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Courses Courses for Kids Free study material Offline Centres More Store The following pie-chart represents different types of sweets sold in a shop on a particular day.Read the pie-chart and answer the following questions:(i) How many kilograms of barfi and gulab jamun are sold if 50 kg sweets are sold on the day?(ii) Find the fraction of the circle representing each of the information. Last updated date: 17th Apr 2024 Total views: 35.1k Views today: 0.35k Verified 35.1k+ views Hint: Here, we will find the amount of the sweets by multiplying the given percentage by the amount sold on the day and then add them to find the net sold. Then we will find the fractions by simplifying the given percentage representing each of the information. Complete step-by-step solution (i) Now we will find the amount of barfi sold when 50 kg sweets are sold on the day. ${\text{Barfi sold}} = 50 \times \dfrac{{30}}{{100}} \\ = 15{\text{ kg}} \\$ We also calculate the amount of gulab jamun sold when 50 kg sweets are sold on the day. ${\text{Gulab Jamun sold}} = 50 \times \dfrac{{25}}{{100}} \\ = \dfrac{{125}}{{10}} \\ = 12.5{\text{ kg}} \\$ Adding the above amount of barfi and gulab jamun sold, we get ${\text{Net sold}} = 15 + 12.5 \\ = 27.5{\text{ kg}} \\$ Thus, the net sold of barfi and gulab jamun is $27.5$ kg. (ii) Since it is given that the pie chart is formed from percentage, so the total number outcomes will be 100. We will now find the fraction of percentage for laddu from the given pie chart. ${\text{Fraction for Laddu}} = \dfrac{{40}}{{100}} \\ = \dfrac{2}{5} \\$ Finding the fraction of percentage for barfi from the given pie chart, we get ${\text{Fraction for Barfi }} = \dfrac{{30}}{{100}} \\ = \dfrac{3}{{10}} \\$ Now we will calculate the fraction of percentage for gulab jamun from the given pie chart. ${\text{Fraction for Gulab Jamun }} = \dfrac{{25}}{{100}} \\ = \dfrac{1}{4} \\$ We will now find the fraction of percentage for jalebi from the given pie chart. ${\text{Fraction for Jalebi }} = \dfrac{5}{{100}} \\ = \dfrac{1}{{20}} \\$ Thus, the fraction for laddu is $\dfrac{2}{5}$, fraction for barfi is $\dfrac{3}{{10}}$, fraction for gulab jamun is $\dfrac{1}{4}$ and fraction for jalebi is $\dfrac{1}{{20}}$. Note: In solving these types of questions, you should be familiar with the conversion of percentage to fractions. Then use the given conditions and values given in the question after examining the pie chart carefully, to find the values. Some students take the amount of one sweet as the required answer instead of adding there amounts to find the net sold.
What is the slope and y-intercept of x+3y=6? Jan 23, 2018 $\text{slope "=-1/3" y-intercept } = 2$ Explanation: $\text{the equation of a line in "color(blue)"slope-intercept form}$ is. •color(white)(x)y=mx+b $\text{where m is the slope and b the y-intercept}$ $\text{rearrange "x+3y=6" into this form}$ $\text{subtract x from both sides}$ $\cancel{x} \cancel{- x} + 3 y = 6 - x$ $\Rightarrow 3 y = - x + 6$ $\text{divide all terms by 3}$ $\Rightarrow y = - \frac{1}{3} x + 2 \leftarrow \textcolor{b l u e}{\text{in slope-intercept form}}$ $\Rightarrow \text{slope "=-1/3" and y-intercept } = 2$ graph{-1/3x+2 [-10, 10, -5, 5]}
# Line of Best Fit We often use a 'best-fitting' line to show any relationship A line of best fit is drawn on the scatter diagram. It is drawn so that the points are evenly distributed on either side of the line. There are various methods for drawing this 'precisely' but you will only be expected to draw the line 'by eye'. When drawing the line of best fit, use a transparent ruler so that you can see how the line fits between all the points before you draw it. ### Example The heights and weights of twenty children in a class are recorded. The results are shown on the scatter diagram below. Katie is 148 cm tall. Draw a line of best fit and use it to estimate her weight. We start by drawing a line of best fit. Katie is 148 cm tall, so we draw a straight line up from 148cm on the horizontal axis until it meets the line of best fit and then along until it meets the vertical axis. Katie weighs approximately 52 kg. As you are only drawing the line of best fit 'by eye', it is unlikely your answer will be exactly the same as someone else's answer. Sometimes you are given the equation of the line of best fit. You can use this in estimation. ### Example The equation of the line of best fit for a set of data is $$w = 1.5h - 170$$. Use this equation to obtain an estimate for the weight of Louise, who is 156 cm tall. We substitute $$h = 156$$ into the equation. $w = 1.5 \times 156 - 170$ $w = 234 - 170$ $w = 64$
Class 7 Maths # Solid Shapes • Shapes with three dimensions are called solid shapes. • The corners of a solid shape are called its vertices. The line segments are called edges and flat surfaces are called faces. • A skeleton outline of a solid that can be folded to make the solid is called a net of that solid. A particular solid can have many types of nets. • An oblique sketch of a solid does not show proportional lengths. An isometric sketch shows proportional measurements of a solid. It is drawn on an isometric dot paper. • Different sections of a solid can be viewed in many ways. • It can be done by cutting or slicing the shape. This would show the cross-section of the solid. • It can be done by observing 2-D shadow of a 3-D shape. • It can be done by looking at the shape from different angles; like front view, side view and top view. ## Exercise 15.1 Question 1: Identify the nets which can be used to make cubes. Answer: (ii), (iii), (iv) and (vi) Question 2: Dice are cubes with dots on each face. Opposite faces of a die always have a total of seven dots on them. Insert suitable numbers in the blanks, remembering that the number on the opposite face should total to 7. Question 3: Can this be a net for a die? Explain your answer. Answer: No, because 4 is opposite 1 and 3 is opposite 6 and they do not add up to 7. Question 4: Here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there in the net here?
# Texas Go Math Grade 2 Lesson 1.5 Answer Key Hundreds, Tens, and Ones Refer to our Texas Go Math Grade 2 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 2 Lesson 1.5 Answer Key Hundreds, Tens, and Ones. ## Texas Go Math Grade 2 Lesson 1.5 Answer Key Hundreds, Tens, and Ones Explore Write the number of hundreds, tens, and ones. Then draw a quick picture. FOR THE TEACHER • Read the following to children. Steven has 243 yellow blocks. How many hundreds, tens, and ones are in this number? Repeat for 423 red blocks. The number of yellow blocks Steven has=243 Now we have to write the digits of place values. The place value of 243 is: 2 is in the hundreds place 4 is in the tens place 3 is in the one’s place. Likewise, we have to repeat for red blocks. The number of red blocks steven has=423 Now we have to write the place values for 423: 4 is in the hundreds place 2 is in the tens place 3 is in the one’s place Math Talk Mathematical Processes Describe how the two numbers are alike. Describe how they are different. Answer: The only difference is two numbers are place values. In Mathematics, place value charts help us to make sure that the digits are in the correct places. To identify the positional values of numbers accurately, first, write the digits in the place value chart and then write the numbers in the usual and the standard form. For example: Model and Draw Write how many hundreds, tens, and ones there are in the model. What are two ways to write this number? Place value: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. There are 2 blocks of hundreds it is equivalent to 200 so it was 2100=200. There are 4 blocks of tens it is equivalent to 10 so it was 410=40 There are 7 blocks of ones it is equivalent to 1 so it was 71=7 The total blocks:200+40+7 (expanded form) The total blocks:247 (standard form) Word form: Two hundred forty-seven. Share and Show Write how many hundreds, tens, and ones are in the model. Write the number in two ways. Question 1. Place value: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. There is 1 block of hundreds it is equivalent to 100 so it was 1100=300. There are 6 blocks of tens it is equivalent to 10 so it was 610=60 There are 0 blocks of ones it is equivalent to 1 so it was 01=0 The total blocks:100+60+0 (expanded form) The total blocks:160 (standard form) Word form: One hundred sixty. Question 2. Place value: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. There are 2 blocks of hundreds it is equivalent to 100 so it was 2100=200. There are 2 blocks of tens it is equivalent to 10 so it was 210=20 There are 5 blocks of ones it is equivalent to 1 so it was 51=5 The total blocks:200+20+5 (expanded form) The total blocks:225 (standard form) Two hundred twenty-five (word form) Question 3. Place value: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. There are 2 blocks of hundreds it is equivalent to 100 so it was 2100=200. There are 3 blocks of tens it is equivalent to 10 so it was 310=30 There are 9 blocks of ones it is equivalent to 1 so it was 91=9 The total blocks:200+30+9 (expanded form) The total blocks:239 (standard form) Two hundred thirty-nine (word form) Problem Solving Write how many hundreds, tens, and ones are in the model. Write the number in two ways. Question 4. Explanation: Place value: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. There are 7 blocks of hundreds it is equivalent to 100 so it was 3100=300. There are 5 blocks of tens it is equivalent to 10 so it was 010=0 There are 4 blocks of ones it is equivalent to 1 so it was 41=4 The total blocks:300+0+4 (expanded form) The total blocks:304 (standard form) Three hundred four (word form) Question 5. Place value: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. There are 7 blocks of hundreds it is equivalent to 100 so it was 3100=300. There are 5 blocks of tens it is equivalent to 10 so it was 710=70 There are 4 blocks of ones it is equivalent to 1 so it was 81=8 The total blocks:300+70+8 (expanded from) The total blocks:378 (standard form) Three hundred seventy-eight (word form) Solve. Write or draw to explain. Question 6. H.O.T. A model for my number has 4 ones blocks, 5 tens blocks, and 7 hundreds blocks. What number am I? Place value: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. There are 7 blocks of hundreds it is equivalent to 100 so it was 7100=700. There are 5 blocks of tens it is equivalent to 10 so it was 510=50 There are 4 blocks of ones it is equivalent to 1 so it was 41=4 Question 7. H.O.T. Multi-Step The hundreds digit of my number is greater than the tens digit. The one’s digit is less than the tens digit. What could my number be? Write it in two ways. ___________ + ___________ + ____________ _____________ Answer: In Mathematics, place value charts help us to make sure that the digits are in the correct places. To identify the positional values of numbers accurately, first, write the digits in the place value chart and then write the numbers in the usual and the standard form. The 10 digits we used to represent the numbers are: 0 1 2 3 4 5 6 7 8 9 – In the hundreds digit the number should be greater than the tens digit. Assume that 9. – The one’s digit should be less than the tens digit. Assume that 7. – The tens digit should be less than the hundreds digit and greater than the ones digit. Assume that 8 The total number is 978 The expanded form of the number is 900+80+7. The word form is nine hundred eighty-seven. Question 8. Representations Daniel has these bags of marbles. How many marbles does Daniel have? (A) 110 (B) 320 (C) 230 Explanation: There are 2 100’s bags and 3 10’s bags. 2 100’s is equivalent to 2100=200 3 tens are equivalent to 310=30 So the place value will be 2 is hundreds place, 3 is in the tens place and 0 is in one’s place. Question 9. Analyze Gina uses pennies to buy a drum. The number of pennies she uses has 7 tens. Which number could it be? (A) 274 (B) 167 (C) 728 Explanation: The number of tens of pennies she uses=7 Now check out the options which number is having 7 in the tens place and this is the right option. Place value: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. Question 10. What is another way to write 2 hundreds, 3 ones, and 8 tens? (A) 280 (B) 823 (C) 283 Explanation: We have to write the standard form. The standard form of 2 hundreds, 3 ones, 8 tens is 283. Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. Question 11. TEXAS Test Prep Karen has 279 pictures. How many hundreds are in this number? (A) 9 hundred (B) 2 hundred (C) 7 hundred Explanation: The number of pictures Karen has=279 Here we have to write the place value of hundred. Place value: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. There are 2 blocks of hundreds it is equivalent to 100 so it was 2100=200. There are 7 blocks of tens it is equivalent to 10 so it was 710=70 There are 9 blocks of ones it is equivalent to 1 so it was 91=9 TAKE-HOME ACTIVITY • Say a 3-digit number, such as 546. Have your child draw a quick picture for that number. Answer: Explanation is shown in the below representation. Explanation: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. There are 5 blocks of hundreds it is equivalent to 100 so it was 5100=500. There are 4 blocks of tens it is equivalent to 10 so it was 410=40 There are 6 blocks of ones it is equivalent to 1 so it was 61=6 The total block: 500+40+6 The total block:546. ### Texas Go Math Grade 2 Lesson 1.5 Homework and Practice Answer Key Write how many hundreds, tens, and ones are in the model. Write the number in two ways. Question 1. 3 blocks of hundreds and 3 blocks of ones and 8 blocks of tens. Explanation: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. Question 2. 2 blocks of hundreds and 7 blocks of ones and 0 tens. Explanation: Place value in Maths describes the position or place of a digit in a number. Each digit has a place in a number. When we represent the number in general form, the position of each digit will be expanded. Those positions start from a unit place or we also call it one’s position. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on. Problem Solving Solve. Write or draw to explain Question 3. A model for my number has 6 one’s blocks, 2 tens blocks, and 8 hundred blocks. What number am I? Explanation: The number of one’s blocks=6 The number of tens blocks=2 The number of hundred blocks=8 Here we write the first hundred places, second tens place and third one’s place. The number is 826. Question 4. Multi-Step The hundreds digit of my number is less than the tens digit. The ones digit is greater than the tens digit. What could my number be? _________ + _________ + ___________ __________ Answer: In Mathematics, place value charts help us to make sure that the digits are in the correct places. To identify the positional values of numbers accurately, first, write the digits in the place value chart and then write the numbers in the usual and the standard form. The 10 digits we used to represent the numbers are: 0 1 2 3 4 5 6 7 8 9 – In the hundreds digit the number should be less than the tens digit. Assume that 1 – The one’s digit should be greater than the tens digit. Assume that 3. – The tens digit should be greater than the hundreds digit and less than the ones digit. Assume that 2 The total number is 123 The expanded form of the number is 100+20+3. The word form is one hundred twenty-three. Lesson Check Question 5. The number of coins needed to buy a notepad has 3 tens. Which number could it be? (A) 432 (B) 173 (C) 593 Explanation: 3 tens are nothing but 310=30 The value of digits of 30: 3 is in the tens place 0 is in one’s place. In option A 3 is in tens place so I prefer that option. Question 6. Tim has one pail of 100 coins. He has a second pail of 5 coins. He has the third pail of 40 coins. How many coins does he have? (A) 154 (B) 451 (C) 145 The number of one pail of coins Tim has=100 The number of the second pail of coins Tim has=5 The number of the third pail of coins Tim has=40 The total number of coins he has=X Add the coins of 1st pail, second pail, and third pail X=100+40+5 X=145 Therefore, the total coins are 145. Question 7. Sara wants to buy a toy. She needs 175 coins. Which model shows the number of coins? Explanation: The value of digits of 175: 1 is in the hundreds place 7 is in the tens place 5 is in one’s place. Question 8. Kevin has 438 marbles. How many hundreds are in this number? (A) 8 hundred (B) 3 hundred (C) 4 hundred
Courses Courses for Kids Free study material Offline Centres More Store # Three resistors $(2\Omega ,4\Omega ,4\Omega )$ are combined to achieve ${{R}_{\max }}$ (maximum resistance) and ${{R}_{\min }}$ (minimum resistance). Then the average and ratio of ${{R}_{\max }}$ and ${{R}_{\min }}$ shall be respectively:A. 5.5, 1:10B. 5.5, 11:1$C.\dfrac{22}{4},10:1$D. Both 1 and 3 are correct Last updated date: 13th Sep 2024 Total views: 429.3k Views today: 10.29k Verified 429.3k+ views Hint: We will make use of the formula for calculating the resistors connected in parallel to find the value of the minimum resistance and the formula for calculating the resistors connected in series to find the value of the maximum resistance. Then, we will find the average and the ratio of these values. Formula used: $\dfrac{1}{{{R}_{P}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}+......+\dfrac{1}{{{R}_{n}}}$ ${{R}_{S}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+......+{{R}_{n}}$ The formula used to find the resistance of the resistors connected in parallel is as follows. $\dfrac{1}{{{R}_{P}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}+......+\dfrac{1}{{{R}_{n}}}$ Similarly, the formula used to find the resistance of the resistors connected in series is as follows. ${{R}_{S}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+......+{{R}_{n}}$ From given, we have the values of the resistors to be equal to$(2\Omega ,4\Omega ,4\Omega )$. As we are given with the resistance values of the 3 resistors, firstly, we will derive the formula for the equivalent resistance of the 3 resistors connected in parallel. So, we have, $\dfrac{1}{{{R}_{P}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}$ Now, substitute the values of the resistance of the 3 resistors given in the above equation. \begin{align} & \dfrac{1}{{{R}_{P}}}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4} \\ & \Rightarrow \dfrac{1}{{{R}_{P}}}=\dfrac{1}{2}+\dfrac{1}{2} \\ \end{align} Continue the further calculation. \begin{align} & \dfrac{1}{{{R}_{P}}}=1 \\ & \Rightarrow {{R}_{P}}=1 \\ \end{align} Therefore, the value of the minimum resistance is$1\,\Omega$. As we are given with the resistance values of the 3 resistors, firstly, we will derive the formula for the equivalent resistance of the 3 resistors connected in series. So, we have, ${{R}_{S}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}$ Now, substitute the values of the resistance of the 3 resistors given in the above equation. \begin{align} & {{R}_{S}}=2+4+4 \\ & \Rightarrow {{R}_{S}}=10 \\ \end{align} Therefore, the value of the maximum resistance is$1\,0\,\Omega$. The average value of the minimum resistance $1\,\Omega$ and the maximum resistance $1\,0\,\Omega$ is calculated as follows. \begin{align} & \text{Avg}=\dfrac{{{R}_{\max }}+{{R}_{\min }}}{2} \\ & \Rightarrow \text{Avg}=\dfrac{10+1}{2} \\ \end{align} Upon further continuing the calculation, we get, $\text{Avg}=5.5$ Therefore, the average value of the maximum and minimum resistance is 5.5 or $\dfrac{22}{4}$. The ratio of the minimum resistance $1\,\Omega$and the maximum resistance $1\,0\,\Omega$ is calculated as follows. \begin{align} & R=\dfrac{{{R}_{\max }}}{{{R}_{\min }}} \\ & \Rightarrow R=\dfrac{10}{1} \\ \end{align} Therefore, the ratio of the maximum and minimum resistance is 10:1. As the values the average and ratio of ${{R}_{\max }}$ and ${{R}_{\min }}$ are $\dfrac{22}{4},10:1$. So, the correct answer is “Option C”. Note: The average of the values of the maximum and minimum resistance can be found by first adding the maximum and minimum values and then dividing this value by 2. The ratio of the values of the maximum and minimum resistance can be found by dividing the values of the maximum and minimum resistance.
# Find the volume generated by revolving about the line y = 6 the area bounded by y = 1/x and 3x + 2y = 7. lemjay \| Certified Educator To start, we need to determine the area bounded by the given equations. To do so, plot the two equations. As shown below, the graph of y=1/x is the red curve and the graph of 3x+2y=7 is the blue line. Hence, the bounded region is the graph at the right. Then, solve the intersection between the blue line and the red curve using substitution method. So, substitute y=1/x to 3x+2y=7. `3x+2y=7` `3x + 2(1/x) = 7` Multiply both sides by x to simplify. `x(3x+2/x)=7x` `3x^2+2=7x` Express the equation in quadratic form `ax^2+bx+c=0` to be able to factor. `3x^2-7x+2=0` `(x-2)(3x-1)=0` Set each factor to zero and solve for x. `x-2=0` and `3x-1=0` `x=2` `x=1/3` Substitute values of x to y=1/x. `x=1/3 ` , `y=1/(1/3) = 3` `x=2` , `y=1/2` Hence the intersection between the two equations are (1/3,3) and (2,1/2). Next, apply the formula of disk method to determine the volume which is: `V = pi int_a^b (r_o^2 - r_i^2) dx` The limits of the integral a and b are the x-coordinates of the intersection of the red curve and the blue line. So, a=1/3 and b=2. `r_o` and `r_i` are the outer and inner radius of the disk formed when the bounded region is revolved around a certain line/axis. So radius is the distance between the axis of rotation and the curve. To determine `r_o` and `r_i` , let's plot the bounded region and the axis of rotation which is y=6. Base on the graph above, the outer radius is the difference between y=6 and y=1/x. `r_o = 6 - 1/x` Then, take the square of `r_o` . `r_o^2= (6-1/x)^2 = 36-12/x + 1/x^2` And the inner radius is the difference between y=6 and y=(7-3x)/2. Note that y=(7-3x)/2 is base on the equation of the blue line which is 3x+2y=7. So, `r_i = 6 - (7-3x)/2= 12/2 - (7-3x)/2 = (3x +5)/2` Take the square of `r_i` . `r_i^2 = ((3x+5)/2^2 = 9/4x^2 +15/2x+25/4` Substitute `r_o^2` and `r_i^2` to the formula of volume. `V= pi int_(1/3)^2 [(36-12/x + 1/x^2) - (9/4x^2+15/2x+25/4)]dx` `V= pi int_(1/3)^2 ( 119/4 -12/x+1/x^2 - 9/4x^2 - 15/2x)dx` `V= pi (119/4x -12lnx -1/x -3/4x^3 -15/4x^2)` `\| _(1/3)^2` `V = pi (1135/36 - 12ln6) = 31.5` Hence, volume is 31.5 cubic units. sciencesolve \| Certified Educator You may use the following formula to determine the volume of the solid generated revolving the region bounded by the curves `y=1/x` and the line `3x+2y=7` around the line y=6 such that: `V = int_a^b pi(R^2(y) - r^2(y))dy` You need to find the limits of integration a and b substituting `1/y` for x in equation `3x+2y=7` such that: `3/y + 2y = 7 => 3 + 2y^2 - 7y = 0` `2y^2 - 7y + 3 = 0` `y_(1,2) = (7+-sqrt(49 - 24))/4 => y_(1,2) = (7+-sqrt25)/4` `y_(1,2) = (7+-5)/4 => y_1 = 3 ; y_2 = 1/2` You need to determine the radii of the washer such that: `R(y) = 6 and r(y) = 6 - 1/` y `V = int_(1/2)^3 pi(36 - (6 - 1/y)^2)dy` `V = pi int_(1/2)^3(36 - 36 + 12/y - 1/y^2)dy` `V = pi int_(1/2)^3 12/y dy - pi int_(1/2)^3 1/y^2 dy` `V = 12piln y\|_(1/2)^3+ pi/y\|_(1/2)^3` `V = 12pi(ln 3 - ln(1/2)) + pi/3 - 2pi` `V = 12pi*ln6 - 5pi/3 => V = pi(36ln6 - 5)/3` Hence, evaluating the volume of solid of revolution, under the given conditions, yields `V = pi(36ln6 - 5)/3` . You may evaluate the volume such that: Using the fundamental theorem of calculus yields: Hence, evaluating the volume of solid generated by revolving about the line the area bounded by and yields
Volume of Cylinders, Cones & Spheres \| Math for Kids Grade 6, 7, 8 1% It was processed successfully! HOW DO YOU FIND THE VOLUME OF CYLINDERS, CONES & SPHERES? Volume is the space inside a 3D solid. Volume is measured in cubic units. Formulas are rules that are written using mathematical symbols that relate quantities. You have already used volume formulas to find the volume of rectangular prisms and cubes. You can also use volume formulas to find the volumes of cones, cylinders, and spheres. To better understand how to find the volume of cylinders, cones, and sphere… HOW DO YOU FIND THE VOLUME OF CYLINDERS, CONES & SPHERES?. Volume is the space inside a 3D solid. Volume is measured in cubic units. Formulas are rules that are written using mathematical symbols that relate quantities. You have already used volume formulas to find the volume of rectangular prisms and cubes. You can also use volume formulas to find the volumes of cones, cylinders, and spheres. To better understand how to find the volume of cylinders, cones, and sphere… ## LET’S BREAK IT DOWN! ### Review Volume You can find the volume of a rectangular prism by seeing how many cubic units fit inside it. The number of unit cubes that fit inside tells you the volume. You can also use a formula. The volume of a rectangular prism is length times width times height. If the length is 3 units, the width is 2 units, and the height is 4 units, the volume is 3 × 2 × 4 = 24 cubic units. Try this one yourself: A rectangular prism has length 5 units, width 6 units, and height 2 units. What is the volume of the rectangular prism? Review Volume You can find the volume of a rectangular prism by seeing how many cubic units fit inside it. The number of unit cubes that fit inside tells you the volume. You can also use a formula. The volume of a rectangular prism is length times width times height. If the length is 3 units, the width is 2 units, and the height is 4 units, the volume is 3 × 2 × 4 = 24 cubic units. Try this one yourself: A rectangular prism has length 5 units, width 6 units, and height 2 units. What is the volume of the rectangular prism? ### Volume of a Cylinder How can you find the volume of a cake in the shape of a cylinder? Think about this shape by comparing it to a rectangular prism. The formula for the volume of a rectangular prism is length times width times height. The length times the width is the area of the base. So, another way to think of this formula is the area of the base times the height. You can think of the volume of a cylinder in the same way: the area of the base times the height. The base is a circle. A circle has area πr2. So, the formula for the volume of a cylinder is πr2h. Amari’s cake is a cylinder with diameter 44 cm and height 50 cm. What is the volume of the cake? The radius is half the diameter, so the radius is 22 cm. Use the formula. 3.14 × 222 × 50 = 75,988 cubic centimeters. Remember that when you use 3.14 to approximate pi, your answer is an approximation. Try this one yourself: A cylinder has height 20 cm and diameter 16 cm. What is the volume of the cylinder? Volume of a Cylinder How can you find the volume of a cake in the shape of a cylinder? Think about this shape by comparing it to a rectangular prism. The formula for the volume of a rectangular prism is length times width times height. The length times the width is the area of the base. So, another way to think of this formula is the area of the base times the height. You can think of the volume of a cylinder in the same way: the area of the base times the height. The base is a circle. A circle has area πr2. So, the formula for the volume of a cylinder is πr2h. Amari’s cake is a cylinder with diameter 44 cm and height 50 cm. What is the volume of the cake? The radius is half the diameter, so the radius is 22 cm. Use the formula. 3.14 × 222 × 50 = 75,988 cubic centimeters. Remember that when you use 3.14 to approximate pi, your answer is an approximation. Try this one yourself: A cylinder has height 20 cm and diameter 16 cm. What is the volume of the cylinder? ### Volume of a Cone How can you find the volume of a plastic cup in the shape of a cone? Compare the cone to a cylinder with the same diameter and same height. If you fill the cone with water and pour it into the cylinder, it takes exactly 3 cones full of water to fill the cylinder. The volume of the cone is one third the volume of the cylinder. Remember, the formula for the volume of a cylinder is πr2h. So, the formula for the volume of a cone is [ggfrac]1/3[/ggfrac]πr2h. Emily has a cone cup. The height is 10 cm and the diameter is 6 cm. What is the volume of the cup? The radius is half the diameter, or 3 cm. Substitute the numbers into the volume formula. The volume of the cone cup is approximately 94.2 cubic centimeters. Try this one yourself: A cone has height 14 cm and diameter 12 cm. What is the volume of the cone? Volume of a Cone How can you find the volume of a plastic cup in the shape of a cone? Compare the cone to a cylinder with the same diameter and same height. If you fill the cone with water and pour it into the cylinder, it takes exactly 3 cones full of water to fill the cylinder. The volume of the cone is one third the volume of the cylinder. Remember, the formula for the volume of a cylinder is πr2h. So, the formula for the volume of a cone is [ggfrac]1/3[/ggfrac]πr2h. Emily has a cone cup. The height is 10 cm and the diameter is 6 cm. What is the volume of the cup? The radius is half the diameter, or 3 cm. Substitute the numbers into the volume formula. The volume of the cone cup is approximately 94.2 cubic centimeters. Try this one yourself: A cone has height 14 cm and diameter 12 cm. What is the volume of the cone? ### Volume of a Sphere How can you find the volume of a sphere? Compare the sphere to a cone with the same radius and the same height. If you fill the cone with water and pour it into the sphere, it takes exactly 2 cones full of water to fill the sphere. The volume of the sphere is twice the volume of the cone. Remember, the formula for the volume of a cone was [ggfrac]1/3[/ggfrac]πr2h. So, [ggfrac]1/3[/ggfrac]πr2h + [ggfrac]1/3[/ggfrac]πr2h is the volume of a sphere. You can make this formula simpler. The height of the sphere, h, is equal to the diameter, or 2r. Now you have [ggfrac]1/3[/ggfrac]πr2(2r) + [ggfrac]1/3[/ggfrac]πr2(2r). You can simplify this expression. Since you can multiply in any order, move the 2s to the front. Remember, r2 times r is r3. Now you have [ggfrac]2/3[/ggfrac]πr3 + [ggfrac]2/3[/ggfrac]πr3. Add these parts together: [ggfrac]4/3[/ggfrac]πr3. So, the formula for the volume of a sphere in its simplest form is [ggfrac]4/3[/ggfrac]πr3. You could use another version of the formula, but this version is simplest. Use this formula to solve a problem. A fishbowl in the shape of a sphere has diameter 20 cm. What is the volume of the fishbowl? The radius of the fishbowl is half the diameter, so it is 10 cm. Substitute that number into the formula. The volume of the fishbowl is approximately 4,187 cm3 Try this one yourself: A sphere has diameter 18 cm. What is the volume of the sphere? Volume of a Sphere How can you find the volume of a sphere? Compare the sphere to a cone with the same radius and the same height. If you fill the cone with water and pour it into the sphere, it takes exactly 2 cones full of water to fill the sphere. The volume of the sphere is twice the volume of the cone. Remember, the formula for the volume of a cone was [ggfrac]1/3[/ggfrac]πr2h. So, [ggfrac]1/3[/ggfrac]πr2h + [ggfrac]1/3[/ggfrac]πr2h is the volume of a sphere. You can make this formula simpler. The height of the sphere, h, is equal to the diameter, or 2r. Now you have [ggfrac]1/3[/ggfrac]πr2(2r) + [ggfrac]1/3[/ggfrac]πr2(2r). You can simplify this expression. Since you can multiply in any order, move the 2s to the front. Remember, r2 times r is r3. Now you have [ggfrac]2/3[/ggfrac]πr3 + [ggfrac]2/3[/ggfrac]πr3. Add these parts together: [ggfrac]4/3[/ggfrac]πr3. So, the formula for the volume of a sphere in its simplest form is [ggfrac]4/3[/ggfrac]πr3. You could use another version of the formula, but this version is simplest. Use this formula to solve a problem. A fishbowl in the shape of a sphere has diameter 20 cm. What is the volume of the fishbowl? The radius of the fishbowl is half the diameter, so it is 10 cm. Substitute that number into the formula. The volume of the fishbowl is approximately 4,187 cm3 Try this one yourself: A sphere has diameter 18 cm. What is the volume of the sphere? ### Relationship Between Cones, Cylinders, and Spheres A restaurant serves smoothies in containers in three different shapes: cones, cylinders, and spheres. All three containers have the same radius and height. All of them are the same price. Which shape container should you order if you want to get the greatest amount of smoothie? Remember, 1 cylinder has the same volume as 3 cones. 1 sphere has the same volume as 2 cones. The cylinder container gives you the greatest amount of smoothie. Try this one yourself: A cylinder and a cone have the same radius. They hold the same amount of water. How many times as tall as the cylinder is the cone? Relationship Between Cones, Cylinders, and Spheres A restaurant serves smoothies in containers in three different shapes: cones, cylinders, and spheres. All three containers have the same radius and height. All of them are the same price. Which shape container should you order if you want to get the greatest amount of smoothie? Remember, 1 cylinder has the same volume as 3 cones. 1 sphere has the same volume as 2 cones. The cylinder container gives you the greatest amount of smoothie. Try this one yourself: A cylinder and a cone have the same radius. They hold the same amount of water. How many times as tall as the cylinder is the cone? ## VOLUME OF CYLINDERS, CONES & SPHERES VOCABULARY Volume The amount of space inside a 3D object. Cone A 3D shape with one circular base. Cylinder A 3D shape with two circular bases. Sphere A 3D shape in the shape of a ball. Pi (π) The ratio of the circumference of a circle to its diameter. This is always the same number: approximately 3.14. The actual number is an irrational number (a decimal that never ends or repeats). The measurement from the center of a circle or sphere to its edge. The radius is half the diameter. The distance straight across a circle or sphere. The diameter always passes through the center point of the circle. ## VOLUME OF CYLINDERS, CONES & SPHERES DISCUSSION QUESTIONS ### What is the formula for the volume of a cylinder? What does each part of the formula mean? V = πr2h. r is the radius of the circular base of the cylinder and h is the height of the cylinder (the distance between the two circular ends). ### What is the formula for the volume of a cone? What does each part of the formula mean? V = [ggfrac]1/3[/ggfrac]πr2h. r is the radius of the circle at the base and h is the distance between the center point of the circular base and the point at the top of the cone. ### What is the formula for the volume of a sphere? What does each part of the formula mean? V = [ggfrac]4/3[/ggfrac]πr3. If I cut the sphere exactly in half I could see a circle on the face. r is the radius of the circle at the very center of the sphere. ### [Draw a cylinder with radius 6 cm and height 10 cm.] What is the volume of this shape? 360π cm3 or about 1,130.4 cm3. The cylinder. X ## Success We’ve sent you an email with instructions how to reset your password. Ok ## Choose Your Free Trial Period ### 7 Days Continue to Lessons ### 90 Days Get 90 days free by inviting other teachers to try it too. Share with Teachers ## Get 90 Days Free ### By inviting 4 other teachers to try it too. 4 required Skip, I will use a 7 days free trial ## Thank You! Enjoy your free 90 days trial
Courses Courses for Kids Free study material Offline Centres More Store # Express the following number as a sum of two odd primes, $84$ Last updated date: 22nd Jul 2024 Total views: 349.2k Views today: 9.49k Verified 349.2k+ views Hint: Prime numbers are the numbers which have only two factors, that is $1$and the number itself. For example, the first five prime numbers are $2,3,5,7\& 11$ Odd prime numbers are the numbers which are odd numbers. A number can be represented as a sum of two numbers like $c = a + b$. As we know that a number can be represented as a sum of two numbers, i.e. $c = a + b$ In this case we have $c = 84$ and $a\& b$ has to be of prime numbers. An odd prime number that is nearest to the number $84$ is $83$ but there is no other odd prime number which will give $84$ by adding it to $83$. So, we cannot use $83$. For instance, let us take the least odd prime number $3$, $83 + 3 = 86$, which is not equal to $84$ Let us consider another odd prime number $79$ which is nearest to $83$. Now consider the least odd prime number 3. $79 + 3 = 82$ which is not equal to $84$. Let us take the next least odd prime number $5$, $79 + 3 = 82$, Now we got the results. Thus $84$ can be represented as a sum of two odd prime numbers that is $79$and $5$. Note: The numbers other than prime numbers are called composite numbers. $1$ is neither a prime number nor a composite number because by the definition of prime number it should have two factors but $1$ does not have any other factor other than itself so $1$ is not a prime number and by the definition of composite number it should have more than two factors again $1$ has no other factor other than itself so, $1$ is not a composite number.
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ### Course: Arithmetic>Unit 7 Lesson 5: Multiply 2-digit numbers with area models # Multiplying with area model: 16 x 27 Sal uses an area model to multiply 16x27. Created by Sal Khan. ## Want to join the conversation? • Is it permissible to use this area concept by subtracting a 'negative' area? For example, using our problem of 16 x 27, we could actually make an area of 20 x 27. We can then subtract an area of 4 x 27 to yield our answer. We can easily calculate 20 x 27 = 540, and then subtract a similarly easy calculation of 4x27=108 → 540-108=432. • Good point, I think you are absolutely right. We can even set up a diagram like Sal has done, using negative numbers in one or both sides of the multiplication. Eg. Using (30 - 3) for 27, and (20 - 4) for 16 in the diagram below We get 600 - 120 - 60 + 12 = 432 `` 30 -3 .______________. \| \| \| \| \| \|20 \| +600 \| -60 \| \| \| \| \|________\|_____\| \| \| \|-4 \| -120 \| +12 \| \|________\|_____\|`` • 30 -3 .____________. \| \| \| \| \| \| 20 \| +600 \| -60 \| \| \| \| \|______\|_____\| \| \| \| -4 \| -120 \| +12 \| \|______\|_____\| this is my area model khan sir • This is an interesting way of using area models to multiply. This is most useful for multiplying numbers that have high digits. • Can you explain me how to do the area model negative numbers if it is possible? Also what do you mean when you say subtracting a 'negative' area? • I don't know what it means by subtracting a negative area, but the question basically says to multiply 16 by 27 which is 432. This probably doesn't help but I hope it does. • What would an aria model for 95 / 6 = ? / = divided by • An area model is shown. Which expressions show how to multiply 4 1/4×2 3/5 using partial products? Label the model with the correct expressions. • I did not know you got energy points for watching a video • You can get 850 energy points if you don't skip any part of the video. • I think the points are to hipe you up
How To Multiply Fractions Question Whenever getting a fractional answer, it is always best to _____________ if possible. 00:00 00:00 Let us say you want to multiply two fractions together like three sevens and two elevens. At first this looks quite complicated but actually multiplying fractions is really easy. All you have to remember is that you multiply the numerator and the denominator separately.Let us start with the numerator. 3 multipled by 2 is 6 and that is our new numerator. 7 multipled by 11 is 77 and this is our new denomiator.It really is as simple as that. Let us try another example.Let us try 8 over 12 multipled by 3 over 5. Once again multiply the numerators together. It gives us 24, that is what 8 times 3 gives us. Then we find our denominator, 12 times 5 is 60 and we are given our answer.However, there is another step here.If you remember whenever we are given a fractional answer it is always best practice to simplify this.24 and 60 both have a common factor of 6 and this means 24 divided by 60 can simplify to 4 over 10.4 and 10 have a common factor of 2 which means we have two fifths. So the simplest answer to 8/12 multipled by 3/5 is two fifths and this is how to multiply fractions together.
# Exponent Law - Part 2 Lesson Objective This lesson shows you the basics behind the next three exponent laws. You'll learn how these laws are actually formulated... This lesson is a continuation from the lesson Exponent Laws - Part 1. The next three laws that you are going to learn can be derived from the first three laws shown in Part 1. So, the ideas behind these laws are not completely new. Again, it is important for you to understand these laws before using them. This helps to minimize mistakes. # Study Tips Tip #1 Let's recall the first three exponent laws before proceeding further. The picture below shows the 1st law. Here is an example on how to use it: 6a10 x 3a-5 = 6 x 3 x a10 x a-5 = 18 x a10 + (-5) = 18 x a10 -5 = 18a5 Tip #2 As for the 2nd law of exponent, the picture below shows the law. Here is an example on how to use it: 6a10 ÷ 3a-5 = 2a10 ÷ a-5 = 2a10 - (-5) = 2a10 + 5 = 2a15 Tip #3 Finally, for the 3rd law of exponent, the picture below shows the law. Here is an example on how to use it: 2(a10)-5 = 2 x a10 x (-5) = 2a-50 Tip #4 You may come across terms like a1 or 21. These terms simply mean that: a1 = a 21 = 2 We can explain them using the following observation: 23 = 2 x 2 x 2 22 = 2 x 2 Therefore: 21 = 2 Now, watch the following math video to learn the next three laws. # Math Video Click play to watch video Math Video Transcript # 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 exponent laws - Part 2 or pick your choice of question below. 1. Question 1 on the basics of the next three exponent laws More Lessons Here are more lessons that you might be interested:
# Identify Number 10 Worksheet on identify number 10 helps the kids to recognize the number. Here the kids need to circle all the number 10’s. Kid’s needs to recognize number 10 from the bunch of numbers which increase the skills to visualize the numbers. This worksheet helps the preschoolers to discriminate the correct number and also help to learn counting by identifying the correct sets of object. Circle all the number 10’s: 8 10 7 10 1 6 10 4 10 9 10 5 2 10 4 3 10 7 1 10 1 6 10 8 10 10 5 10 Count and color the sets of 10 colorfully: Preschoolers and homeschoolers can enjoy practicing these colorful printable math preschool learning activities on numbers to recognize the numbers. Parents and teachers can take the printouts of these worksheets to identify number 10 for kids and help the kids to practice these worksheets which are absolutely free. Take as many printouts you want so that kids can enjoy the practice of counting and coloring the objects as many they want. These printable math worksheets for kids can be access by anyone from anywhere. After practicing identification of number 10 kids can understand the concept of these worksheets on identification of numbers from 1 to 10. Math Only Math is based on the premise that children do not make a distinction between play and work and learn best when learning becomes play and play becomes learning. However, suggestions for further improvement, from all quarters would be greatly appreciated. ## Recent Articles 1. ### Successor and Predecessor \| Successor of a Whole Number \| Predecessor May 24, 24 06:42 PM The number that comes just before a number is called the predecessor. So, the predecessor of a given number is 1 less than the given number. Successor of a given number is 1 more than the given number… 2. ### Counting Natural Numbers \| Definition of Natural Numbers \| Counting May 24, 24 06:23 PM Natural numbers are all the numbers from 1 onwards, i.e., 1, 2, 3, 4, 5, …... and are used for counting. We know since our childhood we are using numbers 1, 2, 3, 4, 5, 6, ……….. 3. ### Whole Numbers \| Definition of Whole Numbers \| Smallest Whole Number May 24, 24 06:22 PM The whole numbers are the counting numbers including 0. We have seen that the numbers 1, 2, 3, 4, 5, 6……. etc. are natural numbers. These natural numbers along with the number zero 4. ### Math Questions Answers \| Solved Math Questions and Answers \| Free Math May 24, 24 05:37 PM In math questions answers each questions are solved with explanation. The questions are based from different topics. Care has been taken to solve the questions in such a way that students
What does 2 to the power of 5 mean? Tracie Meola asked, updated on December 31st, 2021; Topic: the power of two π 194 π 4 βββββ4.3 the power of 4 just means that number multiplied by itself 4 times. e.g 2 to the power of 4 = 2x2x2x2 = 16. 2 to the power of 5 = 2x2x2x2x2 = 32 and so on. So anyway, how do you write 2 to the 5th power? Applying Powers with Superscript For example, to write "2 raised to the 5th power," type "25" with no space between the numbers. Highlight just the "5" and press "Control-Shift-Equals" or click the superscript icon on the "Home" tab, which looks like an "x squared." The "5" will shrink and shift upward. Quite so, what is with the power of 2? A power of two is a number of the form 2n where n is an integer, that is, the result of exponentiation with number two as the base and integer n as the exponent. Never mind, what does 5 with a little 2 mean? Understanding and Computing Square Roots As you saw earlier, 52 is called βfive squared.β βFive squaredβ means to multiply five by itself. In mathematics, we call multiplying a number by itself βsquaringβ the number. We call the result of squaring a whole number a square or a perfect square. What is 10 to the O power? When n is less than 0, the power of 10 is the number 1 n places after the decimal point; for example, 10β2 is written 0.01. When n is equal to 0, the power of 10 is 1; that is, 100 = 1. What is a power of 5? In arithmetic and algebra, the fifth power of a number n is the result of multiplying five instances of n together: n5 = n Γ n Γ n Γ n Γ n. Fifth powers are also formed by multiplying a number by its fourth power, or the square of a number by its cube. What are all the real square roots of 100? 1 Answer. The square root of a number x is, by definition, a number y such that y2=x . In your case, it is immediate to find out that 102=100 , which fits the definition, and thus the square root of 100 is 10. What is a perfect fifth power? Ask for a perfect fifth power of a number from 1 to 50. Then give the fifth root of that number! Note that when you take a fifth power of any number, the units digit of the number and the fifth power are the same. So, if the person gives you a number less than 100,000, your answer is just the last digit. PowerValue 664 7128 8256 9512 Is there a power 2 coming out? Power aired on Starz in the US, and is available on Netflix in the UK. Power Book II: Ghost premieres on Starz in the US on Sunday, September 6, and will air on Starzplay, available through Amazon Prime Video, in the UK. What's the square of 5? Table of Squares and Square RootsNUMBERSQUARESQUARE ROOT 4162.000 5252.236 6362.449 7492.646 What does 2 with a little 3 mean? The Cube Root Symbol This is the special symbol that means "cube root", it is the "radical" symbol (used for square roots) with a little three to mean cube root. What does 10 to the power of 8 mean? 108. 100,000,000. "ten to the eight" billion. Why is 0 to the 0 power undefined? I assume you are familiar with powers. The problem is similar to that with division by zero. No value can be assigned to 0 to the power 0 without running into contradictions. Thus 0 to the power 0 is undefined! What is the value of 6 power 2? So, 62 = 12. When you take 6 and square it (raise it to the power of 2), you are taking 6 and multiplying it by itself. So, 62= 66 = 36. ο»Ώ
# Find the cement bags, sand(CFT) & aggregate in M20, M15 concrete ## Calculate the Quantity of Cement, Sand & Aggregate in M20 Concrete First of all, I want to assure you that it is very easy to determine the quantity of sand, cement, and aggregate in M20, M15, and M10 Concrete. So take a look carefully and will easily be understood. Let’s take some examples, Example: 1 Let’s assume that the volume of total concrete(wet concrete) required for a particular work is 2 m3. Now find out the quantity of cement, sand, and aggregate. Solution: In the case of M 20 concrete, the ratio of cement, sand, and aggregate is 1: 1.5: 3. Dry volume of concrete = wet volume of concrete × 1.54 [ You can take this factor from 1.54 to 1.57 ] Hence, the dry volume of concrete is = 2 × 1.54 = 3.08 m3 The required quantity of cement is = (Cement ratio ÷ Sum of all ratio) × Total volume of concrete = $\frac{1}{(1 + 1.5 + 3)} \times 3.08$ = 0.56 m3 ### How do you calculate cement in bags? First of all, we need to calculate it in kg. We know that the density of cement is 1440 kg/ m3. Therefore, 0.56 m3 cement means 0.56 × 1440 = 806.4 kg cement. In general, each bag of cement weight is 50 kg. So, the required number of cement bags are = 806.4 ÷ 50 = 16.12 ≈ 17 No.s. Quantity of sand required = ( Sand ratio ÷ Sum of all ratio) × Total volume of concrete = $\frac{1.5}{(1 + 1.5 + 3)} \times 3.08$ = 0.84 m3 ### How do you calculate sand in CFT? Our total quantity of sand is 0.84 in m3, but in CFT(cubic feet) it will be ( 0.84 × 35.31) = 29.66 ≈ 30 CFT( cubic feet) [ Note: 1 m3 = 35.31 CFT or cubic feet ] Quantity of aggregate is = ( Aggregate ratio ÷ Sum of all ratio) × Total volume of concrete = $\frac{3}{(1 + 1.5 + 3)} \times 3.08$ = 1.68 m3. ### How do you calculate aggregates in CFT? We know, 1 m3 is equal to 35.31 CFT. Therefore, 1.68 m3 aggregate means 1.68 × 35.31 = 59.32 ≈ 60 CFT. Example: 2 A section of footing is given below. Find out the quantity of cement sand and aggregates required for this footing. Solution: At first, we need to calculate the total volume of concrete required for this footing. The total volume of concrete is equal to the total volume of footing. So, the total volume of concrete is = 1 × 7 × 6 = 42 cubic feet. Dry volume of concrete will be = 42 × 1.54 = 64.68 cubic feet. In the case of M20 concrete, The Quantity of cement required for this footing is = $\frac{1}{(1 + 1.5 + 3)} \times 64.68$ = 11.76 cubic feet. Or, 0.33 in m3 [ Note: 1 m3 = 35.31 CFT or cubic feet] The weight of cement = 0.33 × 1440 = 475.2 kg. Or, 475.2/50 = 9.50 ≈ 10 No.s bags. Quantity of sand required = $\frac{1.5}{(1 + 1.5 + 3)} \times 64.68$ = 17.64 cubic feet or CFT . The Quantity of aggregate required = $\frac{3}{(1 + 1.5 + 3)} \times 64.68$ = 35.25 CFT. ## Calculate the Quantity of Cement, Sand & Aggregate in M15 Concrete. Now, we will calculate the quantity of cement and aggregate in M15 Concrete. Let’s take an example, Example: Let’s assume that the volume of total concrete(wet concrete) required for a particular work is 1 m3. Now find out the quantity of cement, sand, and aggregate. Solution: In the case of M15 concrete, the ratio of cement, sand, and aggregate is 1: 2: 4. Dry volume of concrete = wet volume of concrete × 1.54 Hence, the dry volume of concrete is = 1 × 1.54 = 1.54 m3 The required quantity of cement is = (Cement ratio ÷ Sum of all ratio) × Total volume of concrete = $\frac{1}{(1 + 2+ 4)} \times 1.54$ = 0.22 m3 0.22 m3 cement means 0.22 × 1440 = 316.8 kg cement. In general, each bag of cement weight is 50 kg. So, the required number of cement bags are = 316.8 ÷ 50 = 6.33 ≈ 7 No.s. Quantity of sand required = ( Sand ratio ÷ Sum of all ratio) × Total volume of concrete = $\frac{2}{(1 + 2 + 4)} \times 1.54$ = 0.44 m3 Our total quantity of sand is 0.44 in m3, but in CFT(cubic feet) it will be ( 0.44 × 35.31) = 15.53 ≈ 16 CFT( cubic feet) [ Note: 1 m3 = 35.31 CFT or cubic feet ] Quantity of aggregate is = ( Aggregate ratio ÷ Sum of all ratio) × Total volume of concrete = $\frac{4}{(1 + 2 + 4)} \times 1.54$ = 0.88 m3. We know, 1 m3 is equal to 35.31 CFT. Therefore, 0.88 m3 aggregate means 0.88 × 35.31 = 31.07 ≈ 32 CFT. ## Calculate the Quantity of Cement, Sand & Aggregate in M10 Concrete. Now, we will calculate the quantity of cement and aggregate in M10 Concrete. Let’s take an example, Example: Let’s assume that the volume of total concrete(wet concrete) required for a particular job is 3 m3. Now find out the quantity of cement, sand, and aggregate. Solution: In the case of M10 concrete, the ratio of cement, sand, and aggregate is 1: 3: 6. Dry volume of concrete = wet volume of concrete × 1.54 Hence, the dry volume of concrete is = 3 × 1.54 = 4.62 m3 The required quantity of cement is = (Cement ratio ÷ Sum of all ratio) × Total volume of concrete = $\frac{1}{(1 + 3+ 6)} \times 4.62$ = 0.462m3 0.22 m3 cement means 0.462× 1440 = 665.28 kg cement. In general, each bag of cement weight is 50 kg. So, the required number of cement bags are = 665.28 ÷ 50 = 13.3 ≈ 14 No.s. Quantity of sand required = ( Sand ratio ÷ Sum of all ratio) × Total volume of concrete = $\frac{3}{(1 + 3 + 6)} \times 4.62$ = 1.386 m3 Our total quantity of sand is 1.386 m3 , but in CFT(cubic feet) it will be ( 1.386 × 35.31) = 48.93 ≈ 49 CFT( cubic feet) [ Note: 1 m3 = 35.31 CFT or cubic feet ] Quantity of aggregate is = ( Aggregate ratio ÷ Sum of all ratio) × Total volume of concrete = $\frac{6}{(1 + 3 + 6)} \times 4.62$ = 2.77 m3. We know, 1 m3 is equal to 35.31 CFT. Therefore, 2.77 m3 aggregate means 2.77 × 35.31 = 97.8 ≈ 98 CFT.
# Math Statistics: Bar Charts/Bar Graphs These lessons cover how to create and read bar charts and using bar charts to solve problems. A bar chart represents the data as horizontal or vertical bars. The length of each bar is proportional to the amount that it represents. There are 3 main types of bar charts. • Vertical bar chart • Horizontal bar chart • Double bar chart When constructing a bar chart it is important to choose a suitable scale to represent the frequency. Example: The following table shows the number of visitors to a park for the months January to March. Month January February March Number of visitors 150 300 250 a) Construct a vertical and a horizontal bar chart for the table. b) What is the percentage of increase of visitors to the park in March compared to January? c) What percentage of visitors came in February compared with total number of visitors over the three months? Solution: a) If we choose a scale of 1:50 for the frequency then the vertical bar chart and horizontal bar chart will be as shown. b) Increase in March compared to January is c) Percentage of visitors in February compared to the total number of visitors is How to create bar graphs using given data, and answer questions based on given bar graphs? A bar graph is a visual way to display and compare numerical data. The bars of a bar graph are drawn in relation to a horizontal axis and a vertical axis. A bar graph can have either vertical or horizontal bars. Example: Use the bar graph below to find the difference between the speed limit on a state highway and a suburban street? Examples: 1. Some students travel by train, airplane or boat during the summer. This bar graph shows how many students used each type of travel. How many more students traveled by airplane than boat? 2. Sunny, Rover and Buster buried bones in the yard. This bar graph shows how many bones each dog buried. How many fewer bones did Rover bury than Sunny? 3. The weather in November was sunny, cloudy, or rainy. This bar graph shows the number of days of each kind of weather. How many rainy days were there in November? How to solve problems with bar graphs? Examples: 1. Lola asked her classmates their favorite type of movie and graphed the results. Which of the following types of movies were picked by fewer than 14 people? 2. Puppy Party Place graphed the number of dogs that came in each weekday. On which day did the same number of dogs come to Puppy Party Place as on Monday and Wednesday combined? 3. Desert Zone asked its customers about their favorite ice cream flavor and graphed the results. How many more customers picked the most popular flavor than the least popular flavor? ### Double Bar Chart The double bar chart is used when we want to represent two sets of data on the same chart. We can put the bars side by side or we may put the bars of one set of data on top of the bars of the other set of data. The choice of the forms of double bar chart – side by side or stacked, depends on the main purpose of the chart. A side by side chart is more useful when we compare the two sets of data (example: the number of adult visitors as compared to the number of child visitors); whereas the stacked chart emphasizes the totals of the two sets of data (example: total number of visitors). Example: The following frequency graph shows the number of adult visitors and child visitors to a park. Construct a side by side double bar chart and a stacked double bar chart for the frequency table. Month April May June Number of adult visitors 300 500 700 Number of child visitors 200 600 600 Solution: How to read Double Bar Graphs? How to solve problems using double bar graphs? Example: Based on the data below, which student’s score improved the most between the midterm and final exams? Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
## Precalculus (6th Edition) domain: $(−\infty,+\infty)$ range: $(−\infty,+\infty)$ Refer to the graph below. The given line can be graphed using the x and y-intercepts. RECALL: (1) The x-intercept can be found by setting y=0 then solving for x. (2) The y-intercept can be found by setting x=0 then solving for y. Find the x-intercept of the given equation. Set y=0 then solve for x to obtain: \begin{array}{ccc} &3x+2y&=&0 \\&3x+2(0)&=&0 \\&3x&=&=0 \\&\frac{3x}{3}&=&\frac{0}{3} \\&x&=&0\end{array} The x-intercept is $(0,0)$. Find the y-intercept of the given equation. Set x=0 then solve for y to obtain: \begin{array}{ccc} &3x+2y&=&0 \\&3(0)+2y&=&0 \\&2y&=&=0 \\&\frac{2y}{2}&=&\frac{0}{2} \\&y&=&0\end{array} The y-intercept is $(0,0)$. The intercepts are the same so one more point is needed to graph the line. Set $x=2$ then solve for y to obtain: $3x+2y=0 \\3(2)+2y=0 \\6+2y=0 \\6+2y-6=0-6 \\2y=-6 \\\frac{2y}{2}=\frac{-6}{2} \\y=-3$ The line contains the point (2, -3). Graph the line by plotting $(0,0)$ and $(2, -3)$ and connecting them using a line. (Refer to the graph in the answer part above.) The graph covers all x-values therefore the domain is $(−\infty,+\infty)$. The graph covers all y-values therefore the range is $(−\infty,+\infty)$.
