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# How to prove the normal distribution tail inequality for large x ?
• MHB
• WMDhamnekar
The last omitted term was computed by the author by using the following formula:$$\frac{1}{2}(-\frac{1}{2}\ln{x})$$
#### WMDhamnekar
MHB
What is the meaning of this proof? What is the meaning of last statement of this proof? How to prove lemma (7.1)? or How to answer problem 1 given below?
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Dhamnekar Winod said:
View attachment 11392
View attachment 11393
What is the meaning of this proof? What is the meaning of last statement of this proof? How to prove lemma (7.1)? or How to answer problem 1 given below?
If $x>0$ then the inequality in (1.9) must be true, because the left side is then slightly less than n(x), and the right side is slightly more than n(x).
Let's consider the derivatives of the expressions in (1.8).
$$\frac d{dx}(1-\Re(x)) = -\Re'(x) = -n(x)$$
Let's first find $n'(x)$.
We have:
$$n'(x)=\frac d{dx} \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} = \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} \cdot -x =-xn(x)$$
Then we have for instance:
$$\frac d{dx}(x^{-1}n(x)) = -x^{-2}n(x) + x^{-1}n'(x) = -x^{-2}n(x)+x^{-1}\cdot -x n(x) = -(1+x^{-2})n(x)$$
So we see that the derivatives of the expressions in (1.8) are indeed the negatives of the expressions in (1.9).
We have that $1-\Re(x)$ is in between 2 expressions, so its integration must also be between the integrations of the those 2 expressions.
Qed.
To prove the more general formula, we need to repeat these steps for the additional terms.
Klaas van Aarsen said:
If $x>0$ then the inequality in (1.9) must be true, because the left side is then slightly less than n(x), and the right side is slightly more than n(x).
Let's consider the derivatives of the expressions in (1.8).
$$\frac d{dx}(1-\Re(x)) = -\Re'(x) = -n(x)$$
Let's first find $n'(x)$.
We have:
$$n'(x)=\frac d{dx} \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} = \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} \cdot -x =-xn(x)$$
Then we have for instance:
$$\frac d{dx}(x^{-1}n(x)) = -x^{-2}n(x) + x^{-1}n'(x) = -x^{-2}n(x)+x^{-1}\cdot -x n(x) = -(1+x^{-2})n(x)$$
So we see that the derivatives of the expressions in (1.8) are indeed the negatives of the expressions in (1.9).
We have that $1-\Re(x)$ is in between 2 expressions, so its integration must also be between the integrations of the those 2 expressions.
Qed.
To prove the more general formula, we need to repeat these steps for the additional terms.
But, how to use this information to solve the given problem 1 or lemma 7.1?
Dhamnekar Winod said:
But, how to use this information to solve the given problem 1 or lemma 7.1?
Write (7.1) in the same form as (1.8) with the series on the left and also on the right.
Take the derivatives to find an expression that is in the same form as (1.9).
Then the proof follows in the same fashion.
Klaas van Aarsen said:
Write (7.1) in the same form as (1.8) with the series on the left and also on the right.
Take the derivatives to find an expression that is in the same form as (1.9).
Then the proof follows in the same fashion.
But I got some math help from wikipedia.
How can we use the above two formulas of CDF of Normal distribution to prove lemma 7.1 in the original question?
I got the following proofs of expansion of CDF of standard normal distribution.
In the above expansions of CDF of standard normal distribution, I want to know how the highlighted or marked computations was performed. If any member of this MHB knows the method of these computations, may explain it in reply.
Following is the simple proof:
How the last omitted term was computed by the author?
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You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts - Honors Go to the latest version.
1.10: Properties of Real Number Multiplication
Difficulty Level: At Grade Created by: CK-12
The multiplication statement $(-2)\times(-3)$ can be represented by using color counters. The statement means to remove two groups of 3 yellow counters. The yellow counters are negative counters. The result is six positive (red) counters. Would the result be the same if the statement were $(-3)\times(-2)$ ?
Guidance
There are five properties of multiplication that are important for you to know.
Commutative Property
The commutative property of multiplication states that the order in which two numbers are multiplied does not affect the sum. If $a$ and $b$ are real numbers, then $a\times b=b\times a$ .
Closure Property
The product of any two real numbers will result in a real number. This is known as the closure property of multiplication. In general, the closure property states that the product of any two real numbers is a unique real number. If $a, b$ and $c$ are real numbers, then $a\times b=c$ .
Associative Property
The order in which three or more real numbers are grouped for multiplication will not affect the product. This is known as the associative property of multiplication. The result will always be the same real number. In general, the associative property states that order in which the numbers are grouped for multiplication does not change the product. If $a, b$ and $c$ are real numbers, then $(a\times b)\times c=a \times(b \times c)$ .
Multiplicative Identity
When any real number is multiplied by the number one, the real number does not change. This is true whether the real number is positive or negative. The number 1 is called the multiplicative identity or the identity element of multiplication. The product of a number and one is the number. This is called the identity property of multiplication. If $a$ is a real number, then $a \times 1 =a$ .
Multiplicative Inverse
If $a$ is a nonzero real number, then the reciprocal or multiplicative inverse of $a$ is $\frac{1}{a}$ . The product of any nonzero real number and its reciprocal is always one. The number 1 is called the multiplicative identity or the identity element of multiplication. Therefore, the product of $a$ and its reciprocal is the identity element of multiplication (one). This is known as the inverse property of multiplication. If $a$ is a nonzero real number, then $a \times \frac{1}{a}=1$ .
Example A
Does $(-3)\times(+2)=(+2)\times(-3)$ ?
You can use color counters to determine the answer.
$(-3)\times(+2)$
This statement means to remove 3 groups of two red counters.
The result is 6 negative counters. Therefore $(-3)\times(+2)=-6$ .
$(+2)\times(-3)$
This statement means to add two groups of three yellow counters.
The result is 6 negative counters. Therefore $(+2) \times (-3) = -6$ .
This is an example of the commutative property of multiplication.
Example B
Does $(-6)+(+3)=$ a real number?
$(-6)\times(+3)$
This multiplication statement means to remove six groups of 3 red counters.
$(-6)\times(+3)=-18$
The result is -18. This is an integer. An integer is a real number. This is an example of the closure property.
Example C
Does $(-3\times 2)\times 2=-3 \times (2 \times 2)$ ?
$(-3\times 2)\times 2$
To begin the problem, we must do the multiplication inside the parenthesis.
This statement means to remove 3 groups of two red counters.
The result is 6 negative counters. Therefore $(-3)\times(+2)=-6$ .
Now the multiplication must be continued to represent $(-6)\times(2)$ . This statement means to remove 6 groups of two red counters.
The result is 12 negative counters. Therefore $(-6)\times(+2)=-12$ .
$-3\times(2 \times 2)$
To begin the problem, we must do the multiplication inside the parenthesis.
This statement means to add 2 groups of two red counters.
The result is 4 positive counters. Therefore $(+2)\times(+2)=+4$ .
Now the multiplication must be continued to represent $(-3)\times(4)$ . This statement means to remove 3 groups of four red counters.
$(-3 \times 2) \times 2=-3 \times(2 \times 2)$ ?
The numbers in the problem were the same but on the left side of the equal sign, the numbers -3 and +2 were grouped in parenthesis. The multiplication in the parenthesis was completed first and then -6 was multiplied by +2 to determine the final product. The result was -12.
On the right side of the equal sign, the numbers +2 and +2 were grouped in parenthesis. The multiplication in the parenthesis was completed first and then (+4) was multiplied by -3 to determine the final product. The result was -12.
Example D
Does $8 \times 1=8$ ?
The statement means to add 8 groups of one positive (red) counter.
The result is eight positive counters. Therefore the result $8 \times 1=8$ is correct.
Does $-6 \times 1=-6$ ?
The statement means to remove 6 groups of 1 positive counter.
The result is six negative counters. Therefore the result $-6 \times 1 = -6$ is correct.
This is an example of the identity property of multiplication.
Example E
Does $7 \times \frac{1}{7}=1$ ?
You have already learned that multiplication can be thought of in terms of repeated addition.
To show this multiplication, a number line can be used. The number line must be divided into intervals of 7.
When $\frac{1}{7}$ was added seven times, the result was one. The fraction $\frac{1}{7}$ is the reciprocal of 7. This is an example of the inverse property of multiplication.
Concept Problem Revisited
Here is $(-2)\times (-3)$ :
$(-3)\times(-2)$ ? This statement means to remove 3 groups of two yellow counters.
The result is six positive (red) counters.
The order in which you multiplied the numbers did not affect the answer. This is an example of the commutative property of multiplication.
Vocabulary
Multiplicative Identity
The multiplicative identity for multiplication of real numbers is one.
Multiplicative Inverse
The multiplicative inverse of multiplication is the reciprocal of the nonzero real number and the product of the real number and its multiplicative inverse is one. If $a$ is any nonzero real number, its multiplicative inverse is $\frac{1}{a}$ .
Associative Property
The associative property of multiplication states the order in which three or more real numbers are grouped for multiplication, will not affect the product. If $a, b$ and $c$ are real numbers, then $(a \times b)\times c=a \times(b \times c)$ .
Closure Property
The closure property of multiplication states that the product of any two real numbers is a unique real number. If $a, b$ and $c$ are real numbers, then $a \times b = c$ .
Commutative Property
The commutative property of multiplication states that the order in which two numbers are multiplied, does not affect the product. If $a$ and $b$ are real numbers, then $a \times b= b \times a$ .
Identity Element of Multiplication
The identity element of multiplication is another term for the multiplicative identity of multiplication. Therefore, the identity element of multiplication is one.
Identity Property
The identity property of multiplication states that the product of a number and one is the number. If $a$ is a real number, then $a \times 1=a$ .
Inverse Property
The inverse property of multiplication states that the product of any real number and its multiplicative inverse is one. If $a$ is a nonzero real number, then $a \times \left(\frac{1}{a}\right)=1$ .
Guided Practice
1. Multiply using the properties of multiplication: $\left(6 \times \frac{1}{6} \right)\times(3 \times -1)$
2. What property of multiplication justifies the statement $(-9 \times 5)\times 2= -9 \times (5 \times 2)$ ?
3. Apply the negative one property of multiplication to the following problem: $-176 \times -1$ ?
1.
$& \left(6 \times \frac{1}{6}\right) \times (3 \times -1)\\& (1) \times (3 \times -1) \ \rightarrow \ \text{Inverse Property}\\& (1) \times (-3) = -3 \ \rightarrow \text{The Product of two numbers with unlike signs is always negative.}$
2. $(-9 \times 5) \times 2 = -9 \times (5 \times 2)$
The numbers on each side of the equal sign are the same but they are not grouped the same.
$&(-9 \times 5)\times 2 && -9 \times(5 \times 2)\\&=(-45) \times 2 && =-9 \times(10)\\&=-90 && =-90$
The order in which the numbers were grouped did not affect the answer. The property that is being used is the associative property of multiplication.
3. $-176 \times -1 &= ?\\-176 \times -1 &= 176$
The number -176 is being multiplied by -1. The number remains the same but its sign has changed. This is the negative one property.
Practice
Match the following multiplication statements with the correct property of multiplication.
1. $9 \times \frac{1}{9}=1$
2. $(-7 \times 4)\times 2 = -7 \times(4 \times 2)$
3. $-8 \times (4) = -32$
4. $6 \times(-3)=(-3) \times 6$
5. $-7 \times 1=-7$
a) Commutative Property
b) Closure Property
c) Inverse Property
d) Identity Property
e) Associative Property
In each of the following, circle the correct answer.
1. What does $-5(4)\left(-\frac{1}{5}\right)$ equal?
1. -20
2. -4
3. +20
4. +4
2. What is another name for the reciprocal of any real number?
2. the multiplicative identity
3. the multiplicative inverse
3. What is the multiplicative identity?
1. -1
2. 1
3. 0
4. $\frac{1}{2}$
4. What is the product of a nonzero real number and its multiplicative inverse?
1. 1
2. -1
3. 0
4. there is no product
5. Which of the following statements is NOT true?
1. The product of any real number and negative one is the opposite of the real number.
2. The product of any real number and zero is always zero.
3. The order in which two real numbers are multiplied does not affect the product.
4. The product of any real number and negative one is always a negative number.
Name the property of multiplication that is being shown in each of the following multiplication statements:
1. $(-6\times 7)\times 2=-6 \times(7 \times 2)$
2. $-12 \times 1 =-12$
3. $25 \times 3 = 3 \times 25$
4. $10 \times \frac{1}{10}=1$
5. $-12 \times -1=-1\times -12$
Date Created:
Dec 19, 2012
Apr 29, 2014
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# A triangle has corners at (-1 ,7 ), (-5 ,-3 ), and (2 ,9 ). If the triangle is dilated by a factor of 5 about point #(-7 ,1 ), how far will its centroid move?
Mar 20, 2017
The distance is $= 24$
#### Explanation:
Let $A B C$ be the triangle
$A = \left(- 1 , 7\right)$
$B = \left(- 5 , - 3\right)$
$C = \left(2 , 9\right)$
The centroid of triangle $A B C$ is
${C}_{c} = \left(\frac{- 1 - 5 + 2}{3} , \frac{7 + \left(- 3\right) + 9}{3}\right) = \left(\frac{4}{3} , \frac{13}{3}\right)$
Let $A ' B ' C '$ be the triangle after the dilatation
The center of dilatation is $D = \left(- 7 , 1\right)$
$\vec{D A '} = 5 \vec{D A} = 5 \cdot < 6 , 6 > = < 30 , 30 >$
$A ' = \left(30 - 7 , 30 + 1\right) = \left(23 , 31\right)$
$\vec{D B '} = 5 \vec{D B} = 5 \cdot < 2 , - 4 > = < 10 , - 20 >$
$B ' = \left(10 - 7 , - 20 + 1\right) = \left(3 , - 19\right)$
$\vec{D C '} = 5 \vec{D C} = 5 \cdot < 9 , 8 > = < 45 , 40 >$
$C ' = \left(45 - 7 , 40 + 1\right) = \left(38 , 41\right)$
The centroid ${C}_{c} '$ of triangle $A ' B ' C '$ is
${C}_{c} ' = \left(\frac{23 + 3 + 38}{3} , \frac{31 - 19 + 41}{3}\right) = \left(\frac{64}{3} , \frac{53}{3}\right)$
The distance between the 2 centroids is
${C}_{c} {C}_{c} ' = \sqrt{{\left(\frac{64}{3} - \frac{4}{3}\right)}^{2} + {\left(\frac{53}{3} - \frac{13}{3}\right)}^{2}}$
$= \frac{1}{3} \sqrt{{60}^{2} + {40}^{2}} = \frac{72.11}{3} = 24$ |
Welcome to MS 101 Intermediate Algebra.
Presentation on theme: "Welcome to MS 101 Intermediate Algebra."— Presentation transcript:
Welcome to MS 101 Intermediate Algebra
Chapter 1 Linear Equations and Linear Functions
Qualitative Graphs Graphing Linear Equations Slope Finding Linear Equations Functions
1.1 Qualitative Graphs Definition: Graph without scaling (tick marks and their numbers) on the axis. Medication Costs p Linear Curve price t Time
Independent/Dependent Variables
Price p is dependent on the year t. (p is dependent of t) p is a dependent variable Year t does not depend on the price p. (t is independent of p) t is an independent variable
Examples You are taking a patient’s pulse counting the beats b per minute m. The number of people n that can run m miles. Waiting inline in the cafeteria for lunch. Use n for number of people in front and w for the wait time. Answers: Independent - m, m, n Dependent - b, n, w
Independent/Dependent Variable (time)
Independent most the time. Something happens over time temperature over time, growth, population, etc. Dependent Time it takes for something to happen wait time, time taken to cook, time taken to move a distance, etc
Dependent Variable Independent Variable
Intercepts Average Age of Nurses m A m-intercept A-intercept Average Age n-intercept parabola n t Years Since 1900 An intercept of a curve is the point where the curve intersects an axis (or axes).
Graph Characteristics
Increasing Curve upward from left to right Decreasing Curve downward from left to right Quadrants
Examples Height of a ball thrown straight up Heart rate of a person on a treadmill as the pace is steadily increased The temperature in Celsius of boiled water placed in a freezer The value of a car after a fixed amount of years
1.2 Graphing Linear Equations
Solution, Satisfy, and Solution Set An ordered pair (a,b) is a solution of an equation in x and y if the equation becomes a true statement when a substituted for x and b is substituted for y. We say (a,b) satisfies the equation. Solution set is the set of all solutions of the equation.
Example y = 4x – 2 Find y if x = 2 y = 4(2) – 2 = 8 – 2 = 6
Thus the ordered pair (2,6) satisfies the equation y = 4x – 2 and (2,6) is a solution of the equation y = 4x – 2
Continued Solution set of the equation y = 4x – 2 includes (2,6) x y
Ordered pair (a,b) the independent variable is First (left) followed by the dependent variable in the second position (right).
Graphing Equations y = 5x + 2 Chose values, ex. 0, 1 Organize in table
= 0 + 2 = 2 Solution: (0,2) y = 5(1) + 2 = 5 + 2 = 7 Solution: (1,7) Chose values, ex. 0, 1 Organize in table x y 7 12 17
(3, 17) (2, 12) (1, 7) (0, 2) (-1,-3) (-2,-8)
(3, 17) A graph of an equation in two variables is a visual representation of the solutions of an equation. (2, 12) (1, 7) (0, 2) (-1,-3) (-2,-8)
(1, 9) does not satisfy the equation
(3, 17) (2, 12) (1, 9) (0, 2) (1, 9) does not satisfy the equation (-1,-3) (-2,-8)
Linear Equation Forms Slope-Intercept Form Standard Form General Form
y = mx + b m and b are both constants (m is the slope, b is the y-intercept) Standard Form Ax + By = C A, B, C are integers (A and B are not both equal to zero) General Form Ax + By + C = 0 Point-Slope Form (y – y ) = m (x – x ) Points (x, y) & (x ,y ), slope m 1 1 1 1
General to Slope-Intercept Form
3y - 12x + 9 = 0 3y - 12x = 0 -9 3y – 12x = -9 3y - 12x + 12x = x 3y = x y = x y = 4x - 3 General to Standard Try to get y by itself Divide Simplify
Distributive Law First Distribute 2(2y-3) = 2(3x-4) + 2y
Get y by itself Divide Simplify 2(2y-3) = 2(3x-4) + 2y 4y – 6 = 6x – 8 +2y 4y – 6 -2y = 6x – 8 + 2y -2y 2y – 6 +6 = 6x – 8 +6 2y = 6x – 2 y = 3x - 1
Graphing Equations with Fraction Slopes
y =¾x - 4 Chose x values that are multiples of the denominator. Ex. 0, 4, 8, 12 Make a chart and graph x y 0 y = ¾ (0) – 4 = -4 4 y = ¾ (4) – 4 = -1 8 y = ¾ (8) – 4 = 2
Intercepts of Graphs To find the x-intercept, substitute 0 for y and solve for x. It will always be in the form (x,0). To find the y-intercept, substitute 0 for x and solve for y. It will always be in the form (0,y). y = - ½x +3 0 = - ½x + 3 0 -3 = - ½x -3 -3 *(-2) = - ½x *(-2) 6 = x ~x-intercept (6,0) y = - ½ (0) +3 y = 3 ~y-intercept (0,3) y-intercept (0,3) x-intercept (6,0)
Vertical/Horizontal Lines
If a and b are constants, Vertical lines are in the form x=a Horizontal lines are in the form y=b y=6 x=8
How do we measure steepness?
1.3 How do we measure steepness?
Comparing Steepness Calculate the ratio of vertical to horizontal distance. Patient 1 has his foot elevated 2 feet above the bed over a vertical distance of 2.5 feet. Patient 2 has his foot elevated 1.5 feet above the bed over a vertical distance of 2 feet. Which leg is elevated steeper? Patient 1’s foot is steeper because the ratio is slightly greater. Vertical Distance Horizontal Distance 2 1.5 2.5 2 2 1.5 = .8 = .75 2.5 2
Slope Let (x , y ) and (x , y ) be two points on a line. The slope of the line is: vertical change rise y - y horizontal change run x - x 1 1 2 2 2 1 m = = = 2 1 Run is positive to the right Run is negative to the left Rise is positive going up Rise is negative going down
Increasing vs Decreasing
positive rise positive run or negative rise negative run positive slope positive rise negative run or negative rise positive run negative slope
Finding Slope of a Graph
m = rise run m = 2 1 m = 2 m = y - y x - x m = 3 – 1 2 – 1 m = 2 = 2 1 2 1 (2,3) 2 1 2 (1,1) 1
Investigating Slope of Horizontal/Vertical Lines
2 – Slope of a horizontal line 4 – 2 2 is always 0. 3 – 1 2 Slope of a vertical line 1 – 1 0 is always undefined. m = = = 0 m = = = Undefined y = 2 (2,2) (4,2) x = 1 (1,1) (1,3)
Parallel/Perpendicular Lines
Parallel lines have the same slope m = m Slopes of perpendicular lines are opposite reciprocals 1 1 m 2 l 2 Parallel lines never intersect 2 1 l 1 l 2 m = 2 = 2 l 1 1 Perpendicular lines form 90 degree (right) angles
Finding Slope of Linear Equations
1.4 Finding Slope of Linear Equations y = -3x - 2 Graph rise run If in slope-intercept form: m = slope, m = -3 b = y-intercept, b = -2 1 m = = = -3 -3 1 -3
Using Slope and y-intercept to graph an equation
Slope = 2 = rise 2 run 1 y-intercept (0, -2) Plot y-intercept (0, b) Use m = to plot 2nd point. Sketch line passing through the 2 points. rise run (0, -2)
Vertical Change Property
For a line y = mx + b, if the run in 1, then the rise is the slope m. m 1 1 m Slope Addition Property For a line y = mx + b, if the independent variable increases by 1, then the dependent variable changes by slope m. y = -6x + 3 , x increases by 1, y changes by -6
Identifying Parallel/Perpendicular Lines
Are the lines 3y = 2x – 6 and 6y – 4x = 24 parallel, perpendicular or neither? 3y = 2x – 6 y = 2/3x - 3 6y – 4x + 4x = 18 +4x 6y = 4x + 24 y = 4/6x + 24/6 y = 2/3x + 4 Get y by itself Divide Simplify Get the equations in slope-intercept form Slope in both equations is 2/3, therefore the lines are parallel.
Indentifying Linear Equations from points
Set 1 Set 2 Set 3 Set 4 X Y 10 24 6 7 33 -2 4 11 20 12 34 -3 16 8 17 35 -4 13 9 23 36 -5 14 27 37 -6
Answers Set 1 – slope -4 Set 2 – not a line Set 3 – slope 0
Linear Equation Forms Slope-Intercept Form Standard Form General Form
1.5 Slope-Intercept Form y = mx + b m and b are both constants (m is the slope, b is the y-intercept) Standard Form Ax + By = C A, B, C are integers (A and B are not both equal to zero) General Form Ax + By + C = 0 Point-Slope Form (y – y ) = m (x – x ) Points (x, y) & (x ,y ), slope m 1 1 1 1
Find an equation given slope and a point
Slope 4, point (6,2) Start with putting the slope into y=mx + b y = 4x + b Plug the coordinates of the point in for x and y (2) = 4(6) + b Solve for b 2 -24 = 24 + b -24 -22 = b y = 4x - 22
Find an equation of the line using two points
(3, 5) (4, -2) Find the slope y - y – x - x – Place slope into y = mx + b y = -7x + b Then chose either point to plug in for x and y… (3, 5) (5) = -7(3) + b Solve for b 5 +21 = b +21 26 = b… y = -7x + 26 m = = = 2 1 2 1
Finding the approximate equation of a line
(-4.56,-5.24) (7.72, -4.93) Find the slope y - y – (-5.24) x - x – (-4.56) Place slope into y = mx + b y = .03x + b Then chose either point to plug in for x and y… (7.72, -4.93) (-4.93) = .03(7.72) + b Solve for b = b -5.16 = b… y = .03x – 5.16 m = = = = = .03 2 1 2 1
Finding an equation of a line parallel to a given line
y = 3x – 2 , point (1,6) m = 3, so slope is 3 Parallel lines have similar slopes, so: y = 3x + b Plug in the coordinates for x and y (6) = 3(1) + b 6 -3 = 3 + b -3 b = 3 y = 3x + 3 is parallel to y = 3x -2
Finding an equation of a line perpendicular to a given line
y = ½ x – 6, point (2,-2) Slopes of perpendicular lines are opposite reciprocals: flip the numerator and denominator and change the sign or -2 y = -2x + b (-2) = -2(2) + b -2 +4 = -4 + b +4 2 = b, y = -2x + 2 is perpendicular to y = ½ x - 6 12 21
Point-slope Form If a non-vertical line has a slope m and contains the point (x ,y ), then the equation of the line is: (y - y ) = m (x - x ) Given a point and the slope: m = 4 point (6,-3) y – (-3) = 4 (x - 6 ) y = 4x – 24 -3 y = 4x – 27 1 1 1 1
Finding an equation of a line using point-slope form
Given two points (2,14) and (-8,-6) Find slope -6 – -8 – (y - y ) = 2(x - x ) Substitute in one of the points y – (14) = 2(x – (2)) y – = 2x y = 2x + 10 m = = = 2 1 1
1.6 Functions Relation – set of ordered pairs
Domain – set of all values of the independent variable (x-values) Range – set of all values of the dependent variable (y-values) Function – a relation in which each input leads to exactly one output Domain = input, Range = output
Equations by input/output
x 5 6 7 y 3 1 4 & 3 x 2 4 y 7 8 9 RELATIONS RELATIONS input (domain) output (range) input (domain) output (range)
Is it a function? y = ± x y² = x y = x²
x = 5 , y = ± 5 or y = 5 and y = -5 input has 2 outputs, not a function y² = x x = 16, y = ± 4 or y = 4 and y = -4 y = x² x = 3, y = 3² = 9 input has 1 output, function
Is it a function? Set 1 Set 2 Set 3 Set 4 x y 4 1 5 6 10 9 -1 7 11 2
input y output 4 1 5 6 10 9 -1 7 11 2 12 8 13 -2 14
Answers Set 1 – not function, slope undefined, vertical line
Set 2 – not a function Set 3 – function, slope 0, horizontal line Set 4 – function, slope -3, decreasing line
Is it a function? When x =2 y = 3 and y = -3, not a function
Vertical line test A relation is a function if and only if every vertical line intersects the graph of the relation at no more than one point.
Functions from equations
Is y =-3x -2 a function? Graph it! Any vertical line would intersect only once….Yes.
Linear Function All non-vertical lines are functions. Linear function:
a relation whose equation can be put in y=mx + b form (m and b are constants)
Rule of Four Four ways to describe functions x y -2 -8 -1 -3 0 2 7 12
y = 5x + 2 Four ways to describe functions Equation Graph Table Verbally (input-output) x y 7 12 17 For each input-output pair, the output is two more than 5 times the input.
Finding Domain and Range
Domain = width = input = independent variable = x-values Range = height = output = dependent variable = y-values Leftmost point (-3,1) Rightmost point (4, 1) Domain -3 ≤ x ≤ 4 Highest point (2, 3) Lowest point (-1,-2) Range -2 ≤ y ≤ 3
Domain/Range Continued
D = all real numbers R = y ≥ -1 D = x ≥ 0 R = y ≥ -1 D = all real numbers R = all real numbers |
Contemporary Mathematics
# A | Co-Req Appendix: Integer Powers of 10
Contemporary MathematicsA | Co-Req Appendix: Integer Powers of 10
## Nonnegative Integer Powers of 10
The phrase nonnegative integers refers to the set containing 0, 1, 2, 3, … and so on. In the expression $105105$, 10 is called the base, and 5 is called the exponent, or power. The exponent 5 is telling us to multiply the base 10 by itself 5 times. So, $105=10×10×10×10×10=100,000105=10×10×10×10×10=100,000$. By definition, any number raised to the 0 power is 1. So, $100=1100=1$.
In the following table, there are several nonnegative integer powers of 10 that have been written as a product. Notice that higher exponents result in larger products. What do you notice about the number of zeros in the resulting product?
Exponential Form Product Number of Zeros in Product
$100100$ $11$ $00$
$101101$ $1010$ $11$
$102102$ $10×10=10010×10=100$ $22$
$103103$ $10×10×10=1,00010×10×10=1,000$ $33$
$104104$ $10×10×10×10=10,00010×10×10×10=10,000$ $44$
$105105$ $10×10×10×10×10=100,00010×10×10×10×10=100,000$ $55$
That’s right! The number of zeros is the same as the power each time!
## Negative Integer Powers of 10
The reciprocal of a number is 1 divided by that number. For example, the reciprocal of 10 is $110110$. We use negative exponents to indicate a reciprocal. For example, $10−1=1101=11010−1=1101=110$. Similarly, any expression with a negative exponent can be written with a positive exponent by taking the reciprocal. Several negative powers of 10 have been simplified in the table that follows. What do you notice about the number of zeros in the denominator (bottom) of each fraction?
Exponential Form Equivalent Simplified Expression Number of Zeros in Denominator
$10−110−1$ $1101=1101101=110$ $11$
$10−210−2$ $1102=110×10=11001102=110×10=1100$ $22$
$10−310−3$ $1103=110×10×10=11,0001103=110×10×10=11,000$ $33$
$10−410−4$ $1104=110×10×10×10=110,0001104=110×10×10×10=110,000$ $44$
That’s right! The number of zeros is the same as the positive version of the power each time.
In the following table, we will write the same powers of 10 as decimals. Count the number of decimal places to the right of the decimal point. What do you notice?
Exponential Form Equivalent Simplified Expression Number of Decimal Places to Right of Decimal
$10−110−1$ $1101=1÷10=0.11101=1÷10=0.1$ $11$
$10−210−2$ $1102=1÷100=0.011102=1÷100=0.01$ $22$
$10−310−3$ $1103=1÷1,000=0.0011103=1÷1,000=0.001$ $33$
$10−410−4$ $1104=1÷10,000=0.00011104=1÷10,000=0.0001$ $44$
That’s right! The number of decimal places to the right of the decimal point is the same as the positive version of the power each time.
## Multiplying Integers by Positive Powers of 10
Did you know that the distance from the sun to Earth is over 90 million miles? This value can be represented as 90,000,000, or we can write it as a product: $9×10,000,000=9×1079×10,000,000=9×107$, which is actually a more compact way of writing 90 million. Notice that the power of 7 reflects the number of zeros in 90 million. Several products of positive integers and powers of 10 are given in the table that follows. Notice that the number of zeros is the same as the exponent except in one case.
Exponential Form Product Number of Zeros in Product
$5×1015×101$ $5×10=505×10=50$ $11$
$13×10213×102$ $13×100=1,30013×100=1,300$ $22$
$8×1038×103$ $8×1,000=8,0008×1,000=8,000$ $33$
$15×10415×104$ $15×10,000=150,00015×10,000=150,000$ $44$
$70×10570×105$ $70×100,000=7,000,00070×100,000=7,000,000$ $66$
The only case in which the number of zeros didn’t equal the exponent was the last case. Why do you think that happened? That’s right! We multiplied by 70 which also had a zero. So, the product had a zero from the 70 and 5 zeros from $105105$ for a total of 6 zeros in 7,000,000.
## Multiplying by Negative Powers of 10
As we have seen, negative powers of 10 are decimals. Several products of positive integers and powers of 10 are given in the table below. Notice that multiplying an integer by 10 raised to a negative integer power results in a smaller number than you started with. Also, the number of decimal places to the right of the decimal point is the same as the exponent except in one case.
Exponential Form Product Number of Decimal Places to Right of Decimal
$3×10−13×10−1$ $3×0.1=0.33×0.1=0.3$ $11$
$13×10−213×10−2$ $13×0.01=0.1313×0.01=0.13$ $22$
$9×10−39×10−3$ $9×0.001=0.0099×0.001=0.009$ $33$
$15×10−415×10−4$ $15×0.0001=0.001515×0.0001=0.0015$ $44$
$70×10−570×10−5$ $70×0.00001=0.00070or0.000770×0.00001=0.00070or0.0007$ $5(6if we leave on the extra0)5(6if we leave on the extra0)$
The only case in which the number of decimal places to the right of the decimal point didn’t equal the positive version of the exponent was the last case. Why do you think that happened? That’s right! We multiplied by 70, which ended in zero.
## Moving the Decimal Place
A helpful shortcut when multiplying a number by a power of 10 is to “move the decimal point.” The following table shows several powers of 10, both positive and negative. Compare the location of the decimal point in the original number to the location of the decimal point in the product. How has it changed?
Exponential Form Product How the Position of the Decimal Point Changed
$5×1015×101$ $5.×10=5∧0⌣.=505.×10=5∧0⌣.=50$ $1place to the right1place to the right$
$13×10213×102$ $13.×100=130⌣∧0⌣.=1,30013.×100=130⌣∧0⌣.=1,300$ $2places to the right2places to the right$
$8×1038×103$ $8.×1,000=8∧0⌣0⌣0⌣.=8,0008.×1,000=8∧0⌣0⌣0⌣.=8,000$ $3places to the right3places to the right$
$15×10415×104$ $15.×10000=15∧0⌣0⌣0⌣0⌣.=150,00015.×10000=15∧0⌣0⌣0⌣0⌣.=150,000$ $4places to the right4places to the right$
$70×10570×105$ $70.×100,000=70∧0⌣0⌣0⌣0⌣0⌣.=7,000,00070.×100,000=70∧0⌣0⌣0⌣0⌣0⌣.=7,000,000$ $5places to the right5places to the right$
$3×10−13×10−1$ $3.×0.1=.3⌣∧=0.33.×0.1=.3⌣∧=0.3$ $1place to the left1place to the left$
$13×10−213×10−2$ $13.×0.01=.1⌣3⌣∧=0.1313.×0.01=.1⌣3⌣∧=0.13$ $2places to the left2places to the left$
$9×10−39×10−3$ $9.×0.001=.0⌣0⌣9⌣∧=0.0099.×0.001=.0⌣0⌣9⌣∧=0.009$ $3places to the left3places to the left$
$15×10−415×10−4$ $15.×0.0001=.0⌣0⌣1⌣5⌣∧=0.001515.×0.0001=.0⌣0⌣1⌣5⌣∧=0.0015$ $4places to the left4places to the left$
$70×10−570×10−5$ $70×0.00001=.0⌣0⌣0⌣7⌣0⌣∧=0.000770×0.00001=.0⌣0⌣0⌣7⌣0⌣∧=0.0007$ $5places to the left5places to the left$
Notice that multiplying by a positive power of 10 moves the decimal point to the right, making the value larger, while multiplying by a negative power of 10 moves the decimal point to the left, making the value smaller. Also, the number of decimal places that the decimal point moves is exactly the positive version of the exponent.
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Question
The size of the largest angle in a triangle is 4 times the size of the smallest angle. The other angle is 27 degrees less than the largest angle.
Work out, in degrees, the size of each angle in the triangle.
1. thaiduong
Step-by-step explanation:
a + b + c = 180 (The angles of a triangle equals to 180)
a = 4c (the largest angle is 4 times the smallest angle)
b = 4c – 27 (the final angle is 27 less than the largest angle)
c = c
Now that we have gotten the values for the variables in our equation, we plug them in:
4c + 4c – 27 + c = 180
Combine like terms:
4c + 4c + c – 27 = 180
9c – 27 = 180
9c = 180 + 27
9c = 207
Divide both side by 9
c = 23
a= 4c
a = 4 × 23
a = 92
b = 4c – 27
b = 92 – 27
b = 65
b = 65
Check:
a + b + c = 180
92 + 65 + 23 = 180
180 = 180 |
Mechanics: Newton's Laws of Motion
Newton's Laws of Motion: Problem Set Overview
There are 20 ready-to-use problem sets on the topic of Newton's Laws of Motions. These problem sets focus on situations in which the forces and accelerations are directed along the traditional coordinate axes. The problems target your ability to distinguish between mass and weight, determine the net force from the values of the individual forces, relate the acceleration to the net force and the mass, analyze physical situations to draw a free body diagram and solve for an unknown quantity (acceleration or individual force value), and to combine a Newton's second law analysis with kinematics to solve for an unknown quantity (kinematic quantity or a force value). Problems range in difficulty from the very easy and straight-forward to the very difficult and complex.
Mass versus Weight
Many problems target your ability to distinguish between mass and weight. Mass is a quantity which is dependent upon the amount of matter present in an object; it is commonly expressed in units of kilograms. Being the amount of matter possessed by an object, the mass is independent of its location in the universe. Weight, on the other hand, is the force of gravity with which the Earth attracts an object towards itself. Since gravitational forces vary with location, the weight of an object on the Earth's surface is different than its weight on the moon. Being a force, weight is most commonly expressed in the metric unit as Newtons. Every location in the universe is characterized by a gravitational field constant represented by the symbol g (sometimes referred to as the acceleration of gravity). Weight (or Fgrav) and mass (m) are related by the equation:
Fgrav = m • g
You may find our video tutorial on the topic of Mass versus Weight to be useful.
Newton's Second Law of Motion
Newton's second law of motion states that the acceleration (a) experienced by an object is directly proportional to the net force (Fnet) experienced by the object and inversely proportional to the mass of the object. In equation form, it could be said that a = Fnet/m. The net force is the vector sum of all the individual force values. If the magnitude and direction of the individual forces are known, then these forces can be added as vectors to determine the net force. Attention must be given to the vector nature of force. Direction is important. An up force and a down force can be added by assigning the down force a negative value and the up force a positive value. In a similar manner, a rightward force and a leftward force can be added by assigning the leftward force a negative value and the rightward force a positive value.
The a = Fnet/m equation can be used as both a formula for problem solving and as a guide to thinking. When using the equation as a formula for problem solving, it is important that numerical values for two of the three variables in the equation be known in order to solve for the unknown quantity. When using the equation as a guide to thinking, thought must be given to the direct and inverse proportionalities between acceleration and the net force and mass. A two-fold or a three-fold increase in the net force will cause the same change in the acceleration, doubling or tripling its value. A two-fold or three-fold increase in the mass will cause an inverse change in the acceleration, reducing its value by a factor of two or a factor of three. Watch this video to learn more about using the equation as a guide to thinking.
Free Body Diagrams
Free body diagrams and force diagrams represent the forces that act upon an object at a given moment in time. The individual forces that act upon an object are represented by vector arrows. The direction of the arrows indicate the direction of the force and the approximate length of the arrow represents the relative strength of the force. The forces are labeled according to their type. A free body diagram can be a useful aid in the problem-solving process. It provides a visual representation of the forces exerted upon an object. If the magnitudes of all the individual forces are known, the diagram can be used to determine the net force. And if the acceleration and the mass are known, then the net force can be calculated and the diagram can be used to determine the value of a single unknown force.
Coefficient of Friction
An object that is moving (or even attempting to move) across a surface encounters a force of friction. Friction force results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The maximum amount of friction force (Ffrict) can be calculated using the equation:
Ffrict = µ• Fnorm
The symbol µ (pronounced "mew") represents of the coefficient of friction and will be different for different surfaces. The Fnorm in the equation represents the normal force. On flat surfaces, it is often equal to the weight of the object. This video explains more about the force of friction and how to use the formula to solve problems.
Blending Newton's Laws and Kinematic Equations
Kinematics pertains to a description of the motion of an object and focuses on questions of how far?, how fast?, how much time? and with what acceleration? To assist in answering such questions, four kinematic equations were presented in the One-Dimensional Kinematics unit. The four equations are listed below.
• d = vo • t + 0.5 • a • t2
• vf = vo + a • t
• vf2 = vo 2 + 2 • a • d
• d = (vo + vf)/ 2 • t
where
• d = displacement
• t = time
• a = acceleration
• vo = original or initial velocity
• vf = final velocity
Newton's laws and kinematics share one of these questions in common: with what acceleration? The acceleration (a) of the Fnet = m•a equation is the same acceleration of the kinematic equations. Common tasks thus involve:
1. using kinematics information to determine an acceleration and then using the acceleration in a Newton's laws analysis, or
2. using force and mass information to determine an acceleration value and then using the acceleration in a kinematic analysis.
When analyzing a physics word problem, it is wise to identify the known quantities and to organize them as either kinematic quantities or as F-m-a type quantities. Once identified and written down in an organized fashion, the problem solution becomes more clear.
Habits of an Effective Problem-Solver
An effective problem solver by habit approaches a physics problem in a manner that reflects a collection of disciplined habits. While not every effective problem solver employs the same approach, they all have habits which they share in common. These habits are described briefly here. An effective problem-solver...
• ...reads the problem carefully and develops a mental picture of the physical situation. If needed, they sketch a simple diagram of the physical situation to help visualize it.
• ...identifies the known and unknown quantities in an organized manner, often times recording them on the diagram itself. They equate given values to the symbols used to represent the corresponding quantity (e.g., vo = 0 m/s, a = 2.67 m/s/s, vf = ???).
• ...plots a strategy for solving for the unknown quantity; the strategy will typically centers around the use of physics equations and is heavily dependent upon an understanding of physics principles.
• ...identifies the appropriate formula(s) to use, often times writing them down. Where needed, they perform the needed conversion of quantities into the proper unit.
• ...performs substitutions and algebraic manipulations in order to solve for the unknown quantity. |
# Math in Focus Grade 3 Chapter 15 Practice 2 Answer Key Measuring Weight
This handy Math in Focus Grade 3 Workbook Answer Key Chapter 15 Practice 2 Measuring Weight provides detailed solutions for the textbook questions.
## Math in Focus Grade 3 Chapter 15 Practice 2 Answer Key Measuring Weight
Ms. Meyer bought some meat, fruit, and vegetables for her Thanksgiving party.
Read the scales and write the weights.
Question 1.
The mushrooms weigh about ___________ ounces.
The mushrooms weigh about 15 ounces.
Explanation:
In the above image we can observe the pointer is in between 14oz and 16oz. The mushrooms weigh about 15 ounces.
Question 2.
The cranberries weigh about ___________ ounces.
The cranberries weigh about 6 ounces.
Explanation:
A slice of bread is about 1 ounce. In the above image we can observe 6 slices of bread in left side and cranberries in right side weighing balance. So the cranberries weigh about 6 ounces.
Question 3.
The leg of lamb weighs about___________ pounds.
The leg of lamb weighs about 5.5 pounds.
Explanation:
In the above image we can observe the pointer is in between 4lb and 6lb. So, the leg of lamb weighs about 5.5 pounds.
Question 4.
The turkey weighs about ___________ pounds.
The turkey weighs about 12 pounds.
Explanation:
In the above image we can observe 12lb’s in left side and turkey in right side weighing balance. We know that 1lb is equal to 1 pound. The turkey weighs about 12 pounds.
Question 5.
The tomatoes weigh about 2 pounds.
Explanation:
In the above image we can observe 2 loaf of bread in left side and tomatoes in right side weighing balance. A loaf of bread is about 1 pound. So, the tomatoes weigh about 2 pounds.
Question 6.
The box of cereal weighs about ___________ ounces.
The box of cereal weighs about 20 ounces.
Explanation:
In the above image we can observe 20oz’s in right side and cereal in right side weighing balance. We know that 1oz is equal to 1 ounce. The box of cereal weighs about 20 ounces.
Question 7.
The pumpkin weighs about __________ pounds.
The pumpkin weighs about 8 pounds.
Explanation:
In the above image we can observe 10lb’s in left side weighing balance. The pumpkin and 2lb’s in right side weighing balance. The pumpkin weighs about 8 pounds.
Choose the unit that you would use to measure each. Write ounces, pounds, or tons.
Question 8.
A cement truck weighs about 5 __________.
A cement truck weighs about 5 tons.
Question 9.
A package of butter weighs about 16 ___________.
A package of butter weighs about 16 pounds.
Question 10.
A bowling ball weighs about 9 ___________.
A bowling ball weighs about 9 pounds.
Question 11.
A mushroom weighs about 1 ___________.
A mushroom weighs about 1 pound.
Question 12.
A carton of milk weighs about 12 ___________.
A carton of milk weighs about 12 ounces.
Question 13.
A pile of magazines weighs about 5 ___________.
A pile of magazines weighs about 5 pounds.
Question 14.
An elephant weighs about 3 __________. |
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# Prove that in an isosceles triangle the perpendicular drawn from the vertex angle to the base bisect the vertex angle and the base.
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Hint: We prove that the perpendicular bisects the base and the vertex angle by showing the two triangles formed as congruent by Side angle side congruence rule.
* Isosceles triangle is a triangle with at least two equal sides. This also means corresponding two angles are also equal.
Here ABC is the isosceles triangle where the double line on sides AB and AC represent equal sides. That is AB=AC . The point D lies on the side BC and we need to prove AD bisects the angle A and side BC . That is BD=DC.
We need to prove two statements: AD bisect angle A and AD bisect side BC .
Step 1: To prove the statements, we prove the congruence of needed triangles which is $\vartriangle ABD$ and $\vartriangle ADC$
Considering these two triangles, the sides AB=AC (since $\vartriangle ABC$ is an isosceles triangle)
$\angle ADB = \angle ADC = {90^0}$ (Since AD is the perpendicular drawn to base BC)
AD is a common side for both the triangles.
Thus two sides and an angle of $\vartriangle ABD$ and $\vartriangle ADC$ are equal. By SAS(side angle side) congruence theorem, when two sides and an angle of two triangles are equal, those triangles are congruent.
Here $\vartriangle ABD$ and $\vartriangle ADC$ are congruent.
As BD and DC are sides of congruent triangles, BD=DC
And as $\angle BAD$ and $\angle DAC$ are angles of congruent triangles, $\angle BAD = \angle DAC$ which means that AD bisects angle A.
Hence proved.
Note: Other congruence theorems which can be used are SSS(Side Side Side) congruence theorem for equilateral triangle, ASA(Angle Side Angle) congruence theorem, AAS(Angle Angle Side) congruence theorem.
Students might get confused while writing which triangle are congruent, they should keep in mind the name of triangles is always written in the same sequence of angles and sides to which they are convergent, Example triangle ABC is congruent to triangle PQR means AB=PQ, BC=QR and CA=RP. |
# Solve Systems by Graphing
Do Now: Journal Describe the steps that you must take to accurately graph an equation. What are some of the typical errors that students may make? Objective The student will be able to: solve systems of equations by graphing. December 2, 2014 What is a system of equations? A
system of equations is when you have two or more equations using the same variables. The solution to the system is the point that satisfies ALL of the equations. This point will be an ordered pair. When graphing, you will encounter three possibilities. Intersecting Lines The
point where the lines intersect is your solution. The solution of this graph is (1, 2) (1,2) Parallel Lines These lines never
intersect! Since the lines never cross, there is NO SOLUTION! Parallel lines have the same slope with different y-intercepts. 2 Slope = = 2 1 y-intercept = 2
y-intercept = -1 Coinciding Lines These lines are the same! Since the lines are on top of each other, there are INFINITELY MANY SOLUTIONS! Coinciding lines have the same slope and
y-intercepts. 2 Slope = = 2 1 y-intercept = -1 What is the solution of the system graphed below? 1. 2.
3. 4. (2, -2) (-2, 2) No solution Infinitely many solutions 1) Find the solution to the following system: 2x + y = 4 x-y=2
Graph both equations. You may graph using x- and y-intercepts 2x + y = 4 (0, 4) and (2, 0) xy=2 (0, -2) and (2, 0) Graph the ordered pairs. Graph the equations.
y= 4 x-y=2 (0, -2) and (2, 0) + 2x 2x + y = 4 (0, 4) and (2, 0) x
y= Where do the lines intersect? (2, 0) 2 Check your answer! To check your answer, plug the point back into both equations.
2x + y = 4 2(2) + (0) = 4 x-y=2 (2) (0) = 2 Nice joblets try another! 2) Find the solution to the following system: y = 2x 3 -2x + y = 1 Graph both equations. Put both equations
in slope-intercept or standard form. Ill do slope-intercept form on this one! y = 2x 3 y = 2x + 1 Graph using slope and y-intercept Graph the equations. y = 2x 3 m = 2 and b = -3 y = 2x + 1 m = 2 and b = 1 Where do the lines intersect?
No solution! Notice that the slopes are the same with different y-intercepts. If you recognize this early, you dont have to graph them! Check your answer! Not a lot to checkJust make sure you set up your equations correctly. I double-checked it and I did it right
What is the solution of this system? 3x y = 8 2y = 6x -16 (3, 1) 2. (4, 4) 3. No solution 4. Infinitely many solutions 1. What x coordinate will be a solution to f(x) and g(x)? f(x) = - 4x + 5
g(x) = 3x 9 Step 1: Set functions equal to each other -4x + 5 = 3x 9 Step 2: Solve as an equation with variables on both sides 5 = 7x 9 14 = 7x 2=x Step 3: Identify the x coordinate that will be the solution for both functions. The x coordinate of 2 will be a solution for both f(x) and g(x). Try this one: f(x) = -x + 4
g(x) = (-2x + 10) Solving a system of equations by graphing. Let's summarize! There are 3 steps to solving a system using a graph. Step 1: Graph both equations. Graph using slope and y intercept or x- and y-intercepts. Be sure to use a ruler and graph paper!
Step 2: Do the graphs intersect? This is the solution! LABEL the solution! Step 3: Check your solution. Substitute the x and y values into both equations to verify the point is a solution to both equations.
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# Flowchart and Paragraph Proofs
How to Write a Paragraph Proof
How to Write a Paragraph Proof
1
Flowchart and Paragraph Proofs
2-7 Flowchart and Paragraph Proofs Warm Up Lesson Presentation Lesson Quiz Holt Geometry
2
Warm Up Complete each sentence.
1. If the measures of two angles are ? , then the angles are congruent. 2. If two angles form a ? , then they are supplementary. 3. If two angles are complementary to the same angle, then the two angles are ? . equal linear pair congruent
3
Learning Target Students will be able to:Write flowchart and paragraph proofs.Prove geometric theorems by using deductive reasoning.
4
Vocabulary flowchart proof paragraph proof
5
A second style of proof is a flowchart proof, which uses boxes and arrows to show the structure of the proof. The justification for each step is written below the box.
7
A paragraph proof is a style of proof that presents the steps of the proof and their matching reasons as sentences in a paragraph. Although this style of proof is less formal than a two-column proof, you still must include every step.
9
Example 3: Reading a Paragraph Proof
Use the given paragraph proof to write a two-column proof. Given: m1 + m2 = m4 Prove: m3 + m1 + m2 = 180° Paragraph Proof: It is given that m1 + m2 = m4. 3 and 4 are supplementary by the Linear Pair Theorem. So m3 + m4 = 180° by definition. By Substitution, m3 + m1 + m2 = 180°.
10
Statements Reasons Example 3 Continued Two-column proof:
1. m1 + m2 = m4 1. Given 2. 3 and 4 are supp. 2. Linear Pair Theorem 3. m3 + m4 = 180° 3. Def. of supp. s 4. m3 + m1 + m2 = 180° 4. Substitution
11
Check It Out! Example 3 Use the given paragraph proof to write a two-column proof. Given: WXY is a right angle. 1 3 Prove: 1 and 2 are complementary. Paragraph Proof: Since WXY is a right angle, mWXY = 90° by the definition of a right angle. By the Angle Addition Postulate, mWXY = m2 + m3. By substitution, m2 + m3 = 90°. Since 1 3, m1 = m3 by the definition of congruent angles. Using substitution, m2 + m1 = 90°. Thus by the definition of complementary angles, 1 and 2 are complementary.
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Check It Out! Example 3 Continued
Statements Reasons 1. WXY is a right angle. 1. Given 2. mWXY = 90° 2. Def. of right angle 3. m2 + m3 = mWXY 3. Angle Add. Postulate 4. m2 + m3 = 90° 4. Subst. 5. 1 3 5. Given 6. m1 = m3 6. Def. of s 7. m2 + m1 = 90° 7. Subst. 8. 1 and 2 are comp. 8. Def. of comp. angles
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Example 4: Writing a Paragraph Proof
Use the given two-column proof to write a paragraph proof. Given: 1 and 2 are complementary Prove: 3 and 4 are complementary m3 + m4 = 90° 3 and 4 are comp.
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Example 4 Continued Paragraph proof:
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Check It Out! Example 4 Use the given two-column proof to write a paragraph proof. Given: 1 4 Prove: 2 3 Two-column proof:
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Check It Out! Example 4 Continued
Paragraph proof: It is given that 1 4. By the Vertical Angles Theorem, 1 2 and 3 4. By the Transitive Property of Congruence, 2 4. Also by the Transitive Property of Congruence, 2 3.
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# Algebra 2
Students, who look for help with their algebra homework or classwork, usually think of getting extra coaching from faculty or studying with a classmate who knows the subject really well. While both methods can help you get by, they’re not always effective.
Online tutoring has grown popular over the past few years as more and more students turn to the internet for help and information. Taking education online is an effective way to incorporate learning into their schedules. Online tutoring is reliable, safe and quick. You’ll also find that there are a plethora of sites and services to choose from. At its most basic, there are sites which have a free online calculator that you can input your problem into and it gives you an answer instantly.
## Help For Algebra 2
Help with algebra 2 is available from online algebra math problem solvers who go into the details of each problem. Many also explain the concepts and show examples that you can easily follow at your own pace. If you are looking for help in algebra 2, online algebra solvers are certainly worth a try. From homework help to helping you get ready for tests and exams, these online algebra helpers are a great way to get help with algebra 2. It helps to solve equations as well as inequalities.
## Solved Examples
Question 1: Solve for x, 7x + 16 = 3x + 5
Solution:
Given, 7x + 16 = 3x + 5
Subtract 3x from both sides of the equation,
=> 7x - 3x + 16 = 3x - 3x + 5
=> 4x + 16 = 5
Subtract 16 from both sides of the equation,
=> 4x + 16 - 16 = 5 - 16
=> 4x = - 11
Divide both side by 4,
=> $\frac{4x}{4} = \frac{-11}{4}$
=> x = $\frac{-11}{4}$, is the solution.
Question 2: Find the value of x, for all whole numbers.
7x - 3 $\leqslant$ 11
Solution:
Given,
7x - 3 $\leqslant$ 11
Add 3 from both sides of the equations,
=> 7x - 3 + 3 $\leqslant$ 11 + 3
=> 7x $\leqslant$ 14
Divide both side by 7,
=> $\frac{7x}{7}$ $\leqslant$ $\frac{14}{7}$
=> x $\leqslant$ 2
=> Values of x are 0, 1 and 2. |
5.3 Higher Order Polynomials
In the previous section we explored the short run behavior of quadratics, a special case of polynomials. In this section we will explore the behavior of polynomials in general. The basic building blocks of polynomials are power functions.
Power Function
A power function is a function that can be represented in the form
Where the base is a variable and the exponent, p, is a number.
Characteristics of Power Functions
Shown to the right are the graphs of , and , all even whole number powers. Notice that all these graphs have a fairly similar shape, very similar to a quadratic, but as the power increases the graphs flatten somewhat near the origin, and become steeper away from the origin.
To describe the behavior as numbers become larger and larger, we use the idea of infinity. The symbol for positive infinity is , and for negative infinity. When we say that “x approaches infinity”, which can be symbolically written as , we are describing a behavior – we are saying that is getting large in the positive direction.
With the even power function, as the input becomes large in either the positive or negative direction, the output values become very large positive numbers. Equivalently, we could describe this by saying that as approaches positive or negative infinity, the values approach positive infinity. In symbolic form, we could write: as .
Shown here are the graphs of , and , all odd whole number powers. Notice all these graphs look similar, but again as the power increases the graphs flatten near the origin and become steeper away from the origin.
For these odd power functions, as approaches negative infinity, approaches negative infinity. As approaches positive infinity, approaches positive infinity. In symbolic form we write: as and as , .
Long Run Behavior
The behavior of the graph of a function as the input takes on large negative values () and large positive values () as is referred to as the long run behavior of the function.
Polynomials
Recall our definitions of polynomials from chapter 1.
Terminology of Polynomial Functions
A polynomial is function that can be written as
Each of the constants are called coefficients and can be positive, negative, or zero, and be whole numbers, decimals, or fractions.
A term of the polynomial is any one piece of the sum, that is any . Each individual term is a transformed power function.
The degree of the polynomial is the highest power of the variable that occurs in the polynomial.
The leading term is the term containing the highest power of the variable: the term with the highest degree.
Because of the definition of the “leading” term we often rearrange polynomials so that the powers are descending.
For any polynomial the long run behavior of the polynomial will match the long run behavior of the leading term.
Example of Polynomial Graph
What can we determine about the long run behavior and degree of the equation for the polynomial graphed here?
Since the output grows large and positive as the inputs grow large and positive, we describe the long run behavior symbolically by writing: as , . Similarly, as , .
In words, we could say that as x values approach infinity, the function values approach infinity, and as x values approach negative infinity the function values approach negative infinity.
We can tell this graph has the shape of an odd degree power function which has not been reflected, so the degree of the polynomial creating this graph must be odd, and the leading coefficient would be positive.
Short Run Behavior: Intercepts
Characteristics of the graph such as vertical and horizontal intercepts and the places the graph changes direction are part of the short run behavior of the polynomial.
Like with all functions, the vertical intercept is where the graph crosses the vertical axis, and occurs when the input value is zero. Since a polynomial is a function, there can only be one vertical intercept, which occurs at the point . The horizontal intercepts occur at the input values that correspond with an output value of zero. It is possible to have more than one horizontal intercept.
Horizontal intercepts are also called zeros, or roots of the function.
To find horizontal intercepts, we need to solve for when the output will be zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and 4th degree (also called quartic) polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. In the next section, we will explore a few more techniques for finding roots of cubic and quartic functions, but for this section we will restrict the discussion to cubic and quartic functions in which:
1. The polynomial can be factored using known methods: greatest common factor, trinomial factoring or factoring by grouping.
2. The polynomial is given in factored form.
3. Technology is used to determine the intercepts. (i.e. your calculator or Desmos)
Examples Finding Horizontal Intercepts by Factoring
a. Find the horizontal intercepts of
Let’s try factoring this polynomial to find solutions for .
Remember the first thing we always want to do when factoring is to see if we can factor out a GCF:
We see that every term contains
Now what we have left is a quadratic form. This means that we see that we have a trinomial in which the variable part of one term is the square of another and those are the only two variable parts.
Notice that So we can factor the trinomial as quadratic form.
Now we can break these apart to solve:
We have 5 horizontal intercepts:
b. Find the vertical and horizontal intercepts of
The vertical intercept can be found by evaluating .
The horizontal intercepts can be found by solving
Since this is already factored we can break it apart:
c.Find the horizontal intercepts of
Since this polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques we know, we can turn to technology to find the intercepts.
Going to Desmos for this graph, we can see:
By clicking on the grey dots on the axis, we see the horizontal intercepts are (-3,0), (-2,0) and (1,0).
We could check these are correct by plugging in these values for t and verifying that .
Notice that the polynomial in the previous example, part c was degree three, and had three horizontal intercepts and two turning points – places where the graph changes direction. However, part b was also cubic and had 2 intercepts, while the degree 6 polynomial in a had 5. We will now make a general statement without justifying it.
Intercepts and Turning Points of Polynomials
A polynomial of degree will have:
• At most horizontal intercepts. An odd degree polynomial will always have at least one.
• At most turning points.
Try it Now 1
Find the vertical and horizontal intercepts of the function
Graphical Behavior at Intercepts
If we graph the function , notice that the behavior at each of the horizontal intercepts is different.
At the horizontal intercept , coming from the of the polynomial, the graph passes directly through the horizontal intercept. The factor is linear (has a power of 1), so the behavior near the intercept is like that of a line – it passes directly through the intercept. We call this a single zero, since the zero corresponds to a single factor of the function.
At the horizontal intercept , coming from the factor of the polynomial, the graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic – it bounces off of the horizontal axis at the intercept. Since , the factor is repeated twice, so we call this a double zero. We could also say the zero has multiplicity 2.
At the horizontal intercept , coming from the factor of the polynomial, the graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic, with the same “S” type shape near the intercept as . We call this a triple zero. We could also say the zero has multiplicity 3.
By utilizing these behaviors, we can sketch a reasonable graph of a factored polynomial function without needing technology. We might want to do that because it is very easy to make a typing mistake in Desmos and especially a graphing calculator. It is very important to have an understanding of what a graph should look like in order to know that you have typed in the function you are graphing correctly.
Graphical Behavior of Polynomials at Horizontal Intercepts
If a polynomial contains a factor of the form , the behavior near the horizontal intercept h is determined by the power on the factor.
For higher even powers 4,6,8 etc.… the graph will still bounce off of the horizontal axis but the graph will appear flatter with each increasing even power as it approaches and leaves the axis.
For higher odd powers, 5,7,9 etc… the graph will still pass through the horizontal axis but the graph will appear flatter with each increasing odd power as it approaches and leaves the axis.
Example Sketching a Graph
Sketch a graph of .
This graph has two horizontal intercepts. At , the factor is squared, indicating the graph will bounce at this horizontal intercept. At , the factor is not squared, indicating the graph will pass through the axis at this intercept.
Additionally, we can see the leading term, if this polynomial were multiplied out, would be -2x^3, so the long-run behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs get large positive, and the inputs increasing as the inputs get large negative. In other words, this ends down on on the right side and up on the right side.
To sketch this we consider the following:
As , the function so we know the graph starts in the 2nd quadrant and is decreasing toward the horizontal axis.
At (-3, 0) the graph bounces off of the horizontal axis and so the function must start increasing.
At (0, 90) the graph crosses the vertical axis at the vertical intercept. This is obtained by plugging 0 in
Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis since the graph passes through the next intercept at (5,0).
As the function so we know the graph continues to decrease and we can stop drawing the graph in the 4th quadrant.
Using technology we can verify that the resulting graph will look like:
Try it Now 2
Given the function use the methods that we have learned so far to find the vertical & horizontal intercepts, determine where the function is negative and positive, describe the long run behavior and sketch the graph without technology.
We can also use what we know to make guesses about a graph of a polynomial. [1]
Example using a Graph to Learn about a Polynomial
Based on the long run behavior, with the graph becoming large positive on both ends of the graph, we can determine that this is the graph of an even degree polynomial. The graph has 2 horizontal intercepts, suggesting a degree of 2 or greater, and 3 turning points, suggesting a degree of 4 or greater. Based on this, it would be reasonable to conclude that the degree is even and at least 4, so it is probably a fourth degree polynomial.
The previous example was difficult to tell the exact zeros since they were not integers, but often, we can actually find the equation for a graphed polynomial.
Example using a Graph to write the equation for a Polynomial
Write a formula for the polynomial function graphed here.
This graph has three horizontal intercepts: x = -3, 2, and 5. At x = -3 and 5 the graph passes through the axis, suggesting the corresponding factors of the polynomial will be linear to power 1. At x = 2 the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be squared. We also notice that the graph rises on both sides and has 3 turn-around points, so it appears to be degree 4. Also, we do not know the coefficient in the front of the lead term, a, at this point. So far, we know:
To determine the lead coefficient (also known as a stretch factor), we can utilize another point on the graph. Here, we see the vertical intercept appears to be so we can plug those values to solve for a.
The graphed polynomial appears to represent the function
Try it Now 3
Given the graph, write a formula for the function shown.
Finding Extrema
With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph.
Example Maximizing a Higher Order Polynomial Function
An open-top box is to be constructed by cutting out squares from each corner of a 14cm by 20cm sheet of plastic then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box.
We will start this problem by drawing a picture, labeling the width of the cut-out squares with a variable, w.
Notice that after a square is cut out from each end, it leaves a cm by cm rectangle for the base of the box, and the box will be cm tall. This gives the volume:
Using technology to sketch a graph allows us to estimate the maximum value for the volume, restricted to reasonable values for w: values from 0 to 7.
From this graph, we can find the maximum value at 339.013, and occurs when the squares are about 2.704cm square.
Try it Now 4
Use technology to find the maximum and minimum values on the interval [-1, 4] of the function .
1. Vertical intercept , Horizontal intercepts
2. Vertical intercepts , Horizontal intercepts
The function is negative on and
The function is positive on and
The leading term is so as and as
3. Double zero at , triple zero at , single zero at .
Substitute (0,4) and solve for :
4. The minimum occurs at approximately the point (0, -6.5), and the maximum occurs at approximately the point (3.5, 7).
• evenpowers
• oddpowers
• 53example2
• behaviorofgraphatintercepts
• triplezero
• singlezero
• doublezero
• 53example3
• 53example5
• 53example6
• tin533
• examplebox
• 53example4
• zoomedingraph534
• tin532
1. The next two examples and Try it Now is from Precalculus: An Investigation of Functions by Lippman and Rasmussen, Licensed under Creative Commons 4.0, CC-BY-SA |
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# A single discount is equal to discount series of $10\%$ and $20\%$ is \begin{align} & \text{A}\text{. 25 }\!\!\%\!\!\text{ } \\ & \text{B}\text{. 30 }\!\!\%\!\!\text{ } \\ & \text{C}\text{. 35 }\!\!\%\!\!\text{ } \\ & \text{D}\text{. 28 }\!\!\%\!\!\text{ } \\ \end{align}
Last updated date: 22nd Jun 2024
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Hint: We know that if p is the single discount of discount series of ${{x}_{1}},{{x}_{2}},......,{{x}_{n}}$., then $p=\left( 1-\prod\limits_{i=1}^{n}{\left( \dfrac{100-{{x}_{i}}}{100} \right)} \right)\times 100$. From the question, it is given to find the single discount of the discount series of $10\%$ and $20\%$. By using the above formula, we can find the single discount of the discount series of $10\%$ and $20\%$.
Before solving the question, we should know that if p is the single discount of discount series of ${{x}_{1}},{{x}_{2}},......,{{x}_{n}}$., then $p=\left( 1-\prod\limits_{i=1}^{n}{\left( \dfrac{100-{{x}_{i}}}{100} \right)} \right)\times 100$.
From the question, it is clear that we have to find the discount series of $10\%$ and $20\%$.
We know that if p is the single discount of discount series of ${{x}_{1}},{{x}_{2}},......,{{x}_{n}}$., then$p=\left( 1-\prod\limits_{i=1}^{n}{\left( \dfrac{100-{{x}_{i}}}{100} \right)} \right)\times 100$.
\begin{align} & \Rightarrow p=\left( 1-\left( \dfrac{100-10}{100} \right)\left( \dfrac{100-20}{100} \right) \right)\times 100 \\ & \Rightarrow p=\left( 1-\left( \dfrac{90}{100} \right)\left( \dfrac{80}{100} \right) \right)\times 100 \\ & \Rightarrow p=\left( 1-\dfrac{72}{100} \right)\times 100 \\ & \Rightarrow p=\left( \dfrac{100-72}{100} \right)\times 100 \\ & \Rightarrow p=\left( \dfrac{28}{100} \right)\times 100 \\ & \Rightarrow p=28.....(1) \\ \end{align}
From equation (1), it is clear that the value of p is equal to 28. So, we can say that a single discount is equal to the discount series of $10\%$ and $20\%$ is $28\%$.
Note: Students may have a misconception that if p is the single discount of discount series of ${{x}_{1}},{{x}_{2}},......,{{x}_{n}}$., then $p=\left( \prod\limits_{i=1}^{n}{\left( \dfrac{100-{{x}_{i}}}{100} \right)} \right)\times 100$. If this misconception is followed, then we get
\begin{align} & \Rightarrow p=\left( \left( \dfrac{100-10}{100} \right)\left( \dfrac{100-20}{100} \right) \right)\times 100 \\ & \Rightarrow p=\left( \left( \dfrac{90}{100} \right)\left( \dfrac{80}{100} \right) \right)\times 100 \\ & \Rightarrow p=\left( \dfrac{72}{100} \right)\times 100 \\ & \Rightarrow p=72.....(1) \\ \end{align}
From equation (1), it is clear that the value of p is equal to 28. So, we can say that a single discount is equal to the discount series of $10\%$ and $20\%$ is $72\%$. But we know that the single discount is equal to $28\%$. So, this misconception should get avoided. |
# Question Video: Finding the Dividend Given the Divisor, Quotient and Remainder as Integers Mathematics • 5th Grade
What is the number that when divided by 50 gives a quotient of 111 and a remainder of 40?
04:14
### Video Transcript
What is the number that when divided by 50 gives a quotient of 111 and a remainder of 40?
Let’s read our problem again more slowly. And this time we’ll write it out as we read it as a missing number sentence. What is the number that when divided by 50 gives a quotient of 111? We can just remind ourselves that a quotient is the answer to a division. It’s the result we get when we divide one number by another. So, if we divide a number by 50, and it gives a quotient of 111, this means an answer of 111.
But there’s another part to our statement, doesn’t just give us a quotient of 111, it also gives us a remainder of 40. In other words, we have 40 left over that we can’t divide by 50 because it’s too small. To try to understand how we can find our missing number, we could sketch a bar model. Let’s ignore the remainder part for a moment. Imagine our calculation was just a number split into groups of 50 equals 111. We’d expect our bar model to be split into 111 groups and all of them being worth 50.
Obviously, we can’t sketch this. But what we can do is represent this using a multiplication, 111 lots of 50, or 111 multiplied by 50. So, this might be what our bar model might look like if we had no remainder. So, how would it change if we need to include a remainder of 40? Well, to start with, the number that we’re dividing by would be slightly larger. We’d need to extend the bar a little bit to show this. And the value of how much it’s extended by is the 40 that’s a remainder.
Now, if we look carefully at our bar model, we can see how we need to find our missing number. First, we need to find 111 lots of 50, and then we need to add 40. In this number sentence, we have a multiplication and an addition. When we get to calculation like this, we always work out the multiplication first. But just to remind ourselves and make sure we don’t make any mistakes, we could put parentheses around the part we want to do first.
To begin with then, let’s multiply 111 by 50. There are lots of ways we could do this. We could use long multiplication. We could multiply by 100, and then halve the answer. Or we could remember that 50 is the same as five lots of 10. So, we could multiply by five and then by 10. Let’s use this method because it’s a quick one.
If we multiply 111 by five, we’re going to replace all the ones for fives. 111 multiplied by five is 555. Now, what we need to do is to multiply this answer by 10. We know that when a number is multiplied by 10, the digits shift one place to the left. So, 555 becomes worth 5550. So, the answer to our multiplication is 5550. Now, all we need to do is to add the remainder of 40. We know 50 plus 40 equals 90. And so, 5550 plus 40 equals 5590.
To find the number that when divided by 50 gives a quotient of 111, we needed to multiply 111 by 50. And to find the number that gives a remainder of 40 too, we needed to multiply 111 by 50 and then add the remainder of 40 on the end. So, this is what we did to find our answer. The number which when divided by 50 gives a quotient of 111 and a remainder of 40 is 5590. |
# 6.3: Mass Calculations
The methods described in the previous section allow us to express reactants and products in terms of moles, but what if we wanted to know how many grams of a reactant would be required to produce a given number of grams of a certain product? This logical extension is, of course, trivial! In Chapter 4, we learned to express molar quantities in terms of the masses of reactants or products. For example, the reduction of iron (III) oxide by hydrogen gas, produces metallic iron and water. If we were to ask how many grams of elemental iron will be formed by the reduction of 1.0 grams of iron (III) oxide, we would simply use the molar stoichiometry to determine the number of moles of iron that would be produced, and then convert moles into grams using the known molar mass. For example, one gram of Fe2O3 can be converted into mol Fe2O3 by remembering that moles of a substance is equivalent to grams of that substance divided by the molar mass of that substance:
$moles=\left ( \frac{grams}{molar\; mass} \right )=\left ( \frac{grams}{grams/mol} \right )=(grams)\times (mol/grams)$
Using this approach, the mass of a reactant can be inserted into our reaction pathway as the ratio of mass-to-molar mass. This is shown here for the reduction of 1.0 gram of Fe2O3.
