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# Chapter 6: Exploring Data: Relationships Lesson Plan Displaying Relationships: Scatterplots Making Predictions: Regression Lines Correlation Least-Squares.
## Presentation on theme: "Chapter 6: Exploring Data: Relationships Lesson Plan Displaying Relationships: Scatterplots Making Predictions: Regression Lines Correlation Least-Squares."— Presentation transcript:
Chapter 6: Exploring Data: Relationships Lesson Plan Displaying Relationships: Scatterplots Making Predictions: Regression Lines Correlation Least-Squares Regression Interpreting Correlation and Regression 1 Mathematical Literacy in Today’s World, 8th ed. For All Practical Purposes © 2009, W.H. Freeman and Company
Chapter 6: Exploring Data: Distributions Displaying Relationships 2 Relationship Between Two Variables Examine data for two variables to see if there is a relationship between the variables. Does one influence the other? Study both variables on the same individual. If a relationship exists between variables, typically one variable influences or causes a change in another variable. Explanatory variable explains, or causes, the change in another variable. Response variable measures the outcome, or response to the change. Response variable – A variable that measures an outcome or result of a study (observed outcome). Explanatory variable – A variable that explains or causes change in the response variable.
Chapter 6: Exploring Data: Distributions Displaying Relationships: Scatterplots Data to Be Used for a Scatterplot A scatterplot is a graph that shows the relationship between two numerical variables, measured on the same individual. Explanatory variable, x, is plotted on the horizontal axis, (x). Response variable, y, is plotted on the vertical axis (y). Each pair of related variables (x, y) is plotted on the graph. 3 Example: A study done to see how the number of beers that a student drinks predicts his/her blood alcohol content (BAC). Results of 16 students: Student12345678 Beers52983735 BAC0.100.030.190.120.040.0950.070.06 Student910111213141516 Beers35465714 BAC0.020.050.070.100.130.090.010.05 Explanatory variable, x = beers drunk Response variable, y = BAC level
Chapter 6: Exploring Data: Distributions Displaying Relationships: Scatterplots Scatterplot Example continued: The scatterplot of the blood alcohol content, BAC, (y, response variable) against the number of beers a young adult drinks (x, explanatory variable). The data from the previous table are plotted as points on the graph (x, y). 4 Examining This Scatterplot… 1. What is the overall pattern (form, direction, and strength)? Form – Roughly a straight-line pattern. Direction – Positive association (both increase). Strength – Moderately strong (mostly on line). 2.Any striking deviations (outliers)? Not here. BAC vs. number of beers consumed Outliers – A deviation in a distribution of a data point falling outside the overall pattern.
Chapter 6: Exploring Data: Distributions Regression Lines Regression Line A straight line that describes how a response variable y changes as an explanatory variable x changes. Regression lines are often used to predict the value of y for a given value of x. 5 A regression line has been added to be able to predict blood alcohol content from the number of beers a student drinks. Graphically, you can predict that if x = 6 beers, then y = 0.095 BAC. (Legal limit for driving in most states is BAC = 0.08.) BAC vs. number of beers consumed
Chapter 6: Exploring Data: Distributions Regression Lines Using the Equation of the Line for Predictions It is easier to use the equation of the line for predicting the value of y, given the value of x. Using the equation of the line for the previous example: predicted BAC = −0.0127 + (0.01796)(beers) y = −0.127 + 0.01796 (x) For a young adult drinking 6 beers (x = 6): predicted BAC = −0.0127 + 0.01796 (6) = 0.095 Straight Lines A straight line for predicting y from x has an equation of the form: predicted In this equation, m is the slope, the amount by which y changes when x increases by 1 unit. The number b is the y-intercept, the value of y when x =0. 6
Chapter 6: Exploring Data: Distributions Correlation 9 Correlation, r Measures the direction and strength of the straight-line relationship between two numerical variables. A correlation r is always a number between −1 and 1. It has the same sign as the slope of a regression line. r > 0 for positive association (increase in one variable causes an increase in the other). r < 0 for negative association. (increase in one variable causes a decrease in the other) Perfect correlation r = 1 or r = −1 occurs only when all points lie exactly on a straight line. The correlation moves away from 1 or −1 (toward zero) as the straight-line relationship gets weaker. Correlation r = 0 indicates no straight-line relationship.
Chapter 6: Exploring Data: Distributions Correlation Correlation Correlation is strongly affected by a few outlying observations. (Also, the mean and standard deviation are affected by outliers.) 10 Equation of the Correlation To calculate the correlation, suppose you have data on variable x and y for n individuals. From the data, you have the values calculated for the means and standard deviations x and y. The means and standard deviations for the two variables are ¯ and s x, for the x-values, and ¯ and s y for the y-values. The correlation r between x and y is: 1 (x 1 − ¯) (y 1 − ¯) + (x 2 − ¯) (y 2 − ¯) + … + (x n − ¯) (y n − ¯) n – 1 s x s y s x s y s x s y () r = xyxyxy y x
Chapter 6: Exploring Data: Distributions Correlation Correlation The scatterplots below show examples of how the correlation r measures the direction and strength of a straight-line association. 11
Chapter 6: Exploring Data: Distributions Least Squares Regression Least-Squares Regression Line A line that makes the sum of the squares of the vertical distances of the data points from the line as small as possible. Equation of the Least-Squares Regression Line From the data for an explanatory variable x and a response variable y for n individuals, we have calculated the means ¯, ¯, and standard deviations s x, s y, as well as their correlation r. 12 This equation was used to calculate the line for predicting BAC for number of beers drunk. Predicted y = −0.0127 + 0.01796 x The least-squares regression line is the line: Predicted With slope … And y-intercept … xy
Fuel Economy ModelCity Mileage (x) Highway Mileage (y) Acura RL1624 BMW 550i1523 Chev Malibu2230 Dodge Avenger2130 Hyundai Elantra2433 Lexus ES 3501927 Mercury Milan2029 Mitsu galant2027 Nissan Sentra2129 Nissan Versa2733 Pont Gran Prix1828 Toyota Camry2131
Chapter 6: Exploring Data: Distributions Interpreting Correlation and Regression A Few Cautions When Using Correlation and Regression Both the correlation r and least-squares regression line can be strongly influenced by a few outlying points. Always make a scatterplot before doing any calculations. Often the relationship between two variables is strongly influenced by other variables. Before conclusions are drawn based on correlation and regression, other possible effects of other variables should be considered. A strong association between two variables is not enough to draw conclusions about cause and effect. Sometimes an observed association really does reflect cause and effect (such as drinking beer causes increased BAC). Sometimes a strong association is explained by other variables that influence both x and y. Remember, association does not imply causation. 15
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## Intermediate Algebra (12th Edition)
$(7x+5q)(2x-5q)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $14x^2-25xq-25q^2 ,$ use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ In the trinomial expression above, $a= 14 ,b= -25 ,\text{ and } c= -25 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 14(-25)=-350$ and whose sum is $b$ are $\left\{ 10,-35 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 14x^2+10xq-35xq-25q^2 .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} (14x^2+10xq)-(35xq+25q^2) \\\\= 2x(7x+5q)-5q(7x+5q) \\\\= (7x+5q)(2x-5q) .\end{array} |
# How do you find all the zeros of X^3-3x^2+4x-2?
May 2, 2016
This cubic polynomial has zeros: $x = 1$, $x = 1 - i$, $x = 1 + i$
#### Explanation:
First note that the sum of the coefficients is zero. That is:
$1 - 3 + 4 - 2 = 0$
Hence $x = 1$ is a zero and $\left(x - 1\right)$ a factor:
${x}^{3} - 3 {x}^{2} + 4 x - 2 = \left(x - 1\right) \left({x}^{2} - 2 x + 2\right)$
The remaining quadratic factor only has Complex zeros, which we can find by completing the square and using the difference of squares identity:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
with $a = \left(x - 1\right)$ and $b = i$ as follows:
${x}^{2} - 2 x + 2$
$= {\left(x - 1\right)}^{2} - 1 + 2$
$= {\left(x - 1\right)}^{2} + 1$
$= {\left(x - 1\right)}^{2} - {i}^{2}$
$= \left(\left(x - 1\right) - i\right) \left(\left(x - 1\right) + i\right)$
$= \left(x - 1 - i\right) \left(x - 1 + i\right)$
Hence zeros $x = 1 \pm i$ |
# Triangles
## Types of Triangles for Class 5 Math
There are different types of triangles in mathematics. Here students will learn about the different types of triangles.
In this learning concept, the students will also learn to
• Classify types of triangles based on sides.
• Identify types of triangles on the basis of angles.
• Evaluate types of triangles and their properties.
Each concept is explained to class 5 maths students using illustrations, examples, and mind maps. Students can assess their learning by solving the two printable worksheets given at the page’s end.
Download the class 5 maths types of triangles worksheet and check the solutions to the types of triangle questions for class 5 provided in PDF format.
### How Many Types of Triangles Are There?
• A triangle is a two-dimension closed figure.
• It has three sides and three interior angles.
• The point where two sides intersect each other is called the vertices of the triangle.
• A triangle has three vertices.
### Types of Triangles
1. Types of Triangles Based on Sides:
A triangle is of three types:
• Equilateral triangle
• Isosceles triangle
• Scalene triangle
Equilateral Triangle
In an equilateral triangle, the length of all the sides of the triangle is equal.
Isosceles Triangle
In an isosceles triangle, the length of any two sides of the triangle is equal.
Scalene Triangle
In a scalene triangle, the length of all the sides of the triangle is unequal.
### 2. Types of Triangles on the Basis of Angles:
There are three types of triangles:
• Acute angled triangle
• Obtuse angled triangle
• Right-angled triangle
Acute angled Triangle
When each of the angles of a triangle is less than 90° is called an acute-angled triangle or acute triangle.
Obtuse angled Triangle
When any one of the angles of a triangle is more than 90° is called an obtuse-angled triangle or obtuse triangle.
Right angled Triangle
When any one of the angles of a triangle is equal to 90° is called a right-angled triangle or right-angled triangle or right triangle.
### Angles in Triangle
• There are three angles in a triangle.
• The sum of the angles of a triangle is equal to 180°.
The sum of the angle = 45° + 90° + 45° = 180°
Therefore, the given figure is a triangle.
The sum of the angle = 75° + 56° + 88° = 219°
Therefore, the given figure is not a triangle.
Find the Missing Angle:
#### Example:
Find the missing angle of the given triangle
#### Solution:
The sum of the given angles = 60° + 70° = 130°
The total angle of the triangle = 180°
Missing angle = 180° – 130 °
= 50° |
# Magick squares, sigils and planetary seals
These pages detail an ancient form of magick that have been updated for modern practitioners. It incorporates the use of magick squares and sigils for a more focused intent in spell work. Magick squares are used to draw in the influences of their corresponding planet. Bear in mind that with any work of magick the inclusion of universal forces should be in harmony with your intent.
Magick Squares
Each Magick Square represents a matrix of planetary energy. Magick squares are based on the original work done by ancient mathematicians in their description of numbers. Magical practitioners expanded on this to carry over the correlation between a number and its corresponding planet, therefore representing planetary energy in a mathematical format.