## Infosys Practice Problems On Pins & Dimensions Dear Reader, Below are three problems where you will be asked to find number of pins/dimensions of cardboards based on data in questions. Question 1 M/S. GRT Grand Hotels, Salem built their hotel recently. They wanted to decorate the Enquiry counter in a grand manner. They used square plates for this purpose. The square board is of size 41” x 41”. The carpenter in charge of decoration uses nails all along the edges of the square such that there are 42 nails on each side of the square. Each nail is at the same distance from the neighbouring nails. How many nails does the carpenter use for each square decoration board? a) 124 b)164 c)204 d)168 Answer : b) 164 Solution : To start with one nail can be fitted on each of the corners. Between two nails on any edge 40 nails have to be fitted so that the count becomes 42 when counted on any edge. Therefore total nails = 4 edge nails + 4 x 40 nails in between corners = 4 + 160 = 164 (Note. though this is a simple question, you could be mislead to work out wrongly 42 x 4 = 168.) Question 2 Ms. Ganesh Mahal, Coimbatore wanted to present each of its customers with a momento rectangle in size. The momentos were of size 25 x 20 cm. The company requested the carpenter to make it in such a way that both 25 cm sides should have 24 nails each and the shorter sides should have 10 nails each. How many decorative nails the carpenter would have used in each momento? a) 56 b)64 c)68 d)60 Answer : b) 64 Solution : To begin with there should be one nail in each of the corners. There should be 22 nails on each of the two longer sides so that the count comes to 24 nails on each of those sides. There should be 8 nails on each of the shorter sides so that the count becomes 10 nails on each of the shorter sides. Total nails = 4 corner nails + 2 x 22 nails (longer sides) + 2 x 8 (shorter sides) = 4 + 44 + 16 = 64 nails. Question 3 Ramnath and Company was celebrating Silver Jubilee celebrations and the authorities wanted to present a Square cardboard momento fitted with decorative pins on all the four sides. They fitted 36 decorative pins with 1 cm distance between any two pins and with equal number of pins when counted on any side. Find the dimension of the cardboard. a) 9 cm X 9 cm b) 10 cm X 10 cm c) 12 cm X 12 cm d) None of these Answer : a) 9 cm X 9 cm Solution : To find the dimension, we have to find the number of pins on each of the sides of the square cardboard. Similar to solutions of previous questions, there will be 4 corner pins. Corner pins when added with pins on edges should amount to 36 as given. Let the number of pins in between the corner pins on each of the edges be x. Total Pins = 4 corner pins + 4 x number of pins on each of the edges in between corner pins 36 = 4 + 4x 32 = 4x or 8 = x Therefore number of pins on any of the edge = 2 corner pins + 8 pins in between the corner pins = 10 pins. Distance between each of the pins is 1 cm. Therefore the length of each of the sides = 9cm. Dimension of the square = 9 cm X 9 cm • Fresher IT Jobs
Fibonacci Ratios The last three sections have done much to explain Golden geometry, but haven't explained how they relate to the topic of the Fibonacci series. Actually, the Golden Ratio and the Fibonacci numbers are two concepts that have a lot to do with each other. The most important appearance of the Golden Ratio in Fibonacci mathematics is the Fibonacci ratio sequence. The Fibonacci ratio sequence is the series of numbers you get when you divide each Fibonacci number with the one that precedes it. Let's look at a few terms of this sequence. In order to create the first few terms of the sequence, we divide each Fibonacci number by the one before it. Since there is no Fibonacci number before f(1), we start with f(2). Dividing f(2) by f(1) gives a value of 1, so 1 is the first number of the Fibonacci ratio sequence, or r(1). Then we divide f(3) by f(2) to get the value of r(2), 2. Continuing in this fashion, we get: r(1) = 1 / 1 = 1 r(2) = 2 / 1 = 2 r(3) = 3 / 2 = 1.5 r(4) = 5 / 3 = 1.67 r(5) = 8 / 5 = 1.6 r(6) = 13 / 8 = 1.625 r(7) = 21 / 13 = 1.615 r(8) = 1.6190476 r(9) = 1.6176471 r(10) = 1.6181818 r(11) = 1.6179775 r(12) = 1.6180555 r(13) = 1.6180258 r(14) = 1.6180371 r(15) = 1.6180328 Do you notice a pattern beginning to emerge? We already proved that the decimal approximation of the Golden Ratio () is about 1.6180339. The farther we go in the Fibonacci ratio sequence, the closer we come to . However, since the real value of is an irrational number (can not be expressed as the ratio of two whole numbers), we can never really get exactly to in the sequence. Note that the first, third, fifth, and seventh terms of the sequence are all less than , and the second, fourth, and sixth terms are all greater than . For a more visual demonstration of this relationship, see the In Action section.
# Absolute Value of a Complex Number The complex number is defined as the number in the form a+ib, where a is the real part while ib is the imaginary part of the complex number in which i is known as iota and b is a real number. The value of i is √(-1). Or in other words, a complex number is a combination of real and imaginary numbers. For example, 5+11i, 10+20i, etc. • Real number: A real number is a number that is present in the number system which can be positive, negative, integer, rational irrational, etc. For example, 23, -3, 3/6. • Imaginary number: Imaginary numbers are those numbers that are not real numbers. For example, √3, √11, etc. • Zero complex number: A zero complex number is a number that has its real and imaginary parts both equal to zero. For example, 0+0i. Graphical representation of complex number: The graphical representation of the complex number is as shown in the below image: Here, the real part of the complex number is represented on the horizontal axis while the imaginary part of the complex number is represented on the vertical axis. ### Absolute value The absolute value(Modulus) of a number is the distance of the number from zero. Absolute value is always represented in the modulus(\|z\|) and its value is always positive. So, the absolute value of the complex number Z = a + ib is \|z\| = √ (a2 + b2) So, the absolute value of the complex number is the positive square root of the sum of the square of real part and the square of the imaginary part, i.e., Proof: Let us consider the mode of the complex number z is extended from 0 to z and the mod of a, b real numbers is extended from a to 0 and b to 0. So these values create a right angle triangle in which 0 is the vertex of the acute angle So, using Pythagoras theorem, we get, \|z\|2 = \|a\|2 + \|b\|2 \|z\| = √ a2 + b2 Now, in the sets of complex numbers z1 > z2 or z1 < z2 has no meaning but \|z1\| > \|z2\| or \|z1\| < \|z2\| has meaning because \|z1\| and \|z2\| is a real number. Properties of modulus of a complex number: 1. \|z\| = 0 <=> z = 0i, i.e., Re(z) = 0 and Im(z) = 0 2. \|z\| = \|\| = \|-z\| 3. -\|z\| ≤ Re(z) ≤ \|z\|, -\|z\| ≤ Im(z) ≤ \|z\| 4. z. = \|z2\| 5. \|z1z2\| = \|z1\|\|z2\| 6. \|z1 / z2\| = \|z1\|/\|z2\| 7. \|z1 + z2\|2 = \|z1\| + \|z2\| + 2Re(z1) 8. \|z1 – z2\|2 = \|z1\| + \|z2\| – 2Re(z1) 9. \|z1 + z2\|2 ≤ \|z1\| + \|z2\| 10. \|z1 – z2\|2 ≥ \|z1\| – \|z2\| 11. \|az1 – bz2\|2 + \|bz1 + az2\|2 = (a2 + b2)(\|z1\|2 + \|z2\|2) Or \|z1 – z2\|2 + \|z1 + z2\|2 = 2(\|z1\|2 + \|z2\|2) 12. \|zn\| = \|z\|n 13. 1/z = a – ib/a2 + b2 =/\|z\|2 Example: (i) z = 3 + 4i \|z\| = √(32 + 42) = √(9 + 16) = √25 = 5 (ii) z = 5 + 6i \|z\| = √(52 + 62) = √(25 + 36) = √61 ### Angle of Complex numbers The angle of a complex number or argument of the complex number is the angle inclined from the real axis in the direction of the complex number that represents on the complex plane or argand plan. θ = tan-1(b/a) or arg(Z) = tan-1(b/a) Here, Z = a + ib Properties of the angle or argument of a complex number: • arg(Zn) = n arg(Z) • arg (Z1/ Z2) = arg (Z1) – arg (Z2) • arg (Z1 Z2) = arg (Z1) + arg (Z2) Examples: (i) z = 2 + 2i θ = tan-1(2/2) = tan-1(1) = 45° (ii) z = -4 + 4i θ = tan-1(4/-4) = tan-1(-1) = -45° It is important to note here that the angle θ =-45° is in 4th quadrant, while we always measure angle with the positive x-axis. So, we will have to add 180° to the answer to obtain the real opposite angle. So, θ = 180° + (-45°) = 135° So , the above complex number will make an angle of 135° with the positive x-axis. ### Polar Form of Complex number The polar form of complex number is also a way to represent a complex number. Generally, we represent complex number like Z = a + ib, but in polar form, complex number is represented in the combination of modulus and argument. Here, the modulus of the complex number is known as the absolute value of the complex number, and the argument is known as the angle of the complex number. Z = r(cos θ + isin θ) Proof: Let us considered we have a complex number Z = a+ib. So we draw an argand plane, it is a plane where we can represent complex numbers it is also known as a complex plan. In the argand plan, the horizontal line represents the real axis and the vertical line represents the imaginary axis. Now we plot a on the real axis and b on the imaginary axis. Now we represent vector Z as a position vector that starts at 0 and the tip is at coordinate (a, b). Suppose θ be the angle made by Z over the real axis and the distance between 0 to Z is r(it is also known as the magnitude/absolute value of vector Z). Here, (r, θ) pair is known as the polar coordinates of Z. According to the diagram, we have a right angle triangle So, using Pythagoras’s theorem, we get, r = \|z\|2 = \|a\|2 + \|b\|2 r = \|z\| = √ a2 + b2 This is the modulus Now we find the value of θ tan θ = (a/b) θ = tan-1(a/b) This is the argument Now we find the polar form of the complex number: So using the trigonometric formula, we get cos θ = a/r Now multiply both sides with r we get rcos θ = a sin θ = b/r Now multiply both sides with r we get rsin θ = b So, we get Z = rcos θ + irsin θ Z = r(cos θ + isin θ) Here, r is the absolute value of the complex number and θ is the argument of the complex number. The exponential form of complex numbers: The exponential form of complex numbers uses both the trigonometric ratios of sine and cosine to define the complex exponential as a rotating plane in exponential form. The exponential form of a complex number is generally given by Euler’s Identity, named after famous mathematician Leonhard Euler. It is given as follows: Z = re Example: (i) Given that r = 5 and θ = 45°. Find the polar form of the complex number z = rcosA + (rsinA)i z = 5cos45° + (5sin45°)i = 5(1/√2) + (5 (1/√2))i = 5/√2 + (5/√2)i (ii) Given that r = 6 and θ = 30°. Find the polar form of the complex number z = rcosA + (rsinA)i z = 6cos30° + (6sin30°)i = 6(√3/2) + (6(1/2))i = 3√3 + 3i Convert rectangular form to polar form: Let us discuss this concept with the help of an example: Suppose we have a complex number, i.e., Z = 3 + 4i It is in rectangular form so now we have to convert it to a polar form. Step 1: So, first, we will calculate the modulus of the complex numbers and then the angle. \|Z\|= √(32 + 42) = √(9 + 16) = √25 = 5 Step 2: Now we find the angle of the complex number, θ = tan-1(y/x) = tan-1(4/3) = 53.1° Step 3: As we know that the formula of polar form is: Z = r(cos θ + isin θ) Now put the value of r and θ in this equation, we get Z = 5(cos 53.1 + isin 53.1) Hence, the polar form of 3 + 4i is 5(cos 53.1 + isin 53.1) ### Sample Problems Question 1. Find the absolute value of z = 4 + 8i Solution: Given complex number is z = 4 + 8i As we know that the formula of absolute value is \|z\| = √ (a2 + b2) So, a = 4, and b = 8, we get \|z\| = √(42 + 82) \|z\| = √80 Question 2. Find the absolute value of z = 2 + 4i Solution: Given complex number is z = 2 + 4i As we know that the formula of absolute value is \|z\| = √ (a2 + b2) So, a = 2, and b = 4, we get \|z\| = √(22 + 42) \|z\| = √20 Question 3. Find the angle of the complex number: z = √3 + i Solution: Given complex number is z = √3 + i As we know, that θ = tan-1(b/a) So, a = √3 , and b = 1, we get θ = tan-1(1/ √3 ) θ = 30° Question 4. Find the angle of the complex number: z = 6 + 6i Solution: Given complex number is z = 6 + 6i As we know, that θ = tan-1(b/a) So, a = 6 , and b = 6, we get θ = tan-1(6/6) θ = 45° Question 5. Convert z = 5 + 5i into polar form Solution: Given complex number is z = 5 + 5i As we know that Z = r(cos θ + isin θ) …(1) Now, we find the value of r r = √(52 + 52) r = √(25 + 25) r = √50 Now we find the value of θ θ = tan-1(5/5) θ = tan-1(1) θ = 45° Now put all these values in eq(1), we get Z = √50(cos 45° + isin 45°) Question 6. Convert z = 2 + √3i into polar form Solution: Given complex number is z = 2 + 2√3i As we know that Z = r(cos θ + isin θ) …(1) Now, we find the value of r r = √(22 + (2√3)2) r = √(4 + 12) r = √16 r = 4 Now we find the value of θ θ = tan-1(2√3/2) θ = tan-1(√3) θ = 60° Now put all these values in eq(1), we get Z = 4(cos 60° + isin 60°)
# Lorem Ipsum Lorem ipsum dolor sit amet, consectetur adipiscing elit # Introduction Recently, my significant other approached me with a probability puzzle: Roll a die repeatedly until the sum over all rolls exceeds twelve. What sum is the most probable to occur? Below, I’ll answer the puzzle by intuition first and then verify my answer via simulation in Python. Classical probability theory defines the probability of an event as the total number of its favorable outcomes divided by the number of all possible results. The most likely sum to occur should hence be the sum that can arise in most ways. If the sum over all dice rolls has to exceed twelve, then there are only six possible outcomes of this “experiment”: Anything between thirteen and eighteen is possible. Now, how should we count the number of ways each sum of dice can arise? Since the number of dice rolled is not known beforehand, we cannot take comfort in exact equations. Regardless, we can reason our way through. To that end, notice that there is only one way to end the experiment at 18: You roll three sixes in a row. In contrast, there are at least two ways to complete the dice rolls at 17. First, the dice sum to twelve, and you get a five on your last roll. Alternatively, your dice sum to eleven and the next roll results in a six. Already, a pattern emerges: Sums closer to twelve seem to occur in more ways. • s18 = {(12, 6)} • s17 = {(12, 5); (11, 6)} • s16 = {(12, 4); (11, 5); (10, 6)} • s15 = {(12, 3); (11, 4); (10, 5); (9, 6)} • s14 = {(12, 2); (11, 3); (10, 4); (9, 5); (8, 6)} • s13 = {(12, 1); (11, 2); (10, 3); (9, 4); (8, 5); (7, 6)} The list above shows the set of all possible outcomes and the ways to achieve them on the last roll. Thirteen should be the most probable outcome. However, we cannot yet put a probability on it. Each prior sum is itself a compound outcome: Twelve, eleven, etc., may arise in numerous ways. Hence, let’s simulate the entire process in Python and verify my intuition. ## Simulation in Python The listing below shows a simple Python program that simulates the entire process. The function throw_die() returns a random, uniformly distributed integer between 1 and 6. The function run_experiment() does the actual dice rolling and returns the required sum. Finally, main() runs the experiment 10,0000 times and prints the result to screen. When done, the program output supports my intuition. ```from random import randint from collections import Counter def throw_die(sides=6): return randint(1, sides) def run_experiment(cut=12): sum_dice = 0 while sum_dice <= cut: sum_dice += throw_die() return sum_dice def main(): NTRIALS = 100000 results = [run_experiment() for _ in range(NTRIALS)] print(Counter(results)) ``` Three variations of this program are immediately apparent. First, you could alter the number of sides of the die (six by default). Second, you could change the cut-off point (twelve by default). Finally, you could increase the number of trials. The latter increases the precision of the result. The former two options are more interesting. Playing around with them, first, proves the result independent of the size of the die, and, second, shows that the most probable outcome of this experiment is always the cut-off value + 1. ## Conclusion This post has shown how to combine intuition with simulation in Python to answer a probability puzzle. Let’s see what puzzle makes it into my inbox next. ## Introduction Today I read an interesting post on R-Bloggers. In the post, Method Matters outlines a solution to a typical data scientist interview problem: FizzBuzz. ```In pseudo-code or whatever language you would like: write a program that prints the numbers from 1 to 100. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz”. ``` The problem probes your knowledge of basic programming concepts. More specifically, it asks: “Are you able to identify and deal with multiple cases?” Method Matters introduces two simple solutions in R and Python. The first solution uses a for-loop. It iterates over a set of if-else-statements. I take no issue with this approach even though more elegant implementations are possible. The second solution uses a function to do the heavy lifting, which strikes me as more interesting, more powerful, and more questionable given its implementation by Method Matters. ## 99 problems but for() ain’t one Method Matters defines a function which takes a numeric input to pass through the set of if-else-statements defined in the first solution. It exemplifies how working code can be recycled in more complex settings. Take a moment to study the original code, I’ll detail my reservations below. ``````# define the function fizz_buzz <- function(number_sequence_f){ if(number_sequence_f%%3 == 0 & number_sequence_f%%5 == 0) { print('FizzBuzz') } else if(number_sequence_f%%3 == 0) { print('Fizz') } else if (number_sequence_f%%5 == 0){ print('Buzz') } else { print(number_sequence_f) } } # apply it to the numbers 1 to 100 sapply(seq(from = 1, to = 100, by = 1), fizz_buzz) `````` So what’s not to like about it? Two things bug me in particular. 1. The function does not abstract from the particular instance of the problem. Note how values 3, 5, Fizz, Buzz, and FizzBuzz have been hard-coded into the function’s body. Given that the function recycles a for-loop this is not surprising. However, functions (should) encapsulate algorithms which solve entire classes of problems. 2. More importantly, the function is not vectorized. In R, control statements such as if() or while() expect logical vectors of length one. Whatever condition you pass to them must evaluate to a single Boolean statement.1 Since `fizz_buzz()` consists entirely of such statements, it expects you to pass each sequence element individually. The concluding call to `sapply()` does just that. I don’t mean to offend, but `sapply(seq(from = 1, to = 100, by = 1), fizz_buzz)` is `for()` in disguise. ## fizz_buzz() refactored The refactored solution shown below generalizes from FizzBuzz, and it is vectorized. The function expects two arguments: (1) a numeric vector such as an integer sequence from 1 to 100, and (2) a “dictionary” of named values to evaluate, e.g., `c("Fizz" = 3, "Buzz" = 5)`. The first argument is checked for multiples of each entry in the dictionary. Those TRUE/FALSE evaluations are saved in a logical matrix with rows equal to `length(x)` and columns equal to `length(dictionary)`. Next, each column of that matrix is used to logically index the argument `<x>` and its selected elements are replaced by the corresponding name from the dictionary. Finally, the function returns a vector of strings. ``````replace_sequence_elements <- function(x, dictionary){ # The function searches <x> for multiples of each element # in <dictionary>. Multiples are replaced by the name of # the corresponding dictionary element. If some element # of <x> is a multiple of all elements in dictionary, # then it will be replaced by concatenated dictionary # names. # x ... numeric vector # dictionary ... named, numeric vector # The function returns a vector of type character. stopifnot("names" %in% names(attributes(dictionary))) out <- as.character(x) K <- length(dictionary) tests <- matrix( FALSE, nrow = length(x), ncol = length(dictionary) ) for(k in seq(K)) { tests[, k] <- x %% dictionary[k] == 0 out[tests[, k]] <- names(dictionary[k]) } out[rowSums(tests) == K] <- paste( names(dictionary), collapse = "" ) return(out) } `````` Let’s see if this works: ``````replace_sequence_elements(seq(15), c("Fizz" = 3, "Buzz" = 5)) `````` ```## [1] "1" "2" "Fizz" "4" "Buzz" "Fizz" ## [7] "7" "8" "Fizz" "Buzz" "11" "Fizz" ## [13] "13" "14" "FizzBuzz" ``` ## Conclusion This post responds to and refactors a publicly available solution of a common data scientist interview problem: FizzBuzz. The refactored solution generalizes from the original and can now take any number of cases to evaluate. Moreover, it is vectorized. That said, several pain points remain. For instance, a for-loop is still doing the heavy lifting. As the number of test cases increases this may cause performance issues, at least in R. Also, the refactored solution makes specific assumptions about the structure of its arguments. There must be no missing elements, and test cases must be passed in the form of a named vector. A different approach would be needed to avoid or at least reduce the number of such implicit assumptions. Two instructive examples may be found in the comments to Method Matters’ original post. ## Introduction It’s that time of the academic year again: The semester has just started and every class is immediately overwhelmed by student demand. On top of it, coordination among the teaching staff could have been better. Classes cluster on Mondays and Wednesdays, limiting our students’ flexibility when building their schedule. Put some strategic behavior on the part of your students into the mix and you’ll easily end up with fifty or more applications for a twenty-five to thirty students class. It’s a mess. So, how do you decide who gets to participate without stepping onto everyone’s toes? Alternatively, how do you democratize access to scarce resources? ## Hard criteria are few and far between Students who endure hardships1 are usually admitted, but their numbers are small. Also, and at the risk of kicking in an open door, what counts as a hardship? Some students must take my class because their study regulations demand two seminars from the same area to conclude their electives module. They have no alternatives as long as my class is the only one offered in that module. Does TINA count as a hardship? What about good old “first come, first serve”? In fact, several students argued for their admission based on the fact that they registered on five past twelve a.m. right when admissions started. Fun fact: Our campus management software does not tell me who signed up first. More importantly, technology can be tricky. An occasional glitch in your internet connection may easily move you from first to the last position on the list. In my opinion, “First come, first serve” is neither fair – So, you stayed up late? – nor robust. What if students were to rank their preferences? At Potsdam University, students may express priorities when they sign up for class. In principle, lecturers should admit priority students first and distribute any remaining seats among everyone else. Again, there is a serious drawback: How do you select within each group? For instance, 49 students registered for my class – every single one of them prioritized it. So, I am back to square one. ## Are fair procedures the answer? Apparently, you can’t just look at students to make the call. Maybe there are procedures which lead to fair results? Cues, auctions, and lotteries are on the table. Some colleagues rely on cueing. They first admit students late in their studies. Any remaining seats are then filled with earlier semesters. Justifications of this method boil down to urgency: Older students must transition into the labor market as soon as possible. But why should an eighth-semester student feel more pressure than a seventh-semester student? If you are unable to justify this distinction, then you won’t be able to justify any other. Moreover, I believe cueing perpetuates the problem. If all students have to wait their turn, then all of them will eventually be eighth at some point. Auctions would be preferable. After all, students who are most motivated to participate in my class should place the highest bids. In the ideal world, I would have the honors to teach a group of self-selected geeks. That is, if – and, unfortunately, only if – I could elicit sincere offers. I could require students to bid from their own resources and -rightly – face the wrath of the university’s ethics committee. Alternatively, I provide them with benefits to burn and – rightly – face my wife’s wrath. In short, without institutional backing auctions may seem ideal but they are not a realistic option. That leaves casting lots. Lotteries have been around for ages. In ancient Greece, for instance, many public officials were chosen by lot based on a firm belief in the equality of all citizens. Lots make decisions without consideration of person or intent. In that regard, they are uniquely fair. On the downside, students are not all equal. After all, hardships are all about significant inequalities. Lotteries discard such circumstantial information. ## Moving forward This is the upshot so far: Fair admissions cannot be based on a single criterion or procedure. Rather, some combination of criteria and procedures is required to avoid discrimination against any group of students. Here is how I tried to solve the problem. 1. I decided to accept 35 students. That is 10 more than originally planned and puts me at 140 percent course load. 2. To fill those 35 seats I set up a lottery. It assigns a higher probability to older students. After all, I do subscribe to the argument that older students experience enormous social and economic pressures. However, given my ignorance about their course of studies and given the mounting pressure from younger cohorts, age does not justify deterministic admission.2 3. Students who claimed exceptional circumstances, e.g., small children, were additionally admitted.3 In the end, 39 students were admitted to class. That’s quite a lot, but still manageable based on my past experience. Attrition will do its work and eventually reduce class size to about 30. ## Conclusion Every faculty member fights the admissions battle on her own, no solution is ideal. By combining different admissions criteria and processes I strove to democratize admissions as much as reasonable. The result is unavoidably imperfect. However, in light of my own limited ressources, I believe it constitutes a compromise that will be endurable for everyone. ## Introduction This post demonstrates how small multiples can be used to highlight different parts of a distribution. In principle, ggplot2 offers many, easy to use options that partial out different groups in your data. The aesthetics color, size, and shape come to mind. Moreover, the purpose of small multiples is to show the same (bivariate) association across different groups in your data. Notwithstanding, either approach has drawbacks. When mapping categories to an aesthetic like color, all groups remain on the same canvas. The result may be wanting, especially when you work with big datasets. Small multiples, in contrast, draw out each group but deemphasize the grand picture. Wouldn’t it be nice to find some middle ground? ## Motivation We are going to work with the diamonds data which is available from the ggplot2 package. The goal is to highlight each cut in a scatter plot of price against carat without falling into either of the extremes mentioned above. Here is what the data look like: ``````rm(list = ls()) library("tidyverse") data(diamonds) select(diamonds, price, carat, cut) `````` ```## # A tibble: 53,940 x 3 ## price carat cut ## <int> <dbl> <ord> ## 1 326 0.23 Ideal ## 3 327 0.23 Good ## 5 335 0.31 Good ## 6 336 0.24 Very Good ## 7 336 0.24 Very Good ## 8 337 0.26 Very Good ## 9 337 0.22 Fair ## 10 338 0.23 Very Good ## # … with 53,930 more rows ``` ## Recipe As always, smart layering is the answer. We are going to plot the diamonds data twice using different colors: Once for all diamonds in the data, and once for each cut. The code also includes minor finishing touches (opacity and color). ``````alpha_lvl <- .4 ggplot(data = diamonds, aes(x = carat, y = price)) + geom_point( data = select(diamonds, -cut), # Dropping <cut> plots all our data. colour = "#3288bd", alpha = alpha_lvl ) + geom_point(colour = "#d53e4f", alpha = alpha_lvl) + scale_y_log10() + facet_wrap(vars(cut)) `````` ## Conclusion This post demonstrates how small multiples can highlight different segments within a distribution without losing sight of its overall shape. The key is smart layering. Plot the data twice: Once ignore and once highlight your facets.
# Use the distance formula to find the distances between the given points:(5, 2) and (0,-4) sciencesolve \| Certified Educator The problem provides the information that you need to use distance formula to evaluate the distance between the given points, such that: `d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)` Identifying `(x_1,y_1) = (5,2)` and `(x_2,y_2) = (0,-4)` yields: `d = sqrt((0 - 5)^2 + (- 4 - 2)^2)` `d = sqrt(25 + 36) => d = sqrt 61` Hence, evaluating the distance between the given points, using distance formula, yields `d = sqrt 61.` giorgiana1976 \| Student To determine the distance between 2 given points in the rectangular plane, we'll apply the Pythagorean theorem in the right angle triangle formed by the projections of the given points. We'll note the points as A(5, 2) and B(0,-4). The right angle triangle is ACB, where <C = 90 degrees and AB is the hypothenuse. We'll calculate the cathetus AC: AC = xA - xC AC = 5 - 0 AC = 5 BC = yB - yC = 4 + 2 = 6 The hypothenuse AB: AB^2 = AC^2 + BC^2 AB^2 = 5^2 + 6^2 AB^2 = 25 + 36 AB^2 = 61 AB = sqrt 61 units We'll keep just the positive value, since AB represents a distance.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> Exterior Angles in Convex Polygons Angles on the outside of a polygon formed by extending a side. Estimated6 minsto complete % Progress Practice Exterior Angles in Convex Polygons Progress Estimated6 minsto complete % Exterior Angles in Convex Polygons What if you were given a seven-sided regular polygon? How could you determine the measure of each of its exterior angles? After completing this Concept, you'll be able to use the Exterior Angle Sum Theorem to solve problems like this one. Watch This CK-12 Exterior Angles in Convex Polygons Watch the second half of this video. James Sousa: Angles of Convex Polygons Guidance An exterior angle is an angle that is formed by extending a side of the polygon. As you can see, there are two sets of exterior angles for any vertex on a polygon, one going around clockwise (\begin{align}1^{st}\end{align} hexagon), and the other going around counter-clockwise (\begin{align}2^{nd}\end{align} hexagon). The angles with the same colors are vertical and congruent. The Exterior Angle Sum Theorem states that the sum of the exterior angles of ANY convex polygon is \begin{align}360^\circ\end{align}. If the polygon is regular with \begin{align}n\end{align} sides, this means that each exterior angle is \begin{align}\frac{360}{n}^\circ\end{align}. Example A What is \begin{align}y\end{align}? \begin{align}y\end{align} is an exterior angle and all the given angles add up to \begin{align}360^\circ\end{align}. Set up an equation. \begin{align}70^\circ + 60^\circ + 65^\circ + 40^\circ + y & = 360^\circ\\ y & = 125^\circ\end{align} Example B What is the measure of each exterior angle of a regular heptagon? Because the polygon is regular, the interior angles are equal. It also means the exterior angles are equal. \begin{align}\frac{360^\circ}{7} \approx 51.43^\circ\end{align} Example C What is the sum of the exterior angles in a regular 15-gon? The sum of the exterior angles in any convex polygon, including a regular 15-gon, is \begin{align}360^\circ\end{align}. CK-12 Exterior Angles in Convex Polygons --> Guided Practice Find the measure of each exterior angle for each regular polygon below: 1. 12-gon 2. 100-gon 3. 36-gon For each, divide \begin{align}360^\circ\end{align} by the given number of sides. 1. \begin{align}30^\circ\end{align} 2. \begin{align}3.6^\circ\end{align} 3. \begin{align}10^\circ\end{align} Explore More 1. What is the measure of each exterior angle of a regular decagon? 2. What is the measure of each exterior angle of a regular 30-gon? 3. What is the sum of the exterior angles of a regular 27-gon? Find the measure of the missing variables: 1. The exterior angles of a quadrilateral are \begin{align}x^\circ, 2x^\circ, 3x^\circ,\end{align} and \begin{align}4x^\circ.\end{align} What is \begin{align}x\end{align}? Find the measure of each exterior angle for each regular polygon below: 1. octagon 2. nonagon 3. triangle 4. pentagon To view the Explore More answers, open this PDF file and look for section 6.2. Vocabulary Language: English Spanish exterior angle exterior angle An angle that is formed by extending a side of the polygon. regular polygon regular polygon A polygon in which all of its sides and all of its angles are congruent. Exterior Angle Sum Theorem Exterior Angle Sum Theorem Exterior Angle Sum Theorem states that the exterior angles of any polygon will always add up to 360 degrees.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Checking for Solutions to Systems of Linear Inequalities ## Substitute a given value and simplify Estimated21 minsto complete % Progress Practice Checking for Solutions to Systems of Linear Inequalities Progress Estimated21 minsto complete % Checking for Solutions to a System of Linear Inequalities Three times Aubrey's age minus twice Dakar's age is less than or equal to 25. Four times Aubrey's age plus three times Dakar's age is greater than or equal to 90. Which could be their ages? A. Aubrey is 10 and Dakar is 15. B. Aubrey is 16 and Dakar is 8. C. Aubrey is 17 and Dakar is 13. ### Guidance A linear system of inequalities has an infinite number of solutions. Recall that when graphing a linear inequality the solution is a shaded region of the graph which contains all the possible solutions to the inequality. In a system, there are two linear inequalities. The solution to the system is all the points that satisfy both inequalities or the region in which the shading overlaps. #### Example A Given the system of linear inequalities shown in the graph, determine which points are solutions to the system. a) (0, -1) b) (2, 3) c) (-2, -1) d) (3, 5) Solution: a) The point (0, -1) is not a solution to the system of linear inequalities. It is a solution to $y \le \frac{2}{3}x+3$ (graphed in ${\color{blue}\mathbf{blue}}$ ), but it lies on the line $y=-\frac{4}{5}x-1$ which is not included in the solution to $y>-\frac{4}{5}x-1$ (shown in ${\color{red}\mathbf{red}}$ ). The point must satisfy both inequalities to be a solution to the system. b) The point (2, 3) lies in the overlapping shaded region and therefore is a solution to the system. c) The point (-2, -1) lies outside the overlapping shaded region and therefore is not a solution the system. d) The point (3, 5) lies on the line $y=\frac{2}{3}x+3$ , which is included in the solution to $y \le \frac{2}{3}x+3$ . Since this part of the line is included in the solution to $y>-\frac{4}{5}x-1$ , it is a solution to the system. #### Example B Determine whether the following points are solutions to the system of linear inequalities: $3x+2y & \ge 4\\x+5y &< 11$ a) (3, 1) b) (1, 2) c) (5, 2) d) (-3, 1) Solution: This time we do not have a graph with which to work. Instead, we will plug the points into the equations to determine whether or not they satisfy the linear inequalities. A point must satisfy both linear inequalities to be a solution to the system. a) Yes, $3(3)+2(1) \ge 4$ and $(3)+5(1)<11$ are both valid inequalities. Therefore, (3, 1) is a solution to the system. b) No, $3(1)+2(2) \ge 4$ is true, but $(1)+5(2)=11$ , so the point fails the second inequality. c) No, $3(5)+2(2) \ge 4$ is true, but $(5)+5(2)>11$ , so the point fails the second inequality. d) No, $3(-3)+2(1)=-9+2=-7<4$ , so the point fails the first inequality. There is no need to check the point in the second inequality since it must satisfy both to be a solution. #### Example C Is (-9, 0) a solution to the system below? $y &> 3x -11 \\x + 2y \le 4$ Solution: Substitute the point into each equation and see if the inequalities hold true. $0 &> 3(-9) -11 \\0 &> 38$ The first inequality is true, let's test the second. $-9 + 2(0) \le 4 \\-9 \le 4$ This inequality is also true, therefore (-9, 0) is a solution. Intro Problem Revisit We don't have a graph to work with so we must plug the ages into the system of inequalities to determine whether or not they satisfy both. The system of linear inequalities represented by this situation is: $3A - 2D \le 25\\4A + 3D \ge 90$ Now we test each of the posibilities. For A = 10 and D = 15: $3(10) - 2(15) \le 25$ is fulfilled. $4(10) + 3(15) \ge 90$ is NOT fulfilled. For B = 16 and D = 8: $3(16) - 2(8) \le 25$ is NOT fulfilled and we can stop here. For B = 17 and D = 13: $3(17) - 2(13) \le 25$ is fulfilled. $4(17) + 3(13) \ge 90$ is fulfilled. Therefore, of the answer choices given only C could be Aubrey and Dakar's ages. ### Guided Practice 1. Determine whether the given points are solutions to the systems shown in the graph: a) (-3, 3) b) (4, 2) c) (3, 2) d) (-4, 4) 2. Determine whether the following points are solutions to the system: $y &< 11x-5\\7x-4y & \ge 1$ a) (4, 0) b) (0, -5) c) (7, 12) d) (-1, -3) 1. a) (-3, 3) is a solution to the system because it lies in the overlapping shaded region. b) (4, 2) is not a solution to the system. It is a solution to the red inequality only. c) (3, 2) is not a solution to the system because it lies on the dashed blue line and therefore does not satisfy that inequality. d) (-4, 4) is a solution to the system since it lies on the solid red line that borders the overlapping shaded region. 2. a) Yes, $0<11(4)-5$ , and $7(4)-4(0) \ge 1$ . b) No, $-5=11(0)-5$ so the first inequality is not satisfied. c) Yes, $12<11(7)-5$ , and $7(7)-4(12) \ge 1$ . d) No, $-3>11(-1)-5$ so the first inequality is not satisfied. ### Practice Given the four linear systems graphed below, match the point with the system(s) for which it is a solution. A. B. C. D. 1. (0, 3) 2. (2, -1) 3. (4, 3) 4. (2, -1) 5. (-3, 0) 6. (4, -1) 7. (-1, -2) 8. (2, 1) 9. (2, 5) 10. (0, 0) Given the four linear systems below, match the point with the system(s) for which it is a solution. A. $5x+2y & \le 10\\3x-4y &> -12$ B. $5x+2y &< 10\\3x-4y & \le -12$ C. $5x+2y &> 10\\3x-4y &< -12$ D. $5x+2y & \ge 10\\3x-4y & \ge -12$ 1. (0, 0) 2. (4, 6) 3. (0, 5) 4. (-3, 4) 5. (4, 3) 6. (0, 3) 7. (-8, -3) 8. (1, 6) 9. (4, -5) 10. (4, -2)
The quotient rule states that one radical divided by another is the same as dividing the numbers and placing them under the same radical symbol. Our examples will be using the index to be 2 (square root). No denominator contains a radical. The quotient rule states that a … Adding and Subtracting Radical Expressions, $$a) \sqrt{\color{red}{6}} \cdot \sqrt{\color{blue}{5}} = \sqrt{\color{red}{6} \cdot \color{blue}{5}} = \sqrt{30}$$, $$b) \sqrt{\color{red}{5}} \cdot \sqrt{\color{blue}{2ab}} = \sqrt{\color{red}{5} \cdot \color{blue}{2ab}} = \sqrt{10ab}$$, $$c) \sqrt[4]{\color{red}{4a}} \cdot \sqrt[4]{\color{blue}{7a^2b}} = \sqrt[4]{\color{red}{4a} \cdot \color{blue}{7a^2b}} = \sqrt[4]{28a^3b}$$, $$a) \sqrt{\frac{\color{red}{5}}{\color{blue}{36}}} = \frac{ \sqrt{\color{red}{5}} } { \sqrt{\color{blue}{36}} } Example $$\PageIndex{10}$$: Use Rational Exponents to Simplify Radical Expressions. Problem. In this examples we assume that all variables represent positive real numbers. Simplifying Radical Expressions. Simplify the radicals in the numerator and the denominator. \sqrt{18} = \sqrt{\color{red}{9} \cdot \color{blue}{2}} = \sqrt{\color{red}{9}} \cdot \sqrt{\color{blue}{2}} = 3\sqrt{2} . For all of the following, n is an integer and n ≥ 2. mathhelp@mathportal.org, More help with radical expressions at mathportal.org,$$ \color{blue}{\sqrt5 \cdot \sqrt{15} \cdot{\sqrt{27}}} $$,$$ \color{blue}{\sqrt{\frac{32}{64}}} $$,$$ \color{blue}{\sqrt[\large{3}]{128}} $$. Example Problem #1: Differentiate the following function: y = 2 / (x + 1) Solution: Note: I’m using D as shorthand for derivative here instead of writing g'(x) or f'(x):. Actually, I'll generalize. Important rules to simplify radical expressions and expressions with exponents are presented along with examples. Use Product and Quotient Rules for Radicals When presented with a problem like √4 , we don’t have too much difficulty saying that the answer 2 (since 2 × 2 = 4). If we converted every radical expression to an exponential expression, then we could apply the rules for … = \frac{\sqrt{5}}{6} Come to Algbera.com and read and learn about inverse functions, expressions and plenty other math topics ELEMENTARY ALGEBRA 1-1 0 0 0. 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Using the Quotient Rule to Simplify Square Roots. Another such rule is the quotient rule for radicals. Product Rule for Radicals Often, an expression is given that involves radicals that can be simplified using rules of exponents. The logical and step-bystep approach to problem solving has been a boon to me and now I love to solve these equations. product and quotient rule for radicals, Product Rule for Radicals: Write the radical expression as the quotient of two radical expressions. Use the rule to create two radicals; one in the numerator and one in the denominator. Step 1: Now, we need to find the largest perfect cube that divides into 24. Simplify. The quotient rule can be used to differentiate tan(x), because of a basic quotient identity, taken from trigonometry: tan(x) = sin(x) / cos(x). \sqrt{108} = \sqrt{\color{red}{36} \cdot \color{blue}{3}} = \sqrt{\color{red}{36}} \cdot \sqrt{\color{blue}{3}} = 6\sqrt{3} , No perfect square divides into 15, so \sqrt{15} cannot be simplified. Simplify each radical. Quotient Rule for Radicals Example . If it is not, then we use the product rule for radicals Given real numbers A n and B n, A ⋅ B n = A n ⋅ B n. and the quotient rule for radicals Given real numbers A n … Welcome to MathPortal. If n is even, and a ≥ 0, b > 0, then. Example 1 - using product rule That is, the radical of a quotient is the quotient of the radicals. Finding the root of product or quotient or a fractional exponent is simple with these formulas; just be sure that the numbers replacing the factors a and b are positive. When written with radicals, it is called the quotient rule for radicals. Given a radical expression, use the quotient rule to simplify it. A perfect square fraction is a fraction in which both the numerator and the denominator are perfect squares. More simply, you can think of the quotient rule as applying to functions that are written out as fractions, where the numerator and the denominator are both themselves functions. Go down deep enough into anything and you will find mathematics. Rules for Exponents. b \ne 0 and n is a natural number, then Simplify radical expressions using the product and quotient rule for radicals. If \sqrt[n]{a} and \sqrt[n]{b} are real numbers and n is a natural number, then Example 4: Use the quotient rule to simplify. Then the quotient rule tells us that F prime of X is going to be equal to and this is going to look a little bit complicated but once we apply it, you'll hopefully get a little bit more comfortable with it. Example: Simplify: (7a 4 b 6) 2. To begin the process of simplifying radical expression, we must introduce the product and quotient rule for radicals Product and quotient rule for radicals In this example, we are using the product rule of radicals in reverse to help us simplify the square root of 200. We can take the square root of the 25 which is 5, but we will have to leave the 3 under the square root. Use the quotient rule to divide variables : Power Rule of Exponents (a m) n = a mn. A Radical Expression Is Simplified When the Following Are All True. It has been 20 years since I have even thought about Algebra, now with my daughter I want to be able to help her. Show Step-by-step Solutions. If not, we use the following two properties to simplify them. The radicand has no fractions.$$ \large{\color{blue}{\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{ab}}} $$. f (x) = 5 is a horizontal line with a slope of zero, and thus its derivative is also zero. Finding the root of product or quotient or a fractional exponent is simple with these formulas; just be sure that the numbers replacing the factors a and b are positive. Simplify the radical expression. That is, the product of two radicals is the radical of the product.$$ \color{blue}{\frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[\large{n}]{\frac{a}{b}}} . It isn't on the same level as product and chain rule, those are the real rules. The Quotient Rule A quotient is the answer to a division problem. These include the constant rule, power rule, constant multiple rule, sum rule, and difference rule. More simply, you can think of the quotient rule as applying to functions that are written out as fractions, where the numerator and the denominator are both themselves functions. Quotient Rule for Radicals? Quotient Rule for Radicals . Step 1: Name the top term f(x) and the bottom term g(x). Solution. 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# » Parents ### Play games such as: 1. Any board game that uses dice and cards 2. Dot card and 10 frame games and activities 3. Set 4. Tangos 5. ‘Make 10 Go Fish’ 6. Mastermind 7. Dominoes 8. Mancala 9. Chess/Checkers 10. Cribbage 11. Yahtzee 1. Cribbage 2. Yahtzee 3. Set 4. Tangos 5. Mastermind 6. Dominoes 7. Mancala 8. Chess/Checkers 9. For younger students: anything with Dice and cards (Snakes and Ladders, board games, card games –‘ Make 10 Go Fish’) ### Strike It Out Power of 10 (partitioning): knowing the pairs of numbers that add to make 10 to partition numbers. Examples: 7+4 = 7+3+1 = 10+1= 11 or 8+5 = 8+2+3=10+3=13 Near Doubles: once students know their doubles this can be used to find near doubles. Examples: 7+8 = 7+7+1 = 14+1 = 15 or 6+7 = 6+6+1=12+1=13 Adding 9: we want students to see that adding 9 is the same as adding 10, subtract 1. Example: 9+6 = 10+6-1 = 16-1 = 15 or 9+8 = 10+8-1=18-1=17 Partitioning into place value: this also reinforces the universal “rule” to adding any numbers (whole numbers, decimals, fractions, variables) that you add “like terms”. Example: 324+68 à 300 + (20+60) + (4+8) = 300+80+12 = 392 ## Subtraction Strategies Subtract 9: is just like subtracting 10 +1. Connect this to adding 9. Example: 13-9 = 13-10+1 = 3+1 = 4 Subtract 8: subtract 10 +2. Example: 12-8 = 12-10+2 = 2+2 = 4 Power of 10 or partitioning: break up the numbers to make it a multiple of 10. Example: 13-5 = 13-3-2 = 10-2=8. Think of it as addition and count up. Example: 14-6 = 6 + ? = 14 àthen they could do 6 + 4=10 + 4 =14 so ? = 8. This strategy can be very useful for larger number subtractions. ## Multiplication Strategies For meaning: use arrays, area models, manipulatives and “GROUPS OF” (if they can’t give you a story problem that would use multiplication it means they likely don’t have conceptual understanding) Most students find the 0,1,2,5,10 multiplication facts easier to learn, and so use strategies based on what they know. Using strategies builds number sense and are good pre-algebra thinking activities for students. 3’s: double a number and add another group (if they have become proficient at adding, this will be easier for them). Example: 3 x 7 = 2 x 7 +7= 14 + 7 = 21 4’s: double twice. Example: 4 x 8 = 2 x 8 x 2 = 16 x 2 = 32 6’s: double the 3’s. Example: 6 x 6 = 3 x 6 x 2= 18 x 2 = 36 OR do the # x5 + another group of the #. Example: 6 x 6 = 5 x 6 = 30 + 6 = 36 7’s: do the #x5 + the # x2. Example: 7 x 8 = 5 x 8 + 2 x 8 = 40+16 = 56 8’s: double three times. Example: 8 x 8 = 8 x 2 x 2 x 2= 16 x 2 x 2 = 32 x 2 = 64 OR double the 4’s if they know them. Example: 8 x 6 = 4 x 6 x 2 = 24 x 2 = 48 OR “jump off” what they know by adding or subtracting groups. Example: 8 x 7 (if the student knew 7 x7 = 49, then they can add another group of 7 = 56) 9’s: multiply by 10 and then subtract one group of the number. Example: 9 x 7 = 70-7 = 63 OR use the knowledge that the digits will always add to 9, the pattern of tens and ones 11’s: Pattern found up to 9 x 11. Example: 11 x 4 = 44 OR Multiply by 10 and add another group. Example: 11 x 7 = 10 x 7 + 7 = 70+7=77 12’s: do 10 x the # +2 x the number. Example: 12 x 7 = 10 x 7 + 2 x 7 = 70+14=84 OR add another group to 11’s. Example: 12 x 8 = 11 x8 + 8 = 88+8 = 96 ## Multi-Digit Multiplication Strategies Use box method or Area Model, distributive property, traditional with meaning (multiplying numbers not digits, so no “carrying over”). All of these methods will have future applications like multiplying binomials. Examples: ## Division strategies Most students find it easiest to think about as the opposite of multiplication. For meaning: use arrays, area models, manipulatives and explore both equal sharing and equal grouping and how they are similar and different. Equal Sharing: 12÷4 = 3 means 12 divided into 4 equal groups = 3 in each group: Multi-Digit Division: Repeated subtraction or Partial Quotient method for multi-digit division can be a welcome strategy to use rather than the traditional long division algorithm: Long division algorithm with meaning: 2 Options: #### Parents – How to Help Your Children to Learn & Enjoy Math Part 1 In today’s post we will be exploring how to connect reading to math, math in the home, and math out ... #### Parents – How to Help Your Children to Learn & Enjoy Math Part 2 Most adults use estimation and calculators to do their daily math, rarely do they get out a piece of paper ... #### Parents – How to Help Your Children to Learn & Enjoy Math Part 3 I saved the best for last! Who doesn’t love playing games?! This last post for parents on how to support ... ## Tips for supporting your child at home: Avoid endorsing math anxiety or being “bad at math”. Students who have these attitudes towards math have more difficulties learning math than those who approach it positively. This is the “growth mindset” versus “fixed mindset” and has been proven to really affect learning. Another way to encourage a Growth Mindset is to encourage perseverance through frustration and understanding that mistakes actually make the brain grow – so are not bad but rather are very useful for developing understanding. Math is literally all around us….if we look for it. It doesn’t have to be just computations but rather looking at relationships, patterns and sizes. Problem solving, deciding between choices are also mathematical processes, as are playing games and solving logic puzzles. For younger students: grouping and organizing toys, practicing adding and subtracting using toys, blocks, etc. and looking for and making patterns. Comparing more than and less than and by how much. For example: a child has 4 dolls and 12 stuffed animals, you could ask, “how many more stuffed animals than dolls do you have and how do you know?” Read books that involve math or find the math in nighttime stories. Encourage creativity and ask your child to find the math (shapes, counting, comparing, categorizing, estimating, etc.) in stories, shows, movies etc. This helps develop mathematical habits of mind. We know you want to support your children as best as you can. That can be tough in an age where there is almost too much information and conflicting information! This is particularly true with our new math curriculum. This page is here to help provide you with some information about our math curriculum. It also provides you with some activities you can do at home, and links to other great websites that can help you support your child best. Math can be fun, easy and challenging – but in a good way. It is all about how we approach it. ## Minimize Math Anxiety For many of us math was no fun in school. It was difficult and often a complete mystery. It is easy to think that math is ‘hard’ because of our experiences with it. But for many of us, that is because of how it was taught. Our new methods overcome that. Avoid endorsing math anxiety or being “bad at math”. Students who have these attitudes towards math have more difficulties learning math than those who approach it positively. ## Encourage a Growth Mindset We all have either a “growth mindset” or “fixed mindset” when it comes to math. These mindsets have been proven to really affect learning.To develop a Growth Mindset encourage your child to persevere through frustration and understand that mistakes are a really important part of the learning process – so are not bad but rather are very useful when we reflect on them and understand what type of mistake it was (carless mistake, misunderstanding, etc.). Also we need to remember that learning is challenging and when things get difficult that doesn’t mean there is anything wrong but rather it is a part of learning! New concepts are often difficult at first, while we are trying to connect them to what we already know and make sense of them, and then with time and practice, they are not so difficult anymore. Embrace the struggle – without great struggle there is no great learning! Please see the helpful websites for more information on how to help your child develop a Growth Mindset and for the 10 minute TED talk by Dr. Carol Dweck, the researcher who has researched and written about this extensively. ## Helping with Homework: Many parents don’t understand why we use multiple strategies to solve the problem when only one strategy is needed. We are teaching students WHY the math works when we use different strategies so even though you may think that the strategies are inefficient there is a very important purpose to using them. Some strategies are inefficient but are an important stepping stone towards deeper understanding. Imagine an Olympic diver attempting a complex dive off of a high board without first practicing multiple small steps in a foam pit or supported trampoline. It is the same learning approach. We use manipulatives (blocks, tiles, fraction circles) and pictures to help students make sense of what the math actually means. For example 4 x 3 means 4 groups of 3, or 4 rows of 3. Most humans gather 70%-90% of their information visually, so we are making math more visual to help more people understand it. We also use these visuals to help us find the generalizations in math that become those ‘rules’ you might have memorized. This way students actually understand why the rules work as they do and they often construct the rules themselves which means they are way more likely to remember them! It is great role modeling for your children to see you as a lifelong learner who is open minded to trying new things. We encourage you to try to understand these different methods and visuals as this can help improve your number sense too! If you are really stuck and can’t help your child with their homework, please send along a note to the teacher explaining so. This is more helpful than teaching them a shortcut or the traditional algorithm before they have enough deeper understanding to actually understand the algorithm or short cut. If they can’t explain why something works, then they don’t have the understanding we’re aiming for. ## Involve your child in “real-life” math (adjust for the child’s age) ### Grocery shopping: 1. Practice rounding items to the nearest dollar or dime (example: if a jar of peanut butter is \$4.69 – is this closer to \$4 or \$5? How do you know?) 2. Estimate the value of each item and create a total estimate. This can be a game to see how close you come to the real value. 3. Compare prices to see which is the better deal. This can be done using estimation or using the unit prices listed on most prices. 4. If you buy enough groceries to last 4 or 5 days, estimate the cost per day 5. Determine how much you save if you buy things on sale. ### Cooking/Baking: 1. Just using measuring spoons and cups and reading recipes helps! 2. When doubling or halving recipes, determine how much of each ingredient is needed. 3. Looking at the measurements on measuring spoons and cups, determine how many teaspoons in a tablespoon, how many tablespoons in a ¼ cup, etc. 4. Ask them to do the measuring and show when you estimate (1/2 teaspoon in your palm etc). ### In the kitchen: 1. What holds more – juice container or milk container? How do you know? How much more? 2. Find containers that have similar volumes but different shapes (like a tall skinny container compared to a short fat container). Ask them to determine by looking which has more, then read the volumes 3. Estimate weights of ingredients (a potato, carrots, etc) and then weigh them on a scale to see how close you came. 4. Chop vegetables into fractions! If I chop a celery stalk into 4 pieces, what fraction of the stalk is each piece? ### In the car: (keep in mind that the working memory is not fully developed so don’t surpass 2-digit numbers unless your child can work with larger numbers mentally) 1. Make 20 – they can make 20 by adding, subtracting, multiplying, dividing, or a combination. This can be done with any number and the more ways a child can find the better! For K – make 5, for Gr. 1/2 – make 10, for Gr. 3/4 – Make 20, 25, 30,etc. for Gr. 5- up to make 100 2. Estimate how far 1 km is (use your odometer). Estimate 2 km. Use Siri to get directions and then make a game out of when to turn (she’ll say “turn left in 500m”). It helps students to know how far 500 m or 200 m is. 3. Ask your child an addition (subtraction, multiplication, division) question and then ask them HOW they got it. For example: 15 + 7 (they might say 10 +5+7 = 10+12 =22). 4. Ask your child to create a story problem or real-life problem to match a question. For example: 3 x 4 à A story problem that might work. I have 3 friends over and I give each friend 4 candies from my Halloween stash, how many candies did I give away? 5. Give a question and ask your child to estimate only and then see who is closest (you can play too). For example: 43 x 37 (estimate 40 x 40 = 1600).