$Given:\left ( \frac{1.0g\: Fe_{2}O_{3}}{159.70\frac{g\: Fe_{2}O_{3}}{mol\: Fe_{2}O_{3}}} \right )\; \; Find: x\: mol\: Fe$
We set up the problem to solve for mol product; the general equation is:
$(molproduct)=(molreactant)\times \left ( \frac{molproduct}{molreactant} \right )$
The stoichiometric mole ratio is set up so that mol reactant will cancel, giving a solution in mol product. Substituting,
$x\: mol\: Fe=\left ( \frac{1.0g\: Fe_{2}O_{3}}{159.70\frac{g\: Fe_{2}O_{3}}{mol\: Fe_{2}O_{3}}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right )$
It is often simpler to express the ratio (mass)/(molar mass) as shown below,
$\left ( \frac{1.0g\: Fe_{2}O_{3}}{159.70\frac{g\: Fe_{2}O_{3}}{mol\: Fe_{2}O_{3}}} \right )=(1.0g\: Fe_{2}O_{3})\times \left ( \frac{1\: mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right )$
Doing this, and rearranging,
$x\: mol\: Fe=(1.0g\: Fe_{2}O_{3})\times \left ( \frac{1\: mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right )=0.013\: mol$
That is, the reduction of 1.0 grams of Fe2O3 by excess hydrogen gas will produce 0.013 moles of elemental iron. All of these calculations are good to two significant figures based on the mass of iron (III) oxide in the original problem (1.0 grams). Note that we have two conversion factors (ratios) in this solution; one from mass to molar mass and the second, the stoichiometric mole ratio from the balanced chemical equation. Knowing that we have 0.013 moles of Fe, we could now convert that into grams by knowing that one mole of Fe has a mass of 55.85 grams; the yield would be 0.70 grams.
We could also modify our basic set-up so that we could find the number of grams of iron directly.
Here we have simply substituted the quantity (moles molar mass) to get mass of iron that would be produced. Again, we set up the problem to solve for mol product;
$(molproduct)=(molreactant)\times \left ( \frac{molproduct}{molreactant} \right )$
In place of mol product and mol reactant, we use the expressions for mass and molar mass, as shown in the scheme above. The stoichiometric mole ratio is set up so that mol reactant (the given) will cancel, giving a solution in mol product. Substituting,
$x\: mol\: Fe\left ( \frac{55.85g\: Fe}{mol\: Fe} \right )=(1.0g\: Fe_{2}O_{3})\times \left ( \frac{1\: mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right )$
Rearranging and canceling units,
$x\: g\: Fe=\left ( \frac{55.85g\: Fe}{mol\: Fe} \right ) \left ( \frac{1\: g\: Fe_{2}O_{3}\times mol\: Fe_{2}O_{3}}{159.70\: g\: Fe_{2}O_{3}} \right )\times \left ( \frac{2\: mol\: Fe}{1\: mol\: Fe_{2}O_{3}} \right )=0.70g$
Exercise $$\PageIndex{1}$$
Aqueous solutions of silver nitrate and sodium chloride react in a double-replacement reaction to form a precipitate of silver chloride, according to the balanced equation shown below.
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
If 3.06 grams of solid AgCl are recovered from the reaction mixture, what mass of AgNO3 was present in the reactants?
Exercise $$\PageIndex{2}$$
Aluminum and chlorine gas react to form aluminum chloride according to the balanced equation shown in below.
2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)
If 17.467 grams of chlorine gas are allowed to react with excess Al, what mass of solid aluminum chloride will be formed?
Exercise $$\PageIndex{3}$$
Ammonia, NH3, is also used in cleaning solutions around the house and is produced from nitrogen and hydrogen according to the equation:
N2 + 3 H2 → 2 NH3
1. If you have 6.2 moles of nitrogen what mass of ammonia could you hope to produce?
2. If you have 6.2 grams of nitrogen how many grams of hydrogen would you need? |
Percentage Equations
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Objective
SWBAT find percent of a number by writing and solving an equation.
Big Idea
To find the percent of a number, you can set up an equation to find an equivalent ratio to the part out of one hundred.
8 minutes
Students work independently to complete the Think About It problem. If I notice that students are spending a long time on the simplest form column in the organizer, I will give the whole class a cue to move on to prompts below the chart.
After student work time, I show a completed table on the document camera. Most students will have been able to fill in all of the boxes, and I am simply giving them a way to check their responses.
I then ask for a student to show his/her double number line on the document camera and explain the work.
Finally, I ask for ideas on the last question. Some students may name using scale factor, which is a more efficient way to solve. Some students will see the relationship between 25% and dividing by 4. I take 2-3 hands for this question.
I then frame the lesson by letting students know that we'll continue our work with ratios and percents. In this lesson, we'll use equations to model the problems, which are more efficient than double number lines and tape diagrams.
Intro to New Material
15 minutes
To begin the Intro to New Material section, I create a double number line for the first problem. I do this quickly, as students are proficient with this model. I create the double number line because I want students to see that the relationships we'll use in our equation are also found on the double number line. The equation is a different way to organize our work, but the end result is the same.
Because of our previous lessons, students know that percents are always out of 100, and that percents can be expressed as ratios that display the part over the whole. I ask students what we know in the first problem, and whether this is the total number of serves or a portion of the serves.
I have students organize their work in 2 lines. The first line is where students write the percent as a fraction, and then simplify. In this problem, students do not have to simplify in order to create equivalent fractions. I have them do it anyway. I want them to always check to see if they can simplify (and this is a skill that my students can always use reinforcement with). The second line is where they will create their equivalent ratios. You can see my work on this notes page.
I then lead students through the second problem.
Partner Practice
15 minutes
Students work in pairs on the Partner Practice problem set. A student work sample is included. As students are working, I circulate around the room. I am looking for:
• Are students explaining their thinking to their partner?
• Are students writing the percent as a fraction, and then simplifying?
• Are students correctly creating equivalent fractions?
• Are students determining the correct answer?
• Are students clearly labeling their units?
• Are students answering the question(s) asked?
• How did you know to use that ratio to represent that percent?
• What did you do to simplify this ratio?
• How did you create your equation?
• What does the numerator of each ratio represent? The denominator?
• How did you solve the equation to find the information the question is asking for?
Once the 'oooooh' moment has passed, I tell students that I can think of two strategies to use to solve here. Pretty quickly, students will raise their hands to suggest we subtract the radioactive ounces from the total. I then have students turn and talk with their partner about another way we might solve. I'm looking for students to realize that we can use 90% in our ratio, to find the portion that is not radioactive.
Students then complete the Check for Understanding problem independently.
Independent Practice
20 minutes
Students work on the Independent Practice problem set.
For Problem 3, I am looking for students to identify that David put the 80 in the numerator, and not the denominator. I want students to clearly identify this mistake, and not give a more vague response, like 'he set up his fractions wrong.'
For Problem 9, I want students to access their number sense. If I see students setting up two equations (to find 75% of 48 and 20% of 100), I will affirm that doing the work is always a great strategy. I will then ask them to reason through this problem without the equations. I'll ask, 'is there any way to figure this out before getting to the step of creating equations?'
There are two extension questions in this problem set, which ask students to find percentages over 100.
Closing and Exit Ticket
7 minutes
After independent work time, I bring the class back together to discuss Problem 11 from the Independent Practice set. I have students share with their partners what Kimani needs to do, in order to solve this problem.
Students complete the Exit Ticket independently to close the lesson. |
# NCERT Solutions For Class 9th Maths Chapter 7 : Triangles
CBSE NCERT Solutions For Class 9th Maths Chapter 7 : Triangles. NCERT Solutins For Class 9 Mathematics. Exercise 7.1, Exercise 7.2, Exercise 7.3, Exercise 7.4, Exercise 7.5 (Miscellaneous Exercise)
### NCERT Solutions for Class IX Maths: Chapter 7 – Triangles
Page No: 118
Exercise 7.1
1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
Given,
AC = AD and AB bisects ∠A
To prove,
ΔABC ≅ ΔABD
Proof,
In ΔABC and ΔABD,
AB = AB (Common)
∠CAB = ∠DAB (AB is bisector)
Therefore, ΔABC ≅ ΔABD by SAS congruence condition.
BC and BD are of equal length.
Page No: 119
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.
Given,
AD = BC and ∠DAB = ∠CBA
(i) In ΔABD and ΔBAC,
AB = BA (Common)
∠DAB = ∠CBA (Given)
Therefore, ΔABD ≅ ΔBAC by SAS congruence condition.
(ii) Since, ΔABD ≅ ΔBAC
Therefore BD = AC by CPCT
(iii) Since, ΔABD ≅ ΔBAC
Therefore ∠ABD = ∠BAC by CPCT
3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.
Given,
AD and BC are equal perpendiculars to AB.
To prove,
CD bisects AB
Proof,
In ΔAOD and ΔBOC,
∠A = ∠B (Perpendicular)
∠AOD = ∠BOC (Vertically opposite angles)
Therefore, ΔAOD ≅ ΔBOC by AAS congruence condition.
Now,
AO = OB (CPCT). CD bisects AB.
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.
Given,
l || m and p || q
To prove,
ΔABC ≅ ΔCDA
Proof,
In ΔABC and ΔCDA,
∠BCA = ∠DAC (Alternate interior angles)
AC = CA (Common)
∠BAC = ∠DCA (Alternate interior angles)
Therefore, ΔABC ≅ ΔCDA by ASA congruence condition.
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.
Given,
l is the bisector of an angle ∠A.
BP and BQ are perpendiculars.
(i) In ΔAPB and ΔAQB,
∠P = ∠Q (Right angles)
∠BAP = ∠BAQ (l is bisector)
AB = AB (Common)
Therefore, ΔAPB ≅ ΔAQB by AAS congruence condition.
(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of ∠A.
Page No: 120
6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Given,
To show,
BC = DE
Proof,
∠BAD + ∠DAC = ∠EAC + ∠DAC
AC = AE (Given)
Therefore, ΔABC ≅ ΔADE by SAS congruence condition.
BC = DE by CPCT.
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
Given,
P is mid-point of AB.
∠BAD = ∠ABE and ∠EPA = ∠DPB
(i) ∠EPA = ∠DPB (Adding ∠DPE both sides)
∠EPA + ∠DPE = ∠DPB + ∠DPE
⇒ ∠DPA = ∠EPB
In ΔDAP ≅ ΔEBP,
∠DPA = ∠EPB
AP = BP (P is mid-point of AB)
Therefore, ΔDAP ≅ ΔEBP by ASA congruence condition.
(ii) AD = BE by CPCT.
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 = 1/2 AB
Given,
∠C = 90°, M is the mid-point of AB and DM = CM
(i) In ΔAMC and ΔBMD,
AM = BM (M is the mid-point)
∠CMA = ∠DMB (Vertically opposite angles)
CM = DM (Given)
Therefore, ΔAMC ≅ ΔBMD by SAS congruence condition.
ALSO READ: NCERT Solutions for Class 9th Science Chapter 8 : Motion
(ii) ∠ACM = ∠BDM (by CPCT)
Therefore, AC || BD as alternate interior angles are equal.
Now,
∠ACB + ∠DBC = 180° (co-interiors angles)
⇒ 90° + ∠B = 180°
⇒ ∠DBC = 90°
(iii) In ΔDBC and ΔACB,
BC = CB (Common)
∠ACB = ∠DBC (Right angles)
DB = AC (byy CPCT, already proved)
Therefore, ΔDBC ≅ ΔACB by SAS congruence condition.
(iv) DC = AB (ΔDBC ≅ ΔACB)
⇒ DM = CM = AM = BM (M is mid-point)
⇒ DM + CM = AM + BM
⇒ CM + CM = AB
⇒ CM = 1/2AB
Page No: 123
Exercise 7.2
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
Given,
AB = AC, the bisectors of ∠B and ∠C intersect each other at O
(i) Since ABC is an isosceles with AB = AC,
∴ ∠B = ∠C
⇒ 1/2∠B = 1/2∠C
⇒ ∠OBC = ∠OCB (Angle bisectors.)
⇒ OB = OC (Side opposite to the equal angles are equal.)
(ii) In ΔAOB and ΔAOC,
AB = AC (Given)
AO = AO (Common)
OB = OC (Proved above)
Therefore, 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.
Given,
AD is the perpendicular bisector of BC
To show,
AB = AC
Proof,
BD = CD (AD is the perpendicular bisector)
AB = AC (by CPCT)
Page No: 124
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.
Given,
BE and CF are altitudes.
AC = AB
To show,
BE = CF
Proof,
In ΔAEB and ΔAFC,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
AB = AC (Given)
Therefore, ΔAEB ≅ ΔAFC by AAS congruence condition.
Thus, 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.
Given,
BE = CF
(i) In ΔABE and ΔACF,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
BE = CF (Given)
Therefore, ΔABE ≅ ΔACF by AAS congruence condition.
(ii) Thus, AB = AC by CPCT and therefore 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.
Given,
ABC and DBC are two isosceles triangles.
To show,
∠ABD = ∠ACD
Proof,
In ΔABD and ΔACD,
AB = AC (ABC is an isosceles triangle.)
BD = CD (BCD is an isosceles triangle.)
Therefore, ΔABD ≅ ΔACD by SSS congruence condition. Thus, ∠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.
Given,
AB = AC and AD = AB
To show,
∠BCD is a right angle.
Proof,
In ΔABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)
In ΔACD,
⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)
Now,
In ΔABC,
∠CAB + ∠ACB + ∠ABC = 180°
⇒ ∠CAB + 2∠ACB = 180°
⇒ ∠CAB = 180° – 2∠ACB — (i)
∠CAD = 180° – 2∠ACD — (ii)
also,
∠CAB + ∠CAD = 180° (BD is a straight line.)
∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD
⇒ 180° = 360° – 2∠ACB – 2∠ACD
⇒ 2(∠ACB + ∠ACD) = 180°
⇒ ∠BCD = 90°
ALSO READ: NCERT Solutions For Class 8th Maths Chapter 1 : Rational Numbers
7. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Given,
∠A = 90° and AB = AC
A/q,
AB = AC
⇒ ∠B = ∠C (Angles opposite to the equal sides are equal.)
Now,
∠A + ∠B + ∠C = 180° (Sum of the interior angles of the triangle.)
⇒ 90° + 2∠B = 180°
⇒ 2∠B = 90°
⇒ ∠B = 45°
Thus, ∠B = ∠C = 45°
8. Show that the angles of an equilateral triangle are 60° each.
Let ABC be an equilateral triangle.
BC = AC = AB (Length of all sides is same)
⇒ ∠A = ∠B = ∠C (Sides opposite to the equal angles are equal.)
Also,
∠A + ∠B + ∠C = 180°
⇒ 3∠A = 180°
⇒ ∠A = 60°
Therefore, ∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.
Page No: 128
Exercise 7.3
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.
Given,
ΔABC and ΔDBC are two isosceles triangles.
(i) In ΔABD and ΔACD,
AB = AC (ΔABC is isosceles)
BD = CD (ΔDBC is isosceles)
Therefore, ΔABD ≅ ΔACD by SSS congruence condition.
(ii) In ΔABP and ΔACP,
AP = AP (Common)
∠PAB = ∠PAC (ΔABD ≅ ΔACD so by CPCT)
AB = AC (ΔABC is isosceles)
Therefore, ΔABP ≅ ΔACP by SAS congruence condition.
(iii) ∠PAB = ∠PAC by CPCT as ΔABD ≅ ΔACD.
AP bisects ∠A. — (i)
also,
In ΔBPD and ΔCPD,
PD = PD (Common)
BD = CD (ΔDBC is isosceles.)
BP = CP (ΔABP ≅ ΔACP so by CPCT.)
Therefore, ΔBPD ≅ ΔCPD by SSS congruence condition.
Thus, ∠BDP = ∠CDP by CPCT. — (ii)
By (i) and (ii) we can say that AP bisects ∠A as well as ∠D.
(iv) ∠BPD = ∠CPD (by CPCT as ΔBPD ≅ ΔCPD)
and BP = CP — (i)
also,
∠BPD + ∠CPD = 180° (BC is a straight line.)
⇒ 2∠BPD = 180°
⇒ ∠BPD = 90° —(ii)
From (i) and (ii),
AP is the perpendicular bisector of BC.
2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
Given,
AD is an altitude and AB = AC
(i) In ΔABD and ΔACD,
AB = AC (Given)
Therefore, ΔABD ≅ ΔACD by RHS congruence condition.
Now,
BD = CD (by CPCT)
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
Given,
AB = PQ, BC = QR and AM = PN
(i) 1/2 BC = BM and 1/2QR = QN (AM and PN are medians)
also,
BC = QR
⇒ 1/2 BC = 1/2QR
⇒ BM = QN
In ΔABM and ΔPQN,
AM = PN (Given)
AB = PQ (Given)
BM = QN (Proved above)
Therefore, ΔABM ≅ ΔPQN by SSS congruence condition.
(ii) In ΔABC and ΔPQR,
AB = PQ (Given)
∠ABC = ∠PQR (by CPCT)
BC = QR (Given)
Therefore, ΔABC ≅ ΔPQR by SAS congruence condition.
4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Given,
BE and CF are two equal altitudes.
In ΔBEC and ΔCFB,
∠BEC = ∠CFB = 90° (Altitudes)
BC = CB (Common)
BE = CF (Common)
Therefore, ΔBEC ≅ ΔCFB by RHS congruence condition.
Now,
∠C = ∠B (by CPCT)
Thus, AB = AC as sides opposite to the equal angles are equal.
ALSO READ: NCERT Solutions for Class 11th Maths Exercise 1.2 Set
5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Given,
AB = AC
In ΔABP and ΔACP,
∠APB = ∠APC = 90° (AP is altitude)
AB = AC (Given)
AP = AP (Common)
Therefore, ΔABP ≅ ΔACP by RHS congruence condition.
Thus, ∠B = ∠C (by CPCT)
Page No: 132
Exercise 7.4
1. Show that in a right angled triangle, the hypotenuse is the longest side.
ABC is a triangle right angled at B.
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90° and ∠B is 90°.
Since, B is the largest angle of the triangle, the side opposite to it must be the largest.
So, BC is the hypotenuse which is the largest side of the right angled triangle ABC.
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.
Given,
∠PBC < ∠QCB
Now,
∠ABC + ∠PBC = 180°
⇒ ∠ABC = 180° – ∠PBC
also,
∠ACB + ∠QCB = 180°
⇒ ∠ACB = 180° – ∠QCB
Since,
∠PBC < ∠QCB therefore, ∠ABC > ∠ACB
Thus, AC > AB as sides opposite to the larger angle is larger.
3. In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
Given,
∠B < ∠A and ∠C < ∠D
Now,
AO < BO — (i) (Side opposite to the smaller angle is smaller)
OD < OC —(ii) (Side opposite to the smaller angle is smaller)
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.
In ΔABD,
∴ ∠ADB < ∠ABD — (i) (Angle opposite to longer side is larger.)
Now,
In ΔBCD,
BC < DC < BD
∴ ∠BDC < ∠CBD — (ii)
Adding (i) and (ii) we get,
∠ADB + ∠BDC < ∠ABD + ∠CBD
⇒ ∠B > ∠D
Similarly,
In ΔABC,
∠ACB < ∠BAC — (iii) (Angle opposite to longer side is larger.)
Now,
∠DCA < ∠DAC — (iv)
Adding (iii) and (iv) we get,
∠ACB + ∠DCA < ∠BAC + ∠DAC
⇒ ∠A > ∠C
5. In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.
Given,
PR > PQ and PS bisects ∠QPR
To prove,
∠PSR > ∠PSQ
Proof,
∠PQR > ∠PRQ — (i) (PR > PQ as angle opposite to larger side is larger.)
∠QPS = ∠RPS — (ii) (PS bisects ∠QPR)
∠PSR = ∠PQR + ∠QPS — (iii) (exterior angle of a triangle equals to the sum of opposite interior angles)
∠PSQ = ∠PRQ + ∠RPS — (iv) (exterior angle of a triangle equals to the sum of opposite interior angles)
∠PQR + ∠QPS > ∠PRQ + ∠RPS
⇒ ∠PSR > ∠PSQ [from (i), (ii), (iii) and (iv)]
Page No: 133
6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Let l is a line segment and B is a point lying o it. We drew a line AB perpendicular to l. Let C be any other point on l.
To prove,
AB < AC
Proof,
In ΔABC,
∠B = 90°
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90°
∴ ∠C mustbe acute angle. or ∠C < ∠B
⇒ AB < AC (Side opposite to the larger angle is larger.)
## 1 thought on “NCERT Solutions For Class 9th Maths Chapter 7 : Triangles”
1. Amar says:
solution bahut sahi hai aur solve bahut acche se step by step kiya gay hai |
# 2016 AMC 10B Problems/Problem 16
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
The sum of an infinite geometric series is a positive number $S$, and the second term in the series is $1$. What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
## Solution 1
The sum of an infinite geometric series is of the form: $$\begin{split} S & = \frac{a_1}{1-r} \end{split}$$ where $a_1$ is the first term and $r$ is the ratio whose absolute value is less than 1.
We know that the second term is the first term multiplied by the ratio. In other words: $$\begin{split} a_1 \cdot r & = 1 \\ a_1 & = \frac{1}{r} \end{split}$$
Thus, the sum is the following: $$\begin{split} S & = \frac{\frac{1}{r}}{1-r} \\\\ S & =\frac{1}{r-r^2} \end{split}$$
Since we want the minimum value of this expression, we want the maximum value for the denominator, $-r^2+r$. The maximum x-value of a quadratic with leading coefficient $-a$ is $\frac{-b}{2a}$. $$\begin{split} r & = \frac{-(1)}{2(-1)} \\\\ r & = \frac{1}{2} \end{split}$$
Plugging $r$ $=$ $\frac{1}{2}$ into the quadratic yields: $$\begin{split} S & = \frac{1}{\frac{1}{2} -\left(\frac{1}{2}\right)^2} \\\\ S & = \frac{1}{\frac{1}{4}} \end{split}$$
Therefore, the minimum sum of our infinite geometric sequence is $\boxed{\textbf{(E)}\ 4}$. (Solution by akaashp11)
As an extension to find the maximum value for the denominator we can find the derivative of $-r^2+r$ to get $1-2r$. we know that this changes sign when $r = \frac{1}{2}$ so plugging it in into the original equation we find the answer is $\boxed{\textbf{(E)}\ 4}$.
## Solution 2
After observation we realize that in order to minimize our sum $\frac{a}{1-r}$ with $a$ being the reciprocal of r. The common ratio $r$ has to be in the form of $\frac{1}{x}$ with $x$ being an integer as anything more than $1$ divided by $x$ would give a larger sum than a ratio in the form of $\frac{1}{x}$.
The first term has to be $x$, so then in order to minimize the sum, we have minimize $x$.
The smallest possible value for $x$ such that it is an integer that's greater than $1$ is $2$. So our first term is $2$ and our common ratio is $\frac{1}{2}$. Thus the sum is $\frac{2}{\frac {1}{2}}$ or $\boxed{\textbf{(E)}\ 4}$. Solution 2 by No_One
(edited)
## Solution 3
We can see that if $a$ is the first term, and $r$ is the common ratio between each of the terms, then we can get $$S=\frac{a}{1-r} \implies S-Sr=a$$ Also, we know that the second term can be expressed as $a\cdot r$ notice if we multiply $S-Sr=a$ by $r$, we would get $$r(S-Sr)=ar \implies Sr-Sr^2=1 \implies Sr^2-Sr+1=0$$ This quadratic has real solutions if the discriminant is greater than or equal to zero, or $$S^2-4\cdot S \cdot 1 \ge 0$$ This yields that $S\le 0$ or $S\ge 4$. However, since we know that $S$ has to be positive, we can safely conclude that the minimum possible value of $S$ is $\boxed{\textbf{(E)}\ 4}$.
## Solution 4 (Quick Method)
Let the first term of the geometric series $x$. Since it must be decreasing, we have $x>1$ and the third term is $\frac{1}{x}$. Realize that by AM-GM inequality $x+\frac{1}{x} \ge 2$ with equality if $x = 1$. However, we established that $x>1$ so that means $x+\frac{1}{x} > 2$. So the sum of the first three terms of the sequence $x + \frac{1}{x} + 1$ is greater than $3$, and the geometric series keeps continuing infinitely. This means the sum continues increasing. The only answer choice greater than $3$ is $\boxed{\textbf{(E)}\ 4}$. ~skyscraper
## Solution 5 (Clever Algebra)
Let the first term be $k.$ The sum of the series is $\frac{k}{1- \frac{1}{k}} =\frac{k^2}{k-1}.$ Rewrite this as $\frac{k^2-2k+1}{k-1} +\frac{2k-1}{k-1} = k-1+\frac{2k-2}{k-1} +\frac{1}{k-1} = (k-1) + \left(\frac{1}{k-1}\right) + 2.$ By AM-GM we know that $(k-1) + \left(\frac{1}{k-1}\right) \ge 2$ so the minimum is $2+2 = \boxed{\textbf{(E)}\ 4}.$
## Solution 6 (Calculus)
Set the first term is $a.$, the common ratio should be $\frac{1}{a}.$
The sum to infinity of the series is $S=\frac{a}{1-\frac{1}{a}}=\frac{a^2}{a-1}.$
Since $S$ is positive, we have $a>1.$ Define the function $f(a)=\frac{a^2}{a-1}$ , the domain of this function is $a>1.$
Let $f^{'}(a)=\frac{2a^2-2a-a^2}{(a-1)^2}=\frac{a(a-2)}{(a-1)^2}=0.$ We solve that $a=2.$
It's easy to find that when $1 when $a>2, f^{'}(a)>0.$ Thus $f(a)$ has a minimum value when $a=2.$, which is $f(2)=4.$ Choose $\boxed{\textbf{(E)}\ 4}.$
~PythZhou |
# 2sqrt(28)3sqrt(63)-sqrt(49)
<-- Enter expression
Evaluate 2√283√63-√49
Term 1 has a square root, so we evaluate and simplify:
sqtot = 2
We have a product of 2 square root terms
The product of square roots is equal to the square root of the products
2√283√63 = 2√28*63
2√283√63 = 2√1764
Simplify 2√1764.
If you use a guess and check method, you see that 412 = 1681 and 432 = 1849.
Since 1681 < 1764 < 1849 the next logical step would be checking 422.
422 = 42 x 42
422 = 1764 <--- We match our original number!!!
Multiplying by our outside constant, we get 2 x 42 = 84
Therefore, 2√1764 = ±84
Term 2 has a square root, so we evaluate and simplify:
Simplify -1√49.
If you use a guess and check method, you see that 62 = 36 and 82 = 64.
Since 36 < 49 < 64 the next logical step would be checking 72.
72 = 7 x 7
72 = 49 <--- We match our original number!!!
Multiplying by our outside constant, we get -1 x 7 = -7
Therefore, -1√49 = ±-7
Group constants
84 - 7 = 77 |
# Comparing Fractions
In this lesson, you will learn about comparing fractions. Before you start this lesson, I recommend that you study or review my lesson about fractions
Before I show you two ways to compare fractions, you will need to learn the following.
Cross product: The answer obtained by multiplying the numerator of one fraction by the denominator of another
For instance, to get the cross products of the fractions below:
2 / 5 and 3 / 6
We can do 2 × 6 = 12 and 3 × 5 = 15.
Meaning of inequality sign:
The sign (>) means greater or bigger than.
For instance, 6 > 4
The sign (<) means smaller than
For instance, 4 < 6
Common denominator: When two or more fractions have the same denominator, we say that the fractions have a common denominator.
2 / 6 , 1 / 6 and 5 / 6 all have a common denominator
Now, here are two ways to compare fractions.
The first way is to do a cross product.
Let us compare 2 / 3 and 3 / 4
Start your cross product by multiplying the numerator of the fraction on the left by the denominator of the fraction on the right. You get 2 × 4 = 8
Then, multiply the numerator of the fraction on the right by the denominator of the fraction on the left. You get 3 × 3 = 9
Put 8 beneath 2 / 3 and 9 beneath 3 / 4
2 / 3 3 / 4 8 9
Since 8 is smaller than 9, then 2 / 3 < 3 / 4
Let us compare 5 / 6 and 6 / 8
5 × 8 = 40 and 6 × 6 = 36
Put 40 beneath 5 / 6 and 36 beneath 6 / 8
5 / 6 6 / 8 40 36
Since 40 is bigger than 36, then 5 / 6 > 6 / 8
A second method to use when comparing fractions is to first get a common denominator.
Let us compare again 5 / 6 and 6 / 8
Notice that you can multiply the denominator for the first fraction, which is 6 by 8 and multiply the denominator for the second fraction, which is 8 by 6 to get your common denominator.
Warning! Whatever you multiply the denominator, you have to multiply your numerator by the same thing so that you are in fact getting equivalent fractions.
5 / 6 becomes 40 / 48
6 / 8 becomes 36 / 48
Since 40 is bigger than 36, then 40 / 48 > 36 / 48
Therefore, 5 / 6 > 6 / 8
Rule #1
When two fractions have the same denominator, the bigger fraction is the one with the bigger numerator. Does that make sense?
Let's again use our Pizza in the lesson about fractions as an example. If your pizza has 10 slices and you eat 5.
That is
5 / 10
If you eat one more, that is 6 slices
That is
6 / 10
Now it has become obvious that
6 / 10 > 5 / 10
Rule # 2
When comparing fractions that have the same numerator, the bigger fraction is the one with the smaller denominator.
Once again, let us use our pizza as an example. Say that you bought two large pizzas and they are the same size.
Let's say that the first pizza was cut into 10 slices and the second was cut into 15 slices. No doubt if the second pizza is cut into 15 slices, slices will be smaller.
If you grab 2 slices from the first, the expression for the fraction
2 / 10
If you grab 2 slices from the second, the expression for the fraction
2 / 15
Slices for the latter will definitely smaller.
Therefore, 2 / 10 > 2 / 15
This lesson about comparing fractions is over.
Before I show you two ways to compare fractions, you will need to learn the following.
Cross product: The answer obtained by multiplying the numerator of one fraction by the denominator of another
For instance, to get the cross products of the fractions below:
2 / 5 and 3 / 6
We can do 2 × 6 = 12 and 3 × 5 = 15.
Meaning of inequality sign:
The sign (>) means greater or bigger than.
For instance, 6 > 4
The sign (<) means smaller than
For instance, 4 < 6
Common denominator: When two or more fractions have the same denominator, we say that the fractions have a common denominator.
2 / 6 , 1 / 6 and 5 / 6
all have a common denominator
Now, here are two ways to compare fractions.
The first way is to do a cross product.
Let us compare
2 / 3 and 3 / 4
Start your cross product by multiplying the numerator of the fraction on the left by the denominator of the fraction on the right. You get 2 × 4 = 8
Then, multiply the numerator of the fraction on the right by the denominator of the fraction on the left. You get 3 × 3 = 9
Put 8 beneath 2 / 3 and 9 beneath 3 / 4
2 / 3 3 / 4 8 9
Since 8 is smaller than 9, then
2 / 3 < 3 / 4
Let us compare
5 / 6 and 6 / 8
5 × 8 = 40 and 6 × 6 = 36
Put 40 beneath
5 / 6 and 36 beneath 6 / 8
5 / 6 6 / 8 40 36
Since 40 is bigger than 36, then
5 / 6 > 6 / 8
A second method to use when comparing fractions is to first get a common denominator.
Let us compare again
5 / 6 and 6 / 8
Notice that you can multiply the denominator for the first fraction, which is 6 by 8 and multiply the denominator for the second fraction, which is 8 by 6 to get your common denominator.
Warning! Whatever you multiply the denominator, you have to multiply your numerator by the same thing so that you are in fact getting equivalent fractions.
5 / 6 becomes 40 / 48
6 / 8 becomes 36 / 48
Since 40 is bigger than 36, then
40 / 48 > 36 / 48
Therefore, 5 / 6 > 6 / 8
Rule #1
When two fractions have the same denominator, the bigger fraction is the one with the bigger numerator. Does that make sense?
Let's again use our Pizza in the lesson about fractions as an example. If your pizza has 10 slices and you eat 5.
That is
5 / 10
If you eat one more, that is 6 slices
That is
6 / 10
Now it has become obvious that
6 / 10 > 5 / 10
Rule # 2
When comparing fractions that have the same numerator, the bigger fraction is the one with the smaller denominator.
Once again, let us use our pizza as an example. Say that you bought two large pizzas and they are the same size.
Let's say that the first pizza was cut into 10 slices and the second was cut into 15 slices. No doubt if the second pizza is cut into 15 slices, slices will be smaller.
If you grab 2 slices from the first, the expression for the fraction
2 / 10
If you grab 2 slices from the second, the expression for the fraction
2 / 15
Slices for the latter will definitely smaller.
Therefore, 2 / 10 > 2 / 15
This lesson about comparing fractions is over.
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# 1.4 Applications of exponentials
## 1.4 Applications of exponentials (EMBFC)
There are many real world applications that require exponents. For example, exponentials are used to determine population growth and they are also used in finance to calculate different types of interest.
## Worked example 16: Applications of exponentials
A type of bacteria has a very high exponential growth rate at $$\text{80}\%$$ every hour. If there are $$\text{10}$$ bacteria, determine how many there will be in five hours, in one day and in one week?
### Exponential formula
$\text{final population} = \text{initial population} \times (1 + \text{growth percentage})^{\text{time period in hours}}$
Therefore, in this case: $\text{final population} = \text{10}\left(\text{1,8}\right)^n$ where $$n =$$ number of hours.
### In $$\text{5}$$ hours
final population = $$\text{10}\left(\text{1,8}\right)^{\text{5}} \approx \text{189}$$
### In $$\text{1}$$ day = $$\text{24}$$ hours
final population = $$\text{10}\left(\text{1,8}\right)^{\text{24}} \approx \text{13 382 588}$$
### In $$\text{1}$$ week = $$\text{168}$$ hours
final population = $$\text{10}\left(\text{1,8}\right)^{\text{168}} \approx \text{7,687} \times \text{10}^{\text{43}}$$
Note this answer is given in scientific notation as it is a very big number.
## Worked example 17: Applications of exponentials
A species of extremely rare deep water fish has a very long lifespan and rarely has offspring. If there are a total of $$\text{821}$$ of this type of fish and their growth rate is $$\text{2}\%$$ each month, how many will there be in half of a year? What will the population be in ten years and in one hundred years?
### Exponential formula
$\text{final population} = \text{initial population} \times (1+\text{growth percentage})^{\text{time period in months}}$
Therefore, in this case: $\text{final population} = \text{821}(\text{1,02})^n$ where $$n =$$ number of months.
### In half a year = $$\text{6}$$ months
$$\text{final population} = \text{821}(\text{1,02})^6 \approx \text{925}$$
### In $$\text{10}$$ years = $$\text{120}$$ months
$$\text{final population} = \text{821}(\text{1,02})^{\text{120}} \approx \text{8 838}$$
### In $$\text{100}$$ years = $$\text{1 200}$$ months
$$\text{final population} = \text{821}(\text{1,02})^{\text{1 200}} \approx \text{1,716} \times \text{10}^{\text{13}}$$
Note this answer is also given in scientific notation as it is a very big number.
temp text
## Applications of exponentials
Textbook Exercise 1.7
Nqobani invests $$\text{R}\,\text{5 530}$$ into an account which pays out a lump sum at the end of $$\text{6}$$ years. If he gets $$\text{R}\,\text{9 622,20}$$ at the end of the period, what compound interest rate did the bank offer him? Give answer correct to one decimal place.
\begin{align*} A &= \text{9 622,20} \\ P &= \text{5 530} \\ n &= 6 \end{align*} \begin{align*} A &= P(1 + i)^n \\ \text{9 622,20} &= \text{5 530} (1 + i)^{6}\\ \frac{\text{9 622,20}}{\text{5 530}} &= (1 + i)^{6} \\ \sqrt[6]{\frac{\text{9 622,20}}{\text{5 530}}} &= 1 + i \\ \sqrt[6]{\frac{\text{9 622,20}}{\text{5 530}}} - 1 &= i \\ \therefore i &= \text{0,096709} \ldots \\ &= \text{9,7}\% \end{align*}
The current population of Johannesburg is $$\text{3 885 840}$$ and the average rate of population growth in South Africa is $$\text{0,7}\%$$ p.a. What can city planners expect the population of Johannesburg to be in $$\text{13}$$ years time?
\begin{align*} P &= \text{3 885 840} \\ i &= \text{0,007} \\ n &= 13 \end{align*} \begin{align*} A &= P(1 + i)^n \\ &= \text{3 885 840} (1 + \text{0,007} )^{13} \\ &= \text{4 254 691} \end{align*}
Abiona places $$\text{3}$$ books in a stack on her desk. The next day she counts the books in the stack and then adds the same number of books to the top of the stack. After how many days will she have a stack of $$\text{192}$$ books?
$$3; 6; 12; 24; 48; \ldots$$
\begin{align*} 3 \times 2^{n-1} &= \text{192} \\ 2^{n-1} &= 64 \\ &= 2^6 \\ \therefore n - 1 &= 6 \\ \therefore n &= 7 \end{align*}
A type of mould has a very high exponential growth rate of $$\text{40}\%$$ every hour. If there are initially $$\text{45}$$ individual mould cells in the population, determine how many there will be in $$\text{19}$$ hours.
\begin{align*} \text{Population } &= \text{Initial population } \times \left( 1 + \text{growth percentage } \right)^{\text{time period in hours} } \\ &= 45 (1 + \text{0,4} )^{19} \\ &= \text{26 893} \end{align*} |
Proof 1 uses the fact that the alternate interior angles formed by a transversal with two parallel lines are congruent. Since X and, $$\angle J$$ are remote interior angles in relation to the 120° angle, you can use the formula. Displaying top 8 worksheets found for - Sum Of Interior Angles In A Triangle. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"Answer: No, and here's an example to illustrate.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-is-the-sum-of-any-two-angles-of-a-triangle-always-greater-than-the-third-angle'\u003eread more\u003c/a\u003e"},"name":"🗯Is the sum of any two angles of a triangle always greater than the third angle? If the angles of a triangle are in the ratio 2:3:4, determine the three angles. 4: Interactive fol Angle sum property of a triangle Theorem 1: The angle sum property of a triangle states that the sum of interior angles of a triangle is 180°. The sides opposite to equal angles in a triangle are equal. 1.65 mJacquees/Height. A hyperbolic triangle, whose sides are arcs of these semicircles, has angles that add up to less than 180 degrees. Angles correspond to their opposite sides. The exterior angles, taken one at each vertex, always sum up to 360°. The sum of two sides of a triangle is greater than the third side. Q.2. What is the sum of interior angles of a Nonagon? By interior angle sum property of triangles, ∠A + ∠B + ∠C = 180 0 ⇒ ∠A + 60 0 + 45 0 = 180 0 ⇒ ∠A + 105 0 = 180 0 ⇒ ∠A = 180 -105 0 ⇒ ∠A = 75 0. 60 B. How old is Kylie? Tags: Question 6 . Angles in a triangle worksheets contain a multitude of pdfs to find the interior and exterior angles with measures offered as whole numbers and algebraic expressions. And again, try it for the square: Every triangle has six exterior angles (two at each vertex are equal in measure). You can do this. Label the angles A, B, and C. An exterior angle is supplementary to its adjacent triangle interior angle. Q. The sum of the interior angles is always 180° implies, ∠ x + ∠y + ∠z = 180°. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"Two angles are called complementary when their measures add to 90 degrees.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-are-complementary-angles'\u003eread more\u003c/a\u003e"},"name":"🗯What are complementary angles? The angles inside a triangle are called interior angles. 90. Because the sum of the angles is not equal 180°, the given three angles can not be the angles of a triangle. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"And the inside angle at each vertex of a polygon is always equal to (180° - deflection angle), because (new direction - old direction) + (reverse direction - new direction) = (reverse direction - old direction), which is always 180°.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-why-is-the-sum-of-the-interior-angles-of-a-triangle-equal-to-180'\u003eread more\u003c/a\u003e"},"name":"🗯Why is the sum of the interior angles of a triangle equal to 180? 100 C. 130 D. 150 The angles in any given triangle have two key properties: 1. This one is z. From the definition of an exterior angle, its sums up to the straight angle with the interior angles. This property is known as exterior angle property. The sum of all of the interior angles can be found using the formula S = (n – 2)*180. You can solve for Y. The measure of this angle is x. Unit 5 Section 6 : Finding Angles in Triangles. In a triangle, the exterior angle is always equal to the sum of the interior opposite angle. Step 3 : Arrange the vertices of the triangle around a point so that none of your corners overlap and there are no gaps between them. BLINDING LIGHTS. This may be one the most well known mathematical rules- The sum of all 3 interior angles in a triangle is 180 ∘ . incenter, Note The interior angles only add to 180° when the triangle is planar, meaning it is lying on a flat plane. This is also called angle sum property of a triangle . Therefore, straight angle ABD measures 180 degrees. For an ideal triangle, a generalization of hyperbolic triangles, this sum is equal to zero. Angle sum of a triangle, which appears to be more common and is more concise; Triangle postulate, which is the technical name of this topic, and is how Wolfram MathWorld refers to it; Sum of angles of a triangle, the current name. Since X and, $$\angle J$$ are remote interior angles in relation to the 120° angle, you can use the formula. Proof. Daniel L. A right triangle has only one right angle. Finding the Number of Sides of a Polygon. Exit Ticket (5 minutes) Lesson Summary The sum of the measures of the remote interior angles of a triangle is equal to the measure of the related exterior angle. 180 seconds . The exterior angles, taken one at each vertex, always sum up to 360°. 80. Sum of the Interior Angles of a Triangle Worksheet 1 RTF Sum of the Interior Angles of a Triangle Worksheet 1 PDF View Answers. List of trigonometric identities § Angle sum and difference identities, https://en.wikipedia.org/w/index.php?title=Sum_of_angles_of_a_triangle&oldid=997636359, Short description is different from Wikidata, Articles to be expanded from November 2013, Creative Commons Attribution-ShareAlike License, This page was last edited on 1 January 2021, at 14:37. Step 4 : What do you notice about how the angles fit together around a point ? You know the sum of interior angles is 900 °, but you have no The sum of the angles can be arbitrarily small (but positive). Solution : We know that, the sum of the three angles of a triangle = 180 ° 90 + (x + … In a triangle, the angle opposite to longer side is larger and vice-versa. You can do this. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"A hyperbolic triangle, whose sides are arcs of these semicircles, has angles that add up to less than 180 degrees.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-triangle-does-not-add-up-to-180-degrees'\u003eread more\u003c/a\u003e"},"name":"🗯What triangle does not add up to 180 degrees? How do I get pictures from iCloud to my iPhone? Sum of the Angle Measures in a Triangle is 180 ° - Justify. “SSS” is when we know three sides of the triangle, and want to find the missing angles….To solve an SSS triangle:use The Law of Cosines first to calculate one of the angles.then use The Law of Cosines again to find another angle.and finally use angles of a triangle add to 180° to find the last angle. How do you find an angle of a triangle with 3 sides? Why is the sum of the interior angles of a triangle equal to 180? A triangle has three angles, one at each vertex, bounded by a pair of adjacent sides. Try it first with our equilateral triangle: (n - 2) × 180 °(3 - 2) × 180 °Sum of interior angles = 180 ° Sum of angles of a square. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"The sum of the three angles of any triangle is equal to 180 degrees.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-is-the-sum-of-3-angles-of-a-triangle'\u003eread more\u003c/a\u003e"},"name":"🗯What is the sum of 3 angles of a triangle? "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"Angles around a point add up to 360°.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-what-is-the-sum-of-angles-at-a-point'\u003eread more\u003c/a\u003e"},"name":"🗯What is the sum of angles at a point? Which country has the hardest exams? 90 degrees. One can also consider the sum of all three exterior angles, that equals to 360°[7] in the Euclidean case (as for any convex polygon), is less than 360° in the spherical case, and is greater than 360° in the hyperbolic case. Corresponding and Alternate Angles are also congruent angles. Did you ever work on a jigsaw puzzle, devoting hours and hours to putting it together, only to get almost to the end and find out a piece is missing? The sum of the three angles of a triangle equals 180°. The influence of this problem on mathematics was particularly strong during the 19th century. Proof: • Rotate ΔABC about the midpoint of . So, the sum of three exterior angles added to the sum of three interior angles always gives three straight angles. Proof: Sum of all the angles of a triangle is equal to 180° this theorem can be proved by the below-shown figure. Proof 2 uses the exterior angle theorem. B. This postulate is equivalent to the parallel postulate. Here are three proofs for the sum of angles of triangles. The sum of the three angles of any triangle is equal to 180 degrees. Interior Angles of a Triangle Rule. where f is the fraction of the sphere's area which is enclosed by the triangle. Therefore, straight angle ABD measures 180 degrees. For ∆ABC, Angle sum property of triangle declares that. The diagram below shows the interior and exterior angles of a triangle.. It follows that a 180-degree rotation is a half-circle. So if the measure of this angle is a, the measure of this angle over here is b, and the measure of this angle is c, we know that a plus b plus c is equal to 180 degrees. Calculate the sum of interior angles of… 10 sides, so 8 triangles, so 8 x 180 degrees = 1440 degrees How many degrees in one interior angle… 8 sides, so 6 triangles, so 6 x 180 degrees = 1080 degrees in… Grab each of the points, and move them to increase or decrease their respective angles. So, the three angles of a triangle are 30°, 60° and 90°. a triangle have 3 sides n=3 Why Is The Sky Always Lighter Inside A Rainbow. Try it first with our equilateral triangle: (n - 2) × 180 °(3 - 2) × 180 °Sum of interior angles = 180 ° Sum of angles of a square. Thus, the sum of the interior angles of a triangle is 180°. Sum of the Interior Angles of a Triangle. The measure of one interior angle of a triangle is 45°. pg. …. It is also possible to calculate the measure of each angle if the polygon is regular by dividing the sum by the number of sides. Question: How Do I Download My Pictures From ICloud To My IPhone? Sum of the Interior Angles of a Triangle Worksheet 2 - This angle worksheet features 12 different triangles. Step 5 : What do you notice about how the angles fit together around a point ? This one's y. 100. Though Euclid did offer an exterior angles theorem specific to triangles, no Interior Angle Theorem exists. Example 3 : In a triangle, If the second angle is 5° greater than the first angle and the third angle is 5° greater than second angle, find the three angles of the triangle. I've drawn an arbitrary triangle right over here. A. Step 1 : Draw a triangle and cut it out. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"\"SSS\" is when we know three sides of the triangle, and want to find the missing angles.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-how-do-you-find-an-angle-of-a-triangle-with-3-sides'\u003eread more\u003c/a\u003e"},"name":"🗯How do you find an angle of a triangle with 3 sides? Answer:proved. Thus, the sum of the interior angles of a triangle is 180°.. Do all Triangle angles equal 180? Proof 1 How tall is Jacquees? The interior angles of any plane triangle can together form a straight line. And again, try it for the square: Therefore, a complete rotation is 360 degrees. Interior Angles and Polygons: The measures of the interior angles of any triangle must add up to {eq}180^\circ {/eq}. One can easily see how hyperbolic geometry breaks Playfair's axiom, Proclus' axiom (the parallelism, defined as non-intersection, is intransitive in an hyperbolic plane), the equidistance postulate (the points on one side of, and equidistant from, a given line do not form a line), and Pythagoras' theorem. Instead, you can use a formula that mathematically describes an interesting pattern about polygons and their interior angles. So, the measure of the third angle of the given triangle is 75 0. 2. [1] In the presence of the other axioms of Euclidean geometry, the following statements are equivalent:[2], The sum of the angles of a hyperbolic triangle is less than 180°. which is the center of … Angles B and B’ are corresponding angles, … So, the angle sum ∠A + ∠B + ∠C is equal to the angle sum ∠A’ + ∠B’ + ∠C’. Label the angles A, B, and C. Step 2 : Tear off each “corner” of the triangle. Answer: No, and here’s an example to illustrate. SURVEY . How are we supposed … Note that spherical geometry does not satisfy several of Euclid's axioms (including the parallel postulate.). And what I want to prove is that the sum of the measures of the interior angles of a triangle, that x plus y plus z is equal to 180 degrees. An exterior angle of a triangle is equal to the sum of the opposite interior angles. How many right angles does a triangle have? Sum of all the interior angles of the triangle … That is, Sum of the Three Angles in Any Triangle = 180 ° In the next part, we are going to justify this relationship. Sum of Interior Angles … Question 5: Do interior angles add up to 180º? Your first step should be to set up an equation where the sum of the angles adds up to 180̊. Step 1 : Draw a triangle and cut it out. Please indicate which you prefer. Angles between adjacent sides of a triangle are referred to as interior angles in Euclidean and other geometries. ... its interior angles add up to 3 × 180° = 540° And when it is regular (all angles the same), then each angle is 540 ° / 5 = 108 ° (Exercise: make sure each triangle here adds up to 180°, and check that the pentagon's interior angles add up to 540°) The Interior Angles of a Pentagon add up to 540° In a Euclidean space, the sum of angles of a triangle equals the straight angle (180 degrees, π radians, two right angles, or a half-turn). What is the sum of interior angles of a Heptagon? Sum of all the interior angles of the triangle … If two of the angles of a triangle are 30 and 70 degrees, the third angle measures... answer choices . The two sides of the triangle that are by the right angle are called the legs... and the side opposite of the right angle is called the hypotenuse. Quick Answer: Is Taxi Expensive In Abu Dhabi? 2: Interior angles of triangles pg. Angles between adjacent sides of a triangle are referred to as interior angles in Euclidean and other geometries. So, we all know that a triangle is a 3-sided figure with three interior angles. An exterior angle of a triangle is equal to the sum of the opposite interior angles. Triangle angle sum theorem: Which states that, the sum of all the three interior angles of a triangle is equal to 180 degrees. The three angles A’, B’, and C’ form together a straight angle (they are along the line l). right triangleThe right triangle has one 90 degree angle and two acute (< 90 degree) angles. Exterior angle is defined as the angle formed between a side of triangle and an adjacent side extending outward. "},{"@type":"Question","acceptedAnswer":{"@type":"Answer","text":"To prove the above property of triangles, draw a line \\overleftrightarrow {PQ} parallel to the side BC of the given triangle.\u003ca href='https://aahanaledlights.com/qa/question-why-is-the-sum-of-interior-angles-of-a-triangle-180.html#qa-how-do-you-prove-that-the-sum-of-all-angles-in-a-triangle-is-180'\u003eread more\u003c/a\u003e"},"name":"🗯How do you prove that the sum of all angles in a triangle is 180? 60 degrees. This is illustrated by the blue triangle, which has one vertex fixed at the origin and its two other vertices shaded in yellow and green. Angles in a triangle worksheets contain a multitude of pdfs to find the interior and exterior angles with measures offered as whole numbers and algebraic expressions. To Prove :- ∠4 = ∠1 + ∠2 Proof:- From If, $\alpha$, $\beta$ and $\gamma$ are three interior angles in a triangle, then $\alpha+\beta+\gamma = 180^\circ$ This basic geometrical property of a triangle is often used as a formula in geometry in some special cases. It is not to be confused with, "Angle sum theorem" redirects here. The sum of the measures of the interior angles of all triangles is 180°. For example, ∠ + ∠ = ∠. GoodRx makes money, What makes a double rainbow? Interior angle is defined as the angle formed between two adjacent sides of a triangle. a + b + c = 180º Triangle exterior angle theorem: Which states that, the exterior angle is equal to the sum of two opposite and non-adjacent interior angles. Answer:To Prove:-Sum of interior angles of a triangle is 180°.Steps:-Let us draw a ∆ABC.Then, draw a || line. A circle[5] cannot have arbitrarily small curvature,[6] so the three points property also fails. The sum of the interior angles of any triangle is 180°. )? In hyperbolic geometry, the interior angle sum of a triangle is less than $180^{\circ}$, whilst in elliptic geometry, the interior angle sum is more than $180^{\circ}$. Parallel postulate. ), for which this sum is sum of interior angles of a triangle to 180 degrees triangle... If two of the angles of a triangle is 45° 4: what is the sum of the points and! Of two sides of a triangle is equal to the base of the angle. Maybe it 's a piece you 'd been looking for on and off for a while old lady knives. Properties of interior angles midpoint of its difference from 180° is a of!, see, defined as the angle formed between two adjacent sides of a triangle Rule uses the fact the... 180° – 360° = 180° one at each vertex, always sum up to 180 degrees,... ( < 90 degree angle and two acute ( < 90 degree ).... The influence of this problem on mathematics was particularly strong during the 19th century 3 angles of triangle... Concerning sums of angles of a triangle is 180° the Sky always Lighter inside a triangle is 180 ° Justify. One right angle for the sum of the interior angles of a triangle is 180° of Euclid 's axioms including! Points, and are positioned to form a straight line and measures 180 degrees inside a rainbow 45° x! Are 30°, 60° and 90° tend to be confused with, angle sum of. No interior angle B, and the triangle 3 angles of triangles for - sum of two sides a. Sides of one triangle are equal in measure ) two right angles proven by Johann Heinrich.. Are all interior angles of the other two angles are also congruent angles right angle graphic organizer ( )... Triangles is 180° for which this sum is different } parallel to sum..., defined as the exterior angle theorem exists vertex are equal 8 worksheets found for this concept - =. Also called angle sum theorem '' redirects here six exterior angles of a triangle.... Are in the ratio 2:3:4, determine the three angles of a polygon 8.5! Parallel to the sum of the triangle … thus, the side BC of ∆ABC is.... Sum of the angle measures... answer choices formula that mathematically describes an interesting pattern about polygons and their angles. Baseball in the ratio 2:3:4, determine the three interior angles in triangles spherical..., [ 6 ] so the three angles of a triangle does sum of interior angles of a triangle make their money spherical triangle a... This theorem can be also defined, and the Euclidean triangle postulate can be hailed, how does make... Midpoint of the relation between angular defect and the Euclidean triangle postulate states that the sum of two sides a! Interior and exterior angles of a triangle has six exterior angles, see, defined as the formed. Property also fails of points at the fixed where ∠PRS is exterior angle is defined as the angle opposite equal... Goodrx make their money the gaokao is, Who is the sum of interior of... 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AMPS | THARC | KE8QZC | SFW | TSW
Back to the class
Section 3.5 #13: Find the exact value of $\cos \left( \dfrac{7\pi}{12} \right)$.
Solution: We will rewrite $\dfrac{7\pi}{12}$ as a difference of two angles we can handle using the unit circle. Notice that $$\dfrac{7\pi}{12} = \dfrac{9\pi}{12} - \dfrac{2\pi}{12} = \dfrac{3\pi}{4} - \dfrac{\pi}{6}.$$ Therefore using the difference identity $$\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$$ to compute $$\begin{array}{ll} \cos \left( \dfrac{7\pi}{12} \right) &= \cos \left( \dfrac{3\pi}{4} - \dfrac{\pi}{6} \right) \\ &= \cos \left( \dfrac{3\pi}{4} \right) \cos \left( \dfrac{\pi}{6} \right) + \sin \left( \dfrac{3\pi}{4} \right) \sin \left( \dfrac{\pi}{6} \right) \\ &= \left( - \dfrac{\sqrt{2}}{2} \right) \left( \dfrac{\sqrt{3}}{2} \right) + \left( \dfrac{\sqrt{2}}{2} \right) \left( \dfrac{1}{2} \right) \\ &= - \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} \\ &= \dfrac{\sqrt{2}-\sqrt{6}}{4}. \end{array}$$ note: the solution that Wolfram Alpha calculates, $-\dfrac{\sqrt{3}-1}{2\sqrt{2}}$, is equivalent to this -- just multiply by $\dfrac{\sqrt{2}}{\sqrt{2}}$ and you will see it
Section 3.5 #29: Find the exact value of $$\sin \left( \dfrac{\pi}{12} \right) \cos \left( \dfrac{7\pi}{12} \right) - \cos \left( \dfrac{\pi}{12} \right) \sin\left( \dfrac{7\pi}{12} \right).$$
Solution: Recall the difference identity for sine: $$\sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta).$$ Ths given problem looks like the right hand side of that identity with $\alpha=\dfrac{\pi}{12}$ and $\beta=\dfrac{7\pi}{12}$. So we may compute $$\begin{array}{ll} \sin \left( \dfrac{\pi}{12} \right) \cos \left( \dfrac{7\pi}{12} \right) - \cos \left( \dfrac{\pi}{12} \right) \sin\left( \dfrac{7\pi}{12} \right) &=\sin \left( \dfrac{\pi}{12} - \dfrac{7\pi}{12} \right) \\ &=\sin \left(-\dfrac{6\pi}{12} \right) \\ &=\sin \left( -\dfrac{\pi}{2} \right)\\ &=-1 \end{array}$$
Section 3.5 #77: Find the exact value of $$\cos \left( \tan^{-1} \left( \dfrac{4}{3} \right) + \cos^{-1} \left( \dfrac{5}{13} \right) \right).$$ Solution: Recall the sum identity for cosine: $$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta).$$ Therefore we see that $$\cos \left( \tan^{-1} \left( \frac{4}{3} \right) + \cos^{-1} \left( \frac{5}{13} \right) \right) = \cos \left( \tan^{-1} \left( \frac{4}{3} \right) \right) \cos \left( \cos^{-1} \left( \frac{5}{13} \right) \right) - \sin \left( \tan^{-1} \left( \frac{4}{3} \right) \right) \sin \left( \cos^{-1} \left( \frac{5}{13} \right) \right).$$ Now compute each piece.
$\underline{\cos \left( \tan^{-1} \left( \dfrac{4}{3} \right) \right)}$
Let $\theta=\tan^{-1} \left( \dfrac{4}{3} \right)$ so $\theta$ lies in quadrant I or IV. We see that $\tan(\theta)=\dfrac{4}{3}$ and so we can conclude that $\theta$ lies in quadrant I or III. Therefore $\theta$ lies in quadrant I (and so $\cos(\theta)>0$).
Since $\tan(\theta)=\dfrac{4}{3}$ and $\cos(\theta)>0$ we take $y=4$ and $x=3$. This means that $r^2=3^2+4^2=9+16=25$ so we can conclude that $r=\pm \sqrt{25}=\pm 5$, but a radius is never negative so $r=5$. Now using the idea that $\cos(\theta) = \dfrac{x}{r}$, we may compute $$\begin{array}{ll} \cos \left( \tan^{-1} \left( \dfrac{4}{3} \right) \right) &= \cos (\theta) \\ &= \dfrac{3}{5}. \end{array}$$
$\underline{\cos \left( \cos^{-1} \left( \dfrac{5}{13} \right) \right)}$
Using the inverse property, we may compute $$\cos \left( \cos^{-1} \left( \dfrac{5}{13} \right) \right) = \dfrac{5}{13}.$$
$\underline{\sin \left( \tan^{-1} \left( \dfrac{4}{3} \right) \right)}$
Let $\theta = \tan^{-1} \left(\dfrac{4}{3} \right)$. The analysis we did above for compute $\cos \left( \tan^{-1} \left( \dfrac{4}{3} \right) \right)$ all works the same and we have the same values $x=3$, $y=4$, and $r=5$. Therefore using the idea that $\sin(\theta) = \dfrac{y}{r}$ we compute $$\sin \left( \tan^{-1} \left( \dfrac{4}{3} \right) \right) = \sin (\theta) = \dfrac{4}{5}.$$
$\underline{\sin \left( \cos^{-1} \left( \dfrac{5}{13} \right) \right)}$
Let $\theta = \cos^{-1} \left( \dfrac{5}{13} \right)$ so we may conclude that $\theta$ lies in quadrant I or II. From this we have $\cos(\theta) =\dfrac{5}{13}$ from which we can conclude that $\theta$ lies in quadrant I or IV. Therefore $\theta$ lies in quadrant I and so we can conclude that $\sin(\theta)>0$.
Using $\cos(\theta) = \dfrac{5}{13}$ and the idea that $\cos(\theta) = \dfrac{x}{r},$ we can take $x=5$ and $r=13$. We can find the value of $y$ using the equation $x^2+y^2=r^2$. Plugging in our values yields $5^2+y^2=13^2$, or equivalently $$y^2 = 169 - 25 = 144,$$ so $$y = \pm \sqrt{144}=\pm 12$$ Since $\theta$ lies in quadrant I, we must take the positive value for $y$: $y=12$. Now we may use the idea that $\sin(\theta) = \dfrac{y}{r}$ to compute $$\begin{array}{ll} \sin \left( \cos^{-1} \left( \dfrac{5}{13} \right) \right) &= \sin(\theta)= \dfrac{12}{13} \end{array}$$
Now we put all this information together (continuing our original calculation) to compute $$\begin{array}{ll} \cos \left( \tan^{-1} \left( \dfrac{4}{3} \right) + \cos^{-1} \left( \dfrac{5}{13} \right) \right) &= \left(\dfrac{3}{5} \right) \left( \dfrac{5}{13} \right) - \left( \dfrac{4}{5} \right)\left( \dfrac{12}{13} \right) \\ &=\dfrac{15}{65} - \dfrac{48}{65} \\ &= -\dfrac{33}{65} \\ \end{array}.$$
Section 3.5 #85: Write as an algebraic expression containing $u$ and $v$: $$\cos(\cos^{-1}(u)+\sin^{-1}(v)).$$
Solution: Recall the sum identity for cosine: $$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta).$$ Applying this to the given problem gives us $$\begin{array}{ll} \cos(\cos^{-1}(u)+\sin^{-1}(v)) &=\cos(\cos^{-1}(u))\cos(\sin^{-1}(v))-\sin(\cos^{-1}(u))\sin(\sin^{-1}(v)) \\ &=u \cos(\sin^{-1}(v)) - v \sin(\cos^{-1}(u)). \end{array}.$$
$\underline{\cos(\sin^{-1}(v)}$
Let $\theta=\sin^{-1}(v)$, so $\theta$ lies in quadrant I or IV. From this we may conclude that $\cos(\theta)>0$. Also $\sin(\theta)=v$. Therefore using the fact that $\sin^2(\theta)+\cos^2(\theta)=1$ rewritten as $\cos(\theta)= \pm \sqrt{1-\sin^2(\theta)}$ and using the fact that $\cos(\theta)>0$ we discovered above, we may compute $$\cos(\sin^{-1}(v))=\cos(\theta) = \sqrt{1-\sin^2(\theta)}=\sqrt{1-v^2}.$$
$\underline{\sin(\cos^{-1}(u))}$
Let $\theta=\cos^{-1}(u)$ so $\theta$ lies in quadrant I or II, implying that $\sin(\theta)>0$. Also $\cos(\theta)=u$. Therefore using the fact that $\sin(\theta)=\sqrt{1-\cos^2(\theta)}$ and the formula $\cos(\theta)=u$ we may compute $$\sin(\cos^{-1}(u))=\sin(\theta)=\sqrt{1-\cos^2(\theta)}=\sqrt{1-u^2}.$$
Putting this all together, we can compute $$\cos(\cos^{-1}(u)+\sin^{-1}(v))=u\sqrt{1-v^2}-v\sqrt{1-u^2}.$$
Section 3.6 #27: Find an exact value of $\sin \left( - \dfrac{\pi}{8} \right)$.
Solution: Notice that this is the same as $\sin \left( -\dfrac{1}{2} \cdot \dfrac{\pi}{4} \right)$. So we use the half-angle identity, taking the negative value because $\dfrac{\pi}{8}$ lies in QIV to get $$\sin \left( -\dfrac{\pi}{8} \right) = -\sqrt{ \dfrac{1-\cos(\frac{\pi}{4})}{2}} = -\sqrt{\dfrac{1-\frac{\sqrt{2}}{2}}{2}}=\dfrac{\sqrt{2-\sqrt{2}}}{2}.$$
Section 3.6 #49: Establish the identity $$\cot(2\theta) = \dfrac{\cot^2(\theta)-1}{2\cot(\theta)}.$$
Solution: Compute $$\begin{array}{ll} \cot(2\theta) &= \dfrac{\cos(2\theta)}{\sin(2\theta)} \\ &= \dfrac{\cos^2(\theta)-\sin^2(\theta)}{2\sin(\theta)\cos(\theta)} \\ &= \dfrac{\cos^2(\theta)-\sin^2(\theta)}{2\sin(\theta)\cos(\theta)} \left( \dfrac{\frac{1}{\sin^2(\theta)}}{\frac{1}{\sin^2(\theta)}} \right) \\ &=\dfrac{\cot^2(\theta)-1}{2\cot(\theta)}, \end{array}$$ as was to be shown.
Section 3.6 #79: Find the exact value of $$\sin \left( 2 \sin^{-1} \left( \dfrac{1}{2} \right) \right).$$
Solution: By the double angle identity, $$\sin \left(2\sin^{-1} \left( \dfrac{1}{2} \right) \right)=2 \sin \left( \sin^{-1} \left( \dfrac{1}{2} \right) \right) \cos \left( \sin^{-1} \left( \dfrac{1}{2} \right) \right) = \cos \left( \sin^{-1} \left( \dfrac{1}{2} \right) \right).$$ Let $\theta=\sin^{-1}\left( \dfrac{1}{2} \right)$ then $\theta$ lives in QI or QIV. Also $\sin(\theta) = \dfrac{1}{2}$ so $\theta$ lives in QI or QII. Therefore $\theta$ lives in QI.
Since $\sin(\theta) = \dfrac{y}{r}$ we will take $y=1$ and $r=2$. Using $x^2+y^2=r^2$ we plug in our values and solve for $x$ to get $x= \pm \sqrt{4-1}=\pm\sqrt{3}$. Since $\theta$ is in quadrant I we take $x=\sqrt{3}$. Therefore we compute $$\sin \left( 2 \sin^{-1} \left( \dfrac{1}{2} \right) \right)=\cos\left( \sin^{-1} \left( \dfrac{1}{2} \right) \right)= \dfrac{\sqrt{3}}{2}.$$ |
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# A cubic vessel of height $2m$ is full of water. The work done in pumping the water out of the vessel is?A) $72.3kJ$B) $78.4kJ$C) $64.5kJ$D) $57.9kJ$
Last updated date: 19th Apr 2024
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Hint:Recall that the work done is defined as the measure of the amount of transfer of energy. It is said to be done when an external force is applied on an object or particle in order to move it over a distance in a specific direction. It is a vector quantity because it has both magnitude and direction.
Complete step by step solution:
Given that the height of the cubical vessel is, $h = 2m$
Suppose That the length of the cubic vessel is ‘l’
So its volume will be $v = {l^3}$
The centre of mass of the water in the cubical vessel from the ground is $h = \dfrac{l}{2}$
It is known that the density of the substance is the ratio of mass per unit volume. So, mass of the cubical vessel can be written as
$Mass = Density \times Volume$
$m = v\rho$
Substituting the value of volume, the mass becomes
$m = {l^3}\rho$
Work done in pumping out the water will be equal to the potential energy required to pump the water to the required height. So work done is given by
$w = mgh$
$w = mg\dfrac{l}{2}$
$\Rightarrow w = {l^3}\rho g\dfrac{l}{2}$
$\Rightarrow w = \dfrac{{\rho g{l^4}}}{2}$
Substituting the given values in the above equation, and solving
$\Rightarrow w = \dfrac{1}{2} \times 1000 \times 10 \times {(2)^4}$
$\Rightarrow w = 4900 \times 16$
$\Rightarrow w = 78400J = 78.4kJ$
The work done in pumping the water out of the vessel is $= 78.4kJ$`
Option B is the right answer.
Note: It is to be noted that the centre of mass is defined as the point, where the whole mass of the particles of the body or system is said to be concentrated. When there are a lot of particles in motion that are to be dealt with, then this concept is used. |
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# A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with?
Last updated date: 28th May 2024
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Hint: In this problem the lady spends money sequentially at different spaces. She spends a fraction of what is left on next purchase. This is the problem that will take us into a fractional equation giving the last solution that is money left one rupee.
Try to solve the problem along with the journey of the lady.
Consider the amount lady had initially $= x\,\,Rupees$
First, she spent half she had for hankie and gave one rupee to the beggar
Money spent:
= $\dfrac{x}{2} + 1$
Money left:
$= x - \left( {\dfrac{x}{2} + 1} \right) \\ \Rightarrow \dfrac{x}{2} - 1 \;$
Second, she spent half left on lunch with two rupee tip
Money spent:
$\Rightarrow \dfrac{{\dfrac{x}{2} - 1}}{2} + 2 \\ \Rightarrow \dfrac{x}{4} + \dfrac{3}{2} \;$
Money left:
$\Rightarrow \dfrac{x}{2} - 1 - \left( {\dfrac{x}{4} + \dfrac{3}{2}} \right) \\ \Rightarrow \dfrac{x}{4} - \dfrac{5}{2} \;$
Third, purchase she spent half of what was left on a book and 3 rupees on bus fare
Money Spent:
$\Rightarrow \left( {\dfrac{{\dfrac{x}{4} - \dfrac{5}{2}}}{2} + 3} \right) \\ \Rightarrow \dfrac{x}{8} + \dfrac{7}{4} \;$
Money left:
$\Rightarrow \dfrac{x}{4} - \dfrac{5}{2} - \left( {\dfrac{x}{8} + \dfrac{7}{4}} \right) \\ \Rightarrow \dfrac{x}{8} - \dfrac{{17}}{4} \;$
Money she left with is one rupee
$\Rightarrow \dfrac{x}{8} - \dfrac{{17}}{4} = 1 \\ \Rightarrow \dfrac{x}{8} = \dfrac{{21}}{4} \\ \Rightarrow x = 42\,Rupees \;$
Hence, she had 42 rupees at the start of the journey.
So, the correct answer is “42 rupees”.
Note: The concept of fraction is widely used in many areas like calculating age, estimating profit loss, getting the share of profit of one among the three, etc. The simplification involves all mathematical operations on one variable. The use of multiple operations makes the problem tricky. |
# Difference between revisions of "2015 AMC 8 Problems/Problem 2"
Point $O$ is the center of the regular octagon $ABCDEFGH$, and $X$ is the midpoint of the side $\overline{AB}.$ What fraction of the area of the octagon is shaded?
$\textbf{(A) }\frac{11}{32} \quad\textbf{(B) }\frac{3}{8} \quad\textbf{(C) }\frac{13}{32} \quad\textbf{(D) }\frac{7}{16}\quad \textbf{(E) }\frac{15}{32}$
$[asy] pair A,B,C,D,E,F,G,H,O,X; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot("A",A,dir(45)); dot("B",B,dir(90)); dot("C",C,dir(135)); dot("D",D,dir(180)); dot("E",E,dir(-135)); dot("F",F,dir(-90)); dot("G",G,dir(-45)); dot("H",H,dir(0)); dot("X",X,dir(135/2)); dot("O",O,dir(0)); draw(E--O--X); [/asy]$
## Solution 1
Since octagon $ABCDEFGH$ is a regular octagon, it is split into 8 equal parts, such as triangles $\bigtriangleup ABO, \bigtriangleup BCO, \bigtriangleup CDO$, etc. These parts, since they are all equal, are $\frac{1}{8}$ of the octagon each. The shaded region consists of 3 of these equal parts plus half of another, so the fraction of the octagon that is shaded is $\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\boxed{\text{(D) }\dfrac{7}{16}}.$
## Solution 2
$[asy] pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); a=midpoint(B--C); b=midpoint(C--D); c=midpoint(D--E); d=midpoint(E--F); e=midpoint(F--G); f=midpoint(G--H); g=midpoint(H--A); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot("A",A,dir(45)); dot("B",B,dir(90)); dot("C",C,dir(135)); dot("D",D,dir(180)); dot("E",E,dir(-135)); dot("F",F,dir(-90)); dot("G",G,dir(-45)); dot("H",H,dir(0)); dot("X",X,dir(135/2)); dot("O",O,dir(0)); draw(E--O--X); draw(B--F); draw(A--O); draw(D--H); draw(C--G); draw(a--e); draw(b--f); draw(c--g); draw(d--O); [/asy]$
The octagon has been divided up into 16 identical triangles (and thus they each have equal area). Since the shaded region occupies 7 out of the 16 total triangles, the answer is $\boxed{\textbf{(D)}~\dfrac{7}{16}}$.
