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Whole Numbers Definition
If zero is also included in the natural number( 1, 2, 3, 4, 5, 6, 7 …. тИЮ ), then those numbers are Whole numbers.ex – 0, 1, 2, 3, 4, 5, 6, 7, 8 ……. тИЮ
If W is removed from the word whole, the word becomes a hole word. And we know. That the hole size looks like zero. So from the hole, we will remember zero. Numbers from 0 to infinity are called hole numbers.
Whole Numbers Examples
555, 687, 999, 0, 1800, 1520, 888 etc.
The predecessor of the whole number
The predecessor of the whole number is that number. Which comes 1 digit before the given number. That is, 1 is subtracted from that number to find the predecessor of a whole number.
Ex – What will be the predecessor of 555?
We know To find the predecessor of any number, one is subtracted from that number. Hence the predecessor of 555 is (555-1) = 554.
What will be the predecessor of 2?
Ans – (2-1) = 1
Subtract one of them to find the predecessor of two.
The successor of the whole number
The successor of a whole number is that number. Which comes after 1 digit of the given number. Or to find the successor of a hole number, one is added to that number.
Ex – What will be the successor of 15?
Ans – To find the successor of 15 can be obtained by adding a digit to it. 15 + 1 = 16
What will be the successor of 120?
120+1 = 121
Property of Whole Numbers
Closure Property
If two whole numbers are multiplied and added. So the number received is the whole number. On the other hand, if two whole numbers are divided and subtracted, the number obtained may or may not be a whole number.
The sum of two whole numbers is the whole number.
Whole Number + Whole Number = Whole Number
Ex – 2+2 = 4, 0+1 = 1, 5+8 = 13
Multiplication
Multiplication of two whole numbers is obtained as a whole number.
Whole Number ├Ч Whole Number = Whole Number
Ex – 2├Ч2 = 4 , 5├Ч8 = 40, 12├Ч2 = 24
Subtraction
If the smaller whole number is subtracted from the larger whole number then the number obtained will be the whole number. Conversely, if the whole number is subtracted from the smaller whole number, the number received will not be the whole number.
Whole Number “x” – Whole Number “y” = Whole Number ( x>y)
Ex: 4 – 2 = 2, 10 – 5 = 5, 110 – 20 = 90, 999 – 111 = 888
Whole Number “x” – Whole Number ‘y” = Whole Number (x=y)
Ex: 4 – 4 = 0, 15 – 15 = 0, 50 – 50 = 0, 80 – 80 = 0
Whole Number “x” – Whole Number “y” = integer ( x<y)
Ex: 15 – 20 = -5, 40 – 50 = -10, 130 – 140 = -10
Division
The division of two whole numbers will be the whole number only if the dividend is completely divisible by the divisor.
In the above situation, only the whole number can be obtained in case of division. If the quotient is the point or negative. So the whole number will not be received.
Commutative Property
The whole number follows the Commutative Property in addition and multiplication. On the contrary, it does or may not follow.
Two whole numbers can be added in any case, the whole number is obtained. According to the Whole Numbers Definition, these numbers range from zero to infinity
Whole Number “a” + whole Number “b” = Whole Number “b” + Whole Number “a”
a + b = b + a
Ex: 5 + 4 = 4 + 5 In this case, if 5 and 4 are added or 4 and 5 are added, the same answer will come in the two cases.
Multiplication
According to the Whole Numbers Definition, these numbers range from zero to infinity
Whole Number “a” x whole Number “b” = Whole Number “b” x Whole Number “a”
a x b = b x a
Ex: 5 x 4 = 4 x 5| In this case, if 5 and 4 or 4 and 5 are multiplied, the same answer will be given in two cases.
Subtraction
The whole number does not follow the Commutative Property in case of subtraction.
Ex: a – b тЙа b -a
Division
In the case of division, the whole number does not follow the Commutative Property.
Associative Property
If three whole numbers are added or reduced in any case, then, in that case, the whole number is obtained. The whole number in addition and subtraction follows the Associative Property.
In the case of Yoga, the Whole Number follows the Associative Property.
Ex: (a + b) + c = a + (b + c)
Multiplication
In the case of multiplication, the Whole Number follows the Associative Property.
Ex: (a x b) x c = a x (b x c)
Subtraction
In the event of subtraction, the Whole number does not follow the Associative Property.
( a – b ) – c тЙа a – ( b – c )
Division
The whole number does not follow the Associative Property in the position of the part.
Distributive Property
In the case of addition and multiplication, the Whole number follows the Distributive Property.
FAQ
Is 10 a whole number?
True
Is 9 a whole number?
True
What are the first 5 whole numbers?
0, 1, 2, 3, 4
Is seven a whole number?
True
Is 18 a whole number?
True
Is 100 a whole number?
True
Is 13 a whole number?
True
What is a whole number between 1 and 20?
2, 3, 4,5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19
Is 0 a whole number?
Zero is a whole number. It is neither positive nor negative integer. This is the link between Positive and Negative.
Can whole numbers be negative?
No
What are the properties of whole numbers?
The whole number has four properties. 1 – Closure Property, 2 – Commutative Property, 3 – Associative Property, 4 – Distributive Property
Are whole numbers closed under subtraction?
Ans – The whole number is closed under the Subtraction.
Are whole numbers closed under addition?
Ans – The whole number is closed under the Subtraction.
Are whole numbers also natural numbers?
All whole numbers except zero are natural numbers. All-natural numbers are whole numbers. But not all whole numbers are natural numbers.
Are whole numbers rational numbers?
According to the Whole Numbers Definition, these numbers range from zero to infinity. Every whole number can be written in a rational number.
Are whole numbers closed under multiplication?
Yes
Are whole numbers associative under subtraction?
The subtraction of whole numbers is not associative
Which the whole number has no predecessor?
According to the Whole Numbers Definition. The whole zero (0) has no predecessor.
Which of the whole number is not a natural number?
According to the Whole Numbers Definition. Zero (0) is not a natural number.
Which the whole number doesn’t have a successor?
All whole numbers have successors.
Which the whole number is not a rational number?
According to the Whole Numbers Definition. The rational numbers whose values are completely positive. The same number is the whole number.
Quiz
Definition of Natural Number and property, for example, Sum part-1
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Activities to Teach Students About Fractions on a Number Line
Fractions can be one of the most challenging concepts for students to grasp in mathematics. Teaching fractions using a number line is one of the most effective and engaging methods for students to understand the concept of fractions. A number line provides a visual representation of where fractions fit within a whole and makes the idea of fractions much easier to understand. Here are some activities to teach students about fractions on a number line.
1. Fraction Line-up
Begin the activity by dividing the class into teams of 4-5. Give each group a strip of paper with fractions written on it. Ask the groups to line up in order from smallest to largest fractions on a number line. Walk around the room, checking and correcting the groups, and once they have all correctly placed their fractions, see which team is the first to stand completed.
2. Fraction Walk
Draw a number line on the floor using masking tape. Use different colors to mark whole numbers and fractions. For instance, mark whole numbers in red and fraction bars in blue and put them in the appropriate places. Then ask the students what fraction they think various points represent on the number line, and ask them to explain why. Use this simple, yet enjoyable activity to help students understand the concept of fractions and their placement on the number line.
3. Fraction Concentration
Fraction concentration activity is a fun way of learning about fractions and is an excellent option for students who quickly lose interest in traditional classroom activities. To play this activity, you will need cards with fractions and number lines. Cut the cards and mix them up, and then ask the students to match the fractions with its appropriate placement on the number line.
4. Fraction Builder
Give the students fraction bars and number lines to work with. Ask the learners to create different fractions (unit fractions, equivalent fractions, etc.) by placing the fraction bars on the number line in its proper positional value. By visually marking the position of the numerator and denominator, children will create a mental image of how the fractions work, helping elementary students form a lasting understanding of essential math concepts.
5. Fraction Scavenger Hunt
Conduct a scavenger hunt activity using fractions and number lines. Make a treasure hunt map that includes clues leading to the locations of different fraction bars placed on the number lines around the classroom. Give each student a copy of the treasure hunt map, and they will use the clues to find the different fractions before their team-mates. This activity helps to engage students in learning and is also a fun way to apply math in the real-life world.
In conclusion, teaching fractions using a number line can be a fun, interactive, and beneficial way to help students understand and visualize the concept of fractions. The above activities are helpful tools for familiarizing students with fractions on a number line and are designed to assist learners in developing fluency with fractions, numerator and denominator relationships, and equivalent fractions, all of which are crucial to understanding more complex topics in mathematics.
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NCERT Notes Mathematics for Class 12 Chapter 1:- Relations and Functions
Let A and B be .two non-empty sets, then a function f from set A to set B is a rule whichassociates each element of A to a unique element of B.
It is represented as f: A ? B and function is also called mapping.
f : A ? B is called a real function, if A and B are subsets of R.
Domain and Codomain of a Real Function
Domain and codomain of a function f is a set of all real numbers x for which f(x) is a real number. Here, set A is domain and set B is codomain.
Range of a Real Function
Range of a real function, f is a set of values f(x) which it attains on the points of its domain.
Classification of Real Functions
Real functions are generally classified under two categories algebraic functions and transcendental functions.
1. Algebraic Functions
Some algebraic functions are given below
(i) Polynomial Functions If a function y = f(x) is given by
where, a0, a1, a2,
, an are real numbers and n is any non -negative integer, then f (x) is called a polynomial function in x.
If a0 ? 0, then the degree of the polynomial f(x) is n. The domain of a polynomial function is the set of real number R.
e.g., y = f(x) = 3x5 4x2 2x +1
is a polynomial of degree 5.
(ii) Rational Functions If a function y = f(x) is given by f(x) = ?(x) / ?(x)
where, ?(x) and ?(x) are polynomial functions, then f(x) is called rational function in x.
(iii) Irrational Functions The algebraic functions containing one or more terms having nonintegral rational power x are called irrational functions.
e.g., y = f(x) = 2?x 3?x + 6
2. Transcendental Function
A. function, which is not algebraic, is called a transcendental function. Trigonometric, Inverse trigonometric, Exponential, Logarithmic, etc are transcendental functions.
Explicit and Implicit Functions
(i) Explicit Functions A function is said to be an explicit function, if it is expressed in the form y = f(x).
(ii) Implicit Functions A function is said to be an implicit function, if it is expressed in the form f(x, y) = C, where C is constant.
e.g., sin (x + y) cos (x + y) = 2
Intervals of a Function
(i) The set of real numbers x, such that a ? x ? b is called a closed interval and denoted by [a, b] i.e., {x: x ? R, a ? x ? b}.
(ii) Set of real number x, such that a < x < b is called open interval and is denoted by (a, b) i.e., {x: x ? R, a < x < b}
(iii) Intervals [a,b) = {x: x ? R, a ? x ? b} and (a, b] = {x: x ? R, a < x ? b} are called semiopen and semi-closed intervals.
Graph of Real Functions
1. Constant Function Let c be a fixed real number.
The function that associates to each real number x, this fixed number c is called a constant function i.e., y = f{x) = c for all x ? R.
Domain of f{x) = R
Range of f{x) = {c}
2. Identity Function
The function that associates to each real number x for the same number x, is called the identity function. i.e., y = f(x) = x, ? x ?
R. Domain of f(x) = R
Range f(x) = R
3. Linear Function
If a and b be fixed real numbers, then the linear function is defmed as y = f(x) = ax + b, where a and b are constants.
Domain of f(x) = R
Range of f(x) = R
The graph of a linear function is given in the following diagram, which is a straight line with slope a.
If a, b and c are fixed real numbers, then the quadratic function is expressed as y = f(x) = ax2 + bx + c, a ? 0 ? y = a (x + b / 2a)2 + 4ac b2 / 4a
which is equation of a parabola in downward, if a < 0 and upward, if a > 0 and vertex at ( b / 2a, 4ac b2 / 4a).
Domain of f(x) = R
Range of f(x) is [ ?, 4ac b2 / 4a], if a < 0 and [4ac b2 / 4a, ?], if a > 0 5. Square Root Function Square root function is defined by y = F(x) = ?x, x ? 0.
5. Square Root Function
Square root function is defined by y = F(x) = ?x, x ? 0.
Domain of f(x) = [0, ?)
Range of f(x) = [0, ?)
6. Exponential Function
Exponential function is given by y = f(x) = ax, where a > 0, a ? 1.
7. Logarithmic Function
A logarithmic function may be given by y = f(x) = loga x, where a > 0, a ? 1 and x > 0.
The graph of the function is as shown below. which is increasing, if a > 1 and decreasing, if 0 < a < 1.
Domain of f(x) = (0, ?)
Range of f(x) = R
8. Power Function
The power function is given by y = f(x) = xn ,n ? I,n? 1, 0. The domain and range of the graph y = f(x), is depend on n.
(a) If n is positive even integer.
i.e., f(x) = x2, x4 ,
.
Domain of f(x) = R
Range of f(x) = [0, ?)
(b) If n is positive odd integer.
i.e., f(x) = x3, x5 ,
.
Domain of f(x) = R
Range of f(x) = R
(c) If n is negative even integer.
i.e., f(x) = x- 2, x 4 ,
.
Domain of f(x) = R {0}
Range of f(x) = (0, ?)
(d) If n is negative odd integer.
i.e., f(x) = x- 1, x 3 ,
.
Domain of f(x) = R {0}
Range of f(x) = R {0}
9. Modulus Function (Absolute Value Function)
Modulus function is given by y = f(x) = |x| , where |x| denotes the absolute value of x, that is
|x| = {x, if x ? 0, x, if x < 0
Domain of f(x) = R
Range of f(x) = [0, &infi;)
Domain of f(x) = R
Range of f(x) = {-1, 0, 1}
11. Greatest Integer Function
The greatest integer function is defined as y = f(x) = [x]
where, [x] represents the greatest integer less than or equal to x. i.e., for any integer n, [x] = n, if n ? x < n + 1 Domain of f(x) = R Range of f(x) = I
Properties of Greatest Integer Function
(i) [x + n] = n + [x], n ? I
(ii) x = [x] + {x}, {x} denotes the fractional part of x.
(iii) [- x] = [x], -x ? I
(iv) [- x] = [x] 1, x ? I
(v) [x] ? n ? x ? n,n ? I
(vi) [x] > n ? x ? n+1, n ? I
(vii) [x] ? n ? x < n + 1, n ? I
(viii) [x] < n ? x < n, n ? I
(ix) [x + y] = [x] + [y + x [x}] for all x, y ? R
(x) [x + y] ? [x] + [y]
(xi) [x] + [x + 1 / n] + [x + 2 / n] +
+ [x + n 1 / n] = [nx], n ? N
12. Least Integer Function
The least integer function which is greater than or equal to x and it is denoted by (x). Thus, (3.578) = 4, (0.87) = 1, (4) = 4, (- 8.239) = 8, (- 0.7) = 0
In general, if n is an integer and x is any real number between n and (n + 1).
i.e., n < x ? n + 1, then (x) = n + 1
? f(x) = (x)
Domain of f = R
Range of f= [x] + 1
13. Fractional Part Function
It is denoted as f(x) = {x} and defined as
(i) {x} = f, if x = n + f, where n ? I and 0 ? f < 1
(ii) {x} = x [x]
i.e., {O.7} = 0.7, {3} = 0, { 3.6} = 0.4
(iii) {x} = x, if 0 ? x ? 1
(iv) {x} = 0, if x ? I
(v) { x} = 1 {x}, if x ? I
Graph of Trigonometric Functions
1. Graph of sin x
(i) Domain = R
(ii) Range = [-1,1]
(iii) Period = 2?
2. Graph of cos x
(i) Domain = R
(ii) Range = [-1,1]
(iii) Period = 2?
3.Graph of tan x
(i) Domain = R ~ (2n + 1) ? / 2, n ? I
(ii) Range = [- &infi;, &infi;]
(iii) Period = ?
4. Graph of cot x
(i) Domain = R ~ n?, n ? I
(ii) Range = [- &infi;, &infi;]
(iii) Period = ?
5. Graph of sec x
(i) Domain = R ~ (2n + 1) ? / 2, n ? I
(ii) Range = [- &infi;, 1] ? [1, &infi;)
(iii) Period = 2?
6. Graph of cosec x
(i) Domain = R ~ n?, n ? I
(ii) Range = [- &infi;, 1] ? [1, &infi;)
(iii) Period = 2?
Operations on Real Functions
Let f: x ? R and g : X ? R be two real functions, then
(i) Sum The sum of the functions f and g is defined as f + g : X ? R such that (f + g) (x) = f(x) + g(x).
(ii) Product The product of the functions f and g is defined as fg : X ? R, such that (fg) (x) = f(x) g(x) Clearly, f + g and fg are defined only, if f and g have the same domain. In case, the domain of f and g are different. Then, Domain of f + g or fg = Domain of f ? Domain of g.
(iii) Multiplication by a Number Let f : X ? R be a function and let e be a real number .
Then, we define cf: X ? R, such that (cf) (x) = cf (x), ? x ? X.
(iv) Composition (Function of Function) Let f : A ? B and g : B ? C be two functions. We define gof : A ? C, such that got (c) = g(f(x)), ? x ? A
Alternate There exists Y ? B, such that if f(x) = y and g(y) = z, then got (x) = z
Periodic Functions
A function f(x) is said to be a periodic function of x, provided there exists a real number T > 0, such that F(T + x) = f(x), ? x ? R
The smallest positive real number T, satisfying the above condition is known as the period or the fundamental period of f(x) ..
Testing the Periodicity of a Function
(i) Put f(T + x) = f(x) and solve this equation to find the positive values of T independent of x.
(ii) If no positive value of T independent of x is obtained, then f(x) is a non-periodic function.
(iii) If positive val~es ofT independent of x are obtained, then f(x) is a periodic function and the least positive value of T is the period of the function f(x).
Important Points to be Remembered
(i) Constant function is periodic with no fundamental period.
(ii) If f(x) is periodic with period T, then 1 / f(x) and. ?f(x) are also periodic with f(x) same period T.
{iii} If f(x) is periodic with period T1 and g(x) is periodic with period T2, then f(x) + g(x) is periodic with period equal to LCM of T1 and T2, provided there is no positive k, such that f(k + x) = g(x) and g(k + x) = f(x).
(iv) If f(x) is periodic with period T, then kf (ax + b) is periodic with period T / |a| where a, b ,k ? R and a, k ? 0.
(v) sin x, cos x, sec x and cosec x are periodic functions with period 2?.
(vi) tan x and cot x are periodic functions with period ?.
(vii) |sin x|, |cos x|, |tan x|, |cot x|, |sec x| and |cosec x| are periodic functions with period ?.
(viii) sinn x, cosn x, secn x and cosecnx are periodic functions with period 2? when n is odd, or ? when n is even .
(ix) tann x and cotnx are periodic functions with period ?.
(x) |sin x| + |cos x|, |tan x| + |cot x| and |sec x| + |cosec x| are periodic with period ? / 2.
Even and Odd Functions
Even Functions A real function f(x) is an even function, if f( -x) = f(x).
Odd Functions A real function f(x) is an odd function, if f( -x) = f(x).
Properties of Even and Odd Functions
(i) Even function ± Even function = Even function.
(ii) Odd function ± Odd function = Odd function.
(iii) Even function * Odd function = Odd function.
(iv) Even function * Even function = Even function.
(v) Odd function * Odd function = Even function.
(vi) gof or fog is even, if anyone of f and g or both are even.
(vii) gof or fog is odd, if both of f and g are odd.
(viii) If f(x) is an even function, then d / dx f(x) or ? f(x) dx is odd and if dx .. f(x) is an odd function, then d / dx f(x) or ? f(x) dx is even.
(ix) The graph of an even function is symmetrical about Y-axis.
(x) The graph of an odd function is symmetrical about origin or symmetrical in opposite quadrants.
(xi) An even function can never be one-one, however an odd function mayor may not be oneone.
Different Types of Functions (Mappings)
1. One-One and Many-One Function
The mapping f: A ? B is a called one-one function, if different elements in A have different images in B. Such a mapping is known as injective function or an injection.
Methods to Test One-One
(i) Analytically If x1, x2 ? A, then f(x1) = f(x2) => x1 = x2 or equivalently x1 ? x2 => f(x1) ? f(x2)
(ii) Graphically If any .line parallel to x-axis cuts the graph of the function atmost at one point, then the function is one-one.
(iii) Monotonically Any function, which is entirely increasing or decreasing in whole domain, then f(x) is one-one.
Number of One-One Functions Let f : A ? B be a function, such that A and B are finite sets having m and n elements respectively, (where, n > m).
The number of one-one functions n(n 1)(n 2)
(n m + 1) = { nPm, n ? m, 0, n < m
The function f : A ? B is called many one function, if two or more than two different elements in A have the same image in B.
2. Onto (Surjective) and Into Function
If the function f: A ? B is such that each element in B (codomain) is the image of atleast one element of A, then we say that f is a function of A onto B.
Thus, f: A ? B, such that f(A) = i.e., Range = Codomain Note Every polynomial function f: R ? R of degree odd is onto.
Number of Onto (surjective) Functions Let A and B are finite sets having m and n elements respectively, such that 1 ? n ? m, then number of onto (surjective) functions from A to B is n?r = 1 (- 1)n r nCr rm = Coefficient ofn in n! (ex 1)r If f : A ? B is such that there exists atleast one element in codomain which is not the image of
Thus, f : A ? B, such that f(A) ? B
i.e., Range ? Codomain
Important Points to be Remembered
(i) If f and g are injective, then fog and gof are injective.
(ii) If f and g are surjective, then fog is surjective.
(iii) Iff and g are bijective, then fog is bijective.
Inverse of a Function
Let f : A ? B is a bijective function, i.e., it is one-one and onto function.
We define g : B ? A, such that f(x) = y => g(y) = x, g is called inverse of f and vice-versa.
Symbolically, we write g = f-1
Thus, f(x) = y => f-1(y) = x
PART 2
Ncert Math Notes For Class 12 Chapter 1 Relations and Functions Download PDf
Ncert Math Notes For Class 12 Chapter 1 Relations and Functions Download PDf
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11th NCERT/CBSE Introduction to Introduction to Conic section Exercise 11.1 Questions 15
Hi
Question (1)
Find the equation of the circle with centre (0, 2) and radius 2
Solution
The equation of a circle with centre (h, k) and radius r is given as (x . h)2 + (y . k)2 = r2 It is given that centre (h, k) = (0, 2) and radius (r) = 2. Therefore, the equation of the circle is (x – 0)2 + (y – 2)2 = 22 x2 + y2 + 4 – 4 y = 4 x2+ y2 – 4y = 0
Question (2)
Find the equation of the circle with centre (–2, 3) and radius 4
Solution
The equation of a circle with centre (h, k) and radius r is given as (x. h)2 + (y . k)2 = r2 It is given that centre (h, k) = (–2, 3) and radius (r) = 4. Therefore, the equation of the circle is (x + 2)2 + (y – 3)2= (4)2 x2 + 4x + 4 + y2 – 6y + 9 = 16 x2 + y2 + 4x – 6y – 3 = 0
Question (3)
Find the equation of the circle with centre $\left( {\frac{1}{2},\frac{1}{4}} \right)$ and radius $\frac{1}{{12}}$
Solution
The equation of a circle with centre (h, k) and radius r is given as (x. h)2 + (y – k)2 = r2
It is given that centre $\left( {h,k} \right) = \left( {\frac{1}{2},\frac{1}{4}} \right)$ and radius $\left( r \right) = \frac{1}{{12}}$
Therefore, the equation of the circle is ${\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{4}} \right)^2} = {\left( {\frac{1}{{12}}} \right)^2}$
${x^2} - x + \frac{1}{4} + {y^2} - \frac{y}{2} + \frac{1}{{16}} = \frac{1}{{144}}$
${x^2} - x + \frac{1}{4} + {y^2} - \frac{y}{2} + \frac{1}{{16}} - \frac{1}{{144}} = 0$
144x2 - 144x +36+144y2 - 72y +9-1 = 0
144x2 - 144x +144y2 - 72y + 44 = 0
36x2 - 36x +36y2 - 18y + 11 = 0
36x2 +36y2 - 36x - 18y + 11 = 0
Question (4)
Find the equation of the circle with centre (1, 1) and radius √2
Solution
The equation of a circle with centre (h, k) and radius r is given as (x.h)2 + (y . k)2 = r2
It is given that centre (h, k) = (1, 1) and radius (r) = √2.
Therefore, the equation of the circle is
(x-1)2 + (y-1)2 = (√2)2
x2 - 2x + 1 + y2 - 2y +1 = 2
x2 + y2 - 2y - 2x = 0
Question (5)
Find the equation of the circle with centre (–a, –b) and radius $\sqrt {{a^2} - {b^2}}$
Solution
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (–a, –b) and radius $\left( r \right) = \sqrt {{a^2} - {b^2}}$
Therefore, the equation of the circle is
${\left( {x + a} \right)^2} + {\left( {y + b} \right)^2} = {\left( {\sqrt {{a^2} - {b^2}} } \right)^2}$
x2 +2ax +a2 + y2 + 2by + b2 = a2 - b2
x2 + y2 + 2ax +2by +2b2 = 0
Question (6)
Find the centre and radius of the circle (x + 5)2 + (y – 3)2 = 36
Solution
The equation of the given circle is (x + 5)2 + (y – 3)2 = 36.
(x + 5)2 + (y – 3)2 = 36
⇒ {x – (–5)}2 + (y – 3)2 = 62, which is of the form (x – h)2 + (y – k)2 = r2, where h = –5, k = 3, and r = 6.
Thus, the centre of the given circle is (–5, 3), while its radius is 6.
Question (7)
Find the centre and radius of the circle x2 + y2 – 4x – 8y – 45 = 0
Solution
Given equation for circle x2 + y2 – 4x – 8y – 45 = 0
Comparing it to standard form x2 + y2 + 2gx + 2fy + c = 0 we get
g = -2, f = -4 and c = 45
centre = (-g, -f) = (2, 4)
radius $\left( r \right) = \sqrt {{g^2} + {f^2} - c}$
$\left( r \right) = \sqrt {4 + 16 + 45} = \sqrt {65}$
Question (8)
Find the centre and radius of the circle x2 + y2 – 8x + 10y – 12 = 0
Solution
Given equation for circle x2 + y2 – 8x + 10y – 12 = 0
Comparing it to standard form x2 + y2 + 2gx + 2fy + c = 0 we get
g = -4, f = 5 and c = -12
centre C(-g, -f) = (4, -5)
radius $\left( r \right) = \sqrt {{g^2} + {f^2} - c}$
$\left( r \right) = \sqrt {16 + 25 + 12} = \sqrt {53}$
Question (9)
Find the centre and radius of the circle 2x2 + 2y2 – x = 0
Solution
Given equation for circle is 2x2 + 2y2 – x = 0
or ${x^2} + {y^2} - \frac{1}{2}x = 0$
Comparing it to standard form x2 + y2 + 2gx + 2fy + c = 0 we get
centre C(-g, -f) = $\left( {\frac{1}{4},0} \right)$
radius $\left( r \right) = \sqrt {{g^2} + {f^2} - c}$
$\left( r \right) = \sqrt {\frac{1}{{16}} + 0 + 0} = \frac{1}{4}$
Question (10)
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Solution
circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Let equation of circle be x2 + y2 + 2gx + 2fy + c = 0
∴ centre of Circle, c = (-g, -f)
c∈ line 4x+y=16
∴ -4g -f - 16 = 0 ---(1)
(4,1) belongs to circle
∴ 42 + (1)2 + 2g(4) + 2f(1) + c =
16+ 1 + 8g + 2f + c = 0
8g + 2f + c +17 = 0 ---(2)
(6, 5) ∈ circle
∴ 36 +25 +2g(6) +2 f(5) + c = 0
12g + 10f + c + 61 = 0
---(3)
(2) - (3)
⇒ -4g - 8f - 44 = 0 ---(4)
(1)-(4)
⇒ 7f + 28 = 0
f = -4
Substitute f = -4 in (1) we get
-4g + 4 -16 = 0
-4g = 12
g = -3
Substitute f = -4 and g = -3 in (2)
8(-3) + 2(-4) + c + 17 = 0
-24-8+c 17 = 0
c = 15
∴ equation of circle will be
x2 + y2 - 6x - 8y +15 = 0
Question (11)
Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.
Solution
circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.
Let equation of circle be x2 + y2 + 2gx + 2fy + c = 0
centre of circle c = (-g, -f)
c ∈ x - 3y-11 = 0
∴ -g+3f - 11 = 0 ---(1)
(2, 3) ∈ circle
22 + 32 +2g(2) +2f(3) + c = 0
4g + 6f +c +13 = 0 ---(2)
(-1, 1) ∈ circle
(-1)2 + (1)2 + 2g(-1) +2f(1) + c = 0
-2g +2f + c + 2 = 0 ---(3)
(2) - (3)
6g + 4f + 11 = 0 ----(3A)
6(1) + (3A)
-6g +18f -66 + 6g +4f + 11 = 0
22f = 55
$f = \frac{5}{2}$
Replace f in (1)
$- g + \frac{{15}}{2} - 11 = 0$
$g = - \frac{7}{2}$
Replace $g = - \frac{7}{2} \;and \;f = \frac{5}{2}$ in (1)
$4\left( { - \frac{7}{2}} \right) + 6\left( {\frac{5}{2}} \right) + c + 13 = 0$
c - 14 +15 + 13 = 0
c = -14
∴ Equation of circle will be
${x^2} + {y^2} + 2\left( {\frac{{ - 7}}{2}} \right)x + 2\left( {\frac{5}{2}} \right)y - 14 = 0$
x2 + y2 -7x + 5y - 14 = 0
Question (12)
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Solution
r=5, c ∈ x-axis, passing through (2, 3) Since centre ∈ x-axis
Let p(2,3) ∈ circle
∴ cp = r
∴ (CP)2 = r2 ∴ (a-2)2 + ( 0 - 3)2 = 52
a2 - 4a + 4 + 9 -25 = 0
a2 - 4a -12 = 0
(a - 6) ( a + 2) = 0
a = 6 or a = -2
If a = 6 → c(6, 0), r = 5
The equation of circle will be
(x-6)2 + (y - 0)2 = 25
x2 + y2 -12x + 11 = 0
if a = -2, c = (a, 0) = (-2, 0), r = 5
The equation of circle will be
(x+2)2 + (y-0)2 = 52
x2 + y2 + 4x - 21 = 0
Question (13)
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Solution
circle passing through (0, 0) and making intercepts a and b
So coordinate of A(a, 0) and B(0, b)
Let equation of circle be x2 + y2 + 2gx + 2fy + c = 0
It passes through (0, 0)
0 + 0 + 0 + 0 + c = 0 → c = 0
It passes through A(a, 0)
∴ a2 + 0 +2ga + 0 + c = 0
a2 + 2ga + 0 = 0 (Because c = 0)
g = -a/2
It passes through (0, b)
0 + b2 + 0 + 2 +b + 0 = 0
f = -b/2
Replacing value of g, f and c we get
x2 + y2 - ax - by = 0
Question (14)
Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Solution
The centre of the circle is given as (h, k) = (2, 2).
Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).
$r = \sqrt {{{\left( {2 - 4} \right)}^2} + {{\left( {2 - 5} \right)}^2}}$
$r = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 3} \right)}^2}} = \sqrt {4 + 9} = \sqrt {13}$
Thus, the equation of the circle is
(x - h)2 + (y - k)2 = r2
(x - 2)2 + (y - 2)2 = (√13)2
x2 - 4x + 4 + y2 - 4y + 4 = 13
x2 + y2 -4x - 4y -5 = 0
Question (15)
Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
Solution
The equation of the given circle is x2 + y2 = 25. x2 + y2 = 25 ⇒ (x – 0)2 + (y – 0)2 = 52, which is of the form (x – h)2 + (y – k)2 = r2, where h = 0, k = 0, and r = 5. ∴Centre = (0, 0) and radius = 5 Distance between point (–2.5, 3.5) and centre (0, 0) $= \sqrt {{{\left( { - 2.5 - 0} \right)}^2} + {{\left( {3.5 - 0} \right)}^2}}$
$= \sqrt {6.25 + 12.25}$
$= \sqrt {18.5}$
$= 4.3\left( {approx.} \right) < 5$
Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (–2.5, 3.5) lies inside the circle.
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Chapter 2 - Equations, Inequalities, and Problem Solving - 2.2 Using the Principles Together - 2.2 Exercise Set - Page 94: 61
$x=11$
Work Step by Step
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\dfrac{1}{3}(2x-1)=7 ,$ use the properties of equality to isolate the variable. Do checking of the solution. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{3}(2x-1)=7 \\\\ 3\cdot\dfrac{1}{3}(2x-1)=3\cdot7 \\\\ 2x-1=21 \\\\ 2x=21+1 \\\\ 2x=22 \\\\ x=\dfrac{22}{2} \\\\ x=11 .\end{array} Checking: If $x=11,$ then \begin{array}{l}\require{cancel} \dfrac{1}{3}(2x-1)=7 \\\\ \dfrac{1}{3}(2\cdot11-1)=7 \\\\ \dfrac{1}{3}(22-1)=7 \\\\ \dfrac{1}{3}(21)=7 \\\\ 7=7 \text{ (TRUE) } .\end{array} Hence, the solution is $x=11 .$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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Practice With Points
PRACTICE WITH POINTS
LESSON READ-THROUGH
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
Thanks for your support!
As discussed in Locating Points in Quadrants and on Axes,
an ordered pair $\,(x,y)\,$ is a pair of numbers, separated by a comma, and enclosed in parentheses.
The number that is listed first is called the first coordinate or the $\,{x}$-value.
The number that is listed second is called the second coordinate or the $\,y$-value.
The coordinate plane (also called the $\,xy$-plane) is a device to ‘picture’ ordered pairs.
Each ordered pair corresponds to a point in the coordinate plane,
and each point in the coordinate plane corresponds to an ordered pair.
For this reason, ordered pairs are often called points.
Recall that the origin is the point $\,(0,0)\,$.
This section gives you practice with the connection between ordered pairs and movement in the coordinate plane.
Moving Up/Down/Left/Right
Left/right movement is controlled by the $\,x$-value (the first coordinate).
Moving to the right increases the $\,x$-value; moving to the left decreases it.
For example, if a point $\,(a,b)\,$ is moved two units to the right, then the new coordinates are $\,(a+2,b)\,$.
If a point $\,(a,b)\,$ is moved three units to the left, then the new coordinates are $\,(a-3,b)\,$.
Up/down movement is controlled by the $\,y$-value (the second coordinate).
Moving up increases the $\,y$-value; moving down decreases it.
For example, if a point $\,(a,b)\,$ is moved four units up, then the new coordinates are $\,(a,b+4)\,$.
If a point $\,(a,b)\,$ is moved five units down, then the new coordinates are $\,(a,b-5)\,$.
Recall that the origin is the point $\,(0,0)\,$.
Reflecting a Point about a Line
Suppose you have a line drawn on a piece of paper.
On this same piece of paper, you have a point.
For the moment, suppose that the point does not lie on the line.
To ‘reflect the point about the line’ means, roughly,
that you want the point that is the same distance from the line,
but on the other side of the line.
This idea is illustrated below:
reflecting about an arbitrary line: the distance from P to Q is the same as the distance from P' to Q. (the coordinates of the reflected pointare not so easy to find) reflecting about the $\,x$-axis (the $x$-value stays the same; take the opposite of the $\,y$-value) reflecting about the $\,y$-axis (the $y$-value stays the same; take the opposite of the $\,x$-value)
Notice that if you fold the piece of paper along the line of reflection,
then the original point and its reflection will land right on top of each other.
It's kind of like reflecting in a mirror —except instead of the mirror reflecting ‘back at you’,
it instead projects to the other side.
If a point actually lies on the line that you're reflecting about,
then the reflection of the point is itself.
EXAMPLES:
Question: Start at the origin.
Move to the right $\,2\,$, and down $\,4\,$.
At what point are you?
Solution: $\,(2,-4)\,$
Why?
$\cssId{s58}{(0,0)} \ \ \cssId{s59}{\overset{\text{right 2}}{\rightarrow}}\ \ \cssId{s60}{(0+2,0) = (2,0)} \ \ \cssId{s61}{\overset{\text{down 4}}{\rightarrow}}\ \ \cssId{s62}{(2,0-4) = (2,-4)}$
Question: Start at the point $\,(1,-3)\,$.
Move to the left $\,4\,$, and up $\,2\,$.
At what point are you?
Solution: $\,(-3,-1)\,$
Why?
$\cssId{s70}{(1,-3)} \ \ \cssId{s71}{\overset{\text{left 4}}{\rightarrow}}\ \ \cssId{s72}{(1-4,-3) = (-3,-3)} \ \ \cssId{s73}{\overset{\text{up 2}}{\rightarrow}}\ \ \cssId{s74}{(-3,-3+2) = (-3,-1)}$
Question: Start at the point $\,(-2,5)\,$.
Stay in the same quadrant,
but double your distance from the $\,x$-axis,
and triple your distance from the $\,y$-axis.
At what point are you?
Solution: $\,(-6,10)\,$
Why?
The up/down info (the $\,y$-value) gives distance from the $\,x$-axis.
The left/right info (the $\,x$-value) gives distance from the $\,y$-axis.
Thus:
$(-2,5)$ original point: currently $\,5\,$ units from the $\,x$-axis (above) and $\,2\,$ units from the $\,y$-axis (to the left) $(-2,5\cdot 2) = (-2,10)$ double distance from $\,x$-axis; stay above $(-2\cdot 3,10) = (-6,10)$ triple distance from $\,y$-axis, stay to the left
Question: Start at the point $\,(-3,7)\,$.
Reflect about the $\,x$-axis.
At what point are you?
Solution: $\,(-3,-7)\,$
To reflect about the $\,x$-axis,
the $\,x$-value stays the same,
and you take the opposite of the $\,y$-value.
Question: Start at the point $\,(-3,7)\,$.
Reflect about the $\,y$-axis.
At what point are you?
Solution: $\,(3,7)\,$
To reflect about the $\,y$-axis,
the $\,y$-value stays the same,
and you take the opposite of the $\,x$-value.
Question: Start at the point $\,(-3,7)\,$.
Reflect about the $\,x$-axis, and then reflect about the $\,y$-axis.
At what point are you?
Solution: $\,(3,-7)\,$
Why?
$\cssId{s120}{(-3,7)} \ \ \cssId{s121}{\overset{\text{reflect about x-axis}}{\rightarrow}}\ \ \cssId{s122}{(-3,-7)} \ \ \cssId{s123}{\overset{\text{reflect about y-axis}}{\rightarrow}}\ \ \cssId{s124}{(3,-7)}$
Question: Start at the point $\,(-3,1)\,$.
Reflect about the vertical line that passes through the $\,x$-axis at $\,2\,$.
At what point are you?
Solution: $\,(7,1)\,$
Why?
On this problem, you need to stop and think.
It might be helpful to make a sketch (see below).
• Since you're reflecting about a vertical line, the $\,y$-value won't change.
It was $\,1\,$, and it will remain $\,1\,$.
• Figure out the distance from the point to the line: $2 - (-3) = 5$
You need to go this far on the other side of the vertical line: $2 + 5 = 7$
So, the new $\,x$-value is $\,7\,$.
Question: Start at the point $\,(-3,1)\,$.
Reflect about the horizontal line that passes through the $\,y$-axis at $\,4\,$.
At what point are you?
Solution: $\,(-3,7)\,$
Why?
Again, you need to stop and think.
Again, a sketch may be helpful (see below).
• Since you're reflecting about a horizontal line, the $\,x$-value won't change.
It was $\,-3\,$, and it will remain $\,-3\,$.
• Figure out the distance from the point to the line: $4 - 1 = 3$
You need to go this far on the other side of the horizontal line: $4 + 3 = 7$
So, the new $\,y$-value is $\,7\,$.
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
the Pythagorean Theorem
CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5 6 7 8 9 10 11 12
AVAILABLE MASTERED IN PROGRESS
(MAX is 12; there are 12 different problem types)
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# CONVERTING CUSTOMARY UNITS WORD PROBLEMS WORKSHEET
Converting Customary Units Word Problems Worksheet :
Worksheet given in this section will be much useful for the students who would like to practice solving word problems on converting customary units.
## Converting Customary Units Word Problems Worksheet - Problems
Problem 1 :
A standard elevator in a mid rise building can hold a maximum weight of about 1.5 tons. Assuming an average adult weight of 150 pounds, what is the maximum number of adults who could safely ride the elevator ?
Solution :
First let us convert 1.5 tons into pounds
1.5 tons = 1.5 2000 = 3000 pounds
So, the elevator can hold 3000 pounds of weight.
If the average weight of an adult is 150 pounds,
Maximum no. of adults could safely ride the elevator is
= 3000/150
= 20
So, maximum number of adults who could safely ride the elevator is 20.
Problem 2 :
David prepares 60 pints of juice in two hours. At the same rate, How many cups of juice will he prepare in one minute ?
Solution :
No. of pints prepared in 2 hours = 60
No. of pints prepared in 1 hour = 30
We know that,
1 hour = 60 minutes
1 pint = 2 cups
1 hour -----> 30 pints
60 minutes -----> 30 2 cups
60 minutes -----> 60 cups
So, no. of cups prepared in 60 minutes is 60.
No. of cups prepared in in one minute is
= 60 / 60
= 1 cup
So, 1 cup of juice is prepared in 1 minute.
Problem 3 :
Kemka's little sister needs to take a bubble bath. The package says to put in a drop of bubble bath for every half gallon of water in the bath tub. If bathtub has 12 gallons of water, how many drops can she put into the bath for her sister?
Solution :
Half gallon of water -----> 1 drop of bubble bath
1 gallon of water -----> 2 drops of bubble bath
12 gallons of water -----> 12 2 drops of bubble bath
12 gallons of water -----> 24 drops of bubble bath
So, Kemka can put into 24 drops of bubble bath for her sister with 12 gallons of water.
Problem 4 :
Ivan needs gas for his truck. He knows his truck holds 40 gallons of gas. If he is allowed to fill up 8 quarts of gas once in a time, how many times will he have to fill up his gas can to get his truck full of gas ?
Solution :
1 gallon = 4 quarts
40 gallons = 40 4 quarts = 160 quarts
So, he needs 160 quarts of gas to make his truck full of gas.
Once in a time, he can fill up 8 quarts of gas.
No. of times of filling to make the truck full of gas is
= 160/8
= 20
So, Ivan has to fill up his gas can 20 times to get his truck full of gas.
Problem 5 :
Mrs. Moore took 4 hours 30 minutes to complete a work. How many seconds will Mrs. Moore take to complete the same work ?
Solution :
4 hours 30 minutes = 4 ⋅ 60 min + 30 min
4 hours 30 minutes = 240 min + 30 min
4 hours 30 minutes = 270 minutes
4 hours 30 minutes = 270 60 seconds
4 hours 30 minutes = 16200 seconds
So, Mrs. Moore will take 16200 seconds to complete the same work.
Problem 6 :
Tommy takes 10 minutes time for each pizza he makes. How many seconds will he take to make 4 pizzas ?
Solution :
1 pizza -----> 10 minutes
4 pizzas -----> 4 10 minutes
4 pizzas ------> 40 minutes
4 pizzas ------> 40 60 seconds
4 pizzas ------> 2400 seconds
So, Tommy will take 2400 seconds to make 4 pizzas.
Problem 7 :
A piece of work can be done by Mr. David in 9 days working 10 hours per day. How many hours will be taken by Mr. David to complete another work which is 4 times the first one ?
Solution :
Total number of hours required for David to complete the given work is
= No. of days ⋅ No. of hours per day
= 9 ⋅ 10 hours
= 90 hours
Total number of hours required for David to complete a work which is 4 times the first one is
= 4 ⋅ 90 hours
= 360 hours
So, the required time is 360 hours.
Problem 8 :
Jose needs 6 hours to complete a work. But Jacob need 3/4 of time taken by Jose to complete the same work. In how many minutes will Jose complete the work ?
Solution :
Time required for Jose = 6 hours
Time required for Jacob = 3/4 of time taken Jose
Time required for Jacob = 3/4 6 hours
Time required for Jacob = 3/4 6 60 minutes
Time required for Jacob = 270 minutes
So, Jacob needs 270 minutes to complete the work.
After having gone through the stuff given above, we hope that the students would have understood,
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
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Word problems on direct variation and inverse variation
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Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
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Profit and loss word problems
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Word problems on fractions
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One step equation word problems
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Word problems on sets and venn diagrams
Word problems on ages
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Percent of a number word problems
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Word problems on average speed
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OTHER TOPICS
Profit and loss shortcuts
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Times table shortcuts
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Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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# Fluid dynamics – problems and solutions
Fluid dynamics – problems and solutions
Torricelli’s theorem
1. A container filled with water and there is a hole, as shown in the figure below. If acceleration due to gravity is 10 ms-2, what is the speed of water through that hole?
Known :
Height (h) = 85 cm – 40 cm = 45 cm = 0.45 meters
Acceleration due to gravity (g) = 10 m/s2
Wanted : The speed of water (v)
Solution :
Torricelli’s theorem states that the water leaves the hole with the same speed as an object free fall from the same height. Height (h) = 85 cm – 40 cm = 45 cm = 0.45 meters
Velocity of water is calculated using the equation of the free fall motion :
vt2 = 2 g h
vt2 = 2 g h = 2(10)(0.45) = 9
vt = √9 = 3 m/s
2. A container filled with water and there is a hole, as shown in figure below. If acceleration due to gravity is 10 ms-2, what is the speed of water through that hole.
Known :
Height (h) = 1.5 m – 0.25 m = 1.25 meters
Acceleration due to gravity (g) = 10 m/s2
Wanted : The speed of water (v)
Solution :
vt2 = 2 g h = 2(10)(1.25) = 25
vt = √25 = 5 m/s
3. A container filled with water and there is a hole, as shown in figure below. If acceleration due to gravity is 10 ms-2, what is the speed of water through that hole.
Known :
Height (h) = 1 m – 0.20 m = 0.8 meter
Acceleration due to gravity (g) = 10 m/s2
Wanted : The speed of water (v)
Solution :
vt2 = 2 g h = 2(10)(0.8) = 16
vt = √16 = 4 m/s
4. A container filled with water and there is a hole, as shown in figure below. If acceleration due to gravity is 10 ms-2, what is the speed of water through that hole.
Known :
Height (h) = 20 cm = 0.2 meters
Acceleration due to gravity (g) = 10 m/s2
Wanted: The speed of water (v)
Solution :
5. A container filled with water and there are two holes, as shown in the figure below. What is the ratio of x1 to x2?
Solution
Time interval of the water free fall from hole 1 :
h = 1/2 a t2
0.8 = 1/2 (10) t2
0.8 = 5 t2
t2 = 0.8 / 5 = 0.16
t = 0.4 seconds
Time interval of the water free fall from hole 2 :
h = 1/2 a t2
0.5 = 1/2 (10) t2
0.5 = 5 t2
t2 = 0.5 / 5 = 0.1
t = √0.1 second
The horizontal distance (x) :
x1 = v1 t1 = (2)(0.4) = 0.8 meters
x2 = v2 t2 = (√10)(√0.1) = (10)(0.1) = 1 meter
The ration of x1 to x2 :
x1 : x2 = 0.8 : 1 = 8 : 10 = 4 : 5
The equation of continuity
6. Water flows through a pipe of varying diameter, A to B and then to C. The ratio of A to C is 8 : 3. If the speed of water in pipe A is v, what is the speed of water in pipe C.
Known :
Area of A (AA) = 8
Area of C (AC) = 3
The speed of water in pipe A (vA) = v
Wanted: The speed of water in pipe C (vC)
See also Rounding a flat curve – dynamics of circular motion problems and solutions
Solution :
The equation of continuity :
AA vA = AC vC
8 v = 3 vC
vC = 8/3 v
7. If the speed of water in pipe with a diameter of 12 cm is 10 cm/s, what is the speed of water in a pipe with a diameter of 8 cm?
Known :
Diameter 1 (d1) = 12 cm, radius 1 (r1) = 6 cm
Diameter 2 (d2) = 8 cm, radius 2 (r2) = 4 cm
The speed of water 1 (v1) = 10 cm/s
Wanted : The speed of water 2 (v2)
Solution :
Area 1 (A1) = π r2 = π 62 = 36π cm2
Area 2 (A2) = π r2 = π 42 = 16π cm2
The equation of continuity :
A1 v1 = A2 v2
(36π)(10) = (16π) v2
(36)(10) = (16) v2
360 = (16) v2
v2 = 360/16
v2 = 22.5 cm/s
8. Water flows through a pipe of varying diameter, as shown in figure below. If area 1 (A1) = 8 cm2, A2 = 2 cm2 and the speed of water in pipe 2 = v2 = 2 m/s then what is the speed of water in pipe 1 = v1.
Known :
Area 1 (A1) = 8 cm2
Area 2 (A2) = 2 cm2
Speed of water in pipe 2 (v2) = 2 m/s
Wanted : the speed of water in pipe 1 (v1)
Solution :
The equation of continuity :
A1 v1 = A2 v2
8 v1 = (2)(2)
8 v1 = 4
v1 = 4 / 8 = 0.5 m/s
9. If the diameter of the larger pipe is 2 times the diameter of smaller pipe, what is the speed of fluid at the smaller pipe.
Known :
Diameter of the larger pipe (d1) = 2
Radius of the larger pipe (r1) = ½ d1 = ½ (2) = 1
Area of the larger pipe (A1) = π r12 = π (1)2 = π (1) = π
Diameter of the smaller pipe (d2) = 1
Radius of the smaller pipe (r2) = ½ d2 = ½ (1) = ½
Area of the smaller pipe (A2) = π r22 = π (1/2)2 = π (1/4) = ¼ π
The speed of fluid at the larger pipe (v1) = 4 m/s
Wanted : The speed of fluid at the smaller pipe (v2)
Solution :
The equation of continuity :
A1 v1 = A2 v2
π 4 = ¼ π (v2)
4 = ¼ (v2)
v2 = 8 m/s
Bernoulli’s principle and equation
10. Water is pumped with a 120 kPa compressor entering the lower pipe (1) and flows upward at a speed of 1 m/s. Acceleration due to gravity is 10 m/s and water density is 1000 kg/m-3. What is the water pressure on the upper pipe (II).
Known :
Radius of the lower pipe (r1) = 12 cm
Radius of the lower pipe (r2) = 6 cm
Water pressure in the lower pipe (p1) = 120 kPa = 120,000 Pascal
The speed of water in the lower pipe (v1) = 1 m.s-1
The height of the lower pipe (h1) = 0 m
The height of the upper pipe (h2) = 2 m
Acceleration due to gravity (g) = 10 m.s-2
Density of water = 1000 kg.m-3
Wanted: Water pressure in pipe 2 (p2)
Solution :
The speed of water in pipe 2 is calculated with the equation of continuity :
Water pressure in pipe 2 is calculated using the equation of Bernoulli :
11. A large pipe 5 meters above the ground and a small pipe 1 meter above the ground. The velocity of the water in a large pipe is 36 km/h with a pressure of 9.1 x 105 Pa, while the pressure in the small pipe is 2.105 Pa. What is the water velocity in the small pipe? Water density = 103 kg/m3
Known :
Water pressure in the large pipe (p1) = 9.1 x 105 Pascal = 910,000 Pascal
Water pressure in the small pipe (p2) = 2 x 105 Pascal = 200,000 Pascal
Water speed in the large pipe (v1) = 36 km/h = 36(1000)/(3600) = 36000/3600 =10 m/s
The height of the large pipe (h1) = -4 meters
The height of the small pipe (h2) = 0 meter
Acceleration due to gravity (g) = 10 m.s-2
Density of water = 1000 kg/m3
Wanted: The speed of water in the small pipe (v2)
Solution :
The speed of water in the small pipe (v2) is calculated using the equation of Bernoulli :
12. A pipe with a radius of 15 cm connected with another pipe with a radius of 5 cm. Both are in a horizontal position. The velocity of the water flow in the large pipe is 1 m/s at a pressure of 105 N/m2. What is the water pressure on the small pipe (1 g cm-3)
Known :
Radius of the large pipe (r1) = 15 cm = 0.15 m
Radius of the small pipe (r2) = 5 cm = 0.05 m
The water pressure in the large pipe (p1) = 105 N m-2 = 100.000 N m-2
The speed of water in the large pipe (v1) = 1 m s-1
Acceleration due to gravity (g) = 10 m.s-2
Water density = 1 gr cm-3 = 1000 kg m-3
Height difference (Δh) = 0.
Wanted: Pressure in the small pipe (p2)
Solution :
The speed of water in pipe 2 is calculated using the equation of continuity :
The water pressure in the small pipe (p2) is calculated using the equation of Bernoulli :
1. What is fluid dynamics?
• Answer: Fluid dynamics is the branch of physics that studies the motion of fluids (liquids and gases) and the forces acting on them. It encompasses the principles and equations that describe how fluids flow, interact with solid boundaries, and affect one another.
2. What’s the difference between laminar and turbulent flow?
• Answer: Laminar flow is characterized by smooth, parallel layers of fluid moving in orderly paths. Turbulent flow, on the other hand, is chaotic, with eddies, swirls, and rapid fluctuations. Turbulence generally occurs at high velocities or in irregularly shaped channels.
3. How is the concept of viscosity important in fluid dynamics?
• Answer: Viscosity measures a fluid’s resistance to shear or flow. High-viscosity fluids (like honey) resist flow more than low-viscosity fluids (like water). In fluid dynamics, viscosity plays a crucial role in determining the nature of fluid flow, energy dissipation, and drag forces.
4. What is Bernoulli’s principle?
• Answer: Bernoulli’s principle states that in a steady flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant. Specifically, where the fluid velocity is high, the pressure is low, and vice versa.
5. How does the principle of lift in aerodynamics relate to fluid dynamics?
• Answer: The lift on an aircraft wing can be explained using Bernoulli’s principle and Newton’s third law. As air flows over the wing, it moves faster over the curved top surface than the bottom, creating a pressure difference. This difference in pressure, combined with the downward deflection of air by the wing, results in an upward force or lift.
6. What is the equation of continuity in fluid dynamics?
• Answer: The equation of continuity states that the product of the cross-sectional area (A) of a flow and its velocity (v) remains constant along a streamline in a steady flow. Mathematically, , where and are cross-sectional areas and and are the velocities at two points along the streamline.
7. What role does the Reynolds number play in fluid dynamics?
• Answer: The Reynolds number is a dimensionless quantity that helps predict the flow regime (laminar, transitional, or turbulent) in fluid dynamics. It’s defined as the ratio of inertial forces to viscous forces and depends on factors like fluid velocity, characteristic length, and fluid properties.
8. How does the drag force act on objects moving in a fluid?
• Answer: Drag force opposes the motion of an object through a fluid. It arises due to the viscous resistance of the fluid and the pressure differences around the object. The magnitude and nature of drag depend on factors like the object’s shape, roughness, speed, and the properties of the fluid.
9. What is the Venturi effect?
• Answer: The Venturi effect refers to the decrease in fluid pressure that occurs when a fluid flows through a constricted section of a pipe. As the fluid’s velocity increases in the constricted section (due to the conservation of mass), its pressure decreases according to Bernoulli’s principle.
10. Why does fluid speed up when flowing through a narrow section of a pipe or channel?
• Answer: This behavior can be explained by the principle of conservation of mass. In a steady flow, the volume of fluid entering a section of a pipe must equal the volume leaving. If the pipe narrows, the fluid must speed up to allow the same volume to pass through in a given time.
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# Lesson 15Quartiles and Interquartile Range
Let's look at other measures for describing distributions.
### Learning Targets:
• I can use IQR to describe the spread of data.
• I know what quartiles and interquartile range (IQR) measure and what they tell us about the data.
• When given a list of data values or a dot plot, I can find the quartiles and interquartile range (IQR) for data.
## 15.1Notice and Wonder: Two Parties
Here are two dot plots including the mean marked with a triangle. Each shows the ages of partygoers at a party.
What do you notice and wonder about the distributions in the two dot plots?
## 15.2The Five-Number Summary
Here are the ages of a group of the 20 partygoers you saw earlier, shown in order from least to greatest.
7 8 9 10 10 11 12 15 16 20 20 22 23 24 28 30 33 35 38 42
1. Find and mark the median on the table, and label it “50th percentile.” The data is now partitioned into an upper half and a lower half.
2. Find and mark the middle value of the lower half of the data, excluding the median. If there is an even number of values, find and write down the average of the middle two. Label this value “25th percentile.”
3. Find and mark the middle value of the upper half of the data, excluding the median. If there is an even number of values, find and write down the average of the middle two. Label the value “75th percentile.”
4. You have now partitioned the data set into four pieces. Each of the three values that “cut” the data is called a quartile.
• The first (or lower) quartile is the 25th percentile mark. Write “Q1” next to “25th percentile.”
• The second quartile is the median. Write “Q2” next to that label.
• The third (or upper) quartile is the 75th percentile mark. Write “Q3” next to that label.
5. Label the least value in the set “minimum” and the greatest value “maximum.”
1. Record the five values that you have just identified. They are the five-number summary of the data.
Minimum: _____ Q1: _____ Q2: _____ Q3: _____ Maximum: _____
2. The median (or Q2) value of this data set is 20. This tells us that half of the partygoers are 20 or younger, and that the other half are 20 or older. What does each of the following values tell us about the ages of the partygoers?
1. Q3
2. Minimum
3. Maximum
### Are you ready for more?
Here is the five-number summary of the age distribution at another party of 21 people.
Minimum: 5 years Q1: 6 years Q2: 27 years Q3: 32 years Maximum: 60 years
1. Do you think this party has more or fewer children than the other one in this activity? Explain your reasoning.
2. Are there more children or adults at this party? Explain your reasoning.
## 15.3Range and Interquartile Range
1. Here is a dot plot you saw in an earlier task. It shows how long Elena’s bus rides to school took, in minutes, over 12 days.
Write the five-number summary for this data set by finding the minimum, Q1, Q2, Q3, and the maximum. Show your reasoning.
2. The range of a data set is one way to describe the spread of values in a data set. It is the difference between the greatest and least data values. What is the range of Elena’s data?
3. Another number that is commonly used to describe the spread of values in a data set is the interquartile range (IQR), which is the difference between Q1, the lower quartile, and Q3, the upper quartile.
1. What is the interquartile range (IQR) of Elena’s data?
2. What fraction of the data values are between the lower and upper quartiles? Use your answer to complete the following statement:
The interquartile range (IQR) is the length that contains the middle ______ of the values in a data set.
4. Here are two dot plots that represent two data sets.
Without doing any calculations, predict:
a. Which data set has the smaller IQR? Explain your reasoning.
b. Which data set has the smaller range? Explain your reasoning.
1. Check your predictions by calculating the IQR and range for the data in each dot plot.
## Lesson 15 Summary
Earlier we learned that the mean is a measure of the center of a distribution and the MAD is a measure of the variability (or spread) that goes with the mean. There is also a measure of spread that goes with the median called the interquartile range (IQR).
Finding the IQR involves partitioning a data set into fourths. Each of the three values that cut the data into fourths is called a quartile
• The median, which cuts the data into a lower half and an upper half, is the second quartile (Q2).
• The first quartile (Q1) is the middle value of the lower half of the data.
• The third quartile (Q3) is the middle value of the upper half of the data.
Here is a set of data with 11 values.
12 19 20 21 22 33 34 35 40 40 49 Q1 Q2 Q3
• The median (Q2) is 33.
• The first quartile (Q1) is 20, the median of the numbers less than 33.
• The third quartile (Q3) is 40, the median of the numbers greater than 33.
The difference between the minimum and maximum values of a data set is the range.
The difference between Q1 and Q3 is the interquartile range (IQR). Because the distance between Q1 and Q3 includes the middle two-fourths of the distribution, the values between those two quartiles are sometimes called the middle half of the data
The bigger the IQR, the more spread out the middle half of the data are. The smaller the IQR, the closer the middle half of the data are. We consider the IQR a measure of spread for this reason.
A five-number summary, which includes the minimum, Q1, Q2, Q3, and the maximum, can be used to summarize a distribution.
The five numbers in this example are 12, 20, 33, 40, and 49. Their locations are marked with diamonds in the following dot plot.
Different data sets could have the same five-number summary. For instance, the following data has the same maximum, minimum, and quartiles as the one above.
## Glossary Terms
interquartile range (IQR)
The interquartile range is one way to measure how spread out a data set is. We sometimes call this the IQR. To find the interquartile range we subtract the first quartile from the third quartile.
22 29 30 31 32 43 44 45 50 50 59 Q1 Q2 Q3
For example, the IQR of this data set is 20 because .
quartile
Quartiles are the numbers that divide a data set into four sections that each have the same number of values.
For example, in this data set the first quartile is 20. The second quartile is the same thing as the median, which is 33. The third quartile is 40.
12 19 20 21 22 33 34 35 40 40 49 Q1 Q2 Q3
range
The range is the distance between the smallest and largest values in a data set. For example, for the data set 3, 5, 6, 8, 11, 12, the range is 9, because .
## Lesson 15 Practice Problems
1. Suppose that there are 20 numbers in a data set and that they are all different.
1. How many of the values in this data set are between the first quartile and the third quartile?
2. How many of the values in this data set are between the first quartile and the median?
2. In a word game, 1 letter is worth 1 point. This dot plot shows the scores for 20 common words.
1. What is the median score?
2. What is the first quartile (Q1)?
1. What is the third quartile (Q3)?
2. What is the interquartile range (IQR)?
3. Here are five dot plots that show the amounts of time that ten sixth-grade students in five countries took to get to school. Match each dot plot with the appropriate median and IQR.
1. Median: 17.5, IQR: 11
2. Median: 15, IQR: 30
3. Median: 8, IQR: 4
4. Median: 7, IQR: 10
5. Median: 12.5, IQR: 8
4. Mai and Priya each played 10 games of bowling and recorded the scores. Mai’s median score was 120, and her IQR was 5. Priya’s median score was 118, and her IQR was 15. Whose scores probably had less variability? Explain how you know.
5. Draw and label an appropriate pair of axes and plot the points. , , ,
6. There are 20 pennies in a jar. If 16% of the coins in the jar are pennies, how many coins are there in the jar?
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# Standard Index form OCR Module 8.
## Presentation on theme: "Standard Index form OCR Module 8."— Presentation transcript:
Standard Index form OCR Module 8
Why is this number very difficult to use?
999,999,999,999,999,999,999,999,999,999 Too big to read Too large to comprehend Too large for calculator To get around using numbers this large, we use standard index form.
But it still not any easier to handle!?!
Look at this 100,000,000,000,000,000,000,000,000,000 At the very least we can describe it as 1 with 29 noughts. But it still not any easier to handle!?!
Let’s investigate! Converting large numbers
How could we turn the number 800,000,000,000 into standard index form? We can break numbers into parts to make it easier, e.g. 80 = 8 x 10 and 800 = 8 x 100 800,000,000,000 = 8 x 100,000,000,000 Size given by first number 8 and the index 11 And 100, 000,000,000 = 1011 So, 800,000,000,000 = 8 x 1011 in standard index form
So 30,000 = 3 x 104 in standard index form
Try it out! How can we convert 30,000 into standard index form? Break into easier parts: And, 10,000 = 104 30000 = 3 x 10,000 So 30,000 = 3 x 104 in standard index form The number is now easier to use
Now it’s your turn: 500 = 5 x 100 = 5 x 102 4000 60,000
Copy down the following numbers, and convert them into standard index form. 500 4000 60,000 900,000 7000,000 = 5 x 100 = 5 x 102 = 4 x = 4 x 103 = 6 x 10,000 = 6 x 104 = 9 x 100,000 = 9 x 105 = 7 x 1000,000 = 7 x 106
The first number must be a value between
One of the most important rules for writing numbers in standard index form is: The first number must be a value between 1 and 10 But NOT 10 itself!! For example, 39 x 106 does have a value but it’s not written in standard index form. The first number, 39, is greater than 10. Pupils should copy down rule in green
How could we convert 350,000,000 into standard index form?
Again, we can break the number into smaller, more manageable parts. 350,000,000 = 3.5 x 100,000,000 100,000,000 = 108 3.5 x 100 = 350, x by 1,000,000 makes 350,000,000 350,000,000 = 3.5 x 108 in standard index form
Try it out! How can we convert 67,000 into standard index form?
10,000 = 104 67,000 = 6.7 x 104 in standard index form
Now it’s your turn: Copy out the following numbers and convert them into standard index form. 940 8,600 34, 000 570,000 1,200,000 = 9.4 x 100 = 9.4 x 102 = 8.6 x 1000 = 8.6 x 103 = 3.4 x 10,000 = 3.4 x 104 = 5.7 x 100,000 = 5.7 x 105 = 1.2 x 1000,000 = 1.2 x 106
Quick method of converting numbers to standard form
For example, Converting 45,000,000,000 to standard form Place a decimal point after the first digit Count the number of digits after the decimal point. Pupil book G3 page 137 10 This is our index number (our power of 10) So, 45,000,000,000 = 4.5 x 1010
And numbers less than 1? How can we convert into standard index form? 0.067 = 6.7 x 0.01 0.01 = 10-2 0.067 = 6.7 x 10-2
And numbers less than 1? How can we convert into standard index form? = 2.13 x = 10-4 = 2.13 x 10-4
And numbers less than 1? How can we convert into standard index form? = 6.03 x 0.001 = 10-3 = 6.03 x 10-3
Now it’s your turn: Copy out the following numbers and convert them into standard index form. 0.094 0.903 = 9.4 x 0.01 = 9.4 x 10-2 = 3.9 x = 3.9 x 10-4 = 9.03 x 0.1 = x 10-1 = 4.52 x = x 10-4 = 6.24 x 0.001 = x 10-3
Calculations
Multiply two numbers 4 x 1018 x 3 x 104 4 x 3 x 1018 x 104 = 12 x 1022
Powers of 10 4 x 3 x 1018 x 104 ADD powers = 12 x 1022 NOT Std Form! = 1.2 x 101 x 1022 = 1.2 x 1023
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Note: When slope $(m)$ and a point $(x_1, y_1)$ is given, the equation of the line is $y - y_1 = m ( x-x_1)$
Question 1: Find the equation of the straight line passing through the point $(6,2)$ and having slope $-3$.