# Fractions in Lowest Terms: Definition with Examples Home » Math Vocabulary » Fractions in Lowest Terms: Definition with Examples ## What Are Fractions in Lowest Terms? A fraction is said to be in lowest terms if the numerator and denominator have no common factors other than 1. A fraction in lowest terms is also termed as a fraction in the simplest form Consider an example: Suppose Oliver ate 3 out of 6 slices of a pizza. That means he had $\frac{3}{6}$ fraction of the pizza. Let’s visualize this. Take a look at the portion of the pizza he had. However, it looks like Oliver had a $\frac{1}{2}$ pizza. How come the fractions $\frac{3}{6}$ and $\frac{1}{2}$ represent the same value? $\frac{3}{6}$ and $\frac{1}{2}$ have the same value because $\frac{1}{2}$ is the lowest form of $\frac{3}{6}$. $\frac{3}{6}$ simplified gives $\frac{1}{2}$. In other words, we say that these are equivalent fractions. ## Fractions in the Lowest Term Definition A fraction is said to be in lowest terms if the GCD of the numerator and denominator is 1. Example: $\frac{7}{8}$ is a fraction in lowest terms since GCD(7,8) $= 1$ However, $\frac{6}{8}$ is not in its lowest terms since GCD(6,8) $= 2$ You might wonder what is $\frac{1}{2}$ in lowest terms? What is $\frac{1}{2}$ simplified? The answer is straightforward! $\frac{1}{2}$ itself is in the simplified form. ## How to Reduce Fractions in Lowest Terms Reducing fractions to lowest terms refers to simplifying the fractions such that the numerator and denominator become coprime (relatively prime). In other words, reducing fractions to lowest terms in math means to reducing fractions to a form $\frac{a}{b}(b \neq 0)$ such that GCD(a,b) $= 1$. Let’s learn two simple ways to reduce the fractions in lowest terms. ## Dividing by Common Factors Step 1: Divide both numerator and denominator by their lowest common factor (or any common factor). Step 2: Repeat this step for the resulting fraction. Step 3: Continue this process until the only common factor is 1. This method is simple but time-consuming. It uses several steps to get to the answer. Example 1: Reduce $\frac{48}{60}$ in lowest terms. $\frac{48 \div 2}{60 \div 2} = \frac{24}{30}$ …2 is the common factor of 48 and 60. $\frac{24 \div 2}{30 \div 2} = \frac{12}{15}$ …2 is the common factor of 24 and 30. $\frac{12 \div 3}{15 \div 3} = \frac{4}{5}$ …3 is the common factor of 48 and 60. 4 and 5 have no common factors other than 1. 4860 in its simplest form is 45. Example 2: Write each fraction in lowest terms. i) $\frac{6}{8}$ in lowest terms $= \frac{6 \div 2}{8 \div 2} = \frac{3}{4}$ ii) $\frac{135}{180}$ simplified $= \frac{135 \div 45}{180\div45} = \frac{3}{4}$ ## Using the Greatest Common Divisor Here, we divide the numerator and the denominator by the greatest common factor. The greatest common factor (GCF) method of reducing fractions involves only two simple steps. Step 1: Find the GCD of the numerator and the denominator. Step 2: Divide both the numerator and the denominator by the GCD. Example 1: $\frac{48}{60}$ Let’s first find the GCD of 48 and 60. $48 = 2 \times 2 \times 2 \times 2 \times 3$ $60 = 2 \times 2 \times 3 \times 5$ GCD(48, 60) $= 2 \times 2 \times 3 = 12$ Dividing both numerator and denominator by GCD, we get $\frac{48 \div 12}{60 \div 12} = \frac{4}{5}$ Example 2: $\frac{16}{50}$ $16 = 2 \times 2 \times 2 \times 2$ $50 = 2 \times 5 \times 5$ GCD(16, 50) $= 2$ $\frac{16 \div 2}{50 \div 2} = \frac{8}{25}$ ## Reducing Algebraic Fractions to Lowest Terms Lowest terms in algebra refers to simplifying a fraction with polynomials as numerator and denominator to its lowest form. Step 1: Factorize polynomials in the numerator and denominator. Step 2: Cancel out the common factors. Step 3: Simplify. SImplify $\frac{x^2\;-\; 5x + 6}{x^2\;-\;9}$: Factorize the expressions given in numerator and denominator. $x^2\;-\;5x + 6 = x^2\;-\;2x\;-\;3x + 6$ $= x (x\;-\;2)\;-\;3(x\;-\;2)$ $= (x\;-\;2)(x\;-\;3)$ and $x^2\;-\;9 = (x\;-\;3)(x + 3)$ Cancel the common terms. $\frac{x^2\;-\; 5x + 6}{x^2\;-\;9} = \frac{(x\;-\;2)(x\;-\;3)}{(x\;-\;3)(x + 3)} = \frac{(x\;-\;2)}{(x + 3)}$ ## Solved Examples on Fractions in Lowest Terms 1. Express $\frac{121}{187}$ in lowest terms. Solution: $121 = 11\times11$ $187 = 11\times17$ GCD$(121,\; 187) = 11$ $\frac{121\div 11}{187 \div11} = \frac{11}{17}$ 2. Simplify and write $\frac{96}{108}$ in lowest terms. Solution: $96 = 2\times2\times2\times2\times2\times3$ $108= 2\times2\times3\times3\times3$ GCD$(96,\;108) = 2\times2\times3 = 12$ $\frac{96 \div12}{108 \div12} = \frac{8}{9}$ Thus, $\frac{96}{108} = \frac{8}{9}$ 3. Express $\frac{x^2 + 4x + 4}{x^2\;-\;4}$ in lowest terms. Solution: Using identities: $(a + b)^2 = a^2 + 2ab + b^2$ $a^2\;-\;b^2 = (a + b)(a\;-\;b))$ $\frac{x^2 + 4x + 4}{x^2\;-\;4} = \frac{x^2 + (2\times2\times x) + 2^2}{x^2\;-\;2^2} = \frac{(x + 2)(x + 2)}{(x + 2)(x\;-\;2)} = \frac{x + 2}{x \;-\; 2}$ 4. Reduce to lowest terms: $\frac{180}{354}$ Solution: $180 = 2\times2\times3\times3\times5$ $354 = 2\times3\times59$ GCD$(180,\; 354) = 2\times3 = 6$ $\frac{180\div 6}{354 \div 6} = \frac{30}{59}$ 5. Express $\frac{105}{945}$ in lowest terms. Solution: $105 = 3\times5\times7$ $945 = 3\times3\times3\times5\times7$ GCD$(180,\; 354) = 3\times5\times7 = 105$ $\frac{105\div 105}{945 \div 105} = \frac{1}{9}$ 6. What is the lowest term of $\frac{4}{8}$? Solution: GCD$(4,\; 8) = 2$ Dividing both numerator and denominator by 2, we get $\frac{4}{8} = \frac{4 \div2}{8 \div2} = \frac{1}{2}$ The fraction $\frac{4}{8}$ in its lowest term is $\frac{1}{2}$. 7. Is $\frac{3}{4}$ in the lowest terms? Solution: Here, 3 and 4 are coprime numbers. GCD$(3,\; 4) = 1$ Thus, $\frac{3}{4}$ is in lowest terms. ## Practice Problems on Fractions in Lowest Terms 1 ### Simplify to the lowest terms: $\frac{45}{75}$ $\frac{2}{5}$ $\frac{3}{5}$ $\frac{7}{5}$ $\frac{4}{7}$ CorrectIncorrect Correct answer is: $\frac{3}{5}$ GCD$(45,\; 75) = 15$ $\frac{45\div15}{75\div15} = \frac{3}{5}$ 2 ### Which of the following fractions are not in their lowest terms? $\frac{13}{91}$ $\frac{11}{27}$ $\frac{13}{44}$ $\frac{3}{5}$ CorrectIncorrect Correct answer is: $\frac{13}{91}$ GCD$(13,\;91) = \frac{1}{3}$ $\frac{13}{91} = \frac{1}{7}$ 3 ### On expressing $\frac{8x + 24}{x + 3}$ in lowest term, we get: $8(x + 2)$ $\frac{1}{x + 3}$ $8(x + 1)$ $8$ CorrectIncorrect Correct answer is: $8$ $\frac{8x + 24}{x + 3} = \frac{8(x + 3)}{x + 3} = 8$ 4 ### The fraction $\frac{1}{4}$ represents the lowest form of $\frac{2}{4}$ $\frac{4}{12}$ $\frac{8}{32}$ $\frac{4}{64}$ CorrectIncorrect Correct answer is: $\frac{8}{32}$ $\frac{8}{32} = \frac{8 \div 8}{32 \div 8} = \frac{1}{4}$ 5 ### The fraction ab is said to be in lowest terms when LCM(a, b) $= 1$ GCD(a, b) $= 1$ GCD(a, b) $= 0$ LCM(a, b) $= 0$ CorrectIncorrect Correct answer is: GCD(a, b) $= 1$ A fraction is in lowest terms when the greatest common factor (GCF) of its numerator and denominator is one. 6 ### What is the lowest term of $\frac{4}{10}$? $\frac{2}{5}$ $\frac{2}{10}$ $\frac{2}{6}$ $\frac{8}{20}$ CorrectIncorrect Correct answer is: $\frac{2}{5}$ $\frac{4}{10} = \frac{4\div2}{10\div2} = \frac{2}{5}$ 7 ### What is $\frac{2}{6}$ in its lowest term? $\frac{1}{6}$ $\frac{1}{3}$ $\frac{4}{12}$ $\frac{6}{36}$ CorrectIncorrect Correct answer is: $\frac{1}{6}$ $\frac{2}{6} = \frac{2\div2}{6\div2} = \frac{1}{3}$ ## Frequently Asked Questions on Fractions in Lowest Terms The GCD will be equal to 1. A mixed number is said to be in its simplest form if the highest common factor, i.e., the HCF of its fractional part is 1. A ratio is said to be in the lowest term if antecedent and consequent of the ratio have no common factor other than 1. Fractions whose numerators and denominators are different, but they still represent the same value. Equivalent fractions of $\frac{1}{2} = \frac{2}{4} = \frac{3}{6} = \frac{4}{8}$ and so on. Lowest terms refers to a form of fraction that cannot be simplified further since the numerator and denominator have no factor in common except 1.
Courses Courses for Kids Free study material Offline Centres More Store # A man has $3$ coins A, B & C. A is fair coin. B is biased such that the probability of occurring head on it is $\dfrac{2}{3}$ C is also biased with the probability of occurring head as$\dfrac{1}{3}$. If one coin is selected and tossed three times, giving two heads and one tail, find the probability that the chosen coin was A.$A.\,\dfrac{9}{{25}} \\ B.\;\dfrac{3}{5} \\ C.\;\dfrac{{27}}{{125}} \\ D.\;\dfrac{1}{3} \\$ Last updated date: 13th Jun 2024 Total views: 411k Views today: 10.11k Verified 411k+ views Hint: Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the outcomes. Probability is the state of being probable and the extent to which something is likely to happen in the particular favourable situations. Here, we will find the probability of each of the three coins in the favourable pattern of Head, Head and Tail. Complete step by step solution: Give that: Coin A is the fair coin. Therefore, the probability of occurring head for coin A is $P{(H)_A} = \dfrac{1}{2}$ (By definition – as out of the total two outcomes Head and Tail, our favourable outcomes Head will occur only once.) When coin A is tossed, the probability of getting two heads and one tail is $P{(HHT)_A} = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}$ Therefore, $P{(HHT)_A} = \dfrac{1}{8}\;{\text{ }}.......{\text{(1)}}$ Coin B is the biased coin and The probability of occurring head for coin A is $P{(H)_B} = \dfrac{2}{3}$ (Given) Therefore, the probability of getting tail is equal to one minus the probability of getting Head $P{(T)_B} = 1 - P(H){}_B \\ P{(T)_B} = 1 - \dfrac{2}{3} \\ P{(T)_B} = \dfrac{1}{3} \\$ Now, by substituting the values - When coin B is tossed, the probability of getting two heads and one tail is $P{(HHT)_B} = \dfrac{2}{3} \times \dfrac{2}{3} \times \dfrac{1}{3}$ Therefore, $P{(HHT)_B} = \dfrac{4}{{27}}\,{\text{ }}.......{\text{(2)}}$ Similarly, for coin C The probability of occurring head for coin C is $P{(H)_C} = \dfrac{1}{3}$ (Given) Therefore, the probability of getting tail is equal to one minus the probability of getting Head $P{(T)_B} = 1 - P(H){}_B \\ P{(T)_B} = 1 - \dfrac{1}{3} \\ P{(T)_B} = \dfrac{2}{3} \\$ Now, by substituting the values - When coin C is tossed, the probability of getting two heads and one tail is $P{(HHT)_C} = \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{2}{3}$ Therefore, $P{(HHT)_C} = \dfrac{2}{{27}}\;{\text{ }}........{\text{(3)}}$ From the Equations $(1),\,{\text{(2) \& (3)}}$ Now, combined probability of all the three coins is – $P{(HHT)_T} = P{(HHT)_A} + P{(HHT)_B} + P{(HHT)_C} \\ P{(HHT)_T} = \dfrac{1}{8} + \dfrac{4}{{27}} + \dfrac{2}{{27}} \\$ Simplify by taking LCM $P{(HHT)_T} = \dfrac{{27 + 32 + 16}}{{216}} \\ P{(HHT)_T} = \dfrac{{75}}{{216}} \\$ (Take from both the numerator and the denominator; also same number from numerator and denominator cancels each other) $P{(HHT)_T} = \dfrac{{25}}{{72}}$ Now, the probability that the chosen coin was A, giving two heads and one tail when tossed is – $P(for{\text{ the coin A) = }}\dfrac{{P{{(HHT)}_A}}}{{P(HHT){}_T}} \\ P(for{\text{ the coin A)}} = \dfrac{{\dfrac{1}{8}}}{{\dfrac{{25}}{{72}}}} \\$ Use the property, denominator or numerator goes in denominator and denominator’s denominator goes in numerator. $\therefore P(for{\text{ the coin A) = }}\dfrac{{1 \times 72}}{{8 \times 25}}$ Now, simplify the right hand side of the equation – $\therefore P(for{\text{ the coin A) = }}\dfrac{9}{{25}}$ Thus, the required answer is the probability that the chosen coin was A, giving two heads and one tail when tossed is – $\therefore P(for{\text{ the coin A) = }}\dfrac{9}{{25}}$ Hence, from the given multiple choices, option A is the correct answer. Note: For this type of probability problems, just follow the general formula for probability and basic simplification properties for the fractions. Always remember the probability of any event lies between zero and one.
# INTERSECTING BOX AND CYLINDER In this example we will calculate the three dimensional coordinates of holes through a precast box. We can think of this as a cylinder intersecting the wall of a box while laying on its side. These calculations are possible using cross products to develop a generic mathematical formula but we can simplify this by using a definite cylinder height of 1000 inches. You will see in the diagrams below details for the points we are going to calculate. By the end of this we will create a formula to layout round, arch or elliptical holes at any rotation through a box wall. All dimensions will be in inches. The First step is to plot our hole on the plane y=0. The x,y,z can be expressed using the following parametric equation where r=hole radius. $0<=t<=2\pi$ $x(t) = r * \cos(t)$ $y(t) = 0 (On Plane y=0)$ $z(t) = r * \sin(t)$ Next we want to derive an equation to work if the hole is rotated at an angle phi about the z axis. $0<=t<=2\pi$ $\begin{pmatrix} x_0(t) \\y_0(t) \\z_0(t) \end{pmatrix}=\begin{pmatrix} cos(\phi)&-sin(\phi)&0 \\sin(\phi)&cos(\phi)&0 \\0&0&1 \end{pmatrix}\begin{pmatrix} r * \cos(t)\\0 \\r * \sin(t)\end{pmatrix}$ $\begin{pmatrix} x_0(t) \\y_0(t) \\z_0(t) \end{pmatrix}=\begin{pmatrix} r * \cos(t)\cos(\phi)\\r\cos(t)\sin(\phi)\\r \sin(t)\end{pmatrix}$ Next we want to derive an equation for the hole initially located on the y=-1000 plane and rotated at an angle phi about the z axis. $0<=t<=2\pi$ $\begin{pmatrix} x_1(t) \\y_1(t) \\z_1(t) \end{pmatrix}=\begin{pmatrix} \cos(\phi)&-sin(\phi)&0 \\sin(\phi)&cos(\phi)&0 \\0&0&1 \end{pmatrix}\begin{pmatrix} r * \cos(t)\\-1000 \\r * \sin(t)\end{pmatrix}$ $\begin{pmatrix} x_1(t) \\y_1(t) \\z_1(t) \end{pmatrix}=\begin{pmatrix} r * \cos(t)\cos(\phi)+1000\sin(\phi)\\r\cos(t)\sin(\phi)-1000\cos(\phi)\\r \sin(t)\end{pmatrix}$ Now we need to find an equation for the line between the circle at the origin and the point 1000 units away through points $P_1=(r\cos(t)cos(\phi),r\cos(t)sin(\phi),rsin(\phi))$ and $P_2=(r\cos(t)cos(\phi)+1000sin(\phi),r\cos(t)sin(\phi)-1000\cos(\phi),rsin(t))$ and now we find the vector r(t) between P1 and P2. The formula for a line in parametric form has is r(t)=P+tD where P is a point on the line and D is a direction vector for the line. In our case the direction vector D will be P2-P1 and we know a point P on the line will be P2. $r(\tau)=P_1+\tau(P_2-P_1)$ $r(\tau)=(r\cos(t)cos(\phi),r\cos(t)sin(\phi),rsin(t)) + \tau(r\cos(t)cos(\phi)+1000sin(\phi)-r\cos(t)cos(\phi),r\cos(t)sin(\phi)-1000\cos(\phi)-r\cos(t)sin(\phi),rsin(\phi)-rsin(\phi))$ $r(\tau)=(r\cos(t)cos(\phi),r\cos(t)sin(\phi),rsin(t))+\tau(1000sin(\phi),-1000\cos(\phi),0)$ $r(\tau)=(r\cos(t)cos(\phi)+\tau1000sin(\phi),r\cos(t)sin(\phi)-\tau1000\cos(\phi),rsin(t)$ The next step is to calculate the intersection point of this parametric line and the plane at each wall 0,90,180 and 270. We will use substitution in the above parametric equation to solve. 0 Degree Plane Intersection y=-w/2 We start with the y term of the parametric line since the 0 degree wall is on the plane y=-w/2. $\frac{-w}{2}=r\cos(t)sin(\phi)-\tau1000\cos(\phi)$ $\frac{-w}{2}-r\cos(t)sin(\phi)=-\tau1000\cos(\phi)$ $\tau=\frac{\frac{w}{2}+r\cos(t)sin(\phi)}{1000\cos(\phi)}$ Now that we have Tau we can substitute back into the parametric equation to get equations for x,y and z on the 0 degree wall/plane. L= Box Length and w = Box Width $r(\tau)=(r\cos(t)cos(\phi)+\tau1000sin(\phi),r\cos(t)sin(\phi)-\tau1000\cos(\phi),rsin(t)$ Solve for x: $x_i=(r\cos(t)cos(\phi)+\frac{\frac{w}{2}+r\cos(t)sin(\phi)}{1000\cos(\phi)}1000sin(\phi)$ $x_i=r\cos(t)cos(\phi)+\frac{\frac{w}{2}\sin(\phi)+r\cos(t)sin^2(\phi)}{cos(\phi)}$ Solve for y: $y_i=r\cos(t)sin(\phi)-\frac{\frac{w}{2}+r\cos(t)sin(\phi)}{1000\cos(\phi)}1000\cos(\phi)$ $y_i=r\cos(t)sin(\phi)-\frac{w}{2}-r\cos(t)sin(\phi)$ Solve for z: $z_i=rsin(t)$ 90 Degree Plane Intersection x=-L/2 We start with the x term of the parametric line since the 90 degree wall is on the plane x=-L/2. $\frac{-L}{2}=r\cos(t)cos(\phi)+\tau1000sin(\phi)$ $\frac{-L}{2}-r\cos(t)cos(\phi)=\tau1000\sin(\phi)$ $\tau=\frac{\frac{-L}{2}-r\cos(t)\cos(\phi)}{1000\sin(\phi)}$ Now that we have Tau we can substitute back into the parametric equation to get equations for x,y and z on the 90 degree wall/plane. L= Box Length and w = Box Width $r(\tau)=(r\cos(t)cos(\phi)+\tau1000sin(\phi),r\cos(t)sin(\phi)-\tau1000\cos(\phi),rsin(t)$ Solve for x: $x_i=r\cos(t)cos(\phi)+\frac{\frac{-L}{2}-r\cos(t)\cos(\phi)}{1000\sin(\phi)}1000sin(\phi)$ $x_i=r\cos(t)cos(\phi)-.5L-r\cos(t)cos(\phi)$ Solve for y: $y_i=r\cos(t)sin(\phi)-\frac{\frac{-L}{2}-r\cos(t)\cos(\phi)}{1000\sin(\phi)}1000\cos(\phi)$ $y_i=r\cos(t)sin(\phi)+\frac{\frac{w}{2}\cos(\phi)+r\cos(t)cos^2(\phi)}{sin(\phi)}$ Solve for z: $z_i=rsin(t)$ 180 Degree Plane Intersection y=w/2 We start with the y term of the parametric line since the 180 degree wall is on the plane y=w/2. $\frac{w}{2}=r\cos(t)sin(\phi)-\tau1000\cos(\phi)$ $\frac{-w}{2}+r\cos(t)sin(\phi)=\tau1000\cos(\phi)$ $\tau=\frac{\frac{-w}{2}+r\cos(t)\sin(\phi)}{1000\cos(\phi)}$ Now that we have Tau we can substitute back into the parametric equation to get equations for x,y and z on the 180 degree wall/plane. L= Box Length and w = Box Width $r(\tau)=(r\cos(t)cos(\phi)+\tau1000sin(\phi),r\cos(t)sin(\phi)-\tau1000\cos(\phi),rsin(t)$ Solve for x: $x_i=(r\cos(t)cos(\phi)+\frac{\frac{-w}{2}+r\cos(t)\sin(\phi)}{1000\cos(\phi)}1000sin(\phi)$ $x_i=r\cos(t)cos(\phi)-\frac{\frac{w}{2}\sin(\phi)+r\cos(t)\sin^2(\phi)}{cos(\phi)}$ Solve for y: $y_i=r\cos(t)sin(\phi)-\frac{\frac{-w}{2}+r\cos(t)\sin(\phi)}{1000\cos(\phi)}1000\cos(\phi)$ $y_i=r\cos(t)\sin(\phi)+\frac{w}{2}-r\cos(t)sin(\phi)$ Solve for z: $z_i=rsin(t)$ 270 Degree Plane Intersection x=L/2 We start with the x term of the parametric line since the 270 degree wall is on the plane x=L/2. $\frac{L}{2}=r\cos(t)cos(\phi)+\tau1000sin(\phi)$ $\frac{L}{2}-r\cos(t)cos(\phi)=\tau1000\sin(\phi)$ $\tau=\frac{\frac{L}{2}-r\cos(t)\cos(\phi)}{1000\sin(\phi)}$ Now that we have Tau we can substitute back into the parametric equation to get equations for x,y and z on the 270 degree wall/plane. L= Box Length and w = Box Width $r(\tau)=(r\cos(t)cos(\phi)+\tau1000sin(\phi),r\cos(t)sin(\phi)-\tau1000\cos(\phi),rsin(t)$ Solve for x: $x_i=r\cos(t)cos(\phi)+\frac{\frac{L}{2}-r\cos(t)\cos(\phi)}{1000\sin(\phi)}1000sin(\phi)$ $x_i=r\cos(t)cos(\phi)+.5L-r\cos(t)cos(\phi)$ Solve for y: $y_i=r\cos(t)sin(\phi)-\frac{\frac{L}{2}-r\cos(t)\cos(\phi)}{1000\sin(\phi)}1000\cos(\phi)$ $y_i=r\cos(t)sin(\phi)-\frac{\frac{w}{2}\cos(\phi)+r\cos(t)cos^2(\phi)}{sin(\phi)}$ Solve for z: $z_i=rsin(t)$ Now we can solve for the three dimensional coordinates at any angle. If the plotted coordinates fall outside the box or are extended past the wall we need to clip those points off. To clip those points we can quickly trim points using the following ranges. 0 and 180 degree wall: $\frac{-L}{2}\leq x\leq\frac{L}{2}$ 90 and 270 degree wall: $\frac{-w}{2}\leq y\leq\frac{w}{2}$ These ranges can be applied to all points where L=length of box and w = width of box. All points must fall in the ranges below or should be excluded. $\frac{-L}{2}\leq x_i\leq\frac{L}{2}\\\\ or \\\\\frac{-w}{2}\leq y_i\leq\frac{w}{2}$ $x_i=r\cos(t)cos(\phi)+.5L-r\cos(t)cos(\phi)$ $y_i=r\cos(t)sin(\phi)-\frac{\frac{w}{2}\cos(\phi)+r\cos(t)cos^2(\phi)}{sin(\phi)}$ $z_i=r*sin(t)$
## Section2.5Cyclic Subgroups Often a subgroup will depend entirely on a single element of the group; that is, knowing that particular element will allow us to compute any other element in the subgroup. ###### Example2.32. Suppose that we consider $3 \in {\mathbb Z}$ and look at all multiples (both positive and negative) of $3\text{.}$ As a set, this is \begin{equation} 3 {\mathbb Z} = \{ \ldots, -3, 0, 3, 6, \ldots \}\text{.} \end{equation} It is easy to see that $3 {\mathbb Z}$ is a subgroup of the integers. This subgroup is completely determined by the element $3$ since we can obtain all of the other elements of the group by taking multiples of $3\text{.}$ Every element in the subgroup is “generated” by $3\text{.}$ ###### Example2.33. If $H = \{ 2^n : n \in {\mathbb Z} \}\text{,}$ then $H$ is a subgroup of the multiplicative group of nonzero rational numbers, ${\mathbb Q}^\text{.}$ If $a = 2^m$ and $b = 2^n$ are in $H\text{,}$ then $ab^{-1} = 2^m 2^{-n} = 2^{m-n}$ is also in $H\text{.}$ By Proposition 2.31, $H$ is a subgroup of ${\mathbb Q}^$ determined by the element $2\text{.}$ ###### Proof. The identity is in $\langle a \rangle$ since $a^0 = e\text{.}$ If $g$ and $h$ are any two elements in $\langle a \rangle \text{,}$ then by the definition of $\langle a \rangle$ we can write $g = a^m$ and $h = a^n$ for some integers $m$ and $n\text{.}$ So $gh = a^m a^n = a^{m+n}$ is again in $\langle a \rangle \text{.}$ Finally, if $g = a^n$ in $\langle a \rangle \text{,}$ then the inverse $g^{-1} = a^{-n}$ is also in $\langle a \rangle \text{.}$ Clearly, any subgroup $H$ of $G$ containing $a$ must contain all the powers of $a$ by closure; hence, $H$ contains $\langle a \rangle \text{.}$ Therefore, $\langle a \rangle$ is the smallest subgroup of $G$ containing $a\text{.}$ ###### Remark2.35. If we are using the “+” notation, as in the case of the integers under addition, we write $\langle a \rangle = \{ na : n \in {\mathbb Z} \}\text{.}$ For $a \in G\text{,}$ we call $\langle a \rangle$ the cyclic subgroup generated by $a\text{.}$ If $G$ contains some element $a$ such that $G = \langle a \rangle \text{,}$ then $G$ is a cyclic group. In this case $a$ is a generator of $G\text{.}$ If $a$ is an element of a group $G\text{,}$ we define the order of $a$ to be the smallest positive integer $n$ such that $a^n= e\text{,}$ and we write $\|a\| = n\text{.}$ If there is no such integer $n\text{,}$ we say that the order of $a$ is infinite and write $\|a\| = \infty$ to denote the order of $a\text{.}$ ###### Example2.36. Notice that a cyclic group can have more than a single generator. Both $1$ and $5$ generate ${\mathbb Z}_6\text{;}$ hence, ${\mathbb Z}_6$ is a cyclic group. Not every element in a cyclic group is necessarily a generator of the group. The order of $2 \in {\mathbb Z}_6$ is $3\text{.}$ The cyclic subgroup generated by $2$ is $\langle 2 \rangle = \{ 0, 2, 4 \}\text{.}$ The groups ${\mathbb Z}$ and ${\mathbb Z}_n$ are cyclic groups. The elements $1$ and $-1$ are generators for ${\mathbb Z}\text{.}$ We can certainly generate ${\mathbb Z}_n$ with 1 although there may be other generators of ${\mathbb Z}_n\text{,}$ as in the case of ${\mathbb Z}_6\text{.}$ ###### Example2.37. The group of units, $U(9)\text{,}$ in ${\mathbb Z}_9$ is a cyclic group. As a set, $U(9)$ is $\{ 1, 2, 4, 5, 7, 8 \}\text{.}$ The element 2 is a generator for $U(9)$ since \begin{align} 2^1 & = 2 \qquad 2^2 = 4\\ 2^3 & = 8 \qquad 2^4 = 7\\ 2^5 & = 5 \qquad 2^6 = 1\text{.} \end{align} ###### Example2.38. Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle $S_3\text{.}$ The multiplication table for this group is Figure 2.7 . The subgroups of $S_3$ are shown in Figure 2.39 . Notice that every subgroup is cyclic; however, no single element generates the entire group. ###### Proof. Let $G$ be a cyclic group and $a \in G$ be a generator for $G\text{.}$ If $g$ and $h$ are in $G\text{,}$ then they can be written as powers of $a\text{,}$ say $g = a^r$ and $h = a^s\text{.}$ Since \begin{equation} g h = a^r a^s = a^{r+s} = a^{s+r} = a^s a^r = h g\text{,} \end{equation} $G$ is abelian. ### Subsection2.5.1Subgroups of Cyclic Groups We can ask some interesting questions about cyclic subgroups of a group and subgroups of a cyclic group. If $G$ is a group, which subgroups of $G$ are cyclic? If $G$ is a cyclic group, what type of subgroups does $G$ possess? ###### Proof. The main tools used in this proof are the division algorithm and the Principle of Well-Ordering. Let $G$ be a cyclic group generated by $a$ and suppose that $H$ is a subgroup of $G\text{.}$ If $H = \{ e \}\text{,}$ then trivially $H$ is cyclic. Suppose that $H$ contains some other element $g$ distinct from the identity. Then $g$ can be written as $a^n$ for some integer $n\text{.}$ Since $H$ is a subgroup, $g^{-1} = a^{-n}$ must also be in $H\text{.}$ Since either $n$ or $-n$ is positive, we can assume that $H$ contains positive powers of $a$ and $n \gt 0\text{.}$ Let $m$ be the smallest natural number such that $a^m \in H\text{.}$ Such an $m$ exists by the Principle of Well-Ordering. We claim that $h = a^m$ is a generator for $H\text{.}$ We must show that every $h' \in H$ can be written as a power of $h\text{.}$ Since $h' \in H$ and $H$ is a subgroup of $G\text{,}$ $h' = a^k$ for some integer $k\text{.}$ Using the division algorithm, we can find numbers $q$ and $r$ such that $k = mq +r$ where $0 \leq r \lt m\text{;}$ hence, \begin{equation} a^k = a^{mq +r} = (a^m)^q a^r = h^q a^r\text{.} \end{equation} So $a^r = a^k h^{-q}\text{.}$ Since $a^k$ and $h^{-q}$ are in $H\text{,}$ $a^r$ must also be in $H\text{.}$ However, $m$ was the smallest positive number such that $a^m$ was in $H\text{;}$ consequently, $r=0$ and so $k=mq\text{.}$ Therefore, \begin{equation} h' = a^k = a^{mq} = h^q \end{equation} and $H$ is generated by $h\text{.}$ ###### Proof. First suppose that $a^k=e\text{.}$ By the division algorithm, $k = nq + r$ where $0 \leq r \lt n\text{;}$ hence, \begin{equation} e = a^k = a^{nq + r} = a^{nq} a^r = e a^r = a^r\text{.} \end{equation} Since the smallest positive integer $m$ such that $a^m = e$ is $n\text{,}$ $r= 0\text{.}$ Conversely, if $n$ divides $k\text{,}$ then $k=ns$ for some integer $s\text{.}$ Consequently, \begin{equation} a^k = a^{ns} = (a^n)^s = e^s = e\text{.} \end{equation} ###### Proof. We wish to find the smallest integer $m$ such that $e = b^m = a^{km}\text{.}$ By Proposition 2.43 , this is the smallest integer $m$ such that $n$ divides $km$ or, equivalently, $n/d$ divides $m(k/d)\text{.}$ Since $d$ is the greatest common divisor of $n$ and $k\text{,}$ $n/d$ and $k/d$ are relatively prime. Hence, for $n/d$ to divide $m(k/d)$ it must divide $m\text{.}$ The smallest such $m$ is $n/d\text{.}$ ###### Example2.46. Let us examine the group ${\mathbb Z}_{16}\text{.}$ The numbers $1\text{,}$ $3\text{,}$ $5\text{,}$ $7\text{,}$ $9\text{,}$ $11\text{,}$ $13\text{,}$ and $15$ are the elements of ${\mathbb Z}_{16}$ that are relatively prime to $16\text{.}$ Each of these elements generates ${\mathbb Z}_{16}\text{.}$ For example, \begin{align} 1 \cdot 9 & = 9 & 2 \cdot 9 & = 2 & 3 \cdot 9 & = 11\\ 4 \cdot 9 & = 4 & 5 \cdot 9 & = 13 & 6 \cdot 9 & = 6\\ 7 \cdot 9 & = 15 & 8 \cdot 9 & = 8 & 9 \cdot 9 & = 1\\ 10 \cdot 9 & = 10 & 11 \cdot 9 & = 3 & 12 \cdot 9 & = 12\\ 13 \cdot 9 & = 5 & 14 \cdot 9 & = 14 & 15 \cdot 9 & = 7\text{.} \end{align} ###### 1. Explain what it means for a group to be cyclic. Your explanation should use the word generator at least once. ###### 2. Find two different generators for the group $U(9) = \{1,2,4,5,7,8\}\text{.}$ What does this say about whether $U(9)$ is cyclic? (Compare to the previous question.) ###### 3. Suppose $H$ is a subgroup of a cyclic group $G\text{.}$ Must $H$ be abelian? Explain. ###### 4. After reading the section, what questions do you still have? Write at least one well formulated question (even if you think you understand everything).
# Solve augmented matrices In this blog post, we will show you how to Solve augmented matrices. Our website will give you answers to homework. ## Solving augmented matrices In this blog post, we will take a look at how to Solve augmented matrices. The most common type of function is the linear function. A linear function is a function in which the input and output are related by a straight line. College algebra is the study of linear functions and their properties. It investigates how these functions can be used to model real-world situations. In addition, college algebra also covers topics such as graphing, solving equations, and manipulating algebraic expressions. As a result, college algebra is an important course for any student who plans on pursuing a career in mathematics or another field that uses mathematics. Solving for exponents can be a tricky business, but there are a few basic rules that can help to make the process a bit easier. First, it is important to remember that any number raised to the power of zero is equal to one. This means that when solving for an exponent, you can simply ignore anyterms that have a zero exponent. For example, if you are solving for x in the equation x^5 = 25, you can rewrite the equation as x^5 = 5^3. Next, remember that any number raised to the power of one is equal to itself. So, in the same equation, you could also rewrite it as x^5 = 5^5. Finally, when solving for an exponent, it is often helpful to use logs. For instance, if you are trying to find x in the equation 2^x = 8, you can take the log of both sides to get Log2(8) = x. By using these simple rules, solving for exponents can be a breeze. There are two methods that can be used to solve quadratic functions: factoring and using the quadratic equation. Factoring is often the simplest method, and it can be used when the equation can be factored into two linear factors. For example, the equation x2+5x+6 can be rewritten as (x+3)(x+2). To solve the equation, set each factor equal to zero and solve for x. In this case, you would get x=-3 and x=-2. The quadratic equation can be used when factoring is not possible or when you need a more precise answer. The quadratic equation is written as ax²+bx+c=0, and it can be solved by using the formula x=−b±√(b²−4ac)/2a. In this equation, a is the coefficient of x², b is the coefficient of x, and c is the constant term. For example, if you were given the equation 2x²-5x+3=0, you would plug in the values for a, b, and c to get x=(5±√(25-24))/4. This would give you two answers: x=1-½√7 and x=1+½√7. You can use either method to solve quadratic functions; however, factoring is often simpler when it is possible. In mathematics, "solving for x" refers to the process of finding the value of an unknown variable in an equation. In most equations, the variable is represented by the letter "x." Fractions can be used to solve for x in a number of ways. For example, if the equation is 2x + 1 = 7, one can isolated the x term by subtracting 1 from each side and then dividing each side by 2. This would leave x with a value of 3. In some cases, more than one step may be necessary to solve for x. For example, if the equation is 4x/3 + 5 = 11, one would first need to multiply both sides of the equation by 3 in order to cancel out the 4x/3 term. This would give 12x + 15 = 33. From there, one could subtract 15 from each side to find that x = 18/12, or 1.5. As these examples demonstrate, solving for x with fractions is a matter of careful algebraic manipulation. With a little practice, anyone can master this essential math skill. ## We solve all types of math problems My math teacher isn't fantastic, but is a nice lady. This all has probably taught me everything I could ever care to know with the premium upgrade. Good purchase all around to maintain my grade. Sarah Morgan it has been very helpful as long as I plug in enough info. would be better if I could write the type of equation I am trying to solve. i.e., least common multiple, greatest common factor, addition and subtraction of rational numbers etc. which may help the app. understand the form I am writing in. but pretty awesome app. and the breakdown helps me understand the how and why of the problem Mila Davis Formula solver How to solve for y Math word problem solver step by step Fraction word problem solver Infinite series solver
# Leslie and Corey are brother and sister. Leslie is 5 years older than Corey, and the sum of their ages is 51. Find their ages. The fact that Leslie and Corey are brother and sister is irrelevant information. Let us start by assigning variables to the values we are trying to find, Leslie's and Corey's ages: L = Leslie's age C = Corey's age Leslie is 5 years older than Corey. This means that if we have Corey's age and add 5, then we get Leslie's age. We can express this mathematically as: L = C + 5 (Equation 1) We also know that the sum of Leslie's and Corey's ages is 51. This can be expressed mathematically as: L + C = 51 (Equation 2) We now have two equations (Equation 1 and Equation 2) and two unknowns (L and C). We can use substitution to solve for both unknowns. We know that L = C + 5 from Equation 1. Substitute this for L in Equation 2 to get: ` ` (C+5) + C = 51 Rearranging terms, we can also express this as: C + C + 5 = 51 Combining variables, we get: 2C + 5 = 51 Subtract 5 from both sides of the equation to isolate C: 2C + 5 - 5 = 51 - 5 2C = 46 Now divide both sides by 2: 2C/2 = 46/2 C = 23 Now that we have solved for Corey's age, we can use Equation 1 to determine Leslie's: L = C + 5 L = 23 + 5 = 28 Therefore, Leslie is 28 and Corey is 23. A quick double-check confirms that Leslies is 5 years older than Corey (28 - 5 = 23) and that the sum of the ages is 51 (28 + 23 = 51). Answer: Leslie is 28 years old and Corey is 23 years old. Approved by eNotes Editorial Team
# Solving Absolute-Value Equations ## Presentation on theme: "Solving Absolute-Value Equations"— Presentation transcript: Solving Absolute-Value Equations 2-Ext Solving Absolute-Value Equations Holt Algebra 1 Lesson Presentation Objective Solve equations in one variable that contain absolute-value expressions. The absolute-value of a number is that numbers distance from zero on a number line. For example, \|–5\| = 5. 5 units 6 5 4 3 2 1 1 2 3 4 5 6 Both 5 and –5 are a distance of 5 units from 0, so both 5 and –5 have an absolute value of 5. To write this using algebra, you would write \|x\| = 5. This equation asks, “What values of x have an absolute value of 5?” The solutions are 5 and –5. Notice this equation has two solutions. To solve absolute-value equations, perform inverse operations to isolate the absolute-value expression on one side of the equation. Then you must consider two cases. Example 1A: Solving Absolute-Value Equations Solve each equation. Check your answer. \|x\| = 12 Think: What numbers are 12 units from 0? \|x\| = 12 Case 1 x = 12 Case 2 x = –12 Rewrite the equation as two cases. The solutions are 12 and –12. Check \|x\| = 12 \|12\| \|12\| 12 Example 1B: Solving Absolute-Value Equations Since \|x + 7\| is multiplied by 3, divide both sides by 3 to undo the multiplication. Think: What numbers are 8 units from 0? \|x + 7\| = 8 Case 1 x + 7 = 8 Case 2 x + 7 = –8 – 7 –7 – 7 – 7 x = 1 x = –15 Rewrite the equations as two cases. Since 7 is added to x subtract 7 from both sides of each equation. The solutions are 1 and –15. Example 1B Continued 3\|x + 7\| = 24 The solutions are 1 and –15. Check 3\|x + 7\| = 24 3\|8\| 24 3\|1 + 7\| 24 3(8) 24 3\|15 + 7\| 3\|8\| 3(8) Check It Out! Example 1a Solve each equation. Check your answer. \|x\| – 3 = 4 Since 3 is subtracted from \|x\|, add 3 to both sides. \|x\| – 3 = 4 \|x\| = 7 Think: what numbers are 7 units from 0? Case 1 x = 7 Case 2 –x = 7 –1(–x) = –1(7) x = –7 x = 7 Rewrite the case 2 equation by multiplying by –1 to change the minus x to a positive.. The solutions are 7 and –7. Check It Out! Example 1a Continued Solve the equation. Check your answer. \|x\|  3 = 4 The solutions are 7 and 7. Check \|x\| 3 = 4 7  \|7\|  \| 7\|  7  Check It Out! Example 1b Solve the equation. Check your answer. \|x  2\| = 8 Think: what numbers are 8 units from 0? \|x  2\| = 8 Case 1 x  2 = 8 x = 10 x = 6 Case 2 x  2 = 8 Rewrite the equations as two cases. Since 2 is subtracted from x add 2 to both sides of each equation. The solutions are 10 and 6. Check It Out! Example 1b Continued Solve the equation. Check your answer. \|x  2\| = 8 The solutions are 10 and 6. Check \|x 2\| = 8 10 2\| 8 \|10 2\| 8 \| 6 + (2)\| 8 Not all absolute-value equations have two solutions Not all absolute-value equations have two solutions. If the absolute-value expression equals 0, there is one solution. If an equation states that an absolute-value is negative, there are no solutions. Example 2A: Special Cases of Absolute-Value Equations Solve the equation. Check your answer. 8 = \|x + 2\|  8 Since 8 is subtracted from \|x + 2\|, add 8 to both sides to undo the subtraction. 8 = \|x + 2\|  8 0 = \|x +2\| There is only one case. Since 2 is added to x, subtract 2 from both sides to undo the addition. 0 = x + 2  2 2 = x Example 2A Continued Solve the equation. Check your answer. 8 = \|x +2\|  8 Solution is x = 2 Check 8 =\|x + 2\|  8 To check your solution, substitute 2 for x in your original equation. 8 8 8 \|2 + 2\|  8 8 \|0\|  8   8 Example 2B: Special Cases of Absolute-Value Equations Solve the equation. Check your answer. 3 + \|x + 4\| = 0 Since 3 is added to \|x + 4\|, subtract 3 from both sides to undo the addition. 3 + \|x + 4\| = 0  3 \|x + 4\| = 3 Absolute values cannot be negative. This equation has no solution. Remember! Absolute value must be nonnegative because it represents distance. Check It Out! Example 2a Solve the equation. Check your answer. 2  \|2x  5\| = 7 Since 2 is added to \|2x  5\|, subtract 2 from both sides to undo the addition. 2  \|2x  5\| = 7  2  \|2x  5\| = 5 Since \|2x  5\| is multiplied by a negative 1, divide both sides by negative 1. \|2x  5\| = 5 Absolute values cannot be negative. This equation has no solution. Check It Out! Example 2b Solve the equation. Check your answer. 6 + \|x  4\| = 6 Since 6 is subtracted from \|x  4\|, add 6 to both sides to undo the subtraction. 6 + \|x  4\| = 6 \|x  4\| = 0 x  4 = 0 x = 4 There is only one case. Since 4 is subtracted from x, add 4 to both sides to undo the addition. Check It Out! Example 2b Continued Solve the equation. Check your answer. 6 + \|x  4\| = 6 The solution is x = 4. 6 + \|x  4\| = 6 To check your solution, substitute 4 for x in your original equation. 6 + \|4  4\| 6 6 +\|0\| 6  6 6 6
# Maharashtra Board 10th Class Maths Part 2 Practice Set 6.2 Solutions Chapter 6 Trigonometry ## Practice Set 6.2 Geometry 10th Std Maths Part 2 Answers Chapter 6 Trigonometry Question 1. A person is standing at a distance of 80 m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church. Solution: Let AB represent the height of the church and point C represent the position of the person. BC = 80 m Angle of elevation = ∠ACB = 45° In right angled ∆ABC, tan 45° = ABBC … [By definition] ∴ 1 = AB80 ∴ AB = 80m ∴ The height of the church is 80 m. Question 2. From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. ( 3–√ = 1.73) Solution: Let AB represent the height of lighthouse and point C represent the position of the ship. AB = 90 m Angle of depression = ∠PAC = 60° Now, ray AP \|\| seg BC ∴ ∠ACB = ∠PAC … [Alternate angles] ∴ ∠ACB = 60° In right angled ∆ABC, tan 60° = ABBC … [By definition] ∴ The ship is 51.90 m away from the lighthouse. Question 3. Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building? Solution: Let AB and CD represent the heights of the two buildings, and BD represent the width of the road. Draw seg AM ⊥ seg CD. Angle of elevation = ∠CAM = 60° AB = 10 m BD= 12 m In ꠸ABDM, ∠B = ∠D = 90° ∠M = 90° … [seg AM ⊥ seg CD] ∴ ∠A = 90° … [Remaining angle of ꠸ABDM] ∴ ꠸ABDM is a rectangle …. [Each angle is 90°] ∴ AM = BD = 12 m opposite sides DM = AB = 10 m of a rectangle In right angled ∆AMC, tan 60° = CMAM …[By definition] ∴ 3–√ = CM12 ∴ CM = 123–√ m Now, CD = DM + CM … [C – M – D] ∴ CD = (10 + 123–√)m = 10 + 12 × 1.73 = 10 + 20.76 = 30.76 ∴ The height of the second building is 30.76 m. Question 4. Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops is 22 metre. Find the angle made by the wire with the horizontal. Solution: Let AB and CD represent the heights of two poles, and AC represent the length of the wire. Draw seg AM ⊥ seg CD. Angle of elevation = ∠CAM = θ AB = 7 m CD = 18 m AC = 22 m In ꠸ABDM, ∠B = ∠D = 90° ∠M = 90° …[seg AM ⊥ seg CD] ∴ ∠A = 90° … [Remaining angle of ꠸ABDM] ∴ □ABDM is a rectangle. … [Each angle is 90°] ∴ DM = AB = 7 m … [Opposite sides of a rectangle] Now, CD = CM + DM … [C – M – D] ∴ 18 = CM + 7 ∴ CM = 18 – 7 = 11 m In right angled ∆AMC, sin θ = CMAC …..[By definition] ∴ sin θ = 1122 = 12 But, sin 30° = 12 ∴ θ = 30° ∴ The angle made by the wire with the horizontal is 30°. Question 5. A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree. Solution: Let AB represent the height of the tree. Suppose the tree broke at point C and its top touches the ground at D. AC is the broken part of the tree which takes position CD such that ∠CDB = 60° ∴ AC = CD …(i) BD = 20 m In right angled ∆CBD, tan 60° = BCBD … [By definition] ∴ 3–√ = BC20 ∴ BC = 203–√ m Also, cos 60° = BCCD … [By definition] ∴ 12 = 20CD ∴ CD = 20 × 2 = 40 m ∴ AC = 40 m …[From(i)] Now, AB = AC + BC ….[A – C – B] = 40 + 203–√ = 40 + 20 × 1.73 = 40 + 34.6 = 74.6 ∴ The height of the tree is 74.6 m. Question 6. A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60° with the ground. Assuming that the string is straight, find the length of the string. (3–√ = 1.73) Solution: Let AB represent the height at which kite is flying and point C represent the point where the string is tied at the ground. ∠ACB is the angle made by the string with the ground. ∠ACB = 60° AB = 60 m In right angled ∆ABC, sin 60° = ABAC … [By definition] ∴ AC = 40 3–√ = 40 × 1.73 = 69.20 m ∴ The length of the string is 69.20 m.
42 Q: # The difference between compound interest and simple interest on a sum for two years at 8% per annum, where the interest is compounded annually is Rs.16. if the interest were compounded half yearly , the difference in two interests would be nearly A) Rs.24.64 B) Rs.21.85 C) Rs.16 D) Rs.16.80 Explanation: For 1st year S.I =C.I. Thus, Rs.16 is the S.I. on S.I. for 1 year, which at 8% is thus Rs.200 i.e S.I on the principal for 1 year is Rs.200 Principle = $Rs.10020081$ = Rs.2500 Amount for 2 years, compounded half-yearly $Rs.25001+41004=Rs.2924.4$ C.I = Rs.424.64 Also, $S.I=Rs.25008*2100=Rs.400$ Hence, [(C.I) - (S.I)] = Rs. (424.64 - 400) = Rs.24.64 Q: A certain sum is invested for 2 years in scheme M at 20% p.a. compound interest (compounded annually), Same sum is also invested for the same period in scheme N at k% p.a. simple interest. The interest earned from scheme M is twice of that earned from scheme N. What is the value of k? A) 7 B) 11 C) 9 D) 13 Explanation: Interest earned in scheme M = Interest earned in scheme N = Now, from the given data, k = 11 0 62 Q: A sum is equally invested in two different schemes on CI at the rate of 15% and 20% for two years. If interest gained from the sum invested at 20% is Rs. 528.75 more than the sum invested at 15%, find the total sum? A) Rs. 7000 B) Rs. 4500 C) Rs. 9000 D) Rs. 8200 Explanation: Let Rs. K invested in each scheme Two years C.I on 20% = 20 + 20 + 20x20/100 = 44% Two years C.I on 15% = 15 + 15 + 15x15/100 = 32.25% Now, (P x 44/100) - (P x 32.25/100) = 528.75 => 11.75 P = 52875 => P = Rs. 4500 Hence, total invested money = P + P = 4500 + 4500 = Rs. 9000. 7 1120 Q: What is the interest rate per annum, if a sum of money invested at compound interest amount to Rs. 2400 in 3 years and in 4 years to Rs. 2,520? A) 3.5% B) 4% C) 5% D) 6.5% Explanation: Let 'R%' be the rate of interest From the given data, Hence, the rate of interest R = 5% per annum. 4 815 Q: A sum of Rs. 8,000 is deposited for 3 years at 5% per annum compound interest (compounded annually). The difference of interest for 3 years and 2 years will be A) Rs. 387 B) Rs. 441 C) Rs. 469 D) Rs. 503 Explanation: Given principal amount = Rs. 8000 Time = 3yrs Rate = 5% C.I for 3 yrs = Now, C.I for 2 yrs = Hence, the required difference in C.I is 1261 - 820 = Rs. 441 4 813 Q: Simple interest on a certain sum at 7% per annum for 4 years is Rs. 2415. What will be the compound interest on the same principal at 4% per annum in two years? A) Rs. 704 B) Rs. 854 C) Rs. 893 D) Rs. 914 Explanation: We know that, From given data, P = Rs. 8625 Now, C.I = 3 1143 Q: Find the compound interest on Rs. 6,500 for 4 years if the rate of interest is 10% p.a. for the first 2 years and 20% per annum for the next 2 years? A) Rs. 3845 B) Rs. 4826 C) Rs. 5142 D) Rs. 4415 Explanation: We know the formula for calculating The compound interest where P = amount, r = rate of interest, n = time Here P = 5000, r1 = 10, r2 = 20 Then C = Rs. 4826. 10 1074 Q: What is the difference between the compound interests on Rs. 5000 for 11⁄2 years at 4% per annum compounded yearly and half-yearly? A) Rs. 1.80 B) Rs. 2.04 C) Rs. 3.18 D) Rs. 4.15 Explanation: Compound Interest for 1 12 years when interest is compounded yearly = Rs.(5304 - 5000) Amount after 112 years when interest is compounded half-yearly Compound Interest for 1 12 years when interest is compounded half-yearly = Rs.(5306.04 - 5000) Difference in the compound interests = (5306.04 - 5000) - (5304 - 5000)= 5306.04 - 5304 = Rs. 2.04 6 1129 Q: The difference between simple interest and compound interest of a certain sum of money at 20% per annum for 2 years is Rs. 56. Then the sum is : A) Rs. 3680 B) Rs. 2650 C) Rs. 1400 D) Rs. 1170
Question Video: Adding Two Whole Numbers Mathematics 2,897 + 5,497 = ＿. 04:05 Video Transcript 2,897 plus 5,497 equals what. In this question, we need to find the total of two four-digit numbers. Now, the way that this calculation is being written, horizontally or across the page, might make us think that that’s how we need to try to find the answer. We’ve just got to look at these two four-digit numbers as they are and try to add them together. But never forget just because we’re shown a calculation like this doesn’t mean we can’t rewrite it in a different way. And when we’re adding two four-digit numbers like this, a really helpful way to rewrite the calculation is vertically, in other words, writing both numbers so that the thousands, the hundreds, the tens, and the ones digits are on top of each other in separate columns. Now we can add each pair of digits separately. And we always start by adding the ones. Do you know why we always start by adding the ones by the way? Before we begin, look at the two digits in the thousands column and make a prediction. How many thousands do you think are going to be in the answer? We’ve got 2,000 in our first number and 5,000 in our second number. Well, we know two plus five equals seven, so we might predict that our answer is going to contain 7,000. Well, we’d come back to that prediction in a moment, but to begin with, let’s find the total of our ones. Both numbers contain seven ones. Seven plus seven equals 14 ones. Now we know we can’t show 14 ones in the ones place because we can only show one digit. So we need to take 10 of our 14 ones and exchange them for one 10. We’ll write the little one underneath like this. So we’re still showing 14 ones, but we’re writing it as one 10 and four ones. If we look at our tens digits, we can see that they’re both the same too. Both numbers have a nine in the tens place. Nine plus nine is 18, so nine 10s plus another nine 10s is 18 10s. We mustn’t forget the 10 that we got when we exchanged either, so that’s 19 10s altogether. Again, this is a two-digit number, so we’re going to need to exchange. We can take 10 of our 19 10s and exchange them for one 100. So we can express 19 10s as one 100 and nine 10s. In our hundreds column, we have eight 100s plus another four 100s. This gives us a total of 12 100s, but we can’t forget the one 100 that we’ve exchanged underneath. So that makes 13 100s. We’re going to need to exchange again. A lot of exchanging in this calculation, isn’t there? We need to take 10 of our 13 100s and exchange them for one 1,000 because one 1,000 and three 100s is the same as 13 100s. Finally, let’s add our two 1,000s and five 1,000s that we talked about at the start. 2,000 plus 5,000 equals 7,000, but we’ve got one extra 1,000 that we’ve exchanged. Instead of seven 1,000s, our answer has eight 1,000s. Good job we didn’t start by adding the thousands, isn’t it? This is why we always start on the right and work to the left. It’s because if we have to regroup in a column and exchange, the next column to the left is going to be affected. We found the total of these two four-digit numbers by using the standard written method. 2,897 plus 5,497 equals 8,394.