## Solution 3
For starters what I find helpful is to divide the whole octagon up into triangles as shown here: $[asy] pair A,B,C,D,E,F,G,H,O,X; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot("A",A,dir(45)); dot("B",B,dir(90)); dot("C",C,dir(135)); dot("D",D,dir(180)); dot("E",E,dir(-135)); dot("F",F,dir(-90)); dot("G",G,dir(-45)); dot("H",H,dir(0)); dot("X",X,dir(135/2)); dot("O",O,dir(0)); draw(E--O--X); draw(C--O--B); draw(B--O--A); draw(A--O--H); draw(H--O--G); draw(G--O--F); draw(F--O--E); draw(E--O--D); draw(D--O--C); [/asy]$
Now it is just a matter of counting the larger triangles remember that $\triangle BOX$ and $\triangle XOA$ are not full triangles and are only half for these purposes. We count it up and we get a total of $\frac{3.5}{8}$ of the shape shaded. We then simplify it to get our answer: $\frac{7}{6}$ or $\textbf{(D)}$.
2015 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions |
1. Equation of hyperbola
write an equation of hyperbola having 3/2 as eccentricity. the point (1,-3) as focus and the line y=2 as its corresponding directrix
2. Equation of hyperbola: $\frac{(y-\beta)^2}{b^2}-\frac{(x-\alpha)^2}{a^2}=1$
Eccentricity: $e=\sqrt{1+\frac{b^2}{a^2}}$
Foci: $(\alpha, \beta\pm ae)$
Equation of directrices: $y-\beta=\pm \frac{a}{e}$
You have to solve $\alpha,\beta,a,b$ from this.
write an equation of hyperbola having 3/2 as eccentricity. the point (1,-3) as focus and the line y=2 as its corresponding directrix
Here comes a slightly different approach:
1. Let P denotes an arbitrary point on the hyperbola, r the distance of P to the focus F and l the distance of P to the directrix. Then
$\dfrac rl = e~\implies~r = e \cdot l$
2. Plug in the values you know:
$\sqrt{(x-1)^2+(y+3)^2} = 1.5(y+2)$
3. Simplify and re-arrange so that you'll get the equation of the hyperbola in standard form. I've got:
$-\dfrac{(x-1)^2}{\frac95}+\dfrac{\left(y+\frac65\right)^2}{\l eft(\frac56\right)^2}} = 1$
4. Originally Posted by earboth
Here comes a slightly different approach:
1. Let P denotes an arbitrary point on the hyperbola, r the distance of P to the focus F and l the distance of P to the directrix. Then
$\dfrac rl = e~\implies~r = e \cdot l$
2. Plug in the values you know:
$\sqrt{(x-1)^2+(y+3)^2} = 1.5(y+2)$
3. Simplify and re-arrange so that you'll get the equation of the hyperbola in standard form. I've got:
$-\dfrac{(x-1)^2}{\frac95}+\dfrac{\left(y+\frac65\right)^2}{\l eft(\frac56\right)^2}} = 1$
Thanks,but i get (y-6)^2/36 -(x-1)^2/45=1 when i used PF= e (PD) I wrong or not?
$\sqrt{(x-1)^2+(y+3)^2}=1.5(y-2)$ |
## Precalculus (6th Edition) Blitzer
The solution set of the given system of equations is $\left( {{e}^{1}},{{e}^{-1}},{{e}^{2}}\text{ and }{{e}^{-3}} \right)$.
The solution set of the given system of equations is $\left( {{e}^{1}},{{e}^{-1}},{{e}^{2}}\text{ and }{{e}^{-3}} \right)$. It is provided that, $\ln w=A,\ln x=B,\ln y=C\text{ and }\ln z=D$ Now, the resulting system of equations is: \begin{align} & A+B+C+D=-1 \\ & -A+4B+C-D=0 \\ & A-2B+C-2D=11 \\ & -A-2B+C+2D=-3 \end{align} First write the augmented matrix for the given system of equations. Therefore, the augmented matrix is, $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ -1 & 4 & 1 & -1 \\ 1 & -2 & 1 & -2 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ 0 \\ 11 \\ -3 \\ \end{matrix} \right]$ Now, reduce the matrix to row echelon form by using the row operation First apply ${{R}_{2}}\to {{R}_{2}}+{{R}_{1}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 1 & -2 & 1 & -2 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 11 \\ -3 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 0 & -3 & 0 & -3 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 12 \\ -3 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+{{R}_{1}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 0 & -3 & 0 & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 12 \\ -4 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{5}{{R}_{2}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & -3 & 0 & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ 12 \\ -4 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+3{{R}_{2}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & \frac{6}{5} & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{57}{5} \\ -4 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+{{R}_{2}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & \frac{6}{5} & -3 \\ 0 & 0 & \frac{12}{5} & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{57}{5} \\ \frac{-21}{5} \\ \end{matrix} \right]$ ${{R}_{3}}\to \frac{5}{6}{{R}_{3}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & \frac{12}{5} & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ \frac{-21}{5} \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}-\frac{12}{5}{{R}_{3}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & 0 & 9 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ -27 \\ \end{matrix} \right]$ ${{R}_{4}}\to \frac{1}{9}{{R}_{4}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ -3 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+\frac{5}{2}{{R}_{4}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ 2 \\ -3 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-{{R}_{4}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 0 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} 2 \\ \frac{-1}{5} \\ 2 \\ -3 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}-\frac{2}{5}{{R}_{1}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} 2 \\ -1 \\ 2 \\ -3 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-{{R}_{3}}$, to get $\left[ \left. \begin{matrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} 0 \\ -1 \\ 2 \\ -3 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$, to get $\left[ \left. \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} 1 \\ -1 \\ 2 \\ -3 \\ \end{matrix} \right]$ By the Gauss Jordan Elimination Method, $A=1,B=-1,C=2,D=-3$ Now, \begin{align} & \ln w=A \\ & w={{e}^{A}} \\ & w={{e}^{1}} \end{align} And \begin{align} & \ln x=B \\ & x={{e}^{B}} \\ & x={{e}^{-1}} \end{align} And \begin{align} & \ln y=C \\ & y={{e}^{C}} \\ & y={{e}^{2}} \end{align} And \begin{align} & \ln z=D \\ & z={{e}^{D}} \\ & z={{e}^{-3}} \end{align} Hence, the solution set of the given system of equations is $\left( {{e}^{1}},{{e}^{-1}},{{e}^{2}}\text{ and }{{e}^{-3}} \right)$. |
# How do you find the limit of (2x-8)/(sqrt(x) -2) as x approaches 4?
Oct 8, 2016
$8$
#### Explanation:
As you can see, you will find an indeterminate form of $\frac{0}{0}$ if you try to plug in $4$. That is a good thing because you can directly use L'Hospital's Rule, which says
$\mathmr{if} {\lim}_{x \to a} \frac{f \left(x\right)}{g \left(x\right)} = \frac{0}{0} \mathmr{and} \frac{\infty}{\infty}$
all you have to do is to find the derivative of the numerator and the denominator separately then plug in the value of $x$.
=>lim_(x->a)(f'(x))/(g'(x)
$f \left(x\right) = {\lim}_{x \to 4} \frac{2 x - 8}{\sqrt{x} - 2} = \frac{0}{0}$
$f \left(x\right) = {\lim}_{x \to 4} \frac{2 x - 8}{{x}^{\frac{1}{2}} - 2}$
$f ' \left(x\right) = {\lim}_{x \to 4} \frac{2}{\frac{1}{2} {x}^{- \frac{1}{2}}} = {\lim}_{x \to 4} \frac{2}{\frac{1}{2 \sqrt{x}}} = \frac{2}{\frac{1}{4}} = 8$
Hope this helps :)
Oct 8, 2016
${\lim}_{x \to 4} \frac{2 x - 8}{\sqrt{x} - 2} = 8$
#### Explanation:
As an addition to the other answer, this problem can be solved by applying algebraic manipulation to the expression.
${\lim}_{x \to 4} \frac{2 x - 8}{\sqrt{x} - 2} = {\lim}_{x \to 4} 2 \cdot \frac{x - 4}{\sqrt{x} - 2}$
$= {\lim}_{x \to 4} 2 \cdot \frac{\left(x - 4\right) \left(\sqrt{x} + 2\right)}{\left(\sqrt{x} - 2\right) \left(\sqrt{x} + 2\right)}$
$= {\lim}_{x \to 4} 2 \cdot \frac{\left(x - 4\right) \left(\sqrt{x} + 2\right)}{x - 4}$
$= {\lim}_{x \to 4} 2 \left(\sqrt{x} + 2\right)$
$= 2 \left(\sqrt{4} + 2\right)$
$= 2 \left(2 + 2\right)$
$= 8$ |
Home Math Numbers and Counting as much as 20 | Numbers From 11 to twenty
### Numbers and Counting as much as 20 | Numbers From 11 to twenty
We are going to be taught numbers and counting as much as 20 to acknowledge the numerals 11 by 20. Counting numbers are essential to know in order that we will perceive that numbers have an order and likewise have the ability to rely numbers simply. In our actual life we will relate numbers to portions.
### Let’s find out about numbers and counting as much as 20:
Numbers 11 to twenty:
Allow us to rely roses and write.
Oh! Sure, these are TEN roses and yet one more.
We name yet one more than 9 as TEN.
Identical means,
One multiple ten makes ELEVEN.
ELEVEN is written as 11.
Allow us to be taught extra numbers.
One Ten and One extra make ELEVEN. We write it as 11.
One Ten and Two extra make TWELVE. We write it as 12.
One Ten and Three extra make THIRTEEN. We write it as 13.
One Ten and 4 extra make FOURTEEN. We write it as 14.
One Ten and 5 extra make FIFTEEN. We write it as 15.
One Ten and Six extra make SIXTEEN. We write it as 16.
One Ten and Seven extra make SEVENTEEN. We write it as 17.
One Ten and Eight extra make EIGHTEEN. We write it as 18.
One Ten and 9 extra make NINETEEN. We write it as 19.
One Ten and Ten extra make TWENTY. We write it as 20.
Write numbers from 11 to twenty. One is finished for you.
T O Quantity Title 10 and 1 extra make 1 1 Eleven 10 and a couple of extra make _____ _____ __________ 10 and three extra make _____ _____ __________ 10 and 4 extra make _____ _____ __________ 10 and 5 extra make _____ _____ __________ 10 and 6 extra make _____ _____ __________ 10 and seven extra make _____ _____ __________ 10 and eight extra make _____ _____ __________ 10 and 9 extra make _____ _____ __________ Two teams of 10 is _____ _____ __________
### It’s essential to find out about numbers;
1. To acknowledge the numerals.
2. To know that numbers have an order.
3. To narrate or specific numbers to portions.
Studying numbers and counting as much as 20.
## You may like these
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• ### Counting Earlier than, After and Between Numbers as much as 10 | Quantity Counting
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• ### What’s Zero? | Idea of 0 | Addition Property of ‘Zero’ | Train
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Youngsters let’s benefit from the story about seasons. Right here we’ll talk about concerning the 4 seasons and the length. Some months are too sizzling and a few are too chilly. The interval of sizzling months is named the recent
• ### Counting Ahead and Backward upto 10 | Counting Numbers from 10 to 1
Studying of counting ahead and backward upto 10 ought to by no means be rushed by. Counting ahead and backward is a basic idea within the main stage that must be dealt with properly. As soon as the kid understands the essential idea, there can be no problem
• ### One Lower than Numbers upto 10 | Counting One Much less | Be taught 1 Much less Than
What’s one lower than? 1 lower than means we have to subtract or rely one much less variety of the given numbers. Right here, we’ll be taught counting one lower than upto quantity 10. Examples of counting 1 lower than as much as quantity 10 are given as follows.
• ### One Greater than Numbers upto 10 | Counting One Extra | Be taught 1 extra Than
1 greater than means we have to add or rely yet one more quantity to the given numbers. Right here, we’ll be taught counting yet one more than upto quantity 10. Examples of counting 1 greater than as much as quantity 10 are given as follows.
● Time
Kindergarten Math Actions |
# Story Problems
It’s a story– not a problem. Don’t create some problem you ain’t got. You can use algebra or not. Look.
Story
Jack and Diane are playing poker. After one hand, Diane has 2 more chips than Jack. After the next hand, Diane wins 5 more chips from Jack. At this point, how many more chips does Diane have than Jack?
Explanation Using Algebra
Notice the word more in the question. We will never know how many chips each person has. The question asks what is the difference between Jack’s amount and Diane’s amount. So, the equation is (Diane) – (Jack) = (difference).
Now make some mathematical terms.
After the first hand, Diane has d chips. Jack has (d – 2). But then they play another hand, and Diane wins 5, which makes her term (d + 5). Jack loses 5 more, so Jack’s term is now (d – 7).
Subtract these two terms to find the difference.
(d + 5) – (d – 7) =
When the negative 1 in front of the second set of parentheses is distributed, the 7 becomes positive. The ds cancel, and the answer is 12.
Diane has 12 more chips than Jack.
Simple Explanation
Play god. Pretend you are some Greek god of fortune. Give Jack and Diane the same number of chips.
Jack = 10 , Diane = 10
After one hand, Diane has 2 more chips.
Jack = 9 , Diane = 11
After another hand, Diane wins 5 more.
Jack = 4, Diane =16
Now Diane has 12 more chips than Jack. |
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# Prime numbers are a mathematical mystery.
## Despite many prominent mathematicians
deriving equations to produce prime numbers, there is really no concrete way to find all prime
numbers. From Math Apps this past semester and from reading Math Girls, I have become very
interested in prime numbers. I have gathered what I learned from Math Apps and Math Girls as
well as conducted some other research to provide insight into the quest to solve the mystery of
prime numbers and why they are so useful in our modern world.
## A Very Impractical Way to Find Prime Numbers:
You can find primes using the Sieve of Eratosthenes. We practiced this in Math Apps and I read
about it in Math Girls. Here is an example of it that we did in Math Apps, sifting out all of the
multiples each number until we found all the primes up to 432. The prime numbers are
highlighted in blue.
Figure 1
Math Girls illustrated the idea of the sieve beautifully, if another visual helps you to imagine this
way of finding primes. The prime numbers can fall all the way down through the sieve because
they do not have any factors.
Figure 2
The issue with this method is that it is slow and impractical for finding HUGE prime numbers.
Many mathematicians have created equations to make finding prime numbers painless.
## Mathematician Attempts to Find Primes:
n
Marin Mersenne (French mathematician, 1588-1648) found that Mn = 2 - 1 generates
prime numbers, where n is any integer. It is not known whether Mersenne primes are
infinite. However, this equation does not work for all integers. Heres an example:
Figure 3
As you can see, this works when integers 2, 3, and 5 are inputted into the equation, however
not when 4 is inputted. That is why some primes are categorized as Mersenne primes, because
it does not generate all primes.
Sophie Germain (French mathematician, 1776-1831) said that if p is a prime number,
and 2p+1 also generates a prime number, then p is a Sophie Germain prime. This is a
way that prime numbers are categorized. Here are some examples:
Figure 4
This seems to work until you input 7 and get 15 , which is not a prime number. This is the
reason some primes are Sophie Germain primes and others are not, making Sophie Germain
primes special.
Leonhard Euler (Swiss mathematician, 1707-1783) derived the equation
P(n) = n2 + n + 41 to generate primes, where n an integer. But yet again it does not work
for all integers. Here is a chart of this equation for all positive integers up to 74. The
composite (not prime) results are underlined in red.
Figure 5
Stanislaw Ulam (Polish-American mathematician, 1909-1984) created the Ulam Spiral,
which is a visual representation of prime numbers that sort of creates a pattern. Fun fact,
he found this pattern while doodling during a boring lecture! The pattern can be found
when overlaying Eulers equation: P(n) = n2 + n + 41.
Figure 6
Obviously, not all primes are included in this spiral (the lighter gray dots are also primes), but it
is cool that there is some visible pattern and that it appears to not be totally random.
There is no Conclusion:
The above mathematicians are not the only people who have embarked on solving the
prime mystery, and there are many, many more that I did not even mention (Fermat and Escott
are also famous, just to name a few). The real takeaway is that we have not come up with a way
to generate every single prime number, and that is why it is a mystery. Even though this is very
frustrating, I think it is really cool that there are mysteries in a system that seems so predictable
and concrete. Who knows, maybe you will see me in the newspaper 40 years from now with a
new way to find prime numbers.
## Why Primes are Important in the Real World:
Prime numbers are used in cryptography. Now I admit, I dont know very much about
cryptography, so I have asked my friend Rob Vitali to help me out. I interviewed him and asked
him to explain how prime numbers are used in cryptography. I am going to try and show my
general understanding of the subject and explain what Rob taught me below:
The reason that prime numbers are useful is because there are only two factors, the number
and one. When you have composite numbers, there are multiple factors so these keys can
repeat. So someone can unintentionally decode something because the factors are available.
When you privatize something, it is privatized in a piece of code that can be repeated. But when
you have a prime number the only way to repeat it is to find the prime number.
I mostly understood this explanation, besides his mention of keys. He tried to explain this
key concept to me, but it is very complicated. I think I got the gist though. So computer codes
are made up of keys. Lets say one key is 1, 2, and 3 or a, b, and c. The numbers or letters in
the keys get replaced with other numbers and letters, and thats how people hack. My
understanding is that if you use a prime number in your key, it is very difficult to figure out the
factors to plug in that break the key open.
The advantage of using prime numbers is the factors dont correlate with an unintentional
number that someone just tries to plug in. The reason you dont want to use a non-prime
number is because theres repetition in code that can decode something unintentionally. So you
are trying to make something private and you use the number 12 for example, which has factors
of 2, 2, and 3. So lets say a hacker is trying to decode something you have so they use the
number 6. Since 6 has factors of 2 and 3, certain things are going to be able to be decoded. Not
everything will be decoded correctly, but certain things will be unencrypted because it has the
same factors. But since prime numbers dont have the other factors, you are going to have to
find that exact prime number in order to decode.
My next question was about the enormous primes that people get paid to find. We talked
in Math Apps about the contests offering hundreds of thousands of dollars to the person who
discovers the largest prime number. The largest known prime number currently is 22,338,618
digits, or 274,207,281
1, and was found on January 25th in the Great Internet Mersenne Prime
Search. The reason these prime numbers are valuable is, Youre not going to think of a
gigantic prime number thats in the trillions. Someone would have to create a program to go
from one prime number to the next and check them all and do all of the steps.
Big Take Away: My next quest should be to find gigantic primes and sell them -- I would
make a ton of money.
Works Cited
"Largest known prime number." Wikipedia. Wikimedia Foundation, n.d. Web. 01 Feb. 2017.
https://en.wikipedia.org/wiki/Largest_known_prime_number.
## "Leonhard Euler." Wikipedia. Wikimedia Foundation, n.d. Web. 29 Jan. 2017.
https://en.wikipedia.org/wiki/Leonhard_Euler.
## "Mersenne Prime." Wikipedia. Wikimedia Foundation, n.d. Web. 29 Jan. 2017.
https://en.wikipedia.org/wiki/Mersenne_prime.
"Sophie Germain Prime." Wikipedia. Wikimedia Foundation, n.d. Web. 29 Jan. 2017.
https://en.wikipedia.org/wiki/Sophie_Germain_prime.
## "Stanislaw Ulam." Wikipedia. Wikimedia Foundation, n.d. Web. 29 Jan. 2017.
https://en.wikipedia.org/wiki/Stanislaw_Ulam.
Yki, Hiroshi, and Tony Gonzalez. Math Girls Talk About Integers: Fundamental Skills for
Advanced Mathematics. Austin, TX: Bento , 2014. Print. |
# Which Diagram Could Be Used To Prove △abc ~ △dec Using Similarity Transformations?
Similarity transformations are a powerful tool for proving that two figures are similar. By using similarity transformations, one can prove that two figures are similar without needing to calculate any angles or side lengths. This article will discuss which diagrams can be used to prove that two triangles are similar using similarity transformations.
## Understanding Similarity Transformations
Similarity transformations are a type of transformation that preserves the shape of a figure, but not necessarily the size. There are two types of similarity transformations: dilation and congruence. A dilation is a transformation that enlarges or reduces the size of a figure while preserving the shape. A congruence is a transformation that preserves both the shape and size of a figure.
When two figures are similar, they can be related by a similarity transformation. This means that one figure can be transformed into the other by a dilation or a congruence.
## Proving △abc ~ △dec Using Diagrams
The best way to prove that two triangles are similar is to use a diagram. A diagram can be used to show how the two triangles can be related by a similarity transformation.
One way to prove that △abc ~ △dec is to draw a diagram showing the two triangles and the similarity transformation that relates them. The diagram should show the two triangles, the similarity transformation, and the points of correspondence between the two triangles. This will show that the two triangles are related by a similarity transformation and thus are similar.
Another way to prove that △abc ~ △dec is to draw a diagram showing the two triangles and the similarity transformation that relates them. The diagram should show the two triangles, the similarity transformation, and the points of correspondence between the two triangles. This will show that the two triangles are related by a similarity transformation and thus are similar.
Finally, a third way to prove that △abc ~ △dec is to draw a diagram showing the two triangles and the similarity transformation that relates them. The diagram should show the two triangles, the similarity transformation, and the points of correspondence between the two triangles. This will show that the two triangles are related by a similarity transformation and thus are similar.
In conclusion, similarity transformations are a powerful tool for proving that two figures are similar. By using diagrams, one can prove that two triangles are similar using similarity transformations without needing to calculate any angles or side lengths. |
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# Area of a Circle
## Pi times the radius squared.
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Area of a Circle
Do you know the area of a circle is? Well to think about this, let's look once again at the small round table at Jillian's house.
In an earlier Concept, you learned how to measure around the edge of the circumference using the diameter or the radius. The diameter of this table is 4 feet.
Now, you are going to learn how to measure the flat top of the circle. The space inside the circumference of a circle is known as the area of the circle.
Given the diameter, what is the area of this table?
This Concept is all about calculating area. By the end of it, you will know how to solve this problem.
### Guidance
In an earlier Concept, you learned about the parts of a circle and about finding the circumference of a circle. This Concept is going to focus on the inside of the circle. The inside of a circle is called the area of the circle.
What is the area of a circle?
Remember back to working with quadrilaterals? The area of the quadrilateral is the surface or space inside the quadrilateral. Well, the area of a circle is the same thing. It is the area inside the circle that we are measuring.
In this picture, the area of the circle is yellow. How do you measure the area of a circle?
To figure out the area of a circle, we are going to need a couple of different measurements. The first one is pi. We will need to use the numerical value for pi, or 3.14, to represent the ratio between the diameter and the circumference.
The next measurement we need to use is the radius. Remember that the radius is the distance from the center of a circle to the edge, or is 1/2 of the circle's diameter.
We calculate the area of a circle by multiplying the radius squared (multiplied by itself) by pi (3.14).
Here is the formula.
\begin{align*}A = \pi r^2\end{align*}
Find the area of the circle.
Next, we use our formula and the given information.
\begin{align*}A & = \pi r^2\\ A & = (3.14)(3^2)\\ A & = (3.14)(9)\\ A & = 28.26 \ mm^2\end{align*}
What about if you have been given the diameter and not the radius?
If you have been given the diameter and not the radius, you can still figure out the area of the circle. You start by dividing the measure of the diameter in half since the radius is one-half the measure of the diameter.
Then you use the formula and solve for area.
Notice that the diameter is 6 inches. We can divide this in half to find the radius.
6 \begin{align*}\div\end{align*} 2 \begin{align*}=\end{align*} 3 inches \begin{align*}=\end{align*} radius
Next, we substitute the given values into the formula and solve for the area of the circle.
\begin{align*}A & = \pi r^2\\ A & = (3.14)(3^2)\\ A & = (3.14)(9)\\ A & = 28.26 \ in^2\end{align*}
Now it’s time for you to try a few on your own. Find the area of the circle using the given radius or diameter.
#### Example A
\begin{align*}r = 12 \ cm\end{align*}
Solution: 452.16 sq. cm.
#### Example B
\begin{align*}r = 18 \ cm\end{align*}
Solution: 1017.36 sq. cm
#### Example C
\begin{align*}d = 14 \ in\end{align*}
Solution: 153.86 sq. in.
Now let's go back to the original problem from the beginning of this Concept.
Do you know the area of a circle is? Well to think about this, let's look once again at the small round table at Jillian's house.
In an earlier Concept, you learned how to measure around the edge of the circumference using the diameter or the radius. The diameter of this table is 4 feet.
Now, you are going to learn how to measure the flat top of the circle. The space inside the circumference of a circle is known as the area of the circle.
Given the diameter, what is the area of this table?
If you have been given the diameter and not the radius, you can still figure out the area of the circle. You start by dividing the measure of the diameter in half since the radius is one-half the measure of the diameter.
The diameter of the table is 4 feet, so the radius is 2 feet.
Then you use the formula and solve for area.
Next, we substitute the given values into the formula and solve for the area of the circle.
\begin{align*}A & = \pi r^2\\ A & = (3.14)(2^2)\\ A & = (3.14)(4)\\ A & = 12.56 \ ft^2\end{align*}
### Vocabulary
Here are the vocabulary words in this Concept.
Area
the surface or space of the figure inside the perimeter.
the measure of the distance halfway across a circle.
Diameter
the measure of the distance across a circle
Squaring
uses the exponent 2 to show that a number is being multiplied by itself. \begin{align*}3^2 = 3 \times 3\end{align*}
Pi
the ratio of the diameter to the circumference. The numerical value of pi is 3.14.
### Guided Practice
Here is one for you to try on your own.
The diameter of a circle is 13 feet. What is the area of the circle?
To work on this problem, we must first divide the diameter in half to find the radius.
\begin{align*}13 \div 2 = 6.5\end{align*}
Next, we substitute the given values into the formula for area and solve.
\begin{align*}A & = \pi r^2\\ A & = (3.14)(6.5^2)\\ A & = (3.14)(42.25)\\ A & = 132.665 \ ft^2\end{align*}
Next, we can round up the nearest hundredth.
The answer is 132.67 sq. feet.
### Video Review
Here are videos for review.
### Practice
Directions: Find the area of the following circles given the radius.
1. \begin{align*}r = 4 \ in\end{align*}
2. \begin{align*}r = 5 \ cm\end{align*}
3. \begin{align*}r = 8 \ in\end{align*}
4. \begin{align*}r = 2 \ cm\end{align*}
5. \begin{align*}r = 7 \ m\end{align*}
6. \begin{align*}r = 9 \ in\end{align*}
7. \begin{align*}r = 10 \ ft\end{align*}
8. \begin{align*}r = 11 \ cm\end{align*}
9. \begin{align*}r = 20 \ ft\end{align*}
10. \begin{align*}r = 30 \ miles\end{align*}
Directions: Find the area of the following circles given the diameter.
11. \begin{align*}d = 10 \ in\end{align*}
12. \begin{align*}d = 12 \ m\end{align*}
13. \begin{align*}d = 14 \ cm\end{align*}
14. \begin{align*}d = 16 \ ft\end{align*}
15. \begin{align*}d = 18 \ in\end{align*}
16. \begin{align*}d = 22 \ ft\end{align*}
17. \begin{align*}d = 24 \ cm\end{align*}
18. \begin{align*}d = 28 \ m\end{align*}
19. \begin{align*}d = 30 \ m\end{align*}
20. \begin{align*}d = 36 \ ft\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes |
# Ideas in Geometry/Analytic Geometry
Analytic Geometry is covered in section 1.2.3 of our textbook.
1. What is analytic geometry?
2. What is it used for?
3. What do I need to know and be able to do for the homework and the exam?
Below are the responses I came up with to answer my questions.
## Definition
Initially, I looked at this section and found no clear definition as to what Analytic Goemetry was actually about. I saw a nice little story about Sir Isaac Newton, a plane with a line and a curve, and a formula that seemed a little complicated, but no idea what we were talking about.
After a little research, my definition of Analytic Geometry is this: The study of geometric objects within the coordinate plane, or you could call it Coordinate Geometry.
This means that any geometric object which can be drawn on a coordinate plane is part of this realm. This includes all of your basic shapes - circles, quadrilaterals, triangles, etc.
## Why Do We Use This Stuff?[1]
Analytic Geometry can be used for many things in math:
### Coordinates
The most basic form of this would be the coordinate plane. On which we can plot and manipulate points on a plane using a values of x and y.
### Distance and Angles
By using our coordinate planes, we can determine distance between points along with the angles created by line in relationship to other lines.
This linked picture show us the relationship between angles, distance, and coordinate planes.
### Equations of Curves
Analytic Geometry can also be used to plot curves of equations that use polynomials and other equations that require an input of x and which results in an output of y (x and y act would be coordinates for a point on a plane).
This picture shows up a curve in black with a line tangent to the curve:
### Other Themes
Below are some other important themes of analytical geometry, some of which we need not worry about right now:
Many of these problems involve a good deal of linear algebra and Lesson Four: Algebraic Geometry.
## What Do We Need To Know For This Class?
As you can see, much of what we see in Analytic Geometry will be used throughout this class, but for the purposes of this particular chapter and section, we will focus on the concept of slope of a tangent line at a given point on a curve (In Calculus it is known as the concept of a derivative).
For our example in the book we are given a picture similar to this:
Where:
h is a small number which approaches 0 (For Example: 0.1, 0.01, 0.001...).
x is our input number.
f(x) is our function f with the x value plugged into it (represented by the green curve).
Is this making sense? If not think of the picture in terms of basic slope:
Slope (m) = Rise (Change in y)/Run (Change in x)
So this pictures really shows that our formula to solve this is: m = f(x+h) - f(h)/h Because: f(x+h) - f(h) is the change in the y value and h is the change in x.
### Solving for the Slope of a Tangent Line Given an Equation and a Point
On page 18 of the .pdf textbook, you see the process for solving this kind of problem. To sum it up:
• Set up the formula for slope of a tangent line with the appropriate values inserted for the variables.
• Gather a selection of numbers approaching 0 (0.1, 0.01, 0.001...)
• Substitute these numbers in for h.
• All of the numbers should approach the slope of the tangent line.
### The Calculus Derivative Trick
There is actually an easier process for to find the slope of a tangent line at a point on a curve.
So, let's look at the first example shown on page 18.
On this particular problem, we have taken the derivative of f(x). The derivative of this equation gives us another equation which will tell us the slope of the tangent line at any value of x. The derivative is found by multiplying the power of the exponent of a polynomial by the number in front of the varialbe, then reducing the exponent by 1. Here is the general formula.
$f(x) = x^r\,$,
where r is any real number, then
$f'(x) = rx^{r-1}\,$,
wherever this function is defined.
So by this idea we arrive at :$f'(x) = 2x\,$.
And after substituting 2 for x, our answer for this example the slope is 4.
#### Why does this work?
Let's look at an example when we are solving for an unknown x value (xo) for the same f(x) as in our example.
When this formula is broken down algebraically it becomes clear how this works out. If you were to look at the original method for our example and reduce the formula before substituting smaller values of h, you would notice that the process is similar to the one above.
# DISCLAIMER
However, now that we have a nice method for solving for the slope of the tangent line, I must remind you that this is a shortcut. The key to this section for this class is to be able to explain why all of this makes sense in terms of the coordinate plan and values of x and h. Once you move on to Calculus, you will do much more work with this shortcut, but the key to doing well in MATH 119 is to understand these concepts in geometric terms.
# References
Analytic Geometry from Wolfram Math
Analytic Geometry formulas from Dr. Math
MATH 119 Wiki: Topic:Ideas_in_Geometry |
# Recamán Sequence: Definition & Creating Terms
Share on
## What is a Recamán sequence?
There are two different sequences attributed to Columbian puzzle maker Bernardo Recamán Santos:
Recamán sequence #1 (entry A5132 in the OEIS) is usually the one referred to as “The” Recamán Sequence. It is defined by [1]:
• an = an-1 – n, if an-1 – n > 0 and an-1 has already occurred in the sequence.
• Otherwise, an = an-1 + n.
The first few numbers are:
{1, 3, 6, 2, 7, 13, 20, …}.
Recamán sequence #2 (entry A8336 in the OEIS) is defined by:
• an+1 = an / n if n divides an
The first few numbers are:
{1, 1, 2, 6, 24, 120, …}.
## Intuitive way to Create the Terms of the Recamán Sequence
Neil Sloane of ATT Labs (creator of OEIS) shed a little light on how the sequence is created in a 2008 Math Factor podcast [2].
Step 1: Write down the numbers
1 2 3 4 5 6 7 8 9
with gaps (we’re going to write the terms in those gaps).
Step 2: Write down the first term, which is 1, between the 1 and the 2:
1 (1) 2 3 4 5 6 7 8 9.
Step 2: Calculate the second term.
Ask yourself the question: Can we subtract 2 (the second number in our list) from 1 (the first term)?. We can’t (negative numbers or zero are not allowed), so we have to add 2 instead. 2 + 1 = 3, so:
1 (1) 2 (3) 3 4 5 6 7 8 9.
Step 2: Calculate the third term in the same way.
Ask yourself the question: Can we subtract 3 (the third number in our list from 3 (the second term). We can’t, so we have to add: 3 + 3 = 6, so:
1 (1) 2 (3) 3 (6) 4 5 6 7 8 9.
Continue in this way, obeying the basic rules:
Numbers can’t be zero, negative, or something already in the sequence.
If any of these conditions appear, just add instead of subtract. Note that if you add and get a number that has appeared before, that’s OK. It’s only not allowed if you subtract.
## Deciphering Recaman’s sequence
Although 10230 terms of Recaman’s sequence have been computed, but it still remains a mystery [2] despite the OEIS’s calculation of an enormous number of terms. No one knows if every number will eventually appear; Those dedicated to deciphering the sequence have dubbed it “How to Recamán’s life” [3].
As of the January 2018, Prime Curious! reported that the smallest missing prime in the Sequence is 966727 (as of January 2018).
## References
[1] Ding, C. (2012). Sequences and Their Applications. Springer London.
[2] Math Factor Podcast. Retrieved April 9, 2021 from: http://mathfactor.uark.edu/2008/05/dw-the-online-encyclopedia-of-integer-sequences/
[2] Myers, J. et al. (2020). Three Cousins of Recaman’s Sequence. Retrieved April 9, 2021 from: https://ui.adsabs.harvard.edu/abs/2020arXiv200414000M/abstract
[3] Roberts, S. (2015). Genius At Play: The Curious Mind of John Horton Conway. Bloomsbury Publishing.
CITE THIS AS:
Stephanie Glen. "Recamán Sequence: Definition & Creating Terms" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/recaman-sequence-definition/
------------------------------------------------------------------------------
Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free! |
# What is the vertex form of y = 3x^2 − 50x+300 ?
Apr 22, 2018
$y = 3 {\left(x - \frac{25}{3}\right)}^{2} + \frac{275}{3}$
#### Explanation:
$\text{the equation of a parabola in "color(blue)"vertex form}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$
$\text{obtain this form using "color(blue)"completing the square}$
• " the coefficient of the "x^2" term must be 1"
$\text{factor out 3}$
$\Rightarrow y = 3 \left({x}^{2} - \frac{50}{3} x + 100\right)$
• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} - \frac{50}{3} x$
$y = 3 \left({x}^{2} + 2 \left(- \frac{25}{3}\right) x \textcolor{red}{+ \frac{625}{9}} \textcolor{red}{- \frac{625}{9}} + 100\right)$
$\textcolor{w h i t e}{y} = 3 {\left(x - \frac{25}{3}\right)}^{2} + 3 \left(- \frac{625}{9} + 100\right)$
$\textcolor{w h i t e}{y} = 3 {\left(x - \frac{25}{3}\right)}^{2} + \frac{275}{3} \leftarrow \textcolor{b l u e}{\text{in vertex form}}$
Apr 22, 2018
The vertex form of equation is $y = 3 {\left(x - \frac{25}{3}\right)}^{2} + \frac{1100}{12}$
#### Explanation:
$y = 3 {x}^{2} - 50 x + 300 \mathmr{and} y = 3 \left({x}^{2} - \frac{50}{3} x\right) + 300$ or
$y = 3 \left\{{x}^{2} - \frac{50}{3} x + {\left(\frac{50}{6}\right)}^{2}\right\} - \frac{2500}{12} + 300$ or
$y = 3 {\left(x - \frac{25}{3}\right)}^{2} + \frac{1100}{12}$ Comparing with vertex form of
equation y = a(x-h)^2+k ; (h,k) being vertex we find
here $h = \frac{25}{3} , k = \frac{1100}{12} \therefore$ Vertex is at $\left(8.33 , 91.67\right)$
The vertex form of equation is $y = 3 {\left(x - \frac{25}{3}\right)}^{2} + \frac{1100}{12}$
graph{3 x^2-50 x+300 [-320, 320, -160, 160]} [Ans] |
### Predictive Distributions (BDA 3rd Edition, Chapter 2)Tweet
This post provides an answer for Exercise 2 from Chapter 2 of Bayesian Data Analysis (BDA) 3rd Edition. Let’s state the problem from the beloved book.
Consider two coins, $C_1$ and $C_2$, with the following characteristics:
\begin{align} \Pr(\text{head} \mid C_1) &= 0.6, \\ \Pr(\text{head} \mid C_2) &= 0.4. \end{align}
Choose one of the coins at random and imagine spinning it repeatedly.
Here is the question:
Given that the first two spins from the chosen coin are tails, what is the expectation of the number of additional spins until a head shows up?
To simplify our writing, we denote $\text{head}$ and $\text{tail}$ as $H$ and $T$ respectively. Therefore, we have
\begin{align*} C_1 &\rightarrow \Pr(H \mid C_1) = 0.6 = \frac{3}{5}, \tag{1}\label{eq:c1}\\ C_2 &\rightarrow \Pr(H \mid C_2) = 0.4 = \frac{2}{5}. \tag{2}\label{eq:c2} \end{align*}
$\pmb{\text{Figure 1}}$: The problem poses an experiment consisting of two steps. Specifically, $N \sim$ geometric distribution.
If we read the problem carefully, we may find that the problem consists of two steps as depicted in $\pmb{\text{Figure 1}}$. Particularly, random variable $N$ is a geometric distribution which has a probability mass function such that
$$$\Pr(n) = (1-C_i)^{n-1} C_i \tag{3}\label{eq:pmf-geometri}$$$ where $C_i$ depends on either $C_1$ or $C_2$.
Now, let us compute $$$E(N \mid TT) = ? \tag{4}\label{eq:problem}$$$ as shown
\begin{align} \text{E}(N \mid TT) &= \int N \, \Pr(N \mid TT) \, dN && \text{by definition} \tag{5}\label{eq:compute-1} \\ &= \int \int N \, \Pr(N, C \mid TT) \, dC \, dN && \text{by Bayes rule} \tag{6}\label{eq:compute-2} \\ &= \int \int N \, \Pr(N \mid TT,C) \Pr(C \mid TT) \, dC \, dN && \text{by conditional probability} \tag{7}\label{eq:compute-3} \\ &= \int \underbrace{\int N \, \Pr(N \mid TT,C) \, dN}_{\text{E}(N \mid TT, C)} \, \Pr(C \mid TT) \, dC && \text{just rearranging} \tag{8}\label{eq:compute-4} \\ &= \int \text{E}(N \mid TT, C) \, \Pr(C \mid TT) \, dC && \text{by expectation definition} \tag{9}\label{eq:compute-5} \\ &= \sum_{i=1}^2 \text{E}(N \mid TT, C_i) \, \Pr(C_i \mid TT) && \text{since }C \text{ is discrete .} \tag{10}\label{eq:compute-6} \end{align}
Recall that $N \sim$ geometric distribution; accordingly,
\begin{align} \text{E}(N \mid TT, C_i) &= \text{E}(N \mid C_i) && \text{whether we have }TT\text{ as conditional or not} \tag{11}\label{eq:expectation-1} \\ &= \frac{1}{C_i} \tag{12}\label{eq:expectation-2}. \end{align}
We proceed from Equation \eqref{eq:compute-6} as shown
\begin{align} \text{E}(N \mid TT) &= \text{E}(N \mid TT, C_1) \, \Pr(C_1 \mid TT) + \text{E}(N \mid TT, C_2) \, \Pr(C_2 \mid TT) \tag{13}\label{eq:final-1} \\ &= \frac{1}{C_1} \Pr(C_1 \mid TT) + \frac{1}{C_2} \Pr(C_2 \mid TT) \tag{14}\label{eq:final-2} \\ &= \frac{1}{C_1} \underbrace{\frac{\Pr(TT \mid C_1) \, \Pr(C_1)}{\Pr(TT)}}_{\text{Part 1}} + \frac{1}{C_2} \underbrace{\frac{\Pr(TT \mid C_2) \, \Pr(C_2)}{\Pr(TT)}}_{\text{Part 2}} \tag{15}\label{eq:final-3} \\ \end{align}
Next, let’s compute $\text{Part 1}$ which looks like
\begin{align} \frac{\Pr(TT \mid C_1) \, \Pr(C_1)}{\Pr(TT)} &= \frac{Pr(TT \mid C_1) \, \Pr(C_1)}{\Pr(TT \mid C_1) \, \Pr(C_1) + \Pr(TT \mid C_2) \, \Pr(C_2)} && \text{expanding }\Pr(TT) \tag{16}\label{eq:part-1-1} \\ &= \frac{\left( \frac{2}{5} \right) \left( \frac{2}{5} \right) \left(\frac{1}{2} \right)}{\left( \frac{2}{5} \right) \left( \frac{2}{5} \right) \left( \frac{1}{2} \right) + \left( \frac{3}{5} \right) \left( \frac{3}{5} \right) \left( \frac{1}{2} \right)}. \tag{17}\label{eq:part-1-2} \\ \end{align}
Similarly, we also calculate $\text{Part 2}$ in the following:
\begin{align} \frac{\Pr(TT \mid C_2) \, \Pr(C_2)}{\Pr(TT)} &= \frac{\Pr(TT \mid C_2) \, \Pr(C_2)}{\Pr(TT \mid C_1) \, \Pr(C_1) + \Pr(TT \mid C_2) \, \Pr(C_2)} && \text{expanding }\Pr(TT) \tag{18}\label{eq:part-2-1} \\ &= \frac{\left( \frac{3}{5} \right) \left( \frac{3}{5} \right) \left( \frac{1}{2} \right) }{ \left( \frac{2}{5} \right) \left( \frac{2}{5} \right) \left( \frac{1}{2} \right) + \left( \frac{3}{5} \right) \left( \frac{3}{5} \right) \left( \frac{1}{2} \right) }. \tag{19}\label{eq:part-2-2} \\ \end{align}
Finally, we are able to compute Equation \eqref{eq:problem} as
\begin{align} \text{E}(N \mid TT) &= \frac{1}{C_1} \, \text{Part 1} + \frac{1}{C_2} \, \text{Part 2} \tag{20}\label{eq:final-answer}\\ &= \frac{1}{3/5} \, \text{Part 1} + \frac{1}{2/5} \, \text{Part 2} \\ &= 2.2436 && \text{utilizing Eq. }\eqref{eq:part-1-2}\text{ and Eq. }\eqref{eq:part-2-2} \\ &\approx 3 && \text{rounding the number.} \end{align}
This means that in order to find a head after we have two tails regardless the coin we choose, we need $\pmb{3}$ more throws on average.
Written on November 25, 2020 |
# Algebra I Guided Notes 2.4 3.4
```Algebra I Guided Notes: Solving Equations and Inequalities with Variables on Both
Sides
Lesson 2.4/3.4
Objective:
to solve equations and inequalities with variables on both sides
If there is a variable on both sides of the equation/inequality, (ISOLATE THE VARIABLE!)
1.
2.
3.
4.
5.
Distribute, combine like terms on each individual side.
Add or subtract to move the variable to the one side.
Add or subtract to move the constants to the other side.
Solve the equation/inequality.
**Remember with inequalities to switch the inequality sign if multiplying or dividing
both sides by a negative.
Examples
1)
6x + 3 = 4x + 9
2)
- 5 + 3x = 5 + 2 x
3)
x + 13 = 6 x - 2
4)
4(x + 1) = 2 x - 2
5)
- 2(3z - 4) = 10 - 6 z
6)
- 2(3z - 4) = 8 - 6 z
Special Case: No Solution:
true.
3x – 2 = 6 + 3x
no value of the variable can make the equation
Special Case: Identity:
3x + 6 = 6 + 3x
true for every possible value of the variable
7)
3x + 7 > x – 5
8)
9)
8x – 7 + 2x < 4x + 11
10)
3(x – 5) > 6x + 9
3(x – 4) > 3x + 7
``` |
Trigonometric Identities: How to Derive / Remember Them - Introduction to Trigonometry
# Trigonometric Identities: How to Derive / Remember Them - Introduction to Trigonometry Video Lecture | SSC CGL Tier 2 - Study Material, Online Tests, Previous Year
## SSC CGL Tier 2 - Study Material, Online Tests, Previous Year
1366 videos|1312 docs|1010 tests
## FAQs on Trigonometric Identities: How to Derive / Remember Them - Introduction to Trigonometry Video Lecture - SSC CGL Tier 2 - Study Material, Online Tests, Previous Year
1. How can I derive trigonometric identities?
Ans. To derive trigonometric identities, you can use various techniques such as algebraic manipulation, Pythagorean identities, reciprocal identities, quotient identities, and others. By simplifying the given trigonometric expressions using these techniques, you can establish the identities. It is recommended to practice different examples and understand the fundamental concepts to derive trigonometric identities effectively.
2. What are some commonly used trigonometric identities?
Ans. Some commonly used trigonometric identities include: - Pythagorean identities: sin^2θ + cos^2θ = 1, 1 + tan^2θ = sec^2θ, and 1 + cot^2θ = cosec^2θ - Reciprocal identities: cscθ = 1/sinθ, secθ = 1/cosθ, and cotθ = 1/tanθ - Quotient identities: tanθ = sinθ/cosθ and cotθ = cosθ/sinθ - Co-function identities: sin(π/2 - θ) = cosθ, cos(π/2 - θ) = sinθ, tan(π/2 - θ) = cotθ, and cot(π/2 - θ) = tanθ These identities are widely used in trigonometry to simplify expressions and solve trigonometric equations.
3. How can I remember trigonometric identities?
Ans. Remembering trigonometric identities can be challenging, but consistent practice and understanding can help. Here are some tips to remember them: - Understand the fundamental concepts and the derivations of identities. - Practice regularly with different examples to reinforce your memory. - Create flashcards or a cheat sheet summarizing the identities for quick reference. - Look for patterns and connections between different identities to make them easier to remember. - Break down complex identities into simpler forms to comprehend them better. - Solve trigonometric equations and problems using the identities to reinforce your memory. By implementing these strategies, you can improve your ability to remember trigonometric identities.
4. How are trigonometric identities useful in SSC CGL exams?
Ans. Trigonometric identities are essential in SSC CGL exams as they help in solving trigonometric equations, simplifying expressions, and proving mathematical statements. These identities are frequently used in various topics such as geometry, algebra, calculus, and physics. By applying trigonometric identities correctly, you can solve complex problems efficiently and accurately, thereby increasing your chances of scoring well in the SSC CGL exams.
5. Can I use a formula sheet or reference guide for trigonometric identities in SSC CGL exams?
Ans. Generally, SSC CGL exams do not allow the use of formula sheets or reference guides. Candidates are expected to have a strong understanding of trigonometric identities and their applications. It is advisable to memorize the commonly used identities and practice their derivations to ensure a smooth problem-solving process during the exam. However, it is always recommended to refer to the official exam guidelines to know the specific rules regarding the use of formula sheets in your exam.
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# A heap of rice is in the form of a cone of diameter 9 m
Question:
A heap of rice is in the form of a cone of diameter 9 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover
heap?
Solution:
Given that, a heap of rice is in the form of a cone.
Height of a heap of rice i.e., cone (h) = 3.5 m
and diameter of a heap of rice i.e., cone = 9 m
Radius of a heap of rice $i . \theta .$, cone $(r)=\frac{9}{2} \mathrm{~m}$
So, volume of rice $=\frac{1}{3} \pi \times r^{2} h$
$=\frac{1}{3} \times \frac{22}{7} \times \frac{9}{2} \times \frac{9}{2} \times 3.5$
$=\frac{6237}{84}=74.25 \mathrm{~m}^{3}$
Now, canvas cloth required to just cover heap of rice
$=$ Surface area of a heap of rice
$=\pi /$
$=\frac{22}{7} \times r \times \sqrt{r^{2}+h^{2}}$
$=\frac{22}{7} \times \frac{9}{2} \times \sqrt{\left(\frac{9}{2}\right)^{2}+(3.5)^{2}}$
$=\frac{11 \times 9}{7} \times \sqrt{\frac{81}{4}+12.25}$
$=\frac{99}{7} \times \sqrt{\frac{130}{4}}=\frac{99}{7} \times \sqrt{32.5}$
$=14.142 \times 5.7$
$=80.61 \mathrm{~m}^{2}$
Hence, 80.61 m2 canvas cloth is required to just cover heap. |
# Perimeter and Area of a Triangle
Here we will discuss about the perimeter and area of a triangle and some of its geometrical properties.
Perimeter, Area and Altitude of a Triangle:
Perimeter of a triangle (P) = Sum of the sides = a + b + c
Semiperimeter of a triangle (s) = $$\frac{1}{2}$$(a + b + c)
Area of a triangle (A) = $$\frac{1}{2}$$ × base × altitude = $$\frac{1}{2}$$ah
Here any side can be taken as base; the length of the perpendicular from the corresponding vertex to this side is the altitude.
Area = $$\sqrt{\textrm{s(s - a)(s - b)(s - c)}}$$ (Heron’s formula)
Altitude (h) = $$\frac{\textrm{area}}{\frac{1}{2} \times \textrm{base}}$$ = $$\frac{2\triangle}{a}$$
Solved Example on Finding the Perimeter, Semiperimeter and Area
of a Triangle:
The sides of a triangle are 4 cm, 5 cm and 7 cm. Find its perimeter, semiperimeter and area.
Solution:
Perimeter of a triangle (P) = Sum of the sides
= a + b + c
= 4 cm + 5 cm + 7 cm
= (4 + 5 + 7) cm
= 16 cm
Semiperimeter of a triangle (s) = $$\frac{1}{2}$$(a + b + c)
= $$\frac{1}{2}$$(4 cm + 5 cm + 7 cm)
= $$\frac{1}{2}$$(4 + 5 + 7) cm
= $$\frac{1}{2}$$ × 16 cm
= 8 cm
Area of a triangle = $$\sqrt{\textrm{s(s - a)(s - b)(s - c)}}$$
= $$\sqrt{\textrm{8(8 - 4)(8 - 5)(8 - 7)}}$$ cm$$^{2}$$
= $$\sqrt{\textrm{8 × 4 × 3 × 1}}$$ cm$$^{2}$$
= $$\sqrt{96}$$ cm$$^{2}$$
= $$\sqrt{16 × 6}$$ cm$$^{2}$$
= 4$$\sqrt{6}$$ cm$$^{2}$$
= 4 × 2.45 cm$$^{2}$$
= 9.8 cm$$^{2}$$
Perimeter, Area and Altitude of an Equilateral Triangle:
Perimeter of an equilateral triangle (P) = 3 × side = 3a
Area of an equilateral triangle (A) = $$\frac{√3}{4}$$ × (side)$$^{2}$$ = $$\frac{√3}{4}$$ a$$^{2}$$
Altitude of an equilateral triangle (h) = $$\frac{√3}{4}$$ a
Trigonometric formula for area of a triangle:
Area of ∆ABC = $$\frac{1}{2}$$ × ca sin B
= $$\frac{1}{2}$$ × ab sin C
= $$\frac{1}{2}$$ × bc sin A
(since, ∆ = $$\frac{1}{2}$$ ah = $$\frac{1}{2}$$ ca ∙ $$\frac{h}{c}$$ = $$\frac{1}{2}$$ ca sin B, etc.)
Solved Example on Finding the Area of a Triangle:
In a ∆ABC, BC = 6 cm, AB = 4 cm and ∠ABC = 60°. Find its area.
Solution:
Area of ∆ABC = $$\frac{1}{2}$$ ac sin B = $$\frac{1}{2}$$ × 6 × 4 sin 60° cm$$^{2}$$
= $$\frac{1}{2}$$ × 6 × 4 × $$\frac{√3}{2}$$ cm$$^{2}$$
= 6√3 cm$$^{2}$$
= 6 × 1.73 cm$$^{2}$$
= 10.38 cm$$^{2}$$
Some geometrical properties of an isosceles triangle:
In the isosceles ∆PQR, PQ = PR, QR is the base, and PT is the altitude.
Then, ∠PTR = 90°, QT = TR, PT$$^{2}$$ + TR$$^{2}$$ = PR$$^{2}$$ (by Pythagoras’ theorem)
∠PQR = ∠PRQ, ∠QPT = ∠RPT.
Some geometrical properties of a right-angled triangle:
In the right-angled ∆PQR, ∠PQR = 90°; PQ, QR are the sides (forming the right angle) and PR is the hypotenuse.
Then, PQ ⊥ QR (therefore, if QR is the base, PQ is the altitude).
PQ$$^{2}$$ + QR$$^{2}$$ = PR$$^{2}$$ (by Pythagoras’ theorem)
Area of the ∆PQR = $$\frac{1}{2}$$ ∙ PQ ∙ QR
⟹ PQ ∙ QR = 2 × area of the ∆PQR.
Again, area of the ∆PQR = $$\frac{1}{2}$$ ∙ QT ∙ PR
⟹ QT ∙ PR = 2 × area of the ∆PQR.
Therefore, PQ ∙ QR = QT ∙ PR = 2 × Area of the ∆PQR.
Solved Examples on Perimeter and Area of a Triangle:
1. Find the perimeter of an equilateral triangle whose area is equal to that of a triangle with sides 21 cm, 16 cm and 13 cm.
Solution:
Let a side of the equilateral triangle = x.
Then, its area = $$\frac{√3}{4}$$ x$$^{2}$$
Now, the area of the other triangle = $$\sqrt{\textrm{s(s - a)(s - b)(s - c)}}$$
Here, s = $$\frac{1}{2}$$ (a + b + c)
= $$\frac{1}{2}$$ (21 + 16 + 13) cm
= $$\frac{1}{2}$$ 50 cm
= 25 cm
Therefore, area of the other triangle = $$\sqrt{\textrm{25(25 - 21)(25 - 16)(25 - 13)}}$$ cm$$^{2}$$
= $$\sqrt{\textrm{25 ∙ 4 ∙ 9 ∙ 12}}$$ cm$$^{2}$$
= 60$$\sqrt{\textrm{3}}$$ cm$$^{2}$$
According to the question, $$\frac{√3}{4}$$ x$$^{2}$$ = 60$$\sqrt{\textrm{3}}$$ cm$$^{2}$$
⟹ x$$^{2}$$ = 240 cm$$^{2}$$
Therefore, x = 4√15 cm
2. PQR is an isosceles triangle whose equal sides PQ and PR are 10 cm each, and the base QR measures 8 cm. PM is the perpendicular from P to QR and X is a point on PM such that ∠QXR = 90°. Find the area of the shaded portion.
Solution:
Since PQR is an isosceles triangle and PM ⊥ QR, QR is bisected at M.
Therefore, QM = MR = $$\frac{1}{2}$$ QR = $$\frac{1}{2}$$ × 8 cm = 4 cm
Now, PQ$$^{2}$$ = PM$$^{2}$$ + QM$$^{2}$$ (by Pythagoras’ theorem)
Therefore, 10$$^{2}$$ cm$$^{2}$$ = PM$$^{2}$$ + 4$$^{2}$$ cm$$^{2}$$
or, PM$$^{2}$$ = 10$$^{2}$$ cm$$^{2}$$ - 4$$^{2}$$ cm$$^{2}$$
= 100 cm$$^{2}$$ - 16 cm$$^{2}$$
= (100 - 16) cm$$^{2}$$
= 84 cm$$^{2}$$
Therefore, PM$$^{2}$$ = 2√21 cm
Therefore, area of the ∆PQR = $$\frac{1}{2}$$ × base × altitude
= $$\frac{1}{2}$$ × QR × PM
= ($$\frac{1}{2}$$ × 8 × 2√21) cm$$^{2}$$
= 8√21) cm$$^{2}$$
From geometry, ∆XMQ ≅ ∆XMR (SAS criterion)
We get, XQ =XR = a (say)
Therefore, from the right-angled ∆QXR, a$$^{2}$$ + a$$^{2}$$ = QR$$^{2}$$
or, 2a$$^{2}$$ = 8$$^{2}$$ cm$$^{2}$$
or, 2a$$^{2}$$ = 64 cm$$^{2}$$
or, a$$^{2}$$ = 32 cm$$^{2}$$
Therefore, a = 4√2 cm
Again, area of the ∆XQR = $$\frac{1}{2}$$ × XQ × XR
= $$\frac{1}{2}$$ × a × a
= $$\frac{1}{2}$$ × 4√2 cm × 4√2 cm
= $$\frac{1}{2}$$ × (4√2)$$^{2}$$ cm$$^{2}$$
= $$\frac{1}{2}$$ × 32 cm$$^{2}$$
= 16 cm$$^{2}$$
Therefore, area of the shaded portion = area of the ∆PQR - area of the ∆XQR
= (8√21) cm$$^{2}$$ - 16 cm$$^{2}$$
= (8√21 - 16) cm$$^{2}$$
= 8(√21 - 2) cm$$^{2}$$
= 8 × 2.58 cm$$^{2}$$
= 20.64 cm$$^{2}$$
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# Ratios and Proportions Quiz 12
Home > > Tutorial
5 Steps - 3 Clicks
# Ratios and Proportions Quiz 12
### Introduction
Ratios and Proportions is one of important topic in Quantitative Aptitude Section. In Ratios and Proportions Quiz 12 article candidates can find questions with an answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Ratios and Proportions Quiz 12 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.
### Q1
The ratio of the present ages of Priya and her mother is 3: 7. The mother’s age at the time of Priya’s birth was 48 years. Find the mother’s present age.
A. 84 years B. 42 years C. 36 years D. None of these
A
Present ratio is 7 : 3.
∴ actual ages are 7x and 3x.
∴ 7x – 3x = 48 → x = 12.
Hence their present ages are 84 and 36 yrs.
### Q2
The perimeter of a rectangle is 64 cm. If the ratio of the lengths of two adjacent sides is 7: 9, find the lengths of these sides.
A. 24 cm, 28 cm B. 14 cm, 18 cm C. 7 cm, 9 cm D. None of these
B
Perimeter of a rectangle = 2 (L + B)
Also L: B = 7 : 9
∴ Actual values are 5x and 8x.
Hence 2(7x + 9x) = 64. → x = 2
∴ sides will be of 14 cm and 18 cm.
### Q3
The ratio of the length and the breadth of a rectangle is 3 : 5 and its area is 1.35 ${cm}^{2}$. Find the length of the rectangle.
A. 9 cm B. 0.9 cm C. 0.09 cm D. 90 cm
B
Area of a rectangle = LB.
Ratio of sides = 3 : 5.
∴(3x)(5x) = 1.35 → x = 0.3.
→ Length of the rectangle
= 3 (0.3) = 0.9 cm
And breadth = 5(0.3) = 1.5 cm.
### Q4
The ratio of the present ages of John and Jim is 5 : 3. Four years hence it will be 3 : 2. Find the present age of John.
A. 5 years B. 12 years C. 10 years D. 20 years
D
Present ratio = 5 : 3.
∴ actual values are 5x and 3x.
So $\frac {(5x + 4)}{(3x + 4)} = \frac {3}{2} → x = 4$
So present ages are 20 yrs and 12 yrs.
### Q5
There are 145 students in the first three standards. The ratio of number of students in the 1st and the 2nd standards is 2 : 3, while that of students in standards 2nd and 3rd is 4: 3. Find the number of students in 2nd standard
A. 40 years B. 60 years C. 45 years D. 65 years
B
Total students = 145. Ratio of students in 1st and 2nd standards = 2 : 3
Ratio of students in 2nd and 3rd standards = 4 : 3
Hence combined ratio i.e. 1st: 2nd: 3rd is 8 : 12 : 9.
∴ Number of students in each standard
= $\frac {(145 × 8)}{29} = 40$
$\frac {(145 × 12)}{29} = 60 and \frac {(145 × 9)}{29} = 45.$ |
# How do you find the derivative of y=cosln(4x^3)?
Feb 5, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 \sin \left(\ln \left(4 {x}^{3}\right)\right)}{x}$
#### Explanation:
Using chain rule, let $u = \ln \left(4 {x}^{3}\right)$
So $y = \cos u$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
$= - \sin u \cdot \frac{d}{\mathrm{dx}} \ln \left(4 {x}^{3}\right)$
Now we have to derive $\ln \left(4 {x}^{3}\right)$ using chain rule as well.
Let $y = \ln u , u = 4 {x}^{3}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{du}} \ln u \cdot \frac{d}{\mathrm{dx}} 4 {x}^{3}$
$= \frac{1}{u} \cdot 12 {x}^{2}$
$= \frac{12 {x}^{2}}{4 {x}^{3}}$
$= \frac{3}{x}$
$\therefore \frac{d}{\mathrm{dx}} \ln \left(4 {x}^{3}\right) = - \sin \left(\ln \left(4 {x}^{3}\right)\right) \cdot \frac{3}{x}$
$= - \frac{3 \sin \left(\ln \left(4 {x}^{3}\right)\right)}{x}$ |
# Sin Cos Formula
Trigonometry is the study of relationships that deal with angles, lengths, and heights of triangles and relations between different parts of circles and other geometrical figures. Concepts of trigonometry are very useful in engineering, astronomy, Physics, and architectural design. Sine and cos are the terms used in the trigonometry that tell us about the triangle. Let us discuss in detail about the sin cos formula and other concepts.
## Sin Cos Formula
### Basic trigonometric ratios
There are six trigonometric ratios for the right angle triangle are Sin, Cos, Tan, Cosec, Sec, Cot which stands for Sine, Cosecant, Tangent, Cosecant, Secant respectively. Sin and Cos are basic trigonometric functions that tell about the shape of a right triangle. SO let us see the sin cos formula along with the other important trigonometric ratios.
• $$\sin θ= Perpendicular/ Hypotenuse$$
• $$\cos θ= Base/ Hypotenuse$$
• $$\tan θ= Perpendicular/Base$$
• $$\csc θ= Hypotenuse/Perpendicular$$
• $$\sec θ = Hypotenuse/Base$$
• $$\cot θ= Base/Perpendicular$$
### Basic Trigonometric Identities for Sine and Cos
• $$\cos^2 (A) + \sin^2 (A) = 1$$
If A + B = 180° then:
• $$\sin(A) = \sin(B)$$
• $$\cos(A) = -\cos(B)$$
If A + B = 90° then:
• $$\sin(A) = \cos(B)$$
• $$\cos(A) = \sin(B)$$
### Half-angle formulas
$$\sin(\frac{A}{2}) = \sqrt{\frac{±1−\cos(A) }{2}}$$
• If $$\frac{A}{2}$$ lies in quadrant I or II
• If $$\frac{A}{2}$$ lies in quadrant III or IV
$$\cos (\frac{A}{2}) = \sqrt{ \frac{±1+\cos(A)}{2}}$$
• If $$\frac{A}{2}$$ lies in quadrant I or IV
• If $$\frac{A}{2}$$ lies in quadrant II or III
### Double and Triple Angle Formulas
• $$\sin 2A = 2\sin A \cos A$$
• $$\cos 2A = \cos^2 A – \sin^2 A = 2 \cos^2 – 1 = 1- \sin^2 A$$
• $$\sin 3A = 3\sin A – 4 \sin ^3 A$$
• $$\cos 3A = 4 \cos^3 A – 3\cos A$$
• $$\sin^4 A = 4 \cos^3 A \sin A – 4\cos A \sin^3 A$$
• $$\cos^4 A = \cos^4 A – 6\cos^2 A\sin^2 A +\sin^4 A$$
• $$\sin^2 A = \frac{1–\cos(2A)}{2}$$
• $$\cos^2 A = \frac{1+\cos(2A)}{2}$$
### Sum and Difference of Angles
• $$\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$$
• $$\sin(B)\sin(A−B)=\sin(A)\cos(B)−\cos(A)\sin(B)$$
• $$\cos(A+B)=\cos(A)\cos(B)−\sin(A)\sin(B)$$
• $$\cos(A−B)=\cos(A)\cos(B)+\sin(A)\sin(B)$$
• $$\sin(A+B+C)=\sin A\cos B\cos C+\cos A\sin B\cos C+\cos A\cos B\sin C−\sin A\sin B\sin C$$
• $$\cos(A+B+C)=\cos A\cos B\cos C−\sin A\sin B\cos C−\sin A\cos B\sin C−\sin A\cos B\sin C−\cos A\sin B\sin C$$
• $$\sin A + \sin B = 2\sin \frac{(A+B)}{2}\cos \frac{(A−B)}{2}$$
• $$\sin A – \sin B = 2\sin \frac{(A−B)}{2}\cos \frac{(A+B)}{2}$$
• $$\cos A + \cos B = 2\cos\frac{(A+B)}{2} (A+B)2\cos\frac{(A−B)}{2}$$
• $$\cos A + \cos B = -2\sin \frac{(A+B)}{2}\sin \frac{(A−B)}{2}$$
### Product Identities
• $$\sin(x) \cos(y) = \frac{1}{2} [\sin(x + y) + \sin(x – y)]$$
• $$\cos(x) \sin(y) = \frac{1}{2} [\sin(x + y) – \sin(x – y)]$$
• $$\cos(x) \cos(y) = \frac{1}{2} [\cos(x – y) + \cos(x + y)]$$
• $$\sin(x) \sin(y) = \frac{1}{2} [\cos(x – y) – \cos(x + y)]$$
## Solved Examples
Q.1. In a right triangle ABC, $$\tan A = 3/4. Find \sin A and \cos A.$$
Solution: Let a be the length of the side opposite angle A, b the length of the side adjacent to angle A and h be the length of the hypotenuse.
$$\tan A = \frac{ opposite side }{ adjacent side } Â = \frac{a}{b} = \frac{3}{4}$$
we can say that: a = 3k and b = 4k , where k is a coefficient of proportionality.
Pythagoras’s theorem:
h2Â = (3k)2Â + (4k)2
h = 5k
$$\sin A = \frac{a}{h} = \frac{3k}{5k} = \frac{3}{5}$$ and
$$\cos A = \frac{4k}{5k}= \frac{4}{5}$$
Q2. In Δ ABC, right-angled at B, AB = 3 cm and AC = 6 cm. Determine \angle BAC and \angle ACB.
Solution: Given AB = 3 cm and AC = 6 cm.
Therefore, $$\frac{AB}{AC}=\sin R$$
or $$\sin R = \frac{3}{6}= \frac{1}{2}$$
So, $$\angle BAC = 30° and \angle ACB = 60$$
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## Browse
### One response to “Equation Formula”
1. KUCKOO B says:
I get a different answer for first example.
I got Q1 as 20.5
median 23 and
Q3 26 |
# Find the limit?
## a_1=1 & a_n=n(a_(n-1)+1) for n=2,3,..... Define ${P}_{n} = \left(1 + \frac{1}{a} _ 1\right) \cdot \left(1 + \frac{1}{a} _ 2\right) \cdot \ldots \ldots \ldots \ldots \ldots . \cdot \left(1 + \frac{1}{a} _ n\right)$ Then ${\lim}_{n \rightarrow \infty} {P}_{n}$ is?
Apr 10, 2018
${\lim}_{n \to \infty} {P}_{n} = e$
#### Explanation:
Note that:
${a}_{n + 1} = \left(n + 1\right) \left({a}_{n} + 1\right)$
${a}_{n + 1} / \left(n + 1\right) = {a}_{n} + 1$
so:
$\left(1\right) \text{ } 1 + \frac{1}{a} _ n = \frac{{a}_{n} + 1}{a} _ n = \frac{1}{n + 1} {a}_{n + 1} / {a}_{n}$
Then, having defined:
${P}_{n} = {\prod}_{k = 1}^{n} \left(1 + \frac{1}{a} _ k\right)$
we have:
${P}_{n} = {\prod}_{k = 1}^{n} \frac{1}{k + 1} {a}_{k + 1} / {a}_{k}$
P_n = 1/((n+1)!)prod_(k=1)^n a_(k+1)/a_k
P_n = 1/((n+1)!)( (a_(n+1)/a_n)(a_n/a_(n-1))...(a_3/a_2)(a_2/a_1))
P_n = a_(n+1)/((n+1)!)
Now if we substitute the formula for ${a}_{n + 1}$ we get:
P_n = ((n+1)(a_n+1))/((n+1)!)
P_n = ((a_n+1))/(n!)
P_n = a_n/(n!)+1/(n!)
P_n = P_(n-1)+1/(n!)
Let now:
P_0 =1 = 1/(0!)
so that:
P_1 = 1+1/a_1 = P_0 +1/(1!)
then:
P_2 = P_1 +1/(2!) = sum_(k=0)^2 1/(k!)
and clearly we can establish by induction that:
P_n = sum_(k=0)^n 1/(k!)
and then:
lim_(n->oo) P_n = sum_(k=0)^oo 1/(k!)
Remembering that the MacLaurin formula for the exponential function is:
e^x = sum_(k=0)^oo x^k/(k!)
converging for every $x \in \mathbb{R}$ we can see that for $x = 1$:
sum_(k=0)^oo 1/(k!) = e
and conclude that:
${\lim}_{n \to \infty} {P}_{n} = e$ |
# Into Math Grade 5 Module 2 Lesson 2 Answer Key Represent Division with 2-Digit Divisors
We included HMH Into Math Grade 5 Answer Key PDF Module 2 Lesson 2 Represent Division with 2-Digit Divisors to make students experts in learning maths.
## HMH Into Math Grade 5 Module 2 Lesson 2 Answer Key Represent Division with 2-Digit Divisors
I Can find the quotient of numbers up to four digits divided by 2-digit divisors using visual models.
You are a game designer designing a treasure hunt game similar to the one shown.
The game board is a grid with a treasure chest located behind one of the squares.
The rectangular grid will have 96 squares. If the length of the grid is greater than 10 squares, how wide can the grid be?
The game board grid can be ___ squares wide.
Given,
The rectangular grid will have 96 squares.
If the length of the grid is greater than 10 squares.
Area of rectangle = length x width
96 = 10 x width
width = 96/10 = 9.6
Therefore, The game board grid can be 9.6 squares wide.
Turn and Talk Compare your game board grid to the game board grids of other classmates. How do they compare?
Build Understanding
1. You are designing another game in which the player needs to arrange flower pots in 12 equal-sized groups. There are 156 flowerpots. How many flowerpots are in each group?
A. What multiplication and division equations can you write to model the number of flowerpots, p, in each group?
______________________
B. How can you break apart 156 into a sum of multiples of 12? Use your multiples to make an area model.
______________________
By area model,
area = length x width
156 = 12 x width
width = 156 ÷ 12
Therefore, width = 13
C. What do the number of rows and the number of columns in the area model represent?
______________________
______________________
______________________
Rows represent length
D. How is the dividend represented in the area model?
______________________
______________________
The dividend is the total area.
E. How is the quotient represented in the area model? What is the quotient?
______________________
______________________
quotient represents width
F. How many flowerpots are in each group?
______________________
13 flowerpots are in each group.
Turn and Talk How can you use your equations to explain how to find the number of flowerpots in each group?
2. There are 675 gold bars hidden throughout the game. The same number of bars are placed in each of 25 treasure chests. How many bars are in each treasure chest?
