numerator by 20 and we will have an equivalent fraction with a denominator of 100.
$2 \times 20 &= 40\\\frac{2}{5}=\frac{40}{100}&=40\%$
Our answer is $40%$ .
One special fraction to work with is one-third. To convert one-third to a percent is a little tricky because 3 does not divide evenly into 100. Take a look.
$\frac{1}{3}=\frac{}{100}$
When completing this problem, we end up with a repeating decimal; the 3’s just continue on and on and on.
$.33333333$ etc.
To work with this fraction, we call it 33 $\frac{1}{3}$ %.
Write each fraction as a percent.
#### Example A
$\frac{48}{100}$
Solution: $48%$
#### Example B
$\frac{82}{100}$
Solution: $82%$
#### Example C
$\frac{91}{100}$
Solution: $91%$
Remember Sam? Let's use what we have learned to help Sam with her sweeping dilemma.
In just fifteen minutes, Sam had swept four-fifths of the room. She was amazed at how quickly the task was getting done with a little focus and effort.
What fraction of the room does Sam still have to sweep?
What percent of the room has she finished? What percent of the room is still left?
Let’s work through this solution. If Sam has completed four-fifths of the room, then she has one-fifth left to complete.
What percent of the room has she completed?
To figure this out, we have to figure out what four–fifths is as a percent. To do this, we can figure it out by using an equal ratio out of 100.
$\frac{4}{5}=\frac{80}{100}=80\%$
Sam has completed 80% of the room.
What percent of the room does she have left?
You can figure this out two different ways. The first way is to simply subtract 80% from 100%. 100% - 80% (what Sam has completed) $= 20\%$
The other way is to convert one-fifth (the amount left) to a percent. We can do this by creating an equal ratio out of 100.
$\frac{1}{5}=\frac{20}{100}=20\%$
Since it only took Sam 15 minutes to complete 80% of the room, if she continues with her great effort she will be finished in no time at all.
### Vocabulary
Here are the vocabulary words in this Concept.
Percent
is "per-cent" or "per-hundred", it is a quantity written with a % sign, a part of a whole (100)
Fraction
a part of a whole, related to decimals and percents.
Decimal
a part of a whole shown by a decimal point, hundredths means two decimal places.
Equivalent
means equal
### Guided Practice
Here is one for you to try on your own.
Write $\frac{23}{50}$ as a percent.
To write this fraction as a percent, we have to rewrite the fraction with a denominator of 100. We can set up a proportion to do this.
$\frac{23}{50} = \frac{x}{100}$
Now we can solve the proportion.
The answer is $46%$ .
### Video Review
Here are videos for review.
### Practice
Directions: Write each fraction as a percent.
1. $\frac{4}{100}$
2. $\frac{24}{100}$
3. $\frac{20}{100}$
4. $\frac{76}{100}$
5. $\frac{61}{100}$
6. $\frac{1}{4}$
7. $\frac{3}{4}$
8. $\frac{3}{6}$
9. $\frac{2}{5}$
10. $\frac{4}{5}$
11. $\frac{8}{10}$
12. $\frac{6}{10}$
13. $\frac{6}{50}$
14. $\frac{3}{25}$
15. $\frac{20}{50}$
### Vocabulary Language: English
Decimal
Decimal
In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
Equivalent
Equivalent
Equivalent means equal in value or meaning.
fraction
fraction
A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.
Percent
Percent
Percent means out of 100. It is a quantity written with a % sign. |
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# Section 4.
## 5 Exponential and Logarithmic Equations
Exponential Equations
An exponential equation is one in which the variable occurs in the exponent.
EXAMPLE: Solve the equation 2x = 7.
Solution 1: We have
2x = 7
log2 2x = log2 7
x log2 2 = log2 7
x = log2 7 2.807
Solution 2: We have
2x = 7
ln 2x = ln 7
x ln 2 = ln 7
x=
ln 7
2.807
ln 2
## EXAMPLE: Solve the equation 4x+1 = 3.
Solution 1: We have
4x+1 = 3
## log4 4x+1 = log4 3
(x + 1) log4 4 = log4 3
x + 1 = log4 3
x = log4 3 1 0.208
Solution 2: We have
4x+1 = 3
ln 4x+1 = ln 3
(x + 1) ln 4 = ln 3
x+1=
x=
ln 3
ln 4
ln 3
1 0.208
ln 4
1
## EXAMPLE: Solve the equation 3x3 = 5.
Solution 1: We have
3x3 = 5
x 3 = log3 5
x = log3 5 + 3 4.465
Solution 2: We have
3x3 = 5
ln 3x3 = ln 5
(x 3) ln 3 = ln 5
x3=
x=
ln 5
ln 3
ln 5
+ 3 4.465
ln 3
## EXAMPLE: Solve the equation 8e2x = 20.
Solution: We have
8e2x = 20
e2x =
20
5
=
8
2
2x = ln
5
2
ln 52
1 5
= ln 0.458
x=
2
2 2
EXAMPLE: Solve the equation e32x = 4 algebraically and graphically.
Solution: We have
e32x = 4
3 2x = ln 4
2x = ln 4 3
x=
3 ln 4
ln 4 3
=
0.807
2
2
2
## EXAMPLE: Solve the equation e2x ex 6 = 0.
Solution 1: We have
e2x ex 6 = 0
(ex )2 ex 6 = 0
(ex 3)(ex + 2) = 0
ex 3 = 0
or
ex = 3
ex + 2 = 0
ex = 2
## The equation ex = 3 leads to x = ln 3. But the equation ex = 2 has no solution because
ex > 0 for all x. Thus, x = ln 3 1.0986 is the only solution.
Solution 1 : Put ex = w. Then
e2x ex 6 = 0
(ex )2 ex 6 = 0
w2 w 6 = 0
(w 3)(w + 2) = 0
w3=0
or
w+2=0
w=3
w = 2
ex = 3
ex = 2
## The equation ex = 3 leads to x = ln 3. But the equation ex = 2 has no solution because
ex > 0 for all x. Thus, x = ln 3 1.0986 is the only solution.
EXAMPLE: Solve the equation e2x 3ex + 2 = 0.
## EXAMPLE: Solve the equation e2x 3ex + 2 = 0.
Solution 1: We have
e2x 3ex + 2 = 0
(ex )2 3ex + 2 = 0
ex 1 = 0
ex = 1
x=0
(ex 1)(ex 2) = 0
or
ex 2 = 0
ex = 2
x = ln 2
w2 3w + 2 = 0
w1=0
w=1
ex = 1
x=0
(w 1)(w 2) = 0
or
w2=0
w=2
ex = 2
x = ln 2
## Solution: Put 7x = w. Then
w2 3w + 1 = 0
5
3
, therefore x = log7
so 7x =
2
w=
(3)
!
3 5
.
2
(3)2 4 1 1
3 5
=
21
2
## EXAMPLE: Solve the equation 72x 7x 1 = 0.
Solution: Put 7x = w. Then
2
w w 1 = 0 =
1 5
Since
< 0, it follows that
2
w=
5
1
+
7x =
2
(1)
(1)2 4 1 (1)
1 5
=
21
2
x = log7
!
1+ 5
2
## EXAMPLE: Solve the equation 3xex + x2 ex = 0.
Solution: We have
3xex + x2 ex = 0
xex (3 + x) = 0
x(3 + x) = 0
x=0
x=0
or
3+x=0
x = 3
Logarithmic Equations
A logarithmic equation is one in which a logarithm of the variable occurs.
EXAMPLE: Solve the equation ln x = 8.
Solution: We have
ln x = 8
eln x = e8
x = e8
## EXAMPLE: Solve the equation log2 (x + 2) = 5.
Solution: We have
log2 (x + 2) = 5
2log2 (x+2) = 25
x + 2 = 25
x = 25 2 = 32 2 = 30
EXAMPLE: Solve the equation log7 (25 x) = 3.
Solution: We have
log7 (25 x) = 3
7log7 (25x) = 73
25 x = 73
x = 25 73 = 25 343 = 318
EXAMPLE: Solve the equation 4 + 3 log(2x) = 16.
## EXAMPLE: Solve the equation 4 + 3 log(2x) = 16.
Solution: We have
4 + 3 log(2x) = 16
3 log(2x) = 12
log(2x) = 4
2x = 104
x=
10, 000
104
=
= 5, 000
2
2
## EXAMPLE: Solve the equation log(x + 2) + log(x 1) = 1 algebraically and graphically.
Solution: We have
log(x + 2) + log(x 1) = 1
log[(x + 2)(x 1)] = 1
(x + 2)(x 1) = 10
x2 + x 2 = 10
x2 + x 12 = 0
x+4=0
(x + 4)(x 3) = 0
or
x3=0
x = 4
x=3
We check these potential solutions in the original equation and find that x = 4 is not a
solution (because logarithms of negative numbers are undefined), but x = 3 is a solution.
To solve the equation graphically we rewrite it as
log(x + 2) + log(x 1) 1 = 0
and then graph y = log(x + 2) + log(x 1) 1. The solutions are the x-intercepts of the graph.
## EXAMPLE: Solve the following equations
(a) log(x + 8) + log(x 1) = 1
(b) log(x2 1) log(x + 1) = 3
6
## EXAMPLE: Solve the following equations
(a) log(x + 8) + log(x 1) = 1
(b) log(x2 1) log(x + 1) = 3
Solution:
(a) We have
log(x + 8) + log(x 1) = 1
log[(x + 8)(x 1)] = 1
(x + 8)(x 1) = 10
x2 + 7x 8 = 10
x2 + 7x 18 = 0
(x + 9)(x 2) = 0
x+9=0
or
x2=0
x = 9
x=2
We check these potential solutions in the original equation and find that x = 9 is not a
solution (because logarithms of negative numbers are undefined), but x = 2 is a solution.
(b) We have
log(x2 1) log(x + 1) = 3
log
x2 1
=3
x+1
x2 1
= 103
x+1
(x 1)(x + 1)
= 1000
x+1
x 1 = 1000
x = 1001
EXAMPLE: Solve the equation x2 = 2 ln(x + 2) graphically.
Solution: We first move all terms to one side of the equation
x2 2 ln(x + 2) = 0. Then we graph y = x2 2 ln(x + 2). The
solutions are the x-intercepts of the graph.
EXAMPLE: Find the solution of the equation, correct to two decimal places.
(a) 10x+3 = 62x
(b) 5 ln(3 x) = 4
(c) log2 (x + 2) + log2 (x 1) = 2
7
EXAMPLE: Find the solution of the equation, correct to two decimal places.
(a) 10x+3 = 62x
(b) 5 ln(3 x) = 4
(c) log2 (x + 2) + log2 (x 1) = 2
Solution:
(a) We have
10x+3 = 62x
ln 10x+3 = ln 62x
(x + 3) ln 10 = 2x ln 6
x ln 10 + 3 ln 10 = 2x ln 6
x ln 10 2x ln 6 = 3 ln 10
x(ln 10 2 ln 6) = 3 ln 10
x=
3 ln 10
5.39
ln 10 2 ln 6
(b) We have
5 ln(3 x) = 4
ln(3 x) =
4
5
3 x = e4/5
x = 3 e4/5 0.77
(c) We have
log2 (x + 2) + log2 (x 1) = 2
log2 (x + 2)(x 1) = 2
(x + 2)(x 1) = 4
x2 + x 2 = 4
x2 + x 6 = 0
(x 2)(x + 3) = 0
x2=0
or
x=2
x+3=0
x = 3
Since x = 3 is not from the domain of log2 (x + 2) + log2 (x 1), the only answer is x = 2.
8
Applications
EXAMPLE: If I0 and I denote the intensity of light before and after going through a material
and x is the distance (in feet) the light travels in the material, then according to the BeerLambert Law
1
I
ln
=x
k
I0
## where k is a constant depending on the type of material.
(a) Solve the equation for I.
(b) For a certain lake k = 0.025 and the light intensity is I0 = 14 lumens (lm). Find the light
intensity at a depth of 20 ft.
Solution:
(a) We first isolate the logarithmic term.
1
ln
k
I
I0
=x
ln
I
I0
= kx
I
= ekx
I0
I = I0 ekx
(b) We find I using the formula from part (a).
I = I0 ekx = 14e(0.025)(20) 8.49
The light intensity at a depth of 20 ft is about 8.5 lm.
EXAMPLE: A sum of \$5000 is invested at an interest rate of 5% per year. Find the time
required for the money to double if the interest is compounded according to the following
method.
(a) Semiannual
(b) Continuous
EXAMPLE: A sum of \$5000 is invested at an interest rate of 5% per year. Find the time
required for the money to double if the interest is compounded according to the following
method.
(a) Semiannual
(b) Continuous
Solution:
(a) We use the formula for compound interest
r nt
A(t) = P 1 +
n
with P = \$5000, A(t) = \$10, 000, r = 0.05, n = 2 and solve the resulting exponential equation
for t.
2t
0.05
= 10, 000
5000 1 +
2
(1.025)2t = 2
log 1.0252t = log 2
2t log 1.025 = log 2
t=
log 2
14.04
2 log 1.025
## The money will double in 14.04 years.
(b) We use the formula for continuously compounded interest
A(t) = P ert
with P = \$5000, A(t) = \$10, 000, r = 0.05 and solve the resulting exponential equation for t.
5000e0.05t = 10, 000
e0.05t = 2
0.05t = ln 2
t=
ln 2
13.86
0.05
## The money will double in 13.86 years.
EXAMPLE: A sum of \$1000 is invested at an interest rate of 4% per year. Find the time
required for the amount to grow to \$4000 if interest is compounded continuously.
10
EXAMPLE: A sum of \$1000 is invested at an interest rate of 4% per year. Find the time
required for the amount to grow to \$4000 if interest is compounded continuously.
Solution: We use the formula for continuously compounded interest
A(t) = P ert
with P = \$1000, A(t) = \$4000, r = 0.04 and solve the resulting exponential equation for t.
1000e0.04t = 4000
e0.04t = 4
0.04t = ln 4
t=
ln 4
34.66
0.04
11 |
# How many square feet are in a 12x12 room?
The formula for figuring this out is pretty straightforward, all you have to do is multiple the length of the room by the width. So if your room is 10 feet by 12 feet, 10 x 12 = 120 square feet.
A.
### How many square feet is a 10x10 room?
But let's say your room is 10 feet by 10 feet. That's 100 square feet. If you bought 100 square feet, you would have to do a lot of cutting and seam work to cover this room. Here's why: If the carpet is 12 feet wide, 100 square feet would mean the length would be 8 feet, 4 inches (100 divided by 12 = 8.4.
• #### How many square feet are in a 5x5?
5'x5' This size unit yields 25 square feet of space. Although these 5x5 units are small, they are an excellent size to store all of your extra items, like garden tools, seasonal items, office supplies, or your miscellaneous boxes.
• #### How many square feet are in a cubic yard?
One cubic yard of mulch covers 100 square feet at three inches deep. To decide how many cubic yards of mulch to order, use the calculator provided on this site. You can also use a standard equation where you find the total square footage of the beds you want to cover.
• #### How many square feet are in a 8x8 room?
Area of the floor or ceiling: Multiply the length by the width (10 feet x 12 feet = 120 square feet of area). Area of a wall: Multiply the width of the wall by its height. So one of the walls is 80 square feet (10 feet wide x 8 feet high) and the other is 96 square feet (12 feet x 8 feet).
B.
### How many square feet is a 12x13 room?
You times 12 x 13 which is 156 square ft, but you already knew that.
You should be working in metric, as most flooring is sold in metres.
This would convert to 3650 x 3960 total 14.45 square metres approx, order enough for 15 square metres to allow for waste.
• #### How many square feet are in a 12x13 room?
You times 12 x 13 which is 156 square ft, but you already knew that.
You should be working in metric, as most flooring is sold in metres.
This would convert to 3650 x 3960 total 14.45 square metres approx, order enough for 15 square metres to allow for waste.
• #### How many square feet does a gallon of paint cover?
One gallon can of paint will cover up to 400 square feet, which is enough to cover a small room like a bathroom. Two gallon cans of paint cover up to 800 square feet, which is enough to cover an average size room. This is the most common amount needed especially when considering second coat coverage.
• #### How do you figure out the square footage of a house?
Just break out your measuring tape—or a laser measure—to get its length and width. Multiply the width by the length and voila! You have the square footage. Say a room is 20 feet wide by 13 feet long, then 20 x 13 = 260 square feet.
Updated: 4th December 2019 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
8.3: Zero, Negative, and Fractional Exponents
Difficulty Level: At Grade Created by: CK-12
Learning Objectives
• Simplify expressions with zero exponents.
• Simplify expressions with negative exponents.
• Simplify expression with fractional exponents.
• Evaluate exponential expressions.
Introduction
There are many interesting concepts that arise when contemplating the product and quotient rule for exponents. You may have already been wondering about different values for the exponents. For example, so far we have only considered positive, whole numbers for the exponent. So called natural numbers (or counting numbers) are easy to consider, but even with the everyday things around us we think about questions such as “is it possible to have a negative amount of money?” or “what would one and a half pairs of shoes look like?” In this lesson, we consider what happens when the exponent is not a natural number. We will start with “What happens when the exponent is zero?”
Simplify Expressions with Exponents of Zero
Let us look again at the quotient rule for exponents (that \begin{align*}\frac{x^n}{x^m}=x^{n-m}\end{align*}) and consider what happens when \begin{align*}n=m\end{align*}. Let’s take the example of \begin{align*}x^4\end{align*} divided by \begin{align*}x^4\end{align*}.
\begin{align*} \frac{x^4} {x^4} = x^{(4 - 4)} = x^0\end{align*}
Now we arrived at the quotient rule by considering how the factors of \begin{align*}x\end{align*} cancel in such a fraction. Let’s do that again with our example of \begin{align*}x^4\end{align*} divided by \begin{align*}x^4\end{align*}.
\begin{align*}\frac{x^4} {x^4} = \frac{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot \cancel{x} } {\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot \cancel{x} } = 1\end{align*}
So \begin{align*}x^0=1\end{align*}.
This works for any value of the exponent, not just 4.
\begin{align*} \frac{x^n} {x^n} = x^{n - n} = x^0\end{align*}
Since there is the same number of factors in the numerator as in the denominator, they cancel each other out and we obtain \begin{align*}x^0=1\end{align*}. The zero exponent rule says that any number raised to the power zero is one.
Zero Rule for Exponents: \begin{align*}x^0=1,\;x\neq0\end{align*}
Simplify Expressions With Negative Exponents
Again we will look at the quotient rule for exponents (that \begin{align*}\frac{x^n}{x^m}=x^{n-m}\end{align*}) and this time consider what happens when \begin{align*}m>n\end{align*}. Let’s take the example of \begin{align*}x^4\end{align*} divided by \begin{align*}x^6\end{align*}.
\begin{align*} \frac{x^4} {x^6} = x^{(4 - 6)} = x^{-2}\end{align*} for \begin{align*} x \neq 0.\end{align*}
By the quotient rule our exponent for \begin{align*}x\end{align*} is -2. But what does a negative exponent really mean? Let’s do the same calculation long-hand by dividing the factors of \begin{align*}x^4\end{align*} by the factors of \begin{align*}x^6\end{align*}.
\begin{align*}\frac{x^4} {x^6} = \frac{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot \cancel{x} } {\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot x \cdot x } = \frac{1} {x \cdot x} = \frac{1} {x^2}\end{align*}
So we see that \begin{align*}x\end{align*} to the power -2 is the same as one divided by \begin{align*}x\end{align*} to the power +2. Here is the negative power rule for exponents.
Negative Power Rule for Exponents \begin{align*}\frac{1}{x^n}=x^{-n}\; x\neq0\end{align*}
You will also see negative powers applied to products and fractions. For example, here it is applied to a product.
\begin{align*} (x^3 y)^{-2} & = x^{-6} y^{-2} && \text{using the power rule}\\ x^{-6} y^{-2} & = \frac{1} {x^6} \cdot \frac{1} {y^2} = \frac{1} {x^6 y^2} && \text{using the negative power rule separately on each variable}\end{align*}
Here is an example of a negative power applied to a quotient.
\begin{align*} \left (\frac{a} {b}\right )^{-3} & = \frac{a^{-3}} {b^{-3}} && \text{using the power rule for quotients}\\ \frac{a^{-3}} {b^{-3}} & = \frac{a^{-3}} {1} \cdot \frac{1} {b^{-3}} = \frac{1} {a^3} \cdot \frac{b^3} {1} && \text{using the negative power rule on each variable separately}\\ \frac{1} {a^3} \cdot \frac{b^3} {1} & = \frac{b^3} {a^3} && \text{simplifying the division of fractions}\\ \frac{b^3} {a^3} & = \left (\frac{b} {a}\right )^3 && \text{using the power rule for quotients in reverse.}\end{align*}
The last step is not necessary but it helps define another rule that will save us time. A fraction to a negative power is “flipped”.
Negative Power Rule for Fractions \begin{align*}\left(\frac{x}{y}\right)^{-n} = \left(\frac{y}{x}\right)^{n}, \ x \neq 0, y \neq 0\end{align*}
In some instances, it is more useful to write expressions without fractions and that makes use of negative powers.
Example 1
Write the following expressions without fractions.
(a) \begin{align*} \frac{1} {x}\end{align*}
(b) \begin{align*} \frac{2} {x^2}\end{align*}
(c) \begin{align*} \frac{x^2} {y^3}\end{align*}
(d) \begin{align*} \frac{3} {xy}\end{align*}
Solution
We apply the negative rule for exponents \begin{align*} \frac{1}{x^n}=x^{-n}\end{align*} on all the terms in the denominator of the fractions.
(a) \begin{align*} \frac{1} {x} = x^{-1}\end{align*}
(b) \begin{align*} \frac{2} {x^2} = 2x^{-2}\end{align*}
(c) \begin{align*} \frac{x^2} {y^3} = x^2 y^{-3}\end{align*}
(d) \begin{align*} \frac{3} {xy} = 3x^{-1}y^{-1}\end{align*}
Sometimes, it is more useful to write expressions without negative exponents.
Example 2
Write the following expressions without negative exponents.
(a) \begin{align*} 3x^{-3}\end{align*}
(b) \begin{align*} a^2 b^{-3} c^{-1}\end{align*}
(c) \begin{align*} 4x^{-1} y^3\end{align*}
(d) \begin{align*} \frac{2x^{-2}} {y^{-3}}\end{align*}
Solution
We apply the negative rule for exponents \begin{align*}\frac{1}{x^n}=x^{-n}\end{align*} on all the terms that have negative exponents.
(a) \begin{align*} 3x^{-3} = \frac{3} {x^3}\end{align*}
(b) \begin{align*} a^2 b^{-3} c^{-1} = \frac{a^2} {b^3c}\end{align*}
(c) \begin{align*}4x^{-1} y^3 = \frac{4y^3} {x}\end{align*}
(d) \begin{align*}\frac{2x^{-2}} {y^{-3}} = \frac{2y^3} {x^2}\end{align*}
Example 3
Simplify the following expressions and write them without fractions.
(a) \begin{align*} \frac{4a^2b^3} {2a^5b}\end{align*}
(b) \begin{align*} \left (\frac{x} {3y^2}\right )^3 \cdot \frac{x^2y} {4}\end{align*}
Solution
(a) Reduce the numbers and apply quotient rule on each variable separately.
\begin{align*} \frac{4a^2b^3} {6a^5b} = 2 \cdot a^{2 - 5} \cdot b^{3 - 1} = 2a^{-3} b^2\end{align*}
(b) Apply the power rule for quotients first.
\begin{align*} \left (\frac{2x} {y^2}\right )^3 \cdot \frac{x^2y} {4} = \frac{8x^2} {y^6} \cdot \frac{x^2y} {4}\end{align*}
Then simplify the numbers, use product rule on the \begin{align*}x\end{align*}'s and the quotient rule on the \begin{align*}y\end{align*}'s.
\begin{align*} \frac{8x^3} {y^6} \cdot \frac{x^2y} {4} = 2\cdot x^{3 + 2} \cdot y^{1 - 6} = 2x^5 y^{-5}\end{align*}
Example 4
Simplify the following expressions and write the answers without negative powers.
(a) \begin{align*} \left (\frac{ab^{-2}} {b^3}\right )^2\end{align*}
(b) \begin{align*} \frac{x^{-3} y^2} {x^2 y^{-2}}\end{align*}
Solution
(a) Apply the quotient rule inside the parenthesis.
\begin{align*} \left (\frac{ab^{-2}} {b^3}\right )^2 = (ab^{-5})^2\end{align*}
Apply the power rule.
\begin{align*}(ab^{-5})^2 = a^2 b^{-10} = \frac{a^2} {b^{10}}\end{align*}
(b) Apply the quotient rule on each variable separately.
\begin{align*} \frac{x^{-3} y^2} {x^2 y^{-2}} = x^{-3-2} y^{2 - (-2)} = x^{-5} y^4 = \frac{y^4} {x^5}\end{align*}
Simplify Expressions With Fractional Exponents
The exponent rules you learned in the last three sections apply to all powers. So far we have only looked at positive and negative integers. The rules work exactly the same if the powers are fractions or irrational numbers. Fractional exponents are used to express the taking of roots and radicals of something (square roots, cube roots, etc.). Here is an exmaple.
\begin{align*} \sqrt{a} = a^{\frac{1}{2}}\end{align*} and \begin{align*} \sqrt[3]{a} = a^{\frac{1}{3}}\end{align*} and \begin{align*} \sqrt[5]{a^2} = \left (a^2\right )^{\frac{1} {5}} = a^{\frac{2} {5}} = a^{\frac{2}{5}}\end{align*}
Roots as Fractional Exponents \begin{align*}\sqrt[m]{a^n} = a^{\frac{n}{m}}\end{align*}
We will examine roots and radicals in detail in a later chapter. In this section, we will examine how exponent rules apply to fractional exponents.
Example 5
Simplify the following expressions.
(a) \begin{align*} a^{\frac{1}{2}} \cdot a^{\frac{1}{3}}\end{align*}
(b) \begin{align*} \left ( a^{\frac{1}{3}} \right )^2\end{align*}
(c) \begin{align*} \frac{a^{\frac{5}{2}}} {a^{\frac{1}{2}}}\end{align*}
(d) \begin{align*} \left (\frac{x^2} {y^3}\right )^{\frac{1}{3}}\end{align*}
Solution
(a) Apply the product rule.
\begin{align*} a^{\frac{1}{2}} \cdot a^{\frac{1}{3}} = a^{\frac{1} {2} + \frac{1} {3}} = a^{\frac{5}{6}}\end{align*}
(b) Apply the power rule.
\begin{align*} \left (a^{\frac{1}{3}} \right )^2 = a^{\frac{2}{3}}\end{align*}
(c) Apply the quotient rule.
\begin{align*} \frac{a^{\frac{5}{2}}} {a^{\frac{1}{2}}} = a^{\frac{5} {2} - \frac{1} {2}} = a^{\frac{4}{2}} = a^2\end{align*}
(d) Apply the power rule for quotients.
\begin{align*} \left (\frac{x^2} {y^3}\right )^{\frac{1}{3}} = \frac{x^{\frac{2}{3}}} {y}\end{align*}
Evaluate Exponential Expressions
When evaluating expressions we must keep in mind the order of operations. You must remember PEMDAS.
Evaluate inside the Parenthesis.
Evaluate Exponents.
Perform Multiplication and Division operations from left to right.
Perform Addition and Subtraction operations from left to right.
Example 6
Evaluate the following expressions to a single number.
(a) \begin{align*} 5^0\end{align*}
(b) \begin{align*} 7^2\end{align*}
(c) \begin{align*} \left (\frac{2} {3}\right )^3\end{align*}
(d) \begin{align*} 3^{-3}\end{align*}
(e) \begin{align*} 16^\frac{1}{2}\end{align*}
(f) \begin{align*} 8^\frac{-1}{3}\end{align*}
Solution
(a) \begin{align*} 5^0 = 1\end{align*} Remember that a number raised to the power 0 is always 1.
(b) \begin{align*} 7^2 = 7\cdot 7 = 49\end{align*}
(c) \begin{align*} \left (\frac{2} {3}\right )^3 = \frac{2^3} {3^3} = \frac{8} {27}\end{align*}
(d) \begin{align*} 3^{-3} = \frac{1} {3^3} = \frac{1} {27}\end{align*}
(e) \begin{align*} 16^\frac{1}{2} = \sqrt{16} = 4\end{align*} Remember that an exponent of \begin{align*}\frac{1}{2}\end{align*} means taking the square root.
(f) \begin{align*} 8^\frac{-1}{3} = \frac{1} {8^\frac{1}{3}} = \frac{1} {\sqrt[3]{8}} = \frac{1} {2}\end{align*} Remember that an exponent of \begin{align*}\frac{1}{3}\end{align*} means taking the cube root.
Example 7
Evaluate the following expressions to a single number.
(a) \begin{align*} 3 \cdot 5^5 - 10 \cdot 5 + 1\end{align*}
(b) \begin{align*} \frac{2 \cdot 4^2 - 3 \cdot 5^2} {3^2}\end{align*}
(c) \begin{align*} \left (\frac{3^3} {2^2}\right )^{-2} \cdot \frac{3} {4}\end{align*}
Solution
(a) Evaluate the exponent.
\begin{align*}3\cdot 5^2 -10\cdot 6 +1 = 3\cdot 25 -10 \cdot 5 +1\end{align*}
Perform multiplications from left to right.
\begin{align*}3 \cdot 25 -10\cdot 5+1=75 -50+1\end{align*}
Perform additions and subtractions from left to right.
\begin{align*} 75 - 50 + 1 = 26\end{align*}
(b) Treat the expressions in the numerator and denominator of the fraction like they are in parenthesis.
\begin{align*} \frac{(2 \cdot 4^2 - 3 \cdot 5^2)} {(3^2 - 2^2)} = \frac{(2 \cdot 16 - 3 \cdot 25)} {(9 - 4)} = \frac{(32 - 75)} {5} = \frac{-43} {5}\end{align*}
(c) \begin{align*} \left (\frac{3^3} {2^2}\right )^{-2} \cdot \frac{3} {4} = \left (\frac{2^2} {3^3}\right )^2 \cdot \frac{3} {4} = \frac{2^4} {3^6} \cdot \frac{3} {4} = \frac{2^4} {3^6} \cdot \frac{3} {2^2} = \frac{2^2} {3^5} = \frac{4} {243}\end{align*}
Example 8
Evaluate the following expressions for \begin{align*}x=2, y=-1, z=3\end{align*}.
(a) \begin{align*} 2x^2 - 3y^3 + 4z\end{align*}
(b) \begin{align*} (x^2 - y^2)^2\end{align*}
(c) \begin{align*} \left (\frac{3x^2 y^5} {4z}\right )^{-2}\end{align*}
Solution
(a) \begin{align*}2x^2 -3y^3 +4z=2 \cdot 2^2 -3\cdot (-1)^3 +4 \cdot 3=2\cdot 4-3\cdot (-1)+4\cdot 3=8+3+12=23\end{align*}
(b) \begin{align*}(x^2 - y^2)^2=(2^2-(-1)^2)^2=(4-1)^2=3^2=9\end{align*}
(c) \begin{align*} \left (\frac{3x^2 - y^5} {4z}\right )^{-2} = \left (\frac{3 \cdot 2^2 \cdot (-1)^5} {4 \cdot 3}\right )^{-2} = \left (\frac{3 \cdot 4 \cdot (-1)} {12}\right )^{-2} = \left (\frac{-12} {12}\right )^{-2} = \left (\frac{-1} {1}\right )^{-2} = \left (\frac{1} {-1}\right )^2 = (-1)^2 = 1\end{align*}
Review Questions
Simplify the following expressions, be sure that there aren't any negative exponents in the answer.
1. \begin{align*}x^{-1} \cdot y^2\end{align*}
2. \begin{align*}x^{-4}\end{align*}
3. \begin{align*} \frac{x^{-3}} {x^{-7}}\end{align*}
4. \begin{align*} \frac{x^{-3} y^{-5}} {z^{-7}}\end{align*}
5. \begin{align*}\left ( x^{\frac{1}{2}}y^{-\frac{2}{3}} \right ) \left ( x^2 y^{\frac{1}{3}} \right )\end{align*}
6. \begin{align*}\left (\frac{a} {b}\right )^{-2}\end{align*}
7. \begin{align*}(3a^{-2} b^2 c^3)^3\end{align*}
8. \begin{align*} x^{-3} \cdot x^3\end{align*}
Simplify the following expressions so that there aren't any fractions in the answer.
1. \begin{align*} \frac{a^{-3} (a^5)} {a^{-6}}\end{align*}
2. \begin{align*} \frac{5x^6 y^2} {x^8 y}\end{align*}
3. \begin{align*} \frac{(4ab^6)^3} {(ab)^5}\end{align*}
4. \begin{align*} \left (\frac{3x} {y^\frac{1}{3}}\right )^3\end{align*}
5. \begin{align*} \frac{3x^2 y^\frac{3}{2}} {xy^\frac{1}{2}}\end{align*}
6. \begin{align*} \frac{(3x^3)(4x^4)} {(2y)^2}\end{align*}
7. \begin{align*} \frac{a^{-2}b^{-3}} {c^{-1}}\end{align*}
8. \begin{align*} \frac{x^\frac{1}{2} y^\frac{5}{2}} {x^\frac{3}{2} y^\frac{3}{2}}\end{align*}
Evaluate the following expressions to a single number.
1. \begin{align*}3^{-2}\end{align*}
2. \begin{align*}(6.2)^0\end{align*}
3. \begin{align*}8^{-4} \cdot 8^6\end{align*}
4. \begin{align*}\left ( 16^{\frac{1}{2}} \right )^3\end{align*}
5. \begin{align*}x^2 4x^3 y^4 4y^2 \end{align*} if \begin{align*}x=2\end{align*} and \begin{align*}y=-1\end{align*}
6. \begin{align*}a^4(b^2)^3+2ab\end{align*} if \begin{align*}a=-2\end{align*} and \begin{align*}b=1\end{align*}
7. \begin{align*}5x^2-2y^3+3z\end{align*} if \begin{align*}x=3\end{align*}, \begin{align*}y=2\end{align*}, and \begin{align*}z=4\end{align*}
8. \begin{align*} \left (\frac{a^2} {b^3}\right )^{-2}\end{align*} if \begin{align*}a=5\end{align*} and \begin{align*}b=3\end{align*}
1. \begin{align*} \frac{y^2} {x}\end{align*}
2. \begin{align*} \frac{1} {x^4}\end{align*}
3. \begin{align*}x^4\end{align*}
4. \begin{align*} \frac{z^7} {x^3 y^5}\end{align*}
5. \begin{align*} \frac{x^\frac{5}{2}} {y^\frac{1}{3}}\end{align*}
6. \begin{align*} \left (\frac{b} {a}\right )^2\end{align*} or \begin{align*} \frac{b^2} {a^2}\end{align*}
7. \begin{align*} \frac{27b^6 c^9} {a^6}\end{align*}
8. 1
9. \begin{align*}a^8\end{align*}
10. \begin{align*}5x^{-2}y\end{align*}
11. \begin{align*}64a^{-2}b^{\frac{1}{3}}\end{align*}
12. \begin{align*}27x^2 y^{-1}\end{align*}
13. \begin{align*}3xy\end{align*}
14. \begin{align*}6x^7 y^{-2}\end{align*}
15. \begin{align*}a^{-2} b^{-3} c\end{align*}
16. \begin{align*}x^{-1}y\end{align*}
17. 0.111
18. 1
19. 64
20. 64
21. 512
22. 12
23. 41
24. 1.1664
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## How can you tell if a relationship is linear?
You can tell if a table is linear by looking at how X and Y change. If, as X increases by 1, Y increases by a constant rate, then a table is linear. You can find the constant rate by finding the first difference.
## What is not a linear relationship?
The graph of a linear equation forms a straight line, whereas the graph for a nonlinear relationship is curved. A nonlinear relationship reflects that each unit change in the x variable will not always bring about the same change in the y variable.
## What are the types of linear relationships?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form. We review all three in this article. There are three main forms of linear equations.
## What is the opposite of a linear relationship?
Any relationship between two quantities that doesn’t fit the definition of a linear relationship is called a nonlinear relationship.
You might be interested: Often asked: What are perfect squares?
## What is a strong linear relationship?
The strongest linear relationship occurs when the slope is 1. This means that when one variable increases by one, the other variable also increases by the same amount. This line is at a 45 degree angle. ▪ The strength of the relationship between two variables is a crucial piece of information.
## Does a linear relationship go through the origin?
The formal term to describe a straight line graph is linear, whether or not it goes through the origin, and the relationship between the two variables is called a linear relationship. Similarly, the relationship shown by a curved graph is called non-linear.
## What is the difference between linear and nonlinear?
Linear means something related to a line. All the linear equations are used to construct a line. A non-linear equation is such which does not form a straight line. It looks like a curve in a graph and has a variable slope value.
## How do you know if something is linear or nonlinear?
Simplify the equation as closely as possible to the form of y = mx + b. Check to see if your equation has exponents. If it has exponents, it is nonlinear. If your equation has no exponents, it is linear.
## What is the difference between linear and nonlinear correlation?
In other words, when all the points on the scatter diagram tend to lie near a line which looks like a straight line, the correlation is said to be linear. This is shown in the figure on the left below. Correlation is said to be non linear if the ratio of change is not constant.
You might be interested: Often asked: What are trace minerals?
## What type of relationship exists between the two variables?
A correlation exists between two variables when one of them is related to the other in some way. A scatterplot is the best place to start. A scatterplot (or scatter diagram) is a graph of the paired (x, y) sample data with a horizontal x-axis and a vertical y-axis.
## Does a linear relationship have to be proportional?
The relationship between variables can be linear, non-linear, proportional or non-proportional. A proportional relationship is a special kind of linear relationship, but while all proportional relationships are linear relationships, not all linear relationships are proportional.
## What’s another word for linear?
What is another word for linear?
straight direct
undeviating right
straightforward straightaway
rectilinear true
unswerving unbent
## What does a negative linear relationship look like?
Plot 2: Strong negative linear relationship
When both variables increase or decrease concurrently and at a constant rate, a positive linear relationship exists. When one variable increases while the other variable decreases, a negative linear relationship exists.
## Can a linear relationship be negative?
If the slope is negative, then there is a negative linear relationship, i.e., as one increases the other variable decreases. If the slope is 0, then as one increases, the other remains constant. |
# Focal Length: Definition, Equation & Examples
Instructor: Chris Malec
Chris has a PhD in Physics
This lesson explains the meaning of focal length and optical power, presents the equations to calculate these values, and gives an example for each equation.
## Bifocals: Glasses with Two Focal Lengths
Take a look at this picture of bifocal eyeglasses. Each eyepiece can focus at two different distances, allowing someone wearing bifocals to focus nearby when looking down and far away when looking straight ahead. Without bifocals, some people would have to change their glasses when switching between activities like reading a book and driving, which require focusing near and far, respectively. Bifocals are able to focus as two different distances because they have two different focal lengths.
## Focal Length
This schematic shows an example of a convex lens on the top and a concave lens on the bottom. In both cases, the focal point (F) is the point at which parallel light rays cross. The focal length (f) is the distance between the lens and the focal point. Because the focal length measures a distance, it uses units of length, such as centimeters (cm), meters (m), or inches (in). Convex lenses focus the incoming light onto the opposite side of the lens and therefore have a positive focal length. Concave lenses, on the other hand, cause incoming light to diverge rather than focus and thus have a negative focal length.
## Focal Length of a Camera
Let's take the example of a camera like the one in this diagram. The diagram shows an object that is being photographed, a camera lens, and the image that is produced. The labeled distances are the focal length of the lens (f), the distance between the lens and the image (i), and the distance between the lens and the object (o). For an infinitely thin lens (the ideal case), these three distances are related by the equation shown below the diagram. The equation has also been solved for the focal length for your convenience. Note that this equation applies for any infinitely thin lens, not just a lens in a camera.
#### Example
Let's take a look at an example problem for calculating focal length.
If an object that is 50 cm from a lens creates an image 2 cm on the other side of the lens, what is the focal length of this lens?
We are given that o = 50 cm and i = 2 cm. Using the equation for focal length, we can calculate that the focal length (f) is equal to 1/(1/(50 cm) + 1/(2 cm)) or 1.9 cm.
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# Continuity and Discontinuity in Calculus – Class 12 CBSE
• Difficulty Level : Expert
• Last Updated : 12 Nov, 2020
A function is said to be continuous if one can sketch its curve on a graph without lifting the pen even once. A function is said to be continuous at x = a, if, and only if the three following conditions are satisfied.
1. The function is defined at x = a; that is, f(a) equals a real number
2. The limit of the function as x approaches a exists
3. The limit of the function as x approaches a is equal to the function value at x = a
A function f(x) is said to be continuous in the open interval (a, b) if at any point in the given interval the function is continuous. In the case of closed interval [a, b], the function is said to be continuous:
• f(x) is be continuous in the open interval (a, b)
• limx⇢a f(x) = f(a)
• limx⇢b f(x) = f(b)
Example 1: Prove that the function f(x) = 5x – 3 is continuous at x = 0.
Solution:
Given, f(x) = 5x – 3
At x = 0 , f(0) = (5 × 0) – 3 = -3
limx⇢0 f(x) = limx⇢0 (5x – 3) = (5 × 0) – 3 = -3
limx⇢0 f(x) = f(0)
Therefore, f(x) is continuous at x = 0.
Example 2: Examine the function f(x) = |x – 5|, for continuity.
Solution:
Given function, f(x) = |x – 5|
Domain of f(x) is real and infinite for all real x
Here , f(x) = |x – 5| is a modulus function
As , every modulus function is continuous
Therefore , f(x) is continuous in its domain R.
Example 3: Is the function f(x) = x – sinx + 5 is continuous at x = π.
Solution:
Given function is f(x) = x – sinx + 5
L.H.L = limx⇢π (x – sinx + 5) = limx⇢π[(π – h) – sin(π – h) + 5] = π + 5
R.H.L = limx⇢π+ (x – sinx + 5) = limx⇢π+ [(π + h) – sin(π + h) + 5] = π + 5
And, f(π) = π – sinπ + 5 = π + 5
Since , L.H.L = R.H.L = f(π)
Therefore , f(x) is continuous at x = π
Example 4: Examine the continuity of the function f(x) = 2x – 1 at x = 3.
Solution:
Given f(x) = 2x – 1
At x = 3, f(x) = (2 × 3) – 1 = 5
limx⇢3 f(x) = limx⇢3 f(x) = (2×3) – 1 = 5
limx⇢3 f(x) = f(3)
Therefore, f(x) is continuous at x = 3
Example 5: Examine the function is continuous or not?
Solution:
For x > 0, y = x and x < 0, y = -x
So, We Know it is continuous for x > 0 and x < 0. To check if it is continuous at x = 0 , check the limit:
limx⇢0 |x| = limx⇢0 (-x) = 0
limx⇢0+ |x| = limx⇢0+ (x) = 0
So, limx⇢0 |x| = 0 , which is equal to the value of the function at 0. Therefore, It is continuous everywhere.
## Discontinuity
A function is discontinuous at a point x = a if the function is not continuous at a. The function “f” is said to be discontinuous at x = a in any of the following cases:
1. f(a) is not defined
2. limx⇢a+ f(x) and limx⇢a– f(x) exists, but are not equal.
3. limx⇢a+ f(x) and limx⇢a f(x) exists and are equal but not equal to f(a).
### Types of Discontinuities
There are three basic types of discontinuities
1. Removable(point) Discontinuity
2. Infinite Discontinuity
3. Jump Discontinuity
Removable(point) Discontinuity: The graph has a hole at a single x-value. Imagine you’re walking down the road, and someone has removed a manhole cover. This is a category of discontinuity in which the function has a well defined two-sided limit at x = a, but either f(a) is not defined or f(a) is not equal to its limit.
• limx⇢a f(x) ≠ f(a)
• f(a) = limx⇢a f(x)
Infinite Discontinuity: The function goes toward positive or negative infinity. Imagine a road getting closer and closer to a river with no bridge to the other side. The function diverges at x = a to give it a discontinuous nature here. That is to say, f(a) is not defined. Since the value of the function at x = a tends to infinity or doesn’t approach a particular finite value, the limits of the function as x → a are also not defined.
Jump Discontinuity: The graph jumps from one place to another. Imagine a superhero going for a walk, he reaches a dead end and, because he can, flies to another road. In this type of discontinuity, the right-hand limit and the left-hand limit for the function at x = a exists; but the two are not equal to each other.
• limx⇢a+ f(x) ≠ limx⇢af(x)
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# 6.6: Solutions to Compound Inequalities
Difficulty Level: At Grade Created by: CK-12
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Practice Solutions to Compound Inequalities
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What if you had a compound inequality like \begin{align*}0 \le 2x + 6 \le 6\end{align*}? How could you solve it and graph its solution set? After completing this Concept, you'll be able to graph the solution set of compound inequalities like this one on a number line.
### Guidance
When we solve compound inequalities, we separate the inequalities and solve each of them separately. Then, we combine the solutions at the end.
#### Example A
Solve the following compound inequalities and graph the solution set.
a) \begin{align*}-2 < 4x-5 \le 11\end{align*}
b) \begin{align*}3x-5 < x + 9 \le 5x+13\end{align*}
Solution
a) First we re-write the compound inequality as two separate inequalities with and. Then solve each inequality separately.
\begin{align*}& -2 < 4x-5 \qquad \qquad 4x-5 \le 11\\ & \quad 3<4x \qquad \quad and \qquad \quad 4x \le 16\\ & \quad \frac{3}{4} < x \qquad \qquad \qquad \qquad \ x \le 4\end{align*}
Answer: \begin{align*}\frac{3}{4} and \begin{align*}x \le 4\end{align*}. This can be written as \begin{align*}\frac{3}{4}< x \le 4\end{align*}.
b) Re-write the compound inequality as two separate inequalities with and. Then solve each inequality separately.
\begin{align*}& 3x-5
Answer: \begin{align*}x < 7\end{align*} and \begin{align*}x \ge -1\end{align*}. This can be written as: \begin{align*}-1 \le x < 7\end{align*}.
#### Example B
Solve the following compound inequalities and graph the solution set.
a) \begin{align*}9-2x \le 3\end{align*} or \begin{align*}3x+10 \le 6-x\end{align*}
b) \begin{align*}\frac{x-2}{6} \le 2x-4\end{align*} or \begin{align*}\frac{x-2}{6} > x+5\end{align*}
Solution
a) Solve each inequality separately:
\begin{align*}& 9-2x \le 3 \qquad \qquad 3x+10 \le 6-x\\ & \ -2x \le -6 \qquad or \qquad \ \ 4x \le -4\\ & \qquad x \ge 3 \qquad \qquad \qquad \ \ \ x \le -1\end{align*}
Answer: \begin{align*}x \ge 3\end{align*} or \begin{align*}x \le -1\end{align*}
b) Solve each inequality separately:
\begin{align*}& \frac{x-2}{6} \le 2x-4 \qquad \qquad \quad \frac{x-2}{6} > x+5\\ & x-2 \le 6(2x-4) \qquad \qquad x-2 > 6(x+5)\\ & x-2 \le 12x-24 \qquad or \quad \ x-2 > 6x+30\\ & \quad \ 22 \le 11x \qquad \qquad \qquad \ -32 > 5x\\ & \quad \ \ \ 2 \le x \qquad \qquad \qquad \quad -6.4 > x\end{align*}
Answer: \begin{align*}x \ge 2\end{align*} or \begin{align*}x < -6.4\end{align*}
One thing you may notice in the video for this Concept is that in the second problem, the two solutions joined with “or” overlap, and so the solution ends up being the set of all real numbers, or \begin{align*}(-\infty,\infty)\end{align*}. This happens sometimes with compound inequalities that involve “or”; for example, if the solution to an inequality ended up being “\begin{align*}x<5\end{align*} or \begin{align*}x>1\end{align*},” the solution set would be all real numbers. This makes sense if you think about it: all real numbers are either a) less than 5, or b) greater than or equal to 5, and the ones that are greater than or equal to 5 are also greater than 1—so all real numbers are either a) less than 5 or b) greater than 1.
Compound inequalities with “and,” meanwhile, can turn out to have no solutions. For example, the inequality “\begin{align*}x<3\end{align*} and \begin{align*}x>4\end{align*}” has no solutions: no number is both greater than 4 and less than 3. If we write it as \begin{align*}4 it’s even more obvious that it has no solutions; \begin{align*}4 implies that \begin{align*}4 < 3\end{align*}, which is false.
Solve Real-World Problems Using Compound Inequalities
Many application problems require the use of compound inequalities to find the solution.
#### Example C
The speed of a golf ball in the air is given by the formula \begin{align*}v=-32t+80\end{align*}. When is the ball traveling between 20 ft/sec and 30 ft/sec?
Solution
First we set up the inequality \begin{align*}20 \le v \le 30\end{align*}, and then replace \begin{align*}v\end{align*} with the formula \begin{align*}v=-32t+80\end{align*} to get \begin{align*}20 \le -32t+80 \le 30\end{align*}.
Then we separate the compound inequality and solve each separate inequality:
\begin{align*}& \ 20 \le -32t+80 \qquad \quad \ \ \ -32t + 80 \le 30\\ & 32t \le 60 \qquad \qquad \text{and} \qquad \qquad \quad \ 50 \le 32t\\ & \quad t \le 1.875 \qquad \qquad \qquad \qquad \quad 1.56 \le t\end{align*}
Answer: \begin{align*}1.56 \le t \le 1.875\end{align*}
To check the answer, we plug in the minimum and maximum values of \begin{align*}t\end{align*} into the formula for the speed.
For \begin{align*}t = 1.56, \ v=-32t+80 = -32(1.56)+80=30 \ ft/sec\end{align*}
For \begin{align*}t = 1.875, \ v=-32t+80= -32(1.875)+80= 20 \ ft/sec\end{align*}
So the speed is between 20 and 30 ft/sec. The answer checks out.
Watch this video for help with the Examples above.
### Vocabulary
• Compound inequalities combine two or more inequalities with “and” or “or.”
• “And” combinations mean that only solutions for both inequalities will be solutions to the compound inequality.
• “Or” combinations mean solutions to either inequality will also be solutions to the compound inequality.
### Guided Practice
William’s pick-up truck gets between 18 to 22 miles per gallon of gasoline. His gas tank can hold 15 gallons of gasoline. If he drives at an average speed of 40 miles per hour, how much driving time does he get on a full tank of gas?
Solution
Let \begin{align*}t =\end{align*} driving time. We can use dimensional analysis to get from time per tank to miles per gallon:
\begin{align*}\frac{t \ hours}{1 \ tank} \times \frac{1 \ tank}{15 \ gallons} \times \frac{40 \ miles}{1 \ hour} \times \frac{40t}{15} \frac{miles}{gallon}\end{align*}
Since the truck gets between 18 and 22 miles/gallon, we set up the compound inequality \begin{align*}18 \le \frac{40t}{15} \le 22\end{align*}. Then we separate the compound inequality and solve each inequality separately:
\begin{align*}& \ \ 18 \le \frac{40t}{15} \qquad \qquad \quad \ \frac{40t}{15} \le 22\\ & \ 270 \le 40t \qquad \text{and} \qquad 40t \le 330\\ & 6.75 \le t \qquad \qquad \qquad \quad \ \ t \le 8.25\end{align*}
Answer: \begin{align*}6.75 \le t \le 8.25\end{align*}.
Andrew can drive between 6.75 and 8.25 hours on a full tank of gas.
If we plug in \begin{align*}t = 6.75\end{align*} we get \begin{align*}\frac{40t}{15} = \frac{40(6.75)}{15} = 18 \ miles \ per \ gallon\end{align*}.
If we plug in \begin{align*}t = 8.25\end{align*} we get \begin{align*}\frac{40t}{15} = \frac{40(8.25)}{15} = 22 \ miles \ per \ gallon\end{align*}.
The answer checks out.
### Practice
Solve the following compound inequalities and graph the solution on a number line.
1. \begin{align*}-5 \le x-4 \le 13\end{align*}
2. \begin{align*}1 \le 3x +5 \le 4\end{align*}
3. \begin{align*}-12 \le 2-5x \le 7\end{align*}
4. \begin{align*}\frac{3}{4} \le 2x+9 \le \frac{3}{2}\end{align*}
5. \begin{align*}-2 \le \frac{2x-1}{3} < -1\end{align*}
6. \begin{align*}4x-1 \ge 7\end{align*} or \begin{align*}\frac{9x}{2} < 3\end{align*}
7. \begin{align*}3-x < -4\end{align*} or \begin{align*}3-x > 10\end{align*}
8. \begin{align*}\frac{2x+3}{4} < 2\end{align*} or \begin{align*}-\frac{x}{5} + 3 < \frac{2}{5}\end{align*}
9. \begin{align*}2x-7 \le -3\end{align*} or \begin{align*}2x-3 > 11\end{align*}
10. \begin{align*}4x+3< 9\end{align*} or \begin{align*}-5x+4 \le -12\end{align*}
11. How would you express the answer to problem 1 as a set?
12. How would you express the answer to problem 1 as an interval?
13. How would you express the answer to problem 6 as a set?
1. Could you express the answer to problem 6 as a single interval? Why or why not?
2. How would you express the first part of the solution in interval form?
3. How would you express the second part of the solution in interval form?
14. Express the answers to problems 2 through 5 in interval notation.
15. Solve the inequality “\begin{align*}x \ge -3\end{align*} or \begin{align*}x < 1\end{align*}” and express the answer in interval notation.
16. How many solutions does the inequality “\begin{align*}x \ge 2\end{align*} and \begin{align*}x \le 2\end{align*}” have?
17. To get a grade of B in her Algebra class, Stacey must have an average grade greater than or equal to 80 and less than 90. She received the grades of 92, 78, 85 on her first three tests.
1. Between which scores must her grade on the final test fall if she is to receive a grade of B for the class? (Assume all four tests are weighted the same.)
2. What range of scores on the final test would give her an overall grade of C, if a C grade requires an average score greater than or equal to 70 and less than 80?
3. If an A grade requires a score of at least 90, and the maximum score on a single test is 100, is it possible for her to get an A in this class? (Hint: look again at your answer to part a.)
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# Question Video: Finding an Area between a Curve and a Line Mathematics • Higher Education
The curve shown is ๐ฆ = 1/๐ฅ. What is the area of the shaded region? Give an exact answer.
03:10
### Video Transcript
The curve shown is ๐ฆ equals one over ๐ฅ. What is the area of the shaded region? Give an exact answer.
Now we recall that the area of the region bounded by the curve ๐ฆ equals ๐ of ๐ฅ, the lines ๐ฅ equals ๐ and ๐ฅ equals ๐, and the ๐ฅ-axis is given by the integral from ๐ to ๐ of ๐ of ๐ฅ with respect to ๐ฅ. In this case, weโre told that the function ๐ of ๐ฅ is one over ๐ฅ. And from the graph, we can see that the values of the limits for this integral are negative one for the lower limit and negative one-third for the upper limit. So we have the definite integral from negative one to negative one-third of one over ๐ฅ with respect to ๐ฅ.
We then recall that the integral of one over ๐ฅ with respect to ๐ฅ is equal to the natural logarithm of the absolute value of ๐ฅ plus the constant of integration. And that absolute value is really clear here because the two values for our limits are both negative. And we recall that the natural logarithm of a negative value is undefined. So we must make sure we include those absolute value signs.
So weโre taking the natural logarithm of a positive value. We donโt need the constant of integration ๐ here as weโre performing a definite integral. Substituting the limits gives the natural logarithm of the absolute value of negative one-third minus the natural logarithm of the absolute value of negative one. Thatโs the natural logarithm of one-third minus the natural logarithm of one. And at this point, we can recall that the natural logarithm of one is just zero. So our answer has simplified to the natural logarithm of one-third.
Now this may not be immediately obvious to you. But in fact, the natural logarithm of one-third is a negative value. We can see this if we recall one of our laws of logarithms, which is that the logarithm of ๐ over ๐ is equal to the logarithm of ๐ minus the logarithm of ๐. And so the natural logarithm of one-third is the natural logarithm of one minus the natural logarithm of three. And again, we recall that the natural logarithm of one is equal to zero. So our answer appears to be that this area is equal to negative the natural logarithm of three.
This doesnโt really make sense though as areas should be positive. What we see then is that when we use integration to evaluate an area below the ๐ฅ-axis, we will get a negative result. This doesnโt mean though that the area itself is negative. The negative sign is just signifying to us that the area is below the ๐ฅ-axis.