Here $m = -3$ and $(x_1, y_1) = ( 6, 2)$
Substituting in $y - y_1 = m ( x-x_1)$ we get
$y - 2 = - 3( x-6)$
$\Rightarrow y - 2 = - 3x + 18$
$\Rightarrow 3x + y - 20 = 0$
Hence the equation of the straight line is $3x + y - 20 = 0$
$\\$
Question 2: Find the equation of the straight line passing through $(-2, 3)$ and inclined at an angle of $45^{\circ}$ with the x-axis.
Here $m = \tan 45^{\circ} = 1$ and $(x_1, y_1) = ( -2, 3)$
Substituting in $y - y_1 = m ( x-x_1)$ we get
$y - 3 = 1( x+2)$
$\Rightarrow y-3 = x + 2$
$\Rightarrow x-y+5=0$
Hence the equation of the straight line is $x-y+5=0$
$\\$
Question 3: Find the equation of the line passing through $(0, 0)$ with slope $m$.
Here $m = m$ and $(x_1, y_1) = ( 0,0)$
Substituting in $y - y_1 = m ( x-x_1)$ we get
$y - 0 = m( x-0)$
$\Rightarrow y = mx$
Hence the equation of the straight line is $y=mx$
$\\$
Question 4: Find the equation of the line passing through $(2, 2\sqrt{3})$ and inclined with x-axis at an angle of $75^{\circ}$.
Here $m = \tan 75^{\circ} = \tan ( 45^{\circ} + 30^{\circ})$
$=$ $\frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \tan 30^{\circ}}$
$=$ $\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}$ $=$ $\frac{\sqrt{3}+1}{\sqrt{3}-1}$ $\times$ $\frac{\sqrt{3}+1}{\sqrt{3}+1}$ $=$ $\frac{3+1 + 2\sqrt{3}}{3-1}$ $=$ $\frac{4+2\sqrt{3}}{2}$ $= 2 + \sqrt{3}$
Also $(x_1, y_1) = ( 2, 2\sqrt{3})$
Substituting in $y - y_1 = m ( x-x_1)$ we get
$y - 2\sqrt{3} = (2+\sqrt{3})( x-2)$
$\Rightarrow y-2\sqrt{3} = (2+\sqrt{3})x - 4 - 2\sqrt{3}$
$\Rightarrow (2+\sqrt{3})x-y-4=0$
Hence the equation of the straight line is $(2+\sqrt{3})x-y-4=0$
$\\$
Question 5: Find the equation of the straight line which passes through the point $(1,2)$ and makes such an angle with the positive direction of x-axis whose sine is $\frac{3}{5}$
Given $\sin \theta =$ $\frac{3}{5}$
We know, $\tan \theta=$ $\frac{\sin \theta}{1 - \sin^2 \theta }$ $=$ $\frac{3/5}{\sqrt{1 - 9/25}}$ $=$ $\frac{3}{\sqrt{16}}$ $=$ $\frac{3}{4}$
Therefore $m =$ $\frac{3}{4}$ and $(x_1, y_1) = ( 1, 2)$
Substituting in $y - y_1 = m ( x-x_1)$ we get
$y - 2 =$ $\frac{3}{4}$ $( x-1)$
$\Rightarrow 3x-4y+5=0$
Hence the equation of the straight line is $3x-4y+5=0$
$\\$
Question 6: Find the equation of the straight line passing through $(3, -2)$ and making an angle of $60^{\circ}$ with the positive direction of y-axis
Slope of line $= m = \tan 30^{\circ} =$ $\frac{1}{\sqrt{3}}$ and $(x_1, y_1) = ( 3, -2)$
Substituting in $y - y_1 = m ( x-x_1)$ we get
$y + 2 =$ $\frac{1}{\sqrt{3}}$ $(x-3)$
$\Rightarrow \sqrt{3} y + 2\sqrt{3}= x-3$
$\Rightarrow x-\sqrt{3} y - 3 - 2\sqrt{3} = 0$
Hence the equation of the straight line is $x-\sqrt{3} y - 3 - 2\sqrt{3}=0$
$\\$
Question 7: Find the lines through the point $(0,2)$ making angle $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ with the x-axis. Also, find the lines parallel to them cutting the y-axis at a distance of $2$ units below the origin.
Given $m_1 = \tan$ $\frac{\pi}{3}$ $= \sqrt{3}$
$m_2 = \tan$ $\frac{2\pi}{3}$ $= - \tan$ $\frac{\pi}{3}$ $= -\sqrt{3}$
Also $(x_1, y_2) = ( 0, 2)$
Therefore the equation passing through $( 0,2)$ and having $m_1$ and $m_2$ slope respectively are
$y - 2 = \sqrt{3}(x-0) \Rightarrow y - 2 = \sqrt{3}x \Rightarrow \sqrt{3}x - y + 2 = 0$
$y - 2 = -\sqrt{3}(x-0) \Rightarrow y - 2 = -\sqrt{3}x \Rightarrow \sqrt{3}x + y - 2 = 0$
Now let us find lines parallel to these lines but passing through $( 0, -2)$
Therefore the equation passing through $( 0,-2)$ and having $m_1$ and $m_2$ slope respectively are
$y + 2 = \sqrt{3}(x-0) \Rightarrow y + 2 = \sqrt{3}x \Rightarrow \sqrt{3}x - y - 2 = 0$
$y + 2 = -\sqrt{3}(x-0) \Rightarrow y + 2 = -\sqrt{3}x \Rightarrow \sqrt{3}x + y + 2 = 0$
$\\$
Question 8: Find the equations of the straight lines which cut off an intercept $5$ from the y-axis and are equally inclined to the axes.
Given the lines are equally inclined to the axes.
Hence, their inclination with positive x-axis are $45^{\circ}$ and $135^{\circ}$
Therefore for Line 1: $m_1 = \tan 45^{\circ} = 1$
and for Line 2: $m_2 = \tan 135^{\circ} = - \tan 45^{\circ} = -1$
Hence the equation of Line 1: $y - 5 = 1 ( x - 0) \Rightarrow x - y + 5 = 0$
And the equation for Line 2: $y - 5 = -1 ( x - 0) \Rightarrow x + y -5 = 0$
$\\$
Question 9: Find the equation of the line which intercepts a length $2$ on the positive direction of the x-axis and is inclined at an angle of $135^{\circ}$ with the positive direction of y-axis.
Here $m = \tan 45^{\circ} = 1$ and $(x_1, y_1) = ( 2, 0)$
Therefore equation of the line is: $y-0 = 1 ( x-2) \Rightarrow x-y - 2 = 0$
$\\$
Question 10: Find the equation of the straight line which divides the join of the points $(2, 3)$ and $(- 5, 8)$ in the ratio $3 : 4$ and is also perpendicular to it.
Given points $A(2, 3)$ and $B(- 5, 8)$
Therefore slope of $AB =$ $\frac{8-3}{-5-2}$ $=$ $\frac{-5}{7}$
$\therefore$ Slope of line $\perp$ to $AB =$ $\frac{-1}{(-5/7)}$ $=$ $\frac{7}{5}$
The coordinate of the point that divides AB in the ratio of $3:4$ is
$(x_1, y_1) = \Big($ $\frac{3(-5) + 4(2)}{3+4}$ $,$ $\frac{3(8)+4(3)}{3+4}$ $\Big) = \Big($ $\frac{-7}{7}$ $,$ $\frac{36}{7}$ $\Big) =\Big( -1,$ $\frac{36}{9}$ $\Big)$
Therefore the equation of the required line:
$y -$ $\frac{36}{7}$ $=$ $\frac{7}{5}$ $( x+1)$
$\Rightarrow 7y - 36 =$ $\frac{7}{5}$ $( 7x + 7)$
$\Rightarrow 35y - 180 = 49x + 49$
$\Rightarrow 49x - 35y + 229 = 0$
$\\$
Question 11: Prove that the perpendicular drawn from the point $(4, 1)$ on the join of $(2, -1)$ and $(6,5)$ divides it in the ratio $5 : 8$.
Given points $A(2, -1)$ and $B(6,5 )$
Slope of $AB =$ $\frac{5-(-1)}{6-2}$ $=$ $\frac{6}{4}$ $=$ $\frac{3}{2}$
$PQ$ is $\perp$ to $AB$ . Therefore slope of $PQ =$ $\frac{-1}{3/2}$ $=$ $\frac{-2}{3}$
Therefore equation of $PQ$ is:
$y - 1 =$ $\frac{-2}{3}$ $( x-4)$
$\Rightarrow 3y - 3 = - 2x + 8$
$\Rightarrow 2x + 3y = 11$ … … … … … i)
Equation of $AB$:
$y - 5 =$ $\frac{3}{7}$ $(x-6)$
$\Rightarrow 2y - 10 = 3x - 18$
$\Rightarrow 3x - 2y = 8$ … … … … … ii)
Point of intersection will be found by solving the two equations.
Multiply i) by $3$ and ii) by $2$ and then subtract ii) from i) we get
${ \hspace{1.0cm} 3 \times (2x + 3y = 11)} \\ \underline {(-1) \ 2 \times ( 3x - 2y = 8)} \\ {\hspace{2.5cm}13y = 17}$
$\Rightarrow y =$ $\frac{17}{13}$
Substituting in i) we get
$2x + 3 \Big($ $\frac{17}{13}$ $\Big ) = 11$
$\Rightarrow x =$ $\frac{46}{13}$
Hence the point of intersection is $\Big($ $\frac{46}{13}$ $,$ $\frac{17}{13}$ $\Big)$ $= \Big($ $\frac{5(6) + 8 ( 2) }{5+8}$ $,$ $\frac{5(5) + 8 ( -1) }{5+8}$ $\Big)$
Therefore we see that $Q \Big($ $\frac{46}{13}$ $,$ $\frac{17}{13}$ $\Big)$ divides AB in the ratio of $5:8$
$\\$
Question 12: Find the equations to the altitudes of the triangle whose angular points ate $A(2,-2), B (1, 1)$ and $C (- 1, 0)$.
Given $A(2,-2), B (1, 1)$ and $C (- 1, 0)$.
Let $AD$, $BE$ and $CF$ are the altitudes as shown in the figure.
Slope of $BC =$ $\frac{0-1}{-1-1}$ $=$ $\frac{1}{2}$
Therefore slope of $AD =$ $\frac{-1}{(1/2)}$ $= -2$
Therefore equation of $AD$:
$y - ( -2) = -2 ( x-2)$
$\Rightarrow y + 2 = - 2x + 4$
$\Rightarrow 2x + y + 2 = 0$
Slope of $AC =$ $\frac{0-(-2)}{-1-2}$ $=$ $\frac{2}{-3}$ $=$ $\frac{-2}{3}$
Therefore slope of $BE =$ $\frac{-1}{(-2/3)}$ $=$ $\frac{3}{2}$
Therefore equation of $BE$:
$y - 1 =$ $\frac{3}{2}$ $( x-1)$
$\Rightarrow 2y - 2 = 3x - 3$
$\Rightarrow 3x-2y-1=0$
Slope of $AB =$ $\frac{1-(-2)}{1-2}$ $=$ $\frac{3}{-1}$ $= -3$
Therefore slope of $CF =$ $\frac{-1}{-3}$ $=$ $\frac{1}{3}$
Therefore equation of $CF$:
$y - 0 =$ $\frac{1}{2}$ $( x+1)$
$\Rightarrow 3y = x + 1$
$\Rightarrow x-3y+1=0$
$\\$
Question 13: Find the equation of the right bisector of the line segment joining the points $(3, 4)$ and $(-1,2)$.
Given points $A(3, 4)$ and $B ( -1, 2)$
Therefore the midpoint of $AB = ($ $\frac{-1+3}{2}$ $,$ $\frac{2+4}{2}$ $) = ( 1, 3)$
Slope of $AB =$ $\frac{2-4}{-1-3}$ $=$ $\frac{-2}{-4}$ $=$ $\frac{1}{2}$
Therefore the slope of the bisector $-$ $\frac{-1}{1/2}$ $= -2$
Hence the equation of the bisector:
$y-3 = -2( x-1)$
$\Rightarrow y - 3 = -2x + 2$
$\Rightarrow 2x + y - 5 = 0$
Hence the equation of the right bisector is $2x + y - 5 = 0$
$\\$
Question 14: Find the equation of the line passing through the point $(- 3, 5)$ and perpendicular to the line joining $(2,5)$ and $(-3, 6)$.
Given points $A( 2, 5)$ and $B ( -3, 6)$
Slope of $AB =$ $\frac{6-5}{-3-2}$ $=$ $\frac{1}{-5}$ $=$ $\frac{=1}{5}$
Therefore the slope of the perpendicular line $=$ $\frac{-1}{(-1/5)}$ $= 5$
Hence the equation of the perpendicular which is passing through $( -3, 5)$ :
$y - 5 = 5 ( x+3)$
$\Rightarrow 5x - y + 20 = 0$
Hence the equation of the line is $5x - y + 20 = 0$
$\\$
Question 15: Find the equation of the right bisector of the line segment joining the points $A(1, 0)$ and $B (2, 3)$.
Given points $A(1,0)$ and $B( 2, 3)$
Mid point of $AB = \Big($ $\frac{2+1}{2}$ $,$ $\frac{3+0}{2}$ $\Big) = \Big($ $\frac{3}{2}$ $,$ $\frac{3}{2}$ $\Big)$
Slope of $AB =$ $\frac{3-0}{2-1}$ $=$ $\frac{3}{1}$ $= 3$
Therefore slope of bisector $=$ $\frac{-1}{3}$
Therefore the equation of the right bisector:
$y -$ $\frac{3}{2}$ $=$ $\frac{-1}{3}$ $\Big ( x -$ $\frac{3}{2}$ $\Big)$
$\Rightarrow 6y - 9 = - 2 \Big( x -$ $\frac{3}{2}$ $\Big)$
$\Rightarrow 6y - 9 = - 2x + 3$
$\Rightarrow 2x + 6y - 12 = 0$
$\Rightarrow x + 3y - 6 = 0$
Hence the equation of the right bisector is $x + 3y - 6 = 0$
|
# Definition 8.1 Two inequalities are equivalent if they have the same solution set. Add or Subtract the same value on both sides of the inequality.
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1 8 Inequalities Concepts: Equivalent Inequalities Linear and Nonlinear Inequalities Absolute Value Inequalities (Sections 4.6 and 1.1) 8.1 Equivalent Inequalities Definition 8.1 Two inequalities are equivalent if they have the same solution set. Operations that Produce Equivalent Inequalities. Add or Subtract the same value on both sides of the inequality. Multiply or Divide by the same positive value on both sides of the inequality. Multiply or Divide by the same negative value on both sides of the inequality AND change the direction of the inequality. Example 8. For each pair of inequalities, determine if the two inequalities are equivalent. x +x 5 and x +x x < 5 and x > 1 x x+ > 3 and x x+ 3 > 0 x > 3 and x > 3(x+) x+ You should NEVER multiply both sides of an inequality by an expression involving x that you don t know the sign of. Why? 1
2 8. Solving a Linear or Nonlinear Inequality Example 8.3 (Linear Inequality) Solve the inequality 5x+3 6 7x. Write the solution set in interval notation. Note 8.4 There are many ways to solve this inequality algebraically. We will begin by using addition and subtraction to move all the nonzero quantities to one side. This is not necessary for this inequality, but it will help us to understand the process needed for solving more complicated inequalities. Graph the equations y = 5x+3 and y = 6 7x. How can you approximate the solutions of an inequality graphically? y x
3 Thinking graphically can help us understand the algebraic procedure for solving Nonlinear Inequalities. Given an expression, such as (x+3) (x 1)(x 5), the expression is positive when the graph of y = (x+3) (x 1)(x 5) is. The expression is negative when the graph of y = (x+3) (x 1)(x 5) is. Example 8.5 (Polynomial Inequality) The graph of y = (x+3) (x 1)(x 5) is shown below. The viewing window is [ 10,10] [ 00, 100]. Use the graph to help you approximate the solutions of the inequality. (x+3) (x 1)(x 5) > 0 The algebraic procedure for solving an inequality is based on the intuition we gain from the graphical solution. Algebraic Procedure for Solving Linear and Nonlinear Inequalities 1. Use addition and subtraction to move all nonzero quantities to one side. If fractional expressions are involved, simplify so the nonzero side is a single fractional expression.. Find the zeros of the expression AND the zeros of all denominators. (If you can factor the expression, this can help. Finding a zero means find when the expression equals zero.) 3. Make a sign chart to determine if the values between the zeros from step lead to positive or negative values of the polynomial. 4. Answer the question. Example 8.6 Use the algebraic approach to solve (x+3) (x 1)(x 5) > 0. Be sure to write you answer in interval notation. 3
4 Example 8.7 (Quadratic Inequality) Solve the inequality below. Be sure to write your answer in interval notation. x +x > 8 Example 8.8 (Rational Inequality) Solve the inequality below. Be sure to write your answer in interval notation. x+3 4 x 1 Example 8.9 (Inequality Application) A computer store has determined that the cost C of ordering and storing x laser printers is given by C = x+ 300,000. If the delivery truck can bring at most 450 printers per order, x how many printers should be ordered at a time to keep the cost below \$1600? 4
5 8.3 Absolute Value Inequalities Number lines can be really insightful when working with absolute value equations and inequalities. Recall that we think of the absolute value as a distance. Example 8.10 (Another Distance Example) Solve x+3 > 5 geometrically. Be sure to write your answer in interval notation. Example 8.11 (Another Distance Example) (a) Write a distance sentence that corresponds to this number line. (b) Write an absolute value equation or inequality that corresponds to this number line. Example 8.1 (The Algebraic Approach to Absolute Values) Solve each inequality algebraically. (a) x+3 7 (b) x+ +1 < 3 5
6 8.4 Inequalities Practice Problems 1. Solve each of the inequalities below. Be sure to write your answer in interval notation.. (a) 7x 3 < 10x+ (b) x +7x 10 (c) (x )(x 3) > 0 (d) ( x)(x 3) > 0 (e) x+5 > (f) x+3 1 x 1 (g) 4 x 6 (h) 5x+7 +4 > 10 (i) x 3 9x 0 1 (j) x (a) Write a distance sentence that corresponds to this number line. (b) Write an algebraic equation or inequality that corresponds to this number line (a) Write a distance sentence that corresponds to this number line. (b) Write an algebraic equation or inequality that corresponds to this number line. 4. (Number 60 in Section 4.6 in your textbook) The length of a rectangle is 6 inches longer than its width. What are the possible widths if the area of the rectangle is at least 667 square inches? 5. (Number 55 in Section 4.6 in your textbook) A sales agent is given a choice of two different compensation plans. The first plan has no salary, but a 10% commission on total sales. The second plan has a salary of \$3000 per month, plus a % commission on total sales. What range of monthly sales will make the first plan a better choice for the sales agent? 6
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How to Take Derivatives
Author Info
Updated: April 30, 2019
The derivative is an operator that finds the instantaneous rate of change of a quantity. Derivatives can be used to obtain useful characteristics about a function, such as its extrema and roots.[1] Finding the derivative from its definition can be tedious, but there are many techniques to bypass that and find derivatives more easily.
Part 1 of 2: Preliminaries
1. 1
Understand the definition of the derivative. While this will almost never be used to actually take derivatives, an understanding of this concept is vital nonetheless.
• Recall that the linear function is of the form ${\displaystyle y=mx+b.}$ To find the slope ${\displaystyle m}$ of this function, two points on the line are taken, and their coordinates are plugged into the relation ${\displaystyle m={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}.}$ Of course, this can only be used with linear graphs.
• For nonlinear functions, the line will be curved, so taking the difference of two points can only give the average rate of change between them. The line that intersects these two points is called the secant line, with a slope ${\displaystyle m={\frac {f(x+\Delta x)-f(x)}{\Delta x}},}$ where ${\displaystyle \Delta x=x_{2}-x_{1}}$ is the change in ${\displaystyle x,}$ and we have replaced ${\displaystyle y}$ with ${\displaystyle f(x).}$ This is the same equation as the one before.
• The concept of the derivatives comes in when we take the limit ${\displaystyle \Delta x\to 0.}$ When this happens, the distance between the two points shrinks, and the secant line better approximates the rate of change of the function. When we do send the limit to 0, we end up with the instantaneous rate of change and obtain the slope of the tangent line to the curve (see animation above).[2] Then, we end up with the definition of the derivative, where the prime symbol denotes the derivative of the function ${\displaystyle f.}$
• ${\displaystyle f^{\prime }(x)=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}}$
• Finding the derivative from this definition stems from expanding the numerator, canceling, and then evaluating the limit, since immediately evaluating the limit will give a 0 in the denominator.
2. 2
Understand the derivative notation. There are two common notations for the derivative, though there are others.
• Lagrange's Notation. In the previous step, we used this notation to denote the derivative of a function ${\displaystyle f(x)}$ by adding a prime symbol.
• ${\displaystyle f^{\prime }(x)}$
• This notation is pronounced "${\displaystyle f}$ prime of ${\displaystyle x.}$" To form higher order derivatives, simply add another prime symbol. When derivatives of fourth or higher order are taken, the notation becomes ${\displaystyle f^{(4)}(x),}$ where this represents the fourth derivative.
• Leibniz's Notation. This is the other commonly used notation, and we will use it in the rest of the article.
• ${\displaystyle {\frac {\mathrm {d} f}{\mathrm {d} x}}}$
• (For shorter expressions, the function can be placed in the numerator.) This notation literally means "the derivative of ${\displaystyle f}$ with respect to ${\displaystyle x.}$" It may be helpful to think of it as ${\displaystyle {\frac {\Delta y}{\Delta x}}}$ for values of ${\displaystyle x}$ and ${\displaystyle y}$ that are infinitesimally different from each other. When using this notation for higher derivatives, you must write ${\displaystyle {\frac {\mathrm {d} ^{2}f}{\mathrm {d} x^{2}}},}$ where this represents the second derivative.
• (Note that there "should" be parentheses in the denominator, but no one ever writes them, since everyone understands what we mean without them anyway.)
Part 2 of 2: Basic Techniques
Using the Definition
1. 1
Substitute into the function. For this example, we will define ${\displaystyle f(x)=2x^{2}+6x.}$
• {\displaystyle {\begin{aligned}f(x+\Delta x)&=2(x+\Delta x)^{2}+6(x+\Delta x)\\&=2(x^{2}+2x\Delta x+(\Delta x)^{2})+6x+6\Delta x\\&=2x^{2}+4x\Delta x+2(\Delta x)^{2}+6x+6\Delta x.\end{aligned}}}
2. 2
Substitute the function into the limit. Then evaluate the limit.
• {\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} x}}f(x)&=\lim _{\Delta x\to 0}{\frac {(2x^{2}+4x\Delta x+2(\Delta x)^{2}+6x+6\Delta x)-(2x^{2}+6x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {4x\Delta x+2(\Delta x)^{2}+6\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\Delta x(4x+2\Delta x+6)}{\Delta x}}\\&=\lim _{\Delta x\to 0}4x+2\Delta x+6\\&=4x+6.\end{aligned}}}
• This is a lot of work for such a simple function. We will see that there are plenty of derivative rules to skirt past this type of evaluation.
• You can find the slope anywhere on the function ${\displaystyle f(x)=2x^{2}+6x.}$ Simply plug in any x value into the derivative ${\displaystyle {\frac {\mathrm {d} f(x)}{\mathrm {d} x}}=4x+6.}$
The Power Rule
1. 1
Use the power rule[3] when is a polynomial function of degree n. Multiply the exponent with the coefficient and bring down the power by one.
• The formula is ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(x^{n})=nx^{n-1}.}$
• Although the intuitive method seems to only apply to natural number exponents, it can be generalized to all real numbers; that is, ${\displaystyle n\in \mathbb {R} .}$
2. 2
Use the previous example. ${\displaystyle f(x)=2x^{2}+6x.}$ Remember that ${\displaystyle x=x^{1}.}$
• {\displaystyle {\begin{aligned}f(x)&=2x^{2}+6x\\{\frac {\mathrm {d} }{\mathrm {d} x}}f(x)&=(2)2x^{2-1}+(1)6x^{1-1}\\&=4x+6.\end{aligned}}}
• We have used the property that the derivative of a sum is the sum of the derivatives (technically, the reason why we can do this is because the derivative is a linear operator). Obviously, the power rule makes finding derivatives of polynomials much easier.
• Before going on, it is important to note that the derivative of a constant is 0, because the derivative measures the rate of change, and no such change exists with a constant.
Higher Order Derivatives
1. 1
Differentiate again. Taking a higher order derivative of a function just means you take the derivative of the derivative (for order of 2). For example, if it asks you to take the third derivative, just differentiate the function three times.[4] For polynomial functions of degree ${\displaystyle n,}$ the ${\displaystyle n+1}$ order derivative will be 0.
2. 2
Take the third derivative of the previous example .
• {\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} x}}f(x)&=4x+6\\{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}f(x)&=4\\{\frac {\mathrm {d} ^{3}}{\mathrm {d} x^{3}}}f(x)&=0\end{aligned}}}
• In most applications of derivatives, especially in physics and engineering, you will at most differentiate twice, or perhaps three times.
The Product and Quotient Rules
1. 1
See this article for a full treatment on the product rule. In general, the derivative of a product does not equal the product of the derivatives. Rather, each function "gets its turn" to differentiate.
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(fg)={\frac {\mathrm {d} f}{\mathrm {d} x}}g+f{\frac {\mathrm {d} g}{\mathrm {d} x}}}$
2. 2
Use the quotient rule to take derivatives of rational functions. As with products in general, the derivative of a quotient does not equal the quotient of the derivatives.
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left({\frac {f}{g}}\right)={\frac {g{\frac {\mathrm {d} f}{\mathrm {d} x}}-f{\frac {\mathrm {d} g}{\mathrm {d} x}}}{g^{2}}}}$
• A useful mnemonic for the numerator of the derivative is "Down-dee-up, up-dee-down," since the minus sign means the order matters.
• For example, consider the function ${\displaystyle f(x)={\frac {x^{2}+2x-21}{x-3}}.}$ Let ${\displaystyle g(x)=x^{2}+2x-21}$ and ${\displaystyle h(x)=x-3.}$ Then use the quotient rule.
• {\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} x}}\left({\frac {g}{h}}\right)&={\frac {(x-3)(2x+2)-(x^{2}+2x-21)(1)}{(x-3)^{2}}}\\&={\frac {x^{2}-6x+15}{(x-3)^{2}}}\end{aligned}}}
• Make sure your algebra is up to par. Derivatives involving quotients like these can quickly become cumbersome in terms of the algebra involved. This means you should be comfortable with factoring out constants and keeping track of negative signs.
The Chain Rule
1. 1
Use the chain rule[5] for nested functions. For example, consider the scenario where ${\displaystyle z(y)}$ is a differentiable function of ${\displaystyle y}$ and ${\displaystyle y(x)}$ is a differentiable function of ${\displaystyle x.}$ Then there is a composite function ${\displaystyle z(y(x)),}$ or ${\displaystyle z}$ as a function of ${\displaystyle x,}$ that we can take the derivative of.
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}z(y(x))={\frac {\mathrm {d} z}{\mathrm {d} y}}{\frac {\mathrm {d} y}{\mathrm {d} x}}}$
• As with the product rule, this works with any number of functions; hence the "chain" rule. Here, an easy way to see how this works is if one imagines a ${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} y}}}$ inserted between ${\displaystyle {\frac {\mathrm {d} z}{\mathrm {d} x}}.}$
2. 2
Consider the function . Notice that this function can be decomposed into two elementary functions, ${\displaystyle g(x)=2x^{4}-x}$ and ${\displaystyle h(g)=g^{3}.}$ Then, we want to find the derivative of the composition ${\displaystyle f(x)=h(g(x)).}$
• Use the chain rule ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}h(g(x))={\frac {\mathrm {d} h}{\mathrm {d} g}}{\frac {\mathrm {d} g}{\mathrm {d} x}}.}$ We have now written the derivative in terms of derivatives that are easier to take. Then,
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}h(g(x))=3(2x^{4}-x)^{2}(8x^{3}-1).}$
• With practice, you will see that applying the chain rule is easiest if you "peel away the onion." The first layer is everything inside the parentheses, cubed. The second layer is the function inside the parentheses. When dealing with more complex functions, this way of thinking helps to keep yourself on track and not get lost in what functions are taken with respect to what variables, etc.
Other Important Derivatives
1. 1
See this article for a full treatment on implicit differentiation. Understanding the chain rule is a must in order to implicitly differentiate.
2. 2
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}e^{x}=e^{x}}$
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}a^{x}=a^{x}\ln a}$
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\ln x={\frac {1}{x}}}$
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\log _{a}x={\frac {1}{x\ln a}}}$
3. 3
Memorize basic trigonometric derivatives and how to derive them.
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\sin x=\cos x}$
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\cos x=-\sin x}$
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\tan x=\sec ^{2}x}$
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\cot x=-\csc ^{2}x}$
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\sec x=\sec x\tan x}$
• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\csc x=-\csc x\cot x}$
Community Q&A
Search
• Question
What is the derivative of x^0.09?
Pimemorized
By the power rule, you would first multiply the whole equation by the exponent, which is 0.09 so you get 0.09x^0.09. Then, you subtract the power by one, so it becomes 0.09x^(-.91).
• Question
What is the second derivative of 1/x?
2 x^-3
• Question
What is the derivative of a logarithmic function?
Pimemorized
The derivative of a log function is the derivative of the function divided by the function itself. For example, the derivative of log(x) would be the derivative of x is 1 divided by x, and so log(x) = 1/x.
• Question
What is the derivative of X power of infinite X?
It is infinity.
200 characters left
Tips
• Every technique outlined in this article on calculating derivatives can be verified by a proper use of the definition of the derivative. If, for example, the power rule seems sketchy to you, try and recover the formula using the definition.
Thanks!
• Know your calculator well; try different functions on your calculator to learn their uses. It is especially useful to know how to use the derivative functions of your calculator if they exist.
Thanks!
• Practice the product rule, chain rule, and especially implicit differentiation, as these are more difficult to differentiate and are widely used outside mathematics.
Thanks!
• Now is the time to start writing down every step in your calculations if you hadn't already been doing that. Not only is this required on some exams, but this is a huge help in clearing confusion. Just one missed negative sign or mistake, which are easy to make with more complex functions, will give you the wrong answer. And if you don't know where you went wrong, it is easier to trace back your steps.
Thanks!
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Warnings
• Some students will be tempted to use programs on their calculators to take derivatives. While these programs are very useful for confirming your answers, you should not rely on these. Make sure you understand the concepts of deriving and are able to do it yourself.
Thanks!
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 28 people, some anonymous, worked to edit and improve it over time. Together, they cited 5 references. This article has also been viewed 538,556 times.
Co-authors: 28
Updated: April 30, 2019
Views: 538,556
Categories: Calculus
Article SummaryX
To take the derivative of a function by using the definition, substitute x plus delta x into the function for each instance of x. Then, substitute the new function into the limit, and evaluate the limit to find the derivative. If you're finding the derivative of a polynomial with a function to the degree of n, use the power rule by multiplying the coefficient by the exponent and subtracting 1 from the exponent to lower the power by one. After that, simplify the limit to find the derivative of the equation. For tips on how to do high-order derivatives and use the product and quotient rule, read on!
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# S k i l l i n A R I T H M E T I C
Lesson 2 Section 2
# Place valuePositional numeration
Back to Section 1
THE ORGANIZING PRINCIPLE in our system of naming and writing numbers are the powers of 10. When we write '2,364' for example, we mean the sum of
2 Thousands + 3 Hundreds + 6 Tens + 4 Ones.
This illustrates that, starting with the Ones on the right, we have chosen the powers of 10 to be the units. There are 2 of that unit (Thousands), plus 3 of that unit (Hundreds), plus 6 of that one (Tens), plus 4 of those (Ones). When we write the number, however, we omit the names of the units and the + signs, and just write
2,364.
After the Ones on the right, each place has for its value the next power of 10. We call such a system positional numeration.
5. To which place does each digit belong? Equivalently, what is the unit at each position? Starting with the Ones on the right, each place belongs to the next power of 10.
Example 1. In this number,
139,072,658
the 0 is in which place?
For, in each class of three digits, there are Ones, Tens, and Hundreds.
0 is in the class of thousands and in the Hundreds place. The power of 10 at that position is Hundred thousands.
Example 2. In this number,
386,214,035
how many Ten millions are there? That is, which digit is in the ten millions place?
Answer. 8. For, on counting from the right, the millions are the third group of three digits, 386. The Tens place is the middle one (Ones, Tens, Hundreds). There are 8 Ten millions.
Place value versus absolute value of a digit
In addition to speaking of a digit being "in" a place, we also speak of the place value of the digit itself. In this number,
6,666
each digit has the same absolute or invariable value 6, but a different place value. 6 on the extreme left has the place value 6000; the next 6 has the value 600; the next, 60; and the last, 6.
Expanded form
The numeral for every whole number stands for a sum.
364 = 3 Hundreds + 6 Tens + 4 Ones.
(Even a single digit stand for a sum: 5 = 1 + 1 + 1 + 1 + 1.)
What is written above is called the expanded form of 364.
6. What does it mean to write a number in expanded form? It means to write the sum that the number indicates, and therefore to name the unit at each digit's place.
Example 3. Write 6,325 in expanded form.
6,325 = 6 Thousands + 3 Hundreds + 2 Tens + 5 Ones.
In practice, however, it is often more useful to expand the number in this way:
6,325 = 6,000 + 300 + 20 + 5.
Example 4. Write the expanded form of 10,000.
Answer. 10,000 = 1 Ten-thousand + 0 Thousands + 0 Hundreds + 0 Tens + 0 Ones.
Example 5. 42 = 40 + 2 -- no matter what the unit. 42 eggs = 40 eggs + 2 eggs 42 tens = 40 tens + 2 tens 42 hundreds = 40 hundreds + 2 hundreds
And so on. For there is no "42" apart from 42 units, even though we do not say the word units.
The following question is to prepare for the standard written methods of addition and subtraction. The answer follows from the fact that each digit
6,325
has a place value ten times the digit to its right.
For, each power of 10 is ten times the one to its right:
1000 100 10 1
1000 is made up of ten 100's.
100 is made up of ten 10's.
10 is made up of ten 1's.
And so on.
7. What is the relationship between units of adjacent place value? 1000 100 10 1 Ten units of lower place value can be composed -- grouped together -- to make one unit of the next higher value. Equivalently: One unit of higher place value can be decomposed -- broken up -- into ten units of the next lower value.
Ten 1's can be composed into one 10.
Ten 10's can be composed into one 100.
Ten 100's can be composed into one 1000.
And so on.
Conversely:
1000 100 10 1
One 1000 can be decomposed into ten 100's.
One 100 can be decomposed into ten 10's.
One 10 can be decomposed into ten 1's.
1000 100 10 1
We will see this when we come to regrouping in addition and subtraction.
Rounding off
8. How do we round off, or approximate, a whole number to a given place? 12671300 Look at the digit to the right of the given place: hundreds for example (2). If the digit to the right is a 5 or greater, add 1 to the given place. If it is less than 5, leave the given place unchanged. In either case, replace all the digits to the right of the given place with 0's.
Example 6. Round off 6,528 to the nearest ten.
(The wavy equal signmeans "is approximately equal to.")
2 is in the tens place. To round off to the nearest ten, look at the digit to the right: 8 (greater than 5). Therefore, add 1 to the tens place. Replace 8 with 0.
Example 7. Round off 6,528 to the nearest hundred.
5 is in the hundreds place. To round off to the nearest hundred, look at the digit to the right: 2 (less than 5). Therefore, leave the hundreds place unchanged. Replace 28 with 00.
Example 8. Round off 6,528 to the nearest thousand.
6 is in the thousands place. To round off to the nearest thousand, look at the digit to the right: 5. Therefore, add 1 to the thousands place. Replace 528 with 000.
Example 9. Round off 79,521 to the nearest thousand.
9 is in the thousands place. To round off to the nearest thousand, look at the digit to the right: 5. Therefore, add 1 to 79 -- it becomes 80. Replace 521 with 000.
To round off decimals, see Lesson 11.
At this point, please "turn" the page and do some Problems.
or
Continue on to the next Section.
Section 1 of this Lesson
Please make a donation to keep TheMathPage online.
Even \$1 will help.
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# Sppose that the probability a new vaccine will protect adults from a certain disease is 0.8. The vaccine is given to 200 adults. What is the probability that more than 170 of those adults will be protected by the vaccine?
This question is in the category of a binomial experiment with n trials equal to 200 and the probability of success p equal to 0.8, and the number of successes r equal to 170. There is a formula for the binomial distribution which can be used to calculate the probability. But this would be very time consuming and easy to make some errors in calculation. But there is a simpler way to solve this problem, and that is by using the normal distribution to approximate the binomial distribution. However, certain conditions must be present in order to use this method.
First, we must consider the binomial distribution with n = number of trials, p = probability of success on a single trial, q = 1 - p = probability of failure on a single trial, and r = number of successes. Then if np > 5 and nq > 5, then r has a binomial distribution which you can approximate with the normal distribution. The mean is approximated by np and the standard deviation is approximated by the square root of npq. This approximation becomes more accurate as the sample size n increases.
Now we can put this all together with an example. Suppose the owner of a hotel needs to install new air conditioner units to 25 of the rooms. From past experience with a noted brand, he knows that the air conditioner unit is guaranteed for 5 years, and the probability that it will last 10 years is 0.35. What is the probability that 12 or more of the units will last more than 10 years?
In this problem, n = 25, p = 0.35, q = 0.65. We want the probability that r is greater than 10. We can use the normal approximation to the binomial since np = 8.75, which is greater than 5. Also, nq = 16.25, which is greater than 5. The mean is 8.75 and the standard deviation is square root of npq, which is 2.38. Using the normal distribution, we calculate the z to be (9.5 - 8.75)/2.38 = 0.32. Note that we use 9.5 instead of 10 because of a continuity correction, which converts r to a continuous normal random variable x by subtracting 0.5, since r in this case is the left-point of the interval. Now we find the probability that z > 0.32, which is 0.3745. This value is found by using a table for the areas of a standard normal distribution.
This guide should help assist students having difficulties understanding the normal approximation to the binomial.
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# equations
http://www.cimt.org.uk/projects/mepres/book8/bk8i12/bk8_12i4.htm
In this section we solve linear equations where more than one step is needed to reach the solution.
Methods of solution vary from one equation to another, but there are two main groups:
(i) equations where the unknown letter only appears once (like 2x + 7 = 13), and
(ii) equations where the unknown letter appears more than once (like 5x – 3 = 2x+3).
### Equations where the unknown letter appears once
Example 1
Look at this equation:
5x + 2 = 17
To solve this, we need to get the x on its own.
Think about what has happened to the x: it has been multiplied by 5 and has had 2 added to it.
To get x on its own, we need to reverse this process, by subtracting 2 and dividing by 5.
Here is the working:
5x + 2 = 17 [subtract 2 from both sides] 5x + 2 – 2 = 17 – 2 [note how the subtract 2 cancels out the add 2] 5x = 15 [divide both sides by 5] 5x ÷ 5 = 15 ÷ 5 [note how the divide by 5 cancels out the multiply by 5] x = 3
Always check your solution works in the original equation: 5 × 3 + 2 = 17
Example 2
Look at this equation: p/2+ 3 = 7
Think about what has happened to the p this time: it has been divided by 2 and has had 3 added to it.
To get p on its own, we again need to reverse this process, by subtracting 3 and multiplying by 2. Here is the working:
p/2+ 3 = 7 [subtract 3 from both sides]
p/2 = 4 [note how the subtract 3 cancels out the add 3
p/2 x 2 = 4 x 2 [multiply both sides by 2]
p= 8 [note how the multiply by 2 cancels out the divide by 2]
Remember to check your solution works in the original equation: 8 ÷ 2 + 3 = 7
Example 3
Look at this equation:
5(7 + 2x) = 65
This time, three things have happened to the x: it has been multiplied by 2, it has had 7 added to it, and it
has then been multiplied by 5. To reverse this process, we divide by 5, subtract 7 and then divide by 2.
Here is the working:
5(7 + 2x) = 65 [divide by 5 on both sides] 5(7 + 2x = 13 [subtract 7 on both sides] 5(7 + 2x = 6 [divide by 2 on both sides] 5(7 + 2x = 3
Always check your solution works in the original equation: 5 × (7 + 2 × 3) = 65
Note that the last example could also have been solved by multiplying out the brackets first.
### Equations where the unknown letter appears more than once
Example 4
Look at this equation:
6x – 2 = 4x + 8
To solve this, we need to get the all the “x“s on one side.
There are more “x“s on the left, so we will aim to get all the “x“s on the left and all the “numbers” on the right.
To remove the “4x” on the right, we will subtract 4x on both sides.
To remove the “– 2” on the left, we will add 2 to both sides.
Here is the working:
6x – 2 = 4x + 8 [subtract 4x from both sides] [6x – 4x = 2x] 2x – 2 = 4x + 8 [add 2 to both sides] [8 + 2 = 10] 2x = 4x + 10 [divide both sides by 2] 2x = 4x + 5
As always, check your solution works in the original equation: 6 × 5 – 2 = 4 × 5 + 8
Note that many equations can be solved in more than one way, but all the methods will give the same solution.
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# Interpret Features of a Function
Videos and lessons to help High School students understand how a function models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship.
Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.
• Distinguish linear, quadratic, and exponential relationships based on equations, tables, and verbal descriptions.
• Given a function in a table or in algebraic or graphical form, identify key features such as x- and y-intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; and end behavior.
• Use key features of an algebraic function to graph the function.
Common Core: HSF-IF.B.4
F.IF.4 - Interpret Key Features of the Graph of a Function
The function f(t) = -5t2 + 20t + 60 models the approximate height of an object t seconds after it is launched.
a) What is its maximum height?
b) How many seconds does it take to reach its maximum height?
c) How many seconds does it take for the object to return to the ground after the launch?
Interpreting intercepts of linear functions
After several days of camping and long distance bicycle riding, you decide to ride straight home at a constant rate. After 2 hours of riding, you are 10 km from home and after 4 hours of riding, you are 5 km from home. Once you begin to ride home, in how many hours would you finish your bike ride and arrice at home?
x and y intercepts of exponential functions
How to find the x and y intercepts of an exponential function from a graph, equation, and word description.
End Behavior of Linear and Exponential Functions
How to find end behaviors of linear and exponential functions from graphs, tables, and equations.
Interpreting Linear Functions (Part 1)
Interpreting graphs of linear functions, including horizontal and vertical asymptotes, and rate of change.
Interpreting Linear Functions (Part 2)
Interpreting graphs of linear functions, including horizontal and vertical asymptotes, and rate of change.
Interpreting features of functions (Example 1).
Interpreting features of functions (Example 2).
Interpreting Graphs of Exponential Functions
In this video one is supplied with a real world situation and a graph corresponding to that situation. The graph is then used to answer a series of questions.
Recognizing features of functions (Example 1).
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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Question 157345
In order to find the vertex, we first need to find the x-coordinate of the vertex.
To find the x-coordinate of the vertex, use this formula: {{{x=(-b)/(2a)}}}.
From {{{y=-2x^2+2x+9}}}, we can see that {{{a=-2}}}, {{{b=2}}}, and {{{c=9}}}.
{{{x=(-(2))/(2(-2))}}} Plug in {{{a=-2}}} and {{{b=2}}}.
{{{x=(-2)/(-4)}}} Multiply 2 and {{{-2}}} to get {{{-4}}}.
{{{x=1/2}}} Reduce.
So the x-coordinate of the vertex is {{{x=1/2}}}. Note: this means that the axis of symmetry is also {{{x=1/2}}}.
Now that we know the x-coordinate of the vertex, we can use it to find the y-coordinate of the vertex.
{{{y=-2(1/2)^2+2(1/2)+9}}} Plug in {{{x=1/2}}}.
{{{y=-2(1/4)+2(1/2)+9}}} Square {{{1/2}}} to get {{{1/4}}}.
{{{y=-1/2+2(1/2)+9}}} Multiply {{{-2}}} and {{{1/4}}} to get {{{-1/2}}}.
{{{y=-1/2+1+9}}} Multiply {{{2}}} and {{{1/2}}} to get {{{1}}}.
{{{y=19/2}}} Combine like terms.
So the y-coordinate of the vertex is {{{y=19/2}}}.
So the vertex is *[Tex \LARGE \left(\frac{1}{2},\frac{19}{2}\right)].
Also, since the leading coefficient is negative, this means that the vertex is at a maximum.
----------------------------------------------------
So using this info we can answer the following:
The x coordinate of the vertex is {{{1/2}}}
The y coordinate of the vertex is {{{19/2}}}
The equation of the line of symmetry is {{{x=1/2}}}
The maximum or minimum of f(x) is {{{f(x)=19/2}}}
The value of f(x)=19/2 is a maximum
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This is an example to help prepare students to tackle the Fundamental Theorem of Calculus (FTC). Use it after the lesson on Riemann sums and the definition of the definite integral, but before the FTC derivation.
Consider the area, A, between the graph of $f\left( x \right)=\cos \left( x \right)$ and the x-axis on the interval $\left[ 0,\tfrac{\pi }{2} \right]$. Set up a Riemann sum using the general partition:
$0={{x}_{0}}<{{x}_{1}}<{{x}_{2}}<{{x}_{3}}<\cdots <{{x}_{n-2}}<{{x}_{n-1}}<{{x}_{n}}=\tfrac{\pi }{2}$
$\displaystyle A=\int_{0}^{{\scriptstyle{}^{\pi }\!\!\diagup\!\!{}_{2}\;}}{\cos \left( x \right)dx}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\cos \left( {{c}_{i}} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right)}$
Since we can choose the value of ${{c}_{i}}$ any way we want, let’s take the same intervals and use ${{c}_{i}}$ the number guaranteed by the Mean Value Theorem for the function $F\left( x \right)=\sin \left( x \right)$ on the each sub-interval interval. That is, on each sub-interval at $x={{c}_{i}}$
$\left. \frac{d}{dx}\sin \left( x \right) \right|_{x={{c}_{i}}}^{{}}=\cos \left( {{c}_{i}} \right)=\frac{\sin \left( {{x}_{i}} \right)-\sin \left( {{x}_{i-1}} \right)}{\left( {{x}_{i}}-{{x}_{i-1}} \right)}$
Then, substituting into the Riemann sum above
$\displaystyle A=\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\cos \left( {{c}_{i}} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right)}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\frac{\sin \left( {{x}_{i}} \right)-\sin \left( {{x}_{i-1}} \right)}{\left( {{x}_{i}}-{{x}_{i-1}} \right)}\left( {{x}_{i}}-{{x}_{i-1}} \right)}$
$\displaystyle A=\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\left( \sin \left( {{x}_{i}} \right)-\sin \left( {{x}_{i-1}} \right) \right)}$
Now writing out the terms we have a telescoping series:
$\displaystyle A=\left( \sin \left( {{x}_{1}} \right)-\sin \left( {{x}_{0}} \right) \right)+\left( \sin \left( {{x}_{2}} \right)-\sin \left( {{x}_{1}} \right) \right)+\left( \sin \left( {{x}_{3}} \right)-\sin \left( {{x}_{2}} \right) \right)+$
$\displaystyle \cdots +\left( \sin \left( {{x}_{n-1}} \right)-\sin \left( {{x}_{n-2}} \right) \right)+\left( \sin \left( {{x}_{n}} \right)-\sin \left( {{x}_{n-1}} \right) \right)$
$\displaystyle A=\sin \left( {{x}_{n}} \right)-\sin \left( {{x}_{0}} \right)$
$\displaystyle A=\sin \left( \tfrac{\pi }{2} \right)-\sin \left( 0 \right)=1$
As you can see, this is really just the derivation of the FTC applied to a particular function. Now the students should have a better idea of what’s going on when you solve the problem in general, i.e. when you prove the FTC.
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# Lesson note #
## Statistical Inference Testing of Hypothesis
Type I Error:
Rejection of the null hypothesis when it is true is called a type I error.
Type II Error:
Acceptance of the null hypothesis when it is false is called a type II error. Decision of the test for the Null The Null Hypothesis is Hypothesis True False Accept Correct decision Incorrect decision
Type II Error
Reject Incorrect decision .Type I Error Correct decision
## Test Concerning Mean
One and Two tailed Tests:
A test procedure is called a one tailed test procedure if the alternative hypothesis is one sided. The test will be two tailed if the alternative hypothesis is two sided.
Example:
Let a specified value of population mean is 45. Construct the null and alternative hypothesis for the following questions; a) Do the sample data provide sufficient evidence to indicate that the population mean is greater than 45. H 0 : = 45 leads to one tailed test (or right tailed test) H A : > 45 b) Do the sample data provide sufficient evidence to indicate that the population mean is less than 45. H 0 : = 45 leads to one tailed test (or left tailed test) H A : < 45 c) Do the sample data provide sufficient evidence to indicate that the population mean is not equal to 45. H 0 : = 45 leads to two tailed test (or both tailed test) H A : 45
## Level of Significance and Power of a Test:
The probability of making type I error is called the level of significance of the test denoted by . The probability of making a type two error is denoted by and (1 - ) is called the power of the test. The Null Hypothesis is True Correct decision P(Correct decision)=1- . Incorrect decision P(type I error)= . = level of significance of the test False Incorrect decision Type two error P(type II error) = Correct decision P(correct decision)=1- =Power of test.
## Rejection Rule and Conclusion:
Points to Note i) Rejection of H 0 indicates that an extremely unlikely sample has been drawn which implies that H 0 is very likely to be false. ii) Failing to reject H 0 does not prove that H 0 is true. It implies that H 0 may be true. iii) In testing hypotheses, the assumption is always made that the sample used in the test process is a random sample. iv) It is assumed that the sampling distribution of the test statistic is known. H 0 , H A and . are determined before the test is carried out. v)
## Formal Testing Procedure:
A hypothesis testing procedure involves the following six steps: Step 1:Set up the null and alternative hypothesis ( H 0 & H A ). The alternative hypothesis decides whether the test is one tailed or two tailed. Step 2:Specify a level of significance ( . ). Step 3:Select an appropriate test statistic (z or t-test) and compute the value of the test statistic using sample data assuming null hypothesis to be true. Step 4:Dtermine the critical values and the critical region of the test (using z or t table). Step 5:State a rule to reject the null hypothesis. Step 6:Decide if the null hypothesis is to be rejected and write the conclusion of the test.
## Testing the Mean of a Population:
1.
Null Hypothesis: H 0 : = 0 ( 0 Alternative Hypothesis: H A : 0 OR H A : > 0 H A : < 0 OR 2. Significance level: . = 0.01 or 0.05 3. Test Statistic: x 0 z= n x 0 z= s OR n x 0 t= s OR n with d. f. = n-1)
is the hypothesized value of ) (Two tailed test) (One tailed test) (One tailed test) or 0.10 etc.
(when
is known)
(when
## is unknown and n 30)
(when
is unknown and n
<
30
The test statistic (z or t ) is decided according to the following table. Normal Population Non-Normal Population Sample Size known unknown known unknown (n) n>30 Z - test Z - test Z - test None large sample n small Z - test t - test None None sample Thus t-test is used only if i) The population is normal is unknown ( but s is known or can be computed) ii) iii) n < 30
4. Critical Region: The critical region shown in the curve of normal distribution (z table) or t-distribution. The value find out by z or t table, and the area right and left to that value according to the alternative hypothesis in the case of one tailed test and the area to either left or right according to the alternative hypothesis in one tailed test to be shaded. These values taken by the table.
If the value of Z or t calculated lie in the shaded region or considered as critical region then null hypothesis will reject otherwise accept. Test Two tail One tail
.
2
## 0.10 1.645 1.28
Z 0
Critical values of t for (n-1) degree of freedom are obtained from the t-table.
5. Rejection Rule:
If Z cal > Z tab ; reject H 0 and accept H A OR t cal > t tab ; reject H 0 and accept H A If
Z cal and t cal are calculated values of the test statistic. Z tab and t tab are critical or tabulated values of the test statistic.
6. Conclusion:
Reject or do not reject the null hypothesis on the basis of the above rejection rule. The test will be significant if H 0 is rejected otherwise the test will be insignificant.
## Test Concerning Variance
In testing hypothesis concerning Variance the basic six steps of hypothesis will remain same only the difference of distribution and critical region defining procedure. In testing concerning variance, we use 2 -distribution (Chi square distribution).
For finding the value from table; the degree of freedom will be = n 1
(n 1) s 2 = 2
2
Note: it is incomplete. Write about critical region and rejection rule from page 320, 3rd paragraph, Walpole
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# Position of a Point with Respect to a Circle
### Position of a Point with Respect to a Circle:
#### 1. Circle Equation:
• The equation of a circle in the Cartesian plane is ${x}^{2}+{y}^{2}={r}^{2}$ where $\left(x,y\right)$ are coordinates and $r$ is the radius.
#### 2. Point-Inside, Outside, or On the Circle:
• Inside the Circle: A point $\left(x,y\right)$ is inside the circle if ${x}^{2}+{y}^{2}<{r}^{2}$.
• Outside the Circle: A point $\left(x,y\right)$ is outside the circle if ${x}^{2}+{y}^{2}>{r}^{2}$.
• On the Circle: A point $\left(x,y\right)$ lies on the circle if ${x}^{2}+{y}^{2}={r}^{2}$.
#### 3. Distance from the Circle's Center:
• Distance Formula: For a point $\left({x}_{1},{y}_{1}\right)$ and the circle's center $\left(h,k\right)$, the distance is calculated as $\sqrt{\left({x}_{1}-h{\right)}^{2}+\left({y}_{1}-k{\right)}^{2}}$.
• Comparison to Radius: If the calculated distance is:
• Less than the radius $r$, the point is inside the circle.
• Greater than $r$, the point is outside the circle.
• Equal to $r$, the point lies on the circle.
#### 4. Using Inequality to Determine Position:
• Inequality Test: For a point $\left({x}_{1},{y}_{1}\right)$, compare ${x}_{1}^{2}+{y}_{1}^{2}$ with ${r}^{2}$.
• Inside or Outside Check: If ${x}_{1}^{2}+{y}_{1}^{2}<{r}^{2}$, the point is inside; if ${x}_{1}^{2}+{y}_{1}^{2}>{r}^{2}$, it's outside.
• On the Circle Check: If ${x}_{1}^{2}+{y}_{1}^{2}={r}^{2}$, the point lies on the circle.
#### 5. Geometric Interpretation:
• Visualizing the Circle: Plotting the circle on the Cartesian plane helps understand the point's position relative to the circle.
• Understanding Quadrants: Analyzing points in relation to the circle's center and quadrants helps determine their positions.
#### 6. Applications:
• Geometry and Trigonometry: Used in understanding geometric relationships and applying trigonometric concepts.
• Real-world Scenarios: Applicable in navigation, mapping, and determining object positions.
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# Fraction calculator
This fraction calculator performs all fraction operations - addition, subtraction, multiplication, division and evaluates expressions with fractions. It also shows detailed step-by-step informations.
## The result:
### 1/3 + 2/4 + 1/2 = 4/3 = 1 1/3 ≅ 1.3333333
The spelled result in words is four thirds (or one and one third).
### How do we solve fractions step by step?
1. Add: 1/3 + 2/4 = 1 · 4/3 · 4 + 2 · 3/4 · 3 = 4/12 + 6/12 = 4 + 6/12 = 10/12 = 2 · 5/2 · 6 = 5/6
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(3, 4) = 12. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 3 × 4 = 12. In the following intermediate step, cancel by a common factor of 2 gives 5/6.
In other words - one third plus two quarters is five sixths.
2. Add: the result of step No. 1 + 1/2 = 5/6 + 1/2 = 5/6 + 1 · 3/2 · 3 = 5/6 + 3/6 = 5 + 3/6 = 8/6 = 2 · 4/2 · 3 = 4/3
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(6, 2) = 6. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 6 × 2 = 12. In the following intermediate step, cancel by a common factor of 2 gives 4/3.
In other words - five sixths plus one half is four thirds.
### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right.
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A+ » VCE Blog » VCE Maths Methods Blog » Polynomial Functions [Video Tutorial]
# Polynomial Functions [Video Tutorial]
This tutorial covers material encountered in chapter 4 of the VCE Mathematical Methods Textbook, namely:
• The turning point form and axis of symmetry of a quadratic polynomial
• The quadratic formula and the discriminant (extremely useful!)
• Dividing polynomials via long division and equating coefficients
• The Remainder and Factor theorem
• Study of polynomials in general
## Worksheet
Q1. Sketch the graph of each of the following polynomials, clearly indicating the axis intercepts and the coordinates of the vertex:
(a) h(x)=2(x-4)^2+1
(b) j(x)=x^2-3x+6
(c) b(x)=x^2-1
(d) c(x)=3x^2+9x-7
Q2. Sketch the graph of each of the following cubic polynomials, clearly indicating the axis intercept(s) and the coordinates of the zero gradient:
(a) f(x)=2(x-1)^3-16.
(b) h(x)=3(x-4)^3+7
(c) g(x)=-2(x+2)^3-9
Q3. Express each of these polynomials in turning point form:
(a) x^2+2x
(b) x^2-6x+8
(c) 2x^2+4x-9
(d) -x^2+3x-4
Q4. Without dividing, find the remainder when the first polynomial is divided by the second (Use the Remainder theorem!)
(a) x^3+2x^2+5x+1,\,x+1
(b) x^3-3x^2-x+6,\,x-2
(c) -2x^3+x^2+5x,\,3x+2
Q5. What is the quotient and remainder when x^4+3x^3+2x^2+x+1 is divided by x^2-2x+2 ?
Q6. For what values of l\in \R does p(x)=2x^2-2lx+l+5 have no real solutions?
Q7. The function f:\R\to\R,\,f(x) is a polynomial function of degree 4. Part of the graph of f is shown below, with its x intercepts labelled. If f(1) = 10 find the rule of f.
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# Generalise Number Patterns
## Generalising Linear Number Patterns
We can use algebra to create a general rule for a number sequence. For linear sequences, this is very easy. A linear sequence is a number sequence that goes up by the same amount every time. So for example 1, 4, 7, 10 (+3). The advantage of finding a rule for a number sequence is that we can determine any number in that sequence. This is how we do it. Look at the first few numbers of the sequence and see by how much it goes up or down each time. Then use this value and put it in front of n. In the example, this is 3. “n” stands for the term in the sequence (eg: 2 is the first number in the sequence and 5 the second in the example). Now calculate any number in the sequence using xn (3n in this example). There might be a difference between the actual number and the number that you get using your formula. Modify the formula accordingly (in the example, we had to take away 1). Now you can find any number in the sequence. The 100th number in this sequence would be 299.
ULTIMATE MATHS
WHERE MATHS IS AT YOUR FINGERTIPS!
Example 1 In this number sequence, the numbers increase by two each time. Consequently, we must write down 2n. However, we need to add 2 to make the formula work.
Example 2 In this number sequence, the numbers increase by four each time. Consequently, we must write down 4n. However, we need to subtract 3 to make the formula work.
Generalising quadratic number patterns can be a bit more challenging as generalising linear ones. Quadratic number patterns do not go up or down by the same amount every time. An example of a quadratic number sequence is: 3, 6, 12 & 18. However, it is still possible to generalise these types of sequences by finding the second difference as shown in the example. First find the first difference as we did with the linear sequences and then solve the second difference (the difference of the differences). Here is a rule for the second differences: If the 2nd difference is 2, your formula starts with n² If the 2nd difference is 4, your formula starts with 2n² If the 2nd difference is 6, your formula starts with 3n² In our example, the second difference is two, so our formula starts with n². Then we need to modify the formula to make it work. In this case, we had to add 3. These formulas can be more complicated to modify (eg: n² + 3n -1 or n² +2n). You need to use your maths skills and common sense to change the formula so that works. These number patterns can get very complicated. You can even have n³ depending on the sequence. However, by knowing how to generalise linear and quadratic number sequences you are at a pretty decent level.
## Presentation
Example 1 In this number sequence, the second difference is +4. Consequently, the first part of the formula is 2n². We had to add 2n in this example in order for the formula to work.
Example 2 In this number sequence, the second difference is +6. Consequently, the first part of the formula is 3n². We had to subtract 1 in this example in order for the formula to work.
## More Algebra
You should try to do some linear equations now as this is another basic algebra skill that you should be able to master. If you want to have a look at all our topics, make sure to visit our library. Please share this page if you like it or found it helpful!
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# 4 simple steps to teach your kids odd and even numbers
Not everyone is a number whiz. Some kids take more time to navigate through numbers, and that is okay. It is better to slow down your learning a little bit than learn bits and pieces in a rush. That does not make for a solid foundation, which is exactly what your kids need when it comes to learning all about numbers. Including odd and even numbers. But do not worry. The fix for it is right here, given below.
## How to introduce odd and even numbers to kids
When it comes to the question of “how to introduce odd and even numbers”, stick to the basics. There is more than a good chance kids will already know their number count up to 100. So for starters you can tell them that between 1 to 10, there will be 5 numbers that are not the multiples of 2. These will be odd numbers, and the rest are even ones. Ask them to guess before you reveal which ones.
Also read: What are consecutive numbers?
## How to explain odd and even numbers to kids
Now let us talk about how to explain odd and even numbers. To take it to the next level, tell your kids that even numbers can always be split into two equal parts, while odd ones cannot be. Tell them to remember this as a general rule of thumb so they know at any given point which number will be odd and which even in a given number bracket. A neat trick for tests and exams.
## How to teach odd and even numbers to kids in 4 easy steps
Before we get to the steps, let’s just list all the odd and even numbers between 1 to 10, shall we?
Odd numbers:
• One
• Three
• Five
• Seven
• Nine
Even numbers:
• Two
• Four
• Six
• Eight
• Ten
Now let us dive deep into how to teach odd and even numbers. Step by step.
## Step #1: Explain the basics through interactive play
Just telling kids what odd and even numbers are is kind of boring, is it not? Spice it up a little bit so they retain it better with interactive play learning sessions.
How do you turn it into a game, you ask? That is simple. You can try any of the n number of math apps available online. And if that is not to your liking, you can use things like fruits or colourful blocks to teach kids odd and even numbers in a rather interactive and fun fashion.
Such simple activities will help kids grasp the basic concept and put it into practice quite easily.
## Step #2: Teach them all about pairing and grouping
So now that kids know the basics, you have to find ways to help them retain what they have learned. And that is where pairing and grouping can come in handy.
Call this crucial step because pairing and grouping can help kids visualise odd and even numbers more effectively. Besides, it also reinforces the practical experience of using these numbers so they can apply their principles in the real world.
You can use everyday objects such as coins, colourful clips, plastic balls etc to teach kids about pairing and grouping. Get kids to pair these objects together and see if anyone is left out so they can know if they have odd or even number objects in their hands.
Also read: What are Composite Numbers
## Step #3: Help them recognise patterns
You know that odd and even numbers follow a pattern but your child does not yet. So for them to learn and recognise patterns, you have got to demonstrate it.
You can do this by preparing some flash cards with each card bearing numbers 1 to 10.
Now on a soft board, create two columns. One for odd numbers. And one for even numbers. Now place the flash cards with the numbers 2, 4, 6, 8, and 10 in the even numbers column and the numbers 1, 3, 5, 7, and 9 in the odd number column so kids can see the pattern.
You can jumble up the flash cards and ask your kids to sort them in the right columns as a fun game!
## Step #4: Make them put odd and even numbers to practical use
Finally, you can use story problems to teach kids all about the practical application of odd and even numbers. Think of it as gradual progression – it is the natural next step in their learning. Especially if you know they have mastered the first three steps.
That said, for story problems you can use everyday situations to further reinforce the concept of odd and even numbers in them.
For example, when you go to buy apples and pick up half a dozen, you can ask your child to divide them equally between three people and watch if they answer correctly.
You can even up the ante by creating problems with odd numbers. As long as you encourage your child to use mental math to solve this problem. And it is okay if they use physical objects at first to do the sorting and the solving. The goal is to arrive at the right answer on their own.
Apart from these four steps, you can also resort to using online games as teaching tools. Worksheets and the days of week are great ways to solidify the concept of odd and even numbers too. For more such clever and informative blogs, visit EuroSchool.
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# A block of mass 30kg is being brought down by a chain. If the block acquires a speed of $40\,cm\,{{s}^{-1}}$ in dropping down $2\,m$, find the work down by the chain during the process.$(g=10m{{s}^{-2}})$ (A). $-297.6J$ (B). $-375J$ (C). $-597.6J$ (D). $-674.6J$
Last updated date: 14th Jul 2024
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Hint: When the block is being brought down by the chain; two forces are acting on it. The force applied by the chain and the gravitational attraction are acting in opposite directions therefore both forces are doing some work on the block. The work done by the chain is equal to the difference in total work done on the block and work done by gravity.