Each magick square is made up using three key numbers. The first is the planetary number itself. The second is the square of the planetary number (or the planetary number multiplied by itself). The third is the sum of the square (or all the incremental numbers starting at 1 that it takes to fill the boxes in the square added together and then divided by the planetary number).
Obviously the best way to understand this concept is by example so for the purposes of illustration we’ll use the planet Saturn.
• Saturn is represented by the number 3.
• We start drawing our magick square with 3 boxes horizontally and vertically for a total of 9 squares (the square of 3 being 9).
• The sum of the square in this case are 1+2+3+4+5+6+7+8+9. This equals 45. 45 divided by Saturn’s planetary number of 3 equals 15.
• Each row of numbers in our magick square both horizontally and vertically need to equal 15 (which is the sum of the square of the planetary number of 3).
Aren’t you glad the ancient mathematicians did all the hard work for you!
When drawing the magick square all of the numbers should be placed in the square in sequence beginning with then number 1. Below is the magick square representing Saturn. Notice, as mentioned above, that our square is 3 by 3 and that all the rows equal 15.
Sigils
A sigil is a signature, if you will. It can be seen a symbol of our intent. Sigils incorporated with other influences can add great direction and focus to spell work. Sigils can be traced in air, carved on candles, drawn on paper and burned etc. Sigils can be drawn and formed using the magick squares that have been described above. We’ll continue with the Saturn example started above.
Saturn is the planet that rules institutions such as banks, real estate transactions, investments, educational institutions etc. Lets say our intent was for successful investing. We want a sigil to represent ourselves at being successful in our investment endeavours. So we pick the word investor which will be our sigil, the visualization of self as a smart investor, surrounded by favourable aspects and clear choices to make our money grow.
Now…how do you determine how to draw your sigil. Using the Western system of numerology, each letter has a numerical equivalent that is detailed in the chart below.
```1 2 3 4 5 6 7 8 9
A B C D E F G H I
J K L M N O P Q R
S T U V W X Y Z```
Referring back to our work with Saturn for a favourable investment strategy, select a word of phrase that you want drawn into a sigil. We take our key word of investor and map it to the chart shown above.
```I N V E S T O R
9 5 4 5 1 2 6 9```
Then by starting with our first corresponding number and continuing through each in order, we draw out our sigil.
So we’re left with the following sigil representing our intention of investor:
Using some of the other correspondences for Saturn, we could carve this sigil into a black candle anointed with bergamot or cypress. We can also surround the candle with onyx and place it on top of the world card from a tarot deck. As we light the candle and focus our intent, not only have, we drawn in many aspects of Saturn with crystals, oils and tarot, and we’ve personalized our intent with our sigil.
Planetary Seals
Just as each magick square brings in planetary energy, each planet has a sigil or seal that is designed to block that planet’s energy. Imagine the magick square as the accelerator and the planetary seal as the brakes. Blocking a planet’s energy can be useful in times of retrograde, or when your desired intent is to block the negative effects of a planet. Lets look at Mercury for this example. Many people are effected dramatically when mercury goes retrograde. One of the most common things most often adversely impacted Mercury retrograde is communication. If for example you had a meeting that was important to your career during a Mercury retrograde, you could block the mercury energy by placing the planetary seal over the magick square of Mercury, thus eliminating the effects of Mercury retrograde for you. Bear in mind that when you block a planet you are blocking ALL of the effects of a planet, so this could work against you. Although sometimes blocking a planet temporarily may be appropriate usually a sigil designed to filter the negative effect may be a better way to handle a situation.
For more detailed information on magick squares, seals and correspondences of the seven visible planets, just click on the links below.
The Sun
The Moon
Mars
Mercury
Jupiter
Venus
Saturn
Image credit: Marco Fedele |
# How do you find the roots, real and imaginary, of y= x- (x-2)(x-1) using the quadratic formula?
How do you find the roots, real and imaginary, of y= x- (x-2)(x-1) using the quadratic formula?
ReportAnswer 1
$x = 2 + \sqrt{2}$
$x = 2 - \sqrt{2}$
#### Explanation:
$y = x - \left(x - 2\right) \left(x - 1\right)$
To write the quadratic formula, your equation first needs to be in standard form:
$y = a {x}^{2} + b x + c$
In order to get this, just simplify your equation.
$y = x - \left(x - 2\right) \left(x - 1\right)$
$y = x - \left({x}^{2} - 3 x + 2\right)$
$y = x - {x}^{2} + 3 x - 2$
$y = - {x}^{2} + 4 x - 2$
Now you can look turn this into the quadratic formula:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
The variables in standard form are the same as the ones in the quadratic formula, which means you can easily transfer them over.
$a = - 1$
$b = 4$
$c = - 2$
All you have to do is plug in and solve!
$x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(- 1\right) \left(- 2\right)}}{\left(2\right) \left(- 1\right)}$
$x = \frac{- 4 \pm \sqrt{16 - 8}}{-} 2$
$x = \frac{- 4 \pm \sqrt{8}}{-} 2$
$x = \frac{- 4 \pm 2 \sqrt{2}}{-} 2$
$x = 2 \pm \sqrt{2}$
$x = 2 + \sqrt{2}$
$x = 2 - \sqrt{2}$
##### Add Answer of: How do you find the roots, real and imaginary, of y= x- (x-2)(x-1) using the quadratic formula?
Similar Homework Help Questions
• ### How do you find the roots, real and imaginary, of y=-x^2-12x -7# using the quadratic formula?
Need Online Homework Help? |
What are prime numbers? A prime number is a number bigger than 1 which can only be divided evenly by 1 or itself.
For example, 13 is a prime number because it cannot be divided by any number except 13 or 1. Whereas 14 is not a prime number, as it can be divided by 2 and 7, as well as by 1 and itself.
These are the prime numbers under 20:
2,3,5,7,11,13,17 and 19.
With the exceptions of 0 and 1, any number which is not a prime number is called a composite number. A composite number is a whole number that can be divided evenly by numbers other than 1 or itself.
0 and 1 are not prime or composite numbers because they can only be divided by themselves.
## When do children learn about prime numbers?
Teachers will introduce children to prime numbers in Year 5 and by Year 6 children should be able to name the prime numbers up to 19.
Children will also be expected to be comfortable working out whether numbers up to 100 are prime or not. To practice their knowledge, children will be given questions like the following:
## The number 51 has 4 factors, which 2 of these factors are prime numbers?
Answer: 3 and 17. The other factors are 1 and 51.
This question requires children to also use their knowledge of factors, which they will be taught about at the same time as prime numbers.
## How to help children with prime numbers?
One way of helping children answer these kind of questions is by creating factor trees. Factor trees are especially useful when children are presented with bigger numbers, as they break down numbers into their prime factors. Here are examples of factor trees for the numbers 51 and 31:
The factor tree for the number 51 shows that it is not a prime number but a composite number, as it can be calculated by multiplying either 1 x 51 or 17 x 3. This factor tree is also helpful to answer the question above, as it shows that 17 and 3 are both prime numbers. They cannot be divided by numbers other than 1 or themselves.
After getting children to work out which numbers multiply together to make 31, they will see that the only option is 1 x 31. So, the factor tree shows that 31 is a prime number, as it can only be divided by 1 and itself.
This exercise can really help children to consolidate their knowledge of factors, prime numbers and multiples.
## How does Learning Street help children with prime numbers?
Learning Street teaches introduces the concept of prime numbers to children and then gradually builds up their knowledge. This regular exposure and revision and development work ensures children make progress in a way that is often impossible just by using worksheet providers or books.
In fact, many tutors and tuition centres struggle to build up knowledge in such a complete way.
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# From the top of a hill, the angles of depression of two consecutive kilometer stones due to east are found to be 30° and 45°. Find the height of the hill (or) what is the height of the hill?
Dec 2, 2017
1.366 km
#### Explanation:
Let AB is the height of the hill and two stones are C and D respectively where depression is 45 degree and 30 degree. The distance between C and D is 1 km.
Here depression and hill has formed right angle triangles with the base. We have to find the height of the hill with this through trigonometry.
In triangle ABC, tan 45 = height/base = AB/BC
or, 1 = AB/BC [ As tan 45 degree = 1]
or, AB = BC ..........(i)
Again, triangle ABD, tan 30 = AB/BD
or, $\frac{1}{\sqrt{3}} = \frac{A B}{B C + C D}$ [tan 30 = $\frac{1}{\sqrt{3}}$ =1/1.732]
or, $\frac{1}{1.732} = \frac{A B}{A B + 1}$ [ As AB = BC from (i) above]
or, 1.732 AB = AB +1
or, 1.732 AB - AB = 1
or, AB(1.732-1) = 1
or, AB * 0.732 = 1
or AB = 1/0.732 = 1.366
Hence height of the hill 1.366 km |
# 1.5 Measurement uncertainty, accuracy, and precision (Page 4/11)
Page 4 / 11
## Multiplication and division with significant figures
Rule: When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).
(a) Multiply 0.6238 cm by 6.6 cm.
(b) Divide 421.23 g by 486 mL.
## Solution
(a) $\begin{array}{l}\begin{array}{l}\text{0.6238 cm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}6.6\phantom{\rule{0.2em}{0ex}}\text{cm}=4.11708\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{result is}\phantom{\rule{0.2em}{0ex}}4.1\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2}\phantom{\rule{0.2em}{0ex}}\left(\text{round to two significant figures}\right)\hfill \\ \text{four significant figures}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{two significant figures}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{two significant figures answer}\hfill \end{array}\hfill \end{array}$
(b) $\begin{array}{l}\frac{\text{421.23 g}}{\text{486 mL}}\phantom{\rule{0.2em}{0ex}}=\text{0.86728... g/mL}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{result is 0.867 g/mL}\phantom{\rule{0.2em}{0ex}}\left(\text{round to three significant figures}\right)\\ \frac{\text{five significant figures}}{\text{three significant figures}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{three significant figures answer}\end{array}$
(a) Multiply 2.334 cm and 0.320 cm.
(b) Divide 55.8752 m by 56.53 s.
(a) 0.747 cm 2 (b) 0.9884 m/s
In the midst of all these technicalities, it is important to keep in mind the reason why we use significant figures and rounding rules—to correctly represent the certainty of the values we report and to ensure that a calculated result is not represented as being more certain than the least certain value used in the calculation.
## Calculation with significant figures
One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters.
## Solution
$\begin{array}{lll}V\hfill & =\hfill & l\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}w\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}d\hfill \\ & =\hfill & \text{13.44 dm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{5.920 dm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{2.54 dm}\hfill \\ & =\hfill & \text{202.09459}...\phantom{\rule{0.2em}{0ex}}{\text{dm}}^{3}\left(\text{value from calculator}\right)\hfill \\ & =\hfill & {\text{202 dm}}^{3}\text{, or 202 L}\phantom{\rule{0.2em}{0ex}}\left(\text{answer rounded to three significant figures}\right)\hfill \end{array}$
What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm 3 ?
1.034 g/mL
## Experimental determination of density using water displacement
A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown.
(a) Use these values to determine the density of this piece of rebar.