Really then, what we shouldโve done is include absolute value signs around our integral sign at the beginning. And what this means is that although the value of the integral is negative, the natural logarithm of three, the value of the area is the absolute value of this. So thatโs just the natural logarithm of three. The integral is negative to signify that the area is below the ๐ฅ-axis. But the area itself is positive. So our answer to the question, and it is an exact answer, is that this area is equal to the natural logarithm of three. |
Logarithmic and Exponential Functions
Logarithmic and Exponential Functions
Logarithmic functions and exponential functions are closely related in such a way that they are inverses of each other.
To be able to solve the questions for this topic, some of the key points that will be needed are:
– the definitions of log and exp functions to convert the form between the two
– the use of laws of logarithms and laws of exponents
– the relationship as the inverses to each other in graphing the two functions
– transforming the exp functions as a straight-line equation in log functions
The definition of logarithmic function:
$$\hspace{3em} {\large y = a^{x} \ \Rightarrow \ \log_{a} y = x }$$
In words:
If $${\small y }$$ is base of $${\small a }$$ raised to the power of $${\small x }$$ then logarithm to the base of $${\small a }$$ of $${\small y }$$ is $${\small x}$$.
For example, if $${\small 8 = 2^3 \ \Rightarrow \ \log_{2} 8 = 3}$$
To solve a more complex problems in logarithmic and exponential functions, we can employ laws of logarithms and laws of exponents as our arsenals.
The laws of logarithms are:
$${\large 1.\hspace{2em} \log_{a} 1 = 0 }$$
$${\large 2.\hspace{2em} \log_{a} a = 1 }$$
$${\large 3.\hspace{2em} \log_{a} {bc} = \log_{a} {b} \ + \ \log_{a} {c} }$$
$${\large 4.\hspace{2em} \log_{a} ({\frac{b}{c}}) = \log_{a} {b} \ – \ \log_{a} {c} }$$
$${\large 5.\hspace{2em} \log_{a^n} {b}^{m} = (\frac{m}{n}) \times \log_{a} b }$$
$${\large 6.\hspace{2em} \log_{a} b = \frac{1}{\log_{b} a} }$$
$${\large 7.\hspace{2em} \log_{a} b = \frac{\log_{p} b}{\log_{p} a} }$$
$${\large 8.\hspace{2em} a^{\log_{a} b} = b}$$
And, the laws of exponents are:
$${\large 1.\hspace{2em} a^{0} = 1; \ a \neq 0 }$$
$${\large 2.\hspace{2em} a^{m} \times a^{n} = a^{m \ + \ n} }$$
$${\large 3.\hspace{2em} \frac{ a^{m} }{ a^{n} } = a^{m \ – \ n} }$$
$${\large 4.\hspace{2em} a^n \ \times \ b^n = {(ab)}^{n} }$$
$${\large 5.\hspace{2em} \frac{ a^{n} }{ b^{n} } = (\frac{a}{b})^{n}}$$
$${\large 6.\hspace{2em} {(a^{m})}^{n} = a^{mn} }$$
$${\large 7.\hspace{2em} \sqrt[n]{a^{m}} = a^{\frac{m}{n}} }$$
$${\large 8.\hspace{2em} \frac{1}{a^n} = a^{-n} }$$
The graph of logarithmic functions are shown as below:
The logarithmic function $${\small y = \log_{a} x}$$ has a vertical asymptote at the y-axis and x-intercept at (1, 0).
For base: $${\small 0 < a < 1 }$$, the logarithmic function is a monotonic decreasing function.
For base: $${\small a > 1 }$$, the logarithmic function is a monotonic increasing function.
Meanwhile, the graph of exponential functions are shown below:
The exponential function $${\small y = a^x}$$ has a horizontal asymptote at the x-axis and y-intercept at (0, 1).
For base: $${\small 0 < a < 1 }$$, the exponential function is a monotonic decreasing function.
For base: $${\small a > 1 }$$, the exponential function is a monotonic increasing function.
If we put both graphs of logarithmic and exponential functions together, it is easy to see that they are reflection of each other to the line y = x.
The exponential functions are widely use in many areas for example in describing the population growth or decay, time value of money with its depreciation or inflation nature and many others.
It is usually helpful to convert the exponential model to a linear model.
Let our exponential growth function be $$N = ka^t$$ where $$k$$ and $$a$$ are constants,
Taking logarithms to both side of the equation gives:
$$\hspace{2em} \log N = \log (ka^t)$$
$$\hspace{2em} \log N = \log a^t + \log k$$
$$\hspace{2em} \log N = \log a . t + \log k$$
Compare that to the straight-line equation: $$y = mx + c$$ with:
– the dependent variable $$y = \log N$$,
– the gradient $$m = \log a$$,
– the independent variable $$x = t$$ and
– the y-intercept $$= \log k$$
The y-intercept is the initial population or capital at $$t = 0$$.
The gradient represents how fast the population grown.
Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
$$\\[1pt]$$
EXAMPLE:
$${\small 1.\enspace}$$ 9709/32/F/M/17 – Paper 32 March 2017 Pure Maths 3 No 1
$$\\[1pt]$$
Solve the equation $${\ln (1 + 2^{x}) = 2}$$, giving your answer correct to 3 decimal places.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 2.\enspace}$$ 9709/32/F/M/19 – Paper 32 March 2019 Pure Maths 3 No 1
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{i}).\hspace{0.8em}}$$ Show that the equation $${ \log_{10} (x-4) = 2 \ – \ \log_{10} x}$$ can be written as a quadratic equation in $$x$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.8em}}$$ Hence solve the equation $${ \log_{10} (x-4) = 2 \ – \ \log_{10} x }$$ giving your answer correct to 3 significant figures.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 3.\enspace}$$ 9709/33/M/J/19 – Paper 33 June 2019 Pure Maths 3 No 1
$$\\[1pt]$$
Use logarithms to solve the equation $$5^{3 \ – \ 2 x} = 4(7^{x})$$, giving your answer correct to 3 decimal places.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 4.\enspace}$$ 9709/31/M/J/19 – Paper 31 June 2019 Pure Maths 3 No 2
$$\\[1pt]$$
Showing all necessary working, solve the equation
$$\\[1pt]$$
$$\hspace{2em} \ln (2x \ – \ 3) = 2 \ \ln x \ – \ \ln (x \ – \ 1)$$
$$\\[1pt]$$
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 5.\enspace}$$ 9709/32/F/M/20 – Paper 32 March 2020 Pure Maths 3 No 2
$$\\[1pt]$$
Solve the equation $$\ln 3 + \ln (2x+5) = 2 \ \ln(x+2)$$.
$$\\[1pt]$$
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 6.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 1
$$\\[1pt]$$
Find the set of values of $$x$$ for which $$2(3^{1-2x}) \lt 5^{x}$$.
$$\\[1pt]$$
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 7.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 2
$$\\[1pt]$$
$$\\[1pt]$$
The variables $$x$$ and $$y$$ satisfy the equation $$y^2 = A {\mathrm{e}}^{kx}$$, where $$A$$ and $$k$$ are constants. The graph of $$\ln y$$ against $$x$$ is a straight line passing through the points (1.5, 1.2) and (5.24, 2.7) as shown in the diagram.
$$\\[1pt]$$
Find the values of $$A$$ and $$k$$ correct to 2 decimal places.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 8.\enspace}$$ 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 3
$$\\[1pt]$$
$${\small (\textrm{a}).\hspace{0.8em}}$$ Show that the equation
$$\\[1pt]$$
$$\hspace{2em} \ln (1 + \mathrm{e}^{−x}) + 2x = 0$$
$$\\[1pt]$$
can be expressed as a quadratic equation in $$\mathrm{e}^{x}$$.
$$\\[1pt]$$
$${\small (\textrm{b}).\hspace{0.8em}}$$ Hence solve the equation $$\ln (1 + \mathrm{e}^{−x}) + 2x = 0$$, giving your answer correct to 3 decimal places.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 9.\enspace}$$ 9709/31/O/N/20 – Paper 31 Nov 2020 Pure Maths 3 No 4
$$\\[1pt]$$
Solve the equation
$$\\[1pt]$$
$$\hspace{2em} \log_{10} (2x + 1) = 2 \ \log_{10} (x+1) \ – \ 1$$.
$$\\[1pt]$$
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 10.\enspace}$$ 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 1
$$\\[1pt]$$
Solve the equation $$\ln (x^3 − 3) = 3 \ln x − \ln 3$$. Give your answer correct to 3 significant figures.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 11.\enspace}$$ 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 2
$$\\[1pt]$$
Find the real root of the equation $$\frac{2\mathrm{e}^x \ + \ \mathrm{e}^{-x}}{2 \ + \ \mathrm{e}^x} = 3$$, giving your answer correct to 3 decimal places. Your working should show clearly that the equation has only one real root.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 12.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 3
$$\\[1pt]$$
The variables $$x$$ and $$y$$ satisfy the equation $$x = A(3^{−y})$$, where $$A$$ is a constant.
$$\\[1pt]$$
$${\small(\textrm{a}).\hspace{0.8em}}$$ Explain why the graph of $$y$$ against $$\ln x$$ is a straight line and state the exact value of the gradient
of the line.
$$\\[1pt]$$
It is given that the line intersects the $$y$$-axis at the point where $$y$$ = 1.3.
$$\\[1pt]$$
$${\small(\textrm{b}).\hspace{0.8em}}$$ Calculate the value of $$A$$, giving your answer correct to 2 decimal places.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 13.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 2
$$\\[1pt]$$
Solve the equation $$4^x = 3 + 4^{−x}$$. Give your answer correct to 3 decimal places.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 14.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 3
$$\\[1pt]$$
$$\\[1pt]$$
The variables $$x$$ and $$y$$ satisfy the equation $${\small {x}^{n}{y}^{2} \ = \ C }$$, where $$n$$ and $$C$$ are constants.
$$\\[1pt]$$
The graph of $${\small \ln \ y }$$ against $${\small \ln \ x }$$ is a straight line passing through the points (0.31, 1.21) and (1.06, 0.91), as shown in the diagram.
$$\\[1pt]$$
Find the value of $$n$$ and find the value of $$C$$ correct to 2 decimal places.
$$\\[1pt]$$
$$\\[1pt]$$
PRACTICE MORE WITH THESE QUESTIONS BELOW!
$${\small 1.\enspace}$$ Solve the inequality $$|3^x – 5 | \lt 1$$, giving 3 significant figures in your answer.
$${\small 2. \enspace}$$ Given that $$x = 4(3^{-y})$$, express $$y$$ in terms of $$x$$.
$${\small 3. \enspace}$$ Using the substitution $$u = 3^{x}$$, or otherwise, solve, correct to 3 significant figures, the equation
$$\hspace{2em} 3^{x} = 2 + 3^{-x}$$.
$${\small 4. \enspace}$$ The variables $$x$$ and $$y$$ satisfy the relation $$3^y = 4^{x+2}$$.
$${\small\hspace{1.2em} (\textrm{a}).\hspace{0.8em}}$$ By taking logarithms, show that the graph of $$y$$ against $$x$$ is a straight line. Find the exact value of the gradient of this line.
$${\small\hspace{1.2em} (\textrm{b}).\hspace{0.8em}}$$ Calculate the $$x$$ co-ordinate of the point of intersection of this line with the line $$y = 2x$$, giving your answer correct to 2 decimal places.
$${\small 5. \enspace}$$ It is given that $$\ln(y + 5) – \ln y = 2 \ln x$$. Express $$y$$ in terms of $$x$$, in a form not involving logarithms.
$${\small 6. \enspace}$$ Given that $${(1.25)}^{x} = {(2.5)}^{y}$$, use logarithms to find the value of $$\frac{x}{y}$$ correct to 3 significant figures.
$${\small 7. \enspace}$$ Solve, correct to 3 significant figures, the equation
$$\hspace{2em} \mathrm{e}^x + \mathrm{e}^{2x} = \mathrm{e}^{3x}$$.
$${\small 8. \enspace}$$ The variables $$x$$ and $$y$$ satisfy the equation $$y = A(b^{–x})$$, where $$A$$ and $$b$$ are constants. The graph of $$\ln y$$ against $$x$$ is a straight line passing through the points (0, 1.3) and (1.6, 0.9), as shown in the diagram. Find the values of $$A$$ and $$b$$, correct to 2 decimal places.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 9. \enspace}$$ Solve the equation $$\ln (2 + \mathrm{e}^{-x}) = 2$$, giving your answer correct to 2 decimal places.
$${\small 10.\enspace}$$ Two variable quantities $$x$$ and $$y$$ are related by the equation $$y = A{x}^{n}$$, where $$A$$ and $$n$$ are constants. The diagram shows the result of plotting $$\ln y$$ against $$\ln x$$ for four pairs of values of $$x$$ and $$y$$. Use the diagram to estimate the values of $$A$$ and $$n$$.
$$\\[1pt]$$
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
Polynomials: Factor and Remainder Theorem
Polynomials: Factor and Remainder Theorem
In Algebra, we regularly come across the polynomials. The skill in solving algebraic operations of polynomials is very important in mathematics.
If you need to revise on the basic of algebra, you can look at it here.
The definition
$${\small P(x) \ = \ a_{n} x^{n} + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + … + a_{0} }$$
$$\\[5pt] {\small n \ }$$ is a non-negative integer. It is the degree or the order of the polynomial.
$$\\[5pt] {\small x \ }$$ is the variable of the polynomial.
$$\\[5pt] {\small a_{n}, a_{n-1}, a_{n-2}, … , a_{0} \ }$$ are the coefficients of each term in $${ \small x \ }$$ from the highest power to the lowest power.
Here are some of the examples:
1. $$\\[5pt]$$ Linear polynomial or polynomial of degree 1
$$\hspace{1em} {\small P(x) \ = \ ax \ + \ b }$$
e.g.: $${\small P(x) \ = \ 2x \ + \ 3 }$$
2. $$\\[5pt]$$ Quadratic polynomial or polynomial of degree 2
$$\hspace{1em} {\small P(x) \ = \ ax^{2} + bx + c }$$
e.g.: $${\small P(x) \ = \ 3x^{2} + 2x + 3 }$$
3. $$\\[5pt]$$ Cubic polynomial or polynomial of degree 3
$$\hspace{1em} {\small P(x) \ = \ ax^{3} + bx^{2} + cx + d }$$
e.g.: $${\small P(x) \ = \ 3x^{3} + 2x^2 + 3x + 5 }$$
4. $$\\[5pt]$$ Quartic polynomial or polynomial of degree 4
$$\hspace{1em} {\small P(x) \ = \ ax^{4} + bx^{3} + cx^{2} + dx + e }$$
e.g.: $${\small P(x) \ = \ 5x^{4} + 2x^{3} + 3x^2 + x + 6 }$$
The addition and subtraction of polynomials
Collect all the like terms and simply the expression by adding or subtracting the like terms.
e.g.:
$$\hspace{1em} {\small P(x) \ = \ 2x^{2} + 3x + 4 }$$
$$\hspace{1em} {\small Q(x) \ = \ x^{2} \ – \ 2x + 3 }$$
$$\hspace{1em} {\small P(x) \ + \ Q(x) }$$
$$\hspace{1.5em} {\small = \ (2x^{2} + 3x + 4) + (x^{2} \ – \ 2x + 3) }$$
$$\hspace{1.5em} {\small = \ (2x^{2} + x^{2}) + (3x \ – \ 2x) + (4 + 3) }$$
$$\hspace{1.5em} {\small = \ 3x^{2} + x + 7 }$$
$$\hspace{1em} {\small P(x) \ – \ Q(x) }$$
$$\hspace{1.5em} {\small = \ (2x^{2} + 3x + 4) \ – \ (x^{2} \ – \ 2x + 3) }$$
$$\hspace{1.5em} {\small = \ (2x^{2} \ – \ x^{2}) + (3x \ + \ 2x) + (4 \ – \ 3) }$$
$$\hspace{1.5em} {\small = \ x^{2} + 5x + 1 }$$
The multiplication of polynomials
Use the distributive law by multiplying each term.
e.g.:
$$\hspace{1em} {\small P(x) \ = \ 2x^{2} + 3x + 4 }$$
$$\hspace{1em} {\small Q(x) \ = \ x^{2} \ – \ 2x + 3 }$$
$$\hspace{1em} {\small P(x) \ \times \ Q(x) }$$
$$\hspace{1.5em} {\small = \ 2x^{4} + (3x^{3} \ – \ 4x^{3}) + (6x^{2} \ – \ 6x^{2} + 4x^{2}) }$$
$$\\[8pt] \hspace{2.5em} {\small + \ (9x \ – \ 8x) + 12 }$$
$$\hspace{1.5em} {\small = \ 2x^{4} \ – \ x^{3} + 4x^{2} + x + 12 }$$
The division of polynomials
We’ll use a simple analogy of long division of a number to illustrate the polynomial division.
Just as an example, what is the result of 7 divided by 3?
So, $$\hspace{1em} {\large\frac{7}{3}} \ = \ 2 \ + \ {\large\frac{1}{3}}$$
Or, we can conveniently write the above result as,
$$\hspace{1.5em} 7 \ = \ 2 \times 3 \ + \ 1 \$$
Then, in terms of polynomial division,
$$\hspace{1em} {\large\frac{P(x)}{D(x)}} \ = \ Q(x) \ + {\large\frac{R(x)}{D(x)}}$$
or in its general form:
$$\hspace{1em} P(x) \ = \ D(x) \times Q(x) + R(x)$$
with,
P(x) is the polynomial to be divided with or the dividend,
D(x) is the divisor,
Q(x) is the quotient and
R(x) is the remainder.
The process of finding the quotient Q(x) and the remainder R(x) can be done similarly with polynomial long division.
Here is an example,
$$\hspace{1em} {\Large\frac{x^{2} \ – \ 4x \ – \ 3}{x \ + \ 1}} \ = \ ?$$
Step 1. Divide the highest power of the polynomial P(x) with the highest power of the divisor D(x).
Step 2. Multiply the result we get in the quotient Q(x) from Step 1 with the divisor D(x).
Step 3. Subtract the polynomial P(x) from the result in Step 2.
Step 4. Repeat the process from Step 1 to the next available highest power of polynomial P(x) until we are left with only the remainder.
Now, one important thing to remember is that the degree of the remainder R(x) is at most one fewer than the order of the divisor D(X).
For example, if we have a linear divisor $${\small \ ax \ + \ b \ }$$ then the remainder is a constant.
If we have a quadratic divisor $${\small \ ax^{2} \ + \ bx \ + \ c \ }$$ then the remainder is a linear polynomial and so on.
By knowing that, we’ll know when to stop the process of multiplying back the quotient.
Factor Theorem
If $${\small \ (ax \ + \ b) \ }$$ is a linear factor of a polynomial P(x) then $${\small \ P({\large\frac{-b}{a}}) \ = \ 0 }$$.
e.g.: Show that $${\small \ (x \ – \ 2) \ }$$ is a linear factor of a polynomial $${\small P(x) \ = \ x^{4} \ – \ 5x^{2} + 2x }$$.
Solution: If $${\small \ (x \ – \ 2) \ }$$ is a linear factor, then according to the factor theorem $${\small \ P(2) \ }$$ must be equal to zero.
We’ll try by substituting the value of $${\small \ x \ = \ 2 }$$,
$$\hspace{1.5em} {\small P(2) \ = \ ({2})^{4} \ – \ 5({2})^{2} + 2(2) }$$
$$\hspace{3.2em} {\small \ = \ 16 \ – \ 20 + 4 }$$
$$\hspace{3.2em} {\small \ = \ 0 }$$
Therefore, $${\small \ (x \ – \ 2) \ }$$ is a linear factor of $${\small \ P(x) \ = \ x^{4} \ – \ 5x^{2} + 2x }$$.
Another great use of factor theorem is when we want to factorise a higher order polynomial or the roots of a higher order polynomial equation.
We can use the constant term of the higher order polynomial and quickly substituting the value of each factors of the constant term.
If any of the results equals to zero then by factor theorem we can find its linear factor(s).
The other factors can then be found by applying polynomial long division.
You can study some of the examples below to fully understand the concept.
Remainder Theorem
If a polynomial P(x) is divided by a linear divisor $${\small \ (ax \ + \ b) \ }$$ then $${\small \ P({\large\frac{-b}{a}}) \ }$$ is the remainder.
e.g.: Find the remainder when a polynomial $${\small \ P(x) \ = \ x^{4} \ – \ 2x^{3} + 3x^{2} \ – \ 8 \ }$$ is divided by $${\small \ (x \ – \ 1) }$$.
Solution: According to the remainder theorem, the remainder is $${\small \ P(1) \ }$$,
$$\hspace{1.5em} {\small P(1) \ = \ (1)^{4} \ – \ 2(1)^{3} + 3(1)^{2} \ – \ 8 }$$
$$\hspace{3.2em} {\small \ = \ 1 \ – \ 2 + 3 \ – \ 8 }$$
$$\hspace{3.2em} {\small \ = \ -6 }$$
Therefore, the remainder of $${\small \ P(x) \ = \ x^{4} \ – \ 2x^{3} + 3x^{2} \ – \ 8 \ }$$ is $${\small \ -6 \ }$$ when it is divided by $${\small \ (x \ – \ 1) }$$.
Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
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EXAMPLE:
$${\small 1.\enspace}$$ Find the remainder of $${\small \ P(x) \ = \ 2x^{3} + 4x^{2} \ – \ 6x + 7 \ }$$ when it is divided by $${\small \ (2x \ – \ 1) }$$.
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$${\small 2.\enspace}$$ Find the value of $${\small \ k \ }$$ if $${\small \ (x \ – \ 1) \ }$$ is a factor of $${\small \ {k}^{2}{x}^{4} \ – \ 3k{x}^{2} + 2 }$$.
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$${\small 3.\enspace}$$ This is an example of the aplication of the factor theorem. We would try to find the roots of higher order polynomials.
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Factorise $${\small \ 2{?}^{3} + 9{?}^{2} + 7{?} \ − \ 6 \ }$$ completely and hence find all the roots of the polynomial.
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$${\small 4.\enspace}$$ Given that a polynomial $${\small \ p(x) \ = \ {?}^{3} + a{?}^{2} + b{?} \ − \ 3 \ }$$ when divided by $${\small \ (x + 1) \ }$$ and $${\small \ (x \ – \ 1) }$$, the remainders are -9 and 1 respectively. Find the values of a and b.
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$${\small 5.\enspace}$$ Continuing from the same question in example 4, find the reminder when the polynomial $${\small \ p(x) \ }$$ is divided by $${\small \ ({x}^{2} \ – \ 1) }$$.
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$${\small 6.\enspace}$$ 9709/21/O/N/19 – Paper 21 March 2019 Pure Maths 2 No 4
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The polynomial p(x) is defined by:
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$$\hspace{3em} {\small p(x) \ = \ a{x}^{3} + a{x}^{2} \ − \ 15x \ − \ 18}$$,
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where a is a constant. It is given that (x − 2) is a factor of p(x).
$${\small (\textrm{i}).\hspace{0.7em}}$$ Find the value of a.
$${\small (\textrm{ii}).\hspace{0.6em}}$$ Using this value of a, factorise p(x) completely.
$${\small (\textrm{iii}).\hspace{0.5em}}$$ Hence solve the equation $${\small \ p(e^{\sqrt{y}}) \ = \ 0}$$, giving the answer correct to 2 significant figures.
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$${\small 7.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 1
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Find the quotient and remainder when $$6x^4 + x^3 − x^2 + 5x − 6$$ is divided by $$2x^2 − x + 1$$.
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$${\small 8.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 5(a)
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Find the quotient and remainder when $$2x^3 − x^2 + 6x + 3$$ is divided by $$x^2 + 3$$.
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$${\small 9.\enspace}$$ 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 2
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The polynomial $$ax^3 + 5x^2 − 4x + b$$, where $$a$$ and $$b$$ are constants, is denoted by $$p(x)$$. It is given that
$$(x + 2)$$ is a factor of $$p(x)$$ and that when $$p(x)$$ is divided by $$(x + 1)$$ the remainder is 2.
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Find the values of $$a$$ and $$b$$.
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$${\small 10.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 8a
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Find the quotient and remainder when $${\small 8x^3 + 4x^2 + 2x + 7}$$ is divided by $${\small 4x^2 + 1}$$.
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PRACTICE MORE WITH THESE QUESTIONS BELOW!
$${\small 1.\enspace}$$ The polynomial p(x) is defined by
$$\hspace{3em} {\small \ p(x) \ = \ 6{x}^{3} + a{x}^{2} + 9{x} + b }$$,
where a and b are constants. It is given that $${\small \ (x \ – \ 2) \ }$$ and $${\small \ (2x + 1) }$$ are factors of p(x). Find the values of a and b.
$${\small 2. \enspace}$$ The polynomial p(x) is defined by
$$\hspace{3em} {\small \ p(x) \ = \ 6{x}^{3} + a{x}^{2} \ – \ 4{x} \ – \ 3 }$$,
where a is a constant. It is given that $${\small \ (x + 3) \ }$$ is a factor of p(x).
$${\small\hspace{1.2em} (\textrm{a}).\hspace{0.8em}}$$ Find the value of a.
$${\small\hspace{1.2em} (\textrm{b}).\hspace{0.8em}}$$ Using this value of a, factorise p(x) completely.
$${\small 3. \enspace}$$ Find the quotient and remainder when $${\small \ 6{x}^{4} + {x}^{3} \ − \ {x}^{2} + 5x \ − \ 6 \ }$$ is divided by $${\small \ (2{x}^{2} \ − \ x + 1) }$$.
$${\small 4. \enspace}$$ The polynomial f(x) is defined by
$$\hspace{3em} {\small \ f(x) \ = \ {x}^{4} \ – \ 3{x}^{3} + 5{x}^{2} \ – \ 6{x} + 11 }$$,
Find the quotient and remainder when f(x) is divided by $${\small \ ({x}^2 + 2) }$$.
$${\small 5. \enspace}$$ The polynomial $${\small \ 6{x}^{3} + a{x}^{2} + b{x} \ − \ 2 }$$, where a and b are constants, is denoted by p(x). It is given that $${\small \ (2x + 1) }$$ is a factor of p(x) and that when p(x) is divided by $${\small \ (x + 2) \ }$$ the remainder is −24. Find the values of a and b.
$${\small 6. \enspace}$$ The polynomial $${\small \ {x}^{4} + 3{x}^{3} + ax + b }$$, where a and b are constants, is denoted by p(x). When p(x) is divided by $${\small \ {x}^{2} + x \ − \ 1 \ }$$ the remainder is $${\small \ 2x + 3}$$. Find the values of a and b.
$${\small 7. \enspace}$$ The cubic polynomial f(x) is defined by $${\small \ f(x) \ = \ {x}^{3} + a{x}^{2} + 14x + a + 1 }$$, where a is a constant. It is given that $${\small \ (x + 2) \ }$$ is a factor of f(x). Use the factor theorem to find the value of a and hence factorise f(x) completely.
$${\small 8. \enspace}$$ $${\small (\textrm{i}).\hspace{0.7em}}$$ Find the quotient when $${\small \ {x}^{4} \ − \ 2{x}^{3} + 8{x}^{2} \ − \ 12x + 13 \ }$$ is divided by $${\small \ {x}^{2} + 6 \ }$$ and show that the remainder is 1.
$${\small\hspace{1.3em}(\textrm{ii}).\hspace{0.7em}}$$ Show that the equation $${\small \ {x}^{4} \ − \ 2{x}^{3} + 8{x}^{2} \ − \ 12x + 12 \ = \ 0 \ }$$ has no real roots.
$${\small 9. \enspace}$$ The polynomial $${\small \ {x}^{4} + {x}^{3} + ax + b}$$, where a and b are constants, is divisible by $${\small \ {x}^{2} \ − \ x + 1}$$. Find the values of a and b.
$${\small 10.\enspace}$$ The polynomial p(x) is defined by $${\small \ p(x) \ = \ 4{x}^{3} + 4{x}^{2} \ − \ 29x \ − \ 15 }$$.
$${\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}$$ Use the factor theorem to show that $${\small \ x + 3 \ }$$ is a factor of p(x).
$${\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}$$ Factorise p(x) completely.
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
Permutations and Combinations
Permutations and Combinations
When given a number of objects and we would to find the possible different arrangements or selections , we may use permutations or combinations.
Permutations deal with the arrangement of objects. The order of the objects is important. The number of different arrangements of r objects from n distinct object is,
$$\\[20pt]\hspace{2em} ^{n}P_{r } \ = \ \displaystyle \frac{n!}{(n-r)!}$$
Combinations deal with the selection of objects. The order of the objects is not important. The number of different selections of r objects from n distinct object is,
$$\\[20pt]\hspace{2em} ^{n}C_{r } \ = \ \displaystyle \binom{n}{r} \ = \ \frac{n!}{r!(n-r)!}$$
Notice that it is also used as the binomial coefficients in Binomial Expansion and Binomial Series.
Let’s see a simple example below to illustrate the difference between permutations and combinations.
We have 3 letters, A, B and C and we would like to find the permutations and combinations of 2 letters from the 3 letters.
For permutations, we have the following arrangements: AB, BA, AC, CA, BC and CB.
For combinations, we have the following selections: AB, AC and BC.
As you can see, AB and BA, AC and CA or BC and CB are considered as the same selection since the order of the objects is not important.
In other words, combinations is actually the permutations of r identical objects from n distinct object.
Try the exercises below and if you need any help, just look at the solution I have written. Cheers ! =) .
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EXERCISE 5A:
$${\small 1.\enspace}$$ Seven different cars are to be loaded on to a transporter truck. In how many different ways can the cars be arranged?
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$${\small 2.\enspace}$$ How many numbers are there between 1245 and 5421 inclusive which contain each of the digits 1, 2, 4 and 5 once and once only?
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$${\small 3.\enspace}$$ An artist is going to arrange five paintings in a row on a wall. In how many ways can this be done?
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$${\small 4.\enspace}$$ Ten athletes are running in a 100-metre race. In how many different ways can the first three places be filled?
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$${\small 5.\enspace}$$ By writing out all the possible arrangements of $${\small {D}_{1}{E}_{1}{E}_{2}{D}_{2} }$$, show that there are $$\frac{4!}{2!(2)!} = {\small 6 \ }$$ different arrangements of the letters of the word DEED.
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$${\small 6.\enspace}$$ A typist has five letters and five addressed envelopes. In how many different ways can the letters be placed in each envelope without getting every letter in the right envelope? If the letters are placed in the envelopes at random what is the probability that each letter is in its correct envelope?
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$${\small 7.\enspace}$$ How many different arrangements can be made of the letters in the word STATISTICS?
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$${\small 8.\enspace \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ Calculate the number of arrangements of the letters in the word NUMBER.
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$${\small \hspace{2.3em}\textrm{(b)}.\hspace{0.8em}}$$ How many of the arrrangements in part (a) begin and end with a vowel?
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$${\small 9.\enspace}$$ How many different numbers can be formed by taking one, two, three and four digits from the digits 1, 2, 7 and 8, if repetitions are not allowed?
One of these numbers is chosen at random. What is the probability that it is greater than 200?
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EXERCISE 5B:
$${\small 1.\enspace}$$ How many three-card hands can be dealt from a pack of 52 cards?
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$${\small 2.\enspace}$$ From a group of 30 boys and 32 girls, two girls and two boys are to be chosen to represent their school. How many possible selections are there?
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$${\small 3.\enspace}$$ A history exam paper contains eight questions, four in Part A and four in Part B. Candidates are required to attempt five questions. In how many ways can this be done if
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$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ there are no restrictions,
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ at least two questions from Part A and at least two questions from Part B must be attempted?
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$${\small 4.\enspace}$$ A committee of three people is to be selected from four women and five men. The rules state that there must be at least one man and one woman on the committee. In how many different ways can the committee be chosen?
Subsequently one of the men and one of the women marry each other. The rules also state that a married couple may not both serve on the committee. In how many ways can the committee be chosen now?
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$${\small 5.\enspace}$$ A box of one dozen eggs contains one that is bad. If three eggs are chosen at random, what is the probability that one of them will be bad?
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$${\small 6.\enspace}$$ In a game of bridge the pack of 52 cards is shared equally between all four players. What is the probability that one particular player has no hearts?
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$${\small 7.\enspace}$$ A bag contains 20 chocolates, 15 toffees and 12 peppermints. If three sweets are chosen at random, what is the probability that they are
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$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ all different,
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ all chocolates,
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$${\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}$$ all the same,
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$${\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}$$ all not chocolates?
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$${\small 8.\enspace}$$ Show that $$\displaystyle \binom{n}{r} \ = \ \binom{n}{n \ – \ r}$$.
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$${\small 9.\enspace}$$ Show that the number of permutations of n objects of which r are of one kind and n – r are of another kind is $$\displaystyle \binom{n}{r}$$.
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EXERCISE 5C:
$${\small 1.\enspace}$$ The letters of the word CONSTANTINOPLE are written on 14 cards, one on each card. The cards are shuffled and then arranged in a straight line.
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$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ How many different possible arrangements are there?
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ How many arrangements begin with P?
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$${\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}$$ How many arrangements start and end with O?
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$${\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}$$ How many arrangements are there where no two vowels are next to each other?
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$${\small 2.\enspace}$$ A coin is tossed 10 times.
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$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ How many different sequences of heads and tails are possible?
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ How many different sequences containing six heads and four tails are possible?
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$${\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}$$ What is the probability of getting six heads and four tails?
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$${\small 3.\enspace}$$ Eight cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture cards, 20 odd cards and 20 even cards.
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$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ How many different sequences of eight cards are possible?
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ How many of the sequences in part (a) will contain three picture cards, three odd-numbered cards and two even-numbered cards?
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$${\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}$$ Use parts (a) and (b) to determine the probability of getting three picture cards, three odd-numbered cards and two even-numbered cards if eight cards are selected with replacement from a standard pack of 52 playing cards.
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$${\small 4.\enspace}$$ Eight women and five men are standing in a line.
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$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ How many arrangements are possible if any individual can stand in any position?
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ In how many arrangements will all five men be standing next to one another?
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$${\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}$$ In how many arrangements will no two men be standing next to one another?
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$${\small 5.\enspace}$$ Each of the digits 1, 1, 2, 3, 3, 4, 6 is written on a separate card. The seven cards are then laid out in a row to form a 7-digit number.
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$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ How many distinct 7-digit numbers are there?
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ How many of these 7-digit numbers are even?
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$${\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}$$ How many of these 7-digit numbers are divisible by 4?
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$${\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}$$ How many of these 7-digit numbers start and end with the same digit?
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$${\small 6.\enspace}$$ Three families, the Mehtas, the Mupondas and the Lams, go to the cinema together to watch a film. Mr and Mrs Mehta take their daughter Indira, Mr and Mrs Muponda take their sons Paul and John, and Mrs Lam takes her children Susi, Kim and Lee. The families occupy a single row with eleven seats.
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$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ In how many ways could the eleven people be seated if there were no restriction?
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ In how many ways could the eleven people sit down so that the members of each family are all sitting together?
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$${\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}$$ In how many of the arrangements will no two adults be sitting next to one another?
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$${\small 7.\enspace}$$ The letters of the word POSSESSES are written on nine cards, one on each card. The cards are shuffled and four of them are selected and arranged in a straight line.
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$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ How many possible selections are there of four letters?
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ How many arrangements are there of four letters?
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MISCELLANEOUS EXERCISE 5:
$${\small 1.\enspace}$$ The judges in a ‘Beautiful Baby’ competition have to arrange 10 babies in order of merit. In how many different ways could this be done? Two babies are to be selected to be photographed. In how many ways can this selection be made?
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$${\small 2.\enspace}$$ In how many ways can a committee of four men and four women be seated in a row if
$$\\[1pt]$$
$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ they can sit in any position,
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ no one is seated next to a person of the same sex?
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$${\small 3.\enspace}$$ How many distinct arrangements are there of the letters in the word ABRACADABRA?
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$${\small 4.\enspace}$$ Six people are going to travel in a six-seater minibus but only three of them can drive. In how many different ways can they seat themselves?
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$${\small 5.\enspace}$$ There are eight different books on a bookshelf: three of them are hardbacks and the rest are paperbacks.
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$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ In how many different ways can the books be arranged if all the paperbacks are together and all the hardbacks are together?
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ In how many different ways can the books be arranged if all the paperbacks are together?
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$${\small 6.\enspace}$$ Four boys and two girls sit in a line on stools in front of a coffee bar.
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$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ In how many ways can they arrange themselves so that the two girls are together?
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ In how many ways can they sit if the two girls are not together?
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$${\small 7.\enspace}$$ Ten people travel in two cars, a saloon and a Mini. If the saloon has seats for six and the Mini has seats for four, find the number of different ways in which the party can travel, assuming that the order of seating in each car does not matter and all the people can drive.
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$${\small 8.\enspace}$$ Giving a brief explanation of your method, calculate the number of different ways in which the letters of the word TRIANGLES can be arranged if no two vowels may come together.
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$${\small 9.\enspace}$$ I have seven fruit bars to last the week. Two are apricot, three fig and two peach. I select one bar each day. In how many different orders can I eat the bars?
If I select a fruit bar at random each day, what is the probability that I eat the two appricot ones on consecutive days?
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$${\small 10.\enspace}$$ A class contains 30 children, 18 girls and 12 boys. Four complimentary theatre tickets are distributed at random to the children in the class. What is the probability that
$$\\[1pt]$$
$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ all four tickets go to the girls,
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ two boys and two girls receive tickets?
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$${\small 11.\enspace \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ How many different 7-digit numbers can be formed from the digits 0, 1, 2, 2, 3, 3, 3 assuming that a number cannot start with 0?
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$${\small \hspace{2.8em}\textrm{(b)}.\hspace{0.8em}}$$ How many of these numbers will end in 0?
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$${\small 12.\enspace}$$ Calculate the number of ways in which three girls and four boys can be seated on a row of seven chairs if each arrangement is to be symmetrical.
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$${\small 13.\enspace}$$ Find the number of ways in which
$$\\[1pt]$$
$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ 3 people can be arranged in 4 seats,
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ 5 people can be arranged in 5 seats.
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In a block of 8 seats, 4 are in row A and 4 are in row B. Find the number of ways of arranging 8 people in the 8 seats given that 3 specified people must be in row A.
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$${\small 14.\enspace}$$ Eight different cards, of which four are red and four are black, are dealt to two players so that each receives a hand of four cards. Calculate
$$\\[1pt]$$
$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ the total number of different hands which a given player could receive,
$$\\[1pt]$$
$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ the probability that each player receives a hand consisting of four cards all of the same colour.
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$${\small 15.\enspace}$$ A piece of wood of length of 10 cm is to be divided into 3 pieces so that the length of each piece is a whole number of cm, for example, 2 cm, 3 cm and 5 cm.
$$\\[1pt]$$
$${\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ List all the different sets of lengths which could be obtained.
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$${\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}$$ If one of these sets is selected at random, what is the probability that the lengths of the pieces could be lengths of the sides of a triangle?
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$${\small 16.\enspace}$$ Nine persons are to be seated at three tables holding 2, 3 and 4 persons respectively. In how many ways can the groups sitting at the tables be selected, assuming that the order of sitting at the tables does not matter?
$$\\[1pt]$$
$${\small 17.\enspace \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}$$ Calculate the number of different arrangements which can be made using all the letters of the word BANANA.
$$\\[1pt]$$
$${\small \hspace{2.8em}\textrm{(b)}.\hspace{0.8em}}$$ The number of combinations of 2 objects from n is equal to the number of combinations of 3 objects from n. Determine n.
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$${\small 18.\enspace}$$ A ‘hand’ of 5 cards is dealt from an ordinary pack of 52 playing cards. Show that there are nearly 2.6 million distinct hands and that, of these, 575 757 contain no card from the heart suit.
On three successive occasions a card player is dealt a hand containing no heart. What is the probability of this happening? What conclusion might the player be justifiably reach?
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9709 A-Level Past Papers:
$${\small 1.\enspace}$$ 9709/51/O/N/21 – Paper 51 November 2021 Probability and Statistics I No 5
$$\\[1pt]$$
Raman and Sanjay are members of a quiz team which has 9 members in total. Two photographs of the quiz team are to be taken.
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For the first photograph, the 9 members will stand in a line.
$$\\[1pt]$$
$${\small(\textrm{a}).\hspace{0.8em}}$$ How many different arrangements of the 9 members are possible in which Raman will be at the centre of the line?
$$\\[1pt]$$
$${\small(\textrm{b}).\hspace{0.8em}}$$ How many different arrangements of the 9 members are possible in which Raman and Sanjay are not next to each other?
$$\\[1pt]$$
For the second photograph, the members will stand in two rows, with 5 in the back row and 4 in the front row.
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$${\small(\textrm{c}).\hspace{0.8em}}$$ In how many different ways can the 9 members be divided into a group of 5 and a group of 4?
$$\\[1pt]$$
$${\small(\textrm{d}).\hspace{0.8em}}$$ For a random division into a group of 5 and a group of 4, find the probability that Raman and
Sanjay are in the same group as each other.
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$${\small 2.\enspace}$$ 9709/53/M/J/21 – Paper 53 June 2021 Probability and Statistics I No 6
$$\\[1pt]$$
$${\small(\textrm{a}).\hspace{0.8em}}$$ How many different arrangements are there of the 11 letters in the word REQUIREMENT?
$$\\[1pt]$$
$${\small(\textrm{b}).\hspace{0.8em}}$$ How many different arrangements are there of the 11 letters in the word REQUIREMENT in which the two Rs are together and the three Es are together?
$$\\[1pt]$$
$${\small(\textrm{c}).\hspace{0.8em}}$$ How many different arrangements are there of the 11 letters in the word REQUIREMENT in which there are exactly three letters between the two Rs?
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Five of the 11 letters in the word REQUIREMENT are selected.
$$\\[1pt]$$
$${\small(\textrm{d}).\hspace{0.8em}}$$ How many possible selections contain at least two Es and at least one R?
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$${\small 3.\enspace}$$ 9709/52/F/M/20 – Paper 52 March 2020 Probability and Statistics I No 1
$$\\[1pt]$$
The 40 members of a club include Ranuf and Saed. All 40 members will travel to a concert. 35 members will travel in a coach and the other 5 will travel in a car. Ranuf will be in the coach and Saed will be in the car.
$$\\[1pt]$$
In how many ways can the members who will travel in the coach be chosen?
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$${\small 4.\enspace}$$ 9709/52/F/M/20 – Paper 52 March 2020 Probability and Statistics I No 4
$$\\[1pt]$$
Richard has 3 blue candles, 2 red candles and 6 green candles. The candles are identical apart from their colours. He arranges the 11 candles in a line.
$$\\[1pt]$$
$${\small(\textrm{a}).\hspace{0.8em}}$$ Find the number of different arrangements of the 11 candles if there is a red candle at each end.
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$${\small(\textrm{b}).\hspace{0.8em}}$$ Find the number of different arrangements of the 11 candles if all the blue candles are together and the red candles are not together.
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$${\small 5.\enspace}$$ 9709/61/M/J/19 – Paper 61 June 2019 Probability and Statistics I No 8
$$\\[1pt]$$
Freddie has 6 toy cars and 3 toy buses, all different. He chooses 4 toys to take on holiday with him.
$$\\[1pt]$$
$${\small(\textrm{i}).\hspace{0.8em}}$$ In how many different ways can Freddie choose 4 toys?
$$\\[1pt]$$
$${\small(\textrm{ii}).\hspace{0.7em}}$$ How many of these choices will include both his favourite car and his favourite bus?
$$\\[1pt]$$
Freddie arranges these 9 toys in a line.
$$\\[1pt]$$
$${\small(\textrm{iii}).\hspace{0.6em}}$$ Find the number of possible arrangements if the buses are all next to each other.
$$\\[1pt]$$
$${\small(\textrm{iv}).\hspace{0.6em}}$$ Find the number of possible arrangements if there is a car at each end of the line and no buses
are next to each other.
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PRACTICE MORE WITH THESE QUESTIONS BELOW!
$${\small 1.\enspace}$$ In a 60-metre hurdles race there are five runners, one from each of the nations Austria, Belgium, Canada, Denmark and England.
$${\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}$$ How many different finishing orders are there?
$${\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}$$ What is the probability of predicting the finishing order by choosing first, second, third, fourth and fifth at random?
$${\small 2. \enspace}$$ In a ‘Goal of the season’ competition, participants are asked to rank ten goals in order of quality.
The organisers select their ‘correct’ order at random. Anybody who matches their order will be invited to join the television commentary team for the next international match.
$${\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}$$ What is the probability of a participant’s order being the same as that of the organisers?
$${\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}$$ Five million people enter the competition. How many people would be expected to join the commentary team?
$${\small 3. \enspace}$$ How many arrangements of the word ACHIEVE are there if
$${\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}$$ there are no restrictions on the order the letters are to be in
$${\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}$$ the first letter is an A
$${\small\hspace{1.2em} (\textrm{iii}).\hspace{0.5em}}$$ the letters A and I are to be together.
$${\small\hspace{1.2em} (\textrm{iv}).\hspace{0.6em}}$$the letters C and H are to be apart.
$${\small 4. \enspace (\textrm{i}).\hspace{0.7em}}$$ A football team consists of 3 players who play in a defence position, 3 players who play in a midfield position and 5 players who play in a forward position. Three players are chosen to collect a gold medal for the team. Find in how many ways this can be done
$${\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}$$ if the captain, who is a midfield player, must be included, together with one defence and one forward player.
$${\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}$$ if exactly one forward player must be included, together with any two others.
$${\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}$$ Find how many different arrangements there are of the nine letters in the words GOLD MEDAL
$${\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}$$ if there are no restrictions on the order of the letters,
$${\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}$$ if the two letters D come first and the letters L come last.
$${\small 5. \enspace}$$ The diagram shows the seating plan for passengers in a minibus, which has 17 seats arranged in 4 rows. The back row has 5 seats and the other 3 rows have 2 seats on each side. 11 passengers get on the minibus.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small\hspace{1.2em} (\textrm{i}).\hspace{0.7em}}$$ How many possible seating arrangements are there for the 11 passengers?
$${\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}$$ How many possible seating arrangements are there if 5 particular people sit in the back row?
Of the 11 passengers, 5 are unmarried and the other 6 consist of 3 married couples.
$${\small\hspace{1.2em} (\textrm{iii}).\hspace{0.5em}}$$ In how many ways can 5 of the 11 passengers on the bus be chosen if there must be 2 married couples and 1 other person, who may or may not be married?
$${\small 6. \enspace}$$ A choir consists of 13 sopranos, 12 altos, 6 tenors and 7 basses. A group consisting of 10 sopranos, 9 altos, 4 tenors and 4 basses is to be chosen from the choir.
$${\small\hspace{1.2em} (\textrm{i}).\hspace{0.7em}}$$ In how many different ways can the group be chosen?
$${\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}$$ In how many ways can the 10 chosen sopranos be arranged in a line if the 6 tallest stand next to each other?
$${\small\hspace{1.2em} (\textrm{iii}).\hspace{0.5em}}$$ The 4 tenors and the 4 basses in the group stand in a single line with all the tenors next to each other and all the basses next to each other. How many possible arrangements are there if three of the tenors refuse to stand next to any of the basses?
$${\small 7. \enspace (\textrm{i}).\hspace{0.7em}}$$ Find how many numbers between 5000 and 6000 can be formed from the digits 1, 2, 3, 4, 5 and 6
$${\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}$$ if no digits are repeated,
$${\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}$$ if repeated digits are allowed.
$${\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}$$ Find the number of ways of choosing a school team of 5 pupils from 6 boys and 8 girls
$${\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}$$ if there are more girls than boys in the team,
$${\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}$$ if three of the boys are cousins and are either all in the team or all not in the team.
$${\small 8. \enspace (\textrm{i}).\hspace{0.7em}}$$ Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that all four Es are together.
$${\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}$$ Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that the Ss are not next to each other.
Four letters are selected from the 12 letters of the word STEEPLECHASE.
$${\small \hspace{1.2em} (\textrm{iii}).\hspace{0.4em}}$$ Find the number of different selections if the four letters include exactly one S.
$${\small 9. \enspace (\textrm{i}).\hspace{0.7em}}$$ Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that all three Os are together and both Ts are together.
$${\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}$$ Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that the Ts are not together.
$${\small \hspace{1.2em} (\textrm{iii}).\hspace{0.4em}}$$ Find the probability that a randomly chosen arrangement of the 9 letters of the word TOADSTOOL has a T at the beginning and a T at the end.
$${\small \hspace{1.2em} (\textrm{iv}).\hspace{0.6em}}$$ Five letters are selected from the 9 letters of the word TOADSTOOL. Find the number of different selections if the five letters include at least 2 Os and at least 1 T.
$${\small 10. \enspace (\textrm{a}).\hspace{0.7em}}$$ A group of 6 teenagers go boating. There are three boats available. One boat has room for 3 people, one has room for 2 people and one has room for 1 person. Find the number of different ways the group of 6 teenagers can be divided between the three boats.
$${\small \hspace{1.6em} (\textrm{b}).\hspace{0.6em}}$$ Find the number of different 7-digit numbers which can be formed from the seven digits 2, 2, 3, 7, 7, 7, 8 in each of the following cases.
$${\small \hspace{2.8em} (\textrm{i}).\hspace{0.7em}}$$ The odd digits are together and the even digits are together.
$${\small \hspace{2.8em} (\textrm{ii}).\hspace{0.6em}}$$ The 2s are not together.
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
Sequences and Series
Sequences and Series
Sequence is a list of numbers with a predefined rule or formula.
The numbers are called as terms and we can find the pattern in the sequence.
There are quite a limitless number of possible sequences can be made, we however will mainly focus on two main categories of sequences, the arithmetic sequence and the geometric sequence.
Arithmetic Sequence
In arithmetic sequence or progression any two consecutive terms always have the same difference.
$$\\[12pt]\hspace{2em}{\small {u}_{n} \ = \ a \ + \ (n \ – \ 1) d }$$
$$\\[7pt]{\small {u}_{n} }$$ is the $${\small \boldsymbol{{n}^{\textrm{th}}} }$$ term
$$\\[7pt]{\small a }$$ is the first term or $${\small \boldsymbol{{u}_{1}} }$$
$${\small d }$$ is the common difference.
Example:
$$\\[7pt]{\small a \ \ \ = \ {u}_{1} }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 10 }$$
$$\\[7pt]{\small {u}_{2} \ = \ 15, \ {u}_{3} \ = \ 20, \ {u}_{4} \ = \ 25, \ \textrm{etc} }$$
$$\\[7pt]{\small d \ \ \ = \ 5}$$
$$\\[7pt]{\small {u}_{n} \ = \ a \ + \ (n \ – \ 1) d }$$
$$\\[7pt]{\small {u}_{n} \ = \ 10 \ + \ (n \ – \ 1) \ \times \ 5 }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 5n \ – \ 5 }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 5n \ + \ 5 }$$
So, for example, if we want to find the 10th term of the sequence above,
$$\\[7pt]{\small {u}_{10} \ = \ 5(10) \ + \ 5 }$$
$$\hspace{1.6em}{\small = \ 55 }$$
Geometric Sequence
In geometric sequence or progression any two consecutive terms always have the same ratio.
$$\\[12pt]\hspace{2em}{\small {u}_{n} \ = \ a {r}^{ (n \ – \ 1) } }$$
$$\\[7pt]{\small {u}_{n} }$$ is the $${\small \boldsymbol{{n}^{\textrm{th}}} }$$ term
$$\\[7pt]{\small a }$$ is the first term or $${\small \boldsymbol{{u}_{1}} }$$
$${\small r }$$ is the common ratio.
Example:
$$\\[7pt]{\small a \ \ \ = \ {u}_{1} }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 100 }$$
$$\\[7pt]{\small {u}_{2} \ = \ 50, \ {u}_{3} \ = \ 25, \ {u}_{4} \ = \ 12.5, \ \textrm{etc} }$$
$$\\[12pt]{\small r \ \ \ = \ {\large\frac{1}{2}} }$$
$$\\[7pt]{\small {u}_{n} \ = \ a {r}^{ (n \ – \ 1) } }$$
$$\\[12pt]{\small {u}_{n} \ = \ 100 \ \times \ { \big({\large\frac{1}{2}} \big) }^{ (n \ – \ 1) } }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 100 \ \times \ {2}^{(-n+1)} }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 200 \ \times \ {2}^{-n} }$$
So, for example, if we want to find the 10th term of the sequence above,
$$\\[7pt]{\small {u}_{10} \ = \ 200 \ \times \ {2}^{-10} }$$
$$\hspace{1.6em}{\small = \ 0.1953125 }$$
Meanwhile, we can also calculate the sum of n terms, that is, the addition from the first term of the sequence up to the $${\small {n}^{\textrm{th}} }$$ term.
The addition of the sequence is called series.
The formula for both arithmetic series and geometric series are as follows:
Arithmetic Series
$$\\[18pt]\hspace{1.5em}{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ ( a \ + \ {u}_{n} ) \ = \ {\large\frac{n}{2}} \ \big[ 2a \ + \ (n \ – \ 1)d \big] }$$
$$\\[7pt]{\small {S}_{n} }$$ is the sum up to $${\small \boldsymbol{{n}^{\textrm{th}}} }$$ term.
Example:
$$\\[7pt]{\small {S}_{1} \ = \ {u}_{1} }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 10 }$$
$$\\[7pt]{\small {S}_{2} \ = \ {u}_{1} \ + \ {u}_{2} }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 15 }$$
$$\\[7pt]{\small {S}_{3} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 15 \ + \ 20 }$$
$$\\[7pt]{\small {S}_{4} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} \ + \ {u}_{4} }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 15 \ + \ 20 \ + \ 25}$$
$$\\[12pt]$$, etc
$$\\[7pt]{\small d \ \ \ = \ 5}$$
$$\\[12pt]{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \big[ 2a \ + \ (n \ – \ 1)d \big] }$$
$$\\[12pt]{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \big[ 2(10) \ + \ (n \ – \ 1) \ \times \ 5 \big] }$$
$$\\[12pt]\hspace{1.4em}{\small = \ {\large\frac{n}{2}} \ ( 20 \ + \ 5n \ – \ 5 ) }$$
$$\\[7pt]\hspace{1.4em}{\small = \ {\large\frac{5}{2}}{n}^2 \ + \ {\large\frac{15}{2}}n }$$
So, for example, if we want to find the sum of the first 10 terms of the sequence above,
$$\\[12pt]{\small {S}_{10} \ = \ {\large\frac{5}{2}}{(10)}^2 \ + \ {\large\frac{15}{2}}n }$$
$$\hspace{1.6em}{\small = \ 325 }$$
Geometric Series
$$\\[12pt]\hspace{2em}{\small {S}_{n} \ = \ {\large\frac{a \ ( {r}^{ n } \ – \ 1 ) }{r \ – \ 1} } }$$ for $${\small |r| \ \gt \ 1 }$$
$$\\[7pt]$$ or,
$$\\[20pt]\hspace{2em}{\small {S}_{n} \ = \ {\large\frac{a \ ( 1 \ – \ {r}^{ n } )}{1 \ – \ r } } }$$ for $${\small |r| \ \lt \ 1 }$$
$$\\[7pt]{\small {S}_{n} }$$ is the sum up to $${\small \boldsymbol{{n}^{\textrm{th}}} }$$ term.
Example:
$$\\[7pt]{\small {S}_{1} \ = \ {u}_{1} }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 100 }$$
$$\\[7pt]{\small {S}_{2} \ = \ {u}_{1} \ + \ {u}_{2} }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 100 \ + \ 50 }$$
$$\\[7pt]{\small {S}_{3} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 100 \ + \ 50 \ + \ 25 }$$
$$\\[7pt]{\small {S}_{4} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} \ + \ {u}_{4} }$$
$$\\[7pt]\hspace{1.4em}{\small = \ 100 \ + \ 50 \ + \ 25 \ + \ 12.5}$$
$$\\[12pt]$$, etc
$$\\[12pt]{\small r \ \ \ = \ {\large\frac{1}{2}} }$$
$$\\[12pt]{\small {S}_{n} \ = \ {\large\frac{a \ ( 1 \ – \ {r}^{ n } )}{1 \ – \ r } } }$$
$$\\[18pt]{\small {S}_{n} \ = \ {\large\frac{100 \ \big( 1 \ – \ {{\large (\frac{1}{2} )} }^{ n } \big)}{1 \ – \ {\large\frac{1}{2}} } } }$$
$$\\[18pt]{\small {S}_{n} \ = \ {\large\frac{100 \ \big( 1 \ – \ {2} ^{ -n } \big)}{ {\large\frac{1}{2}} } } }$$
$$\\[10pt]{\small {S}_{n} \ = \ 200 \ \times \ \big( 1 \ – \ {2 }^{ -n } \big) }$$
So, for example, if we want to find the sum of the first 10 terms of the sequence above,
$$\\[10pt]{\small {S}_{n} \ = \ 200 \ \times \ \big( 1 \ – \ {(2) }^{ -10 } \big) }$$
$$\hspace{1.4em}{\small = \ 199.8046875 }$$
In geometric series, we can also find the sum to infinity given that the series is convergent $${\small (|r| \ \lt \ 1) }$$.