Formulas Used:
$E=W=\dfrac{1}{2}m{{v}^{2}}$
$\text{The work is done by the chain on the block} = \text{Total energy}- \text{Work done by gravity}$
Complete step-by-step solution
A block of mass $30\,kg$ held by a chain is moving down with velocity $40\,cm\,{{s}^{-1}}=0.4m{{s}^{-1}}$
Therefore the total work done on the block is –
$E=W=\dfrac{1}{2}m{{v}^{2}}$
Here, $m$ is the mass of the block
$v$ is the velocity of the block
Substituting values in the above equation, we get,
$W=\dfrac{1}{2}\times 30\times {{(0.4)}^{2}}$
$\Rightarrow W=2.4J$ - (1)
Therefore the total work done on the block is $2.4J$
The work done on the block by the gravity is given by-
${{W}_{g}}=mgh$
Here, $g$ is acceleration due to gravity
$h$ is the height through which it was brought down
Therefore,
\begin{align} & {{W}_{g}}=30kg\times 10ms{}^{-2}\times 2m \\ & {{W}_{g}}=600J \\ \end{align}
$\text{The work is done by the chain on the block} = \text{Total energy}- \text{Work done by gravity}$
\begin{align} & {{W}_{c}}=W-{{W}_{g}} \\ & {{W}_{c}}=2.4-600 \\ & {{W}_{c}}=-597.6J \\ \end{align}
The work done by the chain is $-597.6J$, therefore the correct option is (C).
Note: When work done is negative, the work is being done on the system. When work is done is positive, the work is being done by the system. Work done by gravity is defined as the work done to take an object up to a height $h$. When the block is coming down, its potential energy is converted to kinetic energy.
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# Can you simplify 2 x 1
Math pages overview • back
### Simplify terms
Terms are arithmetic expressions that consist of numbers, variables (i.e. letters that stand for unknown numbers of a certain "type") and arithmetic symbols. An equation is not a term, but a statement that two terms have the same value. A term therefore does not contain an equal sign!
If a term consists only of numbers, it can easily be calculated. You have to follow the usual rules:
* Parentheses come first
* Powers before point calculation
* Dot before line calculation
* calculate from left to right
Example:
3 · (4 - 3,5)² - ((5 + 4) - 3 · 11) = 3 · 0,5² - ( 9 - 33 ) = 3 · 0,25 - (-24) = 0,75 + 24 = 24,75
If you do not understand arithmetic with letters at all, you should consider that our numerical signs, such as the 1, are also only symbols. The fact that you can also write 5 · 1 (or equal to 5) for 1 + 1 + 1 + 1 + 1 will probably not lure a mouse out from behind the stove. Calculating with other symbols is not more complicated, but rather simpler, because you have to calculate less. You simply group similar symbols together by "counting" them.
The five (identical) hearts ♥ + ♥ + ♥ + ♥ + ♥ are together as many as five times a such a heart, so 5 · ♥ or simply 5 ♥.
For the sum ♣ + ♦ + ♠ + ♠ + ♠ + ♥ + ♦ + ♣ + ♦ + ♥ + ♣ + ♠ one can briefly write 3 · ♣ + 4 · ♠ + 2 · ♥ + 3 · ♦ or (here too without Painting points): 3 ♣ + 4 ♠ + 2 ♥ + 3 ♦.
It works with letters in the same way as with the card symbols: You combine the same letters and, so to speak, only take one letter of each type times with the total number available. As a reminder: the summands of a sum can be rearranged (commutative law). Hence, e.g. a + c + b + a + b + a is the same as a + a + a + b + b + c or 3a + 2b + c. If only a Letter of a certain "variety" is included, the 1 is omitted, as in the last example with c.
Of course, the number can also be reduced by subtraction: a + a + a - a corresponds to 3a - a, so "three a minus one a", and that of course results in "two a", so: a + a + a - a = 2a.
Examples
x + x + x = 3x x + x - x + x = 2x a + 2a + 3b + 2b = 3a + 5b x + y + x = 2x + y 3a + 4a = 7a 4g - g = 3g h - 7h = -6h 3m + 1 - 5m = 1 - 2m a - b + a = 2a - b 2x + 4 - x + 5 = x + 9 10x - 2.1x = 7.9x 3a + 6b = 3a + 6b a + 2b + c - a = 2b + c 2e - 4f - 3e + 4f = -e
You can't always simplify everything. In the third to last example, for example, you can't do anything at all, since a and b only appear in one summand and the two Not can be "offset" with each other. (Not even the factors 3 and 6, because they only show the respective numbers of the a and the b !!!) In the penultimate example, the 2b and c simply remained for the same reason.
If a variable is left with zero, as in the penultimate example of a or in the last example of f, then one does not write 0a or 0f, but simply omits the variable entirely.
Instead of -1e you just write -e.
Individual numbers, as in the eighth and tenth examples, are also summarized if possible and, of course, they are calculated immediately (see tenth example: 4 + 5 = 9).
Individual numbers and variables from other summands may not be "offset".
##### Multiplications
The expression 3a means written out a + a + a, as we have just seen, not a · a · a. Therefore there is another abbreviated notation, namely the number of equal factors in a product noted at the top right of the letter: a · a · a = a³. The superscript 3 is also called the exponent or exponent. a³ reads "a to the power of three". In addition to "a to the power of two", a² is also said to be "a squared" or simply "a square".
Is it just about a Product, the variables can be summarized by exponents in a similar way to addition:
a · a · a · a = a4 a * a * b * b * b = a2· B3 x y y x x y z = x2· Y3· Z
In the last example the variables are "mixed up". You can still summarize them, because the commutative law applies, i.e. you can first rearrange the factors. With this intermediate step, the last example would be
x y y x y z = x x x y y y y z = x2· Y3· Z
If the product contains numbers, these are extracted, calculated and written to the front. (The commutative law also applies to this "type of multiplication", i.e. the factors may be rearranged even if numbers and variables are mixed.)
4a * 5a = 4 * 5 * a * a = 20a2 x³ x x² = x x x x x x x x = x5 x³ x 4y x x x 5z² x 7 = 4 x 5 x 7 x x3X x y x z2 = 140x4·Y Z2
In such products, the painting dots are usually left out, i.e. instead of 140x4·Y Z2 you write 140x4Y Z2.
Caution: A very "popular" source of error is confusing factor and exponent. Memorize the difference very well! The number in front of the letter is a number and actually means that the variable so often added becomes. In contrast, the exponent indicates how often the variable multiplied becomes.
As a rule x is7 not the same as 7x, because 7x = x + x + x + x + x + x + x, but x7 = x x x x x x x x x x x. You just have to think of some number for x and calculate it with it to see the difference. For example, take the number 2 for x, then 7x is 14, and x7 = 27 = 2·2·2·2·2·2·2 = 128.
x x x x x Not equals 3x (even if there are three x). In this case, get used to not talking and thinking of "three x", but of "x times x times x" or, better, of "x to the power of three"!
In general, one can check its simplifications by inserting the same numbers for all the same variables and calculating both the given term and the simplified term. If both results agree, the simplification should be correct in most cases (although an example is not a proof !!!!), but if they do not agree, the simplification was certainly wrong.
Products from different variables cannot be simplified any further. Instead, you can at most, or write.
##### Summing up mixed summands
In expressions such as, which have different "combinations" (better products) of variables and exponents in their summands, only those may be summarized that exactly match in all variables and associated exponents.
xy + 3x²y - 5xy + 7xy² + 3xz = -4xy + 3x²y + 7xy² + 3xz from + bc + ac + abc = from + bc + ac + abc (no simplification possible!) a²x + 2a³x² - ax + 2ax + 7a²x = 8a²x + 2a³x² + ax
##### Brackets
In principle, parentheses with variables are resolved in the same way as parentheses that contain only pure numbers. The main difference is that you usually cannot really "calculate" the content of the brackets, but you start with the simplifications in the innermost brackets.
Number or variable or product times brackets
Each term in brackets is multiplied by the number (or the variable / product) before (after) the bracket (distributive law). Note the sign rules!
4 · (a + 5b - 2c²) = 4a + 20b - 8c² x · (a + 5b - 2c²) = ax + 5bx - 2c²x (alphabetical order!) -3a · (a + 5b - 2c²) = -3a² - 15ab + 6ac² (x + 3y - z²) x 2 = 2x + 6y - 2z² (x + 3y - z²) x 2yz = 2xyz + 6y²z - 2yz³
Plus bracket
The brackets can simply be left out if there is a plus directly in front of the bracket (without a number!):
3 + (a + 5b - 2c²) = 3 + a + 5b - 2c² 3a + (a + 5b - 2c²) = 3a + a + 5b - 2c² = 4a + 5b - 2c²
Minus bracket
All signs in the brackets are reversed, the brackets and the minus in front of the brackets are omitted:
3 - (a + 5b - 2c²) = 3 - a - 5b + 2c² 2b - (a + 5b - 2c²) = 2b - a - 5b + 2c² = -a - 3b + 2c² 4a - (-a + 5a² - 7c³ ) = 4a + a - 5a² + 7c³ = 5a - 5a² + 7c³ (Caution: 5a and -5a² do not cancel each other out because of the different exponents!)
Bracket times bracket
Each addend in the first bracket is multiplied by each addend in the second bracket (those with a negative sign are also considered as addends). Attention: Note the sign rules!
(3a + 4) · (x - 7y) = 3ax - 21ay + 4x - 28y (2a - 3b + c²) · (5x³ - 7y) = 10ax³ - 14ay - 15bx³ + 21by + 5c²x³ - 7c²y (2a - b) · (3a + 5b) = 6a² + 10ab - 3ab - 5b² = 6a² + 7ab - 5b² (4x - 5y) * (5y + 4x) = 20xy + 16x² - 25y² - 20xy = 16x² - 25y²
With three brackets you first combine a pair, which you leave in brackets, and then multiply with the third bracket.
(x + 2) * (3a - b) * (2a - x) = (3ax - bx + 6a - 2b) * (2a - x) = 6a²x - 3ax² - 2abx + bx² + 12a² - 6ax - 4ab + 2bx
An exponent in brackets means the same as with numbers: The brackets must be taken as often as the exponent indicates.
(x - y) ² = (x - y) x (x - y) = x² - xy - xy + y² = x² - 2xy + y² (2a + 3b) ² = (2a + 3b) x (2a + 3b) = 4a² + 6ab + 6ab + 9b² = 4a² + 12ab + 9b² (5x - 8y) ² = (5x - 8y) · (5x - 8y) = 25x² - 40xy - 40xy + 64y² = 25x² - 80xy + 64y² (Such expressions can can be solved directly with the binomial formulas) (a - b) ³ = (a - b) · (a - b) · (a - b) = (a² - ab - ab + b²) · (a - b) = ( a² - 2ab + b²) * (a - b) = a³ - a²b - 2a²b + 2ab² + ab² - b³ = a³ - 3a²b + 3ab² - b³
#### Further examples
3x² - 3x - 4x · (3 - x) = 3x² - 3x - 12x + 4x² = 7x² - 15x -a · (2a + 3b) ² - (a - b) ³ = -a · (4a² + 12ab + 9b²) - (a³ - 3a²b + 3ab² - b³) = -4a³ - 12a²b - 9ab² - a³ + 3a²b - 3ab² + b³ = -5a³ - 9a²b - 12ab² + b³ 3 · (x - y) ² - ((5 + y²) - x² · 11) (compare very first example) Calculation not possible. 4 = (x - y) ² * (x - y) ² = (x² - 2xy + y²) * (x² - 2xy + y²) (see above) = x4 - 2x³y + x²y² - 2x³y + 4x²y² - 2xy³ + x²y² - 2xy³ + y4 = x4 - 4x³y + 6x²y² - 4xy³ + y4
© Arndt Brünner, September 29, 2003
Version: December 28, 2004
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Mutually Inclusive Events Examples
# Mutually Inclusive Events Examples
Mutually Inclusive Events are those events which are common outcomes in a set of given events. The best way to explain mutually inclusive events is by taking up an example one event being even numbers we get on throwing a dice and the second event being numbers less than $5$ on throwing a dice. If the first event is called '$A$' and the second event is called '$B$', then $A$ = $\{ 2, 4, 6 \}$ and $B$ = $\{ 1, 2, 3, 4 \}$. Outcome which are common to both the events are $A \cap B$ = $\{ 2, 4 \}$ and $P(A \cap B)$ = $\frac{2}{6}$ = $\frac{1}{3}$. We can even draw Venn diagram to represent the example properly
Formula which can be applied in case of mutually inclusive event is
$P(A\ or\ B)$ = $P(A) + P(B) – P(A\ and\ B)$
$P(A \cup B)$ = $P(A) + P(B) – P(A \cap B)$
Where, $P(A \cup B)$ is the combined probability of occurrence of either $A$ or $B,\ P(A)$ is the probability of occurrence of event $A,\ P(B)$ is the probability of occurrence of event $B$ and $P(A \cap B)$ is the probability of joint occurrence of event $A$ and event $B$
## Examples
Example 1:
A fair dice is tossed. What is the probability of tossing an even number or a number greater than $3$?
Solution:
Let event $A$ be getting an even number. So $A$ = $\{ 2, 4, 6 \}$, $n(A)$ = $3$ and event $B$ be getting numbers greater than $3$. So, $B$ = $\{ 4, 5, 6 \}$, $n(B)$ = $3$. The number of outcomes in the sample space is $n(S)$ = $6$. The probability of event $A$ would be $P(A)$ = $\frac{n(A)}{n(S)}$ = $\frac{3}{6}$ = $\frac{1}{2}$. Similarly, the probability of event $B$ would be $P(B)$ = $\frac{n(B)}{n(S)}$ = $\frac{3}{6}$ = $\frac{1}{2}$. The common outcome from both the events would be the number $4$, so the probability of joint occurrence of event $A$ and event $B$ is $P(A \cap B)$ = $\frac{1}{6}$. Thus, the probability of tossing an even number or a number greater than $3$ is
$P(A \cup B)$ = $P(A) + P(B) – P(A \cap B)$
$P(A \cup B)$ = $\frac{1}{2}$ $+$ $\frac{1}{2}$ $–$ $\frac{1}{6}$
$P(A \cup B)$ = $\frac{5}{6}$
Example 2:
What is the probability of choosing a card from a deck of cards that is a heart or a nine?
Solution:
Let event $A$ be choosing a heart from the deck of cards. As there are $13$ heart cards in a deck of cards, so the probability of selecting a heart card would be $P(A)$ = $\frac{n(A)}{n(S)}$ = $\frac{13}{52}$ = $\frac{1}{4}$. Let event $B$ be choosing a nine from a deck of cards. As there are $4$ nine’s in a deck of cards, so the probability of choosing a nine card shall be $P(B)$ = $\frac{n(B)}{n(S)}$ = $\frac{4}{52}$ = $\frac{1}{13}$. The total number of cards present in a deck of cards is $52$, so the number of elements in the sample space would be $n(S)$ = $52$. The common outcome from both the events that is selecting a heart and that too nine in number is1, so the probability of joint occurrence of event $A$ and event $B$ is $P(A \cap B)$ = $\frac{1}{52}$. Thus, the probability of choosing a card from a deck of cards that is a heart or a nine is
$P(A\ or\ B)$ = $P(A) + P(B) - P(A\ and\ B)$
$P(A\ or\ B)$ = $\frac{1}{4}$ $+$ $\frac{1}{13}$ $-$ $\frac{1}{52}$
$P(A\ or\ B)$ = $\frac{16}{52}$
$P(A\ or\ B)$ = $\frac{4}{13}$
Example 3:
What is the probability of choosing a number from $1$ to $15$ that is less than $10$ or even?
Solution:
Let event $A$ be choosing a number from 1 o 15 that is less than $10$. So, $A$ = $\{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$ and $n(A)$ = $9$. Let event $B$ be choosing an even number from $1$ to $15$. So, $B$ = $\{ 2, 4, 6, 8, 10, 12, 14 \}$ and $n(B)$ = $7$. The number of outcomes in the sample space are the numbers from $1$ to $15$, $n(S)$ = $15$. The probability of event $A$ would be $P(A)$ = $\frac{n(A)}{n(S)}$ = $\frac{9}{15}$. Similarly, the probability of event $B$ would be $P(B)$ = $\frac{n(B)}{n(S)}$ = $\frac{7}{15}$. The common outcome from both the events would be the number $A \cap B$ = $\{ 2, 4, 6, 8 \}$ , so the probability of joint occurrence of event $A$ and event $B$ is $P(A \cap B)$ = $\frac{4}{15}$. Thus, the probability of choosing a number from $1$ to $15$ that is less than $10$ or even is
$P(A\ or\ B)$ = $P(A) + P(B) - P(A\ and\ B)$
$P(A\ or\ B)$ = $\frac{9}{15}$ $+$ $\frac{7}{15}$ $-$ $\frac{4}{15}$
$P(A\ or\ B)$ = $\frac{12}{15}$
$P(A\ or\ B)$ = $\frac{4}{5}$
Example 4:
$2$ fair dice are rolled. What is the probability of getting a sum less than $8$ or a sum equal to $9$?
Solution:
Let us first draw the sum table when two fair dice are rolled
Let event $A$ be getting a sum less than $8$ on rolling two fair dice. In the table above those numbers are shaded in brick red color, so $n(A)$ = $21$. Let event $B$ be getting a sum equal to $9$ on rolling two fair dice. In the table above those numbers are shaded in green color, so $n(B)$ = $4$. The number of outcomes in the sample space is the total sum of numbers we get on rolling two fair dice, $n(S)$ = $36$. The probability of event $A$ would be $P(A)$ = $\frac{n(A)}{n(S)}$ = $\frac{21}{36}$. Similarly, the probability of event $B$ would be $P(B)$ = $\frac{n(B)}{n(S)}$ = $\frac{4}{36}$. As there is no elements in common so $A \cap B$ = $0$. In this type of condition the events are said to be mutually exclusive events. Thus, the probability of getting a sum less than $8$ or a sum equal to $9$
$P(A \cup B)$ = $P(A) + P(B) – P(A \cap B)$
$P(A \cup B)$ = $\frac{21}{36}$ $+$ $\frac{4}{36}$ $-$ $0$
$P(A \cup B)$ = $\frac{25}{36}$
Example 5:
What is the probability of getting a sum less than $5$ and a sum less than $4$ when two fair dice is rolled?
Solution:
Let us first draw the sum table when two fair dice are rolled.
Let event $A$ be getting a sum less than $5$ on rolling two fair dice. In the table above those numbers are shaded in purple color, so $n(A)$ = $6$. Let event $B$ be getting a sum less than $4$ on rolling two fair dice. In the table above those numbers are shaded in pink color, so $n(B)$ = $3$. The number of outcomes in the sample space is the total sum of numbers we get on rolling two fair dice, $n(S)$ = $36$. The probability of event $A$ would be $P(A)$ = $\frac{n(A)}{n(S)}$ = $\frac{6}{36}$. Similarly, the probability of event $B$ would be $P(B)$ = $\frac{n(B)}{n(S)}$ = $\frac{3}{36}$. The common outcome from both the events would be the number $A \cap B$ = $\{ 2, 3, 3 \}$, so the probability of joint occurrence of event $A$ and event $B$ is $P(A \cap B)$ = $\frac{3}{36}$. Thus, the probability of getting a sum less than $5$ or a sum less than $4$.
$P(A\ or\ B)$ = $P(A) + P(B) - P(A and B)$P(A\ or\ B)$=$\frac{6}{36}+\frac{3}{36}-\frac{3}{36}P(A\ or\ B)$=$\frac{6}{36}P(A\ or\ B)$=$\frac{1}{6}\$
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Education
# Applications of Maths Concept ‘Area’
In Maths, you must have come across many concepts and theories, which are applied in mathematical calculations, both in academic and real-life. One of the useful concepts is the concept of the area of different shapes, defined in geometry. There are a number of geometrical shapes which we learn in our academics, such as circle, square, triangle, cylinder, etc. All these shapes have their own properties, areas and volumes.
The area of a given shape signifies the region occupied by the shape in a given 2-D or 3-D plane. Therefore, calculating the area helps us to measure the space occupied by a given object of a certain figure. To each shape which are introduced in geometry, there are formulas to calculate the area for them. These shapes have dimensions defined for them, based on which the surface area for them is calculated. Let us go through some of the real-life applications, where the area is measured for a given object.
• By measuring the area of a wall, which has equal length and height, we can calculate the amount of paint required to cover the surface area. Here we can use the formula of area of square for the given wall because a square has all its edges equal in length.
• To cover a floor with a carpet, we need to know the area of the floor by measuring its dimensions.
• Measuring the agricultural fields or plots for purchasing or selling purpose.
• Even to build a Pyramid, the architect must have required the area covered beneath the structure with the help of formula of the area of a two-dimensional triangle.
• To buy a table cloth, we need to know the surface area of the table.
• To construct overhead tanks in an apartment or independent houses.
• Wrapping a gift which are in rectangular shapes, will require to measure the surface area of the gift to wrap it completely.
• To use rods of cylindrical shapes during the construction of buildings, we can measure its lateral surface by using the formula of area of cylinder, by knowing the diameter of the rod.
‘Area’ is a major part of mensuration topic in Maths just like the volume. We may see many real-life examples, where the surface area of any object or place is measured.
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# Difference between revisions of "2020 AMC 8 Problems/Problem 25"
## Problem
Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
$[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("R_1",(3/2,1/2)); label("S_3",(4,1)); label("S_2",(5/2,3/2)); label("S_1",(1,2)); label("R_2",(7/2,5/2)); [/asy]$
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$
## Solution 1
Let the side length of each square $S_k$ be $s_k$. Then, from the diagram, we can line up the top horizontal lengths of $S_1$, $S_2$, and $S_3$ to cover the top side of the large rectangle, so $s_{1}+s_{2}+s_{3}=3322$. Similarly, the short side of $R_2$ will be $s_1-s_2$, and lining this up with the left side of $S_3$ to cover the vertical side of the large rectangle gives $s_{1}-s_{2}+s_{3}=2020$. We subtract the second equation from the first to obtain $2s_{2}=1302$, and thus $s_{2}=\boxed{\textbf{(A) }651}$.
## Solution 2
Assuming that the problem is well-posed, it should be true in the particular case where $S_1 \cong S_3$ and $R_1 \cong R_2$. Let the sum of the side lengths of $S_1$ and $S_2$ be $x$, and let the length of rectangle $R_2$ be $y$. We then have the system $$\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}$$ which we solve to determine $y=\boxed{\textbf{(A) }651}$.
## Solution 3 (faster version of Solution 1)
Since, for each pair of rectangles, the side lengths have a sum of $3322$ or $2020$ and a difference of $S_2$, the answer must be $\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{\textbf{(A) }651}$.
## Solution 4
Assuming that the problem is well-posed, it should be true in the case where $S_1 \cong S_3$. Let the side length of square $S_1$ be $x$ and the side length of square $S_2$ be $y$. We then have the system $$\begin{dcases}2x-y =2020 \\2x+y =3322\end{dcases}$$ and we solve it to determine that $y=\boxed{\textbf{(A) }651}$.
~savannahsolver
~Interstigation
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# Time Speed and Distance Questions for MAH-CET
Question 1:Â A 260 meter long train crosses a 120 meter long wall in 19 seconds .What is the speed of the train?
a)Â 27 km/hr
b)Â 49 km/hr
c)Â 72 km/hr
d)Â 70 km/hr
e)Â None of these
Solution:
Length of the train is 260 metres
Length of the wall is 120 metres
total is 260+120 = 380 metres
Time taken is 19 seconds.
Hence, the speed is 380/19 = 20 m/s = 72 Km/hr
Answer is option C
Question 2: A boat takes 2 hours to travel from point A to B in still water .To find out it’s speed up-stream ,which of the following information is needed.
i. Distance between point A and B
Ii.Time taken to travel down stream from B to A
iii. Speed of the stream of the water
iv. Effective speed of Boat while traveling Downstream from B to A
a)Â All are required
c)Â Only i,iii, and either ii or iv
d)Â Only i and iii
e)Â None of these
Solution:
Time taken by boat to travel from point A to B in still water = 2 hours
To find the upstream speed, we definitely need the speed of stream, thus statement (iii) is mandatory.
Also, the distance between points A and B or the speed of boat in still water is needed.
Thus, statements (i) and (iii) are required to find the upstream speed of the boat.
=> Ans – (D)
Question 3:Â A train running at speed of 120 kmph crosses a signal in 15 seconds .What is the length of the train in meters?
a)Â 300
b)Â 200
c)Â 500
d)Â Cannot determined
e)Â None of these
Solution:
Speed of train = 120 kmph
= $(120 \times \frac{5}{18})$ m/s = $\frac{100}{3}$ m/s
Let length of the train = $l$ meters
Using, speed = distance/time
=> $\frac{100}{3} = \frac{l}{15}$
=> $l=\frac{100}{3} \times 15$
=> $l=100 \times 5=500$ meters
=> Ans – (C)
Question 4:Â A bus covers a distance of 2,924 km,in 43 hours .what is the bus speed?
a)Â 72 km/hr
b)Â 60 km/hr
c)Â 68 km/hr
d)Â cannot determined
e)Â none of these
Solution:
Let speed of bus = $s$ km/hr
Distance covered = 2924 km
Time taken = 43 hours
Using speed = distance/time
=> $s=\frac{2924}{43}=68$ km/hr
=> Ans – (C)
Question 5:Â A 240-metre long train running at the speed of 60 kmph will take how much time to cross another 270-metre long train running in opposite direction at the speed of 48 kmph?
a)Â 17 seconds
b)Â 3 seconds
c)Â 12 seconds
d)Â 8 seconds
e)Â None of these
Solution:
Length of first train = 240 m and second train = 270 m
Total length of the two trains = 240 + 270 = 510 m
Speed of first train = 60 kmph and second train = 48 kmph
Since, the trains are moving in opposite direction, thus relative speed = 60 + 48 = 108 kmph
= $(108 \times \frac{5}{18})$ m/s = $30$ m/s
Let time taken = $t$ seconds
Using, time = distance/speed
=> $t=\frac{510}{30}=17$ seconds
=> Ans – (A)
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Question 6:Â A man takes 2.2 times as long to row a distance upstream as to row the same distance downstream. If he can row 55 km downstream in 2 hours 30 minutes, what is the speed of the boat in still water? (in km/h)
a)Â 40
b)Â 8
c)Â 16
d)Â 24
e)Â 32
Solution:
Let speed of boat in still water = $x$ km/hr
=> Speed of current = $y$ km/hr
Let distance travelled = $d$ km
Acc. to ques, => $2.2 (\frac{d}{x + y}) = \frac{d}{x – y}$
=> $2.2x – 2.2y = x + y$
=> $2.2x – x = y + 2.2y$
=> $3x = 8y$ ————-(i)
Also, the man takes 2 hrs 30 mins in travelling 55 km downstream.
=> $\frac{55}{x + y} = 2 + \frac{1}{2}$
=> $\frac{55}{x + y} = \frac{5}{2}$
=> $x + y = 22$
Multiplying both sides by 8, and using eqn(i), we get :
=> $8x + 3x = 22 \times 8$
=> $x = \frac{22 \times 8}{11} = 16$ km/hr
Question 7:Â At 60% of its usual speed, a train of length L metres crosses a platform 240 metre long in 15 seconds. At its usual speed, the train crosses a pole in 6 seconds. What is the value of L (in metre)?
a)Â 270
b)Â 225
c)Â 220
d)Â 480
e)Â 240
Solution:
Let speed of the train = $10x$ m/s
Length of train = $l$ m
Time taken to cross the pole = 6 sec
Using, $speed = \frac{distance}{time}$
=> $10x = \frac{l}{6}$
=> $x = \frac{l}{60}$
Now, 60% of the speed = $\frac{60}{100} \times 10x = 6x$ m/s
Length of platform = 240 m
Acc. to ques, => $6x = \frac{240 + l}{15}$
=> $6 \times \frac{l}{60} = \frac{240 + l}{15}$
=> $\frac{l}{10} = \frac{240 + l}{15}$
=> $15l = 2400 + 10l$
=> $15l – 10l = 5l = 2400$
=> $l = \frac{2400}{5} = 480$ m
Question 8:Â A boat takes a total time of twelve hours to travel 105 kms upstream and the same distance downstream. The speed of the boat in still water is six times of the speed of the current. What is the speed of the boat in still water? (in km/hr)
a)Â 12
b)Â 30
c)Â 18
d)Â 24
e)Â 36
Solution:
Let speed of current = $x$ km/hr
=> Speed of boat in still water = $6x$ km/hr
Acc. to ques, => $\frac{105}{7x} + \frac{105}{5x} = 12$
=> $\frac{15}{x} + \frac{21}{x} = 12$
=> $\frac{36}{x} = 12$
=> $x = \frac{36}{12} = 3$
$\therefore$ Speed of boat in still water = $6 \times 3 = 18$ km/hr
Question 9:Â At its usual speed, a train of length L metres crosses platform 300 metre long in 25 seconds. At 50% of its usual speed, the train crosses a vertical pole in 20 seconds. What is the value of L?
a)Â 160
b)Â 260
c)Â 200
d)Â 310
e)Â 350
Solution:
Let usual speed of the train = $10x$ m/s
Now, 50% of the speed = $\frac{50}{100} \times 10x = 5x$ m/s
Length of train = $l$ m
Time taken to cross the pole = 20 sec
Using, $speed = \frac{distance}{time}$
=> $5x = \frac{l}{20}$
=> $x = \frac{l}{100}$
Length of platform = 300 m
Acc. to ques, => $10x = \frac{300 + l}{25}$
=> $10 \times \frac{l}{100} = \frac{300 + l}{25}$
=> $\frac{l}{10} = \frac{300 + l}{25}$
=> $25l = 3000 + 10l$
=> $25l – 10l = 15l = 3000$
=> $l = \frac{3000}{15} = 200$ m
Question 10:Â A boat takes a total time of eight hours to travel 63 kms upstream and the same distance downstream. The speed of the current is ${1 \over 8}$th of the speed of the boat in still water. What is the speed of the boat in still water? (in km/hr)
a)Â 32
b)Â 24
c)Â 16
d)Â 8
e)Â 38
Solution:
Let speed of current = $x$ km/hr
=> Speed of boat in still water = $8x$ km/hr
Acc. to ques, => $\frac{63}{9x} + \frac{63}{7x} = 8$
=> $\frac{7}{x} + \frac{9}{x} = 8$
=> $\frac{16}{x} = 8$
=> $x = \frac{16}{8} = 2$
$\therefore$ Speed of boat in still water = $8 \times 2 = 16$ km/hr
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# How does a calculator know the answer?
## Math and take root
What a root is in mathematics and how to pull the root is explained here in a simple way. Go to content:
• A Explanationwhat you need roots for.
• Examples how to calculate roots (also in writing, without a calculator).
• Tasks / exercises to practice this yourself
• A Video this topic.
• A Question and answer area to this area.
Tip: You should already know what an equation and a power is. If you have no idea about this, please see solving equations and powers.
### Explanation take root
We're looking at the basics of pulling the root here. But if you already know the basics, you are welcome to take a look at these topics (will be linked as soon as available):
• Laws of roots / rules of roots
• Root calculation
• Roots with variables
• Root equations
First of all, let's clarify why you need a root. A little hint: Whenever the school speaks of the root or pulling the root, it usually means the square root. Let's take a quick look at the definition, followed by examples:
Note:
The square root of a number a is the positive number that you have to square to get a.
The symbol for the root is this:
Example 1: Simple roots
Let's start with a few simple examples of how to pull the root. The root of 4 is simply 2, because 2 · 2 = 4. The root of 25 is 5, because 5 · 5 = 25. Here are four simple examples:
Let's look at a few more important terms. First of all, we have a root sign. You will find the radicand under the root sign. If you calculate the root you get the root value.
These were simple examples of how to pull the root. So we'll look at a slightly more difficult task in the next section.
Display:
### Examples take root
How can one calculate the root? Well, in school and also during studies, the calculator is usually used. Alternatively, you can calculate the root without a calculator. Either through an approximation procedure or through the written root.
Example root of 2:
If you take the square root of 2 with the calculator you get this:
How do you get this result without a calculator? One possibility is to slowly approach the result through multiplications. We start like this for this example:
We want 2 as a solution. Therefore 1 is too small and 4 is too big. We therefore go in between once with 1.5 · 1.5.
The 2.25 is still bigger than 2. So let's try 1.4.
With the 1.96 we come very close to the 2. But are still below. So we don't try 1.4 but 1.41:
You can go on and on with this. One approaches more and more the 2 below the square root or the square root value 1.4142 ...
Note:
There are other written procedures for calculating roots. However, these are only dealt with marginally at school - if at all - and are therefore not explained here. With the approximation method, however, you can calculate the root once by hand. Alternatively, there is of course the pocket calculator.
Show:
### Explanation and examples
This video covers the basics of calculating the roots. Let's look at this:
• What is a root in math?
• Why do you need the calculation of the roots?
• Simple examples are calculated.
• Simple examples are explained.
Next video »
### Questions with answers take root and root
In this section we look at typical questions with answers about root pulling.
Q: Are there any rules for root extraction?
A: Yeah, sure. As with pretty much anything in mathematics, there are rules in calculating the roots. However, these would blow up the article here. For this reason you can find them together with examples under Root Rules / Root Laws. There you also learn to simplify roots.
Q: is it possible to calculate roots in the head?
A: At least for simple root problems with square numbers everyone should be able to do this, especially for small numbers. You should know the roots of 2, 4, 9, 16 etc. by heart or calculate them quickly using knowledge of the multiplication table. If the roots are complicated, you should be able to roughly estimate what should be the result of the root. Otherwise, you can of course try to follow the written procedure for loosening roots in your head.
Q: Is there a root sign on the keyboard?
A: A normal keyboard does not have a symbol for a root. Various programs therefore offer the option of inserting the root symbol as a special character. \ Sqrt {} is used in latex.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
7.8: Histograms
Difficulty Level: Basic Created by: CK-12
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Practice Histograms
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Suppose you're working in a card store and you notice that the cards have different prices. You pull 15 cards out of inventory and their prices are 0.75, 0.95, 1.25, 1.65, 1.75, 2.25, 0.95, 1.10, 3.55, 5.00, 1.35, 2.25, 3.75, 4.25, and 5.65. If you divide this data up into 0.50 intervals how can you construct a graph of your findings?
Watch This
First watch this video to learn about histograms.
Then watch this video to see some examples.
Watch this video for more help.
Guidance
An extension of the bar graph is the histogram. A histogram is a type of vertical bar graph in which the bars represent grouped continuous data. The shape of a histogram can tell you a lot about the distribution of the data, as well as provide you with information about the mean, median, and mode of the data set. The following are some typical histograms, with a caption below each one explaining the distribution of the data, as well as the characteristics of the mean, median, and mode. Distributions can have other shapes besides the ones shown below, but these represent the most common ones that you will see when analyzing data. In each of the graphs below, the distributions are not perfectly shaped, but are shaped enough to identify an overall pattern.
a)
Figure a represents a bell-shaped distribution, which has a single peak and tapers off to both the left and to the right of the peak. The shape appears to be symmetric about the center of the histogram. The single peak indicates that the distribution is unimodal. The highest peak of the histogram represents the location of the mode of the data set. The mode is the data value that occurs the most often in a data set. For a symmetric histogram, the values of the mean, median, and mode are all the same and are all located at the center of the distribution.
b)
Figure b represents a distribution that is approximately uniform and forms a rectangular, flat shape. The frequency of each class is approximately the same.
c)
Figure c represents a right-skewed distribution, which has a peak to the left of the distribution and data values that taper off to the right. This distribution has a single peak and is also unimodal. For a histogram that is skewed to the right, the mean is located to the right on the distribution and is the largest value of the measures of central tendency. The mean has the largest value because it is strongly affected by the outliers on the right tail that pull the mean to the right. The mode is the smallest value, and it is located to the left on the distribution. The mode always occurs at the highest point of the peak. The median is located between the mode and the mean.
d)
Figure d represents a left-skewed distribution, which has a peak to the right of the distribution and data values that taper off to the left. This distribution has a single peak and is also unimodal. For a histogram that is skewed to the left, the mean is located to the left on the distribution and is the smallest value of the measures of central tendency. The mean has the smallest value because it is strongly affected by the outliers on the left tail that pull the mean to the left. The median is located between the mode and the mean.
e)
Figure e has no shape that can be defined. The only defining characteristic about this distribution is that it has 2 peaks of the same height. This means that the distribution is bimodal.
While there are similarities between a bar graph and a histogram, such as each bar being the same width, a histogram has no spaces between the bars. The quantitative data is grouped according to a determined bin size, or interval. The bin size refers to the width of each bar, and the data is placed in the appropriate bin.
The bins, or groups of data, are plotted on the x\begin{align*}x\end{align*}-axis, and the frequencies of the bins are plotted on the y\begin{align*}y\end{align*}-axis. A grouped frequency distribution is constructed for the numerical data, and this table is used to create the histogram. In most cases, the grouped frequency distribution is designed so there are no breaks in the intervals. The last value of one bin is actually the first value counted in the next bin. This means that if you had groups of data with a bin size of 10, the bins would be represented by the notation [0-10), [10-20), [20-30), etc. Each bin appears to contain 11 values, which is 1 more than the desired bin size of 10. Therefore, the last digit of each bin is counted as the first digit of the following bin.
The first bin includes the values 0 through 9, and the next bin includes the values 9 through 19. This makes the bins the proper size. Bin sizes are written in this manner to simplify the process of grouping the data. The first bin can begin with the smallest number of the data set and end with the value determined by adding the bin width to this value, or the bin can begin with a reasonable value that is smaller than the smallest data value.
Example A
Construct a frequency distribution table with a bin size of 10 for the following data, which represents the ages of 30 lottery winners:
384129334074664560552552546146515957666232476550392235727749\begin{align*}& 38 \quad 41 \quad 29 \quad 33 \quad 40 \quad 74 \quad 66 \quad 45 \quad 60 \quad 55\\ & 25 \quad 52 \quad 54 \quad 61 \quad 46 \quad 51 \quad 59 \quad 57 \quad 66 \quad 62\\ & 32 \quad 47 \quad 65 \quad 50 \quad 39 \quad 22 \quad 35 \quad 72 \quad 77 \quad 49\end{align*}
Step 1: Determine the range of the data by subtracting the smallest value from the largest value.
Range: 7722=55\begin{align*}\text{Range:} \ 77-22=55\end{align*}
Step 2: Divide the range by the bin size to ensure that you have at least 5 groups of data. A histogram should have from 5 to 10 bins to make it meaningful: 5510=5.56\begin{align*}\frac{55}{10}=5.5 \approx 6\end{align*}. Since you cannot have 0.5 of a bin, the result indicates that you will have at least 6 bins.
Step 3: Construct the table.
Bin Frequency
[2030)\begin{align*}[20-30)\end{align*} 3
[3040)\begin{align*}[30-40)\end{align*} 5
[4050)\begin{align*}[40-50)\end{align*} 6
[5060)\begin{align*}[50-60)\end{align*} 8
[6070)\begin{align*}[60-70)\end{align*} 5
[7080)\begin{align*}[70-80)\end{align*} 3
Step 4: Determine the sum of the frequency column to ensure that all the data has been grouped.
3+5+6+8+5+3=30\begin{align*}3+5+6+8+5+3=30\end{align*}
When data is grouped in a frequency distribution table, the actual data values are lost. The table indicates how many values are in each group, but it doesn't show the actual values.
There are many different ways to create a distribution table and many different distribution tables that can be created. However, for the purpose of constructing a histogram, the method shown works very well, and it is not difficult to complete.
Example B
The numbers of years of service for 75 teachers in a small town are listed below:
1, 6, 11, 26, 21, 18, 2, 5, 27, 33, 7, 15, 22, 30, 831, 5, 25, 20, 19, 4, 9, 19, 34, 3, 16, 23, 31, 10, 42, 31, 26, 19, 3, 12, 14, 28, 32, 1, 17, 24, 34, 16, 1,18, 29, 10, 12, 30, 13, 7, 8, 27, 3, 11, 26, 33, 29, 207, 21, 11, 19, 35, 16, 5, 2, 19, 24, 13, 14, 28, 10, 31\begin{align*}& 1, \ 6, \ 11, \ 26, \ 21, \ 18, \ 2, \ 5, \ 27, \ 33, \ 7, \ 15, \ 22, \ 30, \ 8\\ & 31, \ 5, \ 25, \ 20, \ 19, \ 4, \ 9, \ 19, \ 34, \ 3, \ 16, \ 23, \ 31, \ 10, \ 4\\ & 2, \ 31, \ 26, \ 19, \ 3, \ 12, \ 14, \ 28, \ 32, \ 1, \ 17, \ 24, \ 34, \ 16, \ 1,\\ & 18, \ 29, \ 10, \ 12, \ 30, \ 13, \ 7, \ 8, \ 27, \ 3, \ 11, \ 26, \ 33, \ 29, \ 20\\ & 7, \ 21, \ 11, \ 19, \ 35, \ 16, \ 5, \ 2, \ 19, \ 24, \ 13, \ 14, \ 28, \ 10, \ 31\end{align*}
Using the above data, construct a frequency distribution table with a bin size of 5.
Range: 351345=34=6.87\begin{align*}\text{Range:} \ 35-1 & = 34\\ \frac{34}{5} & = 6.8 \approx 7\end{align*}
You will have 7 bins.
When the number of data values is very large, another column is often inserted in the distribution table. This column is a tally column, and it is used to account for the number of values within a bin. A tally column facilitates the creation of the distribution table and usually allows the task to be completed more quickly. For each value that is in a bin, draw a stroke in the Tally column. To make counting the strokes easier, draw 4 strokes and cross them out with the fifth stroke. This process bundles the strokes in groups of 5, and the frequency can be readily determined.
Bin Tally Frequency
[05)\begin{align*}[0-5)\end{align*} |||| |||| |\begin{align*}\cancel{||||} \ \cancel{||||} \ |\end{align*} 11
[510)\begin{align*}[5-10)\end{align*} |||| ||||\begin{align*}\cancel{||||} \ ||||\end{align*} 9
[1015)\begin{align*}[10-15)\end{align*} |||| |||| ||\begin{align*}\cancel{||||} \ \cancel{||||} \ ||\end{align*} 12
[1520)\begin{align*}[15-20)\end{align*} |||| |||| ||||\begin{align*}\cancel{||||} \ \cancel{||||} \ ||||\end{align*} 14
[2025)\begin{align*}[20-25)\end{align*} |||| ||\begin{align*}\cancel{||||} \ ||\end{align*} 7
[2530)\begin{align*}[25-30)\end{align*} |||| ||||\begin{align*}\cancel{||||} \ \cancel{||||}\end{align*} 10
[3035)\begin{align*}[30-35)\end{align*} |||| |||| ||\begin{align*}\cancel{||||} \ \cancel{||||} \ ||\end{align*} 12
11+9+12+14+7+10+12=75\begin{align*}11+9+12+14+7+10+12 = 75\end{align*}
Now that you have constructed the frequency table, the grouped data can be used to draw a histogram. Like a bar graph, a histogram requires a title and properly labeled x\begin{align*}x\end{align*}- and y\begin{align*}y\end{align*}-axes.
Example C
Use the data from Example A that displays the ages of the lottery winners to construct a histogram. The data is shown again below. What percentage of the winners were 50 years of age or older?
Bin Frequency
[2030)\begin{align*}[20-30)\end{align*} 3
[3040)\begin{align*}[30-40)\end{align*} 5
[4050)\begin{align*}[40-50)\end{align*} 6
[5060)\begin{align*}[50-60)\end{align*} 8
[6070)\begin{align*}[60-70)\end{align*} 5
[7080)\begin{align*}[70-80)\end{align*} 3
Use the data as it is represented in the distribution table to construct the histogram.
From looking at the tops of the bars, you can see how many winners were in each category, and by adding these numbers, you can determine the total number of winners. You can also determine how many winners were within a specific category. For example, you can see that 8 winners were 60 years of age or older. The graph can also be used to determine percentages. For example, it can answer the question, “What percentage of the winners were 50 years of age or older?” as follows:
1630=0.533¯¯¯¯¯(0.533)(100%)5.3%.\begin{align*}\frac{16}{30} = 0.5\overline{33} \qquad (0.533) (100\%) \approx 5.3\%.\end{align*}
Guided Practice
a. Use the data and the distribution table that represent the ages of teachers from Example B to construct a histogram to display the data. The distribution table is shown again below:
Bin Tally Frequency
[05)\begin{align*}[0-5)\end{align*} |||| |||| |\begin{align*}\cancel{||||} \ \cancel{||||} \ |\end{align*} 11
[510)\begin{align*}[5-10)\end{align*} |||| ||||\begin{align*}\cancel{||||} \ ||||\end{align*} 9
[1015)\begin{align*}[10-15)\end{align*} |||| |||| ||\begin{align*}\cancel{||||} \ \cancel{||||} \ ||\end{align*} 12
[1520)\begin{align*}[15-20)\end{align*} |||| |||| ||||\begin{align*}\cancel{||||} \ \cancel{||||} \ ||||\end{align*} 14
[2025)\begin{align*}[20-25)\end{align*} |||| ||\begin{align*}\cancel{||||} \ ||\end{align*} 7
[2530)\begin{align*}[25-30)\end{align*} |||| ||||\begin{align*}\cancel{||||} \ \cancel{||||}\end{align*} 10
\begin{align*}[30-35)\end{align*} \begin{align*}\cancel{||||} \ \cancel{||||} \ ||\end{align*} 12
b. Now use the histogram to answer the following questions.
i. How many teachers teach in this small town?
ii. How many teachers have worked for less than 5 years?
iii. If teachers are able to retire when they have taught for 30 years or more, how many are eligible to retire?
iv. What percentage of the teachers still have to teach for 10 years or fewer before they are eligible to retire?
v. Do you think that the majority of the teachers are young or old? Justify your answer.
a.
b. i. \begin{align*}11+9+12+14+7+10+12=75\end{align*}
In this small town, 75 teachers are teaching.
ii. 11 teachers have taught for less than 5 years.
iii. 12 teachers are eligible to retire.
iv. \begin{align*}\frac{17}{75}=0.22\overline{66} \qquad (0.2266)(100\%) \approx 23\%\end{align*}
Approximately 23% of the teachers must teach for 10 years or fewer before they are eligible to retire.
v. Answers will vary, but one possible answer is that the majority of the teachers are young, because 46 have taught for less than 20 years.
Practice
1. What name is given to a distribution that has 2 peaks of the same height?
1. uniform
2. unimodal
3. bimodal
4. discrete
The following histogram shows data collected during a recent fishing derby. The number of fish caught is being compared to the size of the fish caught. Use the histogram to answer the following questions:
1. How many fish were caught?
2. How many fish caught were over 35 cm in length?
3. How many fish caught were between 20 cm and 29 cm in length?
4. Why is there a blank space between 38 cm and 41 cm on the histogram?
The following histogram displays the heights of students in a classroom. Use the information represented in the histogram to answer the following questions:
1. How many students are in the class?
2. How many students are over 60 inches in height?
3. How many students have a height between 54 in and 62 in?
4. Is the distribution unimodal or bimodal? How do you know?
5. The following data represents the results of a test taken by a group of students: \begin{align*}& 95 \quad 56 \quad 70 \quad 83 \quad 59 \quad 66 \quad 88 \quad 52 \quad 50 \quad 77 \quad 69 \quad 80\\ & 54 \quad 75 \quad 68 \quad 78 \quad 51 \quad 64 \quad 55 \quad 67 \quad 74 \quad 57 \quad 73 \quad 53\end{align*} Construct a frequency distribution table using a bin size of 10 and display the results in a properly labeled histogram.
Vocabulary Language: English Spanish
frequency distribution
frequency distribution
A table that lists all of the classes and the number of data values that belong to each of the classes. A distribution in which most of the data values are located to the right of the mean is called a left-skewed distribution, while a distribution in which most of the data values are located to the left of the mean is called a right-skewed distribution.
bar graph
bar graph
A bar graph is a plot made of bars whose heights (vertical bars) or lengths (horizontal bars) represent the frequencies of each category, with space between each bar.
frequency density
frequency density
The vertical axis of a histogram is labelled frequency density.
Frequency table
Frequency table
A frequency table is a table that summarizes a data set by stating the number of times each value occurs within the data set.
Histogram
Histogram
A histogram is a display that indicates the frequency of specified ranges of continuous data values on a graph in the form of immediately adjacent bars.
Interval
Interval
An interval is a range of data in a data set.
Range
Range
The range of a data set is the difference between the smallest value and the greatest value in the data set.
right-skewed distribution
right-skewed distribution
A right-skewed distribution has a peak to the left of the distribution and data values that taper off to the right.
unimodal
unimodal
If a data set has only 1 value that occurs most often, the set is called unimodal.
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Intermediate Algebra: Functions and Graphs
Section3.2Intercepts, Solutions, and Factors
In the last section, we used extraction of roots to solve quadratic equations of the form
\begin{equation*} ~a(x-p)^2=q~ \end{equation*}
But this technique will not work on quadratic equations that also include a linear term, $$bx\text{.}$$ Recall that the most general type of quadratic equation looks like
\begin{equation*} ax^2+bx+c = 0 \end{equation*}
Here is an example.
Suppose a baseball player pops up, that is, she hits the baseball straight up into the air. The height, $$h\text{,}$$ of the baseball after $$t$$ seconds is given by a formula from physics. This formula takes into account the initial speed of the ball (64 feet per second) and its height when it was hit (4 feet).
\begin{equation*} h=-16t^2+64t+4 \end{equation*}
The graph of this equation is shown below.
We would like to know when the baseball was exactly 52 feet high. To find out, we must solve the equation
\begin{equation*} -16t^2+64t+4 = 52 \end{equation*}
where we have substituted 52 for the height, $$h\text{.}$$ We can use the graph to solve this equation, by finding points with $$h$$-coordinate 52. You can see that there are two such points, with $$t$$-coordinates 1 and 3, so the baseball is 52 feet high at 1 second, and again on the way down at 3 seconds.
Can we solve the equation algebraically? Not with the techniques we know, because there are two terms containing the variable $$t\text{,}$$ and they cannot be combined. We will need a new method. To find this method, we are going to study the connection between:
1. the factors of $$ax^2+bx+c\text{,}$$
2. the solutions of the quadratic equation $$~ax^2+bx+c=0,~$$ and
3. the $$x$$-intercepts of the graph of $$y=~ax^2+bx+c=0\text{.}$$
Note3.2.1.
If you would like to review multiplying binomials (the "FOIL" method) or factoring quadratic trinomials, please see the Algebra Toolkit for this section.
Subsection3.2.1Zero-Factor Principle
The method we will learn now is not like extraction of roots, or solving linear equations, where we "undid" in reverse order each operation performed on the variable, like peeling an onion. This new method will seem less direct. It relies on applying a property of our number sustem.
Can you multiply two numbers together and obtain a product of zero? Only if one of the two numbers happens to be zero. (Try it yourself.)
Zero-Factor Principle.
The product of two factors equals zero if and only if one or both of the factors equals zero. In symbols,
\begin{equation*} ab = 0~~~\text{if and only if}~~~a=0~~~\text{or}~~~b=0~~~\text{(or both)} \end{equation*}
Checkpoint3.2.2.QuickCheck 1.
Fill in the blanks.
1. If the sum of two numbers is zero, the numbers must be
• undefined
• opposites
• reciprocals
• numerator
• zero
.
2. If a fraction equals zero, the
• undefined
• opposites
• reciprocals
• numerator
• zero
must be zero.
3. If the product of two numbers is zero, one of the numbers must be
• undefined
• opposites
• reciprocals
• numerator
• zero
.
4. If the divisor in a quotient is zero, the quotient is
• undefined
• opposites
• reciprocals
• numerator
• zero
.
$$\text{opposites}$$
$$\text{numerator}$$
$$\text{zero}$$
$$\text{undefined}$$
Solution.
1. opposites
2. numerator
3. zero
4. undefined
Checkpoint3.2.3.QuickCheck 1.
Fill in the blanks.
1. If the sum of two numbers is zero, the numbers must be .
2. If a fraction equals zero, the must be zero.
3. If the product of two numbers is zero, one of the numbers must be .
4. If the divisor in a quotient is zero, the quotient is .
Here is the simplest possible application of the Zero-Factor Principle (ZFP): For what value(s) of $$x$$ is the equation $$~3x=0~$$ true? You could divide both sides by 3, but you can also see that the product $$3x$$ can equal zero only if one of its factors is zero, so $$x$$ must be zero!
The ZFP is true even if the numbers $$a$$ and $$b$$ are represented by algebraic expressions, such as $$x-6$$ or $$x+2\text{.}$$ For example, if
\begin{equation*} (x-6)(x+2)=0 \end{equation*}
then it must be true that either $$~x-6=0~$$ or $$~x+2=0.~$$ This is how we can use the ZFP to solve quadratic equations.
Example3.2.4.
Solve the equation $$x^2-4x-12=0$$
Solution.
We can factor the expression $$x^2-4x-12\text{,}$$ and write the equation as
\begin{equation*} (x-6)(x+2) = 0 \end{equation*}
Now it is in the form $$~ab=0~\text{,}$$ with $$~a=x-6~$$ and $$~b=x+2~\text{,}$$ so the ZFP tells us that either $$~x-6=0~$$ or $$~x+2=0~\text{.}$$ We solve each of these equations.
\begin{align*} x-6 = 0~~~~\amp \text{or}~~~~x+2 = 0 \amp \amp \blert{\text{Solve each equation.}}\\ x=6~~~~ \amp \text{or}~~~~x=-2 \end{align*}
Once again we see that a quadratic equation has two solutions. You can check that both of these values satisfy the original equation.
Checkpoint3.2.5.Practice 1.
Solve the equation $$\quad x^2-11x+24=0$$
$$x=$$ Use a comma to separate different solutions.
$$3, 8$$
Solution.
$$x=3$$ or $$x=8$$
Checkpoint3.2.6.Practice 1.
Solve the equation $$\quad x^2-11x+24=0$$
Solution.
$$x=3$$ or $$x=8$$
Subsection3.2.2X-Intercepts of a Parabola
Recall that the $$x$$-intercept of a line is the point where $$y=0\text{,}$$ or where the line crosses the $$x$$-axis. We find the $$x$$-intercept by setting $$y=0$$ in the equation of the line, and solving for $$x\text{.}$$ We can find the $$x$$-intercepts of a parabola the same way.
Example3.2.7.
Find the $$x$$-intercepts of the graph of $$~y=x^2-4x-12$$
Solution.
To find the $$x$$-intercepts of the graph, we set $$y=\alert{0}$$ and solve the equation
But this is the same equation we solved in the last Example, because
\begin{equation*} x^2-4x-12 = (x-6)(x+2) \end{equation*}
The solutions of that equation were $$6$$ and $$-2\text{,}$$ so the $$x$$-intercepts of the graph are $$(6,0)$$ and $$(-2,0)\text{.}$$ You can see this by graphing the equation on your calculator, as shown in the figure.
We can state a general result: The $$x$$-intercepts of the graph of
\begin{equation*} y=ax^2+bx+c \end{equation*}
are the solutions of the equation
\begin{equation*} 0=ax^2+bx+c \end{equation*}
So we can always solve a quadratic equation to find the $$x$$-intercepts of a parabola (if there are any).
And we can use this relationship the other way round, too: If we know the $$x$$-intercepts of the graph of $$~y=ax^2+bx+c,~$$ we also know the solutions of the equation $$ax^2+bx+c=0\text{.}$$
Checkpoint3.2.8.Practice 2.
Use technology to graph the equation
\begin{equation*} \begin{gathered} y=(x-3)(2x+3) \end{gathered} \end{equation*}
and find the $$x$$-intercepts of the graph. Use your answers to solve the equation
\begin{equation*} \begin{gathered} (x-3)(2x+3)=0 \end{gathered} \end{equation*}
Check your solutions by applying the ZFP.
$$x=$$ Use a comma to separate different solutions.
$$3, \frac{-3}{2}$$
Solution.
$$x=3$$ or $$x=\dfrac{-3}{2}$$
Checkpoint3.2.9.Practice 2.
Use technology to graph the equation
\begin{gather*} y=(x-3)(2x+3) \end{gather*}
and find the $$x$$-intercepts of the graph. Use your answers to solve the equation
\begin{gather*} (x-3)(2x+3)=0 \end{gather*}
Check your solutions by applying the ZFP.
Solution.
$$x=3$$ or $$x=\dfrac{-3}{2}$$
Now we’ll consider some other quadratic equations. Before we apply the ZFP, we must write the equation so that one side is zero.
Example3.2.10.
Solve $$~3x(x+1)=2x+2$$
Solution.
First, we write the equation in standard form.
\begin{align*} 3x(x+1) \amp = 2x+2 \amp \amp \blert{\text{Apply the distributive law to the left side.}}\\ 3x^2+3x \amp = 2x+2 \amp \amp \blert{\text{Subtract}~2x+2~\text{from both sides.}}\\ 3x^2+x-2 \amp = 0 \end{align*}
Now we factor the left side to obtain
\begin{align*} (3x-2)(x+1) \amp = 0 \amp \amp \blert{\text{Apply the zero-factor principle.}}\\ 3x-2=0~~\text{or} ~~x+1 \amp =0 \amp \amp \blert{\text{Solve each equation.}}\\ x=\dfrac{2}{3}~~\text{or} ~~x \amp = -1 \end{align*}
The solutions are $$\dfrac{2}{3}$$ and $$-1\text{.}$$
Caution3.2.11.
When we apply the zero-factor principle, one side of the equation must be zero. For example, to solve the equation
\begin{equation*} (x-2)(x-4) = 15 \end{equation*}
it is incorrect to set each factor equal to 15! (There are many ways that the product of two numbers can equal 15; it is not necessary that one of the numbers be 15.)
We must first simplify the left side and write the equation in standard form. (The correct solutions are $$7$$ and $$-1\text{;}$$ check that you can find these solutions.)
We summarize the factoring method for solving quadratic equations as follows.
To Solve a Quadratic Equation by Factoring.
1. Write the equation in standard form.
2. Factor the left side of the equation.
3. Apply the zero-factor principle: Set each factor equal to zero.
4. Solve each equation. There are two solutions (which may be equal).
Checkpoint3.2.12.Practice 3.
Solve by factoring: $$\quad (t-3)^2=3 (9-t)$$
$$t=$$ Use a comma to separate different solutions.
Hint.
Begin by multiplying out each side of the equation.
$$-3, 6$$
Solution.
$$t=-3$$ or $$t=6$$
Checkpoint3.2.13.Practice 3.
Solve by factoring: $$\quad (t-3)^2=3 (9-t)$$
Hint: Begin by multiplying out each side of the equation.
Solution.
$$t=-3$$ or $$t=6$$
Checkpoint3.2.14.QuickCheck 2.
Which technique, extracting roots or factoring, is better-suited to each equation?
1. $$4x^2-12x=0$$
• extracting roots
• factoring
2. $$6(4x-1)^2=18$$
• extracting roots
• factoring
3. $$(x+4)^2=16x$$
• extracting roots
• factoring
4. $$9x^2-42=0$$
• extracting roots
• factoring
$$\text{factoring}$$
$$\text{extracting roots}$$
$$\text{factoring}$$
$$\text{extracting roots}$$
Solution.
Extracting roots applies to (b) and (d).
Checkpoint3.2.15.QuickCheck 2.
Which technique, extracting roots or factoring, is better-suited to each equation?
1. $$\displaystyle 4x^2-12x=0$$
2. $$\displaystyle 6(4x-1)^2=18$$
3. $$\displaystyle (x+4)^2=16x$$
4. $$\displaystyle 9x^2-42=0$$
Now we can use factoring to solve the opening problem in this section.
Example3.2.16.
The height, $$h\text{,}$$ of a baseball $$t$$ seconds after being hit is given by
\begin{equation*} h=-16t^2+64t+4 \end{equation*}
When will the baseball reach a height of 64 feet?
Solution.
We substitute $$\alert{64}$$ for $$h$$ in the formula, and solve for $$t\text{.}$$
\begin{align*} -16t^2+64t+4 \amp = \alert{64} \amp \amp \blert{\text{Write the equation in standard form.}}\\ 16t^2-64t+60 \amp = 0 \amp \amp \blert{\text{Factor 4 from the left side.}}\\ 4(4t^2-16t+15) \amp = 0 \amp \amp \blert{\text{Factor the quadratic expression.}}\\ 4(2t-3)(2t-50) \amp = 0 \amp \amp \blert{\text{Set each variable factor equal to zero.}}\\ 2t-3=0~~\text{or} ~~2t-5 \amp =0 \amp \amp \blert{\text{Solve each equation.}}\\ t=\dfrac{3}{2}~~\text{or} ~~t \amp = \dfrac{5}{2} \end{align*}
There are two solutions. At $$t=\dfrac{3}{2}$$ seconds, the ball reaches a height of 64 feet on the way up, and at $$t=\dfrac{5}{2}$$ seconds, the ball is 64 feet high on its way down.
Caution3.2.17.
In the Example above, the factor of 4 does not affect the solutions of the equation at all. You can understand why this is true by looking at some graphs. Use technology to graph the equation
\begin{equation*} y_1=x^2-4x+3 \end{equation*}
in the window
\begin{gather*} \text{Xmin}=-2~~~~\text{Ymin}=-5\\ \text{Xmax}=8~~~~\text{Ymax}=10 \end{gather*}
Notice that when $$y=0\text{,}$$ $$x=1$$ or $$x=3\text{.}$$ These two points are the $$x$$-intercepts of the graph. Now on the same window graph
\begin{equation*} y_2=4(x^2-4x+3) \end{equation*}
as shown below.
This graph has the same $$x$$-values when $$y=0\text{.}$$ The factor of 4 makes the graph "skinnier," but does not change the location of the $$x$$-intercepts.
Checkpoint3.2.18.Practice 4.
1. Solve by factoring $$4t-t^2=0\text{.}$$
$$t=$$ Use a comma to separate solutions.
2. Solve by factoring $$20t-5t^2=0\text{.}$$
$$t=$$ Use a comma to separate solutions.
3. Graph $$y=4t-t^2$$ and $$y=20t-5t^2$$ together in the window
\begin{equation*} \begin{aligned} \text{Xmin}\amp = -2\amp \text{Ymin}\amp = -20\\ \text{Xmax}\amp = 6\amp \text{Ymax}\amp = 25 \end{aligned} \end{equation*}
and locate the horizontal intercepts on each graph.
$$t=$$ Use a comma to separate values.
$$0, 4$$
$$0, 4$$
$$0, 4$$
Solution.
$$t=0$$ and $$t=4$$
Checkpoint3.2.19.Practice 4.
1. Solve by factoring $$~4t-t^2=0\text{.}$$
2. Solve by factoring $$~20t-5t^2=0\text{.}$$
3. Graph $$~y=4t-t^2~$$ and $$~y=20t-5t^2~$$ together in the window
\begin{align*} \text{Xmin}\amp = -2\amp \text{Ymin}\amp = -20\\ \text{Xmax}\amp = 6\amp \text{Ymax}\amp = 25 \end{align*}
and locate the horizontal intercepts on each graph.
Solution.
$$t=0$$ and $$t=4$$
Checkpoint3.2.20.QuickCheck 3.
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Checkpoint3.2.21.QuickCheck 3.
Match each equation with its solutions.
1. $$\displaystyle 3t(t-2)=0$$
2. $$\displaystyle t^2-t=2$$
3. $$\displaystyle 3t^2=12$$
4. $$\displaystyle t(t-3)=2(t-3)$$
1. $$\displaystyle -1,2$$
2. $$\displaystyle 0,2$$
3. $$\displaystyle 2,3$$
4. $$\displaystyle -2,2$$
Subsection3.2.4An Application
Here is another example of how quadratic equations arise in applications.
Example3.2.22.
The size of a rectangular computer monitor screen is taken to be the length of its diagonal. If the length of the screen should be 3 inches greater than its width, what are the dimensions of a 15-inch monitor?
Solution.
We express the two dimensions of the screen in terms of a single variable:
\begin{gather*} \text{Width of screen:}~~w\\ \text{Length of screen:}~~w+3 \end{gather*}
We apply the Pythagorean theorem to write an equation:
\begin{equation*} w^2+(w+3)^2=15^2 \end{equation*}
To solve this equation, we begin by simplifying the left side.
\begin{align*} w^2+w^2+6w+9 \amp = 225 \amp \amp \blert{\text{Write the equation in standard form.}}\\ 2w^2+6w-216 \amp = 0 \amp \amp \blert{\text{Factor 2 from the left side.}}\\ 2(w^2+3w-108) \amp = 0 \amp \amp \blert{\text{Factor the quadratic expression.}}\\ 2(w-9)(w+12) \amp = 0 \amp \amp \blert{\text{Set each factor equal to zero.}}\\ w-9=0 ~~~ \text{or} ~~~ w+12 \amp = 0 \amp \amp \blert{\text{Solve each equation.}}\\ w = 9 ~~~ ~~~ \text{or} ~~~ w \amp = -12 \end{align*}
Because the width of the screen cannot be a negative number, the width is 9 inches, and the length is $$~w+3=12~$$ inches.
Checkpoint3.2.23.Practice 5.
Francine is designing the layout for a botanical garden. The plan includes a square herb garden, with a path 5 feet wide through the center of the garden, as shown above. To include all the species of herbs, the planted area must be 300 square feet. Find the dimensions of the herb garden.
$$20$$
$$20$$
Solution.
20 feet by 20 feet
Checkpoint3.2.24.Practice 5.
Francine is designing the layout for a botanical garden. The plan includes a square herb garden, with a path 5 feet wide through the center of the garden, as shown at right. To include all the species of herbs, the planted area must be 300 square feet. Find the dimensions of the herb garden.
Solution.
20 feet by 20 feet
As we have seen in ehe examples above, the solutions of the quadratic equation
\begin{equation*} a(x - r_1)(x - r_2) = 0 \end{equation*}
are $$r_1$$ and $$r_2\text{.}$$ This is called the factored form of the quadratic equation. Thus, if we know the two solutions of a quadratic equation, we can work backwards to reconstruct the equation.
Example3.2.25.
Find a quadratic equation whose solutions are $$\dfrac{1}{2}$$ and $$-3\text{.}$$
Solution.