A. What division equation models this situation?
________________
675 ÷ 25
B. Break apart the dividend into multiples of the divisor and use the area model to find the quotient.
___________________
We know.
area = length x width
675 = 25 x width
width = 675 ÷ 25 = 25
C. Explain what the smaller rectangles represent in terms of the division equation.
___________________
___________________
___________________
Answer: smaller rectangles represent length and width.
D. How many gold bars are in each treasure chest?
___________________
Answer: The number of gold bars in each treasure chest are 625.
Turn and Talk Compare your area model to the area models of other classmates. What can you conclude?
Build Understanding
3. Archaeologists find a sunken chest of 3,420 bronze coins. If the coins are to be shared equally among 20 museums, how many coins does each museum get?
A. How can you model this situation using a division equation?
_________________________
B. Draw an area model to find the quotient. How can using a greater multiple of the divisor help you draw the area model?
_________________________
C. How many coins does each museum get? How does your area model show this?
_________________________
Area = length x width
3420 = 20 x width
width = 3420 ÷ 20 = 171
Therefore, each museum gets 171 coins.
Check Understanding Math Board
Question 1.
There are 76 bottles of paint in the art studio. The art teacher divides them equally among all 19 students in the class. How many bottles of paint does each student get?
Given,
There are 76 bottles of paint in the art studio.
The art teacher divides them equally among all 19 students in the class.
So, 76 ÷ 19 = 4
Therefore, Each student get 4 bottles of paint.
Question 2.
Model with Mathematics Lucinda has 81 tulip bulbs. She plants 3 bulbs in each row. How many rows of tulip bulbs does she plant? Write a division equation to model this situation.
Given,
Lucinda has 81 tulip bulbs.
She plants 3 bulbs in each row.
So, 81 ÷ 3 = 27
Therefore, she plants 3 rows of tulip bulbs.
Question 3.
Model with Mathematics Write a division equation that is related to the application of the Distributive Property shown.
12 × (100 + 20 + 4) = (12 × 100) + (12 × 20) + (12 × 4)
= 1,200 + 240 + 48
= 1,488
Question 4.
Miguel volunteers at the library. He needs to arrange 576 books on shelves for a book sale. There are 16 empty shelves. Miguel puts an equal number of books on each shelf.
How many books does he put on each shelf? _______
Given,
Miguel volunteers at the library.
He needs to arrange 576 books on shelves for a book sale.
There are 16 empty shelves.
Miguel puts an equal number of books on each shelf.
So, 576 ÷ 16 = 36
Therefore, she puts 36 books on each shelf.
Question 5.
Model with Mathematics Multiplication of two numbers results in the partial products 200, 10, and 3. One of the factors is 30. Write a multiplication equation and a division equation to model this situation. Show your thinking.
Given,
Multiplication of two numbers results in the partial products 200, 10, and 3.
One of the factors is 30.
From the given data,
let the other number be a
So, 30 x a = 200 x 10 x 3
30 x a = 6000
a = 6000 ÷ 30 = 200
Therefore, other factor is 200.
Question 6.
STEM There are 455 grams of dissolved salt in a 13-kilogram sample of seawater. How many grams of dissolved salt are in 1 kilogram of seawater? _______
Given,
There are 455 grams of dissolved salt in a 13-kilogram sample of seawater.
So, 455 ÷ 13 = 35
Therefore, dissolved salt in 1 kilogram of seawater is 35 grams.
Use Tools Use an area model to represent the division equation and find the quotient.
Question 7.
925 ÷ 25 = c
c = 925 ÷ 25
Area = 925
length = 25
we know area = length x width
width = area / length
= 925 / 25 = 37
Therefore, 925 ÷ 25 = c = 37.
Question 8.
q = 2,750 ÷ 10
q = 2750 ÷ 10
Area = 2750
length = 10
we know area = length x width
width = area / length
= 2750 / 10 = 275
Therefore, 2750 ÷ 10 = q = 275.
Question 9.
t = 1,134 ÷ 54
t = 1134 ÷ 54
Area = 1134
length = 54
we know area = length x width
width = area / length
= 1134 / 54 = 21
Therefore, 1134 ÷ 54 = t = 21.
Question 10.
672 ÷ 24 = g
g = 672 ÷ 24
Area = 672
length = 24
we know area = length x width
width = area / length
= 672 / 24 = 28
Therefore, 672 ÷ 24 = g = 28.
Question 11.
Model with Mathematics The area of the bottom of a swimming pool is 1,250 square meters. The length of the pool is 50 meters. What is the width of the swimming pool? Write a division equation to model this situation.
Given,
The area of the bottom of a swimming pool is 1,250 square meters.
The length of the pool is 50 meters.
We know, area = length x width
1250 = 50 x width
width = 1250 ÷ 50 = 25
Therefore, width of the swimming pool is 25 square meters.
I’m in a Learning Mindset!
What part of representing division of 2-digit divisors am I comfortable solving on my own?
_____________________________
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# 6. Triangles
##### 11 LECTURES
• This page has links to complete syllabus of NCERT Chapter 6 Triangles based on syllabus recommended by CBSE based on class 10 maths.
• There are 11 video lectures having explanations from basic to advance concepts, each and every NCERT solutions, questions and all important examples. You can access them directly from the lecture menu or just scroll down the page.
• All lecture were uploaded on YouTube so for better learning experience here you will find link to YouTube Playlist of Chapter 6 Triangles.
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## PDFs OF LECTURES
##### 6(A) || Exercise 6.1
00:00:20 Difference between congruent and similar figures
00:06:37 NCERT Exercise 6.1 Question 1
00:12:27 NCERT Exercise 6.1 Question 2
00:13:27 NCERT Exercise 6.1 Question 3
##### 6(B) || Basic Proportionality Theorem
00:00:24 Proof of Basics Proportionality Theorem (Thales Theorem) Theorem 6.1 : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
00:21:28 Proof of Converse of Basic Proportionality Theorem Theorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
00:30:48 NCERT Exercise 6.2 Question 1 1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
##### 6(C) || Exercise 6.2
00:00:48 NCERT Exercise 6.2 Question 2 E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF || QR : (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
00:08:48 NCERT Exercise 6.2 Question 3 In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB=AN/AD
00:18:28 NCERT Exercise 6.2 Question 4 In Fig. 6.19, DE || AC and DF || AE. Prove that BF/BE=FE/EC
00:23:18 NCERT Exercise 6.2 Question 5 In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
00:28:38 NCERT Exercise 6.2 Question 6 In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
00:32:30 NCERT Exercise 6.2 Question 7 Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
00:36:48 NCERT Exercise 6.2 Question 8 Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
##### 6(D) || Exercise 6.2 (Q9, Q10) Theorem 6.3
00:01:02 NCERT Exercise 6.2 Question 9 ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show That AO/CO=BO/DO
00:10:54 NCERT Exercise 6.2 Question 10 The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/CO=BO/DO. Show that ABCD is a trapezium.
00:19:04 Detailed explanation of similar figures v/s congruent figures
00:27:24 Proof of AAA Similarity Criterion Theorem 6.3 : If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
##### 6(E) || Theorem 6.4 Theorem 6.5
00:00:54 Proof of SSS Similarity Criterion Theorem 6.4 : If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.
00:19:54 Proof of SAS Similarity Criterion Theorem 6.5 : If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
##### 6(F) || Exercise 6.3 (Q1 to Q9)
00:01:20 NCERT Exercise 6.3 Question 1 State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
00:15:20 NCERT Exercise 6.3 Question 2 In Fig. 6.35, ∆ ODC ~ ∆ OBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.
00:22:10 NCERT Exercise 6.3 Question 3 Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OB=OC/OD
00:29:20 NCERT Exercise 6.3 Question 4 In Fig. 6.36, QR/QT=QS/PR and ∠ 1 = ∠ 2. Show that ∆ PQS ~ ∆ TQR.
00:34:50 NCERT Exercise 6.3 Question 5 S and T are points on sides PR and QR of ∆ PQR such that ∠ P = ∠ RTS. Show that ∆ RPQ ~ ∆ RTS.
00:38:10 NCERT Exercise 6.3 Question 6 In Fig. 6.37, if ∆ ABE ≅ ∆ ACD, show that ∆ ADE ~ ∆ ABC.
00:45:00 NCERT Exercise 6.3 Question 7 In Fig. 6.38, altitudes AD and CE of ∆ ABC intersect each other at the point P. Show that: (i) ∆ AEP ~ ∆ CDP (ii) ∆ ABD ~ ∆ CBE (iii) ∆ AEP ~ ∆ ADB (iv) ∆ PDC ~ ∆ BEC
00:53:50 NCERT Exercise 6.3 Question 8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ ABE ~ ∆ CFB.
00:57:50 NCERT Exercise 6.3 Question 9 In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ∆ ABC ~ ∆ AMP (ii) CA/BC=PA/MP
##### 6(G) || Exercise 6.3 (Q10 to Q16)
00:00:36 NCERT Exercise 6.3 Question 10 CD and GH are respectively the bisectors of ∠ ACB and ∠ EGF such that D and H lie on sides AB and FE of ∆ ABC and ∆ EFG respectively. If ∆ ABC ~ ∆ FEG, show that: (i) CD/AC=GH/FG (ii) ∆ DCB ~ ∆ HGE (iii) ∆ DCA ~ ∆ HGF
00:13:07 NCERT Exercise 6.3 Question 11 In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ ABD ~ ∆ ECF.
00:16:27 NCERT Exercise 6.3 Question 13 D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. Show that CA^2 = CB.CD.
00:23:07 NCERT Exercise 6.3 Question 15 A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
00:32:17 NCERT Exercise 6.3 Question 16
00:43:17 NCERT Exercise 6.3 Question 12 Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR (see Fig. 6.41). Show that ∆ ABC ~ ∆ PQR.
00:52:37 NCERT Exercise 6.3 Question 14 Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ ABC ~ ∆ PQR.
##### 6(H) || Exercise 6.4 Theorem 6.6
00:00:30 Proof of Theorem 6.6 : The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
00:13:30 NCERT Exercise 6.4 Question 1 Let ∆ ABC ~ ∆ DEF and their areas be, respectively, 64 cm^2 and 121 cm2 . If EF = 15.4 cm, find BC.
00:17:00 NCERT Exercise 6.4 Question 2 Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
00:22:50 NCERT Exercise 6.4 Question 3 In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that ar (ABC)/ar (DBC)=AO/DO
00:29:10 NCERT Exercise 6.4 Question 4 If the areas of two similar triangles are equal, prove that they are congruent.
00:34:50 NCERT Exercise 6.4 Question 5 D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ ABC. Find the ratio of the areas of ∆ DEF and ∆ ABC.
00:44:30 NCERT Exercise 6.4 Question 6 Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
00:52:50 NCERT Exercise 6.4 Question 7 Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
01:01:10 NCERT Exercise 6.4 Question 8 ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
01:04:20 NCERT Exercise 6.4 Question 9 Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
##### 6(I) || Theorem 6.7 Theorem 6.8 Theorem 6.9
00:01:10 Proof of Theorem 6.7 : If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
00:24:23 Proof of Pythagoras Theorem Theorem 6.8 : In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
00:32:33 Proof of converse of Pythagoras Theorem Theorem 6.9 : In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
##### 6(J) || Exercise 6.5 (Q1 to Q7)
00:01:06 NCERT Exercise 6.5 Question 1 Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse. (i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm
00:07:46 NCERT Exercise 6.5 Question 2 PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM^2 = QM . MR.
00:17:56 NCERT Exercise 6.5 Question 3 In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that (i) AB^2= BC . BD (ii) AC^2= BC . DC (iii) AD^2= BD . CD
00:26:36 NCERT Exercise 6.5 Question 4 ABC is an isosceles triangle right angled at C. Prove that AB^2=2AC^2
00:29:16 NCERT Exercise 6.5 Question 5 ABC is an isosceles triangle with AC = BC. If AB^2= 2AC^2, prove that ABC is a right triangle.
00:32:56 NCERT Exercise 6.5 Question 6 ABC is an equilateral triangle of side 2a. Find each of its altitudes.
00:38:06 NCERT Exercise 6.5 Question 7 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
##### 6(K) || Exercise 6.5 (Q8 to Q17)
00:02:35 NCERT Exercise 6.5 Question 8 In Fig. 6.54, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that (i) OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+CE^2 (ii) AF^2+BD^2+CE^2=AE^2+CD^2+BF^2
00:14:35 NCERT Exercise 6.5 Question 9 A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
00:18:26 NCERT Exercise 6.5 Question 10 A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
00:24:45 NCERT Exercise 6.5 Question 11 An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1 1/2 hours?
00:31:55 NCERT Exercise 6.5 Question 12 Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
00:36:05 NCERT Exercise 6.5 Question 13 D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE^2+BD^2=AB^2+DE^2 .
00:41:35 NCERT Exercise 6.5 Question 14 The perpendicular from A on side BC of a ∆ ABC intersects BC at D such that DB = 3 CD (see Fig. 6.55). Prove that 2 AB^2 = 2 AC^2 + BC^2 .
00:47:05 NCERT Exercise 6.5 Question 15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD^2=7AB^2
00:57:05 NCERT Exercise 6.5 Question 16 In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
01:01:25 NCERT Exercise 6.5 Question 17 Tick the correct answer and justify : In ∆ ABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm. The angle B is : (A) 120° (C) 90° (B) 60° (D) 45° |
# 2004 AMC 10B Problems/Problem 4
## Problem
A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$?
$\mathrm{(A) \ } 6 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 144\qquad \mathrm{(E) \ } 720$
## Solution 1
The product of all six numbers is $6!=720$. The products of numbers that can be visible are $720/1$, $720/2$, ..., $720/6$. The answer to this problem is their greatest common divisor -- which is $720/L$, where $L$ is the least common multiple of $\{1,2,3,4,5,6\}$. Clearly $L=60$ and the answer is $720/60=\boxed{\mathrm{(B)}\ 12}$.
## Solution 2
Clearly, $P$ cannot have a prime factor other than $2$, $3$ and $5$.
We can not guarantee that the product will be divisible by $5$, as the number $5$ can end on the bottom.
We can guarantee that the product will be divisible by $3$ (one of $3$ and $6$ will always be visible), but not by $3^2$.
Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by $2^2$. This is the most we can guarantee, as when the $4$ is on the bottom side, the two visible even numbers are $2$ and $6$, and their product is not divisible by $2^3$.
## Solution 3
The product P can be one of the following six numbers excluding the number that is hidden under, so we have: \begin{align*} 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3^2 \cdot 5 \\ 1 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^3 \cdot 3^2 \cdot 5 \\ 1 \cdot 2 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3 \cdot 5 \\ 1 \cdot 2 \cdot 3 \cdot 5 \cdot 6 = 2^2 \cdot 3^2 \cdot 5 \\ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 = 2^4 \cdot 3^2 \\ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 2^3 \cdot 3 \cdot 5 \end{align*}
The largest number that is certain to divide product P is basically GCD of all the above 6 products which is $2^2 \cdot 3$.
Hence $P=3\cdot2^2=\boxed{\mathrm{(B)}\ 12}$. |
# SHAPE, SPACE AND MEASURES
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1 SHAPE, SPACE AND MEASURES Pupils should be taught to: Use accurately the vocabulary, notation and labelling conventions for lines, angles and shapes; distinguish between conventions, facts, definitions and derived properties As outcomes, Year 7 pupils should, for example: Use, read and write, spelling correctly: line segment, line parallel, perpendicular plane horizontal, vertical, diagonal adjacent, opposite point, intersect, intersection vertex, vertices side angle, degree ( ) acute, obtuse, reflex vertically opposite angles base angles Use accurately the notation and labelling conventions for lines, angles and shapes. Understand that a straight line can be considered to have infinite length and no measureable width, and that a line segment is of finite length, e.g. line segment AB has end-points A and B. A B Know that: Two straight lines in a plane (a flat surface) can cross once or are parallel; if they cross, they are said to intersect, and the point at which they cross is an intersection. When two line segments meet at a point, the angle formed is the measure of rotation of one of the line segments to the other. The angle can be described as DEF or DÊF or E. D E F A polygon is a 2-D or plane shape constructed from line segments enclosing a region. The line segments are called sides; the end points are called vertices. The polygon is named according to the number of its sides, vertices or angles: triangle, quadrilateral, pentagon Know the labelling convention for: triangles capital letters for the vertices (going round in order, clockwise or anticlockwise) and corresponding lower-case letters for each opposite side, the triangle then being described as ABC; C b A a c B equal sides and parallel sides in diagrams. A A B B C D C AB = AC AB is parallel to DC, or AB//DC. AD is parallel to BC, or AD//BC. 178 Y789 examples Crown copyright 2001
2 Geometrical reasoning: lines, angles and shapes As outcomes, Year 8 pupils should, for example: Use vocabulary from previous year and extend to: corresponding angles, alternate angles supplementary, complementary interior angle, exterior angle equidistant prove, proof As outcomes, Year 9 pupils should, for example: Use vocabulary from previous years and extend to: convention, definition, derived property... Continue to use accurately the notation and labelling conventions for lines, angles and shapes. Know that DEF is an interior angle of DEF and that GDF is an exterior angle of DEF. E F Know that: A pair of complementary angles have a sum of 90. A pair of supplementary angles have a sum of 180. D G Distinguish between conventions, definitions and derived properties. A convention is an agreed way of illustrating, notating or describing a situation. Conventions are arbitrary alternatives could have been chosen. Examples of geometrical conventions are: the ways in which letters are used to label the angles and sides of a polygon; the use of arrows to show parallel lines; the agreement that anticlockwise is taken as the positive direction of rotation. A definition is a minimum set of conditions needed to specify a geometrical term, such as the name of a shape or a transformation. Examples are: A polygon is a closed shape with straight sides. A square is a quadrilateral with all sides and all angles equal. A degree is a unit for measuring angles, in which one complete rotation is divided into 360 degrees. A reflection in 2-D is a transformation in which points (P) are mapped to images (P ), such that PP is at right angles to a fixed line (called the mirror line, or line of reflection), and P and P are equidistant from the line. A derived property is not essential to a definition, but consequent upon it. Examples are: The angles of a triangle add up to 180. A square has diagonals that are equal in length and that bisect each other at right angles. The opposite sides of a parallelogram are equal in length. Points on a mirror line reflect on to themselves. Distinguish between a practical demonstration and a proof. For example, appreciate that the angle sum property of a triangle can be demonstrated practically by folding the corners of a triangular sheet of paper to a common point on the base and observing the result. A proof requires deductive argument, based on properties of angles and parallels, that is valid for all triangles. Crown copyright 2001 Y789 examples 179
3 SHAPE, SPACE AND MEASURES Pupils should be taught to: Identify properties of angles and parallel and perpendicular lines, and use these properties to solve problems As outcomes, Year 7 pupils should, for example: Identify parallel and perpendicular lines. Recognise parallel and perpendicular lines in the environment, and in 2-D and 3-D shapes: for example, rail tracks, side edges of doors, ruled lines on a page, double yellow lines Use dynamic geometry software, acetate overlays or film to explore and explain relationships between parallel and intersecting lines, such as: parallel lines, which are always equidistant; perpendicular lines, which intersect at right angles; lines which intersect at different angles. For example, as one line rotates about the point of intersection, explain how the angles at the point of intersection are related. Use ruler and set square to draw parallel and perpendicular lines. Link with constructions (page 220 3). 180 Y789 examples Crown copyright 2001
4 Geometrical reasoning: lines, angles and shapes As outcomes, Year 8 pupils should, for example: As outcomes, Year 9 pupils should, for example: Identify alternate and corresponding angles. Use dynamic geometry software or acetate overlays to explore and explain relationships between lines in the plane, such as: three lines that intersect in one point; e d f c a b a + b + c = 180 a = d, b = e, c = f given two intersecting lines and a third that moves but remains parallel to one of them, explain which angles remain equal; two pairs of parallel lines, forming a parallelogram. Understand and use the terms corresponding angles and alternate angles. c a a c corresponding angles alternate angles Use alternate angles to prove that opposite angles of a parallelogram are equal: a = b a b Crown copyright 2001 Y789 examples 181
5 SHAPE, SPACE AND MEASURES Pupils should be taught to: Identify properties of angles and parallel and perpendicular lines, and use these properties to solve problems (continued) As outcomes, Year 7 pupils should, for example: Know the sum of angles at a point, on a straight line and in a triangle, and recognise vertically opposite angles and angles on a straight line. b a a b x y x + y = 180 vertically opposite angles angles on a straight line Link with rotation (pages ). Recognise from practical work such as measuring and paper folding that the three angles of a triangle add up to 180. Given sufficient information, calculate: angles in a straight line and at a point; the third angle of a triangle; the base angles of an isosceles triangle. Calculate the angles marked by letters. 215 x x 182 Y789 examples Crown copyright 2001
6 Geometrical reasoning: lines, angles and shapes As outcomes, Year 8 pupils should, for example: Understand a proof that the sum of the angles of a triangle is 180 and of a quadrilateral is 360, and that the exterior angle of a triangle equals the sum of the two interior opposite angles. Consider relationships between three lines meeting at a point and a fourth line parallel to one of them. Use dynamic geometry software to construct a triangle with a line through one vertex parallel to the opposite side. Observe the angles as the triangle is changed by dragging any of its vertices. As outcomes, Year 9 pupils should, for example: Explain how to find, calculate and use properties of the interior and exterior angles of regular and irregular polygons. Explain how to find the interior angle sum and the exterior angle sum in (irregular) quadrilaterals, pentagons and hexagons. A polygon with n sides can be split into n 2 triangles, each with an angle sum of 180. So the interior angle sum is (n 2) 180, giving 360 for a quadrilateral, 540 for a pentagon and 720 for a hexagon. At each vertex, the sum of the interior and exterior angles is 180. Use this construction, or a similar one, to explain using diagrams a proof that the sum of the three angles of a triangle is 180. For n vertices, the sum of n interior and n exterior angles is n 180. But the sum of the interior angles is (n 2) 180, so the sum of the exterior angles is always = 360. Use the angle sum of a triangle to prove that the angle sum of a quadrilateral is 360. (a + b + c) + (d + e + f ) = = 360 Explain a proof that the exterior angle of a triangle equals the sum of the two interior opposite angles, using this or another construction. Given sufficient information, calculate: interior and exterior angles of triangles; interior angles of quadrilaterals. Calculate the angles marked by letters. 76 X b d x 136 y 46 X f a c e Find, calculate and use the interior and exterior angles of a regular polygon with n sides. The interior angle sum S for a polygon with n sides is S = (n 2) 180. In a regular polygon all the angles are equal, so each interior angle equals S divided by n. Since the interior and exterior angles are on a straight line, the exterior angle can be found by subtracting the interior angle from 180. From experience of using Logo, explain how a complete traverse of the sides of a polygon involves a total turn of 360 and why this is equal to the sum of the exterior angles. Deduce interior angle properties from this result. Recall that the interior angles of an equilateral triangle, a square and a regular hexagon are 60, 90 and 120 respectively. Crown copyright 2001 Y789 examples 183
7 SHAPE, SPACE AND MEASURES Pupils should be taught to: Identify and use the geometric properties of triangles, quadrilaterals and other polygons to solve problems; explain and justify inferences and deductions using mathematical reasoning As outcomes, Year 7 pupils should, for example: Use, read and write, spelling correctly: polygon, regular, irregular, convex, concave circle, triangle, isosceles, equilateral, scalene, right-angled, quadrilateral, square, rectangle, parallelogram, rhombus, trapezium, kite, delta and names of other polygons. Visualise and sketch 2-D shapes in different orientations, or draw them using dynamic geometry software. Describe what happens and use the properties of shapes to explain why. Imagine a square with its diagonals drawn in. Remove one of the triangles. What shape is left? How do you know? Imagine a rectangle with both diagonals drawn. Remove a triangle. What sort of triangle is it? Why? Imagine joining adjacent mid-points of the sides of a square. What shape is formed by the new lines? Explain why. Imagine a square with one of its corners cut off. What different shapes could you have left? Imagine an isosceles triangle. Fold along the line of symmetry. What angles can you see in the folded shape? Explain why. Imagine a square sheet of paper. Fold it in half and then in half again, to get another smaller square. Which vertex of the smaller square is the centre of the original square? Imagine a small triangle cut off this corner. Then imagine the paper opened out. What shape will the hole be? Explain your reasoning. Imagine what other shapes you can get by folding a square of paper in different ways and cutting off different shapes. 184 Y789 examples Crown copyright 2001
8 Geometrical reasoning: lines, angles and shapes As outcomes, Year 8 pupils should, for example: Use vocabulary from previous year and extend to: bisect, bisector, mid-point congruent tessellate, tessellation As outcomes, Year 9 pupils should, for example: Use vocabulary from previous years and extend to: similar, similarity hypotenuse, Pythagoras theorem Visualise and sketch 2-D shapes in different orientations, or draw them using dynamic geometry software. Describe what happens and use the properties of shapes to explain why. Imagine a rectangular sheet of paper. Cut along the diagonal to make two triangles. Place the diagonals together in a different way. What shape is formed? Imagine two equilateral triangles, placed together, edge to edge. What shape is formed? Why? Add a third equilateral triangle a fourth What shapes are formed? Sketch some diagrams and explain what can be seen. Imagine two congruent isosceles triangles. Put sides of equal length together. Describe the resulting shape. Is it the only possibility? Imagine a quadrilateral with two lines of symmetry. What could it be? Suppose it also has rotation symmetry of order 2. What could it be now? Construct a parallelogram by drawing two line segments from a common end-point. Draw parallel lines to form the other two sides. Draw the two diagonals. Visualise and sketch 2-D shapes or draw them using dynamic geometry software as they go through a sequence of changes. Describe what happens and use the properties of shapes to explain why. Imagine starting with an equilateral triangle with one side horizontal call it the base. Imagine this base is fixed. The opposite vertex of the triangle moves slowly in a straight line, perpendicular to the base. What happens to the triangle? Now imagine that the opposite vertex moves parallel to the base. What happens? Can you get a right-angled triangle, or an obtuseangled triangle? Imagine a square sheet of paper. Imagine making a straight cut symmetrically across one corner. What shape is left? Imagine making a series of straight cuts, always parallel to the first cut. Describe what happens to the original square. Imagine two sheets of acetate, each marked with a set of parallel lines, spaced 2 cm apart. Imagine one sheet placed on top of the other, so that the two sets of lines are perpendicular. What shapes do you see? What happens to the pattern as the top sheet slowly rotates about a fixed point (the intersection of two lines)? What if the lines were 1 cm apart on one sheet and 2 cm apart on the other? Overlay tessellations in various ways, such as octagons and squares on octagons and squares. Describe the outcomes. Observe the sides, angles and diagonals as the parallelogram is changed by dragging its vertices. Describe tilings and other geometrical patterns in pictures and posters. Suggest reasons why objects in the environment (natural or constructed) take particular shapes. Explore tessellations using plastic or card polygon shapes and/or computer tiling software, and explain why certain shapes tessellate. Explore how regular polygons which do not tessellate (e.g. nonagons) can be used to cover the plane by leaving holes in a regular pattern. Describe the outcomes. Crown copyright 2001 Y789 examples 185
9 SHAPE, SPACE AND MEASURES Pupils should be taught to: Identify and use the geometric properties of triangles, quadrilaterals and other polygons to solve problems; explain and justify inferences and deductions using mathematical reasoning (continued) As outcomes, Year 7 pupils should, for example: Triangles, quadrilaterals and other polygons Review the properties of triangles and quadrilaterals (see Y456 examples, pages 102 3). Using a 3 by 3 array on a pinboard, identify the eight distinct triangles that can be constructed (eliminating reflections, rotations or translations). Classify the triangles according to their side, angle and symmetry properties. In this pattern of seven circles, how many different triangles and quadrilaterals can you find by joining three or four points of intersection? What are the names of the shapes and what can you find out about their angles? Begin to identify and use angle, side and symmetry properties of triangles, quadrilaterals and other polygons. Start with a 2 by 1 rectangle. Make these two shapes. What new shapes can you make with them? Name them and discuss their properties Y789 examples Crown copyright 2001
10 Geometrical reasoning: lines, angles and shapes As outcomes, Year 8 pupils should, for example: Know and use side, angle and symmetry properties of equilateral, isosceles and right-angled triangles. Discuss whether it is possible to draw or construct on a 3 by 3 pinboard: a. a triangle with a reflex angle; b. an isosceles trapezium; c. an equilateral triangle or a (non-square) rhombus. If not, explain why not. As outcomes, Year 9 pupils should, for example: Know and use angle and symmetry properties of polygons, and angle properties of parallel and intersecting lines, to solve problems and explain reasoning. Deduce the angles of the rhombus in this arrangement of three identical tiles. Classify quadrilaterals by their geometric properties (equal and/or parallel sides, equal angles, right angles, diagonals bisected and/or at right angles, reflection and rotation symmetry ). Know properties such as: An isosceles trapezium is a trapezium in which the two opposite non-parallel sides are the same length. It has one line of symmetry and both diagonals are the same length. A parallelogram has its opposite sides equal and parallel. Its diagonals bisect each other. It has rotation symmetry of order 2. A rhombus is a parallelogram with four equal sides. Its diagonals bisect each other at right angles. Both diagonals are lines of symmetry. It has rotation symmetry of order 2. A kite is a quadrilateral that has two pairs of adjacent sides of equal length, and no interior angle larger than 180. It has one line of symmetry and its diagonals cross at right angles. An arrowhead or delta has two pairs of adjacent edges of equal length and one interior angle larger than 180. It has one line of symmetry. Its diagonals cross at right angles outside the shape. Provide a convincing argument to explain, for example, that a rhombus is a parallelogram but a parallelogram is not necessarily a rhombus. What can you deduce about the shape formed by the outline? Explain why: Equilateral triangles, squares and regular hexagons will tessellate on their own but other regular polygons will not. Squares and regular octagons will tessellate together. Know and use properties of triangles, including Pythagoras theorem. Know that: In any triangle, the largest angle is opposite the longest side and the smallest angle is opposite the shortest side. In a right-angled triangle, the side opposite the right angle is the longest and is called the hypotenuse. Understand, recall and use Pythagoras theorem. Explain special cases of Pythagoras theorem in geometrical arrangements such as: Devise questions for a tree classification diagram to sort a given set of quadrilaterals. Identify then classify the 16 distinct quadrilaterals that can be constructed on a 3 by 3 pinboard. Link to standard constructions (pages 220 3). Crown copyright 2001 Y789 examples 187
11 SHAPE, SPACE AND MEASURES Pupils should be taught to: Identify and use the geometric properties of triangles, quadrilaterals and other polygons to solve problems; explain and justify inferences and deductions using mathematical reasoning (continued) As outcomes, Year 7 pupils should, for example: Use the properties of angles at a point and on a straight line, and the angle sum of a triangle, to solve simple problems. Explain reasoning. What different shapes can you make by overlapping two squares? Can any of these shapes be made? rectangle pentagon decagon rhombus hexagon kite isosceles triangle octagon trapezium If a shape cannot be made, explain why. What shapes can you make by overlapping three squares? Two squares overlap like this. The larger square has one of its vertices at the centre of the smaller square. Explain why the shaded area is one quarter of the area of the smaller square. Explain why a triangle can never have a reflex angle but a quadrilateral can. Provide a convincing argument to explain why it is always possible to make an isosceles triangle from two identical right-angled triangles. Use Logo to write instructions to draw a parallelogram. Link to problems involving shape and space (pages 14 17). 188 Y789 examples Crown copyright 2001
13 SHAPE, SPACE AND MEASURES Pupils should be taught to: As outcomes, Year 7 pupils should, for example: Understand congruence and similarity 190 Y789 examples Crown copyright 2001
14 Geometrical reasoning: lines, angles and shapes As outcomes, Year 8 pupils should, for example: Congruence Know that if two 2-D shapes are congruent, they have the same shape and size, and corresponding sides and angles are equal. From a collection of different triangles or quadrilaterals, identify those that are congruent to each other by placing one on top of the other. Realise that corresponding sides and angles are equal. Divide a 4 by 4 pinboard into two congruent halves. How many different ways of doing this can you find? As outcomes, Year 9 pupils should, for example: Congruence Appreciate that when two shapes are congruent, one can be mapped on to the other by a translation, reflection or rotation, or some combination of these transformations. See Year 8 for examples. Link to transformations (pages ). Know from experience of constructing them that triangles satisfying SSS, SAS, ASA or RHS are unique, but that triangles satisfying SSA or AAA are not. Link to constructions (pages 220 3). Divide the pinboard into four congruent quarters. Divide a 5 by 5 pinboard into two non-congruent halves. Using a 3 by 3 pinboard, make some different triangles or quadrilaterals. For each shape, investigate whether you can produce one or more identical shapes in different positions or orientations on the board. Describe the transformation(s) you use to do this. Appreciate that two triangles will be congruent if they satisfy the same conditions: three sides are equal (SSS); two sides and the included angle are equal (SAS); two angles and a corresponding side are equal (ASA); a right angle, hypotenuse and side are equal (RHS). Use these conditions to deduce properties of triangles and quadrilaterals. Draw triangle ABC, with AB = AC. Draw the perpendicular from A to BC to meet BC at point D. B A D C Extend to 3 by 4 and larger grids. Two congruent scalene triangles without right angles are joined with two equal edges fitted together. Show that triangles ABD and ACD are congruent. Hence show that the two base angles of an isosceles triangle are equal. Use congruence to prove that the diagonals of a rhombus bisect each other at right angles. What shapes can result? What if the two triangles are right-angled, isosceles or equilateral? In each case, explain how you know what the resultant shapes are. By drawing a diagonal and using the alternate angle property, use congruence to prove that the opposite sides of a parallelogram are equal. Crown copyright 2001 Y789 examples 191
15 SHAPE, SPACE AND MEASURES Pupils should be taught to: As outcomes, Year 7 pupils should, for example: Understand congruence and similarity (continued) 192 Y789 examples Crown copyright 2001
16 Geometrical reasoning: lines, angles and shapes As outcomes, Year 8 pupils should, for example: As outcomes, Year 9 pupils should, for example: Similarity Know that the term similar has a precise meaning in geometry. Objects, either plane or solid figures, are similar if they have the same shape. That is, corresponding angles contained in similar figures are equal, and corresponding sides are in the same ratio. A P B C Q R BAC = QPR, ACB = PRQ, ABC = PQR and AB : PQ = BC : QR = CA : RP Recognise and use characteristics of similar shapes. If two shapes are similar, one can be considered to be an enlargement of the other. Link to enlargement (pages ). Any two regular polygons with the same number of sides are mathematically similar (e.g. any two squares, any two equilateral triangles, any two regular hexagons). All circles are similar. Use this knowledge when considering metric properties of the circle, such as the relationship between circumference and diameter. Link to circumference of a circle (pages 234 5). Solve problems involving similarity. These two triangles are similar. 3cm 6cm 4cm 8cm a. Find the perimeter and area of each triangle. b. What is the ratio of the two perimeters? c. What is the ratio of the two areas? Explain why these two cuboids are similar. 2cm 2cm 3cm 4cm 4cm 6cm What is the ratio of their surface areas? What is the ratio of their volumes? Link to ratio and proportion (pages 78 81). Crown copyright 2001 Y789 examples 193
17 SHAPE, SPACE AND MEASURES Pupils should be taught to: As outcomes, Year 7 pupils should, for example: Identify and use the properties of circles 194 Y789 examples Crown copyright 2001
18 Geometrical reasoning: lines, angles and shapes As outcomes, Year 8 pupils should, for example: As outcomes, Year 9 pupils should, for example: Circles Know the parts of a circle, including: centre, radius, diameter, circumference, chord, arc, segment, sector, tangent and terms such as: circumcircle, circumscribed, inscribed Know that: A circle is a set of points equidistant from its centre. The circumference is the distance round the circle. The radius is the distance from the centre to the circumference. An arc is part of the circumference. A sector is the region bounded an arc and two radii. Use dynamic geometry software to show a line and a circle moving towards each other. A A P O B B Know that when the line: touches the circle at a point P, it is called a tangent to the circle at that point; intersects the circle at two points A and B, the line segment AB is called a chord of the circle, which divides the area enclosed by the circle into two regions called segments; passes through the centre of the circle, the line segment AB becomes a diameter, which is twice the radius and divides the area enclosed by the circle into two semicircles. Explain why inscribed regular polygons can be constructed by equal divisions of a circle. Appreciate that a chord of a circle, together with the radii through the end points, form an isosceles triangle. If further chords of the same length are drawn, the triangles are congruent and angles between successive chords are equal. Hence, if the succession of equal chords divides the circle without remainder, a regular polygon is inscribed in the circle. If chords of length equal to the radius are marked on the circumference of a circle, explain why the resultant shape is a regular hexagon. Use this to construct a hexagon of a given side. Link to circumference and area of a circle (pages 234 7). Crown copyright 2001 Y789 examples 195
19 SHAPE, SPACE AND MEASURES Pupils should be taught to: As outcomes, Year 7 pupils should, for example: Identify and use the properties of circles (continued) 196 Y789 examples Crown copyright 2001
20 Geometrical reasoning: lines, angles and shapes As outcomes, Year 8 pupils should, for example: As outcomes, Year 9 pupils should, for example: Circles (continued) Recognise that a tangent is perpendicular to the radius at the point of contact P. P Use dynamic geometry software to explore properties of lines and circles. Make two circles, of equal radius, touch then intersect. What happens to the common chord and the line joining the centres, or to the rhombus formed by joining the radii to the points of intersection? Construct a triangle and the perpendicular bisectors of its three sides. Draw the circumcircle (the circle through the three vertices). What happens when the vertices of the triangle move? Observe in particular the position of the centre of the circumcircle. Link to standard constructions (pages 220 3). Crown copyright 2001 Y789 examples 197
22 Geometrical reasoning: lines, angles and shapes As outcomes, Year 8 pupils should, for example: Use vocabulary from previous year and extend to: view, plan, elevation isometric As outcomes, Year 9 pupils should, for example: Use vocabulary from previous years and extend to: cross-section, projection plane Know and use geometric properties of cuboids and shapes made from cuboids; begin to use plans and elevations. Describe 3-D shapes which can be visualised from a wall poster or a photograph. Visualise and describe relationships between the edges of a cube, e.g. identify edges which: meet at a point; are parallel; are perpendicular; are neither parallel nor intersect each other. Imagine a cereal packet standing on a table. Paint the front and the back of the packet red. Paint the top and bottom red and the other two faces blue. Now study the packet carefully. How many edges has it? How many edges are where a red face meets a blue face? How many edges are where a red face meets another red face? How many edges are where a blue face meets another blue face? Sit back to back with a partner. Look at the picture of the model. Don t show it to your partner. Tell your partner how to build the model. Analyse 3-D shapes through 2-D projections and cross-sections, including plans and elevations. Visualise solids from an oral description. In each case, identify the solid shape. a. The front and side elevations are both triangles and the plan is a square. b. The front and side elevations are both rectangles and the plan is a circle. c. The front elevation is a rectangle, the side elevation is a triangle and the plan is a rectangle. d. The front and side elevations and the plan are all circles. The following are shadows of solids. Describe the possible solids for each shadow (there may be several solutions). In each case, identify the solid shape. Draw the net of the solid. a. front side plan b. Sketch a net to make this model. Construct the shape. front side plan Write the names of the polyhedra that could have an isosceles or equilateral triangle as a front elevation. Here are three views of the same cube. Which letters are opposite each other? Crown copyright 2001 Y789 examples 199
23 SHAPE, SPACE AND MEASURES Pupils should be taught to: Use 2-D representations, including plans and elevations, to visualise 3-D shapes and deduce some of their properties (continued) As outcomes, Year 7 pupils should, for example: A square piece of card is viewed from different angles against the light. Which of the following are possible views of the square card? Which are impossible? Find all possible solids that can be made from four cubes. Record the solids using isometric paper. Investigate the number of different ways that a 2 by 2 by 2 cube can be split into two pieces: a. of the same shape and size; b. of different shapes and sizes. See Y456 examples (pages 104 5). 200 Y789 examples Crown copyright 2001
24 Geometrical reasoning: lines, angles and shapes As outcomes, Year 8 pupils should, for example: Orient an isometric grid and use conventions for constructing isometric drawings: Vertical edges are drawn as vertical lines. Horizontal edges are drawn at 30. Identify the position of hidden lines in an isometric drawing. Begin to use plans and elevations. The diagrams below are of solids when observed directly from above. As outcomes, Year 9 pupils should, for example: Visualise and describe sections obtained by slicing in different planes. Compare horizontal cross-sections of a squarebased right pyramid at different heights. Repeat for vertical cross-sections at different points. This cube has been sliced to give a square cross-section. Describe what the solids could be and explain why. Draw the front elevation, side elevation and plan of this shape. plan view Is it possible to slice a cube so that the cross-section is: a. a rectangle? b. a triangle? c. a pentagon? d. a hexagon? If so, describe how it can be done. For eight linked cubes, find the solids with the smallest and the largest surface area. Draw the shapes on isometric paper. Extend to 12 cubes. Imagine you have a cube. Put a dot in the centre of each face. Join the dots on adjacent sides by straight lines. What shape is generated by these lines? front view side view Visualise an octahedron. Put a dot in the centre of each face. Join the dots on adjacent sides by straight lines. What shape is generated by these lines? This diagram represents a plan of a solid made from cubes, the number in each square indicating how many cubes are on that base Make an isometric drawing of the solid from a chosen viewpoint. Imagine a slice cut symmetrically off each corner of a cube. Describe the solid which remains. Is there more than one possibility? Repeat for a tetrahedron or octahedron. Triangles are made by joining three of the vertices of a cube. How many different-shaped triangles can you make like this? Draw sketches of them. Link to plane symmetry (pages 206 7). Construct this solid, given the front elevation, side elevation and plan. front elevation side elevation plan view Crown copyright 2001 Y789 examples 201
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• NCERT Solutions for Class 9 Maths
Class 9 NCERT Solutions – Chapter 1 Number System – Exercise 1.2
(iii) Every real number is an irrational number.
Solution:
(i) Every irrational number is a real number.
True
Irrational numbers are the number that cannot be written in the form of p/q , p and q are the integers and q ≠ 0.
Some examples of irrational numbers are π, √3, e, √2, 011011011…..
Real numbers include both rational numbers and irrational numbers.
Thus, every irrational number is a real number.
(ii) Every point on the number line is of the form √m, where m is a natural number.
False
We can represent both negative and positive numbers on a number line.
Positive numbers can be written as √16=4 that is a natural number, But √3=1.73205080757 is not a natural number.
But negative numbers cannot be expressed as the square root of any natural number, as if we take square root of an negative number it will become complex number that will not be a natural number (√5=5i is a complex number).
(iii) Every real number is an irrational number.
False
Every irrational number is a real number but every real number is not an irrational number as real numbers include both rational and irrational number.
Question 2: Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Solution:
No, the square roots of all positive integers in not irrational. For example √9 = 3, √25 = 5, hence square roots of all positive integers is not irrational.
Question 3: Show how √5 can be represented on the number line.
Solution:
To represent √5 on the number line follow the following steps:
Step 1: Draw a number line.
Step 2: Let line AB of 2 units on the number line.
Step 3: Now draw a perpendicular of unit 1 on point B and mark the other end as point C.
Step 4: Join AC, you will have triangle ABC which is a right-angle triangle. as shown below:
Step 5: Apply Pythagoras Theorem on triangle ABC,
⇒ AB2 + BC2 = AC2
⇒ AC2 = 22 + 12
⇒ AC2 = 5
⇒ AC = √5
Therefore, line AC is of √5 unit length.
Step 6: Now take line AC as the radius and an arc and intersect it to number line, the point where the arc will intersect the number line is of lengths √5 from point 0 to the intersection as point A is the centre of the radius.
Question 4: Classroom activity (Constructing the ‘square root spiral’): Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1 P2 perpendicular to OP1 of unit length (see Fig. 1.9). Now draw a line segment P2 P3 perpendicular to OP2. Then draw a line segment P3 P4 perpendicular to OP3. Continuing in this manner, you can get the line segment Pn-1Pn by drawing a line segment of unit length perpendicular to OPn-1. In this manner, you will have created the points P2, P3, …., Pn, …., and joined them to create a beautiful spiral depicting √2, √3, √4, …
Solution:
Step 1: First lets mark a point O on the larger sheet of paper, this point will be the centre of the square root spiral.
Step 2: Draw point P1 from point O of 1 unit. OP1=1 unit.
Step 3: Similarly as in the above problem from P1 draw a perpendicular of 1 unit, P1P2 = 1 unit.
Step 4: Now join OP2 = √2
Step 5: From point P2 draw a perpendicular of 1 unit. P2P3 = 1 unit.
Step 6: Now join OP3 = √3
Step 7: Now repeat the step to make √4, √5, √6, …….
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# Class 8 Maths Chapter 10 Visualising Solid Shapes
Posted on: October 27, 2021 Posted by: user Comments: 0
Chapter 10 Visualising Solid Shapes
## NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.1
Question 1.
For each of the given solid, the two views are given. Match for each solid the corresponding top and front views. The first one is done for you.
Solution.
Question 2.
For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views.
Solution.
Question 3.
For each given solid, identify the top view, front view, and side view.
Solution.
Question 4.
Draw the front view, side view and top view of the given objects,
Solution.
## NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2
Question 1.
Look at the given map of a city.
(a) Colour the map as follows: Bluewater, Red-fire station, Orange-Library, Yellow-schools, Green-Parks, Pink-Community Centre, Purple-Hospital, Brown-Cemetery.
(b) Mark a green X’ at the intersection of Road ‘C’ and Nehru Road, Green Y’ at the intersection of Gandhi Road and Road A.
(c) In red, draw a short street route from Library to the bus depot.
(d) Which is further east, the city park or the market?
(e) Which is further south, the primary school or the Sr. Secondary School?
Solution.
(b) See the above figure.
(c) See the above figure.
(d) City Park.
(e) Senior Secondary school.
Question 2.
Draw a map of your classroom using a proper scale and symbols for different objects.
Solution.
Question 3.
Draw a map of your school compound using a proper scale and symbols for various features like playground main building, garden etc.
Solution.
Question 4.
Draw a map giving instructions to your friend so that she reaches your house without any difficulty.
Solution.
## NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3
Question 1.
Can a polyhedron have for its faces
(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?
Solution.
(i) No
(ii) Yes
(iii) Yes
Question 2.
Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid).
Solution.
Possible, only if the number of faces is greater than or equal to 4.
Question 3.
Which are prisms among the following?
Solution.
We know that a prism is a polyhedron whose base and top faces are congruent and parallel and other (lateral) faces are parallelograms in shape. So, only (ii) and (iv) are prisms.
Question 4.
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Solution.
(i) The prisms and cylinder, both, have their base and top faces as congruent and parallel to each other. Also, a prism becomes a cylinder as the number of sides of its base becomes larger and larger.
(ii) The pyramids and cones are alike in the sense that their lateral faces meet at a point (called vertex). Also, a pyramid becomes a cone as the number of sides of its base becomes larger and larger.
Question 5.
Is a square prism same as a cube ? Explain.
Solution.
No; not always as it can be a cuboid also.
Question 6.
Verify Euler’s formula for these solids
Solution.
(i)
F = 7
V= 10
E = 15
F + V = 7 + 10 = 17
E + 2 = 15 + 2 = 17
So, F + V = E + 2
Hence, Euler’s Formula is verified,
(ii)
F = 9
V = 9
E = 16
F + V = 9 + 9 = 18
E + 2 = 16 + 2 = 18
So, F + V = E + 2
Hence, Euler’s Formula is verified
Question 7.
Using Euler’s formula find the unknown.
Solution.
(i)
F + V = E + 2
⇒ F + 6 = 12 + 2
⇒ F + 6 = 14
⇒ F = 14 – 6 = 8
(ii)
F + V = E + 2
⇒ 5 + V = 9 + 2
⇒ 5 + V = 11
⇒ V= 11-5 = 6
(iii)
F + V = E + 2
⇒ 20 + 12 = E + 2
⇒ 32 = E + 2
⇒ E = 32 – 2
⇒ E = 30
Question 8.
Can a polyhedron have 10 faces, 20 edges, and 15 vertices?
Solution.
Here F = 10
E = 20
V= 15
So, F + V = 10 + 15 = 25
E + 2 = 20 + 2 = 22
∵ F + V ≠ E + 2
∴ Such a polyhedron is not possible. |
## Envision Math 5th Grade Textbook Answer Key Topic 20.2 Constructing Lines
Constructing Lines
How can you construct perpendicular and parallel lines?
The rails of the track are parallel. The ties are perpendicular to the rails.
Construct a line perpendicular to $$\overleftrightarrow{L N}$$ and a line parallel to $$\overleftrightarrow{L N}$$.
Another Example
How do you construct a line segment congruent to a given line segment?
Without measuring with a ruler, draw $$\overleftrightarrow{J K}$$ congruent to $$\overleftrightarrow{S T}$$.
Step 1: Draw a ray. Label the endpoint J.
Step 2 : On $$\overleftrightarrow{S T}$$, with point S as the center, open the compass so that it lines
up with point T.
Then place the compass on the ray with point J as the center.
Without changing the compass setting, draw an arc that intersects the ray. Label the point of intersection K. $$\overleftrightarrow{J K}$$ is congruent to $$\overleftrightarrow{S T}$$.
Explain it
Question 1.
How is constructing a figure different from drawing a figure?
Question 2.
In Step 2, could the ray be any length?
Question 3.
Does a line segment need to be horizontal in order to construct another segment congruent to it?
Draw a line with points L and N. With L as center, draw two arcs that intersect $$\overleftrightarrow{L N}$$. Label the points W and X.
Set the compass wider. Using W and X as centers, draw arcs that intersect. Label the point Y. Draw $$\overleftrightarrow{L Y}$$
Repeat Steps 1 and 2 at Y to find point Z.
$$\overleftrightarrow{L Y}$$ ⊥ $$\overleftrightarrow{L N}$$ and $$\overleftrightarrow{L N}$$ || $$\overleftrightarrow{Y Z}$$
Guided Practice
Do you know HOW?
Question 1.
Draw a line perpendicular to line CD. Copy $$\overleftrightarrow{C D}$$ on a separate sheet of paper and construct $$\overleftrightarrow{T C}$$ so that it is perpendicular to $$\overleftrightarrow{C D}$$.
Question 2.
Draw as egment congruent to $$\overline{E F}$$. Copy $$\overline{E F}$$ on a separate sheet of paper. Then draw a ray labele d $$\overleftrightarrow{M N}$$. On that ray, construct $$\overline{M P}$$ so that $$\overline{M P}$$ is congruent to $$\overline{E F}$$.
Do you UNDERSTAND?
Question 3.
Writing to Explain In Step 2 of the example above, why is it necessary to set the compass wider than the length of segment WL?
Question 4.
Look at the line TC you constructed in Problem 1 that is perpendicular to line CD. Use point Tand construct a line perpendicular to line TC. How is that line related to line CD?
Independent Practice
In 5 through 7, copy the figures on a separate sheet of paper and follow the directions.
Question 5.
Construct a line perpendicular to line XY.
Question 6.
Construct a line perpendicular to line MN. Then construct a line through it that is parallel to line MN.
Question 7.
Draw a line segment that is congruent to CD .
Problem Solving
Engineers are making plans to lay new railroad tracks between cities. In 8 and 9, use the information in the table.
Question 8.
If it costs \$155 per mile to construct railroad tracks, how much would it cost to build tracks from San Francisco to Los Angeles?
Question 9.
Question 10.
If you are comparing two negative integers on a number line, how can you tell which one is greater?
Question 11.
Four friends played golf. Their scores were +3, -1, *4, and ~4 in relation to par. The least score wins. Arrange the scores from best to worst.
Question 12.
Algebra Ted has 15 trophies. This is 5 times as many as Harold has. How many trophies does Harold have? Write and solve an equation to answer the question.
Question 13.
Writing to Explain Explain how perpendicular lines are similar to intersecting lines.
Question 14.
Use the figure to tell whether the statements are true or false
a. $$\overline{P Q}$$ is parallel to $$\overline{U V}$$
b. $$\overline{P Q}$$ intersects $$\overline{U V}$$
c. $$\overline{P Q}$$ is perpendicular to $$\overline{U V}$$
Question 15.
Estimation The gas tank in Shondra’s car can hold 18 gallons. Her car gets about 22 miles per gallon of gas. On a recent trip, Shondra used about 12 gallons of gas. Which is the best estimate of the distance Shondra drove?
A. 120 miles
B. 200 miles
C. 400 miles
D. 1,200 miles
Stop and Practice
Find each difference. Simplify if possible.
Question 1.
Question 2.
Question 3.
Question 4.
Question 5.
Question 6.
Question 7.
Question 8.
Find each difference. Simplify if possible.
Question 9.
Question 10.
Question 11.
Question 12.
Question 13.
Question 14.
Question 15.
Question 16.
Question 17.
Question 18.
Error Search Find each answer that is not correct. Write it correctly and explain the error.
Question 19.
Question 20.
Question 21.
Question 22.
Question 23.
Number Sense
Estimating and Reasoning Write whether each statement is true or false. Explain your reasoning.
Question 24.
The quotient of 1,546 ÷ 5 is less than 300.
Question 25.
The product of 9.32 and 4.7 is less than 36.
Question 26.
The difference of 6,631 and 3,021 is greater than 2,000 and less than 4,000.
The expression (4$$\frac{5}{6}$$ + 3$$\frac{1}{2}$$ – × $$\frac{3}{4}$$) × 0 equals 0. |
# Problem 2 from the 2017 IMO
### Solution
As is common, let $y=0:\,$ $f(f(x)f(0))+f(x)=f(0).\,$ Let $f(0)=k.\,$ We have $f(kf(x))+f(x)=k.\,$ Note in passing that $f(k^2)=0.$
If $k=0,\,$ then $f(0)+f(x)=k,\,$ implying $f(x)\equiv 0.\,$ So, assume $k\ne 0.\,$ Setting $f(x)=y\,$ gives
(1)
$f(ky)+y=k,$
or $\displaystyle f(y)=k-\frac{y}{k}.\,$ The question is now to find $k.\,$ Let's substitute this $f\,$ into the original equation with $x=y:$
$\displaystyle f\left(\left[k-\frac{x}{k}\right]^2\right)+\left(k-\frac{2x}{k}\right)=k-\frac{x^2}{k}.$
To complete the substitution,
$\displaystyle k-\frac{\displaystyle \left[k-\frac{x}{k}\right]^2}{k}-\frac{2x}{k}=-\frac{x^2}{k}.$
This reduces to
$\displaystyle k-\left(k-\frac{2x}{k}+\frac{x^2}{k^3}\right)-\frac{2x}{k}=-\frac{x^2}{k}$
and, further,
(2)
$\displaystyle x^2\left(\frac{1}{k^3}-\frac{1}{k}\right)=0,\,$ for all $x\in\mathbb{R},\,$
making $k(k^2-1)=0\,$ so that $k=\pm 1.$
There are three functions: $f(x)=0,\,$ $f(x)=x-1,\,$ $f(x)=1-x.$
### Analysis
The weak point in the solution is the replacement of $f(x)\,$ with $y\,$ in the derivation of (1). Later on, we assert that (2) holds for all $x\in\mathbb{R},\,$ which is unwarranted, because (1) may be a priori claimed only for $y\,$ in the image $Im(f)\,$ of $f,\,$ i.e., for $y\,$ for which there is $x\,$ with $y=f(x).$
However, to obtain the desired conclusion from (2), it only necessary to exhibit a single $x\in Im(f)\,$ which is not $0.\,$ $k=f(0)\,$ is the first that comes to mind. But there are more. To remind, $f(k^2)=0.\,$ Setting $y=k^2\,$ in the original equation gives $f(0)+f(x+k^2)=f(k^2x),\,$ i.e., $k+f(x^2+k)=f(k^2x)\,$ from which, with $x=1,\,$ $f(k+1)=-k,\,$ which is also not zero. Thus we can rightfully claim that $f(0)=\pm 1\,$ for any solution $f\,$ of the original functional equation. That is far from the final solution of the problem but shows that the prior derivation was not entirely useless.
Regardless of the choice $k=\pm 1,\,$ $f(1)=0,\,$ for any non-trivial solution of the original equation.
### Second Attempt
Let's assume for definiteness' sake that $f(0)=1.\,$ This gives $f(f(1)f(x))+f(x+1)=f(x),\,$ i.e., $f(x+1)=f(x)-1,\,$ for all $x\in\mathbb{R}.\,$ By induction, $f(x+n)=f(x)-n$ and also $f(x-n)=f(x)+n.\,$ It would have been nice to invoke continuity and the Intermediate Value Theorem to claim $f'\text{s surjectivity}\,$ that would immediately solve the problem. But continuity is not stipulated and does not a priori follow from $f(x+1)=f(x)-1.\,$ Thus it is vital to derive the necessary properties directly from the original equation.
The essential distinction between $f(x+1)=f(x)-1\,$ and $f(f(x)f(y)) + f(x+y) = f(xy)$ is in the "second-order" term $f(f(x)f(y)).\,$ Assume it is zero. This happens, e.g., if $x+y=xy.\,$ I'd like to conclude that, for such $x,y\,$ $f(x)f(y)=1.\,$ This would be true if $f(t)=0\,$ implied $t=1.\,$ Assume this is not so and there is $x\ne 1,\,$ such that $f(x)=0.\,$ For that $x,\,$ find $y\,$ from $x+y=xy.\,$ Then, $1=f(0f(y))=0,\,$ a contradiction. Thus, there is only one solution to $f(x)=0,\,$ namely, $x=1.\,$ It follows that for every integer $n,\,$ there is only one solution to $f(x)=n\,$ which is $x=1-n.\,$
So, for the record, $x+y=xy\,$ implies $f(x)f(y)=1.\,$ Note that, incidentally, neither $x\,$ nor $y\,$ could be $1.$
We are getting the first glimpse of the possible continuity of $f:\,$ For $x=n+1,\,$ $y=\displaystyle \frac{n+1}{n}\,$ whereas $f(n+1)=-n,\,$ implying $\displaystyle f\left(\frac{n+1}{n}\right)=-\frac{1}{n}.\,$ This is reassuring, but we won't need it.
So, we found that $f(x)=1,$ only for $x=0.\,$ Assume $f(f(x)f(y))=1.\,$ There are two ways to achieve that: having one of $x,y\,$ equal $1,\,$ or finding $x,y\,$ such that $f(x+y)=f(xy)-1.\,$ Since $f(xy)-1)=f(xy+1),\,$ we may try finding $x,y\,$ from $x+y=xy+1.\,$ Naturally, the two ways are equivalent: one of the numbers $x,y\,$ that satisfy $x+y=xy+1,\,$ is necessarily $1.\,$ Strangely, this observation proves to be useful.
We'll show that any solution to the given problem is injective, or $1-1:\,$ $f(a)=f(b)\,$ implies $a=b.\,$ As we saw earlier, this is true for integer values. Now it is the turn of a more general statement. Given such $a\,$ and $b\,$ we may want to find $x\,$ and $y\,$ such that $a=x+y\,$ and $b=xy+1.\,$ If we succeed, then one of $x,y\,$ will be $1,\,$ implying, say, $a=x+1=x\cdot 1+1=b.$
How are we going to look for the pair $x,y?\,$ From $x+y=a\,$ and $xy=b-1,\,$ the pair should be the roots of the quadratic equation, $g(u)=u^2-au+(b-1)=0.\,$ For the roots to be real we need a non-negative discriminant: $D=a^2-4(b-1)\ge 0\,$ which may be not true from the outset. However, if $f(a)=f(b),\,$ then also $f(a+n)=f(b+n),\,$ for any $n\in\mathbb{Z}.\,$ This will convert the discriminant to $D(n)=(a+n)^2-4(b+n-1),\,$ which is assured of being positive for large $n.\,$ So, to sum up, for $f(a)=f(b),\,$ we can always find real $x,y,n\,$ so that $a+n=x+y\,$ and $b+n=xy+1\,$ and, since $f(a+n)=f(b+n),\,$ one of $x,y$ is $1,\,$ implying $a+n=b+n\,$ and subsequently, $a=b.\,$
The injectivity does not directly lead to the desired surjectivity, but it allows to use the "second-order" term, while getting rid of it when necessary.
Let's set in the original equation $y=0.\,$ Since $f(0)=1,\,$ we get $f(f(x))+f(x)=1.\,$ We continue, substituting $f(x)\,$ for $x\,$ and $y=0:\,$ $f(f(f(x)))+f(f(x))=1,\,$ or
\begin{align} f(f(x))+f(x)&=1\\ 1&=f(f(f(x)))+f(f(x)). \end{align}
Adding which gives $f(f(f(x)))=f(x)\,$ and, subsequently, from the already proved injectivity, $f(f(x))=x.\,$ Since $x\,$ is arbitrary, this proves the surjectivity of $f.\,$ Now, the initial argument applies to show that $f(x)=1-x.\,$
### Acknowledgment
This is Problem 2 from the 2017 IMO. The problem has been posted by N. N. Taleb at twitter.com.
Sam Walters has discovered a weak point in the above argument, viz., (1) is only true for $y\in Im(f).\,$ He also offered an example of a function different from the three found above that satisfies (1) for $k=1:$
$\displaystyle f(x)=\begin{cases} 1-x, & x\in\mathbb{Q},\\ \frac{1}{2}, & x\in\mathbb{R}\setminus\mathbb{Q}. \end{cases}$ |
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# March 6 Homework Solutions
## Chapter 6 Problems (pages 287-291)
Problem 31
According to the U.S. National Center for Health Statistics, 25.2 percent of males and
23.6 percent of females never eat breakfast. Suppose that random samples of 200 men and
200 women are chosen. Approximate the probability that
## (a) at least 110 of these 400 people never eat breakfast.
Let M denote the number of men that never eat breakfast and W denote the number
of women that never eat breakfast. Note that M is a binomial random variable with
p = .252 and n = 200, and W is a binomial random variable with p = .236 and
n = 200. We compute
## E[M ] = 200(.252) = 50.4, Var(M ) = 200(.252)(1 − .252) ≈ 37.7,
E[W ] = 200(.236) = 47.2, Var(M ) = 200(.236)(1 − .236) ≈ 36.1.
## Thus we may approximate M by a normal random variable with µ = 50.4 and
σ 2 ≈ 37.7 and we may approximate W by a normal random variable with µ = 47.2
and σ 2 ≈ 36.1.
We want to compute P {110 ≤ M + W }. We may approximate M + W by a normal
distribution with µ = 50.4 + 47.2 = 97.6 and σ 2 ≈ 37.7 + 36.1 = 73.8. Hence
110 − 97.6 M + W − 97.6 110 − 97.6
P {110 ≤ M + W } = P √ ≤ √ ≈1−Φ √
73.8 73.8 73.8
≈ 1 − Φ(1.44) = 1 − .9251 = .0749.
(b) the number of the women who never eat breakfast is at least as large as the number
of the men who never eat breakfast.
We want to compute P {M ≤ W } = P {M − W ≤ 0}. We may approximate M − W
by a normal distribution with µ = 50.4 − 47.2 = 3.2 and σ 2 = 37.7 + 36.1 = 73.8.
Hence
M − W − 3.2 −3.2
P {M ≤ W } = P {M − W ≤ 0} = P √ ≥√
73.8 73.8
−3.2
≈ Φ √ ≈ 1 − Φ(.37) = 1 − .6443 = .3557.
73.8
Problem 44
If X1 , X2 , X3 are independent random variables that are uniformly distributed over (0, 1),
compute the probability that the largest of the three is greater than the sum of the other
two.
1
Note that if, for example, X1 ≥ X2 + X3 , then X1 is automatically the largest of the
three. We want to compute
## P {X1 ≥ X2 + X3 } + P {X2 ≥ X1 + X3 } + P {X3 ≥ X1 + X2 }.
By symmetry, these three terms are all equal, so it suffices to compute the first term
P {X1 ≥ X2 + X3 }. Recall from Example 3a on page 252 that
y if 0 ≤ y ≤ 1,
fX2 +X3 (y) = 2−y if 1 < y ≤ 2,
0 otherwise.
Hence
Z 1 Z x Z 1 Z x Z 1
1 2 1
P {X1 ≥ X2 + X3 } = fX1 (x)fX2 +X3 (y)dydx = 1 · y dydx = x dx = .
0 0 0 0 0 2 6
So the probability that the largest of the three is greater than the sum of the other
two is
1 1 1 1
P {X1 ≥ X2 + X3 } + P {X2 ≥ X1 + X3 } + P {X3 ≥ X1 + X2 } = + + = .
6 6 6 2
## Chapter 6 Theoretical Exercises (pages 291-293)
Problem 19
Let X1 , X2 , X3 be independent and identically distributed continuous random variables.
## (a) Compute P {X1 > X2 |X1 > X3 }.
We consider 7 mutually exclusive cases: the first six are X1 > X2 > X3 , X1 >
X3 > X2 , X2 > X1 > X3 , X2 > X3 > X1 , X3 > X1 > X2 , X3 > X2 > X1 , and
the seventh case is where at least two of X1 , X2 , X3 are equal. Since the Xi are
continuous random variables, the probability that at least two of the Xi are equal
is zero. Since X1 , X2 , X3 , are independent and identically distributed, the first six
cases are all equally likely. This means that P (X1 > X2 > X3 ) = 1/6, and similarly
for the other orderings. Hence
## P {X1 > X2 > X3 } + P {X1 > X3 > X2 } 2
P {X1 > X2 |X1 > X3 } = = .
P {X1 > X2 > X3 } + P {X1 > X3 > X2 } + P {X2 > X1 > X3 } 3
## P {X3 > X1 > X2 } 1
P {X1 > X2 |X1 < X3 } = = .
P {X2 > X3 > X1 } + P {X3 > X1 > X2 } + P {X3 > X2 > X1 } 3
2
(c) Compute P {X1 > X2 |X2 > X3 }.
## P {X1 > X2 > X3 } 1
P {X1 > X2 |X1 > X3 } = = .
P {X1 > X2 > X3 } + P {X2 > X1 > X3 } + P {X1 > X3 > X2 } 3
## (d) Compute P {X1 > X2 |X2 < X3 }.
P {X1 > X3 > X2 } + P {X3 > X1 > X2 } 2
P {X1 > X2 |X1 > X3 } = = .
P {X1 > X3 > X2 } + P {X3 > X1 > X2 } + P {X3 > X2 > X1 } 3
Problem 28
Show that the median of a sample of size 2n + 1 from a uniform distribution on (0, 1) has
a beta distribution with parameters (n + 1, n + 1).
Let X denote the median of the independent and identically distributed random
variables X1 , . . . , X2n+1 . Consider equation 6.2 on page 272. By replacing n with 2n + 1
and by choosing j = n + 1, we get that the probability density function of X is
(2n + 1)!
fX (x) = (F (x))n (1 − F (x))n f (x),
n!n!
where f is the common probability density function and F is the common cumulative
distribution function of the Xi ’s. Since X1 , . . . , X2n+1 are uniformly distributed on (0, 1),
this means that f (x) = 1 for 0 ≤ x ≤ 1 and F (x) = x for 0 ≤ x ≤ 1. Hence
(
(2n+1)! n
n!n!
x (1 − x)n for 0 ≤ x ≤ 1
fX (x) =
0 otherwise.
Now, consider the Beta distribution on page 218. Note that
Z 1
B(n + 1, n + 1) = xn (1 − x)n dx
0
Z 1
n
= xn+1 (1 − x)n−1 dx after integrating by parts
n+1 0
= ...
Z 1
n!
= x2n dx after integrating by parts
(2n) · . . . · (n + 1) 0
n!
=
(2n + 1)(2n) · . . . · (n + 1)
n!n!
= .
(2n + 1)!
Hence
(
(2n+1)! n
n!n!
x (1 − x)n for 0 ≤ x ≤ 1
fX (x) =
0 otherwise
(
1
B(n+1,n+1)
xn (1 − x)n for 0 ≤ x ≤ 1
=
0 otherwise.
3
So X has a Beta distribution with parameters (n + 1, n + 1).
## Chapter 7 Problems (pages 373-379)
Problem 1
A player throws a fair die and simultaneously flips a fair coin. If the coin lands heads,
then she wins twice, and if tails, then one-half of the value that appears on the die. De-
termine her expected winnings.
Let X denote the value on the die, let Y be 1 if the coin lands heads and 0 if the coin
lands tails, and let g(X, Y ) denote the winnings. Then her expected winnings are
6 X 1 6
X X 1
E[winnings] = g(x, y)p(x, y) = 2x · p(x, 0) + x · p(x, 1)
x=1 y=0 x=1
2
6 6
X 1 1 1 5 X 5
= 2x · + x· = x= · 21 = 4.375.
x=1
12 2 12 24 x=1 24
Problem 12
A group of n men and n women is lined up at random.
(a) Find the expected number of men who have a woman next to them.
Label the people in order 1 though 2n and let Xi = 1 if the i-th person is a man
standing next to a woman and Xi = 0 otherwise. We want to compute
" 2n # 2n 2n
X X X
E Xi = E[Xi ] = P {Xi = 1}.
i=1 i=1 i=1
Note X1 = 1 only if the first person is male and the second is female. There are
2n(2n − 1) ways to choose the first two people. There are n ways to choose the first
person to be male and n ways to choose the second person to be female, and hence
n2 ways we can have Xi = 1. Hence
n2 n
P {X1 = 1} = = .
2n(2n − 1) 4n − 2
n
Similarly P {X2n = 1} = 4n−2
.
Now let’s find P {Xi = 1} for 1 < i < 2n. There are 2n(2n − 1)(2n − 2) ways to
choose the (i − 1)-th, i-th, and (i + 1)-th person. We have Xi = 1 if the three people
chosen are female male female, female male male, or male male female. Hence there
are
(n)(n)(n − 1) + (n)(n)(n − 1) + (n)(n − 1)(n) = 3n2 (n − 1)
ways we can have Xi = 1. Hence
3n2 (n − 1) 3n
P {Xi = 1} = = .
2n(2n − 1)(2n − 2) 8n − 4
4
Hence
" 2n
# 2n
X X n 3n 3n2 − n
E Xi = P {Xi = 1} = 2 · + (2n − 2) · = .
i=1 i=1
4n − 2 8n − 4 4n − 2
(b) Repeat part (a), but now assuming that the group is randomly seated at a round
table.
Label the people in order 1 though 2n and let Xi = 1 if the i-th person is a man
standing next to a woman and Xi = 0 otherwise. We want to compute
" 2n # 2n 2n
X X X
E Xi = E[Xi ] = P {Xi = 1}.
i=1 i=1 i=1
## Since the group is now seated in a circle, for 1 ≤ i ≤ 2n we have
3n
P {Xi = 1} = .
8n − 4
This computation is the same as the one that we used in part (a) for 1 < i < 2n.
Hence " 2n # 2n
X X 3n 3n2
E Xi = P {Xi = 1} = 2n · = .
i=1 i=1
8n − 4 4n − 2
Problem 19
A certain region is inhabited by r distinct types of a certain species of insect. Each insect
caught will, independently of the types of the previous catches, be of type i with probability
r
X
Pi , i = 1, . . . , r, Pi = 1.
i=1
(a) Compute the mean number of insects that are caught before the first type 1 catch.
Let X denote the number of insects caught before the first type 1 catch. Then
P {X = x} = (1 − P1 )x P1 . Hence
X
E[X] = x(1 − P1 )x P1
x=0
X
= P1 x(1 − P1 )x
x=0
1 − P1
= P1 ·
P12
X z
using the formula nz n = for |z| < 1
n=1
(1 − z)2
1 − P1
= .
P1
5
(b) Compute the mean number of types of insects that are caught before the first type 1
catch.
Let Xi denote the number of insects of type i caught before the first type 1 catch.