# 2nd Grade Fractions Lesson Plan Equivalance ## Discussion/Introduction In first grade our students were briefly introduced to fractions—or, more specifically, halves and quarters– as equal divisions done on rectangles and circles. They identified halves and quarters and divided their own samples into two or four equal parts. The emphasis was on neat, symmetrical divisions that made it easy to check equivalence. In second grade, though, we get to work with the concept of strange looking equivalences— portions which are equal but which are different shapes. This concept will probably be counterintuitive to many of your students, but it is an important concept nonetheless: a crucial part of their understanding of space, area, and sameness. This lesson plan gives a brief review of halves and quarters, introduces the new kid on the block—thirds—and goes on to discuss strange looking equivalences and what it means for portions or fractions to be the same. ## Objective That students would gain familiarity in partitioning circles and rectangles in two, three, and four equal shares, and would understand that equal shares of identical wholes need not have the same shape. (Common Core 2.G3) ## Supplies • One orange • Construction paper, pre-cut into a quantity of identical rectangles and circles: 3 circles and 5 rectangles, preferably in an assortment of different colors • Markers • Scissors • Scotch Tape • Beads or other small math manipulatives (12 per student) ## Methodology/Procedure Start class by reviewing fractions, as learned in first grade. Tell your students that you like eating orange every morning, but you only have one orange for today and tomorrow; ask them how much orange you should eat today (half—one of two equal parts). Ask them how you should divide it in half (across the middle), and ask them divide their first circle by drawing a line. Then ask them how much orange you can eat today if you need to make it last for four days. (one fourth – one of four equal parts). Ask them to draw this on their second construction paper circle cutout. Take the orange out of your desk and tell them that actually, you’ll be able to buy a new orange to eat on the fourth morning. It’s just three days that this orange has to last. Ask how much orange you can eat today. Divide the orange into three equal portions, and tell the class that they are called thirds. Tell them that when you are making thirds out of a circle there’s no middle line to divide on. Show them, or draw on the backboard, a picture of an orange cross section cut into thirds; demonstrate how you can start by making a line through the middle, to the center point, and then continuing it as a (wide mouth) Y. Ask your students to draw lines and divide their last construction paper circles into thirds. Have them cut out the segments and lay them over each other to check their own work. Now tell them to take out their first rectangle, and tell them you’d like this to be divided into thirds. Ask them how they would go about it. After they have had plenty of time to experiment call up any students who have been especially successful to show their techniques. If they are all still fumbling, show them your rectangle, creased to show thirds. Give them a chance to imitate this with their own paper, and have the first successful student show them how folding the two sides over the middle segment and creasing the fold will give three equal sections— three thirds. Ask which is larger, a half or a third. And which is smaller, a fourth or a third. Give each student a pile of twelve beads, tiles, or other math manipulatives, and ask them to divide it into thirds. Let them try to figure this out by themselves before you give them any input. You can continue this exercise to halves and quarters. When your students feel comfortable making divisions in halves, thirds, and quarters, it’s time to go on to strange looking equivalences. If your students had forgotten some of their first grade work, you may want to simply spend the rest of the class period playing with fractions and save this second half of the lesson till your next class. If your students had no difficulties with your preliminary exercises, though, you can go straight on. Ask the students to divide their second rectangle in halves. Choose two students who have halves that are proportioned differently, and have them show their halves to the class. Ask which is larger. If there is the slightest doubt in anyone’s mind, point out how, though one is wider, the other is taller. Demonstrate equivalence by cutting and taping the rectangles till they become the same size and shape. Only after you have demonstrated it tell them that half of an object is always the same size as any other half of the same object, no matter what shape it takes. You don’t want them to accept this as a rule for how halves should behave, but simply as a demonstrated fact. Provide each student with three more identical rectangles, pre-cut out of three different colors of construction paper. Ask the students to draw lines to partition each rectangle in fourths, and tell them you’d like the partitions to be different in each rectangle. Give the students the attached worksheet and ask them to mark the areas that are equivalent. ## Common Core Standards In 2.G.3, 2nd grade geometry item 3, the Common Core State Standards for Mathematics reads: 2.G.3 Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape. ## Web Resources/Further Exploration You may also want to point your students to some of our other fractions resources and links: # M&M Fun: A Bar Chart Lesson for 2nd Graders Sample Bar Graph ## Discussion/Introduction You can download this lesson plan as a pdf After weeks and weeks of arithmetic and mental math that tax my students’ minds to the utmost, I always enjoy getting to the graph section of our curriculum. It’s a breather, almost, and the change of pace is very welcome. At the same time, our unit on graphing is a very important one—as students enjoy making colorful charts and graphs they learn how to condense real-world data into a mathematical format that is easy to understand, concise, and easy to analyze. A 2nd grade bar chart lesson can be just complex enough to be exciting, but there’s no need to go into the complicated situations that will leave your students scratching their heads in confusion. The free chart-maker at Meta-Chart (http://www.meta-chart.com/) is a wonderfully easy way to make professional, streamlined charts and graphs; all you do is plug in your data and—eureka!—out comes the graph. ## Objective Students will learn how to analyze data on a simple single-unit bar chart with four categories. They will learn to solve simple put-together, take-apart, and compare problems using information presented in a bar graph by making use of a free online bar graph maker. ## Supplies • A mug with around 20-30 colored M&Ms, and ~8 M&Ms for each child in the bag • 1 sheet of graph paper for each child • Markers • Graph printouts from http://www.meta-chart.com/bar (or, if you have projector capabilities, meta-chart graphs prepared and saved on your laptop). ## Methodology/Procedure When students are investigators what they learn becomes part of them. Bar graphs lend themselves well to a discovery way of teaching because they are, in a way, a special code that can be discovered. Your students will have had some exposure to graphs in grade 1; some will have even done very basic work with bar graphs. They have all the tools they need to figure out what the bar graph is telling them if you give them the time and encouragement to figure it out by themselves. Teacher: Look, I have a mug here with M&Ms in it. You can’t look inside right now, but here is a graph that tells you what the colors of the M&Ms are. [Give students a chance to look at it and think about what it is saying.] Teacher: Can anyone tell me which color I have the most of? Student: Red! Teacher: Yes! You are correct. How did you know? Student: The bar labeled red is the longest. Teacher: You’re right. Can you tell me how many red M&Ms there are in my cup? If a student gives the right answer, applaud him and ask for an explanation why. If no-one knows, ask some more leading questions. Teacher: How long is the red M&M bar? Is there a number which tells how long it is? What do you think that number might be telling us? Once the students have figured out how many red M&Ms there are, reinforce that interpretative ability by asking questions about the other colored M&Ms in the mug. Then go on to problems related comparing. Teacher: How many more red M&Ms do I have than blue? Lead the students to discover that they can find the difference without doing subtraction by simply noticing how much further the red line sticks out. Ask questions comparing each of the other lines. When your students are comfortable comparing, go on to simple put-together and take-apart problems using the data on the graph. Teacher: If I don’t like red or green M&Ms and decide to throw those ones away, how many will I throw away? Given the opportunity to discuss and brainstorm, your students should have no trouble solving this. If their thinking was anything less than automatic, follow this up with a similar question. If your dog only can eat yellow or brown M&Ms, how many will he have? Then ask a take-apart question: If I actually like green M&Ms and want to have as many green M&Ms as red, how many more green M&Ms do I need to put in the mug? Once the students are comfortable doing a variety of problems, have them each draw two axis on their own graph paper, and label the vertical axis with numbers 1-6. Distribute around eight M&Ms to each child, and have them draw a bar graph. They can color each bar the bars the same color as the M&Ms for easy labeling. Post the graphs at the front of the room, and discuss which child has the most red, the most green, the most yellow, and so on. #### Evaluation You’ll be able to tell how well the students understood the concept of a bar graph by how much help and hand-holding they need when it comes to drawing their own graphs. Quick comparing and put-together questions on the smaller numbers of their own M&M collections and graphs will give you another way of testing their ability to understand the logic behind bar graphs. ###### A Single Unit Bar Chart Lesson in Your Classroom If you use my lesson plan I’d love to hear from you regarding your experiences—did you have fun with your students? What is your preferred way of teaching a 2nd grade bar chart lesson? Please comment! ## Common Core Standards The Common Core recognizes the importance of learning how to use graphs for data representation and manipulation in the early grades. For second grade, under measurement and data [ 2.MD.10 ], the Common Core State Standards for Mathematics reads ‘Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put- together, take-apart, and compare problems using information presented in a bar graph.’ ## Web Resources/Further Exploration The chart maker at http://www.meta-chart.com is a convenient, easy to use, free way of preparing charts or graphs for any of your classes!
Home \| \| Maths 10th Std \| Problems involving Angle of Elevation and Depression Problems involving Angle of Elevation and Depression In this section, we try to solve problems when Angles of elevation and depression are given. Problems involving Angle of Elevation and Depression Let us consider the following situation. A man standing at a top of lighthouse located in a beach watch on aeroplane flying above the sea. At the same instant he watch a ship sailing in the sea. The angle with which he watch the plane correspond to angle of elevation and the angle of watching the ship corresponding to angle of depression. This is one example were one oberseves both angle of elevation and angle of depression. In the Fig.6.26, x° is the angle of elevation and y° is the angle of depression. In this section, we try to solve problems when Angles of elevation and depression are given. Example 6.31 From the top of a 12 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 30°. Determine the height of the tower. Solution As shown in Fig.6.27, OA is the building, O is the point of observation on the top of the building OA. Then, OA = 12 m. PP’ is the cable tower with P as the top and P ' as the bottom. Then the angle of elevation of P, MOP = 60° And the angle of depression of P’ , MOP = 30°. Suppose, height of the cable tower PP ' = h metres. Through O, draw OMPP ' MP = PP MP = h OA = h −12 Hence, the required height of the cable tower is 48 m. Example 6.32 A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point ‘A’ on the ground is 60° and the angle of depression to the point ‘A’ from the top of the tower is 45°. Find the height of the tower. (√3=1.732) Solution Let BC be the height of the tower and CD be the height of the pole. Let ‘A’ be the point of observation. Let BC = x and AB = y. From the diagram, BAD = 60° and XCA = 45° = BAC Hence, height of the tower is 6.83 m. Example 6.33 From a window (h metres high above the ground) of a house in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are θ1 and θ2 respectively. Show that the height of the opposite house is . Solution Let W be the point on the window where the angles of elevation and depression are measured. Let PQ be the house on the opposite side. Then WA is the width of the street. Height of the window = h metres = AQ (WR = AQ) Let PA = x metres. Therefore, height of the opposite house = PA+AQ = x + h Hence Proved. Tags : Solved Example Problems \| Trigonometry \| Mathematics , 10th Mathematics : UNIT 6 : Trigonometry Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail 10th Mathematics : UNIT 6 : Trigonometry : Problems involving Angle of Elevation and Depression \| Solved Example Problems \| Trigonometry \| Mathematics
Linear Transformations Linear Transformations Definition: A transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ (or operator if $T: \mathbb{R}^n \to \mathbb{R}^n$) is defined to be linear if the image $(w_1, w_2, ..., w_m)$ is comprised of only linear equations for every mapping $(x_1, x_2, ..., x_n)$, that is $T(x_1, x_2, ..., x_n) = (w_1, w_2, ..., w_m)$. For any vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ and any scalar $k$ a transformation is linear if $T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})$ and $T(k\vec{u}) = kT(\vec{u})$. Let's first look at an example of a linear transformation. Consider the following linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^3$ defined by the following equations: (1) \begin{align} w_1 = x_1 + 3x_2 \\ w_2 = 2x_1 - x_2 \\ w_3 = -x_1 + 4x_2 \end{align} We note that the equations forming the image, that is $w_1$, $w_2$, and $w_3$ are all linear, so this transformation is also considered linear and that $T(x_1, x_2) = (x_1 + 3x_2, 2x_1 - x_2, -x_1 + 4x_2)$. For example, if we take the vector $\vec{x} = (1, 2)$ and apply our linear transformation, we obtain a resultant vector $\vec{w} = (7, 0, 7)$, and we say that $(7, 0, 7)$ is the image of $(1, 2)$ under the linear transformation $T$. In general, a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is generally defined by the following equations: (2) In matrix notation we can represent this transformation as $w = Ax$. $A$ is called the standard matrix for the linear transformation $T$, though sometimes we use the notation $[ T ]$ instead. Either way, the standard matrix is created from the coefficients from the system of linear equations defining the image of $T$. (3) \begin{align} \quad \begin{bmatrix} w_1\\ w_2\\ \vdots\\ w_m \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots\\ x_n \end{bmatrix} \end{align} This linear transformation $T$ is defined by the standard matrix $A$, so we say that $T$ is multiplication by $A$ and often denote it with the notation $T_A (x) = Ax$. Either way, these transformations will geometrically transform some vector or point in $\mathbb{R}^n$ to some other vector or point in $\mathbb{R}^m$. Properties of Linear Transformations We've already stated the following two properties in the definition of a linear transformation, but now we will prove their existence. Property 1: If $T: \mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation, then for any vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ it follows that $T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})$. • Proof: Suppose that $T$ is a linear transformation and is multiplication by $A$. Thus it follows that: (4) \begin{align} T(\vec{u} + \vec{v}) = A(u + v) \\ T(\vec{u} + \vec{v}) = Au + Av \\ T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v}) \\ \blacksquare \end{align} Property 2: If $T: \mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation, then for any vector $\vec{u} \in \mathbb{R}^n$ and any scalar $k$ it follows that $T(k\vec{u}) + kT(\vec{u})$. • Proof: Suppose that $T$ is a linear transformation and is multiplication by $A$. Thus it follows that: (5) \begin{align} T(k\vec{u}) = A(ku) \\ T(k\vec{u}) = k(Au) \\ T(k\vec{u}) = kT(\vec{u}) \\ \blacksquare \end{align}
Succeed with maths – Part 1 Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available. Free course # 3 Introduction to dividing fractions Dividing fractions is a little more difficult, so you’ll start by considering simple examples. You’ll then attempt to translate that to a more complex question by applying a similar strategy. Suppose that you have three pizzas, and you divide each of them in half (see Figure 5). Figure 5 Three circles, each divided into half How many halves do you have? You can see that there are six halves in three pizzas. Thus this shows how many halves there are in three. This is the same as dividing 3 by a . When you calculate , you can think of this as ‘How many halves are there in three?’ Here are few similar examples to work out how to divide by fractions. There are three pizzas again, but this time the question is 'How many thirds can they be cut up into?' (Figure 6). This is the same as asking what is . Figure 6 Three circles, divided into thirds From this, you can see that there are 9 thirds in 3. So . Now divide these three pizzas into quarters, which gives the answer to . Each pizza will be cut into four parts to give quarters and there are three pizzas, so the total number of slices will be . You also know that . Looking at these three calculations, can you see a way of arriving at the answers without drawing out a picture to help? Look at the whole numbers and the denominators. You may have noticed that if you multiply these together, then you arrive at the answer. So it seems that , and . So put into words, you can say that to divide by a fraction, you have to swap the numerator and the denominator, and then multiply. Or, in mathematical language (so that you can communicate this clearly and concisely to others): when the numerator and denominator change places, the result is called the reciprocal of the original fraction. Thus, dividing by a fraction is the same as multiplying by its reciprocal. In the next section you’ll look at this in practice.
# Tessellating Regular Polygons If we imagine a pearl bracelet projected onto a 2D plane, it looks like a collection of circles arranged, touching, in a larger circle. If, instead of using circles, we used n-sided regular polygons, is it possible to fit these together perfectly (edge-edge) to make a ring, and if so, what is the relationship between the number of sides of each polygon, and the number of polygons that fit into a circle? Yes, it is possible, for example, here is a bracelet constructed using eighteen regular nonagons (9 sided shapes). Each polygon is touching edge-edge with its adjacent two neighbors. ## Perfect Tessellations Before we answer the ring question, let’s take a step back and look at perfect tessellations (Euclidean tilings). Below are representations of the first ten regular polygons; starting with the triangle with three equal sides and angles (and three vertices) through to the Dodecagon with twelve equal sides and angles. Which of these tessellates? Let’s start with the triangle. We can place one on the plane, then add another sharing an edge and common vertex, then another sharing the same vertex, tiling as we go around. After we’ve placed our sixth triangle, we’ve completed an orbit. The last triangle placed fits in perfectly. If these were tiles on the floor they would fit together with no gaps, and be flat on the floor (no bumps). Triangles tesselate on a flat plane. If you know about triangles, you know that all their internal angles add up to 180° As all the angles are equal in regular polygons we know that, for an equilateral triangle, each internal angle is 60°. We can see that there are six triangles that meet at the common vertex and 6 × 60° = 360°, which we know is a complete circle. It’s no surprise that triangles tesselate on a plane. Similarly, we can do the same with squares. Squares have an internal angle of 90° so we can get four of them (4 × 90° = 360°) around in a circle. A pentagon has five vertices. It does not tessellate. Let’s see why. First, we need to calculate the internal angle of a pentagon. A pentagon can be divided into three triangles. We know that the three angles of a triangle add up to 180°, so the sum of all the angles in the pentagon is 3 × 180° = 540°, and since there are five equal angles in a regular pentagon, each internal angle is 540° ÷ 5 = 108° Three lots of 108° is 324°, and so is short of a complete circle. Four lots of 108° is 432° which is over a complete circle. Three regular pentagons is too small, four regular pentagons too large. There is no Goldilocks (integer) number of regular pentagons to make a perfect tessellation. For hexagons, these tesselate. The internal angle for a hexagon is 120°, and it’s easy to compute that three of these fit together in a circle. Heptagon Octagon Nonagon Decagon ## Are there any more solutions? There’s a formula for computing the internal angle of a regular polygon, and it’s really simple. Just like pentagon example above, a regular n-sided polygon can be broken up into triangles. For an n-sided shape, it can be broken up into (n-2) triangles. The total of the internal angles is thus (n-2)×180° and, as each angle is equal, the internal angle is (n-2)×180°/n. Here is a table showing a few examples: SidesTotal AngleInternal Angle 3180°60.000° 4360°90.000° 5540°108.000° 6720°120.000° 7900°128.571° 81,080°135.000° 91,260°140.000° 101,440°144.000° 111,620°147.273° 121,800°150.000° 203,240°162.000° 10017,640°176.400° 1,000179,640°179.640° 10,0001,799,640°179.964° For a shape to tessllate, the internal angle has to be a factor of 360. Only the triangle, square, and hexagon fit this criterion. ## Proof Here's a more formal proof. We know, from above, the internal angle of a regular n-sided polygon is (n-2)×180°/n. Let's define the number of times this shape tessellates around as T times. The product of these two needs to be 360°. Simplifying, then adding 4 to each side, and factoring. The only integer solutions to this are {3,6}, {4,4}, and {6,3} corresponding to the triangle, square, and hexagon. For simple (perfect) tiling, these are the only three solutions. ## Skipping a vertex There's another class of tessellations we can make if we skip a vertex. Rather than connected the edges directly adjacent, we skip one. Here we can see, for example, that heptagons still do not tessllate around, but now octagons do. Are there other regular polygons that now tessellate? This tessellation method leaves a hole (which is also a regular polygon) in the middle, and starts us down the path of making our bracelet. The internal angle of the center hole can be calculated by subtracting two lots of the internal angle of the polygons from a full circle. For this to tessellate into a ring, the whole in the center needs to be a regular polygon, and so there needs to be a value k such that the internal angle is 180(k-2)/k. Equating these two requirements: If we look for values of n that result in integer solutions for k we can find perfect tessellations. To avoid negative and invalid solutions 5 ≤ n ≤ 12. nk 510 66 74.667 84 93.600 103.333 113.143 123 We can see from this that the pentagon, hexagon, octagon, and dodecagon tesselate with one skipped vertex. The corresponding holes are shaped decagon, hexagon, square, and triangle. ## Skipping two vertices We don't have to stop here. We can skip two vertices. Rather than making a regular polygon shaped center hole, we now have a multipointed star shape in the middle. We can, however, apply the same technique to determine if we can get a perfect tessallation. A little bit of math gives the relationship between k and n when we skip two verices to be: nk 714 88 96 105 114.400 124 133.714 143.500 153.333 163.200 173.091 183 ## Skipping three vertices I think you can see where we are going with this … Skipping to the third vertex, the hole in the center is looking more like a circular cog wheel, and the relationship is: nk 918 1010 117.333 126 135.200 144.667 154.286 164 173.778 183.600 193.455 203.333 213.231 223.143 233.067 242 Here's one eample: ## Generic If we skip m vetices when attempting to make a perfect ring, the solutions that work are the integer solutions to the formula k=2n/(n-2(m+1))
Rs Aggarwal 2019 2020 Solutions for Class 8 Maths Chapter 5 Playing With Numbers are provided here with simple step-by-step explanations. These solutions for Playing With Numbers are extremely popular among Class 8 students for Maths Playing With Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 8 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate. #### Question 1: The units digit of a two-digit number is 3 and seven times the sum of the digits is the number itself. Find the number. Let the tens place digit be x. The units place digit is 3. ∴ Number = (10x + 3) ... (1) Given: 7( x + 3) = (10 x + 3) 7 x + 21 = 10 x + 3 ∴ 10 x - 7x = 21 - 3 ⇒ 3 x = 18 or x = 6 Using x = 6 in equation (1): The number is 63. #### Question 2: In a two-digit number, the digit at the units place is double the digit in the tens place. The number exceeds the sum of its digits by 18. Find the number. Let the tens digit be x. The digit in the units place is 2x. Number = 10x + 2x Given: (x + 2x) + 18 = (10x + 2x) ∴ 3x + 18 = 12x 12x - 3x = 18 9x =18 x = $\frac{18}{2}$ = 2 The digit in the tens place is 2. The digit in the units place is twice the digit in the tens place. The digit in the units place is 4. Therefore, the number is 24. #### Question 3: A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number. Let the tens place digit be a and the units place digit be b. Then, number is (10a + b). According to the question: 4(a + b) + 3 = (10 a + b) 4a + 4b + 3 = 10a + b 6a - 3b = 3 3(2a - b) = 3 2a - b =1 ... (1) Given: If 18 is added to the number, its digits are reversed. The reverse of the number is (10b + a). ∴ (10a + b) + 18 = 10b + a 10a - a + b -10b = -18 9a - 9b = -18 9(a - b) = -18 a - b = -2 ... (2) Subtracting equation (2) from equation (1): 2a - b = 1 a - b = -2 - + + a = 3 Using a = 3 in equation (1): 2(3) - b = 1 6 - b = 1 ∴ b = 5 Number = 10a+b = 10 $×$ 3 + 5 = 35 #### Question 4: The sum of the digits of a two-digit number is 15. The number obtained by interchanging its digits exceeds the given number by 9. Find the original number. Let the tens place digit be a and the units place digit be b. Then, the number is (10a + b). Given: a + b = 15 ... (1) When the digits are interchanged the number will be (10 b + a). Given: 10a + b + 9 = 10 b + a ∴ 10a - a + b - 10b = -9 9a - 9b = -9 a - b = -1 ... (2) a + b = 15 a - b = -1 2a = 14 ∴ a = 7 Using a = 7 in equation (2): 7 - b = -1 ∴ b = 8 Original number = 10a+b = 10 $×$ 7 + 8 = 78 #### Question 5: The difference between a 2-digit number and the number obtained by interchanging its digits is 63. What is the difference between the digits of the number? Let the tens place digit be 'x' and the units place digit be 'y'. ∴ Number = (10x + y) Number obtained by interchanging the digits = (10y + x) Given: (10x + y) - (10y + x) = 63 ∴ 10x - x + y - 10 y = 63 9x - 9y = 63 9(x - y) = 63 x - y = 7 Therefore, the difference between the digits of the number is 7. #### Question 6: In a 3-digit number, the tens digit is thrice the units digit and the hundreds digit is four times the units digit. Also, the sum of its digits is 16. Find the number. Let the units place digit be x. Then, the tens place digit will be 3x and the hundreds place digit will be 4x. Given: 4x + 3x + x = 16 or 8x = 16 or x =2 Units place digit = 2 Tens place digit = 3 $×$ 2 = 6 Hundreds place digit = 4 $×$ 2 = 8 Therefore, the number is 862. #### Question 1: Test the divisibility of each of the following numers by 2: (i) 94 (ii) 570 (iii) 285 (iv) 2398 (v) 79532 (vi) 13576 (vii) 46821 (viii) 84663 (ix) 66669 A given number is divisible by 2 only when its unit digit is 0, 2, 4, 6 or 8. (i) 94 The number 94 has '4' at its unit's place so, it is divisible by 2. (ii) 570 The number 570 has '0' at its unit's place so, it is divisible by 2. (iii) 285 The number 285 has '5' at its unit's place so, it is not divisible by 2. (iv) 2398 The number 2398 has '8' at its unit's place so, it is divisible by 2. (v) 79532 The number 79532 has '2' at its unit's place so, it is divisible by 2. (vi) 13576 The number 13576 has '6' at its unit's place so, it is divisible by 2. (vii) 46821 The number 46821 has '1' at its unit's place so, it is not divisible by 2. (viii) 84663 The number 84663 has '3' at its unit's place so, it is not divisible by 2. (ix) 66669 The number 66669 has '9' at its unit's place so, it is not divisible by 2. #### Question 2: Test the divisibility of each of the following numers by 5: (i) 95 (ii) 470 (iii) 1056 (iv) 2735 (v) 55053 (vi) 35790 (vii) 98765 (viii) 42658 (ix) 77990 A given number is divisible by 5 only when its unit digit is 0 or 5. (i) 95 The number 95 has '5' at its unit's place so, it is divisible by 5. (ii) 470 The number 470 has '0' at its unit's place so, it is divisible by 5. (iii) 1056 The number 1056 has '6' at its unit's place so, it is not divisible by 5. (iv) 2735 The number 2735 has '5' at its unit's place so, it is divisible by 5. (v) 55053 The number 55053 has '3' at its unit's place so, it is not divisible by 5. (vi) 35790 The number 35790 has '0' at its unit's place so, it is divisible by 5. (vii) 98765 The number 98765 has '5' at its unit's place so, it is divisible by 5. (viii) 42658 The number 42658 has '8' at its unit's place so, it is not divisible by 5. (ix) 77990 The number 77990 has '0' at its unit's place so, it is divisible by 5. #### Question 3: Test the divisibility of each of the following numbers by 10: (i) 205 (ii) 90 (iii) 1174 (iv) 57930 (v) 60005 A given number is divisible by 10 only when its unit digit is 0. (i) 205 The number 205 has '5' at its unit's place so, it is not divisible by 10. (ii) 90 The number 90 has '0' at its unit's place so, it is divisible by 10. (iii) 1174 The number 1174 has '4' at its unit's place so, it is not divisible by 10. (iv) 57930 The number 57930 has '0' at its unit's place so, it is divisible by 10. (v) 60005 The number 60005 has '5' at its unit's place so, it is not divisible by 10. #### Question 4: Test the divisibility of each of the following numers by 3: (i) 83 (ii) 378 (iii) 474 (iv) 1693 (v) 20345 (vi) 67035 (vii) 591282 (viii) 903164 (ix) 100002 A given number is divisible by 3 only when the sum of its digits is divisible by 3. (i) 83 The sum of the digits is 8 + 3 = 11 which is not divisible by 3. So, 83 is not divisible by 3. (ii) 378 The sum of the digits is 3 + 7 + 8 = 18 which is divisible by 3. So, 378 is divisible by 3. (iii) 474 The sum of the digits is 4 + 7 + 4 = 15 which is divisible by 3. So, 474 is divisible by 3. (iv) 1693 The sum of the digits is 1 + 6 + 9 + 3 = 19 which is not divisible by 3. So, 1693 is not divisible by 3. (v) 20345 The sum of the digits is 2 + 0 + 3 + 4 + 5 = 14 which is not divisible by 3. So, 20345 is not divisible by 3. (vi) 67035 The sum of the digits is 6 + 7 + 0 + 3 + 5 = 21 which is divisible by 3. So, 67035 is divisible by 3. (vii) 591282 The sum of the digits is 5 + 9 + 1 + 2 + 8 + 2 = 27 which is divisible by 3. So, 591282 is divisible by 3. (viii) 903164 The sum of the digits is 9 + 0 + 3 + 1 + 6 + 4 = 23 which is not divisible by 3. So, 903164 is not divisible by 3. (ix) 100002 The sum of the digits is 1 + 0 + 0 + 0 + 0 + 2 = 3 which is divisible by 3. So, 100002 is divisible by 3. #### Question 5: Test the divisibility of each of the following numbers by 9: (i) 327 (ii) 7524 (iii) 32022 (iv) 64302 (v) 89361 (vi) 14799 (vii) 66888 (viii) 30006 (ix) 33333 A given number is divisible by 9 only when the sum of the digits is divisible by 9. (i) 327 The sum of the digits is 3 + 2 + 7 = 12 which is not divisible by 9. So, 327 is not divisible by 9. (ii) 7524 The sum of the digits is 7 + 5 + 2 + 4 = 18 which is divisible by 9. So, 7524 is divisible by 9. (iii) 32022 The sum of the digits is 3 + 2 + 0 + 2 + 2 = 9 which is divisible by 9. So, 32022 is divisible by 9. (iv) 64302 The sum of the digits is 6 + 4 + 3 + 0 + 2 = 15 which is not divisible by 9. So, 64302 is not divisible by 9. (v) 89361 The sum of the digits is 8 + 9 + 3 + 6 + 1 = 27 which is divisible by 9. So, 89361 is divisible by 9. (vi) 14799 The sum of the digits is 1 + 4 + 7 + 9 + 9 = 30 which is not divisible by 9. So, 14799 is not divisible by 9. (vii) 66888 The sum of the digits is 6 + 6 + 8 + 8 + 8 = 36 which is divisible by 9. So, 66888 is divisible by 9. (viii) 30006 The sum of the digits is 3 + 0 + 0 + 0 + 6 = 9 which is divisible by 9. So, 30006 is divisible by 9. (ix) 33333 The sum of the digits is 3 + 3 + 3 + 3 + 3 = 15 which is not divisible by 9. So, 33333 is not divisible by 9. #### Question 6: Test the divisibility of each of the following numbers by 8: (i) 134 (ii) 618 (iii) 3928 (iv) 50176 (v) 39392 (vi) 56794 (vii) 86102 (viii) 66666 (ix) 99918 (x) 77736 A given number is divisible by 4 only when the number formed by its last two digits is divisible by 4. (i) 134 The last two digits of 134 are '34' which is not divisible by 4. Hence, 134 is not divisible by 4. (ii) 618 The last two digits of 618 are '18' which is not divisible by 4. Hence, 618 is not divisible by 4. (iii) 3928 The last two digits of 3928 are '28' which is divisible by 4. Hence, 3928 is divisible by 4. (iv) 50176 The last two digits of 50176 are '76' which is divisible by 4. Hence, 50176 is divisible by 4. (v) 39392 The last two digits of 39392 are '92' which is divisible by 4. Hence, 39392 is divisible by 4. (vi) 56794 The last two digits of 56794 are '94' which is not divisible by 4. Hence, 56794 is not divisible by 4. (vii) 86102 The last two digits of 86102 are '02' which is not divisible by 4. Hence, 86102 is not divisible by 4. (viii) 66666 The last two digits of 66666 are '66' which is not divisible by 4. Hence, 66666 is not divisible by 4. (ix) 99918 The last two digits of 99918 are '18' which is not divisible by 4. Hence, 99918 is not divisible by 4. (x) 77736 The last two digits of 77736 are '36' which is divisible by 4. Hence, 77736 is divisible by 4. #### Question 7: Text the divisibility of each of the following numbers by 5: (i) 95 (ii) 470 (iii) 1056 (iv) 2735 (v) 55053 (vi) 35790 (vii) 98765 (viii) 42658 (ix) 77990 A number is divisible by 8 only when the number formed by its last three digits is divisible by 8. (i) 6132 The last three digits of the given number are 132 which is not divisible by 8. So, 6132 is not divisible by 8. (ii) 7304 The last three digits of the given number are 304 which is divisible by 8. So, 7304 is divisible by 8. (iii) 59312 The last three digits of the given number are 312 which is divisible by 8. So, 59312 is divisible by 8. (iv) 66664 The last three digits of the given number are 664 which is divisible by 8. So, 66664 is divisible by 8. (v) 44444 The last three digits of the given number are 444 which is not divisible by 8. So, 44444 is not divisible by 8. (vi) 154360 The last three digits of the given number are 360 which is divisible by 8. So, 154360 is divisible by 8. (vii) 998818 The last three digits of the given number are 818 which is not divisible by 8. So, 998818 is not divisible by 8. (viii) 265472 The last three digits of the given number are 472 which is divisible by 8. So, 265472 is divisible by 8. (ix) 7350162 The last three digits of the given number are 162 which is not divisible by 8. So, 7350162 is not divisible by 8. #### Question 8: Test the divisibility of each of the following numbers by 11: (i) 22222 (ii) 444444 (iii) 379654 (iv) 1057982 (v) 6543207 (vi) 818532 (vii) 900163 (viii) 7531622 A given number is divisible by 11, if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11. (i) 22222 For the given number, sum of the digits at odd places = 2 + 2 + 2 = 6 sum of digits at even places = 2 + 2 = 4 Difference of the above sums = 6 − 4 = 2 Since the difference is not 0 and neither a number divisible by 11 so, the 22222 is not divisible by 11. (ii) 444444 For the given number, sum of the digits at odd places = 4 + 4 + 4 = 12 sum of digits at even places = 4 + 4 + 4 = 12 Difference of the above sums = 12 − 12 = 0 Since the difference is 0 so the given number 444444 is divisible by 11. (iii) 379654 For the given number, sum of the digits at odd places = 4 + 6 + 7 = 17 sum of digits at even places = 5 + 9 + 3 = 17 Difference of the above sums = 17 − 17 = 0 Since the difference is 0 so the given number 379654 is divisible by 11. (iv) 1057982 For the given number, sum of the digits at odd places = 2 + 9 + 5 + 1 = 17 sum of digits at even places = 8 + 7 + 0 = 15 Difference of the above sums = 17 − 15= 2 Since the difference is not 0 and neither a number divisible by 11 so, the 1057982 is not divisible by 11. (v) 6543207 For the given number, sum of the digits at odd places = 7 + 2 + 4 + 6 = 19 sum of digits at even places = 0 + 3 + 5 = 8 Difference of the above sums = 19 − 8 = 11 Since the difference is 11 which is surely divisible by 11 so the given number 6543207 is divisible by 11. (vi) 818532 For the given number, sum of the digits at odd places = 2 + 5 + 1 = 8 sum of digits at even places = 3 + 8 + 8 = 19 Difference of the above sums = 19 − 8 = 11 Since the difference is 11 which is surely divisible by 11 so the given number 818532 is divisible by 11. (vii) 900163 For the given number, sum of the digits at odd places = 3 + 1 + 0 = 4 sum of digits at even places = 6 + 0 + 9 = 15 Difference of the above sums = 15 − 4 = 11 Since the difference is 11 which is surely divisible by 11 so the given number 900163 is divisible by 11. (viii) 7531622 For the given number, sum of the digits at odd places = 2 + 6 + 3 + 7 = 18 sum of digits at even places = 2 + 1 + 5 = 8 Difference of the above sums = 18 − 8 = 10 Since the difference is 10 which is not divisible by 11 so the given number 7531622 is not divisible by 11. #### Question 9: Test the divisibility of each of the following numbers by 7: (i) 693 (ii) 7896 (iii) 3467 (iv) 12873 (v) 65436 (vi) 54636 (vii) 98175 (viii) 88777 (i) 693 69 − (2 × 3) = 69 − 6 = 63, which is divisible by 7. Hence, 693 is divisible by 7. (ii) 7896 789 − (2 × 6) = 789 − 12 = 777, which is divisible by 7. Hence, 7896 is divisible by 7. (iii) 3467 346 − (2 × 7) = 346 − 14 = 332, which is not divisible by 7. Hence, 3467 is not divisible by 7. (iv) 12873 1287 − (2 × 3) = 1287 − 6 = 1281, which is divisible by 7. Hence, 12873 is divisible by 7. (v) 65436 6543 − (2 × 6) = 6543 − 12 = 6531, which is divisible by 7. Hence, 65436 is divisible by 7. (vi) 54636 5463 − (2 × 6) = 5463 − 12 = 5451, which is not divisible by 7. Hence, 54636 is not divisible by 7. (vii) 98175 9817 − (2 × 5) = 9817 − 10 = 9807, which is divisible by 7. Hence, 98175 is divisible by 7. (viii) 88777 8877 − (2 × 7) = 8877 − 14 = 8863, which is not divisible by 7. Hence, 88777 is not divisible by 7. #### Question 10: Find all possible values of x for which the number 7x3 is divisible by 3. Also, find each such number. For a number to be divisible by 3, the sum of the digits must be divisible by 3. will be divisible by 3 in the following cases: So, the numbers can be 723, 753 or 783. #### Question 11: Find all possible values of y for which the number 53y1 is divisible by 3. Also, find each such number. If a number is divisible by 3, then the sum of the digits is also divisible by 3. Sum of the digits = The sum of the digits is divisible by 3 in the following cases: ∴ y = 0, 3, 6 or 9 The possible numbers are 5301, 5331, 5361 and 5391. #### Question 12: Find the value of x for which the number x806 is divisible by 9. Also, find the number. For a number to be divisible by 9, the sum of the digits must be divisible by 9. Sum of the digits in the given number = The sum of the digits is divisible by 9, only in the following case: Thus, the number x806 is divisible by 9 if $x$ is equal to 4. The number is 4806. #### Question 13: Find the value of z for which the number 471z8 is divisible by 9. Also, find the number. If a number is divisible by 9, then the sum of the digits is also divisible by 9. Sum of the digits of the given number = $4+7+1+z+8=20+z$ 27 is divisible by 9. Therefore, 471z8 is divisible by 9 if $z$ is equal to 7. The number is 47178. #### Question 14: Give five examples of numbers, each one of which is divisible by 3 but not divisible by 9. For a number to be divisible by 3, the sum of the digits should be divisible by 3. And for the number to be divisible by 9, the sum of the digits should be divisible by 9. Let us take the number 21. Sum of the digits is 2 + 1 = 3, which is divisible by 3 but not by 9. Hence, 21 is divisible by 3 not by 9. Similarly, lets check the number 24. Here, 2 + 4 = 6. This is divisible by 3 not by 9. The number 30 will be divisible by 3 not by 9 as 3 + 0 = 3. The number 33 will give 3 + 3 = 6, which is divisible by 3 not by 9. Also 39 has the digits 3 + 9 = 12, which is divisible by 3 not by 9. Hence, the numbers 21, 24, 30, 33 and 39 are divisible by 3 not by 9. #### Question 15: Give five examples of numbers, each one of which is divisible by 4 but not divisible by 8. For a number to be divisible by 4, the number formed by its last two digits should be divisible by 4. And for the number to be divisible by 8, the number formed by its last three digits should be divisible by 8. So, the numbers divisible by 4 and not by 8 will be 28, 36, 44, 52, 60. #### Question 1: Replace A, B, C by suitable numerals. $\begin{array}{ccc}& 5& A\\ +& 8& 7\end{array}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\begin{array}{ccc}C& B& 3\end{array}}}$ 1 is carried over. $\left(1+5+8\right)=14$ 1 is carried over. $B=4$ and $C=1$ #### Question 2: Replace A, B, C by suitable numerals. (1 is carried over) (1 is carried over) (1 is carried over) #### Question 3: Replace A, B, by suitable numerals. (with 1 being carried over) This is satisfied if $A$ is equal to $5$. When $A=5$: $A+A+A=15$ (1 is carried over) Or $B=1$ #### Question 4: Replace A, B by suitable numerals. First look at the left column, which is: $6-A=3$ This implies that the maximum value of A can be 3. $A\le 3$ ... (1) The next column has the following: $A-B=7$ To reconcile this with equation (1), borrowing is involved. We know: $12-5=7$ ∴ #### Question 5: Replace A, B, C by suitable numerals. $5-A=9$ This implies that 1 is borrowed. We know: $15-6=9$ $A=6$ $B-5=8$ This implies that 1 is borrowed. $13-5=8$ But 1 has also been lent $B=4$ $C-2=2$ This implies that 1 has been lent. $C=5$ ∴ #### Question 6: Replace A, B, C by suitable numerals. $\begin{array}{ccc}& A& B\\ & ×& 3\end{array}\phantom{\rule{0ex}{0ex}}\overline{)\overline{)\begin{array}{ccc}C& A& B\end{array}}}$ #### Question 7: Replace A, B, C by suitable numerals. $A×B=B⇒A=1$ In the question: First digit = B+1 Thus, 1 will be carried from 1+B2 and becomes (B+1) (B2 -9) B. ∴ C = B2 -1 Now, all B, B+1 and B2 -9 are one digit number. This condition is satisfied for B=3 or B=4. For B< 3, B2 -9 will be negative. For B>3, B2 -9 will become a two digit number. For B=3 , C = 32 - 9 = 9-9 = 0 For B = 4, C = 42 -9 = 16-9 = 7 A=1, B=3, C = 0 or A=1, B=4, C = 7 #### Question 8: Replace A, B, C by suitable numerals. Also, ∴ $A=7\phantom{\rule{0ex}{0ex}}B=C=6$ #### Question 9: Find two numbers whose product is a 1-digit number and the sum is a 2-digit number. are two numbers, whose product is a single digit number. ∴ $1×9=9$ Sum of the numbers is a two digit number. $1+9=10$ #### Question 10: Find three whole numbers whose product and sum are equal. The three whole numbers are 1, 2 and 3. $1+2+3=6=1×2×3$ #### Question 11: complete the magic square given below, so that the sum of the numbers in each row or in each column or along each diagonal is 15. 6 1 5 Taking the diagonal that starts with 6: 6 1 5 4 Now, taking the first row: 6 1 8 5 4 Taking the last column: 6 1 8 5 3 4 Taking the second column: 6 1 8 5 3 9 4 Taking the second row: 6 1 8 7 5 3 9 4 Taking the diagonal that begins with 8: 6 1 8 7 5 3 2 9 4 #### Question 12: Fill in the numbers from 1 to 6 without repetition, so that each side of the triangle adds up to 12. 6+2+4 = 12 4+3+5 = 12 6+1+5 = 12 #### Question 13: Fibonacci numbers Take 10 numbers as shown below: a, b, (a + b), (a + 2b), (2a + 3b), (3a + 5b), (5a + 8b), (8a + 13b), (13a + 21b), and (21a + 34b). Sum of all these numbers = 11(5a + 8b) = 11 × 7th number. Taking a = 8, b = 13; write 10 Fibonacci numbers and verify that sum of all these numbers = 11 × 7th number. Given: The numbers in the Fibonnaci sequence are arranged in the following manner: The numbers are . Sum of the numbers = $8+13+21+34+55+89+144+233+377+610\phantom{\rule{0ex}{0ex}}$ $=1584$ #### Question 14: Complete the magic square: 14 0 8 6 11 4 . 7 2 1 12 The magic square is completed assuming that the sum of the row, columns and diagonals is 30. This is because the sum of all the number of the last column is 30. 3 14 13 0 8 5 6 11 4 9 10 7 15 2 1 12 #### Question 1: If 5x6 is exactly divisible by 3, then the least value of x is (a) 0 (b) 1 (c) 2 (d) 3 (b) 1 If a number is exactly divisible by 3, the sum of the digits must also be divisible by 3. $5+x+6=11+x$ must be divisible by 3. The smallest value of $x$ is 1. is divisible by 3. #### Question 2: If 64y8 is exactly divisible by 3, then the least value of y is (a) 0 (b) 1 (c) 2 (d) 3 (a) 0 If a number is divisible by 3, then the sum of the digits is also divisible by 3. $6+4+y+8=18+y$ This is divisible by 3 as y is equal to 0. #### Question 3: If 7x8 is exactly divisible by 9, then the least value of x is (a) 0 (b) 2 (c) 3 (d) 5 (c) 3 If a number is exactly divisible by 9, the sum of the digits must also be divisible by 9. $7+x+8=15+x$ 18 is divisible by 9. #### Question 4: If 37y4 is exactly divisible by 9, then the least value of y is (a) 2 (b) 3 (c) 1 (d) 4 (d) 4 A number is divisible by 9 if the sum of the digits is divisible by 9. $3+7+y+4=14+y$ For this sum to be divisible by 9: #### Question 5: If 4xy7 is exactly divisible by 3, then the least value of (x + y) is (a) 1 (b) 4 (c) 5 (d) 7 (a) 1 If a number is divisible by 3, the sum of the digits is also divisible by 3. $4+x+y+7=11+\left(x+y\right)$ For the sum to be divisible by 3: #### Question 6: If x7y5 is exactly divisible by 3, then the least value of (x + y) is (a) 6 (b) 0 (c) 4 (d) 3 (d) 3 When a number is divisible by 3, the sum of the digits must also be divisible by 3. $x+7+y+5=\left(x+y\right)+12$ This sum is divisible by 3 if x+y+12 is 12 or 15. For x+y+12 = 12: x+y=0 But x+y cannot be 0 because then x and y both will have to be 0. Since x is the first digit, it cannot be 0. ∴ x+y+12 = 15 or x+y = 15-12=3 #### Question 7: If x4y5z is exactly divisible by 9, then the least value of (x + y + z) is (a) 3 (b) 6 (c) 9 (d) 0 (c) 9 A number is divisible by 9 if the sum of the digits is divisible by 9. $x+4+y+5+z=9+\left(x+y+z\right)$ The lowest value of for the number x4y5z to be divisible by 9. In this case, all x, y and z will be 0. But x is the first digit, so it cannot be 0. ∴ x+4+y+5+z = 18 or x+y+z+9 = 18 or x+y+z = 9 #### Question 8: If 1A2B5 is exactly divisible by 9, then the least value of (A + B) is (a) 0 (b) 1 (c) 2 (d) 10 (b) 1 For a number to be divisible by 9, the sum of the digits must also be divisible by 9. $1+A+2+B+5=\left(A+B\right)+8$ The number will be divisible by 9 if . #### Question 9: If the 4-digit number x27y is exactly divisible by 9, then the least value of (x + y) is (a) 0 (b) 3 (c) 6 (d) 9 (d) 9 If a number is divisible by 9, then the sum of the digits is divisible by 9. $x+2+7+y=\left(x+y\right)+9$ For this to be divisible by 9, the least value of . But for x+y = 0, x and y both will be zero. Since x is the first digit, it can never be 0. ∴ x + y + 9 = 18 or x + y = 9 #### Question 1: Find all possible values of x for which the 4-digit number 320x is divisible by 3. Also, find the numbers. If a number is divisible by 3, then the sum of the digits is also divisible by 3. $3+2+0+x=5+x$ must be divisible by 3. This is possible in the following cases: #### Question 2: Find all possible values of y for which the 4-digit number 64y3 is divisible by 9. Also, find the numbers. For a number to be divisible by 9, the sum of the digits must also be divisible by 9. $6+4+y+3=13+y$ For this to be divisible by 9: $y=5$ The number will be 6453. #### Question 3: The sum of the digits of a 2-digit number is 6. The number obtained by interchanging its digits is 18 more than the original number. Find the original number. Let the two numbers of the two-digit number be 'a' and 'b'. $a+b=6$ ... (1) The number can be written as $\left(10a+b\right)$. After interchanging the digits, the number becomes $\left(10b+a\right)$. $\left(10a+b\right)+18=\left(10b+a\right)\phantom{\rule{0ex}{0ex}}9a-9b=-18$ $a-b=-2$ ... (2) Using $a=2$ in equation (1): $b=6-a=6-2=4$ Therefore, the original number is 24. #### Question 4: Which of the following numbers are divisible by 9? (i) 524618 (ii) 7345845 (iii) 8987148 A number is divisible by 9 if the sum of the digits is divisible by 9. Number Sum of the digits Divisible by 9? 524618 26 No 7345845 36 Yes 8987148 45 Yes #### Question 5: Replace A, B, C by suitable numerals: $A-8=3$ This implies that 1 is borrowed. Then, $7-B=9$ 1 is borrowed from 7. Further, $5-C=2$ But 1 has been borrowed from 5. ∴ 4 - C = 2 $⇒C=2$ #### Question 6: Replace A, B, C by suitable numerals: Here, $B=3$ Since $7×9=63$$C=9$ ∴ #### Question 7: Find the values of A, B, C when $A×B=B⇒A=1$ $\begin{array}{ccc}& 1& B\\ ×& B& 1\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}& 1& B\\ B& {B}^{2}& ×\end{array}\phantom{\rule{0ex}{0ex}}\begin{array}{ccc}B& \left(1+{B}^{2}\right)& B\end{array}$ Now, $B\ne A=1$ and $\left(1+{B}^{2}\right)$ is a single digit number. ∴ $B=2$ ∴ #### Question 8: Mark (✓) against the correct answer If 7x8 is exactly divisible by 3, then the least value of x is (a) 3 (b) 0 (c) 6 (d) 9 (b) 0 If a number is exactly divisible by 3, the sum of its digits is also divisible by 3. $7+x+8=15+x$ $15+x$ can be divisible by 3 even if x is equal to 0. #### Question 9: Mark (✓) against the correct answer If 6x5 is exactly divisible by 9, then the least value of x is (a) 1 (b) 4 (c) 7 (d) 0 (c) 7 When a number is divisible by 9, the sum of the digits is also divisible by 9. $6+x+5=11+x$ To be divisible by 9: #### Question 10: Mark (✓) against the correct answer If x48y is exactly divisible by 9, then the least value of (x + y) is (a) 4 (b) 0 (c) 6 (d) 7 (c) 6 When a number is divisible by 9, the sum of its digits is also divisible by 9. $x+4+8+y=12+\left(x+y\right)$ For $12+\left(x+y\right)$ to be divisible by 9: #### Question 11: If 4867 is divisible by 9, then the least value of is (a) 0 (b) 1 (c) 3 (d) 2 (d) 2 For a number to be divisible by 9, the sum of its digits must be divisible by 9. $4+8+6++7=25+$ Now, and 27 is divisible by 9) View NCERT Solutions for all chapters of Class 8
Some Arithmetic Operations The four basic arithmetic operations found in mathematics are: addition, subtraction, multiplication and division. The word “addition” is derived from Latin which means “addition by addition”. The operation of addition consists of two steps i.e., addition of one element with another element. The following examples of this type of operation are: addition of two numbers or addition of three things together to form a sum. In addition, other kinds of addition such as subtraction and division are known as the same. The second part of the addition is the definition. The definition consists of an addition by a rule. The rule is denoted by “+” sign. The rule of addition is known as “addition by definition”. The third part of addition is subtraction. The subtraction is also known as “subtraction by rule”. The subtraction by rule is a rule that is generally used when you need to find the difference between two objects. The difference can be defined as either zero or negative zero. The fourth and last part of addition and subtraction is division. The division is the simplest type of operation. It is denoted as “divide by rule”. As we know, these four basic addition and subtraction are found in every school and college. You may ask how we know these. The fact is that these operations were created in every elementary school and college. We all know that we are not able to calculate without knowing the above four arithmetic operations. Addition and subtraction are two of the basic arithmetic operations. But they are not the only ones. Other types of addition and subtraction are the division and multiplication operations. Multiplication is the first type of operation of arithmetics. The multiplication operation of two different numbers produces a third number. When multiplying two different numbers, it is better to use a larger number to produce a larger number, but it is easier to multiply two smaller numbers. Division is the second type of arithmetic operation. Divisions can be done when multiplying a number with a smaller number. For example, if you multiply “4” with “5”, you will get “3”. Division is also used to divide a number into smaller parts. You can divide a number by “minus one”. Subtracting a number from the other is called “minus one division”. Divide and multiply are some of the most useful arithmetic operations. You can perform addition, subtraction, division and multiply any number in an exact way. These arithmetic operations can help you in calculating, analyzing, comparing, and planning the work. Subtraction is the process of separating two or more things. This procedure can be done with a single number or with a number of other things. The difference between two objects is the difference between their quantities. The subtraction process separates them. Subtraction is also called the subtractive method. The process of subtracting something from something else. The process of subtracting something from another thing and then subtracting the same thing back from the second object is called subtraction. Subtraction is useful in order to determine whether something has been subtracted from something else. In the algebra, subtraction is also used in addition and multiplication. Subtraction is also known as a simple method in addition and multiplication. For example, if you multiply a number with a number, you will get a product of the two numbers. Subtraction is used to find out the difference of two numbers. The difference of two numbers is the difference between the actual values of the two numbers. In addition, subtraction is also used to find the difference between two quantities. Subtraction is also used in addition and multiplication. In addition, subtraction is used to find the difference between two quantities. In addition, subtraction is also used to find out the difference between two quantities. Subtraction is also used in the calculation of the difference of the quantity of two quantities. This subtraction can be used in the calculation of the difference between two quantities.