Each solution corresponds to a factor of the equation, so the equation must look like this:
\begin{equation*} \left(x-\dfrac{1}{2}\right)(x-(-3)) = 0 \end{equation*}
or, simplifying:
\begin{equation*} \left(x-\dfrac{1}{2}\right)(x+3) = 0 \end{equation*}
We multiply the factors together to obtain
\begin{equation*} x^2+\dfrac{5}{2}x-\dfrac{3}{2} = 0 \end{equation*}
This is an equation that works, but we can make a "nicer" one if we clear the fractions. We can multiply both sides of the equation by 2. We know that multiplying by a constant does not change the solutions of the equation.
By factoring, we can check that this equation really does have the given solutions.
\begin{equation*} 0 = 2x^2+5x-3 = (2x-1)(x+3) \end{equation*}
From here, you can see that the solutions are indeed $$\dfrac{1}{2}$$ and $$-3\text{.}$$
Checkpoint3.2.26.Practice 6.
Find a quadratic equation with integer coefficients whose solutions are $$\dfrac{2}{3}$$ and $$-5\text{.}$$
$$=0$$
$$3x^{2}+13x-10$$
Solution.
$$3x^2+13x-10=0$$
Checkpoint3.2.27.Practice 6.
Find a quadratic equation with integer coefficients whose solutions are $$\dfrac{2}{3}$$ and $$-5\text{.}$$
Solution.
$$3x^2+13x-10=0$$
A quadratic equation in one variable always has two solutions. In some cases, the solutions may be equal. For example, the equation
\begin{equation*} x^2-2x+1=0 \end{equation*}
can be solved by factoring as follows:
\begin{align*} (x-1)(x-1) \amp = 0 \amp \amp \blert{\text{Apply the zero-factor principle.}}\\ x-1=0~~~~\text{or}~~~~x-1 \amp = 0 \end{align*}
Both of these equations have solution $$x=1\text{.}$$ We say that 1 is a solution of multiplicity two, meaning that it occurs twice as a solution of the quadratic equation.
Checkpoint3.2.28.QuickCheck 4.
True or false.
1. We find the $$x$$-intercepts of a graph by setting $$x=0$$ .
• True
• False
2. If $$z$$ is a solution of a quadratic equation, then $$(x-z)$$ is a factor of the left side in standard form.
• True
• False
3. We can factor a constant from both sides of a quadratic equation without changing its solutions.
• True
• False
4. To solve a quadratic equation by factoring, we should factor each side of the equation.
• True
• False
$$\text{False}$$
$$\text{True}$$
$$\text{True}$$
$$\text{False}$$
Solution.
1. False
2. True
3. True
4. False
Checkpoint3.2.29.QuickCheck 4.
True or false.
1. We find the $$x$$-intercepts of a graph by setting $$x=0\text{.}$$
2. If $$z$$ is a solution of a quadratic equation, then $$(x-z)$$ is a factor of the left side in standard form.
3. We can factor a constant from both sides of a quadratic equation without changing its solutions.
4. To solve a quadratic equation by factoring, we should factor each side of the equation.
Exercises3.2.6Problem Set 3.2
Warm Up
Exercise Group.
For Problems 1-4, write each product as a polynomial in simplest form.
1.
1. $$\displaystyle (b+6)(2b-3)$$
2. $$\displaystyle (3z-8)(4z-1)$$
1. $$\displaystyle 2b^2+9b-18$$
2. $$\displaystyle 12z^2-35z+8$$
2.
1. $$\displaystyle (4z-3)^2$$
2. $$\displaystyle (2d+8)^2$$
3.
1. $$\displaystyle 3p(2p-5)(p-3)$$
2. $$\displaystyle 2v(v+4)(3v-4)$$
1. $$\displaystyle 6p^3-33p^2+45p$$
2. $$\displaystyle 6v^3+16v^2-32v$$
4.
1. $$\displaystyle -50(1+r)^2$$
2. $$\displaystyle 12(1-t))^2$$
5.
Factor if possible.
1. $$\displaystyle x^2-16$$
2. $$\displaystyle x^2-16x$$
3. $$\displaystyle x^2-8x+16$$
4. $$\displaystyle x^2+16$$
1. $$\displaystyle (x-4)(x+4)$$
2. $$\displaystyle x(x-16)$$
3. $$\displaystyle (x-4)(x-4)$$
4. cannot be factored
6.
Solve if possible.
1. $$\displaystyle x^2-16=0$$
2. $$\displaystyle x^2-16x=0$$
3. $$\displaystyle x^2-8x+16=0$$
4. $$\displaystyle x^2+16=0$$
Exercise Group.
For Problems 7-12, factor completely.
7.
$$x^2-7x+10$$
$$(x-5)(x-2)$$
8.
$$x^2-225$$
9.
$$w^2-4w-32$$
$$(w-8)(w+4)$$
10.
$$2z^2+11z-40$$
11.
$$9n^2+24n+16$$
$$(x-5)(x-2)$$
12.
$$4n^2-28n+49$$
Skills Practice
Exercise Group.
For Problems 13–20, solve by factoring.
13.
$$2a^2+5a-3=0$$
$$\dfrac{1}{2}, ~-3$$
14.
$$3b^2-4b-4=0$$
15.
$$2x^2-6x=0$$
$$0,~3$$
16.
$$3y^2-6y=-3$$
17.
$$x(2x-3)=-1$$
$$\dfrac{1}{2}, ~1$$
18.
$$t(t-3)=2(t-3)$$
19.
$$z(3z+2)=(z+2)^2$$
$$-1,2$$
20.
$$(v+2)(v-5)=8$$
Exercise Group.
For problems 21-24, solve by extracting roots.
21.
$$3(8x-7)^2=24$$
$$\dfrac{7}{8} \pm \dfrac{\sqrt{8}}{8}$$
22.
$$81\left(x+\dfrac{1}{3}\right)^2=1$$
23.
$$(ax-b)^2=25$$
$$\dfrac{b \pm 5}{a}$$
24.
$$100=\pi x^2-16 \pi$$
Exercise Group.
For problems 25 and 26, graph the function in the ZInteger window, and locate the $$x$$-intercepts of the graph. Use the $$x$$-intercepts to write the quadratic expression in factored form.
25.
$$y=0.1(x^2-3x-270)$$
$$0.1(x-18)(x+15)$$
26.
$$y=-0.06(x^2-22x-504)$$
27.
Use technology to graph all three equations in the same window. What do you notice about the $$x$$-intercepts?
1. $$\displaystyle y = x^2+2x-15$$
2. $$\displaystyle y = 3(x^2+2x-15)$$
3. $$\displaystyle y = 0.2(x^2+2x-15)$$
All three graphs have the same $$x$$-intercepts.
28.
Write a quadratic equation with the given solutions. Give your answers in standard form with integer coefficients.
1. $$-2$$ and $$1$$
2. $$-3$$ and $$\dfrac{1}{2}$$
3. $$\dfrac{-1}{4}$$ and $$\dfrac{3}{2}$$
Applications
29.
Delbert stands at the top of a 300-foot cliff and throws his algebra book directly upward with a velocity of 20 feet per second. The height of his book above the ground $$t$$ seconds later is given by the equation
\begin{equation*} h=-16t^2+20t+300 \end{equation*}
where $$h$$ is in feet.
1. Use your calculator or graphing utility to make a table of values for the height equation, with increments of 0.5 second.
3. What is the highest altitude Delbert’s book reaches? When does it reach that height? Use the TRACE feature to find approximate answers first. Then use the Table feature to improve your estimate.
4. When does Delbert’s book pass him on its way down? (Delbert is standing at a height of 300 feet.) Use the intersect command.
5. Write and solve an equation to answer the question: How long will it take Delbert’s book to hit the ground at the bottom of the cliff?
c. 306.5 ft at 0.625 sec d. 1.25 sec e. 5 sec
30.
The annual increase $$I$$ in the deer population in a national park is given by the formula
\begin{equation*} I=1.2x-0.0002x^2 \end{equation*}
where $$x$$ is the size of the population that year.
1. Make a table of values for $$I$$ for $$0 \le x \le 7000.$$ Use increments of 500 in $$x\text{.}$$
2. How much will a population of 2000 deer increase? A population of 5000 deer? A population of 7000 deer?
3. Use your calculator to graph the annual increase versus the size of the population, $$x\text{,}$$ for $$0 \le x \le 7000.$$ Use your table from part (b) to help you choose appropriate values for Ymin and Ymax.
4. What do the $$x$$-intercepts tell us about the deer population?
5. Estimate the population size that results in the largest annual increase. What is that increase?
31.
One end of a ladder is 10 feet from the base of a wall, and the other end reaches a window in the wall. The ladder is 2 feet longer than the height of the window.
1. Choose a variable for the height of the window. Make a sketch of the situation described, and label the sides of a right triangle.
3. Solve your equation to find the height of the window.
b. $$h^2+10^2=(h+2)^2$$ c. 24 ft
32.
Irene would like to enclose two adjacent chicken coops of equal size against the henhouse wall. She has 66 feet of chicken wire fencing, and she would like the total area of the two coops to be 360 square feet. What should the dimensions of the chicken coops be?
We’ll use three methods to solve this problem: a table of values, a graph, and an algebraic equation.
1. Make a table by hand that shows the areas of coops of various widths, as shown below.
Width Length Area $$4$$ $$27$$ $$216$$ $$6$$ $$24$$ $$288$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$ $$\hphantom{0000}$$
Continue the table until you find a pair of chicken coops whose total area is 360 square feet. (Be careful computing the length of each chicken coop: look at the diagram above.)
2. Write an expression for the length of each of the two coops if their width is $$x\text{.}$$ Then write an expression for the combined area of the coops if their width is $$x\text{.}$$ Graph the equation for $$A\text{,}$$ and use the graph to find the pair of coops whose combined area is 360 square feet. (Is there more than one solution?)
3. Write an equation for the area $$A$$ of the two coops in terms of their width, $$x\text{.}$$ Solve your equation algebraically for $$A=360.$$
33.
A box is made from a square piece of cardboard by cutting 2-inch squares from each corner and then turning up the edges.
1. If the piece of cardboard is $$x$$ inches square, write expressions for the length, width, and height of the box. Then write an expression for the volume, $$V\text{,}$$ of the box in terms of $$x\text{.}$$
2. Use your calculator to make a table of values showing the volumes of boxes made from cardboard squares of side 4 inches, 5 inches, and so on.
3. Graph your expression for the volume on your calculator. What is the smallest value of $$x$$ that makes sense for this problems
4. Use your table or your graph to find what size cardboard you need to make a box with volume 50 cubic inches.
1. $$\displaystyle l=x-4,~w=x-4,~h=2,~V=2(x-4)^2$$
2. 0 cubic in, 2 cubic in, 8 cubic in, etc.
3. $$\displaystyle x = 4$$
4. 9 in by 9 in
5. $$\displaystyle 2(x-4)^2 = 50;~ x=9$$
34.
A length of rain gutter is made from a piece of aluminum 6 feet long and 1 foot wide.
1. If a strip of width $$x$$ is turned up along each long edge, write expressions for the length, width and height of the gutter. Then write an expression for the volume $$V$$ of the gutter in terms of $$x\text{.}$$
2. Use your calculator to make a table of values showing the volumes of various rain gutters formed by turning up edges of 0.1 foot, 0.2 foot, and so on.
3. Graph your expression for the volume. What happens to $$V$$ as $$x$$ increases?
4. Use your table or your graph to discover how much metal should be turned up along each long edge so that the gutter has a capacity of $$\dfrac{3}{4}$$ cubic foot of rainwater.
a. $$l=6, ~w=1-2x, ~h=x, ~V=6x(1-2x)~~$$ e. $$6x(1-2x)=\dfrac{3}{4};~~\dfrac{1}{4}~$$ ft
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# Measurement of Angles
Last Updated on May 9, 2020 by Alabi M. S.
MATHEMATICS
THEME: Mensuration and Geometry
TOPIC: Measurement of Angles
PERFORMANCE OBJECTIVES
By the end of the lesson, the pupils should have attained the following objectives (cognitive, affective and psychomotor):
1. Measure angles in degrees using clock;
1. Measure angles in degrees using protractor;
1. Measure angles in a plane.
ENTRY BEHAVIOR
The pupils are required to already have taught and learned measurement of angle
INSTRUCTIONAL MATERIALS
The teacher will teach the lesson with the aid of Instructional Mathematical Set, and protractor.
METHOD OF TEACHING:
MAIN REFERENCE MATERIALS
Prime Mathematics book 6 (PMB 273); 2. New Method Mathematics book 6 (NMM 205).
CONTENT OF THE LESSON/LESSON NOTE
LESSON 1
PERIOD:
DATE:
TIME:
A protractor is used to measure angles. A full turn or circle is divided into 360 parts. Each part is called a degree. One degree is written as 1°. Some protractors are semi-circles as shown below.
1. ANGLE 90°
2. ANGLE 72°
3. ANGLE 127°
4. ANGLE 18°
LESSSON EVALUATION – Measure these angles with your protractor.
1.
2.
3.
LESSON 2
PERIOD:
DATE:
TIME:
MEASUREMENT OF ANGLES
EXAMPLES – Using a protractor, draw
1. Angle 75°
2. Angle 145°
LESSON EVALUATION – Using a protractor, draw:
1. Angle 45
2. Angles 66
3. Angles 123
LESSON 3
PERIOD:
DATE:
TIME:
MEASUREMENT OF ANGLESTYPES OF ANGLES
1. An acute angle is less than 90°
2. A right angle Is 90°
3. An obtuse angle is more than 90° but less than 180°
4. A reflex angle is more than 180° but less than 360°
NOTE: Angle on a straight line is 180° and Angle at a point is 360°
LESSON 4
PERIOD:
DATE:
TIME:
MEASUREMENT OF ANGLE
EXAMPLE
28° + < ADC = 180°
< ADC = 180° – 28° = 152°
LESSON EVALUATION – Measure the angle marked x with your protractor. Calculate the angle marked y.
1.
2.
LESSON 5
PERIOD:
DATE:
TIME:
MEASUREMENT OF ANGLE IN A PLANETRIANGLE
Measure the three angles of these triangles
1.
2.
Note – sum of angles in a triangle is 180°
Further exercise will be taken from the listed reference materials.
PRESENTATION
To deliver the lesson, the teacher adopts the following steps:
1. To introduce the lesson, the teacher revises the previous lesson- time. Based on this, he/she asks the pupils some questions;
1. Guides pupils in the correct placement of protractors to measure a given angle in degree;
1. Pupils Activities – Use protractor and ruler to measure correctly the given angle.
1. Guides pupils to indicate the edges, vertices and faces of given three dimensional shapes;
1. Pupils Activities – Identify the number of edges, vertices and faces for different shapes.
1. Guides pupils to measure the size of angle, indicate lines that are parallel and perpendicular in each of the given shapes;
1. Pupils Activities – Use protractors to determine the sizes of angles in a given shape; Identify lines that are parallel and perpendicular in three dimensional shapes.
CONCLUSION
• To conclude the lesson for the week, the teacher revises the entire lesson and links it to the following week’s lesson (polygon).
Great Is Thy Faithfulness…
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Fibonacci Numbers¶
The Fibonacci numbers are a famous sequence of numbers defined by the following recursion rule:
$$\begin{matrix} F_0 &=& 0 &\\ F_1 &=& 1 &\\ F_n &=& F_{n-1} + F_{n-2} &\text{for } n>2 \end{matrix}$$
The first few Fibonacci numbers are: $0,1,1,2,3,5,8,13,21, \dots$
Fibonacci numbers are fascinating because they appear in many areas of mathematics. They are also closely related to the golden ratio:
$$\varphi = \frac{1 + \sqrt{5}}{2} = 1.61803\dots$$
The ratio of successive Fibonacci numbers $F_{n+1}/F_n$ approximates $\varphi$ as $n$ gets larger.
Today, we will see three different ways of computing $F_n$ using Python:
1. Directly, using the formula for $F_n$
2. By taking powers of a matrix $F$
3. By finding the diagonal decomposition of $F$, and then taking powers
Computing using recursion¶
The rule $F_n = F_{n-1} + F_{n-2}$ is sometimes called a recursive rule. This just means that future values depend on past values according to some fixed formula.
Let's implement this in Python, and use this to print the first 10 Fibonacci numbers. Can you understand what the code is doing?
In [1]:
def fib(n):
if n == 0:
return 0
elif n == 1:
return 1
else:
return fib(n-1) + fib(n-2)
for i in range(10):
print(fib(i))
0
1
1
2
3
5
8
13
21
34
Computing using matrix powers¶
As we saw in our lecture, the Fibonacci numbers are also given by the matrix equation
$$\begin{pmatrix} F_{n+1} \\ F_n \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} F_{n} \\ F_{n-1} \end{pmatrix}$$
We can also write this as:
$$\begin{pmatrix} F_{n+1} \\ F_n \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n \begin{pmatrix} F_1 \\ F_0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$
Let $F$ be the matrix $\begin{pmatrix} 1& 1 \\ 1 & 0 \end{pmatrix}$.
Taking powers of $F$ gives us another way to computing $F_n$. We can do this in Python using the command np.linalg.matrix_power(F,n). In our code, the variable Fn_vector is the vector $(F_{n+1}, F_n)$, but since we only want $F_n$, we extract entry $1$ (recall that Python indexing starts at 0). Let's print the first 10 Fibonacci numbers to make sure we've got it right:
In [2]:
import numpy as np
F = np.array([[1,1], [1,0]])
def fib_mat(n):
Fn_vector = np.linalg.matrix_power(F,n) @ np.array([1,0])
return Fn_vector[1]
for i in range(10):
print(fib_mat(i))
0
1
1
2
3
5
8
13
21
34
This is nice. We can even compute negative Fibonacci numbers! (Why does this make sense?)
In [3]:
for i in range(-5, 5):
print(fib_mat(i))
5.0
-3.0
2.0
-1.0
1.0
0
1
1
2
3
But calculating $F^n$ requires calculating the previous powers too, so fib_mat isn't really that different from fib. Here are the first ten powers of $F$:
In [4]:
for i in range(10):
print(np.linalg.matrix_power(F,i), "\n")
[[1 0]
[0 1]]
[[1 1]
[1 0]]
[[2 1]
[1 1]]
[[3 2]
[2 1]]
[[5 3]
[3 2]]
[[8 5]
[5 3]]
[[13 8]
[ 8 5]]
[[21 13]
[13 8]]
[[34 21]
[21 13]]
[[55 34]
[34 21]]
Computing using the diagonal decomposition of $F$¶
It turns out that $F$ can be diagonally decomposed! We can compute the diagonal decomposition, also called the eigendecomposition, using the np.linalg.eig command:
In [5]:
np.linalg.eig(F)
Out[5]:
(array([ 1.61803399, -0.61803399]),
array([[ 0.85065081, -0.52573111],
[ 0.52573111, 0.85065081]]))
This function returns 2 things:
1. A vector $d$
2. A matrix $M$
Let's store these into variables d and M:
In [6]:
d, M = np.linalg.eig(F)
Let's also define a diagonal matrix D that given by d along the diagonal:
In [7]:
D = np.diag(d)
D
Out[7]:
array([[ 1.61803399, 0. ],
[ 0. , -0.61803399]])
Then $F = M D M^{-1}$, as we may verify:
In [8]:
M @ D @ np.linalg.inv(M)
Out[8]:
array([[ 1.00000000e+00, 1.00000000e+00],
[ 1.00000000e+00, -1.11022302e-16]])
Actually, it turns out that $M$ is orthogonal!
In [9]:
M @ M.T
Out[9]:
array([[1., 0.],
[0., 1.]])
So $F$ is given by $M D M^T$:
In [10]:
M @ D @ M.T
Out[10]:
array([[ 1.00000000e+00, 1.00000000e+00],
[ 1.00000000e+00, -5.55111512e-17]])
This week's homework is very simple: use d and M to compute $F_n$.
Homework¶
Copy the following script into Spyder, then modify it so that the function my_fib(n) gives $F_n$. Test your function by printing the first 10 Fibonacci numbers (don't worry about rounding errors).
In [70]:
import numpy as np
F = np.array([[1,1],[1,0]])
d,M = ????? # diagonal decomposition of F, stored into a vector v and a matrix M
def my_fib(n):
dn = ????? # vector with entries of d to the power of n
Dn = ????? # diagonal matrix with dn along the diagonal
Fn = ????? # the matrix F to the power of n. Do not use np.linalg.matrix_power or np.linalg.inv!
Fn_vector = Fn @ np.array([1,0])
return Fn_vector[1]
for i in range(10):
print(my_fib(i))
It may help to trying defining dn,Dn,Fn outside the function first, so you can see what they look like and whether they're correct.
Also remember that Python uses "v**n" to denote powers. This works even when v is a vector!
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# The Classic Crossed Ladders Puzzle
Here is a classic puzzle. A pair of ladders leaning against the sides of an alley form a lopsided cross. Each ladder is propped against the base of one wall and leans against the opposite wall. If one ladder is 30 feet long, the other 20 feet long, and the point where they cross 10 feet above the ground, how wide is the alley?
### Contents
#### Eight variables
The lengths of the ladders are denoted by $a$ and $b$. The height of the crossing is $c$. The heights of the points where the ladders reach the walls are $x$ and $y$. The bases of the right triangles whose common height is the crossing are $u$ and $v$. Finally, the width of the alley is $w$. Here is the picture.
#### The equations
These eight variables are connected by five equations. The Pythagorean Theorem provides two equations.
$$a^2 = x^2 + w^2$$
$$b^2 = y^2 + w^2$$
The ratios between sides of similar triangles provide two more.
$$\frac{x}{w} = \frac{c}{v}$$
$$\frac{y}{w} = \frac{c}{u}$$
The fifth equation is a simple linear relation that ties the two triangles together.
$$w = u + v$$
Five equations involving eight variables leaves three degrees of freedom. In principle we could specify any three of the variables and solve for the other five. But that may or may not be easy. We will have two different situations in what follows. The puzzle posed in the opening paragraph specifies $a$, $b$ and $c$, and asks to find $w$. This is nontrivial. On the other hand, our graphic app is driven by $x$, $y$ and $w$. When these three are pinned down by mouse clicks, the other five follow immediately.
#### Harmonic mean
Combining the last three equations generates an elegant relation. This isn't a new equation; it's a consequence of the ones we already have.
$$\frac{1}{c} = \frac{1}{x} + \frac{1}{y}$$
This says that the height of the crossing is one-half of the harmonic mean of the heights of the two ladders. The equation is a familiar one in optics, where it is known as the thin lens equation. It relates the location of an image to the object distance and focal length.
#### Smallest integer solution
The following picture shows a solution of these equations where all eight variables have integer values. In fact, if we collect all eight variables into a vector $s$.
$$s = [a, b, c, u, v, w, x, y]$$
And use the 1-norm to measure the "size" of a solution.
$$||s||_1 = \sum_{i=1}^8 s_i$$
Then ITOT ("It Turns Out That") this is smallest solution with all integer elements.
#### Puzzle
Combining Pythagoras and optics provides a single nonlinear equation for $w$ in terms of $a$, $b$ and $c$.
$$\frac{1}{\sqrt{a^2 - w^2}}+\frac{1}{\sqrt{b^2 - w^2}}=\frac{1}{c}$$
It is not difficult to use the one-dimensional MATLAB zero finder fzero with fixed values for a, b and c to compute a solution of this equation.
a = 30;
b = 20;
c = 10;
F = @(w) 1./sqrt(a^2 - w.^2) + 1./sqrt(b^2 - w.^2) - 1/c
w = fzero(F,c)
F =
@(w)1./sqrt(a^2-w.^2)+1./sqrt(b^2-w.^2)-1/c
w =
12.3119
So this is the answer to the puzzle. For the prescribed values of the lengths of the ladders and the height of the crossing point, the width of the alley has to be about 12.3 feet.
A graph of F(w) in the vicinity of its zero shows why fzero has no trouble.
ezplot(F,[10,15])
set(gca,'xaxislocation','origin')
line(w,0,'marker','.','markersize',24,'color','k')
#### The gorilla in the room
It is easy to solve our single nonlinear equation numerically to compute $w$, but to find an analytic solution is a formidable challenge, even for computer algebra systems. Historically, the approach has focused on an equivalent quartic polynomial.
Take $a = 40$, $b = 30$ and $c = 20$. Let
$$z = w^2$$
Multiply through by the expressions in the denominators to put everything on one level.
$$10\, \sqrt{400 - z} + 10\, \sqrt{900 - z} = \sqrt{400 - z}\, \sqrt{900 - z}$$
Now square both sides to get rid of the sqrt's on the right. Then rearrange terms and square everything again to eliminate the remaining sqrt's. If you're careful, you will eventually reach a polynomial of degree 4 in $z$.
$$z^4 - 2200\, z^3 + 1630000\, z^2 - 454000000\, z + 38500000000 = 0$$
But we're only halfway there. We still have to solve this quartic. In principle it is possible to do this analytically, but let's again abandon algebraic and resort to numeric techniques.
poly = [ 1, -2200, 1630000, -454000000, 38500000000];
z = roots(poly)
z =
1.0e+02 *
8.4877 + 0.5881i
8.4877 - 0.5881i
3.5087 + 0.0000i
1.5158 + 0.0000i
The repeated squaring has produced extraneous roots. ITOT that we can recover $w$ from the fourth one.
w = sqrt(z(4))
w =
12.3119
It is reassuring to find the same width.
I am having a lot of fun with an interactive graphical experience where you vary any one of four parameters and see how it affects the others. You can change the height of the points where the ladders hit the wall, or change the width of the alley. You will find that dragging any one of these three control points affects only a couple of the other quantities.
You will see more action when you vary the crossing point. This changes all of the values except the width. Of course, it is not physically realistic to expect to alter the lengths of the ladders by changing where they cross, but that would actually have to happen if they were constrained to meet the walls.
The complete program for this app is the subject for my blog in two weeks. The title is "Investigating the Classic Crossed Ladders Puzzle". This link should be good after March 14.
#### Acknowledgement
Thanks to Ned Gulley of MathWorks for suggesting that this classic puzzle warrants another look.
Published with MATLAB® R2016a
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# How to Understand Algebra
A quick and easy guide to learning algebra basics
Understanding algebra can seem tricky at first. But if you build up a strong basic knowledge of beginner math facts and learn some of the “language” of algebra, you can understand it much more easily. The basic steps for solving algebra problems involve performing simple operations in small steps that “cancel” the original problem. Doing these steps carefully and in order should get you to the solution.
## Things You Should Know
• Read problem instructions carefully. Look for key words like “solve,” “simplify,” “factor,” or “reduce" so you know what action to perform.
• Use the order of operations to solve problems in the proper sequence: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
• Remember that an equation has an equal sign and can be solved. An expression can only be factored or simplified.
Part 1
Part 1 of 5:
### Algebra Objectives
1. 1
Read the problem instructions carefully. When you have one or more algebra problems, you must read the instructions carefully. Look for key words in the instructions like “solve,” “simplify,” “factor,” or “reduce.” These are some of the most common instructions (although there are others that you will learn). Many people have problems because they try to “solve” a problem when they really only need to “simplify” it.[1]
2. 2
Perform the operations that are instructed. When you read the problem instructions, you should identify the key words and then perform those operations. Many people feel frustration with algebra when they try doing something that is not really part of the intended problem. The basic operations you will be asked for are:[2]
• Solve. You will need to reduce the problem to an actual numerical solution, such as “x=4.” You need to find a value for the variable that can make the problem come true.
• Simplify. You need to manipulate the problem into some simpler form than before, but you will not wind up with what you might consider “an answer.” You will probably not have a single numerical value for the variable.
• Factor. This is similar to “simplify,” and is usually used with complex polynomials or fractions. You need to find a way to turn the problem into smaller terms. Just as the number 12 can be broken into factors of 3x4, for example, you can factor an algebraic polynomial.
• For example, a simple expression like ${\displaystyle 5x}$ can be broken into factors of ${\displaystyle 5}$ and ${\displaystyle x}$.
• For example, the expression ${\displaystyle x^{2}+3x+2}$ can be factored into the terms ${\displaystyle (x+2)}$ and ${\displaystyle (x+1)}$.
• Reduce. To “reduce” a problem generally involves a combination of factoring and then simplifying. You would break the terms of a numerator and denominator into their factors. Then look for common factors on top and bottom, and cancel them out. Whatever remains is the “reduced” form of the original problem. For example, reduce the expression ${\displaystyle {\frac {6x^{2}}{2x}}}$ as follows:
• 1. Factor the numerator and denominator: ${\displaystyle {\frac {(3)(2)(x)(x)}{(2)(x)}}}$
• 2. Look for common terms. Both the numerator and denominator have factors of 2 and x.
• 3. Eliminate the common terms: ${\displaystyle {\frac {(3)(2)(x)(x)}{(2)(x)}}}$
• 4. Copy down what remains: ${\displaystyle 3x}$
3. 3
Learn the difference between “expression” and “equation.” In algebra, the difference between an “expression” and an “equation” is very important. An expression is any group of numbers and variables, collected together. Some examples of expressions are ${\displaystyle x}$, ${\displaystyle 14xyz}$ and ${\displaystyle {\sqrt {2x+15}}}$. All you can do to an expression is simplify or factor it. An equation, on the other hand, contains an = sign. You can simplify or factor equations, but you can also solve them to get a final answer. It is important to look for the difference.[3]
• If you have an expression, like ${\displaystyle 4x^{2}}$, you can never find a single “answer” or “solution.” You could find out that if ${\displaystyle x=1}$, then the expression would have a value of 4, and if ${\displaystyle x=2}$, then the expression would have a value of ${\displaystyle (4)(2)^{2}}$, which is 16. But you cannot get a single “answer."
Part 2
Part 2 of 5:
### Order of Operations
1. 1
Learn PEMDAS. In algebra, the steps you take must occur in a logical order, which is called the “order of operations.” This is often simplified by the mnemonic device “PEMDAS.” The letters of PEMDAS will help you know which operations to perform in order. The letters of PEMDAS stand for:[4]
• Parentheses.
• Exponents.
• Multiplication.
• Division.
• Subtraction.
2. 2
Perform operations inside parentheses first. When you have an expression or equation that includes terms inside parentheses, you need to do whatever is inside the parentheses first. Consider the difference between ${\displaystyle 5*3+2}$ and ${\displaystyle 5*(3+2)}$.[5]
• Without the parentheses, the first expression , ${\displaystyle 5*3+2}$, would become ${\displaystyle 15+2=17}$.
• With the parentheses, ${\displaystyle 5*(3+2)}$, you perform the (3+2) first, so the simplified expression becomes ${\displaystyle 5*5=25}$.
3. 3
Simplify any exponents next. Exponents need to be performed as the next part of simplifying or solving a problem. Consider the expression ${\displaystyle 3*2^{2}}$. Without the order of operations, you wouldn’t know if you should first multiply ${\displaystyle 3*2}$ and then square the result, so your value is 36, or if you square the 2 first, then multiply by 3. Using PEMDAS, the correct operation is:[6]
• ${\displaystyle 3*2^{2}}$
• ${\displaystyle 3*4}$…..Square the 2 first.
• ${\displaystyle 12}$…..This is the expected result.
4. 4
Multiply or divide, from right to left. M and D are the next two parts of PEMDAS, and they go together. After performing any exponents, you then perform multiplication or division from left to right.[7]
• ${\displaystyle 3+4*2-6/3}$
• ${\displaystyle 3+8-2}$…..4*2=8, and 6/3=2. These can be done in the same step.
5. 5
Add or subtract, from right to left. A and S are the final steps of PEMDAS. These mean that you add or subtract whatever terms remain in the expression. You can perform addition and subtraction in the same step, moving from right to left through the problem. Consider the expression ${\displaystyle 4+2-3-1-5+2}$:[8]
• ${\displaystyle 4+2-3-1-5+2}$
• ${\displaystyle 6-3-1-5+2}$…..(Add 4+2)
• ${\displaystyle 3-1-5+2}$…..(Subtract 6-3)
• ${\displaystyle 2-5+2}$…..(Subtract 3-1)
• ${\displaystyle -3+2}$…..(Subtract 2-5)
• ${\displaystyle -1}$…..(Add -3+1)
• If you perform the steps in any other order, you may come up with a different, incorrect result. For example, suppose you chose to do all the additions first, and then the subtractions:
• ${\displaystyle 4+2-3-1-5+2}$
• ${\displaystyle 6-3-1-7}$…..(Add 4+2 and add 5+2)
• ${\displaystyle 3-1-7}$…..(Subtract 6-3)
• ${\displaystyle 2-7}$…..(Subtract 3-1)
• ${\displaystyle -5}$…..(Subtract 2-7. This gives a result of -5, which is incorrect.)
Part 3
Part 3 of 5:
### Variables
1. 1
Get used to symbols other than numbers. In early math, you worked only with numbers. Learning algebra is about being able to solve problems with unknown terms. These unknown terms are represented in the problems with letters. You need to get used to treating these letters like numbers, although you may not know their actual value yet. Some common examples of variables include:[9]
• Letters, such as ${\displaystyle x}$, ${\displaystyle y}$ or ${\displaystyle z}$
• Greek symbols, such as ${\displaystyle \theta }$, ${\displaystyle \alpha }$ or ${\displaystyle \sigma }$.
• Be aware that some symbols might look like variables but are actually known numbers. For example, the Greek symbol pi, ${\displaystyle \pi }$, stands for the number 3.1415.
2. 2
Consider the variable as an unknown place holder. If you think of the phrase, “Two times some number,” you can express that with a variable as ${\displaystyle 2*x}$. The variable ${\displaystyle x}$ takes the place of the unknown “some number.” Usually, your job in an algebra problem is to find the value of the variable.[10]
• For example, when you start with the equation ${\displaystyle 4+x=9}$, you need to think, “What number added to 4 will make 9?” The solution is 5, which you can write algebraically as ${\displaystyle x=5}$.
3. 3
Combine common variables together. When you learn to treat the variables as numbers, you can combine or simplify them as you do with numbers. This is usually referred to as “combining like terms.”[11]
• For example, ${\displaystyle 2x+3x=10}$ just means that 2 of some variable added to 3 of the same variable will equal 10. If you have 2 of something and 3 of the same thing, you can add them together. Then, ${\displaystyle 2x+3x}$ will become 5x, so your problem is ${\displaystyle 5x=10}$, and the solution is ${\displaystyle x=2}$.
• You can only add or subtract the same variable. Some algebra problems may contain two or more variables. In the problem ${\displaystyle 2x+3y=10}$, you cannot combine the ${\displaystyle x}$ and ${\displaystyle y}$ terms together because the different variables represent different unknown numbers.
Part 4
Part 4 of 5:
### Inverse Operations
1. 1
Learn the concept of inverse functions. One key to being successful in algebra is performing inverse functions. The word “inverse” means opposite. Inverse functions are a way of undoing or untangling a problem. If a chosen problem, for example, contains multiplication, you will use division, which is the inverse of multiplication, to solve the problem.[12]
• The inverse of addition is subtraction.
• The inverse of subtraction is addition.
• The inverse of multiplication is division.
• The inverse of division is multiplication.
• The inverse of an exponent is a root (square root, cube root, etc.).
2. 2
Focus on isolating the variable. If you are asked to “solve” an equation, this means that you want to end up with ${\displaystyle x=}$__, with some number in the blank space. You need to use algebra to move everything else away from the ${\displaystyle x}$ term so it is alone on one side of the equals sign. You will do this with a series of inverse operations.[13]
• The key rule to remember is that any operation you make to one side of the equation, you must also do the same to the opposite side of the equation. This will keep the equation balanced and still equal.
3. 3
Cancel addition by using subtraction (and vice versa). Individual terms in an equation are linked by a combination of plus and minus signs. You can “cancel” these to get the variable alone by doing the opposite function.[14]
• For example, if you start with ${\displaystyle x+3=7}$, you want the ${\displaystyle x}$ alone. The inverse of ${\displaystyle +3}$ is ${\displaystyle -3}$. Remember that you must do everything equally to both sides of the equation. So you will get:
• ${\displaystyle x+3=7}$
• ${\displaystyle x+3-3=7-3}$…..(subtract 3 equally on both sides)
• ${\displaystyle x=4}$…..(the +3 and -3 cancel each other out to leave the solution)
• If you start with a subtraction problem, you will cancel it the same way with addition:
• ${\displaystyle x-8=12}$
• ${\displaystyle x-8+8=12+8}$…..(add 8 to both sides)
• ${\displaystyle x=20}$…..(the +8 and -8 cancel each other out to leave the solution)
4. 4
Cancel multiplication by using division (and vice versa). In the same way, you can perform inverse operations on multiplication and division. A term like ${\displaystyle 3x}$ means ${\displaystyle 3*x}$. To get the variable alone, you will divide. Remember that for an equation, you must divide both sides of the equation equally.[15]
• Consider the problem ${\displaystyle 3x=24}$. Since this is a multiplication problem, you will solve it with division:
• ${\displaystyle 3x=24}$
• ${\displaystyle {\frac {3x}{3}}={\frac {24}{3}}}$…..(Divide both sides equally by 3. Note that the ${\displaystyle \div }$symbol is not usually used in algebra. Instead, show division by writing the terms as a fraction.)
• ${\displaystyle x=8}$…..(the 3s on the left cancel each other out to leave the solution)
• Do the same to cancel a division problem with multiplication. Consider the problem ${\displaystyle {\frac {x}{4}}=9}$:
• ${\displaystyle {\frac {x}{4}}=9}$
• ${\displaystyle {\frac {x}{4}}*4=9*4}$…..(multiply both sides by 4)
• ${\displaystyle x=36}$…..(the 4s on the left cancel each other out to leave the solution)
5. 5
Use a combination of add/subtract and multiply/divide. As problems become more complicated, you may have to perform multiple operations to get to a solution. You will usually use addition and subtraction first, to isolate the variable with its coefficient. Then you will use multiplication or division to find the solution.[16]
• ${\displaystyle 3x+5=23}$
• ${\displaystyle 3x+5-5=23-5}$…..(first, subtract 5 from both sides to leave the x term alone)
• ${\displaystyle 3x=18}$…..(the +5 and -5 cancel out on the left)
• ${\displaystyle {\frac {3x}{3}}={\frac {18}{3}}}$…..(divide both sides by 3)
• ${\displaystyle x=6}$…..(the 3s on the left cancel each other out, leaving the solution)
6. 6
Check your result. In algebra, you can almost always find out if you have done the problem correctly by checking your answer. Take the solution that you found, and insert it back in the original problem in place of the variable. Then simplify the problem, and if you reach a true statement, you solution was correct.
• Try the example you just solved, ${\displaystyle 3x+5=23}$. Put the solution of ${\displaystyle x=6}$ in place of the variable:
• ${\displaystyle 3x+5=23}$
• ${\displaystyle 3(6)+5=23}$…..(Insert the value ${\displaystyle x=6}$.)
• ${\displaystyle 18+5=23}$…..(Simplify the equation.)
• ${\displaystyle 23=23}$….. (This is true, so your solution of ${\displaystyle x=6}$ is correct.)
Part 5
Part 5 of 5:
### Foundational Math
1. 1
Learn the basic math facts. Algebra is a system of manipulating numbers and operations to try to solve problems. When you learn algebra, you will learn the rules to follow for solving problems. But to help make that easier, you need to have a strong understanding of basic math facts. You should know basic addition, subtraction, multiplication and division facts and be able to work with them easily. In particular, you should be able to do the following:[17]
• Know your multiplication tables from 1 through 12.
• Know division and factors for numbers up through 144 (12x12).
2. 2
Practice the rules of fractions. Algebra uses the rules of fractions as much as any other numbering system. You need to be comfortable with finding common denominators, adding and subtracting fractions, multiplying and dividing fractions. When you learn algebra, you will expand this knowledge into working with unknown variables, but you need a strong understanding of the basics first.[18]
• Know the importance of reciprocals. You need to know the concept of reciprocal numbers. The short definition of a reciprocal is that it is a fraction turned upside down. Thus, the reciprocal of ${\displaystyle {\frac {2}{3}}}$ is ${\displaystyle {\frac {3}{2}}}$, and the reciprocal of ${\displaystyle {\frac {4}{5}}}$ is ${\displaystyle {\frac {5}{4}}}$. You use reciprocals as an alternative to division, when the problem is complicated. Instead of dividing by one fraction, you can multiply by its reciprocal.
3. 3
Know how to use negative numbers. You will often be using negative numbers or variables. You should review how to add, subtract, multiply, and divide negatives before starting to learn algebra. Here are some basic rules for working with negatives.[19] You can also see our articles on adding and subtracting negative numbers and dividing and multiplying negative numbers.
• On a number line, a negative number is the same distance from zero as the positive, but in the opposite direction.
• A negative plus a negative will also be negative. Adding two negative numbers together makes the number more negative.
• Two negative signs together cancel each other out. Subtracting a negative number is the same as adding a positive number.
• 4-(-3) is the same as 4+3 = 7.
• Multiplying or dividing two negative numbers gives a positive answer.
• Multiplying or dividing one positive number and one negative number gives a negative answer.
Search
• Question
When do you learn algebra?
Daron Cam
Algebra Tutor
Daron Cam is an Academic Tutor and the Founder of Bay Area Tutors, Inc., a San Francisco Bay Area-based tutoring service that provides tutoring in mathematics, science, and overall academic confidence building. Daron has over eight years of teaching math in classrooms and over nine years of one-on-one tutoring experience. He teaches all levels of math including calculus, pre-algebra, algebra I, geometry, and SAT/ACT math prep. Daron holds a BA from the University of California, Berkeley and a math teaching credential from St. Mary's College.
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You typically learn it in high school. Middle school math is typically pre-algebra, although you may dip your toes in algebra when you're in 8th grade if you're taking an advanced class or something.
• Question
Do I need to know algebra to learn calculus?
Daron Cam
Algebra Tutor
Daron Cam is an Academic Tutor and the Founder of Bay Area Tutors, Inc., a San Francisco Bay Area-based tutoring service that provides tutoring in mathematics, science, and overall academic confidence building. Daron has over eight years of teaching math in classrooms and over nine years of one-on-one tutoring experience. He teaches all levels of math including calculus, pre-algebra, algebra I, geometry, and SAT/ACT math prep. Daron holds a BA from the University of California, Berkeley and a math teaching credential from St. Mary's College.
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Yes, absolutely. The courses even build on one another. It's sort of like a pyramid, and if you want to learn calculus at the top, you'll need the algebra underneath it.
• Question
If b=2, what is 10b squared?
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## Tips
• Work with your teacher. If you have questions or problems, speak to your teacher. Some people can understand algebra very quickly, but other people just need some more time. Your teacher may also have another way to explain things to you. Instead of giving up, go ask for some help.[20]
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• Always check your answers. Whenever you finish a problem, work back through it to see if your solution makes the equation check out correctly.
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• Use good study skills. Attend classes, do the assigned readings, and complete your homework. Understanding algebra requires practice.
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This article was co-authored by Daron Cam. Daron Cam is an Academic Tutor and the Founder of Bay Area Tutors, Inc., a San Francisco Bay Area-based tutoring service that provides tutoring in mathematics, science, and overall academic confidence building. Daron has over eight years of teaching math in classrooms and over nine years of one-on-one tutoring experience. He teaches all levels of math including calculus, pre-algebra, algebra I, geometry, and SAT/ACT math prep. Daron holds a BA from the University of California, Berkeley and a math teaching credential from St. Mary's College. This article has been viewed 299,870 times.
Co-authors: 34
Updated: March 9, 2023
Views: 299,870
Categories: Algebra
Article SummaryX
To understand algebra, start by learning addition, subtraction, multiplication, and division facts, and how to do these operations on fractions and negative numbers. Understand the differences between solving, simplifying, factoring, and reducing. Know that an expression is a collection of numbers and variables which can be simplified or factored, but an equation, which always has an “=” sign, can also be solved. Also, try to memorize algebra’s order of operations, which tells you what steps to do in what order to simplify or solve problems. To learn how to apply algebra’s order of operations, keep reading!
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# Draw The Logic Circuit For Following Boolean Expression F A B C
By | December 26, 2021
# Drawing Logic Circuits forBoolean Expressions F, A, B, and C
People often use Boolean expressions to solve problems related to computer science, engineering, and other fields. These mathematical expressions are used to create logic equations that can be used to describe any algebraic or logical problem. Boolean expressions are typically written using symbols (such as F, A, B, and C) to represent either true or false values. By using these symbols and their relationships to one another, a logic circuit can be drawn to solve the problem. In this article, we will explore how to draw the logic circuit for the Boolean expression F A B C.
## Understanding Boolean Expressions
Boolean expressions are mathematical equations created using symbols to represent true and false values. A true value is represented by 1 and a false value is represented by 0. Boolean expressions are used in many areas of computer science, engineering, and other disciplines to solve complex problems. Boolean expressions are often written using symbols such as F, A, B, and C to represent true or false values.
## Components of a Logic Circuit
In order to draw the logic circuit for the Boolean expression F A B C, one must first have a basic understanding of the components of a logic circuit. These components include logic gates, transistors, and switches. Logic gates are used to control the flow of electrical currents and represent true or false values. Transistors are used to amplify the power of an electrical signal. Lastly, switches are used to control the flow of electricity in the circuit.
## Drawing the Logic Circuit
Now that we have a basic understanding of the components of a logic circuit, we can begin to draw the logic circuit for the Boolean expression F A B C. The first step is to decide how many logic gates will be needed. For this example, we will need five logic gates. Next, draw a circuit diagram with the logic gates arranged in the necessary positions. Once the diagram is complete, we can begin to connect the logic gates together.
For this example, we will connect the logic gates together in the following way: the output of the first logic gate is connected to the input of the other logic gate, the output of the second logic gate is connected to the input of the third logic gate, the output of the third logic gate is connected to the input of the fourth logic gate, and the output of the fourth logic gate is connected to the input of the fifth logic gate. Then, connect the output of the fifth logic gate to the input of the sixth logic gate. Finally, connect the output of the sixth logic gate to the input of the seventh logic gate and connect the output of the seventh logic gate back to the input of the first logic gate.
Once the logic gates have been connected, the next step is to add transistors and switches. To do this, place a transistor between each logic gate and connect it to a switch. This switch will allow us to control the flow of electrical current in the circuit. Finally, label each switch and place a checkmark on the ones that are closed.
Once the logic circuit has been drawn, the final step is to add the necessary logic equations. For this example, the logic equations for the Boolean expression F A B C are as follows: F = A + B + C; A = B + C; B = C; and C = 1.
## Conclusion
Drawing the logic circuit for the Boolean expression F A B C is a fairly simple process once you understand the components of a logic circuit and the logic equations that need to be included. By following the steps outlined in this article, anyone can easily create a logic circuit to solve their problem.
Eng Huda M Dawoud
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# 3.1: The Cartesian Plane
Difficulty Level: At Grade Created by: CK-12
Kaitlyn walked into Math class and saw the following image displayed from the overhead projector. Her teacher asked everyone in the class to duplicate the picture on the blank sheet of paper that she had placed on each student’s desk.
When the teacher felt that the students had completed the drawing, she asked them to share their results with the class. Most of the students had difficulty reproducing the picture. Kaitlyn told the class that she could not make the picture the same size as the one shown. She also said that she had a problem locating the leaves in the same places on the stem. Her teacher said that she could offer a solution to these problems.
### Guidance
The Cartesian plane is a system of four areas or quadrants produced by the perpendicular intersection of two number lines. The two number lines intersect at right angles. The point of intersection is known as the origin. One number line is a horizontal line and this is called the -axis. The other number line is a vertical line and it is called the -axis. The two number lines are referred to as the axes of the Cartesian plane. The Cartesian plane, also known as the coordinate plane, has four quadrants that are labeled counterclockwise.
The value of the origin on the -axis is zero. If you think of the -axis as a number line, the numbers to the right of zero are positive values, and those to the left of zero are negative values. The same can be applied to the -axis. The value of the origin on the -axis is zero. The numbers above zero are positive values and those below zero are negative values.
Every point that is plotted on a Cartesian plane has two values associated with it. The first value represents the -value and the second value represents the -value. These two values are called the coordinates of the point and are written as the ordered pair .
To plot a point on the Cartesian plane:
• Start at zero (the origin) and locate the coordinate on the -axis.
• If the coordinate is positive, move to the right of the origin the number of units displayed by the number. If the coordinate is negative, move to the left of the origin the number of units displayed by the number.
• Once the coordinate (also called the abscissa) has been located, move vertically the number of units displayed by the coordinate (also called the ordinate). If the coordinate is positive, move vertically upward from the coordinate, the number of units displayed by the coordinate. If the coordinate is negative, move vertically downward from the coordinate, the number of units displayed by the coordinate.
• The point is can now be plotted.
Examine the points and that have been plotted on the graph below.
• – From the origin, move to the left four units (along the red line on the -axis). Now, move vertically upward two units. Plot the point .
• – From the origin, move to the left two units (along the red line on the -axis). Now, move vertically downward one unit. Plot the point .
• – From the origin, move to the right three units (along the red line on the -axis). Now, move vertically downward four units. Plot the point .
• – From the origin, move to the right six units (along the red line on the -axis). Now, move vertically upward three units. Plot the point .
#### Example A
For each quadrant, say whether the values of and are positive or negative.
Solution: The graph below shows where and values are positive and negative.
#### Example B
On the following Cartesian plane, draw an axis and plot the following points.
Solution:
#### Example C
Determine the coordinates of each of the plotted points on the following graph.
Solution:
#### Concept Problem Revisited
Now, let us return to the beginning of the lesson to find out the solution that the teacher had for the students.
Now that the students can see the picture on a Cartesian plane, the reproduction process should be much easier.
### Vocabulary
Abscissa
The abscissa is the coordinate of the ordered pair that represents a plotted point on a Cartesian plane. For the point (3, 7), 3 is the abscissa.
Cartesian Plane
A Cartesian plane is a system of four areas or quadrants produced by the perpendicular intersection of two number lines. A Cartesian plane is the grid on which points are plotted.
Coordinates
The coordinates are the ordered pair that represent a point on the Cartesian plane.
Coordinate Plane
The coordinate plane is another name for the Cartesian plane.
Ordinate
The ordinate is the coordinate of the ordered pair that represents a plotted point on a Cartesian plane. For the point (3, 7), 7 is the ordinate
Origin
The origin is the point of intersection of the and axes on the Cartesian plane. The coordinates of the origin are (0, 0).
-axis
The -axis is the horizontal number line of the Cartesian plane.
-axis
The -axis is the vertical number line of the Cartesian plane.
### Guided Practice
1. Draw a Cartesian plane that displays only positive values. Number the and axes to twelve. Plot the following coordinates and connect them in order. Use a straight edge to connect the points. When the word “STOP” appears, begin the next line. Plot the points in the order they appear in each Line row.
LINE 1 (6, 0) (8, 0) (9, 1) (10, 3) (10, 6) (9, 8) (7, 9) (5, 9) STOP
LINE 2 (6, 0) (4, 0) (3, 1) (2, 3) (2, 6) (3, 8) (5, 9) STOP
LINE 3 (7, 9) (6, 12) (4, 11) (5, 9) STOP
LINE 4 (4, 8) (3, 6) (5, 6) (4, 8) STOP
LINE 5 (8, 8) (7, 6) (9, 6) (8, 8) STOP
LINE 6 (5, 5) (7, 5) (6, 3) (5, 5) STOP
LINE 7 (3, 2) (4, 1) (5, 2) (6, 1) (7, 2) (8, 1) (9, 2) STOP
LINE 8 (4, 1) (6, 1) (8, 1) STOP
2. In which quadrant would the following points be located?
i) (3, -8)
ii) (-5, 4)
iii) (7, 2)
iv) (-6, -9)
v) (-3, 3)
vi) (9, -7)
3. State the coordinates of the points plotted on the following Cartesian plane.
1. The following picture is the result of plotting the coordinates and joining them in the order in which they were plotted. Your pumpkin can be any color you like.
2. i) (3, -8) – the coordinate is positive and the coordinate is negative. This point will be located in Quadrant IV.
ii) (-5, 4) – the coordinate is negative and the coordinate is positive. This point will be located in Quadrant II.
iii) (7, 2) – the coordinate is positive and the coordinate is positive. This point will be located in Quadrant I.
iv) (-6, -9) – the coordinate is negative and the coordinate is negative. This point will be located in Quadrant III.
v) (-3, 3) – the coordinate is negative and the coordinate is positive. This point will be located in Quadrant II.
vi) (9, -7) – the coordinate is positive and the coordinate is negative. This point will be located in Quadrant IV.
3.
### Practice
Answer the following questions with respect to the Cartesian plane:
1. What name is given to the horizontal number line on the Cartesian plane?
2. What name is given to the four areas of the Cartesian plane?
3. What are the coordinates of the origin?
4. What name is given to the vertical number line on the Cartesian plane?
5. What other name is often used to refer to the coordinate of a point on the Cartesian plane?
On each of the following graphs, select three points and state the coordinates of these points.
1. With a partner, create a picture on a Cartesian plane that is numbered ten round. Using the coordinates, list the points for at least five lines necessary for a classmate to complete this same picture. (Go back to the directions for the pumpkin).
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# NCERT Solutions for Class 9 Maths Chapter 7 – Triangles
NCERT Solutions for Class 9 Maths Chapter 7 Triangles provides the answers and questions related to the chapter. Triangle, the word itself describes its meaning. “Tri” means “three” so a closed figure formed by three intersecting lines is known as a Triangle. Students must have already studied the angle sum property of a triangle in chapter 6 of NCERT Class 9 Maths. Now, inconsequential to it, Chapter 7 of NCERT Class 9 Maths will further brief students about the congruence of triangles, and rules of congruence. Also, they will learn a few more properties of triangles and inequalities in a triangle.
Here we have provided the complete Solution of NCERT Class 9 Maths Chapter 7 Triangles in PDF format solved by experienced teachers. Students can download the free PDF by clicking on the link below to keep it handy for future reference.
### Download NCERT Solutions for Class 9 Maths Chapter 7 – Triangles
List of Exercises in class 9 Maths Chapter 7
Exercise 7.1 Solution 8 Questions (6 Short Answer Questions, 2 Long Answer Question)
Exercise 7.2 Solution 8 Questions (6 Short Answer Questions, 2 Long Answer Question)
Exercise 7.3 Solution 5 Questions (3 Short Answer Questions, 2 Long Answer Question)
Exercise 7.4 Solution 6 Questions (5 Short Answer Questions, 1 Long Answer Question)
Exercise 7.5 (Optional) Solution 4 Questions
### Access Answers of Maths NCERT Class 9 Chapter 7 – Triangles
Exercise: 7.1 (Page No: 118)
1. In quadrilateral ACBD, AC = AD and AB bisect A (see Fig. 7.16). Show that ΔABC ΔABD. What can you say about BC and BD?
Solution:
It is given that AC and AD are equal i.e. AC = AD and the line segment AB bisects A.
We will have to now prove that the two triangles ABC and ABD are similar i.e. ΔABC ΔABD
Proof:
Consider the triangles ΔABC and ΔABD,
(i) AC = AD (It is given in the question)
(ii) AB = AB (Common)
(iii) CAB = DAB (Since AB is the bisector of angle A)
So, by SAS congruency criterion, ΔABC ΔABD.
For the 2nd part of the question, BC and BD are of equal lengths by the rule of C.P.C.T.
2. ABCD is a quadrilateral in which AD = BC and DAB = CBA (see Fig. 7.17). Prove that
(i) ΔABD ΔBAC
(ii) BD = AC
(iii) ABD = BAC.
Solution:
The given parameters from the questions are DAB = CBA and AD = BC.
(i) ΔABD and ΔBAC are similar by SAS congruency as
AB = BA (It is the common arm)
DAB = CBA and AD = BC (These are given in the question)
So, triangles ABD and BAC are similar i.e. ΔABD ΔBAC. (Hence proved).
(ii) It is now known that ΔABD ΔBAC so,
BD = AC (by the rule of CPCT).
(iii) Since ΔABD ΔBAC so,
Angles ABD = BAC (by the rule of CPCT).
3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.
Solution:
It is given that AD and BC are two equal perpendiculars to AB.
We will have to prove that CD is the bisector of AB
Now,
Triangles ΔAOD and ΔBOC are similar by AAS congruency since:
(i) A = B (They are perpendiculars)
(ii) AD = BC (As given in the question)
(iii) AOD = BOC (They are vertically opposite angles)
∴ ΔAOD ΔBOC.
So, AO = OB (by the rule of CPCT).
Thus, CD bisects AB (Hence proved).
4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ΔCDA.
Solution:
It is given that p q and l m
To prove:
Triangles ABC and CDA are similar i.e. ΔABC ΔCDA
Proof:
Consider the ΔABC and ΔCDA,
(i) BCA = DAC and BAC = DCA Since they are alternate interior angles
(ii) AC = CA as it is the common arm
So, by ASA congruency criterion, ΔABC ΔCDA.
5. Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of A (see Fig. 7.20). Show that:
(i) ΔAPB ΔAQB
(ii) BP = BQ or B is equidistant from the arms of A.
Solution:
It is given that the line “l” is the bisector of angle A and the line segments BP and BQ are perpendiculars drawn from l.
(i) ΔAPB and ΔAQB are similar by AAS congruency because:
P = Q (They are the two right angles)
AB = AB (It is the common arm)
BAP = BAQ (As line l is the bisector of angle A)
So, ΔAPB ΔAQB.
(ii) By the rule of CPCT, BP = BQ. So, it can be said the point B is equidistant from the arms of A.
6. In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.
Solution:
It is given in the question that AB = AD, AC = AE, and ∠BAD = ∠EAC
To prove:
The line segment BC and DE are similar i.e. BC = DE
Proof:
We know that BAD = EAC
Now, by adding DAC on both sides we get,
BAD + DAC = EAC +DAC
Now, ΔABC and ΔADE are similar by SAS congruency since:
(i) AC = AE (As given in the question)
(iii) AB = AD (It is also given in the question)
So, by the rule of CPCT, it can be said that BC = DE.
7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB (see Fig. 7.22). Show that
(i) ΔDAP ΔEBP
Solutions:
In the question, it is given that P is the mid-point of line segment AB. Also, BAD = ABE and EPA = DPB
(i) It is given that EPA = DPB
Now, add DPE on both sides,
EPA +DPE = DPB+DPE
This implies that angles DPA and EPB are equal i.e. DPA = EPB
Now, consider the triangles DAP and EBP.
DPA = EPB
AP = BP (Since P is the mid-point of the line segment AB)
BAD = ABE (As given in the question)
So, by ASA congruency, ΔDAP ΔEBP.
(ii) By the rule of CPCT, AD = BE.
8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:
(i) ΔAMC ΔBMD
(ii) DBC is a right angle.
(iii) ΔDBC ΔACB
(iv) CM = ½ AB
Solution:
It is given that M is the mid-point of the line segment AB, C = 90°, and DM = CM
(i) Consider the triangles ΔAMC and ΔBMD:
AM = BM (Since M is the mid-point)
CM = DM (Given in the question)
CMA = DMB (They are vertically opposite angles)
So, by SAS congruency criterion, ΔAMC ΔBMD.
(ii) ACM = BDM (by CPCT)
∴ AC BD as alternate interior angles are equal.
Now, ACB +DBC = 180° (Since they are co-interiors angles)
⇒ 90° +B = 180°
∴ DBC = 90°
(iii) In ΔDBC and ΔACB,
BC = CB (Common side)
ACB = DBC (They are right angles)
DB = AC (by CPCT)
So, ΔDBC ΔACB by SAS congruency.
(iv) DC = AB (Since ΔDBC ΔACB)
⇒ DM = CM = AM = BM (Since M the is mid-point)
So, DM + CM = BM+AM
Hence, CM + CM = AB
⇒ CM = (½) AB
Exercise: 7.2 (Page No: 123)
1. In an isosceles triangle ABC, with AB = AC, the bisectors of B and C intersect each other at O. Join A to O. Show that:
(i) OB = OC (ii) AO bisects A
Solution:
Given:
AB = AC and
the bisectors of B and C intersect each other at O
(i) Since ABC is an isosceles with AB = AC,
B = C
½ B = ½ C
⇒ OBC = OCB (Angle bisectors)
∴ OB = OC (Side opposite to the equal angles are equal.)
(ii) In ΔAOB and ΔAOC,
AB = AC (Given in the question)
AO = AO (Common arm)
OB = OC (As Proved Already)
So, ΔAOB ΔAOC by SSS congruence condition.
BAO = CAO (by CPCT)
Thus, AO bisects A.
2. In ΔABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ΔABC is an isosceles triangle in which AB = AC.
Solution:
It is given that AD is the perpendicular bisector of BC
To prove:
AB = AC
Proof:
BD = CD (Since AD is the perpendicular bisector)
Thus,
AB = AC (by CPCT)
3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.
Solution:
Given:
(i) BE and CF are altitudes.
(ii) AC = AB
To prove:
BE = CF
Proof:
Triangles ΔAEB and ΔAFC are similar by AAS congruency since
A = A (It is the common arm)
AEB = AFC (They are right angles)
AB = AC (Given in the question)
∴ ΔAEB ΔAFC and so, BE = CF (by CPCT).
4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that
(i) ΔABE ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Solution:
It is given that BE = CF
(i) In ΔABE and ΔACF,
A = A (It is the common angle)
AEB = AFC (They are right angles)
BE = CF (Given in the question)
∴ ΔABE ΔACF by AAS congruency condition.
(ii) AB = AC by CPCT and so, ABC is an isosceles triangle.
5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ABD = ACD.
Solution:
In the question, it is given that ABC and DBC are two isosceles triangles.
We will have to show that ABD = ACD
Proof:
Triangles ΔABD and ΔACD are similar by SSS congruency since
AB = AC (Since ABC is an isosceles triangle)
BD = CD (Since BCD is an isosceles triangle)
So, ΔABD ΔACD.
∴ ABD = ACD by CPCT.
6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that BCD is a right angle.
Solution:
It is given that AB = AC and AD = AB
We will have to now prove BCD is a right angle.
Proof:
Consider ΔABC,
AB = AC (It is given in the question)
Also, ACB = ABC (They are angles opposite to the equal sides and so, they are equal)
Now, consider ΔACD,
Also, ADC = ACD (They are angles opposite to the equal sides and so, they are equal)
Now,
In ΔABC,
CAB + ACB + ABC = 180°
So, CAB + 2ACB = 180°
⇒ CAB = 180° – 2ACB — (i)
CAD = 180° – 2ACD — (ii)
also,
CAB + CAD = 180° (BD is a straight line.)
Adding (i) and (ii) we get,
CAB + CAD = 180° – 2ACB+180° – 2ACD
⇒ 180° = 360° – 2ACB-2ACD
⇒ 2(ACB+ACD) = 180°
⇒ BCD = 90°
7. ABC is a right-angled triangle in which A = 90° and AB = AC. Find B and C.
Solution:
In the question, it is given that
A = 90° and AB = AC
AB = AC
⇒ B = C (They are angles opposite to the equal sides and so, they are equal)
Now,
A+B+C = 180° (Since the sum of the interior angles of the triangle)
∴ 90° + 2B = 180°
⇒ 2B = 90°
⇒ B = 45°
So, B = C = 45°
8. Show that the angles of an equilateral triangle are 60° each.
Solution:
Let ABC be an equilateral triangle as shown below:
Here, BC = AC = AB (Since the length of all sides is same)
⇒ A = B =C (Sides opposite to the equal angles are equal.)
Also, we know that
A+B+C = 180°
⇒ 3A = 180°
⇒ A = 60°
∴ A = B = C = 60°
So, the angles of an equilateral triangle are always 60° each.
Exercise: 7.3 (Page No: 128)
1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that
(i) ΔABD ΔACD
(ii) ΔABP ΔACP
(iii) AP bisects A as well as D.
(iv) AP is the perpendicular bisector of BC.
Solution:
In the above question, it is given that ΔABC and ΔDBC are two isosceles triangles.
(i) ΔABD and ΔACD are similar by SSS congruency because:
AB = AC (Since ΔABC is isosceles)
BD = CD (Since ΔDBC is isosceles)
∴ ΔABD ΔACD.
(ii) ΔABP and ΔACP are similar as:
AP = AP (It is the common side)
PAB = PAC (by CPCT since ΔABD ΔACD)
AB = AC (Since ΔABC is isosceles)
So, ΔABP ΔACP by SAS congruency condition.
(iii) PAB = PAC by CPCT as ΔABD ΔACD.
AP bisects A. — (i)
Also, ΔBPD and ΔCPD are similar by SSS congruency as
PD = PD (It is the common side)
BD = CD (Since ΔDBC is isosceles.)
BP = CP (by CPCT as ΔABP ΔACP)
So, ΔBPD ΔCPD.
Thus, BDP = CDP by CPCT. — (ii)
Now by comparing (i) and (ii) it can be said that AP bisects A as well as D.
(iv) BPD = CPD (by CPCT as ΔBPD ΔCPD)
and BP = CP — (i)
also,
BPD +CPD = 180° (Since BC is a straight line.)
⇒ 2BPD = 180°
⇒ BPD = 90° —(ii)
Now, from equations (i) and (ii), it can be said that
AP is the perpendicular bisector of BC.
2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
Solution:
It is given that AD is an altitude and AB = AC. The diagram is as follows:
(i) In ΔABD and ΔACD,
AB = AC (It is given in the question)
∴ ΔABD ΔACD by RHS congruence condition.
Now, by the rule of CPCT,
BD = CD.
3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see Fig. 7.40). Show that:
(i) ΔABM ΔPQN
(ii) ΔABC ΔPQR
Solution:
Given parameters are:
AB = PQ,
BC = QR and
AM = PN
(i) ½ BC = BM and ½ QR = QN (Since AM and PN are medians)
Also, BC = QR
So, ½ BC = ½ QR
⇒ BM = QN
In ΔABM and ΔPQN,
AM = PN and AB = PQ (As given in the question)
∴ ΔABM ΔPQN by SSS congruency.