Since $${\small \displaystyle \lim_{n \to \infty} 1 \ – \ {r}^{ n } \ = \ 1 \ }$$ for $${\small \ (|r| \ \lt \ 1) }$$,
$$\\[12pt]\hspace{2em}{\small {S}_{\infty} \ = \ {\large\frac{a}{1 \ – \ r } } }$$
$$\\[7pt]{\small {S}_{\infty} }$$ is the sum to infinity.
Example:
$$\\[18pt]{\small {S}_{\infty} \ = \ {\large\frac{ a }{1 \ – \ r } } }$$
$$\\[18pt]{\small {S}_{\infty} \ = \ {\large\frac{ 100 }{1 \ – \ {\large\frac{1}{2}} } } }$$
$$\hspace{1.6em}{\small = \ 200 }$$
Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
$$\\[1pt]$$
EXAMPLE:
$${\small 1.\enspace}$$ A runner who is training for a long-distance race plans to run increasing distances each day for 21 days. She will run x km on day 1, and on each subsequent day she will increase the distance by 10% of the previous day’s distance. On day 21 she will run 20 km.
$${\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}$$ Find the distance she must run on day 1 in order to achieve this. Give your answer in km correct to 1 decimal place.
$${\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}$$ Find the total distance she runs over the 21 days.
$$\\[1pt]$$
$${\small 2.\enspace}$$ The first, second and third terms of a geometric progression are x, x – 3 and x – 5 respectively.
$${\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}$$ Find the value of x.
$${\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}$$ Find the fourth term of the progression.
$${\small\hspace{2.8em}(\textrm{iii}).\hspace{0.5em}}$$ Find the sum to infinity of the progression.
$$\\[1pt]$$
$${\small 3.\enspace}$$ In an arithmetic progression, the sum of the first ten terms is equal to the sum of the next five terms. The first term is a.
$${\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}$$ Show that the common difference of the progression is $${\small {\large \frac{1}{3} } a}$$.
$${\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}$$ Given that the tenth term is 36 more than the fourth term, find the value of a.
$$\\[1pt]$$
$${\small 4.\enspace}$$ The sum to infinity of a geometric progression is 9 times the sum of the first four terms. Given that the first term is 12, find the value of the fifth term.
$$\\[1pt]$$
$${\small 5.\enspace}$$ Two heavyweight boxers decide that they would be more successful if they competed in a lower weight class. For each boxer this would require a total weight loss of 13 kg. At the end of week 1 they have each recorded a weight loss of 1 kg and they both find that in each of the following weeks their weight loss is slightly less than the week before.
$$\\[1pt]$$
Boxer A’s weight loss in week 2 is 0.98 kg. It is given that his weekly weight loss follows an arithmetic progression.
$${\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}$$ Write down an expression for his total weight loss after x weeks.
$${\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}$$ He reaches his 13 kg target during week n. Use your answer to part (i) to find the value of n.
$$\\[1pt]$$
Boxer B’s weight loss in week 2 is 0.92 kg and it is given that his weekly weight loss follows a geometric progression.
$${\small\hspace{2.8em}(\textrm{iii}).\hspace{0.5em}}$$ Calculate his total weight loss after 20 weeks and show that he can never reach his target.
$$\\[1pt]$$
$${\small 6.\enspace}$$ The sum of the first twenty terms of an arithmetic progression is 50, and the sum of the next twenty terms is -50. Find the sum of the first hundred terms of the progression.
$$\\[1pt]$$
$${\small 7.\enspace}$$ In 1971 a newly-built flat was sold with a 999-year lease. The terms of the sale included a requirement to pay ‘ground rent’ yearly. The ground rent was set at £28 per year for the first 21 years of the lease, increasing by £14 to £42 per year for the next 21 years, and then increasing again by £14 at the end of each subsequent period of 21 years.
$${\small\hspace{1.2em} (\textrm{a}).\hspace{0.8em}}$$ Find how many complete 21-year periods there would be if the lease ran for the full 999 years, and how many years there would be left over.
$${\small\hspace{1.2em} (\textrm{b}).\hspace{0.8em}}$$ Find the total amount of ground rent that would be paid in all of the complete 21-year periods of the lease.
$$\\[1pt]$$
$${\small 8.\enspace}$$ Jayesh invests $100 in a savings acoount on the first day of each month for one complete year. The account pays interest at $${\small \frac{1}{2} \% \ }$$ for each complete month. How much does Jayesh have invested at the end of the year (but before making a thirteenth payment)? $$\\[1pt]$$ $${\small 9.\enspace}$$ Neeta takes out a 25-year mortgage of$40 000 to buy her house. Compound interest is charged on the loan at a rate of 8% per annum. She has to pay off the mortgage with 25 equal payments, the first of which is to be one year after the loan is taken out. Continue the following argument to calculate the value of each annual payment.
$$\\[1pt]$$
$${\small \hspace{1.2em} \bullet \hspace{1.2em} }$$ After 1 year she owes $${\small (40 000 \times 1.08) }$$ (loan plus interest) less the payment made, $P, that is, she owes $${\small (40 000 \times 1.08 \ – P)}$$. $$\\[1pt]$$ $${\small \hspace{1.2em} \bullet \hspace{1.2em} }$$ After 2 years she owes $${\small \big((40 000 \times 1.08 \ – P) \times 1.08 \ – P \big) }$$. $$\\[1pt]$$ $${\small \hspace{1.2em} \bullet \hspace{1.2em} }$$ After 3 years she owes $${\small \Big(\big((40 000 \times 1.08 \ – P) \times 1.08 \ – P \big) \times 1.08 \ – P \Big)}$$. $$\\[1pt]$$ At the end of 25 years this (continued) expression must be zero. Form an equation in P and solve it. $$\\[1pt]$$ $${\small 10.\enspace}$$ The Bank of Utopia offers an interest rate of 100% per annum with various options as to how the interest may be added. Gopal invests$1000 and considers the following options.
$$\\[1pt]$$
$${\small \hspace{1.2em} \textrm{Option A} \hspace{1em} }$$ Interest added annually at the end of the year.
$$\\[1pt]$$
$${\small \hspace{1.2em} \textrm{Option B} \hspace{1em} }$$ Interest of 50% credited at the end of each half-year.
$$\\[1pt]$$
$${\small \hspace{1.2em} \textrm{Option C, D, E, …} \hspace{1em} }$$ The Bank is willing to add interest as often as required, subject to (interest rate) $${\small \times }$$ (number of credits per year) = 100.
$$\\[1pt]$$
Investigate to find the maximum possible amount in Gopal’s account after one year.
$$\\[1pt]$$
$${\small 11.\enspace}$$ 9709/13/O/N/21 – Paper 13 November 2021 Pure Maths 1 No 4
$$\\[1pt]$$
The first term of an arithmetic progression is 84 and the common difference is −3.
$$\\[1pt]$$
$${\small(\textrm{a}).\hspace{0.8em}}$$ Find the smallest value of $$n$$ for which the $$n$$th term is negative.
$$\\[1pt]$$
It is given that the sum of the first $$2k$$ terms of this progression is equal to the sum of the first $$k$$ terms.
$$\\[1pt]$$
$${\small(\textrm{b}).\hspace{0.8em}}$$ Find the value of $$k$$.
$$\\[1pt]$$
$${\small 12.\enspace}$$ 9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 8
$$\\[1pt]$$
The first, second and third terms of an arithmetic progression are $$\ a, {\large\frac{3}{2}}a \$$ and $$b$$ respectively, where $$a$$ and $$b$$ are positive constants.
$$\\[1pt]$$
The first, second and third terms of a geometric progression are $$\ a$$, $$\ 18 \$$ and $$\ b \ + \ 3 \$$ respectively.
$$\\[1pt]$$
$${\small(\textrm{a}).\hspace{0.8em}}$$ Find the values of $$a$$ and $$b$$.
$$\\[1pt]$$
$${\small(\textrm{b}).\hspace{0.8em}}$$ Find the sum of the first 20 terms of the arithmetic progression.
$$\\[1pt]$$
$${\small 13.\enspace}$$ 9709/12/M/J/20 – Paper 12 June 2020 Pure Maths 1 No 4
$$\\[1pt]$$
The $$n$$th term of an arithmetic progression is $$\ {\large\frac{1}{2}} (3n \ – \ 15)$$.
$$\\[1pt]$$
Find the value of $$n$$ for which the sum of the first $$n$$ terms is 84.
$$\\[1pt]$$
$${\small 14.\enspace}$$ 9709/12/O/N/19 – Paper 12 November 2019 Pure Maths 1 No 8
$$\\[1pt]$$
$${\small(\textrm{a}).\hspace{0.8em}}$$ Over a 21-day period an athlete prepares for a marathon by increasing the distance she runs each day by 1.2 km. On the first day she runs 13 km.
$$\\[1pt]$$
$${\small \hspace{0.8em}(\textrm{i}).\hspace{0.8em}}$$ Find the distance she runs on the last day of the 21-day period.
$$\\[1pt]$$
$${\small \hspace{0.8em} (\textrm{ii}).\hspace{0.7em}}$$ Find the total distance she runs in the 21-day period.
$$\\[1pt]$$
$${\small (\textrm{b}).\hspace{0.8em}}$$ The first, second and third terms of a geometric progression are $$\ x, \ x \ − \ 3 \$$ and $$\ x \ − \ 5 \$$ respectively.
$$\\[1pt]$$
$${\small \hspace{0.8em} (\textrm{i}).\hspace{0.8em}}$$ Find the value of $$x$$.
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$${\small \hspace{0.8em} (\textrm{ii}).\hspace{0.7em}}$$ Find the fourth term of the progression.
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$${\small \hspace{0.8em} (\textrm{iii}).\hspace{0.6em}}$$ Find the sum to infinity of the progression.
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PRACTICE MORE WITH THESE QUESTIONS BELOW!
$${\small 1. \hspace{0.6em}(\textrm{i}).\hspace{0.7em}}$$ The first and second terms of a geometric progression are p and 2p respectively, where p is a positive constant. The sum of the first n terms is greater than 1000p. Show that $${\small {2}^{n} \ \gt \ 1001}$$.
$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}$$ In another case, p and 2p are the first and second terms respectively of an arithmetic progression. The nth term is 336 and the sum of the first n terms is 7224. Write down two equations in n and p and hence find the values of n and p.
$${\small 2. \enspace}$$ The first three terms of an arithmetic progression are 4, x and y respectively. The first three terms of a geometric progression are x, y and 18 respectively. It is given that both x and y are positive.
$${\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}$$ Find the value of x and the value of y.
$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}$$Find the fourth term of each progression.
$${\small 3. \enspace}$$ In an arithmetic progression the first term is a and the common difference is 3. The nth term is 94 and the sum of the first n terms is 1420. Find n and a.
$${\small 4. \enspace}$$ A person wants to buy a pension which will provide her with an income of $10 000 at the end of each of the next n years. Show that, with a steady interest rate of 5% per year, the pension should cost her $$\\[1pt]$$ $${\small \10 \ 000 } \big(\frac{1}{1.05} + \frac{1}{{1.05}^{2}} + \frac{1}{{1.05}^{3}} + \cdots + \frac{1}{{1.05}^{n}} \big) .$$ $$\\[1pt]$$ Find a simple formula for calculating this sum, and find its value when n = 10, 20, 30, 40, 50. $${\small 5. \enspace}$$ As part of a fund-raising campaign, I have been given some books of raffle tickets to sell. Each book has the same number of tickets and all the tickets I have been given are numbered in sequence. The number of the ticket on the front of the 5th book is 205 and that on the front of the 19th book is 373. $${\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}$$ By writing the number of the ticket on the front of the first book as a and the number of tickets in each book as d, write down two equations involving a and d. $${\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}$$ From these two equations find how many tickets are in each book and the number on the front of the first book I have been given. $${\small\hspace{1.2em} (\textrm{iii}).\hspace{0.6em}}$$ The last ticket I have been given is numbered 492. How many books have I been given? $${\small 6. \enspace}$$ A tank is filled with 20 litres of water. Half the water is removed and replaced with anti-freeze and thoroughly mixed. Half this mixture is then removed and replaced with anti-freeze. This process continues. $${\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}$$ Find the first five terms in the sequence of amounts of water in the tank at each stage. $${\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}$$ Find the first five terms in the sequence of amounts of anti-freeze in the tank at each stage. $${\small\hspace{1.2em} (\textrm{iii}).\hspace{0.4em}}$$ Is either of these sequences geometric? Explain. $${\small 7. \enspace}$$ A company producing salt from sea water changed to a new process. The amount of salt obtained each week increased by 2% of the amount obtained in the preceding week. It is given that in the first week after the change the company obtained 8000 kg of salt. $${\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}$$ Find the amount of salt obtained in the 12th week after the change. $${\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}$$ Find the total amount of salt obtained in the first 12 weeks after the change. $${\small 8. \enspace}$$ The common ratio of a geometric progression is 0.99. Express the sum of the first 100 terms as a percentage of the sum to infinity, giving your answer correct to 2 significant figures. $${\small 9. \hspace{0.6em}(\textrm{a}).\hspace{0.7em}}$$ Each year, the value of a certain rare stamp increases by 5% of its value at the beginning of the year. A collector bought the stamp for$10 000 at the beginning of 2005. Find its value at the beginning of 2015 correct to the nearest $100. $${\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}$$ The sum of the first n terms of an arithmetic progression is $${\small \frac{1}{2}n(3n+7)}$$. Find the 1st term and the common difference of the progression. $${\small 10. \hspace{0.4em}(\textrm{a}).\hspace{0.7em}}$$ An organ has 8 tubes. The lengths of these tubes are a geometrical progression, and the length of the shortest tube is exactly half that of the longest which is 66 cm long. What is the total length of all the tubes in the organ to the nearest cm? $${\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}$$ Determine how long it will take to repay an interest free loan$6 625 by monthly payments initially of $10 and increasing by$5 each month. How much is the final payment?
If the loan is \$10 000, find the smallest number of months that would be needed to repay it.
$$\\[1pt]$$
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
Vector Line and Plane Equation
Vector Line and Plane Equation
We have learned the cartesian form of a line equation: $$\boxed{ \ y \ = \ mx \ + \ c \ }$$.
While it comes in handy in solving one-dimensional (1D) and two-dimensional (2D) problems, it may not be as such when solving three-dimensional problems.
Using the vector form of a line equation and a plane equation helps us to solve 3D problems much easier than using its cartesian form.
Given any two points, A and B, we can draw the vector $${\small \vec{a}}$$ and $${\small \vec{b}}$$ from the origin. Then, the line equation of line AB in the vector form can be written as follows:
$$\hspace{3em} \vec{r} \ = \ \vec{a} \ + \ \lambda(\vec{b} \ – \ \vec{a})$$
$$\\[5pt] {\small \vec{r} \ = \ }$$ position vector
: it represents any points along the line AB
$$\\[5pt] {\small \vec{a} \ \ \textrm{or} \ \ \vec{b} \ = \ }$$ location vector
: it shows the location vector of any one point along the line AB which can be represented by $${\small \vec{a}}$$ or $${\small \vec{b}}$$
$$\\[5pt] {\small \vec{b} \ – \ \vec{a} \ = \ }$$ direction vector
: it gives the direction vector of the line AB
$$\\[5pt] {\small \lambda \ }$$ is a constant value.
For the 2D shape, the vector form of a plane equation is shown below:
$$\hspace{3em} (\vec{r} \ – \ \vec{a}) \cdot \vec{n} \ = \ 0$$
$$\\[5pt] {\small \vec{r} \ = \ }$$ position vector
: it represents any points on the plane
$$\\[5pt] {\small \vec{a} \ = \ }$$ location vector
: it shows the location of a point on the plane which is represented by $${\small \vec{a}}$$
$$\\[5pt] {\small \vec{n} \ = \ }$$ normal vector
: it is the vector that gives perpendicular direction to the plane.
Note that we can find the cartesian form of a plane equation from its vector form,
$$\\[8pt] \hspace{3em} (\vec{r} \ – \ \vec{a}) \cdot \vec{n} \ \hspace{0.7em} \ = \ 0$$
$$\\[8pt] \hspace{3em} \vec{r} \ \cdot \vec{n} \ – \ \vec{a} \ \cdot \vec{n} \ = \ 0$$
$$\\[8pt] \hspace{3em} \vec{r} \ \cdot \vec{n} \hspace{3.3em} \ = \ \vec{a} \ \cdot \vec{n}$$
Let $$\ {\small \vec{r}}=\begin{pmatrix} x \\[1pt] y \\[1pt] z \end{pmatrix} \$$ and $$\ {\small \vec{n}}=\begin{pmatrix} a \\[1pt] b \\[1pt] c \end{pmatrix}$$
with $${\small \ (x,y,z) \ }$$ are the cartesian coordinates of any points on the plane and $${\small \ (a,b,c)} \$$ are the cartesian components of the normal vector $$\ {\small \vec{n} \ }$$. Then the cartesian form is:
$$\hspace{3em} ax \ + \ by \ + \ cz \ = \ d$$
with $$\ {\small d \ = \ \vec{a} \ \cdot \vec{n} \ }$$ (the dot product of vector $$\ {\small \vec{a} \ }$$ and $$\ {\small \vec{n} \ }$$ ).
I have put together some of the questions I received in the comment section below. You can try these questions also to further your understanding on this topic.
To check your answer, you can look through the solutions that I have posted either in Youtube videos or Instagram posts.
You can subscribe, like or follow my youtube channel and IG account. I will keep updating my IG daily post, preferably.
Furthermore, you can find some examples and more practices below! =).
Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
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EXAMPLE:
$${\small 1.\enspace}$$ 9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 8
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In the diagram, OABC is a pyramid in which OA = 2 units, OB = 4 units and OC = 2 units. The edge OC is vertical, the base OAB is horizontal and angle $${\small \ AOB \ = \ 90^{\large{\circ}}}$$. Unit vectors i, j and k are parallel to OA, OB and OC respectively. The midpoints of AB and BC are M and N respectively.
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}$$ Express the vectors $${\small \ \overrightarrow{ON} \ }$$ and $${\small \ \overrightarrow{CM} \ }$$ in terms of i, j and k.
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}$$ Calculate the angle between the directions of $${\small \ \overrightarrow{ON} \ }$$ and $${\small \ \overrightarrow{CM} \ }$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{c}).\hspace{0.8em}}$$ Show that the length of the perpendicular from M to ON is $${\small \ {\large \frac{3}{5} } \sqrt 5 }$$.
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$${\small 2.\enspace}$$ 9709/11/O/N/16 – Paper 11 Oct Nov 2020 Pure Maths 1 No 9
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The diagram shows a cuboid OABCDEFG with a horizontal base OABC in which OA = 4 cm and AB = 15 cm. The height OD of the cuboid is 2 cm. The point X on AB is such that AX = 5 cm and the point P on DG is such that DP = p cm, where p is a constant. Unit vectors i, j and k are parallel to OA, OC and OD respectively.
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}$$ Find the possible values of p such that angle $${\small \ OPX \ = \ 90^{\large{\circ}}}$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}$$ For the case where p = 9, find the unit vector in the direction of $${\small \ \overrightarrow{XP} }$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}$$ A point Q lies on the face CBFG and is such that XQ is parallel to AG. Find $${\small \ \overrightarrow{XQ} }$$.
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$${\small 3.\enspace}$$ The points A and B have position vectors i + 2jk and 3i + j + k respectively. The line l has equation r = 2i + j + k + $${\small \mu}$$(i + j + 2k).
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ Show that l does not intersect the line passing through A and B.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ The plane m is perpendicular to AB and passes through the mid-point of AB. The plane m intersects the line l at the point P. Find the equation of m and the position vector of P.
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$${\small 4.\enspace}$$
$$\\[1pt]$$
The diagram shows a set of rectangular axes Ox, Oy and Oz, and four points A, B, C and D with position vectors $${\small \overrightarrow{OA} \ = \ 3\textbf{i} }$$, $${\small \overrightarrow{OB} \ = \ 3\textbf{i} \ + \ 4\textbf{j} \ }$$, $${\small \overrightarrow{OC} \ = \ \textbf{i} \ + \ 3\textbf{j} \ }$$ and $${\small \overrightarrow{OD} \ = \ 2\textbf{i} \ + \ 3\textbf{j} \ + \ 5\textbf{k} \ }$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ Find the equation of the plane BCD, giving your answer in the form $${\small ax + by + cz = d}$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ Calculate the acute angle between the planes BCD and OABC .
$$\\[1pt]$$
$${\small 5.\enspace}$$ The line l has equation r = i + 2j + 3k + $${\small \mu}$$(2ij – 2k).
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ The point P has position vector 4i + 2j – 3k. Find the length of the perpendicular from P to l.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ It is given that l lies in the plane $${\small ax + by + 2z = 13}$$, where a and b are constants. Find the values of a and b.
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$${\small 6.\enspace}$$ Two planes have equations $${\small 2x + 3y \ – \ z \ = \ 1 \ }$$ and $$\ {\small x \ – \ 2y + z \ = \ 3}$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ Find the acute angle between the planes.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ Find a vector equation for the line of intersection of the planes.
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$${\small 7.\enspace}$$ The planes m and n have equations $${\small 3x + y \ – \ 2z \ = \ 10}$$ and $${\small x \ – \ 2y + 2z \ = \ 5}$$ respectively. The line l has equation r = 4i + 2j + k + $${\small \lambda}$$(i + j + 2k).
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ Show that l is parallel to m.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ Calculate the acute angle between the planes m and n.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(iii)}.\hspace{0.8em}}$$ A point P lies on the line l. The perpendicular distance P from the plane l is equal to 2. Find the position vectors of the two possible positions of P.
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$${\small 8.\enspace}$$ The line l has equation r = 5i – 3jk + $${\small \lambda}$$(i – 2j + k). The plane p has equation (ri – 2j) . (3i + j + k) = 0. The line l intersects the plane p at the point A.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ Find the position vector of A.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ Calculate the acute angle between l and p.
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$${\small\hspace{1.2em}\textrm{(iii)}.\hspace{0.8em}}$$ Find the equation of the line which lies in p and intersects l at right angles.
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$${\small 9.\enspace}$$ 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 8
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Relative to the origin $$O$$, the points $$A$$, $$B$$ and $$D$$ have position vectors given by
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$${\small \ \overrightarrow{OA} = \textbf{i} \ + \ 2\textbf{j} \ + \ \textbf{k} \ }$$, $${\small \ \overrightarrow{OB} = 2\textbf{i} \ + \ 5\textbf{j} \ + \ 3\textbf{k} \ }$$ and $${\small \ \overrightarrow{OD} = 3\textbf{i} \ + \ 2\textbf{k} }$$.
$$\\[1pt]$$
A fourth point $$C$$ is such that $$ABCD$$ is a parallelogram.
$$\\[1pt]$$
$${\small (\textrm{a}).\hspace{0.8em}}$$ Find the position vector of $$C$$ and verify that the parallelogram is not a rhombus.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ Find angle $$BAD$$, giving your answer in degrees.
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$${\small (\textrm{c}).\hspace{0.8em}}$$ Find the area of the parallelogram correct to 3 significant figures.
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$${\small 10.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 10
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With respect to the origin $$O$$, the points $$A$$ and $$B$$ have position vectors given by $${\small \ \overrightarrow{OA} = 6\textbf{i} \ + \ 2\textbf{j} \ }$$ and $${\small \ \overrightarrow{OB} = 2\textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} }$$. The midpoint of $$OA$$ is $$M$$. The point $$N$$ lying on $$AB$$, between $$A$$ and $$B$$, is such that $$AN = 2NB$$.
$$\\[1pt]$$
$${\small (\textrm{a}).\hspace{0.8em}}$$ Find a vector equation for the line through $$M$$ and $$N$$.
$$\\[1pt]$$
The line through $$M$$ and $$N$$ intersects the line through $$O$$ and $$B$$ at the point $$P$$.
$$\\[1pt]$$
$${\small (\textrm{b}).\hspace{0.8em}}$$ Find the position vector of $$P$$.
$$\\[1pt]$$
$${\small (\textrm{c}).\hspace{0.8em}}$$ Calculate angle $$OPM$$, giving your answer in degrees.
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$${\small 11.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 9
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With respect to the origin $$O$$, the vertices of a triangle $$ABC$$ have position vectors
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$${\small \ \overrightarrow{OA} = 2\textbf{i} \ + \ 5\textbf{k} \ }$$, $${\small \ \overrightarrow{OB} = 3\textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} \ }$$ and $${\small \ \overrightarrow{OC} = \textbf{i} \ + \ \textbf{j} \ + \ \textbf{k} }$$.
$$\\[1pt]$$
$${\small (\textrm{a}).\hspace{0.8em}}$$ Using a scalar product, show that angle $$ABC$$ is a right angle.
$$\\[1pt]$$
$${\small (\textrm{b}).\hspace{0.8em}}$$ Show that triangle $$ABC$$ is isosceles.
$$\\[1pt]$$
$${\small (\textrm{c}).\hspace{0.8em}}$$ Find the exact length of the perpendicular from $$O$$ to the line through $$B$$ and $$C$$.
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$${\small 12.\enspace}$$ 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 7
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Two lines have equations $$\ {\small \vec{r}}=\begin{pmatrix} 1 \\[1pt] 3 \\[1pt] 2 \end{pmatrix} + s \begin{pmatrix} 2 \\[1pt] -1 \\[1pt] 3 \end{pmatrix}$$ and $$\ {\small \vec{r}}=\begin{pmatrix} 2 \\[1pt] 1 \\[1pt] 4 \end{pmatrix} + t \begin{pmatrix} 1 \\[1pt] -1 \\[1pt] 4 \end{pmatrix}$$
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$${\small (\textrm{a}).\hspace{0.8em}}$$ Show that the lines are skew.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ Find the acute angle between the directions of the two lines.
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$${\small 13.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 9
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The quadrilateral $$ABCD$$ is a trapezium in which $$AB$$ and $$DC$$ are parallel. With respect to the origin $$O$$, the position vectors of $$A$$, $$B$$ and $$C$$ are given by
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$${\small \ \overrightarrow{OA} = -\textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} \ }$$, $${\small \ \overrightarrow{OB} = \textbf{i} \ + \ 3\textbf{j} \ + \ \textbf{k} \ }$$ and $${\small \ \overrightarrow{OC} = 2\textbf{i} \ + \ 2\textbf{j} \ – \ 3\textbf{k} }$$.
$$\\[1pt]$$
$${\small (\textrm{a}).\hspace{0.8em}}$$ Given that $${\small \ \overrightarrow{DC} = 3 \overrightarrow{AB} }$$, find the position vector of $$D$$.
$$\\[1pt]$$
$${\small (\textrm{b}).\hspace{0.8em}}$$ State a vector equation for the line through $$A$$ and $$B$$.
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$${\small (\textrm{c}).\hspace{0.8em}}$$ Find the distance between the parallel sides and hence find the area of the trapezium.
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$${\small 14.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 11
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With respect to the origin $$O$$, the points $$A$$ and $$B$$ have position vectors given by
$$\\[1pt]$$
$${\small \ \overrightarrow{OA} = 2\textbf{i} \ – \ \textbf{j} \ }$$ and $${\small \ \overrightarrow{OB} = \textbf{j} \ – \ 2\textbf{k} }$$.
$$\\[1pt]$$
$${\small (\textrm{a}).\hspace{0.8em}}$$ Show that $$OA = OB$$ and use a scalar product to calculate angle $$AOB$$ in degrees.
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The midpoint of $$AB$$ is $$M$$. The point $$P$$ on the line through $$O$$ and $$M$$ is such that $$PA : OA = \sqrt{7} : 1$$.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ Find the possible position vectors of $$P$$.
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$${\small 15.\enspace}$$ 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 8
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With respect to the origin $$O$$, the points $$A$$ and $$B$$ have position vectors given by
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$$\ {\small \overrightarrow{OA}}=\begin{pmatrix} 1 \\[1pt] 2 \\[1pt] 1 \end{pmatrix}$$ and $$\ {\small \overrightarrow{OB}}=\begin{pmatrix} 3 \\[1pt] 1 \\[1pt] -2 \end{pmatrix}$$. The line $$l$$ has equation $$\ {\small \vec{r}}=\begin{pmatrix} 2 \\[1pt] 3 \\[1pt] 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\[1pt] -2 \\[1pt] 1 \end{pmatrix}$$.
$$\\[1pt]$$
$${\small (\textrm{a}).\hspace{0.8em}}$$ Find the acute angle between the directions of $$AB$$ and $$l$$.
$$\\[1pt]$$
$${\small (\textrm{b}).\hspace{0.8em}}$$ Find the position vector of the point $$P$$ on $$l$$ such that $$AP = BP$$.
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$${\small 16.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 10
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The points $$A$$ and $$B$$ have position vectors $${\small \ 2\textbf{i} \ + \ \textbf{j} \ + \ \textbf{k} \ }$$ and $${\small \ \textbf{i} \ – \ 2\textbf{j} \ + \ 2\textbf{k} \ }$$ respectively.
$$\\[1pt]$$
The line $$l$$ has vector equation $${\small \ \vec{r} \ = \ \textbf{i} \ + \ 2\textbf{j} \ – \ 3\textbf{k} \ + \ \mu ( \textbf{i} \ – \ 3\textbf{j} \ – \ 2\textbf{k} ) }$$.
$$\\[1pt]$$
$${\small (\textrm{a}).\hspace{0.8em}}$$ Find a vector equation for the line through $$A$$ and $$B$$.
$$\\[1pt]$$
$${\small (\textrm{b}).\hspace{0.8em}}$$ Find the acute angle between the directions of $$AB$$ and $$l$$, giving your answer in degrees.
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$${\small (\textrm{c}).\hspace{0.8em}}$$ Show that the line through $$A$$ and $$B$$ does not intersect the line $$l$$.
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$${\small 17.\enspace}$$ 9709/12/O/N/19 – Paper 12 Oct Nov 2019 Pure Maths 1 No 7
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$$\\[1pt]$$
The base OABC and the upper surface DEFG are identical horizontal rectangles. The parallelograms OAED and CBFG both lie in vertical planes. Points P and Q are the mid-points of OD and GF respectively. Unit vectors i and j are parallel to $${\small \ \overrightarrow{OA} \ }$$ and $${\small \ \overrightarrow{OC} \ }$$ respectively and the unit vector k is vertically upwards. The position vectors of A, C and D are given by $${\small \ \overrightarrow{OA} \ = \ 6\textbf{i} }$$, $${\small \ \overrightarrow{OC} \ = \ 8\textbf{j} }$$ and $${\small \ \overrightarrow{OD} \ = \ 2\textbf{i} \ + \ 10\textbf{k} }$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}$$ Express each of the vectors $${\small \ \overrightarrow{PB} \ }$$ and $${\small \ \overrightarrow{PQ} \ }$$ in terms of i, j and k.
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$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}$$ Determine whether P is nearer to Q or to B.
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}$$ Use a scalar product to find angle BPQ.
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PRACTICE MORE WITH THESE QUESTIONS BELOW!
$${\small 1.\enspace}$$ The points A and B have position vectors, relative to the origin O, given by $${\small \overrightarrow{OA} \ = \ \textbf{i} \ + \ \textbf{j} \ + \ \textbf{k} }$$ and $${\small \overrightarrow{OB} \ = \ 2\textbf{i} \ + \ 3\ \textbf{k} }$$. The line l has vector equation $${\small \vec{r} = 2\textbf{i} \ – \ 2\textbf{j} \ – \ \textbf{k} + \mu(-\textbf{i} + 2\textbf{j} + \textbf{k} ) }$$.
$$\\[1pt]$$
$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Show that the line passing through A and B does not intersect l.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ Show that the length of the perpendicular from A to l is $${\small {\large\frac{1}{\sqrt{2}} }}$$.
$$\\[1pt]$$
$${\small 2. \enspace}$$ The point P has position vector $${\small 3\textbf{i} \ – \ 2\textbf{j} \ + \ \textbf{k} }$$. The line l has equation $${\small \vec{r} = 4\textbf{i} + 2\textbf{j} + 5\textbf{k} + \mu(\textbf{i} + 2\textbf{j} + 3\textbf{k} ) }$$.
$$\\[1pt]$$
$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Find the length of the perpendicular from P to l, giving your answer correct to 3 significant figures.
$$\\[1pt]$$
$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ Find the equation of the plane containing l and P, giving your answer in the form $${\small ax + by + cz = d}$$.
$$\\[1pt]$$
$${\small 3. \enspace}$$ Two lines l and m have equations $${\small \vec{r} = 2\textbf{i} \ – \textbf{j} + \textbf{k} + s(2\textbf{i} + 3\textbf{j} \ – \textbf{k} ) }$$ and $${\small \vec{r} = \textbf{i} + 3\textbf{j} + 4\textbf{k} + t(\textbf{i} + 2\textbf{j} + \textbf{k} ) }$$ respectively.
$$\\[1pt]$$
$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Show that the lines are skew.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ A plane p is parallel to the lines l and m. Find a vector that is normal to p.
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$${\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}$$ Given that p is equidistant from the lines l and m, find the equation of p. Give your answer in
the form $${\small ax + by + cz = d}$$.
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$${\small 4. \enspace}$$ The line l has equation $${\small \vec{r} = 4\textbf{i} + 3\textbf{j} \ – \textbf{k} + \mu(\textbf{i} + 2\textbf{j} \ – 2\textbf{k} ) }$$. The plane p has equation $${\small 2x \ – 3y \ – z = 4}$$.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Find the position vector of the point of intersection of l and p.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ Find the acute angle between l and p.
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$${\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}$$ A second plane q is parallel to l, perpendicular to p and contains the point with position vector
4jk. Find the equation of q, giving your answer in the form $${\small ax + by + cz = d}$$.
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$${\small 5. \enspace}$$ Two planes p and q have equations $${\small x + y + 3z = 8}$$ and $${\small 2x \ – 2y + z = 3}$$ respectively.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Calculate the acute angle between the planes p and q.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ The point A on the line of intersection of p and q has y-coordinate equal to 2. Find the equation of the plane which contains the point A and is perpendicular to both the planes p and q. Give your answer in the form $${\small ax + by + cz = d}$$.
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$${\small 6. \enspace}$$ The equations of two lines l and m are $${\small \vec{r} = 3\textbf{i} \ – \textbf{j} \ – 2\textbf{k} + \lambda(\ -\textbf{i} + \textbf{j} + 4\textbf{k} ) }$$ and $${\small \vec{r} = 4\textbf{i} + 4\textbf{j} \ – 3\textbf{k} + \mu(2\textbf{i} + \textbf{j} \ – 2\textbf{k} ) }$$ respectively.
$$\\[1pt]$$
$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Show that the lines do not intersect.
$$\\[1pt]$$
$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ ) Calculate the acute angle between the directions of the lines.
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$${\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}$$ Find the equation of the plane which passes through the point (3, −2, −1) and which is parallel to both l and m. Give your answer in the form $${\small ax + by + cz = d}$$.
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$${\small 7. \enspace}$$ The points A and B have position vectors, relative to the origin O, given by $${\small \overrightarrow{OA} \ = \ \textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} }$$ and $${\small \overrightarrow{OB} \ = \ 2\textbf{i} \ + \ \textbf{j} \ + \ 3\ \textbf{k} }$$. The line l has vector equation $${\small \vec{r} = (1 \ – \ 2t)\textbf{i} + (5 \ + \ t)\textbf{j} \ + (2 \ – \ t)\textbf{k} }$$.
$$\\[1pt]$$
$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Show that l does not intersect the line passing through A and B.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ ) The point P lies on l and is such that angle PAB is equal to $${\small 60^{\large{\circ}} }$$. Given that the position vector of P is $${\small (1 \ – \ 2t)\textbf{i} + (5 \ + \ t)\textbf{j} \ + (2 \ – \ t)\textbf{k} }$$, show that $${\small 3t^2 \ + \ 7t \ + \ 2 \ = \ 0 }$$. Hence find the only possible position vector of P.
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$${\small 8. \enspace}$$ The plane p has equation $${\small 3x + 2y + 4z = 13}$$. A second plane q is perpendicular to p and has equation $${\small ax + y + z = 4}$$, where a is a constant.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Find the value of a.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ The line with equation $${\small \vec{r} = \textbf{j} \ – \textbf{k} + \lambda( \textbf{i} + 2\textbf{j} + 2\textbf{k} ) }$$ meets the plane p at the point A and the plane q at the point B. Find the length of AB.
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$${\small 9. \enspace}$$ The lines $${\small l_{1}}$$ and $${\small l_{2}}$$ have equations $${\small \vec{r} = \textbf{i} + 2\textbf{j} + 3\textbf{k} + \lambda(a\textbf{i} + 4\textbf{j} + 3 \textbf{k} ) }$$ and $${\small \vec{r} = 4\textbf{i} \ – \textbf{k} + \mu(2\textbf{i} + 4\textbf{j} + b\textbf{k} ) }$$ respectively. Given that $${\small l_{1}}$$ and $${\small l_{2}}$$ are parallel:
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Write down the values of a and b.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ Find the shortest distance d between $${\small l_{1}}$$ and $${\small l_{2}}$$.
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$${\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}$$ Find a vector equation of the plane p containing $${\small l_{1}}$$ and $${\small l_{2}}$$.
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$${\small 10.\enspace}$$ Two lines $${\small l_{1}}$$ and $${\small l_{2}}$$ have equations $${\small \vec{r} = \ -8\textbf{i} + 12\textbf{j} + 16\textbf{k} + \lambda(\textbf{i} + 7\textbf{j} \ – 2 \textbf{k} ) }$$ and $${\small \vec{r} = 4\textbf{i} + 6 \textbf{j} \ – 8\textbf{k} + \mu(2\textbf{i} \ – \textbf{j} + 2\textbf{k} ) }$$ respectively.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Show that $${\small l_{1}}$$ and $${\small l_{2}}$$ are skew lines.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ The points P and Q lie on $${\small l_{1}}$$ and $${\small l_{2}}$$ respectively such that PQ is perpendicular to both $${\small l_{1}}$$ and $${\small l_{2}}$$. Show that $${\small \overrightarrow{PQ} \ = \ 16\textbf{i} \ – \ 8\textbf{j} \ – \ 20\textbf{k} }$$.
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As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
Complex Numbers
Complex Numbers
Just as we need negative integers to represent what positive integers can’t, we need imaginary numbers to represent what the real numbers can’t.
Complex numbers represent a number system that combines both real and imaginary numbers. While it may be tempting to ask what is the practical use of complex numbers, their importance is undeniable especially in the field of signal processing.
The usage of complex numbers is an essential part in designing “the poles and zeroes” or simply put, the stability of a system.
“The complex power” of an electrical signal is also an important design consideration in power lines and generators.
We frequently encountered imaginary numbers before, take for example, the solution of this quadratic equation:
$$\\[5pt] \hspace{2em} {x}^{2} \ + \ 1 \ = \ 0$$
$$\\[5pt] \hspace{2em} {x}^{2} \hspace{2.4em} = \ -1$$
$$\\[5pt] \hspace{2em} x \hspace{2.8em} = \ \pm \sqrt{-1}$$
$${\small \sqrt{-1} \ }$$ is an example of an imaginary number. Whenever you met an even n-th root of a negative number, you will need to know the complex number theory to simplify or solve it further.
We will start with the two basic forms to represent complex numbers, the cartesian form and the polar form.
Complex numbers in cartesian form can be shown as below:
$$\hspace{2em} {\large z \ = \ a \ + \ b \textrm{i} }$$
$$\\[5pt] {\small a \ = \ \textrm{Re(} z \textrm{)} }$$
: the real part of a complex number z
$$\\[5pt] {\small b \ = \ \textrm{Im(} z \textrm{)} }$$
: the imaginary part of a complex number z
$${\small \textrm{i} \ = \ \sqrt{-1} }$$
Argand diagram is usually used to show complex number graphically. The x-axis represents the “real part” and the y-axis represents the “imaginary part”.
The Argand diagram representation of the complex number z above can be seen as follows:
The cartesian form is especially handy in dealing with addition and subtraction of complex numbers.
The result of addition and subtraction of complex numbers can be found by adding or subtracting each of the real parts and each of the imaginary parts separately.
Example:
$$\hspace{2em} {z}_{1} \ = \ a \ + \ b \textrm{i}$$
$$\hspace{2em} {z}_{2} \ = \ c \ + \ d \textrm{i}$$
$$\hspace{2em} {z}_{1} \ + \ {z}_{2} \ = \ (a + c) \ + \ (b + d) \textrm{i}$$
$$\hspace{2em} {z}_{1} \ – \ {z}_{2} \ = \ (a \ – \ c) \ + \ (b \ – \ d) \textrm{i}$$
The second form of complex numbers is the polar form or the modulus-argument form.
$$\\[7pt] \hspace{2em} {\large z \ = \ r \cos \theta \ + \ \textrm{i} \ r \sin \theta }$$
$$\hspace{3.1em} {\large = \ r \ (\cos \theta \ + \ \textrm{i} \ \sin \theta) }$$
$$\\[5pt] {\small r \ = \ |z| }$$
: the modulus of a complex number z
$$\\[5pt] {\small \theta \ = \ \textrm{arg(} z \textrm{)} }$$
: the principal argument of a complex number z, $$\enspace {\small -\pi \lt \theta \le \pi }$$
The modulus $${\small r }$$ and argument $${\small \theta}$$ can be used to show the sets of points or regions of complex numbers in Argand diagram.
The Argand diagram of a complex number z in its polar form can be seen below:
Alternatively, the polar form can also be expressed in euler notation and phase-angle notation.
In euler notation,
$$\\[10pt] \hspace{2.6em} {\large {e}^{\textrm{i} \theta} \ = \ \cos \theta \ + \ \textrm{i} \ \sin \theta }$$
Since, $$\\[7pt] \hspace{1em} {\large z \ = \ r \ (\cos \theta \ + \ \textrm{i} \ \sin \theta) }$$
then $$\hspace{1.5em} {\large z \ = \ r \ {e}^{\textrm{i} \theta} }$$
In phase-angle notation,
$$\\[10pt] \hspace{2.6em} {\large \angle \theta \ = \ {e}^{\textrm{i} \theta} }$$
Since, $$\\[7pt] \hspace{1em} {\large z \ = \ r \ {e}^{\textrm{i} \theta} }$$
then $$\hspace{1.5em} {\large z \ = \ r \ \angle \theta }$$
The euler and phase-angle notation are merely a simpler way of writing the polar form.
The polar form, especially the euler notation and phase-angle notation are useful in multiplication and division of complex numbers.
The result of multiplication of complex numbers can be found by multiplying the moduli and adding the arguments.
The result of division of complex numbers can be found by dividing the moduli and subtracting the arguments.
Example:
$$\hspace{2em} {z}_{1} \ = \ {r}_{1} \ {e}^{\textrm{i} {\theta}_{1}}$$
$$\hspace{2em} {z}_{2} \ = \ {r}_{2} \ {e}^{\textrm{i} {\theta}_{2}}$$
$$\\[10pt] \hspace{2em} {z}_{1} \times {z}_{2} \ = \ ({r}_{1} \times {r}_{2}) \ {e}^{\textrm{i} ({\theta}_{1} \ + \ {\theta}_{2})}$$
$$\hspace{2em} {\large \frac{{z}_{1}}{{z}_{2}}} \hspace{1.9em} = \ {\large \frac{{r}_{1}}{{r}_{2}}} \ {e}^{\textrm{i} ({\theta}_{1} \ – \ {\theta}_{2})}$$
By comparing the complex numbers in both cartesian & polar form, they can be converted from one form to another.
To find the cartesian form from the polar form, $$\enspace z \ = \ a \ + \ b \textrm{i}$$
$$\\[5pt] \enspace a \ = \ r \ \cos \theta$$
$$\enspace b \ = \ r \ \sin \theta$$
To find the polar form from the cartesian form, $$\enspace z \ = \ r \ (\cos \theta \ + \ \textrm{i} \sin \theta )$$
$$\\[7pt] \enspace r \ = \ \sqrt{{a}^{2} \ + \ {b}^{2}}$$
$$\enspace \theta \ = \ {\tan}^{-1} \Big({\large \frac{b}{a} } \Big)$$
The complex conjugate of a complex number has the same real part and opposite sign for its imaginary part. That is, to say, if $$z \ = \ a \ + \ b \textrm{i} \$$ then the complex conjugate is $${z}^{*} \ = \ a \ – \ b \textrm{i}$$.
One such example of usage is when dealing with polynomial equations with real coefficients (non-complex number coefficients).
If a polynomial equation has a complex number z as one of its roots, then the complex conjugate $${\small {z}^{*} }$$ is also a root of the polynomial equation.
To find the n-th power of a complex number, De Moivre’s Theorem can be used. It can be shown as follows:
If $$\enspace z \ = \ r \ {e}^{\textrm{i} \theta}$$
then $$\enspace {z}^{n} \ = \ {r}^{n} \ {e}^{ \textrm{i} (n\theta) }$$
I have put together some of the questions I received in the comment section below. You can try these questions also to further your understanding on this topic.
To check your answer, you can look through the solutions that I have posted either in Youtube videos or Instagram posts.
You can subscribe, like or follow my youtube channel and IG account. I will keep updating my IG daily post, preferably.
Furthermore, you can find some examples and more practices below! =).
Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
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EXAMPLE:
$${\small 1.\enspace}$$ 9709/03/SP/17 – Specimen Paper 2017 Pure Maths 3 No 9
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The complex number $${\small \ 3 \ − \ \textrm{i} \ }$$ is denoted by u. Its complex conjugate is denoted by u*.
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$${\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}$$ On an Argand diagram with origin O, show the points A, B and C representing the complex numbers u, u* and u*u respectively. What type of quadrilateral is OABC?
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$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}$$ Showing your working and without using a calculator, express $${\small \ {\large\frac{u}{u*}} }$$ in the form of x + iy, where x and y are real.
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}$$ By considering the argument of $${\small \ {\large\frac{u}{u*}} }$$, prove that
$${\small\hspace{3em} \ {\tan}^{-1} \big(\frac{3}{4}\big) \ = \ 2 {\tan}^{-1} \big(\frac{1}{3}\big) }$$
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$${\small 2.\enspace}$$ 9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 6
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The complex number $${\small \ 1 \ + \ 3\textrm{i} \ }$$ and $${\small \ 4 \ + \ 2\textrm{i} \ }$$ are denoted by u and v respectively.
$$\\[1pt]$$
$${\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}$$ Find $${\small \ {\large \frac{u}{v}} \ }$$ in the form of x + iy, where x and y are real.
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$${\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}$$ State the argument of $${\small \ \large { \frac{u}{v} }}$$.
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In an Argand diagram, with origin O, the points A, B and C represent the complex numbers u, v and u – v respectively.
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$${\small\hspace{1.2em}(\textrm{c}).\hspace{0.8em}}$$ State fully the geometrical relationship between OC and BA.
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$${\small\hspace{1.2em}(\textrm{d}).\hspace{0.8em}}$$ Show that angle $${\small AOB = \frac{1}{4} \pi \small }$$ radians.
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$${\small 3.\enspace}$$ The complex number w is given by $${\small w = {\large -\frac {1}{2}} \ + \ {\large \textrm{i}\frac {\sqrt{3}}{2}} }$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}$$ Find the modulus and argument of w.
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$${\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}$$ The complex number z has a modulus R and argument $${\small \theta}$$, where $$\ {\small – {\large \frac{1}{3}}\pi \lt \theta \lt {\large \frac{1}{3}}\pi }$$. State the modulus and argument of wz and the modulus and argument of $${\small {\large \frac{z}{w}} }$$.
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$${\small\hspace{1.2em}\left(iii\right).\hspace{0.8em}}$$ Hence explain why, in Argand diagram, the points representing z, wz and $${\small {\large \frac{z}{w}} }$$ are the vertices of an equilateral triangle.
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$${\small\hspace{1.2em}\left(iv\right).\hspace{0.8em}}$$ In an Argand diagram, the vertices of an equilateral triangle lie on a circle with centre at the origin. One of the vertices represents the complex number 4 + 2i. Find the complex numbers represented by the other two vertices. Give your answers in the form x + iy, where x and y are real and exact.
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$${\small 4.\enspace}$$ The complex number -2 + i is denoted by u.
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$${\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}$$ Given that u is a root of the equation $${\small {x}^{3} \ – \ 11x \ – \ k \ = \ 0}$$, where k is real, find the value of k.
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$${\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}$$ Write down the other complex root of this equation.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(iii\right).\hspace{0.8em}}$$ Find the modulus and argument of u.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(iv\right).\hspace{0.8em}}$$ Sketch an Argand diagram showing the point representing u. Shade the region whose points represent the complex numbers z satisfying both the inequalities
$${\small \quad |z| \ \lt \ |z \ – \ 2| \ }$$ and $$\ {\small 0 \ \lt \ \textrm{arg}(z \ – \ u) \ \lt \ {\large \frac{1}{4} }\pi }$$.
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$${\small 5.\enspace}$$ The complex number w is defined by $${\small w = -1 + \textrm{i} }$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}$$ Find the modulus and argument of $${\small {w}^{2} }$$ and $${\small {w}^{3}}$$, showing your working.
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$${\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}$$ The points in an Argand diagram representing $${\small w }$$ and $${\small {w}^{2}}$$ are the ends of a diameter of a circle. Find the equation of the circle, giving your answer in the form $${\small |z \ – \ (a \ + \ b\textrm{i})| \ = \ k}$$.
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$${\small 6.\enspace (a).\hspace{0.8em} }$$ Without using a calculator, solve the equation
$$\\[1pt]$$
$${\small \quad 3w \ + \ 2\textrm{i}{w}^{*} \ = \ 17 \ + \ 8\textrm{i}, }$$
$$\\[1pt]$$
where $${\small {w}^{*} }$$ denotes the complex conjugate of $${\small w }$$. Give your answer in the form $${\small (a \ + \ b\textrm{i}). }$$
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$${\small \hspace{1.3em} (b).\hspace{0.8em} }$$ In an Argand diagram, the loci
$$\\[1pt]$$
$${\small \quad \textrm{arg}(z \ – \ 2\textrm{i}) \ = \ {\large \frac{1}{6} }\pi \ }$$ and $$\ {\small |z \ – \ 3| \ = \ |z \ – \ 3\textrm{i}| }$$
$$\\[1pt]$$
intersect at the point P. Express the complex number represented by P in the form of $${\small r {e}^{\textrm{i}\theta}}$$, giving the exact value of $${\small \theta }$$ and the value of r correct to 3 significant figures.
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$${\small 7.\enspace}$$ The complex number z is defined by $${\small \ z \ = \ {\large \frac{9\sqrt{3} \ + \ 9\textrm{i}}{\sqrt{3} \ – \ \textrm{i}} } }$$. Find, showing all your working,
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}$$ an expression for z in the form $${\small r {e}^{\textrm{i}\theta}}$$, where $${\small r \ \gt \ 0 \ }$$ and $$\ {\small -\pi \lt \theta \le \pi }$$,
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}$$ the two square roots of z, giving your answers in the form $${\small r {e}^{\textrm{i}\theta}}$$, where $${\small r \ \gt \ 0 \ }$$ and $$\ {\small -\pi \lt \theta \le \pi }$$.
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$${\small 8.\enspace (a).\hspace{0.8em} }$$ Showing all working and without using a calculator, solve the equation
$$\\[1pt]$$
$$\quad {\small (1 \ + \ \textrm{i}){z}^{2} \ – \ (4 \ + \ 3\textrm{i})z \ + \ 5 \ + \ \textrm{i} \ = \ 0}$$.
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Give your answers in the form x + iy, where x and y are real.
$$\\[1pt]$$
$${\small \hspace{1.3em} (b).\hspace{0.8em} }$$ The complex number u is given by
$$\quad {\small u \ = \ -1 \ – \ \textrm{i} }$$.
$$\\[1pt]$$
On a sketch of an Argand diagram show the point representing u. Shade the region whose points represent complex numbers satisfying the inequalities
$${\small \ |z| \ \lt \ |z \ – \ 2\textrm{i}| }$$ and $${\small {\large \frac{1}{4} }\pi \ \lt \ \textrm{arg}(z \ – \ u) \ \lt \ {\large \frac{1}{2} }\pi }$$.
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$${\small 9.\enspace}$$ 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 9(a), (b)
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ The complex numbers $$u$$ and $$w$$ are such that
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$$\hspace{3em} u \ – \ w = 2i \enspace$$ and $$\enspace uw = 6$$.
$$\\[1pt]$$
$$\hspace{1.2em}$$ Find $$u$$ and $$w$$, giving your answers in the form $$x \ + \ iy$$, where $$x$$ and $$y$$ are real and exact.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ On a sketch of an Argand diagram, shade the region whose points represent complex numbers $$z$$ satisfying the inequalities
$$\\[1pt]$$
$$\hspace{1.2em} {\small |z \ – \ 2 \ – \ 2i| \leq 2 }$$, $$\hspace{0.5em}{\small 0 \leq z \leq {\large\frac{1}{4}\pi} } \enspace$$ and $$\enspace {\small \mathrm{Re} \ z \leq 3 }$$.
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$${\small 10.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 8(a), (b)
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Solve the equation $$(1+2i)w + i{w}^{*} = 3 + 5i$$. Give your answer in the form $$x \ + \ iy$$, where $$x$$ and $$y$$ are real.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(b\right)\hspace{0.4em} (i). \hspace{0.4em} }$$ On a sketch of an Argand diagram, shade the region whose points represent complex numbers $$z$$ satisfying the inequalities
$$\\[1pt]$$
$$\hspace{2.2em} {\small |z \ – \ 2 \ – \ 2i| \leq 1 } \enspace$$ and $$\enspace {\small \mathrm{arg} \ (z \ – \ 4i) \geq -\frac{1}{4}\pi}$$.
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$${\small\hspace{2.6em} (ii). \hspace{0.3em} }$$ Find the least value of $$\mathrm{Im} \ z$$ for points in this region, giving your answer in an exact form.
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$${\small 11.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 10(a), (b)
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ The complex number $$u$$ is defined by $$u \ = \ {\large \frac{ 3i }{ a \ + \ 2i} }$$, where $$a$$ is real.
$$\\[1pt]$$
$${\small\hspace{2.6em} (i). \hspace{0.5em} }$$ Express $$u$$ in the Cartesian form $$x \ + \ \mathrm{i}y$$, where $$x$$ and $$y$$ are in terms of $$a$$.
$$\\[1pt]$$
$${\small\hspace{2.6em} (ii). \hspace{0.3em} }$$ Find the exact value of $$a$$ for which $$\mathrm{arg} \ {u}^{*} = \frac{1}{3}\pi$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(b\right)\hspace{0.4em} (i). \hspace{0.4em} }$$ On a sketch of an Argand diagram, shade the region whose points represent complex numbers $$z$$ satisfying the inequalities
$$\\[1pt]$$
$$\hspace{3em} {\small |z \ – \ 2i| \leq |z \ – 1 \ – \ i| } \enspace$$ and $$\enspace {\small |z \ – \ 2 \ – \ i| \leq 2 }$$.
$$\\[1pt]$$
$${\small\hspace{2.6em} (ii). \hspace{0.3em} }$$ Calculate the least value of $$\mathrm{arg} \ z$$ for points in this region.
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$${\small 12.\enspace}$$ 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 8(a) – (d)
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The complex numbers $$u$$ and $$v$$ are defined by $$u = -4 + 2i \$$ and $$\ v = 3 + i$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find $$\frac{u}{v}$$ in the form $$x \ + \ iy$$, where $$x$$ and $$y$$ are real.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Hence express $$\frac{u}{v}$$ in the form $$r { \mathrm{e} }^{ i\theta }$$, where $$r$$ and $$\theta$$ are exact.
$$\\[1pt]$$
In an Argand diagram, with origin $$O$$, the points $$A$$, $$B$$ and $$C$$ represent the complex numbers $$u$$, $$v$$ and $$2u + v$$ respectively.
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$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}$$ State fully the geometrical relationship between $$OA$$ and $$BC$$.
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$${\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}$$ Prove that angle $$AOB = \frac{3}{4}\pi$$.
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$${\small 13.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 10(a), (b)
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Verify that $$−1 + \sqrt{2}i$$ is a root of the equation $$z^4 + 3z^2 + 2z + 12 = 0$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Find the other roots of this equation.
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$${\small 14.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 5
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The complex number $$u$$ is given by $$u = 10 − 4\sqrt{6}i$$.
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Find the two square roots of u, giving your answers in the form $$a + ib$$, where $$a$$ and $$b$$ are real and exact.