(b) Rebar is mostly iron. Does your result in (a) support this statement? How?
## Solution
The volume of the piece of rebar is equal to the volume of the water displaced:
$\text{volume}=\text{22.4 mL}-\text{13.5 mL}=\text{8.9 mL}={\text{8.9 cm}}^{3}$
(rounded to the nearest 0.1 mL, per the rule for addition and subtraction)
The density is the mass-to-volume ratio:
$\text{density}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass}}{\text{volume}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{69.658 g}}{{\text{8.9 cm}}^{3}}={\text{7.8 g/cm}}^{3}$
(rounded to two significant figures, per the rule for multiplication and division)
From [link] , the density of iron is 7.9 g/cm 3 , very close to that of rebar, which lends some support to the fact that rebar is mostly iron.
An irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown.
(a) Use these values to determine the density of this material.
(b) Do you have any reasonable guesses as to the identity of this material? Explain your reasoning.
(a) 19 g/cm 3 ; (b) It is likely gold; the right appearance for gold and very close to the density given for gold in [link] .
## Accuracy and precision
Scientists typically make repeated measurements of a quantity to ensure the quality of their findings and to know both the precision and the accuracy of their results. Measurements are said to be precise if they yield very similar results when repeated in the same manner. A measurement is considered accurate if it yields a result that is very close to the true or accepted value. Precise values agree with each other; accurate values agree with a true value. These characterizations can be extended to other contexts, such as the results of an archery competition ( [link] ).
what is hybridization
the mixing of atomic orbitals to form molecular of similar energy called hybrid orbitals
Cffrrcvccgg
who are the alchemist?
alchemy science of transmutation. typically it is aim at tranforming lead to or other base metals to gold and the creation of the philosophers stone which in reality isn't a stone it's something priceless something we all need for coming times. don't be fooled
Kendrick
read Corinthians 5 verses 50 to the end of the chapter then read revelations chapter 2 verse 17
Kendrick
The word "Alchemy" comes from the forgotten name for Ancient Egypt, Khemmet. Khem was the name for the Egyptian Empire, but the actual land of Egypt was called Khemmet because the "T" on the end of a word denoted a physical location on Earth and not just an idea.
Michael
Wow!
mendie
What's the mass number of carbon
Akinbola
mass number of carbon is 12.
Nnenna
wat d atomic number of oxygen
safiya
atomic number of oxygen is 8
Nnenna
which quantum number divides shell into orbitals?
azimuthal
Emmanuel
hi
Charlie
azimuthal
reinhard
azimuthal
Charlie
what is atom
an atom is a smallest indivisible part of an element
Henry
an atom is the smallest part of an element that takes part in a chemical reaction
Nana
wat is neutralization
when any acid reacts with base to decrease it's acidity or vice-versa to form salt and solvent.. which is called neutralization
Santosh
explain buffer
Organic
buffer is a solution which resists changes in pH when acid or alkali added to it..
Santosh
hello, who is online
UTHMAN
buffer is the solution which resist the change in pH by addition of small amount of acid or alkali to it
KAUSIK
neutralisation is the process of mixing of a acid and a base to form water and corresponding salt
KAUSIK
how to solve equation on this
what are the elent of ionic and covalent bonding
Princewill
what is gases
Its one of the fundamental sate of matter alone side with liquid, solid and plasma
John
What is chemical bonding
John
To my own definitions. It's a unit of measurement to express the amount of a chemical substance.
What is mole
It's the unit of measurements used to express the amount of chemical substance.
Ozoaniehe
What is pressure
force over area
Jake
force applied per unit area
john
force applied per unit area
Prajapati
Why does carbonic acid don't react with metals
Why does carbonic acid don't react with metal
Some metals will react depending on their Standard Electrode Potential. Carbonic acid is a very weak acid (i.e. a low hydrogen ion concentration) so the rate of reaction is very low.
Paul
sample of carbon-12 has a mass of 6.00g. How many atoms of carbon-12 are in the sample
a sample of carbon-12 has a mass of 6.00g. How many atoms of carbon-12 are in the sample
an object of weight 10N immersed in a liquid displaces a quantity of d liquid.if d liquid displaced weights 6N.determine d up thrust of the object |
# How does cardinality work?
1. Jun 22, 2011
### Greg Bernhardt
In the https://www.physicsforums.com/showthread.php?t=507003" [Broken], we have looked at various types of infinities. In the last section of that post, we have said when we regarded two sets to have equal size (or equal cardinality). We will now flesh out this concept a bit.
Comparing sizes of sets.
In the last part of the previous post, we proposed a definition deciding when two sets have the same size. This definition was
Two sets A and B have the same cardinality if there is a one-to-one correspondence between A and B. We denote this by |A|=|B|.
Note that we have not yet given meaning to |A| and |B|. Right now, we will just introduce |A|=|B| as a notation. We will give a deeper meaning behind this notation later on.
A set of the same cardinality as {1,...,n} is called finite and is said to have n elements. For example {{1},{2,3}} has 2 elements. The empty set is also called finite and is said to have 0 elements. A set which is not of the same cardinality as a finite set, is called infinite.
An important infinite set is the set of natural numbers $\mathbb{N}=\{0,1,2,3,4,5,...\}$. A set of the same cardinality as of $\mathbb{N}$ is called countably infinite. Infinite sets that are not countable are called uncountable. Intuitively, a set is countably infinite if we can give every element of the set a number. I.e. there is a zeroth element, there is a first element, there is a second element,... The crucial part is that the natural numbers suffice for this process.
The integers are countable
Many important sets are countably infinite. Take $\mathbb{Z}$, the integers, for example. At first, it might not be obvious why this is countable. In fact, we can try numbering it as follows:
The zeroth element is 0, the first element is 1, the second element is 2,..., the n'th element is n,...
but then we have not given every element of $\mathbb{Z}$ a number. For example, we have not given -1 a number. However, with some clever reasoning, we can see that following numbering suffices:
The zeroth element is 0, the first element is 1, the second element is -1, the third element is 2, the fourth element is -2,...
In general, if n is odd then the n'th element is (n+1)/2 and if n is even then the n'th element is -n/2.
So the integers form a countable set. What about the rational numbers $\mathbb{Q}$?
The rational numbers are countable
Before we show this, we need some more terminology. We already know how to say that sets have an equal cardinality, but we can not yet say what it means for a set A to have smaller cardinality than a set B. We will do this now:
We say that a set A has a smaller cardinality than a set B if there exists a subset $B^\prime\subseteq B$ such that |A|=|B'|. We will denote this as $|A|\leq |B|$.
Again, when we write $|A|\leq |B|$, then this is just a notation. We have not yet given meaning to |A| and |B|.
Now, to show that the rational numbers are countable, we can limit our self to the positive rational numbers (just apply the reasoning of above to prove the general case). To give an easy proof that the positive rational numbers are countable, we will need the following fundamental result:
The theorem of Cantor-Bernstein-Schroder: Say that we have sets A, B and C such that $|A|\leq |B|\leq |C|$. If $|A|=|C|$, then also $|A|=|B|$.
This theorem is trivial in the finite case, but requires some ingenious method in the infinite case. The proof of this theorem can be found in any good book on set theory.
We will apply this theorem on the natural numbers (which we will take as A) and the positive rational numbers (that we shall take as B). The set that we will pick as C is the set of all couples of natural numbers, i.e. the set
$$\begin{array}{cccccc} (0,0) & (0,1) & (0,2) & (0,3) & (0,4) & ...\\ (1,0) & (1,1) & (1,2) & (1,3) & (1,4) & ...\\ (2,0) & (2,1) & (2,2) & (2,3) & (2,4) & ...\\ (3,0) & (3,1) & (3,2) & (3,3) & (3,4) & ...\\ (4,0) & (4,1) & (4,2) & (4,3) & (4,4) & ...\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots\\ \end{array}$$
To apply the theorem of Cantor-Bernstein-Schroder, we need to show that this set is countable. But how in earth can one show such a thing? It was Cantor that was the first to have found a truly marvelous proof of this fact. What he did is to look at the diagonals of the above set, and this would supply him with an ordering. So what he did was:
The first element is (0,0), the second element is (0,1), the third element is (1,0), the fourth element is (0,2), the fifth element is (1,1), the sixth element is (2,0), the seventh element is (0,3),...
So this indeed shows that C is countable and that |A|=|C|. So by applying Cantor-Bernstein-Schroder, we have that the rational numbers are countable.
The real numbers are uncountable
A set which is truly bigger then $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{Q}$ are the real numbers (denoted by $\mathbb{R}$). This was again shown by Cantor and it is one of the most famous proofs in entire mathematics. What he actually shows is that [0,1] is uncountable, but the theorem of Cantor-Bernstein-Schroder shows that this implies $\mathbb{R}$to be uncountable.
Now, how does Cantor show [0,1] to be uncountable? It goes as follows, say we have a numbering of all the real numbers as:
• The zeroth element is $0.d_{0,0}d_{0,1},d_{0,2}d_{0,3}d_{0,4}...$
• The first element is $0.d_{1,0}d_{1,1},d_{1,2}d_{1,3}d_{1,4}...$
• The second element is $0.d_{2,0}d_{2,1},d_{2,2}d_{2,3}d_{2,4}...$
• The third element is $0.d_{3,0}d_{3,1},d_{3,2}d_{3,3}d_{3,4}...$
• The fourth element is $0.d_{4,0}d_{4,1},d_{4,2}d_{4,3}d_{4,4}...$
• ..........
• The n'th element is $0.d_{n,1},d_{n,2}d_{n,3}d_{n,4}...$
• ..........
The idea of Cantor was to take the diagonal number $0.d_{0,0}d_{1,1}d_{2,2}d_{3,3}d_{4,4}...$. What we do is change this number slightly: if $d_{i,i}$ is not 4, then we change it to a 4 and if it is a 4 then we change it to a 5.
For example, the number 0.12345654321... is changed to 0.44454445444. This new number, which we denote by $0.c_0c_1c_2c_3c_4...$, is not contained in our numbering. Indeed, it does not equal our zeroth element, since by definition $c_0\neq d_{0,0}$. It does not equal our first element, since $c_1\neq d_{1,1}$. In general, it does not equal our n'th element since $c_n\neq d_{n,n}$, etc. But this is a contradiction since we have assumed all the reals to be numbered.
The continuum hypothesis
So, what we have established is
$$|\mathbb{N}|=|\mathbb{Z}|=|\mathbb{Q}|<|\mathbb{R}|$$
A natural question is whether there exists a set C between the rationals and the reals such that
$$|\mathbb{Q}|<|C|<|\mathbb{R}|$$
For years, this was one of the most important unsolved questions in mathematics, but it was finally resolved by Kurt Gödel and Paul Cohen. The answer is not what one should expect. The answer is that it cannot be proved that there is such a set C and it cannot be proved that there isn't such a set C. It's not that we're not smart enough to prove it, it's that the usual axioms of set theory are too weak to prove or disprove the existence of C.
If we want, we can make an axiom that says that such a set C does exist, but it is as valid to make an axiom that says that the set C does not exist. These two axioms lead to two very different kinds of mathematics, but they are both equally valid.