(ii) In ΔABC and ΔPQR,
AB = PQ and BC = QR (As given in the question)
ABC = PQR (by CPCT)
So, ΔABC ΔPQR by SAS congruency.
4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
.
Solution:
It is known that BE and CF are two equal altitudes.
Now, in ΔBEC and ΔCFB,
BEC = CFB = 90° (Same Altitudes)
BC = CB (Common side)
BE = CF (Common side)
So, ΔBEC ΔCFB by RHS congruence criterion.
Also, C = B (by CPCT)
Therefore, AB = AC as sides opposite to the equal angles is always equal.
5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that B = C.
Solution:
In the question, it is given that AB = AC
Now, ΔABP and ΔACP are similar by RHS congruency as
APB = APC = 90° (AP is altitude)
AB = AC (Given in the question)
AP = AP (Common side)
So, ΔABP ΔACP.
∴ B = C (by CPCT)
Exercise: 7.4 (Page No: 132)
1. Show that in a right-angled triangle, the hypotenuse is the longest side.
Solution:
It is known that ABC is a triangle right angled at B.
We know that,
A +B+C = 180°
Now, if B+C = 90° then A has to be 90°.
Since A is the largest angle of the triangle, the side opposite to it must be the largest.
So, AB is the hypotenuse which will be the largest side of the above right-angled triangle i.e. ΔABC.
2. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB.
Solution:
It is given that PBC < QCB
We know that ABC + PBC = 180°
So, ABC = 180°-PBC
Also,
ACB +QCB = 180°
Therefore ACB = 180° -QCB
Now, since PBC < QCB,
∴ ABC > ACB
Hence, AC > AB as sides opposite to the larger angle is always larger.
3. In Fig. 7.49, B < A and C < D. Show that AD < BC.
Solution:
In the question, it is mentioned that angles B and angle C is smaller than angles A and D respectively i.e. B < A and C < D.
Now,
Since the side opposite to the smaller angle is always smaller
AO < BO — (i)
And OD < OC —(ii)
By adding equation (i) and equation (ii) we get
AO+OD < BO + OC
4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).
Show that A > C and B > D.
Solution:
In ΔABD, we see that
So, ADB < ABD — (i) (Since angle opposite to longer side is always larger)
Now, in ΔBCD,
BC < DC < BD
Hence, it can be concluded that
BDC < CBD — (ii)
Now, by adding equation (i) and equation (ii) we get,
ADB + BDC < ABD + CBD
B > D
Similarly, In triangle ABC,
ACB < BAC — (iii) (Since the angle opposite to the longer side is always larger)
DCA < DAC — (iv)
By adding equation (iii) and equation (iv) we get,
ACB + DCA < BAC+DAC
∴ A > C
5. In Fig 7.51, PR > PQ and PS bisect QPR. Prove that PSR > PSQ.
Solution:
It is given that PR > PQ and PS bisects QPR
Now we will have to prove that angle PSR is smaller than PSQ i.e. PSR > PSQ
Proof:
QPS = RPS — (ii) (As PS bisects ∠QPR)
PQR > PRQ — (i) (Since PR > PQ as angle opposite to the larger side is always larger)
PSR = PQR + QPS — (iii) (Since the exterior angle of a triangle equals to the sum of opposite interior angles)
PSQ = PRQ + RPS — (iv) (As the exterior angle of a triangle equals to the sum of opposite interior angles)
PQR +QPS > PRQ +RPS
Thus, from (i), (ii), (iii) and (iv), we get
PSR > PSQ
6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:
First, let “l” be a line segment and “B” be a point lying on it. A line AB perpendicular to l is now drawn. Also, let C be any other point on l. The diagram will be as follows:
To prove:
AB < AC
Proof:
In ΔABC, B = 90°
Now, we know that
A+B+C = 180°
∴ A +C = 90°
Hence, C must be an acute angle which implies C < B
So, AB < AC (As the side opposite to the larger angle is always larger)
Also Access NCERT Exemplar for class 9 Maths Chapter 7 CBSE Notes for class 9 Maths Chapter 7
Triangle is a part of a Geometry. However, the complete Geometry of Class 9 constitutes a weightage of 22 marks out of 80 marks. Have a look at the types of questions that are expected from Geometry in the annual exam of class 9 Maths paper.
NCERT Class 9 Maths Geometry – Marks Distribution Topics Multiple Choice Questions Short Questions Long Questions Introduction to Euclid’s Geometry, Lines and Angles, Triangles, Quadrilaterals, Areas, Circles, Constructions 4 Qs of 1 mark each 2 Qs of 3 marks each 2 Qs of 6 marks each Total Marks 4 Marks 6 Marks 12 Marks
Students should practice all the questions from the exercise to score high marks in class 9 Maths paper. The step by step solution to all the exercise of NCERT Class 9 Chapter 7 is provided below:
In previous classes, students must have used the properties of triangles for solving the questions, but now in NCERT Chapter 7 of Class 9, Maths students will learn how to prove these properties. For solving the questions related to this topic, it is very necessary that student should know the congruence rule. So, firstly go through the theory and then look at the solved examples that are already there in the book. After that, start solving the exercise problems.
Students can also have a look at the NCERT Solutions of Class 9 for Science subject to know the answers of all the chapters with a detailed explanation.
### Important Concepts Learned in NCERT Class 9 Maths Chapter 7 – Triangles
The aim of including this Triangle chapter in class 9 Maths NCERT textbook is to make students know the following concepts:
1. Congruence of triangles.
2. Criteria for congruence of triangles (SAS congruence rule, ASA congruence rule, SSS congruence rule, RHS congruence rule).
3. Properties of triangles.
4. Inequalities of triangle.
We hope this information on “NCERT Solution Class 9 Maths Chapter 7 ” is useful for students. Click on NCERT Solutions to get the solved answers of NCERT book for all the classes. Stay tuned for further updates on CBSE and other competitive exams. To access interactive Maths and Science Videos download CoolGyan’S App and subscribe to YouTube Channel.
## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 7
### List out the important topics covered in NCERT Solutions for Class 9 Maths Chapter 7.
The important concepts covered in NCERT Solutions for Class 9 Maths Chapter 7 are
1. Congruence of triangles.
2. Criteria for congruence of triangles (SAS congruence rule, ASA congruence rule, SSS congruence rule, RHS congruence rule).
3. Properties of triangles.
4. Inequalities of triangles.
### What is the meaning of congruence of triangles in NCERT Solutions for Class 9 Maths Chapter 7?
According to NCERT Solutions for Class 9 Maths Chapter 7congruence of triangles states that that two triangles are said to be congruent if they are copies of each other and when superposed, they cover each other exactly. In other words, two triangles are congruent if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.
### How NCERT Solutions for Class 9 Maths Chapter 7 helpful for board exam preparation?
Practising NCERT Solutions for Class 9 Maths Chapter 7 help the students to top the final exams and ace a subject. These solutions are devised, based on the most updated syllabus, covering all the crucial topics of the respective subjects. Hence, solving these questions will make the students more confident to face the board exams. Topics given in these solutions form the basis for top scores. It also helps the students to get familiar with answering questions of all difficulty levels. These solutions are highly recommended to the students for referencing and to practice for the board exams.
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# Inversion. Chapter Constructing The Inverse of a Point: If P is inside the circle of inversion: (See Figure 7.1)
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## Transcription
1 Chapter 7 Inversion Goal: In this chapter we define inversion, give constructions for inverses of points both inside and outside the circle of inversion, and show how inversion could be done using Geometer s Sketchpad. A cartesian coordinate representation and a number of fascinating applications of inversion are also presented. Definition Let O be the center of a fixed circle of radius r in the Euclidean plane. Let P be any point in the plane other than O. An inversion in circle C(O, r), I(O, r), is a function such that if I (O,r) (P ) = P then P OP and (OP )(OP ) = r 2. Here P is called the inverse of P, O is called the center of inversion, and r is called the radius of inversion, and r 2 is called its power. It follows from the above definition that to each point P of the plane, other than O, there corresponds a unique inverse point P. To make the inversion a transformation of the plane, we add to the plane a single ideal point Ω defined to be the inverse of the center of inversion. The point Ω is considered to lie on every line in the plane. 7.1 Constructing The Inverse of a Point: Given the circle of inversion C (O,r) and a point P, how do you construct the inverse of P? If P is inside the circle of inversion: (See Figure 7.1) Draw the ray OP. Draw a perpendicular to OP at P. This intersects the circle of inversion in two points, label one of them Q. Connect the center of the circle O to Q. Draw a perpendicular to OQ from Q. The intersection of this perpendicular with OP is P, the inverse of P. 52
2 Q O P P. Figure 7.1: Inverse of a Point Inside the Circle of Inversion Note that OP Q OQP. (OQ) 2 = r 2. Hence OQ OP = OP OQ. Therefore, OP OP = If P is outside the circle of inversion: (See Figure 7.2) Let M be the midpoint of the segment OP. Construct a circle centered at M of radius MO. It intersects the circle of inversion C (O,r) in two points, Q and R and goes through O and P. Construct the segment QR. The intersection of QR and OP is the point P, the inverse of P. Q... P O.... M P. R Figure 7.2: Inverse of a point Outside the Circle of Inversion Note that OP Q OQP. (OQ) 2 = r 2. Hence OQ OP = OP OQ. Therefore, OP OP = 53
3 Using Sketchpad: Inverting a Line: To invert a straight line in the circle of inversion C (O,r) follow the following steps: 1. Construct the circle of inversion C and the line l. 2. Construct an arbitrary point P on the line l. 3. Construct the ray OP. 4. Construct the intersection of OP and the circle, call it Q. 5. Construct the segment OP. 6. Mark O as the center of dilation. 7. Mark the ratio r. OP 8. Dilate Q by the ratio r centered at O. The image of Q is P the inverse OP image of P. 9. Hide everything except the circle, its center, the point determining its radius, P and the straight line l and the two point determining it. 10. Select everything and create a tool and call it invcirc. 11. Apply the tool a few times and use arc through three points to determine the image of the line. Inverting a Circle: To invert a circle C 1 in the circle of inversion C (O,r), replace l by C 1 in the steps above. 7.2 Inversion Using Coordinates: Theorem An inversion about x 2 + y 2 = r 2 is given by (x, y) (x, y xr 2 ) = ( x 2 + y, yr 2 2 x 2 + y ) 2 Proof. Since (x, y), (x, y ) and (0, 0) are collinear, we have y = y. Now x x d((0, 0), (x, y)) d((0, 0), (x, y )) = r 2, hence x 2 + y 2 x 2 + y 2 = r 2 and (x 2 +y 2 ) = r4 2 y2 x (x 2 + y 2 ) x 2 x 2 + y 2 (x 2 +y 2 ) = r 4. Hence x 2 = r4 x 2 + y 2 y 2 and x 2 = r4 y 2 (x 2 + y 2 ) x 2 + y 2 = r4 x 2 + y y2 x 2 2 x. Hence 2 x 2 + y2 x 2 = r4 x 2 x 2 + y and x 2 (x 2 + y 2 ) = r4 2 x 2 x 2 + y. Hence 2 x 2 = r 4 x 2 (x 2 + y 2 ) 2 and x = r2 x x 2 + y 2. Similarly, we can show that y = r2 y x 2 + y 2. Exercise 1: What is the image of (x 1) 2 + y 2 x 2 + y 2 = 1? = 1 under an inversion in 54
4 Y X Figure 7.3: Inverse of (x 1) 2 + y 2 = 1 in x 2 + y 2 = 1 Answer. Well, x = x and y = y. Hence x x = = 1 and x 2 +y 2 x 2 +y 2 x 2 +1 (x 1) 2 2 y = y. Hence the image of a circle going through the center of the circle of 2x inversion is a line going through the points of intersection of the two circles. Exercise 2: What is the image of x = 1 2 under an inversion in x2 + y 2 = 1? Y X Figure 7.4: Inverse of x = 1 2 in x2 + y 2 = 1 Answer: (x 1) 2 + y 2 = 1 Exercise: What is the image of x = 1 under an inversion in x 2 + y 2 = 1? Answer: (x 1/2) 2 + y 2 = 1/4. Exercise 3: What is the image of x 2 +2x+y 2 = 0 under an inversion in x 2 +y 2 = 1? Answer: x = 1/2. 55
5 Theorem If two circles are orthogonal, (their tangents at the points of intersection are perpendicular), and if a diameter AB of one circle meets the other circle in the points C and D, then OP 2 = OC OD. P A O C B O D Figure 7.5: Orthogonal Circles are Inverses Proof. OP O is a right triangle, hence (OP ) 2 + (P O ) 2 = (OO ) 2. But OO = OC + CO = OC + P O, hence (OP ) 2 + (P O ) 2 = (OC + P O ) 2. Hence (OP ) 2 = (OC) OC P O, which implies that (OP ) 2 = OC (OC + 2P O ). Hence (OP ) 2 = OC OD. Theorem A circle orthogonal to the circle of inversion inverts into itself, and, a circle through a pair of inverse points is orthogonal to the circle of inversion. T P O P A Figure 7.6: Orthogonal Circles are Inverses Proof. Given the circle of inversion C (O,r) and an orthogonal circle centered at A. Let T be one of the points of intersection of the two circles. Now if a line through O meets this orthogonal circle at P and P then OP OP = OT 2 = R 2. Hence P and P are inverse points. Theorem If P P and Q, Q are pairs of inverse points with respect to some circle C (O,r), Then P Q = P Q r2 OP OQ. Proof. If O, P, Q are noncollinear then, OP = OQ and P OQ = Q OP, OQ OP hence OP Q OQ P. Hence P Q = OQ. Hence P Q = OQ P Q OQ = r2 P Q. P Q OP OP OQ OP OQ 56
6 Q. Q... O P P Figure 7.7: P Q P Q = r2 OP OQ 7.3 Applications of Inversion: Given three non-coaxial concurrent circles, construct a circle C tangent to all three circles. Figure 7.8: A Circle Tangent To Three Non-Coaxial Circles Solution: Invert the circles about a unit circle centered at the point of concurrency of the circles creating a triangle. Now construct the inscribed circle and invert this circle in the circle of inversion to create the required circle. Ptolemy s Theorem: In a cyclic convex quadrilateral, the product of the diagonals is equal to the sum of the products of the two pairs of opposite sides. Proof: Invert the circle and the convex quadrilateral about a circle centered at one of the vertices of the quadrilateral, say A. Now B D = B C + C D. Hence, r 2 BC AB AC + CD AC AD = BD AB AD 57 r2 r2
7 B B A C C D D Figure 7.9: Ptolemy s Theorem Hence, BC AD + CD AB = BD AC. Homework Find the image of the objects below under the specified inversion. (See Figures 7.10 and 7.11 ) 2. Prove that the inverse of the circumcircle C c of a triangle ABC with respect to the incircle C i, as a circle of inversion, is the nine point circle of the triangle XY Z determined by the points of contact of C i with the sides of ABC. 58
8 Figure 7.10: Inversion - HW 59
9 Figure 7.11: Inversion - HW 60
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This is a follow-up to Monday’s post about the smart way to do the binomial expansion. In this one, we’re going to look at how to do C4 binomial expansions - ones with crazy powers like $-3$ or $\frac{3}{2}$.
This bit is very important: you should COMPLETELY ignore the formula in the book. It will give you a headache and make all of your minus signs vanish. It’s also mildly poisonous.
The good news? You can use the same table as you did for the C2 stuff on Monday. The bad news? Pascal’s Triangle doesn’t have a $-3$rd or $\frac{3}{2}$th row, so you need another way of working those numbers out.
### Getting the C numbers
Let’s think about Pascal’s Triangle for a moment. The seventh row (as we saw on Monday) was 1, 7, 21, 35, 35, 21, 7, 1. It turns out, there’s a simple way to generate that series without your calculator. Here’s how:
Start with 1. Multiply it by 7 (the power) and divide by 1 to get 7. Multiply this by 6 and divide by 2 to get 21. Then $21 \times 5 \div 3 = 35$, and so on. Eventually you get to zero and the series stops.
Did you get the recipe there? Multiply by the power divided by one, then for each new number, drop the top and bump the bottom up.
I wouldn’t give you the recipe if it wasn’t useful. It works just the same for numbers that aren’t positive integers. So, if $n$ was $-3$, you’d have:
• $1$
• $1 \times -3 \div 1 = -3$
• $-3 \times -4 \div 2 = 6$
• $6 \times -5 \div 3 = -10$
• … and so on.
If $n$ was $\frac{3}{2}$:
• $1$
• $1 \times \frac{3}{2} \div 1 = \frac{3}{2}$
• $3/2 \times \frac{1}{2} \div 2 = \frac{3}{8}$
• $3/8 \times \frac{-1}{2} \div 3 = \frac{-1}{16}$
• … and so on.
Other than that, the table works just the same way as before (only it goes on forever… normally they only ask you for the first few terms).
(Notice, if your system is strong enough, that these numbers are the same as the $\frac{n(n-1)(n-2)…(n-r)}{r!}$ in the big formula.)
### An example of the binomial expansion
So, let’s say we need to work out $(4 - x)^{\frac{1}{2}}$ up to the term in $x^3$. If you’ve done binomial expansion in class at school, you’d probably groan at that because the method you’ve learned is ridiculous. If not, ask your teacher how s/he would do it and watch him/her groan.
For us, though, it’s easy! We work out the C numbers (we only need the first four):
• $1$
• $1 \times \frac{1}{2} \div 1 = \frac{1}{2}$
• $1/2 \times \frac{-1}{2} \div 2 = \frac{-1}{8}$
• $-1/8 \times \frac{-3}{2} \div 3 = \frac{1}{16}$
… and throw them in the table. The A column starts at $4^{\frac{1}{2}} = 2$ and you divide by 4 each time (dropping the power by 1); the B column starts at 1 and multiplies by $-x$ each time.
C A B CAB $1$ $2$ $1$ $2$ $\frac{1}{2}$ $\frac{1}{2}$ $-x$ $\frac{-1}{4}x$ $\frac{-1}{8}$ $\frac{1}{8}$ $x^2$ $\frac{-1}{64} x^2$ $\frac{1}{16}$ $\frac{1}{32}$ $-x^3$ $-\frac{1}{512} x^3$
… so in about four short lines of working, you get $(4-x)^{1/2} = 2 - \frac{1}{4} x - \frac{1}{64} x^2 - \frac{1}{512} x^3$
### In conclusion
I can barely count the number of ways this is better than the ‘traditional’ method, but here are a few:
• You get at most two minus signs to deal with in any sum;
• This method deals brilliantly with the first number not being one;
• The C numbers are the same every time you have to do the same power (i.e., if you’re doing two separate $\frac{1}{2}$ powers, you only need to work out the C numbers once
• You can use this method for any power $n$
Can you think of any others?
* Edited 2020-01-02 to fix some LaTeX and an arithmetic error that had gone unnoticed for nearly eight years. Thanks, Rob!
* Edited 2021-06-15 to fix a table and LaTeX. Thanks, Adam!
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A
# moment
Figure 1. The principle of moments is demonstrated by a children's teeter-totter (see-saw).
Figure 2. The law of moments
A moment is a tendency to cause rotation about a point or axis, i.e., the turning effect of a force. It is equal to the product of the force F and the perpendicular distance r between the point of axis of rotation (known as the fulcrum) and the line of action of the force, that is
T = Fr
Imagine resting your arm on a table, with your palm up and with your elbow on the table. Imagine a book resting in the palm of your hand. Now lift up the book slightly above the table (say the book weighs 2 kilograms), pivoting your arm slightly above the table, but keeping your elbow on the table. Say the length of your arm from your elbow to your hand is 0.5 meter. By definition, the book is exerting a moment about your elbow equal to the force (the weight of the book) times the distance to the pivot point (your elbow). In this case, (force) × (distance) = 2 × 0.5 = 1 newton·meter. The moment exerted by the book about your elbow is 1 N·m.
Consider another example. If you push on a door as far away from the hinge as possible, it is very easy to push it open. Pushing nearer to the hinge, though, requires a greater effort. This is because a small force some distance away from the fulcrum can have the same moment as a large force near to the fulcrum.
## Clockwise and anticlockwise moments
The turning effect of a force which tends to turn an object in a clockwise direction is called a clockwise moment and the turning effect of an opposing force is called an anticlockwise moment. When the two moments exactly balance each other then the object is said to be in equilibrium. The simplest way of achieving equilibrium is to use two opposing moments of equal size, but it is quite possible to have one large moment balanced by several smaller moments (Figure 2). Here the sum of the smaller moments is equal in size to the larger moment. Equilibrium can also ne attained by making several opposing moments balance each other. For the object to be in equilibrium the sum of the clockwise moments must equal the sum of the anticlockwise moments. This is known as the law of moments.
Values of moment are given a +/- value depending on whether they tend to produce clockwise or anticlockwise rotation. The international standard (S.I.) unit of moment is the newton·meter. The dimensions of moment are [M ] [L ] 2 [T ]–2.
See also torque. Compare with couple.
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# Problem of the Week Problem C and Solution Intersecting Triangles
## Problem
$$\triangle ABC$$ and $$\triangle PQR$$ are equilateral triangles with vertices $$B$$ and $$P$$ on line segment $$MN$$. The triangles intersect at two points, $$X$$ and $$Y$$, as shown.
If $$\angle NPQ = 75\degree$$ and $$\angle MBA = 65\degree$$, determine the measure of $$\angle CXY$$.
## Solution
In any equilateral triangle, all sides are equal in length and each angle measures $$60\degree$$.
Since $$\triangle ABC$$ and $$\triangle PQR$$ are equilateral, $$\angle ABC = \angle ACB = \angle CAB = \angle QPR = \angle PRQ = \angle RQP = 60\degree$$.
Since the angles in a straight line sum to $$180\degree$$, we have $$180\degree = \angle MBA + \angle ABC + \angle YBP = 65\degree + 60\degree + \angle YBP$$.
Rearranging, we have $$\angle YBP = 180\degree - 65\degree - 60\degree = 55\degree$$.
Similarly, since angles in a straight line sum to $$180\degree$$, we have $$180\degree = \angle NPQ + \angle QPR + \angle YPB = 75\degree + 60\degree + \angle YPB$$.
Rearranging, we have $$\angle YPB = 180\degree - 75\degree - 60\degree = 45\degree$$.
Since the angles in a triangle sum to $$180\degree$$, in $$\triangle BYP$$ we have $$\angle YPB + \angle YBP + \angle BYP = 180\degree$$, and so $$45\degree + 55\degree + \angle BYP = 180\degree$$.
Rearranging, we have $$\angle BYP = 180\degree - 45\degree - 55\degree = 80\degree$$.
When two lines intersect, vertically opposite angles are equal. Since $$\angle XYC$$ and $$\angle BYP$$ are vertically opposite angles, we have $$\angle XYC = \angle BYP = 80\degree$$.
Again, since angles in a triangle sum to $$180\degree$$, in $$\triangle XYC$$ we have $$\angle XYC + \angle XCY + \angle CXY = 180\degree$$. We have already found that $$\angle XYC = 80\degree$$, and since $$\angle XCY = \angle ACB$$, we have $$\angle XCY = 60\degree$$. So, $$\angle XYC + \angle XCY + \angle CXY = 180\degree$$ becomes $$80\degree + 60\degree + \angle CXY = 180\degree$$. Rearranging, we have $$\angle CXY = 180\degree - 80\degree - 60\degree = 40\degree$$.
Therefore, $$\angle CXY = 40\degree$$.
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Square Rectangle Triangle
Introduction:
• In geometry, a polygon is a plane figure which is bounded by a closed path or closed circuit which is composed of finite sequence of straight line segments
• Square,rectangleand triangle are the types of polygons and the basic important shapes in geometry. In this article, we are going to discuss some basic informations and formulas for square rectangle and triangle.
Is this topic Square Definition hard for you? Watch out for my coming posts.
Square:
Properties:
• The diagonals of a square bisect each other
• The diagonals of a square are perpendicular.
• Opposite sides of a square are both parallel and equal.
• The diagonals of a square are equal.
Formulas:
• The area of square = (a)2
• The perimeter of square = 4 * a
where a ---> side length
Example problems:
1) Find the area of the square with the side length of 12 cm.
Solution:
Area of square = ( a ) 2
= (12)2
= 12 * 12
Area of square = 144 cm2
2) Find the perimeter of the square with side length of 31 feet.
Solution:
Perimeter of square = 4 * a
= 4 * 31
= 124
Perimeter of square = 124 feet.
Rectangle:
Properties:
• A rectangle is a four-sided polygon.
• The opposite sides are parallel and of equal length.
• Opposite angles are equal to 90 degree.
• Diagonals are equal and bisect each other.
Formulas:
• Area of rectangle = l * w
• Perimeter of rectangle = 2( l + w )
Where, l is the length and b is the width.
Examples:
1) Find the perimeter of rectangle with the length and width are 9 cm and 5 cm.
Solution:
p = 2( l+ w )
= 2 (9+5)
= 2(14)
= 28 cm
2) Find the area of rectangle given the length is 21 m and width is 13 m.
Solution:
Area of a Rectangle = l * w
= 21 * 13
= 273 m2
Between, if you have problem on these topics Triangle Construction,please browse expert math related websites for more help on answer my math questions.
Triangle:
Properties:
• Triangles are formed using straight line segments
• The line segments connect three points that are not in a straight line.
• If the all sides are equal, then the triangle is known as equilateral triangle.
Formulas:
• Area of Triangle = ½ b h
b ---> base
h ---> vertical height
• Perimeter of equilateral triangle = 3 * a
Example problems:
1) Find the area of a triangle through base of 21 m and a height of 7 m.
Solution:
Area of a triangle = ½ b h
= ½ (21) (7)
= 73.5 m2
2) Find the perimeter of equilateral triangle with the side length 9 cm
Solution:
p =3 * a
= 3*9
= 27 cm
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# 4.2 Modeling with linear functions (Page 7/9)
Page 7 / 9
If the function $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ is graphed, find and interpret the slope of the function.
When will the population reach 100,000?
Ten years after the model began
What is the population in the year 12 years from the onset of the model?
For the following exercises, consider this scenario: The weight of a newborn is 7.5 pounds. The baby gained one-half pound a month for its first year.
Find the linear function that models the baby’s weight $\text{\hspace{0.17em}}W\text{\hspace{0.17em}}$ as a function of the age of the baby, in months, $\text{\hspace{0.17em}}t.$
$W\left(t\right)=0.5t+7.5$
Find a reasonable domain and range for the function $\text{\hspace{0.17em}}W.$
If the function $\text{\hspace{0.17em}}W\text{\hspace{0.17em}}$ is graphed, find and interpret the x - and y -intercepts.
$\left(-15,\text{}0\right)\text{\hspace{0.17em}}$ : The x -intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth. $\text{\hspace{0.17em}}\left(0,\text{7}.\text{5}\right)\text{\hspace{0.17em}}$ : The baby weighed 7.5 pounds at birth.
If the function W is graphed, find and interpret the slope of the function.
When did the baby weight 10.4 pounds?
At age 5.8 months
What is the output when the input is 6.2?
For the following exercises, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by 205 each year from 2005 until 2010. In 2005, 12,025 people were inflicted.
Find the linear function that models the number of people inflicted with the common cold $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ as a function of the year, $\text{\hspace{0.17em}}t.$
$C\left(t\right)=12,025-205t$
Find a reasonable domain and range for the function $\text{\hspace{0.17em}}C.$
If the function $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ is graphed, find and interpret the x - and y -intercepts.
$\left(\text{58}.\text{7},\text{}0\right):\text{\hspace{0.17em}}$ In roughly 59 years, the number of people inflicted with the common cold would be 0. $\text{\hspace{0.17em}}\left(0,\text{12},0\text{25}\right)\text{\hspace{0.17em}}$ Initially there were 12,025 people afflicted by the common cold.
If the function $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ is graphed, find and interpret the slope of the function.
When will the output reach 0?
2063
In what year will the number of people be 9,700?
## Graphical
For the following exercises, use the graph in [link] , which shows the profit, $\text{\hspace{0.17em}}y,$ in thousands of dollars, of a company in a given year, $\text{\hspace{0.17em}}t,$ where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ represents the number of years since 1980.
Find the linear function $\text{\hspace{0.17em}}y,$ where $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ depends on $\text{\hspace{0.17em}}t,$ the number of years since 1980.
$y=-2t\text{+180}$
Find and interpret the y -intercept.
Find and interpret the x -intercept.
In 2070, the company’s profit will be zero.
Find and interpret the slope.
For the following exercises, use the graph in [link] , which shows the profit, $\text{\hspace{0.17em}}y,$ in thousands of dollars, of a company in a given year, $\text{\hspace{0.17em}}t,$ where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ represents the number of years since 1980.
Find the linear function $\text{\hspace{0.17em}}y,$ where $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ depends on $\text{\hspace{0.17em}}t,$ the number of years since 1980.
$y=\text{3}0t-\text{3}00$
Find and interpret the y -intercept.
$\left(0,-300\right);$ In 1980, the company lost $300,000. Find and interpret the x -intercept. Find and interpret the slope. $y=30t-300\text{\hspace{0.17em}}$ of form $\text{\hspace{0.17em}}y=mx+b,\text{m}=30.$ For each year after 1980, the company's profits increased$30,000 per year
## Numeric
For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in [link] . Assume that the house values are changing linearly.
Year Mississippi Hawaii
1950 $25,200$74,400
2000 $71,400$272,700
In which state have home values increased at a higher rate?
How look for the general solution of a trig function
stock therom F=(x2+y2) i-2xy J jaha x=a y=o y=b
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined?
what is the value of x in 4x-2+3
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
the indicated sum of a sequence is known as
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JH
Numerade Educator
# (a) Use integration by parts to show that, for any positive integer $n$,$$\int \frac{dx}{(x^2 + a^2)^n}\ dx = \frac{x}{2a^2 (n - 1)(x^2 + a^2)^{n - 1}} + \frac{2n - 3}{2a^2 (n - 1)} \int \frac{dx}{(x^2 + a^2)^{n - 1}}$$(b) Use part (a) to evaluate$\displaystyle \int \frac{dx}{(x^2 + 1)^2}$ and $\displaystyle \int \frac{dx}{(x^2 + 1)^3}$
## (a) $$\int \frac{d x}{\left(x^{2}+a^{2}\right)^{n}}=\frac{x}{2 a^{2}(n-1)\left(x^{2}+a^{2}\right)^{n-1}}+\frac{2 n-3}{2 a^{2}(n-1)} \cdot \int d x\left(x^{2}+a^{2}\right)^{n-1}$$(b) $$\begin{array}{l}{\int \frac{d x}{\left(x^{2}+1\right)^{2}}=\frac{x}{2\left(x^{2}+1\right)}+\frac{1}{2} \tan ^{-1} x} \\ {\int \frac{d x}{\left(x^{2}+1\right)^{3}}=\frac{x}{4\left(x^{2}+1\right)^{2}}+\frac{3 x}{8\left(x^{2}+1\right)}+\frac{3}{8} \tan ^{-1} x}\end{array}$$
#### Topics
Integration Techniques
### Discussion
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##### Heather Z.
Oregon State University
##### Kristen K.
University of Michigan - Ann Arbor
Lectures
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### Video Transcript
Let's use integration by parts toe Prove the following formula and let's go ahead and denote the integral on the left. But I am so we see the end shows up on the right. Is the exponents in the denominator? So for our integration, my parts let's make this choice for you then do you? Is negative to an ex that you think the chain rule from calculus TV is the ex. So then, by using immigration my parts we know the formula. You ve todo So in our case, lets go out and write that out. Look, no, cancel those double negatives. Pull off the two one and that here are always the following trick. I'LL rewrite this as x squared plus a squared minus a squared And then I'll split it this up into two in a girls. So this is then we have X squared plus a square. Now we see that we could cancel one of these terms Run And then for the last in a girl Let's pull out that a square. We still have that too in from here. Good. And then in a girl one over x squared, plus a square. Now notice that we can rewrite these remaining two in a girls in terms of this interval right here. So this integral one over X square, plus a square to the end, is just I am. And here would you have him and plus one that will be I and plus one. So let's go to the next patient. Write this out in the first term here. This is from our UV two and to end I am minus two and a squared I and plus one that's good and soft Fur I n plus one the smiles over here. So we have this one minus to end. Yeah, no. And what's good and simplify this. Cancel all those double those minus signs and then here. Okay, so now there's new formula. This hole's for any end. So instead of using the n plus one on the left, let's just replacing within. So here, just since this works for in the end, let's replace and with and minus one that here is the left hand side just becomes I am the right hand side. Just ride it out. So this is the numerator and then on the denominator to a square. But this time and minus one going on to the next page. Simplify each term here, that's our first term. And then our second term Teo So remember, on the left this's all equal toe I end And then here on the left, I could just rewrite that first her. That's the term that we wanted from the formula. We're basically done here. The last step is it. Just replace Pull off this constant here and then replace i n minus one with the definition as thie Integral. So that's a DX X squared plus a squared all to thee and minus one power. And that's the formula that we attended on proving Now let's go ahead and start on part B. So let me go to the next page here. So we'LL plug in for actually let me just write out two Step the integral here for part B one over X squared plus one square DX. So here we see that take was one and then we have an equals two. So let's fight and use this in our formula. So we get X over to X squared plus one and then going to the second term here we plug in and equals two again. So we get one half and then in a girl, one over X squared plus one. Now, if the Senate rules bother you feel free to do it. You some here, our strengths up. Excuse me? No. And we could simplify this one half and then we have our Tana Vicks plus our constancy. Now there's two intervals in part B. This is our answer for the first animal. And we could actually use this answer from the next interval. So we're still in part be here. But the next integral this time we have X squared plus one. Cute. So using the formula from part eh? This time and equals three equals one. So we have X over We'LL have two times two x square plus once where and then the second fraction two times one times two one over one plus x clear square My notice this integral That just popped out. That's the interval we just evaluated on the last stage. So here was just plug in our answer from the other the previous part of part B And then we have three over four and then we can replace this by previous age X to X squared plus one and then we also had one half. It's an inverse X and let's got and add that constant of integration, see, and very in. And that would be our answer for the second in a girl party and that used the earlier part of the same problem. B. But it also used the formula from party, and that's our final answer.
JH
#### Topics
Integration Techniques
##### Heather Z.
Oregon State University
##### Kristen K.
University of Michigan - Ann Arbor
Lectures
Join Bootcamp
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New SAT Math Workbook
# Example 372 461 372 22320 148800 171492 exercise 3
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Unformatted text preview: hen we multiply by 7, we are really multiplying by 70 and therefore leave a 0 at the extreme right before we proceed with the multiplication. Example: 537 × 72 1074 + 37590 38664 If we multiply by a three-digit number, we leave one zero on the right when multiplying by the tens digit and two zeros on the right when multiplying by the hundreds digit. Example: 372 × 461 372 22320 + 148800 171492 Exercise 3 Find the following products. 1. 526 multiplied by 317 (A) 156,742 (B) 165,742 (C) 166,742 (D) 166,748 (E) 166,708 8347 multiplied by 62 (A) 517,514 (B) 517,414 (C) 517,504 (D) 517,114 (E) 617,114 705 multiplied by 89 (A) 11,985 (B) 52,745 (C) 62,705 (D) 62,745 (E) 15,121 4. 437 multiplied by 607 (A) 265,259 (B) 265,219 (C) 265,359 (D) 265,059 (E) 262,059 798 multiplied by 450 (A) 358,600 (B) 359,100 (C) 71,820 (D) 358,100 (E) 360,820 5. 2. 3. www.petersons.com Operations with Whole Numbers and Decimals 5 4. DIVISION OF WHOLE NUMBERS The number being divided is called the dividend. The number we are dividing by is called the divisor. The answer to the division is called the quotient. When we divide 18 by 6, 18 is the dividend, 6 is the divisor, and 3 is the quotient. If the quotient is not an integer, we have a remainder. The remainder when 20 is divided by 6 is 2, 2 because 6 will divide 18 evenly, leaving a remainder of 2. The quotient in this case is 6 6 . Remember that in writing the fractional part of a quotient involving a remainder, the remainder becomes the numerator and the divisor the denominator. When dividing by a single-digit divisor, no long division procedures are needed. Simply carry the remainder of each step over to the next digit and continue. Example: 9724 6 5 84 31 4 2 4 ) Exercise 4 1. Divide 391 by 23. (A) 170 (B) 16 (C) 17 (D) 18 (E) 180 Divide 49,523,436 by 9. (A) 5,502,605 (B) 5,502,514 (C) 5,502,604 (D) 5,502,614 (E) 5,502,603 3. Find the remainder when 4832 is divided by 15. (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 Divide 42,098 by 7. (A) 6014 (B) 6015 (C) 6019 (D) 6011 (E) 6010 Which of the following is the quotient of 333,180 and 617? (A) 541 (B) 5...
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## This note was uploaded on 08/15/2010 for the course MATH a4d4 taught by Professor Colon during the Spring '10 term at Embry-Riddle FL/AZ.
Ask a homework question - tutors are online
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# By using the formula, find the amount and compound interest on:
Question:
By using the formula, find the amount and compound interest on:
Rs 31250 for 3 years at 8% per annum compounded annually.
Solution:
Principal amount, $P=$ Rs. 31250
Rate of interest, $R=8 \%$ per annum.
Time, $n=3$ years.
The formula for the amount including the compound interest is given below :
$A=$ Rs. $P\left(1+\frac{R}{100}\right)^{\mathrm{n}}$
$\Rightarrow A=$ Rs. $31250\left(1+\frac{8}{100}\right)^{3}$
$\Rightarrow A=$ Rs. $31250\left(\frac{100+8}{100}\right)^{3}$
$\Rightarrow A=$ Rs. $31250\left(\frac{108}{100}\right)^{3}$
$\Rightarrow A=$ Rs. $31250(1.08 \times 1.08 \times 1.08)^{3}$
$\Rightarrow A=$ Rs. 39366
i. e., the amount including the compound interest is Rs 39366 .
$\therefore$ Compound interest $=$ Rs. $(39366-31250)=$ Rs. 8116
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# What is the estimated value of 7.94/2.01 to 1 significant figure?
May 29, 2017
$4$
#### Explanation:
Start by performing the division
$\frac{7.94}{2.01} = 3.950248756$
Now, you're dealing with division, so the answer can only have as many significant figures as the value with the least number of sig figs.
In this case, you have
• $7.94 \to$ three non-zero digits $=$ three sig figs
• $2.01 \to$ two non-zero digits + a zero sandwitched between two non-zero digits $=$ three sig figs
This means that usually, the result of this division should be rounded to three sig figs. However, you must round this value to one significant figure.
Start with the first digit and count two digits, one more than the number of sig figs that you must have for the answer. You should stop at $9$, the second digit
$3.9 \textcolor{red}{\cancel{\textcolor{b l a c k}{50248756}}}$
You now have
$\frac{7.94}{2.01} = 3.9$
Next, compare the second digit to $\textcolor{red}{5}$.
• $\text{second digit} \ge \textcolor{red}{5} \implies$ you add $\textcolor{b l u e}{1}$ to the first digit
• $\text{Second digit} < \textcolor{red}{5} \implies$ you leave the first digit unchanged
In both cases, you must drop the second digit after you make the comparison.
$9 \ge \textcolor{red}{5}$
so you must add $1$ to the first digit to get
$3 + \textcolor{b l u e}{1} = 4$
$\frac{7.94}{2.01} = 4 \to$ rounded to one significant figure
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S k i l l
i n
A R I T H M E T I C
Lesson 3 Section 2
# HOW TO NAME OR READ A DECIMAL
Back to Section 1
6. How do we name or read a decimal? .038 Ignore the decimal point and read the number as a whole number; then say the decimal unit where the last digit falls.
Example 1. .038 "38 thousandths"
Ignore the decimal point and read 038 as the whole number "Thirty-eight." The last digit, 8, falls in the thousandths place.
When we read .038 as "Point 0, 3, 8," that is "spelling" the number, which is often convenient. But its name is "Thirty-eight thousandths."
Example 2. .002135 "2,135 millionths"
Ignore the decimal point, and read 002135 as the whole number 2,135 ("Two thousand one hundred thirty-five" Lesson 2, Question 4). The last digit 5 is in the millionths place.
Example 3. 14.0029 "14 and 29 ten-thousandths."
This is called a mixed number. The decimal point separates the whole number 14 on the left, from the decimal fraction on the right. In a mixed number, we read the decimal point as "and."
Example 4. Write these in numerals:
a) Two hundred four thousand
b) Two hundred four thousandths
c) Two hundred and four thousandths
a) 204,000 This is a purely whole number. b) .204 This is a purely decimal number. The thousandths are the 3rd decimal place. Therefore the 4 must fall in that last place. c) 200.004 This is a mixed number. The word "and" will always signify the decimal point.
Example 5. Write in words: \$607.08
Answer. Six hundred seven dollars and eight cents.
Save "and" for the decimal point.
Note that cents means hundredths. (Centum in Latin means 100.) 1 cent is the hundredth part of one dollar. We write 1 cent either as
\$.01 or 1¢.
When we write the cent sign ¢, we do not write a decimal point.
Example 6. Write "eighty cents" using the dollar sign \$ and using the cent sign ¢.
At this point, please "turn" the page and do some Problems.
or
Continue on to the next Section.
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# Quick Answer: How Do You Find The Sum Of Two Numbers?
## What is the sum of all positive numbers?
Two physicists explain: The sum of all positive integers equals −1/12.
Their viral video introduces mathematics that laymen find preposterous, but physicists find useful..
## What does the sum of two numbers mean?
What is a sum? The sum of two numbers is the answer you get when you add them both together. So the sum of 5 and 4 is 9.
## How do you find the sum of whole numbers?
Using the Formula We can put what Gauss discovered into an easy-to-use formula, which is: (n / 2)(first number + last number) = sum, where n is the number of integers. Let’s use the example of adding the numbers 1-100 to see how the formula works.
## What is the sum of numbers 1 to 50?
The number series 1, 2, 3, 4, . . . . , 49, 50. Therefore, 1275 is the sum of positive integers upto 50.
## How do you find the sum of triangular numbers?
nth triangular number is the sum of n consecutive natural numbers from starting which is simply n(n+1)/2. You want sum of first n triangular numbers. Just take the sum Σni=1i(i+1)2.
## What is the sum of n terms?
Sum of N Terms Formula It is equal to n divided by 2 times the sum of twice the first term – ‘a’ and the product of the difference between second and first term-‘d’ also known as common difference, and (n-1), where n is numbers of terms to be added. Sum of n terms of AP = n/2[2a + (n – 1)d]
## How do you find the sum of n terms?
The Sum Formula The formula says that the sum of the first n terms of our arithmetic sequence is equal to n divided by 2 times the sum of twice the beginning term, a, and the product of d, the common difference, and n minus 1. The n stands for the number of terms we are adding together.
## What is the sum of the first 56 numbers?
The number series 1, 2, 3, 4, . . . . , 55, 56. Therefore, 1596 is the sum of positive integers upto 56.
## What is the sum of all numbers?
For those of you who are unfamiliar with this series, which has come to be known as the Ramanujan Summation after a famous Indian mathematician named Srinivasa Ramanujan, it states that if you add all the natural numbers, that is 1, 2, 3, 4, and so on, all the way to infinity, you will find that it is equal to -1/12.
## What is the sum of the first 100 whole numbers?
Clearly, it is an Arithmetic Progression whose first term = 1, last term = 100 and number of terms = 100. Therefore, the sum of first 100 natural numbers is 5050.
## What is the sum of first 20 natural numbers?
The number series 1, 2, 3, 4, . . . . , 19, 20. Therefore, 210 is the sum of positive integers upto 20.
## What is the sum of 6841?
6,841 (six thousand eight hundred forty-one) is an odd four-digits prime number following 6840 and preceding 6842. In scientific notation, it is written as 6.841 × 103. The sum of its digits is 19.
## How do you find the sum of the first 20 terms?
Exercise. Calculate the sum of the first 20 terms of the arithmetic sequence whose formula for the nth term is: un=1+(n−1)×4. … Answers Without Working.Formula 2. Given an arithmetic sequence, we can calculate the sum of its first n terms, Sn, using the formula: Sn=n2(2.
## What is the sum of the number from 1 to 10?
The number series 1, 2, 3, 4, . . . . , 9, 10. Therefore, 55 is the sum of positive integers upto 10.
## What is the sum of first 50 odd numbers?
The number series 1, 3, 5, 7, 9, . . . . , 99. Therefore, 2500 is the sum of first 50 odd numbers.
## What does calculate the sum of numbers mean?
In mathematics, sum can be defined as the result or answer we get on adding two or more numbers or terms. Here, for example, addends 8 and 5 add up to make the sum 13.
## How do you find the sum of something?
The result of adding two or more numbers. (because 2 + 4 + 3 = 9).
## What is the sum of 3 numbers?
NumberRepeating Cycle of Sum of Digits of Multiples3{3,6,9,3,6,9,3,6,9}4{4,8,3,7,2,6,1,5,9}5{5,1,6,2,7,3,8,4,9}6{6,3,9,6,3,9,6,3,9}8 more rows
## What is the sum of a series?
The sum of the terms of a sequence is called a series . If a sequence is arithmetic or geometric there are formulas to find the sum of the first n terms, denoted Sn , without actually adding all of the terms.
## What is the sum of natural numbers?
A Sum of natural numbers from 1 to n. The answer is n(n+1)/2. Atleast, this is what we were taught all throughout our schooling. So, if ‘n’ were to tend to infinity, summation should tend to infinity.
## What is the sum of 1 to N?
Sum of the First n Natural Numbers. We prove the formula 1+ 2+ … + n = n(n+1) / 2, for n a natural number. There is a simple applet showing the essence of the inductive proof of this result.
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# Triangular number
The first six triangular numbers
A triangular number or triangle number counts objects arranged in an equilateral triangle. The nth triangular number is the number of dots in the triangular arrangement with n dots on a side, and is equal to the sum of the n natural numbers from 1 to n. The sequence of triangular numbers (sequence A000217 in the OEIS), starting at the 0th triangular number, is
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666...
## Formula
Derivation of triangular numbers from a left-justified Pascal's triangle
The triangle numbers are given by the following explicit formulas:
${\displaystyle T_{n}=\sum _{k=1}^{n}k=1+2+3+\dotsb +n={\frac {n(n+1)}{2}}={n+1 \choose 2},}$
where ${\displaystyle \textstyle {n+1 \choose 2}}$ is a binomial coefficient. It represents the number of distinct pairs that can be selected from n + 1 objects, and it is read aloud as "n plus one choose two".
The first equation can be illustrated using a visual proof.[1] For every triangular number ${\displaystyle T_{n}}$, imagine a "half-square" arrangement of objects corresponding to the triangular number, as in the figure below. Copying this arrangement and rotating it to create a rectangular figure doubles the number of objects, producing a rectangle with dimensions ${\displaystyle n\times (n+1)}$, which is also the number of objects in the rectangle. Clearly, the triangular number itself is always exactly half of the number of objects in such a figure, or: ${\displaystyle T_{n}={\frac {n(n+1)}{2}}}$. The example ${\displaystyle T_{4}}$ follows:
${\displaystyle 2T_{4}=4(4+1)=20}$ (green plus yellow) implies that ${\displaystyle T_{4}={\frac {4(4+1)}{2}}=10}$ (green).
The first equation can also be established using mathematical induction.[2] Since the sum of the first (one) natural number(s) is clearly equal to one, a basis case is established. Assuming the inductive hypothesis for some ${\displaystyle n}$ and adding ${\displaystyle n+1}$ to both sides immediately gives
${\displaystyle \sum _{k=1}^{n}k+(n+1)={\frac {n(n+1)}{2}}+(n+1)\rightarrow \sum _{k=1}^{n+1}k={\frac {(n+1)((n+1)+1)}{2}}}$
In other words, since the proposition ${\displaystyle P(n)}$ (that is, the first equation, or inductive hypothesis itself) is true when ${\displaystyle n=1}$, and since ${\displaystyle P(n)}$ being true implies that ${\displaystyle P(n+1)}$ is also true, then the first equation is true for all natural numbers. The above argument can be easily modified to start with, and include, zero.
Carl Friedrich Gauss is said to have found this relationship in his early youth, by multiplying n/2 pairs of numbers in the sum by the values of each pair n + 1.[3] However, regardless of the truth of this story, Gauss was not the first to discover this formula, and some find it likely that its origin goes back to the Pythagoreans 5th century BC.[4] The two formulae were described by the Irish monk Dicuil in about 816 in his Computus.[5]
The triangular number Tn solves the handshake problem of counting the number of handshakes if each person in a room with n + 1 people shakes hands once with each person. In other words, the solution to the handshake problem of n people is Tn−1.[6] The function T is the additive analog of the factorial function, which is the products of integers from 1 to n.
The number of line segments between closest pairs of dots in the triangle can be represented in terms of the number of dots or with a recurrence relation:
${\displaystyle L_{n}=3T_{n-1}=3{n \choose 2};~~~L_{n}=L_{n-1}+3(n-1),~L_{1}=0.}$
In the limit, the ratio between the two numbers, dots and line segments is
${\displaystyle \lim _{n\to \infty }{\frac {T_{n}}{L_{n}}}={\frac {1}{3}}.}$
## Relations to other figurate numbers
Triangular numbers have a wide variety of relations to other figurate numbers.
Most simply, the sum of two consecutive triangular numbers is a square number, with the sum being the square of the difference between the two (and thus the difference of the two being the square root of the sum). Algebraically,
${\displaystyle T_{n}+T_{n-1}=\left({\frac {n^{2}}{2}}+{\frac {n}{2}}\right)+\left({\frac {\left(n-1\right)^{2}}{2}}+{\frac {n-1}{2}}\right)=\left({\frac {n^{2}}{2}}+{\frac {n}{2}}\right)+\left({\frac {n^{2}}{2}}-{\frac {n}{2}}\right)=n^{2}=(T_{n}-T_{n-1})^{2}.}$
This fact can be demonstrated graphically by positioning the triangles in opposite directions to create a square:
6 + 10 = 16
10 + 15 = 25
There are infinitely many triangular numbers that are also square numbers; e.g., 1, 36, 1225. Some of them can be generated by a simple recursive formula:
${\displaystyle S_{n+1}=4S_{n}\left(8S_{n}+1\right)}$ with ${\displaystyle S_{1}=1.}$
All square triangular numbers are found from the recursion
${\displaystyle S_{n}=34S_{n-1}-S_{n-2}+2}$ with ${\displaystyle S_{0}=0}$ and ${\displaystyle S_{1}=1.}$
A square whose side length is a triangular number can be partitioned into squares and half-squares whose areas add to cubes. This shows that the square of the nth triangular number is equal to the sum of the first n cube numbers.
Also, the square of the nth triangular number is the same as the sum of the cubes of the integers 1 to n. This can also be expressed as
${\displaystyle \sum _{k=1}^{n}k^{3}=\left(\sum _{k=1}^{n}k\right)^{2}.}$
The sum of the first n triangular numbers is the nth tetrahedral number:
${\displaystyle \sum _{k=1}^{n}T_{k}=\sum _{k=1}^{n}{\frac {k(k+1)}{2}}={\frac {n(n+1)(n+2)}{6}}.}$
More generally, the difference between the nth m-gonal number and the nth (m + 1)-gonal number is the (n − 1)th triangular number. For example, the sixth heptagonal number (81) minus the sixth hexagonal number (66) equals the fifth triangular number, 15. Every other triangular number is a hexagonal number. Knowing the triangular numbers, one can reckon any centered polygonal number; the nth centered k-gonal number is obtained by the formula
${\displaystyle Ck_{n}=kT_{n-1}+1}$
where T is a triangular number.
The positive difference of two triangular numbers is a trapezoidal number.
## Other properties
Triangular numbers correspond to the first-degree case of Faulhaber's formula.
Alternating triangular numbers (1, 6, 15, 28, ...) are also hexagonal numbers.
Every even perfect number is triangular (as well as hexagonal), given by the formula
${\displaystyle M_{p}2^{p-1}={\frac {M_{p}(M_{p}+1)}{2}}=T_{M_{p}}}$
where Mp is a Mersenne prime. No odd perfect numbers are known, hence all known perfect numbers are triangular.
For example, the third triangular number is (3 × 2 =) 6, the seventh is (7 × 4 =) 28, the 31st is (31 × 16 =) 496, and the 127th is (127 × 64 =) 8128.
In base 10, the digital root of a nonzero triangular number is always 1, 3, 6, or 9. Hence every triangular number is either divisible by three or has a remainder of 1 when divided by 9:
0 = 9 × 0
1 = 9 × 0 + 1
3 = 9 × 0 + 3
6 = 9 × 0 + 6
10 = 9 × 1 + 1
15 = 9 × 1 + 6
21 = 9 × 2 + 3
28 = 9 × 3 + 1
36 = 9 × 4
45 = 9 × 5
55 = 9 × 6 + 1
66 = 9 × 7 + 3
78 = 9 × 8 + 6
91 = 9 × 10 + 1
There is a more specific property to the triangular numbers that aren't divisible by 3, that is, they either have a remainder 1 or 10 when divided by 27. Those that are equal to 10 mod 27, are also equal to 10 mod 81.
The digital root pattern for triangular numbers, repeating every nine terms, as shown above, is "1, 3, 6, 1, 6, 3, 1, 9, 9".
The converse of the statement above is, however, not always true. For example, the digital root of 12, which is not a triangular number, is 3 and divisible by three.
If x is a triangular number, then ax + b is also a triangular number, given a is an odd square and b = a − 1/8
b will always be a triangular number, because 8Tn + 1 = (2n + 1)2, which yields all the odd squares are revealed by multiplying a triangular number by 8 and adding 1, and the process for b given a is an odd square is the inverse of this operation.
The first several pairs of this form (not counting 1x + 0) are: 9x + 1, 25x + 3, 49x + 6, 81x + 10, 121x + 15, 169x + 21, … etc. Given x is equal to Tn, these formulas yield T3n + 1, T5n + 2, T7n + 3, T9n + 4, and so on.
The sum of the reciprocals of all the nonzero triangular numbers is
${\displaystyle \!\ \sum _{n=1}^{\infty }{1 \over {{n^{2}+n} \over 2}}=2\sum _{n=1}^{\infty }{1 \over {n^{2}+n}}=2.}$
This can be shown by using the basic sum of a telescoping series:
${\displaystyle \!\ \sum _{n=1}^{\infty }{1 \over {n(n+1)}}=1.}$
Two other formulas regarding triangular numbers are
${\displaystyle T_{a+b}=T_{a}+T_{b}+ab}$
and
${\displaystyle T_{ab}=T_{a}T_{b}+T_{a-1}T_{b-1},}$
both of which can easily be established either by looking at dot patterns (see above) or with some simple algebra.
In 1796, German mathematician and scientist Carl Friedrich Gauss discovered that every positive integer is representable as a sum of three triangular numbers (possibly including T0 = 0), writing in his diary his famous words, "ΕΥΡΗΚΑ! num = Δ + Δ + Δ". This theorem does not imply that the triangular numbers are different (as in the case of 20 = 10 + 10 + 0), nor that a solution with exactly three nonzero triangular numbers must exist. This is a special case of the Fermat polygonal number theorem.
The largest triangular number of the form 2k − 1 is 4095 (see Ramanujan–Nagell equation).
Wacław Franciszek Sierpiński posed the question as to the existence of four distinct triangular numbers in geometric progression. It was conjectured by Polish mathematician Kazimierz Szymiczek to be impossible and was later proven by Fang and Chen in 2007.[7][8]
Formulas involving expressing an integer as the sum of triangular numbers are connected to theta functions, in particular the Ramanujan theta function.[9][10]
## Applications
A fully connected network of n computing devices requires the presence of Tn − 1 cables or other connections; this is equivalent to the handshake problem mentioned above.
In a tournament format that uses a round-robin group stage, the number of matches that need to be played between n teams is equal to the triangular number Tn − 1. For example, a group stage with 4 teams requires 6 matches, and a group stage with 8 teams requires 28 matches. This is also equivalent to the handshake problem and fully connected network problems.
One way of calculating the depreciation of an asset is the sum-of-years' digits method, which involves finding Tn, where n is the length in years of the asset's useful life. Each year, the item loses (bs) × ny/Tn, where b is the item's beginning value (in units of currency), s is its final salvage value, n is the total number of years the item is usable, and y the current year in the depreciation schedule. Under this method, an item with a usable life of n = 4 years would lose 4/10 of its "losable" value in the first year, 3/10 in the second, 2/10 in the third, and 1/10 in the fourth, accumulating a total depreciation of 10/10 (the whole) of the losable value.
## Triangular roots and tests for triangular numbers
By analogy with the square root of x, one can define the (positive) triangular root of x as the number n such that Tn = x:[11]
${\displaystyle n={\frac {{\sqrt {8x+1}}-1}{2}}}$
which follows immediately from the quadratic formula. So an integer x is triangular if and only if 8x + 1 is a square. Equivalently, if the positive triangular root n of x is an integer, then x is the nth triangular number.[11]
## References
1. ^ "Triangular Number Sequence". Math Is Fun.
2. ^ Andrews, George E. Number Theory, Dover, New York, 1971. pp. 3-4.
3. ^ Hayes, Brian. "Gauss's Day of Reckoning". American Scientist. Computing Science. Retrieved 2014-04-16.
4. ^ Eves, Howard. "Webpage cites AN INTRODUCTION TO THE HISTORY OF MATHEMATICS". Mathcentral. Retrieved 28 March 2015.
5. ^ Esposito,M. An unpublished astronomical treatise by the Irish monk Dicuil. Proceedings of the Royal Irish Academy, XXXVI C. Dublin, 1907, 378-446.
6. ^ http://www.mathcircles.org/node/835
7. ^ Chen, Fang: Triangular numbers in geometric progression
8. ^ Fang: Nonexistence of a geometric progression that contains four triangular numbers
9. ^ Liu, Zhi-Guo (2003-12-01). "An Identity of Ramanujan and the Representation of Integers as Sums of Triangular Numbers". The Ramanujan Journal. 7 (4): 407–434. doi:10.1023/B:RAMA.0000012425.42327.ae. ISSN 1382-4090.
10. ^ Sun, Zhi-Hong (2016-01-24). "Ramanujan's theta functions and sums of triangular numbers". arXiv:1601.06378 [math.NT].
11. ^ a b Euler, Leonhard; Lagrange, Joseph Louis (1810), Elements of Algebra, 1 (2nd ed.), J. Johnson and Co., pp. 332–335
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# Reflection
Here we will learn how to reflect 2D shapes on grid paper and how to describe reflections.
There are also reflection worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What is reflection?
Reflection is a type of transformation that flips a shape in a mirror line (also called a line of reflection) so that each point is the same distance from the mirror line as its reflected point.
E.g.
Triangle P has been reflected in the line x=4 to give Triangle Q .
Triangle P is the object and Triangle Q is the image.
When an object is reflected the object and the image are mirror images of each other.
We can see that each point of triangle P is the same distance from the mirror line as the corresponding point on triangle Q .
The two triangles are congruent because they are the same shape and the same size.
## How to use reflections
In order to reflect a shape on a grid:
1. Reflect the first point.
2. Reflect the other points.
3. Finish the diagram.
## Reflection maths examples
### Example 1: reflect a shape on a grid
Reflect the shape in the line:
1. Reflect the first point.
Choose the first point to reflect. It is easier to start with a point which is closest to the mirror line (the line of reflection). The new point will be exactly the same distance away from the mirror line as the original point.
2Reflect the other points.
Here is a second point being reflected to give its image. The remaining two points have also been added.
3Finish the diagram.
To finish the diagram we can join up the reflected points.
### Example 2: reflect a shape on a grid
Reflect the shape in the line:
1. Reflect the first point.
Choose the first point to reflect. It is easier to start with a point which is closest to the mirror line (the line of reflection). The new point will be exactly the same distance away from the mirror line as the original point.
2Reflect the other points.
Here is a second and a third point being reflected to give its image. The remaining point is on the mirror line and does not move (it is an invariant point).
3Finish the diagram.
To finish the diagram we can join up the reflected points.
## How use reflections on a coordinate grid
In order to reflect a shape on a coordinate grid:
1. Draw the mirror line.
2. Reflect the first point.
3. Reflect the other points.
4. Finish the diagram.
## Reflections on a coordinate grid examples
### Example 3: reflect a shape on a coordinate grid
Reflect Triangle P in the line x = 4 :
1. Draw the mirror line.
The mirror line is x = 4 (the line of reflection). This is a vertical line. Draw this on the diagram.
2Reflect the other points.
Choose the first point to reflect. It is easier to start with a point which is closest to the mirror line (the line of reflection). Let’s reflect the point (3,1) . The new point will be exactly the same distance away from the mirror line as the original point.
3Reflect the other points.
Here is a second point (1,4) being reflected to give its image.
We can reflect the third point (1,1) .
4Finish the diagram.
To finish the diagram we can join up the reflected points.
Here Triangle Q is the image of Triangle P .
### Example 4: reflect a shape
Reflect Triangle P in the line y = 3 :
1. Draw the mirror line.
The mirror line is y = 3 (the line of reflection). This is a horizontal line. Draw this on the diagram.
2Reflect the other points.
Choose the first point to reflect. It is easier to start with a point which is closest to the mirror line (the line of reflection). Let’s reflect the point (3,2) . The new point will be exactly the same distance away from the mirror line as the original point.
3Reflect the other points.
Here is a second point (1,4) being reflected to give its image.
We can reflect the third point (4,0) .
4Finish the diagram.
To finish the diagram we can join up the reflected points.
Here Triangle Q is the image of Triangle P .
## How to describe reflections
In order to describe a reflection of a shape on a coordinate grid:
1. Pair up the points.
2. Identify the midpoints.
3. Join the midpoints.
4. State the equation of the line.
## Describing reflections examples
### Example 5: describe a reflection
Describe the transformation of Shape A to Shape B
1. Pair up the points.
Try to match up the corresponding point on the two shapes and draw a connecting line. Try to do this with more than one set of points.
2Identify the midpoints.
Identify the midpoints of these connecting lines. This will help us to draw in the line of reflection.
3Join the midpoints.
Draw a straight line through the midpoints.
4State the equation of the line.
The equation of the line is y = 4 .
The transformation is a reflection in the line y = 4
### Example 6: describe a reflection
Describe the transformation of Shape A to Shape B
1. Pair up the points.
Try to match up the corresponding point on the two shapes and draw a connecting line. Try to do this with more than one set of points.
2Identify the midpoints.
Identify the midpoints of these connecting lines. This will help us to draw in the line of reflection.
3Join the midpoints.
Draw a straight line through the midpoints.
4State the equation of the line.
The equation of the line is y = x .
The transformation is a reflection in the line y = x
### Common misconceptions
• Overlapping shapes
The original shape (the object) and its reflection (the image) are allowed to overlap each other.
E.g.
• Diagonal mirror line (the line of reflection)
Diagonal mirror lines can be tricky. It is worth turning the diagram so that the mirror line is either horizontal or vertical to make the reflections easier to carry out.
• Describe transformations fully
When you are asked to describe a transformation make sure you state which kind of transformation it is and any other details. So for reflections we need to state that it is a reflection and give the equation of the line of reflection.
### Practice reflection maths questions
1. Reflect the shape in the line of reflection:
The corresponding points on the object and the image must be equidistant (the same distance) from the mirror line.
2. Reflect the shape in the line of reflection:
The corresponding points on the object and the image must be equidistant (the same distance) from the mirror line. The object and the image should be congruent – the same shape and the same size.
3. Reflect the shape in the line y = 4 :
The line y = 4 is a horizontal line going through 4 on the y-axis. The corresponding points on the object and the image must be equidistant (the same distance) from the mirror line.
4. Reflect the shape in the y axis:
The y-axis is the vertical axis. The corresponding points on the object and the image must be equidistant (the same distance) from the mirror line. The object and the image should be congruent – the same shape and the same size.
5. Describe the transformation of Shape P to Shape Q
reflection in x=6
reflection in x=5
reflection in y=5
transformation in x=5
You must state that the transformation is a reflection. The line of reflection is a vertical line, so the equation is x=a , where a is a number. The corresponding points on the object and the image must be equidistant (the same distance) from the mirror line.
6. Describe the reflection of Shape P to Shape Q
reflection in x-axis
reflection in y-axis
reflection in y=-x
reflection in y=x
The line of reflection is a diagonal line. The corresponding points on the object and the image must be equidistant (the same distance) from the mirror line. The mirror line goes through the points (1,-1), (2,-2), (3,-3) and so on.
### Reflection maths GCSE questions
1. Reflect the shaded shape in the mirror line.
(2 marks)
for at least 2 correct vertices
(1)
for correct reflection in the correct position
(1)
2. On the grid below, reflect the shape in the line with the equation x=1
(2 marks)
for drawing the correct mirror line
(1)
for correct reflection in the correct position
(1)
3. Describe fully the single transformation that maps shape A onto shape B .
(2 marks)
Reflection in x-axis (or y=0 )
stating reflection
(1)
for giving the correct line of reflection
(1)
## Learning checklist
You have now learned how to:
• Reflect 2D shapes including polygons on grid
• Reflect 2D shapes including polygons on coordinate grid with x-axis and y-axis
• Identify properties of, and describe the results of reflections applied to given figures
## Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
Find out more about our GCSE maths tuition programme.
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# 2004 AMC 12B Problems/Problem 16
## Problem
A function $f$ is defined by $f(z) = i\overline{z}$, where $i=\sqrt{-1}$ and $\overline{z}$ is the complex conjugate of $z$. How many values of $z$ satisfy both $|z| = 5$ and $f(z) = z$?