# Question Video: Finding the Position of a Shape before a Reflection given Its Image Mathematics • 11th Grade Below is an image of a trapezoid after it was reflected across the dotted line. Which of these figures shows how the trapezoid appeared before the reflection? [A] Figure A [B] Figure B [C] Figure C [D] Figure D 02:29 ### Video Transcript Below is an image of a trapezoid after it was reflected across the dotted line. Which of these figures shows how the trapezoid appeared before the reflection? The image on the left of the screen shows how the trapezoid appears after the reflection. And we need to decide which of the images (A), (B), (C), or (D) shows how the trapezoid appeared before. A key property of reflections is that they preserve the perpendicular distance of all points from the mirror line. To determine how the trapezoid appeared before the reflection, we can therefore consider the perpendicular distance of each vertex of the trapezoid after the reflection and then find the point on the opposite side of the mirror line at the same perpendicular distance. We note that the trapezoid has been drawn on a squared grid and the mirror line is horizontal. Therefore, the perpendicular distance of each point from the mirror line is its vertical distance. Let’s consider each vertex of the trapezoid. The pink vertex is three squares above the mirror line. So its original position would have been three squares below the mirror line at the same horizontal position. The red vertex is two squares below the mirror line. So its original position would be two squares above. The yellow and green vertices are each two squares vertically from the mirror line. So, in fact, they have swapped places under the reflection. Connecting the four crosses shows how the trapezoid would’ve looked before the reflection took place. Looking at the four answer options, we can rule out options (A) and (D) as in both cases the sloping side below the mirror line slopes upwards from left to right instead of downwards. Then, looking at option (B), we can see that whilst the orientation is correct, this trapezoid is in the wrong position on the grid. The bottom-left vertex should be two squares below the mirror line, but it is only one square below. In option (C), however, each of the vertices is in the same position as on our diagram. So option (C) shows how the trapezoid appeared before the reflection.
Courses Courses for Kids Free study material Offline Centres More Store # Find x, if $x,15,36,27$ are in proportion. Last updated date: 13th Sep 2024 Total views: 439.2k Views today: 8.39k Verified 439.2k+ views Hint: Two numbers are said to be proportional to each other if one number has a constant ratio to another number. Proportional relationships are relationships between two variables where their ratios are equivalent. Another way to think about them is that, in a proportional relationship, one variable is always a constant value times the other. That constant is known as the "constant of proportionality". If the ratio $\dfrac{y}{x}$ of two variable $$x$$ and $$y$$ is equal to a constant k then the variable in the numerator of the ratio (y) is the product of the other variable and the constant $y = k \times x$ . In this case y is said to be directly proportional to x with proportionality constant k. If $a,b,c,d$ are proportional then they will have the same proportionality constant. i.e.$\dfrac{a}{b} = \dfrac{c}{d}$ We will apply the same that is mentioned above in our given proportion and we will get an equation. Then by cross multiplication of the terms, we will get the value of x. It is given that,$\;x,15,36,27$ are in proportion to each other. Since $x,15,36,27$ are in proportion so we can write them as follows, $\dfrac{x}{{15}} = \dfrac{{36}}{{27}}$ Let us now solve the above term by the method of cross multiplication, That is let us multiply by 15 on both sides of the above given equation so that we get, $x = \dfrac{{15 \times 36}}{{27}}$ On solving the terms in numerator and denominator we get, $x = \dfrac{{15 \times 4}}{3}$ $x = 5 \times 4$ $x = 20$ Hence we have found the value of$$x$$ as 20. Note: If $x,y,z$ are in proportion then$\dfrac{x}{y} = \dfrac{y}{z}$ . That is if three terms are said to be in proposition we get $\dfrac{x}{y} = \dfrac{y}{z}$ If $a,b,c,d$ are proportional then they will have the same proportionality constant$\dfrac{a}{b} = \dfrac{c}{d}$. That is if four terms are in proportion then we get$\dfrac{a}{b} = \dfrac{c}{d}$. Here while solving the proportion we are just comparing the fractions on both sides of the equation whenever two fractions are said to be equal they both have the same values after the reduction. That is we can say $\dfrac{x}{{15}} = \dfrac{{36}}{{27}} = \dfrac{{12}}{9} = \dfrac{4}{3}$ that is we should get $\dfrac{x}{{15}} = \dfrac{4}{5}$ on cross multiplying we get, $x = \dfrac{{4 \times 15}}{3} = 4 \times 5 = 20$, here the value of x is found easily.
# Do two lines lay in the same 4-dimensional plane I don't know how quite to phrase this, but I'll try. Because two point are co-linear and two lines cannot always be used to define a plane and aren't always in the same plane, are two lines always co-something? 2 0 dimensional shapes make one 1 dimensional shape. 3 0 dimensional shapes make 1 two dimensional shape. x-1 0 dimensional shapes make a x dimensional shape. What do x-1 1 dimensional shapes make? If you allow one of your points to be the origin (or one of your lines to pass through the origin), then you can formulate and answer this question using linear algebra. For instance, given two points (the origin $O$ and one other point $P$), you have a single vector $\overrightarrow{OP}$, and the span of this vector is $1$-dimensional vector space, a line. Given three points (the origin, $P$, and $Q$), they are contained in the vector space spanned by the two vectors $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ which is at most a $2$-dimensional vector space, a plane. Now, given two lines, the first through the origin and another point $P$, and the second through $Q$ and $R$, these lines are contained in the vector space spanned by the vectors $\overrightarrow{OP}$, $\overrightarrow{OQ}$, and $\overrightarrow{QR}$, which is at most $3$-dimensional. So two lines are always contained in a $3$-dimensional space. In the same way, if we have $x-1$ lines, then we can choose that the first goes through the origin and hence is determined by a single vector (one other point). The other lines can each be determined by two vectors (a point on the line and the vector from that point to another on the line). So they will be contained in a space of dimension $2x-3$. • Using the same kind of counting, it will span a subspace that is $x(y+1)-1$ dimensional (at most). For each of the $x$ objects (except one), you get $y+1$ vectors, and $y$ for the remaining one.
Addition is the basic operation of the basic operations in arithmetic. If we add two numbers, we basically count the values of both numbers together and present the result as a new number. In Figure 1 two numbers are added: 6 + 4 = 10, verbally: "six plus four equals ten" or "six added to four is ten" or "six and four makes ten"... Adding any two natural numbers under ten is a basic skill anyone should be able to perform fluently, almost without thinking. In Figure 2 addition under ten is ordered in an accessible table. In fact, students start by learning all the additions that result in 10 or less than 10. Then, for larger sums, they learn to first "make 10" and add what is left. "Making 10" is first adding the complement to 10. For example, 1 is called the complement to 10 of 9 and 7 is called the complement to 10 of 3: 9 + 8 = (9 + 1) + 7 = 10 + 7 = 17 3 + 8 = (3 + 7) + 1 = 10 + 1 = 11 The elements that are added together are called the terms of the addition, and the result is called the sum. Addition is commutative, meaning that changing the order of the terms does not change the sum: 3 + 4 = 4 + 3 2 + 3 + 4 = 4 + 2 + 3 What effect does this property have on the table in Figure 2? Adding zero to a number (or a number to zero) has no effect: 3 + 0 = 3. ## Adding terms > 10 If we master addition under ten and understand the principles of the decimal numeral system, we can try some more difficult additions: In 50 + 30 both digits 5 and 3 are in the tens position of its number, thus the digits can be added: 5 + 3 = 8 and in the tens position this makes 80. 50 + 30 = (5+3) × 10 = 8 × 10 = 80 60 + 40 = (6+4) × 10 = 10 × 10 = 100. 700 + 500 = (7+5) × 100 = 12 × 100 = 1200. So, we constantly need to regroup ones by ones, ten by tens, hundreds by hundreds, etc: 55 + 4 = 50 + 5 + 4 = 50 + 9 = 59. 46 + 6 = 40 + 6 + 6 = 40 + 12 = 40 + 10 + 2 = 50 + 2 = 52. 219 + 23 = 200 + 10 + 20 + 9 + 3 = 200 + 30 + 12 = 200 + 40 + 2 = 242. 77 + 45 = 70 + 40 + 7 + 5 = 110 + 12 = 100 + 10 + 12 = 100 + 22 = 122. So far, you should have been able to calculate the given examples fluently in your head (if not, practice here). More difficult additions can be executed using pen and paper and the standard addition algorithm, also called long addition or column addition. In the above example the addition algorithm is used to calculate 127 + 95 = 222. The recipe is as follows: We write the digits of the involved numbers in a table with horizontal rows and vertical columns. From right to left the first column is for the ones positions, the second column for the tens positions, the third column for the hundreds positions etc. Then we write the numbers to be added (127 and 95) under each other with their digits in the corresponding columns. Then we add the digits in each column, starting with the right column. Each time we write the result in the bottom row under the horizontal line: 7 + 5 = 12. We write the 2 in the bottom row in the ones column and the 1 (= 10) needs to be "carried" to the tens column, and in the next step to be added to the other tens digits in this column. Thus: 1 + 2 + 9 = 12 (the first digit "1" is the "carry"). And again we write the 2 in the result row, and carry the 1 (= 100) to the next column. This example ends with the third column: 1 + 1 = 2 (the first "1" is again the "carry"). Thus: 127 + 95 = 222. If we want, we can write the "carries" in the corresponding columns in a top row, as done in the above example. This compact and widely used algorithm always works for all additions of natural numbers, also if we need to add up more than two numbers. Some more examples: ### Practice application With the next app you can practice with the standard addition algorithm. Use the algorithm on a piece of paper to add-up the numbers. Click the exercise or click "CHECK" to see if you did it right. Exercises: =
# Video: KS2-M18 • Paper 2 • Question 4 These diagrams show three equivalent fractions. Write the missing values. 3/4 = 9/＿ = ＿/24 03:42 ### Video Transcript These diagrams show three equivalent fractions. Write the missing values. Three-quarters equals nine over what equals what twenty-fourths. The first part of the question tells us that the three diagrams show equivalent fractions. Each large rectangle may be divided into different numbers of pieces. But the shaded areas are all the same the fractions have the same value. They are equivalent. That’s why there are equal signs in between them. And the three fractions underneath refer to the three diagrams above. But two of the numbers are missing and we’re asked to write the missing values. The first fraction shows three-quarters. Let’s see where we get this from. Each row of the first rectangle contains one piece: one, two, three, four, four parts altogether. This is where the denominator or the bottom number in the fraction comes from. It’s the total number of equal parts. But how many parts have been shaded? In the first diagram, the shading comes as far as here. So we can see that there are three parts shaded. And this is where we get the numerator or the top number in our fraction from, the number of shaded parts. Three shaded parts out of a possible four equals three-quarters. Our second fraction is equivalent to three-quarters. How many parts has a rectangle been divided into? Three, six, nine, 12. The rectangle has been split into twelfths. So the missing value in our fraction, the denominator, equals 12, the total number of equal parts. We can see that nine out of 12 parts have been coloured in. So nine twelfths is the same as three-quarters. Let’s look at our final fraction. This time, the numerator is missing. This is the number of shaded parts. The denominator 24 tells us that the shape has been split into 24 equal parts: six, 12, 18 — 24 altogether. But how many out of those 24 parts have been shaded? We can see that three rows is the same as 18 parts. So eighteen twenty-fourths is the same as nine twelfths, which is also the same as three-quarters. We found the two missing values by thinking carefully about what the numerator and the denominator stand for in a fraction. We counted the total number of the equal parts and the number of the shaded parts. And we used these to find the missing values. Three-quarters equals nine twelfths equals eighteen twenty-fourths.
Math Relative position of lines Determining point of intersection # Intersecting lines Two lines can intersect and then have a point of intersection. ! ### Requirements 1. Direction vectors are not collinear 2. When equating the linear equations, we receive a distinct result ## Calculating point of intersection To calculate the point of intersection, we follow the scheme above. The result from step 2 for $r$ or $s$ is then inserted into the corresponding linear equation. ### Example $\text{g: } \vec{x} = \begin{pmatrix} 10 \\ 9 \\ 7 \end{pmatrix} + r \cdot \begin{pmatrix} 9 \\ 8 \\ 7 \end{pmatrix}$ $\text{h: } \vec{x} = \begin{pmatrix} 4 \\ 4 \\ 6 \end{pmatrix} + s \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ 1. #### Direction vectors not collinear? First, we examine if the direction vectors are multiples of each other (=collinear). $\vec{a}=t\cdot\vec{b}$ $\begin{pmatrix} 9 \\ 8 \\ 7 \end{pmatrix}=t\cdot\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ Calculate $t$ for every row 1. $9=t\cdot1$ 2. $8=t\cdot1$ 3. $7=t\cdot2$ 1. $t=9$ 2. $t=8$ 3. $t=3.5$ $t$ does not have the same value everywhere. Thus the direction vectors are not collinear. Therefore the lines are skew or have a point of intersection. 2. #### Equate $g$ and $h$ Now equate the linear equations and calculate $r$ and $s$. $\begin{pmatrix} 10 \\ 9 \\ 7 \end{pmatrix} + r \cdot \begin{pmatrix} 9 \\ 8 \\ 7\end{pmatrix}$ $= \begin{pmatrix} 4 \\ 4 \\ 6 \end{pmatrix} + s \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ Now we set up an equation system and solve it. 1. $10+9r=4+s$ 2. $9+8r=4+s$ 3. $7+7r=6+2s$ Equation method: Equate I. and II. $4+s=4+s$ $10+9r=9+8r\quad\|-8r$ $10+r=9\quad\|-10$ $r=-1$ Compute $s$ with e.g. I. (for that insert $r=-1$) $10+9\cdot(-1)=4+s$ $1=4+s\quad\|-4$ $s=-3$ For validation insert $r=-1$ and $s=-3$ into III. $7+7\cdot(-1)=6+2\cdot(-3)$ $0=0$ $r=-1$ and $s=-3$ result from all equations => The lines intersect. 3. #### Determining point of intersection To determine the point of intersection we insert $r=-1$ into $g$. $\begin{pmatrix} 10 \\ 9 \\ 7 \end{pmatrix} + (-1) \cdot \begin{pmatrix} 9 \\ 8 \\ 7 \end{pmatrix}$ $=\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ => The lines intersect in point $S(1\|1\|0)$.
# Range and Mean Deviation The field of statistics has practical applications in almost all fields of life. In finance and investing and manufacturing and various other fields. Whether you want to launch a rocket or calculate a students performance we take the help of statistics. Now let us learn the concepts of range and mean deviation. ## Range and Mean Deviation For a given series of data, statistics aims at analysis and drawing conclusions. The various measures of central tendency – mean, median and mode represent the values in a series. However, we can further implement this analytical claim of statistics, by measuring the scattering and dispersion of data around these measures of central tendency. For example, the range, in a series of men distributed by their ages, gives us an idea of how much the ages can vary from the central value. So without further ado, let us jump right into the concepts of range and mean deviation. ## Range The range can be simply understood as the value that tells us about the scattering of data. This gives us an idea of how much the data can vary. Consequently, it is related to the maximum and minimum values in a distribution. The range is the difference between the maximumÂ and the minimum value in a distribution. Notably, it only gives us the idea of the spread of data. It does not tell us about the dispersion of values from a measure of central tendency. Range = Maximum value – Minimum value ## Mean Deviation To understand the dispersion of data from a measure of central tendency, we can use mean deviation. It comes as an improvement over the range. It basically measures the deviations from a value. This value is generally mean or median. Hence although mean deviation about mode can be calculated, mean deviation about mean and median are frequently used. Note that the deviation of an observation from a value a isÂ d= x-a.Â To find out mean deviation we need to take the mean of these deviations. However, when this value of a is taken as mean, the deviations are both negative and positive since it is the central value. This further means that when we sum up these deviations to find out their average, the sum essentially vanishes. Thus to resolve this problem we use absolute values or the magnitude of deviation. The basic formula for finding out mean deviation is : Mean deviation= Sum of absolute values of deviations from ‘a’ Ã· The number of observations ## Solved Example for You Q:Â The sum of squares of deviation of variates fromÂ their A.M. is always 1. Zero 2. Minimum 3. Maximum 4. Cannot be said Sol: The correct option is “B”. It is a fundamental concept that the sum of squares of deviation of any variate from their arithmetic mean is always minimum. Share with friends ## Customize your course in 30 seconds ##### Which class are you in? 5th 6th 7th 8th 9th 10th 11th 12th Get ready for all-new Live Classes! Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes. Ashhar Firdausi IIT Roorkee Biology Dr. Nazma Shaik VTU Chemistry Gaurav Tiwari APJAKTU Physics Get Started
# Adding Mixed Numbers 13 teachers like this lesson Print Lesson ## Objective SWBAT add mixed numbers with like denominators by changing them to improper fractions. They validate their answers with a visual model. #### Big Idea A mixed number can be solved by changing it to an improper fraction. ## Whole Class Discussion 15 minutes In today's lesson, the students learn to add mixed numbers with like denominators. This relates to 4.NF.B3c because the students add mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction. Before we begin the lesson, I give the students a review by asking a series of questions. First, I ask, What is a mixed number? Student response: A mixed numbers is when you have a whole number and a fraction. I let the students know that today, we are adding mixed numbers with like denominators. What are like denominators? Student response: When the denominators are the same. If the denominators are the same number, what must we do to solve the problem? Student response: Add the numerators. I explain to the students that we are not working with unlike denominators today, so we do not have to find a common denominator. The Adding Mixed Numbers with Like denominators.pptx power point is displayed on the Smart board. The students are sitting at their desk with paper and pencil. I let the students know that we are learning two ways to add mixed numbers. I let the students know that they can solve the problem in either way that is easiest for them. I instruct them to write down the sample problem from the power point, 5 1/5 + 3 2/5. I want the students to work with me as we discuss the power point. (I find that this method works well for my students because it keeps them focused on what is being said in class.) Be sure to line up the fractions under the fractions and the whole numbers under the whole numbers. We learned in addition that we line numbers up according to place value. This also applies to fractions because fractions can be written as decimals. When you add, always begin with the fraction. I remind the students that when they learned to add in earlier grades, they learned to start with the place farthest to the right. Since the denominators are the same, we can begin by adding the numerators. The students add 1 + 2 to get 3. On their papers, they write the 3 as the numerator. What is the denominator in this problem? 5. I was curious to see if the students remembered why we do not add denominators. Therefore, I asked, Can someone explain to me why the denominator did not change? I asked 3 students before I could get one student to tell me that the denominator tells us how many pieces the whole was divided into. (At this point, I changed my original plans for this lesson. I originally planned to have the students just solve the problems by adding and using multiplication to help with the division - if they chose to use improper fractions. However, I decided at this point to have the students draw models because a lot of them were still not connecting the model to the problems.) I reminded the students that in our previous lesson on changing mixed numbers to improper fractions, we drew the models to show the mixed numbers and improper fractions. We learned that the denominator told us how to divide the whole. I explain to them that in the problem that they are currently working on, the denominator stays fifths because the whole is cut into fifths. Back to the problem, next we add the whole number. What is 5 + 3? 8. We have 8 3/5 as our answer. I instruct the students to draw 5 1/5. What did we learn about the wholes? Student response: They have to be the same size. As the students draw their models, I draw models on the board as well. (A teacher modeling for the students can help give those lost students a starting point if they are completely lost.) Together we shade 5 whole and 1 out of 5 for our 5 1/5. Next, the students draw a model of 3 2/5, then shade 3 whole and 2 out of 5 to show 3 2/5. Now, let's see if our model connects to our answer. We said that the answer to the problem was 8 3/5. Let's count to see how many "wholes" are shaded. The students count to get 8. Next, count the boxes that represent the fraction. The students count 3/5 shaded. Therefore, our model and answers connect. Another Way: I remind the students that this is not a new skill. We have learned this skill in a previous lesson. The students write the problem down again, 5 1/5 + 3 2/5. We can change a mixed number to an improper fraction. To change a mixed number to an improper fraction, you multipy the denominator by the whole number, then add the numerator. The students and I work together to change 5 1/5 to 26/5. Next, we change 3 2/5 to 17/5. I tell the students that now we can add the numerators because we have like denominators. The numerator is 43. Our fraction is 43/5. I ask, "Can we leave our answer as an improper fraction?" No. "No, we can not. We must change our improper fraction to a whole or mixed number by using division. We have learned that multiplication can help us with division. We should divide 43 divided by 5. "How many times can 5 go into 43?" 8. "What multiplication fact can help you with this?" 8 x 5 = 40. "What is the remainder?" 3. The quotient is your whole number. The remainder is your numerator. Your divisor is your denominator. We write our mixed number as 8 3/5. So that the students see the connection between the mixed number and improper fraction, I have the students go back to the model that they drew earlier. I have them count the total number of pieces shaded. The students see that there are 43/5 shaded in their model of 5 1/5 + 3 2/5. ## Skill Building/Exploration 20 minutes For this activity, I let the students work independently. I give each student an Adding Mixed Numbers with Like Denominators.docx activity sheet. The students must add mixed numbers with like denominators in one of two ways: 1) by adding the mixed numbers, then simplifying the fraction, or 2) by changing the mixed numbers to improper fractions. The students must draw a model of their mixed numbers (MP4). The students work on real-world scenarios to add mixed numbers with like denominators. The students are guided to the conceptual understanding through questioning by me. As I walk around the classroom, I am questioning the students and looking for common misconceptions among the students. Any misconceptions are addressed at that point, as well as whole class at the end of the activity. Any student that finishes the assignment early, can go to the computer to practice the skill at the following site until we are ready for the whole group sharing: http://studyjams.scholastic.com/studyjams/jams/math/fractions/add-sub-mixed-numbers.htm My Findings: In today's lesson, the students learned to add mixed numbers in two ways. As I monitored and answered questions from the students, I noticed that some of the students still had a hard time drawing the models. I encouraged those students to solve the problem first, then draw a model last. Most of the students could solve the problems, but as stated earlier, the models were giving a few of them a problem. After I kept driving home the point that the denominator tells you how to divide the whole, it became better for most of the ones having problems. I was surprised that several of the students used equivalent fractions to add the mixed numbers. I expected more of the students to just add the fractions and whole numbers. Overall, I was pleased with the lesson. ## Closure 15 minutes To close the lesson, we review the answers to the problems as a whole class. This gives those students who still do not understand another opportunity to learn it. I like to use my document camera to show the students' work during this time. Some students do not understand what is being said, but understand clearly when the work is put up for them to see. I feel that by closing each of my lessons by having students share their work is very important to the success of the lesson. Students need to see good work samples, as well as work that may have incorrect information. From the Video - Adding Mixed Numbers, you can hear a student explain how she solved a problem. More than one student may have had the same misconception. During the closing of the lesson, all misconceptions that were spotted during the activity will be addressed whole class. Misconception(s): -Drawing models correctly (as stated earlier)
## How do you solve an integral with a fraction? And then we’re going to have the integral of b over the other linear factor x minus two. And since it’s x minus 2 squared we’re going to have another a fraction this one’s going to be c. Can you integrate a fraction by parts? Answer: If someone asks you to integrate a fraction, you must try to multiply or divide the top and bottom of the fraction by a number. Occasionally it will be of help if you split a fraction up prior to making an attempt to integrate it. You can use the method of partial fractions for this. ### How is integration useful to a mathematician? The process of finding integrals is called integration. Along with differentiation, integration is a fundamental, essential operation of calculus, and serves as a tool to solve problems in mathematics and physics involving the area of an arbitrary shape, the length of a curve, and the volume of a solid, among others. Why do we use integration by partial fractions? Integration by partial fractions is a method used to decompose and then integrate a rational fraction integrand that has complex terms in the denominator. By using partial fraction, we calculate and decompose the expression into simpler terms so that we can easily calculate or integrate the expression thus obtained. ## What are the rules of integration? The important rules for integration are: • Power Rule. • Sum Rule. • Different Rule. • Multiplication by Constant. • Product Rule. How do you do integration by parts? So we followed these steps: 1. Choose u and v. 2. Differentiate u: u’ 3. Integrate v: ∫v dx. 4. Put u, u’ and ∫v dx into: u∫v dx −∫u’ (∫v dx) dx. 5. Simplify and solve. ### How do you differentiate fractions? How To Find The Derivative of a Fraction – Calculus – YouTube What is use of integration in real life? In real life, integrations are used in various fields such as engineering, where engineers use integrals to find the shape of building. In Physics, used in the centre of gravity etc. In the field of graphical representation, where three-dimensional models are demonstrated. Was this answer helpful? ## Why should we learn integration? Why do we need to study Integration? Often we know the relationship involving the rate of change of two variables, but we may need to know the direct relationship between the two variables. How do you remember integration rules? TRICK TO MEMORIZE INTEGRATION FORMULAS \|\| HOW TO LEARN … ### What is the formula for integration? Basically, integration is a way of uniting the part to find a whole. It is the inverse operation of differentiation. Thus the basic integration formula is ∫ f'(x) dx = f(x) + C. How do you integrate easily? Integration Tricks (That Teachers Won’t Tell You) for Integral Calculus ## How do you memorize integration formulas? Is dy dx a fraction? Show activity on this post. While I do know that dydx isn’t a fraction and shouldn’t be treated as such, in many situations, doing things like multiplying both sides by dx and integrating, cancelling terms, doing things like dydx=1dxdy works out just fine. ### How do you integrate? Basic Integration… How? (NancyPi) – YouTube Why do we learn integration? ## Why do we need integration? Integration ensures that all systems work together and in harmony to increase productivity and data consistency. In addition, it aims to resolve the complexity associated with increased communication between systems, since they provide a reduction in the impacts of changes that these systems may have. What is a real life example of integration? In real life, integrations are used in various fields such as engineering, where engineers use integrals to find the shape of building. In Physics, used in the centre of gravity etc. In the field of graphical representation, where three-dimensional models are demonstrated. ### What should I learn first integration or differentiation? It is very important to focus on differentiation before you start integration. A strong understanding of differentiation makes integration more natural. Is learning integration difficult? Integration is hard! Integration is generally much harder than differentiation. This little demo allows you to enter a function and then ask for the derivative or integral. You can also generate random functions of varying complexity. ## How do you understand integration? In Maths, integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. This method is used to find the summation under a vast scale. Is integration easy? ### What is basics of integration? Basic Rules And Formulae Of Integration BASIC INTEGRATION FORMULAE 01. ∫xndx=xn+1n+1+C;n≠−1∗ 11. 03. ∫exdx=ex+C 13. 04. ∫axdx=axlna+C ∫ a x d x = a x ln ⁡ 14. 05. ∫sinxdx=−cosx+C ∫ sin x d x = − cos ⁡ 15. Why is integration so hard? The problem is that differentiation of elementary functions always involves elementary functions; however, integration (anti-derivative) of elementary function may not involve elementary functions. This is the reason why the process of integration is, in general, harder.
Lesson 4 Using Technology to Work with Sequences Let’s use technology to create a sequence. Problem 1 Technology required. Open a blank spreadsheet. In A1, type 2 and enter. 1. What should you type into cell A2 to generate the sequence 2, 4, 8, 16, 32, . . . when you fill down the column? 2. What should you type into cell A2 to generate the sequence 2, 4, 6, 8, 10, . . . when you fill down the column? Problem 2 Technology required. Open a blank spreadsheet. In A1, type 400 and enter. 1. What should you type into cell A2 to generate the sequence 400, 200, 100, 50, 25, . . . when you fill down the column? 2. What should you type into cell A2 to generate the sequence 400, 325, 250, 175, 100, . . . when you fill down the column? Problem 3 Technology required. Open a blank spreadsheet. 1. If cell A1 = 5 and cell A2 = A1 * 3 + 2, what are the first 5 terms of the sequence? 2. If cell A1 = 1 and cell A2 = (A1 + 2) * 3, what are the first 5 terms of the sequence? 3. If cell A1 = 2 and cell A2 = (A1 + 2) * 3, what are the first 5 terms of the sequence? Problem 4 Technology required. Open a blank spreadsheet. 1. Find the first 5 terms of a geometric sequence that starts with -5 and has a growth factor of -1. 2. Find the first 5 terms of a geometric sequence that starts with -20 and has a growth factor of 0.5. 3. Find the first 5 terms of an arithmetic sequence that starts with -20 and has an rate of change of 5. 4. Find the first 5 terms of an arithmetic sequence that starts with 43 and has an rate of change of -7. Problem 5 Here is the graph of a sequence. 1. Explain how you know this sequence is arithmetic. 2. Explain how you know this sequence is not geometric. (From Unit 1, Lesson 3.) Problem 6 The first two terms of a geometric sequence are 6 and 3. 1. Explain why there is only one geometric sequence with these first two terms. 2. What are the next 3 terms of this geometric sequence? (From Unit 1, Lesson 2.)
## FANDOM 1,168 Pages The Quadratic Formula is a mathematical formula used to find the x-intercepts for a quadratic function, or parabola. When $f(x) = ax^2 + bx + c = 0,\quad a \ne 0$ With $x$ being the variable and $a$, $b$ and $c$ being constant, the quadratic equation is: $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ This is one possible method to determine the x-intercepts. They can also be found by completing the square, factoring or graphing. For the simpler formula where $a = 0,\quad x = \frac{-c}{b}$. ## Discriminant The number of x-intercepts, or solutions, can be determined by the value of the discriminant: $b^2 - 4ac$ If the value is positive, there are always two real solutions. If the value is zero, there is one solution. If it is negative, there are no real solutions but two complex conjugate solutions. ## Derivation by completing the square Given the quadratic equation $ax^2+bx+c=0$ Divide the equation by a: $x^2+\frac{b}{a}x+\frac{c}{a}=0$ Subtract the constant term c/a from both sides: $x^2+\frac{b}{a}x=-\frac{c}{a}$ Add the square of b/2a to both sides to get a perfect square at the left hand side: $x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}{a}$ Factor the left hand side and simplify the right hand side: $\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$ Take the square root of both sides: $x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}$ Solving for x gives the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ Community content is available under CC-BY-SA unless otherwise noted.
Singular Point: Regular and Irregular Examples Singular Point in Differential Equations A differential equation of the form y′′ + p(x)y′ + q(x)y = 0 has a singular point at x0 if either of the following limits do not exist [1]: What this means for second order differential equations is that an initial value problem will not have a unique solution. Alternatively, it may not have any solution, or its solution or derivatives might be discontinuous. For linear homogeneous differential equations, a singular point happens when at least one coefficient is either undefined (i.e. discontinuous) or multivalued at that point [2]. Regular and Irregular Singular Point Singular points can be regular or irregular. Regular singular points are well-behaved, defined in terms of ratios of the differential equation’s polynomial coefficients Q(x)/P(x) and R(x)/P(x) [3]. Irregular singular points exhibit bizarre behavior and cannot easily be pinned down or defined, other than to say that if a point isn’t regular, then it is irregular. To put this more concretely, a regular singular point can be defined as follows: It is where a singularity of P(x) is no worse than 1 / (x – x0) and the singularity of Q(x) is no worse than 1 / (x – x0)2. In other words, it’s where both of the following limits exist: Otherwise, the point is an irregular singular point. Examples of Regular and Irregular Singular Points 1. Example of a regular single point x0 [4]: Here, p2(x) is singular but xp0(x) = -1 is analytic* is x0 = 0 (and for all x). 2. Example of an irregular single point x0: Here, x0 = -1 is an irregular singular point because (x + 1)p1(x) is singular at x = -1. Example question: Are the singular points of (x3 − 3x2)y′′ + y′ + 2y = 0 regular or irregular? Step 1: Find the singular points. As every coefficient is a polynomial, the singular points (0 and 3) are the roots of the leading coefficient, x3 – 3x2. Step 2: Find the limits of each point. x = 0 is an irregular singular point because the limit is undefined: x = 0 is a regular singular point because both limits exist: *Most functions you come across are “analytic.” All polynomial functions, rational functions, exponential functions, logarithmic functions, and trigonometric functions are analytic away from their singularities. Singular Point in Complex Analysis A singular point, also called a singularity, is a point where a complex function isn’t analytic. In other words, it’s an obstacle to analytic continuation where the function can’t be expressed as an infinite series of powers of z. Singular points can be classified as regular points or irregular points (also called essential singularities). A singular point may be an isolated point, or a point on the curve (e.g. a cusp). If there aren’t any other singular points in the neighborhood of z, the point is called an isolated singularity. In some cases, you might be able to assign a value to the discontinuity to fill in the “gap”. If that’s the case, the point is called a removable singularity. References [1] Binegar, B. Lecture 19: Regular Singular Points and Generalized Power Series. Retrieved August 11, 2021 from: https://math.okstate.edu/people/binegar/4233/4233-l19.pdf [2] Dobrushkin, V. MATHEMATICA TUTORIAL for the First Course. Part V: Singular and ordinary points. Retrieved August 11, 2021 from: https://www.cfm.brown.edu/people/dobrush/am33/Mathematica/ch5/singular.html [3] 9.2. Classifying Singular Points as Regular or Irregular. Retrieved August 11, 2021 from: https://www.oreilly.com/library/view/differential-equations-workbook/9780470472019/9780470472019_classifying_singular_points_as_regular_o.html [4] Bertherton, C. Regular and Irregular Singular Points of ODEs. Retrieved August 11, 2021 from: https://atmos.washington.edu/~breth/classes/AM568/lect/lect15.pdf
Statistics # 4.4Geometric Distribution (Optional) Statistics4.4 Geometric Distribution (Optional) There are three main characteristics of a geometric experiment: 1. Repeating independent Bernoulli trials until a success is obtained. Recall that a Bernoulli trial is a binomial experiment with number of trials n = 1. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bull's-eye until you hit the bull's-eye. The first time you hit the bull's-eye is a success so you stop throwing the dart. It might take six tries until you hit the bull's-eye. You can think of the trials as failure, failure, failure, failure, failure, success, stop. 2. In theory, the number of trials could go on forever. There must be at least one trial. 3. The probability, p, of a success and the probability, q, of a failure do not change from trial to trial. p + q = 1 and q = 1 − p. For example, the probability of rolling a three when you throw one fair die is $1 6 1 6$. This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is q = $5 6 5 6$, the probability of a failure. The probability of getting a three on the fifth roll is $( 5 6 )( 5 6 )( 5 6 )( 5 6 )( 1 6 ) ( 5 6 )( 5 6 )( 5 6 )( 5 6 )( 1 6 )$ = .0804. X = the number of independent trials until the first success. p = the probability of a success, q = 1 – p = the probability of a failure. There are shortcut formulas for calculating mean μ, variance σ2, and standard deviation σ of a geometric probability distribution. The formulas are given as below. The deriving of these formulas will not be discussed in this book. $μ= 1 p , σ 2 =( 1 p )( 1 p −1),σ= ( 1 p )( 1 p −1) μ= 1 p , σ 2 =( 1 p )( 1 p −1),σ= ( 1 p )( 1 p −1)$ ## Example 4.16 Suppose a game has two outcomes, win or lose. You repeatedly play that game until you lose. The probability of losing is p = 0.57. If we let X = the number of games you play until you lose (includes the losing game), then X is a geometric random variable. All three characteristics are met. Each game you play is a Bernoulli trial, either win or lose. You would need to play at least one game before you stop. X takes on the values 1, 2, 3, . . . (could go on indefinitely). Since we are measuring the number of games you play until you lose, we define a success as losing a game and a failure as winning a game. The probability of a success $p=.57 p=.57$ and the probability of a failure q = 1 – p = 1 – 0.57 = 0.43. Both p and q remain the same from game to game. If we want to find the probability that it takes five games until you lose, then the probability could be written as P(x = 5). We will explain how to find a geometric probability later in this section. ## Try It 4.16 You throw darts at a board until you hit the center area. Your probability of hitting the center area is p = 0.17. You want to find the probability that it takes eight throws until you hit the center. What values does X take on? ## Example 4.17 A safety engineer feels that 35 percent of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions. If we let X = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions, then X is a geometric random variable. All three characteristics are met. Each accident report she reads is a Bernoulli trial: the accident was either caused by failure of employees to follow instructions or not. She would need to read at least one accident report before she stops. X takes on the values 1, 2, 3, . . . (could go on indefinitely). Since we are measuring the number of reports she needs to read until one that shows an accident caused by failure of employees to follow instructions, we define a success as an accident caused by failure of employees to follow instructions. If an accident was caused by another reason, the report is defined as a failure. The probability of a success p = .35 and the probability of a failure $q=1−p=1−.35=.65 q=1−p=1−.35=.65$. Both p and q remain the same from report to report. If we want to find the probability that the safety engineer will have to examine at least three reports until she finds a report showing an accident caused by employee failure to follow instructions, then the probability could be written as $p=.35 p=.35$. If we want to find how many reports, on average, the safety engineer would expect to look at until she finds a report showing an accident caused by employee failure to follow instructions, we need to find the expected value E(x). We will explain how to solve these questions later in this section. ## Try It 4.17 An instructor feels that 15 percent of students get below a C on their final exam. She decides to look at final exams (selected randomly and replaced in the pile after reading) until she finds one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least 10 exams until she finds one with a grade below a C. What is the probability question stated mathematically? ## Example 4.18 Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55 percent of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people? This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student). ### Problem a. Let X = the number of ________ you must ask ________ one says yes. ### Problem b. What values does X take on? ### Problem c. What are p and q? ### Problem d. The probability question is P(_______). ## Try It 4.18 You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10 percent of them carry the special ink. You randomly call each store until one has the ink you need. What are p and q? ## Notation for the Geometric: G = Geometric Probability Distribution Function X ~ G(p) Read this as X is a random variable with a geometric distribution. The parameter is p; p = the probability of a success for each trial. ## Example 4.19 Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective? Let X = the number of computer components tested until the first defect is found. X takes on the values 1, 2, 3, . . . where p = .02. X ~ G(.02) Find P(x = 7). There is a formula to define the probability of a geometric distribution $P(x) P(x)$. We can use the formula to find $P(x=7) P(x=7)$. But since the calculation is tedious and time consuming, people usually use a graphing calculator or software to get the answer. Using a graphing calculator, you can get $P(x=7)=.0177 P(x=7)=.0177$. The instruction of TI83, 83+, 84, 84+ is given below. ## Using the TI-83, 83+, 84, 84+ Calculator Go into 2nd DISTR. The syntax for the instructions are as follows: To calculate the probability of a value P(x = value), use geometpdf(p, number). Here geometpdf represents geometric probability density function. It is used to find the probability that a geometric random variable is equal to an exact value. p is the probability of a success and number is the value. To calculate the cumulative probability P(x ≤ value), use geometcdf(p, number). Here geometcdf represents geometric cumulative distribution function. It is used to determine the probability of “at most” type of problem, the probability that a geometric random variable is less than or equal to a value. p is the probability of a success and number is the value. To find $P(x=7) P(x=7)$, enter 2nd DISTR, arrow down to geometpdf(. Press ENTER. Enter .02,7). The result is $P(x=7)=.0177 P(x=7)=.0177$. If we need to find $P(x≤7) P(x≤7)$ enter 2nd DISTR, arrow down to geometcdf(. Press ENTER. Enter .02,7). The result is $(x≤=7)=.1319 (x≤=7)=.1319$. The graph of X ~ G(.02) is Figure 4.2 The previous probability distribution histogram gives all the probabilities of X. The x-axis of each bar is the value of X = the number of computer components tested until the first defect is found, and the height of that bar is the probability of that value occurring. For example, the x value of the first bar is 1 and the height of the first bar is 0.02. That means the probability that the first computer components tested is defective is .02. The expected value or mean of X is $E(X)=μ= 1 p = 1 .02 =50 E(X)=μ= 1 p = 1 .02 =50$. The variance of X is $σ 2 =( 1 p )( 1 p −1)=( 1 .02 )( 1 .02 −1)=(50)(49)=2,450 σ 2 =( 1 p )( 1 p −1)=( 1 .02 )( 1 .02 −1)=(50)(49)=2,450$ The standard deviation of X is $σ= σ 2 = 2,450 =49.5 σ= σ 2 = 2,450 =49.5$ Here is how we interpret the mean and standard deviation. The number of components that you would expect to test until you find the first defective one is 50 (which is the mean). And you expect that to vary by about 50 computer components (which is the standard deviation) on average. ## Try It 4.19 The probability of a defective steel rod is .01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer. ## Example 4.20 ### Problem The lifetime risk of developing pancreatic cancer is about one in 78 (1.28 percent). Let X = the number of people you ask until one says he or she has pancreatic cancer. Then X is a discrete random variable with a geometric distribution: X ~ G$( 1 78 ) ( 1 78 )$ or X ~ G(.0128). 1. What is the probability that you ask 10 people before one says he or she has pancreatic cancer? 2. What is the probability that you must ask 20 people? 3. Find the (i) mean and (ii) standard deviation of X. ## Try It 4.20 The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12 percent. Let X = the number of Afghani women you ask until one says that she is literate. 1. What is the probability distribution of X? 2. What is the probability that you ask five women before one says she is literate? 3. What is the probability that you must ask 10 women? 4. Find the (i) mean and (ii) standard deviation of X.
Next: Monoids and Groups Up: Fundamentals Previous: Fundamentals # Algebraic structures We will attempt to give a brief explanation of the following concepts: • N is a monoid • Z is an integral domain • Q is a field • in the field R the order is complete • the field C is algebraically complete If you have been asked by a child to give them arithmetic problems, so they could show off their newly learned skills in addition and subtraction I'm sure that after a few problems such as: 2 + 3, 9 - 5, 10 + 2 and 6 - 4, you tried tossing them something a little more difficult: 4 - 7 only to be told ``That's not allowed.'' What you may not have realized is that you and the child did not just have different objects in mind (negative numbers) but entirely different algebraic systems. In other words a set of objects (they could be natural numbers, integers or reals) and a set of operations, or rules regarding how the numbers can be combined. We will take a very informal tour of some algebraic systems, but before we define some of the terms, let us build a structure which will have some necessary properties for examples and counterexamples that will help us clarify some of the definitions. We know that any number that is divided by six will either leave a remainder, or will be divided exactly (which is after all the remainder 0). Let us write any number by the remainder n it leaves after division by six, denoting it as [ n ]. This means that, 7, 55 and 1 will all be written [1], which we call the class to which they all belong: i.e. , , or, a bit more technically, they are all equivalent to 1 modulo 6. The complete set of class will contain six elements, and this is called partitioning numbers into equivalent classes because it separates (or partitions) all of our numbers into these classes, and any one number in a class is equivalent to any other in the same class. One interesting thing we can do with these classes is to try to add or to multiply them. What can [1] + [3] mean? We can, rather naively try out what they mean in ``normal'' arithmetic: [1] + [3] = [1+3] = [4]. So far so good, let us try a second example and , their sum is 70 which certainly belongs to [4]. Here we see what we meant above by equivalence, 25 is equivalent to 1 as far as this addition is concerned. Of course this is just one example, but fortunately it can be proven that the sum of two classes is always the class of the sums. Now this is the kind of thing we all do when we add hours for example, 7 (o' clock) plus 6 hours is 1 (o' clock), and all we are really doing is adding hours (modulo 12). The neat part comes with multiplication, as we will see later on. But for now just remember, it can be proven that something like will work: the product of two classes is the class of the product. Now for some of the necessary terminology. Next: Monoids and Groups Up: Fundamentals Previous: Fundamentals Alex Lopez-Ortiz Mon Feb 23 16:26:48 EST 1998
# 3.5 divided by 4 #### Understand the Problem The question is asking for the result of dividing the number 3.5 by 4, which is a straightforward mathematical calculation. The result of dividing 3.5 by 4 is $$0.875$$. The answer is ( 0.875 ) or ( \frac{7}{8} ). #### Steps to Solve 1. Set up the division problem We begin by writing down the division problem we need to solve: $$\frac{3.5}{4}$$ 2. Perform the division Next, we divide 3.5 by 4. This can be done by converting 3.5 into a fraction: $$\frac{35}{10}$$ Now we can rewrite the division as: $$\frac{35}{10} \div 4$$ This can be further expressed as: $$\frac{35}{10} \times \frac{1}{4}$$ 3. Multiply the fractions Now, we multiply the numerators and denominators: $$\frac{35 \times 1}{10 \times 4} = \frac{35}{40}$$ 4. Simplify the fraction We can simplify the fraction $\frac{35}{40}$ by finding the greatest common divisor (GCD) of 35 and 40, which is 5. Thus: $$\frac{35 \div 5}{40 \div 5} = \frac{7}{8}$$ 5. Convert to decimal (if needed) If we want the decimal form, we can divide 7 by 8: $$7 \div 8 = 0.875$$ The answer is ( 0.875 ) or ( \frac{7}{8} ).
### Algebra-Solutions Ex-11.2 CBSE Class –VI Mathematics NCERT Solutions Chaper 11 Algebra (Ex. 11.2) Question 1. The side of an equilateral triangle is shown by $l.$ Express the perimeter of the equilateral triangle using $l.$ Answer: Side of equilateral triangle = $l$ Therefore, Perimeter of equilateral triangle = 3 x side = $3l$ Question 2. The side of a regular hexagon is denoted by $l.$ Express the perimeter of the hexagon using $l.$ (Hint: A regular hexagon has all its six sides in equal length) Answer: Side of hexagon = $l$ Therefore, Perimeter of Hexagon = 6 x side = $6l$ Question 3. A cube is a three-dimensional figure. It has six faces and all of them are identical squares. The length of an edge of the cube is given by $l.$ find the formula for the total length of the edges of a cube. Answer: Length of one edge of cube = $l$ Number of edges in a cube = 12 Therefore, total length = $12×l=12l$ Question 4. The diameter of a circle is a line, which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure AB is a diameter of the circle; C is its centre). Express the diameter of the circle $\left(d\right)$ in terms of its radius $\left(r\right).$ Therefore, $d=2r$ (Here r is the radius of the circle) This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on Whole Numebrs, in a general way, by using variables $a,b$ and $c.$
# Difference between revisions of "2011 AMC 12B Problems/Problem 25" ## Problem For every $m$ and $k$ integers with $k$ odd, denote by $\left[\frac{m}{k}\right]$ the integer closest to $\frac{m}{k}$. For every odd integer $k$, let $P(k)$ be the probability that $$\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]$$ for an integer $n$ randomly chosen from the interval $1 \leq n \leq 99!$. What is the minimum possible value of $P(k)$ over the odd integers $k$ in the interval $1 \leq k \leq 99$? $\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{50}{99} \qquad \textbf{(C)}\ \frac{44}{87} \qquad \textbf{(D)}\ \frac{34}{67} \qquad \textbf{(E)}\ \frac{7}{13}$ ## Solution Answer: $(D) \frac{34}{67}$ First of all, you have to realize that if $\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]$ then $\left[\frac{n - k}{k}\right] + \left[\frac{100 - (n - k)}{k}\right] = \left[\frac{100}{k}\right]$ So, we can consider what happen in $1\le n \le k$ and it will repeat. Also since range of $n$ is $1$ to $99!$, it is always a multiple of $k$. So we can just consider $P(k)$ for $1\le n \le k$. LET $\text{fpart}(x)$ be the fractional part function This is an AMC exam, so use the given choices wisely. With the given choices, and the previous explanation, we only need to consider $k = 99$, $87$, $67$, $13$. $1\le n \le k$ For $k > \frac{200}{3}$, $\left[\frac{100}{k}\right] = 1$. 3 of the $k$ that should consider lands in here. For $n < \frac{k}{2}$, $\left[\frac{n}{k}\right] = 0$, then we need $\left[\frac{100 - n}{k}\right] = 1$ else for $\frac{k}{2}< n < k$, $\left[\frac{n}{k}\right] = 1$, then we need $\left[\frac{100 - n}{k}\right] = 0$ For $n < \frac{k}{2}$, $\left[\frac{100 - n}{k}\right] = \left[\frac{100}{k} - \frac{n}{k}\right]= 1$ So, for the condition to be true, $100 - n > \frac{k}{2}$ . ( $k > \frac{200}{3}$, no worry for the rounding to be $> 1$) $100 > k > \frac{k}{2} + n$, so this is always true. For $\frac{k}{2}< n < k$, $\left[\frac{100 - n}{k}\right] = 0$, so we want $100 - n < \frac{k}{2}$, or $100 < \frac{k}{2} + n$ $100 <\frac{k}{2} + n < \frac{3k}{2}$ For k = 67, $67 > n > 100 - \frac{67}{2} = 66.5$ For k = 69, $69 > n > 100 - \frac{69}{2} = 67.5$ etc. We can clearly see that for this case, $k = 67$ has the minimum $P(k)$, which is $\frac{34}{67}$. Also, $\frac{7}{13} > \frac{34}{67}$ . So for AMC purpose, answer is (D). Now, let's say we are not given any answer, we need to consider $k < \frac{200}{3}$. I claim that $P(k) \ge \frac{1}{2} + \frac{1}{2k}$ If $\left[\frac{100}{k}\right]$ got round down, then $1 \le n \le \frac{k}{2}$ all satisfy the condition along with $n = k$ because if $\text{fpart}\left(\frac{100}{k}\right) < \frac{1}{2}$ and $\text{fpart} \left(\frac{n}{k}\right) < \frac{1}{2}$, so must $\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}$ and for $n = k$, it is the same as $n = 0$. , which makes $P(k) \ge \frac{1}{2} + \frac{1}{2k}$. If $\left[\frac{100}{k}\right]$ got round up, then $\frac{k}{2} \le n \le k$ all satisfy the condition along with $n = 1$ because if $\text{fpart}\left(\frac{100}{k}\right) > \frac{1}{2}$ and $\text{fpart} \left(\frac{n}{k}\right) > \frac{1}{2}$ Case 1) $\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}$ -> $\text{fpart}\left(\frac{100}{k}\right) = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)$ Case 2) $\text{fpart} \left(\frac{100 - n}{k}\right) > \frac{1}{2}$ -> $\text{fpart}\left(\frac{100}{k}\right) + 1 = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)$ and for $n = 1$, since $k$ is odd, $\left[\frac{99}{k}\right] \neq \left[\frac{100}{k}\right]$ -> $99.5 = k (p + .5)$ -> $199 = k (2p + 1)$, and $199$ is prime so $k = 1$ or $k =199$, which is not in this set , which makes $P(k) \ge \frac{1}{2} + \frac{1}{2k}$. Now the only case without rounding, $k = 1$. It must be true. $blacksquare$ 2011 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 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 AMC 12 Problems and Solutions
Study Guide # The Fundamental Theorem of Calculus - Using the FTC to Evaluate Integrals ## Using the FTC to Evaluate Integrals If f is the derivative of F, then we call F an antiderivative of f. We already know how to find antiderivatives–we just didn't tell you that's what they're called. It's like when you realize what all of the subtle signs in the M. Night Shyamalan movie mean. Seriously, like whoa. Whenever we're given a derivative and we "think backwards" to find a possible original function, we're finding an antiderivative. ### Sample Problem Let f(x) = 3x2. Find an antiderivative of f. We think backwards: what could we take the derivative of to get 3x2? This derivative looks like it came from the power rule, so the original function must involve x3. Since the derivative of x3 is 3x2, the function F(x) = x3 is an antiderivative of f(x) = 3x2. Any other antiderivative of 3x2 will have the form x3 + C where C is a constant. We generally take C = 0. For the FTC it won't matter which antiderivative we use, so we might as well use the simplest one. These exercises should be mostly review, and help you remember how thinking backwards works. You might want to review the rules for taking derivatives first. To check an answer for this sort of problem, take the derivative of your answer. If you take the derivative of your answer F and get the f given in the problem, then F is an antiderivative of f and you did the problem correctly. Gold stars all around. Now that we know what antiderivatives are, we can use them along with the FTC to evaluate some integrals we didn't know how to evaluate before. The FTC says that if f is continuous on [a, b] and is the derivative of F, then This means if we want to know , we 1) find an antiderivative F of f, 2) evaluate F at the limits of integration, and 3) subtract to find F(b) – F(a). When evaluating definite integrals for practice, you can use your calculator to check the answers. If you don't know how to use your calculator to find integrals you can look in the manual, look online, ask a friend, or ask your teacher. But practice doing integrals by hand until they're so easy you don't even mind anymore. Here are some reasons to practice doing integrals by hand. 1) At some point you'll probably need to pass a test involving integration, without being allowed to have a calculator. Midterm, anyone? 2) Even when you are allowed a calculator, your teacher will probably want to see the steps you took to get your answer. 3) If you're asked to integrate something that uses letters instead of numbers, the calculator won't help much (some of the fancier calculators will, but see the first two points). 4) Later in Calculus you'll start running into problems that expect you to find an integral first and then do other things with it. It will sometimes be easier to find the integral by hand than it will be to distract yourself by putting the integral into your calculator. Before asking you to find too many definite integrals, we should share a nice notational shortcut. To abbreviate F(b) – F(a), write This expression is read "F of x evaluated from a to b." Using this shortcut, our work to find would look like this: This is a nice shortcut because it saves us from having to mess with letters like f and F. We just found the antiderivative x3 put it in brackets [x3] drew a vertical line on the right-hand side and wrote the limits of integration Then expand the shortcut [(2)3] – [(0)3] and simplify to get the answer. Remember that when we expand the shortcut, we use the upper limit of integration first: • ### Integrating with Letters No, we're not talking about snail mail. If we have a constant instead of a number for a limit of integration, not much changes. We apply the FTC, and write a constant instead of a number where it's appropriate to do so. ### Sample Problem If c is a constant greater than 0, find . We apply the FTC like always, but use c for the upper limit of integration instead of a number. Now we can answer questions like this: ### Sample Problem If , what is c? We just worked out that So if the integral is equal to 2, it means Solving, we get c3 = 6 When there's a constant in the integrand, you have to take it into account while finding the antiderivative. If there's a constant in the integrand, that constant will also show up in the antiderivative. A word of warning: when there are constants in the integrand, it can be easy to get mixed up when it comes time to put in the limits of integration. Do they plug into the a or the x? The answer is that the limits of integration plug into the variable of integration. In this example, the limits of integration (1 and 5) went into the variable of integration x, not into the constant a. Be Careful: Plug the limits of integration into the variable of integration. If there are constants in the integrand, leave those alone. DO NOT plug the limits of integration into any constants. • ### Order of Limits of Integration Integrals like to flip-flop on their stance from time to time. Seriously, they're as bad as politicians sometimes. Sometimes you think they're left, sometimes you think they're right, sometime the upper limit is smaller than the lower limit... When we originally stated the FTC we said that if f is continuous on [a, b], then where F ' = f. We can still evaluate integrals this way if the upper limit of integration is smaller than the lower limit. Suppose this is the case, so b < a. By properties of integrals, Since b < a we can use the FTC to say Then Practically speaking, this means you can evaluate integrals without worrying which limit of integration is bigger. The integrand should still be continuous on the interval between the limits of integration, though. • ### Why the Choice of Antiderivative Doesn't Matter Some say po-TAY-to, some say po-TAH-to. Some say to-MAY-to, some say to-MAH-to. Some choose to not include a constant when finding an antiderivative. Some do include the constant. Here's why it doesn't matter what we do with the constant. ### Sample Problem Find . The simplest antiderivative of 4x3 is x4. Using the FTC with that antiderivative, we get Now let's try the FTC with a different antiderivative. How about x4 + 3? Notice how the extra "+ 3"s canceled each other out and we got 15 again. If we used some other antiderivative of 4x3, the same sort of thing would happen. The moral of the story is that when evaluating a definite integral with the FTC, no matter which antiderivative you use, you should get the same answer every time. Since it doesn't matter which antiderivative you use, you may as well use the simplest one. • ### Average Values The average value of the function f on the interval [a,b] is the integral of the function on that interval divided by the length of the interval. Since we know how to find the exact values of a lot of definite integrals now, we can also find a lot of exact average values. What's the average value of an "A" in Calculus class? You tell us. ### Sample Problem Find the average value of f(x) = sin x on the interval . The average value of f(x) = sin x on this interval is Since we know how to evaluate the integral, we know how to find the average value. First let's simplify that stuff out in front of the integral: Now we can rewrite the average value to be a little more tidy. It's tempting to go off and compute the integral in a corner of your paper, then come back and multiply by at the end. Unfortunately, that's dangerous. After working out a long integral, it's very easy to forget to come back and do that last step. Don't do it.