The following forum members have contributed to this FAQ:
bcrowell
Hurkyl
micromass
Redbelly98
Last edited: May 5, 2017
2. Jun 30, 2011 |
# Proportion
A proportion is a statement of equality between two or more ratios. For example,
This proportion is read as, “3 is to 4 as 6 is to 8.”
#### Extremes and Means
The first and last terms of the proportion (the 3 and 8 in this example) are called the extremes. The second and third terms (the 4 and 6 in this example) are called the means. In any proportion, the product of the extremes is equal to the product of the means.
In the proportion 2:3 = 4:6, the product of the extremes, 2 × 6, is 12; the product of the means, 3 × 4, is also 12. An inspection of any proportion will show this to be true.
#### Solving Proportions
Normally when solving a proportion, three quantities will be known, and the fourth will be unknown. To solve for the unknown, multiply the two numbers along the diagonal and then divide by the third number.
Example: Solve for X in the proportion given below.
First, multiply 65 × 100: 65 × 100 = 6500
Next, divide by 80: 6500 ÷ 80 = 81.25
Therefore, X = 81.25.
Example: An airplane flying a distance of 300 miles used 24 gallons of gasoline. How many gallons will it need to travel 750 miles?
The ratio here is: “miles to gallons;” therefore, the proportion is set up as:
Therefore, to fly 750 miles, 60 gallons of gasoline will be required. |
## Sunday, July 25, 2021
### Mathematics test class 10th │ Maths test │ Coordinate Geometry test
Mathematics test class 10th │ Maths test │ Coordinate Geometry test │ Class 10th Maths test
# Class 10 Maths MCQs Chapter 7 Coordinate Geometry
1. The distance of the point P(2, 3) from the x-axis is
(a) 2
(b) 3
(c) 1
(d) 5
Explaination: Reason: The distance from x-axis is equal to its ordinate i.e., 3
2. The distance between the point P(1, 4) and Q(4, 0) is
(a) 4
(b) 5
(c) 6
(d) 3√3
Explaination: Reason: The required distance = $$\sqrt{(4-1)^{2}+(0-4)^{2}}=\sqrt{9+16}=\sqrt{25}=5$$
3. The points (-5, 1), (1, p) and (4, -2) are collinear if
the value of p is
(a) 3
(b) 2
(c) 1
(d) -1
Explaination: Reason: The points are collinear if area of Δ = 0
= $$\frac{1}{2}$$[-5(p + 2) +l(-2 -1) + 4(1 – p)] – 0
⇒ -5 p -10-3 + 4-4p = 0
⇒ -9p = +9
∴ p = -1
4. The area of the triangle ABC with the vertices A(-5, 7), B(-4, -5) and C(4, 5) is
(a) 63
(b) 35
(c) 53
(d) 36
Explaination: Reason: Area of ΔABC = $$\frac{1}{2}$$ [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= $$\frac{1}{2}$$[-5(-5 – 5) -4(5 – 7) + 4(7 – (-5))] = $$\frac{1}{2}$$[-5(-10) -4(-2) + 4(12)]
= $$\frac{1}{2}$$[50 + 8 + 48] = $$\frac{1}{2}$$ × 106 = 53 sq. units
5. The distance of the point (α, β) from the origin is
(a) α + β
(b) α² + β²
(c) |α| + |β|
(d) $$\sqrt{\alpha^{2}+\beta^{2}}$$
Explaination: Reason: Distance of (α, β) from origin (0, 0) = $$\sqrt{(\alpha-0)^{2}+(\beta-0)^{2}}=\sqrt{\alpha^{2}+\beta^{2}}$$
6. The area of the triangle whose vertices are A(1, 2), B(-2, 3) and C(-3, -4) is
(a) 11
(b) 22
(c) 33
(d) 21
Explaination: Reason: Required area= $$\frac{1}{2}$$[1(3 + 4) -2(-4 – 2) -3(2 – 3)]
= $$\frac{1}{2}$$[7 + 12 + 3]
= $$\frac{1}{2}$$ × 22 = 11
7. The line segment joining the points (3, -1) and (-6, 5) is trisected. The coordinates of point of trisection are
(a) (3, 3)
(b) (- 3, 3)
(c) (3, – 3)
(d) (-3,-3)
Explaination: Reason: Since the line segment AB is trisected
8. The line 3x + y – 9 = 0 divides the line joining the points (1, 3) and (2, 7) internally in the ratio
(a) 3 : 4
(b) 3 : 2
(c) 2 : 3
(d) 4 : 3
Explaination: Reason: Let the line 3x + y – 9 = 0 divide the line segment joining A(l, 3) ad B(2, 7) in the ratio K : 1 at point C.
9. The distance between A (a + b, a – b) and B(a – b, -a – b) is
Explaination:
10. If (a/3, 4) is the mid-point of the segment joining the points P(-6, 5) and R(-2, 3), then the value of ‘a’ is
(a) 12
(b) -6
(c) -12
(d) -4
Explaination:
11. If the distance between the points (x, -1) and (3, 2) is 5, then the value of x is
(a) -7 or -1
(b) -7 or 1
(c) 7 or 1
(d) 7 or -1
Explaination: Reason: We have $$\sqrt{(x-3)^{2}+(-1-2)^{2}}=5$$
⇒ (x – 3)² + 9 = 25
⇒ x² – 6x + 9 + 9 = 25
⇒ x² -6x – 7 = 0
⇒ (x – 7)(x + 1) = 0
⇒ x = 7 or x = -1
12. The points (1,1), (-2, 7) and (3, -3) are
(a) vertices of an equilateral triangle
(b) collinear
(c) vertices of an isosceles triangle
(d) none of these
Explaination: Reason: Let A(1, 1), B(-2, 7) and C(3, 3) are the given points, Then, we have
13. The coordinates of the centroid of a triangle whose vertices are (0, 6), (8,12) and (8, 0) is
(a) (4, 6)
(b) (16, 6)
(c) (8, 6)
(d) (16/3, 6)
Explaination: Reason: The co-ordinates of the centroid of the triangle is
14. Two vertices of a triangle are (3, – 5) and (- 7,4). If its centroid is (2, -1), then the third vertex is
(a) (10, 2)
(b) (-10,2)
(c) (10,-2)
(d) (-10,-2)
Explaination: Reason: Let the coordinates of the third vertex be (x, y)
15. The area of the triangle formed by the points A(-1.5, 3), B(6, -2) and C(-3, 4) is
(a) 0
(b) 1
(c) 2
(d) 3/2
Explaination: Reason: Area of ΔABC = $$\frac{1}{2}$$ [-1.5(-2 – 4) + 6(4 – 3) + (-3) (3 + 2)] = $$\frac{1}{2}$$ [9 + 6 – 15] = 0. It is a straight line.
16. If the points P(1, 2), B(0, 0) and C(a, b) are collinear, then
(a) 2a = b
(b) a = -b
(c) a = 2b
(d) a = b
Explaination: Reason: Area of ΔPBC = 0
⇒ $$\frac{1}{2}$$[1(0 – b) + 0(6 – 1) + a(2 – 0)] = 0
⇒ $$\frac{1}{2}$$[-6 + 2a] = 0
⇒ -b + 2a = 0
∴ 2a = b
Please do not paste any url.... |
# RD Sharma Class 12 Solutions Chapter 15 Exercise 15.1 (Updated for 2021-22)
RD Sharma Solutions Class 12 Maths Chapter 15 Exercise 15.1 is mentioned in this article. The key topics covered in exercise 15.1 are Rolle’s theorem, the geometrical theorem of Rolle’s theorem of given functions, and illustrative examples. Our experts have created exercise-wise RD Sharma Class 12 Solutions with explanations for each step to make problem-solving easier for students. It allows them to assess their understanding of the concepts covered in this chapter. Students can utilise RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Exercise 15.1 as reference materials to analyse Maths.
## Download RD Sharma Solutions Class 12 Maths Chapter 15 Exercise 15.1 Free PDF
RD Sharma Solutions Class 12 Maths Chapter 15 Exercise 15.1
### Access answers to Maths RD Sharma Solutions For Class 12 Chapter 15 – Mean Value Theorems Exercise 15.1 Important Questions With Solution
1. Discuss the applicability of Rolle’s Theorem for the following functions on the indicated intervals:
Solution:
(ii) f (x) = [x] for -1 ≤ x ≤ 1, where [x] denotes the greatest integer not exceeding x
Solution:
Solution:
(iv) f (x) = 2x2 – 5x + 3 on [1, 3]
Solution:
Given function is f (x) = 2x2 – 5x + 3 on [1, 3]
Since given function f is a polynomial. So, it is continuous and differentiable everywhere.
Now, we find the values of function at the extreme values.
⇒ f (1) = 2(1)2–5(1) + 3
⇒ f (1) = 2 – 5 + 3
⇒ f (1) = 0…… (1)
⇒ f (3) = 2(3)2–5(3) + 3
⇒ f (3) = 2(9)–15 + 3
⇒ f (3) = 18 – 12
⇒ f (3) = 6…… (2)
From (1) and (2), we can say that, f (1) ≠ f (3)
∴ Rolle’s Theorem is not applicable for the function f in interval [1, 3].
(v) f (x) = x2/3 on [-1, 1]
Solution:
Solution:
2. Verify the Rolle’s Theorem for each of the following functions on the indicated intervals:
(i) f (x) = x2 – 8x + 12 on [2, 6]
Solution:
(ii) f(x) = x2 – 4x + 3 on [1, 3]
Solution:
(iii) f (x) = (x – 1) (x – 2)2 on [1, 2]
Solution:
Mean Value Theorems Ex 15.1 Q2(iv)
Mean Value Theorems Ex 15.1 Q2(v)
Mean Value Theorems Ex 15.1 Q2(vi)
Mean Value Theorems Ex 15.1 Q2(vii)
Mean Value Theorems Ex 15.1 Q2(viii)
Mean Value Theorems Ex 15.1 Q3(i)
Mean Value Theorems Ex 15.1 Q3(ii)
Mean Value Theorems Ex 15.1 Q3(iii)
Mean Value Theorems Ex 15.1 Q3(iv)
Mean Value Theorems Ex 15.1 Q3(v)
Mean Value Theorems Ex 15.1 Q3(vi)
Mean Value Theorems Ex 15.1 Q3(vii)
Mean Value Theorems Ex 15.1 Q3(viii)
Mean Value Theorems Ex 15.1 Q3(ix)
Mean Value Theorems Ex 15.1 Q3(x)
Mean Value Theorems Ex 15.1 Q3(xi)
Mean Value Theorems Ex 15.1 Q3(xii)
Mean Value Theorems Ex 15.1 Q3(xiii)
Mean Value Theorems Ex 15.1 Q3(xiv)
Mean Value Theorems Ex 15.1 Q3(xv)
Mean Value Theorems Ex 15.1 Q3(xvi)
Mean Value Theorems Ex 15.1 Q3(xvii)
Mean Value Theorems Ex 15.1 Q3(xvii)
Mean Value Theorems Ex 15.1 Q3(xviii)
Mean Value Theorems Ex 15.1 Q7
Mean Value Theorems Ex 15.1 Q8(i)
Mean Value Theorems Ex 15.1 Q8(ii)
Mean Value Theorems Ex 15.1 Q8(iii)
Mean Value Theorems Ex 15.1 Q9
Mean Value Theorems Ex 15.1 Q10
Mean Value Theorems Ex 15.1 Q11
## RD Sharma Class 12 Solutions Chapter 15 Exercise 15.1 | Important Concepts
The following are some of the most essential topics of this exercise.