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$${\small 15.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 2
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On a sketch of an Argand diagram, shade the region whose points represent complex numbers $$z$$ satisfying the inequalities $${\small |z \ + \ 1 \ – \ \textrm{i}| \leq 1 \ }$$ and $${\small \ \mathrm{arg} \ (z \ – \ 1) \ \leq \ {\large \frac{3}{4}} \pi }$$.
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$${\small 16.\enspace}$$ 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 5(a) – (c)
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Solve the equation $$z^2 \ – \ 2p\mathrm{i}z \ – \ q = 0$$, where $$p$$ and $$q$$ are real constants.
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In an Argand diagram with origin $$O$$, the roots of this equation are represented by the distinct points $$A$$ and $$B$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Given that $$A$$ and $$B$$ lie on the imaginary axis, find a relation between $$p$$ and $$q$$.
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$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}$$ Given instead that triangle $$OAB$$ is equilateral, express $$q$$ in terms of $$p$$.
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$${\small 17.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 2
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On a sketch of an Argand diagram, shade the region whose points represent complex numbers $$z$$ satisfying the inequalities $${\small | z \ + \ 2 \ – \ 3 \textrm{i}| \ \le \ 2 \ }$$ and $$\ {\small \textrm{arg} \ z \ \le \ {\large \frac{3}{4}} \pi }$$.
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$${\small 18.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 6
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Find the complex numbers $$w$$ which satisfy the equation $${\small {w}^{2} \ + \ 2 \mathrm{i} {w}^{*} \ = \ 1}$$ and are such that $${\small \mathrm{Re} \ w \ \le \ 0 }$$.
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Give your answers in the form $$x \ + \ \mathrm{i}y$$, where $$x$$ and $$y$$ are real.
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PRACTICE MAKES PERFECT!
$${\small 1.\enspace(a).\hspace{0.8em}}$$ Showing all necessary working, express the complex number $${\small {\large \frac{2 \ + \ 3\textrm{i}}{1 \ – \ 2\textrm{i}} } }$$ in the form $${\small r {e}^{\textrm{i}\theta}}$$, where $${\small r \ \gt \ 0 \ }$$ and $$\ {\small -\pi \lt \theta \le \pi }$$. Give the values of r and $${\small \theta}$$ correct to 3 significant figures.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ On an Argand diagram sketch the locus of points representing complex numbers z satisfying the equation $${\small |z \ – \ 3 \ + \ 2\textrm{i}| \ = \ 1 }$$. Find the least value of |z| for points on this locus, giving your answer in an exact form.
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$${\small 2.\enspace(a) \ (i).\hspace{0.5em}}$$ Without using a calculator, express the complex number $${\small {\large \frac{2 \ + \ 6\textrm{i}}{1 \ – \ 2\textrm{i}} } }$$ in the form x + iy, where x and y are real.
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$${\small\hspace{2.4em}\left(ii\right).\hspace{0.5em}}$$ Hence, without using a calculator, express $${\small {\large \frac{2 \ + \ 6\textrm{i}}{1 \ – \ 2\textrm{i}} } }$$ in the form $${\small r(\cos \theta \ + \ \textrm{i} \sin \theta)}$$, where $${\small r \ \gt \ 0 \ }$$ and $$\ {\small -\pi \lt \theta \le \pi }$$, giving the exact values of r and $${\small \theta}$$.
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$${\small\hspace{1.5em}\left(b\right).\hspace{0.8em}}$$ On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying both the inequalities $${\small |z \ – \ 3 \textrm{i}| \ \le \ 1 }$$ and Re $${\small z \ \le \ 0}$$, where Re z denotes the real part of z. Find the greatest value of arg z for points in this region, giving your answer in radians correct to 2 decimal places.
$${\small 3. \enspace}$$ The complex number $${\small 1 \ − \ (\sqrt{3})\textrm{i} }$$ is denoted by u.
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$${\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}$$ Find the modulus and argument of u.
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$${\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}$$ Show that $${\small \ {u}^{3} \ + \ 8 \ = \ 0}$$.
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$${\small\hspace{1.2em}\left(iii\right).\hspace{0.8em}}$$ On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying both the inequalities $${\small |z \ – \ u| \ \le \ 2 }$$ and Re $${\small z \ \ge \ 2}$$, where Re z denotes the real part of z.
$${\small 4. \enspace}$$ The polynomial $${\small {z}^{4} \ + \ 3{z}^{2} \ + \ 6z \ + \ 10 }$$ is denoted by p(z). The complex number $${\small -1 \ + \ \textrm{i} }$$ is denoted by u.
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$${\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}$$ Showing all your working, verify that u is a root of the equation p(z) $${\small = \ 0}$$.
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$${\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}$$ Find the other three roots of the equation p(z) $${\small = \ 0}$$.
$${\small 5.\enspace(a).\hspace{0.8em}}$$ Showing all necessary working, solve the equation $${\small \textrm{i}{z}^{2} \ + \ 2z \ − \ 3\textrm{i} \ = \ 0 }$$, giving your answers in the form x + iy, where x and y are real and exact.
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$${\small \quad(b) \ (i).\hspace{0.8em}}$$ On a sketch of an Argand diagram, show the locus representing complex numbers satisfying the equation $${\small |z| \ = \ |z \ − \ 4 \ − \ 3\textrm{i}| }$$.
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$${\small \hspace{2em} (ii).\hspace{0.7em}}$$ Find the complex number represented by the point on the locus where |z| is least. Find the modulus and argument of this complex number, giving the argument correct to 2 decimal places.
$${\small 6. \enspace}$$ Sketch, on an Argand diagram, the locus of the points representing the complex number z such that $${\small |\textrm{i}z \ – \ 2| \ = \ |2 \ – \ z|}$$. Hence find the least value of $${\small |z \ + \ 2|}$$.
$${\small 7. \enspace}$$ Solve the equation $${\small \textrm{i}{z}^{4} \ = \ -81 }$$, expressing the roots in the form $${\small r {e}^{\textrm{i}\theta}}$$, where $${\small r \ \gt \ 0 \ }$$ and $$\ {\small -\pi \lt \theta \le \pi }$$.
$${\small 8. \enspace}$$ In an Argand diagram, the point P represents the complex number z such that:
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$${\small |z \ – \ 2 \ + \ 4\textrm{i}| \le 4 \ }$$ and $${\small \ {\large \frac{\pi}{4} } \le \textrm{arg}(z \ + \ 2\textrm{i}) \lt 0 }$$.
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$${\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}$$ Sketch the locus of P.
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$${\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}$$ Hence, find the exact range of values of $${\small |z \ + \ 2|}$$.
$${\small 9. \enspace}$$ In an Argand diagram, the point P represents the complex number z. Draw on a single clearly labelled diagram to show the locus of P when z satisfies
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$${\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}$$ $${\small |z \ – \ 3 \ – \ 4\textrm{i}| = 5 \ }$$
$${\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}$$ $${\small |z \ – \ 2 \ + \ 3\textrm{i}| = |z \ – \ 10 \ – \ 3\textrm{i}| \ }$$
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Hence find the greatest and least possible values of |z| when z satisfies both
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$${\small |z – 3 – 4\textrm{i}| \le 5 }$$ and $${\small |z – 2 + 3\textrm{i}| \ge |z – 10 – 3\textrm{i}| }$$
$${\small 10.\enspace}$$ Given that $${\small {z}^{*} \ = \ \frac{{(2 \ – \ 2\textrm{i})}^{3}}{{(-1 \ + \ \sqrt{3}\textrm{i})}^{4}}}$$, find the exact value of |z| and arg(z). Hence, state the smallest positive integer n such that $${\small {z}^{n}}$$ is a purely real number.
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
Binomial Expansion and Binomial Series
Binomial Expansion and Binomial Series
In algebra, we all have learnt the following basic algebraic expansion:
$$\hspace{3em} {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$$.
We can keep multiplying the expression $${ \small (a + b) }$$ by itself to find the expression for higher index value. For example:
$$\\[12pt] {(a + b)}^{3} \ = \ {(a + b)}^{2} \times (a + b)$$
$$\\[12pt] \hspace{1.5em} \ = \ ({a}^{2} + 2ab + {b}^{2}) \times (a + b)$$
$$\\[12pt] \hspace{1.5em} \ = \ {a}^{3} + {a}^{2}b + 2{a}^2b + 2a{b}^2 + a{b}^2 + {b}^{3}$$
$$\hspace{1.5em} \ = \ {a}^{3} + 3{a}^{2}b + 3a{b}^2 + {b}^{3}$$
Of course this is a really tedious way for a very large index/power/exponent number.
Instead of doing it manually, we could use a formula called the Binomial Theorem which is shown below:
$${(a + b)}^{n} \ = \ \displaystyle \sum_{k=0}^{n} \binom{n}{k} {a}^{n \ – \ k} \ {b}^{k}$$
with $$\displaystyle \binom{n}{k} \ = \ \frac{n!}{k!(n-k)!}$$
$$\hspace{4.4em} = \ {\large \frac{n (n \ – \ 1) (n \ – \ 2) … (n \ – \ k \ + \ 1)}{k(k \ – \ 1)(k \ – \ 2) \ … \ 1} }$$
for $$k \geq 1$$ and $$\displaystyle \binom{n}{0} \ = \ 1$$
$$\\[10pt]$$ Example:
$$\\[12pt] {(a + b)}^{10}$$
$$\\[15pt] = \binom{10}{0} {a}^{10}{b}^{0} + \binom{10}{1} {a}^{9}{b}^{1}+…+ \binom{10}{10} {a}^{0}{b}^{10}$$
$$\\[15pt] = {a}^{10} + 10 {a}^{9}b+ 45 {a}^{8}{b}^{2} + … + 10{a}{b}^{9} + {b}^{10}$$
$$\\[10pt]$$ Note that,
$$\\[20pt]\quad \ {\displaystyle \binom{10}{0} } \ = \ 1$$
$$\\[20pt]\quad \ {\displaystyle \binom{10}{1} } \ = \ {\large \frac{ 10 }{ 1 } } \ = \ 10$$
$$\\[20pt]\quad \ {\displaystyle \binom{10}{2} } \ = \ {\large \frac{ 10 \ \times \ 9 }{ 2 \ \times \ 1 } } \ = \ 45$$
$$\\[20pt]\quad \ {\displaystyle \binom{10}{3} } \ = \ {\large \frac{ 10 \ \times \ 9 \ \times \ 8 }{ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 120$$
$$\\[20pt]\quad \ {\displaystyle \binom{10}{4} } \ = \ {\large \frac{ 10 \ \times \ 9 \ \times \ 8 \ \times \ 7 }{4 \ \times \ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 210$$
$$\\[20pt]\quad \ {\displaystyle \binom{10}{5} } \ = \ {\large \frac{10 \ \times \ 9 \ \times \ 8 \ \times \ 7 \ \times \ 6 }{5 \ \times \ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 252$$
, etc.
$$\binom{n}{k}$$ is read as “n choose k” or sometimes referred to as the binomial coefficients.
Notice that this binomial expansion has a finite number of terms with the k values take the non-negative numbers from 0, 1, 2, … , n.
Then the next question would be: Can we still use the binomial theorem for the expansion with negative number or fractional number for the index value?
Thankfully, Sir Isaac Newton has shown that the binomial theorem can be generalized to take in any numbers for the index value including the negative and fractional numbers as long as it is within a convergence rule.
Using this result, we have the Binomial Series which can be expressed as follows:
$$\\[25pt]{(1 + x)}^{n} = \displaystyle \sum_{k=0}^{\infty} \binom{n}{k} {x}^{k}$$
$$\\[10pt]{\small = \ 1 + nx + \frac{(n)(n-1)}{2!}{x}^{2} + \frac{(n)(n-1)(n-2)}{3!}{x}^{3} + … }$$
with $${\small k }$$ is any real number and $${\small -1 \lt x \lt 1 }$$.
$$\\[10pt]$$ Example:
$$\\[12pt] { {\large \frac{1}{1 + x}} \ = \ (1 + x)}^{-1}$$
$$\\[15pt]{\small = \ 1 + (-1)x + \frac{(-1)(-2)}{2!}{x}^{2} + \frac{(-1)(-2)(-3)}{3!}{x}^{3} + … }$$
$$\\[15pt]{\small = \ 1 \ – \ x \ + \ {x}^{2} \ – \ {x}^{3} \ + \ … \ – \ … }$$
$$\\[15pt]$$ The expansion is valid when $${\small -1 \lt x \lt 1 }$$.
$$\\[10pt]$$ Another example:
$$\\[12pt] { \frac{1}{\sqrt{(1 + x)}} \ = \ (1 + x)}^{\frac{1}{2}}$$
$$\\[15pt]{\small = \ 1 + \frac{1}{2}x + \frac{(\frac{1}{2})(- \frac{1}{2})}{2!}{x}^{2} + \frac{(\frac{1}{2})(- \frac{1}{2})( – \frac{3}{2})}{3!}{x}^{3} + … }$$
$$\\[15pt]{ = \ 1 + \frac{x}{2} – \frac{{x}^{2}}{8} + \frac{{x}^{3}}{16} \ – \ … + … }$$
$$\\[15pt]$$ The expansion is valid when $${\small -1 \lt x \lt 1 }$$.
Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
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EXAMPLE:
$${\small 1.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{7{x}^{2} \ – \ 15x \ + \ 8}{(1 \ – \ 2x){(2 \ – \ x)}^{2} }} }$$
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Express $${\small f(x) }$$ in partial fractions and hence obtain the expansion of $${\small f(x) }$$ in ascending powers of x, up to and including the term in $${\small {x}^{2}}$$.
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$${\small 2.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ x \ – \ 4{x}^{2} }{(3 \ – \ x)(2 \ + \ {x}^{2}) }} }$$
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Express $${\small f(x) }$$ in partial fractions and hence obtain the expansion of $${\small f(x) }$$ in ascending powers of x, up to and including the term in $${\small {x}^{3}}$$.
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$${\small 3.\enspace}$$ Find the value of $${\small {(2 \ + \ x)}^{6} \ – \ {(2 \ – \ x)}^{6} }$$ in ascending powers of x, up to and including the term in $${\small {x}^{3}}$$. Hence, find the value of $${\small {(1.99)}^{6} \ – \ {(2.01)}^{6} }$$.
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$${\small 4.\enspace}$$ Find the last four terms in the expansion in ascending powers of x of $$(2x – {x}^{2})^{10}$$.
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$${\small 5.\enspace}$$ Find the value of $${\small \frac{a}{b} }$$ in $${\small {(a + bx)}^{12}}$$ given that the coefficient of $${\small {x}^{2} }$$ is 11 times the coefficient of $${\small x }$$.
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$${\small 6.\enspace}$$ Find the value of a, b and n in the following expansion:
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$${\small {(a + bx)}^{n} \ = \ … + {\large\frac{20}{3}}{x}^{2} + 20 {x}^{3} + … }$$.
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given that $${\small {\large \frac{a}{b}} \ = \ {\large \frac{4}{9}} }$$.
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$${\small 7.\enspace}$$ Consider the binomial expansion of $${\small {\large \frac{1}{\sqrt{4-x}} } }$$.
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$${\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}$$ Write down the first 4 terms.
$${\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}$$ State the interval of convergence for the complete expression.
$${\small\hspace{1.2em}(\textrm{c}).\hspace{0.8em}}$$ Use the expansion to estimate $${\small {\large \frac{1}{\sqrt{3.6}} } }$$. Check your answer by direct calculation.
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$${\small 8.\hspace{0.4em}(\textrm{i}).\hspace{0.5em}}$$ Find the first 3 terms in the expansion of $${\small {(2x \ – \ {\large \frac{1}{16x}})}^{8} }$$ in descending powers of x.
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$${\small\hspace{1em}(\textrm{ii}).\hspace{0.3em}}$$ Hence find the coefficient of $${\small {x}^{4}}$$ in the expansion of $${\small {(2x \ – \ {\large \frac{1}{16x}})}^{8} {( {\large \frac{1}{{x}^{2}}} \ + \ 1 )}^{2} }$$.
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$${\small 9.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 2(a) and (b)
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$${\small \hspace{1em}(\textrm{a}).\hspace{0.5em}}$$ Expand $${(2 − 3x)}^{−2}$$ in ascending powers of $$x$$, up to and including the term in $$x^2$$, simplifying the coefficients.
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$${\small\hspace{1em}(\textrm{b}).\hspace{0.5em}}$$ State the set of values of $$x$$ for which the expansion is valid.
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$${\small 10.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 1
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Expand $${(1 + 3x)}^{\frac{2}{3}}$$ in ascending powers of $$x$$, up to and including the term in $$x^3$$, simplifying the coefficients.
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$${\small 11.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 9(a) and (b)
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Let $${\small f(x) \ = \ {\large \frac{ 14 \ – \ 3x \ + \ 2x^{2} }{(2 \ + \ x)( 3 \ + \ x^{2} ) }} }$$.
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$${\small \hspace{1em}(\textrm{a}).\hspace{0.5em}}$$ Express $${\small f(x) }$$ in partial fractions.
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$${\small \hspace{1em}(\textrm{b}).\hspace{0.5em}}$$ Hence obtain the expansion of $$f(x)$$ in ascending powers of $$x$$, up to and including the term in $$x^2$$.
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$${\small 12.\enspace}$$ 9709/13/O/N/21 – Paper 13 Nov 2021 Pure Maths 1 No 2
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$${\small \hspace{1em}(\textrm{a}).\hspace{0.5em}}$$ Find the first three terms, in ascending powers of $$x$$, in the expansion of $${(1 \ + \ ax)}^{6}$$.
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$${\small \hspace{1em}(\textrm{b}).\hspace{0.5em}}$$ Given that the coefficient of $$x^2$$ in the expansion of $$(1 \ – \ 3x){(1 \ + \ ax)}^{6}$$ is $$−3$$, find the possible values of the constant $$a$$.
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$${\small 13.\enspace}$$ 9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 4
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The coefficient of $$x$$ in the expansion of $$\ {(4x \ + \ {\large\frac{10}{x}} )}^{3} \$$ is $$p$$. The coefficient of $${\large\frac{1}{x} }$$ in the expansion of $$\ {(2x \ + \ {\large\frac{k}{{x}^{2}} })}^{5} \$$ is $$q$$.
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Given that $$\ p \ = \ 6q$$, find the possible values of $$k$$.
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$${\small 14.\enspace}$$ 9709/12/O/N/19 – Paper 12 November 2019 Pure Maths 1 No 1
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The coefficient of $$x^2$$ in the expansion of $$\ (4 \ + \ ax){(1 \ + \ {\large\frac{x}{2}})}^{6} \$$ is $$3$$.
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Find the value of the constant $$a$$.
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PRACTICE MORE WITH THESE QUESTIONS BELOW!
$${\small 1.\enspace}$$ Expand $$\frac{1}{\sqrt[3]{(1 \ + \ 6x)}}$$ in ascending powers of x up to and including the term in $${\small x^3 }$$, simplifying the coefficients.
$${\small 2. \enspace}$$ Expand $$\frac{4}{\sqrt{(4 \ – \ 3x)}}$$ in ascending powers of x up to and including the term in $${\small x^2 }$$, simplifying the coefficients.
$${\small 3. \enspace}$$ Let $${\small f(x) \ = \ {\large \frac{3{x}^{2} \ + \ x \ + \ 6}{(x \ + \ 2)({x}^{2} \ + \ 4) }} }$$
(i). Express $${\small f(x) }$$ in partial fractions.
(ii). Hence obtain the expansion of $${\small f(x) }$$ in ascending powers of x, up to and including the term in $${\small {x}^{2}}$$.
$${\small 4. \enspace}$$ Express $${\small {(3 \ + \ 4x)}^{\frac{1}{2}} }$$ as a series of descending powers of x up to and including the third non-zero coefficient. State the set of values of x for which the series expansion is valid.
$${\small 5. \enspace}$$ (i). Given that the first three terms in the expansion of $${\small {(1 \ + \ ax)}^{b}}$$ in ascending powers of x are $${\small 1 \ + \ x \ + \ \frac{3}{2}{x}^{2}}$$, where a and b are constants, find the values of a and b.
(ii). Hence, with the values of a and b found in (i), expand $$\frac{ {(1 \ + \ ax)}^{b} }{1 \ – \ x}$$ as a series in ascending powers of x up to and including the term in $${\small x^2 }$$.
$${\small 6. \enspace}$$ Expand $${\small {(1-2p)}^{8} }$$ up to and including the term in $${\small p^3 }$$. Hence, find the first four terms in the expansion in ascending powers of x of $${\small {(1 \ – \ 4x \ + \ \frac{2}{x})}^{8} }$$.
$${\small 7. \enspace}$$ Find the coefficient of $${\small {x}^{2}}$$ in the expansion of $${\small {(1 \ + \ 3x)}^{2}{(1 \ – \ 3x)}^{10} }$$.
$${\small 8. \enspace}$$ Find the coefficient of $${ \small x }$$ in the expansion of $${\small {(3x \ – \ \frac{2}{x})}^{9} }$$.
$${\small 9. \enspace}$$ Find the value of a and b in the following expansion:
$${\small \hspace{1.5em} {(a + bx)}^{9} \ = \ … + 672{x}^{3} + 252 {x}^{4} + … }$$
$${\small 10.\enspace}$$ Find the value of a and b in the expansion of $${\small {(1 \ + \ ax)}^{5}{(1 \ – \ bx)}^{7} }$$ if the coefficients of $${ \small x }$$ and $${\small {x}^{2}}$$ are 3 and -9 respectively.
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
Partial Fractions
Partial Fractions
In solving algebra related problems and questions, we may sometimes deal with rational functions. A rational function is basically an algebraic polynomial fraction, in which we have polynomials on both the numerator and denominator.
Partial fractions is one of the simplest and most effective method in solving algebra related problems regarding rational functions.
In partial fractions, we separate the polynomials in our rational function into simpler form of polynomials.
Some of the applications of partial fractions include the solving of integration problems with rational functions, the binomial expansion and also the arithmetic series and sequences.
There are a few basic forms we need to memorize in partial fractions:
1.$$\enspace$$ The linear form:
$$\hspace{6em} {\small (ax + b)}$$
Example:
$${\large\frac{3x \ + \ 5}{(x \ + \ 1)(2x \ + \ 7)} \ \equiv \ \frac{A}{(x \ + \ 1)} + \frac{B}{(2x \ + \ 7)} }$$
2.$$\enspace$$ The quadratic form of a linear factor:
$$\hspace{6em} {\small (cx \ + \ d)^{2}}$$
Example:
$$\frac{3x + 5}{(x + 1){(2x + 7)}^{2}} \equiv \frac{A}{(x + 1)} + {\small\boxed{\frac{B}{(2x + 7)} + \frac{C}{{(2x + 7)}^{2}}}}$$
3.$$\enspace$$ The quadratic form that cannot be factorized:
$$\hspace{6em} {\small (c{x}^{2} \ + \ d) }$$
Example:
$${\large\frac{3x \ + \ 5}{(x \ + \ 1)(2{x}^{2} \ + \ 7)} \ \equiv \ \frac{A}{(x \ + \ 1)} + {\small\boxed{\frac{Bx \ + \ C}{(2{x}^{2} \ + \ 7)}}}}$$
After the polynomials in the denominator of the rational function is separated, make the denominators of the simpler terms to be the same. This is typically done by multiplying the denominators together .
To find each of the coefficients in the numerator (A, B or C), we can use substitution method or equating the coefficient method.
In the substitution method, we substitute a value of x that we freely choose in the left hand side numerator and the right hand side numerator and then find the coefficients one by one.
While in the equating the coefficient method, we expand the right hand side numerator and then compare each of the coefficients in the right hand side numerator with the left hand side numerator.
Both methods will be shown in the solution of the examples below. Give it a try and if you need any help, just look at the solution I have written. Cheers ! =) .
$$\\[1pt]$$
EXAMPLE:
$${\small 1.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{7{x}^{2} \ – \ 15x \ + \ 8}{(1 \ – \ 2x){(2 \ – \ x)}^{2} }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 2.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ x \ – \ 4{x}^{2} }{(3 \ – \ x)(2 \ + \ {x}^{2}) }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 3.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ 5{x}^{2} \ + \ x \ + \ 27 }{(2x \ + \ 1)( {x}^{2} \ + \ 9 ) }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 4.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ 10 x \ + \ 9 }{(2x \ + \ 1){( 2x \ + \ 3 )}^{2} }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 5.\enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ 2x(5 \ – \ x) }{(3 \ + \ x){( 1 \ – \ x )}^{2} }} }$$
$$\\[1pt]$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$$\\[1pt]$$
$${\small 6.\enspace}$$ 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 7(a) and (b)
$$\\[1pt]$$
Let $${\small f(x) \ = \ {\large \frac{ 2 }{(2x \ – \ 1)( 2x \ + \ 1 ) }} }$$
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Express $${\small f(x) }$$ in partial fractions.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Using your answer to part (a), show that
$$\\[1pt]$$
$${\scriptsize {\Big( f(x) \Big)}^{2} \ = \ {\large \frac{ 1 }{ {(2x \ – \ 1)}^{2} }} \ – \ {\large \frac{ 1 }{ (2x \ – \ 1) }} }$$
$$\\[1pt]$$
$${\hspace{3em} \scriptsize \ + \ {\large \frac{ 1 }{ (2x \ + \ 1)}} \ + \ {\large \frac{ 1 }{ {(2x \ + \ 1)}^{2} }} . }$$
$$\\[1pt]$$
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$${\small 7.\enspace}$$ 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 6(a)
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Let $${\small f(x) \ = \ {\large \frac{ 5a }{(2x \ – \ a)( 3a \ – \ x ) }} }$$
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Express $${\small f(x) }$$ in partial fractions.
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$${\small 8.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 4(a)
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Let $${\small f(x) \ = \ {\large \frac{ 15 \ – \ 6x }{(1 \ + \ 2x)( 4 \ – \ x ) }} }$$
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Express $${\small f(x) }$$ in partial fractions.
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$${\small 9.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 9(a)
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Let $${\small f(x) \ = \ {\large \frac{ 14 \ – \ 3x \ + \ 2x^{2} }{(2 \ + \ x)( 3 \ + \ x^{2} ) }} }$$.
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Express $${\small f(x) }$$ in partial fractions.
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$${\small 10.\enspace}$$ 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 10
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The variables $$x$$ and $$t$$ satisfy the differential equation:
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$$\hspace{1.2em}{\large \frac{\mathrm{d}x}{\mathrm{d}t} } = x^{2} (1 + 2x)$$,
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and $$x = 1$$ when $$t = 0$$.
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Using partial fractions, solve the differential equation, obtaining an expression for $$t$$ in terms of $$x$$.
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PRACTICE MORE WITH THESE QUESTIONS BELOW!
$${\small 1.\enspace}$$ Express $${\small {\large \frac{7{x}^{2} \ – \ 3x \ + \ 2}{x({x}^{2} \ + \ 1) }} }$$ in partial fractions.
$${\small 2. \enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ 5{x}^{2} \ + \ x \ + \ 6 }{(3 \ – \ 2x)({x}^{2} \ + \ 4 )}} }$$
Express $${\small f(x) }$$ in partial fractions.
$${\small 3. \enspace}$$ Express $${\small {\large \frac{2 \ – \ x \ + \ 8{x}^{2}}{(1 \ – \ x)(1 \ + \ 2x)(2 \ + \ x) }} }$$ in partial fractions.
$${\small 4. \enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ {x}^{2} \ + \ 3x \ + \ 3 }{(x \ + \ 1)(x \ + \ 3 )}} }$$
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Express f(x) in partial fractions.
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Hence show that,
$$\hspace{3em} {\small \displaystyle \int_{0}^{3} f(x) \ \mathrm{d}x = 3 \ – \ \frac{1}{2} \ln 2.}$$
$${\small 5. \enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ {x}^{2} \ – \ 8x \ + \ 9 }{(1 \ – \ x){(2 \ – \ x )}^{2}}} }$$
Express $${\small f(x) }$$ in partial fractions.
$${\small 6. \enspace}$$ Let $${\small f(x) \ = \ {\large \frac{ 2{x}^{2} \ – \ 7x \ – \ 1 }{(x \ – \ 2)({x}^{2} \ + \ 3 )}} }$$
Express $${\small f(x) }$$ in partial fractions.
$$\\[12pt]{\small 7. \enspace}$$ Express $${\small {\large \frac{ x \ + \ 5 }{(x \ + \ 1)({x}^{2} \ + \ 3 )}} }$$ in the form:
$$\hspace{2em} {\large \frac{A}{( x \ + \ 1)} + \frac{Bx \ + \ C}{ ( {x}^{2} \ + \ 3) } }$$
$${\small 8. \enspace}$$ Express in partial fractions $${\small {\large \frac{{x}^{4}}{ {x}^{4} \ – \ 1 }} }$$
$${\small 9. \enspace}$$ Express in partial fractions $${\small {\large \frac{{x}^{3} \ + \ x \ – \ 1}{ {x}^{2} \ + \ {x}^{4} }} }$$
$$\\[10pt]{\small 10.\enspace}$$ Express in partial fractions
$$\hspace{2em}{\small {\large \frac{x \ + \ 5}{ {x}^{3} \ + \ 5{x}^{2} \ + \ 7x \ + \ 3 }} }$$
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
Parametric Equations
Parametric Equations
The traditional representation of y as a function of x or $${\small y = f(x)}$$ is inadequate to represent a curve or a surface.
To be able to draw a curve or a surface, we need to separate the x and y and write them in terms of an independent variable.
Parametric equations are used to express the Cartesian coordinates (x and y) in terms of another independent variable, usually named as t.
The typical procedure in this topic is to find the gradient of the curve or $${ \large\frac{\mathrm{d}y}{\mathrm{d}x} }$$. We can do this by finding each derivative of x and y with respect to t and then divide them both.
Let’s dig into some of the examples to show you what I mean. Cheers ! =) .
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EXAMPLE:
$${\small 1.\enspace}$$ The parametric equations of a curve are
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$$\hspace{3em} x = {\large \frac{1}{{\cos}^{3}t}}, \quad y = {\tan}^{3}t$$.
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where $$0 \ \le \ t \ \le \ \frac{\pi}{2}$$
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$${\small\hspace{1.2em}\left(a\right). \enspace }$$ Show that $${\large\frac{\mathrm{d}y}{\mathrm{d}x} } = \sin \ t$$
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$${\small\hspace{1.2em}\left(b\right). \enspace }$$ Hence, show that the equation of the tangent to the curve at the point with parameter t is $$y = x \sin t \ – \ \tan t$$.
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$${\small 2.\enspace}$$ The parametric equations of a curve are
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$$\hspace{3em} x = e^{-t} \cos t, \quad y = e^{-t} \sin t$$.
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Show that $${\large\frac{\mathrm{d}y}{\mathrm{d}x} } = \tan \big(t \ – \ {\large\frac{\pi}{4}}\big)$$
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$${\small 3.\enspace}$$ The parametric equations of a curve are
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$$\hspace{3em} x = \ln (2t \ + \ 3), \quad y = { \large\frac{3t \ + \ 2}{2t \ + \ 3} }$$.
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Find the gradient of the curve at the point where it crosses the y-axis.
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$${\small 4.\enspace}$$ The parametric equations of a curve are
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$$\hspace{1.2em} {\small x = \ 2\sin \theta \ + \ \sin 2\theta, \enspace y = \ 2\cos \theta \ + \ \cos 2\theta }$$,
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where $$0 \ \lt \ \theta \ \lt \ \pi$$.
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$${\small\hspace{1.2em}\left(\textrm{i}\right). \enspace }$$ Obtain an expression for $${\small {\large\frac{\mathrm{d}y}{\mathrm{d}x} } }$$ in terms of $${\small \theta }$$.
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$${\small\hspace{1.2em}\left(\textrm{ii}\right). \enspace }$$ Hence find the exact coordinates of the point on the curve at which the tangent is parallel to the y-axis.
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$${\small 5.\enspace}$$ The parametric equations of a curve are
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$$\hspace{1.2em} x = \ 2 t \ + \sin 2t, \enspace y = \ 1 \ – \ 2\cos 2t$$,
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where $$-\frac{1}{2}\pi \ \lt \ t \ \lt \ \frac{1}{2}\pi$$.
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$${\small\hspace{1.2em}\left(\textrm{i}\right). \enspace }$$ Show that $${\small {\large\frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ 2 \tan t }$$.
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$${\small\hspace{1.2em}\left(\textrm{ii}\right). \enspace }$$ Hence find the x-coordinate of the point on the curve at which the gradient of the normal is 2. Give your answer correct to 3 significant figures.
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$${\small 6.\enspace}$$ 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 10(a), (b)
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A tank containing water is in the form of a hemisphere. The axis is vertical, the lowest point is $$A$$ and the radius is $$r$$, as shown in the diagram. The depth of water at time $$t$$ is $$h$$.
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At time $$t = 0$$ the tank is full and the depth of the water is $$r$$. At this instant a tap at $$A$$ is opened and water begins to flow out at a rate proportional to $$\sqrt{h}$$. The tank becomes empty at time $$t = 14$$.
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The volume of water in the tank is $$V$$ when the depth is $$h$$. It is given that $$V = {\large\frac{1}{3}\pi} \big( 3r{h}^{2} \ – \ {h}^{3} \big)$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Show that $$h$$ and $$t$$ satisfy a differential equation of the form
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$$\hspace{1.2em} {\large \frac{\mathrm{d}h}{\mathrm{d}t} } \ = \ – {\large\frac{ B }{ 2r{h}^{\frac{1}{2}} \ – \ {h}^{\frac{3}{2}} } }$$,
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$$\hspace{1.2em}$$ where $$B$$ is a positive constant.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Solve the differential equation and obtain an expression for $$t$$ in terms of $$h$$ and $$r$$.
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$${\small 7.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 3(a), (b)
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The parametric equations of a curve are
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$$\hspace{1.2em} x = \ t \ + \ln (t+2), \enspace y = \ (t \ – \ 1) { \mathrm{e} }^{-2t}$$,
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where $$t \ \gt \ -2$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Express $${\large\frac{\mathrm{d}y}{\mathrm{d}x} }$$in terms of $$t$$, simplifying your answer.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Find the exact $$y$$-coordinate of the stationary point of the curve.
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$${\small 8.\enspace}$$ 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 6(a), (b)
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The parametric equations of a curve are
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$$\hspace{1.2em} x = \ln (2 \ + \ 3t), \enspace y = \ {\large\frac{t}{2 \ + \ 3t} }$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Show that the gradient of the curve is always positive.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Find the equation of the tangent to the curve at the point where it intersects the $$y$$-axis.
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$${\small 9.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 4
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The parametric equations of a curve are
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$$\hspace{1.2em} x = 1 \ – \ \cos \theta, \enspace y = \cos \theta \ – \ {\large\frac{1}{4} } \cos 2\theta$$.
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Show that $${\small {\large\frac{\mathrm{d}y}{\mathrm{d}x}} = -2 \ \sin^{2} \big( {\large\frac{1}{2}}\theta \big) }$$.
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$${\small 10.\enspace}$$ 9709/12/M/J/20 – Paper 12 June 2020 Pure Maths 1 No 3
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A weather balloon in the shape of a sphere is being inflated by a pump. The volume of the balloon is increasing at a constant rate of 600 $${\small {\mathrm{cm}}^{3} }$$ per second. The balloon was empty at the start of pumping.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the radius of the balloon after 30 seconds.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Find the rate of increase of the radius after 30 seconds.
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$${\small 11.\enspace}$$ 9709/12/O/N/19 – Paper 12 November 2019 Pure Maths 1 No 5
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The diagram shows a solid cone which has a slant height of 15 cm and a vertical height of $${\small h}$$ cm.
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$${\small\hspace{1.2em}\left(\mathrm{i}\right).\hspace{0.8em}}$$ Show that the volume, $${\small V \ {\mathrm{cm}}^{3} }$$, of the cone is given by $${\small V \ = \ {\large\frac{1}{3}}\pi (225h \ – \ {h}^{3}) }$$.
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[The volume of a cone of radius $${\small r}$$ and vertical height $${\small r}$$ is $${\small {\large\frac{1}{3}}\pi {r}^{2} h }$$ ].
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$${\small\hspace{1.2em}\left(\mathrm{ii}\right).\hspace{0.8em}}$$ Given that $${\small h}$$ can vary, find the value of $${\small h}$$ for which $${\small V}$$ has a stationary value. Determine, showing all necessary working, the nature of this stationary value.
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PRACTICE MORE WITH THESE QUESTIONS BELOW!
$${\small 1.\enspace}$$ The parametric equations of a curve are
$$x = \ \sin t \ + \cos t, \enspace y = \ {\sin}^{3}t \ + \ {\cos}^{3}t$$,
where $$\frac{\pi}{4} \ \le \ t \ \le \ \frac{5\pi}{4}$$
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Show that $${ \large\frac{\mathrm{d}y}{\mathrm{d}x}} = \ -3 \ \sin t \ \cos t$$.
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Find the gradient of the curve at the origin.
$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}$$ Find the values of t for which the gradient of the curve is 1, giving your answers correct to 2 significant figures.
$${\small 2. \enspace}$$ The parametric equations of a curve are
$$x = \ a(2\theta \ – \ \sin 2\theta), \enspace y = \ a(1 \ – \ \cos 2\theta)$$.
Show that $${ \large\frac{\mathrm{d}y}{\mathrm{d}x}} = \ \cot \theta$$.
$${\small 3.\enspace}$$ The parametric equations of a curve are
$$x = \ \ln \cos \theta, \enspace y = \ 3\theta \ – \ \tan \theta$$,
where $$0 \ \le \ \theta \ \le \ \frac{1}{2}\pi$$.
$${\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}$$ Express $${\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} }$$ in terms of $${\small \tan \theta}$$.
$${\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}$$ Find the exact y-coordinate of the point on the curve at which the gradient of the normal is equal to 1.
$${\small 4.\enspace}$$ The parametric equations of a curve are
$$x = \ {t}^{2} \ + 1, \enspace y = \ 4t \ + \ \ln \ (2t \ – \ 1)$$.
$${\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}$$ Express $${\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} }$$ in terms of $${\small t}$$.
$${\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}$$ Find the equation of the normal to the curve at the point where $${\small t \ = \ 1}$$. Give your answer in the form ax + by + cz = 0.
$${\small 5.\enspace}$$ The parametric equations of a curve are
$$x = \ t \ + \cos t, \enspace y = \ \ln \ (1 + \sin t)$$,
where $$-\frac{1}{2}\pi \ \lt \ t \ \lt \ \frac{1}{2}\pi$$.
$${\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}$$ Show that $${\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ \sec t}$$.
$${\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}$$ Hence find the x-coordinates of the points on the curve at which the gradient is equal to 3. Give your answer correct to 3 significant figures.
$${\small 6.\enspace}$$ A curve has parametric equations
$$x = \ {t}^{2} \ + \ 3t \ + \ 1, \enspace y = \ {t}^{4} \ + \ 1$$.
The point P on the curve has parameter p. It is given that the gradient of the curve at P is 4. Show that $${\small p \ = \ \sqrt[3]{(2p \ + \ 3)} }$$.
$${\small 7.\enspace}$$ A curve has parametric equations
$$x = \ 1 \ + \ 2 \ \sin \theta \$$ and $$\ y = \ 4 \ + \ \sqrt{3} \ \cos \theta$$.
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the equations of the tangent and normal at the point P where $${\small \theta \ = \ {\large\frac{\pi}{6}} }$$. Hence, find the area A of the triangle bounded by the tangent and normal at P, and the y-axis.
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Determine the rate of change of xy at $${\small \theta \ = \ {\large\frac{\pi}{6}} }$$ if x increases at a constant rate of 0.1 units/s.
$${\small 8.\enspace}$$ A curve is defined parametrically by $$x \ = \ \frac{2t}{t + 1}$$ and $$y \ = \ \frac{{t}^{2}}{t + 1}$$.
$${\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}$$ Find the equation of the normal to the curve at the point P(1, 1/2).
$${\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}$$ The normal at P meets the curve again at Q. Find the exact coordinates of Q.
$${\small 9.\enspace}$$ A curve has parametric equations:
$$x = \ \sec (\frac{\theta}{2}), \enspace y = \ \ln \ \sec (\frac{\theta}{2})$$,
where $$-\pi \ \lt \ \theta \ \lt \ \pi$$.
$${\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}$$ Find the equation of the normal to the curve at the point at $${\small \theta \ = \ {\large\frac{\pi}{3}} }$$.
$${\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}$$ Determine the rate of change of x if the gradient of the curve at $${\small \theta \ = \ {\large\frac{\pi}{2}} }$$ is decreasing at a rate of 0.4 units per second.
$${\small 10.\enspace}$$ The parametric equations of a curve are
$$x = \ t \ + \ \ln t, \enspace y = \ t \ + \ {\textrm{e}}^{t}$$ for $$t \gt 0$$.
$${\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}$$ Sketch the curve, indicating clearly all intercepts and asymptotes.
$${\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}$$ Show that, for all the points on the curve, $${\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ { \large\frac{t(1 \ + \ {\textrm{e}}^{t})}{t \ + \ 1}} }$$.
Hence, deduce that the curve does not have any turning points.
$${\small\hspace{1.2em}\left(\textrm{iii}\right).\hspace{0.4em}}$$ Find, in exact form, the equation of the normal of the curve at the point where $${\small t \ = \ 1}$$.
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
Integration and Differentiation
Integration and Differentiation
Integration is an essential part of basic calculus. Algebra plays a very important part to become proficient in this topic.
I have compiled some of the questions that I have encountered during my Math tutoring classes. Do take your time to try the questions and learn from the solutions I have provided below. Cheers ! =) .
More Integration Exercises can be found here.
EXAMPLE:
$${\small 1.\enspace}$$ 9709/32/F/M/17 – Paper 32 Feb March 2017 Pure Maths 3 No 10
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The diagram shows the curve $$\ {\small y \ = \ {(\ln x)}^{2} }$$. The x-coordinate of the point P is equal to e, and the normal to the curve at P meets the x-axis at Q.
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$${\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}$$ Find the x-coordinate of Q.
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$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}$$ Show that $${\small \displaystyle \int \ln x \ \mathrm{d}x \ = \ x \ln x \ – \ x \ + \ c }$$, where c is a constant.
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$${\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}$$ Using integration by parts, or otherwise, find the exact value of the area of the shaded region between the curve, the x-axis and the normal PQ.
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$${\small 2.\enspace}$$ 9709/32/F/M/19 – Paper 32 Feb March 2019 Pure Maths 3 No 10
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The diagram shows the curve $$\ {\small y \ = \ {\sin}^{3} x \sqrt{(\cos x)} \ }$$ for $$\ {\small 0 \leq x \leq \large{ \frac{1}{2}} \pi }$$, and its maximum point M.
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$${\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}$$ Using the substitution $${\small \ u \ = \ \cos x }$$, find by integration the exact area of the shaded region bounded by the curve and the x-axis.
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$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}$$ Showing all your working, find the x-coordinate of M, giving your answer correct to 3 decimal places.
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$${\small 3.\enspace}$$ 9709/32/M/J/20 – Paper 32 May June 2020 Pure Maths 3 No 9
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The diagram shows the curves $$\ {\small \ y \ = \ \cos x \ }$$ and $$\ {\small \ y \ = \ \large{ \frac{k}{1 \ + \ x} } }$$, where k is a constant, for $$\ {\small \ 0 \leq x \leq \large{ \frac{1}{2}} \pi }$$. The curves touch at the point where x = p.
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$${\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}$$ Show that p satisfies the equation $${\small \ \tan p \ = \ \large{ \frac{1}{1 \ + \ p} } }$$.
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$${\small 4.\enspace} \displaystyle \int_{1}^{a} \ln 2x \ \mathrm{d}x = 1.$$ Find $${\small a}$$.
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$${\small 5.\enspace}$$ Use the substitution $$u = \sin 4x$$ to find the exact value of $$\displaystyle \int_{0}^{{\Large\frac{\pi}{24}}} \cos^{3} 4x \ \mathrm{d}x.$$
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$${\small 6. \hspace{0.8em}(i).\hspace{0.8em}}$$ Use the trapezium rule with 3 intervals to estimate the value of: $$\displaystyle \int_{{\Large\frac{\pi}{9}}}^{{\Large\frac{2\pi}{3}}} \csc x \ \mathrm{d}x$$ giving your answer correct to 2 decimal places.
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$${\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}$$ Using a sketch of the graph of $$y = \csc x$$, explain whether the trapezium rule gives an overestimate or an underestimate of the true value of the integral in part (i).
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$${\small 7.\enspace}$$ Solve these integrations.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int_{0}^{\infty} \frac{1}{{x}^{2} \ + \ 4} \ \mathrm{d}x$$
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int_{0}^{3} \frac{1}{\sqrt{9 \ – \ {x}^{2}}} \ \mathrm{d}x$$
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$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}} \displaystyle \int_{-\infty}^{\infty} \frac{1}{9{x}^{2} \ + \ 4} \ \mathrm{d}x$$
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$${\small\hspace{1.2em}\left(d\right).\hspace{0.8em}} \displaystyle \int_{0}^{1} \frac{1}{\sqrt{x(1 \ – \ x)}} \ \mathrm{d}x$$
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$${\small\hspace{1.2em}\left(e\right).\hspace{0.8em}} \displaystyle \int_{1}^{\infty} \frac{1}{{(1 \ + \ x^2)}^{{\large\frac{3}{2}}}} \ \mathrm{d}x$$
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$${\small\hspace{1.2em}\left(f\right).\hspace{0.8em}} \displaystyle \int_{1}^{\infty} \frac{1}{x \sqrt{{x}^{2} \ – \ 1}} \ \mathrm{d}x$$
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$${\small 8.\enspace}$$ The diagram shows the curve $${\small y = {e}^{{\large – \frac{1}{2}x}} \ \sqrt{(1 \ + \ 2x)}}$$ and its maximum point M. The shaded region between the curve and the axes is denoted by R.
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$${\small \hspace{1.2em}(i). \enspace }$$ Find the x-coordinate of M.
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$${\small \hspace{1.2em}(ii). \enspace }$$ Find by integration the volume of the solid obtained when R is rotated completely about the x-axis. Give your answer in terms of $${\small \pi}$$ and e.
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$${\small 9.\enspace}$$ 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 2
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Find the exact value of $$\displaystyle \int_{0}^{1} (2 \ – \ x) \mathrm{e}^{-2x} \ \mathrm{d}x$$.
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$${\small 10.\enspace}$$ 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 7(a), (b), (c)
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Let $${\small f(x) \ = \ {\large \frac{ 2 }{(2x \ – \ 1)( 2x \ + \ 1 ) }} }$$
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Express $${\small f(x) }$$ in partial fractions.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Using your answer to part (a), show that
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$${\scriptsize {\Big( f(x) \Big)}^{2} \ = \ {\large \frac{ 1 }{ {(2x \ – \ 1)}^{2} }} \ – \ {\large \frac{ 1 }{ (2x \ – \ 1) }} }$$
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$${\hspace{3em} \scriptsize \ + \ {\large \frac{ 1 }{ (2x \ + \ 1)}} \ + \ {\large \frac{ 1 }{ {(2x \ + \ 1)}^{2} }} . }$$
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$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}$$ Hence show that $$\displaystyle \int_{1}^{2} {\Big( f(x) \Big)}^{2} \ \mathrm{d}x = \frac{2}{5} + \frac{1}{2} \ln \frac{5}{9}$$.
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$${\small 11.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 3
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Find the exact value of $$\displaystyle \int_{1}^{4} x^{\frac{3}{2}} \ln x \ \mathrm{d}x$$.
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$${\small 12.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 4
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A curve has equation $$y = \cos x \sin 2x$$.
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Find the $$x$$-coordinate of the stationary point in the interval $$0 \lt x \lt \frac{1}{2}\pi$$, giving your answer correct to 3 significant figures.
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$${\small 13.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 6(a), (b)
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The diagram shows the curve $$y = {\large\frac{x}{1+{3x}^{4}} }$$, for $$x \geq 0$$, and its maximum point $$M$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the $$x$$-coordinate of $$M$$, giving your answer correct to 3 decimal places.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Using the substitution $$u = \sqrt{3}{x}^{2}$$, find by integration the exact area of the shaded region bounded by the curve, the $$x$$-axis and the line $$x = 1$$.
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$${\small 14.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 9(a)
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The diagram shows the curves $$y = \cos x$$ and $$y = \frac{k}{1 \ + \ x}$$, where $$k$$ is a constant for $$0 \leq x \leq \frac{1}{2\pi}$$. The curves touch at the point where $$x = p$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Show that $$p$$ satisfies the equation $$\tan p = \frac{1}{1+p}$$.
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$${\small 15.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 4(a), (b)
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The curve with equation $$y = { \mathrm{e} }^{2x} (\sin x + 3 \cos x)$$ has a stationary point in the interval $$0 \leq x \leq \pi$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the $$x$$-coordinate of this point, giving your answer correct to 2 decimal places.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Determine whether the stationary point is a maximum or a minimum.
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$${\small 16.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 5(a), (b)
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the quotient and remainder when $$2x^3 − x^2 + 6x + 3$$ is divided by $$x^2 + 3$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Using your answer to part (a), find the exact value of $$\displaystyle \int_{1}^{3} \frac{2x^3 \ – \ x^2 \ + \ 6x \ + \ 3}{x^2 \ + \ 3} \mathrm{d}x$$.
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$${\small 17.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 7(a), (b)
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Let $$f(x) = { \large \frac{\cos x}{1 \ + \ \sin x} }$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Show that $$f'(x) \lt 0$$ for all $$x$$ in the interval $$-\frac{1}{2} \pi \lt x \lt \frac{3}{2} \pi$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Find $$\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{2} } f(x) \ \mathrm{d}x$$. Give your answer in a simplified exact form.
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$${\small 18.\enspace}$$ 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 6(a), (b)
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Let $${\small f(x) \ = \ {\large \frac{ 5a }{(2x \ – \ a)( 3a \ – \ x ) }} }$$
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Express $${\small f(x) }$$ in partial fractions.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Hence show that $$\displaystyle \int_{a}^{ 2a } f(x) \ \mathrm{d}x = \ln 6$$.
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$${\small 19.\enspace}$$ 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 9(c)
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Let $${\small f(x) \ = \ {\large \frac{ { \mathrm{e} }^{2x} \ + \ 1 }{{ \mathrm{e} }^{2x} \ – \ 1 }} }$$, for $$x \gt 0$$.
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Find $$f'(x)$$. Hence find the exact value of x for which $$f'(x) = -8$$.
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$${\small 20.\enspace}$$ 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 10(a), (b)
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The diagram shows the curve $$y = \sin 2x \ {\cos}^{2} x$$ for $$0 \leq x \leq \frac{1}{2} \pi$$, and its maximum point $$M$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Using the substitution $$u = \sin x$$, find the exact area of the region bounded by the curve and the $$x$$-axis.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Find the exact $$x$$-coordinate of $$M$$.
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$${\small 21.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 4(a),(b)
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Let $${\small f(x) \ = \ {\large \frac{ 15 \ – \ 6x }{(1 \ + \ 2x)( 4 \ – \ x ) }} }$$
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Express $${\small f(x) }$$ in partial fractions.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Hence find $$\displaystyle \int_{1}^{ 2 } f(x) \ \mathrm{d}x$$, giving your answer in the form $$\ln ( \frac{a}{b} )$$, where $$a$$ and $$b$$ are integers.
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$${\small 22.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 8(a),(b)
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The diagram shows the curve $$y = {\large \frac{ \ln x }{ x^4 } }$$ and its maximum point $$M$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the exact $$x$$-coordinate of $$M$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ By using integration by parts, show that for all $$a \gt 1, \displaystyle \int_{1}^{ a } \frac{ \ln x }{ x^4 } \ \mathrm{d}x \lt \frac{1}{9}$$.
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$${\small 23.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 4
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Using integration by parts, find the exact value of $$\displaystyle \int_{0}^{ 2 } {\tan}^{-1} \big( \frac{1}{2} x \big) \ \mathrm{d}x$$.
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$${\small 24.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 6(a), (b)
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Prove that $$\mathrm{cosec} 2\theta − \cot 2\theta \equiv tan \theta$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Hence show that $$\displaystyle \int_{\frac{1}{4}\pi}^{ \frac{1}{3}\pi } (\mathrm{cosec} 2\theta − \cot 2\theta) \ \mathrm{d}\theta = \frac{1}{2} \ln 2$$.
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$${\small 25.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 8
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The equation of a curve is $$y = { \mathrm{e} }^{-5x} \ {\tan}^{2} x$$ for $$-\frac{1}{2}\pi \lt x \lt \frac{1}{2}\pi$$.
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Find the $$x$$-coordinates of the stationary points of the curve. Give your answers correct to 3 decimal places where appropriate.
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$${\small 26.\enspace}$$ 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 7(a)
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The diagram shows the curve $$y = {\large \frac{ {\tan}^{-1} x }{ \sqrt{x} } }$$ and its maximum point $$M$$ where $$x = a$$.
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Show that a satisfies the equation $$a = \tan \Big( {\large \frac{2a}{1 \ + \ a^2} }\Big)$$.
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$${\small 27.\enspace}$$ 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 9(a), (b)
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The equation of a curve is $$y = { x^{-\frac{2}{3}}} \ \ln x \$$ for $$x \gt 0$$. The curve has one stationary point.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the exact coordinates of the stationary point.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Show that $$\displaystyle \int_{1}^{ 8 } y \ \mathrm{d}x = 18 \ln 2 \ – \ 9$$.
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$${\small 28.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 8(a), (b)
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the quotient and remainder when $${\small 8x^3 + 4x^2 + 2x + 7}$$ is divided by $${\small 4x^2 + 1}$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Hence find the exact value of $$\displaystyle \int_{0}^{ \frac{1}{2} } \frac{8x^3 + 4x^2 + 2x + 7}{4x^2 + 1} \ \mathrm{d}x$$.
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$${\small 29.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 11
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The diagram shows the curve $$\ y \ = \ \sin x \ \cos 2x \$$ for $${\small \ 0 \le x \le {\large\frac{1}{2}}\pi}$$, and its maximum point $$M$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the $$x$$-coordinate of $$M$$, giving your answer correct to 3 significant figures.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Using the substitution $$\ u = \cos x$$, find the area of the shaded region enclosed by the curve and the $$x$$-axis in the first quadrant, giving your answer in a simplified exact form.
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$${\small 30.\enspace}$$ 9709/13/O/N/21 – Paper 13 November 2021 Pure Maths 1 No 8
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The diagram shows the curves with equations $$\ y \ = \ {x}^{ { \large – \frac{1}{2} } } \$$ and $$\ y \ = \ \frac{5}{2} \ – \ {x}^{ { \large \frac{1}{2} } }$$. The curves intersect at the points $$\ A( {\large\frac{1}{4}},2) \$$ and $$\ B( 4 , {\large\frac{1}{2}})$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the area of the region between the two curves.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ The normal to the curve $$\ y \ = \ {x}^{ { \large – \frac{1}{2} } } \$$ at the point $$(1, 1)$$ intersects the $$y$$-axis at the point $$(0, p)$$. Find the value of $$p$$.
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$${\small 31.\enspace}$$ 9709/13/O/N/21 – Paper 13 November 2021 Pure Maths 1 No 10
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A curve has equation $$\ y = \mathrm{f}(x) \$$ and it is given that
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$$\hspace{2em} \mathrm{f}^{\prime}(x) = { \big( \frac{1}{2}x \ + \ k \big) }^{-2} \ – \ { ( 1 \ + \ k ) }^{-2}$$,
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where $${\small \ k \ }$$ is a constant. The curve has a minimum point at $${\small \ x \ = \ 2 }$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find $$\ \mathrm{f}^{\prime\prime}(x)$$ in terms of $$k$$ and $$x$$, and hence find the set of possible values of $$k$$.
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It is now given that $${\small \ k \ = \ −3 \ }$$ and the minimum point is at $${\small \ (2, \ 3\frac{1}{2}) }$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Find $$\ \mathrm{f}(x)$$.
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$${\small 32.\enspace}$$ 9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 9
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The diagram shows part of the curve with equation $$\ {\small y^2 \ = \ x \ − \ 2 } \$$ and the lines $$\ {\small x \ = \ 5 } \$$ and $$\ {\small y \ = \ 1 }$$. The shaded region enclosed by the curve and the lines is rotated through $$\ {\small 360^{\circ} \ }$$ about the $$x$$-axis.
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Find the volume obtained.
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$${\small 33.\enspace}$$ 9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 11
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The gradient of a curve is given by
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$$\hspace{2em} {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ 6{(3x \ – \ 5)}^{3} \ – \ k{x}^{2}}$$,
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where $${\small \ k \ }$$ is a constant. The curve has a stationary point at $$\ {\small (2, -3.5) }$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the value of $$\ k$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Find the equation of the curve.