Video: Comparing Objects Up To 10 Using Symbols In this video, we will learn how to compare numbers up to 10 by using the symbols for “less than,” “greater than,” and “equal to.” 03:44 Video Transcript Comparing Objects up to 10 Using Symbols. In this video, we’re going to use these symbols less than, greater than, or equal to to help us compare groups of objects. Here we have two groups of ones bricks. We’re going to compare them. The first thing we need to do is to count how many are in each group. Let’s start by counting this group: one, two, three. Now let’s count the second group. This group has one block. The group with three blocks has the most, and this group has the least number of blocks. Now we can use one of the symbols less than, greater than, or equal to to compare these two groups of objects. We can think of this symbol like a mouth. It’s open, always hungry, ready to eat the largest amount. This group has the most number of bricks, so the mouth is facing towards the greatest number or the largest number. To read this number sentence, we would say three is greater than one. Let’s practice with an example question. Is 10 less than, equal to, or greater than three? Here we have a group of 10 counters and another group with three counters. Which has the most? The group with 10 counters has the most or the greatest number of counters. The open end of the symbol always points towards the greatest number. The number sentence tells us that 10 is greater than three. The correct symbol to use in this sentence is greater than. Let’s try another example. This question asks us is 10 less than, equal to, or greater than 10. Here we have two groups, which both have 10 blocks. Both groups have an equal amount. So our sentence reads 10 is equal to 10. Let’s try one last example. Is one less than, equal to, or greater than three? In this question, we have to compare one and three. So we have a group with one brick and a group with three bricks. The group with three bricks has the greatest number, and we know that the open end of the symbol always points towards the greatest number. So our number sentence reads one is less than three. The correct symbol in this sentence is less than. In this video, we’ve learned how to compare groups of objects using the symbols less than, greater than, and equal to.
# What Is A Dividend? Get Our Analysis, Insights Event Invites & More By using the above form you agree with the storage and handling of your data by this website. Addition, subtraction, multiplication, and division are the most basic math operations. A number can be divided through repeated subtraction. Example, the result of 12 divided by 3 can be obtained by repeatedly subtracting 3 from 12 till you get zero. 3 had to be subtracted consecutively 4 times to get the result. i.e. 12 divided 3 = 4. It can be termed through the following expression Dividend divided by divisor = quotient The dividend is the number that is divided, the number that divides the divided is the divisor and the quotient is the result. Simply put if you take a fraction 5/6, 4 is the dividend and also the numerator of a fraction and 6 is the divisor and is the denominator of the fraction. The result of this is a decimal value and is the quotient. To learn the division facts and to learn how to divide large dividends it is essential to learn 2 digit dividends with 1 digit divisors through long division. How to Find a Dividend? Division of a number can be done using the below formula Dividend / divisor = quotient with dividend being the number that you divide up. The above formula can be used to find the dividend. Dividend – quotient multiplied by divisor. Ex: x/30 = 5 Using the formula Dividend/Divisor = quotient where x is the dividend Dividend = quotient × divisor So x = 30 × 5 = 150 If there is a reminder: Dividend / Divisor = Dividend or Quotient ÷ Divisor = Quotient Reminder – The number left over when one number is divided by another number: The remainder added to the product of the quotient multiplied by the divisor equals the dividend. If there is a remainder, then Dividend = Quotient x Divisor + Remainder Ex 1: 55/9 = 6 and 1 Where 55 is the dividend and 9 is the divisor. The quotient is 6 and 1 is the reminder. Ex 2: x/7 = 3 and 1 is the reminder. Where 7 is the divisor, 3 is the quotient and 1 is the reminder. To find the dividend use Dividend = Quotient X Divisor + Remainder = 3 X 7+ 1 = 22. So the dividend is 22 ##### Related Articles Division Notations The division notation has a bearing on the location of the dividend and the divisor. When “÷” or “/” is used to represent division, the dividend is on the left and the divisor is on the right of the symbol. Ex: 21/7 where 21 is the numerator and also the dividend and 7 is the denominator and the divisor. But if the problem has to be solved using long division then the location of the dividend and divisor are revised. The divisor appears to the right and the dividend to the left or below the division bracket. The divisor, dividend, quotient, and remainder will help you to verify the answer that you get from a division. If any reminder is left, then add that to the product of quotient and divisor, the sum of it is the dividend.
Describing Triangles Say a question describes a triangle as ‘triangle ABC’, with a right angle at B. How do you draw a diagram of this triangle? First of all, you should realise that you’ve only been given a little bit of information in the question, so different people will probably draw different triangles. But they must have a few things in common: · The triangles must have 3 sides and 3 corners. The corners are what the ‘ABC’ letters represent – one corner is labelled ‘A’, one is labelled ‘B’ and one is labelled ‘C’. · You need to have a right angle in the ‘B’ corner. So that’s all the restrictions there are on what triangle you draw. This means that you could draw any of these triangles and it would be fine: Usually it’s easiest to draw one which is sort of in a ‘normal’ orientation, like one of the top two, which have one horizontal and one vertical side. Specifying the hypotenuse side Sometimes you’ll be asked to work out which side in a right-angled triangle is the hypotenuse. It is always the side opposite the right angle. So for the triangles in the previous diagram, side AC is the hypotenuse because it is the side opposite the 90° angle in corner B. Checking whether a triangle is right-angled Sometimes you’ll get given a triangle and be asked to use Pythagoras’ Theorem to check whether the triangle is a right-angled triangle. How do you do this? Well, you take the side lengths and put them into the equation, and if the equation works, this means it is a right-angled triangle. Often the question will have the triangle drawn as if it is a right-angled triangle even when it isn’t, just to try and trick you. Don’t be fooled! Take this triangle for instance: How do we test this triangle using Pythagoras’ Theorem? Well, the theorem is that: Remember that ‘a’ is always the longest side in the triangle, and ‘b’ and ‘c’ are the two other sides. So what we can do is calculate what each side of this equation equals for our triangle, and then see if the two sides of the equation equal each other. If they do, then it is a right-angled triangle. First, the left hand side: The longest side in our triangle is ‘7’. So this becomes: Now we can do the right hand side of the equation: Now, if the triangle is right-angled, the equation’s left hand side has to equal the right hand side. However, for our triangle, our L.H.S. works out to 49, but our R.H.S. is only 41, so they’re not equal. This means our triangle is not a right-angled triangle. In your solution you’d probably want to write something like this: If a = 7, b = 4 and c = 5 then for this triangle, which means that the triangle is not a right-angled triangle. Proving Pythagoras’ Theorem There are lots and lots of ways of proving that Pythagoras’ Theorem is correct for any right-angled triangle. Most of them involve playing with triangles and re-arranging them around to form more complicated shapes. Here’s one of the simpler proofs. It’s good to understand how a theorem is provenunderstanding how something works is always much better than just memorising the theorem. Especially in an exam under pressure – if you understand it you’re less likely to forget it than if you have just memorised some formula. So on to the proof, what we need to do is start with 4 identical right-angled triangles. It doesn’t matter what shape they are as long as they are right-angled, and as long as each of the 4 are identical to each other. Easy so far right? Now, this is the tricky bit. When mathematicians prove anything in mathematics, it generally involves playing around with a whole lot of things before they finally work out a good proof. It’s rare that they get it first go. So for this proof, people had to play around with arranging these triangles in lots of different layouts trying to find a layout which would help them prove the theorem. Finally, they came up with this arrangement: The four triangles are arranged to form a sort of boundary around a central space. The central space is a square. We can label the side lengths in the diagram with letters, using ‘a’ to represent the hypotenuse of the triangles, and ‘b’ and ‘c’ to represent the other sides: To prove theorems in mathematics, you often have to use some stuff you already know to help you prove it. For instance, in this case, the diagram has triangles and squares in it. We already know how to work out the areas of these triangles and squares. We can use our knowledge of areas to help us prove Pythagoras’ Theorem. Now, there are actually two squares in this diagram. There’s the square between all the triangles, but there’s also the larger square formed by the outer boundary of all the shapes: We can write a relationship involving the area of the larger square, the area of the smaller square, and the areas of the 4 triangles. What do you think the area of the triangles is equal to? Well, it’s equal to the area of the large square minus the area of the smaller square. You could even write an equation using the shapes like this: We can work out what each of these areas is and write a proper equation with letters instead of shapes. Area of the four triangles OK, what’s the area of the four triangles? Well, how about one triangle first? We know that the area of a triangle is one half times the base times the height: For our triangle, let’s use ‘b’ as the base, and ‘c’ as the height: But we’ve got four triangles, so their total area is: Area of large square The area of the large square is easy to find, it’s just the side length squared. What is its side length? Well, a side of the big square looks like this: So its length is ‘b + c’. Area of small square It’s a square, so it’s also easy to find out the area for. What is its side length? The side length is just ‘a’, shown in the diagram: Putting it all together So now we’ve got the three areas, we can put them into our equation: Now when you’re trying to prove something, you’ve got to always keep in mind what you’re trying to get to in the end. In our case, we’re trying to get to . We’ve already got an equation with ‘a’s, ‘b’s and ‘c’s in it. We’ve just got to try and rearrange it so that it ends up looking like . So let’s start: First let’s multiply out those brackets: Now we’ve got a ‘2bc’ term on both sides – these will cancel out, leaving us with: One more step to get it into the final form, let’s add ‘a2’ to both sides: Yeeha! We’ve got it into the exact same form as Pythagoras’ Theorem. Let’s check that the ‘a’s, ‘b’s and ‘c’s correspond to the right sides. If this is exactly the same as Pythagoras’ Theorem, then the ‘a’ should represent the longest side of the triangles, the hypotenuse. Check the diagrams – it does! And ‘b’ and ‘c’ represent the other sides of the triangles, like in Pythagoras’ Theorem. So by using some simple knowledge of how areas are calculated, we have managed to show how you get Pythagoras’ Theorem. This is called deriving a proof – working through some calculations to end up with the equation or theorem you’re trying to prove. Using Pythagoras’ Theorem to work out side lengths Okay, so a very common question you’ll get is a triangle with only two of its three sides labelled with a length, and you’ll have to work out the length of the third side. Often the unknown third side’s length will be labelled ‘x’ or ‘a’ or something else. So you’ll be asked something like, “Find x.” There are two main situations you’ll encounter: Find x: Solution In this case, all you do is take your theorem, , and plug the two smaller sides (5 and 3) in as ‘b’ and ‘c’. Also, our hypotenuse is labelled ‘x’, so instead of ‘a’ we’ll write x: Some teachers might like you to do something like this before you start calculating as well, just to tell them exactly what you’re doing: The other type of question gives you a triangle with a known hypotenuse, but an unknown other side: Find x: Solution In this case, you start with the theorem, but this time you know what ‘a’, the hypotenuse is – it’s ‘9’ in this case. The other side you know can go in as either ‘b’ or ‘c’. If the ‘7’ goes in as ‘b’, then ‘x’ is going to be ‘c’. If the ‘7’ goes in as ‘c’, then the ‘x’ is going to be ‘b’. In this case, let’s put ‘7’ in as ‘b’: Significant figures and diagrams How many significant figures should you write your answer to? Well, if the question asks for a specific number of significant figures, then just go by the question. For instance, it may ask for two significant figures, in which case your answer for the last question would be: However, sometimes the question won’t tell you how many significant figures are required. In this sort of situation, you need to look at how many significant figures are given in the diagram. ‘9’ and ‘7’ are numbers with only one significant figure. This would suggest you give your answer to only one significant figure: However, you’d actually have a reasonable chance of guessing an answer to this sort of accuracy. To show the teacher that you’ve actually calculated the answer properly, I usually give two significant figures or the number of significant figures presented in the diagram, whichever is greater. So in this case, although there’s only one significant figure in the diagram numbers, I’d still give two significant figures in my answer: It’s often good to make an estimate of what you think the answer will be, which you can compare with the answer you actually calculate. When you’re working with triangles and Pythagoras’ Theorem, you can use estimates to help check your answers. There are always a few simple things you can check: - What side length are you trying to find? If you’re trying to find the length of the hypotenuse, then you should get an answer which is larger than either of the other two sides. If you’re trying to find one of the other sides, then you should get an answer which is smaller than the length of the hypotenuse. - If you’re trying to find the length of the hypotenuse, it will always be smaller than the sum of the other two sides. - Same goes for the reverse – if you’re trying to find the length of a side which isn’t the hypotenuse, then its length plus the other short side should add up to more than the length of the hypotenuse. Say we have a right-angled triangle like this one, and we need to find ‘a’: First of all, which side is ‘a’ – is it the hypotenuse? It’s opposite the corner with the 90 degree angle, so this means that it is the hypotenuse, the longest side in the triangle. So first up, we should expect ‘a’ to be larger than either ‘5’ or ‘9’, so larger than ‘9’ in other words. Secondly, the length of the hypotenuse will always be smaller than the sum of the other two sides. 5 plus 9 is 14, so we expect our hypotenuse length to be smaller than 14. This means that we should expect our answer for the length of the hypotenuse to be larger than 9, but smaller than 14. If you use the actual theorem, you can work out your answer: Does this check with what we estimated? Well, 10.3 is definitely larger than 9, and it’s also smaller than 14. This tells us that our answer is a ‘reasonable’ one and gives us some confidence that we’ve got it right. More complicated Pythagoras’ Theorem questions Some of the more complicated questions don’t just have triangles in them, but more complicated shapes as well. For instance, you might get triangles within other triangles, or other shapes like squares or rectangles. Here’s an example of a slightly harder question you might get: Pythagorean Triples You may notice that most of the time, when you use Pythagoras’ Theorem to calculate the length of a side, you get an answer which isn’t a whole number – it has decimal places. However, there are a few right-angled triangles where the lengths of all the sides are whole numbers. The sets of three side lengths for these triangles are known as Pythagorean Triples or Pythagorean Triads. The simplest triple is a triangle with side lengths 3, 4 and 5: You can check that this works in Pythagoras’ Theorem: Now, you can scale the size of the triangle and its sides will still form a Pythagorean Triple. For instance, I could make this triangle twice as big: The only way to actually work out all the different triangles that make up the Pythagorean Triples is to use trial and error and go through lots of possible combinations of side lengths. Because triangles of this type often turn up on exams, it’s worth your while to memorise some of the most common triples: 3, 4, 5 5, 12, 13 12, 16, 20 8, 15, 17 If you know these off by heart often in a question you can save yourself any calculating if you recognise that two of the sides form part of a Pythagorean Triple. For instance, check out this triangle: ‘16’ and ‘20’ form part of the Pythagorean Triple 12, 16, 20. This means that without having to do any calculation at all, we can work out that ‘u’ is going to equal 12. Be careful though that you’re using the right bits of the Pythagorean Triple – if 20 hadn’t been the hypotenuse side, we wouldn’t have been able to use that triple, since 20 is the largest of the three numbers in the triple. Answering using a square root sign Some questions will ask you to give the length of the unknown triangle side exactly. Say you get down to the point where you’re working out the length of the unknown side ‘x’: We could find the square root of 26, but this gives us an endless number of decimal places, and our calculator can only display 9 of them: This is not an exact answer, it’s very accurate, but it’s not exact. The only way you can give the answer exactly is to leave the answer in the form before you actually calculated the square root: Some questions may want you to answer using this exact form, others may want you to write a decimal number. If the question doesn’t make it clear, you can always do both: Or using the ‘approximately equals’ symbol: This may seem impossible to do – try calculating what the square root of 29 is using your calculator. You should get something like this: 5.385164807 Although this answer is very accurate, it’s not exactly right – you could plug this question into a computer and you might be able to get the answer with say 500 decimal places instead of just 9. But the answer would still not be exactly correct, because the square root of 29 is a number with an endless number of decimal places. So how can you draw a line that is exactly units long? Well, since we’re in a section on Pythagoras’ Theorem, this should tip you off about how to do it. Notice how the line we want to draw is the square root of 29 long. For right-angled triangles, when we go to find the length of the longest side, the hypotenuse, we usually square the lengths of the other two sides, and then take the square root of that. So if we can draw a right-angled triangle where the squares of the two short sides add up to 29, the hypotenuse side will be exactly units long. So we need to think of two numbers, that when squared, add up to 29: If , then we’ll know that the length of side a is exactly units long. So what are two numbers that when squared, add up to 29? This is where you need to try a few different numbers: What if ‘c’ was ‘1’, then: This isn’t much good – now we have an inexact value for the length of ‘b’, which means we can’t draw the ‘b’ line exactly the right length. What about if ‘c’ was ‘2: Bingo! If we have c = 2 and b = 5, then together their squares add up to 29. So this is how we can go about drawing a line which is exactly units long. Draw two connected lines at right angles to each other, with one line 2 units long and one line 5 units long: Now, according to the mighty Pythagoras’ Theorem, if we draw a straight line to complete the triangle, the length of that line will equal the square root of the squares of the other two sides: So once we’ve drawn the third side in, the length of ‘a’ can be calculated like this: So the hypotenuse of this triangle is a line exactly units long. It’s good to remember this procedure just in case a question comes up in an exam asking you to draw a length like this exactly. Of course, because you’re not perfect with your pen or pencil and because you can’t measure exactly with your ruler, it’s never going to be perfect, plus you probably won’t be able to draw exactly a 90 degree angle in the triangle. But this is at least a theoretical way you could draw the line exactly the right length. Practical Pythagoras’ Theorem questions There are lots of types of questions involving Pythagoras’ Theorem and real life situations which you can get on an exam. Jason and Bob are on a reality TV show. They’re the final two contestants, and it’s come down to the final contest. They start at the corner of a large, rectangular, artificial lake. At the far corner of the lake, there is a 1 million dollar prize. Jason is only allowed to swim. Bob is only allowed to jump - to slow him down he’s had his legs tied together. How much faster will Bob have to jump in order to get to the money before Jason does? Solution First things first, we need to work out what path each person is going to take. Jason’s only allowed to swim, so he’s most likely going to take the most direct route to the other corner. Bob has to stay on land, so he’s going to have to go around the boundary of the lake. He can go either way round, they’re both the same length: So the length of Bob’s total journey is easy to calculate – we just need to add up the two side lengths: For Jason’s route we need Pythagoras’ Theorem (surprise surprise): So obviously, Jason’s route is shorter because he has the most direct path to the money. The question is asking how much faster than Jason Bob is going to have to cover his distance, to get to the money first. If Bob had twice as far to travel, then he’d have to travel just over twice as fast to get there first. In this case, Bob has times as far to travel as Jason, or 1.3 times as far. So in order to get to the money first, Bob is going to have to jump his way to the money at just over 1.3 times as fast a speed as Jason swims.
How to Calculate the Radius of a Circle Download Article Download Article The radius of a circle is the distance from the center of the circle to any point on its circumference.[1] The easiest way to find the radius is by dividing the diameter in half. If you don’t know the diameter but you know other measurements, such as the circle’s circumference (${\displaystyle C=2\pi r}$) or area (${\displaystyle A=\pi r^{2}}$), you can still find the radius by using the formulas and isolating the ${\displaystyle r}$ variable. Method 1 Method 1 of 4: Using the Circumference 1. 1 Write down the circumference formula. The formula is ${\displaystyle C=2\pi r}$ , where ${\displaystyle C}$ equals the circle’s circumference, and ${\displaystyle r}$ equals its radii[2] • The symbol ${\displaystyle \pi }$ ("pi") is a special number, roughly equal to 3.14. You can either use that estimate (3.14) in calculations, or use the ${\displaystyle \pi }$ symbol on a calculator. 2. 2 Solve for r. Use algebra to change the circumference formula until r (radius) is alone on one side of the equation: Example ${\displaystyle C=2\pi r}$ ${\displaystyle {\frac {C}{2\pi }}={\frac {2\pi r}{2\pi }}}$ ${\displaystyle {\frac {C}{2\pi }}=r}$ ${\displaystyle r={\frac {C}{2\pi }}}$ Advertisement 3. 3 Plug the circumference into the formula. Whenever a math problem tells you the circumference C of a circle, you can use this equation to find the radius r. Replace C in the equation with the circumference of the circle in your problem: Example If the circumference is 15 centimeters, your formula will look like this: ${\displaystyle r={\frac {15}{2\pi }}}$ centimeters 4. 4 Round to a decimal answer. Enter your result in a calculator with the ${\displaystyle \pi }$ button and round the result. If you don't have a calculator, calculate it by hand, using 3.14 as a close estimate for ${\displaystyle \pi }$. Example ${\displaystyle r={\frac {15}{2\pi }}=}$ about ${\displaystyle {\frac {7.5}{2*3.14}}=}$ approximately 2.39 centimeters 5. Advertisement Method 2 Method 2 of 4: Using the Area 1. 1 Set up the formula for the area of a circle. The formula is ${\displaystyle A=\pi r^{2}}$ , where ${\displaystyle A}$ equals the area of the circle, and ${\displaystyle r}$ equals the radius.[3] 2. 2 Solve for the radius. Use algebra to get the radius r alone on one side of the equation: Example Divide both sides by ${\displaystyle \pi }$: ${\displaystyle A=\pi r^{2}}$ ${\displaystyle {\frac {A}{\pi }}=r^{2}}$ Take the square root of both sides: ${\displaystyle {\sqrt {\frac {A}{\pi }}}=r}$ ${\displaystyle r={\sqrt {\frac {A}{\pi }}}}$ 3. 3 Plug the area into the formula. Use this formula to find the radius when the problem tells you the area of the circle. Substitute the area of the circle for the variable ${\displaystyle A}$. Example If the area of the circle is 21 square centimeters, the formula will look like this: ${\displaystyle r={\sqrt {\frac {21}{\pi }}}}$ 4. 4 Divide the area by . Begin solving the problem by simplifying the portion under the square root (${\displaystyle {\frac {A}{\pi }})}$. Use a calculator with a ${\displaystyle \pi }$ key if possible. If you don't have a calculator, use 3.14 as an estimate for ${\displaystyle \pi }$. Example If using 3.14 for ${\displaystyle \pi }$, you would calculate: ${\displaystyle r={\sqrt {\frac {21}{3.14}}}}$ ${\displaystyle r={\sqrt {6.69}}}$ If your calculator allows you to enter the whole formula on one line, that will give you a more accurate answer. 5. 5 Take the square root. You will likely need a calculator to do this , because the number will be a decimal. This value will give you the radius of the circle. Example ${\displaystyle r={\sqrt {6.69}}=2.59}$. So, the radius of a circle with an area of 21 square centimeters is about 2.59 centimeters. Areas always use square units (like square centimeters), but the radius always uses units of length (like centimeters). If you keep track of units in this problem, you'll notice that ${\displaystyle {\sqrt {cm^{2}}}=cm}$. 6. Advertisement Method 3 Method 3 of 4: Using the Diameter 1. 1 Check the problem for a diameter. If the problem tells you the diameter of the circle, it's easy to find the radius. If you are working with an actual circle, measure the diameter by placing a ruler so its edge passes straight through the circle's center , touching the circle on both sides.[4] • If you're not sure where the circle center is, put the ruler down across your best guess. Hold the zero mark of the ruler steady against the circle, and slowly move the other end back and forth around the circle's edge. The highest measurement you can find is the diameter. • For example, you might have a circle with a diameter of 4 centimeters. 2. 2 Divide the diameter by two. A circle's radius is always half the length of its diameter. [5] • For example, if the diameter is 4 cm, the radius equals 4 cm ÷ 2 = 2 cm. • In math formulas, the radius is r and the diameter is d. You might see this step in your textbook as ${\displaystyle r={\frac {d}{2}}}$. 3. Advertisement Method 4 Method 4 of 4: Using the Area and Central Angle of a Sector 1. 1 Set up the formula for the area of a sector. The formula is ${\displaystyle A_{sector}={\frac {\theta }{360}}(\pi )(r^{2})}$ , where ${\displaystyle A_{sector}}$ equals the area of the sector, ${\displaystyle \theta }$ equals the central angle of the sector in degrees, and ${\displaystyle r}$ equals the radius of the circle.[6] 2. 2 Plug the sector’s area and central angle into the formula. This information should be given to you. Make sure you have the area of the sector, not the area for the circle. Substitute the area for the variable ${\displaystyle A_{sector}}$ and the angle for the variable ${\displaystyle \theta }$. Example If the area of the sector is 50 square centimeters, and the central angle is 120 degrees, you would set up the formula like this: ${\displaystyle 50={\frac {120}{360}}(\pi )(r^{2})}$. 3. 3 Divide the central angle by 360. This will tell you what fraction of the entire circle the sector represents. Example ${\displaystyle {\frac {120}{360}}={\frac {1}{3}}}$. This means that the sector is ${\displaystyle {\frac {1}{3}}}$ of the circle. Your equation should now look like this: ${\displaystyle 50={\frac {1}{3}}(\pi )(r^{2})}$ 4. 4 Isolate . To do this, divide both sides of the equation by the fraction or decimal you just calculated. Example ${\displaystyle 50={\frac {1}{3}}(\pi )(r^{2})}$ ${\displaystyle {\frac {50}{\frac {1}{3}}}={\frac {{\frac {1}{3}}(\pi )(r^{2})}{\frac {1}{3}}}}$ ${\displaystyle 150=(\pi )(r^{2})}$ 5. 5 Divide both sides of the equation by . This will isolate the ${\displaystyle r}$ variable. For a more precise result, use a calculator. You can also round ${\displaystyle \pi }$ to 3.14. Example ${\displaystyle 150=(\pi )(r^{2})}$ ${\displaystyle {\frac {150}{\pi }}={\frac {(\pi )(r^{2})}{\pi }}}$ ${\displaystyle 47.7=r^{2}}$ 6. 6 Take the square root of both sides. This will give you the radius of the circle. Example ${\displaystyle 47.7=r^{2}}$ ${\displaystyle {\sqrt {47.7}}={\sqrt {r^{2}}}}$ ${\displaystyle 6.91=r}$ So, the radius of the circle is about 6.91 centimeters. 7. Advertisement Community Q&A Search Add New Question • Question How do I find the radius of a circle when I know the chord length? Community Answer It is possible to have quite a few circles, all with different radii, in which one could draw a chord of a given, fixed length. Hence, the chord length by itself cannot determine the radius of the circle. • Question How do I find the radius of a circle when I know the arc length and the central angle? Donagan Top Answerer Divide the central angle into 360°. Multiply the resulting number by the arc length. That gives you the circumference of the circle. Divide the circumference by pi. That's the diameter. Half of the diameter is the radius of the circle. • Question How do I calculate the radius of a circle when no other values are known? Community Answer Technically you can't "calculate" the radius in such a situation. However, it is possible, by construction, to locate the center of such a circle, and then, simply by physically measuring, determine the radius. To do the construction, draw any two chords and construct their perpendicular bisectors; their point of intersection is the center of the circle. Then draw in any radius and measure it with a ruler. Not technically a "calculation." See more answers Ask a Question 200 characters left Include your email address to get a message when this question is answered. Advertisement Tips • The number ${\displaystyle \pi }$ actually comes from circles. If you measure the circumference C and diameter d of a circle very precisely, then calculate ${\displaystyle C\div d}$, you always get ${\displaystyle \pi }$. ⧼thumbs_response⧽ Advertisement About This Article Co-authored by: Math Instructor, City College of San Francisco This article was co-authored by Grace Imson, MA. Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. This article has been viewed 3,361,939 times. 21 votes - 64% Co-authors: 97 Updated: February 3, 2023 Views: 3,361,939 Categories: Geometry Article SummaryX To calculate the radius of a circle by using the circumference, take the circumference of the circle and divide it by 2 times π. For a circle with a circumference of 15, you would divide 15 by 2 times 3.14 and round the decimal point to your answer of approximately 2.39. Be sure to include the units in your answer. To learn more, such as how to calculate the radius with the area or diameter, keep reading the article! Did this summary help you? Thanks to all authors for creating a page that has been read 3,361,939 times. Reader Success Stories • Akhter Hussain Oct 6, 2016 "I had to help my son re-visit his basic skills on geometry while preparation for entrance for an undergraduate..." more Did this article help you? Advertisement
## Thursday, 28 June 2012 ### Steps to Factor Trinomials In the previous post we have discussed about Polynomial Long Division and In today's session we are going to discuss about Steps to Factor Trinomials and How To Factor Trinomials Step By Step, 1. Firstly we are going to compare the given trinomial with the standard form of the trinomial i.e. ax>2 + bx + c and recognize the values of a, b, and c. 2. Now we are going to first look for the factor which is common in all the three terms. Once the common factor is recognized, we will bring it out of the three terms. 3. In the next step, we will split the middle term in such a way that the product of the two splitted terms will be equal to the product of the first and the third term and the sum will be equal to the middle term. 4. Further we will take out the common terms and make the factors. Let us take the following trinomial : 18x>2 + 48xy + 32y>2 Here we will first take out the factor 2, from all the three terms and we get : = 2 * (9x>2 + 24 * x * y + 16y>2) = 2 * (3x)>2 + 2 * 3 * 4 * x * y + (4y )>2 = 2 * 3x + 4y>2 Thus we come to the observation that the step by step procedure must be followed in order to get the factorization of the trinomial. If we are able to recognize the factors directly, relating it to some of the identity, then it becomes more easy for us to factorize. In case the trinomial is 4x>2 + 12x + 9 = (2x)>2 + 2 * 2 * 3x + 3>2 = (2x + 3 )>2 = (2x + 3 ) (2x + 3 ) homework help online is available in math tutorials. Online cbse class 8 books is also available. ## Wednesday, 20 June 2012 ### Polynomial Long Division In the previous post we have discussed about and In today's session we are going to discuss about Polynomial Long Division. In this blog we are going to discuss the Polynomial Long Division. An operation that is used to dividing a polynomial value with the polynomial value is called as polynomial long division. The process in which a value is dividing by the same value or lower degree is also known as polynomial Long Division. Polynomial long division is denoted the term that is added, subtract and multiplied. This equation 7xy2 + 3x – 11 is the representation of the polynomial long division. The polynomial word came from the two words the first one is ‘poly’ and second one is ‘nomial’. Poly means 'many' and nomial means 'term'. By combining both the meaning we get the word I. e. many terms. Polynomial may be denotes constant, variables, and exponent values and we can combine them with addition, subtraction and multiplication operation. (know more about Polynomial long division, here) Lets consider a polynomial p (n), D (n) where degree (D) < degree (p), then the quotient polynomial Q(n) and remainder polynomial R(n) with degree(R) < degree(D), P(n) = Q(n) + R(n) ⇒ P(n) = D(n) Q(n) + R(n), D(n) D(n) By following some steps we can easily find the polynomial long division. Step1: Firstly we need to focus on the higher coefficient term which is present in the equation. Step2: We need to multiply the divisor with the leading term by doing that we can get coefficient term that will be exact. Step3: After getting the coefficient term we only have to change the sign of the variable. If negative sign is present, then we change it into positive sign and vice-versa. Step4 : At last cancel the term of same coefficient or variable. In differential equation solver we use some methods and somewhere it is related to the concept of how compound interest works. Indian Certificate of Secondary Education is a type of exam that is comes under the Council for the Indian School Certificate Examinations. ## Tuesday, 19 June 2012 ### Degree of Polynomial In the previous post we have discussed about and In today's session we are going to discuss about Degree of Polynomial. We know that the polynomial is the combination of the terms joined together with the sign of addition or subtraction. (know more about Polynomial, here) If the Polynomial has only one term we call it a monomial. If there are two terms in the polynomial, then we say that the polynomial is called the binomial and the polynomial with three terms is called trinomial. The polynomial with more than three terms is simply called the polynomial. By the term degree of polynomial, we mean the highest power of the term among all the terms in the given expression. If we have the polynomial 2x + 3x>2 + 5 x>4, then we say that the term 5x>4 has the highest power. SO we say that the degree of the polynomial 2x + 3x>2 + 5 x>4, is 4. On the other hand if we have the polynomial 4x + 3, here the degree of thee polynomial is 1 as the power of x in 4x is maximum, which is equal to 1. We must remember that if the degree of the polynomial is 1, then we call the polynomial as the linear polynomial. In case the degree is 2, then the polynomial is quadratic polynomial and if the degree of the polynomial is 3, then the polynomial is called the conical polynomial. Here we write 2x + 5 is a linear polynomial, 5x>2 + 2x +5 is a quadratic polynomial; and 2x>3 + 5x>2 + 4x + 8 is a cubical polynomial. To learn more about the Green s Theorem, we can take the help from the online math tutor and understand the concept of the lesson clearly. The sample papers of Andhra Pradesh board of secondary education are available online to learn about the patterns of the Question paper which will be very useful to prepare about the exams. ## Monday, 18 June 2012 ### How to deal with Polynomials In the previous post we have discussed about and In today's session we are going to discuss about How to deal with Polynomials. In mathematics, there is a term algebra in which we studied about the statements that express the relationship between the things. In the algebraic notation, relationship between the things can be described as a relationship between the variables and operators that are vary over time. In the same aspect polynomials also consider as a part of algebra that deals with the real numbers and variables. Through the polynomials we can simply perform the basic operation like addition, subtraction and multiplication with the variables. In a more appropriate way we can say that polynomial is a combination of different types of terms with different mathematical operators. (know more about Polynomial , here) In the standard definition we can say that polynomial is an expression that contains the combination of number and variable into it with basic operations and positive integer exponents. In the below we show how to we represent the polynomials into algebraic expression: 3ab2 – 4a + 6 In the above given algebraic notation 3ab2, 4a, 6 can be consider as a terms through which we perform the basic operation that is addition and subtraction. In the above 3, 4 and 6 can be consider as a constants and ‘ab, a’ can be consider as variables. The power of two with the variable ab can be consider as positive exponents. In the other aspect exponents value describe the degree of the term. It means in above notation the term 3ab2 has a degree of 2. In the absence of exponent value we can take the degree of term as a one. Basically in mathematics polynomials are used for describing relationship between the numbers, variables and operations that generate some values. The output of polynomial expression helps the students to get the value of unknown variables. In the study of algebraic expression the concept of polynomial can be categorized into three categories that is monomial, binomial and trinomial. In mathematics Definite Integral can be consider as part of calculus which is used to integrate the function's values between upper limit to lower limit. The ICSE board books help the students to make their study according to their syllabus that are conducted by the Indian certificate of secondary education. ## Saturday, 16 June 2012 ### Factoring Trinomials We know that the trinomial is the polynomial formed by three terms. To learn about Factoring Trinomials, we say that the standard form of the trinomial is ax>2 + bx + c. To find the factors of the above given polynomial , we say that we will either write it in form of some or the other identity or we will try to factorize it by the splitting method. In the splitting method, we say that the middle term of the trinomial is split in such a way that the sum of the two terms is equal to bx (i.e. the second term) and the product of the two split term is equal to the product of the first and the third term of the trinomial. Let us look at the following examples: If we have the polynomial: 4x>2 + 12x + 9 Here we can write the above given polynomial as ( 2x )>2 + 2 2x * 3 + 3>2 We observe that the above given polynomial is in the form of the identity (a + b) >2 = a>2 + 2 * a * b + b>2 So it can be written as (2x + 3) >2 If we solve the polynomial by splitting the second term we say it can be written as : 4x>2 + 6x + 6x + 9 = 2x * ( 2x + 3 ) + 3 * ( 2x + 3 ) = ( 2x + 3 ) * ( 2x + 3 ) = ( 2x + 3 ) >2 Ans We will learn about Function Notation, by visiting the online math tutors and understand the concept related to this topic. We can also download Previous Year Question Papers Of CBSE for the particular subject and it will help us to have a look on the pattern of the previous year question papers. ### Factoring Trinomials We know that the trinomial is the polynomial formed by three terms. To learn about Factoring Trinomials, we say that the standard form of the trinomial is ax>2 + bx + c. To find the factors of the above given polynomial , we say that we will either write it in form of some or the other identity or we will try to factorize it by the splitting method. In the splitting method, we say that the middle term of the trinomial is split in such a way that the sum of the two terms is equal to bx (i.e. the second term) and the product of the two split term is equal to the product of the first and the third term of the trinomial. Let us look at the following examples: If we have the polynomial: 4x>2 + 12x + 9 Here we can write the above given polynomial as ( 2x )>2 + 2 * 2x * 3 + 3>2 We observe that the above given polynomial is in the form of the identity (a + b) >2 = a>2 + 2 * a * b + b>2 So it can be written as (2x + 3) >2 If we solve the polynomial by splitting the second term we say it can be written as : 4x>2 + 6x + 6x + 9 = 2x * ( 2x + 3 ) + 3 * ( 2x + 3 ) = ( 2x + 3 ) * ( 2x + 3 ) = ( 2x + 3 ) >2 Ans We will learn about Function Notation, by visiting the online math tutors and understand the concept related to this topic. We can also download Previous Year Question Papers Of CBSE for the particular subject and it will help us to have a look on the pattern of the previous year question papers. ## Thursday, 7 June 2012 ### How to use Gcd Calculator In the previous post we have discussed about How to use LCM Calculator and In today's session we are going to discuss about How to use Gcd Calculator, By GCD, we mean greatest common divisor. If we want to learn about the divisor of the given numbers, we mean that the greatest divisor, which is divisible by all the given numbers. To learn about gcd, we will first find the factors of the given numbers. We must ensure that the factors of the given number should be all prime factors. Now we will pick the factors of the given number such that they are common to both the numbers whose gcf is to be calculated. In case the gcf is one, the two numbers are not having any of the common factors. We use gcf to find the lowest form of the fraction numbers, rational numbers or the ratios. It simply signifies that the two numbers whose gcf is 1, cannot be divided by the same number. We must remember that the gcf of two prime numbers is always 1 Suppose we want to find the gcf of 9 and 12 Here we will first find the prime factors of 9 and 12. So we say that the prime factors of 9 = 3 * 3 * 1 and the prime factors of 12 = 2 * 2 * 3 * 1 In both the prime factors we find that the numbers 1 and 3 are the prime factors of both the numbers. So 1 * 3 is the gcf of 9 and 12. We can also use gcd calculator to practice the problems based on gcd and understand its logics clearly. If we want to learn about the Central Tendency, we can take online help from the math tutor which is available online every time. To know more about the contents of CBSE syllabus we will visit the site of CBSE and collect the recent information to update our self. ### How to use LCM Calculator Lcm stands for the least common multiple. By the term Lcm, we mean the process of finding the number, which is the least common multiple of all the given numbers. Let’s first see what is a multiple. By the word multiple, we mean that the number which exactly divides the given number. Thus if we say that 12 is the multiple of 3, it means that the number 3 completely divides the number 12. Now if we have to find the Lcm of the two numbers say 3 and 6, then we will first write the multiples of 3 and the multiples of 6, which are as follows : Multiples of 3 = 3 , 6 , 9, 12, 15, 18, 21, 24, . . . . . . Similarly, we have the multiples of 6 = 6, 12, 18, 24, . . . . . . Now we will observe the common multiples of the two numbers 3 and 6 as 6, 12, 18, 24, 30 . . . . which means that these numbers divides both the numbers completely. Out of these numbers we say that the number 6 is the smallest number. So we come to the conclusion that 6 is the L.C. M of the numbers 3 and 6. In the same way we can find the LCM of 3 or more numbers also. We must remember that the LCM of 1 and any number n is n itself. We can also download Lcm Calculator or use it online to understand the concepts of Lcm. To understand the concept of the Circle Graph, we can study and learn from the books of CBSE or take the help of online math tutor. We also have CBSE Sample Paper online which the students can download and use as a guidance tool for the students preparing for the exams and in the next session we will discuss about ## Monday, 4 June 2012 ### Least Common Denominator In today's session we are going to discuss about Least Common Denominator, By Least Common Denominator, we mean the LCM of the denominators of given fractions. Before we learn about Least Common Denominator, let us quickly recall the terms LCM, fractions & denominator; we are already familiar with. By LCM, we mean lowest common denominator. It can be calculated for 2 or more numbers by long division, listing of multiples of the given numbers or in the best way, by prime factorization. Now, fractions are numbers in the form a/b . A fraction, as we can see in the expression, has two parts, a & b. Here, a is the numerator & b is the denominator of the fraction. Thus, we have recalled denominator also that it is the lower part in the fraction. Now coming back to the topic of discussion, i.e. , Least Common Denominator or LCD ; as stated above is the LCM of the denominators of two or more fractions . But why do we need to find such LCD. As we know that the fractions may be like or unlike depending on whether their denominator is same or not & also that to add or subtract fractions; the fractions must be like fractions. If the fractions to be added or subtracted are like, we can add or subtract them easily. But for unlike fractions, we need to make them like by changing their denominators to Least Common Denominator. This is done by finding the LCM of the denominators of all unlike fractions. As for example; if we have to add 3/7+2/5+4/3. Here the fractions are unlike. So we’ll find the LCM of their denominators which comes out 105. Now we can change the fractions & make their denominator 105 by finding equivalent fractions. Thus, finally we have 3/7+2/5+4/3 = (45+42+140)/105=227/105. You can find the Independent Variable Definition at different places online . Also look for ICSE class 10 syllabus online and in next session we will discuss about
# What is the meaning of modular arithmetic? ## What is the meaning of modular arithmetic? : arithmetic that deals with whole numbers where the numbers are replaced by their remainders after division by a fixed number in a modular arithmetic with modulus 5, 3 multiplied by 4 is 2. ## How do you do modulo arithmetic? The modulus is another name for the remainder after division. For example, 17 mod 5 = 2, since if we divide 17 by 5, we get 3 with remainder 2. Modular arithmetic is sometimes called clock arithmetic, since analog clocks wrap around times past 12, meaning they work on a modulus of 12. How do you explain modulo? The modulo operation (abbreviated “mod”, or “%” in many programming languages) is the remainder when dividing. For example, “5 mod 3 = 2” which means 2 is the remainder when you divide 5 by 3. ### What is the point of modular arithmetic? Modular arithmetic is used extensively in pure mathematics, where it is a cornerstone of number theory. But it also has many practical applications. It is used to calculate checksums for international standard book numbers (ISBNs) and bank identifiers (Iban numbers) and to spot errors in them. ### How do you use arithmetic clock? Every time we go past 12 on the clock we start counting the hours at 1 again. If we add numbers the way we add hours on the clock, we say that we are doing clock arithmetic. So, in clock arithmetic 8 + 6 = 2, because 6 hours after 8 o’clock is 2 o’clock. Why is modulo useful? Since any even number divided by 2 has a remainder of 0, we can use modulo to determine the even-ess of a number. This can be used to make every other row in a table a certain color, for example. #### Which one of these is floor division 1 point? Which one of these is floor division? Explanation: When both of the operands are integer then python chops out the fraction part and gives you the round off value, to get the accurate answer use floor division. This is floor division. #### What is modulo and exponentiation? Modular exponentiation is a type of exponentiation performed over a modulus. It is useful in computer science , especially in the field of public-key cryptography . The operation of modular exponentiation calculates the remainder when an integer b (the base) raised to the e th power (the exponent), b e , is divided by a positive integer m (the modulus). What is mod mathematics? Mod in maths is an operation in mathematics that has a great significance in a branch of mathematics called number theory Mod actually stands for remainder as any number can be represented as d=gq+r where d is the number and g is divisor and q is quotient but mod is just a more complex mathematical form where remainder can be negative. ## What is the definition of MoD in math? Mod in maths is an operation in mathematics that has a great significance in a branch of mathematics called number theory. Mod actually stands for remainder as any number can be represented as d=gq+r where d is the number and g is divisor and q is quotient but mod is just a more complex mathematical form where remainder can be negative.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 7 Go to the latest version. # 7.5: Simplify Products or Quotients of Single Variable Expressions Difficulty Level: At Grade Created by: CK-12 Estimated2 minsto complete % Progress Practice Simplify Products or Quotients of Single Variable Expressions Progress Estimated2 minsto complete % Have you ever had a stamp collection? Marc has twice as many stamps in his collection as his Grandfather has in his. Write an expression to represent , the number of stamps in his Grandfather's collection. To solve this problem, you will need to know how to write a single variable expression. Pay attention to this Concept, and you will know how to do this by the end of the Concept. ### Guidance Recall that when you add and subtract terms in an expression, you can only combine like terms. However, you can multiply or divide terms whether they are like terms or not. For example, and are like terms because both terms include the variable . We can multiply them to simplify an expression like this. . However, even though and 3 are not like terms, we can still multiply them, like this. . The Commutative and Associative Properties of Multiplication may help you understand how to multiply expressions with variables. Remember, the Commutative property states that factors can be multiplied in any order. The Associative property states that the grouping of factors does not matter. Let's apply this information. We can take these two terms and multiply them together. First, we multiply the number parts. Next, we multiply the variables. Here is another one. Even though these two terms are different, we can still multiply them together. First, we multiply the number parts. Next, we multiply the variables. Find the product . and are not like terms, however, you can multiply terms even if they are not like terms. Use the commutative and associative properties to rearrange the factors to make it easier to see how they can be multiplied. According to the commutative property, the order of the factors does not matter. So, . According to the associative property, the grouping of the factors does not matter. Group the factors so that the numbers are multiplied first. So, . Now, multiply. The product is . Remember that the word PRODUCT means multiplication and the word QUOTIENT means division. Here is one that uses division. Find the quotient . It may help you to rewrite the problem like this . Then separate out the numbers and variables like this. Now, divide 42 by 7 to find the quotient. The quotient is . Now it's your turn. Find each product or quotient. Solution: Solution: #### Example C Solution: Here is the original problem once again. Marc has twice as many stamps in his collection as his Grandfather has in his. Write an expression to represent , the number of stamps in his Grandfather's collection. To write this, we simply use the variable and the fact that Marc has twice as many stamps. This term represents Marc's stamps. ### Vocabulary Here are the vocabulary words in this Concept. Expression a number sentence without an equal sign that combines numbers, variables and operations. Simplify to make smaller by combining like terms Product the answer in a multiplication problem. Quotient the answer in a division problem. Commutative Property of Multiplication states that the product is not affected by the order in which you multiply factors. Associative Property of Multiplication states that the product is not affected by the groupings of the numbers when multiplying. ### Guided Practice Here is one for you to try on your own. Find the quotient . It may help you to rewrite the problem like this . Then separate out the numbers and variables like this. Now, divide 50 by 10 and divide by to find the quotient. Since any number over itself is equal to 1, you know that . The quotient is 5. ### Video Review Here is a video for review. ### Practice Directions: Simplify each product or quotient. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. ### Vocabulary Language: English Associative property Associative property The associative property states that the order in which three or more values are grouped for multiplication or addition will not affect the product or sum. For example: $(a+b) + c = a + (b+c) \text{ and\,} (ab)c = a(bc)$. Commutative Property Commutative Property The commutative property states that the order in which two numbers are added or multiplied does not affect the sum or product. For example $a+b=b+a \text{ and\,} (a)(b)=(b)(a)$. Expression Expression An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols. Product Product The product is the result after two amounts have been multiplied. Simplify Simplify To simplify means to rewrite an expression to make it as "simple" as possible. You can simplify by removing parentheses, combining like terms, or reducing fractions. Nov 30, 2012 Jan 26, 2016
# Cubic Feet to Liters Volume Conversion Example Problem A volume conversion can be difficult to understand if you try to grasp the problem all in one step. Many volume conversion problems give the student a series of linear distances with one set of units, but want the volume in a different set of units. At first glance, this should be a simple conversion problem. The difficulty comes from students not applying the conversion to each of the dimension measurements. This example problem shows a good way to avoid simple errors by trying to accomplish too much in one step. The example is for the cubic feet to liters volume conversion. ### Volume Conversion Example How many liters of water does it take to fill a swimming pool 11.0 feet by 11.0 feet and 8.00 feet deep? Given: 1 foot = 12 inches 1 inch = 2.54 centimeters 1 Liter = 103 cm3 ### Solution: Our swimming pool’s measurements are given in feet. We need to convert these measurements into something we can use to find the volume measurement of liters. Looking at the given unit conversions, we can convert feet to inches and then to centimeters. 11.0 feet = 335 cm Now the 8.00 feet measurement. 8.00 feet = 243 cm Now we can multiply these together to get the volume of the swimming pool. Volume of swimming pool = 11.0 feet ⋅ 11.0 feet ⋅ 8.00 feet Volume of swimming pool = 335 cm ⋅ 335 cm ⋅ 243 cm Volume of swimming pool = 27,270,675 cm3 = 2.7 × 107 cm3 Now we can use the final conversion to get the volume in liters. step 6 Volume of swimming pool = 2.7 × 104 Liters
### A garden is 30 meters wide and 40 meters long. There is a pathway that runs along the perimeter inside the garden. The area of the pathway is half of the area of the garden. What is the width of the pathway? 5 m Step by Step Explanation: 1. The following figure shows the garden which is 30 meters wide and 40 meters long with a pathway that runs along the perimeter, inside the garden. Here, AB = DC = 40 meters, AD = BC = 30 meters. 2. Let us assume that w is the width of the pathway. If we subtract twice the width of the pathway from the length and width of the garden, then it is equal to the length and width of the garden without the pathway, respectively. Length of the garden without the pathway = 40 - 2w Width of the garden without the pathway = 30 - 2w Area of the garden without the pathway = (40 - 2w) × (30 - 2w) = 1200 - 80w - 60w + 4w2 = 1200 - 140w + 4w2 3. Area of the garden = 40 × 30 = 1200 m2 4. Area of the pathway = Area of the garden - Area of the garden without the pathway = 1200 - (1200 - 140w + 4w2) = 1200 - 1200 + 140w - 4w2 = 140w - 4w2 ----(1) 5. It is given that the area of the pathway is half the area of the garden. Therefore, the area of the pathway = 1200 × 1 2 = 600 m2 ----(2) 6. Comparing equations (1) and (2), we get, 140w - 4w2 = 600 ⇒ 4w2 - 140w + 600 = 0 ⇒ w2 - 35w + 150 = 0 ⇒ w2 - 30w - 5w + 150 = 0 ⇒ w(w - 30) - 5(w - 30) = 0 ⇒ (w - 30)(w - 5) = 0 either, w - 30 = 0 or w - 5 = 0 ⇒ w = 30 ⇒ w = 5 7. Since, w has to be less than the width of the garden, hence the width of the pathway is 5 meters.
Introductory Statistics 2e # Chapter Review ## 3.1Terminology In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive. ## 3.2Independent and Mutually Exclusive Events Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent. In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other. ## 3.3Two Basic Rules of Probability The multiplication rule and the addition rule are used for computing the probability of A and B, as well as the probability of A or B for two given events A, B defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events A and B are mutually exclusive events when they do not have any outcomes in common. ## 3.4Contingency Tables There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables help display data and are particularly useful when calculating probabilites that have multiple dependent variables. ## 3.5Tree and Venn Diagrams A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize. A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. A Venn diagram is especially helpful for visualizing the OR event, the AND event, and the complement of an event and for understanding conditional probabilities.
Home » Math Theory » Geometry » Kites in Mathematics: A Comprehensive Guide for Students # Kites in Mathematics: A Comprehensive Guide for Students ## Introduction A kite is a simple yet interesting quadrilateral shape often appearing in various mathematical problems and concepts. This article is designed to give students an in-depth understanding of kites, their properties, and how they can be applied to real-life situations. We will cover grade appropriateness, math domain, common core standards, definition, key concepts, illustrative examples, real-life applications, practice tests, and FAQs related to kites. Kites are generally introduced to students around 4th to 6th grade as they start learning about different quadrilateral shapes and their properties. However, the complexity of problems involving kites can vary, making them relevant for students in higher grades. ## Math Domain Kites belong to the domain of Geometry, specifically the subdomain of Quadrilaterals, which deals with studying different types of four-sided polygons. ## Applicable Common Core Standards The concept of kites aligns with the following Common Core Standards: 4.G.A.2: Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines or the presence or absence of angles of a specified size. 5.G.B.3: Understand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category. 6.G.A.1: Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing them into rectangles or decomposing them into triangles and other shapes. ## Definition A kite is a type of quadrilateral having two pairs of consecutive, non-overlapping sides that are congruent (equal in length). The vertices where the congruent sides meet are called the non-adjacent or opposite vertices. The figure below represents a kite. ## Key Concepts Diagonals: A kite’s diagonals are perpendicular to one another, and one diagonal is bisected by the other. Angles: The angles between the congruent sides of a kite are equal. Perimeter: The perimeter of a kite is the total or sum of all the lengths of the sides. Area: The area of a kite is one-half the product of its diagonals and can be calculated using the formula: Area = $\frac{d_1 × d_2}{2}$, where d1 and d2 are the lengths of the diagonals. ## Properties of Kites • A kite consists of two pairs of congruent sides that are adjacent. Side AB is congruent to side AD, and side BC is congruent to side DC. • Two angles are equal where the two pairs of sides meet. • Diagonals form a cross at a right angle (90°). • The longer diagonal cuts the other diagonal equally into two parts. ## Discussion with Illustrative Examples Example 1 Consider a kite ABCD, with AB = BC and AD = CD. The diagonals AC and BD intersect at point E. Also, the diagonals are perpendicular, so ∠BEC = 90°. Example 2 In the kite ABCD, the angle between the congruent sides is equal, so ∠ABC = ∠ADC. Example 3 Find the perimeter of a kite with its pairs of equal sides as two and five units. Solution The figure below shows a sample illustration of the kite in this example. The perimeter of a kite is the sum of all the sides of the kite. You can calculate the perimeter by adding the sides of each pair. Perimeter=2+2+5+5 Perimeter=14 units Therefore, the perimeter of the kite is 14 units. Example 4 Find the area of the kite given below. Solution Lengths of the diagonals are: d1=2+2=4 d2=6+2=8 A kite’s area is equal to half of the product of its diagonals. Hence, we have, Area=½ (diagonal 1)(diagonal 2) Area=½(4)(8) Area=½(3)(2) Area=16 square units Therefore, the area of the kite is 16 square units. ## Examples with Solutions Example 1 True or false The diagonals of a kite are always equal in length. Solution False; a kite’s two diagonals are not the same length. Example 2 Given a kite with diagonals 8 cm and 12 cm, calculate its area. Solution Area=½ (diagonal 1)(diagonal 2) Area=½ (8)(12) Area=½ (96) Area=48 cm2 Therefore, the area of the kite is 48 cm2. Example 3 The lengths of a kite’s three sides are three ft., 5 ft, and 3 ft. a. Find the length of the fourth side. b. Find the perimeter of the kite. Solution a. A kite has two pairs of adjacent equal sides, then the length of the fourth side is 5 ft. b. To calculate its perimeter, we have, Perimeter=3+3+5+5 Perimeter=16 ft. Hence, the perimeter of the kite is 16 ft. ## Real-life Application with Solution A park is shaped like a kite with 100 meters and 60 meters diagonals. What is the area of the park? Solution The lengths of the diagonals are: diagonal 1=100 meters diagonal 2=60 meters Finding the area, we have, Area=½ (diagonal 1)(diagonal 2) Area=½ (100)(60) Area=½ (6000) Area=3000 m2 Therefore, the area of the park is 3000 m2. ## Practice Test A. Tell whether the following objects resemble a kite. B. Calculate the perimeter and area of the given kite. ## Frequently Asked Questions (FAQs) ### How to tell if a quadrilateral is a kite? A kite has two pairs of consecutive, non-overlapping sides that are congruent (equal in length). The vertices where the congruent sides meet are called the non-adjacent or opposite vertices. A kite also has perpendicular diagonals, where one bisects the other. ### What is the total of a kite’s internal angles? A kite’s internal angles add up to 360°. ### How many sides does a kite have? A kite has a total of four sides. The two pairs of sides have equal lengths. ### How many pairs of equal angles does a kite have? There is only one pair of equal angles in a kite. Two angles are equal where the two pairs of sides meet. ### How do we calculate the perimeter and area of a kite? The perimeter of a kite can be calculated by adding all the lengths of the sides. Let us say we have the sides m, n, m, and n; the perimeter of a kite is given by Perimeter=m+n+m+n=2m+2n. The area of the kite can be calculated using the formula: Area= $\frac{d_1 × d_2}{2}$, where d1 and d2 are the lengths of the diagonals.