• Rolle’s theorem
• Geometrical interpretation of Rolle’s theorem
• Algebraic interpretation of Rolle’s theorem
• Checking the applicability of Rolle’s theorem
• Verification of Rolle’s theorem for a given function defined on a given interval
• Miscellaneous applications of Rolle’s theorem
RD Sharma Solutions for Class 12 Maths Chapter 15 Exercise 15.1 are immensely helpful while doing your homework and also while preparing for CBSE finals. If you have any queries, feel free to ask us in the comment section below.
## FAQs on RD Sharma Solutions Class 12 Maths Chapter 15 Exercise 15.1
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There are a total of 10 questions in RD Sharma Solutions Class 12 Maths Chapter 15 Exercise 15.1.
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Yes, it is important to attend all of the questions offered in RD Sharma Solutions for Class 12 Maths Chapter 15 Exercise 15.1 in order to answer all types of questions that may appear in examinations.
### 2 thoughts on “RD Sharma Class 12 Solutions Chapter 15 Exercise 15.1 (Updated for 2021-22)”
1. Thanks Monica Maam for providing all the sol. |
# How do you differentiate y =x /sec ^2x^3 using the chain rule?
Jul 4, 2018
$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos {x}^{3} \left[\cos {x}^{3} - 6 {x}^{3} \sin {x}^{3}\right]$
#### Explanation:
Here,
$y = \frac{x}{\sec} ^ 2 \left({x}^{3}\right)$
$\implies y = x \cdot {\cos}^{2} \left({x}^{3}\right)$
Diff.w.r.t. $x$ , using Product rule and Chain Rule:
$\frac{\mathrm{dy}}{\mathrm{dx}} = x \cdot \frac{d}{\mathrm{dx}} \left({\cos}^{2} {x}^{3}\right) + {\cos}^{2} {x}^{3} \cdot \frac{d}{\mathrm{dx}} \left(x\right)$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = x \cdot 2 \cos {x}^{3} \left(- \sin {x}^{3}\right) \frac{d}{\mathrm{dx}} \left({x}^{3}\right) + {\cos}^{2} {x}^{3} \cdot 1$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x \sin {x}^{3} \cos {x}^{3} \cdot 3 {x}^{2} + {\cos}^{2} {x}^{3}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 6 {x}^{3} \sin {x}^{3} \cos {x}^{3} + {\cos}^{2} {x}^{3}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} {x}^{3} - 6 {x}^{3} \sin {x}^{3} \cos {x}^{3.} . . \to \left(1\right)$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \cos {x}^{3} \left[\cos {x}^{3} - 6 {x}^{3} \sin {x}^{3}\right]$
Note :
From $\left(1\right)$ we can simplify in the form :
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} {x}^{3} - 6 {x}^{3} \sin {x}^{3} \cos {x}^{3}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} {x}^{3} - 3 {x}^{3} \left\{2 \sin {x}^{3} \cos {x}^{3}\right\}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} {x}^{3} - 3 {x}^{3} \cdot \sin 2 \left({x}^{3}\right)$ |
Solve 3x^3-7x+3=0 | Uniteasy.com
# Solve the cubic equation:
## $$3x^3-7x+3=0$$
Since the discriminant $$\Delta <0$$, the cubic equation has three distinct real roots.
$$\Delta=-0.22050754458162$$
$$\begin{cases} x_1=\dfrac{2}{3}\sqrt{7}\cos \bigg[\dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)\bigg] \\ x_2=\dfrac{2}{3}\sqrt{7} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\dfrac{2\pi}{3}\bigg] \\ x_3=\dfrac{2}{3}\sqrt{7} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\dfrac{4\pi}{3} \bigg] \end{cases}$$
In decimals,
$$\begin{cases} x_1=1.2341129259511 \\ x_2=-1.7084121785118 \\ x_3=0.47429925256064 \end{cases}$$
Detailed Steps on Solution
## 1. Cardano's solution
Divide each term by $$3$$, the coefficient of the cubic term.
Then the cubic equation is converted to,
Let $$x=u-v$$
Cube both sides and extract common factor from two middle terms after expanding the bracket.
\begin{aligned} \\x^3&=(u-v)^3\\ & =u^3-3u^2v+3uv^2-v^3\\ & =-3uv(u-v)+u^3-v^3\\ \end{aligned}
Since $$u-v=x$$, substitution gives a linear term for the equation. Rearrange terms.
$$x^3+3uvx-u^3+v^3=0$$
Compare the cubic equation with the original one (1)
$$\begin{cases} 3uv=-\dfrac{7}{3}\quad\text{or}\quad v=-\dfrac{7}{9u}\\ v^3-u^3=1\\ \end{cases}$$
$$v=-\dfrac{7}{9u}$$ gives relationship between the two variables. Substitute the value of $$v$$ to the second equation
$$\Big(-\dfrac{7}{9u}\Big)^3-u^3=1$$
Simplifying gives,
$$u^3+\dfrac{343}{729}\dfrac{1}{u^3}+1=0$$2
Let $$m=u^3$$, then the equation is transformed to a quadratic equation in terms of $$m$$. Once the value of $$m$$ is determined, $$v^3$$ could be determined by $$v^3=1+u^3$$.
$$m^2+1m+\dfrac{343}{729}=0$$
Sovling the quadratic euqation will give two roots (some may be equal). Here we only cosider one case with positive sign before the square root radical since the negative case will produce the same result.
\begin{aligned} \\u^3=m&=-\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\Big(1^2\Big)-4\cdot \dfrac{343}{729}}\\ & =-\dfrac{1}{2}+\dfrac{1}{2}\sqrt{1-\dfrac{1372}{729}}\\ & =-\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\dfrac{643}{729}}i\\ & =-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i\\ \end{aligned}
$$v^3$$ can be determined by the equation we deduced $$v^3-u^3=1$$. Then,
\begin{aligned} \\v^3&=1+u^3\\ & =1-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i\\ & =\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i\\ \end{aligned}
Now we have,
$$u^3=-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i$$ and $$v^3=\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i$$
Evaluating the simplest cubic equation $$x^3-A=0$$, it has 3 roots, in which the first root is a real number or a complex number. The second and third are expressed in the product of cubic root of unity and the first one.
If $$ω = \dfrac{-1+i\sqrt{3}}{2}$$, then its reciprocal is equal to its conjugate, $$\dfrac{1}{ω}=\overline{ω}$$.
$$\begin{cases} r_1=\sqrt[3]{A}\\ r_2=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{A}\\ r_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{A}\\ \end{cases}$$
Similary, taking cubic root for $$u^3$$ and $$v^3$$ also gives 3 roots.
$$\begin{cases} u_1=\sqrt[3]{-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}\\ u_2=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}\\ u_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}\\ \end{cases}$$
For $$v_2$$ and $$v_3$$, the complex numbers before radicals are the conjugates of those for $$u_2$$ and $$u_3$$, which can be verified by the reciprocal property of the cubic root of unity from the equation $$v=-\dfrac{7}{9u}$$. The radicand can be taken as the negative conjugate of that in $$u_1$$, $$u_2$$ and $$u_3$$, which is the same in value.
$$\begin{cases} v_1=\sqrt[3]{\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}\\ v_2=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}\\ v_3=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}\\ \end{cases}$$
Since $$x=u-v$$, the firt root $$x_1$$ can be expressed as the sum of cubic root of two conjugate complex numbers
$$x_1=\sqrt[3]{-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}+\sqrt[3]{-\dfrac{1}{2}-\dfrac{\sqrt{643}}{54}i}$$
Let $$u^3_1=z=-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i$$, and $$z$$ can be expressed in trigonomic form $$r(\cos θ + i \sin θ)$$, where $$r$$ and $$θ$$ are the modulus and principle argument of the complex number.
Then $$-v^3_1$$ is the conjugate $$\overline{z}=-\dfrac{1}{2}-\dfrac{\sqrt{643}}{54}i$$, or $$\overline{z} = r(\cos θ - i \sin θ)$$
Now let's calculate the value of $$r$$ and $$θ$$.
\begin{aligned} \\r&=\sqrt{\Big(-\dfrac{1}{2}\Big)^2+\Big(\dfrac{\sqrt{643}}{54}\Big)^2}\\ & =\dfrac{7}{27}\sqrt{7}\\ \end{aligned}
$$\cosθ=\dfrac{-\dfrac{1}{2}}{\dfrac{7}{27}\sqrt{7}}=-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}$$
The argument is the inverse function of the cosθ
$$θ=\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)$$
Using de Moivre’s formula, the cubic root of $$z$$ could be determined.
\begin{aligned} \\u_1=\sqrt[3]{z}&=\sqrt[3]{r}(\cos\dfrac{θ}{3}+i\sin\dfrac{θ}{3})\\ & =\sqrt[3]{\dfrac{7}{27}\sqrt{7}}\Big[\cos\dfrac{1}{3}\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\sin\dfrac{1}{3}\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)\Big]\\ & =\dfrac{\sqrt{7}}{3}\Big[\cos\dfrac{1}{3}\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\sin\dfrac{1}{3}\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)\Big]\\ \end{aligned}
Since $$-v^3$$ is the conjugate of $$z$$ as we mentioned above,
\begin{aligned} \\v_1=-\sqrt[3]{\overline{z}}&=-\sqrt[3]{r}(\cos\dfrac{θ}{3}-i\sin\dfrac{θ}{3})\\ & =\dfrac{\sqrt{7}}{3}\Big[-\cos\dfrac{1}{3}\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\sin\dfrac{1}{3}\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)\Big]\\ \end{aligned}
The first root is the difference of $$u_1$$ and $$v_1$$.
\begin{aligned} \\x_1&=u_1-v_1\\ & =2\cdot \dfrac{\sqrt{7}}{3}\cos\dfrac{1}{3}\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)\\ & =\dfrac{2}{3}\sqrt{7}\cos\dfrac{1}{3}\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)\\ \end{aligned}
The second root is the difference of $$u_2$$ and $$v_2$$, in which $$u_2$$ is the product of the cubic root of $$z$$ and the cubic root of unity, $$v_2$$ is the product of the negative of the conjugate of $$z$$ and the conjugate of the cubic root of unity.