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 2 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 8$
## Solutions
### Solution 1
Let $z = a+bi$, so $\overline{z} = a-bi$. By definition, $z = a+bi = f(z) = i(a-bi) = b+ai$, which implies that all solutions to $f(z) = z$ lie on the line $y=x$ on the complex plane. The graph of $|z| = 5$ is a circle centered at the origin, and there are $2 \Rightarrow \mathrm{(C)}$ intersections.
### Solution 2
Let $z=a+bi$, like above. Therefore, $z = a+bi = i\overline{z} = i(a-bi) = ai+b$. We move some terms around to get $bi-b = ai-a$. We factor: $b(i-1) = a(i-1)$. We divide out the common factor to see that $b = a$. Next we put this into the definition of $|z| = a^2 + b^2 = a^2 + a^2 = 2a^2 = 25$. Finally, $a = \pm\sqrt{\frac{25}{2}}$, and $a$ has two solutions.
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# NCERT Exemplar Class 7 Maths Chapter 1 Integers
In this chapter, we provide NCERT Exemplar Problems Solutions for Class 7 Maths Chapter 1 Integers for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 7 Maths Chapter 1 Integers pdf, free NCERT Exemplar Problems Solutions for Class 7 Maths Chapter 1 Integers book pdf download. Now you will get step by step solution to each question.
## NCERT Exemplar Class 7 Maths Chapter 1 Integers Solutions
Multiple Choice Questions (MCQs)
Question 1:
When the integers 10, 0, 5, -5,-7 are arranged in descending or ascending order, then find out which of the following integers always remains in the middle of the arrangement.
(a) 0 (b) 5 (c) – 7 (d) -5
Solution:
(a) To arrange these integers in ascending or descending order, first we locate these points on number line.
As we know, if a point or number lies on the right side to the other number, then the number is greater. Then,
Ascending order -7, – 5,0, 5,10
Middle term = 0
Descending order → 10, 5, 0, – 5, – 7
Middle term = 0
Hence, zero always remains in the middle of the arrangement.
Question 2:
By observing the number line, state which of the following statements is not true?
(a) B is greater than -10 (b) A is greater than 0 (c) B is greater than A (d) 6 is smaller than 0
Solution:
(c) As we know that, if a point or number lies on the right side to the other number, then the number is greater.
Here, B is greater than -10 but smaller than 0 and A is greater than 0 but smaller than 10. Also, 6 is smaller than A.
Question 3:
By observing the above number line, state which of the following statements is true?
(a) B is 2 (b) A is – 4 (c) S is -13 (d) B is – 4
Solution:
(d) Since, 8 lies at the left side of 0, so it will be negative and it is at 4th place.
So, 8 =-4
Similarly, A lies at the right side of 0, so it will be positive and it is at 7th place.
So, A = 7. Hence, the value of A – 7 and value of 8 = – 4.
Question 4:
Next three consecutive numbers in the pattern 11, 8, 5, 2,______ ,__ ,__ are
(a) 0, – 3, – 6 (b)-1,-5,-8 (c) – 2, — 5, – 8 (d)-1,-4,-7
Solution:
(d) By observing the series, difference between two consecutive numbers is 3,
Question 5:
The next number in the pattern – 62,- 37,- 12 is_________________ .
(a) 25 (b) 13 (c) 0 (d) -13
Solution:
(b) By observing the series, difference between two consecutive numbers is 25, i.e.
-37 -(-62) = -37 + 62 =25
– 12 – (- 37) = -12 + 37 =25
So, next number will be
-12 + 25 = 13
Question 6:
Which of the following statements is not true? ;
(a) When two positive integers are added, we always get a positive integer.
(b) When two negative integers are added, we always get a negative integer.
(c) When a positive integer and a negative integer are added, we always get a negative integer.
(d) Additive inverse of an integer 2 is (-2) and additive inverse of (-2) is 2.
Solution:
(c)
(a) True, when two positive integers are added, the resultant number is also a positive integer.
(b) True, while adding integers, if both the numbers have same sign, the resultant number also get that sign.
(c) False, while adding the integers of different signs, the resultant number get the sign of greater number.
(d) True, additive inverse of an integer is the same integer value, with opposite sign.
Question 7:
On the following number line value, ‘zero’ is shown by the point
(a) X (b) Y (c) Z (d) W
Solution:
(c) All the points are equally spaced.
One division = 5 units
So, X = -15 + 5 = -10
Y=-10+ 5 = – 5
Z=- 5 +5 = 0
Hence, zero is shown by the point Z.
Question 8:
Solution:
(c) Descending order in number line, is from right to left.
Accordingly,
Question 9:
On the number line, the value of (-3) x 3 lies on right hand side of
(a) -10 (b) – 4 (c) 0 (d) 9
Solution:
(a) (-3) x 3 equals to – 9.
On the number line, it is shown as
So, as we can see – 9 lies on the right hand side of -10.
Question 10:
The value of 5 + (- 1) does not lie between
(a) 0 and -10 (b) 0 and 10 (c) – 4 and -15 (d) – 6 and 6
Solution:
(b) 5÷ (-1) equals to – 5.
On the number line, it is placed as
Now, as we can see, -5 lies between (0 and -10) , (-4 and -15) and (-6 and 6). But it does not lie between 0 and 10.
Question 11:
Water level in a well was 20 m below ground level. During rainy season, rainwater collected in different water tanks was drained into the well and the water level rises 5 m above the previous level. The wall of the well is lm 20cm high and a pulley afixed at a height of 80 cm. Raghu wants to draw water from the well. The minimum length of the rope, that he can use is
Solution:
(a) Details given in the question, can be described in the figure shown below
Question 12:
(-11) x 7 is not equal to
(a) 11 x (-7) (b) -(11 x 7) (c) (- 11) x (- 7 ) (d) 7 x (-11)
Solution:
(c) (— 11) x 7 = (-77) (we know, in multiplication, if sign of both numbers are different, then the sign of the resultant is negative and if sign of both numbers are same, then the sign of the resultant is positive.)
Option (a), 11 x (-7) =-77
Option (b), – (11 x 7) = – 77
Option (c), (-11) x (- 7) = 77
Option (d),7 x(-11) = -77
Question 13:
(- 10) x (- 5) + (- 7) is equal to
(a) -57 (b) 57 (c) -43 (d) 43
Solution:
(d) (-10) x (- 5) + (- 7) = {(- 10) x (- 5)} + (- 7)= 50 + (- 7) = 50- 7 = 43
Question 14:
Which of the following is not the additive inverse of a?
(a) – (-a) (b) a x (-1) (c) — a (d) a + (-1)
Solution:
(a) Additive inverse of a is (- a).
[additive inverse of an integer is the same integer value, with opposite sign]
So,
Option (a), – (- a) = a
Option (b),ax(-1) = -a
Option (c), – a
Option (d), a + (-1) = – a
Question 15:
Which of the following is the multiplicative identity for an integer o?
(a) a (b) 1 (c) 0 (d) -1
Solution:
(b) Multiplicative identity for an integer a is 1.
[∵ a multiplicative identity is that identity in which any number is multiplied by that identity, it gives out the same number.]
Question 16:
[(- 8) x (- 3)] x (- 4)] is not equal to
(a) (- 8) x [(- 3) x (- 4)] (b) [(- 8) x (- 4)] x (- 3)
(c) [(- 3) x (- 8)] x (- 4) (d) (- 8) x (- 3) – (- 8) x (- 4)
Solution:
Question 17:
(- 25) x [6 + 4] is not same as
(a) (-25) x 10 (b) (-25) x 6 + (- 25) x 4
(c) -25 x 6 x 4 (d) – 250
Solution:
(c) (- 25) x [6 + 4] = (- 25) x 10
Also, (- 25) x [6 + 4] = – 25 x 6 + (- 25) x 4
[using distributive property, i.e. ax(b + c) = axb + axc] = -150-100=-250
Hence, (-25) x (6 + 4) is not same as -25 x 6 x 4.
Question 18:
– 35 x 107 is not same as
(a) – 35 x (100 + 7) (b) (- 35) x 7 + (- 35) x 100
(c) – 35 x 7 + 100 (d) (-30 -5) x 107
Solution:
(c) – 35 x 107 =-35 x (100+7)= (- 35) x 100 + (- 35) x 7 [using distributive property, i.e. a x (b + c) = a x b + a x c]
= (- 35) x 7 + (- 35) x 100 [as addition is commutative, i.e. a + b = b + a)
Also, – 35 x 107 =(- 30 – 5) x 107 [∴ (-30 – 5) = (- 35)]
Hence, – 35 x 107 is not same as -35 x 7 + 100.
Question 19:
(- 43) x (- 99)+ 43 is equal to
(a) 4300 (b) – 4300 (c) 425 (d) -4214
Solution:
(a) (- 43) x (- 99) + 43 = (- 1) (43) x (- 99) + 43
= 43 {-(- 99) + 1} [taking 43 as common]
= 43 (99 + 1)= 43 x 100= 4300
Question 20:
(- 16) ÷ 4 is not same as
(a) (-4) ÷ 16 (b) -(16 ÷ 4) (c) 16 ÷ (-4) (d) – 4
Solution:
Question 21:
Which of the following does not represent an integer?
(a) 0 ÷ (- 7) (b) 20 ÷ (- 4) (c) (-9) ÷ 3 (d) (-12) ÷ 5
Solution:
(d) An integer is a whole number (not a fractional number) that can be positive, negative or zero. So,
Question 22:
Which of the following is different from the others?
(a) 20 + (-25) (b) (-37) -(-32)
(c) (-5) x (— 1) (d) 45 ÷ (- 9)
Solution:
(c) Option (a), 20 + (- 25) = 20 – 25 = – 5
Option (b), (- 37) – (- 32)= – 37 + 32 = – 5
Option (c), (- 5) x (- 1)= 5
Option (d), (45) ÷ (- 9)= 45/-9 = – 5
Question 23:
Which of the following shows the maximum rise in temperature?
(a) 23° to 32° (b)-10°to 1°
(c) -18° to-11° (d) -5° to 5°
Solution:
(b) Rise in temperature,
(a) 32° – 23° = 9° (b) 1°-(-10)° = 1° + 10° = 11° (maximum)
(c) -11°-(-18)°=-11°+18° = 7° (d) 5° -(-5°) = 5° + 5° = 10°
Question 24:
If a and b are two integers, then which of the following may not be an integer?
(a) a+b (b) a-b (c) a x b (d) a+b
Solution:
(d) Addition, subtraction and multiplication of two or more integers is always an integer. But, division of integers may or may not be an integer.
e.g. 2 + 3 = 2 ⁄ 3 (not an integer)
3 + 3 = 1 (integer)
Question 25:
For a non-zero integer a, which of the following is not defined?
(a) a ÷ 0 (b) 0 ÷ a (c) a ÷1 (d) 1 ÷ a
Solution:
(a) Division of any number by zero is not defined,
a + 0 = not defined
In questions 26 to 30, encircle the odd one of the following:
Question 26:
(a) (-3,3)
(b) (-5,5)
(c) (-6,1)
(d) (-8,8)
Solution:
(c) By observation, we can say that both the values are same in options (a), (b) and (d).
So, odd one is option (c).
Question 27:
(a) (-1,-2)
(b) (-5,2)
(c) (- 4,1)
(d) (- 9,7)
Solution:
(d) By observation, we can say that the sum of both values are same in options (a), (b) and (c).
So, odd one is option (d).
Question 28:
(a) (- 9) x 5 x 6 x (- 3)
(b) 9 x (-5) x 6 x (-3)
(c) (- 9) x (- 5) x (- 6) x 3
(d) 9 x (- 5) x (- 6) x 3
Solution:
Question 29:
(a) (-100)+ 5 (b) (-81)+ 9
(c) (- 75) + 5 (d) (-32)+ 9
Solution:
Here, option (a), (b) and (c) are the negative integers, but option (d) is not the negative integer. So, odd one is option (d).
Question 30:
(a) (- 1) x (- l)
(b) (- 1) x (- l) x (- 1)
(c) (-1) x (-1) x (-1) x (-1)
(d) (-1) x (-1) x (-1) x (-1) x (-1) x (-1)
Solution:
(a) (-1) x (-1) = 1
(b) (- 1)x(-1)x(-1) = — 1
(c) (- 1) x (- 1) x (- 1) x (- 1) = 1
(d) (-1) x (-1) x (-1) x (- 1) x {-1) x(-1) = 1
Hence, value of options (a), (c), (d) are same but value of option (b) is different.
Fill in the Blanks
In questions 31 to 71, fill in the blanks to make the statements true.
Question 31 :
(- a) + b = b + additive inverse of ___ .
Solution:
Additive inverse is the negation of a number.
As we know, addition is commutative for integers, i.e. – a + b = b + (-a)
Now ‘- a’ is the additive inverse of a. So, a will be the answer.
Question 32:
__________ ÷ (- 10) = 0
Solution:
Division of 0 by any number, results as zero. So, the answer is 0.
Question 33:
(- 157) x (— 19) + 157 =___________ .
Solution:
(-157) x (-19) + 157 = (-1) x (157) x (-19) + 157
= 157 {- (-19) + 1} [taking 157 as common]
= 157 {19+1} = 157 x20= 3140
Question 34:
[(- 8)+ ___ ]+ ___ = ___ + [(-3) + ___ ] = – 3
Solution:
[(- 8) + (- 3)] + 8 = (- 8) + [(- 3) + 8]
[ addition is associative, i.e. a + (to + c) = (a + b) +c]
= – 8 + 5 = – 3
Question 35:
On the following number line, (- 4) x 3 is represented by the point_____ .
Solution:
Question 36:
If x, y and z are integers, then (x +____ ) + z =____ + (y +____ )
Solution:
Addition is associative for integers, i.e. (a + b) + c = a + (b + c)
=> (x + y) + z = x + (y + z)
Question 37:
(-43)+ __________ =(-43)
Solution:
Zero (0) is an additive identity for integers, i.e. a+0 = 0+ a = afor any integer a.
So, (- 43) + 0 = – 43
Question 38:
(- 8) + (- 8) + (- 8) =_______ x (- 8)
Solution:
Let x be the missing number.
Question 39:
11 x (- 5) = – (_____ x____ ) =_____
Solution:
We can write the equation as,
11 x (-5) =-(11×5) = -55
Question 40:
40 (- 9) x 20 =_______
Solution:
(-9) x 20 = -180 [ in multiplication of integers, if both the numbers have different signs, then the result is a negative number]
Question 41:
(- 23) x (42) = (- 42) x______
Solution:
(- 23) x (42)= (- 1) x (23) x (42)= (- 1) x (42) x (23)
[multiplication is commutative, i.e. a x b = b x a] = (- 42) x (23)
Question 42:
While multiplying a positive integer and a negative integer, we multiply them as ___ numbers and put a ___ sign before the product.
Solution:
When multiplying a positive integer and a negative integer, we multiply them as whole numbers and put a negative sign before the product.
Question 43:
If we multiply___ number of negative integers, then the resulting integer is positive.
Solution:
If we multiply even numbers of negative integers, then the resulting integer is positive
Question 44:
If we multiply six negative integers and six positive integers, then the resulting integer is___ .
Solution:
If we multiply six negative integers and six positive integers, then the resulting integer is positive, because even numbers of negative integers, in multiplication becomes positive.
Question 45:
If we multiply five positive integers and one negative integer, then the resulting integer is ___ .
Solution:
If we multiply 5 positive integers and one negative integer, then the resulting integer is negative.
Question 46:
________ is the multiplicative identity for integers.
Solution:
1 is the multiplicative identity for integers, i.e. a x 1 = 1 x a = a for any integer a.
Question 47:
We get additive inverse of an integer a, when we multiply it by ___ .
Solution:
Additive inverse of an integer is the same integer value, with opposite sign. So, we get additive inverse of integer a, when we multiply it by (– 1).
Question 48:
(- 25) x (- 2) =______.
Solution:
Two negative integers make the resultant integer, positive.
(- 25) x (- 2) = 50
Question 49:
(- 5) x (- 6) x (- 7) =______.
Solution:
Odd negative integers make the resultant integer, negative.
(- 5) x (- 6) x (- 7) = 30 x (- 7) = – 210
Question 50:
3 x (- 1) x (- 15) =_______.
Solution:
Two negative integers and one positive integer make the resultant integer, positive.
3 x (-1) x (-15) = (-3) x ( – 15)= 45c
Question 51:
[12 x (- 7)] x 5 =_____ x [(- 7) x ____ ]
Solution:
Multiplication is associative for integers, i.e.
(a x b) x c = a x (b x c)
So, [12 x (- 7)] x 5 = 12 x [(- 7) x 5]
Question 52:
23 x (- 99) = ____ x (- 100 + ____ ) = 23 x ____ + 23 x ____
Solution:
We can write the equation as,
23 x (- 99) = 23 x (- 100 + 1)= 23 x (- 100) + 23 x 1
[ integers show distributive property of multiplication over addition, i.e. a x (b + c) = a x b + a x c]
Question 53:
35 x (- 1) = – 35
Solution:
– 35 x (-1) = – 35 $left[ because left( frac{-35}{-1} right)=35 right]$
Question 54:
____ x (- 1) = 47
Solution:
(- 47) x (- 1) = 47 $left[because quad frac{47}{-1}=left( -47 right) right]$
Question 55:
88 x ____ = – 88
Solution:
88 x (- 1) = – 88 $left[ because frac{-88}{88}=left( -1 right) right]$
Question 56:
____ x (- 93) = 93
Solution:
(- 1) X (- 93) = 93 $left[ because frac{93}{-93}=left( -1 right) right]$
Question 57:
(- 40} x ___ =80
Solution:
(- 40) x (-2) = 80 $left[ because frac{93}{-40}=left( -2 right) right]$
Question 58:
____ x (-23) = – 920
Solution:
(40) x (-23) = – 920 $left[ because left( frac{-920}{-23} right)=40 right]$
Question 59:
When we divide a negative integer by a positive integer, we divide them as whole numbers and put a ____ sign before quotient.
Solution:
When we divide a negative integer by a positive integer or a positive integer by a negative integer, we divide them as whole numbers and put a negative sign before quotient.
Question 60:
When (-16) is divided by ____ the quotient is 4.
Solution:
When (-16) is divided by negative integer, i.e. -4 the quotient is 4 as both signs are cancelled out.
Question 61:
Division is the inverse operation of ____ .
Solution:
Division is the inverse operation of multiplication.
Question 62:
65 ÷ (- 13) =_____.
Solution:
65 ÷ (-13) = 65 x $frac{1}{left( -13 right)}$ [ division is inverse of multiplication]
= -5
Question 63:
(-100) ÷ (-10) =_____.
Solution:
(-100) ÷ (-10) = (-100) x $frac{1}{left( -10 right)}$ [ division is inverse of multiplication]
= (-10)
Question 64:
(-225) ÷ 5 = _____.
Solution:
(-225) ÷ 5 = -225 x $frac{1}{5}$ [ division is inverse of multiplication]
= -45
Question 65:
_____ ÷ (-1) = (- 83)
Solution:
83 ÷ (-1) = – 83 $left[ because frac{-83}{-1}=83 right]$
Question 66:
____ ÷ (-1) = 75
Solution:
75 ÷(-1) = 75 $left[ because frac{75}{-1}=(-75) right]$
Question 67:
51 ÷ ____ =(-51)
Solution:
51 ÷ (-1) = (-51) $left[ because frac{51}{-51}=(-1) right]$
Question 68:
113 ÷ ____ = (- 1)
Solution:
113 ÷ (-113) = (-1) $left[ because frac{113}{-1}=(-113) right]$
Question 69:
-95 ÷ (-1) = 95
Solution:
-95 ÷ (-1) =95 $left[ because frac{-95}{95}=(-1) right]$
Question 70:
(-69) ÷ 69 =_____.
Solution:
(-69) ÷ 69 = (- 1)
Question 71:
(-28) ÷ (-28) =
Solution:
(-28) ÷ (-28) = 1
True/False
In questions 72 to 108, state whether the statements are True or False.
Question 72:
5 – (-8) is same as 5 + 8.
Solution:
True
5 – (-8) = 5 + 8
Question 73:
(-9) + (-11) is greater than (-9) – (- 11).
Solution:
False
(-9)+ (-11) = – 9- 11 = -20
and (-9)-(-11) =-9+11 = 2
So, (-9) – (-11) is greater than (-9) + (-11).
Question 74:
Sum of two negative integers always gives a number smaller than both the integers.
Solution:
True
e.g. Taking two negative integers, i.e. (-5) and (-3).
(-5) + (-3)=-5-3=-8
= – 8 < – 5 and – 8 < – 3
Question 75:
Difference of two negative integers cannot be a positive integer.
Solution:
False
e.g. Taking two negative integers, i.e. -4 and -5.
=> – 4 – (-5) = – 4+5=1 [positive integer]
Question 76:
We can write a pair of integers, whose sum is not an integer.
Solution:
False
Because, sum of two integers, is always be an integer.
Question 77:
Integers are closed under subtraction.
Solution:
True
Because, if we subtract two integers we get another integer.
Question 78:
(- 23) + 47 is same as 47 + (- 23).
Solution:
True
Because, addition is commutative, i.e. a + b = b + a
=> (-23) + 47 = 47 + (-23)
Question 79:
When we change the order of integers their sum remains the same.
Solution:
True
Because, sum of two integers is commutative, i.e. a + b = b + a for two integers a and b.
Question 80:
When we change the order of integers, their difference remains the same.
Solution:
False
Subtraction of two integers is not commutative, i.e. a – b ≠b – a for two integers a and b.
Question 81:
Going 500 m towards East first and then 200 m back, is same as going 200 m towards West first and then going 500 m back.
Solution:
True
Case I Going 500 m towards East first, i.e. point A to B and then 200 m back, i.e. B to C.
As per the above figure shown, final position is C, i.e. 300 m in East.
Question 82:
(-5) x (33) = 5 x (- 33)
Solution:
True
∴ LHS = (-5) x 33 = (-165)
and RHS = 5 x (-33) = (-165)
Hence, LHS = RHS
Question 83:
(-19) x (-11) = 19 x 11
Solution:
True
Product of two negative integers is a positive integer, i.e. (-a) x (-b) = a x b where, a and b are positive integers.
=> LHS = (-19) x (-11) = 209
RHS = 19×11 = 209 Hence,
LHS = RHS
Question 84:
(-20) x (5 – 3) = (-20) x (-2)
Solution:
False
LHS = (-20) x (5 – 3) = (-20) x 2 = (-40)
RHS = (-20) x (-2) = 40
Hence, LHS = RHS
Question 85:
4 x (-5) = (-10) x (-2)
Solution:
False
LHS = 4 x (-5) = – 20
RHS = (-10) x (-2) = 20 Hence,
LHS = RHS
Question 86:
(-1) x (-2) x (-3) = 1 x 2x 3
Solution:
False
LHS = (-1) x (-2) x (-3) = (-6)
RHS = 1 x 2 x 3 = 6
Hence, LHS = RHS
Question 87:
(-3) x 3 = (-12) – (-3)
Solution:
True
LHS = (-3) x 3 = (- 9)
RHS = (-12) – (-3) = (-12) + 3 = (-9)
Hence, LHS = RHS
Question 88:
Product of two negative integers is a negative integer.
Solution:
False
Product of two negative integers is a positive integer, i.e. (-a) x (-b) = ab where, a and b are two positive integers.
Question 89:
Product of three negative integers is a negative integer.
Solution:
True
Product of three negative integers is a negative integer, i.e.
(-a) x (-0) x (-c) = (- abc)
where, a,b and c are three positive integers.
Question 90:
90 Product of a negative integer and a positive integer is a positive integer.
Solution:
False
Product of a negative integer and a positive integer is a negative integer, i.e. a x (-b) = -ab
where, a and b are two positive integers.
Question 91:
When we multiply two integers their product is always greater than both the integers.
Solution:
False
e.g. Let two integers are (-5) and 2.
So, (-5) x2 = -10
=> (— 10) < (- 5) and (-10) <2.
Question 92:
Integers under multiplication.
Solution:
True
If we multiply two integers, we get an integer.
Question 93:
(-237) x 0 is same as 0 x (-39).
Solution:
True
When we multiply a number with 0, we always get 0.
=> (-237) x 0 = 0
Question 94:
Multiplication is not commutative for integers.
Solution:
False
Multiplication is commutative for integers, i.e. axb=bxa for any two integers a and b.
Question 95:
(-1) is not a multiplicative identity of integers.
Solution:
True
1 is multiplicative identity for integers, i.e, a x 1 = 1 x a = a for any integer a.
Question 96:
99 x 101 can be written as (100 – 1) x (100 + 1).
Solution:
True
99 x 101 = 9999
and (100 -1) x (100 + 1) = 100 x (100 + 1) -1 x (100 + 1)
= 100 x 100+1 x 100-1 x 100 -1 x 1 [using distributive property]
= 10000 + 100 – 100 – 1 = 9999
Question 97:
If a, b and c are integers and b ≠ 0, then ax(b-c) = a x b – a x c
Solution:
True
Multiplication can be distributive over subtraction,
i.e. a x (b-c) = a x b-a x c
Question 98:
(a + b) x c=a x c + a x b
Solution:
False
Integers show distributive property of multiplication over addition, i.e. ax(b + c) = a x b + a x c, where a, b and c are integers.
Question 99:
a x b = b x a
Solution:
True
Multiplication is commutative for integers, i.e. a x b = b x a where, a and b are integers.
Question 100:
a ÷ b = b ÷ a
Solution:
False
Division is not commutative for integers, i.e. a÷b ≠ b÷a where, a and b are integers.
Question 101:
a-b = b-a
Solution:
False
Subtraction is not commutative for integers, i.e. a – b ≠ b – a where, a and b are integers.
Question 102:
a + (- b) = – (a + b)
Solution:
True
Division of a negative integer and a positive integer is always a negative integer
i.e $frac{a}{-b}=frac{-b}{a}=-left( frac{a}{b} right)$ where, a and b are integers.
Question 103:
a ÷ (-l) = – a
Solution:
True
a + (-1) = $frac{a}{left( -1 right)}$= – a [as division of a negative and positive integer is always negative]
Question 104:
Multiplication fact (-8) x (-10) = 80 is same as division fact 80 ÷ (-8) = (-10).
Solution:
Question 105:
Integers are closed under division.
Solution:
False
Because, when we divide two integers, we may or may not get an integer.
e.g. ² ⁄ ¹ = 2 (integer) and ² ⁄ ³ (not an integer).
Question 106:
[(-32) ÷ 8] ÷ 2 = – 32 ÷ [8 ÷ 2]
Solution:
False
Question 107:
The sum of an integer and its additive inverse is zero (0).
Solution:
True
Additive inverse is the number, that when added to a given number yields zero.
Question 108:
The successor of 0 x (-25) is 1 x (-25).
Solution:
False
We know that, successor means adding 1 to the given number.
Here, given number is 0 x (-25) = 0 [on multiplying by 0 to any number the result is zero]
Hence, the successor of 0 = 0 + 1 = 1 but 1 ≠ 1 x (-25).
Question 109:
Observe the following patterns and fill in the blanks to make the statements true:
Solution:
Question 110:
Science Application An atom consists of charged particles called electrons and protons. Each proton has a charge of +1 and each electron has a charge of -1. Remember number of electrons is equal to number of protons, while answering these questions:
(a) What is the charge on an atom?
(b) What will be the charge on an atom, if it loses an electron?
(c) hat will be the charge on an atom, if it gains an electron?
Solution:
(a) Let a be the number of electrons in an atom.
Number of protons in the atom, will also be equal to a. Since, an atom has equal number of protons and electrons.
Charge on one electron = (-1)
Total charge in a electrons = a x (-1) = – a
Charge on one proton = (+1)
Total charge in a protons = a x (+1) = + a
Hence, total charge on the atom = Charge of electrons + Charge of protons
= – a + a = 0
(b) If an atom loses an electron, it will have (a -1) electrons and a protons.
Charge in one electron = (-1)
Charge in (a-1) electrons = (a -1) x (-1) = – (a -1) = (1 – a)
Charge in one proton = (+1)
Charge in a protons = (+1) x a = (+a)
Hence, total charge on the atom = Charge of electrons + Charge of protons
= 1 – a + a= + 1
(c) If an atom gains an electron, it will have (a + 1) electrons and a protons
Charge in one electron = -1
Charge in (a +1) electrons = -1 x (a + 1) = – (a + 1)
Charge in one proton = (+1)
Charge in a protons = (+1) x a = (+ a)
Hence, total charge on the atom = Charge of electrons + Charge of protons
= a – (a + 1) = (-1)
Question 111:
An atom changes to a charged particle called ion, if it loses or gains electrons. The charge on an ion is the charge on electrons plus charge on protons. Now, write the missing information in the table given below:
Solution:
(a) For Hydroxide ion,
Proton charge + Electron charge = Ion charge
Electron charge = Ion charge – Proton charge
Electron charge = -1 – 9 = -10
Hence, the electron charge in a Hydroxide ion is -10.
(b) For Sodium ion,
Electron charge = Ion charge – Proton charge = + 1 -11 = -10
Hence, the electron charge in a Sodium ion is -10.
(c) For Aluminium ion,
Ion charge = Proton charge + Electron charge
Ion charge = 13-10 = 3
Hence, the ion charge in an Aluminium ion is 3,
(d) For Oxide ion,
Ion charge = Proton charge + Electron charge
Ion charge = 8 -10 = – 2
Hence, the ion charge in an Oxide ion is -2.
Question 112:
Social Studies Application remembering that 1AD came immediately after 1 BC, while solving following problems take 1BC as -1 and 1AD as + 1.
(a) The Greeco-Roman era, when Greece and Rome ruled Egypt, started in the year 330 BC and ended in the year 395 AD. How long did this era last?
(b) haskaracharya was born in the year 1114 AD and died in the year 1185 AD. What was his age when he died?
(c) Turks ruled Egypt in the year 1517 AD and Queen Nefertis ruled . Egypt about 2900 years, before the Turks ruled. In what year did she rule?
(d) Greek Mathematician Archimedes lived between 287 BC and 212 BC and Aristotle lived between 380 BC and 322 BC. Who lived during an earlier period?
Solution:
Question 113:
The table shows the lowest recorded temperatures for each continent. Write the continents in order from the lowest recorded temperature to the highest recorded temperature.
Solution:
Lowest to heights (ascending order) in a negative number, the number that has greater value of actually smaller and vice-versa.
So, accordingly, we arrange them in ascending order
Question 114:
Write a pair of integers whose product is -12 and there lies seven integers between them (excluding the given integers).
Solution:
For a pair of integers, whose product is -12 and there lies seven integers between them, Two solutions are possible, i.e. (-6and 2) and (-2 and 6).
Question 115:
From given integers in Column I, match an integer of Column II, so that their product lies between -19 and -6.
Solution:
-5 x 3= (-15) which lies between -19 and -6.
6 x (-2)= (-12) which lies between -19 and -6.
– 7 x 1 = (- 7) which lies between -19 and -6.
8 x (-1)= (- 8) which lies between -19and -6.
Question 116:
Write a pair of integers, whose product is -36 and whose difference is 15.
Solution:
For a pair of integers, whose product is -36 and difference is 15, one possible solution is (-3,12).
So, first integer = – 3 and second integer =12
Their product = (-3) x 12 = – (3 x 12) = – 36
and the difference between these two integers is 12 – (- 3) = 15.
Question 117:
Match the following:
Solution:
Question 118:
You have Rs. 500 in your saving account at the beginning of the month.
The record below, shows all of your transactions during the month. How much money is in your account after these transactions?
How much money is in your account after these transactions?
Solution:
According to the question,
Already available amount = Rs. 500
On 4/9 with cheque number 384102 withdraw Rs. 120.
Also, with cheque number 275146 on 12/9 deposited amount was Rs. 200.
In the same way, on 22/9 with cheque number 384103, Rs. 240 paid to LIC of India, also.
On 29/9 with cheque number 801351, deposited amount was Rs. 150.
Thus, net amount available in bank account will be
= Already saved amount + Deposited amount – Debited amount (paid amount)
= 500 + 200+ 150-120-240
= 850 + (- 360) = Rs. 490
Question 119:
(a) Write a positive integer and a negative integer whose sum is a negative integer.
Solution:
A number of solutions can be possible.
e.g. Let first integer = 4and second integer = (-6)
Sum = 4 + (- 6) = – 2 [negative integer]
(b) Write a positive integer and a negative integer whose sum is a positive integer.
Solution:
A number of solutions can be possible.
e.g. Let first integer =8and second integer = – 2
Sum = 8 + (- 2) = 6 [positive integer]
(c) Write a positive integer and a negative integer whose difference is a negative integer.
Solution:
A number of solutions can be possible.
e.g. Let first integer = (- 7) and second integer = 2
Difference = (- 7 – 2) = (- 9) [negative integer]
(d) Write a positive integer and a negative integer whose difference is a positive integer.
Solution:
A number of solutions can be possible.
e.g. Let first integer = 4 and
second integer = (- 3) [positive integer]
(e) Write two integers, which are smaller than -5 but their difference is -5.
Solution:
For two integers, which are smaller than -5 but their difference is -5.
Let first integer = -11 and second integer = (- 6) [ -11 < (-5) and -6 < (-5)]
Difference = -11 – (- 6) = -11 + 6 = (- 5)
(f) Write two integers which are greater than -10 but their sum is smaller than -10.
Solution:
For two integers which are greater than -10 but their sum is smaller than -10.
Let first integer = – 4 and second integer = – 7 [ – 4 > (-10) and -7 > -10]
Sum = – 4 + (-7)= -11 < -10
(g) Write two integers which are greater than – 4 but their difference is smaller than -4.
Solution:
For two integers which are greater than -4 but their difference is smaller than -4.
Let first integer = (-1) and second integer = 4 [ -1 > – 4 and 4 > – 4]
Difference = -1-4 = -5<(-4)
(h) Write two integers which are smaller than – 6 but their difference is greater than -6.
Solution:
For two integers which are smaller than -6 but their difference is greater than -6.
e.g. Let first integer = (- 8) and second integer = (- 9) [ — 8 < — 6 and – 9 < – 6]
Difference = – 8 – (- 9) = -8 + 9 = 1 > (- 6)
(i) Write two negative integers whose difference is 7.
Solution:
A number of solutions can be possible.
e.g. Let first integer = (- 3) and second integer = (-10)
Difference = – 3 – (-10) = 7
( j) Write two integers, such that one is smaller than – 11 and other is greater than -11 but their difference is -11.
Solution:
For two integers, such that one is smaller than – 11 and other is greater than -11.
Let first integer = – 20and second integer = – 9 [-20< -11 and —9> — 11]
Difference = – 20 – (- 9) = (-11)
(k) Write two integers whose product is smaller than both the integers.
Solution:
A number of solutions can be possible.
e.g. Let first integer = – 3 and second integer = 5
Product = -3×5 = -15 [—15 < – 3 and — 15 < 5]
(l) Write two integers, whose product is greater than both the integers.
Solution:
A number of solutions can be possible.
e.g. Let first integer = 4 and second integer = 6
Product = 6 x 4 = 24 [-24 > 6 and 24 > 4]
Question 120:
What’s the error? Ramu evaluated the expression – 7 – (-3) and came up with the answer – 10 . What did Ramu do wrong?
Solution:
Ramu went wrong in solving – (- 3) and took it as – 3 only.
Correct answer = – 7 – ( – 3)= – 7 + 3= – 4
Question 121:
What’s the error? Reeta evaluated -4 + d for d = — 6 and gave an answer of 2. What might Reeta have done wrong?
Solution:
Reeta went wrong in solving + (- 6) and took it as + 6.
Correct answer = -4 + d= -4+(-6) = -4-6 = -10
Question 122:
The table given below, shows the elevations relative to sea level of four locations. Taking sea level as zero (0), answer the following questions.
(a) Which location is closest to sea level?
(b) Which location is farthest from sea level?
(c) Arrange the locations from the least to the greatest elevation.
Solution:
(a) From the adjacent figure, we can clearly see that C is closest to sea level.
(b) D is farthest from sea level.
(c) Locations from the least to the greatest elevation will be in the order A,Sand D.
Question 123:
You are at an elevation 380 m above sea level as you start a motor ride. During the ride, your elevation changes by the following metres 540 m, -268 m, 116 m, -152 m, 490 m, -844 m, 94 m. What is your elevation relative to the sea level at the end of the ride?
Solution:
As per the given information, initial position of motor was 380 m.
During the ride, change in elevation was 540 m, -268 m, 116 m, -152 m, 490 m, -844 m and 94 m.
Net change in position = 540 + (- 268) + (116) + (-152) + (490) + (- 844) + 94 = – 24 m Initial position was 380 m. So, at the end of the ride the position would be = 380+ (-24) = 356 m
Question 124:
Evaluate the following, using distributive property.
(i) -39 x 99 (ii) (-85) x 43 +43 x (-15)
(iii) 53 x (-9) – (-109) x 53 (iv) 68 x (-17) + (-68) x 3
Solution:
Question 125:
If ‘*’ is an operation have, such that for integers a and b. We have a*b=a x b+(a x a+b x b), then find
(i) (-3)*(-5) (ii) (-6)*2
Solution:
(i) We have, a* b = a x b +(a x a+b x b)
Now, put a = (-3) and b = (-5)
(-3)* (-5) = (- 3)x(- 5)+ [(- 3)x(- 3)+ (- 5)x(- 5)]
= 15+(9+25)= 15+34= 49
(ii) Now, put a = – 6 and b = 2
(-6)*2 = (-6×2)+{(-6)x(-6)+2×2}
= -6×2 + (36+4)= -12 + 40= 28
Question 126:
Solution:
Question 127:
Below u, v, w and x represent different integers, where u = (-4) and x ≠ 1. By using following equations, find each of the values
u x v=u, x x w =w and u+x = w
(a) v (b) w (c) x
Explain your reason, using the properties of integers
Solution:
Question 128:
Height of a place A is 1800 m above sea level. Another place B is 700 m below sea level. What is the difference between the levels of these two places?
Solution:
As per the given information, we can draw the diagram,
Let O be the point of level of sea.
Difference between these two points, A and B
= Height between sea level and point A+Height between point B and sea level =AO + OB=1800+ 700 = 2500 m
Question 129:
Solution:
Question 130:
Sana and Fatima participated in an apple race. The race was conducted in 6 parts. In the first part, Sana won by 10 seconds. In the second part, she lost by 1 min, then won by 20 seconds in the third part and lost by 25 seconds in the fourth part, she lost by 37 seconds in the fifth part and won by 12 seconds in the last part. Who won the race finally?
Solution:
Let difference in time denoted by positive, when Sana wins the race and negative, when Sana loses the race.
Total difference in time taken by Sana in all the six parts
=10-60+20-25-37+12 =-80s [ 1 min=60s]
Hence, Fatima won the race by 80 s.
Question 131:
131 A green grocer had a profit of Rs. 47 on Monday, a loss of Rs. 12 on Tuesday and loss of Rs. 8 on Wednesday. Find his net profit or loss in 3 days.
Solution:
As per the given information,
Profit on Monday = Rs. 47 and loss on Tuesday = Rs. 12
and loss on Wednesday =Rs. 8
Net profit = Total profit – Total loss
Now,total profit = Rs. 47 and total loss =12 + 8 = t 20
Net profit =47-20 = Rs. 27
Question 132:
In a test, +3 marks are given for every correct answer and -1 mark are given for every incorrect answer. Sona attempted all the questions and scored +20 marks, though she got 10 correct answers.
(i) How many incorrect answers has she attempted?
(ii) How many questions were given in the test?
Solution:
Let x be the correct answers and y be the incorrect answers, given by Sona.
It is given that, if she gives 10 correct answers and her score is 20. Since, for every correct answer, +3 is given and for every incorrect answer, -1 is given.
Hence,
Question 133:
In a true-false test containing 50 questions, a student is to be awarded 2 marks for every correct answer and -2 for every incorrect answer and 0 for not supplying any answer. If Yash scored 94 marks in a test, what are the possibilities of his marking correct or wrong answer?
Solution:
Since, Yash scored 94 marks.
Hence, there are two possibilities:
(i) 47 correct answers and 3 unattempted.
Question 134:
A multistory building has 25 floors above the ground level each of height 5 m. It also has 3 floors in the basement each of height 5m. A lift in building moves at a rate of lm/s. If a man starts from 50m above the ground, how long will it take him to reach at 2nd floor of basement?
Solution:
Man covers the distance above the ground = 50 m
and man covers the distance below the ground = 2 x 5 = 10 m
Question 135:
Taking today as zero on the number line, if the day before yesterday is 17 January, what is the date 3 days after tomorrow?
Solution:
If we take today as zero, then two days before today is 17 January. Hence, 3 days after tomorrow will be at 4th place from zero on the number line.
So, required date will be (17+6) January =23 January
Question 136:
The highest point measured above sea level is the summit of Mt. Everest, which is 8848 m above sea level and the lowest point is challenger deep at the bottom of Mariana Trench which is 10911 m below sea level. What is the vertical distance between these two points?
Solution:
As per the given information, we can draw the diagram,
Let A be the point above the sea level and B be the point below the sea level.
Vertical distance between points A and B = Distance between point A and sea level
+ Distance between point B and sea level
= AO+OB = 8848+10911 = 19759 m
All Chapter NCERT Exemplar Problems Solutions For Class 7 maths
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• #### 1. Find the compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually.
1. Rs. 610
2. Rs. 612
3. Rs. 614
4. Rs. 616
Explanation:
\begin{aligned}
Amount = [7500 \times (1+ \frac{4}{100})^2] \\
= (7500 \times \frac{26}{25} \times \frac{26}{25}) \\
= 8112 \\
\end{aligned}
So compound interest = (8112 - 7500) = 612
• #### 2. Albert invested amount of 8000 in a fixed deposit for 2 years at compound interest rate of 5 % per annum. How much Albert will get on the maturity of the fixed deposit.
1. Rs. 8510
2. Rs. 8620
3. Rs. 8730
4. Rs. 8820
Explanation:
\begin{aligned}
=> (8000 \times(1+\frac{5}{100})^2) \\
=> 8000 \times \frac{21}{20}\times \frac{21}{20} \\
=> 8820
\end{aligned}
• #### 3. What will be the compound interest on Rs. 25000 after 3 years at the rate of 12 % per annum
1. Rs 10123.20
2. Rs 10123.30
3. Rs 10123.40
4. Rs 10123.50
Explanation:
\begin{aligned}
(25000 \times (1 + \frac{12}{100})^3) \\
=> 25000\times\frac{28}{25}\times\frac{28}{25}\times\frac{28}{25} \\
=> 35123.20 \\
\end{aligned}
So Compound interest will be 35123.20 - 25000
= Rs 10123.20
• #### 4. A man saves Rs 200 at the end of each year and lends the money at 5% compound interest. How much will it become at the end of 3 years.
1. Rs 662
2. Rs 662.01
3. Rs 662.02
4. Rs 662.03
Explanation:
\begin{aligned}
[200(1+\frac{5}{100})^3 + 200(1+\frac{5}{100})^2+ \\ 200(1+\frac{5}{100})]
= [200(\frac{21}{20} \times \frac{21}{20} \times \frac{21}{20})\\
+ 200(\frac{21}{20}\times\frac{21}{20})+200(\frac{21}{20})] \\
= 662.02
\end{aligned}
• #### 5. Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually
1. Rs 312
2. Rs 412
3. Rs 512
4. Rs 612
Explanation:
\begin{aligned}
Amount = P(1+\frac{R}{100})^n \\
\text{C.I. = Amount - P}
\end{aligned}
• #### 6. Find the compound interest on Rs.16,000 at 20% per annum for 9 months, compounded quarterly
1. Rs 2520
2. Rs 2521
3. Rs 2522
4. Rs 2523
Explanation:
Please remember, when we have to calculate C.I. quarterly then we apply following formula if n is the number of years
\begin{aligned}
Amount = P(1+\frac{\frac{R}{4}}{100})^{4n}
\end{aligned}
Principal = Rs.16,000;
Time=9 months = 3 quarters;
Rate = 20%, it will be 20/4 = 5%
So lets solve this question now,
\begin{aligned}
Amount = 16000(1+\frac{5}{100})^3 \\
= 18522\\
C.I = 18522 - 16000 = 2522
\end{aligned}
• #### 7. The present worth of Rs.169 due in 2 years at 4% per annum compound interest is
1. Rs 155.25
2. Rs 156.25
3. Rs 157.25
4. Rs 158.25
Explanation:
In this type of question we apply formula
\begin{aligned}
Amount = \frac{P}{(1+\frac{R}{100})^n} \\
Amount = \frac{169}{(1+\frac{4}{100})^2} \\
Amount = \frac{169 * 25 * 25}{26*26} \\
Amount = 156.25
\end{aligned}
631.525
• #### naresh aswani 5 years ago
Why use this formula in quetion number 7.
As this question is as similar as other question like we have principal, rate ,interest.
Plzz help me ..
• #### Nitesh Bibra 6 years ago
can any1 xpalin 8th question how does it solve and get 6.....
• #### AKsingh 6 years ago
can any one explain 15
• #### Sheenu Mehra 7 years ago
in 4th question, 200 rupee deposite every year, so amount is calculated first 200 for 3 years, then 200 for 2 years which is deposite in 2nd year, then 200 for 1 year which is deposite in 3rd year, at last all amount is added
• #### ramya 7 years ago
can ay one explain the process for the 4th problem
#### Sahana 5 years ago replied
Think n do use brains
• #### hemanth 8 years ago
Helpful. Very good questions. Easy to understand for beginners.
|
+0
# algebra
0
117
1
Complete the equation of the line through (-6,-5) (-4,-4)
Guest Jul 28, 2017
Sort:
#1
+1493
+1
To figure out the equation of a line that passes through the given points (-6, -5) and (-4, -4), you must first know the standard form of a line. It is the following:
$$y=mx+b$$
Let m = slope of the line
Let b = the y-intercept (the point where the line touches the y-axis)
The first step is to figure out the slope of the line. How do we do that, you may ask? All you do is remember the slope formula.
$$m=\frac{y_2-y_1}{x_2-x_1}$$
We already have enough information to calculate the slope, m. We do this by substituting the given points into the formula.
$$m=\frac{-5-(-4)}{-6-(-4)}$$ Simplify the fraction into simplest terms by evaluating the numerator and denominator separately. $$m=\frac{-5+4}{-6+4}$$ Of course, subtracting a negative is the same as adding a positive. $$m=\frac{-1}{-2}$$ The negatives in the numerator and denominator cancel each other out. $$m=\frac{1}{2}$$
Great! We know the slope! Now, the only variable to figure out next is b, the y-intercept. We can do this by plugging in points of points on the line in the equation.
$$y=\frac{1}{2}x+b$$
In other words, to solve for b, you must plug in a point we know is one the line (either (-6,-5) or (-4,-4)) for x and y. I'll choose (-4,-4):
$$y=\frac{1}{2}x+b$$ Plug in the coordinate (-4,-4) in its appropriate spots and then solve for b. $$-4=\frac{1}{2}*-4+b$$ Now, solve for b. $$-4=-2+b$$ Add 2 on both sides. $$-2=b$$
Now that we have solved for both m and b, the equation that passes through the points (-6,-5) and (-4,-4) is $$y=\frac{1}{2}x-2$$.
Do you need your answer in point-slope form? No problem! Remember the point-slope form
$$y-y_1=m(x-x_1)$$
Of course, m is the slope again. We have already calculated that. Let's substitute that in.
$$y-y_1=\frac{1}{2}(x-x_1)$$
$$y_1\hspace{1mm}\text{and}\hspace{1mm}x_1$$ represent a point on the line. You can either substitute the first or the second set of coordinates. It doesn't matter. However, in the end, your answer should be one of these:
$$y+5=\frac{1}{2}(x+6)$$
$$y+4=\frac{1}{2}(x+4)$$
TheXSquaredFactor Jul 28, 2017
edited by TheXSquaredFactor Jul 28, 2017
edited by TheXSquaredFactor Jul 28, 2017
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## More Percent Change Questions
Start Quiz
"Prices are down 25%."
"Wages have increased by 13%."
You may hear phrases like this all the time, but could you calculate the new figure? To succeed on the GED Math test, you'll need to know how.
In some cases, you will be given the original number and the new number, and asked to calculate the percent change. To do this, subtract the old number from the new number, and then divide the difference by the old number. Multiply by 100 (or move the decimal point two places to the right) to convert the decimal into a percentage.
Let's look at an example. The price of a shirt has increased from \$20 to \$24. What is the percent change? First, we subtract the old price from the new: 24 − 20 = 4. We divide this difference by the original price, 4 ÷ 20 = 0.2, and then multiply by 100 to reach the solution: 20%.
But what if the number is decreasing? The process remains the same. Returning to our shirt, imagine the price fell by \$4 instead. In this case, the percent change would be calculated like this:
• 16 − 20 = -4 (Don't worry about the negative in this problem, it just means that the change is a decrease.)
• -4 ÷ 20 = -0.2, or 20%
Notice that the percent change is the same no matter whether the original amount increases or decreases, if it does so by the same amount.
Another way you might see this concept on the GED is when the percent change and either the original or new value are given, and you must calculate the missing value. Consider this example: a book that previously cost \$14 has been discounted by 25%. How much does it cost now? To solve, convert 25% into a decimal and multiply it by the original cost: 0.25 × 14 = 3.5. A book that was \$14 has been discounted \$3.50. So, to find the new price, just subtract the amount of the discount from the original price: \$14 − \$3.50 = \$10.50. Simple, right?
Here are a few practice questions to help you get the hang of it!
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# 7.4 Area and Arc Length in Polar Coordinates
### Learning Objectives
• Derive the formula for area of a region in polar coordinates.
• Determine the arc length of a polar curve.
In the rectangular coordinate system, the definite integral provides a way to calculate the area under a curve. In particular, if we have a function defined from to where on this interval, the area between the curve and the x-axis is given by We can also find the arc length of this curve using the formula In this section, we study formulas for area and arc length in the polar coordinate system.
### Areas of Regions Bounded by Polar Curves
Consider a polar curve defined by the function where We want to derive a formula for the area of the region bounded by the curve and between the radial lines and , see Figure 1 below. When defining areas in rectangular coordinates, we approximated the regions with the union of rectangles, and here we are going to use sectors of a circle. Our first step is to partition the interval into n equal-width subintervals. The width of each subinterval is and the ith partition point is given by the formula Each partition point defines a line with slope passing through the pole as shown in the following graph.
The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to obtain a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.
Recall that the area of a circle is A fraction of a circle can be measured by the central angle The full angle is , and so the fraction of the circle is given by The area of the sector is this fraction multiplied by the total area:
Since the radius of a typical sector in Figure 1 above is given by the area of the ith sector can be found as
Summing the areas of sectors for , we obtain a Riemann sum that approximates the polar area:
We take the limit as to get the exact area:
This gives the following theorem.
### Area of a Region Bounded by a Polar Curve
Suppose is continuous and nonnegative on the interval with The area of the region bounded by the graph of between the radial lines and is
### Finding an Area of a Polar Region
Find the area of one petal of the rose defined by the equation
#### Solution
The graph of is shown below.
Each petal starts and ends at the pole, so to find the endpoints of the interval that corresponds to one petal, we need to solve the equation . We have and the solutions are since the zeros of the sine function are of the form , where is integer. It follows that the petal in the first quadrant corresponds to . To find the area inside this petal, use (*) from the above theorem with and
To evaluate this integral, use the formula with
Find the area inside the cardioid defined by the equation
#### Hint
Use (*). Be sure to determine the correct limits of integration before evaluating.
In the previous example, we were looking for the area inside one curve. We can also use (*) to find the area between two polar curves. For this, we need to find the points of intersection of the curves, determine which function defines the outer curve and which defines the inner curve, and then subtract the corresponding polar areas. This process is illustrated in the example below.
### Finding the Area between Two Polar Curves
Find the area outside the cardioid and inside the circle
#### Solution
First draw a graph containing both curves as shown below.
To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for
This gives the solutions and in the interval , which are the limits of integration since from the picture we see that on . The circle is the red graph, which is the outer function, and the cardioid is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between and then subtract the area inside the cardioid between and
## Analysis
Notice that equating the formulas and solving for only yielded two solutions: and However, in the graph there are three intersection points. The third intersection point is the pole. The reason why this point did not show up as a solution is because the pole is on both graphs but corresponds to different values of Indeed, for the cardioid we get
so the values for that solve this equation are where k is an integer. For the circle we get
The solutions to this equation are of the form for any integer value of n. These two solution sets have no points in common. Regardless of this fact, the curves intersect at the pole. This case must always be taken into consideration, although in this partcular example it did not affect the calculations.
Find the area inside the circle and outside the circle
#### Hint
Use (*) and take advantage of symmetry.
### Arc Length for Polar Curves
Here we derive a formula for the arc length of a curve defined in polar coordinates.
In rectangular coordinates, the arc length of a parameterized curve for is given by
In polar coordinates we define the curve by the equation where In order to adapt the arc length formula for a polar curve, we use the equations
Differentiating, we obtain
Applying the known arc length formula, we get
This gives us the following theorem.
### Arc Length of a Curve Defined by a Polar Function
Let be a function whose derivative is continuous on an interval The length of the polar curve from to is
### Finding the Arc Length of a Polar Curve
Find the arc length of the cardioid
#### Solution
As goes from to the cardioid is traced out exactly once. Therefore we can use those as the limits of integration in the formula from the above theorem to obtain
Next, the identity implies that Substituting gives so the integral becomes
When , we have that and since cosine is non-negative on this interval, the absolute value can be dropped and we obtain
Find the total arc length of
#### Hint
To determine the correct limits, make a table of values.
### Key Concepts
• The area of the region bounded by the polar curve and between the radial lines and is given by the integral
• To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtract the corresponding areas.
• The arc length of a polar curve defined by the equation with is given by the integral
### Key Equations
• Area of a region bounded by a polar curve
• Arc length of a polar curve
### Exercises
For the following exercises, determine a definite integral that represents the area of the given region.
1. Region enclosed by
2. Region in the first quadrant within the cardioid
3. Region enclosed by one petal of .
4. Region enclosed by one petal of .
5. Region below the polar axis and enclosed by
6. Region in the first quadrant enclosed by
7. Region enclosed by the inner loop of
(Hint: inner loop corresponds to .)
8. Region enclosed by the inner loop of
9. Region enclosed by and outside the inner loop.
10. Region common to
(Hint: divide the region into several parts so that each of the parts is determined by one of the curves.)
11. Region common to
(Hint: divide the region into several parts so that each of the parts is determined by one of the curves.)
12. Region common to
(Hint: divide the region into several parts so that each of the parts is determined by one of the curves.)
For the following exercises, find the area of the described region.
13. Region enclosed by
14. Region above the polar axis and enclosed by
15. Region below the polar axis and enclosed by
16. Region enclosed by one petal of
17. Region enclosed by one petal of
18. Region enclosed by
19. Region enclosed by the inner loop of
20. Region enclosed by and outside the inner loop.
21. Common interior of
22. Common interior of
23. Common interior of
For the following exercises, find a definite integral that represents the arc length.
24.
25. ,
26.
27.
For the following exercises, find the length of the curve over the given interval.
28.
29.
30.
31.
32
32.
For the following exercises, use the integration capabilities of a calculator to approximate the length of the curve.
33. [T]
6.238
34. [T]
35. [T]
2
36. [T]
37. [T]
4.39
For the following exercises, rewrite the equation of the given polar curve in rectangular coordinates and use the familiar formula from geometry to find the area enclosed by the curve. Confirm your answer by using the definite integral.
38.
39.
For the following exercises, use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.
40.
|
# Factors of 19683 in pair
Factors of 19683 in pair are (1, 19683) , (3, 6561) , (9, 2187) , (27, 729) and (81, 243)
#### How to find factors of a number in pair
1. Steps to find factors of 19683 in pair 2. What is factors of a number in pair? 3. What are Factors? 4. Frequently Asked Questions 5. Examples of factors in pair
### Example: Find factors of 19683 in pair
Factor Pair Pair Factorization
1 and 19683 1 x 19683 = 19683
3 and 6561 3 x 6561 = 19683
9 and 2187 9 x 2187 = 19683
27 and 729 27 x 729 = 19683
81 and 243 81 x 243 = 19683
Since the product of two negative numbers gives a positive number, the product of the negative values of both the numbers in a pair factor will also give 19683. They are called negative pair factors.
Hence, the negative pairs of 19683 would be ( -1 , -19683 ) .
#### What does factor pairs in mathematics mean?
In mathematics, factor pair of a number are all those possible combination which when multiplied together give the original number in return. Every natural number is a product of atleast one factor pair. Eg- Factors of 19683 are 1 , 3 , 9 , 27 , 81 , 243 , 729 , 2187 , 6561 , 19683. So, factors of 19683 in pair are (1,19683), (3,6561), (9,2187), (27,729), (81,243).
#### What is the definition of factors?
In mathematics, factors are number, algebraic expressions which when multiplied together produce desired product. A factor of a number can be positive or negative.
#### Properties of Factors
• Every number is a factor of zero (0), since 19683 x 0 = 0.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Every factor of a number is an exact divisor of that number, example 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683 are exact divisors of 19683.
• Factors of 19683 are 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683. Each factor divides 19683 without leaving a remainder.
#### Steps to find Factors of 19683
• Step 1. Find all the numbers that would divide 19683 without leaving any remainder. Starting with the number 1 upto 9841 (half of 19683). The number 1 and the number itself are always factors of the given number.
19683 ÷ 1 : Remainder = 0
19683 ÷ 3 : Remainder = 0
19683 ÷ 9 : Remainder = 0
19683 ÷ 27 : Remainder = 0
19683 ÷ 81 : Remainder = 0
19683 ÷ 243 : Remainder = 0
19683 ÷ 729 : Remainder = 0
19683 ÷ 2187 : Remainder = 0
19683 ÷ 6561 : Remainder = 0
19683 ÷ 19683 : Remainder = 0
Hence, Factors of 19683 are 1, 3, 9, 27, 81, 243, 729, 2187, 6561, and 19683
• Is 19683 a composite number?
Yes 19683 is a composite number.
• Is 19683 a perfect square?
No 19683 is not a perfect square.
• Write all odd factors of 19683?
The factors of 19683 are 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683.
Odd factors of 19683 are 1 , 3 , 9 , 27 , 81 , 243 , 729 , 2187 , 6561 , 19683.
• What is the mean of all prime factors of 19683?
Factors of 19683 are 1 , 3 , 9 , 27 , 81 , 243 , 729 , 2187 , 6561 , 19683. Prime factors of 19683 are 3 , 3 , 3 , 3 , 3 , 3 , 3 , 3 , 3. Therefore mean of prime factors of 19683 is (3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3) / 9 = 3.00.
• What do you mean by proper divisors?
A number x is said to be the proper divisor of y if it divides y completely, given that x is smaller than y.
#### Examples of Factors
Can you help Rubel in arranging 19683 blocks in order to form a rectangle in all possible ways? For arranging 19683 blocks in order to form a rectangle in all possible ways we need to find factors of 19683 in pair.
Factors of 19683 are 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683. So, factors of 19683 in pair are (1,19683), (3,6561), (9,2187), (27,729), (81,243).
Ariel has been asked to write all factor pairs of 19683 but she is finding it difficult. Can you help her find out?
Factors of 19683 are 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683. So, factors of 19683 in pair are (1,19683), (3,6561), (9,2187), (27,729), (81,243).
How many total number of factors of 19683 in pair are possible?
Factors of 19683 are 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683. Hence, the factors of 19683 in pair are (1,19683), (3,6561), (9,2187), (27,729), (81,243). Therefore, in total 5 pairs of factors are possible.
Sammy wants to write all the negative factors of 19683 in pair, but don't know how to start. Help Sammy in writing all the factor pairs.
Negative factors of 19683 are -1, -3, -9, -27, -81, -243, -729, -2187, -6561, -19683. Hence, factors of 19683 in pair are (-1,-19683), (-3,-6561), (-9,-2187), (-27,-729), (-81,-243).
Help Deep in writing the positive factors of 19683 in pair and negative factor of 19683 in pair.
Factors of 19683 are 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683. Positive factors of 19683 in pair are (1,19683), (3,6561), (9,2187), (27,729), (81,243). Negative factors of 19683 in pair are (-1,-19683), (-3,-6561), (-9,-2187), (-27,-729), (-81,-243).
Find the product of all factors of 19683.
Factors of 19683 are 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683. So the product of all factors of 19683 would be 1 x 3 x 9 x 27 x 81 x 243 x 729 x 2187 x 6561 x 19683 = 2.954312706550834e+21.
Find the product of all prime factors of 19683.
Factors of 19683 are 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683. Prime factors are 3, 3, 3, 3, 3, 3, 3, 3, 3. So, the product of all prime factors of 19683 would be 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 = 19683.
A student has been assigned the following tasks by the teacher:
- Finding out all positive factors of 19683.
- Writing all prime factors of 19683.
- Writing all the possible factors of 19683 in pair.
Help him in writing all these.
Positive factors of 19683 are 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683.
Prime factors of 19683 are 3, 3, 3, 3, 3, 3, 3, 3, 3.
Factors of 19683 in pair are (1,19683), (3,6561), (9,2187), (27,729), (81,243).
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# Converting Rational Numbers to Decimals Using Long Division
Related Topics:
Lesson Plans and Worksheets for Grade 7
Lesson Plans and Worksheets for all Grades
Videos, examples, lessons, and solutions to help Grade 7 students learn how to convert rational numbers to decimals using long division.
### New York State Common Core Math Grade 7, Module 2, Lesson 14
Lesson 14 Student Outcomes
• Students understand that every rational number can be converted to a decimal.
• Students represent fractions as decimal numbers that either terminate in zeros or repeat, and students represent repeating decimals using, a bar over the shortest sequence of repeating digits.
• Students interpret word problems and convert between fraction and decimal forms of rational numbers.
Lesson 14 Summary
The real world requires that we represent rational numbers in different ways depending on the context of a situation. All rational numbers can be represented as either terminating decimals or repeating decimals using the long division algorithm. We represent repeating decimals by placing a bar over the shortest sequence of repeating digits.
Example 1: Can All Rational Numbers Be Written as Decimals?
a. Using the division button on your calculator, explore various quotients of integers 1 through 11. Record your fraction representations and their corresponding decimal representations in the space below.
b. What two types of decimals do you see?
Example 2: Decimal Representations of Rational Numbers
In the chart below, organize the fractions and their corresponding decimal representation listed in Example 1 according to their type of decimal.
Example 3: Converting Rational Numbers to Decimals Using Long-Division
Use the long division algorithm to find the decimal value of - 3/4
Exercise 1
Convert each rational number to its decimal form using long division.
a. - 7/8
b. 3/16
Example 4: Converting Rational Numbers to Decimals Using Long-Division
Use long division to find the decimal representation of 1/3.
Exercise 2
Calculate the decimal values of the fraction below using long division. Express your answers using bars over the shortest sequence of repeating digits.
a. - 4/9
b. - 1/11
c. 1/7
d. - 5/6
Example 5: Fractions Represent Terminating or Repeating Decimals
How do we determine whether the decimal representation of a quotient of two integers, with the divisor not equal to zero, will terminate or repeat?
Example 6: Using Rational Number Conversions in Problem Solving
a. Eric and four of his friends are taking a trip across the New York State Thruway. They decide to split the cost of tolls equally. If the total cost of tolls is \$8, how much will each person have to pay?
b. Just before leaving on the trip, two of Eric’s friends have a family emergency and cannot go. What is each person’s share of the \$8 tolls now?
Lesson 14 Problem Set
2. Chandler tells Aubrey that the decimal value of - 1/17 is not a repeating decimal. Should Aubrey believe him?
Explain.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Perimeter of Triangle
Perimeter of Triangle: The perimeter of any two-dimensional figure is defined as the distance around the figure. We can calculate the perimeter of any closed shape just adding up the length of each of the sides. In this article, you will first learn about what is the perimeter, how to find the perimeter of different types of triangles when all side lengths are known. Furthermore, the solved examples will help you to get more views on the topic.
### What is the Perimeter of a Triangle?
The sum of the lengths of the sides is the perimeter of any polygon. In the case of a triangle,
Perimeter = Sum of the three sides
Always include units in the final answer. If the sides of the triangle are measured in centimetres, then the final answer should also be in centimetres.
### Perimeter of Triangle Formula
The formula for the perimeter of a closed shape figure is usually equal to the length of the outer line of the figure. Therefore, in the case of a triangle, the perimeter will be the sum of all the three sides. If a triangle has three sides a, b and c, then,
Perimeter, P = a + b +c
## Perimeter of an Isosceles, Equilateral and Scalene Triangle
Below table helps us to understand how to find the perimeter of different triangles- Equilateral triangle, Isosceles triangle and Scalene triangle.
Where a, b, c and l are the side lengths and P = Perimeter.
This formula implies to find the perimeter of a triangle, add the lengths of all of its 3 sides together. If A, B and C are the side measures, and X is perimeter then
### Perimeter of Right Triangle
A right triangle has a base(b), hypotenuse(h) and perpendicular(p) as its sides, By the Pythagoras theorem, we know,
h2 = b2 + p2
Therefore, the Perimeter of a right angle triangle= b + p + h
## Examples
Let us consider some of the examples on the perimeter of a triangle:
Example 1: Find the perimeter of a polygon whose sides are 5 cm, 4 cm and 2 cm.
Solution: Let,
a = 5 cm
b = 4 cm
c = 2 cm
Perimeter = Sum of all sides = a + b + c = 5 + 4 + 2 = 11
Therefore, the answer is 11 cm.
Example 2: Find the perimeter of a triangle whose each side is 10 cm.
Solution: Since all three sides are equal in length, the triangle is an equilateral triangle.
i.e. a = b = c = 10 cm
Perimeter = a + b + c
= 10 + 10 + 10
= 30
Perimeter = 30 cm.