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $f(0)=1 ,\\\\ f(-1)=\sqrt{2} ,\\\\ f(-10)=\sqrt{101}$ $\bf{\text{Solution Outline:}}$ Substitute the given function value in $f(t)=\sqrt{t^2+1} .$ $\bf{\text{Solution Details:}}$ If $t=0 ,$ then \begin{array}{l}\require{cancel} f(t)=\sqrt{t^2+1} \\\\ f(0)=\sqrt{0^2+1} \\\\ f(0)=\sqrt{1} \\\\ f(0)=1 .\end{array} If $t=-1 ,$ then \begin{array}{l}\require{cancel} f(t)=\sqrt{t^2+1} \\\\ f(-1)=\sqrt{(-1)^2+1} \\\\ f(-1)=\sqrt{2} .\end{array} If $t=-10 ,$ then \begin{array}{l}\require{cancel} f(t)=\sqrt{t^2+1} \\\\ f(-10)=\sqrt{(-10)^2+1} \\\\ f(-10)=\sqrt{101} .\end{array} Hence, \begin{array}{l}\require{cancel} f(0)=1 ,\\\\ f(-1)=\sqrt{2} ,\\\\ f(-10)=\sqrt{101} .\end{array}
# Systems of Equations Problems by Alicia Solve this problem by using a two order system. A man bought 42 stamps, some 13¢ and some 18¢. How many of each kind did he buy if the cost was \$6.66? ### Comments for Systems of Equations Problems Average Rating Mar 19, 2010 Rating Systems of Equations Problems by: Karin Hi Alicia, When you write a system of equations, you have to write two equations. The first thing that you want to ask yourself, is what two different pieces of information to I know? In this problem, you know how many stamps (42) and you know the cost of the stamps. So, we are going to write one equation based on the number of stamps and the other equation based on the cost of the stamps. The next thing you want to do is to define your variables. What pieces of information don't you know that you need to know? You don't know how many 13 cent stamps and how many 18 cent stamps you bought. So, let's define those variables. Let x = the number of 13 cent stamps. Let y = the number of 18 cent stamps. Now let's write our two equations. The number of stamps. x + y = 42 The cost of stamps. .13x + .18y = 6.66 Now that you've written the two equations, we need to solve. You can use substitution or linear combinations. I would use substitution. Let's solve x + y = 42 for y. x -x + y = 42 - x y = -x + 42 Now substitute this into the second equation for y. .13x + .18y = 6.66 .13x + .18(-x+42) = 6.66 - Substitute .13x - .18x + 7.56 = 6.66 - Distribute -.05x + 7.56 = 6.66 - Combine like terms -.05x + 7.56 -7.56 = 6.66 - 7.56 - Subtract 7.56 -.05x = -.90 - Simplify -.05x/-.05 = -.90/-.05 - Divide by -.05 x = 18 X = 18, so there were 18 13 cent stamps. x + y = 42 18 + y = 42 y = 24 There were 24 18 cent stamps. Let's check: .13(18) + .18(24) =6.66 6.66 = 6.66 The final answer is: There were 18 13 cent stamps and 24 18 cent stamps. Need More Help With Your Algebra Studies?
# Subtracting Whole Numbers with Subtraction in Parentheses - Examples, Exercises and Solutions Subtraction of whole numbers with subtractions in parentheses refers to a situation where we perform the mathematical operation of subtraction on the difference of some terms that are in parentheses. For example: $12 - (3-2) =$ One way to solve this exercise will be to distribute the parentheses. To do this, we must remember that according to the law of signs of addition/ subtraction, after removing parentheses, the expressions that were inside them change their sign. That is, in our example: $12 - (3-2) =$ $12 - 3 + 2 =$ $9 + 2 = 11$ When distributing the parentheses, we will place a $-$ in front of the number $3$ and a $+$ before the $2$. As you can see, in both cases the sign that was inside the parentheses has switched to the opposite sign. Another way to solve this exercise is to use the order of operations, that is to say: $12 - (3-2) =$ We will start by solving the expression in parentheses by using the order of operations and we will get: $12 - 1 = 11$ ## examples with solutions for subtracting whole numbers with subtraction in parentheses ### Exercise #1 $38-(18+20)=$ ### Step-by-Step Solution According to the order of operations, first we solve the exercise within parentheses: $18+20=38$ Now, the exercise obtained is: $38-38=0$ ### Answer $0$ ### Exercise #2 $8-(2+1)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $2+1=3$ Now we solve the rest of the exercise: $8-3=5$ ### Answer $5$ ### Exercise #3 $22-(28-3)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $28-3=25$ Now we obtain the exercise: $22-25=-3$ ### Answer $-3$ ### Exercise #4 $12:(2\times2)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $2\times2=4$ Now we divide: $12:4=3$ ### Answer $3$ ### Exercise #5 $100-(30-21)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $30-21=9$ Now we obtain: $100-9=91$ ### Answer $91$ ## examples with solutions for subtracting whole numbers with subtraction in parentheses ### Exercise #1 $80-(4-12)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $4-12=-8$ Now we obtain the exercise: $80-(-8)=$ Remember that the product of plus and plus gives us a positive: $-(-8)=+8$ Now we obtain: $80+8=88$ ### Answer $88$ ### Exercise #2 $7-(4+2)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $4+2=6$ Now we solve the rest of the exercise: $7-6=1$ ### Answer $1$ ### Exercise #3 $37-(4-7)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $4-7=-3$ Now we obtain: $37-(-3)=$ Remember that the product of a negative and a negative results in a positive, therefore: $-(-3)=+3$ Now we obtain: $37+3=40$ ### Answer $40$ ### Exercise #4 $28-(4+9)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $4+9=13$ Now we obtain the exercise: $28-13=15$ ### Answer $15$ ### Exercise #5 $13-(7+4)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $7+4=11$ Now we subtract: $13-11=2$ ### Answer $2$ ## examples with solutions for subtracting whole numbers with subtraction in parentheses ### Exercise #1 $55-(8+21)=$ ### Step-by-Step Solution According to the order of operations, we first solve the exercise within parentheses: $8+21=29$ Now we obtain the exercise: $55-29=26$ ### Answer $26$ ### Exercise #2 $60:(10\times2)=$ ### Step-by-Step Solution We write the exercise in fraction form: $\frac{60}{10\times2}=$ Let's separate the numerator into a multiplication exercise: $\frac{10\times6}{10\times2}=$ We simplify the 10 in the numerator and denominator, obtaining: $\frac{6}{2}=3$ ### Answer $3$ ### Exercise #3 $60:(5\times3)=$ ### Step-by-Step Solution We write the exercise in fraction form: $\frac{60}{5\times3}$ We break down 60 into a multiplication exercise: $\frac{20\times3}{5\times3}=$ We simplify the 3s and obtain: $\frac{20}{5}$ We break down the 5 into a multiplication exercise: $\frac{5\times4}{5}=$ We simplify the 5 and obtain: $\frac{4}{1}=4$ ### Answer $4$ ### Exercise #4 $73-(22-(-11))=$ ### Step-by-Step Solution According to the order of operations, first we solve the exercise within parentheses: Remember that the product of a negative by a negative gives a positive result, therefore: $-(-11)=+11$ Now we obtain the exercise: $73-(22+11)=$ We solve the exercise within parentheses: $22+11=33$ We obtain: $73-33=40$ ### Answer $40$ ### Exercise #5 $-45-(8+10)=$ ### Step-by-Step Solution According to the order of operations, first we solve the exercise within parentheses: $8+10=18$ Now we obtain the exercise: $-45-(18)=$ We open the parentheses, remember to change the corresponding sign: $-45-18=-63$ ### Answer $-63$
# How do you write f(x)=2x^2-7x-4 in vertex form? Aug 14, 2016 $y = 2 {\left(x - \frac{7}{4}\right)}^{2} - \frac{81}{8}$ #### Explanation: Given - $y = 2 {x}^{2} - 7 x - 4$ Find the vertex $\left(x , y\right)$ $x = \frac{- b}{2 a} = \frac{- \left(- 7\right)}{2 \times 2} = \frac{7}{4}$ At $x = \frac{7}{4}$ $y = 2 {\left(\frac{7}{4}\right)}^{2} - 7 \left(\frac{7}{4}\right) - 4 = - \frac{81}{8}$ $x , y$ coordinates of the vertex are $\left(\frac{7}{4} , - \frac{81}{4}\right)$ The vertex form of the quadratic equation is - $y = a {\left(x - h\right)}^{2} + k$ We need the values of $a , h , k$ $a = 2$ [coefficient of ${x}^{2}$] $h = \frac{7}{4}$ [x-coordinate of the vertex] $k = \frac{- 81}{8}$[ y-coordinate of the vertex] Now plug in the values in the equation $y = 2 {\left(x - \frac{7}{4}\right)}^{2} - \frac{81}{8}$
Module 3: Probability # Two Basic Rules of Probability Barbara Illowsky & OpenStax et al. When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not. ## The Multiplication Rule If A and B are two events defined on a sample space, then: P(A AND B) = P(B)P(A\|B). This rule may also be written as $displaystyle{P}{({A}{mid}{B})}=frac{{{P}{({A}text{ AND } {B})}}}{{{P}{({B})}}}$ (The probability of A given B equals the probability of A and B divided by the probability of B.) If A and B are independent, then P(A\|B) = P(A). Then P(A AND B) = P(A\|B)P(B) becomes P(A AND B) = P(A)P(B). If A and B are defined on a sample space, then: P(A OR B) = P(A) + P(B) – P(A AND B). If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) – P(A AND B) becomes P(A OR B) = P(A) + P(B). ### Example Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska • Klaus can only afford one vacation. The probability that he chooses A is P(A) = 0.6 and the probability that he chooses B isP(B) = 0.35. • P(A AND B) = 0 because Klaus can only afford to take one vacation • Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) = P(A) + P(B) = 0.6 + 0.35 = 0.95. Note that the probability that he does not choose to go anywhere on vacation must be 0.05. ### Example Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90. 1. What is the probability that he makes both goals? 2. What is the probability that Carlos makes either the first goal or the second goal? 3. Are A and B independent? 4. Are Aand B mutually exclusive? Solution: 1. The problem is asking you to find P(A AND B) = P(B AND A). Since P(B\|A) = 0.90: P(B AND A) = P(B\|A) P(A) = (0.90)(0.65) = 0.585Carlos makes the first and second goals with probability 0.585. 2. The problem is asking you to find P(A OR B). P(A OR B) = P(A) + P(B) – P(A AND B) = 0.65 + 0.65 – 0.585 = 0.715Carlos makes either the first goal or the second goal with probability 0.715. 3. No, they are not, because P(B AND A) = 0.585. P(B)P(A) = (0.65)(0.65) = 0.4230.423 ≠ 0.585 = P(B AND A)So, P(B AND A) is not equal to P(B)P(A). 4. No, they are not because P(A and B) = 0.585. To be mutually exclusive, P(A AND B) must equal zero. Watch this video for another example about first determining whether a series of events are mutually exclusive, then finding the probability of a specific outcome. ### Try it Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.75. D = the event Helen makes the second shot. P(D) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws? P(D\|C) = 0.85 P(C AND D) = P(D AND C) P(D AND C) = P(D\|C)P(C) = (0.85)(0.75) = 0.6375 Helen makes the first and second free throws with probability 0.6375. ### Example A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly. 1. What is the probability that the member is a novice swimmer? 2. What is the probability that the member practices four times a week? 3. What is the probability that the member is an advanced swimmer and practices four times a week? 4. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not? 5. Are being a novice swimmer and practicing four times a week independent events? Why or why not? Solution: 1. $displaystylefrac{{28}}{{150}}$ 2. $displaystylefrac{{80}}{{150}}$ 3. $displaystylefrac{{40}}{{150}}$ 4. P(advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time. 5. No, these are not independent events. P(novice AND practices four times per week) = 0.0667P(novice)P(practices four times per week) = 0.09960.0667 ≠ 0.0996 ### try it A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year? $displaystyle{P}=frac{{{200}-{140}-{40}}}{{200}}=frac{{20}}{{200}}={0.1}$ ### Example Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25. Let M = math class, S = speech class, M\|S = math given speech 1. What is the probability that Felicity enrolls in math and speech?Find P(M AND S) = P(M\|S)P(S). 2. What is the probability that Felicity enrolls in math or speech classes?Find P(M OR S) = P(M) + P(S) – P(M AND S). 3. Are M and S independent? Is P(M\|S) = P(M)? 4. Are M and S mutually exclusive? Is P(M AND S) = 0? Solution: 1. 0.1625 2. 0.6875 3. No 4. No ### try it A student goes to the library. Let events B = the student checks out a book and D = the student check out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D\|B) = 0.5. 1. Find P(B AND D). 2. Find P(B OR D). 1. P(B AND D) = P(D\|B)P(B) = (0.5)(0.4) = 0.20. 2. P(B OR D) = P(B) + P(D) − P(B AND D) = 0.40 + 0.30 − 0.20 = 0.50 ### Example Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random. 1. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative? 2. Given that the woman has breast cancer, what is the probability that she tests negative? 3. What is the probability that the woman has breast cancer AND tests negative? 4. What is the probability that the woman has breast cancer or tests negative? 5. Are having breast cancer and testing negative independent events? 6. Are having breast cancer and testing negative mutually exclusive? Solution: 1. P(B) = 0.143; P(N) = 0.85 2. P(N\|B) = 0.02 3. P(B AND N) = P(B)P(N\|B) = (0.143)(0.02) = 0.0029 4. P(B OR N) = P(B) + P(N) – P(B AND N) = 0.143 + 0.85 – 0.0029 = 0.9901 5. No. P(N) = 0.85; P(N\|B) = 0.02. So, P(N\|B) does not equal P(N). 6. No. P(B AND N) = 0.0029. For B and N to be mutually exclusive, P(B AND N) must be zero. ### Try it A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports? Let A = student is a senior going to college. Let B = student plays sports. $displaystyle{P}{({B})}=frac{{140}}{{200}}$$displaystyle{P}{({B}{mid}{A})}=frac{{50}}{{140}}$P(A AND B) = P(B\|A)P(A) $displaystyle{P}{({A} text{ AND } {B})}={(frac{{140}}{{200}})}{(frac{{50}}{{140}})}=frac{{1}}{{4}}$ ### Example Refer to the information in Example 5. P = tests positive. 1. Given that a woman develops breast cancer, what is the probability that she tests positive. Find P(P\|B) = 1 – P(N\|B). 2. What is the probability that a woman develops breast cancer and tests positive. Find P(B AND P) = P(P\|B)P(B). 3. What is the probability that a woman does not develop breast cancer. Find P(B′) = 1 – P(B). 4. What is the probability that a woman tests positive for breast cancer. Find P(P) = 1 – P(N). Solution: 1. 0.98 2. 0.1401 3. 0.857 4. 0.15 ### try it A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D\|B) = 0.5. 1. Find P(B′). 2. Find P(D AND B). 3. Find P(B\|D). 4. Find P(D AND B′). 5. Find P(D\|B′). 1. P(B′) = 0.60 2. P(D AND B) = P(D\|B)P(B) = 0.20 3. P(B\|D)=$displaystylefrac{{{P}{({B}text{ AND } {D})}}}{{{P}{({D})}}}=frac{{{0.20}}}{{{0.30}}}={0.66}$ 4. P(D AND B′) = P(D) – P(D AND B) = 0.30 – 0.20 = 0.10 5. P(D\|B′) = P(D AND B′)P(B′) = (P(D) – P(D AND B))(0.60) = (0.10)(0.60) = 0.06 ## References DiCamillo, Mark, Mervin Field. “The File Poll.” Field Research Corporation. Available online at http://www.field.com/fieldpollonline/subscribers/Rls2443.pdf (accessed May 2, 2013). Rider, David, “Ford support plummeting, poll suggests,” The Star, September 14, 2011. Available online at http://www.thestar.com/news/gta/2011/09/14/ford_support_plummeting_poll_suggests.html (accessed May 2, 2013). “Mayor’s Approval Down.” News Release by Forum Research Inc. Available online at http://www.forumresearch.com/forms/News Archives/News Releases/74209_TO_Issues_-_Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013). “Roulette.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Roulette (accessed May 2, 2013). Shin, Hyon B., Robert A. Kominski. “Language Use in the United States: 2007.” United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/data/acs/ACS-12.pdf (accessed May 2, 2013). Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013). Data from U.S. Census Bureau. Data from the Wall Street Journal. Data from The Roper Center: Public Opinion Archives at the University of Connecticut. Available online at http://www.ropercenter.uconn.edu/ (accessed May 2, 2013). Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed May 2,2 013). ## Concept Review The multiplication rule and the addition rule are used for computing the probability of A and B, as well as the probability of A or B for two given events A, B defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events A and B are mutually exclusive events when they do not have any outcomes in common. ## Formula Review The multiplication rule: P(A AND B) = P(A\|B)P(B) The addition rule: P(A OR B) = P(A) + P(B) – P(A AND B)
Conic Sections Hyperbola and Line Hyperbola and line relationships The equation of the tangent at the point on the hyperbola Hyperbola and line relationships Let examine relationships between a hyperbola and a line passing through the center of the hyperbola, i.e., the origin. A line y = mx intersects the hyperbola at two points if the slope \|m\| < b/a but if \|m\| > b/a then, the line y = mx does not intersect the hyperbola at all. The diameters of a hyperbola are straight lines passing through its center. The asymptotes divide these two pencils of diameters into, one which intersects the curve at two points, and the other which do not intersect. A diameter of a conic section is a line which passes through the midpoints of parallel chords. Conjugate diameters of the hyperbola (or the ellipse) are two diameters such that each bisects all chords drawn parallel to the other. As the equation of a hyperbola can be obtained from the equation of an ellipse by changing the sign of b2 that is, a2(1 - e2) = -a2(e2 - 1) = -b2 this way, we can use other formulas relating to the ellipse to obtain corresponding formulas for the hyperbola. Therefore, when we examine conditions which determine position of a line in relation to a hyperbola that is, when solve the system of equations, y = mx + c b2x2 - a2y2 = a2b2 then if, a2m2 - b2 > c2 the line intersects the hyperbola at two points, a2m2 - b2 = c2 the line is the tangent of the hyperbola, a2m2 - b2 < c2 the line and the hyperbola do not intersect. Condition for a line to be the tangent to the hyperbola - tangency condition A line is the tangent to the hyperbola if a2m2 - b2 = c2. Regarding the asymptotes, to which c = 0, this condition gives \|m\| = b/a and that is why we can say that the hyperbola touches the asymptotes at infinity. From the tangency condition it also follows that the slopes of the tangents will satisfy the condition if That is, the tangents to the hyperbola can only be parallel to a line belonging to the pencil of lines that do not intersect the hyperbola. The equation of the tangent at the point on the hyperbola As we already mentioned, the points of contact of a line and the hyperbola can be obtained from the corresponding formula for the ellipse by changing b2 with -b2 thus the tangency point or the point of contact. So, the intercept and slope of the tangent or b2x1x - a2y1y = a2b2 the equation of the tangent at a point P1(x1, y1) on the hyperbola. Polar and pole of the hyperbola If from a point A(x0, y0), exterior to the hyperbola, drawn are tangents, then the secant line passing through the contact points, is the polar of the point A. The point A is called the pole of the polar. The equation of the polar is derived the same way as for the ellipse, b2x0x - a2y0y = a2b2 the equation of the polar of the point A(x0, y0). Construction of the tangent at the point on the hyperbola The tangent at the point P1(x1, y1) on the hyperbola is the bisector of the angle F1P1F2 subtended by focal radii, r1 and r2 at P1 . The proof shown for the ellipse applied to the hyperbola gives, or See the title ' The angle between the focal radii at a point of the ellipse'. Construction of tangents from a point outside the hyperbola With A as center draw an arc through F2, and from F1as center, draw an arc of radius 2a. These arcs intersect at points S1 and S2. Tangents are the perpendicular bisectors of the line segments F2S1 and F2S2. Tangents can also be drawn as lines through A and the intersection points of lines through F1S1 and F1S2, with the hyperbola. These intersections are at the same time the points of contact D1 and D2. Hyperbola and line examples Example: Determine the semi-axis a such that the line 5x - 4y - 16 = 0 be the tangent of the hyperbola 9x2 - a2y2 = 9a2. Solution: Rewrite the equation 9x2 - a2y2 = 9a2 \| ¸ 9a2 and the equation of the tangent 5x - 4y - 16 = 0 or Then, plug the slope and the intercept into tangency condition, Therefore, the given line is the tangent of the hyperbola Example: Find the angle between the ellipse, which passes through points, A(Ö5, 4/3) and B(1, 4Ö2/3), and the hyperbola whose asymptotes are y = ± x/ 2 and the linear eccentricity or half of the focal distance c = Ö5. Solution: Find the equation of the ellipse by solving the system of equations, thus, the equation of the ellipse The equation of the hyperbola by solving the system of equations, Therefore, equation of the hyperbola Angle between curves is the angle between tangents at intersection of the curves. By solving the system of equations of the curves we obtain the points of intersection, (1) 4x2 + 9y2 = 36 (2) => (1) 4(4y2 + 4) + 9y2 = 36, 25y2 = 20, y1,2 = ±2/Ö5, x1,2 = ±6/Ö5 (2) x2 - 4y2 = 4 => x2 = 4y2 + 4 Tangent of the ellipse and the hyperbola at the intersection S1(6/Ö5, 2/Ö5), S1(6/Ö5, 2/Ö5) => te :: b2x1x + a2y1y = a2b2, 4x + 3 = 6/Ö5 or y = (-4/3)x + 2Ö5, S1(6/Ö5, 2/Ö5) => th :: b2x1x - a2y1y = a2b2, 3x - 4 = 2/Ö5 or y = (3/4)x - Ö5/2. fulfilled is the perpendicularity condition. Therefore, the angle between curves j = 90°. Conic sections contents
Sign up or log in to Magoosh AP Calc Prep. The disk and washer methods are useful for finding volumes of solids of revolution. In this article, we’ll review the methods and work out a number of example problems. By the end, you’ll be prepared for any disk and washer methods problems you encounter on the AP Calculus AB/BC exam! ## Solids of Revolution The disk and washer methods are specialized tools for finding volumes of certain kinds of solids — solids of revolution. So what is a solid of revolution? Starting with a flat region of the plane, generate the solid that would be “swept out” as that region revolves around a fixed axis. For example, if you start with a right triangle, and then revolve it around a vertical axis through its upright leg, then you get a cone. The cone generated as a solid of revolution by revolving a right triangle around a vertical axis Here’s another cool example of a solid of revolution that you might have seen hanging up as a decoration! Tissue paper decorations that unfold from flat to round are examples of solids of revolution. Watch the next few seconds of the video below to see how it unfolds in real time. ## The Disk and Washer Methods: Formulas So now that you know a bit more about solids of revolution, let’s talk about their volumes. Suppose S is a solid of revolution generated by a region R in the plane. There are two related formulas, depending on how complicated the region R is. ### Disk Method The simplest case is when R is the area under a curve y = f(x) between x = a and x = b, revolved around the x-axis. Now imagine cutting the solid into thin slices perpendicular to the x-axis. Each slice looks like a disk or cylinder, except that the outer surface of the disk may have a curve or slant. Let’s approximate each slice by a cylinder of height dx, where dx is very small. In fact, I like to think of each disk as being generated by revolving a thin rectangle around the x-axis. Then you can see that the height of the rectangle, y, is the same as the radius of the disk. Now let’s compute the volume of a typical disk located at position x. The radius is y, which itself is just the function value at x. That is, r = y = f(x). The height of the disk is equal to dx (think of the disk as a cylinder standing on edge). Therefore, the volume of a single cylindrical disk is: V = π r2h = π f(x)2dx. This calculation gives the approximate volume of a thin slice of S. Next, to approximate the volume of the entirety of S, we have to add up all of the disk volumes throughout the solid. For simplicity, assume that the thickness of each slice is constant (dx). Also, for technical reasons, we have to keep track of the various x-values along the interval from a to b using the notation xk for a “generic” sample point. Finally, by letting the number of slices go to infinity (by taking a limit as n → ∞), we develop a useful formula for volume as an integral. ### Example 1: Disk Method Let R be the region under the curve y = 2x3/2 between x = 0 and x = 4. Find the volume of the solid of revolution generated by revolving R around the x-axis. #### Solution Let’s set up the disk method for this problem. The volume of the solid is 256π (roughly 804.25) cubic units. ### Washer Method Now suppose the generating region R is bounded by two functions, y = f(x) on the top and y = g(x) on the bottom. This time, when you revolve R around an axis, the slices perpendicular to that axis will look like washers. No, we’re not talking about clothes washers or dishwashers… A washer is like a disk but with a center hole cut out. The formula for the volume of a washer requires both an inner radius r1 and outer radius r2. We’ll need to know the volume formula for a single washer. V = π (r22r12) h = π (f(x)2g(x)2) dx. As before, the exact volume formula arises from taking the limit as the number of slices becomes infinite. ### Example 2: Washer Method Determine the volume of the solid. Here, the bounding curves for the generating region are outlined in red. The top curve is y = x and the bottom one is y = x2 #### Solution This is definitely a solid of revolution. We’ll set up the formula with f(x) = x (top) and g(x) = x2 (bottom). But what should we use as a and b? Well, just as in some area problems, you may have to solve for the bounds. Clearly the region is bounded by the two curves between their common intersection points. Set f(x) equal to g(x) and solve to locate these points of intersection. x = x2 → xx2 = 0 → x(1 – x) = 0. We find two such points: x = 0 and 1. So set a = 0 and b = 1 in the formula. ### Example 3: Different Axes Set up an integral that computes the volume of the solid generated by revolving he region bounded by the curves y = x2 and x = y3 around the line x = -1. #### Solution Be careful not to blindly apply the formula without analyzing the situation first! This time, the axis of rotation is a vertical line x = -1 (rather than the horizontal x-axis). The radii will be horizontal segments, so think of x1 and x2 (rather than y-values). Furthermore, because everything is turned on its side compared to previous problems, we have to make sure both boundary functions are solved for x. The thickness of the washer is now dy (instead of dx). Finally, because the axis of revolution is one unit to the left of the y-axis, that adds another unit to each radius. (The further away the axis, the longer the radius must be to reach the figure, right?) Take a look at the graph below to help visualize what’s going on. • Inner Radius: x = y3 + 1 • Outer Radius: x = y1/2 + 1 As before, set the functions equal and solve for points of intersection. Those are again at x = 0 and 1. Using the Washer Method formula for volume, we obtain: The problem only asks for setup, so we are done at this point. ##### About Shaun Ault Shaun earned his Ph. D. in mathematics from The Ohio State University in 2008 (Go Bucks!!). He received his BA in Mathematics with a minor in computer science from Oberlin College in 2002. In addition, Shaun earned a B. Mus. from the Oberlin Conservatory in the same year, with a major in music composition. Shaun still loves music -- almost as much as math! -- and he (thinks he) can play piano, guitar, and bass. Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed! 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# How do you graph f(x)=-2/(x+3) using holes, vertical and horizontal asymptotes, x and y intercepts? Jul 30, 2017 The vertical asymptote is $x = - 3$ The horizontal asymptote is $y = 0$ See the graph below #### Explanation: To calculate the vertical asymptote, we perform ${\lim}_{x \to - {3}^{+}} f \left(x\right) = {\lim}_{x \to - {3}^{+}} - \frac{2}{x + 3} = - \infty$ ${\lim}_{x \to - {3}^{-}} f \left(x\right) = {\lim}_{x \to - {3}^{+}} - \frac{2}{x + 3} = + \infty$ The vertical asymptote is $x = - 3$ To calculate the horizontal asymptote, we perform ${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} - \frac{2}{x + 3} = {0}^{-}$ ${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} - \frac{2}{x + 3} = {0}^{+}$ The horizontal asymptote is $y = 0$ The general form of the graph is $\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a a}$$- 3$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$ $\textcolor{w h i t e}{a a a a}$$- \left(x + 3\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a}$$\| \|$$\textcolor{w h i t e}{a a a}$$-$ $\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a}$↗$\textcolor{w h i t e}{a a a a a a}$$\| \|$$\textcolor{w h i t e}{a a a}$↗ The intercept with the y-axis is when $x = 0$ $f \left(0\right) = - \frac{2}{3}$ So the intercept is $\left(0 , - \frac{2}{3}\right)$ graph{-2/(x+3) [-18.02, 18.03, -9.01, 9.01]}
# Networks Prim’s Algorithm ## Presentation on theme: "Networks Prim’s Algorithm"— Presentation transcript: Networks Prim’s Algorithm Decision Maths Networks Prim’s Algorithm Prim`s Algorithm In Lesson 1 we learnt about Kruskal`s algorithm, which was used to solve minimum connector problems. Another method that can be used is Prim`s algorithm. Step 1 – Select any node Step 2 – Connect it to the nearest node Step 3 – Connect one node already selected to the nearest unconnected node. Step 4 – Repeat 3 until all nodes are connected. Prim`s Algorithm Consider the example we looked at last lesson. Select any node you like. Lets select F. Prim`s Algorithm Connect it to the nearest node. C and D are both 3 away so we can choose either. Lets select C. Prim`s Algorithm The nearest node to either of F or C is D, which is only 3 away from F. So connect D to F. Prim`s Algorithm The nearest to D, F or C is E which is 2 from D. So connect E to D. Prim`s Algorithm The nearest to any of these four nodes is A which is 5 away from F. Connect A to F. Prim`s Algorithm We now need to connect the last node, B. The shortest arc is AB, which is 2. Connect B to A. Prim`s Algorithm All the nodes are now connected so this is the minimum connector or minimal spanning tree. Distance Table The Network can also be represented as a table. The infinity symbol (∞) means there is no edge between the two nodes. Prim`s on a Distance Table We are going to apply Prim`s algorithm to the distance table. This demonstrates how a computer could apply the algorithm. Prim`s is more suitable than Kruskal`s as computers have a problem recognising loops. As you go through the algorithm, see if you can relate the procedure to the last example. Prim`s on a Distance Table Step 1 – Select any arbitrary node. Step 2 – Delete the row and loop the column that correspond to the node selected. Step 3 – Choose the smallest number in the loop. Step 4 – Delete the row that this smallest number is in. Step 5 – Loop the column that corresponds to the row just deleted. Step 6 – Choose the smallest number in any loop. Step 7 – Repeat steps 4, 5 and 6 until all rows have been deleted and columns looped. Prim`s on a Distance Table Here I have chosen F. Delete the row. Loop the column. Select the smallest number in the loop. Prim`s on a Distance Table Delete row C. Loop column C. Select the smallest number in any loop that is not crossed out. Prim`s on a Distance Table Delete row D. Loop column D. Select the smallest number in any loop that is not crossed out. Prim`s on a Distance Table Delete row E. Loop column E. Select the smallest number in any loop that is not crossed out. Prim`s on a Distance Table Delete row A. Loop column A. Select the smallest number in any loop that is not crossed out. Prim`s on a Distance Table Delete row B. Loop column B. Prim`s on a Distance Table The algorithm is complete when all the columns have been looped and the rows crossed out. The circles show the edges in the minimum connector. Prim`s on a Distance Table In this case they are AB, DE, AF, CF, DF Can you explain why this procedure is exactly the same as applying Prim`s algorithm? Similar presentations
Lesson Explainer: Linear Relations: 𝑎𝑥+𝑏𝑦=𝑐 \| Nagwa Lesson Explainer: Linear Relations: 𝑎𝑥+𝑏𝑦=𝑐 \| Nagwa # Lesson Explainer: Linear Relations: ππ₯+ππ¦=π Mathematics In this explainer, we will learn how to identify and graph linear relations between two variables given the relation in the form and write the ordered pairs that satisfy the given equation. Letβs begin by defining what we mean by a linear relation. ### Definition: Linear Relations If two variables and are related by an equation of the form , where , , and are real numbers, then and are linearly related. Such a relation can be represented by a set of ordered pairs . Suppose Sarah wants to send a birthday card to her cousin. It costs EGP 2.40 to send a card, and since Sarah has left it a little late in the day, the post office has only 20 pt and 50 pt stamps left (where pt stands for piastres and there are 100 piastres in one pound). Letβs consider how many 20 pt and 50 pt stamps she could use to send her card. Since EGP 2.40 is equal to 240 pt, the total value of the stamps must be 240 pt. Letβs call the number of 20 pt stamps and the number of 50 pt stamps . Then, to represent the postage, we have the linear relation We want to know which, if any, pairs of values of and together satisfy this equation. Since we cannot break the stamps up into smaller values, and must be natural numbers, that is, positive whole numbers. In fact, the ordered pair satisfies the equation. That is, and , since Alternatively, the ordered pair also satisfies the linear relation, since . Hence, Sarah could either use stamps and stamps or stamps and stamps. In fact, there are 3 possible combinations of the two stamp values that satisfy the given linear relation, as shown in the table below. 20 pt Stamps 50 pt Stamps Ordered Pair 24 72 120 Note that since each term in the equation of our linear relation has a common factor of 10, we can divide the equation through by 10. This gives and each of our ordered pairs also satisfies this new equation. Since there are now no factors that are common to all of the terms, we call this the simplest form of the linear relation. Letβs look now at an example of how we determine whether or not an ordered pair satisfies a particular linear relation. ### Example 1: Determining Which Ordered Pairs Satisfy a Given Linear Relation Which of the following satisfies the relation ? To answer this question, we try each pair of values in turn in the given equation. To do this, from each pair, we substitute the first value for and the second for , and if the right- and left-hand sides are equal, then we can say the pair satisfies the relation . We see that only one of the given ordered pairs satisfies the relation , that is, option D. Hence, the ordered pair satisfies the relation. In our next example, we see how to find the missing value in an ordered pair, given a linear relation. ### Example 2: Completing a Table of Values for a Linear Relation In the table below some of the - and -values are missing from the ordered pairs that satisfy the linear relation . Find the missing and values. π₯ π¦ Ordered Pair β4 0 3 β13 32 (β4,β―) (β―,β13) (0,β―) (3,β―) (β―,32) We can find the missing value in each ordered pair by substituting the known - or -value into the equation for the linear relation. We then solve for the missing - or -value. For our first ordered pair, since we want to solve for , letβs rearrange the equation so that is the subject. We can do this by adding to both sides, giving Now, substituting , we have . That is, . Our first ordered pair is therefore . Note that we could have approached this slightly differently by first substituting the known -value into the given equation and then solving for . Since for our second ordered pair we know the -value, letβs make the subject of our relation . We can do this by subtracting from both sides and dividing through by . This gives us Now, substituting into this equation, we have This evaluates to . Hence, our second ordered pair is . For our third ordered pair, we know that and we have . Hence, the ordered pair is . In our fourth ordered pair, we know that , hence, and the ordered pair is . For the last ordered pair, we know that , hence, and the ordered pair is . The missing values from the ordered pairs are therefore , , , , and , and we can complete the table as shown. π₯ π¦ Ordered Pair β4 β2 0 3 7 β23 β13 β3 12 32 (β4,β23) (β2,β13) (0,β3) (3,12) (7,32) In our next example, we see how to find an unknown coefficient given a linear relation and an ordered pair satisfying that relation. ### Example 3: Finding a Coefficient in a Linear Relation given an Ordered Pair Given that satisfies the relation , find the value of . To find the value of , we substitute the - and -values from the given ordered pair into the linear relation. We then solve the resulting equation for . With and , the linear relation is which evaluates to To solve for , we add 21 to both sides then divide both sides by so that The value of that satisfies the given linear relation is therefore . A linear relation can be represented graphically as a straight line, hence the term linear. If we know at least two ordered pairs that satisfy a specific linear relation, to represent the relation graphically, we plot the points represented by the ordered pairs and draw the line that passes between those points. For example, the ordered pairs and satisfy the linear relation We can represent this relation graphically by plotting the points with coordinates , and , and drawing a line through those points. Note that while we have plotted the line corresponding to the linear relation using only two points, in fact, every point on the line is represented by an ordered pair that satisfies the linear relation , where and are the coordinates of the point. With this in mind, looking once again at our postage stamps example, we note that the linear relation , which we can write equivalently as , can be represented by the graph below. In this case, however, while the linear relation is represented completely by the plotted line, we know that there are only three points on the line that correspond to the actual scenario of buying stamps with a value of 240 pt. Remember, is the number of 20 pt stamps and is the number of 50 pt stamps, and these cannot be broken down into smaller units. Hence, since the solutions must be positive whole number values, there are only three possibilities: , , and . In the linear relations discussed so far, that is, relations of the form , the coefficients and have been nonzero. Letβs now look at the special cases where either or is equal to 0. • If , our linear relation is of the form . Dividing through by the coefficient , we have . This means that for every value of , is equal to the constant . Graphically, this is a horizontal line through the point . • If , our linear relation is of the form . Dividing through by the coefficient , we have . This means that for every value of , is equal to the constant . Graphically, this is a vertical line through the point . In our next two examples, we are asked to plot the graphs of linear relations of this type. ### Example 4: Drawing the Graph of a Linear Relation Given the relation , sketch its graph if , , and . To sketch the graph of the relation , where the coefficients take the values given, we first substitute the given values into the equation. This gives That is, . To make the subject of this new equation, we divide through by 3. Hence, . Another way of looking at this is to recall that if in the linear relation , then . This is exactly what we found by substituting the given coefficients into the relation. To sketch the graph of the relation , we note that the interpretation of this relation is that for every value of , is equal to the constant value . The graph of the relation is therefore a horizontal line through the point , which we can sketch as shown below. ### Example 5: Drawing the Graph of a Linear Relation Given the relation , sketch the graph of this relation if , , and . To sketch the graph of the relation , where the coefficients take the values given, we first substitute the given values into the equation. This gives That is, , and to make the subject of this new equation, we divide through by 5. We then have . We can look at this another way by recalling that if in the linear relation , then . In our case, with and , this is exactly what we found. To sketch the graph of the relation , we note that we interpret this relation as follows: for every value of , is equal to . The graph of the relation is therefore a vertical line through the point , which we can sketch as shown below, noting that as a decimal, . In our final example, we are given a linear relation and must find the missing values in a table of ordered pairs. We then use these values to sketch a graph of the relation. ### Example 6: Completing a Table of Values for a Linear Relation and Then Graphing the Relation • Given the linear relation , complete the table of values below. π₯ π¦ 0 1 5 β7 β1 3 • Using the completed table, sketch a graph to represent the relation. Each column in the table represents an ordered pair whose - and -values together satisfy the given relation . To find the missing values, we substitute the known value of or into the equation for the relation and solve for the unknown value. For the first ordered pair, we are given that . To find the corresponding -value, we substitute into the given relation as follows: Now, to make the subject, we add 14 to both sides and divide by , giving Our first missing value is therefore , which we enter into our table. π₯ π¦ β2 0 1 5 β7 β1 3 The next missing value is the -value corresponding to . To find this -value, we substitute into our relation, giving Now, to make the subject, we divide both sides by 2, and we find Our second missing value is therefore . The third missing value is the -value corresponding to . Substituting into our relation gives Solving this for , we find . Our final missing value is the -value corresponding to , and substituting into our relation gives Solving this for , we find . We can now complete the table with the missing values we have found. π₯ π¦ β2 0 1 3 5 β7 β3 β1 3 7 We are asked to use the completed table to sketch the graph of the given relation. To do this, we note the ordered pairs of - and -values, as these are the coordinates of points on the line representing the linear relation. These are , , , , and . Plotting these points and sketching the line that passes through them gives us the graph of the linear relation , as shown below. Note that since each term in the given equation has a common factor of 2, the equation can be divided through by 2 without changing the ordered pairs that satisfy the equation. This gives , and since there are now no common factors, we say that this is the simplest form of the linear relation. The graph is the same as that of the original relation. We complete this explainer by recapping the key points covered. ### Key Points • A linear relation of the form can be represented by a set of ordered pairs , where each pair of values satisfies the equation of the given linear relation. • We may find the value of an unknown coefficient in a linear relation given an ordered pair satisfying the linear relation. • To sketch the graph of a linear relation, we plot the coordinates corresponding to two or more ordered pairs that satisfy the relation. The line that passes through these points represents the linear relation. • If the equation of a linear relation is multiplied by a constant, both the graph of the relation and the ordered pairs satisfying the relation remain unchanged. If there are no common factors in the terms of the equation of the relation, then the linear relation is said to be in its simplest form. • If in the linear relation , then the graph of the relation is the horizontal line passing through the point . • If in the linear relation , then the graph of the relation is the vertical line passing through the point .