\begin{aligned} \\x_2&=u_2-v_2\\ & =\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}+\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{2}-\dfrac{\sqrt{643}}{54}i}\\ & =\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{r}(\cos\dfrac{θ}{3}+i\sin\dfrac{θ}{3})+\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{r}(\cos\dfrac{θ}{3}-i\sin\dfrac{θ}{3})\\ & =-\sqrt[3]{r}(\cos\dfrac{ θ}{3} + \sqrt{3} \sin\dfrac{ θ}{3} ) \\ & =2\sqrt[3]{r}\cos\Big(\dfrac{θ}{3}+ \dfrac{4\pi}{3}\Big)\\ & =\dfrac{2}{3}\sqrt{7}\cos\Big(\dfrac{1}{3}\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+ \dfrac{4\pi}{3}\Big)\\ \end{aligned}
The third root is the difference of $$u_3$$ and $$v_3$$, in which $$u_3$$ is the product of the cubic root of $$z$$ and the conjugate of the cubic root of unity, $$v_3$$ is the product of the negative of the conjugate of $$z$$ and the cubic root of unity.
\begin{aligned} \\x_3&=u_3-v_3\\ & =\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}+\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{2}-\dfrac{\sqrt{643}}{54}i}\\ & =\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{r}(\cos\dfrac{θ}{3}+i\sin\dfrac{θ}{3})+\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{r}(\cos\dfrac{θ}{3}-i\sin\dfrac{θ}{3})\\ & =\sqrt[3]{r}(-\cos\dfrac{θ}{3}+\sqrt{3} \sin \dfrac{ θ}{3})\\ & =2\sqrt[3]{r}\cos\Big(\dfrac{θ}{3}+ \dfrac{2\pi}{3}\Big)\\ & =\dfrac{2}{3}\sqrt{7}\cos\Big(\dfrac{1}{3}\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+ \dfrac{2\pi}{3}\Big)\\ \end{aligned}
## 2. Vieta's Substitution
In Cardano' solution, $$x$$ is defined as the difference of $$u$$ and $$v$$. If we substitute the value of $$v$$ (4) into (2), we get the equation. $$x=u+\dfrac{7}{9u}$$. And then substitute the equation to the cubic equation $$x^3-\dfrac{7}{3}x+1=0$$. This method is called Vieta's Substitution for solving a cubic equation, which simplied the Cardano' solution. The substitution expression can be obtained by the following formula directly.
$$x=u-\dfrac{p}{3u}$$
Substitute the expression $$x=u+\dfrac{7}{9u}$$ to the cubic equation
$$+3\Big(u+\dfrac{7}{9u}\Big)^3-7\Big(u+\dfrac{7}{9u}\Big)+3=0$$
Expand brackets and cancel the like terms
$$u^3+\cancel{\dfrac{7}{3}u^2\dfrac{1}{u}}+\cancel{\dfrac{49}{27}u\dfrac{1}{u^2}}+\dfrac{343}{729}\dfrac{1}{u^3}-\cancel{\dfrac{7}{3}u}-\cancel{\dfrac{49}{27}\dfrac{1}{u}}+1=0$$
Then we get the same equation as (2)
$$u^3+\dfrac{343}{729}\dfrac{1}{u^3}+1=0$$
The rest of the steps will be the same as those of Cardano's solution
## $$x^3-\dfrac{7}{3}x+1=0$$
Move the linear term and constant of (1) to its right hand side. We get the following form of the equation.
$$x^3=\dfrac{7}{3}x-1$$3
Let the root of the cubic equation be the sum of two cubic roots
$$x=\sqrt[3]{r_1}+\sqrt[3]{r_2}$$4
in which $$r_1$$ and $$r_2$$ are two roots of a quadratic equation
$$z^2-\alpha z+ β=0$$5
Using Vieta's Formula, the following equations are established.
$$r_1+r_2 = \alpha \quad \text{and} \quad r_1r_2 = β$$
To determine $$\alpha$$, $$β$$, cube both sides of the equation (4)
$$x^3=3\sqrt[3]{r_1r_2}(\sqrt[3]{r_1}+\sqrt[3]{r_2})+r_1+r_2$$
Substituting, the equation is simplified to
$$x^3=3\sqrt[3]{β}x+\alpha$$
Compare the cubic equation with (3), the following equations are established
$$\begin{cases} 3\sqrt[3]{β}=\dfrac{7}{3}\\ \alpha=-1\\ \end{cases}$$
Solving for $$β$$ gives
$$β=\dfrac{343}{729}$$
So the quadratic equation (5) is determined as
$$z^2+z+\dfrac{343}{729}=0$$6
$$\begin{cases} r_1=-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i\approx-0.5+0.46958230863355i\\ r_2=-\dfrac{1}{2}-\dfrac{\sqrt{643}}{54}i\approx-0.5-0.46958230863355i\\ \end{cases}$$
Therefore, one of the roots of the cubic equation could be obtained from (4).
$$x_1=\sqrt[3]{-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}+\sqrt[3]{-\dfrac{1}{2}-\dfrac{\sqrt{643}}{54}i}$$
in decimals,
$$x_1=1.2341129259511$$
However, since the cube root of a quantity has triple values,
The other two roots could be determined as,
$$x_2=\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}+\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{2}-\dfrac{\sqrt{643}}{54}i}$$
$$x_3=\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{2}+\dfrac{\sqrt{643}}{54}i}+\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{2}-\dfrac{\sqrt{643}}{54}i}$$
Since the expression involes cubic root of complex number, the final result can be deduced by using trigonometric method as shown in Cardano's solution.
## 4. Cubic Root Formula
The equation $$3x³ - 7x + 3=0$$ has no quadratic term, compared with the general cubic equation. We can use the root formula to calculate the roots direvtly.
$$x^3 +px+q=0$$
For the equation , we have $$p=-\dfrac{7}{3}$$ and $$q = 1$$
### Calculate the discriminant
The nature of the roots are determined by the sign of the discriminant.
Since $$p$$ is negative, the discriminant will be less than zero if the absolute value of $$p$$ is large enough.
\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{1^2}{4}+\dfrac{\Big(-\dfrac{7}{3}\Big)^3}{27}\\ & =\dfrac{1}{4}-\dfrac{343}{729}\\ & =\dfrac{1\cdot 729-343\cdot 4}{2916}\\ & =-0.22050754458162\\ \end{aligned}
### 4.1 Use the root formula directly
If $$\Delta < 0$$, then there are 3 distinct real roots for the cubic equation
$$\begin{cases} x_1 &= 2\sqrt[3]{r} \cos\dfrac{ θ}{3} \\ x_2 & = 2\sqrt[3]{r}\cos\Big( \dfrac{ θ}{3}+\dfrac{2\pi}{3} \Big) \\ x_3&= 2\sqrt[3]{r}\cos\Big( \dfrac{ θ}{3}+\dfrac{4\pi}{3} \Big) \end{cases}$$
where
$$\theta = \arccos(\dfrac{3q}{2p}\sqrt{-\dfrac{3}{p} } )$$ and $$\sqrt[3]{r} = \sqrt{\dfrac{-p}{3} }$$
Substitute the values of $$p$$ and $$q$$ to determine the value of $$\theta$$
\begin{aligned} \\\theta&= \arccos(\dfrac{3q}{2p}\sqrt{-\dfrac{3}{p} } )\\ & =\arccos\Big(\dfrac{3\cdot 1}{2 \cdot -\dfrac{7}{3}}\sqrt{-\dfrac{3}{-\dfrac{7}{3}} }\Big)\\ & = \arccos\Big(-\dfrac{1}{2}\cdot 3\cdot 1\cdot \dfrac{3}{7}\sqrt{3\cdot \dfrac{3}{7}}\Big)\\ & =\arccos\Big(-\dfrac{1}{2}\cdot 3\cdot 1\cdot \dfrac{3}{7}\cdot 3\sqrt{\dfrac{1}{7}}\Big)\\ & = \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)\\ \end{aligned}
Then we can determine one third of $$\theta$$
$$\dfrac{\theta}{3} = \dfrac{1}{3} \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)$$
Substitute the value of $$p$$ to determine the cubic root of $$r$$
\begin{aligned} \\\sqrt[3]{r}& =\sqrt{\dfrac{-p}{3}}\\ & =\sqrt{\frac{-(-7)}{3 \cdot 3}} \\ & =\sqrt{\dfrac{7}{9}}\\ & =\dfrac{\sqrt{7}}{3}\\ \end{aligned}
Substitute the value of $$\dfrac{\theta}{3}$$ and $$\sqrt[3]{r}$$ to the root formulas. Then we get
\begin{aligned} \\x_1&= 2\sqrt[3]{r} \cos\dfrac{ θ}{3}\\ & =2\cdot \dfrac{\sqrt{7}}{3}\cdot \cos \bigg[ \dfrac{1}{3}\cdot\arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)\bigg]\\ & =\dfrac{2}{3}\sqrt{7}\cos \bigg[\dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)\bigg]\\ & \approx \dfrac{2}{3}\sqrt{7} \cos 0.79585205265092\\ & \approx 1.2341129259511\\ \end{aligned}
\begin{aligned} \\x_2&=\dfrac{2}{3}\sqrt{7} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\dfrac{2\pi}{3}\bigg]\\ & \approx -1.7084121785118\\ \end{aligned}
\begin{aligned} \\x_3&=\dfrac{2}{3}\sqrt{7} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\dfrac{4\pi}{3} \bigg]\\ & \approx 0.47429925256064\\ \end{aligned}
### 4.2 Trigonometric Method
For a depressed cubic equation, another method is to convert the equation to the form of triple angle identity. By comparing the trigonometric value of the triple angle, the value of the single angle is determined. And then the roots of the cubic equation can be found.
Compare the given equation with the standard depressed cubic equation $$x^3 +px+q=0$$, we get $$p=-\dfrac{7}{3}$$ and $$q = 1$$
Using the formula
$$x= 2\sqrt{\dfrac{-p}{3}}u$$
to introduce an intermediate variable $$u$$ so that the given equation is transformed to a new equation in terms of $$u$$, which is analogous to a trignometric triple angle identity.
Then,
$$x= \dfrac{2}{3}\sqrt{7}u$$
Substitute to the cubic equation and simplify.
$$3\Big(\dfrac{2}{3}\sqrt{7}u\Big)^3 -7\Big(\dfrac{2}{3}\sqrt{7}u\Big)+3=0$$
Expand and simplify coefficients
$$3\cdot \dfrac{8}{27}\cdot7\sqrt{7}u^3 -\dfrac{14}{3}\sqrt{7}u+3=0$$
Continute simplifying
$$56\sqrt{7}u^3-42\sqrt{7}u+27=0$$
Dividing the equation by $$14\sqrt{7}$$ gives
$$4u^3-3u+\dfrac{27}{14}\sqrt{\dfrac{1}{7}}=0$$
The equation becomes the form of a triple angle identity for cosine function.
$$4\cos^3θ-3\cos θ-\cos3θ =0$$
Comparing the equation with triple angle identity gives
$$u =\cos θ$$
and
$$\cos3θ =-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}$$
The fact that the cosine function is periodic with period 2π implies the following equation.
$$\cos(3θ-2πk) =-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}$$
Solving for θ yields
$$θ =\dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\dfrac{2πk}{3} ,$$
in witch $$k=0,1,2$$.