Example 3: What is the missing side length of a triangle whose perimeter is 40 cm and two sides are 10 cm each?
Solution: Given,
Perimeter = 40 cm
Length of two sides is the same i.e. 10 cm.
Thus, the triangle is an isosceles triangle.
Using formula: P = 2l + b
40 = 2 * 10 + b
40 = 20 + b
or b = 20
Missing side length is 20 cm.
### What does the Perimeter of a Triangle Mean?
The perimeter of a triangle is the total distance around the edges of a triangle. In other words, the length of the boundary of a triangle is its perimeter.
### How to Calculate the Perimeter of a Triangle?
To calculate the perimeter of a triangle, add the length of its sides. For example, if a triangle has sides a, b, and c, then the perimeter of that triangle will be P = a + b + c.
### Calculate the Perimeter of a Right Triangle with Base as 3 cm and height as 4 cm.
First, using the Pythagorean theorem, calculate the hypotenuse of the right triangle.
h =√(base2+perpendicular2)
h = √(32+42)
h = √(9 + 16)
h = √25
Or, h = 5 cm
So, the perimeter of the triangle = 3 + 4 + 5 = 12 cm.
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Question
# how to calculate 42*54 in vedic math
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Solution
## Dear Student 42*54 $42×54=_ _ _\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒4×5=20\phantom{\rule{0ex}{0ex}}⇒write20inthefirstblank\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒42×54=20 _ _\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒2×4=8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒write8inthelastblank\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒42×54=20 _ 8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒now\phantom{\rule{0ex}{0ex}}⇒\left(2×5\right)+\left(4×4\right)=10+16=26\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒write6inthemiddleblankandadd2tothefirstblank\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒42×54=22 6 8\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒Sotheansweris2268\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Regards\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
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# Parametrize the portion of the sphere of radius 4 centred at the origin that lies inside a cylinder above the xy-plane
Parametrize the portion of the surface of the sphere of radius 4 centred at the origin that lies inside the cylinder determined by $$x^2 + y^2 = 12$$ and above the xy-plane.
Solution: The radius of the sphere and cylinder are 4 and $$2\sqrt{3}$$. Using the spherical coordinates, a parametrization is , $$\vec G(\phi, \theta) =(4 sin\, \phi \,\,cos \,\theta, 4 sin \,\phi\,\, sin \theta, 4 cos \,\phi)$$ where $$\theta\in [0,2\pi]$$. Please help me find the range of $$\phi$$.
Pick a point on the intersection of the sphere and cylinder. Call this point $$w$$. Let $$A$$ be the line from $$w$$ to the origin. Let $$B$$ be the vertical line contained by the cylinder from $$w$$ to $$z=0$$. Let $$C$$ be the line from the origin to the endpoint of $$B$$ at $$z=0$$. Then, we get the following right triangle,
Now, $$B$$ is a line in the surface of the cylinder, and the curve in the diagram is the surface of the sphere in the triangle's plane. Thus, $$A$$ must be the radius of the sphere, and $$C$$ must be the radius of the cylinder. Furthermore, the angle $$\psi = \frac{\pi}2-\phi$$.
Thus $$|A|=4$$ and $$|C|=2\sqrt{3}$$. Finding the angle $$\gamma$$, we expect that $$\sin(\gamma) = \frac{|C|}{|A|}=\frac{2\sqrt{3}}4=\frac{\sqrt{3}}2$$. Of course, $$\sin(\gamma)=\frac{\sqrt{3}}2$$ when $$\gamma = \frac{\pi}3$$. Since $$ABC$$ is a right angle triangle, then $$\delta = \frac{\pi}2$$. Thus, $$\gamma+\delta+\psi = \frac{\pi}3+\frac{\pi}2+\psi = \pi$$. So, $$\psi=\frac{\pi}6$$. Finally, $$\psi = \frac{\pi}6= \frac{\pi}2-\phi$$, so $$\phi = \frac{\pi}3$$.
Now, consider the radius of the sphere sweeping down from the $$z$$-axis toward the line $$A$$. As the radius sweeps downward, $$\phi$$ increases from $$0$$ to $$\frac{\pi}3$$. Thus, $$0\le\phi\le\frac{\pi}3$$.
• In the answer, however, it is given as $\phi\in[0,\pi/3]$ Sep 18, 2023 at 5:11
• Sorry, got my notation wrong. Usually, $\phi$ is represented as the downward sweep from the $z$-axis, in which case $0\le\phi\le\pi/3$ is correct, but I have represented it as the upward sweep from the $z$-axis. I will edit the question later today to reflect this. Sep 20, 2023 at 5:28
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### Grade 6 - Mathematics1.21 Prime Factorization of Numbers up to 1000
Method:
1. Factorize the given number
2. The factors chosen need to be a prime number
Example: Find the prime factors of 504.
Solution:
Factorize the number 504
Correct Incorrect Incorrect 504 = 2 x 252 252 = 2 x 126 126 = 2 x 63 63 = 3 x 21 21 = 3 x 7 504 = 2 x 2 x 2 x 3 x 3 x 7 504 = 2 x 252 252 = 4 x 63 63 = 3 x 21 21 = 3 x 7 504 = 2 x 4 x 3 x 3 x 7 504 = 2 x 252 252 = 2 x 126 126 = 2 x 63 63 = 9 x 7 504 = 2 x 2 x 2 x 9 x 7 2, 3 and 7 are prime numbers 4 is not a prime number 9 is not a prime number
Answer: 2 x 2 x 2 x 3 x 3 x 7
Directions: Find the prime factors of the following. Also write at least ten examples of your own.
Name: ___________________Date:___________________
### Grade 6 - Mathematics1.21 Prime Factorization of Numbers up to 1000
Q 1: What are the factors of the number: 502? 1x2x2511x2x2x2x2511x2x2x2511x2x3x251 Q 2: What are the factors of the number: 944? 1x2x2x2x2x591x2x2x2x2x2x591x2x2x2x2x3x591x2x2x2x2x2x2x59 Q 3: What are the factors of the number: 246? 1x2x3x3x411x2x2x2x3x411x2x2x3x411x2x3x41 Q 4: What are the factors of the number: 771? 1x3x3x2571x2x3x2571x2x2x3x2571x3x257 Q 5: What are the factors of the number: 890? 1x2x5x891x2x2x2x5x891x2x3x5x891x2x2x5x89 Question 6: This question is available to subscribers only! Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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# Understanding probability and long term expectations exercise solutions
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CHAPTER 16 SOLUTIONS AND MINI-PROJECT NOTES CHAPTER 16 UNDERSTANDING PROBABILITY AND LONG TERM EXPECTATIONS EXERCISE SOLUTIONS 16.1 a. This statement is an example of a personal probability because it is based on an individual's belief of his/her chances of getting the flu. b. The relative frequency interpretation applies here because this probability is based on repeated observations of the frequency of flu cases each winter. 16.2 a. Assuming that the traits of having red hair and having blond hair are mutually exclusive, Rule 2 can be applied. So the probability of having red hair should be 23% − 14% = 9%. b. The probability that a child is living with one parent is the probability that a child is living with just its mother or just its father. Because these two possibilities are mutually exclusive, the probabilities can be added using Rule 2. The answer is 0.247. c. Note that the probability of a birth resulting in more than one child is 0.0301 + 0.0019 = 0.032. So by Rule 1, the probability that a birth will result in only one child is 1 − 0.032 = 0.968. 16.3 a. This contradicts Rule 1. There are only two possibilities: a person either is wearing a seat belt or is not. So the two probabilities should sum to 1. b. A probability cannot exceed 1. c. This contradicts Rule 4. Red sports cars are a subset of red cars, so the probability that a car is red cannot be less than the probability that it is both red and a sports car. 16.4 a. The probability of not being born on Friday the 13th is 1 − 1/214 = 213/214, by Rule 1. b. No. For instance, the probability of being born on Friday the 13th if you are born in a year with two is roughly 2/365, whereas if there are three in the year it would be 3/365. c. "The probability of being born on Friday the 13th is 1/214" means that over the long run (many years) 1 in 214 people will be born on a Friday the 13th. 16.5 Explanation 1 is closer to what it means. The number of heads may be far from half for a small number of tosses, so Explanation 2 is incorrect. 16.6 Not all events are repeatable. An example is the probability that the Earth will be uninhabitable by the year 2050. 16.7 a. Because 1/2 the cards are red and 1/2 black, in the long run 1/2 the cards you pick will be black and 1/2 will be red. So whether you guess black or red you have a probability of 1/2 of being right by chance. b. It is a relative-frequency probability because it is based on knowledge of the physical situation, which can be repeated numerous times. c. This would be a personal probability because it is based neither on physical assumptions nor repeated observations, but on a person's belief. d. In this case the value of 0.60 would be a relative frequency probability because it was arrived at by repeated observations of her "guessing" ability. 16.8 a. You could either go through the whole phone book and count the portion of people listed who had your first name or you could repeatedly randomly select an entry and note the portion of the time that the first name matched yours. b. By either method the resulting probability would be a relative-frequency probability. The resulting probability would be based either on physical assumptions (how many listings you know to have your first name) or on repeated observations. 16.9 a. By Rule 1, it must be 1 − 0.9 = 0.1, or 10%. b. By Rule 3, the probability that customer A and customer B will not pay their bills in one week is (0.1)(0.1)= 0.01, or 1%. The assumption has been made that customer A not paying in one week does not influence whether or not customer B pays in a week. Yes, it seems reasonable to assume that there will, in general, be no connection between these two events. 16.10 No. To say the probability is 5/10, you are assuming that the events are mutually exclusive, meaning that if you got an interesting piece of mail on Monday you could not get one on Tuesday, and so on. But you could easily get an interesting piece of mail on two of these days or all of them. (You would need to subtract the probability that you would receive an interesting piece of mail on two or more of these days.) 16.11 We would have to assume independence, so that if a person belongs to one of these organizations it would not make them more or less likely to belong to the other. This assumption is probably not valid because older people are more likely to belong to AAA, and only those over 50 are allowed to belong to AARP. 16.12 By Rule 4, alternative A must have a higher probability than B because bank tellers who are active in the feminist movement are a subset of all bank tellers. 16.13 This is a relative-frequency probability. It means that over the long run the portion of times luggage will be temporarily lost is 1/176. 16.14 a. This probability was most likely determined by observing the relative frequency. Flights from New York to San Francisco were observed and their punctuality over a long period of time was recorded. b. This probability was determined by repeated observations of how often Americans read a book for pleasure. c. This probability could have been most easily determined by physical knowledge of the situation, assuming all possible poker hands are equally likely and determining the proportion of them that result in four of a kind. 16.15 a. Whatever the birth month of the first person, the probability that the second person was born in that same month is 1/12. By Rule 1 the probability of a different month is therefore 1 − 1/12 = 11/12. You could also notice that for the second person, 11 out of the 12 months would not match the first person's birth month, so the probability of a different month is 11/12. b. Two months have been taken by the first two people, so there are ten left. If the third person was born in any of those ten months, their birth month would be different. Therefore, the probability is 10/12. c. If parts a and b both happen, then all three were born in different months. The probability of this happening can be found by multiplying the probabilities: (11/12)(10/12) = 110/144 or 0.76. d. If part a and part b did not both happen this would mean that at least two people have the same birth month. In part c we found the probability of the complement of this to be 110/144. Therefore, by Rule 1 we know that the probability that parts a and b do not both happen is 1 − 110/144 = 34/144 = 0.236. Hence, the probability that at least two of the people were born in the same month is 34/144 or 0.236. 16.16 Here is an example provided by a student: I think that the probability that I will finish my math homework by the due date is 70%. I also believe that I will receive an A with a probability of 60%. I will need to turn in the assignment on time in order to get an A, so getting an A is a subset of turning in the assignment. This prediction is coherent because 60% is smaller than 70%. 16.17 a. These are not likely to be independent because spouses' political views are very likely to influence each other, and spouses are likely to belong to the same political party. b. The temperature plays a role in whether or not it snows so these can not be independent. c. These will be independent because the numbers are chosen randomly each week and knowledge of one week's numbers will not give you any knowledge of the second week's numbers. d. These events are probably independent because the Dow Jones Industrial Average is not likely to be affected by earthquakes and earthquakes are clearly not affected by the Dow Jones Industrial Average. You might argue that they are not independent because a major earthquake in a city like New York, Los Angeles or San Francisco would probably play havoc with the stock market. This is an acceptable answer if you provide an explanation. 16.18 a. The probability that you do not find money is 1 − 0.1 = 0.9. b. The probability of finding money for the first time on the third try would be (0.9)(0.9)(0.1) = 0.081. c. This is the accumulated probability at the third try, which is the sum of the individual probabilities for tries 1, 2 and 3. It is thus (0.1) + (0.9)(0.1) + (0.9)(0.9)(0.1) = 0.1 + 0.09 + 0.081 = 0.271. 16.19 a. 99/100 b. (99/100)(1/100) = .0099. 16.20 a. The probability of making it 70 years without being struck by lightning would be (684,999/685,0000)70. b. If the average lifetime is 70 years, then Krantz's probability should be 1 − (684,999/685,000)70, or one minus the probability in part a. c. These probabilities do not apply to any particular person. Your chances depend on things like where you live and what you do during storms. d. About (290,000,000) (1/685,000), or approximately 423 people. 16.21 a. The expected commute time is (20 min)(0.2) + (15 min)(0.8) = 16 minutes. b. No, the commute times are always either 15 or 20 minutes. This is just an average over the long run. 16.22 a. You lose \$3 if the births are all girls, so the probability that you lose \$3 is (0.49)(0.49)(0.49) = 0.117649. Your friend loses \$3 if they are all boys, so the probability is (0.51)(0.51)(0.51) = 0.132651. b. Expected value = (+\$1)(0.51) + (\$1)(0.49) = 0.51 0.49 = 0.02 or 2 cents for each birth. c. Making 2 cents on each birth, after 1,000 births you make about (\$0.02)(1,000) = \$20. 16.23 It costs \$1 to play, so the possible outcomes are \$19, \$1 and −\$1. The expected value is (\$19)(0.012) + (\$1)(0.137) + (−\$1)(0.851) = −\$0.486. So your expected loss would be about 49 cents for each dollar played. 16.24 Your expected grade point average would be (4.0)(0.3) + (3.0)(0.7) = 3.3. You would not necessarily get this in any one quarter; expected value gives you only a long-term average. Even four or five years of college may not be enough to reach this long-term average exactly. 16.25 The possible number of parents is 2 with probability 0.72, 1 with combined probability for the mother and father of 0.25, and 0 with probability 0.03. The expected value for the number of parents is (2)(0.72) + (1)(0.25) = 1.69. This is not particularly meaningful for this example because we cannot interpret 1.69 parents. 16.26 The best examples are for discrete data in which the expected value is also the mode. For instance, suppose a small box of raisins (sultanas) contains 19 pieces with probability 1/10, 20 with probability 8/10 and 21 with probability 1/10. Then the expected value is 20 raisins, which is also the most likely value. 16.27 The probability that results in a break-even situation is (cost)/(cost+\$5,000). To figure this out, call the probability of an accident p. Then the insurance company earns your cost with probability (1−p) and loses \$5,000 with probability p. The expected value is: (cost)(1−p) + (−\$5,000)(p) = cost − (cost+\$5,000)(p). Setting this to zero and solving for p gives the answer. In general, you will find that the answer is much higher than your own personal probability. (That's how the insurance companies pay for their expenses and make a profit.) 16.28 It is based on personal probability. The event of an asteroid that large hitting the earth has not occurred often enough to find a long-run relative frequency. NOTES ABOUT THE MINI-PROJECTS FOR CHAPTER 16 Mini-Project 16.1 The purpose of this mini-project is to learn firsthand that the tendency of most people is to ignore Rule 4 in this setting. The report for this project will be fairly short and should explain whether or not it was difficult to convince people of the right answer. Mini-Project 16.2 Make sure the plot uses all flips accumulated up to that point and not just the previous ten. If done correctly, the value on the vertical axis should start settling down to a constant as the number of flips (on the horizontal axis) increases. If flipping is done in an unbiased way that constant should be about 0.50. That is what the relative-frequency interpretation of the probability of a head would predict. Mini-Project 16.3 The purpose of this project is to demonstrate that personal probabilities can be quite variable and still all be correct. In contrast, if we all agree on the physical assumptions, then we should all agree on the relative-frequency probability in this project. The probability of a heart should be given as 1/4 by everyone. Mini-Project 16.4 Make sure to count the cost of playing when computing the expected value for each game. The interpretation is that this is what it costs per play in the long run. There is no correct answer to part b, although if one game has a substantially higher expected value it would probably be the desired game. The psychological impact of potentially winning a large amount must also be considered. Page of
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Cara Pengurangan Pecahan
>Hello Sohib EditorOnline, in this article, we will discuss the methods and techniques on how to reduce fractions or “cara pengurangan pecahan” in Bahasa Indonesia. Fractions are essential in mathematics, and they are used in many daily situations in our lives. Understanding the reduction of fractions is important to simplify them, making it easier to use them in calculations and other mathematical operations.
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Understanding Fractions
Fractions represent a part of a whole or a ratio between two numbers. Fractions consist of a numerator, which is the top part, and a denominator, which is the bottom part. For example, in the fraction 3/4, the number 3 is the numerator, and the number 4 is the denominator. The numerator represents how many parts we have while the denominator represents the total number of parts that make up the whole. Fractions can be proper, improper or mixed, depending on their numerator, and denominator.
Proper Fractions
A proper fraction is a fraction where the numerator is smaller than the denominator. For example, 2/5, 3/7, and 1/3 are all proper fractions. These types of fractions represent a part of a whole that is less than one. Proper fractions can also be simplified or reduced to their lowest terms.
Improper Fractions
An improper fraction has a numerator that is greater than or equal to the denominator. For example, 7/4, 5/3 and 9/8 are all improper fractions. Improper fractions represent a part of a whole that is more than one. Improper fractions can also be simplified or converted into a mixed number.
Mixed Numbers
A mixed number is a combination of a whole number and a proper fraction. For example, 3 1/2, 5 3/4 and 2 2/3 are all mixed numbers. Mixed numbers can also be converted into improper fractions or simplified to their lowest terms.
Reducing Fractions
To reduce fractions, we need to find the greatest common divisor (GCD) of the numerator and denominator, then divide both by the GCD. The resulting fraction will be in its lowest terms or simplified form. There are several methods and techniques to find the GCD.
Method 1: Prime Factorization
One method of finding the GCD is by using prime factorization. Prime factorization involves finding the prime factors of each number and multiplying the common prime factors. For example, to reduce the fraction 12/20 to its lowest terms, we need to find the GCD of 12 and 20. Prime factorization of 12 yields 2 x 2 x 3 while prime factorization of 20 yields 2 x 2 x 5.
Number Prime Factors
12 2 x 2 x 3
20 2 x 2 x 5
The common prime factors between 12 and 20 are 2 and 2. Therefore, the GCD of 12 and 20 is 2 x 2 = 4. To reduce 12/20 to its lowest terms, we divide both numerator and denominator by 4. The resulting fraction is 3/5.
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Method 2: Euclidean Algorithm
Another method for finding the GCD of two numbers is the Euclidean algorithm. It involves dividing the larger number by the smaller number and then taking the remainder. We continue this process until we obtain a remainder of zero. The last non-zero remainder is the GCD of the two numbers.
For example, to reduce 24/36 to its lowest terms, we use the Euclidean algorithm as follows:
Number Division Remainder
36 ÷24 12
24 ÷12 0
The GCD of 24 and 36 is the last non-zero remainder, which is 12. To simplify 24/36, we divide both numerator and denominator by 12. The resulting fraction is 2/3.
1. What is a fraction?
A fraction represents a part of a whole or a ratio between two numbers. Fractions consist of a numerator, which is the top part, and a denominator, which is the bottom part. The numerator represents how many parts we have while the denominator represents the total number of parts that make up the whole.
2. What is a proper fraction?
A proper fraction is a fraction where the numerator is smaller than the denominator. Proper fractions represent a part of a whole that is less than one and can be simplified or reduced to their lowest terms.
3. What is an improper fraction?
An improper fraction has a numerator that is greater than or equal to the denominator. Improper fractions represent a part of a whole that is more than one and can be converted into a mixed number or simplified to their lowest terms.
4. How do you reduce fractions?
To reduce fractions, we need to find the greatest common divisor (GCD) of the numerator and denominator, then divide both by the GCD. The resulting fraction will be in its lowest terms or simplified form. There are several methods and techniques to find the GCD, such as prime factorization and Euclidean algorithm.
5. Why is reducing fractions important?
Reducing or simplifying fractions is important because it makes them easier to use in calculations and other mathematical operations. Simplifying fractions also helps in understanding the relationship between different fractions and their equivalents.
Conclusion
In conclusion, reducing fractions or “cara pengurangan pecahan” is an essential skill in mathematics. Understanding fractions, their types and reducing them to their lowest terms is crucial in simplifying mathematical operations involving fractions. We hope that this article has provided you with the necessary information and techniques to reduce fractions easily and correctly.
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## Introduction to Data Interpretation
#### Data Interpretation
Direction: Study the graph and answer the questions.
1. In which year the sale of cool-sip is minimum?
1. From the given bar diagram , we see
Minimum sales in 1989 = 60 lakh bottles
##### Correct Option: D
From the given bar diagram , we see
Minimum sales in 1989 = 60 lakh bottles
Hence , the sale of cool-sip is minimum in 1989 year .
Direction: The graph shows the result of 10th class students of a school for 4 years. Study the graph and answer the questions :
1. The ratio of students who scored second class to the total students appeared in the year 2000 is
1. According to given bar graph , we have
Number of students who scored second class in the year 2000 = 50
Total students appeared in the year 2000 = 20 + 50 + 90 = 160
Required ratio = Number of students who scored second class in the year 2000 : Total students appeared in the year 2000
##### Correct Option: C
According to given bar graph , we have
Number of students who scored second class in the year 2000 = 50
Total students appeared in the year 2000 = 20 + 50 + 90 = 160
Required ratio = Number of students who scored second class in the year 2000 : Total students appeared in the year 2000
Required ratio = 50 : 160 = 5 : 16
1. The year in which the maximum number of students appeared for the 10th class exam is
1. As per the given bar graph , we have
Number of students in Year 2000 ⇒ 20 + 50 + 90
= 160
Number of students in Year 2001 ⇒ 30 + 60 + 110 = 200
Number of students in Year 2002 ⇒ 15 + 60 + 120 = 195
##### Correct Option: A
As per the given bar graph , we have
Number of students in Year 2000 ⇒ 20 + 50 + 90
= 160
Number of students in Year 2001 ⇒ 30 + 60 + 110 = 200
Number of students in Year 2002 ⇒ 15 + 60 + 120 = 195
Number of students in Year 2003 ⇒ 10 + 40 + 120 = 170
Hence , required answer is 2001 year .
1. The percentage increase of first-class in the year 2003 over the year 2002 is approximately
1. From the given bar diagram , we see
Number of students of first-class in the year 2003 = 120
Number of students of first-class in the year 2002 = 120
Increase = 120 - 120 = 0
Required % increase = Increase × 100 Number of students of first-class in the year 2002
##### Correct Option: B
From the given bar diagram , we see
Number of students of first-class in the year 2003 = 120
Number of students of first-class in the year 2002 = 120
Increase = 120 - 120 = 0
Required % increase = Increase × 100 Number of students of first-class in the year 2002
Required % increase = 0 × 100 = 0 120
1. The number of students appeared for the 10th class exam in the year 2002 is
1. As per the given bar graph , we have
Required number of students in 2002 = 15 + 60 + 120
##### Correct Option: B
As per the given bar graph , we have
Required number of students in 2002 = 15 + 60 + 120 = 195
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# What Is 7/37 as a Decimal + Solution With Free Steps
The fraction 7/37 as a decimal is equal to 0.189.
One of the fundamental operations in mathematics is called “Division,” which can also be expressed mathematically as the fraction, which is sometimes more helpful in solving or simplifying complicated mathematical expressions. A fraction has the form “p/q,” where p is the numerator (top entity), and q is the denominator (bottom entity).
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 7/37. The following figure shows the long division process:
Figure 1
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 7
Divisor = 37
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 7 $\div$ 37
This is when we go through the Long Division solution to our problem.
## 7/37 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 7 and 37, we can see how 7 is Smaller than 37, and to solve this division, we require that 7 be Bigger than 37.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 7, which after getting multiplied by 10 becomes 70.
We take this 70 and divide it by 37; this can be done as follows:
70 $\div$ 37 $\approx$ 1
Where:
37 x 1 = 37
This will lead to the generation of a Remainder equal to 70 – 37 = 33. Now this means we have to repeat the process by Converting the 33 into 330 and solving for that:
330 $\div$ 37 $\approx$ 8
Where:
37 x 8 = 296
This, therefore, produces another Remainder which is equal to 330 – 296 = 34. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 340.
340 $\div$ 37 $\approx$ 9
Where:
37 x 9 = 333
Finally, we have a Quotient generated after combining the three pieces of it as 0.189=z, with a Remainder equal to 7.
Images/mathematical drawings are created with GeoGebra.
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A tank with rectangular base and rectangular sides,
Question:
A tank with rectangular base and rectangular sides, open at the top is to the constructed so that its depth is $2 \mathrm{~m}$ and volume is $8 \mathrm{~m}^{3}$. If building of tank cost 70 per square metre for the base and Rs 45 per square matre for sides, what is the cost of least expensive tank?
Solution:
Let $l, b$ and $h$ be the length, breadth and height of the tank, respectively.
Height, $h=2 \mathrm{~m}$
Volume of the tank $=8 \mathrm{~m}^{3}$
Volume of the tank $=l \times b \times h$
$\therefore l \times b \times 2=8$
$\Rightarrow l b=4$
$\Rightarrow b=\frac{4}{l}$
Area of the base $=1 b=4 \mathrm{~m}^{2}$
Area of the 4 walls, $A=2 h(l+b)$
$\therefore A=4\left(l+\frac{4}{l}\right)$
$\Rightarrow \frac{d A}{d l}=4\left(1-\frac{4}{l^{2}}\right)$
For maximum or minimum values of $A$, we must have
$\frac{d A}{d l}=0$
$\Rightarrow 4\left(1-\frac{4}{1^{2}}\right)=0$
$\Rightarrow l=\pm 2$
However, the length cannot be negative.
Thus,
$I=2 \mathrm{~m}$
$\therefore b=\frac{4}{2}=2 \mathrm{~m}$
Now,
$\frac{d^{2} A}{d l^{2}}=\frac{32}{l^{3}}$
$\mathrm{At} 1=2:$
$\frac{d^{2} A}{d l^{2}}=\frac{32}{8}=4>0$
Thus, the area is the minimum when $/=2 \mathrm{~m}$
We have
$l=b=h=2 \mathrm{~m}$
$\therefore$ Cost of building the base $=$ Rs $70 \times(l b)=$ Rs $70 \times 4=$ Rs 280
Cost of building the walls = Rs $2 h(l+b) \times 45=$ Rs $90(2)(2+2)=$ Rs $8(90)=$ Rs 720
Total cost $=$ Rs $(280+720)=$ Rs 1000
Hence, the total cost of the tank will be Rs 1000 .
Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.
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# Boolean Algebra Rules & Laws
Boolean algebra is a set of rules which are used to simplify the given logic expression without changing its original functionality. So in this article, we are going to learn about Boolean algebra. The Boolean algebra was invented by George Boole; an English mathematician who helped in establishing modern symbolic logic and whose algebra of logic, now popularly known as Boolean algebra. Other names for boolean algebra are binary algebra and logical algebra.
We can also say that Boolean algebra is mathematics which is used to analyze circuits containing gates. Boolean algebra holds real importance in designing the circuits as these help in reducing the number of gates required thus making your system compact and reduce the overall production cost.
Boolean algebra has some rules and laws which we need to know to apply them to reduce boolean expression. But before that let us understand where the Boolean algebra can be used.
Boolean algebra can be used when number of variables are less in a boolean expression. For example: F = A.B+A.B’. Here only two variables are present, so this function can be easily solved by using rules and laws of Boolean algebra.
But in cases where the function contains more than three variables, it is better not to use Boolean algebra as the equation becomes long and you may sometimes end up with the wrong answer. For variables more than three, there are other techniques to simply which are simple like K-map about which you may learn later.
Note: The variables in Boolean algebra can have values either “0” or “1” but expression can have an infinite number of variables. So, after learning where we can use Boolean algebra let us learn about Boolean algebra laws.
The important rules and laws in Boolean algebra are as follows.
1. Rules associated with Complement operation
The complement is the inverse of a variable and is indicated by a over a variable or a bar over a variable. For example, the complement of the variable A is A’.
The complement of 0 is 1
The complement of 1 is 0
If A = 1, then A’ = 0
If A = 0, then A’ = 1
The truth table for this operation is given below.
A Z=A’ 0 1 1 0
Note: (A’)’ = A
2. Rules associated with AND operation
Consider the 2 variable AND operation.
Z = A.B
The truth table for 2 variable AND operation is given below.
A B Z=A.B 0 0 0 0 1 0 1 0 0 1 1 1
Here we see that Z is 1 when both the variables are 1.
Now some of the rules that we can derive from this are
(i) A.A = A
(ii) A.0 = 0
(iii) A.1 = A
(iv) A.A’= 0
3. Rules associated with OR
Consider the 2 variable OR operation.
Z = A+B
The truth table for 2 variable OR operation is given below.
A B Z= A+B 0 0 0 0 1 1 1 0 1 1 1 1
Here we see that Z is 1 when any one variable is 1.
Now some of the rules that we can derive from this are
(i) A+A = A
(ii) A+0 = A
(iii) A+1 = 1
(iv) A+A’= 1
4. Distributive Law
• A.(B+C) = A.B + A.C
• A+(B.C) = (A+B).(A+C)
Example: Y=A +A’B. Simplify.
⇒ (A+A’).(A+B) [using distributive law]
⇒ (A.A+A.B+A’A+A’B)
⇒ A+B [using AND law 1 = A, Here consider A+B as A]
Note: The following result is also used as a property in solving problems.
A+A’B= A+B
5. Commutative Law
• A+B= B+A
• A.B = B.A
This is simple law which states that the order in which the AND/OR is applied in expression doesn’t matter.
6. Associative Law
• A.(B.C) = A.(B.C)
• A+(B+C)=(A+B)+C
This law allows the removal of brackets from an expression and regrouping of the variables.
7. De Morgan’s Law
• (A+B)’=A’.B’
• (A.B)’=A’+B’
De morgan’s law consists of two theorems but at this stage knowing the rule is enough.
With this, we have covered all the important rules which are required to solve the problems using Boolean algebra.
Another question then how to start a problem? It’s simple just always remember the priority pyramid while solving the Boolean algebra.
Just follow the arrow’s direction in the pyramid to give priority to the type of operation while solving a problem.
Let’s tabularize all the rules which we have learnt till now
Compliment (A’)’=A AND A.A=A A.0=0 A.1=A A.A’=0 OR A+A=A A+0=A A+1=1 A+A’=1 Associative A+BC=(A+B).(A+C) A.(B+C)=A.B+A.C Distributive A+A’B=A+B A’+AB=A’+B De Morgan’s Law (A+B)’=A’+B’ (A.B)’=A’+B’
Now let’s take up some examples to help you all understand better
EXAMPLE 1
Y= AB+AB’
• Y=A.B + A.B’ {Taking A common from expression}
• Y=A.[B+B’] {Now using OR law}
• Y=A.1 { Now using AND law}
• Y= A
EXAMPLE 2
Y= AB+AB’C+AB’C’
• Y= AB+AB’C+AB’C’ {Taking A common from expression}
• Y= A [B+B’C+B’C’]
• Y=A[B+B'(C+C’)] {Taking B’ common from expression}
• Y=A[B+B'(1)] {Now using OR law }
• Y=A[B+B’] {Now using OR law}
• Y=A[1]
• Y=A
EXAMPLE 3
Y=(A+B+C)(A+B’+C)(A+B+C’)
• Y=(A+B+C)(A+B’+C)(A+B+C’)
• Let X=A+B
• Y=(X+C)(A+B’+C)(X+C’)
• Y=(X+C)(X+C’)(A+B’+C)
• Y=(X+C.C’)(A+B’+C)
• Y=(X+0)(A+B’+C)
• Y=X.(A+B’+C)
• Y=(A+B).(A+B’+C)
• Y=A+B.(B’+C)
• Y=A+B.B’+B.C
• Y=A+0+B.C
• Y=A+BC
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algebra and geometry ppt1.ppt
1. UNIVERSITY OF GUYANA EMA 3102: BASIC ALGEBRA AND GEOMETRY LECTURER: MR. PETER WINTZ 1
2. GROUP C MEMBERS • Devishwar Bahadur • Keith Damon • Tanna James • Judy Kendall • Shanaz Hosain • Ophela Johnson • Shelana Cornelius • Towana Johnson 2
3. THE SOLUTION SET OF LINEAR AND QUADRATIC INEQUALITIES 3
4. Presentation objectives In this presentation, you will learn about: • Linear inequalities • Solving linear inequalities with one variable • Solving inequalities with two variables • Quadratic Inequalities • Solving Quadratic Inequalities 4
5. Key Terms • Variables:A symbol for a value we do not know as yet. It is usually a letter like x or y. • Inequality:Inequality refers to a relationship that makes a non-equal comparison between two numbers or other mathematical expressions. • Equation:An equation is a mathematical expression that contains an equal sign. • Constant value: A constant is a fixed value. • Linear: relating to, resembling, or having a graph that is a line and especially a straight line 5
6. Linear Inequalities • Linear inequalities are defined as expressions in which two linear expressions are compared using the inequality symbols. • Expressions in which two values are compared using the inequality symbols. 6
7. Symbols to represent linear inequalities Symbol Name Symbol Example Not Equal ≠ x ≠ 5 Less than < x+9 < 15 Greater than > 3x +2 > 2x + 1 Less than or equal to ≤ x ≤ -6 Greater than or equal to ≥ y ≥ 2x+3 7
8. Linear inequalities with one variable • The linear equations in one variable are equations that are written as ax + b = 0, where a, and b are two integers and x is a variable, and there is only one solution. • 8x + 3 = 8, for particular, is a linear equation with only one variable. As a result, there is only one solution to this equation. 8
9. Example 1 Solve and graph the inequality: x > 2 Solution 9
10. Graph 10
11. Example 2 Solve and graph the inequality: x ≤ 3 Solution 11
12. Graph 12
13. Example 3 Solve and graph the inequality: x-5 >1 Solution x-5>1 x-5>1 x > 5+1 x>6 13
14. 14
15. Checking Answer • x-5 >1 • x=8 8-5 >1 3 >1 TRUE 15
16. Example 4 Solve and graph the inequality: 3x+1 ≤ 10 16
17. Solution 3x+1 ≤ 10 3x ≤10-1 3x ≤9 x ≤ 9/3 x ≤3 17
18. 18
19. Check Answer • 3x+1 ≤ 10 • x=1 • 3(1)+1 ≤ 10 • 3+1 ≤ 10 • 4 ≤10 • TRUE 19
20. Linear inequalities with two variable • A linear inequality in two variables is formed when symbols other than equal to, such as greater than or less than are used to relate two expressions, and two variables are involved. 20
21. Example 1 21
22. 22
23. Checking Answer (5,-5) • y>x-3 • -5>5-3 • -5>2 • FALSE (2,5) • y>x-3 • 5>2-3 • 5>2-3 • 5 >-1 • TRUE 23
24. Example 2 24
25. 25
26. Checking Answer (-1,2) • y≤2x-2 • 2≤2(-1)-2 • 2≤-2-2 • 2≤-4 • FALSE (2,-4) • y≤2x-2 • -4≤2(2)-2 • -4≤ 4-2 • -4 ≤ 2 • TRUE 26
27. Quadratic Inequalities • A quadratic inequality is an equation of second degree that uses an inequality sign instead of an equal sign. • The quadratic inequality has been derived from the quadratic equation ax2 + bx + c = 0 • Further if the quadratic polynomial ax2 + bx + c is not equal to zero, then they are either ax2 + bx + c > 0, or ax2 + bx + c < 0, and are referred as quadratic inequalities. 27
28. Quadratic Inequalities • Some examples of quadratic inequalities in one variable are: • x2 + x - 1 > 0 • 2x2 - 5x - 2 > 0 • x2 + 2x - 1 < 0 28
29. Example 1 29 Factors of10 X=10 + = 7 5 2 Negative
30. Solution in interval notation: -5<x<-2 30 x=2 x=-4 x=-6 (x+5) (x+2)< 0 (2+5) (2+2)<0 10x4<0 40<0 False (x+5) (x+2)< 0 (-4+5) (-4+2)<0 1x-2<0 -2<0 True (x+5) (x+2)< 0 (2+5) (2+2)<0 10+4<0 14<0 False
31. Example 2 31 Factors of 16 X=16 + = 10 8 2 Positive
32. ` 32 Solution in interval notation: -8>x>-2 x=-9 x=-5 x=1 (x+8) (x+2)< 0 (-9+8) (-9+2)<0 -1x-7<0 7>0 (x+8) (x+2)< 0 (-5+8) (-5+2)<0 3x-3<0 -9>0 (x+8) (x+2)< 0 (1+8) (1+2)<0 8x2<0 16>0
33. Example 3 33 Factors of 8 X=8 + = 9 8 1 Negative
34. • Solution in interval notation:-8≤x ≤-1 34
35. THE END 35
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# QA – Average Speed Explained
Average speed is simply the total distance travelled divided by the total time taken. Example:
Aragorn, your watchman, took part in a two-stage race where he travelled by car and on foot. His speed in the two stages was 60 km/hr and 15 km/hr respectively. In each stage, he covered 60 kms. Calculate Aragorn’s average speed across the entire race.
In this question, one will be tempted to simply take the average of the two speeds given and calculate the overall average speed as (60 + 15)/2 = 37.5 km/hr.
But, this is not the actual average speed.
Remember: Average speed = Total distance travelled / Total time taken
In this case, time taken in the first stage = 60/ 60 = 1 hr
Time taken in the second stage = 60/ 15 = 4 hrs
Hence, total time taken = 4 + 1 hrs = 5 hrs
Total distance covered = 60 + 60 = 120 kms
Hence, average speed = 120/ 5 = 24 km/hr
Note: If a journey involves two stages where the speeds in the two stages are x & y but the distance covered in the two stages is the same then the average speed = 2xy/(x + y)
Hence, Aragorn’s average speed
= (2*60*15)/(60 + 15)
= 1800/75 = 24 km/hr
So the above formula is applicable when the distance covered in the stages is equal. But is there a formula to calculate average speed if the time taken in each stage is equal?
When the time taken for each of the stages of a journey is the same, you can directly take the average of the individual speeds. This average would be the overall average speed.
Example:
A bus travelled at a speed of 30 km/hr for 2 hours on a busy road. It then reached a highway and travelled at a speed of 60 km/hr for another 2 hours. Calculate the average speed of the bus?
Explanation:
Since the time taken in each stage is the same, we simply find the average of the two speeds. Hence, overall average speed = (30 + 60)/ 2 = 45 km/hr
Let us now discuss a few easy problems involving trains.
Note: When we discuss cars, bikes, etc. overtaking each other, we do not consider their lengths. But, in the case of trains, we must consider their lengths.
A Rajdhani train with a length of 200 metres crosses a station in just 30 seconds. If the speed of the train was 90 km/hr, what was the length of the station platform?
Explanation: We have already seen that we need to consider the length of the train. Similarly, we must also consider the length of platforms, etc.
In this question, speed of the train = 90 * 5/18 = 25 m/s.
Distance covered by the train in 30 seconds = 30 * 25 = 750 m
This 750 metres must be the sum of the length of the platform and the length of the train.
Hence, length of the platform = distance covered – length of the train = 750 – 200 = 550 metres
Let us consider another interesting problem:
A Japanese Maglev train travelling at 380 km/hr takes 15 seconds to pass a man riding his Suzuki GSX bike in the same direction at a speed of 272 km/hr. Find the length of the train.
Explanation:
Here this problem involves relative speed. Since the train and the bike are going in the same direction, we must take the difference of the two speeds to get the relative speed.
Hence, relative speed = 380 – 272 = 108 km/hr
Since the time given is in seconds, let us convert this speed to m/s.
Hence, relative speed = 108 * 5/18 = 30 m/s
As the train takes 15 seconds to pass the bike, the distance covered in this time would be the length of the train.
Therefore, length of the train = distance covered = time * speed
= 15 * 30 = 450 metres
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# Find the values of a and b
$$(x-a)^2=x^2-12x+b$$
Find the values of $a$ and $b$
Can you show me the working out?
-
Open term in LHS, compare both sides – Awesome Mar 16 '14 at 15:36
If you don't know the theorem used implicitly in some of the other answers (that polynomial functions are identical iff they have equal coefficients), you can proceed more simply as follows. Expanding the square and subtracting it from the RHS yields
$$(2a-12) x + b-a^2\, =\, 0$$
Evaluating the above at $\,x = 0\,$ yields $\,b-a^2 = 0\,$ so $\, b = a^2,\,$ so the above becomes
$$(2a-12)x\, =\, 0$$
Evaluating the above at $\,x = 1\,$ yields $\,2a-12=0\,$ so $\, 2a=12,\,$ so $\, a=6,\,$ so $\, b = a^2 = 36.$
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Hints:
\begin{align*}(1)&\;\;(x-a)^2=x^2-12x+b\iff x^2-2ax+a^2=x^2-12x+b\\{}\\(2)&\;\;\text{Two polynomials are identical iff the coefficients of corresponding powers of \;x\; are identical}\end{align*}
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Well, firstly you should expand the expression and get $$x^2-2ax+a^2=x^2-12x+b$$ then cross out $x^2$ $$a^2-2ax=b-12x$$ Which means that \begin{cases} -2ax = -12x \Rightarrow a=6\\ a^2=b \Rightarrow b=a^2=6^2=36\\ \end{cases} So $a=6$ and $b=36$
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The left-hand side equals $x^2-2ax+a^2$. Comparing linear coefficients, we get $-2a = -12$ or $a=6$. Comparing constant coefficients, we get $b=a^2=6^2=36$. Hence $a=6$ and $b=36$.
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# Circle: Circumference
The distance around the outside of a circle is called the circumference (sometimes we think of it as the perimeter of a circle).
The formula for the circumference of a circle can be written in two ways: $d\pi$ or $2r\pi$
These forumulae are identical because $d$ stands for diameter and $r$ stands for radius -- and the radius (the distance from the center to the edge of a circle) is always half of a diameter (the distance, through the center, from edge to edge of a circle). See the lesson Circles (Diameter vs. Radius) for more information.
So, to find the circumference of a circle, you multiple the diameter times pi. If you are given the radius, instead of the diameter, multiply the radius times two to get the diameter, and multiply times pi.
Example:
Find the circumference of the following circle:
The diameter = 10 inches.
Circumference= $d\times \pi=10 \times \pi = 10 \pi$
Some assignments will let you stop at $10\pi$. Other times you have to calculate the radius in inches.
Use $3.14$ or $\dfrac{22}{7}$ for pi.
Circumference = $3.14 \times 10 = 31.4 \text{inches}$ Circumference = $\dfrac{22}{7} \times 10 = \dfrac{220}{7}=31.4 \text{inches}$
• ## Circle: Circumference
1. Find the circumference of a circle with a radius of 12 inches.
2. Find the circumference of a circle with a diameter of 8 inches.
3. Find the circumference of the circle:
4. Find the circumference of the circle:
5. If the circumference of a circle is 942 feet, what is the length of the diameter?
6. Find the circumference of a circular ampitheater if the length across its center is 450 feet.
7. If a circular room has a radius of 25 feet, what is the circumference of the room?
8. If a circular field has a diameter of 25 feet, what is the circumference of the field?
9. If the circumference of a swimming pool is 188.4 feet, how long is the radius of the swimming pool?
10. You want to wrap a ribbon around the bottom of a skirt that forms a circle when laid out flat on the ground. If the measurement from the center of the circle to the edge of the skirt is 16 inches, how long a ribbon do you need?
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# Add Two Numbers II Leetcode Solution
Difficulty Level Medium
## Problem Statement
The Add Two Numbers II LeetCode Solution – “Add Two Numbers II” states that two non-empty linked lists represent two non-negative integers where the most significant digit comes first and each node contains exactly one digit. We need to add the two numbers and return the sum as the linked list.
## Example:
`Input: l1 = [7,2,4,3], l2 = [5,6,4]`
`Output: [7,8,0,7]`
Explanation:
• The most significant digit comes first.
• When we add the above two numbers, we get [7, 8, 0, 7].
`Input: l1 = [2,4,3], l2 = [5,6,4]`
`Output: [8,0,7]`
Explanation:
• [8, 0, 7] is the sum we get on adding the above two numbers.
## Approach
### Idea:
1. The main idea to solve this problem is to use basis maths to add two linked lists.
2. Reverse both the linked list to add the digits starting from the least significant position.
3. Iterate till both the linked list becomes empty.
4. At any position, extract the digit present at the current position for both the linked lists.
5. If the digit is absent in either of the linked lists, then take the digit as 0 for simplicity.
6. A digit that will be added now to the sum linked list will be the remainder obtained by adding a, b, and carry where:
1. a = digit present at current position for a first linked list.
2. b = digit present at current position for a second linked list.
3. carry = digit that needs to be added.
7. Our new carry would become (a + b + carry) / 10.
8. Continue the above steps for each position, till we get the desired sum linked list.
## Code
### Add Two Numbers II Leetcode C++ Solution:
```class Solution {
public:
ListNode *curr = head, *prev = nullptr;
while(curr != nullptr){
ListNode* next_node = curr->next;
curr->next = prev;
prev = curr;
curr = next_node;
}
return prev;
}
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
l1 = reverse(l1);
l2 = reverse(l2);
int carry = 0;
ListNode* dummy = new ListNode(-1);
ListNode* curr = dummy;
while(l1 != nullptr || l2 != nullptr){
int a = l1 == nullptr ? 0 : l1->val;
int b = l2 == nullptr ? 0 : l2->val;
curr->next = new ListNode((a + b + carry)%10);
carry = (a + b + carry)/10;
curr = curr->next;
if(l1 != nullptr){
l1 = l1->next;
}
if(l2 != nullptr){
l2 = l2->next;
}
}
if(carry > 0){
curr->next = new ListNode(carry);
}
return reverse(dummy->next);
}
};```
### Add Two Numbers II Leetcode Java Solution:
```class Solution {
ListNode curr = head, prev = null;
while(curr != null){
ListNode next_node = curr.next;
curr.next = prev;
prev = curr;
curr = next_node;
}
return prev;
}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
l1 = reverse(l1);
l2 = reverse(l2);
int carry = 0;
ListNode dummy = new ListNode(-1);
ListNode curr = dummy;
while(l1 != null || l2 != null){
int a = l1 == null ? 0 : l1.val;
int b = l2 == null ? 0 : l2.val;
curr.next = new ListNode((a + b + carry)%10);
carry = (a + b + carry)/10;
curr = curr.next;
if(l1 != null){
l1 = l1.next;
}
if(l2 != null){
l2 = l2.next;
}
}
if(carry > 0){
curr.next = new ListNode(carry);
}
return reverse(dummy.next);
}
}```
## Complexity Analysis for Add Two Numbers II Leetcode Solution
### Time Complexity
The time complexity of the above code is O(N + M) since we traverse both the linked lists in linear time, and added the two numbers represented as the linked list.
### Space Complexity
The space complexity of the above code is O(max(N, M)).
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# Understanding 4.135 Repeating as a Fraction
Have you ever encountered a decimal number like 4.135 repeating? It looks like a fraction, but you don’t know exactly how to turn it into one. In this article, we’ll explain how to convert 4.135 repeating into a fraction, and some of the other mathematical concepts related to it.
## What Is 4.135 Repeating?
4.135 repeating is a number that can be written as a decimal, but it is actually a rational number. A rational number is any number that can be expressed as a fraction or ratio. 4.135 repeating is simply a way of expressing the same rational number in decimal form. It can also be written as 4.1353, which is an infinite repeating decimal.
## How to Convert 4.135 Repeating to a Fraction?
The easiest way to convert 4.135 repeating into a fraction is to multiply the decimal by 10, 100, 1000, and so on until the repeating portion of the decimal is “canceled out.” For 4.135 repeating, this would look like the following:
4.135 x 10 = 41.35
4.135 x 100 = 413.5
4.135 x 1000 = 4135
This process eliminates the repeating portion of the decimal, leaving only the non-repeating portion. In this case, 4135 is the non-repeating portion of the decimal, so we can now write the decimal as a fraction:
4.135 repeating = 4135 / 1000
## Other Applications of 4.135 Repeating
Now that we know how to convert 4.135 repeating into a fraction, we can use it in other mathematical operations. For example, we can use 4.135 repeating to calculate the area of a square with sides of 4.135 repeating. The area of the square would be 17.3609 repeating.
We can also use it to calculate the volume of a cube with sides of 4.135 repeating. The volume of the cube would be 70.6873 repeating.
##
In this article, we discussed how to convert 4.135 repeating into a fraction and some of the other applications of the number. We hope this article has helped you gain a better understanding of this concept and the ways it can be used in mathematics.
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# If $\alpha ,\beta$ are the roots of the equation ${x^2} + px - q = 0$ and $\gamma ,\delta$ are the roots of the equation ${x^2} + px + r = 0$, then the value of $(\alpha - \gamma )(\alpha - \delta )$ is $A)p + r \\ B)p - r \\ C)q - r \\ D)q + r \\$
Last updated date: 22nd Jul 2024
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Hint: Here to solve this problem we get the given value we need to know the sum of the roots and products of roots of given quadratic equations. They have mentioned the roots of given equations.
We know that if there is any quadratic equation in the form $a{x^2} + bx + c = 0$.Then we know that
Sum of roots $= -\dfrac{{{\text{ coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$
Product of roots $= \dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$
Here given equation is ${x^2} + px - q = 0$ whose roots are $\alpha ,\beta$
Now by using the above concept
Sum of the roots $\left( {\alpha + \beta } \right)$ $= \dfrac{{ - p}}{1} = - p$
Product of roots $\left( {\alpha .\beta } \right) = \dfrac{{ - q}}{1} = - q$
Here it is also mentioned $\gamma ,\delta$ are the root of equation ${x^2} + px + r = 0$.
Sum of the roots $\left( {\gamma + \delta } \right) = \dfrac{{ - p}}{1} = - p$
Product of roots $\left( {\gamma \delta } \right) = \dfrac{r}{1} = r$
Now here we have to find the value of $(\alpha - \gamma )(\alpha - \delta )$
$\Rightarrow \left( {\alpha - \gamma } \right)\left( {\alpha - \delta } \right)$
$\Rightarrow {\alpha ^2} - \alpha (\gamma + \delta ) + \gamma \delta$
Since we the value $\gamma + \delta = - p$ and $\gamma \delta = r$.So now on substituting the value and further simplification we get
$\Rightarrow q + r$
Option D is the correct one .
Note: In this problem before finding the value of given term we have got the required value of to solve the given term. So get the value we have used the sum of roots and products of roots of the given two functions. Here we have to concentrate on finding the sum of roots of the equation and product roots of the equation by using the formulas.
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General
Which Expression Is Correctly Simplified Using The Laws Of Exponents
The laws of exponents simplify expressions when like terms have the same powers, but when a fraction has a negative or positive exponent, the simplified expression must be equal to the original one. The following table outlines the laws of expansion, decomposition, and divisors and how they relate to the process of simplifying expressions. To make use of this table, download Active Learning 1 and distribute the exponents to find a solution.
The laws of exponents simplify expressions that contain more than one operand. You can simplify any complex expression by combining the powers of the exponents in a single operation. For example, consider the example below. It shows an expression with two operands with two different powers. The product of two exponents gives the result of a single operation. If the product is a positive number, then it’s simplified by using the power rule.
In an interactive, students can explore the differences between these two methods. The table provides both the steps used to simplify an expression. In the first step, students should rewrite the negative exponents as positive. The second step involves applying the product rule to the exponents. This interactive provides the same solutions, but in a different order. Whichever method you use, you’ll have a much easier time tackling equations involving exponents.
For example, the two students’ answers are the same. In each case, they used different coefficients and exponents to simplify the expression. In the first case, Frankie got -4 and Marilyn got -64 using the product rule of exponents. However, the second solution is correct. In the second case, the two women used the same formula, but they couldn’t tell which one was right.
The laws of exponents are an essential part of mathematics education. These laws simplify complex mathematical expressions. They are the key to a student’s success. The laws of exponents are helpful in the context of a wide variety of applications. So, if you’re unsure of which way to apply the laws of exponotors, just check out the interactive below.
The laws of exponents allow you to simplify complex expressions without using a calculator. As you can see, there are different ways to use the laws of exponents to simplify an expression. It’s up to you to choose which method you prefer. You can even try both methods in the same context. For instance, if the equation has a fractional component, you’ll need to multiply the fraction by the same number in the opposite direction.
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```MATH 101 Quiz #5 (v.A2)
Last Name:
Friday, March 18
First Name:
Student-No:
Section:
1. 1 mark Does the series
∞
X
n=2
n2
√ converge? (Answer “yes” or “no” in the box.)
3n2 + n
Solution: We have
n2
1
1
√
√
6= 0,
=
lim
=
n→∞ 3n2 +
3
n n→∞ 3 + 1/(n n)
lim
so the series diverges by the Test for Divergence.
Marking scheme: 1 for a correct answer in the box
2. 2 marks Relate the number 0.53̄ = 0.5333333 . . . to the sum of a geometric series, and use
that to represent it as a rational number (a fraction or combination of fractions, with no
3
3
Solution: The number is 0.5 + 100
+ 1000
+ 1034 + · · · = 12 + 1032
series sums to
3
1
3
1
=
=
,
1
102 1 − 10
10(10 − 1)
30
so the fraction is
8
15
P∞
n=0
10−n . The geometric
1
1
8
+
= .
2 30
15
Marking scheme:
• 1 mark for writing the geometric series (either with
• 1 mark for getting the right answer
P
or · · · notation)
3. 2 marks Show that the series
∞
X
n=2
3
converges.
n(log n)5/4
3
. Then f (x) is positive and decreasing, so that the sum
x(log x)5/4
Z ∞
∞
X
f (n) and the integral
f (x) dx either both converge or both diverge, by the Integral
Solution: Let f (x) =
2
2
Test. For the integral, we use the substitution u = log x, du =
Z ∞
Z ∞
3 dx
3 du
=
,
5/4
x(log x)5/4
2
log 2 u
dx
x
to get
which converges by the p-test (5/4 > 1).
Marking scheme:
• 1 point for correctly using the Integral Test. (For this quiz, students can earn this mark
even if they don’t state that f (x) is positive and decreasing; however, on the final exam,
students must show that they know these hypotheses are important.)
• 1 point for showing the integral converges (citing the p-test is OK)
4. 5 marks Find the solution to the differential equation
yy 0
1
= that satisfies y(0) = 3.
x
e − 2x
y
Solve completely for y as a function of x.
Solution: Cross-multiplying, we rewrite the equation as
dy
= ex − 2x
dx
y 2 dy = (ex − 2x) dx.
y2
Integrating both sides, we find
1 3
y = ex − x2 + C.
3
Setting x = 0 and y = 3, we find C = 8; therefore the solution is
1 3
y = ex − x 2 + 8
3
y = (3ex − 3x2 + 24)1/3 .
Marking scheme:
• 1 mark for separating the variables
• 2 marks for integrating both sides of the equation
• 1 mark for finding the value of the constant
• 1 mark for solving for y
```
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Te Kete Ipurangi
Communities
Schools
### Te Kete Ipurangi user options:
Level One > Number and Algebra
# "Teen" and "Ty" Numbers
Keywords:
Specific Learning Outcomes:
Identify all of the numbers in the range 0–100, at least.
Description of mathematics:
Number Framework Stage 4
Required Resource Materials:
Student copies of Hundreds Board (Material Master 4-4)
Tens Frames (Material Master 4-6)
Slavonic abacus
Transparent counters
Activity:
#### Background
“Teen” numbers can be very troublesome for some students because they do not realise that “teen” means 10, and the usual rule of saying tens before ones is broken. For example, sixteen means literally “six and ten”, which really should be “ten and six”. Also 11 and 12 break the “teen” pattern by not being “oneteen” and “twoteen”. Further confusion can occur for the students who fail to realise words ending in “ty” are tens. In particular, many students confuse “sixty” with “sixteen”. Persistence in teaching “teen” and “ty” is needed in overcoming these problems as understanding these numbers is essential by the time part-whole thinking emerges.
#### Activity – “Teen” Numbers
Seat the students in pairs. One partner shows 10 fingers. The other partner shows any number of fingers from one to nine, say six. The “ones” person says “six” and the other partner says “10”, and together they say “is sixteen”.
As a class, record teen numbers as equations on the board or modelling book, and get the students to read them out loud. For example 10 + 4 = 14 is on the board or modelling book. The students say “Ten and four is the same as (equals) 14.”
#### Activity – “Teen” Numbers
In pairs, one student points to a number between 10 and 20 on the hundreds board and the other student reads the number. Then together they show that many fingers. Repeat with the roles reversed.
#### Activity – “Teen” Numbers
On the Slavonic abacus, push across 10 beads on the first row, and then push across extra beads for any amount from 11 to 19, to allow the students to practise recalling the “teen” number facts.
#### Activity – “Ty” Words
Screen a Slavonic abacus from the students’ view while you move complete rows of 10. Turn the abacus around and ask the students to tell you how many beads they can see. Link the number of tens to the structure of the word, for example, “eight tens is eighty”, and its numeral on the hundreds board, for example, 80.
#### Activity –”Teen” and “Ty” Bingo
Every student has a hundreds board and eight transparent counters. Each student places transparent counters on any eight “teen” and “ty” numbers of their choice. You show a succession of “teen” and “ty” numbers on the Slavonic abacus. If any student has a counter on the matching number, they remove that counter. The first player to remove all their counters wins.
## Who wins?
Identify and order decimals to three places.
Order fractions, decimals and percentages.
Order numbers in the range 0–100.
Order the numbers in the range 0–1000.
Order whole numbers in the range 0–1 000 000.
Identify symbols for any fraction, including tenths, hundredths, thousandths, and those greater than 1.
Find equivalent fractions and order fractions.
Know benchmarks for converting between common fractions, decimals and percentages.
Identify and order decimals to three places.
Order fractions, decimals and percentages.
## Squeeze - Guess my Number
Order numbers in the range 0–100.
Order the numbers in the range 0–1000.
Order whole numbers in the range 0–1 000 000.
Read decimals with tenths, count forwards and backwards in tenths, order decimals with tenths.
Identify and order decimals to three places.
## Number Line Flips
Say the forwards and backwards number word sequences in the range 0–10.
Say the forwards and backwards number word sequences in the range 0–20.
Order the numbers in the range 0–20.
Order numbers in the range 0–100.
Order the numbers in the range 0–1000.
Order whole numbers in the range 0–1 000 000.
Read decimals with tenths, count forwards and backwards in tenths, order decimals with tenths.
Identify and order decimals to three places.
## Rocket - Where will I fit?
Order numbers in the range 0–100.
Order the numbers in the range 0–1000.
Order whole numbers in the range 0–1 000 000.
Read decimals with tenths, count forwards and backwards in tenths, order decimals with tenths.
Find equivalent fractions and order fractions.
Identify and order decimals to three places.
Order fractions, decimals and percentages.
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# The line segment joining the points $P (3, 3)$ and $Q (6, -6)$ is trisected at the points $A$ and $B$ such that $A$ is nearer to $P$. If $A$ also lies on the line given by $2x + y + k = 0$, find the value of $k$.
Given:
The line segment joining the points $P (3, 3)$ and $Q (6, -6)$ is trisected at the points $A$ and $B$ such that $A$ is nearer to $P$ and $A$ also lies on the line given by $2x + y + k = 0$.
To do:
We have to find the value of $k$.
Solution:
As line segment $PQ$ is trisected by the points $A$ and $B$.
$PA: AQ = 1:2$
Using section formula, we have
$P( x,\ y)=( \frac{nx_{1}+mx_{2}}{m+n}, \frac{ny_{1}+my_{2}}{m+n})$
Then, co-ordinates of $A$ are:
$A=( \frac{1\times6+2\times3}{1+2}, \frac{1\times(-6)+2\times3}{1+2})$
$\Rightarrow A=( \frac{6+6}{3}, \frac{-6+6}{3})$
$\Rightarrow A=( \frac{12}{3}, \frac{0}{3})$
$\Rightarrow A=( 4, 0)$
The point $A( 4,\ 0)$ lies on the line $2x+y+k=0$.
This implies, point $A( 4,\ 0)$ satisfies the above equation.
$\Rightarrow 2(4)+0+k=0$
$\Rightarrow 8+k=0$
$\Rightarrow k=-8$
The value of $k$ is $-8$.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
50 Views
|
Year 7 Interactive Maths - Second Edition
## Patterns
Patterns often occur in science and mathematics, and they are formed by groups of shapes, objects, diagrams or numbers. Patterns are a part of daily life and are very useful in mathematics as they can help us to solve some problems.
## Square Numbers
A square number is a number that can be arranged in a square pattern. The patterns of the first three square numbers are depicted below.
The first square number, 1, is shown as a square containing a single dot. There is 1 row containing 1 dot.
So, there is 1 × 1 = 1 dot in this square number.
The second square number, 4, is shown as a
square with 2 dots in each side. It is clear that there are 2 rows each containing 2 dots.
So, there are 2 × 2 = 4 dots in this square number.
The third square number, 9, is shown as a square with 3 dots in each side. It is clear that there are 3 rows each containing 3 dots.
So, there are 3 × 3 = 9 dots in this square number.
From the above discussion, we find that the following numbers are all squares.
Following the pattern above, we can find any particular square number.
E.g. The fourth square number will form a square with 4 rows each containing 4 dots.
So, there are 4 × 4 = 16 dots in the square.
Likewise, the fifth square number will form a square with 5 rows each containing 5 dots.
So, there are 5 × 5 = 25 dots in the square.
Note that 52 is often read as '5 to the power 2' and 2 is called the index (or power).
## Index Form
If a number is written with an index, then it is said to be in index form.
E.g. 36 can be written as 62 .
#### Example 1
Write the tenth square number.
##### Solution:
The tenth square number will form a square with 10 rows each containing 10 dots.
#### Example 2
Write down a number between 50 and 100 that is both odd and square.
##### Solution:
We find that 81 is between 50 and 100, and it is both an odd and square number.
So, the required number is 81.
#### Example 3
Write 20 × 20 in index form.
#### Example 4
Write 122 in words.
##### Solution:
122 in words is 'twelve squared'.
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# Free Fractions Subjective Test 01 Practice Test - 6th grade
In a cake eating contest, Saumya got through 34 of a cake before time was called;
Shreya finished just 14 of her cake. How much more cake Saumya ate? [1 MARK]
#### SOLUTION
Solution :
Solution: 1 Mark
The amount of cake Saumya ate more than Shreya was ;
3414=24=12 more cake.
Nimmi was asked to solve all these questions by her teacher. [2 MARKS]
(i) 2+35
(ii) 5+15
(iii) 1+45
Help her solve these questions.
#### SOLUTION
Solution :
Steps: 1 Mark
(i) 2+35=10+35=135.
(ii) 5+15=25+15=265.
(iii) 1+45=5+45=95.
Last week, Suman spent 10 hours watching television. Shobha watched television for 13 as many hours as Suman did. How many hours did Shobha spend watching television? [2 MARKS]
#### SOLUTION
Solution :
Steps: 1 Mark
Shobha spent 13 times of 10 hours watching television.
13×10=103 or 313
Shobha spent 103 hours or 313 hours (which is 3 hours 20 minutes) watching television.
What fraction of a day is passed in 6 hours? Express the fraction in the simplest form. [3 MARKS]
#### SOLUTION
Solution :
Steps: 1 Mark
Fraction: 1 Mark
There are 24 hours in a day.
6 hours of a day
=624
=14 (expressed in the simplest form)
624 and 14 are equivalent fractions.
6 hours of a day represents 14th of a day.
Out of the following fractions which is the smallest and which the greatest? [3 MARKS]
12,34,45,32,63
#### SOLUTION
Solution :
Smallest Fraction: 1 Mark
Greatest Fraction: 1 Mark
Steps: 1 Mark
63=2=21
LCM for the fractions 12,34,45,32,21 is 20
12=1020
34=1520
45=1620
32=3020
21=4020
Now that the denominator is equal, we can compare the fractions. The numerator 40 is the greatest and 10 is the least. Hence, the corresponding original fractions will be the greatest and the least.
Greatest fraction = 63 = 2
Smallest fraction = 12 = 0.5
2 multiplied by 12= ?,
3 multiplied by 14= ?,
7 multiplied by 13= ? and
17 multiplied by 15= ?
Simplify your answer and write it as a proper fraction or as a whole or mixed number. [3 MARKS]
#### SOLUTION
Solution :
Steps: 2 Marks
21×12=1
31×14=34
71×13=73=213
171×15=175=325
Out of the numbers from 0 to 10, what fraction of them are: [4 MARKS]
1) Odd numbers
2) Prime numbers
3) Even numbers
#### SOLUTION
Solution :
Each option: 1 Mark
There are total of 11 numbers from 0 to 10
1) Number of odd numbers (1,3,5,7,9) = 5.
Fraction of odd numbers = 511
2) Number of prime numbers (2,3,5,7) = 4
Fraction of prime numbers 411
3) Number of even numbers (2,4,6,8,10) = 5
Fraction of even numbers 511
(a) Show the fractions 44,24,34 on a number line and simplify.