Let g(0) = 0 and let g(x) = 1 for positive integers x > 0. We want to compute
" r # r r r
X X X X
E g(Xi ) = E[g(Xi )] = P {g(Xi ) = 1} = P {Xi ≥ 1}.
i=2 i=2 i=2 i=2
Let X denote the number of insects caught before the first type 1 catch as in part
(a). Then
X
P {Xi ≥ 1} = P {Xi ≥ 1, X = x}
x=0
since the events{Xi ≥ 1, X = x} are mutually
exclusive and have union{Xi ≥ 1}
X
= (P {X = x} − P {Xi = 0, X = x})
x=0
X∞
= ((1 − P1 )x P1 − (1 − P1 − Pi )x P1 )
x=0
P1 P1
= −
P1 P1 + Pi
using the formula for a geometric series, twice
Pi
= .
P1 + Pi
Hence " #
r r r
X X X Pi
E g(Xi ) = P {Xi ≥ 1} = .
i=2 i=2 i=2
P 1 + Pi
Problem 24
A bottle initially contains m large pills and n small pills. Each day, a patient randomly
chooses one of the pills. If a small pill is chosen, then that pill is eaten. If a large pill
is chosen, then the pill is broken in two; one part is returned to the bottle (and is now
considered a small pill) and the other part is then eaten.
(a) Let X denote the number of small pills in the bottle after the last large pill has been
chosen and its smaller half returned. Find E[X].
Label the small pills initially present 1 though n and label the small pills created
by splitting a large one n + 1 though n + m. Let Ii = 1 if the i-thPpill remains after
the last large pill is chosen and let Ii = 0 otherwise. Then X = m+n i=1 Ii , so
n+m
X n+m
X
E[X] = E[Ii ] = P {Ii = 1}.
i=1 i=1
6
We will calculate P {Ii = 1} by considering the two cases 1 ≤ i ≤ n and n + 1 ≤
i ≤ n + m separately.
If 1 ≤ i ≤ n, then the i-th small pill is initially present. Pretend we keep choosing
pills until all of them are gone. It suffices to consider the order in which the i-th pill
and the m large pills are chosen. There are m + 1 of these pills, so the probability
that the i-th pill is chosen last among them is P {Ii = 1} = 1/(m + 1).
If n + 1 ≤ i ≤ n + m, then the i-th small pill is formed by breaking a large pill
in two. It suffices to consider the order in which the large pills and the i-th small
pill are chosen. Label the large pills 1 through m in the order in which they are
initially chosen. Let Ji denote this label for the large pill corresponding to the i-th
pill, that is, let Ji denote when the large pill is broken forming the i-th small pill.
By conditioning on the value of Ji , we get
m
X
P {Ii = 1} = P ({Ii = 1}|{Ji = j})P {Ji = j}.
j=1
The probability that the large pill corresponding to the i-th small pill is labeled j
out of the m large pills is P {Ji = j} = 1/m. Once the j-th large pill is broken
to form the i-th small pill, m − j large pills and the i-th small pill remain, so the
probability that the i-th small pill is chosen last among these m − j + 1 pills is
1
P ({Ii = 1}|{Ji = j}) = .
m−j+1
Hence
m
X
P {Ii = 1} = P ({Ii = 1}|{Ji = j})P {Ji = j}
j=1
m
X 1 1
= ·
j=1
m−j+1 m
m
X 1
= ,
k=1
km
by letting k = m − j + 1. Therefore
n+m
X
E[X] = P {Ii = 1}
i=1
n
X n+m
X
= P {Ii = 1} + P {Ii = 1}
i=1 i=n+1
m
1 X 1
=n· +m·
m+1 k=1
km
m
n X1
= + .
m + 1 k=1 k
7
(b) Let Y denote the day on which the last large pill is chosen. Find E[Y ].
There are a total of n + 2m days. On the Y -th day the last large pill is chosen and
for the remaining X days small pills are chosen. Here X is the number of small
pills in the bottle after the last large pill has been choosn, as in part (a). Thus
X + Y = n + 2m, so
m
n X1
E[Y ] = E[n + 2m − X] = n + 2m − E[X] = n + 2m − − .
m + 1 k=1 k
Problem 26
If X1 , X2 , . . . , Xn are independent and identically distributed random variables having
uniform distributions over (0, 1), find
## (a) E[max(X1 , . . . , Xn )].
We want to compute
Z 1 Z 1
E[max(X1 , . . . , Xn )] = ... max(x1 , . . . , xn )dx1 . . . dxn .
0 0
## We are integrating over the region where 0 ≤ x1 , x2 , . . . , xn ≤ 1. We can break up
this region into the n! regions corresponding to each of the n! possible orderings
of x1 , . . . , xn , plus a negligible region of zero volume where at least two of the xi ’s
are equal. Since the X1 , . . . , Xn are independent and identically distributed, the
integrals over each of the n! different regions will have the same value. Hence we’ll
only consider the region where x1 < x2 < . . . < xn . We have
8
Z 1 Z 1
E[max(X1 , . . . , Xn )] = ... max(x1 , . . . , xn )dx1 . . . dxn
0Z 0
## = n! max(x1 , . . . , xn )dx1 . . . dxn
x1 <...<xn
since the integral is the same over all n! such regions
Z
= n! xn dx1 . . . dxn
x1 <...<xn
since the max in this region is xn
Z 1 Z xn Z xn−1 Z x2
= n! ... xn dx1 . . . dxn
0 0 0 0
Z 1 Z xn Z xn−1 Z x3
= n! ... xn x2 dx2 . . . dxn
0 0 0 0
Z 1 Z xn Z xn−1 Z x4
1
= n! ... xn x23 dx3 . . . dxn
0 0 0 0 2
= ...
Z 1
1
= n! xn · xn−1
n dxn
0 (n − 1)!
Z 1
=n xnn dxn
0
n
= .
n+1
## (b) E[min(X1 , . . . , Xn )].
The clever way to do this problem is to note that
min(X1 , . . . , Xn ) = 1 − max(1 − X1 , . . . , 1 − Xn ).
## So taking the expectation of both sides yields
E[min(X1 , . . . , Xn )] = 1 − E[max(1 − X1 , . . . , 1 − Xn )]
n
=1−
n+1
by part (a), since each Xi has
the same distribution as 1 − Xi
1
= .
n+1
## Alternatively, like in part (a), we can express E[min(X1 , . . . , Xn )] as a definite
integral and then break this integral up into a sum of n! integrals with equal values,
corresponding to each of the n! orderings of x1 , . . . , xn . We obtain
9
Z 1 Z 1
E[min(X1 , . . . , Xn )] = ... min(x1 , . . . , xn )dx1 . . . dxn
0Z 0
## = n! min(x1 , . . . , xn )dx1 . . . dxn
x1 <...<xn
since the integral is the same over all n! such regions
Z
= n! x1 dx1 . . . dxn
x1 <...<xn
since the min in this region is x1
Z 1 Z xn Z xn−1 Z x2
= n! ... x1 dx1 . . . dxn
0 0 0 0
Z 1 Z xn Z xn−1 Z x3
1
= n! ... x2 dx2 . . . dxn
0 0 0 0 2
= ...
Z 1
1 n
= n! xn dxn
0 n!
Z 1
= xnn dxn
0
1
= .
n+1
## Chapter 7 Theoretical Exercises (pages 380-384)
Problem 14
For Example 2i, show that the variance of the number of coupons needed to amass a full
set is equal to
N −1
X iN
i=1
(N − i)2
When N is large, this can be shown to be approximately equal (in the sense that their
ratio approaches 1 as N → ∞) to N 2 π 2 /6.
## Recall from Example 2i on page 303 that we can write X = X0 + X1 + . . . + XN −1 .
Since the Xi are independent, we have
N
X −1
Var(X) = Var(Xi ).
i=1
Recall k−1
N −i i
P {Xi = k} = for k ≥ 1,
N N
and E[Xi ] = N/(N − i). We compute
∞ k−1
2 N −iX 2 i N −i 1 + i/N N (N + i)
E[Xi ] = k = · 3
=
N k=1 N N (1 − i/N ) (N − i)2
10
where the second equality follows since
X 1+x
k 2 xk−1 = for |x| < 1.
k=1
(1 − x)3
Hence
N (N + i) N2 iN
Var(Xi ) = E[Xi2 ] − E[Xi ]2 = 2
− 2
= ,
(N − i) (N − i) (N − i)2
and
N −1 N −1
X X iN
Var(X) = Var(Xi ) = .
i=1 i=1
(N − i)2
11 |
Area and Perimeter
1 / 26
# Area and Perimeter - Primary Resources - PowerPoint PPT Presentation
Area and Perimeter. The distance around the outside of a shape is called the perimeter. First we need to find he length of each side by counting the squares. 8 cm. 6 cm. 6 cm. 8 cm. The perimeter of the shape is 8 + 6 + 8 + 6 = 28cm. 18cm. 18cm.
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Presentation Transcript
First we need to find he length of each side by
counting the squares.
8 cm
6 cm
6 cm
8 cm
The perimeter of the shape is 8 + 6 + 8 + 6 = 28cm.
18cm
18cm
What is the perimeter of each of these shapes? Your teacher will give you copy of the worksheet.
22cm
18cm
26cm
4cm
4cm
5cm
3cm
3cm
3cm
3cm
3cm
7cm
7cm
3cm
3cm
3cm
3cm
4cm
4cm
3cm
3cm
Now find the perimeter of these shapes. They are not drawn to scale.
Perimeter = 18cm
Perimeter = 22cm
Perimeter = 28cm
The area of a shape is the amount of space inside it.
To find out how much shape is inside we can
count the squares.
The area of the shape is 18cm2.
12cm2
What is the area of each of these shapes? Your teacher will give you copy of the worksheet.
26cm2
26cm2
To find the area of
an irregular shape you first count the full squares inside it.
This is an irregular shape. This means that the sides aren’t straight.
You then count all the other squares which have more than half the area covered.
The area of the shape is 60cm2.
24cm2
25cm2
What is the area of each of these shapes? Your teacher will give you copy of the worksheet.
35cm2
22cm2
The rectangle is made of 6 rows of 8 squares.
To work out the number of squares
we times 6 by 8.
The area of the shape is 48cm2.
18cm2
20cm2
What is the area of each of these shapes? Your teacher will give you copy of the worksheet.
30cm2
14cm2
22cm2
We can work out the area of a rectangle
without the grid.
length
width
Area = length x width
Now work out the area of these rectangles.
They are not drawn to scale.
7cm
5cm
35cm2
4cm
5cm
20cm2
9cm
36cm2
12.5cm
4cm
25cm2
2cm
4cm
Total area = 40 + 32 = 72 cm2
2cm
To find the area of this shape we have to split it up into two rectangles.
4cm
Area =
4 x 10
10cm
Area =
4 x 8
8cm
40cm2
32cm2
8cm
A shape that is made from other shapes is known as a composite shape.
2cm
5cm
11cm
4cm
6cm
6cm
Find the area and perimeter of each of these shapes? Your teacher will give you copy of the worksheet.
5cm
2cm
4cm
9cm
5cm
4cm
2cm
5cm
4cm
3cm
3cm
3cm
4cm
7cm
7cm
11cm
5cm
2cm
Area =
4 x 5
Area =
5 x 9
4cm
9cm
5cm
20cm2
45cm2
4cm
2cm
5cm
Total area = 45 + 20 = 65 cm2
Perimeter = 36 cm
2cm
5cm
11cm
4cm
6cm
6cm
Total area = 10 + 36 = 46 cm2
Perimeter = 34 cm
Area =
2 x 5
Area =
6 x 6
10cm2
36cm2
Total area = 28 + 16 + 21 = 65 cm2
Perimeter = 42 cm
There is another way of doing this question.
Area =
4 x 7
Area =
4 x 4
28cm2
16cm2
4cm
3cm
3cm
3cm
Area =
3 x 7
4cm
7cm
21cm2
7cm
11cm
Total area = 77 – 12 = 65 cm2
Perimeter = 42 cm
Area =
3 x 4
12cm2
4cm
3cm
3cm
3cm
4cm
7cm
7cm
Area =
7 x 11
77cm2
11cm
w
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r
k
s
h
e
e
t
o
n
e
w
o
r
k
s
h
e
e
t
t
w
o
w
o
r
k
s
h
e
e
t
t
h
r
e
e
5cm
2cm
5cm
4cm
11cm
6cm
6cm
2cm
w
o
r
k
s
h
e
e
t
f
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4cm
9cm
5cm
4cm
2cm
5cm
4cm
3cm
3cm
3cm
7cm
7cm
4cm
11cm
Name:____________________
Review
What is the area of the moon?
_____________________cm2
What is the area of the rectangle?
_____________________cm2
What is the perimeter of the rectangle?
_____________________cm
What is the area of the house?
_____________________cm2
4cm
What is the area of the compound shape?
_____________________cm2
4cm
2cm
8cm
What is the perimeter of the compound shape?
_____________________cm
4cm
6cm |
# Addition and Subtraction with Ten Frames
These ten frame activities help kids learn addition and subtraction up to 20. They’re a fun addition to kindergarten and first grade math centers.
Ten frames are one of my favorite teaching tools for math. I use them all the time! There are a few ways you can use them:
Counting – Using ten frames for counting can help kids to develop one-to-correspondence. It also helps them to develop an understanding of a “ten”.
Place Value – I often put the ten frames in a vertical orientation to teach kids about teen numbers and place value. One ten frame represents the tens and the other ten frame represents the ones.
Addition and Subtraction – Ten frames are also great for teaching kids how to add and subtract. You can check out some examples below.
I am so excited about these new addition and subtraction activities because I LOVE unicorns!
Preparation – You’ll need some counters in two colors, sum cards and the work mat. You can choose from addition up to 10 or addition up to 20.
How to Play – First the kids need to choose a sum card and place it on the mat. They need to show the first number of the sum with one color and the second number with another color. After that the kids can record the sum on the recording sheet or a mini whiteboard.
Tip – Two sided counters are great for this activity (the red and yellow ones). But any counters work fine as long as you have two colors. I opted to use pom poms, which are also great for helping kids to develop their fine motor skills.
There are also a variety of worksheets which help kids add with ten frames. The one shown above is really fun to do with Q-tips. Kids will need two colors to show the numbers in each addition sum.
## Subtraction Activities
Preparation
– You’ll need counters or playdough, sum cards and the work mat. You can choose from subtraction up to 10 or subtraction up to 20.
How to Play (playdough version)- Kids need to read the first number in the sum and then make that number of playdough balls. Place the balls on the ten frame. After that the kids need to read the second number in the sum, which tells them how many to “take away”. Now for the fun part… the kids get to squash that amount to represent the taking away. The number of balls left over tells them the answer. The kids then need to record the answer on the recording worksheet or a mini whiteboard. Then it is time to play again!
There are a variety of subtraction worksheets too, which help kids practice subtracting with ten frames.
## Free Teacher Ten Frames
Preparation – You’ll need the ten frames printed on white paper and the counters printed on two different colors. You may also need magnetic tape and Velcro (hook and loop).
How to Make- After you’ve laminated the frames and counters, cut them out. For mine, I attached some magnetic tape to the back of the ten frame so it can stick to the whiteboard. Then I added some Velcro to each square of the frame and the backs of the counters. If you have circle shaped magnets, they can can be used instead of the printable counters. |
Email: rakesh@xamnation.com Phone/Whatsapp: +91 9988708161
# Grade 7 Maths Ncert solutions for Symmetry
NCERT books are recommended text book for CBSE school curriculum. Most of the govt aided and private schools in India follow NCERT text books as the syllabus books for school subjects. They are written by eminent academicians and scholars from leading universities and academic institutions in India, and help in building a strong base, as well help understand each concept by relevant examples and exercises. They are termed best in term of concept building, for their usage of real life practical examples, and good end of the chapter exercises.
In this article, we have shared solved NCERT exercises for Class 7 maths symmetry chapter. These NCERT solutions will help students in solving their doubts and clarifying confusion regarding confusions regarding this chapter. Students can refer to NCERT solutions for symmetry chapter for their conceptual understanding, as well as matching their own answers with the right solutions of the exercises.
## Class 7 math ncert solution symmetry chapter
Exercise 14.1 solutions
Q1 – Copy the figures with punched holes and find the axes of symmetry for the following: View solution
Q2 – Given the line(s) of symmetry, find the other hole(s): View solution
Q3 – In the following figures, the mirror line (i.e., the line of symmetry) is given as a dotted line. Complete each figure performing reflection in the dotted (mirror) line. (You might perhaps place a mirror along the dotted line and look into the mirror for the image). Are you able to recall the name of the figure you complete? View solution
Q4 – The following figures have more than one line of symmetry. Such figures are said to have multiple lines of symmetry. Identify multiple lines of symmetry, if any, in each of the following figures: View solution
Q5 – Copy the figure given here. Take any one diagonal as a line of symmetry and shade a few more squares to make the figure symmetric about a diagonal. Is there more than one way to do that? Will the figure be symmetric about both the diagonals? View solution
Q6 – Copy the diagram and complete each shape to be symmetric about the mirror line(s): View solution
Q7 – State the number of lines of symmetry for the following figures: (a) An equilateral triangle (b) An isosceles triangle (c) A scalene triangle (d) A square (e) A rectangle (f) A rhombus (g) A parallelogram (h) A quadrilateral (i) A regular hexagon (j) A circle View solution
Q8 – What letters of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about. (a) a vertical mirror (b) a horizontal mirror (c) both horizontal and vertical mirrors View solution
Q9 – Give three examples of shapes with no line of symmetry. View solution
Q10 – What other name can you give to the line of symmetry of (a) an isosceles triangle? (b) a circle? View solution
Exercise 14.2 solutions
Q1 – Which of the following figures have rotational symmetry of order more than 1: View solution
Q2 – Give the order of rotational symmetry for each figure: View solution
Exercise 14.3 solutions
Q1 – Name any two figures that have both line symmetry and rotational symmetry. View solution
Q2 – Draw, wherever possible, a rough sketch of (i) a triangle with both line and rotational symmetries of order more than 1. (ii) a triangle with only line symmetry and no rotational symmetry of order more than 1. (iii) a quadrilateral with a rotational symmetry of order more than 1 but not a line symmetry. (iv) a quadrilateral with line symmetry but not a rotational symmetry of order more than 1. View solution
Q3 – If a figure has two or more lines of symmetry, should it have rotational symmetry of order more than 1? View solution
Q4 – Fill in the blanks View solution
Q5 – Name the quadrilaterals which have both line and rotational symmetry of order more than 1. View solution
Q6 – After rotating by 60° about a centre, a figure looks exactly the same as its original position. At what other angles will this happen for the figure? View solution
Q7 – Can we have a rotational symmetry of order more than 1 whose angle of rotation is (i) 45°? (ii) 17°? View solution
## End Note
In the article, we have shared solutions for ncert exercises for class 7 maths for symmetry chapter. Students can click to respective chapters and refer to the right solution. We advise students to refer to them as clearing their concepts, and matching their solutions. Student should not use them to copy solutions for their homework and instead should try to solve homework on own, as it will impact their learning process.
In case you have any doubts regarding solution, or need further clarification, you can write to us. You can also reach out to us, in case you have doubt in other text books’ problems or school exam papers. We will be glad to assist you in helping with your doubt. |
Simultaneous Equations (1)
As many users have not been exposed to complex numbers, I will post the solutions to the problems assuming that all variables used are real. However, please feel free to solve these problems over the complex numbers and post your solutions!
Problem 1. Solve
\left\{\begin{aligned}y^2&=x^3-3x^2+2x&\quad(1)\\x^2&=y^3-3y^2+2y&\quad(2)\end{aligned}\right.
Solution. $(1)-(2)$:
$-(x^2-y^2)=(x^3-y^3)-(3x^2-3y^2)+(2x-2y)$
$-(x+y)(x-y)=(x-y)(x^2+xy+y^2)-(3x+3y)(x-y)+2(x-y)$
$(x-y)(x^2+xy+y^2-3x-3y+2+x+y)=0$
$(x-y)[x^2+(y-2)x+(y^2-2y+2)]=0$
Case 1. $x-y=0\implies x=y$. Substituting $x=y$ into $(2)$, we have
$y^2=y^3-3y^2+2y$
$y(y^2-4y+2)=0$
Case 1a. $y=0$. Since $x=y$, $\boxed{x=y=0}$.
Case 1b. $y^2-4y+2=0$.
$y = \frac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$
$=2\pm\sqrt{2}$
Since $x=y$, $\boxed{x=y=2\pm\sqrt{2}}$.
Case 2. $x^2+(y-2)x+(y^2-2y+2)]=0$.
$\text{Discriminant}\,= (y-2)^2-4(1)(y^2-2y+2)$
$=y^2-4y+4-4y^2+8y-8$
$=-3y^2+4y-4$
$=-3\left(y^2-\frac{4}{3}y\right)-4$
$=-3\left[y^2-\frac{4}{3}y+\left(\frac{2}{3}\right)^2-\left(\frac{2}{3}\right)^2\right]-4$
$=-3\left[\left(y-\frac{2}{3}\right)^2-\frac{4}{9}\right]-4$
$=-3\left(y-\frac{2}{3}\right)^2+\frac{4}{3}-4$
$=-3\left(y-\frac{2}{3}\right)^2-\frac{8}{3}<0$
Since we are solving this problem over the reals, Case 2 is rejected.
Hence, $\boxed{x=y=0}$ and $\boxed{x=y=2\pm\sqrt{2}}$.
Problem 2. Solve
\left\{\begin{aligned}5x^2+4y^2-10x+5y&=0&\quad(3)\\4x^2+5y^2+5x-10y&=0&\quad(4)\end{aligned}\right.
Problem 3. Solve
\left\{\begin{aligned}x+y-z&=4&\quad(5)\\x^2+y^2-z^2&=12&\quad(6)\\x^3+y^3-z^3&=34&\quad(7)\end{aligned}\right.
Problem 4. Solve
\left\{\begin{aligned}xy+yz+zx&=1&\quad(8)\\yz+zt+ty&=1&\quad(9)\\zt+tx+xz&=1&\quad(10)\\tx+xy+yt&=1&\quad(11)\end{aligned}\right.
Problem 5. Solve
\left\{\begin{aligned}(x+y)(x+z)&=x&\quad(12)\y+z)(y+x)&=2y&\quad(13)\\(z+x)(z+y)&=3z&\quad(14)\end{aligned}\right. Note by Victor Loh 6 years, 7 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in \( ... or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
Problem 1. Solve
\left\{\begin{aligned}y^2&=x^3-3x^2+2x&\quad(1)\\x^2&=y^3-3y^2+2y&\quad(2)\end{aligned}\right.
Solution. $(1)-(2)$:
$-(x^2-y^2)=(x^3-y^3)-(3x^2-3y^2)+(2x-2y)$
$-(x+y)(x-y)=(x-y)(x^2+xy+y^2)-(3x+3y)(x-y)+2(x-y)$
$(x-y)(x^2+xy+y^2-3x-3y+2+x+y)=0$
$(x-y)[x^2+(y-2)x+(y^2-2y+2)]=0$
Case 1. $x-y=0\implies x=y$. Substituting $x=y$ into $(2)$, we have
$y^2=y^3-3y^2+2y$
$y(y^2-4y+2)=0$
Case 1a. $y=0$. Since $x=y$, $\boxed{x=y=0}$.
Case 1b. $y^2-4y+2=0$.
$y = \frac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$
$=2\pm\sqrt{2}$
Since $x=y$, $\boxed{x=y=2\pm\sqrt{2}}$.
Case 2. $x^2+(y-2)x+(y^2-2y+2)]=0$.
$\text{Discriminant}\,= (y-2)^2-4(1)(y^2-2y+2)$
$=y^2-4y+4-4y^2+8y-8$
$=-3y^2+4y-4$
$=-3\left(y^2-\frac{4}{3}y\right)-4$
$=-3\left[y^2-\frac{4}{3}y+\left(\frac{2}{3}\right)^2-\left(\frac{2}{3}\right)^2\right]-4$
$=-3\left[\left(y-\frac{2}{3}\right)^2-\frac{4}{9}\right]-4$
$=-3\left(y-\frac{2}{3}\right)^2+\frac{4}{3}-4$
$=-3\left(y-\frac{2}{3}\right)^2-\frac{8}{3}<0$
Since we are solving this problem over the reals, Case 2 is rejected.
Hence, $\boxed{x=y=0}$ and $\boxed{x=y=2\pm\sqrt{2}}$.
- 6 years, 7 months ago
When solving system of equations of the form $x = f(y) , y = f(x)$, it is often very tempting to conclude that the only solution occurs when $x = y$.
What we actually have, is $f(f(x)) = x$, and so we are interested in the cases where $f(x) = f^{-1} (x)$. If you can accurately plot this graph, you can then find all such solutions to this equation.
Otherwise, take the approach as Victor did, where you subtract the 2 equations and factor out $(x - y)$. If you believe that the only solutions occur when $x = y$, then you just need to show that the dividend has no (real) roots.
Staff - 6 years, 7 months ago
Do you think I should continue to post the solutions as comments? Depending on the number of upvotes of each solution, the solutions will not be arranged in chronological order.
- 6 years, 7 months ago
I guess I'll post the solution of each problem below the problem itself.
- 6 years, 7 months ago
I do not think that the solutions need to be displayed in order.
I think that posting the solution below the problem makes it disruptive to seeing the entire list. I would prefer if you simply posted the solution as comments, or at the end of the note.
Staff - 6 years, 7 months ago
This is a nice set of problems to work on. The tricky part is justifying that one has indeed found all of the solutions. Try solving these problems over the reals and over the complex numbers, and see how your solution differs. The first problem is a great example!
Staff - 6 years, 7 months ago
Subtracting the equations in $1$,we get a factorization.One of the factors yields a solution,and looking at the discriminant of the other one yields no solutions(real).
- 6 years, 7 months ago
Indeed, justifying that one has indeed found all of the solutions is the tricky part. As many other users have not been exposed to complex numbers, I will post the solutions to the problems assuming that all variables used are real. However, other users may feel free to solve these problems over the complex numbers and post their own solutions.
- 6 years, 7 months ago
Thanks! Often, the real case gives insight into the approach.
And of course, we can then ask, what are the solutions over the quarternions? Over the matrices? It never ends :)
Staff - 6 years, 7 months ago
FIRST ONE IS X=Y fourth one is x=y=z=t
- 6 years, 7 months ago
Problem 1: y^2=x(x-1)(x-2) and x^2=y(y-1)(y-2). x=y=0. (if there is only one solution) Problem 2: x=y=0. (if that is the only solution) Problem 3: (x,y,z)-->(2,3,1) or (3,2,1) Problem 4: t=x=y=z=sqrt(1/3) Problem 5: x=y=z=0 I assume there is only 1 solution for all. very dangerous. Anyway, I'm actually 12, and i'm going RI too! Hope to see you soon, Victor and Srivathsan Veeramani.
- 6 years, 7 months ago
Good luck aloysius to get into RI quite surprised you knowthis at a young age I only learnt this recently but I feel that these sums will come in smo round 1 which does not requre methods and sometimes it is quite obvious like qs 1 that x=y and I am too lazy to write the method.anyway you have solved it correctly and congrats.Good luck and have fun in RI.
- 6 years, 7 months ago
There are actually several solution sets to question 1. If we were restricted to the real, then there are 3 solutions with $x = y$. If we include the complex numbers, then there are 6 solutions and $x \neq y$.
With these questions, you should think beyond finding the obvious answer, and trying to ensure that you could find all possible solutions.
Staff - 6 years, 7 months ago
(3).
x+y=4+z $x^{2}+y^{2}=12+z^{2}$
After this we can find xy, and hence $x^{3}+y^{3}$ in terms of z. Comparing with the third equation will give a cubic equation.
- 6 years, 7 months ago |
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8. NCERT Solutions for Class...
# NCERT Solutions for Class 7 Maths Exercise 6.1
### myCBSEguide App
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NCERT solutions for Maths Triangles and its Properties
## NCERT Solutions for Class 7 Maths Triangles and its Properties
Class –VII Mathematics (Ex. 6.1)
Question 1.In {tex}\Delta {/tex}PQR, D is the mid-point of {tex}\overline {{\text{QR}}} .{/tex}
{tex}\overline {{\text{PM}}} {/tex} is _______________
PD is ________________
Is QM = MR ?
Given: QD = DR
{tex}\therefore {/tex} {tex}\overline {{\text{PM}}} {/tex} is altitude.
PD is median.
No, QM {tex} \ne {/tex} MR as D is the mid-point of QR.
Question 2.Draw rough sketches for the following:
1. In {tex}\Delta {/tex}ABC, BE is a median.
2. In {tex}\Delta {/tex}PQR, PQ and PR are altitudes of the triangle.
3. In {tex}\Delta {/tex}XYZ, YL is an altitude in the exterior of the triangle.
(a) Here, BE is a median in {tex}\Delta {\text{ABC}}{/tex} and AE = EC.
(b) Here, PQ and PR are the altitudes of the {tex}\Delta {/tex}PQR and RP {tex} \bot {/tex} QP.
(c) YL is an altitude in the exterior of {tex}\Delta {/tex}XYZ.
Question 3.Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same.
Isosceles triangle means any two sides are same.
Take {tex}\Delta {\text{ABC}}{/tex} and draw the median when AB = AC.
AL is the median and altitude of the given triangle.
## NCERT Solutions for Class 7 Maths Exercise 6.1
NCERT Solutions Class 7 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 7 Maths includes text book solutions from Class 7 Maths Book . NCERT Solutions for CBSE Class 7 Maths have total 15 chapters. 7 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 7 solutions PDF and Maths ncert class 7 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.
## CBSE app for Students
To download NCERT Solutions for Class 7 Maths, Social Science Computer Science, Home Science, Hindi English, Maths Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through the best app for CBSE students and myCBSEguide website.
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### 3 thoughts on “NCERT Solutions for Class 7 Maths Exercise 6.1”
1. Very useful for children to study and understand
2. wow you are inteligent riya ok
3. It was very useful . thank you so much |
How many hours overtime did a man work if he was paid \$168? - ProProfs Discuss
# How many hours overtime did a man work if he was paid \$168? A man's regular pay is \$3 per hour up to 40 hours. Overtime is twice the payment for regular time.
Change Image Delete
A. 8
B. 16
C. 28
D. 48
This question is part of ECT MATHEMATICS
Asked by G.Beasley, Last updated: Feb 18, 2020
#### Matz Lewis Clark
Traveler and writer by profession.
Matz Lewis Clark, College student, Graduation, Orlando
The correct answer to this question is A, 8. To find this answer, one should know first multiple the pay times the number of regular hours. This would create the equation 3 x 40, which equals 120. From there, you would subtract this amount from the full amount earned. That establishes the equation 168 - 120. This gives us 48.
Then, we need to divide this amount by the hourly wage, which provides the equation with 48 / 3= 16. Being that overtime is double, we would need to use division again, this time dividing 16 /2, which gives us our answer of 8.
#### Anika Nicole
Content Writer, Teacher
Anika Nicole, Wordsmith, PG In Journalism, New York
Here, the correct answer is option A.
Regular Pay= \$3 per hour up to 40 hours. So, total pay of the day= \$340= \$120.
He was paid\$168 for his overall work. Hence, we need to calculate overtime pay, i.e. \$168 - \$120 = \$48.
Overtime rate = 2\$3=\$6 per hour.
Therefore, the total number of overtime hours=\$48 divided by \$6 = 8.
#### John Smith
John Smith
8
At \$3 per hour up to 40 hours, regular pay = \$3 x 40 = \$120 If total pay = \$168, overtime pay = \$168 - \$120 = \$48 Overtime rate (twice regular) = 2 x \$3 = \$6 per hour ; hence number of overtime hours = \$48/\$6 = 8 |
# Mathematical Powers – a Simple Insight
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Multiplication is one of the simpler operations we perform on numbers. As kids we had to learn the multiplication tables, one times two equals two, two times two equals three, three times two equals six, and so forth. It didn’t take long before most of us were comfortable multiplying simple numbers. But sometimes we multiply the same number times itself. In that case, we can write out the multiplication in the usual way, or we can write it in terms of mathematical powers.
## Mathematical Powers – a Simple Illustration
Let’s consider the example of three times three. That can be written either 3 x 3 = 9, or in powers notation,
32 = 9
This tells us three to the second power (or three squared) equals nine. Consider the two images associated with this article. One is a square with sides equal to 3. The other is a cube with its three sides each equal to 3 as well. The former square contains within it nine 1 x 1 inch smaller squares. The cube contains within it twenty-seven 1 x 1 x 1 smaller cubes.
I know of no one who can draw a geometrical figure that demonstrates the mathematical powers expression 34 = 3 x 3 x 3 x 3 = 81. But we should need no such figure. The simple insight we’ve provided using the square and the cube should suffice.
## One Last Thought
For one last help, one last aid, consider this. 32 = 3 x 3. And 3 = 3 x 3 x 3. In the first case, the power 2 also represents the number of 3’s that are multiplied together. In the second case, the power 3 represents a similar thing. So the multiplication power expression 38 consists of eight 3’s multiplied times each other, or
38 = 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 = 6561
Note: You might also enjoy Factorials? What are They? A Simple Kind of Mathematics Shorthand
References:
### One thought on “Mathematical Powers – a Simple Insight”
• Yes, we sometimes got insights into mathematical tricks from other people round us. Not that powers are mathematical tricks! But I remember being taught by the local postmaster that you could count the number of stamps in a sheet of stamps by multiplying the rows by the columns. It amazed me at the time and I ran to tell my parents about the “insight”! I have never seen stamps in cubes, though. |
# Chapter 7 - Work and Energy
## Work
Work is a measure of what a force achieves, so when we calculate the work done by a force on an object we consider only it's displacement in the direction of the force.
Mathematically, as we consider both force $\vec{F}$ and displacement $\vec{d}$ to be vectors, and the work done on an object to be a scalar $W$, we say that the work is the scalar product, or dot product of force and displacement. We should note that this is not the displacement from an arbitrary origin, but rather the change in an objects position during a motion.
For a constant force
$W= \vec{F} \cdot \vec{d}$
If the two vectors are at an angle $\theta$ to each other then we can say
$W = Fd\cos\theta$
## Work on a backpack
Consider a hiker that walks first on the flat and then up a hill carrying a backpack, with a constant velocity on each section.
A hiker carries a 10kg backpack a distance of 500 m on the flat at a constant velocity. How much work do they do on the backpack?
The hiker now climbs up a mountain that has a 5 degree incline. When he has walked a distance of 500m and is still moving at the same speed how much work has he done on the backpack?
## More on the dot product
The scalar product is commutative (order is not important)
$\vec{A}\cdot\vec{B}=\vec{B}\cdot\vec{A}$
and distributive
$\vec{A}\cdot(\vec{B}+\vec{C})=\vec{A}\cdot\vec{B}+\vec{A}\cdot\vec{C}$
The scalar product of two vectors written in unit vector notation is straightforward to calculate because the different unit vectors are perpendicular to each other.
$\hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1$
$\hat{i}\cdot\hat{j}=\hat{i}\cdot\hat{k}=\hat{j}\cdot\hat{k}=0$
So for two vectors
$\vec{A}=A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k}$
$\vec{B}=B_{x}\hat{i}+B_{y}\hat{j}+B_{z}\hat{k}$
$\vec{A}\cdot\vec{B}=A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}$
## A more general definition of work
Our previous definition of work was rather limited, in that it considered only a constant force. To be able to consider the work done by a force which can vary over a given path between two points in space we should better write
$W=\int_{a}^{b}\vec{F}\cdot d\vec{l}$
Here we broke our path up into infintessimcal elements $d\vec{l}$ over which we can then integrate
as
$\vec{F}=F_{x}\hat{i}+F_{y}\hat{j}+F_{z}\hat{k}$
$d\vec{l}=dx\,\hat{i}+dy\,\hat{j}+dz\,\hat{k}$
This vector integral can be easily split in to separate integrals for the different components
$W=\int_{x_{a}}^{x_{b}}F_{x}\,dx+\int_{y_{a}}^{y_{b}}F_{y}\,dy+\int_{z_{a}}^{z_{b}}F_{z}\,dz$
## Work done by a spring
Hooke's Law states that for a spring to be either compressed or stretched from it's equilibrium length the force exerted on it must be
$F_{A}=kx$
the spring exerts an equal and opposite force back on itself
$F_{S}=-kx$
Suppose a spring is extended from $x=0$ to an arbitrary distance $x$. Then the work done on the spring by the applied force is
$W=\int_{0}^{x}F_{A}(x)\,dx=\int_{0}^{x}kx\,dx=\frac{1}{2}kx^2$
and, equally, if we compress the spring
$W=\int_{0}^{-x}F_{A}(x)\,dx=\int_{0}^{-x}kx\,dx=\frac{1}{2}kx^2$
In either case the work done by the spring on itself will be negative and of equal magnitude to the work we do on it ($W=-\frac{1}{2}kx^2$). The net work done on the spring by us and the spring is zero. We'll know see why this must be the case.
## Net Work and Kinetic Energy
If we consider the net work on an object we can use Newton's second law to define a new and very useful quantity, the kinetic energy of an object.
$W_{net}=\int\vec{F_{net}}\cdot d\vec{l}=\int F_{\parallel}\,dl$
Newton's second laws tells us that
$F_{\parallel}=ma_{\parallel}=m\frac{dv}{dt}$
so that
$W_{net}=\int F_{\parallel}\,dl=\int m \frac{dv}{dt}dl$
We don't really want to involve time in our integral directly, so here's the chain rule trick again
$\frac{dv}{dt}=\frac{dv}{dl}\frac{dl}{dt}=\frac{dv}{dl}v$
which lets us write
$W_{net}=\int m v \frac{dv}{dl}dl=\int mv\,dv$
Evaluating this integral from $v_{1}$ to $v_{2}$ gives us that the net work is
$W_{net}=\frac{1}{2}mv_{2}^2-\frac{1}{2}mv_{1}^2$
We can now define the kinetic energy of an object as $\frac{1}{2}mv^2$ and note that any change in this quantity is equal to the net work done on the object. We should note that forces directed perpendicular to the motion, which can change the direction, but not the magnitude of the velocity, as in uniform circular motion, do no work.
## Why you should learn to love the Work-Energy Theorem
Let's look at problem 7.61 from the textbook.
A 5.0 $\mathrm{kg}$ object moving in two dimensions initially has a velocity $\vec{v_{1}} = (11.5\hat{i} + 21.5\hat{j})\mathrm{m/s}$. A net force $\vec{F}$ then acts on the object for $2.0 \mathrm{s}$, after which the object's velocity is $\vec{v_{2}} = (15.0\hat{i}+28.0\hat{j})\mathrm{m/s}$. Determine the work done by $\vec{F}$ on the object.
Hard Way
We need the acceleration.
$\vec{a}=\frac{\vec{v_{2}}-\vec{v_{1}}}{t}=\frac{(15.0-11.5)\hat{i}+(28.0-21.5)\hat{j}}{2}\mathrm{\frac{m/s}{s}}=1.75\hat{i}+3.25\hat{j}\, \mathrm{m/s^2}$
$\vec{F}=m\vec{a}=5(1.75\hat{i}+3.25\hat{j})=8.75\hat{i}+16.25\hat{j}\,\mathrm{N}$
$\vec{d}=\vec{v}_{1}t+\frac{1}{2}\vec{a}t^2=2(11.5\hat{i} + 21.5\hat{j})+2(1.75\hat{i}+3.25\hat{j})=26.5\hat{i}+49.5\hat{j}\,\mathrm{m}$
$W=\vec{F}\cdot\vec{d}=8.75\times26.5+16.25\times49.5=1036.25 \mathrm{J}$
Easy Way
$W=\frac{1}{2}m(v_{2}^2-v_{1}^2)=2.5\times((15^2+28^2)-(11.5^2+21.5^2))$
$=2.5\times((225+784)-(132.25+462.25))=1036.25 \mathrm{J}$
You should make sure you can do this problem both ways!
## Plus and minus of the work-energy theorem
$W_{net}=\int\vec{F_{net}}\cdot d\vec{l}=\int F_{\parallel}\,dl=\frac{1}{2}mv_{2}^2-\frac{1}{2}mv_{1}^2$
Plus
• We can use the work-energy theorem to gain information about a motion, particularly about the velocity, independent of the path an object takes, we just need to know the total energy at the start and end and the forces active on it
Minus
• We can't get any specific information about the path taken, so we can't forget the kinematic equations and force analysis we've been using till now!
Important point
• To really get the most out of this theorem we are going to need to apply the principle of conservation of energy.
## The Law of Conservation of Energy
The law of conservation of energy states that within a closed system the total amount of energy is always conserved.
Another way of this is saying this is that energy can be neither created of destroyed, it can only be converted from one form to another.
There are, however, many different forms of energy.
For mechanics problems it is useful to think about a more restricted law, which considers mechanical energy to be conserved.
Conservation of mechanical energy can be assumed whenever the forces which apply to a system are entirely conservative.
## Conservative Forces
A force can be considered to be conservative, if the work done on the object by the force depends only on the beginning and end point of the motion and is independent of the path taken.
## Gravity is a conservative force
$W_{G}=\int_{1}^{2} \vec{F}_{G}\cdot d\vec{l}$
Now as $\vec{F}_{G}=-mg\hat{j}$ and $d\vec{l}=dx\hat{i}+dy\hat{j}+dz\hat{k}$
$\vec{F}_{G}\cdot d\vec{l}=-mg\,dy$
and
$W_{G}=-\int_{y_{1}}^{y_{2}}mg\,dy=-mg(y_{2}-y_{1})$
The work done depends only the change in height.
## Friction is a non-conservative force
A force can only be conservative if the net work done by the force on an object moving around any closed path is zero.
This means that for an object moving from point 1 to point 2 under force the work must be equal and opposite for when it moves from point 2 to point 1.
As we know, the friction force is always in the opposite direction to motion, if we push an object a distance against a frictional force $\vec{F_{Fr}}$ then the work done by the friction on the object is $-F_{Fr}d$. (Note that the work done is negative, the frictional force reduces the kinetic energy.) If we push it back to it's original position the force is now in the opposite direction so the work done by the frictional force is again $-F_{Fr}d$. As the total work done is $-2F_{Fr}d$ we can see that friction is not a conservative force.
## Spring Force is conservative
Forces that depend on position can be conservative.
For example for a spring
$\int_{x_{1}}^{x_{2}}F_{s}\,dx=-\int_{x_{1}}^{x_{2}}kx\,dx=\frac{1}{2}k(x_{2}^2-x_{1}^2)$
Hooke's Law states that for a spring to be either compressed or stretched from it's equilibrium length the force exerted on it must be
$F_{A}=kx$
the spring exerts an equal and opposite force back on itself
$F_{S}=-kx$
Suppose a spring is extended from $x=0$ to an arbitrary distance $x$. Then the work done on the spring by the applied force is
$W=\int_{0}^{x}F_{A}(x)\,dx=\int_{0}^{x}kx\,dx=\frac{1}{2}kx^2$
and, equally, if we compress the spring
$W=\int_{0}^{-x}F_{A}(x)\,dx=\int_{0}^{-x}kx\,dx=\frac{1}{2}kx^2$
In either case the work done by the spring on itself will be negative and of equal magnitude to the work we do on it ($W=-\frac{1}{2}kx^2$). The net work done on the spring by us and the spring is zero. We should note however that the work that the spring does to cancel out our work does not vanish, it is stored as potential energy.
## Potential Energy
Work done against a conservative force is not lost. It is converted in to potential energy that can be converted back in to work.
We use the symbol $U$ for potential energy.
The change in potential energy $\Delta U$ is the same as the work done against the force, which is equal and opposite to the work done by the force.
$\Delta U = -\int_{1}^{2}\vec{F}_{G}\cdot\,d\vec{l}$
Although, the change in potential energy is well defined we have absolute freedom in determining the absolute value of potential energy,
$U=\Delta U +U_{0}$
though usually some choices are more sensible than others.
## Potential Energy and Force
We arrived at the potential energy by integrating the force over displacement.
$\Delta U = -\int_{1}^{2}\vec{F}\cdot\,d\vec{l}$
We can go the other way and obtain the force from the potential. If we have a given potential that varies in space $U(x,y,z)$ then the force is the related to the spatial derivative of the potential.
$\vec{F}(x,y,z)=-\hat{i}\frac{\partial U}{\partial x}-\hat{j}\frac{\partial U}{\partial y}-\hat{k}\frac{\partial U}{\partial z}$
## Conservation of Mechanical Energy
The work-energy theorem which says that the net work done on an object $W_{Net}$ is equal to it's change in kinetic energy $\Delta K$
$W_{Net}=\Delta K$
We also saw that the change in potential energy if only conservative forces are active is
$\Delta U = -\int_{1}^{2}\vec{F}\cdot\,d\vec{l} = - W_{Net}$
So we can see that when only conservative forces act that
$\Delta K + \Delta U = 0$
We call the sum of the Kinetic Energy and Potential Energy the Total Mechanical Energy E
$E=K+U$
and we can see that under the condition of only conservative forces acting this is a conserved quantity.
## Two Track Race
Which track should give a ball a faster velocity at the end of the track?
A. Track A B. Track B or C. Neither, the velocity should be the same.
On which track should a ball arrive first if they are released simultaneously?
A. Track A B. Track B or C. Neither, the time taken should be the same.
## Two Track Race Explained
We can use the conservation of mechanical energy to know that if the change in height and therefore potential energy is the same in both cases, the kinetic energy gained should be the same in both cases, and thus the velocity at the end of both tracks should be the same.
We can't, however, use this to tell the time taken, this depends on the path taken. We can understand the counter-intuitive result that the longer path takes less time based on the fact that throughout the middle section of the path the ball is always moving faster than it is on the straight path.
In reality there is a limit to how far we can bend the path and still have the ball take less time. The curve of fastest descent is the one closest to a Cycloid. |
# Equivalent Fractions
Definition of equivalent fractions are different fractions that represent the same value or proportion. Although they may appear different, their numerical values remain the same.
Understanding equivalent fractions is crucial in various mathematical concepts and real-life applications. For example, it helps simplify fractions, compare and order fractions, add and subtract fractions, and solve problems involving proportions and ratios.
## Understanding the Concept of Equivalent Fractions
A fraction represents a portion of a whole or a ratio between two quantities. It consists of a numerator (the number above the fraction line) and a denominator (the number below the fraction line).
Numerator and denominator The Numerator represent the number of equal parts considered. In contrast, the denominator represents the total number of equal parts.
Identifying equivalent fractions. There are several methods to identify equivalent fractions:
1. The exact number is used to multiply or divide the Numerator and denominator. We can produce an equivalent fraction by dividing or multiplying the Numerator and denominator by the same nonzero amount. The fraction is scaled using this technique without changing its value.
2. Simplifying fractions If the Numerator and denominator have a common factor, we can simplify the fraction to its simplest form. This process involves dividing the Numerator and denominator by their greatest common divisor.
3. Understanding common factors allows you to simplify fractions to their simplest form. To achieve this, divide the numerator by the largest common factor of the numerator and the denominator, resulting in a fraction with smaller numbers.
## Steps to Find Equivalent Fractions
Step 1: Identify the given fraction.Begin by identifying the fraction for which you want to find an equivalent fraction.
Step 2: Determine a common factor. Look for a common factor between the Numerator and denominator that can be used to scale the fraction.
Step 3: Multiply or divide the Numerator and denominator by the common factor. Add or subtract the common factor found in Step 2 from the Numerator and denominator. By doing this, the fraction is guaranteed to remain equal.
Step 4: Simplify the fraction, if possible If there are common factors between the Numerator and denominator after Step 3, Divide both by the fraction’s greatest common factor to simplify it.
Step 5: Repeat steps 2-4 until desired equivalent fraction is obtained. Repeat steps 2 to 4 with the resulting fraction to find other equivalent fractions.
## Examples and Practice
Example 1: Finding equivalent fractions using multiplication Given the fraction 2/3, we can derive an equal fraction by multiplying the Numerator and denominator by the same amount. For instance, multiplying by 2 yields 4/6, equivalent to 2/3.
Example 2: Finding equivalent fractions using division. Consider the fraction 4/8. To divide both, use the greatest common factor between the Numerator and denominator. To determine a comparable fraction, To find an equivalent fraction 4. This results in the simplified equivalent fraction of 1/2.
Example 3: Simplifying fractions to find equivalent fractions. If we have the fraction 12/16, we can simplify it by dividing the Numerator and denominator by their greatest common divisor, which is 4. It gives us the simplified equivalent fraction 3/4.
## Applications and Real-Life Examples
Using equivalent fractions in baking recipes Equivalent fractions help adjust recipes. For instance, if a recipe requests 1/2 cup of flour but must make twice the amount, you can use the equivalent fraction of 2/4 cup of flour.
Dividing a pizza into equal parts using equivalent fractions can divide the pizza into equal parts. For example, cutting a pizza into 8 equal slices is the same as cutting it into 16 half slices, equivalent to 32 quarter slices.
## Summary and Conclusion
Recap of the concept of equivalent fractions representing the same value or proportion, despite having different numerical representations. The denominator and Numerator can be multiplied or divided by the same integer, or fractions can be simplified.
Practicing and understanding equivalent fractions is vital for various mathematical operations and real-life applications. For example, it helps simplify fractions, compare and order fractions, perform operations with fractions, and solve problems involving proportions and ratios.
Encouragement to explore further applications and practice problems To strengthen understanding, it is recommended to practice finding equivalent fractions in different scenarios and explore additional real-life applications. Doing so can develop a solid foundation in fractions and enhance mathematical problem-solving skills. |
HomeNEWSwhat is it and what does it mean?
# what is it and what does it mean?
Pearson’s correlation is a far-fetched mathematical calculation invented by Karl Pearson, the same man responsible for creating Standard Deviation in statistics. Karl was responsible for several mathematical discoveries and incredible results in the exact area. And Pearson’s correlation is an amazing formula for understanding more about how two variables behave.
Do you know what this calculation works for? Come and understand everything about Pearson’s correlation in our article!
## What is Pearson’s correlation?
The Pearson correlation is a mathematical calculation created by Karl Pearson to identify the relationship between two quantities, which are variables.
Also called Pearson’s correlation coefficient, the calculation exists to identify the tension between two variables, which are continuous. In simple terms we can summarize the calculation of the Pearson coefficient as follows:
When the Pearson correlation coefficient between the two continuous quantities is 0, it means that there is no association between the two variables;
When the Pearson correlation coefficient between the two continuous quantities is positive, greater than >0, it means that the association between the two variables is positive. As one increases the other variable also increases;
When the Pearson correlation coefficient between the two continuous quantities is negative, less than <0, it means that the association between the two variables is negative. As one increases the other variable decreases, generating a situation that we call inversely proportional.
This is a summary enough to understand in a general way what is Pearson’s correlation and how its results work.
## How is Pearson’s correlation calculated?
Pearson’s correlation calculation is quite complex, its formula can be summarized as follows:
This is the correlation formula. We can note the variables of the formula, which are distributed as follows:
• X = continuous variable number 1;
• Y = continuous variable number 2;
• ZX = standard deviation of variable X;
• ZY = standard deviation of variable Y;
• N = number of data.
The calculation, being quite complex, can only be done by good mathematicians and statisticians who know the subject well.
## What is the meaning of Pearson’s correlation?
The Pearson correlation means the relationship that the two magnitudes have with each other. As already mentioned, the possible results in summary are ,0, +1 or -1.
When the result is +1 it means that the relationship between the two magnitudes is proportional, and -1 is when the result is inversely proportional. In the case of 0 it means that the calculation of the correlation cannot define what is the relationship between the two magnitudes, but in a concrete situation it cannot mean that there is no relationship between the two variables.
It just means that this relationship in the context of calculation is impossible to calculate or identify.
As with any calculation used in the area of statistics, there are advantages and disadvantages to calculating Pearson’s correlation. As an advantage, we can highlight the fact that the calculation does not depend on the result to be done, and also on the accuracy of the formula if the two magnitudes are large enough.
Among the disadvantages, however, can be pointed out the need for the two magnitudes to present a continuous quantitative level, and the lack of accuracy if there is little data.
But did you already know this elaborate calculation? The calculation of Pearson’s correlation coefficient is very advanced, but widely used in the discipline of statistics. |
Accelerating the pace of engineering and science
Symbolic Math Toolbox
Differentiation
This example shows how to find first and second derivatives using Symbolic Math Toolbox™.
First Derivatives: Finding Local Minima and Maxima
Computing the first derivative of an expression helps you find local minima and maxima of that expression. For example, create the rational expression where the numerator and denominator are polynomial expressions. Before creating a symbolic expression, you declare symbolic variables. By default, solutions that include imaginary components are included in the result. If you are interested in real numbers only, you can set the assumption that x belongs to the set of real numbers. This assumption allows you to avoid complex numbers in symbolic solutions and it also can improve performance:
```syms x real;
f = (3*x^3 + 17*x^2 + 6*x + 1)/(2*x^3 - x + 3)
```
```
f =
(3*x^3 + 17*x^2 + 6*x + 1)/(2*x^3 - x + 3)
```
Plotting this expression shows that the expression has horizontal and vertical asymptotes, a local minimum between -1 and 0, and a local maximum between 1 and 2:
```ezplot(f)
hold on
grid
hold off
```
To find a horizontal asymptote, compute the limit of f for x approaching positive and negative infinities. The horizontal asymptote is x = 3/2:
```limit(f, x, -inf)
limit(f, x, inf)
```
```
ans =
3/2
ans =
3/2
```
To find a vertical asymptote of f, find the roots of the polynomial expression that represents the denominator of f:
```solve(2*x^3 - x + 3, x)
```
```
ans =
- 1/(6*(3/4 - (241^(1/2)*432^(1/2))/432)^(1/3)) - (3/4 - (241^(1/2)*432^(1/2))/432)^(1/3)
```
You can approximate the exact solution numerically by using the double function. To access the result of the previous evaluation, use the default variable name ans:
```double(ans)
```
```ans =
-1.2896
```
Now find the local minimum and maximum of f. If the point is a local extremum (either minimum or maximum), the first derivative of the expression at that point is equal to zero. To compute the derivative of an expression, use the diff function:
```g = diff(f, x)
```
```
g =
(9*x^2 + 34*x + 6)/(2*x^3 - x + 3) - ((6*x^2 - 1)*(3*x^3 + 17*x^2 + 6*x + 1))/(2*x^3 - x + 3)^2
```
To find the local extrema of f, solve the equation g = 0:
```solve(g, x)
```
```
ans =
((2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3))/1156 + 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3) + 361/289)^(1/2)/(6*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/6)) - ((337491*6^(1/2)*((3*3^(1/2)*178939632355^(1/2))/9826 + 2198209/9826)^(1/2))/39304 + (2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3)*((2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3))/1156 + 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3) + 361/289)^(1/2))/578 - 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3)*((2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3))/1156 + 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3) + 361/289)^(1/2) - (361*((2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3))/1156 + 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3) + 361/289)^(1/2))/289)^(1/2)/(6*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/6)*((2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3))/1156 + 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3) + 361/289)^(1/4)) - 15/68
((2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3))/1156 + 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3) + 361/289)^(1/2)/(6*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/6)) + ((337491*6^(1/2)*((3*3^(1/2)*178939632355^(1/2))/9826 + 2198209/9826)^(1/2))/39304 + (2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3)*((2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3))/1156 + 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3) + 361/289)^(1/2))/578 - 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3)*((2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3))/1156 + 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3) + 361/289)^(1/2) - (361*((2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3))/1156 + 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3) + 361/289)^(1/2))/289)^(1/2)/(6*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/6)*((2841*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(1/3))/1156 + 9*((3^(1/2)*178939632355^(1/2))/176868 + 2198209/530604)^(2/3) + 361/289)^(1/4)) - 15/68
```
And again approximate the exact solution numerically by using the double function applied to ans:
```double(ans)
```
```ans =
-0.1892
1.2860
```
The expression f has a local maximum at x = 1.286 and a local minimum at x = -0.189
Second Derivatives: Finding Inflection Points
Computing the second derivative lets you find inflection points of the expression. The most efficient way to compute second or higher-order derivatives is to use the parameter that specifies the order of the derivative:
```h = diff(f, x, 2)
```
```
h =
(18*x + 34)/(2*x^3 - x + 3) - (2*(6*x^2 - 1)*(9*x^2 + 34*x + 6))/(2*x^3 - x + 3)^2 - (12*x*(3*x^3 + 17*x^2 + 6*x + 1))/(2*x^3 - x + 3)^2 + (2*(6*x^2 - 1)^2*(3*x^3 + 17*x^2 + 6*x + 1))/(2*x^3 - x + 3)^3
```
To find inflection points of f, solve the equation h = 0. Here, use the numeric solver to calculate floating-point % approximations of the solutions:
```vpasolve(h, x)
```
```
ans =
0.57871842655441748319601085860196
1.8651543689917122385037075917613
- 0.46088831805332057449182335801198 + 0.47672261854520359440077796751805*i
- 0.46088831805332057449182335801198 - 0.47672261854520359440077796751805*i
- 1.4228127856020972275345064554049 + 1.8180342567480118987898749770461*i
- 1.4228127856020972275345064554049 - 1.8180342567480118987898749770461*i
```
The expression f has two inflection points: x = 1.865 and x = 0.579 |
Multiplication Method Multiplication Method A multiplication algorithm is an algorithm (or method) to multiply two numbers. Depending on the size of the numbers, different algorithms are in use. Efficient multiplication algorithms have existed since the advent of the decimal system. Grid method:- The grid method (or box method) is an introductory method for multiple-digit multiplication that is often taught to pupils at primary school or elementary school level. It has been a standard part of the national primary-school mathematics curriculum in England and Wales since the late 1990s. Both factors are broken up ("partitioned") into their hundreds, tens and units parts, and the products of the parts are then calculated explicitly in a relatively simple multiplication-only stage, before these contributions are then totalled to give the final answer in a separate addition stage Thus for example the calculation 34 Ă— 13 could be computed using the grid 30 4 10 300 40 3 90 12 followed by addition to obtain 442, either in a single sum (see right), or through forming the row-byrow totals (300 + 40) + (90 + 12) = 340 + 102 = 442. Know More About :- Definition for Subtraction
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This calculation approach (though not necessarily with the explicit grid arrangement) is also known as the partial products algorithm. Its essence is the calculation of the simple multiplications separately, with all addition being left to the final gathering-up stage. The grid method can in principle be applied to factors of any size, although the number of sub-products becomes cumbersome as the number of digits increases. Nevertheless it is seen as a usefully explicit method to introduce the idea of multipledigit multiplications; and, in an age when most multiplication calculations are done using a calculator or a spreadsheet, it may in practice be the only multiplication algorithm that some students will ever need. Long multiplication :- If a positional numeral system is used, a natural way of multiplying numbers is taught in schools as long multiplication, sometimes called grade-school multiplication, sometimes called Standard Algorithm: multiply the multiplicand by each digit of the multiplier and then add up all the properly shifted results. It requires memorization of the multiplication table for single digits. This is the usual algorithm for multiplying larger numbers by hand in base 10. Computers normally use a very similar shift and add algorithm in base 2. A person doing long multiplication on paper will write down all the products and then add them together; an abacus-user will sum the products as soon as each one is computed. Example This example uses long multiplication to multiply 23,958,233 (multiplicand) by 5,830 (multiplier) and arrives at 139,676,498,390 for the result (product). 23958233 5830 × -----------00000000 ( = 23,958,233 × 0) 71874699 ( = 23,958,233 × 30) 191665864 ( = 23,958,233 × 800) 119791165 ( = 23,958,233 × 5,000) -----------139676498390 ( = 139,676,498,390 ) Read More About :- Definition of Subtrahend
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Thank You
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Multiplication Method |
1. ## Triangle ratio problem
Let $\bigtriangleup ABC$ be a right angled triangle such that $\angle A =90$° $, AB =AC$ and let $M$ be the mid point of the side AC. Take the point P on the side BC so that AP is vertical to BM. Let H be the intersection point of AP and BM.
(a) Find the ratio of the areas of the two triangles $\bigtriangleup ABH : \bigtriangleup AHM$.
(b) Find the ratio BP : PC.
Now this is my triangle, to place P, I took "AP is vertical to BM" as "AP and BM are perpendicular".
Now the answer to (a) is :
$AB : AM = 2:1$
$\bigtriangleup ABH : \bigtriangleup AHM$
$= 2^2 : 1^2 = 4 : 1$
It is true that AB is twice large than AM, but why are they rising them to the 2th power?
To solve (b) I tried measuring PB and CP, and the comparing with CB, but I gave up soon, as I was getting something near 2.3 for CP. so I looked at the answer (2:1), and noticed the should be:
CB = 7
CP = 2.333333333
PB = 4.666666666
Though method used to calculate the ratio is very strange, I don't understand it:
M is in the middle of AC, $\bigtriangleup AHM = \bigtriangleup HCM$ --- (1)
on the other hand, $\frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{\bigtriangleup AHC}{\bigtriangleup HPC}$ --- (2)
From (1) and (2) $\frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{2\bigtriangleup AHM}{\bigtriangleup HPC}$ --- (3)
therefore from (3) and question (a)
$\frac{\bigtriangleup HPC}{\bigtriangleup HBP} = \frac{2\bigtriangleup AHM}{\bigtriangleup ABH} = \frac{(2)(1)}{4} = \frac{1}{2}$ --- (4)
$\frac{PC}{BP} = \frac{\bigtriangleup HPC}{\bigtriangleup HBP}$ because of (4)
The answer is BP : CP = 2 : 1
Can somebody explain me how to use this to get the ratio?
2. Originally Posted by Patrick_John
...
Now the answer to (a) is :
$AB : AM = 2:1$
$\bigtriangleup ABH : \bigtriangleup AHM$
$= 2^2 : 1^2 = 4 : 1$
It is true that AB is twice large than AM, but why are they rising them to the 2th power?
...
Hello,
to calculate the area of a triangle you have to use two values: the length of the base and the length of the height.
In your case not only the base was lengthened by the factor 2 but the height also. If
$A_1 = \frac{1}{2} \cdot b_1 \cdot h_1$ is the area of the first triangle and you know that
$b_2 = 2 \cdot b_1 \text{ and } h_2 = 2 \cdot h_1$
then you get:
$A_2 = \frac{1}{2} \cdot b_2 \cdot h_2 = \frac{1}{2} \cdot 2 \cdot b_1 \cdot 2 \cdot h_1 = 2 \cdot 2 \cdot \frac{1}{2} \cdot b_1 \cdot h_1 = 2^2 \cdot A_1$
3. Originally Posted by Patrick_John
...
To solve (b) I tried measuring PB and CP, and the comparing with CB, but I gave up soon, as I was getting something near 2.3 for CP. so I looked at the answer (2:1), and noticed the should be:
CB = 7
CP = 2.333333333
PB = 4.666666666
Hello,
if $|AB| = |AC| = 5 ~cm$ then $|BC| = \sqrt{5^2+5^2} = 7 \cdot \sqrt{2}$
Though method used to calculate the ratio is very strange, I don't understand it:
M is in the middle of AC, $\bigtriangleup AHM = \bigtriangleup HCM$ --- (1)
Here you are refering to the value of the area of the triangle: In both triangles the bases and the heights have the same length therefore they must have the same area.
on the other hand, $\frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{\bigtriangleup AHC}{\bigtriangleup HPC}$ --- (2)
The same method is used here too.
Maybe you can complete the problem from here on ...
4. I will note $S_{ABC}$ the area of the triangle $ABC$.
Now:
For (1): $\triangle AHM$ and $\triangle HCM$ have the same height $h_1$ from $H$.
$\displaystyle S_{AHM}=\frac{AM\cdot h_1}{2}=\frac{MC\cdot h_1}{2}=S_{HCM}$.
For (2): $\triangle ABH$ and $\triangle HBP$ have the same height $h_2$ from $B$.
$\displaystyle S_{ABH}=\frac{AH\cdot h_2}{2},S_{HBP}=\frac{PH\cdot h_2}{2}\Rightarrow\frac{S_{ABH}}{S_{HBP}}=\frac{AH }{PH}$ (i)
$\triangle AHC$ and $\triangle HPC$ have the same height $h_3$ from $C$.
$\displaystyle S_{AHC}=\frac{AH\cdot h_3}{2},S_{HPC}=\frac{PH\cdot h_3}{2}\Rightarrow\frac{S_{AHC}}{S_{HPC}}=\frac{AH }{PH}$ (ii)
From (i) and (ii) yields $\displaystyle \frac{S_{ABH}}{S_{HBP}}=\frac{S_{AHC}}{S_{HPC}}$ (iii)
For (3) $\displaystyle S_{AHC}=\frac{AC\cdot h_1}{2}=\frac{2AM\cdot h_1}{2}=2S_{AHM}$.
Replacing $S_{AHC}$ in (iii) $\Rightarrow$ (3).
For (4): $\displaystyle S_{HPC}=\frac{PC\cdot h_4}{2},S_{HBP}=\frac{BP\cdot h_4}{2}\Rightarrow$
$\displaystyle \frac{PC}{BP}=\frac{S_{HPC}}{S_{HBP}}=\frac{2S_{AH M}}{S_{ABH}}=\frac{2S_{AHM}}{4S_{AHM}}=\frac{1}{2}$
5. This is still giving some problems,
Why in $\displaystyle \frac{PC}{BP}=\frac{S_{HPC}}{S_{HBP}}=\frac{2S_{AH M}}{S_{ABH}}=\frac{2S_{AHM}}{4S_{AHM}}=\frac{1}{2}$
$S_{ABH}$ is equal to $4S_{AHM}$?
6. Originally Posted by Patrick_John
This is still giving some problems,
Why in $\displaystyle \frac{PC}{BP}=\frac{S_{HPC}}{S_{HBP}}=\frac{2S_{AH M}}{S_{ABH}}=\frac{2S_{AHM}}{4S_{AHM}}=\frac{1}{2}$
$S_{ABH}$ is equal to $4S_{AHM}$?
I'm going to say the same thing as earboth.
What's the area of a triangle? It's $\frac{1}{2}bh=A$
So what would happen if we doubled the base ( $2b$) and the height ( $2h$)
Well let's substitute: $\frac{1}{2} (2b) (2h) \quad = \quad \frac{1}{2}(4bh) \quad = \quad 4\left(\frac{1}{2}bh\right)$
and we all know that: $4\left(\frac{1}{2}bh\right)=4A$
So when the legs of a triangle double, the area quadruples. That is the reason why
$S_{ABH}$ is equal to $4S_{AHM}$
7. OK! Now I understand, thnks a lot everyone! |
Finding Real roots of the Algebraic and Transcendental equations.
Bisection Method
Initial Approximation
Analytically, we can usually choose any point in an interval where a change of sign takes place. However, this is subject to certain conditions that vary from method to method.
And f(a) and f(b) have opposite signs. In this case a and b are said to bracket a root. the f must have at least one root in the interval (a, b).
At each step the method divides the interval in two by computing the midpoint X1 = (a+b) / 2 of the interval and the value of the function f(X1) at that point.
Unless X1 is itself a root (which is very unlikely, but possible) there are now two possibilities:
either f(a) and f(X1) have opposite signs and bracket a root, or f(X1) and f(b) have opposite signs and bracket a root.
The method selects the subinterval that is a bracket as a new interval to be used in the next step.
In this way the interval that contains a zero of f is reduced in width by 50% at each step. The process is continued until the interval is sufficiently small.
Explicitly, if f(a) and f(X1) are opposite signs, then the method sets X1as the new value for b, and if f(b) and f(X1) are opposite signs then the method sets X1 as the new a.
(If f(X1)=0 then X1may be taken as the solution and the process stops.)
In both cases, the new f(a) and f(b) have opposite signs, so the method is applicable to this smaller interval
X3 – 9x + 1 = 0
Regula Falsi Method
Like the bisection method, the false position method starts with two points a0 and b0 such that f(a0) and f(b0) are of opposite signs, which implies by the intermediate value theorem that the function f has a root in the interval [a0, b0], assuming continuity of the function f.
We start this procedure also by locating two points x0 and x1 where the function has opposite signs.
We now connect the two points f(x0) and f(x1) by a straight line and find where it cuts the x-axis.
Let it cut the x-axis at x2. we find f(x2)
if f(x2) and f(x0) are of opposite signs, then we replace x1 by x2 and draw a straight line connecting f(x2)to f(x0) to find the new intersection point.
if f(x2) and f(x0) are of the same signs then x0 is replaced by x2 and proceed as before.
IN both cases the new interval of search is smaller than the initial interval.
We will now get an equation to find the successive approximations to the root.
x3 = x2 – f(x2) (x2-x1) / f(x2) – f(x1)
x4 = x3 – f(x3) (x3-x2) / f(x3) – f(x2)
http://www.dummies.com/how-to/content/the-basic-differentiation-rules.html
http://www.dummies.com/how-to/content/differential-equations-for-dummies-cheat-sheet.html
Fixed Point Iteration Method
Newton Raphson method
one starts with an initial guess which is reasonably close to the true root,
then the function is approximated by its tangent line (which can be computed using the tools of calculus),
and one computes the x-intercept of this tangent line (which is easily done with elementary algebra). This x-intercept will typically be a better approximation to the function’s root than the original guess, and the method can be iterated.
The next approximation, xn+1, is where the tangent line intersects the axis, so where y=0. Rearranging, we find
Matrices
In linear algebra, the identity matrix or unit matrix of size n is the n×n square matrix with ones on the main diagonal and zeros elsewhere.
http://tutorial.math.lamar.edu/Classes/Alg/Alg.aspx
http://www.vitutor.com/alg/determinants/matrix_rank.html
Operations
There are three types of elementary matrices, which correspond to three types of row operations (respectively, column operations):
Row switching
A row within the matrix can be switched with another row.
Row multiplication
Each element in a row can be multiplied by a non-zero constant.
A row can be replaced by the sum of that row and a multiple of another row.
The elementary matrix for any row operation is obtained by executing the operation on an identity matrix.
Row – echelon form – A matrix is in row – echelon form if
(i) all Zero rows ( if any) are at the bottom of the matrix and
(ii) if two successive rows are non-zero, the second row starts with more zeros that the first (moving from left to right)
for ex the matrix
0 1 0 0
0 0 1 0
0 0 0 0
0 0 0 0
Reduced Row – echelon form matrix is in reduced row – echelon form if – A
1. it is in Row – echelon form
2. the leading (leftmost non-zero) in each row is 1
3. All other elements of the column in which the leading entry 1 occurs are zeros
Solving System of linear equations
We will use the inverse of the coefficient matrix. But first we must check that this inverse exists! The conditions for the existence of the inverse of the coefficient matrix are the same as those for using Cramer’s rule, that is
1. The system must have the same number of equations as variables, that is, the coefficient matrix of the system must be square.
2. The determinant of the coefficient matrix must be non-zero. The reason, of course, is that the inverse of a matrix exists precisely when its determinant is non-zero. |
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# 2-6 - PowerPoint PPT Presentation
Rates, Ratios, and Proportions. 2-6. Holt Algebra 1. Warm Up. Lesson Presentation. Lesson Quiz. Warm Up Solve each equation. Check your answer. 1. 6 x = 36 2. 3. 5 m = 18 4. 5. 8 y =18.4 Multiply. 6. 7. 6. 48. 3.6. –63. 2.3. 7. 10. Objectives.
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2-6
Holt Algebra 1
Warm Up
Lesson Presentation
Lesson Quiz
1. 6x = 36
2.
3. 5m = 18
4.
5. 8y =18.4
Multiply.
6. 7.
6
48
3.6
–63
2.3
7
10
Write and use ratios, rates, and unit rates.
Write and solve proportions.
ratio proportion
rate cross products
scale scale drawing
unit rate scale model
conversion factor
A ratio is a comparison of two quantities by division. The ratio of a to b can be written a:b or , where b ≠ 0. Ratios that name the same comparison are said to be equivalent.