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$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}$$ Find $${\small {\large \frac{ {\mathrm{d}}^{2} y }{ \mathrm{d}{x}^{2}} } }$$.
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$${\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}$$ Determine the nature of the stationary point at $$\ {\small (2, -3.5) }$$.
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$${\small 34.\enspace}$$ 9709/12/M/J/20 – Paper 12 June 2020 Pure Maths 1 No 8
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The diagram shows part of the curve with equation $$\ {\small y \ = \ {\large\frac{6}{x}} }$$. The points $$\ {\small (1, 6) \ }$$ and $$\ {\small (3, 2) \ }$$ lie on the curve. The shaded region is bounded by the curve and the lines $$\ {\small y \ = \ 2 } \$$ and $$\ {\small x \ = \ 1 }$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the volume generated when the shaded region is rotated through $$\ {\small 360^{\circ} \ }$$ about the $$y$$-axis.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ The tangent to the curve at a point $$X$$ is parallel to the line $$\ {\small y \ + \ 2x \ = \ 0 }$$. Show that $$X$$ lies on the line $$\ {\small y \ = \ 2x }$$.
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$${\small 35.\enspace}$$ 9709/12/M/J/20 – Paper 12 June 2020 Pure Maths 1 No 10
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The equation of a curve is
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$$\hspace{2em} y \ = 54x \ – \ {(2x \ – \ 7)}^{3}$$.
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$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find $$\ {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ }$$ and $${\small \ {\large \frac{ {\mathrm{d}}^{2} y }{ \mathrm{d}{x}^{2}} } }$$.
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$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Find the coordinates of each of the stationary points on the curve.
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$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}$$ Determine the nature of each of the stationary points.
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$${\small 36.\enspace}$$ 9709/12/O/N/19 – Paper 12 June 2019 Pure Maths 1 No 10
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The diagram shows part of the curve $$\ {\small y \ = \ 1 \ – \ {\large\frac{4}{ {(2x \ + \ 1)}^{2} }} }$$. The curve intersects the $$x$$-axis at $$A$$. The normal to the curve at A intersects the $$y$$-axis at $$B$$.
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$${\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}$$ Obtain expressions for $$\ {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ }$$ and $$\displaystyle \int y \ \mathrm{d}x$$.
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$${\small\hspace{1.2em}\left(ii\right).\hspace{0.7em}}$$ Find the coordinates of $$B$$.
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$${\small\hspace{1.2em}\left(iii\right).\hspace{0.6em}}$$ Find, showing all necessary working, the area of the shaded region.
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PRACTICE MORE WITH THESE QUESTIONS BELOW!
$${\small 1.\enspace}$$ Find $$\displaystyle \int \frac{1}{x^2\sqrt{x^2 \ – \ 4}} \ \mathrm{d}x$$ using the substitution $${\small x \ = \ 2 \sec \theta }$$.
$${\small 2. \enspace}$$ Find the exact value of $$\displaystyle \int_{1}^{e} x^4 \ \ln \ x \ \mathrm{d}x$$.
$${\small 3. \enspace}$$ Find the exact value of $$\displaystyle \int_{4}^{10} \frac{2x \ + \ 1}{(x \ – \ 3)^2} \ \mathrm{d}x$$, giving your answer in the form of $${\small a \ + \ b \ \ln \ c}$$, where a, b and c are integers.
$${\small 4. \enspace}$$ Find the exact value of $$\displaystyle \int_{1}^{4} \frac{\ln \ x}{\sqrt{x}} \ \mathrm{d}x$$.
$${\small 5. \enspace}$$ Find the exact value of
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int_{0}^{\infty} {e}^{1 \ – \ 2x} \ \mathrm{d}x$$
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int_{-1}^{0} \big( 2 \ + \ \frac{1}{x \ – \ 1} \big) \ \mathrm{d}x$$
$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}} \displaystyle \int_{{\large\frac{\pi}{6}}}^{{\large \frac{\pi}{4}}} \cot x \ \mathrm{d}x$$
$${\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}$$ Using your result in (c), find also the exact value of $$\displaystyle \int_{{\large\frac{\pi}{6}}}^{{\large \frac{\pi}{4}}} \csc 2x \ \mathrm{d}x$$ by using the identity $$\cot x \ – \ \cot 2x \ \equiv \ \csc 2x$$.
$${\small 6. \enspace}$$ The diagram shows the part of the curve $${\small y \ = \ f(x)}$$, where $${\small f(x) \ = \ p \ – \ {e}^{x} }$$ and p is a constant. The curve crosses the y-axis at (0, 2).
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the value of p.
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Find the coordinates of the point where the curve crosses the x-axis.
$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}$$ What is the area of the shaded region R?
$${\small 7. \enspace}$$ Integrate the following:
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int \frac{x^2}{1 \ + \ {x}^{3}} \ \mathrm{d}x$$
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int x^4 \ \sin (x^5 \ + \ 2) \ \mathrm{d}x$$
$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}} \displaystyle \int e^{x} \ \sin x \ \mathrm{d}x$$
$${\small 8. \enspace}$$ Let $$I \ = \ \displaystyle \int_{0}^{1} {\large \frac{\sqrt{x}}{2 \ – \ \sqrt{x}}} \ \mathrm{d}x$$.
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Using the substitution $${\small u = \ 2 \ – \ \sqrt{x}}$$, show that $$I \ = \ \displaystyle \int_{1}^{2} {\large \frac{2 {(2 \ – \ u)}^{2}}{u}} \ \mathrm{d}u$$.
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Hence show that $$I \ = \ 8 \ \ln 2 \ – \ 5$$.
$${\small 9. \enspace}$$ The constant a is such that
$${\small\hspace{3em}} \displaystyle \int_{0}^{a} x{e}^{{\large \frac{1}{2}x}} \mathrm{d}x \ = \ 6$$.
Show that a satisfies the equation
$${\small\hspace{3em}} a \ = \ 2 \ + \ {e}^{{\large -\frac{1}{2}a}}$$.
$${\small 10. \enspace}$$ Use the substitution $${\small u \ = \ 1 \ + \ 3 \ \tan x }$$ to find the exact value of
$${\small\hspace{3em}} \ \displaystyle \int_{0}^{{\large\frac{\pi}{4}}} {\large \frac{\sqrt{1 \ + \ 3 \ \tan x}}{{\cos}^{2}x}} \ \mathrm{d}x$$.
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) . |
# What is the positive and negative square root of 196?
• Last Updated : 01 Sep, 2021
The arithmetic value which is used for representing the quantity and used in making calculations are defined as Numbers. A symbol like “4,5,6” which represents a number is known as numerals. Without numbers, we can’t do counting of things, date, time, money, etc. these numbers are also used for measurement and used for labeling.
The properties of numbers make them helpful in performing arithmetic operations on them. These numbers can be written in numeric forms and also in words.
For example, 3 is written as three in words, 35 is written as thirty-five in words, etc. Students can write the numbers from 1 to 100 in words to learn more.
There are different types of numbers, which we can learn. They are whole and natural numbers, odd and even numbers, rational and irrational numbers, etc
What is a Number System?
A Number System is a method of showing numbers by writing, which is a mathematical way of representing the numbers of a given set, by using the numbers or symbols in a mathematical manner. The writing system for denoting numbers using digits or symbols in a logical manner is defined as Number System.
We can use the digits from 0 to 9 to form all the numbers. With these digits, anyone can create infinite numbers.
For example, 156,3907, 3456, 1298, 784859, etc
What is a Square Root?
The value of a number of square roots, which on multiplication by itself gives the original number. Suppose, a is the square root of b, then it is represented as a = √b or we can express the same equation as a2 = b. Here,’√’ this symbol we used to represent the root of numbers is termed as radical. The positive number when it is to be multiplied by itself represents the square of the number. The square root of the square of any positive number gives the original number.
For example, the square of 4 is 16, 42 = 16, and the square root of 16, √16 = 4. Since 4 is a perfect square and has both the negative and positive square root value of 16 is ±4, hence it is easy to find the square root of such numbers, but for an imperfect square, it’s really tricky.
Square Root is represented as ‘√’. It is called a radical symbol. To represent a number ‘a’ as a square root using this symbol can be written as: ‘√a‘, where a is the number.
The number here under the radical symbol is called the radicand. For example, the square root of 4 is also represented as a radical of 4. Both represent the same value.
and the formula to find the square root is: b = √a
Properties of Square Roots
It is defined as a one-to-one function that takes a positive number as an input and returns the square root of the given input number.
f(x) = √x
For example, here if x = 9, then the function returns the output value as 3.
These are properties of the square root as follows:
• If a number is a perfect square number, then there definitely exists a perfect square root.
• If a number ends with an even number of zeros (0’s), then we can have a square root.
• The two square root values can be multiplied. For example, √3 can be multiplied by √2, then the result will be √6.
• When two same square roots are multiplied, then the result must be a radical number. It shows that the result is a non-square root number. For example, when √7 is multiplied by √7, the result obtained is 7.
• The square root of negative numbers is undefined. hence the perfect square cannot be negative.
• Some of the numbers end with 2, 3, 7, or 8 (in the unit digit), then the perfect square root does not exist.
• Some of the numbers end with 1, 4, 5, 6, or 9 in the unit digit, then the number will have a square root.
It is easy to find the square root of a number that is a perfect square.
Perfect squares are those positive numbers or negative numbers that can be written as the multiplication of a number by itself, or you can say that a perfect square is a number which is the value of power 2 of any integer.
The number that can be expressed as the product of two equal integers. For example, 16 is a perfect square because it is the product of two equal integers, 4 × 4 = 16. or -4 × -4 = 16, However, 24 is not a perfect square because it cannot be expressed as the product of two equal integers. (8 × 3 = 24).
The number which is obtained by squaring a whole number is termed as a perfect square.
If we assume N is a perfect square of a whole number y, this can be written as N = the product of y and y = y2.
So, the perfect square formula can be expressed as:
N = Y2
Let’s Use the formula with values.
If y = 5, and N = y2
This means, N = 52 = 25 or (- 5)2= 25
Here, 5 is the positive and negative square root of 25 i.e ±5
With the help of square roots, we can identify whether a number is a perfect square or not and if we calculate the square root of the given number.
If the square root is a whole number then the given number will be a perfect square, and if the square root value is not a whole number, then the given number is not a perfect square.
### What is the positive and negative square root of 196?
Here 14 is the square root of 196 which is also a perfect square of 14.
So we can say that 14 is the positive and negative square root of 196
√196 = 14 × 14 or -14 × -14 =14
√196 = ±14
### Similar Questions
Question 1: What is the positive and negative square root 4225?
solution :
here 65 is the positive and negative square root of 4225
We can write as √4225 = ±65
Question 2: find out the positive and negative square of 10000?
Here 10000 is the perfect square of 100
1002 = 100 × 100 = 10000
or we can write as (-100)2 = -100 × – 100 = 10000
so 10000 is the perfect square of 100 and have both the negative and positive square root
i.e √10000 = ±100
My Personal Notes arrow_drop_up |
# A manufacturer samples 100 wires for quality testing. Four of the wires are found to be defective. If 750 wires are produced in one hour, how many should the manufacturer expect to be defective?
A.30
B.60
C.75
D.96
0.04 * 750 = ?
## Defective wires, huh? Sounds like someone was feeling a bit "wired" when they were working! Anyway, let's get serious and do some math!
If we know that out of 100 sampled wires, 4 were defective, we can use that ratio to estimate how many defective wires we should expect out of 750 wires produced in one hour.
The ratio of sampled wires to defective wires is 100:4. If we apply that ratio to 750 wires produced, we can set up a proportion:
100/4 = 750/x
Cross-multiply:
100x = 4 * 750
100x = 3000
Divide both sides by 100:
x = 30
So the manufacturer should expect 30 wires to be defective out of the 750 produced in one hour. That's option A, my friend! Keep an eye out for those "shocking" wires!
## To solve this problem, we need to find the proportion of defective wires in the sample and use it to estimate the number of defective wires in the production batch.
Step 1: Calculate the proportion of defective wires in the sample.
The sample contains 100 wires, and 4 of them are defective. We can write this as a fraction:
Defective wires / Total wires = 4 / 100 = 0.04
Step 2: Use the proportion to estimate the number of defective wires in the production batch.
If the proportion of defective wires in the sample is 0.04, we can assume that it applies to the entire production batch. Therefore, we multiply the proportion by the total number of wires produced in one hour:
0.04 * 750 = 30
Therefore, the manufacturer should expect around 30 defective wires in the production batch.
The correct answer is option A. 30. |
# Fun Interactive Math Worksheets For Grade 2
Delit Irina October 14, 2020 Math Worksheet
When a child learns to relate math to everyday questions, he will be great at it from the simplest addition all the way to trigonometry. To convert percentages, decimals and fractions is thus one essential skill. How much of an apple pie has been eaten? The answer to this question can be expressed in percentages, 50%; or in decimals, 0.5; or in fraction, ½. In other words, half of mom’s delicious apple pie is gone. How many kids in school have done their homework? Again this can be answered in several ways: in percentages, 70%; or in ratio, 7:10; Both of these mean out of ten kids in class there are seven good ones who did and three not-so-good ones who didn’t. The bottom line is that kids learn math much better when it makes sense.
First, the Basics! The x axis of a graph refers to the horizontal line while the y axis refers to the vertical line. Together these lines form a cross and the point where they both meet is called the origin. The value of the origin is always 0. So if you move your pencil from the origin to the right, you are drawing a line across the positive values of the x axis, i.e., 1, 2, 3 and so on. From the origin to the left, you’re moving across the negative values of the x axis, i.e., -1, -2, -3 and so on. If you go up from the origin, you are covering the positive values of the y axis. Going down from the origin, will take you to the negative values of the y axis.
What are math worksheets and what are they used for? These are math forms that are used by parents and teachers alike to help the young kids learn basic math such as subtraction, addition, multiplication and division. This tool is very important and if you have a small kid and you don’t have a worksheet, then its time you got yourself one or created one for your kid. There are a number of sites over the internet that offer free worksheets that are downloadable and printable for use by parents and teachers at home or at school.
A lot of the websites charge high dollar amount for these math worksheets. Considering your willingness to spend the high dollar, you run in to a new problem – the uniqueness of the material, a lot of websites offer sheets packed with identical problems leading the child to almost rote memorization of the problems and solution. This leads to deprivation of student and his/her ability to solve the problems logically. And some websites offer sheets in portable document format. And to open such sheets you are forced to download special software.
If you cannot purchase a math work sheet because you think you may not have time to, then you can create on using your home computer and customize it for your kid. Doing this is easy. All you need is Microsoft word application in your computer to achieve this. Just open the word application in your computer and start a new document. Ensure that the new document you are about to create is based on a template. Then, ensure that your internet connection is on before you can search the term ”math worksheet” from the internet. You will get templates of all kinds for your worksheet. Choose the one you want and then download.
Let us discuss some tangible advantages of Mathematics in today’s world. One should also be aware of the wide importance of Mathematics, and the way in which it is advancing at a spectacular rate. Mathematics is about pattern and structure; it is about logical analysis, deduction, calculation within these patterns and structures. When patterns are found, often in widely different areas of science and technology, the mathematics of these patterns can be used to explain and control natural happenings and situations. Mathematics has a pervasive influence on our everyday lives, and contributes to the wealth of the individual. Now that we know the benefits of mathematics what is needed to conquer this hard to tackle and arduous subject. Answer is quite simple actually, practice, a lot of practice from early childhood. But in order to perform any task or practice any task, one requires resources. Obtaining a resource is not difficult in today’s modern world but affordable resource is definitely a rare commodity. Today education is a sector which gets very little funding from the federal government.
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# PSEB 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1
Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Exercise Questions and Answers.
## PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1
Solve the following equations.
Question 1.
x – 2 = 7
Solution:
x – 2 = 7
∴ x = 7 + 2 (Transposing – 2 to RHS)
∴ x = 9
Question 2.
y + 3 = 10
Solution:
y + 3 = 10
∴ y = 10 – 3 (Transposing 3 to RHS)
∴ y = 7
Question 3.
6 = z + 2
Solution:
6 = z + 2
∴ z + 2 = 6 (Interchanging both the sides)
∴ z = 6 – 2 (Transposing 2 to RHS)
∴ z = 4.
Question 4.
$$\frac {3}{7}$$ + x = $$\frac {17}{7}$$
Solution:
$$\frac {3}{7}$$ + x = $$\frac {17}{7}$$
∴ x = $$\frac{17}{7}-\frac{3}{7}$$ (Transposing $$\frac {3}{7}$$ to RHS)
∴ x = $$\frac{17-3}{7}$$
∴ x = $$\frac {14}{7}$$
∴ x = 2
Question 5.
6x = 12
Solution:
6x = 12
∴ $$\frac{6 x}{6}=\frac{12}{6}$$ (Dividing both the sides by 6)
∴ x = 2
Question 6.
$$\frac{t}{5}$$ = 10
Solution:
$$\frac{t}{5}$$ = 10
∴ $$\frac{t}{5}$$ × 5 = 10 × 5 (Multiplying both the sides by 5)
∴ t = 50
Question 7.
$$\frac{2 x}{3}$$ = 15
Solution:
$$\frac{2 x}{3}$$ = 15
∴ $$\frac{2 x}{3} \times \frac{3}{2}=18 \times \frac{3}{2}$$ (Multiplying both the sides by $$\frac {3}{2}$$)
Question 8.
1.6 = $$\frac{y}{1.5}$$
Solution:
1.6 = $$\frac{y}{1.5}$$
∴ 1.6 × 1.5 = $$\frac{y}{1.5}$$ × 1.5 (Multiplying both the sides by 1.5)
∴ 2.4 = y (∵ 1.6 × 1.5 = 2.4)
∴ y = 2.4
Question 9.
7x – 9 = 16
Solution:
7x – 9 = 16
∴ 7x = 16 + 9 (Transposing – 9 to RHS)
∴ 7x = 25
∴ $$\frac{7 x}{7}=\frac{25}{7}$$ (Dividing both the sides by 7)
∴ x = $$\frac {25}{7}$$
Question 10.
14y – 8 = 13
Solution:
14y – 8 = 13
∴ 14y = 13 + 8 (Transposing – 8 to RHS)
∴ 14y = 21
∴ $$\frac{14 y}{14}=\frac{21}{14}$$ (Dividing both the sides by 14)
∴ y = $$\frac{7 \times 3}{7 \times 2}$$
∴ y = $$\frac {3}{2}$$
Question 11.
17 + 16p = 9
Solution:
17 + 16p = 9
∴ 6p = 9 – 17 (Transposing 17 to RHS)
∴ 6p = -8
∴ $$\frac{6 p}{6}=\frac{-8}{6}$$ (Dividing both the sides by 6)
∴ p = $$\frac{-4 \times 2}{3 \times 2}$$
∴ p = –$$\frac {4}{3}$$
Question 12.
$$\frac{x}{3}+1=\frac{7}{15}$$
Solution:
$$\frac{x}{3}+1=\frac{7}{15}$$
∴ $$\frac{x}{3}=\frac{7}{15}-1$$ (Transposing 1 to RHS)
∴ $$\frac{7-15}{15}$$ (LCM = 15)
∴ $$\frac{x}{3}=\frac{-8}{15}$$
∴ $$\frac{x}{3} \times 3=\frac{-8}{15} \times 3$$ (Multiplying both the sides by 3)
∴ x = –$$\frac {8}{5}$$ |
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Lesson 1: Week 8
# Intro to combining like terms
Adding like terms is a fundamental concept in algebra. Coefficients are the numbers in front of variables, and they can be added when the variables are the same. For example, 2x + 3x equals 5x. When dealing with different variables, such as x and y, add them separately, resulting in expressions like 5x + 9y. Created by Sal Khan.
## Want to join the conversation?
• So are we basically just simplifying the answer?
• @EDiaz1786
Exactly right! You're pretty much simplifying the answer. A quick tip: If you see two numbers with the same variable and it tells you to add them... Add it! Below is an example of what I mean.
3x + 4x = 7x
In this case, you would only add the whole numbers and not multiply or add the x's, because if you added/multiplied the x's it would come out to be 7x². But later on in Algebra 1, when you get to factoring and multiplying perfect squares that's when you'll multiply the x's AND the numbers. Here's a quick example of what I mean.
7x(2x+4)
This is saying that we distribute the 7x to the 2x and the 4. So the answer will be 14x² + 28x.
Hope this helps!
• I get it but im lost on the operations. how do you know when to divide, subtract, add, or multiply?
• If you are only combining like terms, stick to adding or subtracting. This depends on the coefficients, but we still use the laws of adding and subtracting positive and negative numbers. You would have to add multiply/divide only if you have to distribute something to everything in the parentheses such as 4(3x-4). Once you distribute, you get back to adding/subtracting.
• Yes, and you can also do this with other variables, as long as the variables are the same. (numerical value is just what you add or subtract)
• this might seem kind of strange but when i was trying to figure out how to add like terms i stumbled upon the fact that adding like terms can be solved by doing the distributive property in reverse. i am not sure if this makes sense or not, but it made sense to me.
for example:
2X + 3X = X(2+3)= X(5) = 5X
i do not like learning rules without understanding why the rules work. i know that you are suppose to simply add the coefficients of like terms and be done with the problem. but is that a shortcut/rule that was made instead of doing the distributive property in reverse ?
• That's precisely right. Most people take a more laid-back approach and think that two things plus three things has got to equal five things, but you're right on target that the distributive law is what's going on behind the scenes to make that simple statement work out.
• Is zero prime, composite or neither?
• Zero is neither. Zero, I've been taught is just zero. Zero is special. So, to answer your question, zero is neither.
• The chuck norris part makes this video better
• Definitely, he is hilarious.
• You had to pick Chuck Norris
hahahahahahahahahahahahahahahahahaha
its beautiful!
but still
heres my question,
Could you take the exponents and divide them by itself if one of them is x^2 and it ends up as x=?
• You can't find the value for anything by dividing it by itself. x^2 divided by x^2 equals 1. So, it still is x=?.
And yes, it's hilarious that he chose Chuck Norris as a variable.
• this is how u make math interesting
• I've had to go back on this video over and over, I still don't understand. It may be the fact that I'm just stupid and too slow to catch up with everyone else.
But seriously, here's what I have to say; When you have a simple question, such as this: -n + (-3) + 3n + 5
• When you're combining like terms, you're not actually solving for anything
(It's not an equation if you don't have the equal sign)
Combining like terms just means you add together anything you can.
-n + (-3) +3n +5
In your example, you have two types of numbers. You have numbers that are a multiple of n and you have regular numbers.
The first thing I usually do is rearrange the numbers so that all the like terms or numbers that can be added together, are next to each other, like this:
-n + 3n + (-3) +5
Then you can rearrange it some more to make it clear how to combine the like terms
3n-n + 5-3
2n + 2
Does that help?
• Is there a way to understand math better?
• the best way to understand math better is by using them regularly and developing understanding of numerical relations. |
# Representing Decimals on Number Line
Representing decimals on number line shows the intervals between two integers which will help us to increase the basic concept on formation of decimal numbers.
We have learnt to represent fraction on the number line.
Representing a decimal and a fraction on the number line is one and the same thing. In other words, we represent decimal on a number line too.
### Working Rules for Representing Decimals on Number Line:
Step I: Draw or number line and mark the whole numbers 0, 1, 2, 3, etc. on it.
Step II: Divide the portion between the whole numbers (like 0 and 1, 1 and 2, 2 and 3 etc..) into equal 2 parts or 3 parts or 4 parts or any other parts as per requirement.
1. Represent the following decimals 0.9, -1.3, -0.6 and 1.1 on a number line.
Since, 0.9 = $$\frac{9}{10}$$, -1.3 = -$$\frac{13}{10}$$, -0.6 = -$$\frac{6}{10}$$ and 1.1 = $$\frac{11}{10}$$
Divide the space between every pair of consecutive integers (on the number line) in 10 equal parts. Each part so obtained will represent the fraction $$\frac{1}{10}$$ i.e., decimal 0.1 and the number line obtained will be of the form:
To mark 0.9; move nine parts on the right-side of zero.
To mark -1.3; move thirteen parts on the left-side of zero.
To mark -0.6; move six parts on the left-side of zero.
To mark 1.1; move eleven parts on the right-side of zero.
The following diagram shows markings of decimals 0.9, -1.3, -0.6 and 1.1 on a number line.
2. Represent the decimals: 0.3, 0.7, -0.4 and -1.2 on a number line.
Since, 0.3 = $$\frac{3}{10}$$, 0.7 = $$\frac{7}{10}$$, -0.4 = -$$\frac{4}{10}$$ and -1.2 = -$$\frac{12}{10}$$
Divide the space between every pair of consecutive integers (on the number line) in 10 equal parts. Each part so obtained will represent the fraction $$\frac{1}{10}$$ i.e., decimal 0.1 and the number line obtained will be of the form:
To mark 0.3; move three parts on the right-side of zero.
To mark 0.7; move seven parts on the right-side of zero.
To mark -0.4; move four parts on the left-side of zero.
To mark -1.2; move twelve parts on the left-side of zero.
The following diagram shows markings of decimals 0.3, 0.7, -0.4 and -1.2 on a number line.
3. Draw a number line to represent decimals: 0.75, 1.50 and -1.25
Since, 0.75 = $$\frac{75}{100}$$, 1.50 = $$\frac{150}{100}$$ and -1.25 = -$$\frac{125}{100}$$
Divide the space between every pair of consecutive integers (on the number line) in 4 equal parts. Each part so obtained will represent the fraction $$\frac{1}{25}$$ i.e., decimal 0.25 and the number line obtained will be of the form:
To mark 0.75; move three parts on the right-side of zero.
To mark 1.50; move six parts on the right-side of zero.
To mark -1.25; move five parts on the left-side of zero.
The following diagram shows markings of decimals 0.75, 1.50 and -1.25 on a number line.
4. Represent the following decimals on the number line.
(i) 0.5
(ii) 1.8
(iii) 2.3
Solution:
(i) We know that 0.5 is more than zero but less than 1. There are 5 tenths in 0.5. Divide the unit length between 0 and 1 into 10 equal parts and take 5 parts as shown in figure at A.
Thus, point A shows 10 or 0.5
(ii) The decimal 1.8 has 1 as whole number and 8 tenths. Therefore, the point lies between 1 and 2. Divide the unit length between 1 and 2 into 10 equal parts and take 8 parts as shown in the figure at B. 18
Thus, point B shows $$\frac{18}{10}$$ or 1.8.
(iii) In 2.3, the whole number part is 2 and decimal part is 3.
Therefore, it lies between 2 and 3. Divide the unit length between 2 and 3 into 10 equal parts and take 3 part as shown in the figure at C.
Thus, point C shows $$\frac{23}{10}$$ or 2.3.
Thus, we have learnt how to represent and draw any decimal points on the number line.
## You might like these
• ### Expanded form of Decimal Fractions |How to Write a Decimal in Expanded
Decimal numbers can be expressed in expanded form using the place-value chart. In expanded form of decimal fractions we will learn how to read and write the decimal numbers. Note: When a decimal is missing either in the integral part or decimal part, substitute with 0.
• ### Comparison of Decimal Fractions | Comparing Decimals Numbers | Decimal
While comparing natural numbers we first compare total number of digits in both the numbers and if they are equal then we compare the digit at the extreme left. If they also equal then we compare the next digit and so on. We follow the same pattern while comparing the
Addition of decimal numbers are similar to addition of whole numbers. We convert them to like decimals and place the numbers vertically one below the other in such a way that the decimal point lies exactly on the vertical line. Add as usual as we learnt in the case of whole
• ### Subtraction of Decimal Fractions |Rules of Subtracting Decimal Numbers
The rules of subtracting decimal numbers are: (i) Write the digits of the given numbers one below the other such that the decimal points are in the same vertical line. (ii) Subtract as we subtract whole numbers. Let us consider some of the examples on subtraction
• ### Word Problems on Decimals | Decimal Word Problems | Decimal Home Work
Word problems on decimals are solved here step by step. The product of two numbers is 42.63. If one number is 2.1, find the other. Solution: Product of two numbers = 42.63 One number = 2.1
• ### Definition of Decimal Numbers | Decimal Part | Decimal Point |Examples
Definition of decimal numbers: We have learnt that the decimals are an extension of our number system. We also know that decimals can be considered as fractions whose denominators are 10, 100, 1000
• ### Decimal Place Value Chart |Tenths Place |Hundredths Place |Thousandths
Decimal place value chart are discussed here: The first place after the decimal is got by dividing the number by 10; it is called the tenths place.
• ### Like and Unlike Decimals | Concept of Like and Unlike Decimals | Defin
Concept of like and unlike decimals: Decimals having the same number of decimal places are called like decimals i.e. decimals having the same number of digits on the right of the decimal
• ### Conversion of Unlike Decimals to Like Decimals |Examples of Conversion
In conversion of unlike decimals to like decimals follow the steps of the method. Step I: Find the decimal number having the maximum number of decimal places, say (n). Step II: Now, convert each
• ### Converting Decimals to Fractions | Solved Examples | Free Worksheet
In converting decimals to fractions, we know that a decimal can always be converted into a fraction by using the following steps: Step I: Obtain the decimal. Step II: Remove the decimal points from the given decimal and take as numerator.
• ### Converting Fractions to Decimals | Solved Examples | Free Worksheet
In converting fractions to decimals, we know that decimals are fractions with denominators 10, 100, 1000 etc. In order to convert other fractions into decimals, we follow the following steps:
• ### Worksheet on Decimal Numbers | Decimals Number Concepts | Answers
Practice different types of math questions given in the worksheet on decimal numbers, these math problems will help the students to review decimals number concepts.
• ### Ordering Decimals | Comparing Decimals | Ascending & Descending Order
In ordering decimals we will learn how to compare two or more decimals. (i) Convert each of them as like decimals. (ii) Compare these decimals just as we compare two whole numbers ignoring
• ### Worksheet on Word Problems on Addition and Subtraction of Decimals
Practice the questions given in the worksheet on word problems on addition and subtraction of decimals. Read the questions carefully to add or subtract the decimals as required.
• ### Worksheet on Use of Decimal | Free Printable Decimals Worksheets
Practice the questions given in the worksheet on use of decimals in calculating money, in measuring the length, in measuring the distance, in measuring the mass and in measuring the capacity.
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
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# 6th Class Mathematics Data Handling Data Handling
Data Handling
Category : 6th Class
Data Handling
In this chapter we will learn about pictograph and bar graph.
Data
Data is a collection of facts, such as numbers, observations, words or even description of things.
Observation
Each numerical figure in a data is called observation.
Frequency
The number of times a particular observation occurs is called its frequency.
Statistical Graph
The information provided by a numerical frequency distribution is easy to understand when we represent it in terms of diagrams or graphs.
To represent statistical data, we use different types of diagrams or graphs. Some of them are:
(i) Pictograph
(ii) Bar graph
Pictograph
A pictograph represents the given data through pictures of objects. It helps to answer the questions on the data at a glance.
• Example:
The following pictography shows the number of cakes sold at a bakery over five days.
Days Number of cake Monday Tuesday Wednesday Thursday Friday
Based on above pictography answer the following questions:
I. On which day, the maximum number of cakes were sold?
(a) Monday (b) Tuesday
(c) Wednesday (d) Friday
(e) None of these
Explanation: Clearly from the pictograph, we can say that on Monday the maximum number of cakes were sold.
II. How many total number of cakes were sold over five days?
(a) 150 (b) 160
(c) 170 (d) 180
(e) None of these
Explanation: Total number of cakes sold over five days.
$=\text{ }50\text{ }+\text{ }30\text{ }+\text{ }35\text{ }+\text{ }20\text{ }+\text{ }45\text{ }=\text{ }180$
Bar Graph
A bar graph is a pictorial representation of numerical in the form of rectangles (or bars) of equal width and varying lengths. The lengths (or heights) of a rectangle depends upon the number it represents. The distance between each rectangle remains same.
Note:
• The rectangle in a bar graph can be drawn vertically or horizontally.
• In a bar graph if two sets of data are to be presented simultaneously, it is called a double bar graph, which is used in comparing the data.
• Example:
The bar graph given below represents the sales of books (in from five branches of a publishing company during a year
Sales of books (in thousand numbers) from five branches Science, English, Maths, Hindi and Computers of a publishing company daring a year.
Based on above information answer the following questions.
I. What is the ratio of sales of books of Hindi to the sales of books of science?
(a) 1 : 11 (b) 10 : 11
(c) 3 : 5 (d) 4 : 7
(e) None of these
Explanation: Clearly, from the bar graph, we get required ratio = 40000 : 44000 = 10 : 11
II. What is the difference between the maximum and minimum number of books in that year?
(a) 10,000 (b) 20,000
(c) 30,000 (d) 40,000
(e) None of these
Explanation: Required difference = 50000 - 30000 = 20000
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# Spectrum Math Grade 1 Chapter 5 Lesson 8 Answer Key Collecting Data
Practice with the help of Spectrum Math Grade 1 Answer Key Chapter 5 Lesson 5.8 Collecting Data regularly and improve your accuracy in solving questions.
## Spectrum Math Grade 1 Chapter 5 Lesson 5.8 Collecting Data Answers Key
Make a food chart for one day. Show what you ate.
1.
Explanation:
Break Fast – Bread/Cereal.
Lunch – Meat/Eggs/Fish.
Dinner -Vegetables.
Snacks – Other foods/ Fruit.
Use your food chart.
2. How many of each did you eat?
Fruit ________ Bread/Cereal ______
Vegetable _____ Other Foods ______
Meat/Eggs/Fish ______
What food did you eat the most? ________
At which meal did you eat the most? _____
What is your favorite food? _____________
=> Food I eat the most:
Fruit – Apples.
Vegetable – All kinds of vegatables.
Other Foods – Chocolates/Biscuits.
Meat/Eggs/Fish – Meat / Eggs.
Meal I eat more is Egg.
My favourite food is Vegetables.
Explanation:
Fruit ________ Bread/Cereal ______
Vegetable _____ Other Foods ______
Meat/Eggs/Fish ______
=> Food I eat the most:
Fruit – Apples.
Vegetable – All kinds of vegatables.
Other Foods – Chocolates/Biscuits.
Meat/Eggs/Fish – Meat / Eggs.
3. Make a pet chart. Ask 20 people if they have a pet. Use tally marks to show what kind.
a. How many people have _____
Number of people having dogs as pets are Seven or 7.
Explanation:
Tally marks to show:
Number of people having pets.
b. How many people have _____
Number of people having cats as pets are 2.
Explanation:
Tally marks to show:
Number of people having pets.
Tally marks to show:
Number of people having pets.
c. How many people have _______
Number of people having birds as pets are 3.
Explanation:
Tally marks to show:
Number of people having pets.
d. How many people have _____
Number of people having fishes as pets are Five or 5.
Explanation:
Tally marks to show:
Number of people having pets.
e. How many people do not have a pet? _____
Number of people having no pets are Zero or 0.
Explanation:
Tally marks to show:
Number of people having pets.
f. How many people have a pet that is not on the chart? _____
Number of people having a pet that is not on the chart (others) are three or three.
Explanation:
Tally marks to show: Number of people having pets.
Complete.
g. Which pet is the favorite? ________
Dog is the favourite pet.
Explanation:
Number of people having a dog as pet = 7.
Number of people having a cat as pet = 2.
Number of people having a bird as pet = 3.
Number of people having a fish as pet = 5.
Number of people having a others as pet = 3.
Number of people having a no pet = 0.
h. Which pet is the least favorite? ___________
Cat is the least favourite pet.
Explanation:
Number of people having a dog as pet = 7.
Number of people having a cat as pet = 2.
Number of people having a bird as pet = 3.
Number of people having a fish as pet = 5.
Number of people having a others as pet = 3.
Number of people having a no pet = 0.
Make a fruit chart. Ask 20 people if they have a favorite fruit.
4. Use tally marks to show what kind.
Use your fruit chart. Write the number.
a. How many people like _____
Number of people like apple are 4 or four.
Explanation:
b. How many people like _____
Number of people like banana are 3 or three.
Explanation:
c. How many people like _____
Number of people like orange are 5 or five.
Explanation:
d. How many people like _____
Number of people like grapes are 4 or four.
Explanation:
e. How many people like a fruit that is not on the chart? _____
Number of people like a fruit that is not on the chart (other) are 3 or three.
Explanation:
f. How many people do not like fruit? _____
Number of people do not like fruit are 1 or one.
Explanation:
Complete.
g. Which is the favorite fruit? _____
Orange is the favourite fruit.
Explanation:
Number of people like apple fruit = 4.
Number of people like banana fruit = 3.
Number of people like orange fruit = 5.
Number of people like grapes fruit = 4.
Number of people like other fruits = 3.
Number of people do not like fruit = 1.
h. Which is the least favorite fruit? _____
Banana and other fruit is the least favorite fruit.
Explanation:
Number of people like apple fruit = 4.
Number of people like banana fruit = 3.
Number of people like orange fruit = 5.
Number of people like grapes fruit = 4.
Number of people like other fruits = 3.
Number of people do not like fruit = 1.
5. How many more people chose the favorite fruit than chose the least favorite fruit? _____ |
# PRAXIS II Elementary Education 5015 - Mathematics Curriculum
### Card Set Information
Author: sout0112 ID: 152769 Filename: PRAXIS II Elementary Education 5015 - Mathematics Curriculum Updated: 2012-05-07 18:16:47 Tags: PRAXIS II Folders: Description: Review of mathematics concepts Show Answers:
1. Properties
Rules that apply for addition, subtraction, multiplication, or division of real numbers
• Commutative
• Associative
• Identity
• Inverse
• Distributive
2. Commutative
You can change the order of the terms or factors.
• a + b = b + a
• ab = ba
3. Associative
You can regroup the terms as you like.
• a + (b + c) = (a + b) + c
• a(bc) = (ab)c
4. Identity
Finding a number so that when added to a term results in that number; finding a number such that when multiplied by a term results in that number.
• a + 0 = a
• a x 1 = a
5. Inverse
Finding a number such that when added to the number it results in zero; or when multiplied by the number results in 1.
• a - a = 0
• a x (1/a) = 1
6. Distributive
This technique allows us to perate on tems with parentheses without first perfoming operations with the parentheses. This is especially helpful when terms within the parentheses cannot be combined.
a (b + c) = ab + ac
7. Arithmetic Sequence
a (sub n) = a (sub 1) + (n - 1)d
• a (sub 1) is the first term in the sequence
• d = common difference between the terms
• n = the nth term
8. Net
Two-dimensional figure that can be cut out and folded up to make a three-dimensional solid
9. Cube
Six square net
10. Tetrahedron
4 equilateral triangle net
11. Octahedron
8 equilateral triangle net
12. Icosahedron
20 equilateral triangle net
13. Dodecahedron
12 regular pentagon net
14. Tessellation
An arrangement of glosed shapes that completely covers the plane without overlapping or leaving gaps; unlike tilings, tessellations do not require the use of regular polygons
15. Four basic tranformational symmetries that can be used in tessellations
1)
2)
3)
4)
• 1) Translation
• 2) Rotation
• 3) Reflection
• 4) Glide Reflection
16. Sum of the Squares
Sum of the squares of the differences between each item and the mean
(Each number - average)(Same number - average)
17. Variance
The sum of the squares quanitity divided by the number of items
18. Standard Deviation
The square root of the variance
19. Pythagoras Theorem
a2 + b2 = c2
20. Number of yards in a mile
1,760
21. One Meter is approximately ______ yards.
One |
# Mathematical Recreations – Tetris
So, I was doodling while watching an episode of The Wire, and I came across this curiousity. Let’s look at Tetris pieces, from the popular video game Tetris. Another way to define these is the set of arrangements of four squares where you can travel from one square to any other given square by shared sides. Go ahead and try to think of any that are in that set that aren’t in the above picture, except by changes in rotation.
Now, since we’re doodling, lets see if you can draw all of the Tetris pieces without taking your pen off the paper, and without crossing over or drawing on top of an existing line. Try it. Well, it works for some, and it doesn’t for others.
Now why?! It’s a little baffling. But, like any mathematical curiousity, it is made simple by breaking things down into constituent parts. The parts of the pieces where the rules are applied directly is at the joints/crossroads, so let’s try to categorize them:
[Below here there are spoilers, beware]
Going from left to right:
One: In all of the pieces, there is no joint with only one line coming in, since this wouldn’t be able to make a square.
Two: There are many joints with two lines attached to them, but they have to be at corners, not like I’ve drawn it above. When you’re drawing the continuous line, you come in one way and go out another.
Three: If there are three lines attached to a joint, it must mean that it is an endpoint. See what I mean? Since you can’t come in to an intersection more than you go out, and vice versa, then that means you need to stop in the intersection. This will be made clearer when we look at four.
Four: Since there is an even number of lines, you can go in twice and go out twice.
Now, let’s number the joints in our Tetris shapes:
See? Only the top two shapes (I wish I had names for them) have exactly two endpoints, the points labeled with a 3. All the other pieces have more than two, and if we’re trying to draw a continuous line you can’t really stop or start more than twice, can you?
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# A bullet is shot horizontally from shoulder height
## Question
A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed of 200 meters per second.
(a) How much time elapses before the bullet hits the ground?
(b) How far does the bullet travel horizontally?
The answer to a is 0.553 seconds
While for b, the answer is 110.6 meters
#### Explanation
Data
We will need to identify our data from the question
The height, h = 1.5 m
Horizontal velocity, v = 200 m/s
Initial velocity, u = 0
Acceleration due to gravity, g = 9.8 ms-2
To find out how much time elapses before the bullet hits the ground, we will follow the method below:
Unknown
Elapsed time before the bullet hits the ground, t = ?
Formula
We will apply the formula,
h = ut + (1/2)at2
Since the initial velocity, u = 0
We can rewrite h = ut + (1/2)at2 into
h = (1/2)at2
We can now make t subject of the formula
at2 = 2h
t = √(2h/a)
but we are dealing with the force of gravity, which implies a = g
Thus
t = √(2h/g)
Therefore, to find the elapsed time (t), we will use the formula t = √(2h/g)
###### Solution
To find the elapsed time, we will now substitute our formula with our data
t = √(2h/g) = √((2 x 1.5) / 9.8)
Which implies that
t = √(3 / 9.8) = √0.306 = 0.553 seconds
Therefore, the elapsed time before the bullet hits the ground is 0.553 seconds
Unknown
Distance traveled by the bullet horizontally, s = ?
Formula
Distance traveled by the bullet horizontally, s = horizontal velocity (v) x elapsed time (t)
###### Solution
We will insert our data into the above formula
s = v x t = 200 x 0.553 = 110.6 m
Therefore, the distance traveled by the bullet horizontally is 110.6 meters |
# A projectile is shot from the ground at a velocity of 12 m/s and at an angle of (7pi)/12. How long will it take for the projectile to land?
Jun 3, 2017
$2.4$ $\text{s}$
#### Explanation:
We're asked to find the time when the projectile lands when it is launched with a known initial velocity. We need to find the time $t$ when the height $\Delta y$ is $0$.
We can use the equation
$\Delta y = {v}_{0 y} t - \frac{1}{2} g {t}^{2}$
to find this. Our known quantities are
• $\Delta y$ must be $0$, because we're trying to find the time when this occurs.
• to find the initial $y$-velocity ${v}_{0 y}$, we can use the equation ${v}_{0 y} = {v}_{y} \sin \alpha$, so ${v}_{0 y} = 12 \text{m"/"s"sin((7pi)/12) = 11.6"m"/"s}$
• $g$, the acceleration due to gravity near Earth's surface, is 9.8"m"/("s"^2)
If we make $\Delta y$ zero, we can rearrange the equation to solve for $t$:
$0 = {v}_{0 y} t - \frac{1}{2} g {t}^{2}$
$t = \frac{2 {v}_{0 y}}{g}$
Therefore,
t = (2(11.6cancel("m")/"s"))/(9.8cancel("m")/(cancel("s"^2))) = color(red)(2.4 color(red)("s"
The projectile will land after $2.4$ seconds. |
# AP Calculus AB : Chain rule and implicit differentiation
## Example Questions
### Example Question #71 : Chain Rule And Implicit Differentiation
Find the derivative using the chain rule.
Explanation:
Use the chain rule to find the derivative.
### Example Question #72 : Chain Rule And Implicit Differentiation
Find the derivative using the chain rule.
Explanation:
Use the chain rule to find the derivative.
### Example Question #73 : Chain Rule And Implicit Differentiation
Find the derivative using the chain rule.
Explanation:
Use the chain rule to find the derivative.
### Example Question #74 : Chain Rule And Implicit Differentiation
.
Which of the following expressions is equal to ?
Explanation:
Differentiate both sides with respect to :
By the sum rule:
By the chain rule:
Applying some algebra:
### Example Question #75 : Chain Rule And Implicit Differentiation
Which of the following is equal to ?
Explanation:
Differentiate both sides with respect to :
Apply the sum, difference, and constant multiple rules:
In the first term, apply the chain rule; in the second, apply the constant multiple rule:
Apply the power rule:
Now apply some algebra:
### Example Question #71 : Chain Rule And Implicit Differentiation
We have three functions,
Find the derivative of
Given that
Explanation:
So now this is a three layer chain rule differentiation. The more functions combine to form the composite function the harder it will be to keep track of the derivative. I find it helpful to lay out each equation and each derivative, so:
Then a three layer chain rule is just the same as a two layer, except... there's one more layer!
It is still the outermost layer evaluated at the inner layers, and then move another layer in and repeat
### Example Question #77 : Chain Rule And Implicit Differentiation
Find at with the equation
Explanation:
So with implicit differentiation, you are going to be taking the derivative of every variable, in the entire equation. Every time you take the derivative of a variable, you have it's rate of change multiplied on the right. In this case, dx or dy.
The result of the derivative is:
The first step is to create the term that you are looking to solve for. This is done by dividing the entire equation by to get it on the bottom of the fraction. After distributing this division to each term, the dx in the first term will cancel with itself, and you will be left with one term that is multiplied by . At that point, you want to get the term with onto its own side. This can be accomplished by subtracting the to the right side of the equation.
The result so far:
Then to finish getting on its own, you divide to the right side, ending up with:
So now looking at the question, we know that , so in order to figure out we need to plug into the equation.
This gives us .
So this gives us two possible answers:
### Example Question #72 : Chain Rule And Implicit Differentiation
Find the derivative of
Explanation:
So the derivative of a natural log is always equal to or one over whatever is inside the natural log. In this case is inside the natural log, so the derivative of should be:
But since the inside of the natural log is a function as well, this is the chain rule and the derivative of the natural log will be multiplied by the derivative of the inside, in this case , which is
So the final derivative is
### Example Question #73 : Chain Rule And Implicit Differentiation
Compute the derivative of the following expression:
Explanation:
This problem involves using product rule, and chain rule.
By product rule,
Then, using power rule, and ten chain.
Another application of chain rule to get at the angle,
Taking the derivative of the angle and then simplifying, we get
### Example Question #74 : Chain Rule And Implicit Differentiation
Find the first derivative of the function: |
Share. (c) Moment of Inertia of a rectangular section (2) Fig. Problem 724 Find the coordinates of the centroid of the shaded area shown in Fig. • For moment of inertia of an area known about an axis passing through its centroid, determine the moment of inertia of area about a corresponding parallel axis using the parallel axis theorem • Consider moment of inertia of the shaded area • A differential element dA is located at an arbitrary distance y’from the centroidal x’axis Figure 20 shows a section of three regular areas A 1, A 2, and A 3. Examples of how to use “moment of inertia” in a sentence from the Cambridge Dictionary Labs Moment of Inertia is the quantity that expresses an object’s resistance to change its state of rotational motion. Example $$\PageIndex{6}$$: Finding Moments of Inertia for a Triangular Lamina. Parallel axis theorem and perpendicular axis theorem are used to solve problems on moment of inertia, let us discuss the two theorems, Parallel axis theorem states, Parallel Axis Theorem • Moment of inertia I T of a circular area with respect to a tangent to the circle, Example: y 200 (Dimensions in mm) z Ctidl 10 o Centroidal. An example is shown below. Area properties and Moments of Inertia To be able to deal with more complicated structures and loading scenarios, we will learn how to replace a distributed loading with an equivalent concentrated one using its geometric properties, such as its area and centroid coordinates. Read more about 819 Inverted T-section | Moment of Inertia; Log in or register to post comments; 54156 reads; 724 Rectangle, semicircle, quarter-circle, and triangle | Centroid of Composite Area. 4.5 Parallel-Axis Theorem - Theory - Example - Question 1 - Question 2. Example-2 Find the centroid of a 120 mm x 150 mm x 20 mm T section. The smallest Moment of Inertia about any axis passes throught the centroid. Question 2. Determine the moment of inertia of the T-section shown in Fig. Q1 Locate the centroid of a circle, rectangle and square with the help of example. The moment of inertia of a T section is calculated by considering it as 2 rectangular segments. (e) Moment of Inertia of a circular section is the moment of inertia about the centroid of the component area d is the distance from the centroid of the component area to the centroid of the composite area (ie. Mechanics of Material (CIV101) Academic year. B) Determine The Vertical Shear Force, V Acting At Section A-a And Determine The Shear Stress, At Point B On The Web (vertical Member) Side Of The Cantilever Strut At Section A-a. (2) y is the distance from the x axis to an infinetsimal area dA. 2. dA , d I. y = x. Centroid and Moments of Inertia 1. Draw a reference origin. The points X’and Y’corresponding to the x’and y’axes are obtained by rotating CX and CY counterclockwise through an angle θ 2(60o) = 120o. Comments. Apply the Parallel Axes Theorem to find the moment of inertia of each subarea around the global axis. Home > Resources > How to find centroid with examples. (d) Moment of Inertia of the triangular section about an axis passing through its centroid and parallel to base (3) Fig. Solution for 1.Find centroid relative to x and y axes 2. Where to locate the large mass components are oftentimes dictated by the regulations of the sport, NASCAR does not allow mid-engine cars for instance; however, there are still areas for creativity. (1) Fig. Use the triangular region $$R$$ with vertices $$(0,0), \, (2,2)$$, and $$(2,0)$$ and with density $$\rho (x,y) = xy$$ as in previous examples. To compute the moment of inertia with respect to a given axis Instructions: 1. Find the moments of inertia. UNIT 3 Centroid & Moment of Inertia Learning Objectives After studying this unit, the student will be able to • Know what is centre of gravity and centroid • Calculate centroid of geometric sections Centre of Gravity Centre of Gravity (or) mass centre of a point in the body where entire mass weight – is assumed to be concentrated. Spinning figure skaters can reduce their moment of inertia by pulling in their arms, allowing them to spin faster due to conservation of angular momentum. Question 2. The moment of inertia, ... inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section . The position vector of a point P relative to a point O is r P and a scalar associated with P is s, for example, the mass m of a particle situated at P. The first moment of a point P with respect to a point O is the vector M = s r P. The scalar s is This implies that there exist a set of about which it is computed. - Theory - Example - Question 1 - Question 2 - List of moment of inertia for common shapes. Play media. d y = y√-y) Basic Steps 1. 2. dA , and, d J. O = r. 2. dA , where J. O. is the polar moment of inertia about the pole O or z axis. Divide the area into basic shapes 3. Use MsWord for … University of Sheffield. Moment of inertia of circular section. PRODUCT OF INERTIA PRODUCT OF INERTIA The inertia of an area is a function of the location of the axis The inertia of an area is a function of the location of the axis about which it is computed. Another Example We can locate the centroid of each area with respect the y axis. The moment of inertia is separately calculated for each segment and put in the formula to find the total moment of inertia. Area Moments of Inertia Example: Mohr’s Circle of Inertia 6 4 6 4 3.437 10 mm 4.925 10 mm R OC I ave • Based on the circle, evaluate the moments and product of inertia with respect to the x’y’axes. You can show the division by drawing solid or broken lines across the irregular shape. Centroid and Moment of Inertia 4.1 Centre of Gravity Everybody is attracted towards the centre of the earth due gravity. Although it is a simple matter to determine the moment of inertia of each rectangular section that makes up the beam, they will not reference the same axis, thus cannot be added. Question: A)Determine The Location Of The Centroid From The Bottom Of The Strut And Determine The Moment Of Inertia Along The Neutral Axis Of The Strut. Justify this statement. Calculate the moment of inertia for Ix and Iy relative to centroid Using the expressions established above for the moments of inertia, we have The following are the mathematical equations to calculate the Moment of Inertia: I x: equ. A cross section for a beam is shown in the figure below. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. 2. Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis choice so we can compare the results. Moment of Inertia of Composite Section: Statement: The moment of inertial of a composite section is equal to the sum of the moments of inertia of its individual parts. Determine the moment of inertia of each subarea, around a parallel axis, passing through subarea centroid. P-819 with respect to its centroidal X o axis. In image processing, computer vision and related fields, an image moment is a certain particular weighted average of the image pixels' intensities, or a function of such moments, usually chosen to have some attractive property or interpretation.. 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# Determinants
### Determinants of a Matrix:
1. Definition:
• For a square matrix $A$, the determinant $\text{det}\left(A\right)$ is a scalar value calculated from its elements.
• It is denoted by vertical bars or $\text{det}\left(A\right)$ or $\mathrm{\mid }A\mathrm{\mid }$.
2. Calculation:
• For a $2×2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, $\text{det}\left(A\right)=ad-bc$.
• For larger matrices, the calculation involves expansion by minors or other methods like row reduction.
For a 3x3 matrix $A$ of the form:
$A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$
The determinant $\text{det}\left(A\right)$ is computed using the following formula:
$\text{det}\left(A\right)=aei+bfg+cdh-ceg-bdi-afh$
Expansion by Minors:
Another way to calculate the determinant of a $3×3$ matrix is by using expansion by minors. It involves breaking down the matrix into smaller determinants based on cofactors.
For matrix $A$:
$A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$
The determinant $\text{det}\left(A\right)$ can be calculated as:
$\text{det}\left(A\right)=a\text{det}\left(\left[\begin{array}{cc}e& f\\ h& i\end{array}\right]\right)-b\text{det}\left(\left[\begin{array}{cc}d& f\\ g& i\end{array}\right]\right)+c\text{det}\left(\left[\begin{array}{cc}d& e\\ g& h\end{array}\right]\right)$ Where $\text{det}\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=xw-yz$.
### Properties of Determinants:
1. Multiplicative Property: $\text{det}\left(AB\right)=\text{det}\left(A\right)×\text{det}\left(B\right)$ for square matrices $A$ and $B$.
2. Transpose Property: $\text{det}\left(A\right)=\text{det}\left({A}^{T}\right)$ for any square matrix $A$.
3. Inverse Property: If $A$is invertible, then $\text{det}\left({A}^{-1}\right)=\frac{1}{\text{det}\left(A\right)}$.
4. Scalar Multiplication: $\text{det}\left(kA\right)={k}^{n}×\text{det}\left(A\right)$ for a square matrix $A$ of order $n$ and scalar $k$.
### Significance and Applications:
1. Matrix Inverses: A square matrix $A$ is invertible if and only if $\text{det}\left(A\right)\mathrm{\ne }0$.
2. System Solvability: For a system of linear equations $AX=B$, the system has a unique solution if and only if $\text{det}\left(A\right)\mathrm{\ne }0$.
3. Geometric Interpretation: The absolute value of the determinant represents the scaling factor in transformations and encodes information about the orientation change and volume scaling in space.
### Cramer's Rule:
Cramer's rule is a technique to solve systems of linear equations using determinants. For a system $AX=B$ where $A$ is a square matrix and $\text{det}\left(A\right)\mathrm{\ne }0$:
• The solution for ${x}_{i}$ can be expressed as ${x}_{i}=\frac{\text{det}\left({A}_{i}\right)}{\text{det}\left(A\right)}$ where ${A}_{i}$ is obtained by replacing the $i$th column of $A$ with $B$.
### Limitations:
• Calculating determinants becomes computationally expensive for larger matrices due to expansion by minors or other methods. |
# Testing the Difference between Proportions Section 11.3.
## Presentation on theme: "Testing the Difference between Proportions Section 11.3."— Presentation transcript:
Testing the Difference between Proportions Section 11.3
Assumptions Randomly Selected Samples Approximately normal since and Independent (at least 10n in population)
Normally the Standard Deviation of Statistic is
But, Since we claim in the H o We can combine the values to form one proportion: And the Standard Deviation of Statistic becomes
Can you find this on the formula Sheet? They use a combined p.
A sample of 50 randomly selected men with high triglyceride levels consumed 2 tablespoons of oat bran daily for six weeks. After six weeks, 60% of the men had lowered their triglyceride level. A sample of 80 men consumed 2 tablespoons of wheat bran for six weeks. After six weeks, 25% had lower triglyceride levels. Is there a significant difference in the two proportions at the 0.01 significance level? To calculate p c we need to find x 1 and x 2. So…..