Then we can determine the value of $$u$$, that is $$\cos θ$$
\begin{aligned} \\u&= \cos θ\\ & = \cos\Big[\dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\dfrac{2πk}{3}\Big]\\ \end{aligned}
and subsiquently,
\begin{aligned} \\x&= \dfrac{2}{3}\sqrt{7}u\\ & = \dfrac{2}{3}\sqrt{7} \cos\Big[\dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\dfrac{2πk}{3}\Big]\\ \end{aligned}
Substituting $$k=0,1,2$$ gives the roots of solution set.
\begin{aligned} \\x_1&= \dfrac{2}{3}\sqrt{7}\cos\Big[\dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)\Big]\\ & \approx1.2341129259511\\ \end{aligned}
\begin{aligned} \\x_2&= \dfrac{2}{3}\sqrt{7}\cos\Big[\dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\dfrac{2π}{3}\Big]\\ & \approx-1.7084121785118\\ \end{aligned}
\begin{aligned} \\x_3&= \dfrac{2}{3}\sqrt{7}\cos\Big[\dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\dfrac{4π}{3}\Big]\\ & \approx0.47429925256064\\ \end{aligned}
which shows the trigonometric method gets the same solution set as that by using roots formula.
## 5. Summary
In summary, we have tried the method of trigonometric to explore the solutions of the equation. The cubic equation $$3x³ - 7x + 3=0$$ is found to have three real roots . Exact values and approximations are given below.
$$\begin{cases} x_1=\dfrac{2}{3}\sqrt{7}\cos \bigg[\dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)\bigg] \\ x_2=\dfrac{2}{3}\sqrt{7} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\dfrac{2\pi}{3}\bigg] \\ x_3=\dfrac{2}{3}\sqrt{7} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\Big(-\dfrac{27}{14}\sqrt{\dfrac{1}{7}}\Big)+\dfrac{4\pi}{3} \bigg] \end{cases}$$
Convert to decimals,
$$\begin{cases} x_1=1.2341129259511 \\ x_2=-1.7084121785118 \\ x_3=0.47429925256064 \end{cases}$$
## 6. Graph for the function $$f(x) = 3x³ - 7x + 3$$
Since the discriminat is less than zero, the curve of the cubic function $$f(x) = 3x³ - 7x + 3$$ has 3 intersection points with horizontal axis.
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# About the counting problem last Christmas
Last Christmas, we gave out a free worksheet based on the familiar carol “The Twelve Days of Christmas”, where students are asked to count the number of gifts they will receive on each day and cumulatively the total number of gifts received up to the n-th day. While working through the exercises, we (and our students) gained some interesting insights into the problem of repeated reasoning. Here, we’re sharing what we observed.
The US Common Core for Mathematical Practice outlined eight mathematical practices that should be integrated into daily lessons, to develop long term empowering “habits” in students. The 8th of these standards is titled “Look for and Express Regularity in Repeated Reasoning”. It focuses on recognizing patterns and structure, so as to be able to make generalizations, and apply it to future situation.
For example, in the “Counting the Twelve Days of Christmas” worksheet, the pattern students observed is a series of triangular numbers.
Having understood the basic structure, students are encouraged to look more deeply into the pattern. What if we turn the triangular structure around and lay them out in a rectangular fashion?
What if we double the dots now?
Students can now more easily see how the sequence can be generalized into the number of dots in the n-th step, since they are now dealing with a more tractable rectangular shape. The total number of dots in the original triangular for the n-th step is simply half of the size of the rectangle, which is n times (n+1), i.e. n(n+1)/2.
In this way, students have found a generalization for the n-th triangular number, which they can apply to obtain any number in the sequence, for example in the worksheet, the number of gifts on the 12th day would be 12(12+1)/2 = 12×13 / 2 = 78.
Using this as a model to help kids look for pattern can help them develop their own strategies when dealing with similar situations. While working through the Christmas problems in class, we used a lot of “what if” questions to encourage creativity and curiosity, and help the kids explore what happens when we try different strategies, and had great fun doing it!
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Exponential Function
In Mathematics, an exponential function is a function whose value is a constant which is raised to the power of an argument, especially with the constant “e”. The exponent function is often used in many real-life applications. In this article, let us discuss what is an exponential function formula, properties, derivatives with examples.
Exponential Function Formula
The exponential function is an important mathematical function which is of the form
f(x) = ax
Where a>0 and a is not equal to 1.
x is any real number.
If the variable is negative, the function is undefined for -1 < x < 1.
Here,
“x” is a variable
“a” is a constant, which is the base of the function.
But, mostly the base of the exponential function is encountered by the transcendental number “e”, which is approximately equal to 2.71828.
Exponential Function Graph
The following figure represents the graph of exponents of x. It can be seen that as the exponent increases, the curves get steeper and the rate of growth increases respectively. Thus for x > 1, the value of y = fn(x) increases for increasing values of (n).
From the above discussion, it can be seen that the nature of polynomial functions is dependent on its degree. Higher the degree of any polynomial function, then higher is its growth. A function which grows faster than a polynomial function is y = f(x) = ax, where a>1. Thus, for any of the positive integers n the function f (x) is said to grow faster than that of fn(x).
Thus, the exponential function having base greater than 1, i.e., a > 1 is defined as y = f(x) = ax. The domain of exponential function will be the set of entire real numbers R and the range are said to be the set of all the positive real numbers.
It must be noted that exponential function is increasing and the point (0, 1) always lies on the graph of an exponential function. Also, it is very close to zero if the value of x is largely negative.
Exponential function having base 10 is known as a common exponential function. Consider the following series:
The value of this series lies between 2Â &3. it is represented by e. Keeping e as base the function, we get y = ex, which is a very important function in mathematics known as a natural exponential function.
For a > 1, the logarithm of b to base a is x if ax = b. Thus, loga b = x if ax = b. This function is known as logarithmic function.
For base a = 10, this function is known as common logarithm and for base a = e, it is known as natural logarithm denoted by ln x. Following are some of the important observations regarding logarithmic functions which has base a>1.
• The domain of log function consists of positive real numbers only, as we cannot interpret the meaning of log functions for negative values.
• For the log function, though the domain is only the set of positive real numbers, the range is set of all real values, I.e. R
• When we plot the graph of log functions and move from left to right, the functions show increasing behaviour.
• The graph of log function never cuts x-axis or y-axis, though it seems to tend towards them.
• Logap = α, logbp = β and logba = µ, then aα = p, bβ = p and bµ = a
• Logbpq = Logbp + Logbq
• Logbpy = ylogbp
• Logb (p/q) = logbp – logbq
Exponential Function Derivative
Let us now focus on the derivative of exponential functions.
The derivative of ex with respect to x is ex, I.e. d(ex)/dx = ex
It is noted that the exponential function f(x) =ex  has a special property. It means that the derivative of the function is the function itself.
(i.e) = f ‘(x) = ex = f(x)
Exponential Function Properties
The following are the properties of the exponential functions:
Exponential Function Example
Example 1:Â
Solve 4x = 43
Solution:
Since the bases are the same (i.e. 5), equate the values of powers.
So, the value of x is 3.
Example 2:
Solve 61-x = 64
 Solution:
From the given equation, it is noted that the bases are the same (i.e. 6), equate the values of powers.
It means that
1-x = 4
Now, simplify this equation, we get
-x = 4-1
-x = 3
x =-3
Therefore, the value of x is -3. |
# Difference between revisions of "2006 AIME II Problems/Problem 10"
## Problem
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumilated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
## Solution
### Solution 1
The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$. We let this probability be $p$; then the probability that $A$ and $B$ end with the same score in these give games is $1-2p$.
Of these three cases ($|A| > |B|, |A| < |B|, |A|=|B|$), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases).
There are ${5\choose k}$ ways to $A$ to have $k$ victories, and ${5\choose k}$ ways for $B$ to have $k$ victories. Summing for all values of $k$,
$1-2p = \frac{1}{2^{5} \times 2^{5}}\left(\sum_{k=0}^{5} {5\choose k}^2\right) = \frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \frac{126}{512}.$
Thus $p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}$. The desired probability is the sum of the cases when $|A| \ge |B|$, so the answer is $\frac{126}{512} + \frac{193}{512} = \frac{319}{512}$, and $m+n = \boxed{831}$.
### Solution 2
You can break this into cases based on how many rounds $A$ wins out of the remaining $5$ games.
• If $A$ wins 0 games, then $B$ must win 0 games and the probability of this is $\frac{{5 \choose 0}}{2^5} \frac{{5 \choose 0}}{2^5} = \frac{1}{1024}$.
• If $A$ wins 1 games, then $B$ must win 1 or less games and the probability of this is $\frac{{5 \choose 1}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}}{2^5} = \frac{30}{1024}$.
• If $A$ wins 2 games, then $B$ must win 2 or less games and the probability of this is $\frac{{5 \choose 2}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}}{2^5} = \frac{160}{1024}$.
• If $A$ wins 3 games, then $B$ must win 3 or less games and the probability of this is $\frac{{5 \choose 3}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}}{2^5} = \frac{260}{1024}$.
• If $A$ wins 4 games, then $B$ must win 4 or less games and the probability of this is $\frac{{5 \choose 4}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}}{2^5} = \frac{155}{1024}$.
• If $A$ wins 5 games, then $B$ must win 5 or less games and the probability of this is $\frac{{5 \choose 5}}{2^5} \frac{{5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}+{5 \choose 5}}{2^5} = \frac{32}{1024}$.
Summing these 6 cases, we get $\frac{638}{1024}$, which simplifies to $\frac{319}{512}$, so our answer is $319 + 512 = \boxed{831}$. |
# Exploring Circles: A Comprehensive Overview
1. A-level maths topics
2. Geometry
3. Circles
From the simplest shapes to the most complex, circles have long been a source of fascination and intrigue. As one of the basic geometrical shapes, circles appear in everything from art to architecture and mathematics. But what is a circle exactly? In this comprehensive overview, we'll explore the definition, properties, and applications of circles. A circle is a two-dimensional shape made up of points equidistant from a central point. It has no edges or corners, and all points along its circumference are equidistant from the center.
The properties of a circle make it an important shape in mathematics and design. In mathematics, circles are used to model real-world situations such as arcs, angles, and curves. Circles can also be used to calculate the area of a shape, the circumference of a circle, and other mathematical concepts. In design, circles can be used to create aesthetically pleasing designs and patterns.
In this article, we'll discuss the definition of a circle and its properties. We'll also explore the applications of circles in mathematics and design. By the end of this article, you'll have a comprehensive understanding of circles and their uses. The first thing to consider when discussing circles is their properties.
A circle has no sides, no corners, and the same circumference all around. It's also the only shape whose internal angles all add up to 360°. The two most important properties of circles are their radius and their diameter. The radius is the distance between the centre of the circle and any point on its circumference.
The diameter is twice the length of the radius. Circles also have an equation that helps us calculate information about them. This is known as the circle equation, or the equation of a circle. It's written as x² + y² = r² where x and y are coordinates and r is the radius of the circle. This equation can be used to calculate the area of a circle, which is equal to πr².In addition to this, circles can be divided into different types.