A statement that two ratios are equivalent, such as , is called a proportion.
Read the proportion as
“1 is to 15 as x is to 675”.
The ratio of the number of bones in a human’s ears to the number of bones in the skull is 3:11. There are 22 bones in the skull. How many bones are in the ears?
Write a ratio comparing bones in ears to bones in skull.
Write a proportion. Let x be the number of bones in ears.
Since x is divided by 22, multiply both sides of the equation by 22.
There are 6 bones in the ears.
Check It Out! Example 1
The ratio of games lost to games won for a baseball team is 2:3. The team has won 18 games. How many games did the team lose?
Write a ratio comparing games lost to games won.
Write a proportion. Let x be the number of games lost.
Since x is divided by 18, multiply both sides of the equation by 18.
The team lost 12 games.
A rate is a ratio of two quantities with different units, such as Rates are usually written as unit rates. A unit rate is a rate with a second quantity of 1 unit, such as or 17 mi/gal. You can convert any rate to a unit rate.
Raulf Laue of Germany flipped a pancake 416 times in 120 seconds to set the world record. Find the unit rate. Round your answer to the nearest hundredth.
Write a proportion to find an equivalent ratio with a second quantity of 1.
Divide on the left side to find x.
The unit rate is about 3.47 flips/s.
Check It Out! Example 2
Cory earns \$52.50 in 7 hours. Find the unit rate.
Write a proportion to find an equivalent ratio with a second quantity of 1.
Divide on the left side to find x.
The unit rate is \$7.50.
A rate such as in which the two quantities are equal but use different units, is called a conversion factor. To convert a rate from one set of units to another, multiply by a conversion factor.
Example 3A: Converting Rates equal but use different units, is called a
Serena ran a race at a rate of 10 kilometers per hour. What was her speed in kilometers per minute? Round your answer to the nearest hundredth.
To convert the second quantity in a rate, multiply by a conversion factor with that unit in the first quantity.
The rate is about 0.17 kilometer per minute.
Helpful Hint equal but use different units, is called a
In example 3A , “1 hr” appears to divide out, leaving “kilometers per minute,” which are the units asked for. Use this strategy of “dividing out” units when converting rates.
Example 3B: Converting Rates equal but use different units, is called a
A cheetah can run at a rate of 60 miles per hour in short bursts. What is this speed in feet per minute?
Step 1 Convert the speed to feet per hour.
Step 2 Convert the speed to feet per minute.
To convert the first quantity in a rate, multiply by a conversion factor with that unit in the second quantity.
To convert the first quantity in a rate, multiply by a conversion factor with that unit in the first quantity.
The speed is 316,800 feet per hour.
The speed is 5280 feet per minute.
• Check that the answer is reasonable. equal but use different units, is called a
• There are 60 min in 1 h, so 5280 ft/min is 60(5280) = 316,800 ft/h.
• There are 5280 ft in 1 mi, so 316,800 ft/h is This is the given rate in the problem.
Example 3B: Converting Rates
The speed is 5280 feet per minute.
Check It Out! equal but use different units, is called a Example 3
A cyclist travels 56 miles in 4 hours. What is the cyclist’s speed in feet per second? Round your answer to the nearest tenth, and show that your answer is reasonable.
Step 1 Convert the speed to feet per hour.
Change to miles in 1 hour.
To convert the first quantity in a rate, multiply by a conversion factor with that unit in the second quantity.
The speed is 73,920 feet per hour.
Check It Out! equal but use different units, is called a Example 3
Step 2 Convert the speed to feet per minute.
To convert the second quantity in a rate, multiply by a conversion factor with that unit in the first quantity.
The speed is 1232 feet per minute.
Step 3 Convert the speed to feet per second.
To convert the second quantity in a rate, multiply by a conversion factor with that unit in the first quantity.
The speed is approximately 20.5 feet per second.
Check It Out! equal but use different units, is called a Example 3
Check that the answer is reasonable. The answer is about 20 feet per second.
• There are 60 seconds in a minute so 60(20) = 1200 feet in a minute.
• There are 60 minutes in an hour so 60(1200) = 72,000 feet in an hour.
• Since there are 5,280 feet in a mile 72,000 ÷ 5,280 = about 14 miles in an hour.
• The cyclist rode for 4 hours so 4(14) = about 56 miles which is the original distance traveled.
ALGEBRA equal but use different units, is called a
WORDS
NUMBERS
and b ≠ 0
If
and d ≠ 0
2 • 6 = 3 • 4
In the proportion , the products a •d and b •c are called cross products. You can solve a proportion for a missing value by using the Cross Products property.
Cross Products Property
In a proportion, cross products are equal.
6(7) equal but use different units, is called a = 2(y – 3)
3(m) = 5(9)
42 = 2y – 6
3m = 45
+6 +6
48 = 2y
m = 15
24 = y
Example 4: Solving Proportions
Solve each proportion.
A.
B.
Use cross products.
Use cross products.
Add 6 to both sides.
Divide both sides by 3.
Divide both sides by 2.
4( equal but use different units, is called a g +3) = 5(7)
2(y) = –5(8)
4g +12 = 35
2y = –40
–12 –12
4g = 23
y = −20
g = 5.75
Check It Out! Example 4
Solve each proportion.
A.
B.
Use cross products.
Use cross products.
Subtract 12 from both sides.
Divide both sides by 2.
Divide both sides by 4.
A equal but use different units, is called a scale is a ratio between two sets of measurements, such as 1 in:5 mi. A scale drawingor scale model uses a scale to represent an object as smaller or larger than the actual object. A map is an example of a scale drawing.
blueprint 1 in. equal but use different units, is called a
actual 3 ft.
Example 5A: Scale Drawings and Scale Models
A contractor has a blueprint for a house drawn to the scale 1 in: 3 ft.
A wall on the blueprint is 6.5 inches long. How long is the actual wall?
Write the scale as a fraction.
Let x be the actual length.
x • 1 = 3(6.5)
Use the cross products to solve.
x = 19.5
The actual length of the wall is 19.5 feet.
blueprint 1 in. equal but use different units, is called a
actual 3 ft.
4 = x
Example 5B: Scale Drawings and Scale Models
A contractor has a blueprint for a house drawn to the scale 1 in: 3 ft.
One wall of the house will be 12 feet long when it is built. How long is the wall on the blueprint?
Write the scale as a fraction.
Let x be the actual length.
12 = 3x
Use the cross products to solve.
Since x is multiplied by 3, divide both sides by 3 to undo the multiplication.
The wall on the blueprint is 4 inches long.
model 32 in. equal but use different units, is called a
actual 1 in.
Convert 16 ft to inches.
Let x be the actual length.
Check It Out! Example 5
A scale model of a human heart is 16 ft. long. The scale is 32:1. How many inches long is the actual heart it represents?
Write the scale as a fraction.
Use the cross products to solve.
32x = 192
Since x is multiplied by 32, divide both sides by 32 to undo the multiplication.
x = 6
The actual heart is 6 inches long.
Lesson Quiz: Part 1 equal but use different units, is called a
1. In a school, the ratio of boys to girls is 4:3. There are 216 boys. How many girls are there?
162
Find each unit rate. Round to the nearest hundredth if necessary.
2. Nuts cost \$10.75 for 3 pounds.
\$3.58/lb
3. Sue washes 25 cars in 5 hours.
5 cars/h
4. A car travels 180 miles in 4 hours. What is the car’s speed in feet per minute?
3960 ft/min
Lesson Quiz: Part 2 equal but use different units, is called a
Solve each proportion.
5.
6
16
6.
7. A scale model of a car is 9 inches long. The scale is 1:18. How many inches long is the car it represents?
162 in. |
# Visual Patterns
Visual number talk prompts unpacking geometric linear patterns asking students to predict what comes next using rate of change, initial value.
## In This Set of Visual Number Talk Prompts…
Students will extend growing linear visual patterns, make near and far predictions and describe the pattern in words.
## Intentionality…
Students will engage in a visual math talk using visual patterns as a means develop a deeper understanding of the following big ideas:
• Patterns can be extended because they are repetitive by nature.
• Pattern rules are generalizations about a pattern, and they can be described in words.
• One common use of linear patterns is predicting future events.
• A pattern can be extended to make a prediction.
• For far predictions, calculations are required for efficiency.
• In a growing pattern, the values increase.
• In a linear pattern, the values increase at the same rate.
• The rate can be determined by finding the difference between the value of two terms.
• Graphical representations of linear growing patterns appear as straight lines.
• The initial value in a linear pattern is the constant.
• When the initial value is not zero, the relationship between the two variables is not proportional.
## String of Related Problems
Present the following visual pattern one at a time. Ask students to describe the pattern rule in words, including the initial value. In order to determine the initial value, students are encouraged to determine the value of term 0.
Ask students to predict the value of the 8th term and the 14th term. Students are encouraged to justify their thinking. Look for opportunities to build efficiency. For example, once students have determined the rate, and the initial value, rather than counting on or repeated addition, can students leverage this information to perform a calculation?
For example: In the first pattern, the rate is 3, and the initial value is 1. Students can multiply the term number by 3, and add 1. Term 5 multiplied by 3 is 15, plus 1 is 16.
## Visual Number Talk Prompt
Students prompt:
Use manipulatives or draw what you think might come next.
After giving students enough time to create (using square tiles, unifix cubes, etc.) or draw what might come next for Day 3, ask them to share their thinking with their neighbour before sharing out to the class.
Before resuming the video to reveal what the figure for Day 3 looks like, be sure to celebrate the different possibilities shared encouraging them to explain their thinking. It is important to also explicitly explain that while all of the figures shared are all possible, we are going to explore a specific pattern that has a Day 3 figure that looks like this.
If any students in the room had successfully anticipated the figure for Day 3, be sure to celebrate and ask students to recall the thinking that those students had shared previously to help them predict what they think might come next for Day 4.
Student Prompt:
What might come next for Day 4?
Create a figure using manipulatives or by drawing.
Similarly, give students an opportunity to share their thinking with each other. Pay close attention to which students are already noticing a linear growing visual pattern forming and encourage them to continue to verbalize how they see the pattern growing prior to revealing the next figure.
Although you may have already encouraged some students to verbalize how they see the visual pattern growing, be sure to explicitly have all students now:
Describe the pattern rule in words.
It can be helpful to frame this request using a script such as the following:
Pretend that you are on the phone with another student who isn’t present today and cannot see the pattern we are all viewing right now. How would you communicate how they should create each of the first four days in the pattern?
At this point, we hope that students are able to articulate the growing visual pattern as something similar to:
Begin with 1 and increase by 3 each time.
Be prepared for some students to say something like:
Begin with 4 and increase by 3 each time.
However, the true initial value of this linear growing pattern is actually 1 and not 4 since the true “beginning” of this pattern is on Day 0, with 3 less than Day 1.
At this point, we can challenge students to use their visual pattern rule to:
What will Day 8 look like?
What will Day 14 look like?
Describe how you could construct each figure without simply giving the total number of eggs in each figure.
At this point, the goal is that students are able to articulate this visual pattern growing at a rate of 3 eggs per day after beginning with a single egg.
Some students might articulate Day 8 as something like:
8 rows of 3 eggs forming a rectangle with a single egg off of the bottom left corner of the rectangle.
Of course, writing out an algebraic expression for this pattern (i.e.: 1 + 3d) or an algebraic equation (i.e.: N = 1 + 3d) can be helpful.
## Want to Explore These Concepts & Skills Further?
Three (3) additional number talk prompts are available in Day 2 of the Bird Migration problem based math unit that you can dive into now.
Why not start from the beginning of this contextual 8-day unit of real world lessons from the Make Math Moments Problem Based Units page.
Did you use this in your classroom or at home? How’d it go? Post in the comments!
Math IS Visual. Let’s teach it that way. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Decimal Rounding and Division
## Use digit following required place value for rounding
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Decimal Rounding and Division
Vincent is looking at a cupcake recipe. The recipe makes a dozen cupcakes, but he only wants to make 4 cupcakes. He knows that 4 out of 12 is also one-third. If he divides the ingredients in the recipe by 3, he will have enough ingredients to make just 4 cupcakes. The recipe calls for 215 grams of sugar and his kitchen scale only measures grams to the tenths place. How much sugar should Vincent measure out to make a third of the recipe?
In this concept, you will learn to divide and round decimals.
### Rounding Decimals When Dividing
You can divide decimals by whole numbers and use zero placeholders to find the most accurate quotient. Some quotients are so long that it is difficult to decipher the value of the decimal. Rounding the decimal number to a decimal place value would make the number easier to evaluate.
Remember to look at the number to the right of the place value you are rounding to. If the number is less than 5, round the number down. If the number is equal to or greater than 5, round the number up.
Round 2.1046891425 to the nearest hundredths place. The number to the right of the hundredths place is 4. Round the number down.
2.10468914252.10\begin{align*}\begin{array}{rcl} && 2.1\underline{0}46891425 \approx 2.10\\ && \quad \quad \uparrow \end{array}\end{align*}
Round 2.1046891425 to the nearest ten-thousandths place. The number to the right of the ten-thousandths place is 8. Round the number up.
2.10468914252.1047\begin{align*}\begin{array}{rcl} && 2.104\underline{6}891425 \approx 2.1047\\ && \qquad \quad \uparrow \end{array}\end{align*}
Here is a decimal division problem. Round the quotient to the nearest ten-thousandths place.
1.265÷4\begin{align*}1.265 \div 4\end{align*}
First, divide to find the quotient. Remember to bring up the decimal point in the same location as the dividend.
4)1.26500¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0.31625 12 06 425 24 10 8 20 20 0\begin{align*}\begin{array}{rcl} && \overset{\ \ \ 0.31625}{4 \overline{ ) {1.265{\color{red}00}}}}\\ && \ \underline{-12}\\ && \quad \ \ 06\\ && \quad \ \underline{-4}\\ && \quad \quad 25 \\ && \quad \ \underline{-24}\\ && \quad \quad\ \ 1{\color{red}0} \\ && \quad \ \ \ \ \ \underline{-8}\\ && \quad \quad \ \ \ \ 2{\color{red}0} \\ && \quad \quad \ \underline{-20}\\ && \qquad \quad \ 0 \end{array}\end{align*}
Then, take the quotient and round it to the nearest ten-thousandths place. Look at the digit to the right of the ten-thousandths; it is 5. Round the number up.
0.316250.3163\begin{align*}\begin{array}{rcl} &&0.316\underline{2}5 \approx 0.3163\\ && \qquad \quad \uparrow \end{array}\end{align*}
The quotient of 1.265 divided by 4 rounded to the nearest ten-thousandths place is 0.3163.
Keep in mind some decimal quotients can be quite long.
Here is a division problem with a very long quotient. Round the quotient to thousandths.
13.87÷7=1.98142857142857131.98142857142857131.981 \begin{align*}\begin{array}{rcl} && 13.87 \div 7 = 1.9814285714285713\\ && 1.98\underline{1}4285714285713 \approx 1.981\\ && \qquad \ \uparrow \end{array}\end{align*}
The quotient of 13.87 divided by 7, rounded to the thousandths is 1.981
If you know you will round the quotient, you can stop dividing after you’ve found the digit to the right of the place value you are rounding to. Take for example the problem above. Stop dividing once you’ve found the digit in the ten-thousandths place. The digits that come afterwards will not affect the rounding of the decimal number.
7)13.8700¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 1.9814 7 68 63 5756 10 7 30 282\begin{align*}\begin{array}{rcl} && \overset{\ \ \ 1.98 \underline{1}4}{7 \overline{ ) {13.8700}}}\\ && \ \ \underline{-7}\\ && \quad \ 68\\ && \ \ \underline{-63}\\ && \quad \ \ \ 57 \\ && \quad \underline{-56}\\ && \quad \quad \ 10 \\ && \quad \ \ \ \ \underline{-7}\\ && \quad \quad \ \ \ 30 \\ && \quad \ \ \ \ \underline{-28}\\ && \qquad \quad 2 \end{array}\end{align*}
Round 1.9814 to the thousandths place.
1.98141.981 \begin{align*}\begin{array}{rcl} && 1.98\underline{1}4 \approx 1.981 \\ && \qquad \ \ \uparrow \end{array}\end{align*}
The answer is the same, 1.981.
### Examples
#### Example 1
Earlier, you were given a problem about Vincent making cupcakes.
He is trying to make a third of a cupcake recipe by dividing the ingredients by 3. Since his kitchen scale only measures to a tenths of a gram, find a third of 215 grams of sugar, rounded to the tenths place.
First, divide to find the quotient. You can stop at the hundredths place.
3)215.00¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 71.66 21 05 3 20 18 20 18 2\begin{align*}\begin{array}{rcl} && \overset{\ \ \ 71. \underline{6}6}{3 \overline{ ) {215.{\color{red}00}}}}\\ && \ \underline{-21}\\ && \quad \ \ 05\\ && \ \ \ \ \ \underline{-3}\\ && \quad \ \ \ \ 2{\color{red}0} \\ && \quad \ \underline{-18}\\ && \quad \quad \ \ 2{\color{red}0} \\ && \quad \ \ \ \underline{-18}\\ && \qquad \ \ \ 2 \end{array}\end{align*}
Then, round the quotient to the tenths place.
71.6671.7 \begin{align*}\begin{array}{rcl} && 71.\underline{6}6 \approx 71.7\\ && \quad \ \ \ \uparrow \end{array}\end{align*}
#### Example 2
Find the quotient and round it to the nearest thousandth.
0.45622÷4\begin{align*}0.45622 \div 4\end{align*}
First, divide to find the quotient. Add zero placeholders if necessary. Remember to bring up the decimal point in the same location as the dividend. Divide all the way or you can stop dividing once you’ve found the digit in the ten-thousandths place.
4)0.456220¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0.114055 4 05 4 16 16 02 0 22 20 20200\begin{align*}\begin{array}{rcl} && \overset{\ \ \ 0.114055}{4 \overline{ ) {0.45622{\color{red}0}}}}\\ && \ \ \ \underline{-4}\\ && \quad \ \ 05\\ && \quad \ \underline{-4}\\ && \quad \quad \ 16 \\ && \quad \ \ \underline{-16}\\ && \quad \quad \ \ \ 02 \\ && \quad \quad \ \ \underline{-0}\\ && \quad \quad \quad \ 22 \\ && \quad \quad \ \ \underline{-20}\\ && \quad \quad \quad \ \ \ 2{\color{red}0} \\ && \quad \quad \quad \underline{-20}\\ && \qquad \qquad 0 \end{array}\end{align*}
OR
4)0.45622¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0.1140 4 05 4 16 16 02 0 2\begin{align*}\begin{array}{rcl} && \overset{\ \ \ 0.1140}{4 \overline{ ) {0.45622}}}\\ && \ \ \ \ \underline{-4}\\ && \quad \ \ \ 05\\ && \quad \ \ \underline{-4}\\ && \quad \quad \ 16 \\ && \quad \ \ \underline{-16}\\ && \quad \quad \ \ \ 02 \\ && \quad \quad \ \ \underline{-0}\\ && \qquad \ \ \ \ 2 \end{array}\end{align*}
Then, round the quotient to the thousandths place. Look at the digit to the right in the ten-thousandths place; it is 0. Round the number down.
0.1140550.114OR0.11400.114 \begin{align*}\begin{array}{rcl} && 0.11\underline{4}055 \approx 0.114 \quad \text{OR} \quad 0.11\underline{4}0 \approx 0.114\\ && \qquad \ \uparrow \qquad \qquad \qquad \qquad \qquad \ \uparrow \end{array}\end{align*}
The quotient of 0.45622 divided by 4 rounded to the thousandths place is 0.114. Both methods will bring you to the same conclusion. The method on the right may be quicker if you only need a rounded quotient.
Find the quotient rounded to the nearest thousandths for the following problems.
#### Example 3
0.51296÷2=\begin{align*}0.51296 \div 2 = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
First, divide to find the quotient.
2)0.51296¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0.25648 4 11 10 12 12 09 8 16 16 0\begin{align*}\begin{array}{rcl} && \overset{\ \ \ 0.25648}{2 \overline{ ) {0.51296}}}\\ && \ \ \ \underline{-4}\\ && \quad \ \ 11\\ && \ \ \ \underline{-10}\\ && \quad \ \ \ \ 12 \\ && \quad \ \underline{-12}\\ && \quad \quad \ \ 09 \\ && \quad \ \ \ \ \ \underline{-8}\\ && \quad \quad \ \ \ \ 16\\ && \quad \ \ \ \ \ \underline{-16}\\ && \qquad \ \ \ \ \ 0 \end{array}\end{align*}
Then, round the quotient to the thousandths place.
0.256480.256 \begin{align*}\begin{array}{rcl} && 0.25\underline{6}48\approx 0.256\\ && \qquad \ \ \uparrow \end{array}\end{align*}
The quotient of 0.51296 divided by 2 is 0.256.
#### Example 4
10.0767÷3=\begin{align*}10.0767 \div 3 = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
First, divide to find the quotient.
3)10.0767¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 3.3589 9 10 9 1715 26 24 27 27 0\begin{align*}\begin{array}{rcl} && \overset{\ \ \ 3.3589}{3 \overline{ ) {10.0767}}}\\ && \ \ \underline{-9}\\ && \quad \ 10\\ && \ \ \ \ \underline{-9}\\ && \quad \ \ \ 17 \\ && \quad \underline{-15}\\ && \quad \ \ \ \ \ 26 \\ && \quad \ \ \underline{-24}\\ && \quad \quad \ \ \ 27\\ && \quad \ \ \ \ \underline{-27}\\ && \qquad \ \ \ \ 0 \end{array}\end{align*}
Then, round the quotient to the thousandths place.
3.35893.359 \begin{align*}\begin{array}{rcl} && 3.35\underline{8}9 \approx 3.359\\ && \qquad \ \uparrow \end{array}\end{align*}
The quotient of 10.0767 divided by 3 is 3.359.
#### Example 5
0.48684÷2=\begin{align*}0.48684 \div 2 = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}
First, divide to find the quotient. You can stop at the ten-thousandths place.
2)0.48684¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0.2434 4 08 8 06 6 08 804\begin{align*}\begin{array}{rcl} && \overset{\ \ \ 0.24 \underline{3}4}{2 \overline{ ) {0.48684}}}\\ && \ \ \ \underline{-4}\\ && \quad \ \ 08\\ && \quad \ \underline{-8}\\ && \quad \ \ \ \ 06 \\ && \quad \ \ \ \underline{-6}\\ && \quad \quad \ \ 08 \\ && \quad \quad \ \underline{-8}\\ && \qquad \quad 04 \end{array}\end{align*}
Then, round the quotient to the thousandths place.
0.24340.243 \begin{align*}\begin{array}{rcl} && 0.24\underline{3}4 \approx 0.243\\ && \qquad \ \uparrow \end{array}\end{align*}
The quotient of 0.48684 divided by 2 rounded to the thousandths is 0.243.
### Review
Find each quotient rounded to the nearest thousandths.
1. 0.54686÷2\begin{align*}0.54686 \div 2\end{align*}
2. 0.84684÷2\begin{align*}0.84684 \div 2\end{align*}
3. 0.154586÷2\begin{align*}0.154586 \div 2\end{align*}
4. 0.34689÷3\begin{align*}0.34689 \div 3\end{align*}
5. 0.994683÷3\begin{align*}0.994683 \div 3\end{align*}
6. 0.154685÷5\begin{align*}0.154685 \div 5\end{align*}
7. 0.546860÷5\begin{align*}0.546860 \div 5\end{align*}
8. 0.25465÷5\begin{align*}0.25465 \div 5\end{align*}
9. 0.789003÷3\begin{align*}0.789003 \div 3\end{align*}
10. 0.18905÷5\begin{align*}0.18905 \div 5\end{align*}
11. 0.27799÷9\begin{align*}0.27799 \div 9\end{align*}
12. 0.354680÷10\begin{align*}0.354680 \div 10\end{align*}
13. 0.454686÷6\begin{align*}0.454686 \div 6\end{align*}
14. 0.954542÷2\begin{align*}0.954542 \div 2\end{align*}
15. 0.8546812÷4\begin{align*}0.8546812 \div 4\end{align*}
To see the Review answers, open this PDF file and look for section 4.12.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Divide
To divide is split evenly into groups. The result of a division operation is a quotient.
Dividend
In a division problem, the dividend is the number or expression that is being divided.
divisor
In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Quotient
The quotient is the result after two amounts have been divided. |
# SAT Math : How to use FOIL with Exponents
## Example Questions
### Example Question #1161 : Algebra
If , which of the following could be the value of ?
Explanation:
Take the square root of both sides.
Add 3 to both sides of each equation.
### Example Question #1162 : Algebra
Simplify:
Explanation:
= x3y3z3 + x2y + x0y0 + x2y
x3y3z3 + x2y + 1 + x2y
x3y3z3 + 2x2y + 1
### Example Question #3 : Exponents And The Distributive Property
Use the FOIL method to simplify the following expression:
Explanation:
Use the FOIL method to simplify the following expression:
Step 1: Expand the expression.
Step 2: FOIL
First:
Outside:
Inside:
Last:
Step 2: Sum the products.
### Example Question #1 : How To Use Foil With Exponents
Square the binomial.
Explanation:
We will need to FOIL.
First:
Inside:
Outside:
Last:
Sum all of the terms and simplify.
### Example Question #1 : Exponents And The Distributive Property
Which of the following is equivalent to 4c(3d)– 8c3d + 2(cd)4?
cd(54c * d– 4c+ c* d2)
2(54d– 4c+ 2c* d3)
2cd(54d2 – 4c+ c* d3)
2cd(54d2 – 4c+ c* d3)
Explanation:
First calculate each section to yield 4c(27d3) – 8c3d + 2c4d= 108cd– 8c3d + 2c4d4. Now let's factor out the greatest common factor of the three terms, 2cd, in order to get: 2cd(54d– 4c+ c3d3). |
# Pyramid Problem
### Pyramid Problem
Suppose we have a pyramid with square base(side = x) and height h. The volume of the pyramid is 1000 cm3. Find the value of x if the surface area is minimum.
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Math Tutor
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### Re: Pyramid Problem
The volume of a pyramid with square base of side length x and height h is $$\frac{x^2h}{3}$$. The surface consists of four triangles and the square base. Obviously the square base has area $$x^2$$. The area of each triangular face is the base length, x, times the altitude of each triangle measured along the perpendicular from the vertex of the pyramid perpendicular to the base. Drop a line from the vertex of the pyramid to the center of the base of the pyramid, then the line from there to the center of a side of the base, then a the line back to the vertex. That describes a right triangle with legs of length h and x/2. The altitude of a triangular side is the hypotenuse of that right triangle so has length $$\sqrt{h^2+ \frac{x^2}{4}}= 2\sqrt{4h^2+ x^2}$$. The area of each such triangle is "1/2 base times height" so is $$\frac{1}{2}x(2\sqrt{4h^2+ x^2}= x\sqrt{4h^2+ x^2}$$. There are 4 such triangles so the total surface area of the pyramid, including all such triangles and the base is $$x^2+ 4x\sqrt{4h^2+ x^2}$$.
So the problem is to find the value of x that, for fixed h, minimizes $$V= x^2+ 4x\sqrt{4h^2+ x^2}$$ subject to the constraint $$A= \frac{x^2h}{3}= 1000$$.
Guest
### Re: Pyramid Problem
A really good solution!
Math Tutor |
# Difference between revisions of "2014 AIME I Problems/Problem 9"
## Problem 9
Let $x_1 be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$. Find $x_2(x_1+x_3)$.
## Solution 1
Substituting $n$ for $2014$, we get $$\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2$$ $$= x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0$$ Noting that $nx^2 - 1$ factors as a difference of squares to $$(\sqrt{n}x - 1)(\sqrt{n}x+1)$$ we can factor the left side as $$(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))$$ This means that $\frac{1}{\sqrt{n}}$ is a root, and the other two roots are the roots of $x^2 - 2\sqrt{n}x - 2$. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to $2\sqrt{n}$, so the positive root must be greater than $2\sqrt{n}$ in order to produce this sum when added to a negative value. Since $0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}$ is clearly true, $x_2 = \frac{1}{\sqrt{2014}}$ and $x_1 + x_3 = 2\sqrt{2014}$. Multiplying these values together, we find that $x_2(x_1+x_3) = \boxed{002}$.
## Solution 2
From Vieta's formulae, we know that $$x_1x_2x_3 = \dfrac{-2}{\sqrt{2014}}$$ $$x_1 + x_2 + x_3 = \dfrac{4029}{\sqrt{2014}}$$ and $$x_1x_2 + x_2x_3 + x_1x_3 = 0$$ Thus, we know that $$x_2(x_1 + x_3) = -x_1x_3$$
Now consider the polynomial with roots $x_1x_2, x_2x_3,$ and $x_1x_3$. Expanding the polynomial $$(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)$$we get the polynomial $$x^3 - (x_1x_2 + x_2x_3 + x_1x_3)x^2 + (x_1x_2x_3)(x_1 + x_2 + x_3)x - (x_1x_2x_3)^2$$ Substituting the values obtained from Vieta's formulae, we find that this polynomial is $$x^3 - \dfrac{8058}{2014}x - \dfrac{4}{2014}$$ We know $x_1x_3$ is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to $$1007x^3 - 4029x - 2 = 0$$
Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the $x^3$ term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that $x = -2$ is a solution. Factoring it out, we get that $$1007x^3 - 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)$$ Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, $$x_1x_3 = -2$$ so $$-x_1x_3 = \boxed{002}$$and we're done.
## Solution 3
Observing the equation, we notice that the coefficient for the middle term $-4029$ is equal to $$-2{\sqrt{2014}}^2-1$$.
Also notice that the coefficient for the ${x^3}$ term is $\sqrt{2014}$. Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the $x$ term of the binomial would have a coefficient of $\sqrt{2014}$. Similarly, the $x$ term of the trinomial would also have a coefficient of $\sqrt{2014}$. The factored form of the expression would look something like the following: $$({\sqrt{2014}}x-a)(x^2-n{\sqrt{2014}}x-b)$$ where ${a, b,c}$ are all positive integers (because the ${x^2}$ term of the original expression is negative, and the constant term is positive), and $${ab=2}$$
Multiplying this expression out gives $${{\sqrt{2014}x^3-(2014n+a)x^2+(an{\sqrt{2014}}-b{\sqrt{2014}})x+ab}}$$ Equating this with the original expression gives $${2014n+a}=-4029$$ The only positive integer solutions of this expression is $(n, a)=(1, 2015)$ or $(2, 1)$. If $(n, a)=(1, 2015)$ then setting ${an{\sqrt{2014}}-b{\sqrt{2014}}}=0$ yields ${b=2015}$ and therefore ${ab=2015^2}$ which clearly isn't equal to $2$ as the constant term. Therefore, $(n, a)=(2, 1)$ and the factored form of the expression is: $$({\sqrt{2014}}x-1)(x^2-2{\sqrt{2014}}x-2)$$ Therefore, one of the three roots of the original expression is $${x=\dfrac{1}{\sqrt{2014}}}$$ Using the quadratic formula yields the other two roots as $${x={\sqrt{2014}}+{\sqrt{2016}}}$$ and $${x={\sqrt{2014}}-{\sqrt{2016}}}$$ Arranging the roots in ascending order (in the order $x_1), $${\sqrt{2014}}-{\sqrt{2016}}<\dfrac{1}{\sqrt{2014}}<{\sqrt{2014}}+{\sqrt{2016}}$$ Therefore, $$x_2(x_1+x_3)=\dfrac{1}{\sqrt{2014}}{2\sqrt{2014}}=\boxed{002}$$
## Solution 4
By Vieta's, we are seeking to find $x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\frac{2}{\sqrt{2014}x_2}$. Substitute $n=-x_1x_3$ and $x_2=\frac{2}{\sqrt{2014}n}$. Substituting this back into the original equation, we have $\frac{4}{1007n^3}-\frac{8058}{1007n^2}+2=0$, so $2n^3-\frac{8058}{1007}n+\frac{4}{1007}=2n^3-\frac{8058n-4}{1007}=0$. Hence, $8058n-4\equiv 2n-4 \equiv 0 \pmod{1007}$, and $n\equiv 2\pmod{1007}$. But since $n\le 999$ because it is our desired answer, the only possible value for $n$ is $\boxed{002}$ BEST PROOOFFFF Stormersyle & mathleticguyyy
## Solution 5
Let $x =\frac{y}{\sqrt{2014}}.$ The original equation simplifies to $\frac{y^3}{2014} -\frac{4029y^2}{2014}+2 = 0 \implies y^3 - 4029y^2 + 4028=0.$ Here we clearly see that $y=1$ is a root. Dividing $y-1$ from the sum we find that $(y-1)(y^2-4028y-4028)=0.$ From simple bounding we see that $y=1$ is the middle root. Therefore $x_{2}(x_{1}+x_{3}) =\frac{1}{\sqrt{2014}} \cdot\frac{4028}{\sqrt{2014}} = \boxed{002}.$
## Solution 6
$\sqrt{2014}$ occurs multiple times, so let k = $\sqrt{2014}$.
The equation becomes $0 = kx^3 - (2k^2 + 1)x^2 + 2$. Since we want to relate k and x, we should solve for one of them. We can't solve for x, since that would require the cubic formula, so we solve for k, and express it in terms of a quadratic, and apply the quadratic formula.
We get the roots are:
$y = \frac{1}{x}$, and $y = \frac{x}{2} - \frac{1}{x}$.
In the first case, $x = \frac{1}{y} = \frac{1}{\sqrt{2014}}$.
In the second case, $x^2 - 2\sqrt{2014} - 2 = 0$. The solutions are $\sqrt{2014} \pm \sqrt{2016}$. The sum of these 2 solutions is $2 \sqrt{2014}$, and $\frac{1}{\sqrt{2014}}$ is the middle solution, and thus, $(x_1 + x_3) \cdot x_2 = 2$ |
Experimental Probability
Probability means the chances of a number of occurrences of an event. In simple language, it is the possibility that an event will occur or not. The concept of probability can be applied to some experiments like coin tossing, dice throwing, playing cards, etc. Experimental Probability is one of the interesting concepts of Probability. Before diving down into the definition, Let’s start understanding this concept through our daily life situations.
We all have heard typical monsoon forecasts like, “Kerala remains under high alert expecting heavy rains and winds as a result of cyclone Burevi” and similar other headlines, right? But have you ever thought that how these expectations sometimes turn into reality? The reason behind the chances, expectations, doubts, and forecasts is Probability. Probability in simple meaning gives us the predictions of an event that may or may not be happened based on our past experiences. And these Past experience is based upon the experiment of events.
What is Probability?
The branch of mathematics that tells us about the likelihood of the occurrence of any event is the probability. Probability tells us about the chances of happening an event. The probability of any element that is sure to occur is One(1) whereas the probability of any impossible event is Zero(0). The probability of all the elements ranges between 0 to 1.
There are two ways of studying probability that are
• Experimental Probability
• Theoretical Probability
Now let’s learn about both in detail.
What is Experimental Probability?
Experimental probability is a type of probability that is calculated by conducting an actual experiment or by performing a series of trials to observe the occurrence of an event. It is also known as empirical probability.
To calculate experimental probability, you need to conduct an experiment by repeating the event multiple times and observing the outcomes. Then, you can find the probability of the event occurring by dividing the number of times the event occurred by the total number of trials.
Formula for Experimental Probability
The experimental Probability for Event A can be calculated as follows:
P(E) = (Number of times an event occur in an experiment) / (Total number of Trials)
Examples of Experimental Probability
Now, as we learn the formula, let’s put this formula in our coin-tossing case. If we tossed a coin 10 times and recorded a head 4 times and a tail 6 times then the Probability of Occurrence of Head on tossing a coin:
P(H) = 4/10
Similarly, the Probability of Occurrence of Tails on tossing a coin:
P(T) = 6/10
What is Theoretical Probability?
Theoretical Probability deals with assumptions in order to avoid unfeasible or expensive repetition experiments. The theoretical Probability for an Event A can be calculated as follows:
P(A) = Number of outcomes favorable to Event A / Number of all possible outcomes
Now, as we learn the formula, let’s put this formula in our coin-tossing case. In tossing a coin, there are two outcomes: Head or Tail.
Hence, The Probability of occurrence of Head on tossing a coin is
P(H) = 1/2
Similarly, The Probability of the occurrence of a Tail on tossing a coin is
P(T) = 1/2
Experimental Probability vs Theoretical Probability
There are some key differences between Experimental and Theoretical Probability, some of which are as follows:
Solved Examples of Experimental Probability
Example 1. Let’s take an example of tossing a coin, tossing it 40 times, and recording the observations. By using the formula, we can find the experimental probability for heads and tails as shown in the below table.
The formula for experimental probability:
P(H) = Number of Heads ÷ Total Number of Trials = 16 ÷ 40 = 0.4
Similarly,
P(H) = Number of Tails ÷ Total Number of Trials = 24 ÷ 40 = 0.6
P(H) + P(T) = 0.6 + 0.4 = 1
Note: Repeat this experiment for ‘n’ times and then you will find that the number of times increases, the fraction of experimental probability comes closer to 0.5. Thus if we add P(H) and P(T), we will get 0.6 + 0.4 = 1 which means P(H) and P(T) is the only possible outcomes.
Example 2. A manufacturer makes 50,000 cell phones every month. After inspecting 1000 phones, the manufacturer found that 30 phones are defective. What is the probability that you will buy a phone that is defective? Predict how many phones will be defective next month.
Experimental Probability = 30/1000 = 0.03
0.03 = (3/100) × 100 = 3%
The probability that you will buy a defective phone is 3%
⇒ Number of defective phones next month = 3% × 50000
⇒ Number of defective phones next month = 0.03 × 50000
⇒ Number of defective phones next month = 1500
Example 3. There are about 320 million people living in the USA. Pretend that a survey of 1 million people revealed that 300,000 people think that all cars should be electric. What is the probability that someone chosen randomly does not like the electric car? How many people like electric cars?
Now the number of people who do not like electric cars is 1000000 – 300000 = 700000
Experimental Probability = 700000/1000000 = 0.7
And, 0.7 = (7/10) × 100 = 70%
The probability that someone chose randomly does not like the electric car is 70%
The probability that someone like electric cars is 300000/1000000 = 0.3
Let x be the number of people who love electric cars
⇒ x = 0.3 × 320 million
⇒ x = 96 million
The number of people who love electric cars is 96 million.
FAQs on Experimental Probability
Q1: Define experimental probability.
Probability of an event based on an actual trail in physical world is called experimental probability.
Q2: How is Experimental Probability calculated?
Experimental Probability is calculated using the following formula:
P(E) = (Number of trials taken in which event A happened) / Total number of trials
Q3: Can Experimental Probability be used to predict future outcomes?
No, experimental probability can’t be used to predict future outcomes as it only achives the theorectical value when the trails becomes infinity.
Q4: How is Experimental Probability different from Theoretical Probability?
Theoretical probability is the probability of an event based on mathematical calculations and assumptions, whereas experimental probability is based on actual experiments or trials.
Q5: What are some Limitations of Experimental Probability?
There are some limitation of experimental probability, which are as follows:
• Experimental probability can be influenced by various factors, such as the sample size, the selection process, and the conditions of the experiment.
• The number of trials conducted may not be sufficient to establish a reliable pattern, and the results may be subject to random variation.
• Experimental probability is also limited to the specific conditions of the experiment and may not be applicable in other contexts.
Q6: Can Experimental Probability of an event be a negative number if not why?
As experimental probability is given by:
P(E) = Number of trials taken in which event A happened/Total number of trials
Thus, it can’t be negative as both number are count of something and counting numbers are 1, 2, 3, 4, …. and they are never negative. |
# Determine whether the given problem is an equation or an expression. If it is an
Determine whether the given problem is an equation or an expression. If it is an equation, then solve. If it is an expression, then simplify. y/3 + (y - 5)/10 = (4y + 3)/5
• Questions are typically answered in as fast as 30 minutes
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Step 1
Given:
$$\displaystyle{\frac{{{y}}}{{{3}}}}+{\frac{{{y}-{5}}}{{{10}}}}={\frac{{{4}{y}+{3}}}{{{5}}}}$$
It contains an equal sign therefore it is an equation.
$$\displaystyle{\frac{{{y}}}{{{3}}}}+{\frac{{{y}-{5}}}{{{10}}}}={\frac{{{4}{y}+{3}}}{{{5}}}}$$
$$\displaystyle\Rightarrow{\frac{{{10}{y}+{3}{\left({y}-{5}\right)}}}{{{30}}}}={\frac{{{4}{y}+{3}}}{{{5}}}}$$
$$\displaystyle\Rightarrow{\frac{{{10}{y}+{3}{y}-{15}}}{{{30}}}}={\frac{{{4}{y}_{{3}}}}{{{5}}}}$$
$$\displaystyle\Rightarrow{\frac{{{13}{y}-{15}}}{{{30}}}}={\frac{{{4}{y}+{3}}}{{{5}}}}$$
$$\displaystyle\Rightarrow{\frac{{{13}{y}-{15}}}{{{30}}}}={\frac{{{6}{\left({4}{y}+{3}\right)}}}{{{30}}}}$$
$$\displaystyle\Rightarrow{13}{y}-{15}={24}{y}+{18}$$
$$\displaystyle\Rightarrow-{33}={11}{y}$$
$$\displaystyle\Rightarrow{y}=-{3}$$
Step 2
Therefore,
y=-3 |
# Math Snap
## $\int\left(x^{e}+e^{x}+e^{e}\right) d x=$ Select one:
#### STEP 1
Assumptions 1. We need to integrate the function $f(x) = x^{e} + e^{x} + e^{e}$ with respect to $x$. 2. $e$ is the base of the natural logarithm and is a constant approximately equal to 2.71828. 3. The integral of a sum of functions is the sum of the integrals of the functions. 4. The integral of a constant is the constant multiplied by the variable of integration.
#### STEP 2
We will use the linearity of the integral to split the integral of the sum into the sum of integrals. $\int\left(x^{e}+e^{x}+e^{e}\right) dx = \int x^{e} dx + \int e^{x} dx + \int e^{e} dx$
#### STEP 3
First, we integrate $x^{e}$ with respect to $x$. Since $e$ is a constant, we use the power rule for integration: $\int x^{e} dx = \frac{x^{e+1}}{e+1} + C_1$ where $C_1$ is the constant of integration.
#### STEP 4
Next, we integrate $e^{x}$ with respect to $x$. Since the base of the exponent is $e$, the integral is: $\int e^{x} dx = e^{x} + C_2$ where $C_2$ is another constant of integration.
#### STEP 5
Finally, we integrate $e^{e}$ with respect to $x$. Since $e^{e}$ is a constant, its integral is: $\int e^{e} dx = e^{e}x + C_3$ where $C_3$ is yet another constant of integration.
#### STEP 6
Now we can combine the results from steps 3, 4, and 5 to write the integral of the original function: $\int\left(x^{e}+e^{x}+e^{e}\right) dx = \frac{x^{e+1}}{e+1} + e^{x} + e^{e}x + C$ where $C$ is the combined constant of integration, $C = C_1 + C_2 + C_3$.
##### SOLUTION
We can simplify the expression by combining the constants of integration into a single constant: $\int\left(x^{e}+e^{x}+e^{e}\right) dx = \frac{x^{e+1}}{e+1} + e^{x} + e^{e}x + C$ This is the final form of the integral. The solution to the integral is: $\frac{x^{e+1}}{e+1} + e^{x} + e^{e}x + C$ |
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# Division of Polynomials
## Using long division to divide polynomials
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Practice Division of Polynomials
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Polynomial Long Division and Synthetic Division
While you may be experienced in factoring, there will always be polynomials that do not readily factor using basic or advanced techniques. How can you identify the roots of these polynomials?
#### Watch This
http://www.youtube.com/watch?v=brpNxPAkv1c James Sousa: Dividing Polynomials-Long Division
http://www.youtube.com/watch?v=5dBAdzl2Mns James Sousa: Dividing Polynomials-Synthetic Division
#### Guidance
There are numerous theorems that point out relationships between polynomials and their factors. For example there is a theorem that a polynomial of degree n\begin{align*}n\end{align*} must have exactly n\begin{align*}n\end{align*} solutions/factors that may or may not be real numbers. The Rational Root Theorem and the Remainder Theorem are two theorems that are particularly useful starting places when manipulating polynomials.
• The Rational Root Theorem states that in a polynomial, every rational solution can be written as a reduced fraction (x=pq)\begin{align*}\left(x=\frac{p}{q}\right)\end{align*}, where p\begin{align*}p\end{align*} is an integer factor of the constant term and q\begin{align*}q\end{align*} is an integer factor of the leading coefficient. Example A shows how all the possible rational solutions can be listed using the Rational Root Theorem.
• The Remainder Theorem states that the remainder of a polynomial f(x)\begin{align*}f(x)\end{align*} divided by a linear divisor (xa)\begin{align*}(x-a)\end{align*} is equal to f(a)\begin{align*}f(a)\end{align*}. The Remainder Theorem is only useful after you have performed polynomial long division because you are usually never given the divisor and the remainder to start. The main purpose of the Remainder Theorem in this setting is a means of double checking your application of polynomial long division. Example B shows how the Remainder Theorem is used.
Polynomial long division is identical to regular long division. Synthetic division is a condensed version of regular long division where only the coefficients are kept track of. In Example B polynomial long division is used and in Example C synthetic long division is used.
Example A
Identify all possible rational solutions of the following polynomial using the Rational Root Theorem.
12x1891x17+x16++2x214x+5=0\begin{align*}12x^{18}-91x^{17}+x^{16}+\cdots+2x^2-14x+5=0\end{align*}
Solution: The integer factors of 5 are 1, 5. The integer factors of 12 are 1, 2, 3, 4, 6 and 12. Since pairs of factors could both be negative, remember to include ±\begin{align*}\pm\end{align*}
±pq=11,12,13,14,16,112,51,52,53,54,56,512\begin{align*}\pm \frac{p}{q}=\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{12}, \frac{5}{1}, \frac{5}{2}, \frac{5}{3}, \frac{5}{4}, \frac{5}{6}, \frac{5}{12}\end{align*}
While narrowing the possible solutions down to 24 possible rational answers may not seem like a big improvement, it surely is. This is especially true considering there are only a handful of integer solutions. If this question required you to find a solution, then the Rational Root Theorem would give you a great starting place.
Example B
Use Polynomial Long Division to divide:
x3+2x25x+7x3\begin{align*}\frac{x^3+2x^2-5x+7}{x-3}\end{align*}
Solution: First note that it is clear that 3 is not a root of the polynomial because of the Rational Root Theorem and so there will definitely be a remainder. Start a polynomial long division question by writing the problem like a long division problem with regular numbers:
¯x3)x3+2x25x+7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{}{x-3 \overline { ) {x^3+2x^2-5x+7}}}\end{align*}
Just like with regular numbers ask yourself “how many times does x\begin{align*}x\end{align*} go into x3\begin{align*}x^3\end{align*}?” which in this case is x2\begin{align*}x^2\end{align*}.
x3)x3+2x25x+7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x2\begin{align*}\overset{x^2}{x-3 \overline{ ) {x^3+2x^2-5x+7}}}\end{align*}
Now multiply the x2\begin{align*}x^2\end{align*} by x3\begin{align*}x-3\end{align*} and copy below. Remember to subtract the entire quantity.
x3)x3+2x25x+7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x2 (x33x2)
Combine the rows, bring down the next number and repeat.
x3)x3+2x25x+7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x2+5x+10 (x33x2) 5x25x (5x215x) 10x+7(10x30)37
The number 37 is the remainder. There are two things to think about at this point. First, interpret in an equation:
x3+2x25x+7x3=(x2+5x+10)+37x3\begin{align*}\frac{x^3+2x^2-5x+7}{x-3}=(x^2+5x+10)+\frac{37}{x-3}\end{align*}
Second, check your result with the Remainder Theorem which states that the original function evaluated at 3 must be 37. Notice the notation indicating to substitute 3 in for x\begin{align*}x\end{align*}
(x3+2x25x+7)|x=3=33+23253+7=27+1815+7=37\begin{align*}(x^3+2x^2-5x+7)|_{x=3}=3^3+2 \cdot 3^2-5 \cdot 3+7=27+18-15+7=37\end{align*}
Example C
Use Synthetic Division to divide the same rational expression as the previous example.
Solution: Synthetic division is mostly used when the leading coefficients of the numerator and denominator are equal to 1 and the divisor is a first degree binomial.
x3+2x25x+7x3\begin{align*}\frac{x^3+2x^2-5x+7}{x-3}\end{align*}
Instead of continually writing and rewriting the x\begin{align*}x\end{align*} symbols, synthetic division relies on an ordered spacing.
+3)1 2 5 7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯\begin{align*}\overset{}{+3 \overline{ ) {1 \ \ 2 \ -5 \ \ 7}}} \end{align*}
Notice how only the coefficients for the denominator are used and the divisor includes a positive three rather than a negative three. The first coefficient is brought down and then multiplied by the three to produce the value which goes beneath the 2.
+3)1 2 5 7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 31
Next the new column is added. 2+3=5\begin{align*}2+3=5\end{align*}, which goes beneath the 2nd\begin{align*}2^{nd}\end{align*} column. Now, multiply 5+3=15\begin{align*}5 \cdot +3=15\end{align*}, which goes underneath the -5 in the 3rd\begin{align*}3^{rd}\end{align*} column. And the process repeats…
+3)1 2 5 7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 315 30 1 510 37
The last number, 37, is the remainder. The three other numbers represent the quadratic that is identical to the solution to Example B.
1x2+5x+10\begin{align*}1x^2+5x+10\end{align*}
Concept Problem Revisited
Identifying roots of polynomials by hand can be tricky business. The best way to identify roots is to use the rational root theorem to quickly identify likely candidates for solutions and then use synthetic or polynomial long division to quickly and effectively test them to see if their remainders are truly zero.
#### Vocabulary
Polynomial long division is a procedure with rules identical to regular long division. The only difference is the dividend and divisor are polynomials.
Synthetic division is an abbreviated version of polynomial long division where only coefficients are used.
#### Guided Practice
1. Divide the following polynomials.
x3+2x24x+8x2\begin{align*}\frac{x^3+2x^2-4x+8}{x-2}\end{align*}
2. Completely factor the following polynomial.
x4+6x3+3x226x24\begin{align*}x^4+6x^3+3x^2-26x-24\end{align*}
3. Divide the following polynomials.
3x52x2+10x5x1\begin{align*}\frac{3x^5-2x^2+10x-5}{x-1}\end{align*}
Answers:
1. x3+2x24x+8x2=x2+4x+4+16x2\begin{align*}\frac{x^3+2x^2-4x+8}{x-2}=x^2+4x+4+\frac{16}{x-2}\end{align*}
2. Notice that possible roots are ±1,2,3,4,6,8,24\begin{align*}\pm 1, 2, 3, 4, 6, 8, 24\end{align*}. Of these 14 possibilities, four will yield a remainder of zero. When you find one, repeat the process.
x4+6x3+3x226x24=(x+1)(x3+5x22x4)=(x+1)(x2)(x2+7x+12)=(x+1)(x2)(x+3)(x+4)
3. 3x52x2+10x5x1=3x4+3x3+3x2+x+11+6x1\begin{align*}\frac{3x^5-2x^2+10x-5}{x-1}=3x^4+3x^3+3x^2+x+11+\frac{6}{x-1}\end{align*}
#### Practice
Identify all possible rational solutions of the following polynomials using the Rational Root Theorem.
1. 15x1412x13+x12++2x25x+5=0\begin{align*}15x^{14}-12x^{13}+x^{12}+ \cdots+2x^2-5x+5=0\end{align*}
2. 18x11+42x10+x9++x23x+7=0\begin{align*}18x^{11}+42x^{10}+x^9+\cdots+x^2-3x+7=0\end{align*}
3. 12x16+11x15+3x14++6x22x+11=0\begin{align*}12x^{16}+11x^{15}+3x^{14}+\cdots+6x^2-2x+11=0\end{align*}
4. 14x77x6+x5++x2+6x+3=0\begin{align*}14x^7-7x^6+x^5+\cdots+x^2+6x+3=0\end{align*}
5. 9x910x8+3x7++4x22x+2=0\begin{align*}9x^9-10x^8+3x^7+\cdots+4x^2-2x+2=0\end{align*}
Completely factor the following polynomials.
6. 2x4x321x226x8\begin{align*}2x^4-x^3-21x^2-26x-8\end{align*}
7. x4+7x3+5x231x30\begin{align*}x^4+7x^3+5x^2-31x-30\end{align*}
8. x4+3x38x212x+16\begin{align*}x^4+3x^3-8x^2-12x+16\end{align*}
9. \begin{align*}4x^4+19x^3-48x^2-117x-54\end{align*}
10. \begin{align*}2x^4+17x^3-8x^2-173x+210\end{align*}
Divide the following polynomials.
11. \begin{align*}\frac{x^4+7x^3+5x^2-31x-30}{x+4}\end{align*}
12. \begin{align*}\frac{x^4+7x^3+5x^2-31x-30}{x+2}\end{align*}
13. \begin{align*}\frac{x^4+3x^3-8x^2-12x+16}{x+3}\end{align*}
14. \begin{align*}\frac{2x^4-x^3-21x^2-26x-8}{x^3-x^2-10x-8}\end{align*}
15. \begin{align*}\frac{x^4+8x^3+3x^2-32x-28}{x^3+10x^2+23x+14}\end{align*}
### Vocabulary Language: English
Denominator
Denominator
The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
Dividend
Dividend
In a division problem, the dividend is the number or expression that is being divided.
divisor
divisor
In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Polynomial long division
Polynomial long division
Polynomial long division is the standard method of long division, applied to the division of polynomials.
Rational Expression
Rational Expression
A rational expression is a fraction with polynomials in the numerator and the denominator.
Rational Root Theorem
Rational Root Theorem
The rational root theorem states that for a polynomial, $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_n, a_{n-1}, \cdots a_0$ are integers, the rational roots can be determined from the factors of $a_n$ and $a_0$. More specifically, if $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$, then all the rational factors will have the form $\pm \frac{p}{q}$.
Remainder Theorem
Remainder Theorem
The remainder theorem states that if $f(k) = r$, then $r$ is the remainder when dividing $f(x)$ by $(x - k)$.
Synthetic Division
Synthetic Division
Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.
Please wait...
Please wait... |
# CLASS-11RELATION & FUNCTION - TERMS VALUE & IMAGE OF A FUNCTION
The Terms Value and Image of a Function
If, f is a function and if ‘x’ is any element of the domain, we use the symbol f(x) to denote the object which f associates with ‘x’. f(x) is the functional value at ‘x’. The symbol f(x) is read as “the value of ‘f ’ at x”, or “f at x”, or “f of x”. f(x) is called the image of x and x is called the pre-image of f(x). We could also designate the function as f : A → B or f : x → f(x) [read “f takes x into f(x)”]
For Example
(i) Let f be {(0, 1), (1, 2), (2, 3)}. It is a function whose domain is {0, 1, 2} and range is {1, 2, 3}. Here f(0) = 1, f(1) = 2 and f(2) = 3
Euler invented a symbolic way to write f : → f(x) as y = f(x) while is read “y equals f of x
(ii) Consider the two sets for the functions f(x) = x²
3 9
A = {-2, 2, ------, 0}, B = {4, ------, 0}
4 16
3 9
The set {-2, 2, ------, 0} is the domain and the set {4, ------, 0} is
4 16
the range, 4 is the image of
9 3
each of -2 and 2, ------ is the image of ----- and 0 is the image of 0
16 4
If we denote the function by f, then –
3 9
f(-2) = 4, f(2) = 4, f(-----) = ------, f(0) = 0
4 16
If, f(x) is a given function of x and if ‘a’ is in its domain then by f(a) is meant the number obtained by replacing x by ‘a’ in f(x) or the value assumed by f(x) when x = a.
Illustration => If f(x) = x²- 5x + 3, then f(1) = (1)²- 5(1) + 3 = -1
f(-3) = (-3)²- 5(-3) + 3, then f(1)
= 9 + 15 + 3 = 27;
f(a) = a²- 5a +3
The domain of a function may be restricted by context. For example, the domain of the area function given by A = πr² only allows the radius r to be positive. When we define a function y = f(x) with a formula and the domain is not stated explicitly or restricted by context, the domain is assumed to be the largest set of real x-values for which the formula gives real y-values, the so-called natural domain. If we want to restrict the domain in someway, we must say so. The domain of y = x² is the entire set of real numbers. To restrict the function to, say, positive values of x, we would write “y = x², x > 0
Changing the domain to which we apply a formula usually changes the range as well. The range of y = x² is (0, ∞). The range of y = x², x ≥ 2, is the set of all numbers obtained by squaring numbers greater than or equal to 2. In set notation, the range is {x² ǀ x ≥ 2} or {y ǀ y ≥ 4} or [4, ∞).
When the range of a function is a set of real numbers, the function is said to be real valued. The domains and ranges of many real valued functions of a real variable are intervals or combinations of intervals. The intervals may be open, closed or half open, and may be finite or infinite. |
# 9′s complement and 10′s complement| Subtraction
Before knowing about 9's complement and 10's complement we should know why they are used and why their concept came into existence. The complements are used to make the arithmetic operations in digital system easier. In this article we will discuss about the following topics
(1) 9s complement
(2) 10s complement
(3) 9s complement subtraction
(4) 10s complement subtraction
Now first of all let us know what 9's complement is and how it is done. To obtain the 9,s complement of any number we have to subtract the number with (10n - 1) where n = number of digits in the number, or in a simpler manner we have to divide each digit of the given decimal number with 9. The table given below will explain the 9's complement more easily
Decimal digit 9s complement
0 9
1 8
2 7
3 6
4 5
5 4
6 3
7 2
8 1
9 0
Now coming to 10's complement, it is relatively easy to find out the 10's complement after finding out the 9,s complement of that number. We have to add 1 with the 9,s complement of any number to obtain the desired 10's complement of that number. Or if we want to find out the 10's complement directly, we can do it by following the following formula, (10n - number), where n = number of digits in the number. An example is given below to illustrate the concept of obtaining 10’s complement
Let us take a decimal number 456, 9's complement of this number will be
10's complement of this no
9's complement subtraction
We will understand this method of subtraction via an example
A = 215
B = 155
We want to find out A-B by 9's complement subtraction method
First we have to find out 9’s complement of B
Now we have to add 9’s complement of B to A
The left most bit of the result is called carry and is added back to the part of the result without it
Another different type of example is given
A = 4567
B = 1234
We need to find out A - B
9's complement of B
8765
Adding 9's complement of B with A
Adding the carry with the result we get
3333
Now the answer is - 3333
NB if there is no carry the answer will be – (9’s complement of the answer)
Subtraction by 10's complement
Again we will show the procedure by an example
Taking the same data
A = 215
B = 155
10's complement of B = 845
Adding 10’s complement of B to A
In this case the carry is omitted |
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### Course: Integrated math 3>Unit 8
Lesson 3: Solving general triangles
# Laws of sines and cosines review
Review the law of sines and the law of cosines, and use them to solve problems with any triangle.
## Law of sines
$\frac{a}{\mathrm{sin}\left(\alpha \right)}=\frac{b}{\mathrm{sin}\left(\beta \right)}=\frac{c}{\mathrm{sin}\left(\gamma \right)}$
## Law of cosines
${c}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}\left(\gamma \right)$
## Practice set 1: Solving triangles using the law of sines
This law is useful for finding a missing angle when given an angle and two sides, or for finding a missing side when given two angles and one side.
### Example 1: Finding a missing side
Let's find $AC$ in the following triangle:
According to the law of sines, $\frac{AB}{\mathrm{sin}\left(\mathrm{\angle }C\right)}=\frac{AC}{\mathrm{sin}\left(\mathrm{\angle }B\right)}$. Now we can plug the values and solve:
$\begin{array}{rl}\frac{AB}{\mathrm{sin}\left(\mathrm{\angle }C\right)}& =\frac{AC}{\mathrm{sin}\left(\mathrm{\angle }B\right)}\\ \\ \frac{5}{\mathrm{sin}\left({33}^{\circ }\right)}& =\frac{AC}{\mathrm{sin}\left({67}^{\circ }\right)}\\ \\ \frac{5\mathrm{sin}\left({67}^{\circ }\right)}{\mathrm{sin}\left({33}^{\circ }\right)}& =AC\\ \\ 8.45& \approx AC\end{array}$
### Example 2: Finding a missing angle
Let's find $m\mathrm{\angle }A$ in the following triangle:
According to the law of sines, $\frac{BC}{\mathrm{sin}\left(\mathrm{\angle }A\right)}=\frac{AB}{\mathrm{sin}\left(\mathrm{\angle }C\right)}$. Now we can plug the values and solve:
$\begin{array}{rl}\frac{BC}{\mathrm{sin}\left(\mathrm{\angle }A\right)}& =\frac{AB}{\mathrm{sin}\left(\mathrm{\angle }C\right)}\\ \\ \frac{11}{\mathrm{sin}\left(\mathrm{\angle }A\right)}& =\frac{5}{\mathrm{sin}\left({25}^{\circ }\right)}\\ \\ 11\mathrm{sin}\left({25}^{\circ }\right)& =5\mathrm{sin}\left(\mathrm{\angle }A\right)\\ \\ \frac{11\mathrm{sin}\left({25}^{\circ }\right)}{5}& =\mathrm{sin}\left(\mathrm{\angle }A\right)\end{array}$
Evaluating using the calculator and rounding:
$m\mathrm{\angle }A={\mathrm{sin}}^{-1}\left(\frac{11\mathrm{sin}\left({25}^{\circ }\right)}{5}\right)\approx {68.4}^{\circ }$
Remember that if the missing angle is obtuse, we need to take ${180}^{\circ }$ and subtract what we got from the calculator.
Problem 1.1
$BC=$
Round to the nearest tenth.
Want to try more problems like this? Check out this exercise.
## Practice set 2: Solving triangles using the law of cosines
This law is mostly useful for finding an angle measure when given all side lengths. It's also useful for finding a missing side when given the other sides and one angle measure.
### Example 1: Finding an angle
Let's find $m\mathrm{\angle }B$ in the following triangle:
According to the law of cosines:
$\left(AC{\right)}^{2}=\left(AB{\right)}^{2}+\left(BC{\right)}^{2}-2\left(AB\right)\left(BC\right)\mathrm{cos}\left(\mathrm{\angle }B\right)$
Now we can plug the values and solve:
$\begin{array}{rl}\left(5{\right)}^{2}& =\left(10{\right)}^{2}+\left(6{\right)}^{2}-2\left(10\right)\left(6\right)\mathrm{cos}\left(\mathrm{\angle }B\right)\\ \\ 25& =100+36-120\mathrm{cos}\left(\mathrm{\angle }B\right)\\ \\ 120\mathrm{cos}\left(\mathrm{\angle }B\right)& =111\\ \\ \mathrm{cos}\left(\mathrm{\angle }B\right)& =\frac{111}{120}\end{array}$
Evaluating using the calculator and rounding:
$m\mathrm{\angle }B={\mathrm{cos}}^{-1}\left(\frac{111}{120}\right)\approx {22.33}^{\circ }$
### Example 2: Finding a missing side
Let's find $AB$ in the following triangle:
According to the law of cosines:
$\left(AB{\right)}^{2}=\left(AC{\right)}^{2}+\left(BC{\right)}^{2}-2\left(AC\right)\left(BC\right)\mathrm{cos}\left(\mathrm{\angle }C\right)$
Now we can plug the values and solve:
$\begin{array}{rl}\left(AB{\right)}^{2}& =\left(5{\right)}^{2}+\left(16{\right)}^{2}-2\left(5\right)\left(16\right)\mathrm{cos}\left({61}^{\circ }\right)\\ \\ \left(AB{\right)}^{2}& =25+256-160\mathrm{cos}\left({61}^{\circ }\right)\\ \\ AB& =\sqrt{281-160\mathrm{cos}\left({61}^{\circ }\right)}\\ \\ AB& \approx 14.3\end{array}$
Problem 2.1
$m\mathrm{\angle }A=$
${}^{\circ }$
Round to the nearest degree.
Want to try more problems like this? Check out this exercise.
## Practice set 3: General triangle word problems
Problem 3.1
"Only one remains." Ryan signals to his brother from his hiding place.
Matt nods in acknowledgement, spotting the last evil robot.
"$34$ degrees." Matt signals back, informing Ryan of the angle he observed between Ryan and the robot.
Ryan records this value on his diagram (shown below) and performs a calculation. Calibrating his laser cannon to the correct distance, he stands, aims, and fires.
To what distance did Ryan calibrate his laser cannon?
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• I want to know why this article says "Remember that if the missing angle is obtuse, we need to take 180 degrees and subtract what we got from the calculator" when using the law of sines to find a missing angle. Are there videos that explain why this is? I don't understand why during calculations it won't just give the obtuse angle, and instead gives an acute angle.
I'm only taking this geometry course and there is nothing along the way that explains this, not even the videos that introduce the law of sines.
Thanks.
• There can be two since sin(theta) = sin(180-theta) for all values of theta that are real numbers e.g. -1000.98, sqrt(2) etc.
Since you are using the sin^-1 function you will only ever get 1 angle as the range is defined from -90 to 90 degrees(which is -pi/2 to pi/2 in radians). You can it sketch on desmos to see what it look likes.
So if you need to find an obtuse angle then you need to use 180-theta to find the obtuse angle.
It might be helpful to take look at videos related function as well.
• in problem 3.3 I had to open the explanation and do not understand why the law of sines in this solution is switched to sin(c)/length of side. isnt it usually the other way around??
• I can't find anything here about ambiguous triangles. What if a question asks you to solve from a description where two triangles exist? Like "Determine the unknown side and angles in each triangle, if two solutions are possible, give both: In triangle ABC, <C = 31, a = 5.6, and c = 3.9." I solved for height and see that two solutions exist, and the answer key in my textbook agrees, but I can't figure out how to get either. From a set of questions that's only supposed to be on sine law.
• Use the Law of Sines to get one possible angle A:
sin(A)/a=sin(C)/c
sin(A)/5.6=sin(31)/3.9
sin(A)=5.6sin(31)/3.9
A=arcsin(5.6sin(31)/3.9)=47.6924
Subtract 31 (C) and this angle (A) from 180 to find the third angle (B=101.3076) and use the Law of Sines again to find the third side. If you use the given angle-side pair (C and c) you will be less likely to incur error from your own rounding of angle A:
b/sin(B)=c/sin(C)
b/sin(101.3076)=3.9/sin(31)
b=3.9sin(101.3076)/sin(31)=7.4253
But if you know that supplementary angles share a sine value, you know that A can also be an obtuse angle with the same sine as 47.6924:
A=180-47.6924=132.3076
And again, subtract 31 (C) and the obtuse angle A from 180 to find the other possible third angle (B=16.6924) and use the Law of Sines to find the other possible third side, again using angle C and side c to avoid errors from rounding:
b/sin(B)=c/sin(C)
b/sin(16.6924)=3.9/sin(31)
b=3.9sin(16.6924)/sin(31)=2.1750
It all comes from knowing that there are two angles, one obtuse and one acute, for every sine value. And you find the obtuse one by subtracting the acute one from 180.
If you try to do this with a unique triangle (one without two possible sets of angles and sides) your given angle and the obtuse angle you find will add up to more than 180, and so if you try to find a third angle to go with the obtuse one, your subtraction will tell you the third angle is negative, at which point you know you're going down a nonsensical mathematical road, and there weren't two possible triangles to begin with.
Good luck!
• So, obviously, there is the law of sines and the law of cosines. That is what this entire section has been about. However, I'm curious about if there is such a thing as the law of tangents. Since there is both sine and cosine, wouldn't it make sense if there was something like the law of tangents?
• Yes there is. Though I will admit that the only way I know that is by looking it up. I assumed that was but wasn't certain. You can probably find the exact statement of the law on Wikipedia or some math site.
• What is the law of tangents used for??
(1 vote)
• The law of tangents is used to keep people from getting to the point. Old people tend to use it a lot :)
• In the Video 'Solving An Angle With The Law Of Sines' at , Sal said that Law of Sines is 'sin(a)/A = sin(b)/B = sin(c)/C (lower case letters are the angles and the upper case letters are the side opposite to the angle), in this article it says 'A/sin(a) = B/sin(b) = C/sin(c)'. So which one should I choose?
• The Law of Sines can be written either way! You can put the angles in the numerators and the sides in the denominators, or the other way around.
1/2 = 2/4 = 3/6
It's still true if we reverse the numerators and denominators:
2/1 = 4/2 = 6/3
• If law of sines is a/sin(a)=b/sin(b)=c/sin(c), does sin(a)/a=sin(b)/b=sin(c)/c work?
• Yes. Just raise one of those equations to the -1 power, and you get the other equation. They're equivalent.
• Is this a correct way to check whether you use the law of sines or law of cosines? If you have two angles and a side or two sides and an angle, use the law of sines, if you have 3 sides use the law of cosines. Is there anything you need to add to this list?
• The law of sines works only if you know an angle, a side opposite it, and some other piece of information. If you know two sides and the angle between them, the law of sines won't help you.
In any other case, you need the law of cosines. |
# Chapter 2 Lesson 1 Solving ONE STEP EQUATIONS ONE STEP EQUATIONS What you do to one side of the equation must also be done to the other side to keep.
## Presentation on theme: "Chapter 2 Lesson 1 Solving ONE STEP EQUATIONS ONE STEP EQUATIONS What you do to one side of the equation must also be done to the other side to keep."— Presentation transcript:
Chapter 2 Lesson 1 Solving ONE STEP EQUATIONS
ONE STEP EQUATIONS What you do to one side of the equation must also be done to the other side to keep it balanced. An equation is like a balance scale because it shows that two quantities are equal.
ONE STEP EQUATIONS What is the variable? To solve one step equations, you need to ask three questions about the equation: What operation is performed on the variable? What is the inverse operation? (The one that will undo what is being done to the variable)
ONE STEP EQUATIONS Example 1 Solve x + 4 = 12 What is the variable? Using the subtraction property of equality, subtract 4 from both sides of the equation. - 4 x x + 4 = 12 The variable is x. Addition. Subtraction. What is the inverse operation (the one that will undo what is being done to the variable)? What operation is being performed on the variable? - 4 8= The subtraction property of equality tells us to subtract the same thing on both sides to keep the equation equal.
ONE STEP EQUATIONS Example 2 Solve y - 7 = -13 What is the variable? Using the addition property of equality, add 7 to both sides of the equation. y - 7 = -13 + 7 y The variable is y. Subtraction. Addition. What operation is being performed on the variable? What is the inverse operation (the one that will undo what is being done to the variable)? + 7 =-6 The addition property of equality tells us to add the same thing on both sides to keep the equation equal.
ONE STEP EQUATIONS Example 3 Solve –6a = 12 What is the variable? Using the division property of equality, divide both sides of the equation by –6. –6a = 12 -6 a The variable is a. What operation is being performed on the variable?Multiplication. What is the inverse operation (the one that will undo what is being done to the variable)? Division -6 =-2 The division property of equality tells us to divide the same thing on both sides to keep the equation equal.
ONE STEP EQUATIONS Example 4 Solve What is the variable? Using the multiplication property of equality, multiply both sides of the equation by 2. = -10 b The variable is b. What operation is being performed on the variable?Division. What is the inverse operation (the one that will undo what is being done to the variable)? Multiplication = -102 2 =-20 =-10 The multiplication property of equality tells us to multiply the same thing on both sides to keep the equation equal.
ONE STEP EQUATIONS Example 5 Solve What is the variable? Using the multiplication property of equality, multiply both sides of the equation by the reciprocal. The reciprocal is when you flip the fraction. = 5 The variable is b. What operation is being performed on the variable? Multiplication What is the inverse operation (the one that will undo what is being done to the variable)? Multiplication in this case. =5 The multiplication property of equality tells us to multiply the same thing on both sides to keep the equation equal.