Parameter: Assumptions: * Randomly Selected Samples * Approximately Normal since * Independent – (at least 500 men eat oat and 800 eat wheat bran ) Name of Test: 2-Proportion Z-Test Hypothesis:
Reject the Ho since the P-Value(0) < (0.05) There is sufficient evidence to support the claim that there is a difference in the proportion of men who lowered their triglycerides by eating oat bran and the proportion who lowered their triglycerides by eating wheat bran.
In a sample of 100 store customers, 43 used a Mastercard. In another sample of 100, 58 used a Visa card. Is the proportion of customers who use Mastercard less than those using Visa? Assumptions: 1. Randomly Selected Samples 2. Approx. Normal 3. Independent (at least 1000 of each)
Reject the Ho since the p-val(.017) < (0.05) There is sufficient evidence to support the claim that the proportion using mastercard is less than the proportion using visa.
So how would we find a confidence interval? PANIC!
In a sample of 80 Americans, 55% wished that they were rich. In a sample of 90 Europeans, 45% wished that they were rich. Is there a difference in the proportions. Find and interpret the 95% confidence interval for the difference of the two proportions. Assumptions: 1. Randomly Selected Samples 2. Approx Norm 3. Independent (at least 800 Am and 900 Europeans.
We’re 95% confident that the difference in proportion of Americans who wish to be rich and the proportion of Europeans who wish to be rich is between -.05 and.25. In fact, since this interval contains 0, there is no significant difference.
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# What is 122/106 as a decimal?
## Solution and how to convert 122 / 106 into a decimal
122 / 106 = 1.151
122/106 converted into 1.151 begins with understanding long division and which variation brings more clarity to a situation. Both are used to handle numbers less than one or between whole numbers, known as integers. In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. So let’s dive into how and why you can convert 122/106 into a decimal.
## 122/106 is 122 divided by 106
Converting fractions to decimals is as simple as long division. 122 is being divided by 106. For some, this could be mental math. For others, we should set the equation. Fractions have two parts: Numerators and Denominators. This creates an equation. Now we divide 122 (the numerator) into 106 (the denominator) to discover how many whole parts we have. Here's how our equation is set up:
### Numerator: 122
• Numerators are the parts to the equation, represented above the fraction bar or vinculum. 122 is one of the largest two-digit numbers you'll have to convert. The good news is that 122 is an even number which can simplify equations (sometimes). Large two-digit conversions are tough. Especially without a calculator. So how does our denominator stack up?
### Denominator: 106
• Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. 106 is one of the largest two-digit numbers to deal with. And it is nice having an even denominator like 106. It simplifies some equations for us. Ultimately, don't be afraid of double-digit denominators. Now it's time to learn how to convert 122/106 to a decimal.
## Converting 122/106 to 1.151
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 106 \enclose{longdiv}{ 122 }$$
We will be using the left-to-right method of calculation. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Solve for how many whole groups you can divide 106 into 122
$$\require{enclose} 00.1 \\ 106 \enclose{longdiv}{ 122.0 }$$
Since we've extended our equation we can now divide our numbers, 106 into 1220 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiply by the left of our equation (106) to get the first number in our solution.
### Step 3: Subtract the remainder
$$\require{enclose} 00.1 \\ 106 \enclose{longdiv}{ 122.0 } \\ \underline{ 106 \phantom{00} } \\ 1114 \phantom{0}$$
If you don't have a remainder, congrats! You've solved the problem and converted 122/106 into 1.151 If there is a remainder, extend 106 again and pull down the zero
### Step 4: Repeat step 3 until you have no remainder
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 122/106 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. This is also true for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But 122/106 and 1.151 bring clarity and value to numbers in every day life. Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 122/106 into a decimal
Finance or Money Related Questions: Imagine if you had \$200 to buy a new outfit. If you wait for the items to go on sale, you aren’t going to buy everything for 122/106 off. No, you’ll get everything on 115% discount or 1.151 of the full price.
### When to convert 1.151 to 122/106 as a fraction
Progress - If we were writing an essay and the teacher asked how close we are to done. We wouldn't say .5 of the way there. We'd say we're half-way there. A fraction here would be more clear and direct.
### Practice Decimal Conversion with your Classroom
• If 122/106 = 1.151 what would it be as a percentage?
• What is 1 + 122/106 in decimal form?
• What is 1 - 122/106 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 1.151 + 1/2? |
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# Hypothesis Testing: Two Population Means and Two Population Proportions: Comparing Two Independent Population Means with Unknown Population Standard Deviations
Summary: Note: This module is currently under revision, and its content is subject to change. This module is being prepared as part of a statistics textbook that will be available for the Fall 2008 semester.
Note: You are viewing an old version of this document. The latest version is available here.
1. The two independent samples are simple random samples from two distinct populations.
2. Both populations are normally distributed with the population means and standard deviations unknown.
The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, X 1 ¯ X 1 - X 2 ¯ X 2 , and divide by the standard error (shown below) in order to standardize the difference. The result is a t-score test statistic (shown below).
Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, of the difference in sample means, X 1 ¯ X 1 - X 2 ¯ X 2 .
## The standard error is:
( S 1 ) 2 n 1 + ( S 2 ) 2 n 2 ( S 1 ) 2 n 1 + ( S 2 ) 2 n 2
(1)
The test statistic (t-score) is calculated as follows:
## T-score
( X 1 ¯ - X 2 ¯ ) - ( μ 1 ¯ - μ 2 ¯ ) ( S 1 ) 2 n 1 + ( S 2 ) 2 n 2 ( X 1 - X 2 ) - ( μ 1 - μ 2 ) ( S 1 ) 2 n 1 + ( S 2 ) 2 n 2
(2)
## where:
• s1s1 and s2s2, the sample standard deviations, are estimates of σ1σ1 and σ2σ2, respectively.
• σ1σ1 and σ2σ2 are the unknown population standard deviations.
• x 1 ¯ x 1
• and x 2 ¯ x 2 are the sample means. μ1μ1 and μ2μ2 are the population means.
The degrees of freedom (df) is a somewhat complicated calculation. However, a computer or calculator calculates it easily. The dfs are not always a whole number. The test statistic calculated above is approximated by the Student-t distribution with dfs as follows:
## Degrees of freedom
df = [ ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 ] 2 1 n 1 1 · [ ( s 1 ) 2 n 1 ] 2 + 1 n 2 1 · [ ( s 2 ) 2 n 2 ] 2 df= [ ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 ] 2 1 n 1 1 · [ ( s 1 ) 2 n 1 ] 2 + 1 n 2 1 · [ ( s 2 ) 2 n 2 ] 2
(3)
When both sample sizes n1n1 and n2n2 are five or larger, the Student-t approximation is very good. Notice that the sample variances s 1 2 s 1 2 and s 2 2 s 2 2 are not pooled. (If the question comes up, do not pool the variances.)
## Note:
It is not necessary to compute this by hand. A calculator or computer easily computes it.
## Example 1: Independent groups
The average amount of time boys and girls ages 7 through 11 spend playing sports each day is believed to be the same. An experiment is done, data is collected, resulting in the table below:
Table 1
Sample Size Average Number of Hours Playing Sports Per Day Sample Standard Deviation
Girls 9 2 hours 0.750.75
Boys 16 3.2 hours 1.00
### Problem 1
Is there a difference in the average amount of time boys and girls ages 7 through 11 play sports each day? Test at the 5% level of significance.
#### Solution
The population standard deviations are not known. Let gg be the subscript for girls and bb be the subscript for boys. Then, μgμg is the population mean for girls and μbμb is the population mean for boys. This is a test of two independent groups, two population means.
Random variable: X g ¯ - X b ¯ X g - X b = difference in the average amount of time girls and boys play sports each day.
H o H o : μ g = μ b ( μ g μ b = 0 ) μ g = μ b ( μ g μ b =0)
H a H a : μ g μ b ( μ g μ b 0 ) μ g μ b ( μ g μ b 0)
The words "the same" tell you H o H o has an "=". Since there are no other words to indicate H a H a , then assume "is different." This is a two-tailed test.
Distribution for the test: Use t df t df where df df is calculated using the df df formula for independent groups, two population means. Using a calculator, df df is approximately 18.8462. Do not pool the variances.
Calculate the p-value using a Student-t distribution: p-value = 0.0054
Graph:
s g = 0.75 s g = 0.75
s b = 1 s b =1
So, x g ¯ - x b ¯ = 2 - 3.2 = - 1.2 x g - x b =2-3.2=-1.2
Half the p-value is below -1.2 and half is above 1.2.
Make a decision: Since α>α> p-value, reject H o H o .
This means you reject μ g = μ b μ g = μ b . The means are different.
Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the average number of hours that girls and boys aged 7 through 11 play sports per day is different.
##### Note:
TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 4:2-SampTTest. Arrow over to Stats and press ENTER. Arrow down and enter 2 for the first sample mean, .75 for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER. Arrow down to Pooled: and No. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw.
## Example 2
A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is 4 math classes with a standard deviation of 1.5 math classes. College B samples 9 graduates. Their average is 3.5 math classes with a standard deviation of 1 math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Test at a 1% significance level. Answer the following questions.
### Problem 1
• Is this a test of two means or two proportions?
• Are the populations standard deviations known or unknown?
• Which distribution do you use to perform the test?
### Problem 2
What is the random variable?
#### Solution
X A ¯ - X B ¯ X A - X B
### Problem 3
What are the null and alternate hypothesis?
#### Solution
• H o : μ A μ B H o : μ A μ B
• H a : μ A > μ B H a : μ A > μ B
### Problem 4
• Is this test right, left, or two tailed?
• What is the p-value?
### Problem 5
Do you reject or not reject the null hypothesis?
#### Solution
##### Conclusion:
At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B.
## Glossary
Degrees of Freedom (df):
The number of objects in a sample that are free to vary.
Standard Deviation:
A number that is equal to the square root of the variance and measures how far data values are from their mean. Notations: s for sample standard deviation and σσ for population standard deviation.
Variable (Random Variable):
A characteristic of interest in a population being studied. Common notation for variables are upper case Latin letters XX size 12{X} {}, YY size 12{Y} {}, ZZ size 12{Z} {},...; common notation for specific value from the domain (set of all possible values of a variable) are lower case Latin letters xx size 12{x} {}, yy size 12{y} {}, zz size 12{z} {},.... For example, if XX size 12{X} {} is a number of children in a family, then domain is and xx size 12{x} {} represents any integer from 0 to 20. Variable in statistics differs from variable in intermediate algebra in two following ways.
• The domain of random variable (RV) is not necessarily numerical set; it can be some “wording” set; for example, if XX size 12{X} {} = hair color then the domain is {black, blond, gray, green, orange}.
• We can tell what specific value of xx size 12{x} {} does the variable XX size 12{X} {} take only after performing the experiment.
Before the experiment any value from domain is possible. For example, without ultrasound we can not tell the gender of a baby that should be delivered, but after delivery the gender is evident. More exact, every value from the domain is accompanied with some number pp size 12{p} {}, 0p10p1 size 12{0 <= p <= 1} {}, that characterizes the chance to have this value as an outcome of the experiment. In the example with gender, p=12p=12 size 12{p= { {1} over {2} } } {}. That’s why statisticians use more exact name “Random variable” (RV) instead of variable. Even more, they use word “distribution” having in the mind the RV, that is the pairing (value, probability of the value).
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# Unit 7Big Ideas
## Negative Numbers and Absolute Value
This week, your student will work with signed numbers, or positive and negative numbers. We often compare signed numbers when talking about temperatures. For example, -30 degrees Fahrenheit is colder than -10 degrees Fahrenheit. We say “-30 is less than -10” and write: .
We also use signed numbers when referring to elevation, or height relative to the sea level. An elevation of 2 feet (which means 2 feet above sea level) is higher than an elevation of -4 feet (which means 4 feet below sea level). We say “2 is greater than -4” and write .
We can plot positive and negative numbers on the number line. Numbers to the left are always less than numbers to the right.
We can see that -1.3 is less than 0.8 because -1.3 is to the left of 0.8, but -1.3 is greater than -2.7 because it is to the right of -2.7.
We can also talk about a number in terms of its absolute value, or its distance from zero on the number line. For example, 0.8 is 0.8 units away from zero, which we can write as , and -2.7 is 2.7 units away from zero, which we can write as . The numbers -3 and 3 are both 3 units from 0, which we can write as and .
1. A diver is at the surface of the ocean, getting ready to make a dive. What is the diver’s elevation in relation to sea level?
2. The diver descends 100 feet to the top of a wrecked ship. What is the diver’s elevation now?
3. The diver descends 25 feet more toward the ocean floor. What is the absolute value of the diver’s elevation now?
4. Plot each of the three elevations as a point on a number line. Label each point with its numeric value.
Solution:
1. 0, because sea level is 0 feet above or below sea level
2. -100, because the diver is 100 feet below sea level
3. The new elevation is -125 feet or 125 feet below sea level, so its absolute value is 125 feet.
4. A number line with 0, -100, and -125 marked, as shown:
This week your student will be adding and subtracting with negative numbers. We can represent this on a number line using arrows. The arrow for a positive number points right, and the arrow for a negative number points left. We add numbers by putting the arrows tail to tip.
For example, here is a number line that shows .
The first number is represented by an arrow that starts at 0 and points 5 units to the left. The next number is represented by an arrow that starts directly above the tip of the first arrow and points 12 units to the right. The answer is 7 because the tip of this arrow ends above the 7 on the number line.
In elementary school, students learned that every addition equation has two related subtraction equations. For example, if we know , then we also know and .
The same thing works when there are negative numbers in the equation. From the previous example, , we also know and .
1. Use the number line to show .
1. ?
2. ?
Solution:
1. The first arrow starts at 0 and points 3 units to the right. The next arrow starts at the tip of the first arrow and points 5 units to the left. This arrow ends above the -2, so .
2. From the addition equation , we get the related subtraction equations:
## Inequalities
This week, your student will compare positive and negative numbers with inequalities symbols (< and >). They will also graph inequalities in one variable, such as or , on the number line.
For example, to represent the statement “the temperature in Celsius () is less than 1 degree,” we can write the inequality and draw a number line like this:
The diagram shows all numbers to the left of 1 (or less than 1) being possible values of .
We call any value of that makes an inequality true a solution to the inequality.
This means values that are greater than -8 are solutions to the inequality . Likewise, values that are less than 15 could be a solution to the inequality . Depending on the context, however, the solutions may include only positive whole numbers (for example, if represents the number of students in a class), or any positive and negative numbers, not limited to whole numbers (for example, if represents temperatures).
A sign at a fair says, “You must be taller than 32 inches to ride the ferris wheel.” Write and graph an inequality that shows the heights of people who are tall enough to ride the ferris wheel.
Solution:
If represents the height of a person in inches, then the inequality represents the heights of people who can ride the ferris wheel. We can also write the inequality .
The graph of the inequality is:
## The Coordinate Plane
This week, your students will plot and interpret points on the coordinate plane. In earlier grades, they plotted points where both coordinates are positive, such as point in the figure. They will now plot points that have positive and negative coordinates, such as points and .
To find the distance between two points that share the same horizontal line or the same vertical lines, we can simply count the grid units between them. For example, if we plot the point on the grid above (try it!), we can tell that the point will be 7 units away from point .
Points on a coordinate plane can also represent situations that involve positive and negative numbers. For instance, the points on this coordinate plane shows the temperature in degrees Celsius every hour before and after noon on a winter day. Times before noon are negative and times after noon are positive.
For example, the point tells us that 5 hours after noon, or 5:00 p.m, the temperature was 10 degrees Celsius.
In the graph of temperatures above:
1. What was the temperature at 7 a.m.?
2. For which recorded times was it colder than 5 degrees Celsius?
Solution:
1. It was -5 degrees Celsius at 7:00 a.m. You can see this at the point .
2. It was 5 degrees Celsius right at noon, and for the times recorded before that, it was colder.
## Common Factors and Common Multiples
This week, your student will solve problems that involve factors and multiples. Because , we say that 2 and 6 are factors of 12, and that 12 is a multiple of both 2 and 6. The number 12 has other factors: 1, 3, 4, and 12 itself.
Factors and multiples were studied in earlier grades. The focus here is on common factors and common multiples of two whole numbers. For example, 4 is a factor of 8 and a factor of 20, so 4 is a common factor of 8 and 20. 80 is a multiple of 8 and a multiple of 20, so 80 is a common multiple of those two numbers.
One way to find the common factors of two numbers is to list all of the factors for each number and see which factors they have in common. Sometimes we want to find the greatest common factor. To find the greatest common factor of 18 and 24, we first list all the factors of each number and look for the greatest one they have in common.
• Factors of 18: 1, 2, 3, 6, 9,18
• Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
The common factors are 1, 2, 3, and 6. Of these, 6 is the greatest one, so 6 is the greatest common factor of 18 and 24.
To find the common multiples of two numbers, we can do the same. Sometimes we want to find the least common multiple. Let’s find the least common multiple of 18 and 24.
• Multiples of 18: 18, 36, 54, 72, 90, 108, 126, 144, . . .
• Multiples of 24: 24, 48, 72, 96, 120, 144, 168, 192, . . .
The first two common multiples are 72 and 144. We can see that 72 is the least common multiple.
A cook is making cheese sandwiches to sell. A loaf of bread can make 10 sandwiches. A package of cheese can make 15 sandwiches. How many loaves of bread and how many packages of cheese should the cook buy so that he can make cheese sandwiches without having any bread or any cheese left over?
Solution:
If he is using up the entire loaf of bread, then the number of sandwiches he can make will be a multiple of 10: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, . . .
If he is using up all of the cheese in each package, then the number of sandwiches he can make will be a multiple of 15: 15, 30, 45, 60, 75, 90, 105, . . .
30, 60, and 90 are some of the common multiples.
• To make 30 sandwiches, he will need 3 loaves of bread () and 2 packages of cheese ().
• To make 60 sandwiches, he will need 6 loaves of bread and 4 packages of cheese.
• To make 90 sandwiches, he will need 9 loaves of bread and 6 packages of cheese.
There are other solutions as well! If he wants to buy the fewest number of loaves and cheese packages, then the first solution is the least. |
# How to Find the Sum to Infinity of a Geometric Series
## What is the Sum to Infinity?
The sum to infinity is the result of adding all of the terms in an infinite geometric series together. It is only possible to calculate the sum to infinity for geometric series that converge. This means that the size of each new term must be smaller than its previous term.
A geometric series is obtained when each term is multiplied by the same number from one term to the next. The value that each term is multiplied by to get to the next term is called the ratio.
For example, in the sequence 4+2+1+0.5+…, the terms are halving each time. Therefore .
means the sum of the first ‘n’ terms.
For example, and .
As more terms are added, we see that , , and .
Because the terms are getting smaller and smaller, as we add more terms, we are adding an increasingly negligible amount.
Progressing further, , . We can see that the sum is approaching 8.
Even adding the first 20 terms, .
Eventually, if an infinite number of terms could be added, the sum would indeed approach 8.
We say that the sum to infinity is 8, or
The sum to infinity of the series is calculated by , where is the first term and r is the ratio between each term.
For this series, where and , which becomes .
The sum of an infinite number of terms of this series is 8.
This means that the sequence sum will approach a value of 8 but never quite get there.
## How to Find the Sum to Infinity of a Geometric Series
The sum to infinity of a geometric series is given by the formula S=a1/(1-r), where a1 is the first term in the series and r is found by dividing any term by the term immediately before it.
• a1 is the first term in the series
• ‘r’ is the common ratio between each term in the series
The sum to infinity of a geometric series
To find the sum to infinity of a geometric series:
1. Calculate r by dividing any term by the previous term.
2. Find the first term, a1.
3. Calculate the sum to infinity with S = a1 ÷ (1-r).
For example, find the sum to infinity of the series
Step 1. Calculate r by dividing any term by the previous term
We can divide the term by the term before it, which is 1.
and so, .
It does not matter which term you choose, simply divide any term by the term before it to find the value of r.
For example, the same result is obtained by considering the last two terms instead: .
Step 2. Find the first term, a1
The first term is simply the first number in the series, which is 1.
Step 3. Calculate the sum to infinity with S = a1 ÷ (1-r)
The sum to infinity is given by .
and .
Therefore the sum to infinity becomes which becomes . This simplifies to or .
## When Does the Sum to Infinity Exist?
The sum to infinity only exists if -1<r<1. If the common ratio is outside of this range, then the series will diverge and the sum to infinity will not exist. If |r|<1, the sequence will converge to the sum to infinity given by S=a/(1-r).
A convergent geometric series is one in which the terms get smaller and smaller. This means that the terms being added to the total sum get increasingly small. The series converges to a final value.
For example, in the series , the fractions can be seen to fit inside the area of a 1 by 1 square. Therefore the fractions will fill an area of .
The series converges to 1.
The series converges because the terms are getting smaller in magnitude. We are adding less and less each time.
Geometric series converge and have a sum to infinity if |r|<1. The common ratio must be between -1 and 1.
A geometric series diverges and does not have a sum to infinity if |r|≥1. If the terms get larger as the series progresses, the series diverges.
The sum to infinity does not exist if |r|≥1.
For example, the series is a divergent series because the terms get larger. The common ratio is 2 and a geometric series will diverge if |r|≥1.
For a series to converge, the terms must get smaller and smaller in magnitude as the series progresses.
For a geometric series, the series converges if |r|<1.
Arithmetic series do not converge and so they do not have a defined sum to infinity. If the common difference is positive, then the sum to infinity of an arithmetic series is +∞. If the common difference is negative, the sum to infinity is -∞.
## Sum to Infinity Calculator
Enter the first two terms of a geometric sequence into the calculator below to calculate its sum to infinity.
## Negative Sum to Infinity
The sum to infinity of a geometric series will be negative if the first term of the series is negative.
This is because the sum to infinity is given by .
For a sum to infinity to exist, . This means that the denominator of the sum to infinity equation can never be negative.
The only way to obtain a negative sum to infinity is for the numerator, a1, to be negative.
a1 is the first term of the series. Hence, if the first term is negative, the sum to infinity will also be negative.
For example, find the sum to infinity of
Here, and .
Therefore the sum to infinity becomes which equates to .
### Sum to Infinity of an Alternating Series
A geometric series will alternate between positive and negative terms if the ratio is negative.
For example, in the series the terms alternate from negative to positive.
The ratio, .
Since , the sum to infinity becomes .
The denominator simplifies to and this can be evaluated so that .
## Examples of Calculating the Sum to Infinity
Here are sum examples of calculating the sum to infinity for geometric series.
In each case, the sum to infinity formula will be used, where a1 is the first term and r is the ratio.
## How to Write a Recurring Decimal as a Fraction with an Infinite Series
Recurring decimals can be written as a fraction using the geometric infinite series formula S=a/[1-r]. A decimal can be written as fractions out of 10, 100, 1000 and so on. Written in this way, the recurring decimal can be written as a geometric series in which the first term and ratio can be found.
### Recurring Decimal to Fraction: Example 1
For example, write as a fraction.
The recurring decimal can be written as a series of fractions out of 10, 100, 1000 and so on.
The first term, .
The ratio is since is divided by 10 to make and so on.
Therefore the sum to infinity becomes .
This simplifies so that . Evaluating , the sum to infinity is .
### Recurring Decimal to Fraction: Example 2
Write the recurring decimal as a fraction.
Here
This can be written as
Therefore and .
The sum to infinity becomes .
This simplifies to and is found to be .
## Sum to Infinity Proof
The proof of the sum to infinity formula is derived from the formula for the first n terms of a geometric series: Sn=a[1-rn]/[1-r]. If -1<r<1 then as n→∞, rn→0. Substituting rn with 0, the sum to infinity S=a[1-0]/[1-r], which simplifies to S=a/[1-r].
Here is the derivation of the sum to infinity of a geometric series in steps.
1. The formula for the sum of the first n terms of a geometric series is .
2. Since , as , .
3. Substituting and into becomes .
4. Simplifying, .
This derivation works since the common ratio is defined to be between -1 and 1.
When a number between -1 and 1 is raised to a high power, its size decreases.
For example:
As the power tends to an infinitely large value, the decimal size tends toward zero.
As such as , .
This is why the ratio must be defined as -1<r<1 for a geometric series to have a sum to infinity. |
Temp Directory?: /var/www/html/mathimages/imgUpload/tmp Bedsheet Problem - Math Images
# Bedsheet Problem
(Difference between revisions)
Revision as of 09:51, 27 May 2011 (edit)← Previous diff Revision as of 10:05, 27 May 2011 (edit) (undo)Next diff → Line 38: Line 38: The first summation is for the actual folds. The second is for the amount of sheet loss in a particular fold. The expression $(k-1)$ the $2^{n-1}$ is the radius of the outer folded layer. The first summation is for the actual folds. The second is for the amount of sheet loss in a particular fold. The expression $(k-1)$ the $2^{n-1}$ is the radius of the outer folded layer. - The following are the steps to simplify the equation. + The following are the steps to simplify the equation. we exclude in all steps ${\pi}*t[itex] but the final answer will have this term. [itex]\sum_{k=1}^n k-1$=$\sum_ {k=0}^n-1 k$=$\frac{(n-1)/n}{2}\$. $\sum_{k=1}^n k-1$=$\sum_ {k=0}^n-1 k$=$\frac{(n-1)/n}{2}\$. Line 57: Line 57: S(b-1)=$b^{n+1}-b$ S(b-1)=$b^{n+1}-b$ S=$\frac{b^{n+1}-b}{b-1}$ S=$\frac{b^{n+1}-b}{b-1}$ + + We then apply this general relationship to the two terms. + + $\sum_{i=1}^n \frac{4^i}{8}$=$\frac{4^{n+1}-4}{3*8}$ + + We simplify and get $\frac{4^{n+1}-4}{24}$=$frac{4^{n}-1}{6}$=$frac{2^{2n}-1}{6}$ + + $\sum_{i=1}^n \frac{2^i}{4}$=$\frac{2^{n+1}-2}{1*4}$ + + We simplify and get $\frac{2^{n+1}-2}{4}$=$\frac{2^{n}-1}{2}$=$\frac{3*2^n-3}{6}$ + + We add the two terms and get + $\frac{2^{2n}-1+3*2^n-3}{6}$ + $\frac{2^{2n}+3*2^n-4}{6}$ + $\frac{(2^n+4)(2^n-1)}{6}$ The following equation gives the loss function for folding a paper in half in one direction. The following equation gives the loss function for folding a paper in half in one direction.
## Revision as of 10:05, 27 May 2011
Bedsheet Problem
Take a piece of paper. Now try to fold it in half more than 7 times. Is it possible? This leads into the problem: what is the ultimate number of folds a flat piece of material can achieve?
There is an urban legend that a piece of paper cannot fold more than 7 times. All who claimed the myth was valid could only cite empirical evidence, they could not explain or prove it mathematically. The puzzle was both mysterious and inexplicable. The wonder that limits the folding has been discovered. This image shows a sheet folded 12 times.
# Basic Description
This well-known problem was not solved before for many reason. First, many believed the puzzle was true because no one understood or proved mathematically the geometric limitations. Most people folded a sheet of paper and no matter what material was used people usually got around 5, 6, and 7 folds. Because 7 was the highest number it became the folding limit. Also, there was little creativity in disproving the myth. No one realized that the thickness, width, flexibility of the material, strength of the person and crease lines all play a role in how many folds are achievable. Nevertheless, a sheet of paper was folded 12 times.
The thickness of the sheet of paper limits the amount of times the paper can be folded. After each fold the thickness doubles. For example after the third fold the thickness is the same as eight sheets. This brings up the issue with folding a sheet of paper in half: it ends up being very thick really fast. In fact, the limit of the folding occurs when the thickness is more than the width. At this point the paper is bent and has curved ends. It is no longer flat and much harder to fold. If we fold in the same direction, which we would not usually do in reality, a narrow strip with a constant thickness will form. Then after a certain number of folds, the thickness will double each time and the width is halved.
Now consider that we are folding the sheet of paper in alternate directions. Then the width is halved every two folds. The thickness behaves the same: it doubles every fold. Folding a sheet in alternative direction requires that the sheet is wide enough to make a fold in the other direction. From this the sheet would have a greater volume. Folding in alternative direction is a possible advantage because unlike folding in one direction the sheet will not unravel a previous fold.
#### King's Problem
Overall the paper folding connects to exponential growth. This is the essence of exponential growth: very small amounts rapidly become astronomically large through simple doubling. There is a math folktale about a clever merchant who asked the King to pay him with grains of wheat on a chessboard. The merchant asked the King to place one grain on the first square, two on the second, four on the third, eight on the fourth, sixteen on the fifth, and so on until the 64 boxes were filled. The king was too proud to admit that he could not calculate the sum of the grains. He foolishly granted the wish, not knowing that it would wipe out his stock and he would be in debt. If it isn't clear yet imagine the 20th box where the number 2 is multiplied by itself 20 times. This means that just on the 20th box the merchant would have more than a million grains. This is an example of exponential growth.
The bed sheet problem though becomes a little bit more complex. The thickness does not simply double after each fold. When a sheet of paper is folded, one end of the paper is placed onto the opposite end. The paper begins to crumple and it does not fold smoothly. With each fold the additional layers make it hard to lay one end over the opposite side. The paper begins to curve and puff at the ends. After a certain number of folds (this varies based on the thickness and width of the paper) the paper will reach a limit and any additional folds will no longer keep the paper long and flat but rather it will be a semi-circle. In addition to the thickness doubling, the sheet of paper begins to curve. The circumference of the semicircle at the edges are included in the thickness formula.
# A More Mathematical Explanation
#### Derivations
If you keep folding a sheet of paper the thickness doubles after each fold. With [...]
#### Derivations
If you keep folding a sheet of paper the thickness doubles after each fold. With one fold, you have a thickness of two paper, for two folds it has a thickness of 4 sheets of paper and so on. The layers increase by $2^N$ where N is the number of folds. This is somewhat similar to the chessboard example.
However, the sheet does not remain flat and long. It begins to curve at the edge. Each time a fold takes place the edges become round and the radius of the curved section. The radius of each layer is a half of the thickness.
With each layer the radius is different. The curved section of the folded sheet is noticeable when the thickness of the paper is equal or greater than the width. With an increase in folds, the radius section begins to take up a greater percentage of the paper's volume. The volume and thickness in curved section grows exponentially. The volume squares with each fold while the thickness doubles. The folding limit is reached when there is not enough volume or length of the remaining paper to fill the curved section. Any folds pass the absolute limit the entire sheet will become a semi-circle. To be specific we consider the folded section to be the region that has two times the number of folds of the straight layers. The curved ends are not counted as part of the folded section.
##### Single Direction
The length of paper required to fold a sheet of paper N times in a single direction, meaning the folding occurs in only one direction, includes how much paper has been lost after a certain number of folds. For a single fold there is a minimum length of the thickness times $/pi$, the length of the semicircle.
The following definition are used in the equation.
• L=length of material needed
• t=thickness of one sheet of material
• n=number of folds
• k=dummy variable
• i=dummy variable
L=$\pi t * \sum_{i=1}^n {\sum_{k=1}^{2^{i-1}} 2^{i-1}-(k-1)}$
The first summation is for the actual folds. The second is for the amount of sheet loss in a particular fold. The expression $(k-1)$ the $2^{n-1}$ is the radius of the outer folded layer.
The following are the steps to simplify the equation. we exclude in all steps ${\pi}*t[itex] but the final answer will have this term. [itex]\sum_{k=1}^n k-1$=$\sum_ {k=0}^n-1 k$=Failed to parse (lexing error): \frac{(n-1)/n}{2}\ .
In our case we get $(2^{i-1}-1)(2^{i-1})/2$
Simplifying the expression we get $-2^{2i-3}+2^{i-2}$
$\sum_{i=1}^n \frac{2^{2i}}{8}+\sum_{i=1}^n \frac{2^i}{4}$
$\sum_{i=1}^n \frac{4^i}{8}+\sum_{i=1}^n \frac{2^i}{4}$
S=$\sum_{i=1}^n b^i$ S=$b+b^2+\cdots+b^n$ S*b=$b^2+\cdots+b^{n+1}$ S*b-S=$b^{n+1}-b$ S(b-1)=$b^{n+1}-b$ S=$\frac{b^{n+1}-b}{b-1}$
We then apply this general relationship to the two terms.
$\sum_{i=1}^n \frac{4^i}{8}$=$\frac{4^{n+1}-4}{3*8}$
We simplify and get $\frac{4^{n+1}-4}{24}$=$frac{4^{n}-1}{6}$=$frac{2^{2n}-1}{6}$
$\sum_{i=1}^n \frac{2^i}{4}$=$\frac{2^{n+1}-2}{1*4}$
We simplify and get $\frac{2^{n+1}-2}{4}$=$\frac{2^{n}-1}{2}$=$\frac{3*2^n-3}{6}$
We add the two terms and get $\frac{2^{2n}-1+3*2^n-3}{6}$ $\frac{2^{2n}+3*2^n-4}{6}$ $\frac{(2^n+4)(2^n-1)}{6}$
The following equation gives the loss function for folding a paper in half in one direction.
Eq. 1 $\pi t/6 * (2^n+4)(2^n-1)$
##### Alternative Direction
Alternate direction folding uses both flat directions of the paper, so the width decreases after every fold. The alternative direction derivation is complicated because there are two folding limits. For the last fold to be achieved the previous fold's length has the be $\pi$ times the thickness. This relationship applies for both methods of folding. The sides of the fold are $\pi t 2^{(n-1)}$ and the thickness at this stage is $t 2^{(n-1)}$. The total area of all sheet is the area of a single sheet times the number of sheets.
Area=$\pi^2 t^2 2^{3(n-1)}$.
Taking the square root of the area will give the limiting width of the original sheet.
• W=Limiting width
• t=Thickness of material
• n=Number of folds
Equation 2 refers to the limiting paper width based on the last fold.
Eq. 2 W=$\pi t* 2^{3(n-1)/2}$
The equation though isn't completely accurate because it does not include the materials lost in the radii of previous folds. For odd number of folds there is lost if the paper is folded in the odd fold direction. The old fold do not contribute to the loss in the even fold direction. For large number of folds there is a possible limit that is 60% less than the actual limit. In practice a sheet of material could not fold that tight so equation 2 accurately demonstrates the limiting width.
#### Limitations
In order to compute the length of the paper required and the thickness of each sheet it is important to understand what exactly limits the number of folds.
• Smoothness
In particular, the smoothness of the paper determines if the paper can slide around the curved sections. As the thickness increases so does the stiffness and the resistance to folding. At this point no human or machine effort can produce another fold. The stiffness issue occurs when the folded section is length is less than π times the thickness. In order to do another successful fold the length to thickness ratio has to be greater than π.
Another important factor is the difference in length of layers. At the round ends each layer has a different radius and circumference. This can happen when paper is pushed out the folded section and is not included in the curved section. The radius section then takes the majority of the paper’s volume until it reaches a point from making the folded materials into a semi-circle.
• Folding Technique
After each fold, it becomes more difficult to set each layer flatly on top of each other. This happens when layers are not wrinkle free and lumps begin to form in the inner portions. From this, paper extends beyond the folded section.
• Miscellaneous
1. The strength of a person can affect how many fold are achievable. The stronger the person the flatter the edges.
2. The paper is not uniform.
3. There's a difference in the limited number of folds based on whether the paper is being folded in the single direction or the alternative direction.
#### Related Problem
Rather than how many folds are achievable, a similar problem to the bed sheet problem focuses on the thickness of the sheet of paper. If you were to take a large sheet of paper with thickness 1/400 inch and fold it in half 50 times, how tall would it be (if it was theoretically possible to fold a sheet of paper 50 times)? Most people in their minds would imagine it to be as thick as a phone book. Some might even being daring and say a few feet. After 3 folds the sheet of paper would be as thick as your fingernail. And, if you continue to fold until the paper is at 50 folds then it will be as thick as 40 million miles. The paper would be able to reach the sun and return back to earth. This sounds ridiculous, but it isn't. This is an example in mathematics called geometric progression. Each number in the sequence is multiplied by a fixed number to get the next term.
# Why It's Interesting
This well-known problem is interesting because it was solved by Britney Gallivan, who was at the time a junior in high school. She was asked by her teacher to fold a sheet of paper 12 times and as an incentive she would get extra credit. She failed multiples times. Later she succeeded after using a thin gold sheet and proved the assumption wrong. Gallivan was able to achieve 12 folds by folding a roll of thin toilet paper that stretched over three-fourths of a mile. It took seven hours in a shopping mall with her parents, but Gallivan was able to bust a myth as well as derive a formula relating the width, thickness of a paper and the number of folds achievable. The urban legend of 7 folds was disproved in 2001.
Gallivan showed that any person can do the impossible. She solved a problem that many mathematicians attempted but also failed to solve. It is interesting that a high school student with such fervor demonstrated a person should not accept anything to be true without evidence.
Gallivan has inspired others to break the record for most folds. Late April 2011 some students from Massachusetts claimed the world record in achieving 13 folds. They used the thinnest type of toilet paper that stretched over 2 miles.
# References
http://pomonahistorical.org/12times.htm
http://sciencebits.wordpress.com/2008/09/06/folding-paper/
Gallivan, B. C. "How to Fold Paper in Half Twelve Times: An 'Impossible Challenge' Solved and Explained." Pomona, CA: Historical Society of Pomona Valley, 2002.
http://mathworld.wolfram.com/Folding.html
[[Category:]] |
# 3.1: Introduction to Linear Functions
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Imagine placing a plant in the ground one day and finding that it has doubled its height just a few days later. Although it may seem incredible, this can happen with certain types of bamboo species. These members of the grass family are the fastest-growing plants in the world. One species of bamboo has been observed to grow nearly 1.5 inches every hour.1 In a twenty-four hour period, this bamboo plant grows about 36 inches, or an incredible 3 feet! A constant rate of change, such as the growth cycle of this bamboo plant, is a linear function.
Figure $$\PageIndex{1}$$: A bamboo forest in China (credit: “JFXie”/Flickr)
Recall from Functions and Function Notation that a function is a relation that assigns to every element in the domain exactly one element in the range. Linear functions are a specific type of function that can be used to model many real-world applications, such as plant growth over time. In this chapter, we will explore linear functions, their graphs, and how to relate them to data.
## Footnotes
1 www.guinnessworldrecords.com/...growing-plant/
This page titled 3.1: Introduction to Linear Functions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. |
# Intro to subtraction
### Intro to subtraction
What is subtraction? Subtraction is taking things away. When you take a number away from another you are subtracting. And the number becomes smaller. Start with 5 apples, then subtract 2, you are left with 3 apples. In math, we can write it with numbers: 5 – 2 = 3 In words, ıt’s read as “five minus
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### Intro to addition
In this lesson, you will learn addition. What is addition? Addition is, bringing two or more numbers (or things) together to make a new total. When you add two numbers or things you are counting them together. You use addition every day in life. I have 1 circle.And I add 1 more circle.Now I have
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### Counting to 100
İn this lesson you will learn how to count to 100 Counting with words and numbers up to 100 1 – one 2 – two3 – three4 – four5 – five6 – six7 – seven8 – eight9 – nine10 – ten 11 – eleven12 – twelve13 – thirteen14 – fourteen15 – fifteen16 – sixteen17 –
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### Counting
In this lesson, you will learn how to count to 10 Count to 10 1 2 3 4 5 6 7 8 9 10 one two three four five six seven eight nine ten Count to 20 Count to 100
### Exponents
Before we talk about exponents. Let’s review multiplication. We know that multiplication is (repeated addition).For example: if we multiply 2 x 3 =?İt means (add the number “2” 3 times) or 3 2’s being added together.Like so: 2+2+2 = 6 The exponent of a number says how many times to use the number in a
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### Multiplication Table
Use the interactive multiplication table chart to quickly multiply two numbers. When we think about multiplication, we usually think of the Times Tables. In my opinion, you should not teach your child the Times Tables until he or she understands what it means to multiply two numbers. After your child has understood the concept, then
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# Write the dimensional formula of velocity.
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Hint: In order to find the dimensional formula of velocity, we need to know the formula of velocity i.e., velocity, $v = \dfrac{s}{t}$ where $s =$displacement, $t =$time taken.
Then we need to find units of velocity, displacement and time in the terms of fundamental quantities.
Complete step-by-step solution:
Fundamental quantities: The quantities which are independent of any other quantities are called fundamental quantities.
Dimensional formula: The expression showing the relationship between the fundamental quantities and expressing the powers of fundamental quantities to be raised to obtain one unit of derived quantity is called dimensional formula.
E.g., let $K$be a quantity then it’s dimensional formula=${[M]^a}{[L]^b}{[T]^c}$where $M =$mass, $L =$Length and $T =$time
Now, Velocity, $v = \dfrac{s}{t} - - (i)$
SI unit of displacement, $s = m$(meter), $t =$$s$(Second)
Substituting the above value in equation$(i)$ we get:
$v = \dfrac{m}{s} \\ v = m{s^{ - 1}} \\$
Thus, SI unit of velocity is $m{s^{^{ - 1}}}$
In dimensional formula we represent metre as $M$, second as $S$
So,
Dimensional Formula of velocity=$\left[ M \right]{\left[ T \right]^{ - 1}}$
The dimensional formula of velocity is $\left[ M \right]{\left[ T \right]^{ - 1}}$
Note: There are $7$fundamental quantities that don’t depends on other quantities are as follows: |
## How do you find the equation of a graph?
How To: Given a graph of linear function, find the equation to describe the function.Identify the y-intercept of an equation.Choose two points to determine the slope.Substitute the y-intercept and slope into the slope-intercept form of a line.
## How do you write an equation for a function on a graph?
How To: Given THE graph of A linear function, find the equation to describe the function.Identify the y-intercept from the graph.Choose two points to determine the slope.Substitute the y-intercept and slope into slope-intercept form of a line.
## What is a function equation?
A function is an equation that has only one answer for y for every x. A function assigns exactly one output to each input of a specified type. It is common to name a function either f(x) or g(x) instead of y. f(2) means that we should find the value of our function when x equals 2.
## What is a function on a graph?
The graph of the function is the set of all points (x,y) in the plane that satisfies the equation y=f(x) y = f ( x ) . If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because that x value has more than one output.
## How do you write an equation for a function?
When you are given an equation and a specific value for x, there should only be one corresponding y-value for that x-value. For example, y = x + 1 is a function because y will always be one greater than x. Equations with exponents can also be functions.
## How do you write an equation of a line?
The slope-intercept form of a linear equation is written as y = mx + b, where m is the slope and b is the value of y at the y-intercept, which can be written as (0, b). When you know the slope and the y-intercept of a line you can use the slope-intercept form to immediately write the equation of that line.
## What is plot graph?
A plot is a graphical technique for representing a data set, usually as a graph showing the relationship between two or more variables. The plot can be drawn by hand or by a computer. Graphs of functions are used in mathematics, sciences, engineering, technology, finance, and other areas.
## How do you plot a bar graph?
On a graph, draw two lines perpendicular to each other, intersecting at 0. The horizontal line is x-axis and vertical line is y-axis. Along the horizontal axis, choose the uniform width of bars and uniform gap between the bars and write the names of the data items whose values are to be marked.
## What are the 3 ways to graph a linear equation?
There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y-intercept and slope. The third is applying transformations to the identity function f(x)=x f ( x ) = x .
## Is an equation a formula?
equation: A statement formed by placing an equals sign between two numerical or variable expressions. formula: An equation that states a rule about a relationship. Highly active question.
## Is the equation a function?
A function is a set of ordered pairs where each input (x-value) relates to only one output (y-value). A function may or may not be an equation. Equations are functions if they meet the definition of a function. But, there are equations that are not functions.
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How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
#### H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […] |
### Gambling at Monte Carlo
A man went to Monte Carlo to try and make his fortune. Is his strategy a winning one?
### Marbles and Bags
Two bags contain different numbers of red and blue marbles. A marble is removed from one of the bags. The marble is blue. What is the probability that it was removed from bag A?
### Coin Tossing Games
You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by a head (you win). What is the probability that you win?
# Pay Attention
##### Age 14 to 16 ShortChallenge Level
Using numbers to make the proportions easier to work with:
Let's imagine that there were 100 people in the audience and the talk was 60 minutes long.
Then 6 people heard 0 minutes, 22 people heard 60 minutes.
100 $-$ 6 $-$ 22 = 72, and 72$\div$2 = 36, so 36 people heard 40 minutes ($\frac{2}{3}$ of an hour) and 36 people heard 20 minutes ($\frac{1}{3}$ of an hour).
To find the average number of minutes heard, we need the total number of minutes heard divided by the number of people (which is 100).
The 6 people who heard 0 minutes did not contribute to the total number of minutes heard.
The 22 people who each heard 60 minutes heard a total of 22$\times$60 = 1320 minutes.
The 36 people who each heard 40 minutes heard a total of 36$\times$40 = 1440 minutes.
The 36 people who each heard 20 minutes heard a total of 36$\times$20 = 720 minutes.
That is a total of 1320 + 1440 + 720 = 3480 minutes.
So the average number of minutes heard was 3480$\div$100 = 34.8.
So the average proportion of the talk heard was $\frac{34.8}{60}=\frac{348}{600}=\frac{29}{50}=58\%$
Using 100 people and proportions of the talk:
Let's imagine that there were 100 people in the audience.
To find the average proportion of the talk heard, we need the total proportion of the talk heard divided by 100.
The 6 people who slept for the whole talk heard none of the talk.
The 22 people who heard all of the talk heard 22$\times$ the whole talk.
100 $-$ 6 $-$ 22 = 72, and 72$\div$2 = 36, so 36 people heard $\frac{2}{3}$ of the talk and 36 people heard $\frac{1}{3}$ of the talk.
The 36 people who heard $\frac{2}{3}$ of the talk heard a total of 36$\times\frac{2}{3}$= 24$\times$ the whole talk.
The 36 people who heard $\frac{1}{3}$ of the talk heard a total of 36$\times\frac{1}{3}$= 12$\times$ the whole talk.
That is a total of 22 + 24 + 12 = 58 $\times$ the whole talk.
So on average, $\frac{58}{100}=58\%$ of the talk was heard.
Using percentages and proportions:
If everyone in the audience had heard all of the talk, then the proportion of the talk heard by the audience would be 1. If half of the audience had slept through the talk and half had heard half of it, then the proportion of the talk heard by the audience would be half of a half, which is a quarter. That would also be the average proportion of the talk heard by the audience.
6% of the audience heard none of the talk, so they did not contribute to the proportion of the talk that was heard by the audience.
22% of the audience heard all of the talk, so they contribute 22% to the proportion that was heard.
100 $-$ 6 $-$ 22 = 72, and 72$\div$2 = 36, so 36% of the audience heard $\frac{2}{3}$ of the talk and 36% heard $\frac{1}{3}$ of the talk.
The 36% who heard $\frac{2}{3}$ of the talk contribute 36%$\times\frac{2}{3}$=24% to the proportion of the talk that was heard by the audience.
The 36% who heard $\frac{1}{3}$ of the talk contribute 36%$\times\frac{1}{3}$=12% to the proportion of the talk that was heard.
So the total proportion of the talk that was heard by the audience was:
22% + 24% + 12% = 58%.
So the average proportion of the talk that was heard was 58%, or $\frac{29}{50}$.
You can find more short problems, arranged by curriculum topic, in our short problems collection. |
# What is the perimeter of a triangle with corners at (6 ,0 ), (5 ,2 ), and (5 ,4 )?
Feb 1, 2018
Perimeter of the triangle $P = \textcolor{b r o w n}{8. 5953}$
#### Explanation:
Given : $A \left(6 , 0\right) , B \left(5 , 2\right) , C \left(5 , 4\right)$
Distance between two points, given the two end points' coordinates is given by the formula,
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
$a = \sqrt{{\left(5 - 6\right)}^{2} + {\left(2 - 0\right)}^{2}} = \sqrt{{1}^{2} + {2}^{2}} \approx \textcolor{g r e e n}{2.2361}$
$b = \sqrt{{\left(6 - 5\right)}^{2} + {\left(0 - 4\right)}^{2}} = \sqrt{{1}^{2} + {4}^{2}} \approx \textcolor{g r e e n}{4.1231}$
$c = \sqrt{{\left(6 - 5\right)}^{2} + {\left(0 - 2\right)}^{2}} = \sqrt{{1}^{2} + {2}^{2}} = \textcolor{g r e e n}{2.2361}$
It is an isosceles triangle with sides a & c = 2.2361.
Perimeter of the triangle $P = \frac{a + b + c}{2} = \frac{2.2361 + 4.1231 + 2.2361}{2} = \textcolor{b r o w n}{8. 5953}$ |
# Lesson 13
Median
### Lesson Narrative
In this lesson, students consider another measure of center, the median, which divides the data into two groups with half of the data greater and half of the data less than the median. To find the median, they learn that the data are to be arranged in order, from least to greatest. They make use of the structure of the data set (MP7) to see that the median partitions the data into two halves: one half of the values in the data set has that value or smaller values, and the other half has that value or larger. Students learn how to find the median for data sets with both even and odd number of values.
Students engage in MP2 as they find the median of a numerical data set and interpret it in context. They begin to see that, just like the mean, the median can be used to describe what is typical in a distribution, but that it is interpreted differently than is the mean.
### Learning Goals
Teacher Facing
• Comprehend that the “median” is another measure of center, which uses the middle of all the values in an ordered list to summarize the data.
• Identify and interpret the median of a data set given in a table or on a dot plot.
• Informally estimate the center of a data set and then compare (orally and in writing) the mean and median with this estimate.
### Student Facing
Let's explore the median of a data set and what it tells us.
### Required Preparation
For the Finding the Middle activity, each student will need an index card.
### Student Facing
• I can find the median for a set of data.
• I can say what the median represents and what it tells us in a given context.
Addressing
Building Towards
### Glossary Entries
• median
The median is one way to measure the center of a data set. It is the middle number when the data set is listed in order.
For the data set 7, 9, 12, 13, 14, the median is 12.
For the data set 3, 5, 6, 8, 11, 12, there are two numbers in the middle. The median is the average of these two numbers. $$6+8=14$$ and $$14 \div 2 = 7$$.
### Print Formatted Materials
Teachers with a valid work email address can click here to register or sign in for free access to Cool Down, Teacher Guide, and PowerPoint materials.
Student Task Statements pdf docx Cumulative Practice Problem Set pdf docx Cool Down Log In Teacher Guide Log In Teacher Presentation Materials pdf docx
### Additional Resources
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# How do I find the equation of a perpendicular bisector of a line segment with the endpoints (-2, -4) and (6, 4)?
##### 2 Answers
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### This answer has been featured!
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Jan 23, 2016
#### Answer:
$x + y - 2 = 0$
#### Explanation:
Here's how.
You need to find the slope of the segment since it is the negative reciprocal of the slope of the bisector.
Another one you will have to find is the midpoint of the segment because the bisector will pass trough that point.
If you have those two, you can now determine the equation of the perpendicular bisector using point-slope form of a line.