(b) A 9.85m long rope is divided into 5 equal parts. Find the length of each part. [4 MARKS]
#### SOLUTION
Solution :
Each option: 2 Marks
(a) 44=1
24=12
34 is in its simplest form
Representation of the fractions on the number line is shown below
(b) Length of rope = 9.85m
After dividing the rope into 5 equal parts each part will be 15th of the total part.
15×9.85=9.855=1.97m
(a) Nisha had 55 balloons; she gave 46 balloons to Astha. What fraction of the balloons is left with Nisha? If Ashta loses 6 ballons what fraction of balloons in left with her?
(b)
Sameera purchased 312 kg apples and 434 kg oranges. What is the total weight of fruits purchased by her?
(c) Solve : 223+312 [4 MARKS]
#### SOLUTION
Solution :
(a) Concept: 1 Mark
Solution: 1 Mark
(b) Solution: 1 Mark
(c) Solution: 1 Mark
(a) Total no of balloons = 55
Balloons left with Nisha = 55 - 46 = 9
Fraction of balloon left = 955
Balloons left with Nisha = 46 - 6 = 40
Fraction of balloon left with Nisha = 4046=2023
(b)The total weight of the fruits = 312+434
= 72+194
144+194
334
814
(c) 223+312=83+72
=8×2+7×36
=16+216
=376
=616
(a) Aashu's elder sister divided a watermelon into 6 equal parts and Aarush eats 4 of them. What fraction of watermelon has he eaten hand what fraction is remaining? What is the name given to the fractions obtained?
(b)
Suman studies for 523 hours daily. She devotes 245 hours of her time for Science and Mathematics. How much time does she devote for other subjects? [4 MARKS]
#### SOLUTION
Solution :
(a) Solution: 1 Mark
Name: 1 Mark
(b) Concept: 1 Mark
Steps: 1 Mark
(a) No of parts Arush has eaten = 4
Total no of parts = 6
So fraction of parts he has eaten = 46=23
Fraction of watermelon remaining = 123=323=13
The fractions 23 and 13 are called proper fractions.
(b) Total time of Suman’s study = 523=173 hours
Time devoted by her for Science and Mathematics = 245=145 hours
Thus, time devoted by her for other subjects = 173145
= 17×51514×315
4315
= 21315
|
# Blog
## Magic calculator
By
,
Are you a magician? Or a mathematician! Guess someone’s number with this cool trick.
### What to do
1. Cut out the six rectangles on the magic calculator sheet.
2. Ask your partner to think of a number between 1 and 63.
3. Hand the six rectangles to your partner. Ask them to look for their number and separate the cards that have their number on from the ones that don’t.
4. When your partner is finished, pick up all the cards that have their number on them.
5. Take the first number on each of those cards and add them together. This sum will be their number!
### What’s happening?
This trick might look like magic. But it’s actually a maths trick! It’s also an introduction to the way that computers count.
There’s more than one way to write numbers. The normal way that we use is called decimal. Computers use a different way of writing numbers, called binary. Binary only has two digits: 1 and 0. In a computer, these two digits can correspond to ‘on’ and ‘off’ for electrical switches.
Binary numbers can do everything that regular, decimal numbers can do, but there are some differences. Since there are only two digits in binary, numbers have to be a lot longer. For example, 78 in decimal is written as 1001110 in binary.
Decimal is based around the number ten. As a result, we have a ones column, a tens column, a hundreds column and so on. Each column is ten times the value of the previous column.
Binary is based around the number two. It has a ones column, a twos column, a fours column, an eights column and so on. Each column is double the value of the previous column.
For example, the number 1001 has a 1 in the ones column and a 1 in the eights column. Its value is therefore one plus eight, or nine.
Each of the cards in this trick is focused on one digit of a number’s binary expression. If someone hands you the one and eight cards, then you know their number in binary has 1s in the ones and eights column, and 0s everywhere else. To work out the value of their number, you just have to add them up!
|
Wow, we received loads of solutions here! Let's have a look at a few of them:
#### Charlie's Squares
William and Chris, from Croftlands Junior School, tried to approach Charlie's Squares this way:
We found that the best way to organise our information was to draw a table. Here is the table we drew:
Square x-coordinate y-coordinate 1 2 2 2 5 3 3 8 4 4 11 5 5 14 6
Using the table, we found out that the x-coordinate was going up 3 every time. We added a couple of extra examples in the table. We spotted that the y-coordinate was the number of the square plus 1. Then we tried spotting patterns by working with the y coordinates to find the x coordinates.
A great way to start! They then went on to find the general formula correctly. Callum, Elys, Cerys, Elgan, Cullen, Ethan, Ifan and Twm, from Ysgol Llanegryn, jumped straight in with the following:
We first found a pattern: (2,2), (5,3), (8,4), and we then discussed how to find any centre square. We turned to algebra. The nth term in the pattern is (3n -1, n + 1) so the coordinates of the 20th centre point would be (60-1, 20+1) which is (59,21).
Justin, from the John of Gaunt School, correctly noted:
In the case of going to the left, the process is repeated but instead of increasing the horizontal and vertical coordinates, the coordinates decrease by the same amounts.
Sam, from Fern Avenue, gave the following answer to Alison's Triangles:
I started by comparing triangles 1 and 3, and realised that 'middle' vertices on nearby odd triangles were exactly 8 points horizontally apart. That, with the fact that all odd middle vertices have coordinates of the form (x,10), allows you to find the middle vertex of any odd numbered triangle.
The same applies for all even numbered triangles, except that the vertices are on coordinates that are 4 grid squares to the right of the odd triangles.
After noticing this, I realised that the even isosceles triangles continued to the middle vertex of the odd numbered triangles, and vice-versa. This meant that they either went down or up 5 squares, and right 2 squares.
I now realised that if I knew the coordinates of any triangle, then I knew the middle vertex of the next triangle.
Therefore, triangle 23 will have vertices at (88,5), (90,10), and (92,5).
Joe and Jack, from Springfield Primary School, gave us their formula for the x-coordinate of the top/bottom vertex of the nth triangle, which was 4n-2. From this they plugged in n = 23, which gave them the correct x-coordinate, and noticed that the y-coordinate alternates between 10 (when n was odd) and 0 (when n was even). Great!
#### More Squares from Charlie
Penny, Jacob, Patrick and Cameron, from Inter Lakes, submitted solutions to More Squares from Charlie. Cameron wrote:
The coordinates of the centre of square 22b are (44,-36). To work this out, first I made a chart for the x-coordinates and the y-coordinates. To get from one 'b' square to the next, add 2 to the x-coordinate and subtract 2 from the y-coordinate. My quick and efficient strategy is knowing the x and y axis rules.
Brilliant!
#### A recap of rules for the nth term
Like some of the students featured above, Ivan Ivanov from the 47th High School in Sofia, Bulgaria put his reasoning into algebraic form. He found formulas for the coordinates of the shapes using the nice notation below. Well done to everyone who found similar formulae.
#### Charlie's Squares
1. If $C(n)$ is the centre of square $n$,
then the coordinates of $C(n)$ satisfy the equations: $x(n) = 3n - 1$, and $y(n) = n + 1$.
2. If $L(n)$ is the bottom left hand vertex of square $n$,
then the coordinates of $L(n)$ satisfy the equations: $x(n) = 3n - 2$, and $y(n) = n - 1$.
These formulas both give the correct coordinates even when Charlie goes left with values $n = ....-2, -1, 0, 1, 2, 3....$.
#### Alison's Triangles
1. If $C(n)$ is the top or bottom vertex of triangle $n$ (for odd and even $n$ respectively), then the coordinates of $C(n)$ satisfy the equations: $x(n) = 4n - 2$, $y(n) = 10$ when $n$ is odd and $y(n) = 0$ when $n$ is even.
2. If $L(n)$ is the left-most vertex of triangle $n$, then the coordinates of $L(n)$ satisfy the equations: $x(n) = 4n - 4$, $y(n) = 5$.
3. The right-most vertex of triangle $n$ is the same as the left-most vertex of triangle $n-1$.
Again, these formulas give the correct coordinates when Alison works left with values $n = ....-2, -1, 0, 1, 2, 3....$.
#### More Squares from Charlie
If $B(n)$ is the centre of square $nb$, then the coordinates of $B(n)$ satisfy the equations: $x(n) = 2n$, and $y(n)B = -2n + 8$.
You might like to think if you can find a formula for squares such as 19c or 199a. What about if Charlie continues the pattern and draws squares such as 1d and 1e, etc?
Michael Sena from NSBH sent in a little computer program that could give the coordinates of any of Charlie's first set of squares, even producing a table of the first few. Well done!
Thanks again to everyone for their solutions!
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Mathematics » Rational Expressions and Equations » Solve Rational Equations
Solving a Rational Equation For a Specific Variable
Solving a Rational Equation For a Specific Variable
When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.
We’ll start with a formula relating distance, rate, and time. We have used it many times before, but not usually in this form.
Example
Solve: $$\frac{D}{T}=R\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}T.$$
Solution
Note any value of the variable that would make any denominator zero. Clear the fractions by multiplying both sides ofthe equations by the LCD, T. Simplify. Divide both sides by R to isolate T. Simplify.
The example below uses the formula for slope that we used to get the point-slope form of an equation of a line.
Example
Solve: $$m=\frac{x-2}{y-3}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}y.$$
Solution
Note any value of the variable that would make any denominator zero. Clear the fractions by multiplying both sides ofthe equations by the LCD, $$y-3$$. Simplify. Isolate the term with y. Divide both sides by m to isolate y. Simplify.
Be sure to follow all the steps in the example below. It may look like a very simple formula, but we cannot solve it instantly for either denominator.
Example
Solve $$\frac{1}{c}+\frac{1}{m}=1\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}c.$$
Solution
Note any value of the variable that would make any denominator zero. Clear the fractions by multiplying both sides ofthe equations by the LCD, $$cm$$. Distribute. Simplify. Collect the terms with c to the right. Factor the expression on the right. To isolate c, divide both sides by $$m-1$$. Simplify by removing common factors.
Notice that even though we excluded $$c=0\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}m=0$$ from the original equation, we must also now state that $$m\ne 1$$.
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# Quentin Tarantino And The Stars (02/14/2020)
How will Quentin Tarantino do on 02/14/2020 and the days ahead? Let’s use astrology to complete a simple analysis. Note this is just for fun – take it with a grain of salt. I will first find the destiny number for Quentin Tarantino, and then something similar to the life path number, which we will calculate for today (02/14/2020). By comparing the difference of these two numbers, we may have an indication of how well their day will go, at least according to some astrology practitioners.
PATH NUMBER FOR 02/14/2020: We will consider the month (02), the day (14) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. This is how it’s calculated. First, for the month, we take the current month of 02 and add the digits together: 0 + 2 = 2 (super simple). Then do the day: from 14 we do 1 + 4 = 5. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 2 + 5 + 4 = 11. This still isn’t a single-digit number, so we will add its digits together again: 1 + 1 = 2. Now we have a single-digit number: 2 is the path number for 02/14/2020.
DESTINY NUMBER FOR Quentin Tarantino: The destiny number will calculate the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Quentin Tarantino we have the letters Q (8), u (3), e (5), n (5), t (2), i (9), n (5), T (2), a (1), r (9), a (1), n (5), t (2), i (9), n (5) and o (6). Adding all of that up (yes, this can get tedious) gives 77. This still isn’t a single-digit number, so we will add its digits together again: 7 + 7 = 14. This still isn’t a single-digit number, so we will add its digits together again: 1 + 4 = 5. Now we have a single-digit number: 5 is the destiny number for Quentin Tarantino.
CONCLUSION: The difference between the path number for today (2) and destiny number for Quentin Tarantino (5) is 3. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is of questionable accuracy. If you want to see something that people really do vouch for, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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# Chapter 3:Stresses And Strains¶
## Problem 3.1,Page no.54¶
In [65]:
import math
#Initilization of Variables
P=40 #mm #Force applied to stretch a tape
L=30 #m #Length of steel tape
A=6*1 #mm #Cross section area
E=200*10**9*10**-6 #KN/m**2 #Modulus of Elasticity
#Calculations
sigma_L=(P*L*10**3)*(A*E)**-1 #mm
#Result
print"The Elongation of steel tape is",round(sigma_L,4),"mm"
The Elongation of steel tape is 1.0 mm
## Problem 3.2,Page no.54¶
In [4]:
import math
#Initilization of VAriables
#D=(D_0-2) #cm #Inside Diameter of cyclinder
#A=(pi*(D_0-1)) #cm**2 #Area of cross-section
#L=(pi*(D_0-1)*5400) #N #Crushing load for column
F=6 #Factor of safety
T=1 #cm #wall thickness of cyclinder
#S=L*F**-1
#After Simplifying,we get
S=600*10**3
#Calculations
D_0=(S*F)*(pi*54000)**-1+1 #cm #Outside diameter of cyclinder
#Result
print"The outside Diameter of cyclinder is",round(D_0,2),"cm"
The outside Diameter of cyclinder is 22.22 cm
## Problem 3.3,Page no.56¶
In [25]:
import math
#Initilization of variables
P=800 #N #force applied to steel wire
L=150 #m #Length of steel wire
E=200 #GN/m**2 #Modulus of Elasticity
d=10 #mm #Diameter of steel wire
W=7.8*10**4 #N/m**3 #Weight Density of steel
#A=(pi*4**-1)*(d)**2 #m**2
#After simplifying Area,we get
A=7.85*10**-5 #m**2
#calculation (Part-1)
dell_L_1=(P*L*10**-3)*(A*E*10**9*10**-6)**-1 #mm
#calculation (Part-2)
#Elongation due to Weight of wire
dell_L_2=((pi*4**-1)*150*W*L*10**-3)*(2*A*E*10**7)**-1 #mm
#calculation (Part-3)
#Total Elongation of wire
dell_L_3=dell_L_1+dell_L_2
#Result
print"The Elongation due to 800N Load is",round(dell_L_1,2),"mm"
print"The Elongation due to Weight of wire is",round(dell_L_2,2),"mm"
print"Total Elongation of wire is",round(dell_L_3,2),"mm"
The Elongation due to 800N Load is 7.64 mm
The Elongation due to Weight of wire is 4.39 mm
Total Elongation of wire is 12.03 mm
## Problem 3.4,Page no.55¶
In [29]:
import math
#Intilization of variables
d=10 #mm #Diameter of Punching Hole
t=4 #mm #Thickness of Mild Steel Plate
tou=320 #N/mm**2 #Shear Strength of mild Steel
#Calculations
#Force Required for punching the hole
P=tou*pi*d*t #N
#Area of punch in contact with the plate surface
A=(pi*4**-1*d**2) #mm*2
#Compressive stress
sigma_c=P*A**-1 #N/mm*2
#Result
print"Force Required for punching the hole is",round(P,2),"N"
print"Compressive stress is",sigma_c,"N/mm*2"
Force Required for punching the hole is 40212.39 N
Compressive stress is 512.0 N/mm*2
## Problem 3.6,Page no.57¶
In [67]:
import math
#Initilization of variables
P=200*10**3 #N
L_1=0.10 #mm #Length of of portin AB
L_2=0.16 #mm #Length of of portin BC
L_3=0.12 #mm #Length of of portin CD
E=200*10**9 #N
d_1=0.1 #cm
d_2=0.08 #cm
d_3=0.06 #cm
A_1=(pi*4**-1)*(0.1)**2 #mm**2
A_2=(pi*4**-1)*(0.08)**2 #mm**2
A_3=(pi*4**-1)*(0.06)**2 #mm**2
#Calculations
dell_L_1=(P*L_1*10**3)*(A_1*E)**-1 #mm
dell_L_2=(P*L_2*10**3)*(A_2*E)**-1 #mm
dell_L_3=(P*L_3*10**3)*(A_3*E)**-1 #mm
dell_L=dell_L_1+dell_L_2+dell_L_3 #mm
#Result
print"Total Elongation of steel bar is",round(dell_L,3),"mm"
Total Elongation of steel bar is 0.087 mm
## Problem 3.7,Page no.57¶
In [44]:
import math
#Initilization of variables
#from F.B.D,we get
P_1=50 #KN
P_2=20 #KN
P_3=40 #KN
d=0.02 #mm #Diameter of steel bar
L_1=0.4 #mm
L_2=0.3 #mm
L_3=0.2 #mm
E=210*10**9 #N
#After simplifying Area,we get
A=pi*10**-4 #m**2 #Area of cross section
#Calculations
sigma_AB=P_1*1000*A #N/m**2
sigma_BA=P_2*1000*A #N/m**2
sigma_CD=P_3*1000*A #N/m**2
dell_L=((P_1*L_1+P_2*L_2+P_3*L_3)*(A*E)**-1)*10**6 #mm
#Result
print"Total Elongation of Steel bar is",round(dell_L,3),"mm"
Total Elongation of Steel bar is 0.515 mm
## Problem 3.8,Page no.58¶
In [26]:
import math
import numpy as np
#Initilization of variables
#R_a+R_c=25 #KN #R_a,R_b are reactions at supports A and C respectively
L_ab=2 #m
L_bc=3 #m
#Calculation
#From F.B.D,we get
#dell_L_AB=(R_a*L_AB)*(A*E)**-1 #Elongation of portion AB
#dell_L_BC=(R_c*L_BC)*(A*E)**-1 #Compression of portion BC
#After simplifying above equations we get,
#R_a=(1.5)*R_c #KN
#R_a+R_c=25 #KN
#Solving the above simultaneous equations using matrix method
A=np.array([[1,-1.5],[1,1]]) #Here the coefficients of the first equations of unknowns are setup
B=np.array([0,25]) #Here the RHS of both equations are setup
C=np.linalg.solve(A,B)
#print C[0] #Prints the first element in the vector C
#print C[1] #Prints the second element in the vector C
#Result
print"The reaction at support A is",round(C[0],2),"KN"
print"The reaction at support C is",round(C[1],2),"KN"
The reaction at support A is 15.0 KN
The reaction at support C is 10.0 KN
## Problem 3.9,Page no.59¶
In [17]:
import math
#Initilization of variables
#P is the force acting on the bar BC compressive in nature and force on AB is (100-P) Tensile in nature
E=200*10*9 #N
A_1=3*10**-4 #cm**2 #Area of AB
A_2=4*10**-4 #cm**2 #Area of BC
L=1.5 #cm #Length of bar
#Calculations
#The total elongation of bar
#(((100-P)*10**3*1.5)*(3*10**-4*E)**-1)-((P*10**3*1.5)*(4*10**-4*E)**-1)=0
#The total elongation of bar is limited to 1
#(25-0.4375*P)*10**-4=1*10**-3
#After simplifying above equation we get,
P=-(10-25)*0.4375**-1 #KN #Total elongation of bar
F_AB=100-P #KN #force in AB
F_BC=P #KN #Force in BC
sigma_AB=(((F_AB)*(3*10**-4)**-1)*10**-3) #KN #Stress in AB
sigma_BC=((F_BC)*(4*10**-4)**-1*10**-3) #KN #Stress in Bc
#Result
print"F_AB",round(F_AB,2),"KN"
print"F_BC",round(F_BC,2),"KN"
print"sigma_AB",round(sigma_AB,2),"KN"
print"sigma_BC",round(sigma_BC,2),"KN"
F_AB 65.71 KN
F_BC 34.29 KN
sigma_AB 219.05 KN
sigma_BC 85.71 KN
## Problem 3.12,Page no.61¶
In [11]:
import math
#Initilization of variables
d=20 #mm #steel rod diameter
n=4 #number of steel rod
sigma_c=4 #N/mm**2 #stress in concrete
#E_S*E_c**-1=15
#Calculations
A_s=4*pi*4**-1*d**2 #mm**2 #Area os steel rod
sigma_s=15*sigma_c #N/mm**2 #stress in steel
#P=sigma_s*A_s+sigma_c*A_c
#After substituting and simplifying above equation we get,
A_c=(P*10**3-sigma_s*1256)*(sigma_c)**-1 #mm**2 #Area of the concrete
X=(A_s+A_c)**0.5 #mm #Total cross sectional area
P_s=A_s*sigma_s*10**-3 #KN #Load carried by steel
#Result
print"stress induced in steel is",round(sigma_s,3),"KN"
print"cross sectional area of column is",round(X,2),"mm"
Load carried by steel is 75.4 KN
stress induced in steel is 60.0 KN
cross sectional area of column is 327.74 mm
## Problem 3.13,Page no.62¶
In [1]:
import math
#Initilization of variables
A_s=500 #mm**2
E_s=200000
E_al=80000
A_al=1000
#Calculations
#(P_al*L_al)*(A_al*E_al)**-1+(P_s*L_s)*(A_s*E_s)**-1=1*2**-1
P=1*1000**-1*((A_s*E_s*A_al*E_al)*(A_s*E_s+A_al*E_al)**-1) #N
P_s=P_al=P #N
sigma_t=P_s*A_s**-1 #N/mm**2 #Tensile stress in bolt
sigma_c=P_al*A_al**-1 #N/mm**2 #Compressive stress in Aluminium tube
#result
print"Tensile stress in bolt is",round(sigma_t,2),"N/mm**2"
print"Compressive stress in Aluminium tube is",round(sigma_c,2),"N/mm**2"
Tensile stress in bolt is 88.89 N/mm**2
Compressive stress in Aluminium tube is 44.44 N/mm**2
## Problem 3.14,Page no.63¶
In [1]:
import math
#Initilization of variables
A=1600 #mm**2 #Area of the Bar
dell_L=0.4 #mm #Contraction of metal bar
L=200 #mm #Length of metal bar
sigma_t=0.04 #mm #Guage Length
t=40
#Calculations
sigma_L=dell_L*L**-1
E=((P*L)*(A*dell_L)**-1*10**-3) #N/mm**2 #Young's Modulus
m=t*sigma_t**-1*sigma_L
#Result
print"The value of Young's Modulus is",round(E,2),"N/mm*2"
print"The value of Poissoin's ratio is",round(m,2)
The value of Young's Modulus is 150.0 N/mm*2
The value of Poissoin's ratio is 2.0
## Problem 3.15,Page no.63¶
In [21]:
import math
#Initilization of variables
A_s=0.003848 #m**2 #Area of steel bar
A_al=0.003436 #m**2 #Area of Aluminium tube
E=220*10*9 #N #Young's modulus of steel
E=70*10*9 #N #Young's modulus of aluminium
P=600*10**3 #N #Load applied to the bar
#dell_L_al-dell_L_s=0.00015 #mm #difference between strain in aluminium bar and steel bar
#Calculations
#Let the aluminium tube be compressed by dell_L_al and steel bar by by dellL_s
#dell_L_al=sigma_al*E_al**-1*L_al
#dell_L_s=sigma_s*E_s**-1*L_s
#After substituting and simplifying above equation we get,
#((sigma_al*70**-1)-(sigma_s*220**-1))=300000 #(equation 1)
#After simplifying above equation we get,
#sigma_al=17462.165*10**4-1.1199*sigma_s #(equation 2)
#Now substituting sigma_al in equation(1)
#((17462.165*10**4-1.1199*sigma_s)*(70)**-1)-(sigma_s*220**-1)=300000
#After simplifying above equation we get,
sigma_s=-((300000-249.4594*10**4)*0.0205444**-1)*10**-6 #MN/m**2 #stress developed in steel bar
#sigma_al=17462.165*10**4-1.1199*sigma_s
sigma_al=(17462.165*10**4-1.1199*106822005.02)*10**-6
#Result
print"stress developed in steel bar is",round(sigma_s,2),"MN/m**2"
print"stress developed in aluminium bar is",round(sigma_al,2),"MN/m**2"
stress developed in steel bar is 106.82 MN/m**2
stress developed in aluminium bar is 54.99 MN/m**2
## Problem 3.16,Page no.64¶
In [43]:
import math
#Initilization of variables
E=200 #GN/m**2 #Modulus of elasticity
alpha=11*10**-6 #per degree celsius #coeffecient o flinear expansion of steel bar
L=6 #m #Length of rod
#Calculations
#(Part-1) #IF the walls do not yield
t=58 #degree celsius #Fall in temperature #(t=80-22)
dell=alpha*t #strain
sigma=E*10**9*dell*10**-6 #MN/m**2 #Stress
A=pi*4**-1*6.25*10**-4 #mm**2 #Area of wall and rod
P=sigma*10**6*A*10**-3 #KN #Pull Exerted
#(Part-2) #IF the walls yield together at the two ends is 1.15 mm
L_2=L*(1-alpha*t) #m #Length of rod at 22 degree celsius
L_3=L-L_2 #m #Decrease in Length
#As the walls yield by 1.5 mm, actual decrease in length is
L_4=L_3-0.0015 #m
dell_2=L_4*L**-1 #strain
P_2=E*10**9*dell_2*A*10**-3 #KN
#Result
print"Pull Exerted by the rod:when walls do not yield",round(P,2),"KN"
print" :when total yield together at two ends is 1.5 mm",round(P_2,2),"KN"
Pull Exerted by the rod:when walls do not yield 62.64 KN
:when total yield together at two ends is 1.5 mm 38.09 KN
## Problem 3.17,Page no.65¶
In [54]:
import math
#Initilization of variables
D=4.5 #cm #External Diameter of tube
d=3 #cm #Internal diameter of tube
t=3 #mm #thickness of tube
t_1=30 #degree celsius
t_2=180 #degree celsius #when metal heated
L=30 #cm #Original LEngth
alpha_s=1.08*10**-5 #Per degree celsius #coefficient of Linear expansion of steel tube
alpha_c=1.7*10**-5 #Per degree celsius #coefficient of Linear expansion of copper tube
E_s=210 #GPa #Modulus of Elasticity of steel
E_c=110 #GPA #Modulus of Elasticity of copper
#Calculation
#For Equilibrium of the system, Total tension in steel=Total tension in copper
#sigma_s*A_s=sigma_c*A_c (equation 1)
A_c=pi*4**-1*d**2 #cm**2 #Area of copper
A_s=pi*4**-1*(D**2-d**2) #cm**2 #Area of steel
#simplifying equation 1
#sigma_s=1.785*sigma_c
T=t_2-t_1 #change in temperature
#Actual expansion of steel=Actual expansion of copper
#alpha_s*T*L+sigma_s*E_s**-1*L=alpha_c*T*L-sigma_c*E_c**-1*L
#After substituting values in above equation and simplifying we get
sigma_c=(930*10**5*1.7591**-1)*10**-6 #MN/m**2 #Stress in copper
sigma_s=1.785*sigma_c #MN/m**2 #Stress in steel
#Increase in Length of either component
L_2=(alpha_s*T+sigma_s*10**6*(E_s*10**9)**-1)*L
#Result
print"stress in copper bar is",round(sigma_c,2),"MN/m**2"
print"stress in steel bar is",round(sigma_s,2),"MN/m**2"
print"Increase in Length is",round(L_2,3),"cm"
stress in copper bar is 52.87 MN/m**2
stress in steel bar is 94.37 MN/m**2
Increase in Length is 0.062 cm
## Problem 3.18,Page no.66¶
In [60]:
import math
#Initilization of variables
t_1=15 #degree celsius #temperature of steel bar
t_2=315 #degree celsius #raised temperature
E_s=210 #GPa #Modulus of Elasticity of steel bar
E_c=100 #GPa #Modulus of Elasticity of copper bar
dell_L=0.15 #cm #Increase in Length of bar
#Calculation
#For Equilibrium of the system, Tension in steel bar = Tension in copper bar
#sigma_s*A_s = sigma_c*2*A_c
#sigma_S=2*sigma_c
#Actual expansion of steel = Actual expansion of copper
#L*alpha_s*T+sigma_s*E_s**-1*L = L*alpha_c*T-sigma_c*E_c**-1*L (Equation 1)
T=t_2-t_1 #per degree celsius #change in temperature
#After substituting values in above equation and simplifying we get
sigma_c=(1650*10**5*1.9524**-1)*10**-6 #MN/m**2 #Stress in copper
sigma_s=2*sigma_c #MN/m**2 #Stress in steel
#Actual Expansion of steel bar
#L*alpha_s*T+sigma_s*E_s**-1*L = L*alpha_c*T-sigma_c*E_c**-1*L
#After substituting values in above equation and simplifying we get
L=0.15*10**-2*0.0044048**-1 #m
#Result
print"Stress in copper bar is",round(sigma_c,2),"MN/m**2"
print"Stress in steel bar is",round(sigma_s,2),"MN/m**2"
print"Original Length of bar is",round(L,2),"m"
Stress in copper bar is 84.51 MN/m**2
Stress in steel bar is 169.02 MN/m**2
Original Length of bar is 0.34 m
## Problem 3.19,Page no.67¶
In [64]:
import math
#Initilization of variables
L=100 #m #Length of rod
A=2 #cm**2 #cross sectional area
rho=80 #KN/m**3
#Calculatiom
W=A*10**-4*L*rho #KN
sigma_s=10+1.6 #KN #Rod experiencing max tensile stress when it is at top performing upstroke
sigma_s_2=sigma_s*10**3*200**-1 #N/mm**2 #corresponding stress at this moment
sigma_c=1 #KN ##Rod experiencing max compressive stress at its lower end,free from its own weight
sigma_c_2=sigma_c*10**3*200**-1 #corresponding stress at this moment
#Result
print"Tensile stress in bar",round(sigma_s_2,2),"N/mm**2"
print"Compressive stress in bar",round(sigma_c_2,2),"N/mm**2"
Tensile stress in bar 58.0 N/mm**2
Compressive stress in bar 5.0 N/mm**2
## Problem 3.20,Page no.68¶
In [6]:
import math
#Initilization of variables
sigma=0.012 #strain
P=150 #KN #Total Load on the Post
E=1.4*10**4 #N/mm**2 #modulus of elasticity
#b be the width of the post in mm
#2b is the longer dimension of the post in mm
#Calculations
#We know,
#sigma=(P*(A*E)**-1)
#After substituting values and simplifying, we get
b=((150*10**3)*(0.012*1.4*2*10**4)**-1)**0.5
q=2*b #mm #Longer dimension of post
#Result
print"Width of post is",round(b,2),"mm"
print"Longer dimension of post is",round(q,2),"mm"
Width of post is 21.13 mm
Longer dimension of post is 42.26 mm
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# SAT II Math II : Circles, Ellipses, and Hyperbolas
## Example Questions
### Example Question #1 : Circles, Ellipses, And Hyperbolas
Refer to the above figure. The circle has its center at the origin. What is the equation of the circle?
Explanation:
The equation of a circle with center and radius is
The center is at the origin, or , so . To find , use the distance formula as follows:
Note that we do not actually need to find
We can now write the equation of the circle:
### Example Question #1 : Circles, Ellipses, And Hyperbolas
Refer to the above diagram. The circle has its center at the origin; is the point . What is the length of the arc , to the nearest tenth?
Explanation:
First, it is necessary to determine the radius of the circle. This is the distance between and , so we apply the distance formula:
The circumference of the circle is
Now we need to find the degree measure of the arc. We can do this best by examining this diagram:
The degree measure of is also the measure of the central angle formed by the green radii. This is found using the relationship
Using a calculator, we find that . We can adjust for the location of :
which is the degree measure of the arc.
Now we can calculate the length of the arc:
### Example Question #1 : Circles, Ellipses, And Hyperbolas
On the coordinate plane, the vertices of a square are at the points with coordinates . Give the equation of the circle.
Explanation:
The figure in question is below.
The center of the circle can be seen to be the origin, so, if the radius is , the equation will be
.
The circle passes through the midpoints of the sides, so we will find one of these midpoints. The midpoint of the segment with endpoints and can be found by using the midpoint equations, setting :
The circle passed though this midpoint . The segment from this point to the origin is a radius, and its length is equal to . Using the following form of the distance formula, since we only need the square of the radius:
,
set :
Substituting in the circle equation for , we get the correct response,
### Example Question #1 : Coordinate Geometry
Find the diameter of the circle with the equation .
Explanation:
Start by putting the equation in the standard form of the equation of a circle by completing the square. Recall the standard form of the equation of a circle:
, where the center of the circle is at and the radius is .
From the equation, we know that .
Since the radius is , double its length to find the length of the diameter. The length of the diameter is .
### Example Question #1 : Coordinate Geometry
A triangle has its vertices at the points with coordinates , and . Give the equation of the circle that circumscribes it.
None of these
Explanation:
The circumscribed circle of a triangle is the circle which passes through all three vertices of the triangle.
In general form, the equation of a circle is
.
Since the circle passes through the origin, substitute ; the equation becomes
Therefore, we know the equation of any circle passing through the origin takes the form
for some .
Since the circle passes through , substitute ; the equation becomes
Solving for :
Now we know that the equation takes the form
for some .
Since the circle passes through , substitute ; the equation becomes
Solving for :
The general form of the equation of the circle is therefore
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Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
December 10th
December 10th
# Solving Inequalities with Fractions and Parentheses
After studying this lesson, you will be able to:
• Solve inequalities with fractions and parentheses.
Below are the steps for solving inequalities. Remember we are applying the same rules as we did for equations. If an inequality contains fractions, the fractions can be cleared out by multiplying every term in the inequality by the common denominator. Also, if an inequality contains parentheses, the parentheses can be removed by using the distributive property.
If we multiply or divide an inequality by a negative, we reverse the inequality symbol.
The steps for solving inequalities are the same as those for solving equations:
1. Remove parentheses and clear fractions (if necessary)
2. Collect like terms on each side of the inequality symbol
3. Get the variables together on one side
4. Isolate the variable
5. Check
Example 1
We have a fraction. To clear it, multiply by the common denominator which is 13 Multiply each side by the common denominator x > -78
Check by substituting into the original inequality
Example 2
We have a fraction. To clear it, multiply by the common denominator which is -4 Multiply each side by the common denominator (remember, to reverse the inequality symbol since we're multiplying by a negative) -40
Check by substituting into the original inequality
Example 3
We have a fraction. To clear it, multiply by the common denominator which is 36 Multiply each side by the common denominator 4 (-4) < -5r (3) Reduce 36 and 9 on the left side to get 4 and reduce 12 and 36 on the right side to get 3. Then, do the multiplying. -16 < -15r Divide each side by -15 (remember to reverse the symbol)
Check by substituting into the original inequality
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# Lesson 5: Representing Subtraction
Let's subtract signed numbers.
## 5.1: Equivalent Equations
For the equations in the second and third columns, write two more equations using the same numbers that express the same relationship in a different way. If you get stuck, consider looking at the examples in the first column.
$2+ 3= 5$
$3 + 2 = 5$
$5 - 3 = 2$
$5 - 2 = 3$
$9+ (\text- 1)= 8$
$\text- 11+ x= 7$
## 5.2: Subtraction with Number Lines
1. Here is an unfinished number line diagram that represents a sum of 8.
1. How long should the other arrow be?
2. For an equation that goes with this diagram, Mai writes $3 + {?} = 8$.
Tyler writes $8 - 3 = {?}$. Do you agree with either of them?
3. What is the unknown number? How do you know?
2. Here are two more unfinished diagrams that represent sums.
For each diagram:
1. What equation would Mai write if she used the same reasoning as before?
2. What equation would Tyler write if he used the same reasoning as before?
3. How long should the other arrow be?
4. What number would complete each equation? Be prepared to explain your reasoning.
3. Draw a number line diagram for $(\text-8) - (\text-3) = {?}$ What is the unknown number? How do you know?
1. Match each diagram to one of these expressions:
$3 + 7$
$3 - 7$
$3 + (\text- 7)$
$3 - (\text- 7)$
2. Which expressions in the first question have the same value? What do you notice?
3. Complete each of these tables. What do you notice?
expression value
row 1 $8 + (\text- 8)$
row 2 $8 - 8$
row 3 $8 + (\text-5)$
row 4 $8 - 5$
row 5 $8 + (\text-12)$
row 6 $8 - 12$
expression value
row 1 $\text-5 + 5$
row 2 $\text-5 - (\text-5)$
row 3 $\text-5 + 9$
row 4 $\text-5 - (\text-9)$
row 5 $\text-5 + 2$
row 6 $\text-5 - (\text-2)$
## Summary
The equation $7 - 5 = {?}$ is equivalent to ${?} + 5= 7$. The diagram illustrates the second equation.
Notice that the value of $7 + (\text-5)$ is 2.
We can solve the equation ${?} + 5= 7$ by adding -5 to both sides. This shows that $7 - 5= 7 + (\text- 5)$
Likewise, $3 - 5 = {?}$ is equivalent to ${?} + 5= 3$.
Notice that the value of $3 + (\text-5)$ is -2.
We can solve the equation ${?} + 5= 3$ by adding -5 to both sides. This shows that $3 - 5 = 3 + (\text- 5)$
In general:
$$a - b = a + (\text- b)$$
If $a - b = x$, then $x + b = a$. We can add $\text- b$ to both sides of this second equation to get that $x = a + (\text- b)$
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# Let $a,b$ and $c$ be three real numbers satisfying $\left[ {\begin{array}{*{20}{c}} a&b&c \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&9&7 \\ 8&2&7 \\ 7&3&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&0 \end{array}} \right]......\left( E \right)$ Let $b = 6$, with $a$ and $c$ satisfying $\left( E \right).$ If $\alpha$ and $\beta$ are the roots of the quadratic equation $a{x^2} + bx + c = 0$, then $\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}}$ is:(A) 6(B) 7(C) $\dfrac{6}{7}$ (D) $\infty$
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Hint: The multiplication of two matrices is possible if the no. of columns in matrix A is equal to the no. of rows in matrix B. Here we multiplied the two given matrix and form the equations by comparing the values of both sides.
Since, $a,b$ and $c$ be three real numbers satisfies
$\left[ {\begin{array}{*{20}{c}} a&b&c \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&9&7 \\ 8&2&7 \\ 7&3&7 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&0 \end{array}} \right]$
So, we get the equations
$a + 8b + 7c = 0 \\ 9a + 2b + 3c = 0 \\ 7a + 7b + 7c = 0 \Rightarrow a + b + c = 0 \\$
Since, $b = 6$, so the equations become
$a + 8\left( 6 \right) + 7c = 0 \Rightarrow a + 7c = - 48....(1) \\ 9a + 2\left( 6 \right) + 3c = 0 \Rightarrow 9a + 3c = - 12....(2) \\ a + 6 + c = 0 \Rightarrow a + c = - 6....(3) \\$
On subtracting equation (3) from (1), we get
$a + 7c - \left( {a + c} \right) = - 48 - \left( { - 6} \right)$
$\Rightarrow a + 7c - a - c = - 48 + 6 \\ \Rightarrow 6c = - 42 \\ \Rightarrow c = - 7 \\$
Substitute the value of $c$ in equation (3), we get
$a + \left( { - 7} \right) = - 6 \\ \Rightarrow a - 7 = - 6 \\ \Rightarrow a = - 6 + 7 \\ \Rightarrow a = 1 \\$
So, we have $a = 1,b = 6,c = - 7$
Given quadratic equation is $a{x^2} + bx + c = 0$. After putting the values of $a,b$ and $c$, it becomes ${x^2} + 6x - 7 = 0$.
Since, $\alpha$ and $\beta$ are the roots of this equation, So
Sum of roots, $\alpha + \beta = $$\dfrac{{ - b}}{a} = \dfrac{{ - 6}}{1} = - 6 Multiplication of roots, \alpha \beta = \dfrac{c}{a} = \dfrac{{ - 7}}{1} = - 7 Now, \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}} =\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{{\beta + \alpha }}{{\alpha \beta }}} \right)}^n}} =\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{{ - 6}}{{ - 7}}} \right)}^n}} = \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{6}{7}} \right)}^n}} On expand it, we get- = {\left( {\dfrac{6}{7}} \right)^0} + {\left( {\dfrac{6}{7}} \right)^1} + {\left( {\dfrac{6}{7}} \right)^2} + {\left( {\dfrac{6}{7}} \right)^3} + ........................ + {\left( {\dfrac{6}{7}} \right)^n} = 1 + {\left( {\dfrac{6}{7}} \right)^1} + {\left( {\dfrac{6}{7}} \right)^2} + {\left( {\dfrac{6}{7}} \right)^3} + ........................ + {\left( {\dfrac{6}{7}} \right)^n} This is an infinite Geometric Progression, whose sum of infinite terms is given by {S_\infty } = \dfrac{a}{{1 - r}} Where a is the first term of G.P. and r is the common ratio of G.P. Here we have, a = 1and r = \dfrac{6}{7} \therefore \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}}$$ = \dfrac{1}{{1 - \dfrac{6}{7}}}$
$\Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}} = \dfrac{1}{{\dfrac{{7 - 6}}{7}}}$
$\Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}} = \dfrac{1}{{\dfrac{1}{7}}}$
$\Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}} = 7$
Hence, option (B) is the correct answer.
Note: If $\alpha$ and $\beta$ are the roots of the quadratic equation $a{x^2} + bx + c = 0$, then sum of roots, $\alpha + \beta = \dfrac{{ - b}}{a}$ and multiplication of roots, $\alpha \beta = \dfrac{c}{a}$. Also, the sum of infinite terms of an G.P. is ${S_\infty } = \dfrac{a}{{1 - r}}$.
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# Solution
Prove the following statement holds for every positive integer $n$:
$$\displaystyle{\sum_{i=1}^{n} 2i = n^2+n}$$
One Solution:
Let us argue by induction...
Starting with the basis step, we must show that the statement holds for the smallest value of $n$ that we wish to consider.
Since we seek to prove the statement holds for every positive integer, then the smallest value we wish to consider would be $n=1$.
When $n=1$, the left side collapses to a "sum" containing only a single term: $$\sum_{i=1}^n 2i = 2\cdot1 = 2$$
While on the right side, if $n=1$, $$n^2 + n= 1^2 + 1 = 2$$
Since the left and right sides agree in value, the statement is true when $n=1$
Now we proceed with the inductive step, where we must show that if we know the statement holds for some particular value of $n$, say $n=k$, then the statement must also hold when $n=k+1$.
Equivalently for this problem, we need to show if the following "inductive hypothesis" is true $$\sum_{i=1}^k 2i = k^2+k$$ then it must also be true that $$\sum_{i=1}^{k+1} 2i = (k+1)^2 + (k+1)$$
To this end, we will attempt to manipulate the left side of the above equation until it looks like the right side. We can use the inductive hypothesis to provide the transition from an expression involving $\sum$ to a simpler algebraic expression.
$$\begin{array}{rcl} \sum_{i=1}^{k+1} 2i &=& \left( \sum_{i=1}^{k} 2i \right) + 2(k+1)\\\\ &=& k^2 + k + 2(k+1)\\\\ &=& k^2 + k + 2k + 2 \\\\ &=& (k^2 + 2k + 1) + (k + 1)\\\\ &=& (k+1)^2 + (k+1)\\\ \end{array}$$
Having met with success in both the basis and inductive steps, we are free to conclude by the principle of mathematical induction, that the following is true for every positive integer $n$: $$\sum_{i=1}^n 2i= n^2+n$$
QED.
A Second Solution:
If we are allowed to appeal to the fact that $$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$ then we can argue the result directly with the following: $$\begin{array}{rcl} \sum_{i=1}^{k}2i &=& 2 \left[ \sum_{i=1}^n i \right]\\\\ &=& 2 \left[ \frac{n(n+1)}{2} \right]\\\\ &=& n(n+1)\\\\ &=& n^2 + n\\\\ \end{array}$$ QED.
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## Mathematical Induction -- Second Principle
### Subjects to be Learned
• second principle of mathematical induction
### Contents
There is another form of induction over the natural numbers based on the second principle of induction to prove assertions of the form x P(x) . This form of induction does not require the basis step, and in the inductive step P(n) is proved assuming P(k) holds for all k < n . Certain problems can be proven more easily by using the second principle than the first principle because P(k) for all k < n can be used rather than just P(n - 1) to prove P(n).
Formally the second principle of induction states that
if n [ k [ k < n P(k) ] P(n) ] , then n P(n) can be concluded.
Here k [ k < n P(k) ] is the induction hypothesis.
The reason that this principle holds is going to be explained later after a few examples of proof.
Example 1: Let us prove the following equality using the second principle:
For any natural number n , 1 + 3 + ... + ( 2n + 1 ) = ( n + 1 )2.
Proof: Assume that 1 + 3 + ... + ( 2k + 1 ) = ( k + 1 )2 holds for all k, k < n.
Then 1 + 3 + ... + ( 2n + 1 ) = ( 1 + 3 + ... + ( 2n - 1 ) ) + ( 2n + 1 )
= n2 + ( 2n + 1 ) = ( n + 1 )2
by the induction hypothesis.
Hence by the second principle of induction 1 + 3 + ... + ( 2n + 1 ) = ( n + 1 )2 holds for all natural numbers.
Example 2: Prove that for all positive integer n, i ( i! ) = ( n + 1 )! - 1
Proof: Assume that
1 * 1! + 2 * 2! + ... + k * k! = ( k + 1 )! - 1 for all k, k < n.
Then 1 * 1! + 2 * 2! + ... + ( n - 1 ) * ( n - 1 )! + n * n!
= n! - 1 + n * n!
by the induction hypothesis.
= ( n + 1 )n! - 1
Hence by the second principle of induction i ( i! ) = ( n + 1 )! - 1 holds for all positive integers.
Example 3: Prove that any positive integer n, n > 1, can be written as the product of prime numbers.
Proof: Assume that for all positive integers k, n > k > 1, k can be written as the product of prime numbers.
We are going to prove that n can be written as the product of prime numbers.
Since n is an integer, it is either a prime number or not a prime number. If n is a prime number, then it is the product of 1, which is a prime number, and itself. Therefore the statement holds true.
If n is not a prime number, then it is a product of two positive integers, say p and q. Since both p and q are smaller than n, by the induction hypothesis they can be written as the product of prime numbers (Note that this is not possible, or at least very hard, if the First Principle is being used). Hence n can also be written as the product of prime numbers.
### Test your understanding of second principle of induction :
Indicate which of the following statements are correct and which are not.
Click True or False , then Submit. There is one set of questions.
Next -- Introduction to Relation
Back to Schedule
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# RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-4) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8
## Class 8: RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-4) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8
The document RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-4) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8
#### Answer 1: (i) We have:
(ii) We have:
To find the cube root of 5832, we use the method of unit digits.
Let us consider the number 5832.
The unit digit is 2; therefore the unit digit in the cube root of 5832 will be 8.
After striking out the units, tens and hundreds digits of the given number, we are left with 5.
Now, 1 is the largest number whose cube is less than or equal to 5.
Therefore, the tens digit of the cube root of 5832 is 1.
(iii) We have:
To find the cube root of 2744000, we use the method of factorisation.
On factorising 2744000 into prime factors, we get:
2744000=2×2×2×2×2×2×5×5×5×7×7×72744000=2×2×2×2×2×2×5×5×5×7×7×7
On grouping the factors in triples of equal factors, we get:
2744000={2×2×2}×{2×2×2}×{5×5×5}×{7×7×7}2744000=2×2×2×2×2×2×5×5×5×7×7×7
It is evident that the prime factors of 2744000 can be grouped into triples of equal factors and no factor is left over.
Now, collect one factor from each triplet and multiply; we get:
2×2×5×7=1402×2×5×7=140
This implies that 2744000 is a cube of 140.
Hence,
(iv) We have:
To find the cube root of 753571, we use the method of unit digits.
Let us consider the number 753571.
The unit digit is 1; therefore the unit digit in the cube root of 753571 will be 1.
After striking out the units, tens and hundreds digits of the given number, we are left with 753.
Now, 9 is the largest number whose cube is less than or equal to 753 (93<753<10393<753<103).
Therefore, the tens digit of the cube root 753571 is 9.
(v) We have:
To find the cube root of 32768, we use the method of unit digits.
Let us consider the number 32768.
The unit digit is 8; therefore, the unit digit in the cube root of 32768 will be 2.
After striking out the units, tens and hundreds digits of the given number, we are left with 32.
Now, 3 is the largest number whose cube is less than or equal to 32 (33<32<4333<32<43).
Therefore, the tens digit of the cube root 32768 is 3.
#### Question 2: Show that:
(i) LHS = 3×4=12273×643=3×3×33×4×4×43=3×4=12
RHS =
=3×4=1227×643=3×3×3×4×4×43=3×3×3×4×4×43=3×4=12
Because LHS is equal to RHS, the equation is true.
(ii) LHS =
=4×9=3664×7293=4×4×4×9×9×93=4×4×4×9×9×93=4×9=36
RHS = =4×9=36
Because LHS is equal to RHS, the equation is true.
(iii) LHS =
=
=5×2×3=30
RHS =
=5×(2×3)=30
Because LHS is equal to RHS, the equation is true.
(iv) LHS =
−5×−10=50
RHS =
=5×10=50-1253×-10003=-5×-5×-53×-10×-10×-103=-5×-10=50
Because LHS is equal to RHS, the equation is true.
#### Answer 3: Property:For any two integers a and b,
(i) From the above property, we have:
=2×5=10
(ii) From the above property, we have:
(For any positive integer x,
Cube root using units digit:
Let us consider the number 1728.
The unit digit is 8; therefore, the unit digit in the cube root of 1728 will be 2.
After striking out the units, tens and hundreds digits of the given number, we are left with 1.
Now, 1 is the largest number whose cube is less than or equal to 1.
Therefore, the tens digit of the cube root of 1728 is 1.
=12
On factorising 216 into prime factors, we get:
216=2×2×2×3×3×3216=2×2×2×3×3×3
On grouping the factors in triples of equal factors, we get:
216={2×2×2}×{3×3×3}216=2×2×2×3×3×3
Now, taking one factor from each triple, we get:
= 2×3 = 6
Thus
=12×6=72
(iii) From the above property, we have:
(For any positive integer x,
Cube root using units digit:
Let us consider the number 2744.
The unit digit is 4; therefore, the unit digit in the cube root of 2744 will be 4.
After striking out the units, tens, and hundreds digits of the given number, we are left with 2.
Now, 1 is the largest number whose cube is less than or equal to 2.
Therefore, the tens digit of the cube root of 2744 is 1.
Thus
=3×14=42
(iv) From the above property, we have:
(For any positive integer x
Cube root using units digit:
Let us consider the number 15625.
The unit digit is 5; therefore, the unit digit in the cube root of 15625 will be 5.
After striking out the units, tens and hundreds digits of the given number, we are left with 15.
Now, 2 is the largest number whose cube is less than or equal to 15 ((23<15<33)23<15<33.
Therefore, the tens digit of the cube root of 15625 is 2.
Also
9, because 9×9×9=729
Thus
=9×25=225
#### Answer 4:Property:For any two integers a and b,
(i) From the above property, we have:
=4×6=24
(ii) Use above property and proceed as follows:
=2×17=34
(iii) From the above property, we have:
(∵∵ 700=2×2×5×5×7 and 49=7×7)
=2×5×7=70
(iv) From the above property, we have:
=125×a2(5×a2)
(∵ =a×a=a2 and =5)
=125a25a2=120a2
#### Answer 5: (i) Let us consider the following rational number:
- 125/729
Now
= - 5/9 ( 729=9×9×9 and 125 = 5×5×5729=9×9×9 and 125 = 5×5×5)
(ii) Let us consider the following rational number:
10648/12167
Now
Cube root by factors:
On factorising 10648 into prime factors, we get:
10648=2×2×2×11×11×1110648=2×2×2×11×11×11
On grouping the factors in triples of equal factors, we get:
10648={2×2×2}×{11×11×11}10648=2×2×2×11×11×11
Now, taking one factor from each triple, we get:
=2×11=22106483=2×11=22
Also
On factorising 12167 into prime factors, we get:
12167=23×23×2312167=23×23×23
On grouping the factors in triples of equal factors, we get:
12167={23×23×23}12167=23×23×23
Now, taking one factor from the triple, we get:
Now
= 22/23
(iii) Let us consider the following rational number:
-19683/24389
Now,
Cube root by factors:
On factorising 19683 into prime factors, we get:
19683=3×3×3×3×3×3×3×3×319683=3×3×3×3×3×3×3×3×3
On grouping the factors in triples of equal factors, we get:
19683={3×3×3}×{3×3×3}×{3×3×3}19683=3×3×3×3×3×3×3×3×3
Now, taking one factor from each triple, we get:
=3×3×3=27196833=3×3×3=27
Also
On factorising 24389 into prime factors, we get:
24389=29×29×2924389=29×29×29
On grouping the factors in triples of equal factors, we get:
24389={29×29×29}24389=29×29×29
Now, taking one factor from each triple, we get:
Now
= -27/ 29
(iv) Let us consider the following rational number:
Now
(686 and 3456 are not perfect cubes; therefore, we simplify it as 686/34566863456 by prime factorisation.)
= -7/ 12
(v) Let us consider the following rational number:
Now
Cube root by factors:
On factorising 39304 into prime factors, we get:
39304=2×2×2×17×17×1739304=2×2×2×17×17×17
On grouping the factors in triples of equal factors, we get:
39304={2×2×2}×{17×17×17}39304=2×2×2×17×17×17
Now, taking one factor from each triple, we get:
=2×17=34
Also
On factorising 42875 into prime factors, we get:
42875=5×5×5×7×7×742875=5×5×5×7×7×7
On grouping the factors in triples of equal factors, we get:
42875={5×5×5}×{7×7×7}42875=5×5×5×7×7×7
Now, taking one factor from each triple, we get:
=5×7=35
Now
= 34/35
#### Question 6: Find the cube root of each of the following rational numbers:(i) 0.001728(ii) 0.003375(iii) 0.001(iv) 1.331
0.001728=1728/1000000
Now
On factorising 1728 into prime factors, we get:
1728=2×2×2×2×2×2×3×3×31728=2×2×2×2×2×2×3×3×3
On grouping the factors in triples of equal factors, we get:
1728={2×2×2}×{2×2×2}×{3×3×3}
Now, taking one factor from each triple, we get:
=2×2×3=1217283=2×2×3=12
Also
= 100
12/100 = 0.12
(ii) We have:
0.003375= 3375/ 10000000.003375=33751000000
Now
On factorising 3375 into prime factors, we get:
3375=3×3×3×5×5×53375=3×3×3×5×5×5
On grouping the factors in triples of equal factors, we get:
3375={3×3×3}×{5×5×5}3375=3×3×3×5×5×5
Now, taking one factor from each triple, we get:
=3×5=1533753=3×5=15
Also
=10010000003=100×100×1003=100
= 15/100 = 0.15
(iii) We have:
1/10 = 0.1
(iv) We have:
= 11/10 = 1.1
#### Question 7: Evaluate each of the following
(i) To evaluate the value of the given expression, we need to proceed as follows:
=3+0.2+0.4=3.6 Thus, the answer is 3.6.
(ii) To evaluate the value of the given expression, we need to proceed as follows:
=10+0.20.5=9.7
(iii) To evaluate the value of the given expression, we need to proceed as follows:
(iv) To evaluate the value of the expression, we need to proceed as follows:
1=11=0
(v) To evaluate the value of the expression, we need to proceed as follows:
=13/10 =1.3
#### Question 8: Show that:
9/10
= 9/10
Because LHS is equal to RHS, the equation is true.
(ii)
= -8/7
RHS =
= -8/7
Because LHS is equal to RHS, the equation is true.
#### Question 9: Fill in the blanks:
(i) 5
3×5
=5×3
=3×5=3×5 (Commutative law)
(ii) 8×8=648×8=64
(iii) 3
12=4×3
(iv) 20
(v) 7×8=567×8=56
(vi) 4×5×6=1204×5×6=120
(vii) 3
(viii) 11
9/11
(ix) 13×13×13=2197
13×13×13=2197
#### Answer 10: Volume of a cube is given by:
V=s3V=s3, where s = side of the cube
Now
s3=474.552 cubic metres
To find the cube root of 474552, we need to proceed as follows:
On factorising 474552 into prime factors, we get:
474552=2×2×2×3×3×3×13×13×13474552=2×2×2×3×3×3×13×13×13
On grouping the factors in triples of equal factors, we get:
474552={2×2×2}×{3×3×3}×{13×13×13}474552=2×2×2×3×3×3×13×13×13
Now, taking one factor from each triple, we get:
=2×3×13=78
Also
= 7.8
Thus, the length of the side is 7.8 m.
#### Answer 11: Let the numbers be 2x, 3x and 4x.
According to the question:
(2x)3+(3x)3+(4x)3=0.3341258x3+27x3+64x3=0.3341258x3+27x3+64x3=0.33412599x3=0.334125
Thus, the numbers are:
2×0.15=0.30 3×0.15=0.454×0.15=0.60 Question 12: Find the side of a cube whose volume is
#### Answer 12: Volume of a cube with side s is given by: V=s3V=s3
(By prime factorisation)
29/6
#### (i) 36 and 384 are not perfect cubes; therefore, we use the following property:
for any two integers a and b
(By prime factorisation)
=2×2×2×3=24
(ii) 96 and 122 are not perfect cubes; therefore, we use the following property:
for any two integers a and b
(By prime factorisation)
=2×2×2×3=24 Thus, the answer is 24.
(iii) 100 and 270 are not perfect cubes; therefore, we use the following property:
for any two integers a and b
(By prime factorisation)
=2×3×5=30 Thus, the answer is 30.
(iv) 121 and 297 are not perfect cubes; therefore, we use the following property: for any two integers a and b
(By prime factorisation)
=11×3=33 Thus, the answer is 33.
#### (i) To find the cube root, we use the following property:
for two integers a and b
Now
(By the above property)
(By prime factorisation)
=3×5×9=135 Thus, the answer is 135.
(ii) To find the cube root, we use the following property:
for two integers a and b
Now
(By the above property)
(By prime factorisation)
=3×7×13=273 Thus, the answer is 273.
(iii) To find the cube root, we use the following property:
for two integers a and b
Now
(By the above property)
(By prime factorisation)
=5×7×17=595 Thus, the answer is 595.
(iv) To find the cube root, we use the following property:
for two integers a and b
Now
(By the above property)
(By prime factorisation)
=5×11×7=385 Thus, the answer is 385.
#### Answer 16: (i) Let us consider the number 226981.The unit digit is 1; therefore, the unit digit of the cube root of 226981 is 1.After striking out the units, tens and hundreds digits of the given number, we are left with 226.Now, 6 is the largest number, whose cube is less than or equal to 226 (63<226<7363<226<73).Therefore, the tens digit of the cube root of 226981 is 6.(ii) Let us consider the number 13824.The unit digit is 4; therefore, the unit digit of the cube root of 13824 is 4.After striking out the units, tens and hundreds digits of the given number, we are left with 13.Now, 2 is the largest number, whose cube is less than or equal to 13 (23<13<3323<13<33).Therefore, the tens digit of the cube root of 13824 is 2.(iii) Let us consider the number 571787.The unit digit is 7; therefore, the unit digit of the cube root of 571787 is 3.After striking out the units, tens and hundreds digits of the given number, we are left with 571.Now, 8 is the largest number, whose cube is less than or equal to 571 (83<571<9383<571<93).Therefore, the tens digit of the cube root of 571787 is 8.(iv) Let us consider the number 175616.The unit digit is 6; therefore, the unit digit of the cube root of 175616 is 6.After striking out the units, tens and hundreds digits of the given number, we are left with 175.Now, 5 is the largest number, whose cube is less than or equal to 175 (53<175<6353<175<63).Therefore, the tens digit of the cube root of 175616 is 5.
The document RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-4) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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# HARMONIC MEAN FORMULA
Harmonic mean is used to calculate the average of a set of numbers. The number of elements will be averaged and divided by the sum of the reciprocals of the elements.
It is calculated by dividing the number of observations by the sum of reciprocal of the observation.
The formula to find the harmonic mean is given by:
For Ungrouped Data:
## H.M of X = $\bar{X}$ = $\frac{n}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+...............+\frac{1}{x_{n}}}=\frac{n}{\sum (\frac{1}{x})}$
Where,
n = Total number of numbers or terms.
x1, x2, x3, …. xn = Individual terms or individual values.
Lets Work Out- Example: Find the harmonic mean of the following data {8, 9, 6, 11, 10, 5} ? Solution: Given data: {8, 9, 6, 11, 10, 5} So Harmonic mean = $\frac{6}{\frac{1}{8}+\frac{1}{9}+\frac{1}{6}+\frac{1}{11}+\frac{1}{10}+\frac{1}{5}}$ H =$\frac{6}{0.7936 }$= 7.560 Harmonic mean(H) = 7.560
For Grouped Data:
## H.M of X = $\bar{X}$ = $\bar{X}=\frac{f_{1}+f_{2}+f_{3}+................+f_{n}}{\frac{f_{1}}{x_{1}}+\frac{f_{2}}{x_{2}}+\frac{f_{3}}{x_{3}}+................+\frac{f_{n}}{x_{n}}}=\frac{\sum f}{\sum (\frac{f}{x})}$
Where,
f = Individual entries.
x1, x2, x3, …. xn = Individual terms or individual values.
Lets Work Out-
Example:The table given below represent the frequency-distribution of ages for Standard 1st students.
Ages 4 5 6 7 Number of Students 6 4 10 8
Find the Harmonic Mean of the given class.
Solution:
Here the data given are distributed data. So the ages are the variables and the number of student is considered as the frequency.
Ages (x) Number of Students (f) $\frac{f}{x}$ 4 10 2.5 5 6 1.2 6 8 1.33 7 4 0.57 Total $\sum f$=28 $\sum(\frac{f}{x})$= 5.6
So Harmonic mean = $\frac{\sum f}{\sum(\frac{f}{x})}=\frac{28}{5.6}$ = 5 years
Therefore, Harmonic mean(H) = 5 years
Related Formulas Regular Square Pyramid Formula Profit Formula Octagon Formula Parallel Formula Perimeter of a Triangle Formula Natural Log Formula Perimeter Formulas Surface Area of a Square Pyramid Formula
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Adding Fractions with Unlike Denominators - PowerPoint PPT Presentation
Adding Fractions with Unlike Denominators. 1. 3. 1. 1. 1. 5. 5. 5. 1. 1. 1. 1. 15. 15. 15. 15. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 15. 15. 15. 15. 15. 15. 15. 15. 15. 15. 15. 15. 15. 15. 15. 15. 15. 15.
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Presentation Transcript
with
Unlike Denominators
3
1
1
1
5
5
5
1
1
1
1
15
15
15
15
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
15
15
15
15
15
15
15
15
15
15
15
15
15
15
15
15
15
15
15
15
15
15
15
15
+
1
6
6
1
4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
12
12
12
12
12
12
12
12
12
12
12
12
12
12
+
=
Step 1: Find the LCD of the fractions.
Step 2: Rename the fractions using the LCD.
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
4
4
4
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
Step 3: Add the numerators. Denominator remains.
1
6
Step 4: Simplify, if necessary.
LCD: 12
3
9
3
×
=
3
12
4
?
1
2
2
×
=
6
2
12
?
+
+
11
12
Step 1: Find the LCD of the fractions.
Step 2: Rename the fractions using the LCD.
1
3
Step 3: Add the numerators. Denominator remains.
1
1
1
1
1
1
1
1
9
9
9
9
9
9
9
9
Step 4: Simplify, if necessary.
LCD: 9
3
3
1
×
=
3
9
3
?
1
9
+
+
4
9
Step 1: Find the LCD of the fractions.
1
2
Step 2: Rename the fractions using the LCD.
1
1
5
5
Step 3: Add the numerators. Denominator remains.
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Step 4: Simplify, if necessary.
1
1
1
1
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
10
LCD: 10
2
4
2
×
=
2
10
5
?
1
5
5
×
=
2
5
10
?
+
+
9
10
Step 1: Find the LCD of the fractions.
2
Step 2: Rename the fractions using the LCD.
1
1
3
3
1
1
1
1
1
1
1
1
6
6
6
6
6
6
6
6
Step 3: Add the numerators. Denominator remains.
Step 4: Simplify, if necessary.
LCD: 6
3
3
1
×
=
3
6
2
?
1
+
6
+
2
2
4
÷
=
2
3
6
GCF: 2
Step 1: Find the LCD of the fractions.
1
2
1
1
3
3
Step 2: Rename the fractions using the LCD.
Step 3: Add the numerators. Denominator remains.
1
1
1
1
1
1
1
1
1
1
1
1
1
1
6
6
6
6
6
6
6
6
6
6
6
6
6
6
Step 4: Simplify, if necessary.
LCD: 6
3
3
1
×
=
3
6
2
?
2
2
4
×
=
3
2
6
?
+
+
7
1
1
=
6
6
1 Whole
Step 1: Find the LCD of the fractions.
Step 2: Rename the fractions using the LCD.
Step 3: Add the numerators. Denominator remains.
Step 4: Simplify, if necessary.
3
3
1
×
=
3
12
4
?
LCD: 12
2
6
3
×
=
2
12
6
+
?
3
3
9
÷
=
GCF: 3
12
3
4
Step 1: Find the LCD of the fractions.
Step 2: Rename the fractions using the LCD.
Step 3: Add the numerators. Denominator remains.
Step 4: Simplify, if necessary.
7
7
1
×
=
7
21
3
?
LCD: 21
3
6
2
×
=
3
21
7
+
?
13
21
Step 1: Find the LCD of the fractions.
Step 2: Rename the fractions using the LCD.
Step 3: Add the numerators. Denominator remains.
Step 4: Simplify, if necessary.
4
9
LCD: 9
3
3
1
×
=
3
9
3
+
?
7
9
Step 1: Find the LCD of the fractions.
Step 2: Rename the fractions using the LCD.
Step 3: Add the numerators. Denominator remains.
Step 4: Simplify, if necessary.
3
3
1
×
=
3
24
8
?
LCD: 24
4
12
3
×
=
4
24
6
+
?
3
5
15
÷
=
GCF: 3
24
3
8
Step 1: Find the LCD of the fractions.
Step 2: Rename the fractions using the LCD.
Step 3: Add the numerators. Denominator remains.
Step 4: Simplify, if necessary.
5
10
2
×
=
5
15
3
?
LCD: 15
3
12
4
×
=
3
15
5
+
?
7
22
1
=
15
15
|
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