ONE STEP EQUATIONS Example 5 Solve What is the variable? Using the multiplication property of equality, multiply both sides of the equation by the reciprocal. The reciprocal is when you flip the fraction. = 5 The variable is b. What operation is being performed on the variable? Multiplication What is the inverse operation (the one that will undo what is being done to the variable)? Multiplication in this case. =5 The multiplication property of equality tells us to multiply the same thing on both sides to keep the equation equal.
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Maths Division Questions For Kids
Division is another basic operation of simple Arithmetical problems. For new school goers, the division is a new mathematical concept in which the problem is broken down into a series of easy and simple steps. An obelus symbol (÷ or †) is used to represent the mathematical operation of division. Young learners are made to learn simple mathematical division sums at the elementary level.
Young kids are taught to solve simple and easy division questions with examples like how to divide cookies among themselves, divide single-digit and two-digit numbers, understand long division sums and solve basic Maths division questions like (10÷2 =5). Having a sound knowledge about division sums is an effective step towards kids learning as it prepares young children to gear up for higher classes.
In the image given above, there is a relationship between the various parts of division sums problem which are explained below –
• Dividend: It refers to the number which we divide. In the problem, the dividend is 359.
• Divisor: It refers to the number by which we divide, here the divisor is 6.
• Quotient: This is the result or outcome that we obtain from the problem. In this case, it is 59.
• Remainder: This is the remaining number that is left over after division, here the remainder is 5.
What is Simple Division?
In Maths, the simple division is a method used for dividing small numbers into groups and parts. Simple division sums help in dividing easy and simple division questions into divisible terms. One such example of simple division questions is given in the image attached below:
Some more examples:
What is Long Division?
In Mathematics, long division is a method used for dividing large numbers into small groups or parts. It helps in breaking the long division sums problem into a sequence of simple and easier steps. Just like other long division questions, a large number (dividend) is divided by another number (divisor) which gives a result (quotient) and sometimes the number left over which is known as remainder.
Some more examples:
Young learners are more receptive to learning new things every day. Some division questions include dividing a single or double-digit number, three-digit calculation and long division questions, etc. Meanwhile, you can also download the Disney BYJU’S Early Learn App and make learning an entertaining and fun-filled experience for your child. |
Introduction to Matrices
(Notation & Terminology)
A matrix is a rectangular array of numbers, written between a large pair of parentheses or square brackets, and usually named by a capital letter. For example: $A = \begin{pmatrix} 2 & -1 & 0 \\ 3 & 5 & 6\end{pmatrix}, \quad B = \begin{bmatrix} 2 & 5 \\ -1 & 0 \end{bmatrix}$ $C = \begin{pmatrix} 2 & 3 & -1 \\ 0 & 4 & 5 \\ 1 & 7 & -8 \end{pmatrix}$ are all matrices.
Whether you write a matrix between square brackets or large parenthese makes no difference whatsoever, both notations are widely accepted.
Order (size) of a Matrix
Given a matrix, its order, or size, is written: $m\times n$ where:
• $$m$$ is the number of rows the matrix has
• $$n$$ is the number of columns the matrix has
If a matrix $$A$$ is of order $$m\times n$$ we'll often indicate this using notation: $A_{m,n}$ For example, consider the matrices $$A$$ and $$B$$ shown here: $A = \begin{pmatrix} 2 & 1 & -5 \\ 0 & 7 & 8 \end{pmatrix} \quad B = \begin{pmatrix} -1 & 0 & 7 \\ 6 & 2 & -3 \\ -5 & 1 & 9 \end{pmatrix}$ We say that:
• $$A$$ is a $$2\times 3$$ "two by three" matrix, which we can write $$A_{2,3}$$
• $$B$$ is a $$3\times 3$$ "three by three" matrix, which we can write $$B_{3,3}$$
Square Matrices
If a matrix has same number of rows as columns then we say it is a square matrix. We'll come across square matrices a lot in this course. Here are a few examples: $A_2 = \begin{pmatrix} 3 & 4 \\ -1 & 5 \end{pmatrix}, \quad B_3 = \begin{pmatrix} 1 & -5 & 3 \\ -4 & 2 & 6 \\ 0 & 8 & 9\end{pmatrix}$ $C_4 = \begin{pmatrix}1 & 0 & -1 & 1\\ 3 & 5 & 7 & -9\\ -2 & 4 & -6 & 8 \\ 1 & -1 & 0 & 1\end{pmatrix}$ Notice that when dealing with square
Index Notation
To refer to specific entries inside a matrix we'll often use index notation. With index notation we can refer to any specific entry using the notation: $a_{ij}$ where:
• $$i$$ refers to the row the entry is in
• $$j$$ refers to the column the entry is in
For instance, given the $$3\times 3$$ matrix $$A$$ defined as: $A = \begin{pmatrix}3 & 0 & -1 \\ 2 & 7 & 5 \\ -4 & 6 & 1 \end{pmatrix}$ we can refer to the entry $$7$$, which is on the second row and in the second column, as well as the entry $$6$$ on the thord row and in the second column by writing: $a_{22} = 7, \quad a_{32} = 6$
Generic Matrix Notation (for Square Matrices)
When working with matrices it's important to be familiar/comfortable with matrix notation. In particular when deriving/learning formula we'll work with generic matrices.
$$2\times 2$$ Matrices
When working with $$2\times 2$$ matrices we'll often use the following generic matrix $$A$$: $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $$a$$, $$b$$, $$c$$ and $$d$$ can be any real numbers.
$$3\times 3$$ Matrices
When working with $$3\times 3$$ matrices we'll need to use index notation, $$a_{ij}$$ for each of the entries in the matrix. The typical matrix we'll use is: $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$
$$4\times 4$$ Matrices and more
When working $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{pmatrix}$
Some "Special" Matrices
Zero Matrix
A zero matrix, also called null matrix, has all entries equal to $$0$$, and is often referred to with the notation $$0_{m,n}$$. Here are some zero matrices: $0_{2,2} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}, \quad 0_{1,3} = \begin{pmatrix} 0 & 0 & 0 \end{pmatrix}$ $0_{3,3} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$
Row Matrix
A row matrix only has one row: $A_{1,n} = \begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} \end{pmatrix}$ Here are some examples of row matrices: $A = \begin{pmatrix} -2 & 1 & 5 \end{pmatrix}, \quad B = \begin{pmatrix} 4 & -6 \end{pmatrix}$ Note: row matrices can be thought of as row vectors, in fact we'll often treat them that way as we learn more about matrices.
Column Matrix
A column matrix only has one column: $A_{m,1} = \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{pmatrix}$ Here are some examples of column matrices: $A = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix}, \quad B = \begin{pmatrix} 6 \\ 0 \end{pmatrix}, \quad C = \begin{pmatrix} 1 \\ 3 \\ -1 \\ 2 \end{pmatrix}$ Note: column matrices can be thought of as column vectors, in fact we'll often treat them that way as we learn more about matrices.
Identity Matrix
An identity matrix is a square matrix whose diagonal entries are all equal to $$1$$ and all other entries are equal to $$0$$. Here are some identity matrices: $I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ $I_4 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$ as we'll be seeing, when we multiply a matrix by an identity matrix it doesn't change its value (just like when we multiply a number by $$1$$).
Exercise
1. State the order (size) of each of the following matrices:
1. $$A = \begin{pmatrix} 2 & -3 \\ 0 & 1\end{pmatrix}$$
1. $$B = \begin{pmatrix} 1 & -5 \\ 4 & 6 \\ 0 & 8\end{pmatrix}$$
1. $$C = \begin{pmatrix} 4 & 7 & 0 & -3 \\ 5 & 9 & 2 & 1 \end{pmatrix}$$
1. $$D = \begin{pmatrix} 8 \end{pmatrix}$$
1. $$E = \begin{pmatrix} 4 \\ -2 \\ 6 \end{pmatrix}$$
1. $$F = \begin{pmatrix} 8 & 0 & 2 \\ -2 & 3 & 4 \\ 5 & 1 & 0 \\ 7 & 9 & 10 \end{pmatrix}$$
2. Given the matrix $$A = \begin{pmatrix} 2 & 3 & -5 \\ 0 & 1 & 4 \\ 6 & -8 & 9 \end{pmatrix}$$, state the value of each of the following entries:
1. $$a_{2,3}$$
1. $$a_{1,2}$$
1. $$a_{2,1}$$
1. $$a_{1,3}$$
1. $$a_{3,3}$$
1. $$a_{2,3}$$
3. For each of the matrices below:
1. State its order
2. State whether it is a square matrix, a column matrix, a row matrix or neither of those.
1. $$A = \begin{pmatrix}3 & 5 \\ 1 & -2 \end{pmatrix}$$
1. $$B = \begin{pmatrix}7 \\ 1 \\ -3 \end{pmatrix}$$
1. $$C = \begin{pmatrix}2 & 3 & 9 \\ 1 & 4 & 8 \end{pmatrix}$$
1. $$D = \begin{pmatrix} -4 & 5 \end{pmatrix}$$
1. $$E = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$$
1. $$F = \begin{pmatrix}2 & -1 & 5 \\ 1 & 7 & 9 \\ 8 & 0 & 3 \end{pmatrix}$$
Note: this exercise can be downloaded as a worksheet to practice with: |
Calculus 10th Edition
$y=3x - 2$
You begin by finding the derivative of the given function: $x^{3}$ You can begin by using the limiting process: $\lim\limits_{\Delta x \to 0} \frac {f(\Delta x+x)-f(x)} {\Delta x}$ Substituting the given function in: $\lim\limits_{\Delta x \to 0} \frac {(x+\Delta x)^{3}-x^{3}} {\Delta x}$ This can now be multiplied out: $\lim\limits_{\Delta x \to 0} \frac {x^{3}+3x^{2}\Delta x + 3x \Delta x^{2} + \Delta x^{3} - x^{3}} {\Delta x}$ Delta x can then be divided out and the x cubes will cancel: $\lim\limits_{\Delta x \to 0} 3x^{2} + 3x \Delta x + \Delta x^{2}$ Applying the limit gives you $3x^{2}$ This must then equal the slope of the given equation because the lines are parallel, putting the equation in slope intercept form yields: $y = 3x +1$ therefore the slope is 3 and 3 must equal the derivative of the function. Dividing both sides by 3 and taking the square root yields Positive and negative 1. Only one is required for the problem so positive 1 will be chosen. $f(1)=1^{3}$, therefore, the tangent line passes through (1, 1). Now the equation is left. $y=3x+b$ $1 = 3(1) +b$ $b=-2$ And the solution can now be written as $y=3x -2$ |
### Subtracting Fractions
#### Lesson Objective
In this lesson, we will learn about subtracting fractions. Here, we will be using some examples to explain this lesson.
The basic idea behind subtracting fractions is the same as adding fractions. So, once you know how to add fractions, it is very easy to subtract fractions.
In this lesson, we will learn how to subtract fractions that involve:
• proper fractions with like denominators
• proper fractions with unlike denominators
• proper and mixed fractions with unlike denominators
The study tips and math video below explain more.
### Study Tips
#### Tip #1
The adding fractions lesson covered the basics that you need to know. Feel free to go through that lesson if you are not sure how to add fractions.
### Math Video
#### Lesson Video
You can contribute to the development of this site and keep it free by getting all six video lessons and volume of solids and calculator app for just US\$1.99 from Apple App Store.
I'd like to contribute or to know more about the app
#### Math Video Transcript
00:00:03.030
In this lesson, we will learn about subtracting fractions.
00:00:07.110
Now, the idea behind subtracting fraction is similar to adding fraction.
00:00:13.070
To recall, let's add 3/5, with 1/5.
00:00:18.180
Now, we know that, since the fractions have like denominators, we can just add the numerators together, and keep the denominator the same.
00:00:28.200
Hence, we get 3 plus 1,/5.
00:00:33.150
From this observation, we can see that, when we subtract the fractions, 3/5 with 1/5, we just need to subtract the numerators, and keep the denominators the same.
00:00:47.110
Hence, we get 3 minus 1/5.
00:00:52.110
Next, subtract 3 with 1. This gives 2. Finally, we get the fraction, 2/5.
00:01:02.240
Alright, let's visually see how subtracting fractions work. We can see that, by subtracting these 3 green parts with this 1 green part, we get 2 green parts.
00:01:15.220
The 2 green parts represent the numerator 2, and all the 5 parts in this bar represent the denominator 5.
00:01:26.110
Next example, let's subtract, 1/3, with 2/9.
00:01:32.070
Notice that, these 2 fractions have unlike denominators. This means that, the size of these parts are not the same, as you can see here.
00:01:44.060
Because of this, we can visually see that, we cannot subtract these two fractions as they are.
00:01:51.120
Therefore, the only way to subtract these fractions, is to make all the parts to have the same size. This means that, these fractions must have like denominators.
00:02:03.090
To do so, we need to use Equivalent Fractions.
00:02:08.080
Now, using Equivalent Fractions, we can change this denominator to 9, by multiplying both the numerator and denominator of this fraction with 3.
00:02:18.090
This gives the fraction, 3/9.
00:02:22.230
Now, these fractions have like denominators. This means that, all the parts will have the same size. As you can see right here.
00:02:34.240
With this, we can now subtract these two fractions just like the previous example. By doing so, we get, 3 minus 2/9.
00:02:46.240
Minus 3 with 2. This gives 1. Finally, we have the fraction, 1/9.
00:02:57.130
Next example, let's subtract, 1/3 with 1 1/2.
00:03:04.020
Notice that, this fraction is a mixed fraction. To minimize mistakes, it is advisable to convert it to an improper fraction.
00:03:13.000
Note that, when doing the conversion, we just need to focus on this mixed fraction, and ignore this minus sign. Now, we multiply 2 with 1. This gives 2.
00:03:28.200
Next, we add 2 with 1. This gives 3, which becomes the improper fraction's numerator. Now, we have the improper fraction, 3/2.
00:03:43.000
Notice that, we cannot subtract these 2 fractions because they have unlike denominators.
00:03:49.190
Therefore, the only way to subtract these fractions, is to make them to have like denominators. We can do so, by using equivalent fractions.
00:04:01.240
Here's how. We can make the denominators the same by multiplying the numerator and denominator of 1/3, with the other fraction's denominator 2, and by multiplying the numerator and denominator of 3/2, with the other fraction's denominator 3.
00:04:20.020
Let's do so. Multiplying 1/3 with 2, and multiplying 3/2 with 3. This gives the equivalent fractions, 2/6, and 9/6 respectively. The denominators are now the same.
00:04:33.180
Now, we subtract these two fractions. This gives 2 minus 9/6.
00:04:46.070
Subtracting 2 with 9, gives negative 7.
00:04:50.210
Notice the negative sign here? We can rewrite it this way so that it looks neater. With this, we have the fraction, negative 7/6.
00:05:01.200
Notice that, negative 7/6 is an improper fraction. So, rather than leaving the answer like this, it is recommended to change it to a mixed fraction, using long division.
00:05:14.080
Now, when doing the conversion, we just need to focus on this fraction, and ignore this sign.
00:05:22.040
Let's start. 7/6 is the same as 7 divides 6. Now, this division gives the quotient as 1. This quotient is actually the whole number for the mixed fraction.
00:05:37.130
Next, we multiply 1 with 6. This gives 6. 7 minus 6 gives the remainder as 1.
00:05:47.200
This remainder, 1, is actually the mixed fraction's numerator.
00:05:54.240
So here, we have the final answer as, negative 1 1/6.
00:06:03.160
That is all for this lesson. Try out the practice question to test your understanding.
### Practice Questions & More
#### Multiple Choice Questions (MCQ)
Now, let's try some MCQ questions to understand this lesson better.
You can start by going through the series of questions on subtracting fractions or pick your choice of question below.
• Question 1 on subtracting fractions with like denominators
• Question 2 on subtracting fractions with unlike denominators
#### Site-Search and Q&A Library
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# SAT Math : How to find the equation of a curve
## Example Questions
### Example Question #1 : X And Y Intercept
Solve the equation for x and y.
x² + y = 31
x + y = 11
x = 8, –6
y = 13, 7
x = 6, 15
y = 5, –4
x = 5, –4
y = 6, 15
x = 13, 7
y = 8, –6
x = 5, –4
y = 6, 15
Explanation:
Solving the equation follows the same system as the first problem. However since x is squared in this problem we will have two possible solutions for each unknown. Again substitute y=11-x and solve from there. Hence, x2+11-x=31. So x2-x=20. 5 squared is 25, minus 5 is 20. Now we know 5 is one of our solutions. Then we must solve for the second solution which is -4. -4 squared is 16 and 16 –(-4) is 20. The last step is to solve for y for the two possible solutions of x. We get 15 and 6. The graph below illustrates to solutions.
### Example Question #1 : How To Find The Equation Of A Curve
Solve the equation for x and y.
x² – y = 96
x + y = 14
x = 10, –11
y = 25, 4
x = 15, 8
y = 5, –14
x = 25, 4
y = 10, –11
x = 5, –14
y = 15, 8
x = 10, –11
y = 25, 4
Explanation:
This problem is very similar to number 2. Derive y=14-x and solve from there. The graph below illustrates the solution.
### Example Question #581 : Geometry
Solve the equation for x and y.
5x² + y = 20
x² + 2y = 10
No solution
x = √10/3, –√10/3
y = 10/3
x = 14, 5
= 4, 6
x = √4/5, 7
= √3/10, 4
x = √10/3, –√10/3
y = 10/3
Explanation:
The problem involves the same method used for the rest of the practice set. However since the x is squared we will have multiple solutions. Solve this one in the same way as number 2. However be careful to notice that the y value is the same for both x values. The graph below illustrates the solution.
### Example Question #1 : How To Find The Slope Of Perpendicular Lines
Solve the equation for x and y.
x² + y = 60
x – y = 50
x = 11, –10
y = 40, 61
x = 40, 61
y = 11, –10
x = –40, –61
y = 10, –11
x = 10, –11
y = –40, –61
x = 10, –11
y = –40, –61
Explanation:
This is a system of equations problem with an x squared, to be solved just like the rest of the problem set. Two solutions are required due to the x2. The graph below illustrates those solutions.
### Example Question #61 : Geometry
A line passes through the points and . What is the equation for the line? |
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# A body of 8 kg is moving under the force $F = 3xN$ when $x$ is the distance covered, if the initial position of the particle is $x = 2m$ and final $x = 10m$. What is the speed of the particle?(A) 6 m/s(B) 12 m/s(C) 36 m/s(D) 144 m/s
Last updated date: 13th Jun 2024
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Hint: As seen, force in itself is dependent on the position of the particle hence the acceleration is not a constant. For such a changing force, the equation of motion can be used to solve the problem, with an integral sign before the acceleration.
Formula used: In this solution we will be using the following formulae;
${v^2} = {u^2} + 2\int {adx}$ where $v$ is the final velocity (or instantaneous velocity at a particular position) of a body, $u$ is the initial velocity at the start point, $a$ is the acceleration of the object, $x$ signifies the distance, and $\int {adx}$ signifies an integral of a changing acceleration with respect to the position of the object.
Acceleration $a = \dfrac{F}{m}$ where $F$ is the force acting on a body of mass $m$.
Complete Step-by-Step solution:
The force acting on a body is in itself proportional to the position of the body. Hence, the acceleration of the body is changing. For such a situation, we have the equation of motion to be given as
${v^2} = {u^2} + 2\int {adx}$ where $v$ is the final velocity of a body, $u$ is the initial velocity at the start point, $a$ is the acceleration of the object, $x$ signifies the distance, and $\int {adx}$ signifies an integral of a changing acceleration with respect to the position of the object.
But, generally, acceleration is given by
$a = \dfrac{F}{m}$ where $F$ is the force acting on a body of mass $m$.
Hence,
${v^2} = {u^2} + 2\int {\dfrac{F}{m}dx}$
Assuming the body begins from rest, and inserting known values and limit of integration, we have
${v^2} = 2\int_2^{10} {\dfrac{{3x}}{8}dx} = \dfrac{3}{4}\int_2^{10} {xdx}$
$\Rightarrow {v^2} = \dfrac{3}{4}\left( {\dfrac{{{{10}^2}}}{2} - \dfrac{{{2^2}}}{2}} \right) = \dfrac{3}{8}\left( {100 - 4} \right)$
By computation,
${v^2} = 36$
$v = 6m/s$
Hence, the correct option is A.
Note: For understanding, the equation, ${v^2} = {u^2} + 2\int {adx}$ can be derived from the definition of instantaneous acceleration, which is
$a = \dfrac{{dv}}{{dt}}$
$\Rightarrow dv = adt$
But also, we know that
$v = \dfrac{{dx}}{{dt}}$, hence,
$dt = \dfrac{{dx}}{v}$
Hence, inserting into $dv = adt$, we get
$dv = a\dfrac{{dx}}{v}$
$\Rightarrow vdv = adx$
Integrating both sides from initial to final value, we have
$\dfrac{{{v^2} - {u^2}}}{2} = \int_{{x_o}}^{{x_1}} {adx}$
Rearranging we have,
${v^2} = {u^2} + 2\int_{{x_o}}^{{x_1}} {adx}$ |
GRAPHING CUBIC, SQUARE ROOT AND CUBE ROOT FUNCTIONS.
Presentation on theme: "GRAPHING CUBIC, SQUARE ROOT AND CUBE ROOT FUNCTIONS."— Presentation transcript:
GRAPHING CUBIC, SQUARE ROOT AND CUBE ROOT FUNCTIONS
43210 In addition to level 3.0 and above and beyond what was taught in class, the student may: - Make connection with other concepts in math. - Make connection with other content areas. Students will be able to construct, compare, and analyze function models and interpret and solve contextual problems. Function models: - absolute value - square root - cube root - piecewise Analyze multiple representations of functions using: - Key features - Translations - Parameters/limits of domain Students will be able to construct and compare function models and solve contextual problems. Function models: - linear - exponential - quadratic Illustrate the graphical effects of translations on function models using technology. With help from the teacher, the student has partial success with the unit content. Even with help, the student has no success with the unit content. Focus 11 - Learning Goal: Students will be able to construct, compare and analyze function models and interpret and solve contextual problems.
Use your calculator to create a table of values of the following equations. xy = x 2 4 2 0 -2 -4 x 4 2 0 -2 -4 16 4 4 0 2 1.41 0 Error
xy = x 2 4 2 0 -2 -4 x 4 2 0 -2 -4 16 4 4 0 2 1.41 0 Error
What similarities and differences do you see between these two graphs? Both graphs intersect at (0, 0) and (1, 1) The square root function is a reflection of the part of the quadratic function in the first quadrant, about y = x (when x is positive). One opens up and one goes to the right…. Why do they intersect at (0, 0) and (1, 1)? y = x 2
xy = x 3 -8 -2 0 1 2 8 x -8 -2 0 1 2 8 -512 1 -8 0 512 8 -2 -1.26 0 1 1.26 2
y = x 3
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# A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}. Find P(E|F) and P (F|E),
This is a multi part question answered separately on Clay6.com
Comment
A)
Toolbox:
• In any problem that involves tossing a coin or a die and determing the outcomes, we can write down the sample space and cout the number of favorable outcomes. Then we find P(E), P(F), P(E$\cap$F) using the set of outcomes.
• Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
Given that a fair die is rolled, number of total outcomes = 6.
Given E: {1, 3, 5} $\rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{3}{6} = \frac{1}{2}$
Given F: {2,3} $\rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{2}{6} = \frac{1}{3}$
Given G: {2,3,4,5} $\rightarrow P(G) = \large \frac{\text{Number of favorable outcomes in G}}{\text{Total number of outcomes in S}} = \frac{4}{6} = \frac{2}{3}$
It also follows that we can calculate the various "Union" and "Intersection" of the sets and their probabilities as follows:
$\Rightarrow E \cup F = {1,2,3,5} \rightarrow P( E \cup F) = \large \frac{4}{6} = \frac{2}{3}$
$\Rightarrow E \cap F = {3} \rightarrow P( E \cap F) = \large \frac{1}{6}$
$\Rightarrow E \cup G = {1,2,3,4,5} \rightarrow P( E \cup G) = \large \frac{5}{6}$
$\Rightarrow E \cap G = {3,5} \rightarrow P( E \cap G) = \large \frac{2}{6} = \frac{1}{3}$
Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
Therefore P($\large \frac{E}{F}$) = $\Large \frac{\Large \frac{1}{6}}{\Large \frac{1}{3}}$ = $\large \frac{1}{2}$
and P($\large \frac{F}{E}$) = $\Large \frac{\Large \frac{1}{6}}{\Large \frac{1}{2}}$ = $\large \frac{1}{3}$ |
## Instructions on using the complex number calculator
• This complex number calculator dynamically calculates the operations that are needed to be done on given complex numbers. The calculated values are rounded off to 4 decimal places for accuracy.
• For single complex number, the given inputs are real part of a complex number, imaginary part of the complex number, and power of the complex number.
• For single complex number, the given outputs are modulus of the entered complex number, angle of the entered complex number, and real and imaginary parts after taking the desired power of the entered complex number.
• For single complex number, this calculator can calculate the Modulus/Magnitude/Absolute value of the entered complex number.
• For single complex number, this calculator can calculate the Angle/Argument value of the entered complex number in radians.
• For single complex number, this calculator can calculate the Power of the entered complex number. This power can also yield squareroot and cuberoot of the entered complex number if the entered power value if 0.5 and 0.3334, respectively.
• For two complex numbers, this calculator can calculate the sum of the entered complex numbers.
• For two complex numbers, this calculator can calculate the multiplication of the entered complex numbers.
• For two complex numbers, this calculator can calculate the division of the entered complex numbers.
• After getting the result from sum, multiplication, and disvision, the real and imaginary values of the resultant can be plugged-in to single complex number calculations to do further operations.
## Formulations
The magnitude/modulus/absolute value, and the angle of the given complex number are found by considering the following figure.
The formulas for the multiplication and division of two complex numbers are also mentioned in Fig. 1.
## Example
We have two complex numbers z1 =2-i, and z2 =-5+i2.
a) Find the magnitude and angle of both complex numbers.
b) Calculate their square, square root, and cube root individually.
c) Find their sum, multiplication, and division (z1/z2).
### Solution
a)
$|z_1|=\sqrt{2^2+1^2}=2.2361$ $|\theta_1|=\tan^{-1}\frac{-1}{2}=-0.4636$ $|z_2|=\sqrt{5^2+2^2}=5.3852$ $|\theta_2|=\tan^{-1}\frac{2}{-5}=2.7611$
b)
$z_1^2=|z|^2(\cos 2\theta+i\sin 2\theta)=|2.2361|^2(\cos 2\theta+i\sin 2\theta)=3.0005-i3.9998$ $z_1^{\frac{1}{2}}=|z|^{\frac{1}{2}}(\cos {\frac{1}{2}}\theta+i\sin {\frac{1}{2}}\theta)=|2.2361|^{\frac{1}{2}}(\cos {\frac{1}{2}}\theta+i\sin {\frac{1}{2}}\theta)=1.4554-i0.3435$ $z_1^{\frac{1}{3}}=|z|^{\frac{1}{3}}(\cos {\frac{1}{3}}\theta+i\sin {\frac{1}{3}}\theta)=|2.2361|^{\frac{1}{3}}(\cos {\frac{1}{3}}\theta+i\sin {\frac{1}{3}}\theta)=1.2921-i0.2013$ $z_2^2=|z|^2(\cos 2\theta+i\sin 2\theta)=|5.3852|^2(\cos 2\theta+i\sin 2\theta)=21.0008-i19.9997$ $z_2^{\frac{1}{2}}=|z|^{\frac{1}{2}}(\cos {\frac{1}{2}}\theta+i\sin {\frac{1}{2}}\theta)=|5.3852|^{\frac{1}{2}}(\cos {\frac{1}{2}}\theta+i\sin {\frac{1}{2}}\theta)=0.4388+i2.2787$ $z_2^{\frac{1}{3}}=|z|^{\frac{1}{3}}(\cos {\frac{1}{3}}\theta+i\sin {\frac{1}{3}}\theta)=|5.3852|^{\frac{1}{3}}(\cos {\frac{1}{3}}\theta+i\sin {\frac{1}{3}}\theta)=1.0612+i1.3953$
c)
$z_1+z_2=2-i-5+i2=-3+i$ $z_1*z_2=(2-i)\times(-5+i2)=2\times(-5)-(-1)\times(2)-i[(-1)\times(-5)+(2)\times(2)]=-8+i9$ $\frac{z_1}{z_2}=\frac{2\times(-5)+(-1)\times(2)}{(-5)^2+(2)^2}+i\frac{(-1)\times(-5)-(2)\times(2)}{(-5)^2+(2)^2}=-0.4138+i0.0345$ |
You are asked to expand (x + 7)10. You will need to multiply (x + 7) times itself 10 times. (x + 7) (x + 7) (x + 7) (x + 7) (x + 7) (x + 7)(x + 7) (x + 7) (x + 7) (x + 7) This could take forever - well, at least quite a while! Is there an easier way to arrive at this expansion? Let's investigate!
When the binomial expression (a + b)n is expanded, there are certain patterns that are noticeable.
Take a look at the expansions when the values of n range from 0 to 4.
If you "pull off" the coefficients of the terms (shown in red), you will discover that the coefficients form a triangle known as Pascal's Triangle.
Pascal's Triangle can be generated by following a certain pattern.
Pascal's Triangle Pattern: The two outside edges of the triangle are comprised of ones.
The other terms are each the sum of the two terms immediately above them in the triangle.
Notice the symmetry of the triangle.
The triangle can grow for as many rows as you desire,
but the work becomes more tedious as the rows increase.
In addition to the observation that the coefficients of a binomial expansion
are the entries that create Pascal's triangle, there are
several other interesting patterns
and observations regarding the expansion of (a + b)n.
1.
The expansion is a series (an adding of the terms, a summation).
2.
The number of terms in each expansion is one more than n. (terms = n + 1)
3.
The power of a starts with an and decreases by one in each successive term ending with a0.
The
power of b starts with b0 and increases by one in each successive term ending with bn.
4.
The power of b is always one less than the "number" of the term.
The
power of a is always n minus the power of b.
5.
The sum of the exponents in each term adds up to n.
6.
The coefficients of the first and last terms are each one.
7.
The coefficients of the middle terms form an interesting (but not easily recognizable) pattern where each coefficient can be determined from the previous term. The coefficient is the product of the previous term's coefficient and a's index (power), divided by the "number" of that previous term.
Check it out:
The second term's coefficient is determined by a4:
To Get Coefficient From the Previous Term:
The third term's coefficient is determined by 4a3b:
This pattern will eventually be expressed as a combination of the form n C k.
When these patterns and observations are pulled together, along with some mathematical
syntax, a theorem is formed pertaining to the expansion of binomial terms:
Binomial Theorem (or Binomial Expansion Theorem)
Most of the syntax used in this theorem should look familiar. The notation is another way of writing a combination such as n C k (read "n choose k").
Remember in observation #6 (above) we said there will be a connection between
and n C k .
Here is the connection: Using our coefficient pattern from #6 in a general setting, we get:
Let's examine the coefficient of that fourth term, the one in the box above.
If we write a combination n C k using k = 3, (for the previous term), we see the connection:
The Binomial Theorem in expanded form is: Remember that and that
Expand: (x + 2)5 Let a = x, b = 2, n = 5 and substitute into the Binomial Theorem. Do not substitute a value for k. This is a good time to put your graphing calculator to work to calculate the combinations. See calculator link at the bottom of this page.
Expand: (2x4 - y)3 The "a" value in this problem is the expression "2x4". Also, the sign is negative making the "b" value "-y". ((2x4) + (-y))3 Let a = (2x4), b = (-y), n = 3 and substitute. Be sure to raise the entire parentheses to the indicated power. Watch out for signs. Use your calculator for the combinations (see calculator link at bottom of page).
Finding a Particular Term in a Binomial Expansion
What if you are asked to find just "one" term in a binomial expansion,
such as just the 5th term of (3x - 4)12 ?
Let's call the term we are looking for the rth term. From our observations, we know that the coefficient of this term will be n C r-1, the power of b will be r - 1 and the power of a will be n minus the power of b. Putting this information together gives us a formula for finding the rth term of a binomial expansion.
The rth term of the expansion of (a + b)n is:
Find 5th term of (3x - 4)12 Let r = 5, a = (3x), b = (-4), n = 12 and substitute. Use your calculator for the combination. Be sure to use the parentheses!! Be careful to raise the entire parentheses to the indicated power. |
Z-test calculator for two proportions examples
$Z$-Test for two proportions
In this tutorial we will discuss $z$-test calculator for testing two population proportions with step by step numerical examples on $Z$-test for testing two population proportions.
Z test calculator for two population proportions
The $Z$-test calculator for testing two population proportions makes it easy to calculate the test statistic, $Z$ critical value and the $p$-value given the sample information, level of significance and the type of alternative hypothesis (i.e. left-tailed, right-tailed or two-tailed.)
Z test Calculator for two proportions
Sample 1 Sample 2
Sample size
No. of Successes
Level of Significance ($\alpha$)
Tail Left tailed
Right tailed
Two tailed
Results
sample proportions:
pooled estimate of proportion:
Standard Error of Diff. of prop.:
Test Statistics Z:
Z-critical value(s):
p-value:
How to use $z$-test calculator for testing two proportions?
Step 1 - Enter the sample size for first sample $n_1$ and second sample $n_2$
Step 2 - Enter the no. of successes for first sample $X_1$ and second sample $X_2$
Step 3 - Enter the level of significance $\alpha$
Step 4 - Select the alternative hypothesis (left-tailed / right-tailed / two-tailed)
Step 5 - Click on "Calculate" button to get the result
$Z$-Test for two proportions Example 1
A survey indicate that of 900 women randomly sampled, 345 use smart-phones. For the men, 450 of the 1025 who were randomly sampled use smartphones. Test whether a percentage of women who uses smartphone is less than men. Use $\alpha=0.05$.
Solution
Given that among $n_1 = 900$ women $X_1= 345$ women use smartphones and among $n_2=1025$ men $X_2=450$ men use smartphones.
The sample proportions of women and men who use smartphones are respectively
$\hat{p}_1=\frac{X_1}{n_1}=\frac{345}{900}=0.383$ and
$\hat{p}_2=\frac{X_2}{n_2}=\frac{450}{1025}=0.439$.
The pooled estimate of sample proportion is
$\hat{p} =\frac{X_1+X_2}{n_1+n_2}=\frac{345+450}{900+1025} =0.413$
Step 1 State the hypothesis testing problem
We wish to test the hypothesis that percentage of women who use smartphones is less as compared to percentage of men who use smartphones, (i.e., $p_1<p_2$).
So the hypothesis testing problem can be setup as
$H_0 : p_1 = p_2$ against $H_1 : p_1 < p_2$ ($\textit{left-tailed}$)
Step 2 Define test statistic
The test statistic for testing above hypothesis testing problem is
\begin{aligned} Z & =\frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}. \end{aligned}
The test statistic $Z$ follows standard normal distribution $N(0,1)$.
Step 3 Specify the level of significance $\alpha$
The significance level is $\alpha = 0.05$.
Step 4 Determine the critical value
As the alternative hypothesis is $\textit{left-tailed}$, the critical value of $Z$ $\text{ is }$ $\text{-1.64}$ (From Normal Statistical Table).
The rejection region (i.e. critical region) is $\text{Z < -1.64}$.
Step 5 Computation
The test statistic under the null hypothesis is
\begin{aligned} Z_{obs}&= \frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}\\ &= \frac{(0.383-0.439)-0}{\sqrt{\frac{0.413*(1-0.413)}{900}+\frac{0.413*(1-0.413)}{1025}}}\\ &= -2.476 \end{aligned}
The rejection region (i.e. critical region) is $\text{Z < -1.64}$. The test statistic is $Z_{obs} =-2.476$ which falls $inside$ the critical region, we $\textit{reject}$ the null hypothesis.
OR
Step 6 Decision ($p$-value approach)
The test is $\text{left-tailed}$ test, so the p-value is the area to the $\text{left}$ of the test statistic ($Z_{obs}=-2.476$) is p-value = $0.0066$.
The p-value is $0.0066$ which is $\textit{less than}$ the significance level of $\alpha = 0.05$, we $\textit{reject}$ the null hypothesis.
Interpretation
There is enough evidence to conclude that percetage of women who use smartphones is less than the percentage of men who use smartphones.
$Z$-Test for two proportions Example 2
A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2823 occupants not wearing seat belts, 31 were killed. Among 7765 occupants wearing seat belts, 16 were killed. The claim is that the fatality rate is higher for those not wearing seat belts. Are Seat Belts Effective? Use $\alpha = 0.01$.
Solution
Let $p_1$ be the proportion of occupants not wearing seat belts and $p_2$ be the proportion of occupants wearing seat belts.
Given that number of front-seat occupants not wearing seat belt $n_1 = 2823$ among them $X_1= 31$ were killed and the number of front-seat occupants wearing seat belt $n_2=7765$ among them $X_2=16$ were killed.
The estimated sample proportions are
$\hat{p}_1=\frac{X_1}{n_1}=\frac{31}{2823}=0.011$.
$\hat{p}_2=\frac{X_2}{n_2}=\frac{16}{7765}=0.002$.
The pooled estimate of sample proportion is
$\hat{p} =\frac{X_1+X_2}{n_1+n_2}=\frac{31+16}{2823+7765} =0.004$
The claim that the fatality rate is higher for those not wearing seat belts can be expressed as $p_1 > p_2$.
Step 1 State the hypothesis testing problem
The hypothesis testing problem is
$H_0 : p_1 = p_2$ against $H_1 : p_1 > p_2$ ($\textit{right-tailed}$)
Step 2 Define test statistic
The test statistic for testing above hypothesis testing problem is
\begin{aligned} Z & =\frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}. \end{aligned}
The test statistic $Z$ follows standard normal distribution $N(0,1)$.
Step 3 Specify the level of significance
The significance level is $\alpha = 0.01$.
Step 4 Determine the critical value
As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $Z$ $\text{ is }$ $\text{2.33}$ (From Normal Statistical Table).
The rejection region (i.e. critical region) is $\text{Z > 2.33}$.
Step 5 Computation
The test statistic under the null hypothesis is
\begin{aligned} Z_{obs}&= \frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}\\ &= \frac{(0.011-0.002)-0}{\sqrt{\frac{0.004*(1-0.004)}{2823}+\frac{0.004*(1-0.004)}{7765}}}\\ &= 6.106 \end{aligned}
The rejection region (i.e. critical region) is $\text{Z > 2.33}$. The test statistic is $Z_{obs} =6.106$ which falls $inside$ the critical region, we $\textit{reject}$ the null hypothesis.
OR
Step 6 Decision ($p$-value approach)
The test is $\text{right-tailed}$ test, so the p-value is the area to the $\text{right}$ of the test statistic ($Z_{obs}=6.106$) is p-value = $0$.
The p-value is $0$ which is $\textit{less than}$ the significance level of $\alpha = 0.01$, we $\textit{reject}$ the null hypothesis.
Interpretation
There is enough evidence to support the claim that the fatality rate is higher for those not wearing seat belts.
Thus we conclude that the use of seat belts appears to be effective in saving lives.
$Z$-Test for two proportions Example 3
Two machines used in the same operation are to be compared. A random sample of 80 parts from the first machine yields 6 non-conforming ones. A random sample of 120 parts from the second machine shows 14 non-conforming ones.
Are the two machine differ significantly with respect to the proportion of non-confirming? Use $\alpha= 0.05$.
Solution
Given information
. First Machine Second Machine
Sample size $n_1=80$ $n_2=200$
no.of non-confirming $X_1=6$ $X_2=14$
The sample proportions are
$\hat{p}_1=\frac{X_1}{n_1}=\frac{6}{80}=0.075$.
$\hat{p}_2=\frac{X_2}{n_2}=\frac{14}{200}=0.07$.
The pooled estimate of sample proportion is
$\hat{p} =\frac{X_1+X_2}{n_1+n_2}=\frac{6+14}{80+200} =0.071$
Step 1 State the hypothesis testing problem
The hypothesis testing problem is $H_0:$ Two machines do not differ significantly with respect to proportion of non-confirming against $H_1:$ Two machines differ significantly with respect to proportion of non-confirming.
i.e., $H_0 : p_1 = p_2$ against $H_1 : p_1 \neq p_2$ ($\textit{two-tailed}$)
Step 2 Define test statistic
The test statistic for testing above hypothesis testing problem is
\begin{aligned} Z & =\frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}. \end{aligned}
The test statistic $Z$ follows standard normal distribution $N(0,1)$.
Step 3 Specify the level of significance $\alpha$
The significance level is $\alpha = 0.05$.
Step 4 Determine the critical value
As the alternative hypothesis is $\textit{two-tailed}$, the critical value of $Z$ $\text{ are }$ $\text{-1.96 and 1.96}$ (From Normal Statistical Table).
The rejection region (i.e. critical region) is $\text{Z < -1.96 or Z > 1.96}$.
Step 5 Computation
The test statistic under the null hypothesis is
\begin{aligned} Z_{obs}&= \frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}\\ &= \frac{(0.075-0.07)-0}{\sqrt{\frac{0.071*(1-0.071)}{80}+\frac{0.071*(1-0.071)}{200}}}\\ &= 0.147 \end{aligned}
The rejection region (i.e. critical region) is $\text{Z < -1.96 or Z > 1.96}$. The test statistic is $Z_{obs} =0.147$ which falls $outside$ the critical region, we $\textit{fail to reject}$ the null hypothesis.
OR
Decision ($p$-value approach)
The test is $\text{two-tailed}$ test, so the p-value is the area to the $\text{extreme}$ of the test statistic ($Z_{obs}=0.147$) is p-value = $0.8833$.
The p-value is $0.8833$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.
Interpretation
There is no sufficient evidence to support the alternative hypothesis. Thus we conclude that the two machine do not differ significantly with respect to the proportion of non-confirming.
Endnote
In this tutorial, you learned the about how to solve numerical examples on $Z$-test for testing two population proportions. You also learned about the step by step procedure to apply $Z$-test for testing two population proportions and how to use $Z$-test calculator for testing two population proportions to get the value of test statistic, p-value, and z-critical value.
Let me know in the comments if you have any questions on $Z$-test calculator for two proportions with examples and your thought on this article. |
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#### Page No 9:
Yes, 0 is a rational number.
0 can be expressed in the form of the fraction $\frac{p}{q}$, where $p=0$ and q can be any integer except 0.
#### Page No 9:
(i) $\frac{5}{7}$
(ii) $\frac{8}{3}$
$\frac{8}{3}=2\frac{2}{3}$
(iii) $-\frac{23}{6}=-3\frac{5}{6}$
(iv) 1.3
$1.3=\frac{13}{10}=1\frac{3}{10}$
(v) – 2.4
$-2.4=\frac{-24}{10}=\frac{-12}{5}=-2\frac{2}{5}$
#### Page No 9:
(i)
Let:
x = $\frac{3}{8}$ and y = $\frac{2}{5}$
Rational number lying between x and y:
=
(ii) 1.3 and 1.4
Let:
x = 1.3 and y = 1.4
Rational number lying between x and y:
=
(iii)
Let:
x = $-$1 and y = $\frac{1}{2}$
Rational number lying between x and y:
= $-\frac{1}{4}$
(iv)
Let:
x$-\frac{3}{4}$ and y = $-\frac{2}{5}$
Rational number lying between x and y:
=
(v)
A rational number lying between will be
$\frac{1}{2}\left(\frac{1}{9}+\frac{2}{9}\right)=\frac{1}{2}×\frac{1}{3}=\frac{1}{6}$
#### Page No 9:
n = 3
$d=\frac{\left(y-x\right)}{n+1}=\frac{\frac{7}{8}-\frac{3}{5}}{3+1}=\frac{11}{40}×\frac{1}{4}=\frac{11}{160}$
Rational numbers between will be
$\left(x+d\right),\left(x+2d\right),...,\left(x+nd\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{3}{5}+\frac{11}{160}\right),\left(\frac{3}{5}+2×\frac{11}{160}\right),\left(\frac{3}{5}+3×\frac{11}{160}\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{107}{160}\right),\left(\frac{118}{160}\right),\left(\frac{129}{160}\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{107}{160}\right),\left(\frac{59}{80}\right),\left(\frac{129}{160}\right)$
There are infinitely many rational numbers between two given rational numbers.
#### Page No 9:
n = 4
n + 1 = 4 + 1 = 5
Thus, rational numbers between are $\frac{16}{35},\frac{17}{35},\frac{18}{35},\frac{19}{35}$.
#### Page No 9:
x = 2, y = 3 and n = 6
$d=\frac{y-x}{n+1}=\frac{3-2}{6+1}=\frac{1}{7}$
Thus, the required numbers are
$\left(x+d\right),\left(x+2d\right),\left(x+3d\right),...,\left(x+nd\right)\phantom{\rule{0ex}{0ex}}=\left(2+\frac{1}{7}\right),\left(2+2×\frac{1}{7}\right),\left(2+3×\frac{1}{7}\right),\left(2+4×\frac{1}{7}\right),\left(2+5×\frac{1}{7}\right),\left(2+6×\frac{1}{7}\right)\phantom{\rule{0ex}{0ex}}=\frac{15}{7},\frac{16}{7},\frac{17}{7},\frac{18}{7},\frac{19}{7},\frac{20}{7}$
#### Page No 9:
n = 5
n + 1 = 6
$x=\frac{3}{5},y=\frac{2}{3}$
$d=\frac{y-x}{n+1}=\frac{\frac{2}{3}-\frac{3}{5}}{6}=\frac{10-9}{90}=\frac{1}{90}$
Thus, rational numbers between will be
$\left(x+d\right),\left(x+2d\right),\left(x+3d\right),\left(x+4d\right),\left(x+5d\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{3}{5}+\frac{1}{90}\right),\left(\frac{3}{5}+\frac{2}{90}\right),\left(\frac{3}{5}+\frac{3}{90}\right),\left(\frac{3}{5}+\frac{4}{90}\right),\left(\frac{3}{5}+\frac{5}{90}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{55}{90}\right),\left(\frac{56}{90}\right),\left(\frac{57}{90}\right),\left(\frac{58}{90}\right),\left(\frac{59}{90}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{11}{18}\right),\left(\frac{28}{45}\right),\left(\frac{19}{30}\right),\left(\frac{29}{45}\right),\left(\frac{59}{90}\right)\phantom{\rule{0ex}{0ex}}$
#### Page No 10:
Let:
x = 2.1, y = 2.2 and n = 16
We know:
d = $\frac{y-x}{n+1}=\frac{2.2-2.1}{16+1}=\frac{0.1}{17}=\frac{1}{170}$= 0.005 (approx.)
So, 16 rational numbers between 2.1 and 2.2 are:
(x + d), (x + 2d), ...(x + 16d)
= [2.1 + 0.005], [2.1 + 2(0.005)],...[2.1 + 16(0.005)]
= 2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17, 2.175 and 2.18
#### Page No 10:
(i) Every natural number is a whole number.
True, since natural numbers are counting numbers i.e N = 1, 2,...
Whole numbers are natural numbers together with 0. i.e W = 0, 1, 2,...
So, every natural number is a whole number
(ii) Every whole number is a natural number.
False, as whole numbers contain natural numbers and 0 whereas natural numbers only contain the counting numbers except 0.
(iii) Every integer is a whole number.
False, whole numbers are natural numbers together with a zero whereas integers include negative numbers also.
(iv) Every integer is a rational number.
True, as rational numbers are of the form . All integers can be represented in the form .
(v) Every rational number is an integer.
False, as rational numbers are of the form . Integers are negative and positive numbers which are not in $\frac{p}{q}$ form.
For example, $\frac{1}{2}$ is a rational number but not an integer.
(vi) Every rational number is a whole number.
False, as rational numbers are of the form . Whole numbers are natural numbers together with a zero.
For example, $\frac{5}{7}$ is a rational number but not a whole number.
#### Page No 18:
(i) $\frac{13}{80}$
Denominator of $\frac{13}{80}$ is 80.
And,
80 = 24$×$5
Therefore, 80 has no other factors than 2 and 5.
Thus, $\frac{13}{80}$ is a terminating decimal.
(ii) $\frac{7}{24}$
Denominator of $\frac{7}{24}$ is 24.
And,
24 = 23$×$3
So, 24 has a prime factor 3, which is other than 2 and 5.
Thus, $\frac{7}{24}$ is not a terminating decimal.
(iii) $\frac{5}{12}$
Denominator of $\frac{5}{12}$ is 12.
And,
12 = 22$×$3
So, 12 has a prime factor 3, which is other than 2 and 5.
Thus, $\frac{5}{12}$ is not a terminating decimal.
(iv) $\frac{31}{375}$
Denominator of $\frac{31}{375}$ is 375.
$375={5}^{3}×3$
So, the prime factors of 375 are 5 and 3.
Thus, $\frac{31}{375}$ is not a terminating decimal.
(v) $\frac{16}{125}$
Denominator of $\frac{16}{125}$ is 125.
And,
125 = 53
Therefore, 125 has no other factors than 2 and 5.
Thus, $\frac{16}{125}$ is a terminating decimal.
#### Page No 19:
(i) $\frac{5}{8}$ = 0.625
By actual division, we have:
It is a terminating decimal expansion.
(ii) $\frac{7}{25}$
$\frac{7}{25}$ = 0.28
By actual division, we have:
It is a terminating decimal expansion.
(iii) $\frac{3}{11}$ = $0.\overline{27}$
It is a non-terminating recurring decimal.
(iv) $\frac{5}{13}$ = $0.\overline{384615}$
It is a non-terminating recurring decimal.
(v) $\frac{11}{24}$
$\frac{11}{24}$ =
By actual division, we have:
It is nonterminating recurring decimal expansion.
(vi) $\frac{261}{400}$$=0.6525$
It is a terminating decimal expansion.
(vii) $\frac{231}{625}$$=0.3696$
It is a terminating decimal expansion.
(viii) $2\frac{5}{12}$
2$\frac{5}{12}$ = $\frac{29}{12}$ =
By actual division, we have:
It is non-terminating decimal expansion.
#### Page No 19:
(i) $0.\overline{2}$
Let x = 0.222... .....(i)
Only one digit is repeated so, we multiply x by 10.
10x = 2.222... .....(ii)
Subtracting (i) from (ii) we get
$9x=2\phantom{\rule{0ex}{0ex}}⇒x=\frac{2}{9}$
(ii) $0.\overline{53}$
Let x = 0.5353... .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 53.5353... .....(ii)
Subtracting (i) from (ii) we get
$99x=53\phantom{\rule{0ex}{0ex}}⇒x=\frac{53}{99}$
(iii) $2.\overline{93}$
Let x = 2.9393... .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 293.9393... .....(ii)
Subtracting (i) from (ii) we get
$99x=291\phantom{\rule{0ex}{0ex}}⇒x=\frac{291}{99}=\frac{97}{33}$
(iv) $18.\overline{48}$
Let x = 18.4848... .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 1848.4848... .....(ii)
Subtracting (i) from (ii) we get
$99x=1830\phantom{\rule{0ex}{0ex}}⇒x=\frac{1830}{99}=\frac{610}{33}$
(v) $0.\overline{235}$
Let x = 0.235235... .....(i)
Three digits are repeated so, we multiply x by 1000.
1000x = 235.235235... .....(ii)
Subtracting (i) from (ii) we get
$999x=235\phantom{\rule{0ex}{0ex}}⇒x=\frac{235}{999}$
(vi) $0.00\overline{32}$
Let x = 0.003232... .....(i)
we multiply x by 100.
100x = 0.3232... .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 32.3232... .....(iii)
Subtracting (ii) from (iii) we get
$9900x=32\phantom{\rule{0ex}{0ex}}⇒x=\frac{32}{9900}=\frac{8}{2475}$
(vii) $1.3\overline{23}$
Let x = 1.32323... .....(i)
we multiply x by 10.
10x = 13.2323... .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
1000x = 1323.2323... .....(iii)
Subtracting (ii) from (iii) we get
$990x=1310\phantom{\rule{0ex}{0ex}}⇒x=\frac{131}{99}$
(viii) $0.3\overline{178}$
Let x = 0.3178178... .....(i)
we multiply x by 10.
10x = 3.178178... .....(ii)
Again multiplying by 1000 as there are 3 repeating numbers after decimals we get
10000x = 3178.178178... .....(iii)
Subtracting (ii) from (iii) we get
$9990x=3175\phantom{\rule{0ex}{0ex}}⇒x=\frac{3175}{9990}=\frac{635}{1998}$
(ix) $32.12\overline{35}$
Let x = 32.123535... .....(i)
we multiply x by 100.
100x = 3212.3535... .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 321235.35... .....(iii)
Subtracting (ii) from (iii) we get
$9900x=318023\phantom{\rule{0ex}{0ex}}⇒x=\frac{318023}{9900}$
(x) $0.40\overline{7}$
Let x = 0.40777... .....(i)
we multiply x by 100.
100x = 40.7777... .....(ii)
Again multiplying by 10 as there is 1 repeating number after decimals we get
1000x = 407.777... .....(iii)
Subtracting (ii) from (iii) we get
$900x=367\phantom{\rule{0ex}{0ex}}⇒x=\frac{367}{900}$
#### Page No 19:
Given: $2.\overline{36}+0.\overline{23}$
Let
First we take x and convert it into $\frac{p}{q}$
100x = 236.3636... ...(iii)
Subtracting (i) from (iii) we get
$99x=234\phantom{\rule{0ex}{0ex}}⇒x=\frac{234}{99}$
Similarly, multiply y with 100 as there are 2 decimal places which are repeating themselves.
$100y=23.2323...$ ...(iv)
Subtracting (ii) from (iv) we get
$99y=23\phantom{\rule{0ex}{0ex}}⇒y=\frac{23}{99}$
Adding x and y we get
$2.\overline{36}+0.\overline{23}$$x+y=\frac{234}{99}+\frac{23}{99}=\frac{257}{99}$
#### Page No 19:
x = 0.3838... ...(i)
Multiply with 100 as there are 2 repeating digits after decimals
100x = 38.3838... ...(ii)
Subtracting (i) from (ii) we get
99x = 38
$⇒x=\frac{38}{99}$
Similarly, we take
y = 1.2727... ...(iii)
Multiply y with 100 as there are 2 repeating digits after decimal.
100y = 127.2727... ...(iv)
Subtract (iii) from (iv) we get
99y = 126
#### Page No 23:
A number that can neither be expressed as a terminating decimal nor be expressed as a repeating decimal is called an irrational number. A rational number, on the other hand, is always a terminating decimal, and if not, it is a repeating decimal.
Examples of irrational numbers:
0.101001000...
0.232332333...
#### Page No 23:
(i) $\sqrt{\frac{3}{81}}$
$\sqrt{\frac{3}{81}}=\sqrt{\frac{1}{27}}=\frac{1}{3}\sqrt{\frac{1}{3}}$
It is an irrational number.
(ii) $\sqrt{361}$ = 19
So, it is rational.
(iii) $\sqrt{21}$
(iv) $\sqrt{1.44}$ = 1.2
So, it is rational.
(v) $\frac{2}{3}\sqrt{6}$
It is an irrational number
(vi) 4.1276
It is a terminating decimal. Hence, it is rational.
(vii) $\frac{22}{7}$
(viii) 1.232332333...
(xi) 6.834834... is a rational number because it is repeating.
#### Page No 23:
x be a rational number and y be an irrational number then x + y necessarily will be an irrational number.
Example: 5 is a rational number but $\sqrt{2}$ is irrational.
So, 5 + $\sqrt{2}$ will be an irrational number.
#### Page No 23:
a be a rational number and b be an irrational number then ab necessarily will be an irrational number.
Example: 6 is a rational number but $\sqrt{5}$ is irrational. And 6$\sqrt{5}$ is also an irrational number.
#### Page No 23:
Product of two irrational numbers is not always an irrational number.
Example: $\sqrt{5}$ is irrational number. And $\sqrt{5}×\sqrt{5}=5$ is a rational number. But the product of another two irrational numbers is $\sqrt{6}$ which is also an irrational numbers.
#### Page No 23:
(i) 2 irrational numbers with difference an irrational number will be .
(ii) 2 irrational numbers with difference is a rational number will be
(iii) 2 irrational numbers with sum an irrational number
(iv) 2 irrational numbers with sum a rational number is
(v) 2 irrational numbers with product an irrational number will be
(vi) 2 irrational numbers with product a rational number will be
(vii) 2 irrational numbers with quotient an irrational number will be
(viii) 2 irrational numbers with quotient a rational number will be .
#### Page No 23:
(i) Let us assume, to the contrary, that $3+\sqrt{3}$ is rational.
Then, $3+\sqrt{3}=\frac{p}{q}$, where p and q are coprime and $q\ne 0$.
$⇒\sqrt{3}=\frac{p}{q}-3\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}=\frac{p-3q}{q}$
Since, p and q are are integers.
$⇒\frac{p-3q}{q}$ is rational.
So, $\sqrt{3}$ is also rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $3+\sqrt{3}$ is rational.
Hence, $3+\sqrt{3}$ is irrational.
(ii) Let us assume, to the contrary, that $\sqrt{7}-2$ is rational.
Then, $\sqrt{7}-2=\frac{p}{q}$, where p and q are coprime and $q\ne 0$.
$⇒\sqrt{7}=\frac{p}{q}+2\phantom{\rule{0ex}{0ex}}⇒\sqrt{7}=\frac{p+2q}{q}$
Since, p and q are are integers.
$⇒\frac{p+2q}{q}$ is rational.
So, $\sqrt{7}$ is also rational.
But this contradicts the fact that $\sqrt{7}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $\sqrt{7}-2$ is rational.
Hence, $\sqrt{7}-2$ is irrational.
(iii) As, $\sqrt[3]{5}×\sqrt[3]{25}$
Hence, $\sqrt[3]{5}×\sqrt[3]{25}$ is rational.
(iv) As, $\sqrt{7}×\sqrt{343}$
Hence, $\sqrt{7}×\sqrt{343}$ is rational.
(v) As,
Hence, $\sqrt{\frac{13}{117}}$ is rational.
(vi) As, $\sqrt{8}×\sqrt{2}$
Hence, $\sqrt{8}×\sqrt{2}$ is rational.
#### Page No 23:
As, few rational numbers between 2 and 2.5 are: 2.1, 2.2, 2.3, 2.4, ...
And,
Since,
So, irrational number between 2 ans 2.5 are:
Hence, a rational and an irrational number can be 2.1 and $\sqrt{5}$, respectively.
Disclaimer: There are infinite rational and irrational numbers between any two rational numbers.
#### Page No 23:
There are infinite number of irrational numbers lying between .
As,
So, the three irrational numbers lying between are:
1.420420042000..., 1.505005000... and 1.616116111...
#### Page No 23:
The two rational numbers between 0.5 and 0.55 are: 0.51 and 0.52
The two irrational numbers between 0.5 and 0.55 are: 0.505005000... and 0.5101100111000...
Disclaimer: There are infinite number of rational and irrational numbers between 0.5 and 0.55.
#### Page No 23:
So, the three different irrational numbers are: 0.72020020002..., 0.7515511555111... and 0.808008000...
Disclaimer: There are an infinite number of irrational numbers between two rational numbers.
#### Page No 24:
The rational numbers between the numbers 0.2121121112... and 0.2020020002... are:
Disclaimer: There are an infinite number of rational numbers between two irrational numbers.
#### Page No 24:
The two irrational numbers between 0.16 and 0.17 are 0.161161116... and 0.1606006000...
Disclaimer: There are an infinite number of irrational numbers between two rational numbers.
#### Page No 24:
(i) True
(ii) False
Example:
(iii) True
(iv) False
(v) True
(vi) False
Example:
(vii) False
Real numbers can be divided into rational and irrational numbers.
(viii) True
(ix) True
#### Page No 28:
(i)
(ii)
(iii) ${\left(3-\sqrt{3}\right)}^{2}$
(iv) ${\left(\sqrt{5}-\sqrt{3}\right)}^{2}$
(v)
$\left(5+\sqrt{7}\right)\left(2+\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=5×2+5×\sqrt{5}+\sqrt{7}×2+\sqrt{7}×\sqrt{5}\phantom{\rule{0ex}{0ex}}=10+5\sqrt{5}+2\sqrt{7}+\sqrt{35}$
(vi)
#### Page No 28:
$=3×6+3×4\sqrt{2}+\sqrt{3}×6+\sqrt{3}×4\sqrt{2}\phantom{\rule{0ex}{0ex}}=18+12\sqrt{2}+6\sqrt{3}+4\sqrt{6}$
#### Page No 28:
(i)
Hence, is rational.
(ii) ${\left(\sqrt{3}+2\right)}^{2}$
Since, the sum and product of rational numbers and an irrational number is always an irrational.
$⇒$$7+4\sqrt{3}$ is irrational.
Hence, ${\left(\sqrt{3}+2\right)}^{2}$ is irrational.
(iii) $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$
$=\frac{2\sqrt{13}}{3\sqrt{13×4}-4\sqrt{13×9}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{13}}{\sqrt{13}\left(3\sqrt{4}-4\sqrt{9}\right)}\phantom{\rule{0ex}{0ex}}=\frac{2}{\left(3×2-4×2\right)}$
Hence, $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$ is rational.
(iv) $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$
$=2\sqrt{2}+4×4\sqrt{2}-6\sqrt{2}\phantom{\rule{0ex}{0ex}}=2\sqrt{2}+16\sqrt{2}-6\sqrt{2}\phantom{\rule{0ex}{0ex}}=12\sqrt{2}$
Since, the product of a rational number and an irrational number is always an irrational.
Hence, $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$ is rational.
#### Page No 28:
(i) As,
Hence, the number of chocolates distributed by Reema is 14.
(ii) The moral values depicted here by Reema is helpfulness and caring.
Disclaimer: The moral values may vary from person to person.
#### Page No 28:
(i) $3\sqrt{45}-\sqrt{125}+\sqrt{200}-\sqrt{50}$
$=3\sqrt{9×5}-\sqrt{25×5}+\sqrt{100×2}-\sqrt{25×2}\phantom{\rule{0ex}{0ex}}=3×3\sqrt{5}-5\sqrt{5}+10\sqrt{2}-5\sqrt{2}\phantom{\rule{0ex}{0ex}}=9\sqrt{5}-5\sqrt{5}+5\sqrt{2}\phantom{\rule{0ex}{0ex}}=4\sqrt{5}+5\sqrt{2}$
(ii) $\frac{2\sqrt{30}}{\sqrt{6}}-\frac{3\sqrt{140}}{\sqrt{28}}+\frac{\sqrt{55}}{\sqrt{99}}$
$=\frac{2\sqrt{6×5}}{\sqrt{6}}-\frac{3\sqrt{28×5}}{\sqrt{28}}+\frac{\sqrt{5×11}}{\sqrt{9×11}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}×\sqrt{5}}{\sqrt{6}}-\frac{3\sqrt{28}×\sqrt{5}}{\sqrt{28}}+\frac{\sqrt{5}×\sqrt{11}}{\sqrt{9}×\sqrt{11}}\phantom{\rule{0ex}{0ex}}=2\sqrt{5}-3\sqrt{5}+\frac{\sqrt{5}}{3}$
$=-\sqrt{5}+\frac{\sqrt{5}}{3}\phantom{\rule{0ex}{0ex}}=\frac{-3\sqrt{5}+\sqrt{5}}{3}\phantom{\rule{0ex}{0ex}}=\frac{-2\sqrt{5}}{3}$
(iii) $\sqrt{72}+\sqrt{800}-\sqrt{18}$
$=\sqrt{36×2}+\sqrt{400×2}-\sqrt{9×2}\phantom{\rule{0ex}{0ex}}=6\sqrt{2}+20\sqrt{2}-3\sqrt{2}\phantom{\rule{0ex}{0ex}}=23\sqrt{2}$
#### Page No 35:
To represent $\sqrt{5}$ on the number line, follow the following steps of construction:
(i) Mark points 0 and 2 as O and P, respectively.
(ii) At point A, draw AB $\perp$ OA such that AB = 1 units.
(iii) Join OB.
(iv) With O as centre and radius OB, draw an arc intersecting the number line at point P.
Thus, point represents $\sqrt{5}$ on the number line.
Justification:
In right $∆$OAB,
Using Pythagoras theorem,
$\mathrm{OB}=\sqrt{{\mathrm{OA}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{2}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+1}\phantom{\rule{0ex}{0ex}}=\sqrt{5}$
#### Page No 35:
To represent $\sqrt{3}$
on the number line, follow the following steps of construction:
(i) Mark points 0 and 1 as O and A, respectively.
(ii) At point A, draw AB $\perp$ OA such that AB = 1 units.
(iii) Join OB.
(iv) At point B, draw DB $\perp$ OA such that DB = 1 units.
(v) Join OD.
(vi)
With O as centre and radius OD, draw an arc intersecting the number line at point Q.
Thus, point Q represents $\sqrt{3}$ on the number line.
Justification:
In right $∆$OAB,
Using Pythagoras theorem,
$\mathrm{OB}=\sqrt{{\mathrm{OA}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{1}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+1}\phantom{\rule{0ex}{0ex}}=\sqrt{2}$
Again, in right
$∆$ODB,
Using Pythagoras theorem,
$\mathrm{OD}=\sqrt{{\mathrm{OB}}^{2}+{\mathrm{DB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\sqrt{2}\right)}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{2+1}\phantom{\rule{0ex}{0ex}}=\sqrt{3}$
#### Page No 35:
To represent $\sqrt{10}$
on the number line, follow the following steps of construction:
(i) Mark points 0 and 3 as O and B, respectively.
(ii) At point A, draw AB $\perp$ OA such that AB = 1 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents $\sqrt{10}$ on the number line.
Justification:
In right $∆$OAB,
Using Pythagoras theorem,
$\mathrm{OA}=\sqrt{{\mathrm{OB}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{3}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+1}\phantom{\rule{0ex}{0ex}}=\sqrt{10}$
#### Page No 35:
To represent $\sqrt{8}$
on the number line, follow the following steps of construction:
(i) Mark points 0 and 2 as O and B, respectively.
(ii) At point B, draw AB $\perp$ OA such that AB = 2 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents $\sqrt{8}$ on the number line.
Justification:
In right $∆$OAB,
Using Pythagoras theorem,
$\mathrm{OA}=\sqrt{{\mathrm{OB}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{2}^{2}+{2}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+4}\phantom{\rule{0ex}{0ex}}=\sqrt{8}$
#### Page No 35:
To represent $\sqrt{4.7}$ on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 4.7 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents $\sqrt{4.7}$.
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB $-$ AO = 4.7 $-$ 2.85 = 1.85 units
In a right angled triangle OBD,
#### Page No 35:
To represent $\sqrt{10.5}$ on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 10.5 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents $\sqrt{10.5}$.
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB $-$ AO = 10.5 $-$ 5.75 = 4.75 units
In a right angled triangle OBD,
#### Page No 35:
To represent $\sqrt{7.28}$ on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 7.28 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents $\sqrt{7.28}$.
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB $-$ AO = 7.28 $-$ 4.14 = 3.14 units
In a right angled triangle OBD,
#### Page No 35:
To represent $\left(1+\sqrt{9.5}\right)$ on the number line, follow the following steps of construction:
(i) Mark two points A and B on a given line such that AB = 9.5 units.
(ii) From B, mark a point C on the same given line such that BC = 1 unit.
(iii) Find the mid point of AC and mark it as O.
(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.
(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.
(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.
(vii) From E, mark a point F on the same given line such that EF = 1 unit.
Thus, let us treat the given line as the number line, with B as 0, C as 1, E as $\sqrt{9.5}$ and so on, then point F represents $\left(1+\sqrt{9.5}\right)$.
Justification:
Here, in semi-circle, radii OA = OC = OD =
And, OB = AB $-$ AO = 9.5 $-$ 5.25 = 4.25 units
In a right angled triangle OBD,
#### Page No 35:
3 < 3.765 < 4
Divide the gap between 3 and 4 on the number line into 10 equal parts.
Now, 3.7 < 3.765 < 3.8
In order to locate the point 3.765 on the number line, divide the gap between 3.7 and 3.8 into 10 equal parts.
Further, 3.76 < 3.765 < 3.77
So, to locate the point 3.765 on the number line, again divide the gap between 3.76 and 3.77 into 10 equal parts.
Now, the number 3.765 can be located on the number line. This can be shown as follows:
Here, the marked point represents the point 3.765 on the number line.
#### Page No 35:
$4.\overline{67}=4.6767$ (Upto 4 decimal places)
4 < 4.6767 < 5
Divide the gap between 4 and 5 on the number line into 10 equal parts.
Now, 4.6 < 4.6767 < 4.7
In order to locate the point 4.6767 on the number line, divide the gap between 4.6 and 4.7 into 10 equal parts.
Further, 4.67 < 4.6767 < 4.68
To locate the point 4.6767 on the number line, again divide the gap between 4.67 and 4.68 into 10 equal parts.
Again, 4.676 < 4.6767 < 4.677
To locate the point 4.6767 on the number line, again divide the gap between 4.676 and 4.677 into 10 equal parts.
Now, the number 4.6767 can be located on the number line. This can be shown as follows:
Here, the marked point represents the point $4.\overline{67}$ on the number line up to 4 decimal places.
#### Page No 43:
$\frac{1}{\sqrt{2}+\sqrt{3}}$
$=\frac{1}{\sqrt{3}+\sqrt{2}}×\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}-\sqrt{2}}{1}$
Here, the denominator i.e. 1 is a rational number. Thus, the rationalising factor of the denominator in $\frac{1}{\sqrt{2}+\sqrt{3}}$ is $\sqrt{3}-\sqrt{2}$.