Let:
${m}_{s} \implies$slope of the segment
${m}_{b} \implies$slope of the bisector
$\left({x}_{1} , {y}_{1}\right) \implies \left(- 2 , - 4\right)$
$\left({x}_{2} , {y}_{2}\right) \implies \left(6 , 4\right)$
$M \left({x}_{m} , {y}_{m}\right) \implies$midpoint of segment
Solving for the slope of the segment:
${m}_{s} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{4 - \left(- 4\right)}{6 - \left(- 2\right)} = \frac{8}{8} = 1$
Solving for the slope of the bisector:
${m}_{b} = - \frac{1}{m} _ s = - \frac{1}{1} = - 1$
Solving for the midpoint of the segment:
${x}_{m} = \frac{{x}_{1} + {x}_{2}}{2} = \frac{- 2 + 6}{2} = 2$
${y}_{m} = \frac{{y}_{1} + {y}_{2}}{2} = \frac{4 - 4}{2} = 0$
$M \left(2 , 0\right)$
Solving for the equation of the segment:
POINT-SLOPE FORM
$y - {y}_{m} = {m}_{b} \left(x - {x}_{m}\right)$
$y - 0 = - 1 \left(x - 2\right)$
$y = - x + 2 \text{ } \leftarrow$slope-intercept form
$x + y - 2 = 0 \text{ } \leftarrow$ standard form
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Jul 16, 2016
#### Answer:
$x + y - 2 = 0$
#### Explanation:
Let $\left(x , y\right)$ be any point on the perpendicular bisector. From elementary geometry, we can easily see that this point must be equidistant from the two points $\left(- 2 , - 4\right) \mathmr{and} \left(6 , 4\right)$. Using the Euclidean distance formula gives us the equation
${\left(x + 2\right)}^{2} + {\left(y + 4\right)}^{2} = {\left(x - 6\right)}^{2} + {\left(y - 4\right)}^{2}$
This can be rewritten as
${\left(x + 2\right)}^{2} - {\left(x - 6\right)}^{2} = {\left(y - 4\right)}^{2} - {\left(y + 4\right)}^{2}$
Using ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, this simplifies to
$8 \left(2 x - 4\right) = - 8 \cdot 2 y$ which simplifies to
$x + y - 2 = 0$
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40. Square Root of 5! When the exponent is $$\frac{1}{2}$$, it is called a square root. Let us understand this process with an example. $$\sqrt{12} = \sqrt{2 \times 2\times3} = 2\sqrt{3}$$. Example: $$\sqrt{9}$$ = $$3$$ as ($$3 \times 3 = 9$$), $$\sqrt[\text{n}]{x } = x ^\frac{1}{n}$$. If you're seeing this message, it means we're having trouble loading external resources on our website. There will be $$\frac{n}{2}$$ digits in the square root of an even number with $$n$$ digits. Often the way that the quotient is written out in long division problems creates confusion. Remainder when 17 power 23 is divided by 16. Sample: Calculate square root of 5 using division method. We can use any of the above methods for finding the square root, such as prime factorization, long division, and so on. Now, you will be able to easily solve problems on the square root calculator, square root formula, how to find square root and square root symbol. Long Division Method . Therefore, if we take the square root on both the sides, we get; â144 = â(2x2x2x2x3x3) As we can see here, there are two pairs of 2 and one pair of 3. Know and learn the method or the process from which you can find the approximate value of the square root of 10. The square root of 324 is $${\displaystyle \pm}18$$, ( $$18 \times 18 = 324$$) and How to find the square root of 1444 by long division method Here we will show you how to calculate the square root of 1444 using the long division method. But if the number is not a perfect square, then it is difficult to find the square root of it. Hence, â144 = 2x2x3 â144 = 12. Write the number. inches. Therefore, we will combine here 2 with 2 and multiply by 2 such as 22 x 2 to get 44, Hence, the final quotient we get is 12, which is the answer. He measured the length of the board to be 16 sq. Am I on the right track? In radical form, it is denoted as â144 = 12. Find the product of the factors obtained by taking one factor from each pair. Translating the word problems in to algebraic expressions. Which number, when multiplied by itself, gives $$9$$ as the product? Now, we will take down the other two numbers, i.e. Input: N = 144 Output: 12 12 2 = 144 Find the greatest 3 digit and 4 digit perfect square. We will use the average method to find out the square root of an imperfect square number 120. Let us find the square root of $$16$$ using this method. If a factor cannot be grouped, retain them under the square root symbol. ToThere are main two ways to find square root of a given number. Simply type in 144 followed by âx to get the answer. But, it can only be used if a number is a perfect square. Here is the answer to questions like: Square root of 53824 or what is the square root of 53824? There are two methods to determine the square root value of 144. 1 and leave the rest two digits i.e. Below are the solved steps to take the root of 144. Steps involved in square root by long- division method. Textbook Solutions 5346. But, it can only be used if a number is a perfect square. The square root is nothing but the exponent $$\frac{1}{2}$$. Step 2: We divide the left-most number by the largest number whose square is less than or equal to the number in the left-most pair. Divide the given number into its prime factors. Step 5: Now, we will continue this process further using a decimal point and adding zeros in pairs to the remainder. Use the square root calculator below to find the square root of any imaginary or real number. Help Kate find out the square root of $$529$$ by the prime factorization method. We will have two pairs, i.e. how to find Square Root or How to find Square root by long division method. If you want to know the square root of 10 value approximately (exact value difficult to find), we must use the long division method definitely. \end{align}\]. Long Division Method . The square roots of numbers that are not perfect squares are part of irrational numbers. We have done repeated subtraction $$13$$ times. To the right of the obtained sum, find a suitable number which, together with the result of the sum, forms a new divisor for the new dividend that is carried down. 15 - 3 &=12 \\ 44, as dividend and add 1 to the divisor to get our next divisor, i.e. See also in this web page a Square Root Table from 1 to 100 as well as the Babylonian Method or Hero's Method. Sum of all three digit numbers divisible by 6. For what value of $$x$$ is this statement valid. We can find the exact square root of any given number using this method. Let us find the square root 120 using the average method, the following are the steps. For example: $$\sqrt{(n \times n)} = (\sqrt{n^2})$$ = $$n$$. This is the lost art of how they calculated the square root of 1444 by hand before modern technology was invented. To simplify a square root, we need to find the prime factorization of the given number. Step 1 : x 4 has been decomposed into two equal parts x 2 and x 2.. He decides to give each guest a square piece of cake of side $$1\:\text{inch}$$. Syllabus. The nearest square numbers to $$7$$ are $$4$$ and $$9$$, Therefore, $$\sqrt{7}$$ lies between $$\sqrt{4}$$ and $$\sqrt{9}$$, which are $$2$$ and $$3$$, $$2.5^2 = 6.25$$ , $$2.6^2 = 6.76$$ and $$2.7^2 = 7.29$$, So $$\sqrt{7}$$ lies between $$2.6$$ and $$2.7$$. Proceed as usual. Many times, we surprised and shocked as well. Math, 16.08.2019 18:38, garganjali1985. Since there are 121 guests, number of cake pieces should be $$121$$. ($$-2 \times -2 = 4$$), Pair of Linear Equations in Two Variables, Cue Learn Private Limited #7, 3rd Floor, 80 Feet Road, 4th Block, Koramangala, Bengaluru - 560034 Karnataka, India, $$\therefore$$ Length of the sides of the square cake = $$11\text{ inches}$$. 1+1 =2. You will also be introduced to the square root formula and the square root symbol. Finding the square root by long division method. Worksheet on Square root using the long division method which helps the students to prepare for the exams or any other tests. We did that with our calculator and got the following answer: â144 = 12. For digits after decimal point, pair them from left to right). Evaluate $$\sqrt{\frac{169}{25}} - \sqrt{\frac{64}{49}} +\sqrt{\frac{196}{9}}$$. Help Mary find the square root of 169 by repeated subtraction method. This method works only for perfect square numbers. Calculation of a square root by hand is a little like long-hand division. Therefore, taking 1 as divisor and quotient and as dividend, we get remainder as 0. Step 4: The new number in the quotient will have the same number as selected in the divisor. Also, learn here about square root in a detailed way. CBSE CBSE Class 8. Sum of all three digit numbers divisible by 7 Given an integer X which is a perfect square, the task is to find the square root of it by using long division method. Here is the list of square roots of perfect square numbers and some non-perfect square numbers from $$1$$ to $$10$$. For example, the square root of 16 is 4, because 16 is a perfect square of 4, such as: 4 2 = 16 and â16 = 4. Step 6: The quotient thus obtained will be the square root of the number. Hence. We can find the exact square root of any given number using this method. Generate work with steps for 2 by 1, 3by 2, 3 by 1, 4 by 3, 4by 2, 4 by 1, 5 by 4, 5 by 3, 5 by 2, 6 by 4, 6 by 3 & 6 by 2 digit long division practice or homework exercises. See also in this web page a Square Root Table from 1 to 100 as well as the Babylonian Method or Hero's Method. If the number of digits in it is odd, then the left-most single digit too will have a bar.Thus we have, 7 29.So 1st bar is on 29 and 2nd bar is on 7. When $$n$$ = $$2$$, we call it square root. We can use the subtraction method, prime factorization method, approximation method, and long division method to find the square root of a given number. The two methods to determine the square root value of 144 are Prime Factorization Method and Long Division Method. We make pairs from right. Let's understand why the square root by division method (digit by digit method) works. Step 1: To find the square root of a decimal number we put bars on the integral part (i.e., 21) of the number in the usual manner.And place bars on the decimal part (i.e., 16) on every pair of digits beginning with the first decimal place. This method helps in estimating and approximating the square root of a given number. Square root of 5! Advertisement. Required fields are marked *. units. That product is the square root of the given number. Online calculator which calculates the square root of a given number using Long Division (LD) method. Enter the number for which you need to calculate the square root. Use the square root calculator below to find the square root of any imaginary or real number. Here, we are going to learn the ways to find the square root of 144 without using calculators. L.C.M method to solve time and work problems. This method is very useful and is the fastest method of all to find the root. We will have two pairs, i.e. since a square is either a positive number or zero. How to calculate the square root of 144 with a computer. 529 &= 23 \times 23 \\ We can find the exact square root of any given number using this method. Finding the Square Root of 120 by Average Method. find the square root of 239.421 correct to three decimal places by long division method how did this came Find the square root of the following numbers by division method 1. Group the digits into pairs (For digits to the left of the decimal point, pair them from right to left. Find the least 4 digit perfect square number. For this, we have to know about square root. From left hand side for every two digits keep the bar as shown in fig. Once we understand square roots, we need not use a square root calculator, but we can learn how to calculate it mentally. We know, square root of 4, or 9 or 16 and we can do it instantly but the square root of 5! John is baking a square cake for his friend's birthday party. Hence, we can see the square root value of 144 is 12. 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We will have two pairs, i.e. Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in. This is a very long process and time-consuming. Hence, we then use long division method. $$1$$ and $$80$$ Definition: This describes a "long hand" or manual method of calculating or extracting square roots. Find the Square of the Following by Long Division Method: 12544 Concept: Finding Square Root by Division Method. Let us understand this process with an example. of digits are odd then the left most single digit will also have a bar Your email address will not be published. 16 -1 &= 15 \\ The square root of a number is the number that gets multiplied to itself to give the product. Remainder when 2 power 256 is divided by 17. What should be the length of the sides of his square cake so that he can serve all the guests $$1$$ piece each? Peter is playing with random numbers and thinks of finding out the square root of $$6$$ to $$3$$ decimal places by the long division method. Therefore, if we take the square root on both the sides, we get; As we can see here, there are two pairs of 2 and one pair of 3. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Add the last digit of the quotient to the divisor. Set up a "division" with the number under the radical. ($$-18 \times -18 = 324$$), The square root of 4 is $${\displaystyle \pm}2$$, ( $$2 \times 2 = 4$$) and It can be used to find the root of imperfect squares and of large numbers, which is not possible in the case of prime factorisation. But the square root of 3, â3, is not easy, as 3 is not a perfect square. Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. \begin{align}\sqrt{121} = 11\end{align}. Examples: Input: N = 484 Output: 22 22 2 = 484. \Rightarrow\sqrt{529} &= 23 There are $$121$$ guests attending the party. We can also find the square root of any number using long division method. We will learn both the methods here, which will help students to find the root of any number easily and in the fastest way. Concept Notes & Videos 270. Your email address will not be published. Here are a few problems for you to practice. Check out the interactive simulations to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page. The condition is the same — as being either less than or equal to the dividend. Then $$(e^{2}+ 2ye )\le a^{2}-y^{2}$$ and this is essentially what we do in the long division method. Find the square root of 3844 by division method. Let's find the square root of $$180$$ Step 1: Place a bar over every pair of digits of the number starting from the unitâs place (right-most side). For fractions, there is also a similar rule: $$\frac{\sqrt{x}}{\sqrt{y}} = \sqrt\frac{x}{y}$$, $$\frac{\sqrt{50}}{\sqrt{10}} = \sqrt\frac{50}{10} =\sqrt{5}$$, Square Root of a negative number cannot be a real number. The value of the square root of 144 is equal to 12. Click hereðto get an answer to your question ï¸ Find the square root of 3249 by Division method. Question Bank Solutions 4748. Square root of $$13$$ in the decimal form is $${\displaystyle \pm}3.6055$$ rounded to $$4$$ decimal places. 12 - 5 &= 7 \\ \begin{align} Since each piece of cake has to be $$1\:\text{inch}$$ by $$1\:\text{inch}$$, the area of the square cake should be $$121$$ sq. $$1$$ and $$80$$ Other questions on the subject: Math. Long division method 2. Here is the answer to questions like: Square root of 3364 or what is the square root of 3364? Long Division method When the exponent is $$2$$, it is called a square. For example, 12 ×12= 144. Perform division as per steps shown below: 1. \begin{align} \sqrt{-x} = i\sqrt{x} \ \end{align}, Here, $$i$$ is the square root of $$-1$$, \begin{align} \sqrt{-16} =\sqrt{16} \times \sqrt {-1} &= 4i \end{align}, where $$i$$ is represented as the square root of $$-1$$. Clearly, 144 is a perfect square, thus the prime factors are. To find the square root of a given number through the prime factorization method: Let us find the square root of 144 by this method. Clearly, 144 is a perfect square, thus the prime factors are, 144 = 2x2x2x2x3x3. Suppose you need to find the square root of 66564. Now the square of 1 is 1. Step 2 : Multiplying the quotient (x 2) by 2, so we get 2x 2.Now bring down the next two terms -12x 3 and 42x 2.. By dividing -12x 3 by 2x 2, we get -6x. Question 1 : Find the square root of the following polynomials by division method (i) x 4 â12x 3 + 42x 2 â36x + 9. Square root times square root; What is the square root (sqrt) of 3? As the number 10 is not a perfect square, so we cannot get root 10 value easily. 1. Step 1: Place a bar over every pair of digits starting from the unit digit. Square root of 4096 by long division method: Steps: Draw the two vertical lines. Squares and square roots are special exponents. Use the calculator above to find the square root of any given number. Now take first two digits i.e. Long division calculator with step by step work for 3rd grade, 4th grade, 5th grade & 6th grade students to verify the results of long division problems with or without remainder. For numbers before decimal point, we start from right ⦠Step 1: Place a bar over every pair of digits of the number starting from the unit’s place (right-most side). But complex numbers have the solutions to the square root of a negative number. Help him estimate the square root of $$7$$ to $$2$$ decimal places using the approximation method. \[\begin{align} \end{align}. We will subtract the consecutive odd numbers from the number for which we are finding the square root, till we reach $$0$$. Step 1: Find out the two perfect square numbers which are very close to the given number on either side. Given below are the list of topics that are closely connected to square roots. To calculate the value to root 144 without using a calculator, there are two methods: one is the prime factorisation and another is the long division method. Square Root of 5 â Long Division, Average & Equation Method. Find the square root of 2025 / 4900; how to find square root of 841 by prime factorisation as finding factors of 841 will take so long time in exam; Find the least number that must be added to 6412 to get a perfect square. It is represented by the radical symbol pronounced as the square root. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Step 3: Bring down the number under the next bar to the right of the remainder. (viii) 7921 Rough 167 × 7 = 1169 168 × 8 = 1344 169 × 9 = 1521 Therefore, â7921 = 89 Ex 6.4, 1 Find the square root of each of the following numbers by Division method. Use the square root calculator below to find the square root of any number. In prime factorisation, we have to find the prime factors in the given number. Jack tries different methods of finding the square root of a number. We know that 256 is a square number which is $$16 \times 16$$. units. Form pairs of similar factors such that both factors in each pair are equal. Square Root of 5 â Long Division, Average & Equation Method. 4. Answers: 1 Get. Find the distance between the point (at^2,2at) and (at^2,2at) Answers: 1. continue. Let us understand this process with an example. It is written in the radical form as $\sqrt{144}$ = 12. Select/Type your answer and click the "Check Answer" button to see the result. We hope you enjoyed learning about square root with the simulations and practice questions. Let us use this method to find $$\sqrt {15}$$, Find the nearest perfect square numbers to $$15$$, $$9$$ and $$16$$ are the perfect square numbers nearest to $$15$$, We know that $$\sqrt{16}$$ = $$4$$ and $$\sqrt{9}$$ = $$3$$, This implies that $$\sqrt {15}$$ lies between $$3$$ and $$4$$, Now, we need to see if $$\sqrt {15}$$ is closer to $$3$$ or $$4$$, Thus, $$\sqrt {15}$$ lies between $$3.5$$ and $$4$$ and is closer to $$4$$, Let us find the squares of $$3.8$$ and $$3.9$$, This implies that $$\sqrt {15}$$ lies between $$3.8$$ and $$3.9$$, We can repeat the process and check between $$3.85$$ and $$3.9$$, We can observe that $$\sqrt{15}$$ = $$3.872$$. To determine the long division method, we will first divide the digits of the number into pairs of segments, starting with the digit in the units place. Check that below 40 is there any perfect square number or not. Is it possible to calculate? The number of times we subtract is the square root of the given number. In between write the number 4096. Finding the square root by long division method. Below are the steps for using the long division method. $$1$$ and $$80$$. By continuting in this way, we get the following steps. \begin{align}169-1 &= 168 \\168-3 &= 165\\165 -5 &= 160 \\160 -7 &= 153 \\153 - 9 &= 144\\144 - 11 &= 133\\133 -13 &= 120 \\120 - 15 &= 105 \\105 - 17 &= 88\\88- 19 &= 69 \\69 - 21 &=48\\48- 23 &= 25 \\25 - 25 &=0\end{align}. Many wonder, how can we possibly get our answer using only the basic, easy to follow steps of long division? Let's find the square root of $$180$$ Step 1: Place a bar over every pair of digits of the number starting from the unitâs place (right-most side). If you are using a computer that has Excel or Numbers, then you can enter SQRT (144) in a cell to get the square root of 144. Step 1) Set up 1444 in pairs of two digits from right to left: As per the given expression above, we can define the square root of natural number 144, is a value which on multiplying by itself gives 144. Thus we have, 05. Mathew has a game board of area 256 sq. Division method for finding square roots step 1 : First place a bar over every pair of digits starting from the unit digit , if the no. These topics will also give you a glimpse of how such concepts are covered in Cuemath. But how do we express 16 in terms of 256? This method is easy to apply, as we have learned about prime factors in our earlier classes. Find the Square root Shortcut Trick and Easy Way. Finding square root using long division. Since the last digit is 4, then either the square of 2 or 8 can get the last digit as 4. Math, 16.08.2019 19:40, Simrankanojia. And what more do I need to add to make this proof complete? (vii) 5776 Rough 144 × 4 = 576 145 × 5 = 725 146 × 6 = 876 Therefore, â5776 = 76 Ex 6.4, 1 Find the square root of each of the following numbers by Division method. 7 - 7 &= 0 At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! This method works when the given number is a perfect square number. And ( at^2,2at ) Answers: 1. continue this proof complete web page a root...: square root of a negative number check answer '' button to the. Not use a square number or zero the exponent \ ( 121\ ) guests attending the.. Side for every two digits keep the bar as shown in fig has... 1: find out the square root times square root of a given number using this method works the... Here is the square root of 4096 by long division in 144 followed by to. We did that with our calculator and got the following are the steps for using the approximation method all digit. \Frac { 1 } { 2 } \ ] = 12 ) to \ ( 16\ ) ).... Root in a detailed way this way, we have to find out the square root of square... Starting from the unit digit Concept: Finding square root of 3364 only. Is a little like long-hand division 4: the quotient to the dividend: 22 22 2 = 484:..., â3, is not a perfect square, thus the prime factors in each pair are equal we to! Method Simply type in 144 followed by âx to get our answer using only the basic, easy to steps. Is either a positive number or zero continuting in this web page a square root 144... A little like long-hand division: 12544 Concept: Finding square root of 120 by Average method and as... Of 3, â3, is not a perfect square quotient thus will! We subtract is the fastest method of all three digit numbers divisible by 6 of an imperfect number. Digits into pairs ( for digits to the right of the given number using long division.! Board to be 16 sq radical symbol pronounced as the Babylonian method or Hero 's method number! Get root 10 value easily filter, please make sure that the quotient to the left of square! ) using this method 4 has been decomposed into two equal parts x and... 3364 or what is the square root of 5 the approximation method pieces should be (... Prime factorisation, we call it square root of 66564 method ) works the students computer! By division method Kate find out the square root of any given number using long division ( ). What more do I need to add to make this proof complete grouped, retain them under the root..., i.e step 5: now, we need not use a cake. 44, as we have subtracted \ ( n\ ) = \ ( 121\ ) guests attending the.! Of similar factors such that both factors in our earlier classes is represented by the factorization... Technology was invented make sure that the quotient thus obtained will be the root... Not a perfect square & Equation method are main two ways to find square root of 53824 with our and... Many times, we square root of 144 by long division method the following answer: â144 = 12 less than equal. Of math experts is dedicated to making learning fun for our favorite readers, the teachers all! Like: square root value of 144 is equal to 12 is dedicated to making fun... Not a perfect square number 120 not use a square root of a given number method... Teachers explore all angles of a given number Finding the square root of 4096 by long division method is! ( \sqrt { 144 } \ ) 2 = 484 we understand square roots, we not. Written in the radical symbol pronounced as the Babylonian method or the process from you! Left hand side for every two digits keep the bar as shown in fig follow steps long. Different methods of Finding the square root of \ ( \frac { 1 {. Art of how they calculated the square root of 4, or 9 or and. Of 66564 = 484 value easily and learn the method or Hero 's method he decides to each! Favorite readers, the teachers explore all angles of a negative number the number 10 is not a square... Covered in Cuemath.kasandbox.org are unblocked mathew has a game board of 256. Of 3364 or what is the fastest method of all three digit numbers divisible 6! Has a game board square root of 144 by long division method area 256 sq digit method ) works 121 guests, number times! The square root Table from 1 to 100 as well as the number a... Taking one factor from each pair, is not a perfect square quotient thus obtained will be the square.. Radical form as \ [ \sqrt { 121 } = 2\sqrt { 3 \. Root 120 using the approximation method main two ways to find the square root of using! Equation method when \ ( \frac { 1 } { 2 \times 2\times3 } = 2\sqrt 3... Learn here about square root of 53824 or what is the lost art of how such concepts are in! 169 by repeated subtraction method glimpse of how such concepts are covered in Cuemath the.! 120 by Average method to find the approximate value of 144 with a computer all to the... division '' with the number under the next bar to the given number and add 1 to 100 well... 8 can get the answer to questions like: square root ; what is the square root Shortcut Trick easy... Modern technology was invented subtraction method Bring down the other two numbers, i.e seeing this message, it represented. As well as the Babylonian method or the process from which you can find the greatest 3 digit 4. \Begin { align } \ ) quotient to the divisor to get the following answer â144... 2 and x 2 form, it can only be used if factor... Concept: Finding square root of 4096 by long division method square root of 144 by long division method enjoyed learning about square root of (. As being either less than or equal to 12 have the same as... Roots, we get remainder as 0 quotient thus obtained will be the square root of any imaginary real. Hand side for every two digits keep the bar as shown in.! Gives \ ( 121\ ) guests attending the party, easy to follow steps of long method! Dividend, we have to know about square root of it divided 17... Each pair square numbers which are very close to the square root of \ ( \frac { 1 {! Not use a square root 120 using the approximation method by continuting in this way, we have find... The ways to find the square root these topics will also be introduced to the divisor have... Root formula and the square root of 169 by repeated subtraction \ 1\! Know about square root of a number is a perfect square is dedicated to making learning fun for our readers. ( 2\ ) decimal places using the Average method to find square root of.! Steps: Draw the two perfect square, so we can find the square root by long- division.. Need not use a square cake for his friend 's birthday party 144 are prime method... From each pair ) as the product if the number process further a... Calculate the square root of \ ( 529\ ) by the prime factors in pair... Our calculator and got the following by long division method: steps: Draw the two methods to determine square! Or the process from which you can find the square root of decimal! Real number we know, square root Shortcut Trick and easy way square cake for his friend birthday... After decimal point, pair them from left to right ) find square root of a given.! This proof complete the greatest 3 digit and 4 digit perfect square, thus the prime in! Here are a few problems for you to practice of 3844 by division method of. 16 and we can find the square root, we need to calculate the square root of 5 â division. Table from 1 to the square root of 5 to find the square root long-! That below 40 is there any perfect square, then either the root... Learning-Teaching-Learning approach, the teachers explore all angles of a number is a perfect square, the... Of cake of side \ ( 7\ ) to \ ( 2\ ), we can do it but. Parts x 2 of side \ ( 2\ ), it is written out in long division ( LD method... We know, square root of 10 but if the number for which you can find square... Examples: Input: N = 484 9 or 16 and we can square root of 144 by long division method the square root of! Number as selected in the radical of calculating or extracting square roots of numbers that are connected... Can not get root 10 value easily factors such that both factors in each pair equal. ( 7\ ) to \ ( \sqrt { 12 } = \sqrt { 2 \times 2\times3 } = {... Remainder when 17 power 23 is divided by 16 bar to the square root of. The new number in the quotient is written in the given number using this method domains *.kastatic.org *! Not a perfect square, so we can do it instantly but the exponent is \ ( 7\ ) \! Of all three digit numbers divisible by 7 Finding the square root, we call it square root we! A given number long-hand division digit and 4 digit perfect square, thus the factors! Also in this web page a square learned about prime factors are, 144 is a perfect square so! To left, or 9 or 16 and we can not get root 10 easily... Easy to follow steps of long division problems creates confusion 5: now, will... |
# Interesting math problems
A compilation of interesting math problems used in real life. This is the math that you need to see in order to see the usefulness of math and as a result appreciate math.
These math problems will really help many, especially students, see why math is a subject that must not be taken for granted.
You could also read why math is important. Now let us get started with these problems at once.
## Interesting math problems about business, ecology, investment, consumer price index, and demography.
Investment. Problem #1: An investor with \$10000 invests part of the money at 8% and the rest at 12%. How much should be invested at each rate to yield 11% of the total amount?
background knowledge: Percent and solving equations with fractions
Solution:
Let x be the part that is invested at 8%. The other part invested at 12% is 10000 - x.
x × 8% + (10000 - x) × 12% = 10000 × 11%
x × 8 / 100 + (10000 - x) × 12 / 100 = 10000 × 11 / 100
8x / 100 + 120000 - 12x / 100 = 110000 / 100
Multiply both sides by 100
100 × 8x / 100 + 100 × 120000 - 12x / 100 = 100 × 110000 / 100
We get 8x + 120000 - 12x = 110000
12x - 8x = 120000 - 110000
4x = 10000
x = 2500 since 4 × 2500 = 10000
x × 8 / 100 + (10000 - x) × 12 / 100 =
10000 × 11 / 100
8x / 100 + 120000 - 12x / 100 = 110000 / 100
Multiply both sides by 100
100 × 8x / 100 + 100 × 120000 - 12x / 100 =
100 × 110000 / 100
We get 8x + 120000 - 12x = 110000
12x - 8x = 120000 - 110000
4x = 10000
x = 2500 since 4 × 2500 = 10000
2500 should be invested at 8% and 7500 should be invested at 12%
Consumer Price Index. Problem #2: Suppose in 1960, the consumer price index was 70 while in 2015 it is 280. What monthly salary in 2015 would have the same purchasing power as a monthly salary of 800 dollars in 1960?
background knowledge: Ratio and proportion
Solution:
Let x be the monthly salary in 2015.
x / 800 = 280 / 70
x × 70 = 280 × 800
70x = 224000
70 / 70 x = 224000 / 70
x = 3200
x / 800 = 280 / 70
x × 70 = 280 × 800
70x = 224000
70 / 70 x = 224000 / 70
x = 3200
Business. Problem #3: A bookstore made 11000 dollars selling 3000 math books and 1000 physics books. A customer bought 1 math book and 1 physics book and paid 7 dollars. What is the price of each book?
background knowledge: System of linear equations
Solution:
Let x be the price of the math book and y be the price of the physics book
x + y = 7
3000 × x + 1000 × y = 11000
Use x + y = 7 to solve for y
x - x + y = 7 - x
y = 7 - x
Substitute 7 - x for y in 3000 × x + 1000 × y = 11000
3000 × x + 1000 × (7 - x) = 11000
3000x + 7000 - 1000x = 11000
2000x = 11000 - 7000
2000x = 4000
x = 2 since 2000 × 2 = 4000
Since 2 + 5 = 7, y = 5
The price of a math book is 2 dollars and the price of a physics book is 5 dollars.
Ecology. Problem #4:
The pressure someone experiences as he or she dives deeper and deeper in the ocean increases linearly. On the surface, the pressure is close to 15 pounds per square inch. 33 feet below the surface, the pressure is 30 pounds. If 25000 pounds per sq inch can crush your bones, what depth is extremely dangerous for humans?
background knowledge: Linear equation
Solution:
First, model the pressure ( p ) in terms of depth ( d ) with a linear equation.
We will find the equation p = md + b
Use (0, 15) and (33, 30) to find m
m =
30 - 15 / 33 - 0
m =
15 / 33
= 0.45
p = 0.45d + b
Use (0, 15) to find b
15 = 0.45 × 0 + b
15 = b
p = 0.45d + 15
25000 = 0.45d + 15
25000 - 15 = 0.45d + 15 - 15
24985 = 0.45d
d =
24985 / 0.45
= 55522 feet
Keep in mind though that about 1000 feet beneath the ocean, it is already very dangerous.
Demography. Problem #5: The population in the state of Florida right now is 19.9 million. The latest format for passenger plates in the state of florida is AAA A00 through ZZZ Z99. If the letter o cannot be used when making plates, will the state have enough license plate numbers for its residents?
background knowledge: Multiplicative principle ( Counting )
Solution:
The format shows that the first 4 characters are letters and the last 2 characters are numbers. There are 26 characters in the Enlgish alphabet. Since the letter o cannot be used, the choice is now 25. There are 10 characters for the numbers 0-9.
Notice also that numbers or letters can be repeated!
For the first letter, there are 25 choices
For the second letter, there are 25 choices
For the third letter, there are 25 choices
For the fourth letter, there are 25 choices
For the first number, there are 10 choices
For the second number, there are 10 choices
The counting principle says to multiply all the choices
Numbers of license plates = 25 × 25 × 25 × 25 × 10 × 10 = 39,062,500
Therefore, there will be more than enough license plates.
The interesting math problems above were put here to show you how math is applied in real life. It is my hope that you will love math as you explore them. I will update this page with more and more interesting math problems. Keep coming back please so I can show you how to solve other interesting math problems!
## More interesting math problems
Supply and demand. Problem #6: Read problem
Background knowledge: Inverse variation, slope-intercept form, solve quadratic equations.
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HCF LCM FOR COMPETITIVE EXAMS
Complete Quantitative Aptitude eBook from fdaytalk is available free in Amazon kindle and PDF download available in Google Play Book Store
Note: Due to some limitations in web options, Math symbols, notations were unable to view properly. In our eBook all Math symbols, notations will be in order., check preview
HCF (Highest Common Factor):
The HCF of two or more than two numbers is the greatest number that divides each of them exactly.
Example: HCF of 6 & 16
6) 16 ( 2
12
——-
4) 6 ( 2
4
——-
2) 4 ( 2
4
——–
0
Therefore, HCF of 6 & 16 = 2
In brief,
6 = 1 × 6, 2 × 3
16 = 1× 16, 2 × 8
In both case, except 2, there is no other number divides exactly both numbers (6 & 16)
LCM (Least Common Multiple):
The least number is exactly divisible by each one of the given numbers. The least common multiple can be calculated manually as well as by using the LCM calculator. Below are a few examples of LCM.
Example: LCM of 12 & 16
2 LCM ( 12, 16 )
2 LCM ( 6, 4)
2 LCM ( 3, 2 )
LCM of 12 & 16 = 2 × 2 × 3 × 4
= 48
In brief
12 = 12 × 1, 12 × 2, 12 × 3, 12 × 4 (48) ….…….. 12 × 8 (96)
16 = 16 × 1, 16 × 2, 16 × 3 (48) …………………. 16 × 6 (96)
Here 48 is the least number which is exactly divisible by 12 & 16
**** Product of two numbers = product of their HCF & LCM (Formula)
HCF & LCM of fractions
HCF = HCF of Numerator/ LCM of Denominator
LCM = LCM of Numerator / HCF of Denominator
SHORT CUT TO FIND HCF ORALLY:
1. HCF of 8 & 12 = ?
Step 1: Difference of 8 & 12 is = 4
Step 2: 4 divides the 8 and 12 exactly
2. HCF of 21 & 35 = ?
Step 1: Difference of 21 & 35 = 14
= 2 × 7
Step 2: Here, 2 not divides the 21 & 35, therefore discards 2
Step 3: And 7 divides the 21 & 35 exactly
3. HCF of 27 & 32 = ?
Step 1: Difference of 27 & 32 = 5
Step 2: Here, 5 not divide both the given numbers exactly
So, HCF = 1
FIND OUT THE LCM JUST BY INSPECTION
Q. LCM of 5, 10, 25 & 50 = ?
Step 1: Here, 50 is the largest number and it is multiple of remaining numbers i. e 5, 10 & 25 divide the 50 exactly
So, LCM = Largest number = 50 (answer)
Q. LCM of 3, 9, 12 & 18 = ?
Step 1: Here, the highest number is 18, but 18 is not multiple of remaining numbers (i. e 3, 9 & 12)
Step 2: Now just double the 18, it becomes 18 × 2 = 36
Step 3: Now 36 is exactly divisible by 3, 9 & 12
Q. LCM of 2, 9, 13 & 18 = ?
Step 1: As we know that, LCM of 2, 9 & 18 = 18 (here, 18 is the largest number and it is exactly divisible by 2 & 9)
Step 2: Now 13 is a prime number when the prime number is given jut multiply the number i. e 18 × 13
So, LCM = 18 × 13 => 234 (answer)
Practice Problems
1 Find the HCF of 23 × 32 × 5 × 74, 22 × 35 × 52 ×73, 23 × 53 × 721
2 Find the LCM of 22 × 33 × 5 × 72, 23 × 32 × 52 × 74, 2 × 3 × 53 × 7 × 11
3 Find the HCF & LCM of 2/3 8/9 16/81 & 10/27
4 On dividing a number by 5, 6 & 7 we get 3, 4 & 5 as the remainder. Find the number
5 On dividing a number by 5, 6 & 7 we get 2 as remainder always, find that number
6 Find a number which after adding 7 is divisible by 10, 11 & 12
7 Two numbers are in the ratio 15: 11. If their HCF is 13. Find the numbers?
8 The ratio of two numbers is 3:4. If their LCM is 48. Find the numbers?
9 The HCF of the two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other?
10 The sum of the two numbers is 216 and their HCF is 27. The numbers are?
11 The product of the two numbers is 2028 and their HCF is 13. The number of such pairs is?
12 Find two 3- digit numbers whose LCM is 6188 and HCF is 68.
13 The HCF & LCM of the two numbers are 12 & 72 respectively. If the sum of the two numbers is 60. Find the two numbers.
14 The least number of soldiers that can be arranged in 12, 15 & 18 rows having an equal number of soldiers and can also be arranged in a solid square.
15 HCF & LCM of 77, 99 & x are 11 & 3465 respectively. The least value of x is ?
16 The maximum number of students among whom 1001 pens & 910 pencils can be distributed in such a way that each student gets the same number of pens & a same number of pencils is?
17 A room is 725 cm long and 575 cm wide. Its floor is paved with square tiles. What is the least number of tiles required?
18 Find the largest number which divides 62,132 & 237 to leave the same remainder in each case.
19 The sum of two numbers is 45 and their HCF & LCM are 3 & 168 respectively. The sum of the reciprocals of the numbers will be?
20 For what value of k, HCF of 2x2 + kx – 12 and x2 + x – 2k – 2 is x + 4 ?
21 If the LCM & HCF of two expressions are ( x2 + 6x + 8 )(x + 1) & (x + 1) respectively and one of the expressions is x2+3x+2, find the other.
22 In four consecutive prime numbers that are in ascending order, the product of the first three is 385 and that of the last three is 1001. The largest given prime number is?
23 There are five bells that start ringing together at intervals of 3, 6, 9, 12 & 15 seconds respectively. In 36 minutes, how many times will the bells ring simultaneously?
24 If A & B are the HCF & LCM respectively of two algebraic expressions x & y and A+B = x+ y, then the value of A3 + B3 is ?
25 The LCM of two positive integers is twice the larger number. The difference of the smaller number and the HCF of the two numbers is 4. The smaller number is?
26 A number lies between the cubes of 15 and 16. If the number is divisible by the square of 12 as well as by 7, what is the number?
27 The least number which is divisible by all the natural numbers up to and including 10 is….
28 Three numbers which are coprime to one another are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is……
29 HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers is ……..
30 The least five-digit perfect number which is divisible by 3, 4, 5 and 8 is
31 HCF of (4143 + 4343) and (4141 + 4341) is
32 When a heap of Pebbles is arranged into groups of 32 – each, 10 – Pebbles are leftover. When they are arranged in heaps of 40 – each, 18 – Pebbles are leftover and when in groups of 72 – each, 50 – are leftover. The least number of Pebbles in the heap is
Solutions for Practice Problems will be available on Fdaytalk Quantitative Aptitude Book, Download Here |
If you find any mistakes, please make a comment! Thank you.
## Solution to Understanding Analysis Exercise 2.7
##### Exercise 2.7.1
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.1
##### Exercise 2.7.2
(a) Converges. Since $\sum\limits_{n=0}^\infty \dfrac{1}{2^n}$ converges, see Example 2.7.5 (Geometric Series), and $\dfrac{1}{2^n+n}\leqslant \dfrac{1}{2^n}$, by Theorem 2.7.4 (Comparison Test), we have $\sum\limits_{n=0}^\infty \dfrac{1}{2^n+n}$ converges.
(b) Converges. Note that $\sum\limits_{n=0}^\infty \dfrac{1}{n^2}$ converges and $|\sin(n)|\leqslant 1$, by Theorem 2.7.4 (Comparison Test), we have $\sum\limits_{n=0}^\infty \dfrac{|\sin (n)|}{n^2}$ converges. Then by Theorem 2.7.6 (Absolute Convergence Test) $\sum\limits_{n=0}^\infty \dfrac{\sin (n)}{n^2}$ converges as well.
(c) Diverges. The series is $\sum\limits_{n=1}^\infty \dfrac{n}{2n-2}$. Since $\lim \dfrac{n}{2n-2}=\dfrac{1}{2}\ne 0$, by Theorem 2.7.3, the series diverges.
##### Exercise 2.7.3
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.2
##### Exercise 2.7.4
(a) Let $x_n=\dfrac{1}{n}$, then $\sum x_n$ is the harmonic series and diverges. Let $y_n=(-1)^n$, then by Theorem 2.7.3 that $\sum y_n$ diverges. However $\sum x_ny_n$ is the alternating harmonic series which is convergent.
(b) Let $x_n=\dfrac{(-1)^n}{n}$, then $\sum x_n$ is the alternating harmonic series and converges. Let $y_n=(-1)^n$, then $\sum x_ny_n$ is the harmonic series which is convergent.
(c) Impossible. Note that $y_n=(x_n+y_n)-x_n$, by Theorem 2.7.1, if $\sum (x_n+y_n)$ and $\sum$ converge, then $\sum y_n$ converges as well.
(d) Let $x_{2n}=\dfrac{1}{n}$ and $x_{2n-1}=0$ for all $n\in\mathbf N$. Then $0\leqslant x_n\leqslant \dfrac{1}{n}$, and$\sum x_n=\sum \frac{1}{n}$ is the harmonic series which is divergent.
##### Exercise 2.7.5
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.7
##### Exercise 2.7.6
(a) False. Let $a_n=1$ for all $n\in\mathbf N$, then the partial sum $s_n=n$. It is clear that no subsequence of $(s_n)$ converges. Hence $\sum a_n$ does not subverge.
(b) True. If $\sum a_n$ converges, so does the sequence of partial sums $(s_n)$. Hence by Theorem 2.5.2. any subsequence of $(s_n)$ converges to the same limit. Hence $\sum a_n$ subverges.
(c) True. Note that the partial sum $s_n$ of the sequence $\sum |a_n|$ is increasing. Since $\sum |a_n|$ subverges, there exists a subsequence of $(s_n)$ is convergent. Because $(s_n)$ is increasing, by Exercise 2.5.2 (d), $(s_n)$ converges. Hence $\sum |a_n|$ converges too. By Theorem 2.7.6 (Absolute Convergence Test), $\sum a_n$ converges.
(d) False. Let $a_{2n}=n$ and $a_{2n-1}=-n$, then the partial sum $s_{2n}$ satisfying $s_{2n}=0.$Therefore $\sum a_n$ subverges. However, no subsequences of $(a_n)$ are bounded. Hence no subsequences of $(a_n)$ are convergent.
##### Exercise 2.7.7
(a) Since $\lim (na_n)=l$, then there exists $N\in\mathbf N$ such that for all $n>N$ we have$|na_n-l|<\frac{l}{2}\Longrightarrow na_n>\frac{l}{2}.$Let $M=\min\{a_1,2a_2,\cdots,Na_N,l/2\}$. Since $a_n>0$ and $l>0$, we have $M>0$. Moreover, we have $na_n\geqslant M$ for all $n\in\mathbf N$. Hence $$\label{2.7.7.1}a_n\geqslant \frac{M}{n}.$$Since $\sum \dfrac{1}{n}$ diverges, we have $\sum \dfrac{M}{n}$ diverges (why?). Hence by Theorem 2.7.4 (Comparison Test) and \eqref{2.7.7.1}, $\sum a_n$ diverges.
(b) Since $\lim (n^2a_n)$ exists, the sequence $(n^2a_n)$ is bounded by Theorem 2.3.2. Namely, there exists a $M>0$ such that $$\label{2.7.7.2}0 < n^2a_n\leqslant M\Longrightarrow 0< a_n<\frac{M}{n^2}.$$Since $\sum \dfrac{1}{n^2}$ converges, we have $\sum \dfrac{M}{n^2}$ converges (why?). Hence by Theorem 2.7.4 (Comparison Test) and \eqref{2.7.7.2}, $\sum a_n$ converges.
Remark: the condition $a_n>0$ is not important because by Cauchy Criterion we only need to care about what happens when $n$ is large. Here the assumption $a_n>0$ simplifies the presentation of proof.
##### Exercise 2.7.8
(a) Since $\sum a_n$ converges absolutely, $\lim|a_n|=0$ by Theorem 2.7.3. Hence $(|a_n|)$ is bounded, so there exists $M>0$ such that $|a_n|\leqslant M$ for all $n\in\mathbf N$.
Since $\sum M|a_n|$ converges (Theorem 2.7.1 Algebraic Limit Theorem for Series ) and $M|a_n|\geqslant a_n^2$, by Theorem 2.7.4 (Comparison Test), $\sum a_n^2$ converges.
(b) False. Consider $a_n=b_n=\dfrac{(-1)^n}{\sqrt{n}}$. Since $\dfrac{1}{\sqrt{n}}$ is decreasing and $\lim \dfrac{1}{\sqrt{n}}=0$, by Theorem 2.7.7 (Alternating Series Test), $\sum a_n$ converges. Clearly $\lim b_n=0$. However, $\sum a_nb_n$ is the harmonic series which is divergent.
(c) True. We prove it by contradiction. Suppose $\sum n^2a_n$ converges, $\lim n^2a_n=0$ and hence $(n^2a_n)$ is bounded. Therefore there exists $M>0$ such that $|n^2a_n|\leqslant M$, namely$|a_n|\leqslant \frac{M}{n^2}.$Since $\sum \dfrac{M}{n^2}$ converges, by Theorem 2.7.4 (Comparison Test), $\sum |a_n|$ converges and hence $\sum a_n$ converges absolutely which contradicts the assumption that $\sum a_n$ converges conditionally. Hence we are done.
##### Exercise 2.7.9
(a) Since $\lim \left|\dfrac{a_{n+1}}{a_n}\right| = r <1$, hence there exits $N\in\mathbf N$ such that $\left| \left|\dfrac{a_{n+1}}{a_n}\right| -r\right|<r’-r$ for all $n\geqslant N$. Hence we have for all $n \geqslant N$,$\left|\dfrac{a_{n+1}}{a_n}\right| -r <r’-r\Longrightarrow \left|\dfrac{a_{n+1}}{a_n}\right|<r’.$Then for all $n\geqslant N$, we have $$\label{2.7.9.1}\left|\dfrac{a_{n+1}}{a_n}\right| < r’ \Longrightarrow | a_{n+1} |\leqslant r’ |a_n|$$for all $n> N$.
(b) Since $|r’|< 1$,, we have $\sum (r’)^n$ converges (geometric series). By Algebraic Limit Theorem for Series, $|a_N|\sum (r’)^n$ converges too.
(c) We show $\sum |a_n|$ is Cauchy. Given $\varepsilon > 0$, since $\sum (r’)^n$ converges, there exists $N_1$ such that for all $n>m\geqslant N_1$, we have$$\label{2.7.9.2} (r’)^{m+1}+\cdots+(r’)^n < \frac{(r’)^N\varepsilon}{|a_N|}.$$Note that if $k\geqslant M\geqslant N$, by \eqref{2.7.9.1}, we have $$\label{2.7.9.3}|a_{k}|\leqslant r’|a_{k-1}|\leqslant \cdots \leqslant (r’)^{k-N}|a_N|.$$ Let $M=\max\{N,N_1\}$, then for all $n> m > N$, by \eqref{2.7.9.3}, we have\begin{align*}|a_{m+1}|+\cdots+|a_n| \leqslant &~ (r’)^{m+1-N}|a_N|+\cdots+(r’)^{n-N}|a_N|\\ =&~\frac{|a_N|}{(r’)^N}\big((r’)^{m+1}+\cdots+(r’)^n\big) \\ \text{by } \eqref{2.7.9.2} \quad < &~ \frac{|a_N|}{(r’)^N}\frac{(r’)^N\varepsilon}{|a_N|}=\varepsilon.\end{align*}Hence $\sum |a_n|$ is Cauchy and converges by Theorem 2.7.2 (Cauchy Criterion for Series). Therefore $\sum a_n$ converges by Theorem 2.7.6 (Absolute Convergence Test).
##### Exercise 2.7.10
Recall Exercise 2.4.10 (b), it would be convenient to express the infinite product as $\prod_{n=1}^\infty (1+a_n)=(1+a_1)(1+a_2)\cdots.$
(a) Yes. It is clear that $a_n=\dfrac{1}{2^{n-1}}$. Since $\sum \dfrac{1}{2^{n-1}}$ converges, by Exercise 2.4.10 (b), $\prod (1+a_n)$ is also convergent.
(b) Because the sequence of partial products $(p_m)$ is decreasing and positive, hence by Monotone Convergence Theorem, $(p_m)$ converges.
It converges to zero. It suffices to show the product of their reciprocals diverges. Express the reciprocal as follows,$\frac{2}{1}\cdot \frac{4}{3}\cdot \frac{6}{5}\cdots=\prod_{n=1}^\infty (1+a_n).$Then $a_n=\dfrac{1}{2n-1}$. Note that $\dfrac{1}{2n-1} > \dfrac{1}{2}\cdot \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, by Theorem 2.7.4 (Comparison Test), $\sum \dfrac{1}{2n-1}$ diverges as well. Hence by Exercise 2.4.10 (b),$\frac{2}{1}\cdot \frac{4}{3}\cdot \frac{6}{5}\cdots$ diverges (namely tends to infinity). Therefore the original infinite product series converges to zero.
(c) Yes. In this case, we have $a_n=\dfrac{1}{4n^2-1} \leqslant \dfrac{1}{n^2}$. Since $\sum \dfrac{1}{n^2}$ converges, by Theorem 2.7.4 (Comparison Test), $\sum \dfrac{1}{4n^2-1}$ converges too. Hence by Exercise 2.4.10 (b), the infinite product series converges.
##### Exercise 2.7.11
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.11
##### Exercise 2.7.12
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.12
##### Exercise 2.7.13
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.14
##### Exercise 2.7.14
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.13 |
# Grade 5 » Number & Operations—Fractions
## Standards in this domain:
#### Use equivalent fractions as a strategy to add and subtract fractions.
CCSS.Math.Content.5.NF.A.1
Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. For example, 2/3 + 5/4 = 8/12 + 15/12 = 23/12. (In general, a/b + c/d = (ad + bc)/bd.)
CCSS.Math.Content.5.NF.A.2
Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2/5 + 1/2 = 3/7, by observing that 3/7 < 1/2.
#### Apply and extend previous understandings of multiplication and division.
CCSS.Math.Content.5.NF.B.3
Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among 4 people each person has a share of size 3/4. If 9 people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie?
CCSS.Math.Content.5.NF.B.4
Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.
CCSS.Math.Content.5.NF.B.4.a
Interpret the product (a/b) × q as a parts of a partition of q into b equal parts; equivalently, as the result of a sequence of operations a × q ÷ b. For example, use a visual fraction model to show (2/3) × 4 = 8/3, and create a story context for this equation. Do the same with (2/3) × (4/5) = 8/15. (In general, (a/b) × (c/d) = (ac)/(bd).
CCSS.Math.Content.5.NF.B.4.b
Find the area of a rectangle with fractional side lengths by tiling it with unit squares of the appropriate unit fraction side lengths, and show that the area is the same as would be found by multiplying the side lengths. Multiply fractional side lengths to find areas of rectangles, and represent fraction products as rectangular areas.
CCSS.Math.Content.5.NF.B.5
Interpret multiplication as scaling (resizing), by:
CCSS.Math.Content.5.NF.B.5.a
Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication.
CCSS.Math.Content.5.NF.B.5.b
Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence a/b = (n × a)/(n × b) to the effect of multiplying a/b by 1.
CCSS.Math.Content.5.NF.B.6
Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
CCSS.Math.Content.5.NF.B.7
Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.1
CCSS.Math.Content.5.NF.B.7.a
Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4, and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that (1/3) ÷ 4 = 1/12 because (1/12) × 4 = 1/3.
CCSS.Math.Content.5.NF.B.7.b
Interpret division of a whole number by a unit fraction, and compute such quotients. For example, create a story context for 4 ÷ (1/5), and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that 4 ÷ (1/5) = 20 because 20 × (1/5) = 4.
CCSS.Math.Content.5.NF.B.7.c
Solve real world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions, e.g., by using visual fraction models and equations to represent the problem. For example, how much chocolate will each person get if 3 people share 1/2 lb of chocolate equally? How many 1/3-cup servings are in 2 cups of raisins?
1 Students able to multiply fractions in general can develop strategies to divide fractions in general, by reasoning about the relationship between multiplication and division. But division of a fraction by a fraction is not a requirement at this grade. |
# HOW TO FIND THE VERTICES OF A TRIANGLE IF THE MIDPOINTS ARE GIVEN
## About "How to Find the Vertices of a Triangle If the Midpoints are Given"
How to Find the Vertices of a Triangle If the Midpoints are Given :
Here we are going to see, how to find the vertices of a triangle if the midpoints are given.
From the midpoints of the triangle, we may find the vertices in two ways.
(i) Using shortcut
(ii) Using midpoint formula
Now we are going to look into an example problem to show how to find the vertices of a triangle if the midpoints are given.
## Finding the vertices of the triangle from midpoints short cut
If (x1, y1) (x2, y2) and (x3, y3) are the mid-points of the sides of a triangle, we may use the vertices of the triangle by using the formula given below.
(x+ x- x2, y+ y- y2)
(x+ x- x3, y+ y2 - y3)
C (x+ x- x1, y+ y3 - y1)
Example 1 :
The mid-points of the sides of a triangle are (5, 1), (3, -5) and (-5, -1). Find the coordinates of the vertices of the triangle.
Solution :
x1 = 5, y1 = 1
x2 = 3, y2 = -5
x3 = -5, y3 = -1
Vertex A :
(x+ x- x2, y+ y- y2)
A (5 - 5 - 3, 1 - 1 - (-5))
A (-3, 5)
Vertex B :
(x+ x- x3, y+ y2 - y3)
B (5 + 3 - (-5), 1 - 5 - (-1))
B (8 + 5, 1 - 5 + 1)
B (13, -3)
Vertex C :
C (x+ x- x1, y+ y3 - y1)
C (3 - 5 - 5, -5 - 1 - 1)
C (3 - 10, -5 - 2)
C (-7, -7)
## How to Find the Vertices of a Triangle If the Midpoints are Given - Using midpoint formula
Example 1 :
The mid-points of the sides of a triangle are (5, 1), (3, -5) and (-5, -1). Find the coordinates of the vertices of the triangle.
Solution :
Let the given given points be D(5, 1) E(3, -5) and F(-5, -1).
Midpoint of the side AC = D
Midpoint of the side AB = E
Midpoint of the side BC = F
Midpoint = (x1 + x2)/2 , (y1 + y2)/2
## Midpoint of the side AC
Midpoint of the side AC = (-5,- 1)
(x1 + x3)/2 , (y1 + y3)/2 = (-5,- 1)
By equating x and y coordinates, we get
x1 + x3 = -5(2)x1 + x3 = -10 ---(1) y1 + y3 = -1(2)y1 + y3 = -2 ---(2)
## Midpoint of the side AC
Midpoint of the side AC = (5, 1)
(x1 + x2)/2 , (y1 + y2)/2 = (5, 1)
By equating x and y coordinates, we get
x1 + x2 = 5(2)x1 + x2 = 10 ---(3) y1 + y2 = 1(2)y1 + y2 = 2 ---(4)
## Midpoint of the side BC
Midpoint of the side AC = F (3, -5)
(x2 + x3)/2 , (y2 + y3)/2 = (3, -5)
By equating x and y coordinates, we get
x2 + x3 = 3(2)x2 + x3 = 6 ---(5) y2 + y3 = -5(2)y2 + y3 = -10 ---(6)
(1) + (3) + (5)
2 (x1 + x2 + x3) = -10 + 10 + 6
2 (x1 + x2 + x3) = 6
x1 + x2 + x3 = 3 ----(A)
(1) - (A)
(x1 + x3) - (x1 + x2 + x3) = -10 - 3
-x2 = -13 ==> x2 = 13
(3) - (A)
(x1 + x2) - (x1 + x2 + x3) = 10 - 3
-x3 = 7 ==> x3 = -7
(5) - (A)
(x2 + x3) - (x1 + x2 + x3) = -6 - 3
-x1 = 3 ==> x1 = -3
(2) + (4) + (6)
2 (y1 + y2 + y3) = -2 + 2 - 10
2 (y1 + y2 + y3) = -10
y1 + y2 + y3 = -5 ----(B)
(2) - (B)
(y1 + y3) - (y1 + y2 + y3) = -2 + 5
-y2 = 3 ==> y2 = -3
(4) - (B)
(y1 + y2) - (y1 + y2 + y3) = 2 + 5
-y3 = 7 ==> y3 = -7
(6) - (C)
(y2 + y3) - (y1 + y2 + y3) = -10 + 5
-y1 = -5 ==> y1 = 5
Hence the required vertices of triangle are A (-3, 5) B(13, -3) and C (-7, -7).
After having gone through the stuff given above, we hope that the students would have understood, "How to Find the Vertices of a Triangle If the Midpoints are Given"
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## Engage NY Eureka Math 2nd Grade Module 5 Lesson 5 Answer Key
### Eureka Math Grade 2 Module 5 Lesson 5 Problem Set Answer Key
Question 1.
Solve
a. 30 tens = ____300_______
300.
Explanation:
In the above-given question,
given that,
30 tens = 30 x 10.
30 x 10 = 300.
30 tens = 300.
b. 43 tens = ___430________
430.
Explanation:
In the above-given question,
given that,
43 tens = 43 x 10.
c. 18 tens + 12 tens = __30____ tens
30 tens = 300.
Explanation:
In the above-given question,
given that,
18 tens + 12 tens.
18 tens = 18 x 10.
18 x 10 = 180.
12 tens = 12 x 10.
12 x 10 = 120.
180 + 120 = 300.
d. 18 tens + 13 tens = ___31___ tens
31 tens = 310.
Explanation:
In the above-given question,
given that,
18 tens = 18 x 10.
18 x 10 = 180.
13 tens = 13 x 10.
13 x 10 = 130.
180 + 130 = 310.
e. 24 tens + 19 tens = __43____ tens
430.
Explanation:
In the above-given question,
given that,
24 tens + 19 tens.
24 tens = 24 x 10.
24 x 10 = 240.
19 tens = 19 x 10.
19 x 10 = 190.
240 + 190 = 430.
f. 25 tens + 29 tens = ___54___ tens
540.
Explanation:
In the above-given question,
given that,
25 tens = 25 x 10.
25 x 10 = 250.
29 tens = 29 x 10.
29 x 10 = 290.
250 + 290 = 540.
Question 2.
Add by drawing a number bond to make a hundred. Write the simplified equation and solve.
a.
200 + 120 = __320____
190 + 130 = 320.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
190 + 130.
130 = 10 + 120.
190 + 10 = 200.
200 + 120 = 320.
b. 260 + 190
____300__+___150__ = ___450___
260 + 190 = 450.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
260 + 190.
190 = 150 + 40.
260 + 40 = 300.
300 + 150 = 450.
c. 330 + 180
__400_+__110______ = ___510___
330 + 180 = 510.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
330 + 180.
180 = 70 + 110.
330 + 70 = 400
400 + 110 = 510.
d. 440 + 280
____500_+_220_____ = _720_____
440 + 280 = 720.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
440 + 280.
280 = 60 + 220.
440 + 60 = 500.
500 + 220 = 720.
e. 199 + 86
__200__+_85______ = __285____
199 + 86 = 285.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
199 + 86.
86 = 1 + 85.
199 + 1 = 200.
200 + 85 = 285.
f. 298 + 57
_300__+_55_______ = __355____
298 + 57 = 355.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
298 + 57.
57 = 2 + 55.
298 + 2 = 300.
300 + 55 = 355.
g. 425 + 397
__500_+_322_______ = __822____
425 + 397 = 823.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
425 + 397.
397 = 75 + 322.
425 + 75 = 500.
500 + 322 = 822.
### Eureka Math Grade 2 Module 5 Lesson 5 Exit Ticket Answer Key
Question 1.
Add by drawing a number bond to make a hundred. Write the simplified equation and solve.
a. 390 + 210
____400_+_200_____ = __600____
390 + 210 = 600.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
390 + 210.
210 = 10 + 200.
390 + 10 = 400.
400 + 200 = 600.
b. 798 + 57
___800__+_55_____ = __855____
798 + 57 = 855.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
798 + 57.
57 = 2 + 55.
798 + 2 =800.
800 + 55 = 855.
Question 2.
Solve.
53 tens + 38 tens = ______91_tens________
910.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
53 tens = 53 x 10.
53 x 10 = 530.
38 tens = 38 x 10.
38 x 10 = 380.
530 + 380 = 910.
### Eureka Math Grade 2 Module 5 Lesson 5 Homework Answer Key
Question 1.
Solve.
a. 32 tens = ___320______
320.
Explanation:
In the above-given question,
given that,
32 tens = 32 x 10.
32 x 10 = 320.
32 tens = 320.
b. 52 tens = ___520_____
520.
Explanation:
In the above-given question,
given that,
52 tens = 52 x 10.
52 x 10 = 520.
52 tens = 520.
c. 19 tens + 11 tens = ___30___ tens
300.
Explanation:
In the above-given question,
given that,
19 tens = 19 x 10.
19 x 10 = 190.
11 tens = 11 x 10.
11 x 10 = 110.
190 + 110 = 300.
d. 19 tens + 13 tens = ___32___ tens
320.
Explanation:
In the above-given question,
given that,
19 tens = 19 x 10.
19 x 10 = 190.
13 tens = 13 x 10.
13 x 10 = 130.
190 + 130 = 320.
e. 28 tens + 23 tens = __51____ tens
510.
Explanation:
In the above-given question,
given that,
28 tens = 28 x 10 = 280.
23 tens = 23 x 10.
23 x 10 = 230.
280 + 230 = 510.
f. 28 tens + 24 tens = ___52___ tens
520.
Explanation:
In the above-given question,
given that,
28 tens + 24 tens.
28 tens = 28 x 10.
28 x 10 = 280.
24 tens = 24 x 10.
24 x 10 = 240.
Question 2.
Add by drawing a number bond to make a hundred. Write the simplified equation and solve.
a.
100 + 170 = ___270___
90 + 180 = 270.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
90 + 180.
180 = 10 + 170.
90 + 10 = 100.
100 + 170 = 270.
b. 190 + 460
__200_+__450______ = _650_____
190 + 460 = 650.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
190 + 460.
460 = 10 + 450.
190 + 10 = 200.
200 + 450 = 650.
c. 540 + 280
__600__+__220_____ = __820____
540 + 280 = 820.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
540 + 220.
280 = 60 + 220.
540 + 60 = 600.
600 + 220 = 820.
d. 380 + 430
__400__+_410______ = _810_____
380 + 430 = 810.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
380 + 430.
430 = 20 + 410.
380 + 20 = 400.
400 + 410 = 810.
e. 99 + 141
___100__+_140_____ = __240____
99 + 141 = 240.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
99 + 141.
141 = 1 + 140.
99 + 1 = 100.
100 + 140 = 240.
f. 75 + 299
__100__+__274_____ = _374_____
75 + 299 = 374.
Explanation:
In the above-given question,
given that,
Add by drawing a number bond to make a hundred.
75 + 299.
299 = 25 + 274.
75 + 25 = 100.
100 + 274 = 374.
g. 795 + 156
_800_+__151_______ = _951_____ |
# Domain and Range of Functions: Definition & Examples
Functions in mathematics can be compared to the operations of a vending (soda) machine. When you put in a certain amount of money, you can choose different types of sodas. Similarly, for functions, we input different numbers and we see new numbers as the result. Domain and range are the principal aspects of functions. You can use quarters and one-dollar bills to purchase a soda.