Functions and algebra: Use Mathematical models to investigate linear programming problems # Unit 1: Linear programming Dylan Busa ### Unit outcomes By the end of this unit you will be able to: • Sketch given constraints. • Find the feasible region and objective function. • Use a boundary search to find the vertices of the feasible region. • Optimise the objective function. ## What you should know Before you start this unit, make sure you can: ## Introduction Qhubeka is a wonderful organisation that donates bicycles to children in rural areas to make things such as getting to and from school easier and quicker. Qhubeka makes all the bicycles it donates, thereby also creating employment. Suppose it makes two types of bicycles – the Buffalo and the Cheetah. The Buffalo is rugged and designed for areas where the roads aren’t so good. The Cheetah is designed for areas where there are at least basic roads. Now suppose that each day Qhubeka can produce at most seven Buffalos and six Cheetahs. It takes three technicians to build a Buffalo and two technicians to build a Cheetah. There is a total of $\scriptsize 18$ technicians at the organisation. Each Buffalo costs $\scriptsize \text{R2}\ 400$ to make and each Cheetah costs $\scriptsize \text{R1}\ 200$ to make. If every bicycle that Qhubeka makes is needed by a child, how many of each type should Qhubeka make in order to minimise costs? This is an example of an optimisation problem. There are several constraints that all need to be taken into account to find an optimal solution, in this case the lowest cost. Another example of an optimisation problem would be a farmer working out how much of two different crops to plant, with different input costs and different market values, in order to maximise his profit. In all cases, we want to calculate the maximum or minimum in a specific situation. ## Solve optimisation problems graphically The easiest and most efficient way to solve these kinds of optimisation problems is graphically using a technique called linear programming. Linear programming is a graphical modelling technique in which the optimal solution of a linear function is found by applying various other constraints. Linear programming is, therefore, a special case of the broader area of mathematical programming and is most often used to guide decision making in business, engineering, social sciences, and physical sciences. A detailed worked example of the solution to the problem Qhubeka faces is the best way to demonstrate how linear programming works. ### Example 1.1 This is what we know: • Qhubeka makes two types of bicycles – the Buffalo and the Cheetah. • Each day Qhubeka can produce at most seven Buffalos and four Cheetahs. • Each Buffalo needs three technicians to build it. • Each Cheetah needs two technicians. • There is a total of $\scriptsize 18$ technicians. • Each Buffalo costs $\scriptsize \text{R2}\ 400$ to make and each Cheetah costs $\scriptsize \text{R1}\ 200$ to make. 1. If every bicycle that Qhubeka makes is needed by a child, how many of each type should Qhubeka make per day in order to minimise costs? 2. What is this minimum cost? Solution There are five basic steps to using linear programming to solve optimisation problems. Step 1: Assign variables In this problem, there are two types of bicycles and we want to find out how many of each kind should be produced. We need to assign a variable to each type of bicycle. You can assign any variables you like but, because we will be sketching linear functions, it is often simplest to just use $\scriptsize x$ and $\scriptsize y$. Let the number of Buffalo bicycles be $\scriptsize x$. Let the number of Cheetah bicycles be $\scriptsize y$. In both cases, the values of $\scriptsize x$ and $\scriptsize y$ are limited to natural numbers. We cannot have a fraction of a bicycle. Therefore $\scriptsize \{x\|x\in \mathbb{N}\}\text{ }$ and $\scriptsize \{y\|y\in \mathbb{N}\}\text{ }$. Step 2: Express the constraints In all cases a set of constraints is given to you. The first constraint we are given is that each day at most seven Buffalo bicycles can be built. Therefore, $\scriptsize x\le 7$. We are also told that at most four Cheetah bicycles can be built. Therefore, $\scriptsize y\le 4$. Each Buffalo needs three technicians and each Cheetah needs two technicians. But there are only $\scriptsize 18$ technicians. Therefore, we can say that $\scriptsize 3x+2y\le 18$ where $\scriptsize 3x$ is the total number of technicians needed to build $\scriptsize x$ Buffalos and $\scriptsize 2y$ is the total number of technicians needed to build $\scriptsize y$ Cheetahs. Step 3: Determine the objective function The objective function is the linear function we are trying to find a maximum or minimum solution for. In our case, we need to minimise cost. We are told that each Buffalo costs $\scriptsize \text{R2}\ 400$ to make and each Cheetah costs $\scriptsize \text{R1}\ 200$ to make. Therefore, $\scriptsize 2\ 400x+1\ 200y=C$ where $\scriptsize C$ is the total cost of producing $\scriptsize x$ Buffalo bicycles and $\scriptsize y$ Cheetah bicycles. Step 4: Graph the constraints It is time to start graphing our constraints so that we can optimise the objective function. Here are our constraints: • $\scriptsize x\le 7$ • $\scriptsize y\le 4$ • $\scriptsize 3x+2y\le 18$ Let’s start with $\scriptsize x\le 7$. If we plot the graph of $\scriptsize x=7$ we get a vertical line that cuts the x-axis at $\scriptsize 7$. But our graph needs to show $\scriptsize x\le 7$. Therefore, this is not a line but a region (see Figure 2). Remember that $\scriptsize x\in \mathbb{N}\text{ }$ so we are only including natural numbers between zero and seven. Next, we graph the next constraint, $\scriptsize y\le 4$ (see Figure 3). Again, $\scriptsize y\in \mathbb{N}\text{ }$ so we are only including natural numbers between zero and four. The overlapping region is what we are interested in because this represents all the values for $\scriptsize x$ and $\scriptsize y$ that would satisfy both constraints at the same time. But we have another constraint: $\scriptsize 3x+2y\le 18$. \scriptsize \begin{align}3x+2y & \le 18\\\therefore 2y & \le -3x+18\\\therefore y & \le -\displaystyle \frac{3}{2}x+9\end{align} First, we sketch the line $\scriptsize y=-\displaystyle \frac{3}{2}x+9$ and then we shade the region below this line because our constraint is $\scriptsize y\le -\displaystyle \frac{3}{2}x+9$ (see Figure 4). If the constraint was $\scriptsize y\ge -\displaystyle \frac{3}{2}x+9$, we would shade the region above the line. The region where all three separate regions overlap is called the feasible region. It is in this region that the solution to our problem lies. We could explicitly show that we are only interested in the natural numbers by plotting these points within the feasible region (see Figure 5). Step 5: Graph the objective function Our objective function is $\scriptsize 2\ 400x+1\ 200y=C$. \scriptsize \begin{align}2\ 400x+1\ 200y&=C\\ \therefore 1\ 200y & =-2\ 400x+C\\ \therefore y & =\displaystyle \frac{{-2\ 400}}{{1\ 200}}x+\displaystyle \frac{C}{{1\ 200}}\\ \therefore y&=-2x+\displaystyle \frac{C}{{1\ 200}}\end{align} Now we cannot really graph this linear function because we don’t yet know what the value of $\scriptsize C$ is. But remember, $\scriptsize C$ is what we are trying to minimise, for a maximum number of bicycles produced. In other words, we are trying to find the lowest value of $\scriptsize C$ that will make our objective function pass through one of the possible solution points with the largest coordinates within the feasible region. So, we start with any line with a gradient of $\scriptsize -2$ and move this line up and down to find the smallest y-intercept value that still lets the objective function pass through one of the points in the feasible region with the largest possible $\scriptsize x$ and $\scriptsize y$ coordinates. The objective function must pass through $\scriptsize (3,4)$ for the y-intercept to be a minimum ($\scriptsize 10$), and $\scriptsize x$ and $\scriptsize y$ to be maximum. The solution is shown below in Figure 6. We are now in a position to answer the questions. 1. Qhubeka must make three Buffalo bicycles and four Cheetah bicycles per day to minimise costs. 2. The minimum solution to the objective function results in a y-intercept of $\scriptsize (0,10)$. The objective function is $\scriptsize y=-2x+\displaystyle \frac{C}{{1\ 200}}\text{ }$. Therefore: \scriptsize \begin{align}\displaystyle \frac{C}{{1\ 200}} & =10\\\therefore C & =12\ 000\end{align}The minimum cost is $\scriptsize \text{R}12\ 000$. ### Note If you have an internet connection, visit a simulation of this linear programming solution. There is a slider to change the value of $\scriptsize C$ in order to find the minimum or maximum solution to the objective function. Let’s look at another example. ### Example 1.2 A TVET college class wants to raise funds to help buy new workshop tools. To raise money, they decide to host a cake sale and sell two different types of cakes – chocolate cake and carrot cake. The class estimates that they can sell at most $\scriptsize 60$ chocolate cakes and $\scriptsize 50$ carrot cakes. Each chocolate cake takes $\scriptsize 50\ \min$ to bake and each carrot cake takes $\scriptsize 80\ \min$ to bake. In total, the class calculates that they have $\scriptsize 4\ 400\min$ to bake all the cakes. Each chocolate cake costs $\scriptsize \text{R}30$ to make and each carrot cake costs $\scriptsize \text{R}40$ to make and they have a total of $\scriptsize \text{R2}\ 400$ to spend on ingredients. They can sell each chocolate cake for $\scriptsize \text{R3}0$ profit and each carrot cake for $\scriptsize \text{R6}0$ profit. 1. Assuming they can sell all the cakes they bake, how many of each type of cake should they bake to maximise their profit? 2. What will this maximum profit be? Solution Step 1: Assign variables Let the number of chocolate cakes be $\scriptsize x$. Let the number of carrot cakes be $\scriptsize y$. We cannot have fractions of cakes or negative numbers of cakes. Therefore $\scriptsize \{x\|x\in \mathbb{N}\}\text{ }$ and $\scriptsize \{y\|y\in \mathbb{N}\}\text{ }$. Step 2: Express the constraints • $\scriptsize x\le 60$ (at most $\scriptsize 60$ chocolate cakes can be sold) • $\scriptsize y\le 50$ (at most $\scriptsize 50$ carrot cakes can be sold) • $\scriptsize 50x+80y\le 4\ 400$ (each chocolate cake takes $\scriptsize 50\ \min$ and each carrot cake takes $\scriptsize 80\ \min$ to bake and they have a total of $\scriptsize 4\ 400\ \min$ to bake). • $\scriptsize 30x+40y=2\ 400$ (each chocolate cake costs $\scriptsize \text{R}30$ and each carrot cake costs $\scriptsize \text{R}40$ and they have a total of $\scriptsize \text{R}2\ 400$ to spend) Step 3: Determine the objective function $\scriptsize 30x+60y=P$, where $\scriptsize P$ is profit. Step 4: Graph the constraints \scriptsize \begin{align}50x+80y\le 4\ 400\\\therefore 80y\le -50x+4\ 400\\\therefore y\le -\displaystyle \frac{{50}}{{80}}x+\displaystyle \frac{{4\ 400}}{{80}}\\\therefore y\le -\displaystyle \frac{5}{8}x+55\end{align} \scriptsize \begin{align}30x+40y & \le 2\ 400\\\therefore 40y & \le -30x+2\ 400\\\therefore y & \le -\displaystyle \frac{{30}}{{40}}x+\displaystyle \frac{{2\ 400}}{{40}}\\\therefore y & \le -\displaystyle \frac{3}{4}x+60\end{align} The graphed constraints are shown in Figure 7. Step 5: Graph the objective function \scriptsize \begin{align}30x+60y & =P\\\therefore 60y & =-30x+p\\\therefore y & =-\displaystyle \frac{{30}}{{60}}x+\displaystyle \frac{P}{{60}}\\\therefore y & =-\displaystyle \frac{1}{2}x+\displaystyle \frac{P}{{60}}\end{align} An instance of the objective function is graphed in Figure 8. In this case, we want to maximise the objective function, in other words, we want to find the maximum y-intercept that the graph can have while still passing through one of the possible solution points in the feasible region. From the graph of the feasible region, we can see that this point is $\scriptsize (40,30)$. When the objective function is made to pass through this point its y-intercept is $\scriptsize 50$ (see Figure 9). 1. The class should bake $\scriptsize 40$ chocolate cakes and $\scriptsize 30$ carrot cakes to maximise their profit. 2. $\scriptsize y=-\displaystyle \frac{1}{2}x+\displaystyle \frac{P}{{60}}$ \scriptsize \begin{align}\displaystyle \frac{P}{{60}} & =50\\\therefore P=3\ 000\end{align} The maximum profit they can make is $\scriptsize \text{R3}\ 000$. ### Exercise 1.1 Question taken from Question 2.3 of NC(V) Mathematics Paper 1 March 2014 A clothing company manufactures yellow T-shirts ($\scriptsize x$) and black trousers ($\scriptsize y$) for a day-care school. The following system of inequalities are obtained: $\scriptsize x\ge 200$ $\scriptsize x+y\le 600$ $\scriptsize x+2y\le 900$ $\scriptsize 50x+100y\le 45\ 000$ 1. Sketch the graphs of the given constraints. 2. Indicate the feasible region. 3. Determine the number of t-shirts and pairs of trousers that will yield a maximum profit if $\scriptsize P=30x+40y$. 4. Determine the number of t-shirts and pairs of trousers that will yield a minimum profit if $\scriptsize P=30x+40y$. The full solutions are at the end of the unit. ## Summary In this unit you have learnt the following: • How to express given constraints using variables. • How to plot these constraints on the Cartesian plane. • How to determine the feasible region. • How to determine the objective function. • How to maximise or minimise the objective function. # Unit 1: Assessment #### Suggested time to complete: 15 minutes Question 1 is based on Question 2.5 from NC(V) Mathematics Paper 1 October 2012 1. An entrepreneur sells apples and bananas at the fruit stand. Suppose $\scriptsize x$ apples and $\scriptsize y$ bananas are sold every hour, and that the following constraints are applicable: $\scriptsize 10\le x\le 40$ $\scriptsize x+y\le 60$ $\scriptsize 15\le y\le 45$ $\scriptsize 2y+x\ge 60$ 1. Sketch the graph with the given constraints. 2. Determine the feasible region. 3. Determine how many apples and bananas should be sold every hour so that the maximum profit can be made if $\scriptsize P=3x+5y$. Question 2 is taken from Question 3.3 from NC(V) Mathematics Paper 1 October 2014 1. Lesedi is a small company that makes two types of cards, type A ($\scriptsize x$) and type B ($\scriptsize y$). With the available labour and materials, the following are the constraints inequalities: $\scriptsize x+y\le 200$ $\scriptsize x\ge 40$ $\scriptsize y\ge 10$ $\scriptsize x\le 150$ $\scriptsize y\le 120$ 1. Sketch the graph of the given constraints. 2. Determine the feasible region. 3. Find the values of $\scriptsize x$ and $\scriptsize y$ which will maximise the objective equation $\scriptsize P=5x+10y$. 4. Find the values of $\scriptsize x$ and $\scriptsize y$ which will minimise the objective equation $\scriptsize P=5x+10y$ and state the value of $\scriptsize P$ when this function is minimised. The full solutions are at the end of the unit. # Unit 1: Solutions ### Exercise 1.1 1. $\scriptsize x\in \mathbb{N}\text{ }$ and $\scriptsize y\in \mathbb{N}\text{ }$ Constraints are: $\scriptsize x\ge 200$ . \scriptsize \begin{align}x+y & \le 600\\\therefore y & \le -x+600\end{align} . \scriptsize \begin{align}x+2y & \le 900\\\therefore 2y & \le -x+900\\\therefore y & \le -\displaystyle \frac{1}{2}x+450\end{align} 2. The feasible region is the region bound by points $\scriptsize \text{A},\text{B},\text{C}$ and $\scriptsize \text{D}$. 3. . \scriptsize \begin{align} P&=30x+40y\\ \therefore 40y&=-30x+P\\ \therefore y&=-\displaystyle \frac{3}{4}x+\displaystyle \frac{P}{40} \end{align} The objective function is maximised when passing through $\scriptsize \text{C}(300,300)$. Therefore, $\scriptsize 300$ t-shirts and $\scriptsize 300$ pairs of trousers will yield a maximum profit. 4. The objective function is minimised when passing through $\scriptsize \text{A}(200,0)$. Therefore, $\scriptsize 200$ t-shirts and zero pairs of trousers will yield a minimum profit. Back to Exercise 1.1 ### Unit 1: Assessment 1. . 1. $\scriptsize x\in \mathbb{N}\text{ }$ and $\scriptsize y\in \mathbb{N}\text{ }$ Constraints are: $\scriptsize 10\le x\le 40$ \scriptsize \begin{align}x+y & \le 60\\\therefore y & \le -x+60\end{align} $\scriptsize 15\le y\le 45$ \scriptsize \begin{align}&2y+x\ge 60\\& \therefore y\ge -\displaystyle \frac{1}{2}x+30\quad \text{Remember that because }y\ge \text{ you need to take the region above the line}\end{align} 2. The feasible region is the region bound by points $\scriptsize \text{A},\text{B},\text{C},\text{D},\text{E}$ and $\scriptsize \text{F}$. 3. . \scriptsize \begin{align} P&=3x+5y\\ \therefore 5y&=-3x+P\\ \therefore y&=-\displaystyle \frac{3}{5}x+\displaystyle \frac{P}{5} \end{align} The objective function is maximised when passing through $\scriptsize \text{F}(15,45)$. Therefore, selling $\scriptsize 15$ apples and $\scriptsize 45$ bananas will yield a maximum profit. 2. . 1. $\scriptsize x\in \mathbb{N}\text{ }$ and $\scriptsize y\in \mathbb{N}\text{ }$ Constraints are: \scriptsize \begin{align}x+y & \le 200\\\therefore y & \le -x+200\end{align} $\scriptsize x\ge 40$ $\scriptsize y\ge 10$ $\scriptsize x\le 150$ $\scriptsize y\le 120$ 2. The feasible region is the region bound by points $\scriptsize \displaystyle \text{A},\text{B},\text{C},\text{D}$ and $\scriptsize \text{E}$. 3. . \scriptsize \begin{align} P&=5x+10y\\ \therefore 10y&=-5x+P\\ \therefore y&=-\displaystyle \frac{1}{2}x+\displaystyle \frac{P}{10} \end{align} The objective function is maximised when passing through $\scriptsize \text{E}(80,120)$. Therefore, $\scriptsize x=80$ and $\scriptsize y=120$. 4. The objective function is minimised when passing through $\scriptsize \text{B}(40,10)$. Therefore, $\scriptsize x=40$ and $\scriptsize y=10$. When minimised, the objective function has a y-intercept of $\scriptsize (0,30)$. \scriptsize \begin{align}\displaystyle \frac{P}{{10}} & =30\\\therefore P & =300\end{align} Therefore $\scriptsize P=300$ is the value of $\scriptsize P$ when the function is minimised. Back to Unit 1: Assessment
# How do you solve (x-4)(x+1)>=0 using a sign chart? Dec 1, 2016 The answer is x in ] -oo,-1 [ uu [4, +oo[ #### Explanation: Let $f \left(x\right) = \left(x - 4\right) \left(x + 1\right)$ Let's do the sign chart $\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$4$$\textcolor{w h i t e}{a a a a}$$+ \infty$ $\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$ $\textcolor{w h i t e}{a a a a}$$x - 4$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$ $\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$ So, $f \left(x\right) \ge 0$ when x in ] -oo,-1 [ uu [4, +oo[ graph{(x-4)(x+1) [-12.66, 12.65, -6.33, 6.33]}
# What is 4\times ( 13+ 8)- 8^{2} -:( 2\times 4)? Sep 22, 2016 $76$ #### Explanation: Always count the number of terms first. Each term simplifies to a single answer and these are added or subtracted in the last step. Within each term the normal order of operation applies. Brackets, then: the strongest operations of powers and roots Multiply and divide You may do more than one calculation in a line. $\left[4 \times \textcolor{\to m a \to}{\left(13 + 8\right)}\right] - \left[\textcolor{b l u e}{{8}^{2}} \div \textcolor{\lim e}{\left(2 \times 4\right)}\right] \text{ } \leftarrow$ there are 2 terms =$\text{ "4xx color(tomato)(21)color(white)(xxxxx)- color(blue)(64) div color(lime)(8)" } \leftarrow$ keep the terms separate =$\text{ } 84 \textcolor{w h i t e}{\times \times \times \times} - 8$ =$\text{ } 76$
# How do you find the B-value of a function? ## How do you find the B-value of a function? We’ve learned that in a quadratic function, f(x) = a * x^2 + b * x + c, the b-value is the middle value, the one multiplied by the x. ### What is the B in MX B? y = mx + b is the slope intercept form of writing the equation of a straight line. In the equation ‘y = mx + b’, ‘b’ is the point, where the line intersects the ‘y axis’ and ‘m’ denotes the slope of the line. The slope or gradient of a line describes how steep a line is. What is BX in a quadratic equation? The standard form of a quadratic function is , where a, b, and c are real numbers, and . ax2 is the quadratic term. bx is the linear term. c is the constant term. The coefficient of the quadratic term, a, determines how wide or narrow the graphs are, and whether the graph turns upward or downward. How do you write a perfect square quadratic equation? Steps to Solving Equations by Completing the Square 1. Rewrite the equation in the form x2 + bx = c. 2. Add to both sides the term needed to complete the square. 3. Factor the perfect square trinomial. 4. Solve the resulting equation by using the square root property. ## What is the value of B in the equation YB? The slope-intercept form of a linear equation (Algebra 1, Visualizing linear functions) – Mathplanet. ### How do you find the B value of two points? Steps to find the equation of a line from two points: 1. Find the slope using the slope formula. 2. Use the slope and one of the points to solve for the y-intercept (b). 3. Once you know the value for m and the value for b, you can plug these into the slope-intercept form of a line (y = mx + b) to get the equation for the line. How do you solve for B in MX B? Step 2: Use the formula y = mx + b to determine the y-intercept, b. Replace x and y in the formula with the coordinates of one of the given points, and replace m with the calculated value, (2). Find the equation of the line whose graph contains the points (1,–2) and (6,5). What is B in the slope formula? ## What is squared plus B Squared? The theorem states that if a right triangle has two sides equal to a and b, and a hypotenuse equal to c, then a squared plus b squared equals c squared. Consequently, how do you type squared? ### How do you write C Squared in math? The theorem states that if a right triangle has two sides equal to a and b, and a hypotenuse equal to c, then a squared plus b squared equals c squared. Keeping this in view, how do you type squared? To type a² symbol in Android , type ‘a’ and long press 2. For Windows users, In Notepad ,you can type Alt code for a². How do I use the inequalities section? You can usually find the exact answer or, if necessary, a numerical answer to almost any accuracy you require. The inequalities section lets you solve an inequality or a system of inequalities for a single variable. You can also plot inequalities in two variables.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> Conversions of Length, Mass, Capacity in Metric Units Determine equivalent units of metric measurement for length, mass and capacity. Estimated25 minsto complete % Progress Practice Conversions of Length, Mass, Capacity in Metric Units Progress Estimated25 minsto complete % Conversions of Length, Mass, Capacity in Metric Units Caleb is playing a math game on the computer. He has to compare metric units by filling in the blank with the correct sign. His choices are greater than (>), less than (<), or equal to (=). This is the problem Caleb is working on. What should Caleb choose? In this concept, you will learn to convert metric units of length, mass, or capacity. Guidance This concept combines a couple of different skills. Here you will combine converting metric units of length, mass, and capacity and multiplying decimals using powers of ten. Converting metric units using powers of ten involves moving the decimal point. Use this chart as a guide. Metri Unit kilo- hecto- deka- (base unit) meter/gram/liter deci- centi- milli- Symbol k h da m/g/L d c m Number 1,000 100 10 1 0.1 0.01 0.001 Here is a conversion problem. When you convert a larger unit to a smaller unit, multiply the number by a power of ten. 1 centimeter is equal to 10 millimeter. You multiply by a power of ten once or . Therefore, multiply 150 cm by 10. Move the decimal point one place to the right when multiplying by 10. The decimal point in a whole number is after the number. Add a zero placeholder to 150. Let’s look at this one involving capacity. When you convert a smaller unit to a larger unit, divide the number by a power of ten. 1,000 milliliters is equal to 1 liter. You divide by a power of ten 3 times or . Divide 1250 milliliters by 1,000. Move the decimal point three places to the left when dividing by 1,000. You can convert any unit of measure as long as you know the conversion equivalents and remember how to use powers of ten to move the decimal point to the left or to the right. Guided Practice Convert the units of measurement. First, identify the unit conversion. You are converting a smaller unit to a larger unit so you will divide. You know that a milliliter is one-thousandth of a liter. Then, divide 12,350 by a power of or 1,000. Move the decimal point three places to the left. 12,350 milliliters is equal to 12.35 liters. Examples Covert the units of measurement. Example 1 First, identify the unit conversion. You are converting a smaller unit to a larger unit so you will divide. Then, divide 1,340 mL by a power of or 1,000. Move the decimal point three places to the left. 1,340 milliliters is equal to 1.34 liters. Example 2 First, identify the unit conversion. You are converting a larger unit to a smaller unit so you will multiply. Then, multiply 66 g by a power of or 1,000. Move the decimal point three places to the right. 66 grams is equal to 0.066 milligrams. Example 3 First, identify the unit conversion. You are converting a smaller unit to a larger unit so you will divide. Then, divide 1,123 m by a power of or 1,000. Move the decimal point three places to the left. 1,123 meters is equal to 1.123 kilometers. Remember Caleb’s math problem? Caleb has to compare complete the math problem. First, convert 5.5 g to mg. You are converting a larger unit to a smaller unit so you will multiply. Then, multiply 5.5 by a power of or 1,000. Move the decimal point three places to the right. Next, substitute 5,500 mg for 5.5 g in the original problem and compare. Caleb should answer that 5.5 g is greater than (>) 4,500 mg. Explore More Convert the units of measurement using powers of ten. 1. 5.6 km = ______ m 2. 890 m = ______ km 3. 9230 m = ______ km 4. 40 cm = ______ mm 5. 5000 mm = ______ cm 6. 500 cm = ______ m 7. 7.9 m = ______ cm 8. 99 m = ______ cm 9. 460 cm = ______ m 10. 34 cm = ______ m 11. 4.3 km = ______ m 12. 760 m = ______ km 13. 4300 m = ______ km 14. 5000 g = ______ kg 15. 560 mL = ______ L 16. 6210 mL = ______ L 17. 8900 mL = ______ L 18. 18. 7.5 L = ______ mL 19. 19. 0.5 L = ______ mL To view the Explore More answers, open this PDF file and look for section 4.20. Vocabulary Language: English Capacity Capacity The volume of liquid an object or item can hold. Customary units of capacity include fluid ounces, cups, pints, quarts and gallons. Centimeter Centimeter A centimeter is a very small unit of metric measure. A centimeter is one hundredth of one meter. Kilometer Kilometer A kilometer is a common metric unit of length that is equal to 1,000 meters. Length Length Length is a measurement of how long something is. Examples of customary units of length are inches, feet, yards and miles. Mass Mass Mass is a measurement of the amount of matter in an object. Measurement Measurement A measurement is the weight, height, length or size of something. Meter Meter A meter is a common metric unit of length that is slightly larger than a yard. Metric System Metric System The metric system is a system of measurement commonly used outside of the United States. It contains units such as meters, liters, and grams, all in multiples of ten. Millimeter Millimeter A millimeter is the smallest common metric unit of length. Proportion Proportion A proportion is an equation that shows two equivalent ratios. Ratio Ratio A ratio is a comparison of two quantities that can be written in fraction form, with a colon or with the word “to”.
Home » Posts » The three different ways to solve a quadratic # The three different ways to solve a quadratic In this post I’ll look at the three different ways to solve a quadratic equation Oh yes, Sorry, first post for a while. I spent much of February making videos for my Facebook Group (And yes, that’s a link to it!) Let’s consider the equation x2 – 8x + 15 = 0 The first way we learn is to factorise, but spotting the factorisation isn’t always is easy or straightforward. If you can’t see an immediate factorisation, you might want to reach for the formula, The formula has the advantages that it can always be applied – and can tell you quite quickly if there are real* solutions. Its a good idea to work out the square root part first, as this will tell you if there are solutions and how many. In this case : 64 – 4 x 15 x 1 = 64 – 60 = 4. The square root of 4 is 2. (8 + 2)//2 = 5 and (8 – 2)/2 = 3 The fact that the solutions are integers suggests that the factorisation does exist, and we could have spotted it if we had looked a bit longer. What numbers multiply to make +15? 1 & 15 and 3 & 5; and the negatives. Our numbers need to add to -8, so the ‘answer’ is -3 and -5. The factorisation is (x – 3)(x – 5), and from that we can see the roots are 3 and 5. Roots is just a fancy mathematician way of saying the solution. The last method is called ‘Completing the square’, which isn’t immediately obvious but is actually the way the formula can be found. We take the number before the x (called the co-efficient) and halve it – In this case that gives us -4, which we write like this (x – 4)2 – which can be expanded to x2 – 8x + 16 So how can we get from x2 – 8x + 15 = 0 to x2 – 8x + 16? By adding 1 to each side to give x2 – 8x + 16 = 1, or (x – 4)2 = 1 Taking the square root of both sides gives x – 4 = +-1. Add 4 to both sides gives x = 3 and x = 5 We get the same answer for the three different ways to solve a quadratic – but it would be awkward if that were not so! For help on this and other Maths topics, see my current deals on Lockdown tuition.
# 1995 AIME Problems/Problem 8 ## Problem For how many ordered pairs of positive integers $(x,y),$ with $y are both $\frac xy$ and $\frac{x+1}{y+1}$ integers? ## Solution 1 Since $y\|x$, $y+1\|x+1$, then $\text{gcd}\,(y,x)=y$ (the bars indicate divisibility) and $\text{gcd}\,(y+1,x+1)=y+1$. By the Euclidean algorithm, these can be rewritten respectively as $\text{gcd}\,(y,x-y)=y$ and $\text{gcd}\, (y+1,x-y)=y+1$, which implies that both $y,y+1 \| x-y$. Also, as $\text{gcd}\,(y,y+1) = 1$, it follows that $y(y+1)\|x-y$. [1] Thus, for a given value of $y$, we need the number of multiples of $y(y+1)$ from $0$ to $100-y$ (as $x \le 100$). It follows that there are $\left\lfloor\frac{100-y}{y(y+1)} \right\rfloor$ satisfactory positive integers for all integers $y \le 100$. The answer is $$\sum_{y=1}^{99} \left\lfloor\frac{100-y}{y(y+1)} \right\rfloor = 49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.$$ ^ Another way of stating this is to note that if $\frac{x}{y}$ and $\frac{x+1}{y+1}$ are integers, then $\frac{x}{y} - 1 = \frac{x-y}{y}$ and $\frac{x+1}{y+1} - 1 = \frac{x-y}{y+1}$ must be integers. Since $y$ and $y+1$ cannot share common prime factors, it follows that $\frac{x-y}{y(y+1)}$ must also be an integer. ## Solution 2 We know that $x \equiv 0 \mod y$ and $x+1 \equiv 0 \mod y+1$. Write $x$ as $ky$ for some integer $k$. Then, $ky+1 \equiv 0\mod y+1$. We can add $k$ to each side in order to factor out a $y+1$. So, $ky+k+1 \equiv k \mod y+1$ or $k(y+1)+1 \equiv k \mod y+1$. We know that $k(y+1) \equiv 0 \mod y+1$. We finally achieve the congruence $1-k \equiv 0 \mod y+1$. We can now write $k$ as $(y+1)a+1$. Plugging this back in, if we have a value for $y$, then $x = ky = ((y+1)a+1)y = y(y+1)a+y$. We only have to check values of $y$ when $y(y+1)<100$. This yields the equations $x = 2a+1, 6a+2, 12a+3, 20a+4, 30a+5, 42a+6, 56a+7, 72a+8, 90a+9$. Finding all possible values of $a$ such that $y, we get $49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.$ ## Solution 3 (Rigorous, but straightforward) We use casework: For $y=1$, we have $3,5,\cdots ,99$, or $49$ cases. For $y=2$, we have $8,14,\cdots ,98$, or $16$ cases. For $y=3$, we have $15,27,\cdots ,99$, or $8$ cases. For $y=4$, we have $24,44\cdots ,84$, or $4$ cases. For $y=5$, we have $35,65,95$, or $3$ cases. For $y=6$, we have $48,90$, or $2$ cases. For $y=7$, we have $63$, or $1$ case. For $y=8$, we have $80$, or $1$ case. For $y=9$, we have $99$, or $1$ case. Adding, we get our final result of $\boxed{085}.$ Note: In general, all of the solutions for all $y$ are $y+by(y+1)$, for any $b$ such that $y+by(y+1)=x<100$ ~SirAppel Also, note that $n+1$ just starts with $(y+1)^2$, and adds $y(y+1)$ until it is greater than $100$. ~Yiyj1
1 / 10 # Section 4.2 Binomial Distributions - PowerPoint PPT Presentation Section 4.2 Binomial Distributions. Key Concept. This section presents a basic definition of a binomial distribution along with notation, and it presents methods for finding probability values. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about 'Section 4.2 Binomial Distributions' - truly Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript Binomial Distributions This section presents a basic definition of a binomial distribution along with notation, and it presents methods for finding probability values. Binomial probability distributions allow us to deal with circumstances in which the outcomes belong to two relevant categories such as acceptable/defective or survived/died. A binomial probability distribution results from a procedure that meets all the following requirements: 1. The procedure has a fixed number of trials. 2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.) 3. Each trial must have all outcomes classified into twocategories (commonly referred to as success and failure). 4. The probability of a success remains the same in all trials. Notation for Binomial Probability Distributions S and F (success and failure) denote two possible categories of all outcomes; p and q will denote the probabilities of S and F, respectively, so P(S) = p (p = probability of success) P(F) = 1 – p = q (q = probability of failure) n denotes the number of fixed trials. x denotes a specific number of successes in n trials, so x can be any whole number between 0 and n, inclusive. p denotes the probability of success in one of the n trials. q denotes the probability of failure in one of the n trials. P(x) denotes the probability of getting exactly x successes among the n trials. Probabilities We will now discuss three methods for finding the probabilities corresponding to the random variable x in a binomial distribution. n! P(x) = •px•qn-x (n –x)!x! for x = 0, 1, 2, . . ., n Method 1: Using the Binomial Probability Formula where n = number of trials x = number of successes among n trials p = probability of success in any one trial q = probability of failure in any one trial (q = 1 – p) Table A-1 in Appendix A Part of Table A-1 is shown below. With n = 12 and p = 0.80 in the binomial distribution, the probabilities of 4, 5, 6, and 7 successes are 0.001, 0.003, 0.016, and 0.053 respectively. STATDISK, Minitab, Excel and the TI-83 Plus calculator can all be used to find binomial probabilities. Excel TI-83 Plus calculator It is especially important to interpret results. The range rule of thumb suggests that values are unusual if they lie outside of these limits: Maximum usual values = µ + 2 Minimum usual values = µ– 2 
Graph of the function ${\displaystyle f(x)={\frac {x^{3}+3x^{2}-6x-8}{4}}.}$ In mathematics, the graph of a function ${\displaystyle f}$ is the set of ordered pairs ${\displaystyle (x,y)}$, where ${\displaystyle f(x)=y.}$ In the common case where ${\displaystyle x}$ and ${\displaystyle f(x)}$ are real numbers, these pairs are Cartesian coordinates of points in two-dimensional space and thus form a subset of this plane. In the case of functions of two variables, that is functions whose domain consists of pairs ${\displaystyle (x,y),}$ the graph usually refers to the set of ordered triples ${\displaystyle (x,y,z)}$ where ${\displaystyle f(x,y)=z,}$ instead of the pairs ${\displaystyle ((x,y),z)}$ as in the definition above. This set is a subset of three-dimensional space; for a continuous real-valued function of two real variables, it is a surface. In science, engineering, technology, finance, and other areas, graphs are tools used for many purposes. In the simplest case one variable is plotted as a function of another, typically using rectangular axes; see Plot (graphics) for details. A graph of a function is a special case of a relation. In the modern foundations of mathematics, and, typically, in set theory, a function is actually equal to its graph. [1] However, it is often useful to see functions as mappings, [2] which consist not only of the relation between input and output, but also which set is the domain, and which set is the codomain. For example, to say that a function is onto ( surjective) or not the codomain should be taken into account. The graph of a function on its own doesn't determine the codomain. It is common [3] to use both terms function and graph of a function since even if considered the same object, they indicate viewing it from a different perspective. Graph of the function ${\displaystyle f(x)=x^{4}-4^{x}}$ over the interval [−2,+3]. Also shown are the two real roots and the local minimum that are in the interval. ## Definition Given a mapping ${\displaystyle f:X\to Y,}$ in other words a function ${\displaystyle f}$ together with its domain ${\displaystyle X}$ and codomain ${\displaystyle Y,}$ the graph of the mapping is [4] the set ${\displaystyle G(f)=\{(x,f(x)):x\in X\},}$ which is a subset of ${\displaystyle X\times Y}$. In the abstract definition of a function, ${\displaystyle G(f)}$ is actually equal to ${\displaystyle f.}$ One can observe that, if, ${\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} ^{m},}$ then the graph ${\displaystyle G(f)}$ is a subset of ${\displaystyle \mathbb {R} ^{n+m}}$ (strictly speaking it is ${\displaystyle \mathbb {R} ^{n}\times \mathbb {R} ^{m},}$ but one can embed it with the natural isomorphism). ## Examples ### Functions of one variable Graph of the function ${\displaystyle f(x,y)=\sin \left(x^{2}\right)\cdot \cos \left(y^{2}\right).}$ The graph of the function ${\displaystyle f:\{1,2,3\}\to \{a,b,c,d\}}$ defined by ${\displaystyle f(x)={\begin{cases}a,&{\text{if }}x=1,\\d,&{\text{if }}x=2,\\c,&{\text{if }}x=3,\end{cases}}}$ is the subset of the set ${\displaystyle \{1,2,3\}\times \{a,b,c,d\}}$ ${\displaystyle G(f)=\{(1,a),(2,d),(3,c)\}.}$ From the graph, the domain ${\displaystyle \{1,2,3\}}$ is recovered as the set of first component of each pair in the graph ${\displaystyle \{1,2,3\}=\{x:\ \exists y,{\text{ such that }}(x,y)\in G(f)\}}$. Similarly, the range can be recovered as ${\displaystyle \{a,c,d\}=\{y:\exists x,{\text{ such that }}(x,y)\in G(f)\}}$. The codomain ${\displaystyle \{a,b,c,d\}}$, however, cannot be determined from the graph alone. The graph of the cubic polynomial on the real line ${\displaystyle f(x)=x^{3}-9x}$ is ${\displaystyle \{(x,x^{3}-9x):x{\text{ is a real number}}\}.}$ If this set is plotted on a Cartesian plane, the result is a curve (see figure). ### Functions of two variables Plot of the graph of ${\displaystyle f(x,y)=-\left(\cos \left(x^{2}\right)+\cos \left(y^{2}\right)\right)^{2},}$ also showing its gradient projected on the bottom plane. The graph of the trigonometric function ${\displaystyle f(x,y)=\sin(x^{2})\cos(y^{2})}$ is ${\displaystyle \{(x,y,\sin(x^{2})\cos(y^{2})):x{\text{ and }}y{\text{ are real numbers}}\}.}$ If this set is plotted on a three dimensional Cartesian coordinate system, the result is a surface (see figure). Oftentimes it is helpful to show with the graph, the gradient of the function and several level curves. The level curves can be mapped on the function surface or can be projected on the bottom plane. The second figure shows such a drawing of the graph of the function: ${\displaystyle f(x,y)=-(\cos(x^{2})+\cos(y^{2}))^{2}.}$
# GED Math : Area of a Triangle ## Example Questions ← Previous 1 3 ### Example Question #205 : Geometry And Graphs The above figure shows Square is the midpoint of is the midpoint of . What percent of the square is shaded? Explanation: The answer is independent of the sidelengths of the rectangle, so to ease calculations, we will arbitrarily assign to Square sidelength 4 inches. The shaded region can be divided by a perpendicular segment from to , then another from to that segment, as follows: By the fact that rectangles have opposite sides that are congruent, and by the fact that is the midpoint of and is the midpoint of , the shaded region is a composite of: Square , which has sides of length 2 and therefore area ; Right triangle , which has two legs of length 2 and, thus, area ; and, Right triangle , which has legs of length 4 and 2 and, thus, area . The area of the shaded region is therefore . Square has area , so the shaded region is of the square. ### Example Question #206 : Geometry And Graphs Give the area of the above triangle. Explanation: This triangle is isosceles by the converse of the Isosceles Triangle Theorem. The altitude of the triangle divides it into two triangles that are both right and that are congruent, as follows: Each triangle is a triangle. The altitude is the short leg of each triangle, and, therefore, its length is half that of the common hypotenuse, or 6. The long leg of each right triangle has length times that of the short leg, or , and the base of the entire large triangle is twice this, or The area of the triangle is half the product of the height and the base: ### Example Question #207 : Geometry And Graphs Note: Figure NOT drawn to scale. Refer to the above diagram. In terms of area, is what percent of ? Explanation: The area of a triangle is half the product of its baselength and its height. To find the area of , we can use the lengths of the legs and : To find the area of , we can use the hypotenuse , the length of which is 30, and the altitude perpendicular to it: In terms of area, is of . ### Example Question #208 : Geometry And Graphs Note: Figure NOT drawn to scale. Refer to the figure above. Give the area of the blue triangle. Explanation: The inscribed rectangle is a 20 by 20 square. Since opposite sides of the square are parallel, the corresponding angles of the two smaller right triangles are congruent; therefore, the two triangles are similar and, by definition, their sides are in proportion. The small top triangle has legs 10 and 20; the blue triangle has legs 20 and , where can be calculated with the following proportion: The legs of the blue triangle are 20 and 40; half their product is the area: ### Example Question #209 : Geometry And Graphs Find the area of a triangle with a base of 7in and a height that is two times the base. Explanation: To find the area of a triangle, we will use the following formula: where b is the base and h is the height of the triangle. Now, we know the base of the triangle is 7in. We also know the height of the triangle is two times the base. Therefore, the height is 14in. So, we can substitute. We get ### Example Question #210 : Geometry And Graphs Find the area of a triangle with a base of 8in and a height that is half the base. Explanation: To find the area of a triangle, we will use the following formula: where b is the base and h is the height of the triangle. Now, we know the base of the triangle is 8in. We also know the height of the triangle is half the base. Therefore, the height is 4in. So, we can substitute. We get ### Example Question #1 : Area Of A Triangle What is the area of a triangle with a base of and a height of ? Explanation: Write the formula for the area of a triangle. Substitute the base and height. Simplify the fractions. ### Example Question #2 : Area Of A Triangle Find the area of a triangle with a base of 6cm and a height that is three times the base. Explanation: To find the area of a triangle, we will use the following formula: where b is the base and h is the height of the triangle. Now, we know the base of the triangle is 6cm. We also know the height is three times the base. Therefore, the height is 18cm. So, we substitute. We get ### Example Question #3 : Area Of A Triangle Find the area of a triangle with a base of 10in and a height of 9in. Explanation: To find the area of a triangle, we will use the following formula: where b is the base and h is the height of the triangle. Now, we know the base of the triangle is 10in. We know the height of the triangle is 9in. So, we can substitute. We get ### Example Question #4 : Area Of A Triangle Find the area of a triangle with a base of 8cm and a height of 12cm. Explanation: To find the area of a triangle, we will use the following formula: where b is the base and h is the height of the triangle. We know the base of the triangle is 8cm. We know the height of the triangle is 12cm. Now, we can substitute. We get ← Previous 1 3
# In the present case by looking at the two equations • No School • AA 1 • 48 This preview shows page 29 - 32 out of 48 pages. is to rearrange the equation that is easier to rearrange. In the present case by looking at the two equations it seems that equation (1) might be easier to rearrange. 3 1 6 2 6 2 3 1 2 4 6 4 6 2 (3) 2 2 4 2 2 4 4 x x x y y x y y x - + = = - = = - = - Now let us put (3) into (2): what we have to do here is to replace 3 1 2 2 y by x - : 3 1 9 3 3 9 4 3 3 4 3 3 4 3 4 3 2 2 2 2 2 2 2 ( 4 ) 3 2 3 9 8 3 6 9 2 2 2 2 5 15 15 5 15 3 (4) 2 2 5 x y x x x x x x y x x x x x x x - - = ⇒ - - - = ⇒ - - + = ⇒ - + = + × - + × + - + + = = - = ⇒ - = = = - - 14243 Now putting (4) in (3) we get: 3 1 3 3 6 ( 3) 3 2 2 2 2 2 y y y = - × - = + = = Therefore, the solution to our system of equations is 3 3 x and y = - = . Note : Now if you hate fractions you can write 3 9 1.5 4.5 2 2 x as x and as . So you will have: 7.5 4 1.5 3 4.5 2.5 7.5 3 2.5 x x x x - + = + ⇒ - = = = - - . According to (3), 3 1 1.5 0.5 2 2 y x y x = - = - . 3 we knowthat x = - 1.5 0.5( 3) 1.5 1.5 3 therefore y = - - = + = . Subscribe to view the full document. 30 2.3.2 The elimination method In this method one has to subtract (or add) the equations from (or to) one another. Again let us try to solve our first example using the elimination method. 4 0 3 7 x y x y - = + = As before we have to label these equations: 4 0 (1) 3 7 (2) x y x y - = + = Now we need to match up the numbers in front of the ' ' x s or y s so that when we subtract (or add) one equation from (or to) the other, one of the variable disappear. Again we need to choose the easiest option. Looking at the system we can see that by adding (1) to (2) the ' y s disappear (or they are eliminated) 4 0 (1) 3 7 (2) 7 7 0 7 7 7 1 7 x y x y x x x - = + + = + = = = = We need now to replace x by its value in any of the equations (it does not matter which one; but always choose the easiest one to calculate). Looking at the system the easiest one to calculate is the first equation, (1). 4 0 we found that 1 4 1 0 4 0 4 x y and x y y y - = = × - = - = = As you can see we arrived to the same results as using the substitution method. We can also try our second example using the elimination approach. 2 4 6 4 3 3 x y x y + = - - = Again we have to label these equations:
Aptitude ➤ Decimal Fractions ➤ Set 1 #### Decimal Fractions : Important Facts and Formulae 1. Decimal Fractions: Fractions in which denominators are powers of 10 are known as decimal fractions. Thus, 1/10 = 1 tenth = .1; 1/100 = 1 hundredth = .01; 99/100 = 99 hundredths = .99; 7/1000 = 7 thousandths = .007, etc.; 2. Conversion of a Decimal into Vulgar Fraction: Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove the decimal point and reduce the fraction to its lowest terms. Thus, 0.25 = 25/100 = 1/4 ; 2.008 = 2008/1000 = 251/125. 3. Annexing Zeros and Removing Decimal Signs: a) Annexing zeros to the extreme right of a decimal fraction does not change its value. Thus, 0.8 = 0.80 = 0.800, etc. b) If numerator and denominator of a fraction contain the same number of decimal places, then we remove the decimal sign. Thus, 1.84/2.99 = 184/299 = 8/13; 0.365/0.584 = 365/584 = 5/8 ;. 4. Operations on Decimal Fractions: a) Addition and Subtraction of Decimal Fractions: The given numbers are so placed under each other that the decimal points lie in one column. The numbers so arranged can now be added or subtracted in the usual way. b) Multiplication of a Decimal Fraction By a Power of 10: Shift the decimal point to the right by as many places as is the power of 10. Thus, 5.9632 x 100 = 596.32; 0.073 x 10000 = 730. c) Multiplication of Decimal Fractions: Multiply the given numbers considering them without decimal point. Now, in the product, the decimal point is marked off to obtain as many places of decimal as is the sum of the number of decimal places in the given numbers. Suppose we have to find the product (0.2 x 0.02 x .002). Now, 2 x 2 x 2 = 8. Sum of decimal places = (1 + 2 + 3) = 6. .2 x .02 x .002 = .000008 d) Dividing a Decimal Fraction By a Counting Number: Divide the given number without considering the decimal point, by the given counting number. Now, in the quotient, put the decimal point to give as many places of decimal as there are in the dividend. Suppose we have to find the quotient (0.0204 ÷ 17). Now, 204 ÷ 17 = 12. Dividend contains 4 places of decimal. So, 0.0204 ÷ 17 = 0.0012 e) Dividing a Decimal Fraction By a Decimal Fraction: Multiply both the dividend and the divisor by a suitable power of 10 to make divisor a whole number. Now, proceed as above. Thus, 0.00066/0.11 = (0.00066 x 100)/(0.11 x 100) = 0.066/11 = 0.006 5. Comparison of Fractions: Suppose some fractions are to be arranged in ascending or descending order of magnitude, then convert each one of the given fractions in the decimal form, and arrange them accordingly. Let us to arrange the fractions 3/5 , 6/7 and 7/9 in descending order. Now, 3/5 = 0.6, 6/7 = 0.857, 7/9 = 0.777... Since, 0.857 > 0.777... > 0.6. So, 6/7 > 7/9 > 3/5 . 6. Recurring Decimal: If in a decimal fraction, a figure or a set of figures is repeated continuously, then such a number is called a recurring decimal. n a recurring decimal, if a single figure is repeated, then it is expressed by putting a dot on it. If a set of figures is repeated, it is expressed by putting a bar on the set. Thus, 1/3 = 0.333... = 0.3; 22/7 = 3.142857142857.... = 3.142857 Pure Recurring Decimal: A decimal fraction, in which all the figures after the decimal point are repeated, is called a pure recurring decimal. Converting a Pure Recurring Decimal into Vulgar Fraction: Write the repeated figures only once in the numerator and take as many nines in the denominator as is the number of repeating figures. Thus, 0.5 = 5/9 ; 0.53 = 53/99 ; 0.067 = 67/999 , etc. Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat and some of them are repeated, is called a mixed recurring decimal. Eg. 0.1733333.. = 0.173. Converting a Mixed Recurring Decimal Into Vulgar Fraction: In the numerator, take the difference between the number formed by all the digits after decimal point (taking repeated digits only once) and that formed by the digits which are not repeated. In the denominator, take the number formed by as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits. Thus, 0.16 = (16 - 1)/90 = 15/90 = 1/6 ; 0.2273 = (2273 - 22)/9900 = 2251/9900 . 7. Some Basic Formulae: i. (a + b)(a - b) = (a2 - b2) ii. (a + b)2 = (a2 + b2 + 2ab) iii. (a - b)2 = (a2 + b2 - 2ab) iv. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) v. (a3 + b3) = (a + b)(a2 - ab + b2) vi. (a3 - b3) = (a - b)(a2 + ab + b2) vii. (a3 + b3 + c3 - 3abc) = (a + b + c)(a2 + b2 + c2 - ab - bc - ac) viii. When a + b + c = 0, then a3 + b3 + c3 = 3abc. Question 11 Q11. The value of (0.051x0.051x0.051 + 0.041x0.041x0.041) / (0.051x0.051 - 0.051x0.041 + 0.041x0.041) is ? Question 12 Q12.
# How do you construct a polygon in a circle? ## How do you construct a polygon in a circle? Procedure: Set the compass to the radius of the circle and strike six equidistant arcs about its perimeter. Connect two neighboring intersections to the center of the circle. Bisect the resulting angle. Beginning at the intersection of the bisector and the circle strike six more arcs around the circle. When constructing an inscribed polygon with a compass and straightedge how should you start the construction 5 points? When constructing an inscribed polygon with a compass and straightedge, how should you start the construction? Place a point on your paper and then use a compass to construct a circle. When constructing inscribed polygons and parallel line how are the steps different? When constructing inscribed polygons and parallel lines, how are the steps different? Four right angles are created. A compass is used to copy an angle. ### What is inscribed polygon? An inscribed polygon might refer to any polygon which is inscribed in a shape, especially: A cyclic polygon, which is inscribed in a circle (the circumscribed circle) A midpoint polygon of another polygon. What is the first step in constructing a regular pentagon inscribed in a circle? Draw an arc of a circle, center point D, with the radius BA until this circular arc cuts the other circular arc about point B; there arises the intersection C. Connect the points BCDEA. This results in the pentagon. When constructing inscribed polygons and perpendicular lines How are the steps similar 5 points? When constructing inscribed polygons and perpendicular lines how are the steps similar? Intersecting arcs are created and connected. What is the first step when constructing parallel lines? You just studied 10 terms! ## When constructing an inscribed square by hand which step comes after constructing a circle? When constructing an inscribed square by hand, which step comes after constructing a circle? Set compass to the diameter of the circle. Set compass to the radius of the circle. Use a straightedge to draw a diameter of the circle. Which step is the same when constructing an inscribed square an inscribed equilateral triangle? For the square construction, the diameter will be used, but for the equilateral triangle construction, the radius will be used. Which step is the same when constructing an inscribed regular hexagon and an inscribed equilateral triangle? Set the compass width to the radius of the circle. What does it mean if a polygon is inscribed in a circle? Inscribed Polygon A polygon is inscribed in a circle if all its vertices are points on the circle and all sides are included within the circle. All regular polygons can be inscribed in a circle. The center of an inscribed polygon is also the center of the circumscribed circle. ### What are the tools used in constructing a polygon? Any regular polygon can be constructed with geometry software or the appropriate tools: • Compass: A device that allows you to create a circle with a given radius. • Straightedge: Anything that allows you to produce a straight line. What happens if a circle is inscribed in a polygon? A circle inscribed in a polygon is a circle inside this polygon that touches all its sides. Thus, each side of a polygon is a tangent for an inscribed circle. The center of an inscribed circle is inside the polygon. A polygon containing an inscribed circle is called a circumscribed polygon. What is inscribed center of a regular polygon? In geometry, the incircle or inscribed circle of a polygon is the largest circle contained in the polygon; it touches (is tangent to) the many sides. The center of the incircle is called the polygon’s incenter . ## What is the difference between Polygon and circle? As nouns the difference between circle and polygon is that circle is (lb) a two-dimensional geometric figure, a line, consisting of the set of all those points in a plane that are equally distant from another point while polygon is . As a verb circle is to travel around along a curved path. What is the definition for circumscribed polygons? A circumscribed polygon is a polygon in which each side is a tangent to a circle . The circle is inscribed in the polygon and the polygon is circumscribed about the circle. (It is a circle in a polygon) Inscribed and Circumscribed Polygons
# How do you solve -9=(-4+m)/2? Dec 22, 2016 $m = - 14$ #### Explanation: First, multiply each side of the equation by $2$ to eliminate the fraction and to keep the equation balanced. Eliminating the fraction will make the problem easier to work with: $\textcolor{red}{2} \times - 9 = \textcolor{red}{2} \times \frac{- 4 + m}{2}$ $- 18 = \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times \frac{- 4 + m}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}$ $- 18 = - 4 + m$ Now, we can isolate the $m$ term and solve for $m$: $- 18 + \textcolor{red}{4} = - 4 + m + \textcolor{red}{4}$ $- 14 = - 4 + 4 + m$ $- 14 = 0 + m$ $- 14 = m$ Dec 22, 2016 $m = - 14$ #### Explanation: Given $\textcolor{w h i t e}{\text{XXX}} - 9 = \frac{- 4 + m}{2}$ If we multiply both sides by $2$ we can get rid of the fraction on the right side: $\textcolor{w h i t e}{\text{XXX}} - 9 \textcolor{b l u e}{\times 2} = \frac{- 4 + m}{\cancel{2}} \textcolor{b l u e}{\times \cancel{2}}$ $\textcolor{w h i t e}{\text{XXX}} - 18 = - 4 + m$ Now adding $4$ to both sides will isolate the constant on one side and the variable on the other side: $\textcolor{w h i t e}{\text{XXX}} - 18 \textcolor{red}{+ 4} = \cancel{- 4} + m \textcolor{red}{+ \cancel{4}}$ $\textcolor{w h i t e}{\text{XXX}} - 14 = m$ or (in the more common order) $\textcolor{w h i t e}{\text{XXX}} m = - 14$
# Chapter 8.1 Notes: Find Angle Measures in Polygons Goal: You will find interior and exterior angle measures in polygons. ## Presentation on theme: "Chapter 8.1 Notes: Find Angle Measures in Polygons Goal: You will find interior and exterior angle measures in polygons."— Presentation transcript: Chapter 8.1 Notes: Find Angle Measures in Polygons Goal: You will find interior and exterior angle measures in polygons. Properties of a Polygon In a polygon, two vertices that are endpoints of the same side are called consecutive vertices. A diagonal of a polygon is a segment that joins two nonconsecutive vertices. Finding Interior Angle Measures in Polygons Theorem 8.1 Polygon Interior Angles Theorem: The sum of the measures of the interior angles of a convex n-gon is (n – 2)180. Corollary to Theorem 8.1 Interior Angles of a Quadrilateral: The sum of the measures of the interior angles of a quadrilateral is 360 o. Ex.1: Find the sum of the measures of the interior angles of a convex octagon. GUIDED PRACTICE for Examples 1 and 2 The coin shown is in the shape of a regular 11- gon. Find the sum of the measures of the interior angles. 1. ANSWER 1620° The sum of the measures of the interior angles of a convex polygon is 1440°. Classify the polygon by the number of sides. 2. ANSWER decagon Ex.2: The sum of the measures of the interior angles of a convex polygon is 900 o. Classify the polygon by the number of sides. Ex.3: Find the value of x in the diagram shown. Ex.4: In the diagram below, find and Ex.5: The sum of the measures of the interior angles of a convex polygon is 1440 o. Classify the polygon by the number of sides. Finding Exterior Angle Measures in Polygons Theorem 8.2 Polygon Exterior Angles Theorem: The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex, is 360 o. Exterior Angle Sum Theorem Exterior Angle Sum Theorem If a polygon is convex, then the sum of the measures of the exterior angles, one at each vertex, is 360°. Ex.6: What is the value of x in the diagram shown? Ex.7: A convex hexagon has exterior angles with measures 34 o, 49 o, 58 o, 67 o, and 75 o. What is the measure of an exterior angle at the sixth vertex? Ex.8: The trampoline shown is shaped like a regular dodecagon. Find (a) the measure of each interior angle, and (b) the measure of each exterior angle. Ex.9: A stop sign is shaped like a regular octagon. Find (a) the measure of each interior angle, and (b ) the measure of each exterior angle. Download ppt "Chapter 8.1 Notes: Find Angle Measures in Polygons Goal: You will find interior and exterior angle measures in polygons." Similar presentations
## What is standard deviation formula with example? The standard deviation measures the spread of the data about the mean value. It is useful in comparing sets of data which may have the same mean but a different range. For example, the mean of the following two is the same: 15, 15, 15, 14, 16 and 2, 7, 14, 22, 30. ## What does std dev tell you? What is standard deviation? Standard deviation tells you how spread out the data is. It is a measure of how far each observed value is from the mean. In any distribution, about 95% of values will be within 2 standard deviations of the mean. ## How do you calculate 3 standard deviations from the mean? Third, calculate the standard deviation, which is simply the square root of the variance. So, the standard deviation = √0.2564 = 0.5064. Fourth, calculate three-sigma, which is three standard deviations above the mean. In numerical format, this is (3 x 0.5064) + 9.34 = 10.9. ## How do you find the sample standard deviation? Sample standard deviationStep 1: Calculate the mean of the data—this is xˉx, with, bar, on top in the formula.Step 2: Subtract the mean from each data point. Step 3: Square each deviation to make it positive.Step 4: Add the squared deviations together.Step 5: Divide the sum by one less than the number of data points in the sample. ## What is Z value? The value of the z-score tells you how many standard deviations you are away from the mean. A positive z-score indicates the raw score is higher than the mean average. For example, if a z-score is equal to +1, it is 1 standard deviation above the mean. A negative z-score reveals the raw score is below the mean average. ## How do you find Sigma? The symbol for Standard Deviation is σ (the Greek letter sigma).Say what?Work out the Mean (the simple average of the numbers)Then for each number: subtract the Mean and square the result.Then work out the mean of those squared differences.Take the square root of that and we are done! You might be interested: Equation for angular frequency ## How do you interpret skewness? The rule of thumb seems to be:If the skewness is between -0.5 and 0.5, the data are fairly symmetrical.If the skewness is between -1 and – 0.5 or between 0.5 and 1, the data are moderately skewed.If the skewness is less than -1 or greater than 1, the data are highly skewed. ## What does skewness and kurtosis tell us? Skewness is a measure of symmetry, or more precisely, the lack of symmetry. Kurtosis is a measure of whether the data are heavy-tailed or light-tailed relative to a normal distribution. That is, data sets with high kurtosis tend to have heavy tails, or outliers. ## Why is the mean useful? An important property of the mean is that it includes every value in your data set as part of the calculation. In addition, the mean is the only measure of central tendency where the sum of the deviations of each value from the mean is always zero. ## What percentage is 4 sigma? Five-sigma corresponds to a p-value, or probability, of 3×107, or about 1 in 3.5 million.Don’t be so sure. σConfidence that result is real 3 σ99.87% 3.5 σ99.98% > 4 σ100% (almost) ## What is the 3 standard deviation rule? The Empirical Rule states that 99.7% of data observed following a normal distribution lies within 3 standard deviations of the mean. Under this rule, 68% of the data falls within one standard deviation, 95% percent within two standard deviations, and 99.7% within three standard deviations from the mean. ## Why is standard deviation 3 times? In the empirical sciences the so-called three-sigma rule of thumb expresses a conventional heuristic that nearly all values are taken to lie within three standard deviations of the mean, and thus it is empirically useful to treat 99.7% probability as near certainty. ## What is the symbol for sample mean? You might be interested: Define linear equation ## What is the formula of standard deviation for grouped data? Var = (Mean square) – (Mean)^2 To find the standard deviation, take the square root of the variance. StDev = sqrt(Var) Note that these values are estimates, because with grouped data, you don’t have the exact figures to work with. ### Releated #### Arrhenius equation r What relationship is described by the Arrhenius equation? The Arrhenius equation describes the relationship between the rate constant, k, and the energy of activation, Ea. In this equation, A is an empirical constant, R, is the ideal-gas constant, e is the base of natural logarithms, and T is the absolute temperature. How do you solve […] #### Siri equation How do you calculate body fat percentage with density? Once body density is determined, percent bodyfat (%BF) can be calculated using the Siri equation below : % Body Fat = (495 / Body Density) – 450. What is density of a body? In physics, density is defined as an object’s mass divided by its volume. […]
# How do you simplify 8^(2/3)? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 60 Jul 24, 2016 $= 4$ #### Explanation: ${8}^{\frac{2}{3}}$ can be written as root3(8^2 $= \sqrt[3]{64}$ =root3((4)(4)(4) $= 4$ Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 8 Jun 30, 2017 $4$ #### Explanation: One of the laws of indices states: ${x}^{\frac{m}{n}} = \sqrt[n]{\left({x}^{m}\right)} = {\left(\sqrt[n]{x}\right)}^{m}$ Therefore ${8}^{\frac{2}{3}}$ can be written as $\sqrt[3]{\left({8}^{2}\right)} \mathmr{and} {\left(\sqrt[3]{8}\right)}^{2}$ I prefer to use the second form because it uses smaller numbers - the root is found first and then that is squared. $\sqrt[3]{8} = 2 \mathmr{and} {2}^{2} = 4$ So: ${\left(\textcolor{b l u e}{\sqrt[3]{8}}\right)}^{2} = {\textcolor{b l u e}{2}}^{2} = 4$ Consider a question such as ${32}^{\frac{3}{5}}$ $\sqrt[5]{\textcolor{b l u e}{\left({32}^{3}\right)}}$ would mean finding ${32}^{3}$ first ... ouch! ${\left(\textcolor{b l u e}{\sqrt[5]{32}}\right)}^{3}$ would mean finding $\sqrt[5]{32}$ first. That is $\textcolor{b l u e}{2}$ ${\left(\textcolor{b l u e}{\sqrt[5]{32}}\right)}^{3} = {\textcolor{b l u e}{2}}^{3} = 8$ • 13 minutes ago • 15 minutes ago • 19 minutes ago • 26 minutes ago • 3 minutes ago • 7 minutes ago • 7 minutes ago • 10 minutes ago • 13 minutes ago • 13 minutes ago • 13 minutes ago • 15 minutes ago • 19 minutes ago • 26 minutes ago
# How many standard deviation units at the baseline of a normal curve? ## How many standard deviation units at the baseline of a normal curve? The standard normal distribution always has a mean of zero and a standard deviation of one. ## What is the standard deviation of a normal curve? The standard normal distribution is a normal distribution with a mean of zero and standard deviation of 1. ## How do you calculate standard deviation from a normal curve? First, it is a very quick estimate of the standard deviation. The standard deviation requires us to first find the mean, then subtract this mean from each data point, square the differences, add these, divide by one less than the number of data points, then (finally) take the square root. ## Where can we locate the mean in the normal curve? The mean is in the center of the standard normal distribution, and a probability of 50% equals zero standard deviations. ## What does the normal curve represent? The normal distribution is a continuous probability distribution that is symmetrical on both sides of the mean, so the right side of the center is a mirror image of the left side. The area under the normal distribution curve represents probability and the total area under the curve sums to one. ## What are the characteristics of the normal curve? Characteristics of a Normal Curve All normal curves are bell-shaped with points of inflection at μ ± σ . All normal curves are symmetric about the mean . Therefore, by the definition of symmetry, the normal curve is symmetric about the mean . The area under an entire normal curve is 1. ## What is normal probability curve and its characteristics? A normal curve is a bell-shaped curve which shows the probability distribution of a continuous random variable. Moreover, the normal curve represents a normal distribution. The total area under the normal curve logically represents the sum of all probabilities for a random variable. ## How do you construct a normal curve? To create a normal distribution graph with a specified mean and standard deviation, start with those values in some cells in a worksheet. The example uses a mean of 10 and a standard deviation of 2. Enter those values in cells F1 and H1. Next, set up the x-values for a standard normal curve. ## Why do we standardize the normal distribution? The standard score (more commonly referred to as a z-score) is a very useful statistic because it (a) allows us to calculate the probability of a score occurring within our normal distribution and (b) enables us to compare two scores that are from different normal distributions. ## How do you standardize standard deviation? Typically, to standardize variables, you calculate the mean and standard deviation for a variable. Then, for each observed value of the variable, you subtract the mean and divide by the standard deviation. ## Is it better to have a higher or lower z score? z-score is like percentile. ... Z score shows how far away a single data point is from the mean relatively. Lower z-score means closer to the meanwhile higher means more far away. Positive means to the right of the mean or greater while negative means lower or smaller than the mean. ## What is considered an extreme Z score? An extreme score happens when z value is above 2 or below -2. if x=50, z=(45-45)/2=0 ... ## What are z-scores for? In finance, Z-scores are measures of an observation's variability and can be used by traders to help determine market volatility. The Z-score is also sometimes known as the Altman Z-score. A Z-Score is a statistical measurement of a score's relationship to the mean in a group of scores. 0. ## What is the z-score for the 60th percentile? Percentilez-Score 590. ## How do you find a normal distribution percentage? Example, continued. Consider the normal distribution N(100, 10). To find the percentage of data below 105. ## What is the z score of 5%? The z-score of 0. ## What is Z value for 5 significance level? a z-score less than or equal to the critical value of -1. ## How do you find the area between the mean and the Z score? To find the area between two points we : 1. convert each raw score to a z-score. 2. find the area for the two z-scores. 3. subtract the smaller area from the larger area.

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