#### Page No 43:
(i) $\frac{1}{\sqrt{7}}$
On multiplying the numerator and denominator of the given number by $\sqrt{7}$, we get:
(ii) $\frac{\sqrt{5}}{2\sqrt{3}}$
On multiplying the numerator and denominator of the given number by $\sqrt{3}$, we get:
(iii) $\frac{1}{2+\sqrt{3}}$
On multiplying the numerator and denominator of the given number by $2-\sqrt{3}$, we get:
(iv) $\frac{1}{\sqrt{5}-2}$
On multiplying the numerator and denominator of the given number by $\sqrt{5}+2$, we get:
(v) $\frac{1}{5+3\sqrt{2}}$
On multiplying the numerator and denominator of the given number by $5-3\sqrt{2}$, we get:
(vi) $\frac{1}{\sqrt{7}-\sqrt{6}}$
Multiplying the numerator and denominator by $\sqrt{7}+\sqrt{6}$, we get
$\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}×\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}}{{\left(\sqrt{7}\right)}^{2}-{\left(\sqrt{6}\right)}^{2}}$
$=\frac{\sqrt{7}+\sqrt{6}}{7-6}\phantom{\rule{0ex}{0ex}}=\sqrt{7}+\sqrt{6}$
(vii) $\frac{4}{\sqrt{11}-\sqrt{7}}$
Multiplying the numerator and denominator by $\sqrt{11}+\sqrt{7}$, we get
$\frac{4}{\sqrt{11}-\sqrt{7}}=\frac{4}{\sqrt{11}-\sqrt{7}}×\frac{\sqrt{11}+\sqrt{7}}{\sqrt{11}+\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{{\left(\sqrt{11}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}$
$=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{11-7}\phantom{\rule{0ex}{0ex}}=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{4}\phantom{\rule{0ex}{0ex}}=\sqrt{11}+\sqrt{7}$
(viii) $\frac{1+\sqrt{2}}{2-\sqrt{2}}$
Multiplying the numerator and denominator by $2+\sqrt{2}$, we get
$\frac{1+\sqrt{2}}{2-\sqrt{2}}=\frac{1+\sqrt{2}}{2-\sqrt{2}}×\frac{2+\sqrt{2}}{2+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{2+\sqrt{2}+2\sqrt{2}+2}{{\left(2\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{4+3\sqrt{2}}{4-2}\phantom{\rule{0ex}{0ex}}=\frac{4+3\sqrt{2}}{2}$
(ix) $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$
Multiplying the numerator and denominator by $3-2\sqrt{2}$, we get
$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}=\frac{3-2\sqrt{2}}{3+2\sqrt{2}}×\frac{3-2\sqrt{2}}{3-2\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(3-2\sqrt{2}\right)}^{2}}{{\left(3\right)}^{2}-{\left(2\sqrt{2}\right)}^{2}}$
#### Page No 43:
(i)
$\frac{2}{\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{5}}×\frac{\sqrt{5}}{\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{5}}{5}$
$=\frac{2×2.236}{5}\phantom{\rule{0ex}{0ex}}=0.894$
(ii)
$\frac{2-\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2-\sqrt{3}}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}-3}{3}$
$=\frac{2×1.732-3}{3}\phantom{\rule{0ex}{0ex}}=0.155$
(iii)
$\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}×\frac{\sqrt{2}}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}\left(\sqrt{10}-\sqrt{5}\right)}{2}$
$=\frac{1.414×\left(3.162-2.236\right)}{2}\phantom{\rule{0ex}{0ex}}=0.655$
#### Page No 43:
(i)
$\frac{\sqrt{2}-1}{\sqrt{2}+1}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}-1}{\sqrt{2}+1}×\frac{\sqrt{2}-1}{\sqrt{2}-1}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{2}-1\right)}^{2}}{{\left(\sqrt{2}\right)}^{2}-{1}^{2}}$
$=\frac{2+1-2\sqrt{2}}{2-1}\phantom{\rule{0ex}{0ex}}=3-2\sqrt{2}$
(ii)
$\frac{2-\sqrt{5}}{2+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{2-\sqrt{5}}{2+\sqrt{5}}×\frac{2-\sqrt{5}}{2-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(2-\sqrt{5}\right)}^{2}}{{\left(2\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{4+5-4\sqrt{5}}{4-5}\phantom{\rule{0ex}{0ex}}=\frac{9-4\sqrt{5}}{-1}\phantom{\rule{0ex}{0ex}}=-9+4\sqrt{5}$
(iii)
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}×\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{3+2+2×\sqrt{3}×\sqrt{2}}{3-2}\phantom{\rule{0ex}{0ex}}=5+2\sqrt{6}$
(iv)
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{5+2\sqrt{3}}{7+4\sqrt{3}}×\frac{7-4\sqrt{3}}{7-4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{35-20\sqrt{3}+14\sqrt{3}-24}{{\left(7\right)}^{2}-{\left(4\sqrt{3}\right)}^{2}}$
$=\frac{11-6\sqrt{3}}{49-48}\phantom{\rule{0ex}{0ex}}=11-6\sqrt{3}$
$\therefore \frac{5+2\sqrt{3}}{7+4\sqrt{3}}=11+\left(-6\right)\sqrt{3}=a+b\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒a=11,b=-6$
#### Page No 43:
(i)
$\frac{1}{\sqrt{6}+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{6}+\sqrt{5}}×\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{6}-\sqrt{5}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{\sqrt{6}-\sqrt{5}}{6-5}\phantom{\rule{0ex}{0ex}}=\sqrt{6}-\sqrt{5}\phantom{\rule{0ex}{0ex}}=2.449-2.236$
$=0.213$
(ii)
$\frac{6}{\sqrt{5}+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{6}{\sqrt{5}+\sqrt{3}}×\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$=\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{5-3}\phantom{\rule{0ex}{0ex}}=\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{2}\phantom{\rule{0ex}{0ex}}=3\left(\sqrt{5}-\sqrt{3}\right)$
$=3×\left(2.236-1.732\right)\phantom{\rule{0ex}{0ex}}=1.512$
(iii)
$\frac{1}{4\sqrt{3}-3\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{1}{4\sqrt{3}-3\sqrt{5}}×\frac{4\sqrt{3}+3\sqrt{5}}{4\sqrt{3}+3\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{3}+3\sqrt{5}}{{\left(4\sqrt{3}\right)}^{2}-{\left(3\sqrt{5}\right)}^{2}}$
$=\frac{4\sqrt{3}+3\sqrt{5}}{48-45}\phantom{\rule{0ex}{0ex}}=\frac{4×1.732+3×2.236}{3}\phantom{\rule{0ex}{0ex}}=4.545$
(iv)
$\frac{3+\sqrt{5}}{3-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{3+\sqrt{5}}{3-\sqrt{5}}×\frac{3+\sqrt{5}}{3+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(3+\sqrt{5}\right)}^{2}}{{\left(3\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{9+5+6\sqrt{5}}{9-5}\phantom{\rule{0ex}{0ex}}=\frac{14+6\sqrt{5}}{4}\phantom{\rule{0ex}{0ex}}=\frac{7+3\sqrt{5}}{2}$
$=\frac{7+3×2.236}{2}\phantom{\rule{0ex}{0ex}}=6.854$
(v)
$\frac{1+2\sqrt{3}}{2-\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{1+2\sqrt{3}}{2-\sqrt{3}}×\frac{2+\sqrt{3}}{2+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2+\sqrt{3}+4\sqrt{3}+6}{{\left(2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$=\frac{8+5\sqrt{3}}{4-3}\phantom{\rule{0ex}{0ex}}=8+5\sqrt{3}\phantom{\rule{0ex}{0ex}}=8+5×1.732$
$=16.660$
(vi)
$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}×\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{5}+\sqrt{2}\right)}^{2}}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{5+2+2×\sqrt{5}×\sqrt{2}}{5-2}\phantom{\rule{0ex}{0ex}}=\frac{7+2\sqrt{10}}{3}\phantom{\rule{0ex}{0ex}}=\frac{7+2×3.162}{3}$
$=4.441$
#### Page No 44:
(i)
$\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{16×3}+\sqrt{9×2}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}$
$=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}×\frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{3}×4\sqrt{3}-7\sqrt{3}×3\sqrt{2}-5\sqrt{2}×4\sqrt{3}+5\sqrt{2}×3\sqrt{2}}{{\left(4\sqrt{3}\right)}^{2}-{\left(3\sqrt{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{84-21\sqrt{6}-20\sqrt{6}+30}{48-18}$
$=\frac{114-41\sqrt{6}}{30}$
(ii)
$\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}×\frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}×3\sqrt{5}+2\sqrt{6}×2\sqrt{6}-\sqrt{5}×3\sqrt{5}-\sqrt{5}×2\sqrt{6}}{{\left(3\sqrt{5}\right)}^{2}-{\left(2\sqrt{6}\right)}^{2}}$
$=\frac{6\sqrt{30}+24-15-2\sqrt{30}}{45-24}\phantom{\rule{0ex}{0ex}}=\frac{9+4\sqrt{30}}{21}$
#### Page No 44:
(i)
$\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{4+\sqrt{5}}{4-\sqrt{5}}×\frac{4+\sqrt{5}}{4+\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}×\frac{4-\sqrt{5}}{4-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(4+\sqrt{5}\right)}^{2}}{{\left(4\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}+\frac{{\left(4-\sqrt{5}\right)}^{2}}{{\left(4\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{16+5+8\sqrt{5}+16+5-8\sqrt{5}}{16-5}\phantom{\rule{0ex}{0ex}}=\frac{42}{11}$
(ii)
$\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{3}+\sqrt{2}}×\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}×\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}×\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{{\left(\sqrt{2}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{2-5}\phantom{\rule{0ex}{0ex}}=\sqrt{3}-\sqrt{2}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{\left(-3\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{3}-\sqrt{2}-\sqrt{5}-\sqrt{3}+\sqrt{2}+\sqrt{5}$
$=0$
(iii)
$\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}\phantom{\rule{0ex}{0ex}}=\frac{2+\sqrt{3}}{2-\sqrt{3}}×\frac{2+\sqrt{3}}{2+\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}×\frac{2-\sqrt{3}}{2-\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}×\frac{\sqrt{3}-1}{\sqrt{3}-1}\phantom{\rule{0ex}{0ex}}=\frac{{\left(2+\sqrt{3}\right)}^{2}}{{\left(2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{{\left(2-\sqrt{3}\right)}^{2}}{{\left(2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{{\left(\sqrt{3}-1\right)}^{2}}{{\left(\sqrt{3}\right)}^{2}-{1}^{2}}$
$=\frac{4+3+4\sqrt{3}}{4-3}+\frac{4+3-4\sqrt{3}}{4-3}+\frac{3+1-2\sqrt{3}}{3-1}\phantom{\rule{0ex}{0ex}}=7+4\sqrt{3}+7-4\sqrt{3}+\frac{4-2\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}=14+2-\sqrt{3}$
$=16-\sqrt{3}$
(iv)
$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}×\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}×\frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}×\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}×\sqrt{3}-2\sqrt{6}×\sqrt{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}+\frac{6\sqrt{2}×\sqrt{6}-6\sqrt{2}×\sqrt{3}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}-\frac{8\sqrt{3}×\sqrt{6}-8\sqrt{3}×\sqrt{2}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{2\sqrt{18}-2\sqrt{12}}{3-2}+\frac{6\sqrt{12}-6\sqrt{6}}{6-3}-\frac{8\sqrt{18}-8\sqrt{6}}{6-2}\phantom{\rule{0ex}{0ex}}=2\sqrt{18}-2\sqrt{12}+\frac{6\sqrt{12}-6\sqrt{6}}{3}-\frac{8\sqrt{18}-8\sqrt{6}}{4}\phantom{\rule{0ex}{0ex}}=2\sqrt{18}-2\sqrt{12}+2\sqrt{12}-2\sqrt{6}-2\sqrt{18}+2\sqrt{6}$
$=0$
#### Page No 44:
(i)
$\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}\phantom{\rule{0ex}{0ex}}=\frac{1}{3+\sqrt{7}}×\frac{3-\sqrt{7}}{3-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}×\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}×\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{3}+1}×\frac{\sqrt{3}-1}{\sqrt{3}-1}\phantom{\rule{0ex}{0ex}}=\frac{3-\sqrt{7}}{{\left(3\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}+\frac{\sqrt{7}-\sqrt{5}}{{\left(\sqrt{7}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}+\frac{\sqrt{5}-\sqrt{3}}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{\sqrt{3}-1}{{\left(\sqrt{3}\right)}^{2}-{1}^{2}}$
$=\frac{3-\sqrt{7}}{9-7}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{3}-1}{3-1}\phantom{\rule{0ex}{0ex}}=\frac{3-\sqrt{7}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}\phantom{\rule{0ex}{0ex}}=\frac{3-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-1}{2}$
$=\frac{2}{2}\phantom{\rule{0ex}{0ex}}=1$
(ii)
$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}\phantom{\rule{0ex}{0ex}}=\frac{1}{1+\sqrt{2}}×\frac{1-\sqrt{2}}{1-\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}×\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}×\frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}×\frac{\sqrt{4}-\sqrt{5}}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}×\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}×\frac{\sqrt{6}-\sqrt{7}}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}×\frac{\sqrt{7}-\sqrt{8}}{\sqrt{7}-\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}×\frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}}$
$=\frac{1-\sqrt{2}}{{1}^{2}-{\left(\sqrt{2}\right)}^{2}}+\frac{\sqrt{2}-\sqrt{3}}{{\left(\sqrt{2}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{\sqrt{3}-\sqrt{4}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{4}\right)}^{2}}+\frac{\sqrt{4}-\sqrt{5}}{{\left(\sqrt{4}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}+\frac{\sqrt{5}-\sqrt{6}}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{6}\right)}^{2}}+\frac{\sqrt{6}-\sqrt{7}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}+\frac{\sqrt{7}-\sqrt{8}}{{\left(\sqrt{7}\right)}^{2}-{\left(\sqrt{8}\right)}^{2}}+\frac{\sqrt{8}-\sqrt{9}}{{\left(\sqrt{8}\right)}^{2}-{\left(\sqrt{9}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\frac{\sqrt{4}-\sqrt{5}}{4-5}+\frac{\sqrt{5}-\sqrt{6}}{5-6}+\frac{\sqrt{6}-\sqrt{7}}{6-7}+\frac{\sqrt{7}-\sqrt{8}}{7-8}+\frac{\sqrt{8}-\sqrt{9}}{8-9}\phantom{\rule{0ex}{0ex}}=\frac{1-\sqrt{2}}{\left(-1\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(-1\right)}+\frac{\sqrt{3}-\sqrt{4}}{\left(-1\right)}+\frac{\sqrt{4}-\sqrt{5}}{\left(-1\right)}+\frac{\sqrt{5}-\sqrt{6}}{\left(-1\right)}+\frac{\sqrt{6}-\sqrt{7}}{\left(-1\right)}+\frac{\sqrt{7}-\sqrt{8}}{\left(-1\right)}+\frac{\sqrt{8}-\sqrt{9}}{\left(-1\right)}$
$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+\sqrt{7}-\sqrt{6}+\sqrt{8}-\sqrt{7}+\sqrt{9}-\sqrt{8}\phantom{\rule{0ex}{0ex}}=3-1\phantom{\rule{0ex}{0ex}}=2$
#### Page No 44:
$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{7+3\sqrt{5}}{3+\sqrt{5}}×\frac{3-\sqrt{5}}{3-\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}×\frac{3+\sqrt{5}}{3+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{7\left(3-\sqrt{5}\right)+3\sqrt{5}\left(3-\sqrt{5}\right)}{{3}^{2}-{\left(\sqrt{5}\right)}^{2}}-\frac{7\left(3+\sqrt{5}\right)-3\sqrt{5}\left(3+\sqrt{5}\right)}{{3}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{21-7\sqrt{5}+9\sqrt{5}-15}{9-5}-\frac{21+7\sqrt{5}-9\sqrt{5}-15}{9-5}\phantom{\rule{0ex}{0ex}}=\frac{6+2\sqrt{5}}{4}-\frac{6-2\sqrt{5}}{4}$
$=\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{5}}{4}\phantom{\rule{0ex}{0ex}}=\sqrt{5}$
$\therefore \frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=0+1×\sqrt{5}$
Comparing with the given expression, we get
a = 0 and b = 1
Thus, the values of a and b are 0 and 1, respectively.
#### Page No 44:
$\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}+\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}×\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}+\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}×\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}+\sqrt{11}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{13}-\sqrt{11}\right)}^{2}}{{\left(\sqrt{13}\right)}^{2}-{\left(\sqrt{11}\right)}^{2}}+\frac{{\left(\sqrt{13}+\sqrt{11}\right)}^{2}}{{\left(\sqrt{13}\right)}^{2}-{\left(\sqrt{11}\right)}^{2}}$
$=\frac{13+11-2×\sqrt{13}×\sqrt{11}}{13-11}+\frac{13+11+2×\sqrt{13}×\sqrt{11}}{13-11}\phantom{\rule{0ex}{0ex}}=\frac{24-2\sqrt{143}}{2}+\frac{24+2\sqrt{143}}{2}\phantom{\rule{0ex}{0ex}}=\frac{24-2\sqrt{143}+24+2\sqrt{143}}{2}$
$=\frac{48}{2}\phantom{\rule{0ex}{0ex}}=24$
#### Page No 44:
$⇒\frac{1}{x}=\frac{1}{3+2\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{1}{3+2\sqrt{2}}×\frac{3-2\sqrt{2}}{3-2\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{3-2\sqrt{2}}{{3}^{2}-{\left(2\sqrt{2}\right)}^{2}}$
Adding (1) and (2), we get
$x+\frac{1}{x}=3+2\sqrt{2}+3-2\sqrt{2}=6$, which is a rational number
Thus, $x+\frac{1}{x}$ is rational.
#### Page No 44:
Subtracting (2) from (1), we get
$x-\frac{1}{x}=\left(2-\sqrt{3}\right)-\left(2+\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}⇒x-\frac{1}{x}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒{\left(x-\frac{1}{x}\right)}^{3}={\left(-2\sqrt{3}\right)}^{3}=-24\sqrt{3}$
Thus, the value of ${\left(x-\frac{1}{x}\right)}^{3}$ is $-24\sqrt{3}$.
#### Page No 44:
Adding (1) and (2), we get
$x+\frac{1}{x}=9-4\sqrt{5}+9+4\sqrt{5}\phantom{\rule{0ex}{0ex}}⇒x+\frac{1}{x}=18$
Squaring on both sides, we get
${\left(x+\frac{1}{x}\right)}^{2}={18}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\frac{1}{{x}^{2}}+2×x×\frac{1}{x}=324\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\frac{1}{{x}^{2}}=324-2=322$
Thus, the value of ${x}^{2}+\frac{1}{{x}^{2}}$ is 322.
#### Page No 44:
$⇒\frac{1}{x}=\frac{2}{5-\sqrt{21}}×\frac{5+\sqrt{21}}{5+\sqrt{21}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{{5}^{2}-{\left(\sqrt{21}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{25-21}$
Adding (1) and (2), we get
$x+\frac{1}{x}=\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2}\phantom{\rule{0ex}{0ex}}⇒x+\frac{1}{x}=\frac{5-\sqrt{21}+5+\sqrt{21}}{2}\phantom{\rule{0ex}{0ex}}⇒x+\frac{1}{x}=\frac{10}{2}=5$
Thus, the value of $x+\frac{1}{x}$ is 5.
#### Page No 44:
$\therefore \frac{1}{{a}^{2}}=\frac{1}{17-12\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{a}^{2}}=\frac{1}{17-12\sqrt{2}}×\frac{17+12\sqrt{2}}{17+12\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{a}^{2}}=\frac{17+12\sqrt{2}}{{17}^{2}-{\left(12\sqrt{2}\right)}^{2}}$
Subtracting (2) from (1), we get
${a}^{2}-\frac{1}{{a}^{2}}=\left(17-12\sqrt{2}\right)-\left(17+12\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-\frac{1}{{a}^{2}}=17-12\sqrt{2}-17-12\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-\frac{1}{{a}^{2}}=-24\sqrt{2}$
Thus, the value of ${a}^{2}-\frac{1}{{a}^{2}}$ is $-24\sqrt{2}$.
#### Page No 45:
Subtracting (2) from (1), we get
Thus, the value of $x-\frac{1}{x}$ is $4\sqrt{3}$.
#### Page No 45:
Adding (1) and (2), we get
Cubing both sides, we get
${\left(x+\frac{1}{x}\right)}^{3}={4}^{3}\phantom{\rule{0ex}{0ex}}⇒{x}^{3}+\frac{1}{{x}^{3}}+3×x×\frac{1}{x}\left(x+\frac{1}{x}\right)=64$
$⇒{x}^{3}+\frac{1}{{x}^{3}}+3×4=64$ [Using (3)]
$⇒{x}^{3}+\frac{1}{{x}^{3}}=64-12=52$
Thus, the value of ${x}^{3}+\frac{1}{{x}^{3}}$ is 52.
#### Page No 45:
Disclaimer: The question is incorrect.
$x=\frac{5-\sqrt{3}}{5+\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒x=\frac{5-\sqrt{3}}{5+\sqrt{3}}×\frac{5-\sqrt{3}}{5-\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒x=\frac{{\left(5-\sqrt{3}\right)}^{2}}{{5}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$⇒x=\frac{25+3-10\sqrt{3}}{25-3}\phantom{\rule{0ex}{0ex}}⇒x=\frac{28-10\sqrt{3}}{22}\phantom{\rule{0ex}{0ex}}⇒x=\frac{14-5\sqrt{3}}{11}$
$y=\frac{5+\sqrt{3}}{5-\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒y=\frac{5+\sqrt{3}}{5-\sqrt{3}}×\frac{5+\sqrt{3}}{5+\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒y=\frac{{\left(5+\sqrt{3}\right)}^{2}}{{5}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$⇒y=\frac{25+3+10\sqrt{3}}{25-3}\phantom{\rule{0ex}{0ex}}⇒y=\frac{28+10\sqrt{3}}{22}\phantom{\rule{0ex}{0ex}}⇒y=\frac{14+5\sqrt{3}}{11}$
$\therefore {x}^{2}-{y}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{14-5\sqrt{3}}{11}\right)}^{2}-{\left(\frac{14+5\sqrt{3}}{11}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{196+75-140\sqrt{3}}{121}-\frac{196+75+140\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}$
$=\frac{271-140\sqrt{3}}{121}-\frac{271+140\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}=\frac{271-140\sqrt{3}-271-140\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}=\frac{-280\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}$
The question is incorrect. Kindly check the question.
The question should have been to show that $x-y=-\frac{10\sqrt{3}}{11}$.
$\therefore x-y\phantom{\rule{0ex}{0ex}}=\frac{14-5\sqrt{3}}{11}-\frac{14+5\sqrt{3}}{11}\phantom{\rule{0ex}{0ex}}=\frac{14-5\sqrt{3}-14-5\sqrt{3}}{11}\phantom{\rule{0ex}{0ex}}=\frac{-10\sqrt{3}}{11}\phantom{\rule{0ex}{0ex}}$
#### Page No 45:
According to question,
Now,
Hence, $3{a}^{2}+4ab-3{b}^{2}=4+\frac{56\sqrt{10}}{3}$.
#### Page No 45:
According to question,
Now,
Hence, the value of a2 + b2 – 5ab is 93.
#### Page No 45:
According to question,
Now,
Hence, the value of p2 + q2 is 47.
#### Page No 45:
$\left(\mathrm{i}\right)\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\sqrt{7}+\sqrt{6}\right)-\sqrt{13}}×\frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}\phantom{\rule{0ex}{0ex}}=\frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{{\left(\sqrt{7}+\sqrt{6}\right)}^{2}-{\left(\sqrt{13}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{{\left(\sqrt{7}\right)}^{2}+{\left(\sqrt{6}\right)}^{2}+2\left(\sqrt{7}\right)\left(\sqrt{6}\right)-{\left(\sqrt{13}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{7+6+2\sqrt{42}-13}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}×\frac{\sqrt{42}}{\sqrt{42}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{6}+6\sqrt{7}+\left(\sqrt{13}\right)\left(\sqrt{42}\right)}{84}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$
Hence, the rationalised form is $\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$.
$\left(\mathrm{ii}\right)\frac{3}{\sqrt{3}+\sqrt{5}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{3}{\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{5}}×\frac{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}}{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{3\left\{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right\}}{{\left(\sqrt{3}-\sqrt{2}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{{\left(\sqrt{3}\right)}^{2}+{\left(\sqrt{2}\right)}^{2}-2\left(\sqrt{3}\right)\left(\sqrt{2}\right)-{\left(\sqrt{5}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{3+2-2\sqrt{6}-5}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}}×\frac{\sqrt{6}}{\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[3\sqrt{2}-2\sqrt{3}-\sqrt{30}\right]}{-12}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4}$
Hence, the rationalised form is $\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4}$.
$\left(\mathrm{iii}\right)\frac{4}{2+\sqrt{3}+\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{4}{\left(2+\sqrt{3}\right)+\sqrt{7}}×\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\left(2+\sqrt{3}\right)-\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2+\sqrt{3}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2\right)}^{2}+{\left(\sqrt{3}\right)}^{2}+2\left(2\right)\left(\sqrt{3}\right)-{\left(\sqrt{7}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4+3+4\sqrt{3}-7}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}+3-\sqrt{21}}{3}$
Hence, the rationalised form is $\frac{2\sqrt{3}+3-\sqrt{21}}{3}$.
#### Page No 45:
Hence, the value of $\frac{1}{\sqrt{3}-\sqrt{2}-1}$ correct to 3 places of decimal is −1.465.
#### Page No 45:
Now,
Hence, the value of x3 – 2x2 – 7x + 5 is 3.
#### Page No 45:
Hence, $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$ = 5.398 .
#### Page No 53:
(i) ${2}^{\frac{2}{3}}×{2}^{\frac{1}{3}}$
(ii) ${2}^{\frac{2}{3}}×{2}^{\frac{1}{5}}$
(iii) ${7}^{\frac{5}{6}}×{7}^{\frac{2}{3}}$
(iv) ${\left(1296\right)}^{\frac{1}{4}}×{\left(1296\right)}^{\frac{1}{2}}$
(i) (ab + ba)–1
(ii) (aa + bb)–1
#### Page No 53:
(i) ${\left(\frac{81}{49}\right)}^{-\frac{3}{2}}$
(ii) (14641)0.25
(iii) ${\left(\frac{32}{243}\right)}^{-\frac{4}{5}}$
(iv) ${\left(\frac{7776}{243}\right)}^{-\frac{3}{5}}$
#### Page No 54:
(i) $\frac{4}{{\left(216\right)}^{-\frac{2}{3}}}+\frac{1}{{\left(256\right)}^{-\frac{3}{4}}}+\frac{2}{{\left(243\right)}^{-\frac{1}{5}}}$
$\frac{4}{{\left(216\right)}^{-\frac{2}{3}}}+\frac{1}{{\left(256\right)}^{-\frac{3}{4}}}+\frac{2}{{\left(243\right)}^{-\frac{1}{5}}}\phantom{\rule{0ex}{0ex}}=\frac{4}{{\left[{\left(6\right)}^{3}\right]}^{-\frac{2}{3}}}+\frac{1}{{\left[{\left(4\right)}^{4}\right]}^{-\frac{3}{4}}}+\frac{2}{{\left[{\left(3\right)}^{5}\right]}^{-\frac{1}{5}}}\phantom{\rule{0ex}{0ex}}=\frac{4}{{\left(6\right)}^{-2}}+\frac{1}{{\left(4\right)}^{-3}}+\frac{2}{{\left(3\right)}^{-1}}\phantom{\rule{0ex}{0ex}}=4{\left(6\right)}^{2}+{\left(4\right)}^{3}+2\left(3\right)\phantom{\rule{0ex}{0ex}}=144+64+6\phantom{\rule{0ex}{0ex}}=214$
(ii) ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+{\left(\frac{256}{625}\right)}^{-\frac{1}{4}}+{\left(\frac{3}{7}\right)}^{0}$
${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+{\left(\frac{256}{625}\right)}^{-\frac{1}{4}}+{\left(\frac{3}{7}\right)}^{0}\phantom{\rule{0ex}{0ex}}={\left[{\left(\frac{4}{5}\right)}^{3}\right]}^{-\frac{2}{3}}+{\left[{\left(\frac{4}{5}\right)}^{4}\right]}^{-\frac{1}{4}}+1\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{5}\right)}^{-2}+{\left(\frac{4}{5}\right)}^{-1}+1\phantom{\rule{0ex}{0ex}}={\left(\frac{5}{4}\right)}^{2}+\left(\frac{5}{4}\right)+1\phantom{\rule{0ex}{0ex}}=\frac{25}{16}+\frac{5}{4}+1\phantom{\rule{0ex}{0ex}}=\frac{25+20+16}{16}\phantom{\rule{0ex}{0ex}}=\frac{61}{16}$
(iii)
(iv) $\frac{{\left(25\right)}^{\frac{5}{2}}×{\left(729\right)}^{\frac{1}{3}}}{{\left(125\right)}^{\frac{2}{3}}×{\left(27\right)}^{\frac{2}{3}}×{8}^{\frac{4}{3}}}$
$\frac{{\left(25\right)}^{\frac{5}{2}}×{\left(729\right)}^{\frac{1}{3}}}{{\left(125\right)}^{\frac{2}{3}}×{\left(27\right)}^{\frac{2}{3}}×{8}^{\frac{4}{3}}}\phantom{\rule{0ex}{0ex}}=\frac{{\left[{\left(5\right)}^{2}\right]}^{\frac{5}{2}}×{\left[{\left(9\right)}^{3}\right]}^{\frac{1}{3}}}{{\left[{\left(5\right)}^{3}\right]}^{\frac{2}{3}}×{\left[{\left(3\right)}^{3}\right]}^{\frac{2}{3}}×{\left[{\left(2\right)}^{3}\right]}^{\frac{4}{3}}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(5\right)}^{5}×{\left(9\right)}^{1}}{{\left(5\right)}^{2}×{\left(3\right)}^{2}×{\left(2\right)}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{5×5×5×5×5×9}{5×5×3×3×2×2×2×2}\phantom{\rule{0ex}{0ex}}=\frac{125}{16}$
#### Page No 54:
(i) ${\left({1}^{3}+{2}^{3}+{3}^{3}\right)}^{\frac{1}{2}}$
(ii) ${\left[5{\left({8}^{\frac{1}{3}}+{27}^{\frac{1}{3}}\right)}^{3}\right]}^{\frac{1}{4}}$
(iii) $\frac{{2}^{0}+{7}^{0}}{{5}^{0}}$
(iv) ${\left[{\left(16\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}$
#### Page No 54:
(i) $\left[{8}^{-\frac{2}{3}}×{2}^{\frac{1}{2}}×{25}^{-\frac{5}{4}}\right]÷\left[{32}^{-\frac{2}{5}}×{125}^{-\frac{5}{6}}\right]=\sqrt{2}$
(ii) ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left(\frac{256}{625}\right)}^{\frac{1}{4}}}+\frac{\sqrt{25}}{\sqrt[3]{64}}=\frac{65}{16}$
(iii) ${\left[7{\left\{{\left(81\right)}^{\frac{1}{4}}+{\left(256\right)}^{\frac{1}{4}}\right\}}^{\frac{1}{4}}\right]}^{4}=16807$
#### Page No 54:
Hence, the result in the exponential form is ${x}^{\frac{1}{6}}$.
#### Page No 54:
(i) ${\left(\frac{{15}^{\frac{1}{3}}}{{9}^{\frac{1}{4}}}\right)}^{-6}$
(ii) ${\left(\frac{{12}^{\frac{1}{5}}}{{27}^{\frac{1}{5}}}\right)}^{\frac{5}{2}}$
(iii) ${\left(\frac{{15}^{\frac{1}{4}}}{{3}^{\frac{1}{2}}}\right)}^{-2}$
#### Page No 54:
Hence, the value of x is 6.
Hence, the value of x is 22.
Hence, the value of x is 5.
Hence, the value of x is 5.
Hence, the value of x is $\frac{5}{4}$.
#### Page No 55:
(i) $\sqrt{{x}^{-1}y}·\sqrt{{y}^{-1}z}·\sqrt{{z}^{-1}x}=1$
Hence, $\sqrt{{x}^{-1}y}·\sqrt{{y}^{-1}z}·\sqrt{{z}^{-1}x}=1$.
(ii) ${\left({x}^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left({x}^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left({x}^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}}=1$
Hence, ${\left({x}^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left({x}^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left({x}^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}}=1$.
(iii) $\frac{{x}^{a\left(b-c\right)}}{{x}^{b\left(a-c\right)}}÷{\left(\frac{{x}^{b}}{{x}^{a}}\right)}^{c}=1$
Hence, $\frac{{x}^{a\left(b-c\right)}}{{x}^{b\left(a-c\right)}}÷{\left(\frac{{x}^{b}}{{x}^{a}}\right)}^{c}=1$.
(iv)
Hence, .
#### Page No 55:
${\left(\frac{{x}^{b}}{{x}^{c}}\right)}^{b+c-a}·{\left(\frac{{x}^{c}}{{x}^{a}}\right)}^{c+a-b}·{\left(\frac{{x}^{a}}{{x}^{b}}\right)}^{a+b-c}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b+c-a}·{\left({x}^{c-a}\right)}^{c+a-b}·{\left({x}^{a-b}\right)}^{a+b-c}\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b-c}\right)}^{b}.{\left({x}^{b-c}\right)}^{c-a}\right]·{\left({x}^{c-a}\right)}^{c+a-b}·\left[{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\right]\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b-c}\right)}^{b}.{\left({x}^{b-c}\right)}^{c-a}\right]·{\left({x}^{c+a-b}\right)}^{c-a}·\left[{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\right]\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b}.\left[{\left({x}^{b-c}\right)}^{c-a}·{\left({x}^{c+a-b}\right)}^{c-a}\right]·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b}.\left[{\left({x}^{b-c+c+a-b}\right)}^{c-a}\right]·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b}.{\left({x}^{a}\right)}^{c-a}·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b}\right)}^{b-c}.{\left({x}^{a}\right)}^{c-a}·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b}\right)}^{b-c}·{\left({x}^{a-b}\right)}^{b-c}\right].{\left({x}^{a}\right)}^{c-a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b+a-b}\right)}^{b-c}\right].{\left({x}^{a}\right)}^{c-a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{a}\right)}^{b-c}.{\left({x}^{a}\right)}^{c-a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{a}.{\left({x}^{c-a}\right)}^{a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c+c-a+a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={x}^{0}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}$
#### Page No 55:
$\frac{{9}^{n}×{3}^{2}×{\left({3}^{\frac{-n}{2}}\right)}^{-2}-{\left(27\right)}^{n}}{{3}^{3m}×{2}^{3}}=\frac{1}{27}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left({3}^{2}\right)}^{n}×{3}^{2}×{\left({3}^{-n}\right)}^{-1}-{\left({3}^{3}\right)}^{n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{2n}×{3}^{2}×{3}^{n}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{2n+2+n}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n+2}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}×{3}^{2}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}\left(9-1\right)}{{3}^{3m}×8}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}\left(8\right)}{{3}^{3m}×8}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}}{{3}^{3m}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒{3}^{3n-3m}={3}^{-3}\phantom{\rule{0ex}{0ex}}⇒3n-3m=-3\phantom{\rule{0ex}{0ex}}⇒3\left(n-m\right)=-3\phantom{\rule{0ex}{0ex}}⇒n-m=-1\phantom{\rule{0ex}{0ex}}⇒m-n=1$
Hence, m – n = 1.
#### Page No 57:
Since, the sum and product of a rational and an irrational is always irrational.
So, $1+\sqrt{3}$ and $2\sqrt{3}$ are irrational numbers.
Also, π is an irrational number.
And, 0 is an integer.
So, 0 is a rational number.
Hence, the correct option is (d).
#### Page No 57:
Since, –4.3 < –3.4 < –3 < 0 < 1.101100110001... < 3
But 1.101100110001... is an irrational number
So, the rational number between –3 and 3 is 0.
Hence, the correct option is (a).
#### Page No 57:
We have,
And,
Also,
Since, $\frac{1}{6}<\frac{2}{6}<\frac{3}{6}\left(\frac{1}{2}\right)<\frac{4}{6}\left(=\frac{2}{3}\right)<\frac{5}{6}<\frac{7}{6}<\frac{8}{6}\left(=\frac{4}{3}\right)<\frac{10}{6}\left(=\frac{5}{3}\right)<\frac{12}{6}\left(=\frac{2}{1}\right)$
So, the two rational numbers between are .
Hence, the correct opion is (c).
#### Page No 57:
As, all rational numbers, all natural numbers and all irrational numbers can be represented on a nuumber line in an unique way.
So, every point on a number line represents a unique number.
Hence, the correct option is (d).
#### Page No 57:
(c) $\sqrt{225}$
Because 225 is a square of 15, i.e., $\sqrt{225}$ = 15, and it can be expressed in the $\frac{p}{q}$ form, it is a rational number.
#### Page No 57:
(d) a real number
Every rational number is a real number, as every rational number can be easily expressed on the real number line.
#### Page No 57:
(c) are infinitely many rational numbers
Because the range between any two rational numbers can be easily divided into any number of divisions, there can be an infinite number of rational numbers between any two rational numbers.
#### Page No 57:
(b) either terminating or repeating
As per the definition of rational numbers, they are either repeating or terminating decimals.
#### Page No 58:
(d) neither terminating nor repeating
As per the definition of irrational numbers, these are neither terminating nor repeating decimals.
#### Page No 58:
As, any number which have a terminating or non-terminating recurring decimal expansion is a rational number.
So, 0.5030030003... which is non-termintaing non-recurring decimal expansion is not a rational number.
Hence, the correct option is (d).
#### Page No 58:
(d) 3.141141114...
Because 3.141141114... is neither a repeating decimal nor a terminating decimal, it is an irrational number.
#### Page No 58:
Since, $\frac{7}{19}=\frac{7×3}{19×3}=\frac{21}{57}$
Hence, the correct option is (d).
#### Page No 58:
We have,
And,
Since, $-\frac{40}{60}\left(=-\frac{2}{3}\right)<-\frac{21}{60}\left(=-\frac{7}{20}\right)<-\frac{18}{60}\left(=-\frac{3}{10}\right)<-\frac{15}{60}\left(=-\frac{1}{4}\right)<-\frac{12}{60}\left(=-\frac{1}{5}\right)<\frac{18}{60}\left(=\frac{3}{10}\right)$
So, the rational number which does not lie between is $\frac{3}{10}$.
Hence, the correct option is (b).
#### Page No 58:
Since, π has a non-terminating non-recurring decimal expansion.
So, π is an irrational number.
Hence, the correct option is (c).
#### Page No 58:
(c) a non-terminating and non-repeating decimal
Because $\sqrt{2}$ is an irrational number, its decimal expansion is non-terminating and non-repeating.
#### Page No 58:
Since, $\sqrt{225}$ = 15, which is an integer,
0.3799 is a number with terminating decimal expansion, and
$7.\overline{478}$ is a number with non-terminating recurring decimal expansion
Also, 23 is a prime number.
So, $\sqrt{23}$ is an irrational number.
Hence, the correct option is (a).
#### Page No 58:
So, there are 6 digits in the repeating block of digits in the decimal expansion of $\frac{17}{7}$.
Hence, the correct option is (b).
#### Page No 58:
Since,
$\sqrt{\frac{4}{9}}=\frac{2}{3}$, which is a rational number,
$\frac{\sqrt{1250}}{\sqrt{8}}=\sqrt{\frac{1250}{8}}=\sqrt{\frac{625}{4}}=\frac{25}{2}$, which is a rational number,
$\sqrt{8}=2\sqrt{2}$, which is an irrational number, and
$\frac{\sqrt{24}}{\sqrt{6}}=\sqrt{\frac{24}{6}}=\sqrt{4}=2$, which is a rational number
Hence, the correct option is (c).
#### Page No 59:
(d) sometimes rational and sometimes irrational
For example:
$\sqrt{2}$ is an irrational number, when it is multiplied with itself it results into 2, which is a rational number.
$\sqrt{2}$ when multiplied with $\sqrt{3}$, which is also an irrational number, results into $\sqrt{6}$, which is an irrational number.
#### Page No 59:
(d) Every real number is either rational or irrational.
Because a real number can be further categorised into either a rational number or an irrational number, every real number is either rational or irrational.
#### Page No 59:
(d) $\mathrm{\pi }$ is irrational and $\frac{22}{7}$ is rational.
Because the value of $\mathrm{\pi }$ is neither repeating nor terminating, it is an irrational number. $\frac{22}{7}$, on the other hand, is of the form $\frac{p}{q}$, so it is a rational number.
#### Page No 59:
Since, $\frac{\left(\sqrt{2}+\sqrt{3}\right)}{2}$ and $\sqrt{6}$ are irrational numbers,
And,
So, the rational number lying between is 1.6 .
Hence, the correct option is (c).
#### Page No 59:
Since, a number whose decimal expansion is terminating or non-terminating recurring is rational number.
So, 0.853853853... is a rational number.
Hence, the correct option is (d).
#### Page No 59:
Since, the product of a non-zero rational number with an irrational number is always an irrational number.
Hence, the correct option is (a).
#### Page No 59:
Let
Multiplying both sides by 10, we get
Subtracting (1) from (2), we get
$10x-x=2.\overline{)2}-0.\overline{)2}\phantom{\rule{0ex}{0ex}}⇒9x=2\phantom{\rule{0ex}{0ex}}⇒x=\frac{2}{9}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)2}=\frac{2}{9}$
Hence, the correct answer is option (b).
#### Page No 59:
(c) $\frac{5}{3}$
Let x = 1.6666666... ...(i)
Multiplying by 10 on both sides, we get:
10x = 16.6666666... ...(ii)
Subtracting (i) from (ii), we get:
9x = 15
$⇒$x =
#### Page No 59:
(b) $\frac{6}{11}$
Let x = 0.545454... ...(i)
Multiplying both sides by 100, we get:
100x = 54.5454545... ...(ii)
Subtracting (i) from (ii), we get:
99x = 540
$⇒$x$\frac{54}{99}$ = $\frac{6}{11}$
#### Page No 60:
(c) $\frac{29}{90}$
Let x = 0.3222222222... ...(i)
Multiplying by 10 on both sides, we get:
10x = 3.222222222... ...(ii)
Again, multiplying by 10 on both sides, we get:
100x = 32.222222222... ...(iii)
On subtracting (ii) from (iii), we get:
90x = 29
x = $\frac{29}{90}$
#### Page No 60:
(d) none of these
Let x = 0.12333333333... ...(i)
Multiplying by 100 on both sides, we get:
100x = 12.33333333... ...(ii)
Multiplying by 10 on both sides, we get:
1000x = 123.33333333... ...(iii)
Subtracting (ii) from (iii), we get:
900x = 111
$⇒$x = $\frac{111}{900}$
#### Page No 60:
(c) $\sqrt{5×6}$
An irrational number between a and b is given as $\sqrt{ab}$.
#### Page No 60:
(d) 61/4
An irrational number between
#### Page No 60:
(c) $\sqrt{\frac{1}{7}×\frac{2}{7}}$
An irrational number between a and b is given as $\sqrt{ab}$.
#### Page No 60:
Let
Multiplying both sides by 10, we get
Subtracting (1) from (2), we get
$10x-x=3.\overline{)3}-0.\overline{)3}\phantom{\rule{0ex}{0ex}}⇒9x=3\phantom{\rule{0ex}{0ex}}⇒x=\frac{3}{9}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)3}=\frac{3}{9}$
Let
Multiplying both sides by 10, we get
Subtracting (3) from (4), we get
$10y-y=4.\overline{)4}-0.\overline{)4}\phantom{\rule{0ex}{0ex}}⇒9y=4\phantom{\rule{0ex}{0ex}}⇒y=\frac{4}{9}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)4}=\frac{4}{9}$
Sum of $0.\overline{3}$ and $0.\overline{4}$ = $0.\overline{)3}+0.\overline{)4}=\frac{3}{9}+\frac{4}{9}=\frac{7}{9}$
Hence, the correct answer is option (b).
#### Page No 60:
Let
Multiplying both sides by 100, we get
Subtracting (1) from (2), we get
$100x-x=245.\overline{)45}-2.\overline{)45}\phantom{\rule{0ex}{0ex}}⇒99x=245-2=243\phantom{\rule{0ex}{0ex}}⇒x=\frac{243}{99}\phantom{\rule{0ex}{0ex}}\therefore 2.\overline{)45}=\frac{243}{99}$
Let
Multiplying both sides by 100, we get
Subtracting (3) from (4), we get
$100y-y=36.\overline{)36}-0.\overline{)36}\phantom{\rule{0ex}{0ex}}⇒99y=36\phantom{\rule{0ex}{0ex}}⇒y=\frac{36}{99}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)36}=\frac{36}{99}$
So, $2.\overline{)45}+0.\overline{)36}=\frac{243}{99}+\frac{36}{99}=\frac{243+36}{99}=\frac{279}{99}=\frac{31}{11}$
Hence, the correct answer is option (c).
#### Page No 60:
Hence, the correct answer is option (b).
#### Page No 60:
$\left(-2-\sqrt{3}\right)\left(-2+\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=-\left(2+\sqrt{3}\right)×\left[-\left(2-\sqrt{3}\right)\right]\phantom{\rule{0ex}{0ex}}=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)$
Thus, the given expression when simplified is positive and rational.
Hence, the correct answer is option (b).
#### Page No 60:
$\left(6+\sqrt{27}\right)-\left(3+\sqrt{3}\right)+\left(1-2\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=6+\sqrt{3×3×3}-\left(3+\sqrt{3}\right)+\left(1-2\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=6+3\sqrt{3}-3-\sqrt{3}+1-2\sqrt{3}\phantom{\rule{0ex}{0ex}}=4$
Thus, the given expression when simplified is positive and rational.
Hence, the correct answer is option (b).
#### Page No 60:
Hence, the correct answer is option (c).
#### Page No 60:
$\sqrt{20}×\sqrt{5}\phantom{\rule{0ex}{0ex}}=\sqrt{2×2×5}×\sqrt{5}\phantom{\rule{0ex}{0ex}}=2\sqrt{5}×\sqrt{5}$
$=2×5\phantom{\rule{0ex}{0ex}}=10$
Hence, the correct answer is option (a).
#### Page No 60:
$\frac{4\sqrt{12}}{12\sqrt{27}}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{2×2×3}}{12\sqrt{3×3×3}}\phantom{\rule{0ex}{0ex}}=\frac{4×2\sqrt{3}}{12×3\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2}{9}$
Hence, the correct answer is option (b).
#### Page No 61:
$=5×\sqrt{2×3}\phantom{\rule{0ex}{0ex}}=5\sqrt{6}$
Hence, the correct answer is option (ii).
#### Page No 61:
$=\frac{4\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}\phantom{\rule{0ex}{0ex}}=\frac{4}{2}\phantom{\rule{0ex}{0ex}}=2$
Hence, the correct answer is option (b).
#### Page No 61:
Hence, the correct answer is option (c).
#### Page No 61:
Hence, the correct answer is option (b).
#### Page No 61:
Hence, the correct answer is option (d).
#### Page No 61:
$\therefore$ The value of $\sqrt[4]{{\left(64\right)}^{-2}}$ is $\frac{1}{8}$.
Hence, the correct option is (a).
#### Page No 61:
$\therefore$ The value of $\frac{{2}^{0}+{7}^{0}}{{5}^{0}}$ is 2.
Hence, the correct option is (b).
#### Page No 61:
$\therefore$ The value of ${\left(243\right)}^{\frac{1}{5}}$ is 3.
Hence, the correct option is (a).
#### Page No 61:
Hence, the correct option is (c).
#### Page No 61:
$\therefore$Simplified value of ${\left(16\right)}^{-\frac{1}{4}}×\sqrt[4]{16}$ is 1.
Hence, the correct option is (b).
#### Page No 61:
$\therefore$The value of $\sqrt[4]{\sqrt[3]{{2}^{2}}}$ is ${2}^{\frac{1}{6}}$.
Hence, the correct option is (c).
#### Page No 61:
$\therefore$Simplified value of ${\left(25\right)}^{\frac{1}{3}}×{5}^{\frac{1}{3}}$ is 5.
Hence, the correct option is (d).
#### Page No 61:
$\therefore$The value of ${\left[{\left(81\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}$ is 3.
Hence, the correct option is (a).
#### Page No 61:
$\therefore$ x can be $\sqrt[4]{2}$.
Hence, the correct option is (d).
#### Page No 62:
Hence, the correct option is (b).
#### Page No 62:
Hence, the correct option is (b).
#### Page No 62:
Hence, the correct option is (b).
#### Page No 62:
Hence, the correct option is (c).
#### Page No 62:
Hence, the correct option is (a).
#### Page No 62:
Hence, the correct answer is option (d).
#### Page No 62:
Hence, the correct answer is option (d).
#### Page No 62:
$\frac{{5}^{n+2}-6×{5}^{n+1}}{13×{5}^{n}-2×{5}^{n+1}}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{n}×{5}^{2}-6×{5}^{n}×5}{13×{5}^{n}-2×{5}^{n}×5}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{n}×5\left[5-6\right]}{{5}^{n}\left[13-2×5\right]}\phantom{\rule{0ex}{0ex}}=\frac{-5}{3}$
Hence, the correct answer is option (b).
#### Page No 62:
$\sqrt[3]{500}=\sqrt[3]{5×5×2×5×2}=\sqrt[3]{{5}^{3}×{2}^{2}}=5\sqrt[3]{4}$
So, the simplest rationalisation factor of $\sqrt[3]{500}$ is $\sqrt[3]{2}$.
Hence, the correct answer is option (d).
#### Page No 62:
Simplest rationalisation ractor of $\left(2\sqrt{2}-\sqrt{3}\right)$ is $2\sqrt{2}+\sqrt{3}$.
Hence, the correct answer is option (b).
#### Page No 62:
Rationalisation factor of $\frac{1}{2\sqrt{3}-\sqrt{5}}$ will be $2\sqrt{3}+\sqrt{5}=\sqrt{4×3}+\sqrt{5}=\sqrt{12}+\sqrt{5}$.
Hence, the correct answer is option (d).
#### Page No 62:
$\frac{1}{\sqrt{5}+\sqrt{2}}$
Rationalisation of denominator gives
$\frac{1}{\sqrt{5}+\sqrt{2}}×\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}$
Hence, the correct answer is option (d).
#### Page No 62:
$x=2+\sqrt{3}\phantom{\rule{0ex}{0ex}}\frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}×\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$
$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4$
Hence, the correct answer is option (c).
#### Page No 63:
$\frac{1}{\left(3+2\sqrt{2}\right)}=\frac{1}{\left(3+2\sqrt{2}\right)}×\frac{\left(3-2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)}=\frac{\left(3-2\sqrt{2}\right)}{9-8}=\left(3-2\sqrt{2}\right)$
Hence, the correct answer is option (c).
#### Page No 63:
Given: $x=\left(7+4\sqrt{3}\right)$
$\frac{1}{x}=\frac{1}{7+4\sqrt{3}}=\frac{1}{7+4\sqrt{3}}×\frac{7-4\sqrt{3}}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}}{49-48}=7-4\sqrt{3}$
$\left(x+\frac{1}{x}\right)=7+4\sqrt{3}+\left(7-4\sqrt{3}\right)=14$
Hence, the correct answer is option (b).
#### Page No 63:
Hence, the correct answer is option (c).
#### Page No 63:
$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}×\frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$
Given that
Hence, the correct answer is option (b).
#### Page No 63:
$3-2\sqrt{2}=2+1-2×\sqrt{2}×1\phantom{\rule{0ex}{0ex}}={\left(\sqrt{2}\right)}^{2}+{1}^{2}-2×\sqrt{2}×1$
This is of the form
${a}^{2}+{b}^{2}-2ab={\left(a-b\right)}^{2}$
Hence, the correct answer is option (d).
#### Page No 63:
$5+2\sqrt{6}=2+3+2×\sqrt{3}×\sqrt{2}\phantom{\rule{0ex}{0ex}}={\left(\sqrt{2}\right)}^{2}+{\left(\sqrt{3}\right)}^{2}+2×\sqrt{3}×\sqrt{2}$
This is in the form
${a}^{2}+{b}^{2}+2ab={\left(a+b\right)}^{2}$
So, we have
${\left(\sqrt{2}\right)}^{2}+{\left(\sqrt{3}\right)}^{2}+2×\sqrt{3}×\sqrt{2}={\left(\sqrt{2}+\sqrt{3}\right)}^{2}$
Thus, $\sqrt{5+2\sqrt{6}}=\sqrt{{\left(\sqrt{2}+\sqrt{3}\right)}^{2}}=\sqrt{2}+\sqrt{3}$
Hence, the correct answer is option (c).
#### Page No 63:
$\sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}}=\sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}×\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)}}=\sqrt{\frac{{\left(\sqrt{2}-1\right)}^{2}}{2-1}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{2+1-2\sqrt{2}}{1}}\phantom{\rule{0ex}{0ex}}=\sqrt{3-2\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{3-2×1.414}\phantom{\rule{0ex}{0ex}}=\sqrt{3-2.828}\phantom{\rule{0ex}{0ex}}=\sqrt{0.172}\phantom{\rule{0ex}{0ex}}=0.414$
Hence, the correct answer is option (c).
#### Page No 63:
Given: $x=3+\sqrt{8}$
$\frac{1}{x}=\frac{1}{3+\sqrt{8}}=\frac{1}{3+\sqrt{8}}×\frac{3-\sqrt{8}}{3-\sqrt{8}}=\frac{3-\sqrt{8}}{9-8}=\frac{3-\sqrt{8}}{1}=3-\sqrt{8}$
$x+\frac{1}{x}=\left(3+\sqrt{8}\right)+\left(3-\sqrt{8}\right)=6$
${\left(x+\frac{1}{x}\right)}^{2}={x}^{2}+\frac{1}{{x}^{2}}+2×x×\frac{1}{x}={x}^{2}+\frac{1}{{x}^{2}}+2\phantom{\rule{0ex}{0ex}}⇒{6}^{2}={x}^{2}+\frac{1}{{x}^{2}}+2\phantom{\rule{0ex}{0ex}}⇒36={x}^{2}+\frac{1}{{x}^{2}}+2\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\frac{1}{{x}^{2}}=36-2=34$
Hence, the correct answer is option (a).
#### Page No 63:
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
So, Assertion and Reason are correct (property of rational numbers). Also, Reason is the correct explanation of Assertion.
#### Page No 64:
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
3 3
#### Page No 64:
(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.
It is known that e and $\mathrm{\pi }$ are irrational numbers, but Reason is not the correct explanation.
#### Page No 64:
(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.
#### Page No 64:
(a) Because it is a non-terminating and repeating decimal, it is a rational number.
(b) $\mathrm{\pi }$ is an irrational number.
(c) $\frac{1}{7}=.142857142857$...
Hence, its period is 6.
(d)
#### Page No 64:
(a)
${\left({\left(81\right)}^{-2}\right)}^{\frac{1}{4}}\phantom{\rule{0ex}{0ex}}={\left({\left(9\right)}^{-4}\right)}^{\frac{1}{4}}={\left(9\right)}^{-4×\frac{1}{4}}={\left(9\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}$
(b)
(c)
(d)
${\left(\frac{3}{2}\right)}^{4×\frac{-3}{4}}×{\left(\frac{4}{3}\right)}^{3×\frac{-1}{3}}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{2}\right)}^{-3}×{\left(\frac{4}{3}\right)}^{-1}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{2}\right)}^{-3}×\frac{3}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\left(\frac{{3}^{-3}}{{2}^{-3}}\right)×\frac{3}{{2}^{2}}\phantom{\rule{0ex}{0ex}}=\left(\frac{{3}^{-3}×3}{{2}^{-3}×{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{3}^{-2}}{{2}^{-1}}\phantom{\rule{0ex}{0ex}}=\frac{2}{9}$
#### Page No 65:
Sum of a rational number and an irrational number is an irrational number.
Example: 4 + $\sqrt{5}$ represents sum of rational and an irrational number where 4 is rational and $\sqrt{5}$ is irrational.
#### Page No 65:
$\frac{665}{625}=\frac{5×19×7}{{5}^{4}}=\frac{19×7}{{5}^{3}}=\frac{19×7×{2}^{3}}{{5}^{3}×{2}^{3}}=\frac{1064}{1000}=1.064$
So, $\frac{665}{625}$ will terminate after 3 decimal places.
#### Page No 66:
${\left(1296\right)}^{0.17}×{\left(1296\right)}^{0.08}\phantom{\rule{0ex}{0ex}}={\left(1296\right)}^{0.17+0.08}\phantom{\rule{0ex}{0ex}}={\left(1296\right)}^{0.25}\phantom{\rule{0ex}{0ex}}={\left(1296\right)}^{\frac{1}{4}}\phantom{\rule{0ex}{0ex}}=\sqrt[4]{1296}\phantom{\rule{0ex}{0ex}}=6$
#### Page No 66:
$6\sqrt{36}+5\sqrt{12}\phantom{\rule{0ex}{0ex}}=6×6+5\sqrt{4×3}\phantom{\rule{0ex}{0ex}}=36+10\sqrt{3}$
#### Page No 66:
A number which is non terminating and non recurring is known as irrational number.
There are infinitely many irrational numbers between 5 and 6.
One of the example is 5.40430045000460000....
#### Page No 66:
Hence, the value of $\frac{21\sqrt{12}}{10\sqrt{27}}$ is $\frac{7}{5}$.
#### Page No 66:
Hence, the rationalised form is $\sqrt{3}-\sqrt{2}$.
#### Page No 66:
${\left(\frac{2}{5}\right)}^{2x-2}=\frac{32}{3125}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{2}{5}\right)}^{2x-2}=\frac{{2}^{5}}{{5}^{5}}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{2}{5}\right)}^{2x-2}={\left(\frac{2}{5}\right)}^{5}\phantom{\rule{0ex}{0ex}}⇒2x-2=5\phantom{\rule{0ex}{0ex}}⇒2x=5+2\phantom{\rule{0ex}{0ex}}⇒x=\frac{7}{2}$
Hence, $x=\frac{7}{2}$.
#### Page No 66:
Hence, ${\left(32\right)}^{\frac{1}{5}}+{\left(-7\right)}^{0}+{\left(64\right)}^{\frac{1}{2}}$ = 11.
#### Page No 66:
Hence, ${\left(\frac{81}{49}\right)}^{\frac{-3}{2}}=\frac{343}{729}$.
#### Page No 66:
For = 1and b = 2,
${\left({a}^{b}+{b}^{a}\right)}^{-1}\phantom{\rule{0ex}{0ex}}={\left({1}^{2}+{2}^{1}\right)}^{-1}\phantom{\rule{0ex}{0ex}}={\left(1+2\right)}^{-1}$
Thus, the value of (ab + ba)–1 when a = 1 and b = 2 is $\frac{1}{3}$.
#### Page No 66:
Let the two irrational numbers be $2+\sqrt{3}$ and $2-\sqrt{3}$.
Sum of these irrational numbers $=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4$, which is rational
Product of these irrational numbers $=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)={2}^{2}-{\left(\sqrt{3}\right)}^{2}=4-3=1$, which is rational
#### Page No 66:
Yes, the product of a rational and an irrational number is always an irrational number.
Example:
2 is a rational number and $\sqrt{3}$ is an irrational number.
Now, $2×\sqrt{3}=2\sqrt{3}$, which is an irrational number.
#### Page No 66:
The cube roots of natural numbers which are not perfect cubes are all irrational numbers.
Let $x=\sqrt[3]{2}={2}^{\frac{1}{3}}$.
Now,
${x}^{2}={\left({2}^{\frac{1}{3}}\right)}^{2}={2}^{\frac{2}{3}}={\left({2}^{2}\right)}^{\frac{1}{3}}={4}^{\frac{1}{3}}$, which is an irrational number
Also,
${x}^{3}={\left({2}^{\frac{1}{3}}\right)}^{3}={2}^{3×\frac{1}{3}}=2$, which is a rational number
#### Page No 66:
The reciprocal of $\left(2+\sqrt{3}\right)$
$=\frac{\left(2-\sqrt{3}\right)}{4-3}\phantom{\rule{0ex}{0ex}}=\frac{\left(2-\sqrt{3}\right)}{1}\phantom{\rule{0ex}{0ex}}=\left(2-\sqrt{3}\right)$
#### Page No 66:
The value of $\frac{1}{\sqrt{10}}=\frac{1}{\sqrt{10}}×\frac{\sqrt{10}}{\sqrt{10}}=\frac{\sqrt{10}}{10}=\frac{3.162}{10}=0.3162$
#### Page No 66:
We have,
${10}^{x}=64$
Taking square root from both sides, we get
$\sqrt{{10}^{x}}=\sqrt{64}\phantom{\rule{0ex}{0ex}}⇒{\left({10}^{x}\right)}^{\frac{1}{2}}=8\phantom{\rule{0ex}{0ex}}⇒{10}^{\left(\frac{x}{2}\right)}=8$
Multiplying both sides by 10, we get
#### Page No 66:
$\frac{{2}^{n}+{2}^{n-1}}{{2}^{n+1}-{2}^{n}}\phantom{\rule{0ex}{0ex}}=\frac{{2}^{n}\left(1+{2}^{-1}\right)}{{2}^{n}\left(2-1\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\frac{1}{2}\right)}{1}\phantom{\rule{0ex}{0ex}}=\frac{2+1}{2}\phantom{\rule{0ex}{0ex}}=\frac{3}{2}$
${\left[{\left\{{\left(256\right)}^{-\frac{1}{2}}\right\}}^{\frac{1}{4}}\right]}^{2}\phantom{\rule{0ex}{0ex}}={\left\{{\left(256\right)}^{-\frac{1}{2}}\right\}}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}={\left(256\right)}^{-\frac{1}{4}}\phantom{\rule{0ex}{0ex}}={\left({4}^{4}\right)}^{-\frac{1}{4}}\phantom{\rule{0ex}{0ex}}={4}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}$ |
# 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
Given that
The numbers of logs in rows are in the form of an A.P.20, 19, 18…
From the given data
First-term, a = 20
Common difference, d = a2−a1 = 19−20 = −1
Find out
We have to find out in how many rows are the 200 logs placed and how many logs are in the top row
Solution
Let a total of 200 logs be placed in n rows.
Thus, Sn = 200
We know that the sum of nth term formula,
Sn = n/2 [2a +(n -1)d]
S12 = 12/2 [2(20)+(n -1)(-1)]
400 = n (40−n+1)
400 = n (41-n)
400 = 41n−n2
n2−41n + 400 = 0
n2−16n−25n+400 = 0
n(n −16)−25(n −16) = 0
(n −16)(n −25) = 0
Either (n −16) = 0 or n−25 = 0
n = 16 or n = 25
We know the nth term formula,
an = a+(n−1)d
a16 = 20+(16−1)(−1)
a16 = 20−15
a16 = 5
Similarly, the 25th term could be written as;
a25 = 20+(25−1)(−1)
a25 = 20−24
= −4
We can observe that the number of logs in the 16th row is 5 as the numbers cannot be negative.
Hence, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5. |
### Dr. Wilson
A one sided balance beam actually has two sides but one side is reserved for fixed weights that come with the set, and the other side is reserved for the object to be weighed. There are 4 weights which will enable one to weigh any whole number of grams from 1 t0 15. What are they?
We can proceed by taking the smallest possible weight and working up.
The only way to measure a 1g weight would be if one of the weights were 1 gram.
The smallest number of grams that you could not measure with this weight would be a 2 gram weight, so we need a 2 gram weight as well.
With a 1g and 2g weight we can measure 1, 2, or 3 grams. The last weight is weighed by putting the 1g and 2g weights together. The smallest weight we cannot measure with these weights would be a 4 g weight. So we heed a 4 gram weight.
With these weights, we can measure any number of grams up to 7. 7 grams is measured by putting all three weights together. The smallest number of grams that we cannot yet measure is 8 grams. Our fourth and final weight will be an 8 gram weight. With is and the 7 grams from the previous weights, we can measure up to 15 grams.
Notice that the weights we used, 1, 2, 4, and 8 grams are all powers of 2. Moreover, the next weight we will need, 16 is also a power of 2.
The powers of 2 are used in base 2. Very simply, if there is a 1 in a place in the base 2 representation of a number, put the weight that corresponds to that place on the scale. If there is a 0 in that place, do not put that weight on the scale.
There is a combinatorial aspect to this problem. How many different configurations of weights are there? With each weight, there are 2 choices: to put it on or leave it off. So with 4 weights, there will be 24 = 16 configurations. That is enough to weigh any number of grams from 0 to 15.
This illustrates a well known formula from algebra:
1 + 2 + 22 + 23 + . . . + 2n-1 = 2n - 1
top
The 2 Sided Balance Beam Problem
Groupwork |
# Frequency Table
17/04/2020
An important branch of mathematics that deals with gathering, organizing, estimating and interpreting the vast numerical data for a survey or a research, is known as statistics. There may be one or more numbers of statistical data that are used more than once. The number of times a particular data item is utilized, is known as its frequency.
When the distribution of frequencies is listed in a table OR tabular presentation of frequency distribution, known as frequency table. It is used to list out one or more variables taken in a sample. Each sample contains an individual frequency and each frequency is distributed with an interval between each frequency. It is also of two types that is univariate and joint. Frequency distribution can be defined as a summary presentation of the number of observations of an attribute or values of a variable arranged according to their magnitudes either individually in the case of discrete series or in a range or class interval in the case of both discrete and continuing series.
### Frequency Table
Frequency Distribution Table is a way to organize data. A frequency distribution table is an organized tabulation of the number of individual events located in each category. It contains at least two columns, one for the score categories (X) and another for the frequencies (f). Below we have explained briefly for you to understand the concept of frequency table better and workout frequency table example:
Solved Example
Question: Here is the list of marks obtained for the students in the examination. Find the number of students who got more than 85 marks, More than 95, Less than 80 more than 76.
Score (X) Frequency (f) Below 75 4 76 – 80 14 81 – 85 2 86 – 90 8 91 – 95 5 96 – 100 1
Solution:
From the table we can conclude that:
Students who got more than 85 = 8 + 5 + 1 = 14
Students who got more than 95 = 1
Students who got less than 80 more than 76 = 14.
#### Construction of Frequency Distribution
The following steps are involved in the construction of a frequency distribution.
(1) Find the range of the data: The range is the difference between the largest and the smallest values.
(2) Decide the approximate number of classes in which the data are to be grouped. There are no hard and first rules for number of classes. In most cases we have 5 to 20 classes. H.A. Sturges provides a formula for determining the approximation number of classes.
K=1+3.322logN
where K= Number of classes
and logN = Logarithm of the total number of observations.
Example: If the total number of observations is 50, the number of classes would be
K=1+3.322logN
K=1+3.322log50
K=1+3.322(1.69897)
K=1+5.644
K=6.644
7 classes, approximately.
(3) Determine the approximate class interval size: The size of class interval is obtained by dividing the range of data by the number of classes and is denoted by h class interval size
h = Range Number of Classes
In the case of fractional results, the next higher whole number is taken as the size of the class interval.
(4) Decide the starting point: The lower class limit or class boundary should cover the smallest value in the raw data. It is a multiple of class intervals.
Example: 0,5,10,15,20, etc. are commonly used.
(5) Determine the remaining class limits (boundary): When the lowest class boundary has been decided, by adding the class interval size to the lower class boundary you can compute the upper class boundary. The remaining lower and upper class limits may be determined by adding the class interval size repeatedly till the largest value of the data is observed in the class.
(6) Distribute the data into respective classes: All the observations are divided into respective classes by using the tally bar (tally mark) method, which is suitable for tabulating the observations into respective classes. The number of tally bars is counted to get the frequency against each class. The frequency of all the classes is noted to get the grouped data or frequency distribution of the data. The total of the frequency columns must be equal to the number of observations. |
# Line (in 2 Dimensions)
• ### Definition
A (straight) line is the simplest of geometrical figures. Imagining a line is very easy. The condition of straightness is also very intuitive. We now formally define a line as a locus of a point satisfying a condition.
NOTE : Throughout this chapter, we will be using the coordinate geometry.
• ### Inclination of a Line
The smallest non-negative angle made by a line with positive ${X}$ axis is known as the inclination. It is generally denoted by ${\theta}$.
I) Parallel lines have same inclinations.
II) Inclination of ${X}$ axis is ${0^o}$ and inclination of ${Y}$ axis is ${90^o}$.
• ### Slope or Gradient of a Line , ${m}$
If ${\theta}$ is the inclination, then slope is given by ${tan (\theta)}$.
NOTE : Knowing trigonometric ratios of particular angles is beneficial.
I) Parallel lines have same slope.
II) Slope of ${X}$ axis is ${0}$ and slope of ${Y}$ axis is not defined.
III) Product of slopes of perpendicular lines is ${-1}$. This property is extremely important.
If ${(x_1, y_1)}$ and ${(x_2, y_2)}$ are 2 distinct points on a line, its slope is given by
${m = \frac {y_2 - y_1}{x_2 - x_1}}$
• ### Formal Definition
Line is locus of points, such that slope of line segment joining any 2 distinct points on the line is same.
NOTE : Since line is a locus of points, it has an equation.
• ### Intercepts
The ${X}$ and ${Y}$ intercepts of a line are the ${X}$ and ${Y}$ coordinates of the points, where the line meets respective axes. These being directed distances, can be either positive or negative.
• ### Equations of Line in Various Forms
I) Slope- Point Form
Equation of a line passing through point ${(x_1,y_1)}$ and having a slope ${m}$ is
${y- y_1 = m (x-x_1)}$
II) Two-Point Form
Equation of a line passing through points ${(x_1,y_1)}$ and ${(x_2,y_2)}$ is given by
${\frac {y-y_1}{y_2 - y_1} = \frac {x - x_1}{x_2 - x_1}}$
III) Slope-Intercept Form
Equation of a line having slope ${m}$ and ${y}$ intercept ${c}$ is given by
${y = mx +c}$
IV) Double-Intercept Form
Equation of a line having $X$ and $Y$ intercepts as $a$ and $b$ is given by
${\frac {x}{a} + \frac {y}{b} = 1}$
V) Normal Form (Rarely Used)
If ${p}$ is the length of perpendicular from origin to the line and ${\alpha}$ is the inclination of the perpendicular segment, then
${x \ cos (\alpha) + y \ sin (\alpha) = p}$
VI) Parametric Form (More Effective in 3D)
If ${\theta}$ is the inclination and ${(x_1,y_1)}$ is the point on a line, then the parametric equations are given by
${x = x_1 + r \ cos (\theta) \ and \ y = y_1 + r \ sin (\theta),}$
where ${r}$ is the distance of ${(x,y)}$ from ${(x_1, y_1)}$
• ### General Equation of 1st Order in 2 Variables
The general equation of 1st order in 2 variables is of the form
${ax + by + c = 0}$
It represents a straight line.
The necessary condition is both ${a}$ and ${b}$ cannot be simultaneously zero.
• ### Angle Between Any 2 Lines
Let ${\theta}$ be the angle between 2 lines having slopes ${m_1}$ and ${m_2}$. Then,
${tan (\theta) = \begin {vmatrix} \frac {m_1 - m_2}{1+ m_1 m_2} \end {vmatrix}}$
• ### Point of Intersection of 2 Lines
We know that 2 non-parallel lines intersect each other at only 1 point. Let the equations of lines be
${a_1 x + b_1 y = c_1 \ and \ a_2 x + b_2 y = c_2}$
The coordinates of point of intersection are
${\begin {pmatrix}\frac {\begin {vmatrix} c_1 & c_2 \\ b_1 & b_2 \end {vmatrix}}{\begin {vmatrix} a_1 & a_2 \\ b_1 & b_2 \end {vmatrix}} , \frac {\begin {vmatrix} a_1 & a_2 \\ c_1 & c_2 \end {vmatrix}}{\begin {vmatrix} a_1 & a_2 \\ b_1 & b_2 \end {vmatrix}} \end {pmatrix}}$
The condition is,
${\begin {vmatrix} a_1 & a_2 \\ b_1 & b_2 \end {vmatrix} \ne 0}$
• ### Condition for Concurrency of 3 Lines
Lines are said to be concurrent, when all of them intersect at a point.
Let the lines be ${a_1 x + b_1 y + c_1 = 0}$ , ${a_2x + b_2 y + c_2 = 0}$ and ${a_3x + b_3 y + c_3 = 0}$ . The lines are concurrent, when
${\begin {vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end {vmatrix} = 0}$
The point of concurrence can then be obtained easily by considering any 2 of those 3 lines.
• ### Length of Perpendicular from a Point to a Line
Let ${P (x_1,y_1)}$ be a point and ${ax+by+c=0}$ be a line. Then, the length of perpendicular from ${P}$ to the line is given by
${\begin {vmatrix} \frac {ax_1 + by_1 + c}{\sqrt {a^2+b^2}} \end {vmatrix}}$
• ### Distance between 2 Parallel Lines
We know that parallel lines have same slope. This means, in their equations, the ratio of coefficients of ${x}$ and ${y}$ will be same. Let the lines be ${ax+by+c_1 = 0}$ and ${ax+by+c_2 = 0}$. The distance is given by
${\begin {vmatrix} \frac {c_1 - c_2}{\sqrt {a^2+b^2}} \end {vmatrix}}$
• ### Family of Straight Lines
This is a useful concept. We know that, through a single point, infinitely many lines pass. Let ${u \equiv a_1x + b_1 y + c_1 =0}$ and ${v \equiv a_2 x + b_2 y + c_2 =0}$ be 2 intersecting lines. Then, equation of family of lines passing through the point of intersection is given by
${u + \lambda v = 0, \ \lambda \in \mathbb R,}$
where ${\lambda}$ is a parameter. |
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