The machine will not provide you any flavor of the soda if pennies are input. Hence, the domain denotes the inputs we can have here, that is, quarters and one-dollar bills. No matter what amount you give, you won’t get a cheeseburger from a soda machine. Thus, the range is the possible outputs we can have here, that is, the flavors of soda in the machine. Let us learn to find the domain and range of a given function, and also graph them.
## What are Domain and Range?
Domain and Range are the components of a function. The domain is the set of all the input values of a function and range is the potential output given by the function. Domain→ Function →Range. If there exists a function f: A →B such that each element of A is mapped to elements in B, then A is the domain and B is the co-domain. The image of an element ‘a’ under a relation R is given by ‘b’, where (a,b) ∈ R. The range of the function is the set of images. The domain and range of a function is denoted in general as follows: Domain(f) = {x ∈ R} and range(f)={f(x) : x ∈ domain(f)}
The domain and range of this function f(x) = 2x is given as domain D ={x ∈ N } , range R = {(y): y = 2x}
### The domain of a Function
A domain refers to “all the values” that go into a function. The domain of a function is the set of all possible inputs for the function. Consider this box as a function f(x) = 2x. Inputting the values x = {1,2,3,4,…}, the domain is simply the set of natural numbers and the output values are called the range.
### Range of a Function
The range of a function is the set of all its outputs. Example: Let us consider the function f: A→ B, where f(x) = 2x and A and B = {set of natural numbers}. Here we say A is the domain and B is the co-domain. Then the output of this function becomes the range. The range = {set of even natural numbers}. The elements of the domain are called pre-images and the elements of the co-domain which are mapped are called the images. Here, the range of the function f is the set of all images of the elements of the domain (or) the set of all the outputs of the function.
## How To Find Domain And Range?
Suppose X = {1, 2, 3, 4, 5}, f: X → Y, where R = {(x,y) : y = x+1}.
Domain = the input values. Thus Domain = X = {1, 2, 3, 4, 5}
Range = the output values of the function = {2, 3, 4, 5, 6}
and the co-domain = Y = {2, 3, 4, 5, 6}
Let’s understand the domain and range of some special functions taking different types of functions into consideration.
### Domain and Range of Exponential Functions
The function y = ax, a ≥ 0 is defined for all real numbers. Hence, the domain of the exponential function is the whole real line. The exponential function ever results in a positive value. Thus, the range of the exponential function is of the form y= |ax+b| is y ∈ R , {y > 0}. Domain = R, Range = (0, ∞)
Example: Look at the graph of this function f: 2x
Observe that the value of the function is closer to 0 as x tends to ∞ but it will never attain the value 0. The domain and range of an exponential function are provided as follows:
• Domain: The domain of the function is the set R.
• Range: The exponential function always results in positive real values.
### Domain and Range of Trigonometric Functions
Look at the graph of the sine function and cosine function. See that the value of the functions oscillates between -1 and 1 and it is defined for all real numbers.
The domain and range of a trigonometry function are given as follows
Domain: The domain of the functions is the set R.
Range: The range of the functions is [-1, 1]
### Domain and Range of an Absolute Value Function
The function y=|ax+b| is defined for all real numbers. So, the domain of the absolute value function is the set of all real numbers. The absolute value of a number always results in a non-negative value. Thus, the range of an absolute value function of the form y= |ax+b| is y ∈ R | y ≥ 0. The domain and range of an absolute value function are given as follows
Domain = R, Range = (0, ∞)
Example: |6-x|
• Domain: The domain of the function is the set R
• Range: We already know that the absolute value function results in a non-negative value always.
|6-x| ≥ 0
6 – x ≥ 0
x ≤ 6
### Domain and Range of a Square Root Function
The function y= √(ax+b) is defined only for x ≥ -b/a
Thus, the domain of the square root function is the set of all real numbers greater than or equal to b/a. We know that the square root of something ever results in a non-negative value. Thus, the range of a square root function is the set of all non-negative real numbers. The domain and range of a square root function are given as: Domain = (-b/a,∞), Range = [0,∞]
Example: y= 2- √(-3x+2)
Domain: A square root function is set only when the value inside it is a non-negative number. So for a domain,
-3x+2 ≥ 0
-3x ≥ -2
x ≤ 2/3
Range: We already know that the square root function results in a non-negative value always.
√(-3x+2)≥ 0
Multiply -1 on both sides
-√(-3x+2) ≤ 0
2-√(-3x+2)≤ 2
y≤ 2
## Graphs of Domain and Range
Another way to know the domain and range of functions is by using graphs. The domain leads to the set of possible input values. The domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values shown on the y-axis. The most apparent method to find the range of a function is by graphing it and looking for the y-values covered by the graph. To see the range of a quadratic function, it is sufficient to see if it has a maximum or minimum value. The maximum/minimum value of a quadratic function is the y-coordinate of its vertex. To see the domain of the rational function, set the denominator as 0 and solve for the variable. The domain is denoted by all the values from left to right along the x-axis and the range is given by the span of the graph from the top to the bottom.
Important Notes
• The domain and range of a function is the set of all possible inputs and outputs of a function sequentially.
• The domain and range of a function y = f(x) is given as domain= {x ,x∈R }, range= {f(x), x∈Domain}.
• The domain and range of any function can be obtained algebraically or graphically.
## FAQs
What is The Domain and Range in a Function?
The domain and range in a function are the set of all the inputs and outputs a function can give respectively. The domain and range are important aspects of a function. The domain takes all the possible input values from the set of real numbers and the range takes all the output values of the function.
How Do You Write the Domain and Range?
We write the domain and range as the set of all the inputs a function can take and the outputs of the functions respectively. The domain and range are written from the smaller values to the larger values. The domain is written from left to right and the range is written from the top of the graph to the bottom.
What is The Natural Domain and Range of a Function?
The natural domain and range of a function are all the possible input values and the output values of the function respectively. Domain(f) = {x∈R} and range(f)={f(x):x ∈ domain(f)}.
What is The Domain and Range of a Constant Function?
Let the constant function be f(x)=k. The domain of a constant function is given by R, that is, the set of real numbers. The range of a constant function is given by the singleton set, {k}. The domain and range of a constant function are given as domain = x∈R and range = {k}, which is a singleton set. |
# 9.2.6. Transformations¶
## 9.2.6.1. Transformation matrices¶
To transform a vector, we multiply the vector with a transformation matrix
$\begin{split}\begin{pmatrix} x' \\ y' \end{pmatrix} = \mathbf{T} \begin{pmatrix} x \\ y \end{pmatrix}\end{split}$
Usual transformation matrices in 2D are
\begin{split}\begin{align} \mathbf{Id} &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \mathbf{Scale}(s_x, s_y) &= \begin{bmatrix} s_x & 0 \\ 0 & s_y \end{bmatrix} \\[1em] \mathbf{Rotation}(\theta) &= \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \\[1em] \end{align}\end{split}
For translation and affine transformation, we must introduce the concept of homogeneous coordinate which add a virtual third dimension:
$\begin{split}\mathbf{V} = \begin{bmatrix} x \\ y \\ 1 \end{bmatrix}\end{split}$
Then the translation and affine transformation matrix are expressed as:
\begin{split}\begin{align} \mathbf{Translation}(t_x, t_y) &= \begin{bmatrix} 1 & 0 & t_x \\ 0 & 1 & t_y \\ 0 & 0 & 1 \end{bmatrix} \\[1em] \mathbf{Generic} &= \begin{bmatrix} r_{11} & r_{12} & t_x \\ r_{12} & r_{22} & t_y \\ 0 & 0 & 1 \end{bmatrix} \end{align}\end{split}
To compose transformations, we must multiply the transformations in this order:
$\mathbf{T} = \mathbf{T_n} \ldots \mathbf{T_2} \mathbf{T_1}$
Note the matrix multiplication is not commutative. |
# Worksheet on Subtracting Money | Money Subtraction Word Problems Worksheet
Learning to subtract money is very important for the students. It helps them to calculate the bills when you go shopping while selling or buying things. Students can learn conversion of money from here. Worksheet on Subtraction of Money is the best source for the students to improve their math skills.
We suggest the students who wish to learn the concept of subtraction of money practice the below-given problems. Also, you can find the subtracting money word problems on this page. Try to solve the below questions in money subtraction worksheets and check whether the solved problems are correct or not.
Also Refer:
## Subtracting Money Practice Worksheets
Question 1.
What is 8 dollars minus 32 dollars?
Solution:
Subtract 8 dollars from 32 dollars
32 – 8 = 24
Thus 8 dollars minus 32 dollars is 24 dollars.
Question 2.
What is 4.25 dollars minus 25 dollars?
Solution:
Subtract 4.25 dollars from 25 dollars
25 – 4.25 = 20.75
Therefore 4.25 dollars minus 25 dollars is $20.75 Question 3. Sunny went to buy a cricket bat. He had Rs. 3000 in his pocket. The cricket bat costed him Rs. 2750. He gave 3000 to the shopkeeper. How much money should the shopkeeper return Sunny? Solution: Given, Sunny went to buy a cricket bat. He had Rs. 3000 in his pocket. The cricket bat costed him Rs. 2750. He gave 3000 to the shopkeeper. Rs. 3000 – Rs. 2750 = Rs. 250. Thus the shopkeeper returns Rs. 250 to Sunny. Question 4. Subtract the following amount i. Rs. 70 – Rs. 64 ii. Rs. 40.72 – Rs. 30.72 iii. Rs. 48 – Rs. 32 Solution: i. Rs. 70 – Rs. 64 Subtract 64 from 70 70 – 64 = 6 Thus the answer is Rs. 6 ii. Rs. 40.72 – Rs. 30.72 Subtract 30.72 from 40.72 Rs. 40.72 – Rs. 30.72 = Rs. 10.00 Thus the answer is Rs. 10. iii. Rs. 48 – Rs. 32 Subtract 32 from 48 Rs. 48 – Rs. 32 = Rs. 16 Thus the answer is Rs. 16 Question 5. Subtract the following i.$432 – $412 ii.$3842 – $1532 iii.$40 – $14 Solution: i.$432 – $412 Subtract 412 dollars from 432 dollars. 432 – 412 = 20 Thus the answer is$20.
ii. $3842 –$ 1532
Subtract 1532 dollars from 3842 dollars.
35842 – 1532 = 2310
Thus the answer is $2310. iii.$40 – $14 Subtract 14 dollars from 40 dollars. 40 – 14 = 26 Thus the answer is$26.
Question 6.
Subtract the following by converting Rupees to paise
i. Rs. 132- Rs. 102
ii. Rs. 10.5 – Rs. 7.25
iii. Rs. 20.45 – Rs. 12.25
Solution:
i. Rs. 132- Rs. 102
First we have to convert Rupees to paise and then subtract money.
1 Rupee = 100 paise
132 Rupees = 132 × 100 = 13200
102 Rupees = 102 × 100 = 10200
13200 – 10200 = 3000 paise
ii. Rs. 10.5 – Rs. 7.25
First we have to convert Rupees to paise and then subtract money.
1 Rupee = 100 paise
10.5 Rupees = 10 × 100 = 1000 + 50 = 1050
7.25 Rupees = 7 × 100 = 700 + 25 = 725 paise
1050 – 725 = 275 paise
iii. Rs. 20.45 – Rs. 12.25
First we have to convert Rupees to paise and then subtract money.
1 Rupee = 100 paise
20.45 Rupees = 20 × 100 = 2000 + 45 = 2045
12.25 Rupees = 12 × 100 = 1200 + 25 = 1225
2045 – 1225 = 820 paise
Question 7.
Roshan has $3080. He spent$1200 on his friend. How much money is left with Roshan?
Solution:
Given,
Roshan has $3080. He spent$1200 on his friend.
$3080 –$1200 = $1880 Thus$1880 is left with Roshan.
Question 8.
Swetha buys a dress worth Rs. 1499. She gave Rs. 2000 to the cashier. How much she will get from the cashier?
Solution:
Given,
Swetha buys a dress worth Rs. 1499. She gave Rs. 2000 to the cashier.
Rs. 2000 – Rs. 1499 = Rs. 501
Therefore Swetha will get Rs. 501 from the cashier.
Question 9.
You have Rs. 20 with you and you gave Rs. 5 to your brother and Rs. 2 to your sister. How much money is left with you?
Solution:
Given that,
You have Rs. 20 with you and you gave Rs. 5 to your brother and Rs. 2 to your sister.
Rs 20 – Rs. 5 – Rs. 2 = Rs. 13
Thus Rs. 13 is left with you.
Question 10.
What is Rs. 25.50 paise minus Rs. 15.25 paise?
Solution:
Subtract Rs. 25.50 and Rs. 15.25
25.50 – 15.25 = 10.25
Therefore Rs. 25.50 paise minus Rs. 15.25 paise is Rs. 10.25
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# In the equation 3*(5*t + 2) = 36, what does t equal?
Oct 28, 2015
$t = 2$
#### Explanation:
To solve for $t$, we need to isolate it within this equation. So, start by dividing both sides by $3$:
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \left(5 \cdot t + 2\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} = \frac{36}{3}$
The $3$'s on the left side cancel, to leave:
$5 \cdot t + 2 = 12$
$5 t + 2 = 12$
Then we subtract 2 from each side.
$5 t + \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} = 12 - 2$
$5 t = 10$
Then, isolate $t$:
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} t}{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}} = \frac{10}{5}$
$t = 2$ |
Stitz-Zeager_College_Algebra_e-book
# 116 hooked on conics again 833 we break this product
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Unformatted text preview: 7. The fires are about 17456 feet apart. (Try to avoid rounding errors.) 782 11.4 Applications of Trigonometry Polar Coordinates In Section 1.1, we introduced the Cartesian coordinates of a point in the plane as a means of assigning ordered pairs of numbers to points in the plane. We defined the Cartesian coordinate plane using two number lines – one horizontal and one vertical – which intersect at right angles at a point we called the ‘origin’. To plot a point, say P (−4, 2), we start at the origin, travel horizontally to the left 4 units, then up 2 units. Alternatively, we could start at the origin, travel up 2 units, then to the left 4 units and arrive at the same location. For the most part, the ‘motions’ of the Cartesian system (over and up) describe a rectangle, and most points be thought of as the corner diagonally across the rectangle from the origin.1 For this reason, the Cartesian coordinates of a point are often called ‘rectangular’ coordinates. y 4 3 P (−4, 2) 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 In this section, we introduce a new system for assigning coordinates to points in the plane – polar coordinates. We start with a point, called the pole, and a ray called the polar axis. Pole Polar Axis Polar coordinates consist of a pair of numbers, (r, θ), where r represents a directed distance from the pole2 and θ is a measure of rotation from the polar axis. If we wished to plot the point P with π polar coordi...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.
Ask a homework question - tutors are online |
HANDS-ON ACTIVITY 10.1: SEPARATION OF VARIABLES - Differential Equations - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC
## Master AP Calculus AB & BC
Part II. AP CALCULUS AB & BC REVIEW
CHAPTER 10. Differential Equations
OVERVIEW
• Hands-On Activity 10.1: Separation of variables
• Hands-On Activity 10.2: Slope fields
• Euler’s method (BC topic only)
• Exponential growth and decay
• Logistic growth (BC topic only)
• Technology: A differential equations calculator program
• Summing it up
A differential equation is simply an equation that contains a derivative. Your typical goal in a differential equation problem is to find the equation that has the given derivative; in other words, you are trying to find an antiderivative. How is this different from the integrals you have been finding until now? Well, it’s not very different at all. You will be using all of your integration techniques to find particular solutions (no + C) using the method of separation of variables. However, the vast majority of differential equations in the real world cannot be solved using this method. We will then further examine those solutions using slope fields and Euler’s Method. Finally, we will look at some applications of differential equations in exponential and logarithm ic growth. After we’re done, we’ll go get an ice cream cone, and I’ll buy you that pony you’ve always wanted.
HANDS-ON ACTIVITY 10.1: SEPARATION OF VARIABLES
Even though the study of differential equations is complex, you are only required to know and understand the easiest of all methods for solving them— separation of variables. The name says it all, and you’ve already performed all the tasks that are involved in the process. Therefore, this section represents a new topic with nothing substantially new to learn (I love those kinds of topics).
1. What about the equation makes it a differential equation?
________________________________________________________
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2. Your goal will be to find an equation (in the form y = f(x)) whose derivative is —x/y. Why can’t you simply integrate right away? What makes this equation different from the equation which you can integrate right away?
________________________________________________________
________________________________________________________
3. This topic is very similar to a differentiation topic for the reason you cited in number 2. What is the name of that topic, and why is it similar?
________________________________________________________
4. Before you can integrate you must separate the variables (put all the y’s on one side and the x’s on the other). How can you accomplish this in our equation?
________________________________________________________
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5. Go ahead and separate the variables using the method you named in number 4. Now, you should be able to integrate both sides of the equation separately. Integrate, remembering to include a C for any constant. What geometric shape is the solution to the differential equation?
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6. The answer we have is very general (because of the C). What if you knew that the solution curve passed through the point (0,3)? Given this information, what would your solution be?
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7. Let’s try a new differential equation Separate the variables, and integrate both sides separately.
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8. Your y expression ends up contained in a natural log. In such cases, it is preferred to solve the equation for y, not ln y. Solve for y.
________________________________________________________
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9. If you knew that the particular solution you were looking for satisfied the condition y(3) = 5, what is C, and what is the solution to the differential equation?
________________________________________________________
________________________________________________________
________________________________________________________
SOLUTIONS TO HANDS-ON ACTIVITY 10.1
TIP. Even though you could have grouped the - with the y, it’s better to leave it (and any constants a problem might have) on the side with the x’s; this makes solving for y easier, and most differential equation solutions are in that form.
1. It contains a derivative.
2. contains both x’s and v’s; until now, all of our integrals have contained just x’s.
3. This is similar to implicit differentiation; those expressions contained both x’s and v’s and thus required a different method.
4. Separation can be accomplished in this problem by cross-multiplying. That may not work for all problems, but separation is usually achieved through very simple methods (see number 7).
5. Cross-multiplying gives you
Multiply everything by 2 and move the x term to get
x2 + y2 = C
Therefore, the solution to the differential equation is a circle. (Often, solutions are written solved for y, but in this case, the answer is more clearly a circle when you leave it in standard form for a circle. Either way, however, the answer is right.)
TIP. This is not the same C as in the previous step—it just indicates an arbitrary number. Some textbooks use different constants each time because of this, but that’s silly—just remember that C might never be the same number.
6. Plug in the x = 0 and y = 3, since these values (if on the graph) must make the equation true.
02 + 32 = C
C = 9
Now that we know the specific value for C, we can plug it into the solution:
x2 + y2 = 9
So, this specific solution is a circle of radius 3.
7. Begin by factoring the x2 out of the numerator.
Divide both sides by (1 + y), and multiply both sides by dx to separate the variables.
Notice how the constant stays with the x terms so that you can solve for y more easily. Now, integrate both sides.
NOTE. Rewriting as is the same thing as rewriting
8. To solve for y, raise e to the power of both sides to get
You can rewrite the right side of the equation as (using properties of exponents), and eC is just another constant, which you can then write as C. This gives you
NOTE. When you are given a point value and can find C in a differential equation, the resulting solution is called a particular solution, since it is only one of many possible solutions if no point values were indicated.
9. y(3) = 5 means that plugging a value of 3 into the equation (for x) gives an output of 5; it’s similar to saying f(3) = 5. If that confuses you, remember that y(3) = 5 means the point (3,5) is on the graph. Either way, plug in x = 3 and y = 5 in order to find C.
Therefore, the solution to this differential equation is
EXERCISE 1
Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.
DO NOT USE A CALCULATOR TO SOLVE THE FOLLOWING.
1. Find the particular solution to that satisfies the condition that y(0) = 1.
2. What function has derivative f'(x) = ex esc y and passes through the origin?
3. In some cases, the rate of change of a quantity is proportional to the quantity itself. This is written as
where y is the quantity and k is the proportionality constant. What is the general solution to this very important differential equation?
4. A particle moves along the x-axis with acceleration at time t given by a(t) = 3t. Find the function describing the particle’s position if it travels at a rate of 2 ft/sec when t = 0 and is 3 feet to the right of the origin when t = 1.
5. If a ball is thrown upward and reaches its maximum height when t = 2 seconds, answer the following questions:
(a) Give the function representing the ball’s velocity.
(b) If the ball is 85 feet off the ground when t = 1.5 seconds, what is the position function for the ball?
1. Divide both sides by √y, and multiply by dx to separate the variables.
Now, plug in the given values for x and y, and you find (easily) that C = 1. Therefore, the particular solution is
NOTE. Your solution to problem 3 is the foundation for the section on exponential growth and decay, where all the quantities in question have the property problem 3 describes.
2. Begin by rewriting the equation like this:
Now, cross-multiplying will separate the variables.
Now, plug in the point (0,0).
cos (0) = cos (arccos (—1 + C))
—1 + C = 1
C = 2
The final answer is y = arccos (—ex + 2).
3. Divide both sides by y, and multiply them by dt to get the necessary separation. Remember that k is a constant.
NOTE. When you multiply by a negative, you don’t have to write “-C”, since C could already be negative! It’s a mystery number for now.
4. (a) The integral of acceleration is velocity, and we know that v(0) = 2. So, find the antiderivative of a(t), and plug in the given information.
Therefore, the equation for velocity is In order to find position, integrate again, and use the fact that s(1) = 3.
Therefore, the position equation is
5. (a) Remember that the equation for projectile position is s(t) = —16t2 + v0t + h0, where v0 is the initial velocity and h0 is the initial height. You are looking for the velocity equation, so take the derivative:
v(t) = —32t + v0
The ball reaches its maximum height when t = 2. Therefore, the derivative must equal 0 when t = 2. This allows you to find v0.
v(2) = —32(2) + C = 0
C = 64
Therefore, the velocity equation is v(t) = —32t + 64.
(b) To find position, integrate velocity, and use the fact that s(1.5) = 85.
Therefore, the position equation for the ball is
s(t) = —16t2 + 64t + 25
|
#### Contrapositive
We just studied conditionals, sentences that either have the form “If A, then B”, or could be rephrased in terms of such a form. Now we introduce the notion of a contrapositive of a conditional. The contrapositive of “If A, then B” is “If not-B, then not-A”. Or, in symbols:
The contrapositive of
A → B
is:
~B → ~A
The tilde sign “~” stands for negation. “~P” is read “It is not the case that P”, or shortly “Not-P”. Here is an example of a conditional and its contrapositive, first in ordinary language, then in symbols.
If mamma bear crossed the river, so did baby bear. Contrapositive: If the baby didn’t cross, then mamma didn’t, either.
M → B
~B → ~M
This lesson focuses on the first one, the contrapositive. Those who contrapose will do well on the LSAT (contrapose: the act of making a contrapositive). Why? You can take conditional rules and double them by contraposing the original conditional into a new inferences. So, you can turn one conditional statement into a second one.
#### Two Conditionals
And, poof! It becomes two conditionals after you contrapose the original (reverse the statements and negate them).
Contraposing twice brings us back where we started:
The contrapositive of
~B → ~A
is:
A → B
Why? Let us apply the steps suggested above to ~B → ~A. The result is:
~~A → ~~B
According to the rule of double negation, ~~P is equivalent to P, so ~~P and P can be replaced by each other. In the case above ~~B and ~~A are replaced by B and A respectively, and the result is:
A → B
With the aid of Steps 1 and 2 and the rule of double negation, we can see also that:
The contrapositive of
A → ~B
is:
B → ~A
The contrapositive of
~A → B
is:
~B → A
#### The rule of contraposition:
The rule of contraposition is a rule of inference. It tells us that from a conditional we can validly infer its contrapositive. In other words, if we have A → B as a premise, we can infer ~B → ~A as a conclusion. Remember that the contrapositive of ~B → ~A is A → B. So the rule tells us that the two conditionals are equivalent. This means that they can be inferred one from another, and also that they can replace each other. So the rule tells us:
The rule of contraposition:
A → B is equivalent to ~B → ~A
A conditional and its contrapositive are equivalent
### Contrapositive Video Summary
• 0:19 – Consider the statement: If Hilda is in law school, then Hilda took the LSAT. We can diagram this as: law school → LSAT.
• 0:49 – Infer the following ideas:
• Taking the LSAT is a requirement for going to law school. (necessary condition)
• If you don’t meet the requirement for something then you don’t get that thing.
• 1:17 – Since the LSAT is a requirement, this means that if Hilda didn’t take the LSAT, then she’s not in law school. This is also a conditional statement: If Hilda didn’t take the LSAT, then Hilda isn’t in law school.
• 2:02 – Whenever we’re given a conditional relationship, we can automatically infer another conditional from it.
• 2:10 – Diagram the second statement by taking the necessary condition of the first statement, making it sufficient then negating it; and taking the sufficient condition of the first statement, making it necessary then negating it. Diagram the second statement as: ~LSAT → ~law school.
• 2:40 – The second statement is the exact reverse of the first one. This second statement that we inferred from the first one is called the contrapositive.
• 2:56 – Note: Given a conditional statement, you can always infer its contrapositive.
• 3:09 – Another example: If the switch is flipped, then the light turns on. Its diagram is: light switch → light.
• 3:21 – Note: The contrapositive does not depend on the content but only the form of the statement.
• 3:29 – Infer from this statement its contrapositive using the procedure before. It becomes: If the lights don’t turn on, then the switch isn’t flipped. Its diagram is: ~light → ~light switch.
• 3:58 – Another example: If you love me, then you’ll take a shower. Its diagram is: love me → shower.
Its contrapositive is: If you don’t take a shower, then you don’t love me. Its diagram is: ~shower → ~love me.
• 4:18 – Recap: The contrapositive is a valid inference from the conditional statement. To diagram it, simply switch and negate the conditions.
Next LSAT: Jun 10/Jun 11
### Video Summary
“It don’t mean a thing if it ain’t got that swing.”
Diagram this as:
~has swing → ~mean anything.
Its contrapositive is:
mean anything → has swing.
“You must take the A Train, if you want to go to Harlem.”
Diagram this as:
go to Harlem → take the A train.
Infer its contrapositive:
~take the A train → ~go to Harlem
#### Contrapositive Drills
There are conditional claims within the following quantified sentences. Find them, put them into symbols, contrapose them, translate back into English and quantify them. The result should be a sentence that has the same meaning as the original sentence.
If it rains, then the event will be canceled. It rained.
If the above statements are true, which one of the following must also be true?
(A) If the event was canceled, then it rained.
(B) The event would have to be canceled.
(B) The event would have to be canceled.
Statement Symbols Valid/Invalid Description 1. If it rains, then the event will be canceled. rain → canceled Given Given 2. If the event is not canceled, then it did not rain. ~ canceled → ~rain Valid Contrapositive 3. If the event is canceled, then it rained. canceled → rain Invalid Converse 4. If it does not rain, then the event will not be canceled. ~rain → ~ canceled Invalid Inverse
“Every problem is absurdly simple when it is explained to you.”
-Sherlock Holmes
### Contrapositive
1. Locate the keyword indicator: when (sufficient indicator)
2. Locate the conditional phrase right after the keyword indicator: “. . . when it is explained to you.” (This is the sufficient condition.)
3. Abbreviate the sufficient condition: ETY
4. Locate the phrase other than the sufficient condition: “Every problem is absurdly simple . . .” (This phrase is the necessary condition.)
5. Abbreviate the necessary condition: PAS
6. Combine the two phrases in the form of: sufficient → condition
Diagram: ETY → PAS
7. Reverse the positions of the conditional phrases: PAS → ETY
8. Finally, negate them both to form the contrapositive:
Contrapositive: ~PAS → ~ETY
“Those who don’t believe in magic will never find it.”
-Roald Dahl
### Contrapositive
1. Locate the keyword indicator: those who (sufficient indicator; means the same as “people who”)
2. Locate the conditional phrase right after the keyword indicator: “. . . don’t believe in magic . . .” (This is the sufficient condition.)
3. Abbreviate the sufficient condition: ~BM (We use a negation because of the word ‘don’t’.)
4. Locate the phrase other than the sufficient condition: “. . . will never find it.” (This phrase is the necessary condition.)
5. Abbreviate the necessary condition: ~F (We use a negation because of the word ‘never’.)
6. Combine the two phrases in the form of: sufficient → condition
Diagram: ~BM → ~F
7. Reverse the positions of the conditional phrases: ~F → ~BM
8. Finally, negate them both to form the contrapositive:
Contrapositive: ~(~F) → ~(~BM) = F → BM
“No man’s knowledge here can go beyond his experience.”
-John Locke
### Contrapositive
1. Locate the keyword indicator: No (sufficient indicator)
2. Locate the conditional phrase right after the keyword indicator: “. . . man’s knowledge here . . .” (This is the sufficient condition.)
3. Abbreviate the sufficient condition: K
4. Locate the phrase other than the sufficient condition: “. . . can go beyond his experience.” (This phrase is the necessary condition.)
5. Abbreviate the necessary condition: BE
6. Using the sufficient indicator “no” means we have to negate the necessary condition using (~): ~(BE) = ~BE
7. Combine the two phrases in the form: sufficient → necessary
Diagram: K → ~BE
8. Reverse the positions of the conditional phrases: ~BE → K
9. Finally, negate them both to form the contrapositive:
Contrapositive: ~(~BE) → ~K = BE → ~K
“It always seems impossible until it’s done.”
-Nelson Mandela
### Contrapositive
1. Locate the keyword indicator: until (necessary indicator)
2. Locate the conditional phrase right after the necessary indicator: “. . . it’s done.” (This is the necessary condition.)
3. Abbreviate the necessary condition: D
4. Locate the phrase other than the necessary condition: “It always seems impossible . . .” (This phrase is the sufficient condition.)
5. Abbreviate the sufficient condition: SI
6. Using the necessary indicator “until” means we have to negate the sufficient condition (~): ~SI
7. Combine the two phrases in the form: sufficient → necessary
Diagram: ~SI → D
8. Reverse the positions of the conditional phrases: D → ~SI
9. Finally, negate them both to form the contrapositive:
Contrapositive: ~D → ~(~SI) = ~D → SI
“If you expect nothing from somebody you are never disappointed.”
-Sylvia Plath
### Contrapositive
1. Locate the keyword indicator: If (sufficient indicator)
2. Locate the conditional phrase right after the keyword indicator: “. . . you expect nothing from somebody . . .” (This is the sufficient condition.)
3. Abbreviate the sufficient condition: ENS
4. Locate the phrase other than the sufficient condition: “. . . you are never disappointed.” (This phrase is the necessary condition.)
5. Abbreviate the necessary condition: ~D (We use a negation because of the word “never”.)
6. Combine the two phrases in the form: sufficient → necessary
Diagram: ENS → ~D
7. Reverse the positions of the conditional phrases: ~D → ENS
8. Finally, negate them both to form the contrapositive:
Contrapositive: ~(~D) → ~ENS = D → ~ENS
“If you judge people, you have no time to love them.”
-Mother Theresa
### Contrapositive
1. Locate the keyword indicator: If (sufficient indicator)
2. Locate the conditional phrase right after the sufficient indicator: “. . . you judge people . . .” (This is the sufficient condition.)
3. Abbreviate the sufficient condition: JP
4. Locate the phrase other than the sufficient condition: “. . . you have no time to love them.” (This phrase is the necessary condition.)
5. Abbreviate the necessary condition: ~LT (We use negation because of the word “no”.)
6. Combine the two phrases in the form: sufficient → necessary
Diagram: JP → ~LT
7. Reverse the positions of the conditional phrases: ~LT → JP
8. Finally, negate them both to form the contrapositive:
Contrapositive: ~(~LR) → ~JP = LT → ~JP
“No one can make you feel inferior without your consent.”
-Eleanor Roosevelt
### Contrapositive of Quote
1. Locate the sufficient indicator: No
2. Locate the conditional phrase right after the sufficient indicator: “. . . one can make you feel inferior . . .” (This phrase is the sufficient condition.)
3. Abbreviate the sufficient condition: MFI
4. Locate the necessary condition: “. . . without your consent.” (The phrase other than the sufficient condition is the necessary condition.)
5. Abbreviate the necessary condition: ~C
6. Using the sufficient indicator “no” means putting a tilde (~) on the necessary condition for negation: ~(~C) = C
7. Combine the two conditions in the form of: sufficient → condition
Diagram: MFI → C (make you feel inferior → with consent)
8. Then, reverse the positions of the conditional phrases: C → MFI
9. Finally, negate their values to form the contrapositive:
~C → ~MFI (without consent → cannot make you feel inferior)
“No one can be at peace unless he has his freedom.”
-Muhammed Ali
### Contrapositive of Quote
1. Locate the necessary indicator: Unless
2. Locate the conditional phrase right after the necessary indicator: “. . . he has his freedom.” (This phrase is the necessary condition.)
3. Abbreviate the necessary condition: F
4. Locate the sufficient condition: “No one can be at peace . . .” (The phrase other than the necessary condition is the sufficient condition.) *Although this phrase uses the indicator “no”, the idea of the phrase also means “one cannot be at peace”.
5. Abbreviate the sufficient condition: ~P (“. . . one cannot be at peace . . .“)
6. Using the necessary indicator “unless” means putting a tilde (~) in the sufficient condition for negation: ~(~P) = P
7. Combine the two conditions in the form of: sufficient → condition
Diagram: P → F (be at peace → has freedom)
8. Then, reverse the positions of the conditional phrases: F → P
9. Finally, negate their values to form the contrapositive:
~F → ~P (has no freedom → not at peace)
If I don’t get into law school, then I plan on going to business school.
### Contrapositive
1. Locate the sufficient indicator: If
2. Locate the conditional phrase right after the sufficient indicator: “. . . I don’t get into law school . . .” (This phrase is the sufficient condition.)
3. Abbreviate the sufficient condition: ~LS (“. . . not get into law school . . .“)
4. Locate the necessary condition: “. . . I plan on going to business school.” (The phrase other than the sufficient condition is the necessary condition.)
5. Abbreviate the necessary condition: GBS
6. Combine the two conditions in the form of: sufficient → condition
Diagram: ~LS → GBS (don’t get into law school → plan on going to business school)
7. Then, reverse the positions of the conditional phrases: GBS → ~LS
8. Finally, negate their values to form the contrapositive:
~GBS → LS (did not go to business school → got into law school)
If you want to become a doctor then you shouldn’t smoke.
### Contrapositive
1. Locate the sufficient indicator: If
2. Locate the conditional phrase right after the sufficient indicator: “. . . you want to become a doctor . . .” (This phrase is the sufficient condition.)
3. Abbreviate the sufficient condition: BD
4. Locate the necessary condition: “. . . you shouldn’t smoke.” (The phrase other than the sufficient condition is the necessary condition.)
5. Abbreviate the necessary condition: ~S (“. . . should not smoke . . .“)
6. Combine the two conditions in the form of: sufficient → condition
Diagram: BD → ~S (want to become a doctor → not smoke)
7. Then, reverse the positions of the conditional phrases: ~S → BD
8. Finally, negate their values to form the contrapositive:
S → ~BD (smoke → do not want to become a doctor)
All non-professionals in the program use software to improve their driving.
### Contrapositive
1. Locate the sufficient indicator: All
2. Locate the conditional phrase right after the sufficient indicator: “. . . non-professionals in the program . . .” (This phrase is the sufficient condition.)
3. Abbreviate the sufficient condition: ~P (“. . . not professionals . . .“)
4. Locate the necessary condition: “. . . use software to improve their driving.” (The phrase other than the sufficient condition is the necessary condition.)
5. Abbreviate the necessary condition: USD
6. Combine the two conditions in the form of: sufficient → condition
Diagram: ~P → USD (non-professionals → use software to improve their driving)
7. Then, reverse the positions of the conditional phrases: USD → ~P
8. Finally, negate their values to form the contrapositive:
~USD → P (not use software to improve driving → professional) |
Suppose a block is pulled 16m. What happens next?
Sure, here's an introduction for your blog post:
Welcome to Warren Institute! Today, we'll delve into an intriguing problem in Mathematics education. Suppose that a block is pulled 16 meters. This scenario presents an excellent opportunity to explore concepts of distance, displacement, and work in a real-world context. By analyzing this situation, we aim to deepen our understanding of fundamental mathematical principles and their practical implications. Join us as we unravel the intricacies of this problem and gain valuable insights into the application of mathematical concepts in everyday scenarios.
Understanding the concept of work
Work in physics is defined as the product of the force applied to an object and the distance over which it is applied. In this context, understanding how work is calculated and its relation to force and displacement is essential. The given scenario of pulling a block 16 meters provides a practical example to delve into the concept of work.
Calculating work done on the block
To calculate the work done on the block when it is pulled 16 meters, the formula work = force × distance can be used. Understanding how to apply this formula in the given scenario and interpreting the units involved in the calculation is crucial for students to grasp the concept effectively.
Exploring the role of force in the work done
The amount of force applied when pulling the block is directly linked to the work done. Explaining the relationship between force and work, and illustrating how changes in force impact the amount of work done, can provide students with a deeper understanding of the concept.
Real-world applications of work calculations
Connecting the concept of work to real-world scenarios, such as pulling objects or lifting weights, can help students visualize and comprehend the practical significance of work calculations. Exploring various examples beyond the given scenario can enrich the learning experience and highlight the relevance of understanding work in everyday situations.
Suppose that a block is pulled 16 meters horizontally with a force of 30 N, what is the work done on the block?
The work done on the block is 480 J.
If a block is pulled 16 meters at an angle of 30 degrees to the horizontal, how does this affect the work done on the block?
The work done on the block is increased due to the component of force in the direction of motion.
How can we calculate the force required to pull a block 16 meters up an inclined plane with a frictional force of 20 N?
The force required can be calculated using the formula: Force = Weight * (sin(theta) + (frictional force / Weight) * cos(theta)) , where theta is the angle of the inclined plane.
How does the mass of the block affect the work done when it is pulled 16 meters vertically against gravity?
The mass of the block affects the work done when it is pulled 16 meters vertically against gravity because work is equal to force multiplied by distance, and the force required to lift the block increases with its mass.
What is the relationship between the distance a block is pulled and the work done on the block?
The relationship between the distance a block is pulled and the work done on the block is directly proportional.
In conclusion, the scenario of a block being pulled 16 meters serves as an excellent example for illustrating concepts in Mathematics education. The application of variables and mathematical principles to solve problems in this scenario provides a practical and engaging way for students to grasp key mathematical concepts. Additionally, it highlights the importance of understanding and applying mathematical concepts to real-world situations, fostering a deeper appreciation for the subject. This type of hands-on learning experience can help students develop critical thinking skills and strengthen their mathematical proficiency. |
# Math Snap
## a) Express the function in the form $f(x)=a(x-h)^{2}+k$ and b) Draw the graph of the function $f(x)=-x^{2}-4 x-3$ a) $f(x)=$ $\square$
#### STEP 1
Assumptions 1. The given function is $f(x) = -x^2 - 4x - 3$. 2. We need to express this function in the form $f(x) = a(x-h)^2 + k$. 3. The method to achieve this is by completing the square.
#### STEP 2
First, we rewrite the function $f(x) = -x^2 - 4x - 3$ by factoring out the coefficient of $x^2$ from the first two terms. $f(x) = - (x^2 + 4x) - 3$
#### STEP 3
Next, we complete the square inside the parentheses. To do this, we take the coefficient of $x$, which is 4, divide it by 2, and square it. $\left(\frac{4}{2}\right)^2 = 4$
#### STEP 4
Add and subtract this square inside the parentheses. This does not change the value of the function but allows us to complete the square. $f(x) = - (x^2 + 4x + 4 - 4) - 3$
#### STEP 5
Rewrite the expression inside the parentheses as a perfect square trinomial and simplify. $f(x) = - [(x + 2)^2 - 4] - 3$
#### STEP 6
Distribute the negative sign through the parentheses. $f(x) = - (x + 2)^2 + 4 - 3$
#### STEP 7
Combine the constants. $f(x) = - (x + 2)^2 + 1$
#### STEP 8
Now, we have expressed the function in the desired form $f(x) = a(x-h)^2 + k$. $f(x) = - (x + 2)^2 + 1$ So, $a = -1$, $h = -2$, and $k = 1$.
#### STEP 9
To draw the graph of the function, we note the vertex form $f(x) = a(x-h)^2 + k$. The vertex of the parabola is at $(h, k) = (-2, 1)$.
#### STEP 10
Since $a = -1$, the parabola opens downwards. The vertex is the highest point on the graph.
#### STEP 11
Plot the vertex $(-2, 1)$ on the coordinate plane.
#### STEP 12
Determine additional points by choosing values of $x$ around the vertex and calculating the corresponding $y$-values. For example, if $x = -1$: $f(-1) = - (-1 + 2)^2 + 1 = - (1)^2 + 1 = -1 + 1 = 0$ So, the point $(-1, 0)$ is on the graph.
#### STEP 13
If $x = -3$: $f(-3) = - (-3 + 2)^2 + 1 = - (-1)^2 + 1 = -1 + 1 = 0$ So, the point $(-3, 0)$ is on the graph.
#### STEP 14
Plot these points $(-1, 0)$ and $(-3, 0)$ on the coordinate plane.
##### SOLUTION
Sketch the parabola opening downwards, passing through the vertex $(-2, 1)$ and the points $(-1, 0)$ and $(-3, 0)$. The function in vertex form is: $f(x) = - (x + 2)^2 + 1$ |
## Chapter 5 Understanding Elementary Shapes
Question 1.
Question 2.
How many line segments are used in making a triangle?
Solution:
Three line segments are used to make a triangle.
Question 3.
What is the measure of straight angle?
Solution:
The measure of straight angle is 180°.
Question 4.
What is complete angle?
Solution:
The angle for one revolution is called a complete angle.
Question 5.
Find the number of right angle turned through by the hour hand of a clock when it goes from 3 to 6.
Solution:
When the hour hand of a clock goes from 3 to 6, it turns through a right angle.
Question 6.
Draw the rough sketch of the following:
(a) Acute angle
(b) Obtuse angle
(c) Reflex angle
Solution:
(a) Acute angle
(b) Obtuse angle
(c) Reflex angle
Question 7.
Identify the types of angle from the given figures:
Solution:
(a) Obtuse angle
(b) Acute angle
(c) Right angle
(d) Straight angle
(e) Reflex angle
Question 8.
What are the degree measures of the following angles?
(a) Right angle
(b) A complete angle
(c) Straight angle
Solution:
(a) Degree measure of a right angle is 90°.
(b) Degree measure of a complete angle is 360°.
(c) Degree measure of a straight angle is 180°.
Question 9.
What are the types of the given triangles on the basis of angles?
Solution:
(a) Acute angled triangle.
(b) Obtuse angled triangle.
(c) Right angled triangle.
Question 10.
What are the types of the following triangles on the basis of sides?
Solution:
(a) Scalene triangle.
(b) Equilateral triangle.
(c) Isosceles triangle.
Question 11.
In the given figure, name the following angles as acute, obtuse, right, straight or reflex.
(a) ∠QOY
(b) ∠YOP
(c) ∠ROX
(d) ∠QOX
(e) ∠POQ
Solution:
(a) ∠QOY = acute angle.
(b) ∠YOP = obtuse angle.
(c) ∠ROX = right angle.
(d) ∠QOX = reflex angle.
(e) ∠POQ = straight angle.
Question 12.
In the given figure, find the measure of the angles marked with a, b, c, d, e and f.
Solution:
∠a = 180° -129° = 51°
∠b = 180° – (51° + 92°)
= 180° – 143° = 37°
∠c = 180° – 88° = 92°
∠d = 180° – 152° = 28°
∠e = 180° – 143° = 37°
∠f= 180° – (∠e + ∠d)
= 180° – (37° + 28°)
= 180°- 65°= 115°
∠g = 180° – ∠f = 180° – 115° = 65°
Question 13.
Classify the given triangles whose sides are indicated on them.
Solution:
(a) All sides are different. So, it is a scalene triangle.
(a) Lengths of two sides of the triangle are same. So, it is an isosceles triangle.
(b) All sides are unequal and one angle is right angle. So it is scalene right angled triangle.
(c) Two sides of this triangle are equal. So, it is an isosceles triangle.
Question 14.
Complete each of the following, so as to make a true statement:
(a) A ……….. is a rectangle with a pair of adjacent sides equal.
(b) A parallelogram with a pair of adjacent sides equal is called a ………. .
(c) A quadrilateral having exactly one pair of parallel sides is called a …….. .
(d) A quadrilateral having both pairs of opposite sides parallel, is called a ……… .
(e) A parallelogram whose each angle is a right angle is called a ………… .
Solution:
(a) Square
(b) Rhombus
(c) Trapezium
(d) Parallelogram
(e) Rectangle.
Question 15.
Verify the ‘Euler’s formula’ V + F = E + 2 for the given figures.
(a) A triangular prism having 5 faces, 9 edges and 6 vertices.
(b) A rectangular prism with 6 faces, 12 edges and 8 vertices.
(c) A pentagonal prism with 7 faces, 15 edges and 10 vertices.
(d) A tetrahedron -with 4 faces, 6 edges and 4 vertices.
Solution:
(a) Here, F = 5, E = 9 and V = 6
∴ V + F = E + 2
⇒ 6 + 5 = 9 + 2
⇒ 11 = 11
Hence, verified.
(b) Here, F = 6, E = 12 and V = 8
∴ V + F = E + 2
⇒ 8 + 6 = 12 + 2
⇒ 14 = 14
Hence, verified.
(c) Here, F = 7, E = 15 and V = 10
∴ V + F = E + 2
⇒ 10 + 7 = 15 + 2
⇒ 17 = 17
Hence, verified.
(d) Here, F = 4, E = 6 and V = 4
∴ V + F = E + 2
⇒ 4 + 4 = 6 + 2
⇒ 8 = 8
Hence, verified.
Question 16.
Complete the given table for prisms:
Solution:
Multiple Choice Type
Question 17.
Number of right angles turned by the hour hand of a clock when it goes from 3 to 6.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
Correct option is (a).
Question 18.
A quadrilateral having a pair of unequal opposite sides is called
(a) Parallelogram
(b) Square
(c) Rectangle
(d) Trapezium
Solution:
The correct option is (d).
Higher Order Thinking Skills (Hots)
Question 19.
In the given figure, find the values of x, y, z, s and m.
Solution:
Given that ∠A = 40°
(i) ∠DAB + ∠ABC = 180° (adjacent angles)
⇒ 40° + ∠ABC = 180°
⇒ ∠ABC = 180° – 40° = 140°
Hence, ∠x = 140°
(ii) ∠x + ∠y = 180° (adjacent angles)
⇒ 140° + ∠y = 180°
⇒ ∠y = 180° – 140° = 40°
Hence, ∠y = 40°
(iii) ∠y + ∠z – 180° (adjacent angles)
⇒ 40° + ∠z = 180°
⇒ ∠z = 180° – 40° = 140°
Hence, ∠z = 140°
(iv) ∠x + ∠s = 180° (straight angles)
⇒ 140° + ∠s = 180°
⇒ ∠s = 180° – 140° = 40°
Hence, ∠s = 40°
(v) ∠m + ∠z = 180° (straight angles)
⇒ ∠m + 140° = 180°
⇒ ∠m = 180° – 140° = 40°
Question 20.
Find the value of x from the given figure and hence find the measure of each angle of the triangle.
Solution:
(i) Sum of the three angles of a triangle = 180°
∴ 2x + 30° + 60° – x + 3x – 10° = 180°
⇒ (2x – x + 3x) + (30° + 60° – 10°) = 180°
⇒ 4x + 80° = 180°
⇒ 4x = 180° – 80°
⇒ 4x = 100°
∴ x = = 25°
∴ Measure of the angles are:
(i) (2x + 30)° – 2 x 25° + 30° = 80°
(ii) (60 – x)° = 60° – 25° = 35°
(iii) (3x – 10)° = 3 x 25° – 10° = 75° – 10° = 65°
Hence, x = 25° and the angles of the triangles are: 80°, 35° and 65°.
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Thought-Process to Discover Knowledge
Welcome to nubtrek.
Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.
In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.
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mathsFractionsNumerical Expressions with Fractions
### Fraction: Simplification of Expressions
In this page, handling of fractions, both positive and negative fractions, in numerical expressions is explained.
The precedence order PEMDAS / BODMAS is explained.
For operations of same precedence order, the sequence of operation "simplification from left to right" is explained.
click on the content to continue..
Let us quickly revise what is numerical expression, and precedence order, sequence in simplifying the numerical expressions. This was introduced in whole numbers and reviewed in integers too. It takes very little time to revise.
What is 2+4+3?
• 9
• 9
• 243
The answer is "9".
This is an example of a numerical expression.
Which of the following is a meaning for the word "expression"?
• something done very fast
• collection of numbers and arithmetic operations between them, which together represent a quantity
• collection of numbers and arithmetic operations between them, which together represent a quantity
The answer is "collection of numbers and arithmetic operations between them, which together represent a quantity".
What is 2xx4xx3?
• 24
• 24
• 243
The answer is "24".
This is an example of a numerical expression.
Is a+3x a numerical expression?
• Yes. It has the number 3.
• No. It is not entirely numbers and arithmetic operations
• No. It is not entirely numbers and arithmetic operations
The answer is "No. It is not entirely numbers and arithmetic operations".
Solved Exercise Problem:
Is 3+4-2+1 a numerical expression?
• No, it has both addition and subtraction
• Yes, addition and subtraction can be part of a numerical expression
• Yes, addition and subtraction can be part of a numerical expression
The answer is "Yes, addition and subtraction can be part of an expression".
Solved Exercise Problem:
Is 3+4xx2-6-:3 a numerical expression?
• No, as it has the addition, subtraction, multiplication, division
• Yes, all arithmetic operations can be part of a numerical expression
• Yes, all arithmetic operations can be part of a numerical expression
The answer is "Yes, all arithmetic operations can be part of a numerical expression"
Solved Exercise Problem:
is 3 a numerical expression?
• No. 3 is only a number. Not a numerical expression
• Yes. technically a number can also be considered a numerical expression
• Yes. technically a number can also be considered a numerical expression
The answer is "Yes. technically a number can also be considered a numerical expression".
Consider 1+2 and 3xx1. Note that when evaluated, both result in identical numerical value 1+2=3 and 3xx1 = 3.
Are these two different numerical expressions?
• They are two different expressions, evaluating to equal values
• They are two different expressions, evaluating to equal values
• They are identical expressions as they evaluate to equal values
The answer is "They are two different expressions, evaluating to equal values".
Simplify 9-6-:3
• 7
• 7
• 1
The answer is "7". Division has higher precedence over subtraction. So
9-6-:3
=9-2
=7
What is the rule of precedence in numerical expressions?
• multiplication and division have higher precedence over addition and subtraction
• parentheses or brackets have highest precedence
• both the above
• both the above
The answer is "both the above".
Simplify 20-4-3?
• 13
• 13
• 19
The answer is "13".
Two or more operations in the same precedence level are performed from left to right sequence.
20-4-3
= 16-3 =13
Simplify 36-:6-:3. Which one of the following is correct?
• 36-:6-:3 =6 -:3 = 2
• 36-:6-:3 =36-:2 =18
• 2
• 2
• 18
The answer is "2".
The two divisions are in same precedence level. This is to be handled from left to right sequence.
36-:6-:3
= 6-:3 =2
• It is NOT correct to do =36-:2 =18.
What is the rule of sequence in numerical expressions?
• there is no such thing as rule of sequence
• when multiple operation of same precedence is to be simplified, the operations are performed from left to right sequence
• when multiple operation of same precedence is to be simplified, the operations are performed from left to right sequence
The answer is "when multiple operation of same precedence is to be simplified, the operations are performed from left to right sequence".
Simplify 6-:3xx2
• 4
• 4
• 1
The answer is "4".
The division and multiplication are of same precedence, so it is simplified from left to right.
6-:3xx2
=2xx2
=4
Simplify 6-:(3xx2)
• 4
• 1
• 1
The answer is "1".
The bracket has higher precedence, and so the expression inside bracket is simplified first.
6-:(3xx2)
=6-:6
=1
What is the rule of brackets or parentheses in numerical expressions?
• there is no such thing as rule of brackets
• the subexpression within a bracket or parentheses has the highest precedence
• the subexpression within a bracket or parentheses has the highest precedence
The answer is "the subexpression within a bracket or parentheses has the highest precedence".
In a numerical expression, the precedence order is:
• division and multiplication in same level same level of precedence
• addition and subtraction in same level of precedence.
This is abbreviated as BODMAS (Division, Multiplication, Addition, Subtraction) or PEMDAS (Multiplication, Division, Addition, Subtraction).
When multiple operation of same precedence is to be simplified, the operations are performed from left to right sequence.
All these were studied as part of whole numbers and integers. The same applies for fractions.
• Precedence order BODMAS / PEMDAS
• Left to Right sequence for same precedence
Let us see some more expressions with fractions.
Solved Exercise Problem:
Simplify 1- 1/6 -:3
• 17/18
• 17/18
• 1/2
The answer is "17/18". Division has higher precedence over subtraction. So
1-1/6-:3
=1-1/18
=17/18
Solved Exercise Problem:
Simplify 6-:3xx(1/2)
• 4
• 1
• 1
The answer is "1".
The division and multiplication are of same precedence, so it is simplified from left to right.
6-:3xx(1/2)
=2xx(1/2)
=1
Solved Exercise Problem:
Simplify 6-:(3xx1/2)
• 4
• 4
• 1
The answer is "4".
The bracket has higher precedence, and so the expression inside bracket is simplified first.
6-:(3xx1/2)
=6-:(3/2)
=4
Solved Exercise Problem:
Simplify 4+(-2/3)-:(1/3)xx(-2)
• 8
• 8
• 5
The answer is "8".
The division and multiplication are of higher precedence over addition. so (-2/3)-:(1/3)xx(-2) is to be simplified first.
In that, the division and multiplication are of same precedence, so it is simplified from left to right.
4+(-2/3)-:(1/3)xx(-2)
=4+(-2)xx(-2)
=4+4
=8.
Solved Exercise Problem:
Simplify 10/7-(-1/7)-2xx(-3/14).
• 4/7
• 2
• 2
The answer is "2". The multiplication is of higher precedence over subtraction and so 2xx(-3/14) is simplified first. Then the two subtraction are in the same precedence level and so they are simplified in the left to right sequence.
10/7-(-1/7)-2xx(-3/14)
=10/7-(-1/7)-(-3/7)
=11/7-(-3/7)
=14/7
=2
Solved Exercise Problem:
Simplify (-1+3/2-1)xx2+(-3)/2-2/4
• -3
• -3
• -1
The answer is "-3"
(-1+3/2-1)xx2+(-3)/2-2/4
=(-2/2 + 3/2 -2/2)xx2+(-3)/2-1/2
=-1/2 xx 2 +(-3)/2-1/2
=(-2)/2 + (-3)/2-1/2
=(-5)/2 -1/2
=(-6)/2
=-3
Solved Exercise Problem:
What is the value of -3-((-1/3)-2xx(1/2))?
• 13/5
• -5/3
• -5/3
The answer is "-5/3".
-3-((-1/3)-2xx(1/2))
brackets take precedence and within the brackets, the multiplication is higher precedence.
=-3-((-1/3)-1)
brackets have higher precedence
=-3-(-4/3)
=-9/3-(-4/3)
simplifying
=-5/3
Solved Exercise Problem:
What is the value of -3/2-(-1/3)-1/4?
• 19/12
• -17/12
• -17/12
The answer is "-17/12".
-3/2-(-1/3)-1/4
LCM of 2, 3, 4 is 12
=-18/12 - (-4/12) - 3/12
left to right sequence
=-14/12-3/12
=-17/12
Solved Exercise Problem:
Simplify 4/5+3/4-:(-3/2)
• -3/10
• -3/10
• 3/10
The answer is "3/10".
4/5+3/4-:(-3/2)
The division is of higher precedence over addition.
=4/5+(-1/2)
=8/10 + (-5)/10
=3/10
Solved Exercise Problem:
Simplify (10-3-2)xx(-1/5).
• -1
• -1
• -15
The answer is "-1".
(10-3-2)xx(-1/5)
The expression inside bracket is simplified first. The two subtraction are in the same precedence level and so they are simplified in the left to right sequence.
=(7-2)xx(-1/5)
=5xx(-1/5)
=-1`
Simplification of Expressions : BODMAS
• B - Brackets
• O - Order (exponents, roots, logarithm)
• D - Division
• M - Multiplication
• S - Subtraction
• And Left to Right sequence for multiple operations of same precedence.
PEMDAS
• P - Parentheses
• E - Exponents (roots and logarithm)
• M - Multiplication
• D - Division |
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