There are semicircles, which contain 180° of the full 360° of a circle; quarter circles, which contain 90°; and even eighth circles, which contain 45°. All these types of circles have their own equations that can be used to calculate information about them. Finally, there are some special uses for circles in mathematics. For example, they can be used to construct polygons such as triangles, squares, and pentagons. They can also be used to calculate the circumference of other shapes such as ellipses and rectangles. It's clear that circles are an important shape in mathematics and can be used for many different applications.
## Special Uses for Circles
Circles have many special uses, such as constructing polygons and calculating the circumference of other shapes.
Polygons are shapes consisting of three or more sides, and they can be constructed using a circle. To do so, draw a circle and mark off a number of points on its circumference. Then draw lines between the points to form the polygon. The circumference of a circle can also be used to calculate the circumference of other shapes.
For example, an ellipse can be approximated by drawing a large circle around it, then measuring the circumference of the circle. From this, you can calculate the circumference of the ellipse. The same principle can be applied to other shapes as well. By drawing a larger circle around them and measuring its circumference, you can get an approximate measure of the circumference of the smaller shape.
## Properties of Circles
Circles are a unique shape with several interesting properties.
They have no sides or corners, meaning they are smooth and continuous. Additionally, all circles have the same circumference regardless of their size. The circumference is the distance around the outside of the circle, and it can be calculated using the equation 2πr, where r is the radius of the circle. Circles are also unique in that all points on a circle are equidistant from the center. This means that if you draw a line from one point on the circle to another, it will always be the same length.
Furthermore, the area of a circle can be calculated with the equation πr2, where r is again the radius of the circle. Finally, circles have an angle measurement called the degree measure. This angle measure is equal to 360° and is used to measure angles inside and outside of circles.
## Types of Circles
Circles come in a variety of shapes and sizes, from semicircles to quarter circles to eighth circles. Let’s take a closer look at each type of circle.
#### Semicircles
are exactly what their name implies — half of a circle.
They can be found in nature in the form of arches or as part of a wheel. Mathematically, they are formed when one side of a circle is split in two, creating two equal parts. Semicircles can be used to create all sorts of shapes and designs, including the traditional circle.
#### Quarter circles
are similar to semicircles, but they are formed by cutting a circle into four equal parts.
This can also be referred to as a right angle, as it is formed by intersecting two lines at a 90-degree angle. Quarter circles can be used to create arcs, curves, and other shapes.
#### Eighth circles
are the smallest circles that can be formed. They are created by cutting a circle into eight equal parts.
Eighth circles can be used to form intricate designs, such as mandalas or repeating patterns. This type of circle is often used in art and design.
## Equation of a Circle
The equation of a circle is an algebraic expression that describes the dimensions and position of a circle. It can be used to calculate the center, radius, circumference, and area of a circle. The equation of a circle is typically written as:x2 + y2 = r2,where x and y are the coordinates of the center of the circle, and r is the radius. This equation can be used to find information about any point on the circle.
For example, if you know the coordinates of any point on the circle, you can use the equation to calculate the radius and center. Similarly, if you know the radius, you can use the equation to find the coordinates of any point on the circle. The equation of a circle can also be used to determine whether two circles are the same or not. This can be done by comparing their equations and seeing if they have the same radius and center. In addition, the equation of a circle can be used to calculate the area and circumference of a circle. To do this, we use the formula for the circumference of a circle:C = 2πr where C is the circumference and r is the radius.
Then, we can use this formula to calculate the area of a circle:A = πr2where A is the area and r is the radius. We've explored circles in great detail in this article, uncovering their properties, equations, types, and special uses. This knowledge will provide a strong foundation to build upon and help you further your understanding of this important shape in mathematics. Circles are a fundamental shape found in nature and mathematics, and they can be used for a variety of purposes. From the planets that orbit the sun to the wheels on a car, circles are an essential element that we encounter every day. Whether you’re a student or a professional, having a good understanding of circles is essential to success.
##### Shahid Lakha
Shahid Lakha is a seasoned educational consultant with a rich history in the independent education sector and EdTech. With a solid background in Physics, Shahid has cultivated a career that spans tutoring, consulting, and entrepreneurship. As an Educational Consultant at Spires Online Tutoring since October 2016, he has been instrumental in fostering educational excellence in the online tutoring space. Shahid is also the founder and director of Specialist Science Tutors, a tutoring agency based in West London, where he has successfully managed various facets of the business, including marketing, web design, and client relationships. His dedication to education is further evidenced by his role as a self-employed tutor, where he has been teaching Maths, Physics, and Engineering to students up to university level since September 2011. Shahid holds a Master of Science in Photon Science from the University of Manchester and a Bachelor of Science in Physics from the University of Bath. |
# Question #e355e
Jan 4, 2017
$y = {40}^{\circ}$
#### Explanation:
Given that $A B = A C \mathmr{and} \angle C = {40}^{\circ} , \implies \Delta A B C$ is an isosceles triangle.
An isosceles triangle is a triangle that has two sides of equal length.
The unequal side of an isosceles triangle is usually referred to as the 'base' of the triangle. In the figure, $B C$ is the base.
The base angles of an isosceles triangle are always equal. In the figure, the angles $\angle B$ and $\angle C$ are the same.
Given that $\angle C = {40}^{\circ}$
$\implies \angle B = \angle C = {40}^{\circ}$
$\implies \angle A = 180 - \left(\angle B + \angle C\right) = 180 - \left(40 + 40\right) = {100}^{\circ}$
Given $\angle A = 2 y + 20$
$\implies 2 y + 20 = {100}^{\circ}$
$\implies 2 y = {80}^{\circ}$
$\implies y = {40}^{\circ}$ |
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# The Inverse of a Matrix
A square matrix that has an inverse is invertible (non-singular). Not every square matrix has an inverse.
Non-square matrices do not have an inverse and are singular. To show this we examine two matrices; A and B. If A is of order m×n and B is of order n×m where mn, the products of AB and BA cannot be equal, and by matrix multiplication definition AB and BA cannot be multiplied.
The inverse of a square matrix is easiest to understand if we begin with the equation ax = b where a ≠ 0.
To solve this equation for x we multiply both sides of the equation by a
−1:
ax = b
a−1 ax = a−1 b
x = a−1 b
a−1 is the multiplicative inverse of a because a−1 a = 1. This is similar to the definition of the multiplicative inverse of a matrix:
If we let A be an n×n matrix and let In be the n×n identity matrix then,
AA−1 = In = A−1 A
This identifies A−1 as the multiplicative inverse of A, the A inverse.
## How to Find the Inverse of a Square Matrix Using a System of Linear Equations
By applying matrix multiplication to a square matrix of which we want to find the inverse and using the matrix equation AX = I to solve for X, when operations have been completed the square matrix X is the inverse matrix A−1, X = A−1, and we will have solved AA−1 = In. We show how to find the inverse of a 2×2 matrix; however this scheme applies to any square matrix:
Matrix A
2×2 C1 C2 R1: 1 4 R2: −1 −3
Matrix X
2×2 C1 C2 R1: X11 X12 R2: X21 X22
Identity Matrix
2×2 C1 C2 R1: 1 0 R2: 0 1
Matrix A×X
1X11 + 4X21 1X12 + 4X22
−1X11 − 3X21 −1X12 − 3X22
Identity Matrix
=
1 0 0 1
Next, equate corresponding entries to obtain two systems of linear equations…
Linear Equations System 1
X11 + 4X21 = 1
−X11 − 3X21 = 0
Linear Equations System 2
X12 + 4X22 = 0
−X12 − 3X22 = 1
From the first system we determine that X11 = −3 and X21 = 1.
From the second system we determine −X12 = −4 and X22 = 1.
We now write the inverse of A as:
Matrix X (A−1)
2×2 C1 C2 R1: −3 −4 R2: 1 1
It is recommended that you also understand the Gauss-Jordan Elimination method, especially if you are working with 3×3 matrices or larger. By solving both systems of linear equations simultaneously it is more efficient than solving for the inverse of a matrix.
## Using the Matrix Formula to Find the Inverse
The matrix formula works for 2×2 matrices. The following shows how the formula works.
Matrix A
a b c d
If matrix A is invertible then adbc ≠ 0. Should adbc ≠ 0 the inverse of matrix A is given by:
A−1 = 1 / (ad − bc) × A
The denominator adbc is the determinant of the 2 × 2 matrix.
Matrix A
3 −1 −2 2
ad − bc = (3)(2) − (−2)(−1) = 4
And A−1 = 1 / (ad − bc) × A = ¼ A
The inverse is a scalar multiplication of ¼ by the array elements of matrix A. |
• # Revision:Further Vectors
TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Further Vectors
## Vector equations of a line
Lines can be written in several different ways:
There’s the c4 way:
The other c4 way:
Also the Cartesian form:
## Vector equations of a plane
Similar to the c4 line equation:
Similar again: I used numbers for the sake of notation. 2 i’s won’t look pretty!
Here’s a new one:
Or:
Or:
## Vector product
• The definition of the vector product is where is the unit vector perpendicular to both a and b in the direction given by the right hand screw rule.
• For vectors in component form, the vector product may be calculated using either of the following determinants, the second of which is given in the formula booklet:
or
• The vector product of any 2 parallel vectors is zero
• The vector product is anti-commutative as
• Area of a triangle is , of a parallelogram is , and representing adjacent sides.
## Intersections
#### Point of intersection of a line and a plane
• Express the position vector of a general point on the line as a single vector, e.g.
• substitute this vector into the normal form of the equation of a plane and find , e.g.
etc.
• Substitute your value of back into the equation of the line.
#### Line of intersection of 2 intersecting planes
• Find the direction vector of the line by using the fact that the line is perpendicular to both normal vectors, i.e.
• Find any point , common to both planes by using any value (usually 0) for x, y or z
• Equation of line is
## Angles
• The direction vector of a line is given by d in the equation r= d
• The direction of the normal to a plane is given by n in the equation r.n = a.n
• The angle between two planes is the angle between their 2 normals, n_1 and n_2. This is usually found by using the scalar product, a.b =
• The angle between two lines is the angle between their 2 direction vectors, d_1 and d_2.
• The angle between a line and a plane is found by finding the angle between d and n then subtracting from 90 degrees. Alternatively, if is the angle between d and n then as you may use
## Distances
• The distance of a plane r.n = p from the origin is or where the sign is unimportant.
• To find the distance between two planes, first find their distances from the origin. If the two p-values are the same sign, the planes are on the same side of the origin so subtract the 2 distances. If the two p-values are different signs, the origin lies between them so add the positive distances.
#### Distance between a plane and a point, A
1)*Express the equation in the form
• If the point is , quote the result from the formula booklet giving:
Distance =
2)*Find the equation of the plane through the given point A, parallel to the original plane, using r.n = a.n
• Find the distance between these 2 planes.
3)*Locate any point, P on the plane and find the vector
• If is the angle between and n, then the required distance h is so h =
4)*Find the co-ordinates of F, the foot of the perpendicular from A to the plane and find AF. This method should only really be used if F is specifically asked for because it’s a bit long winded. Anyway, F can be found as follows:
• As AF is perpendicular to the plane, then n, so = a + n
• Proceed as for and intersection of a line and a plane
#### Distance between a point A, from a line L
1)*Locate any point, P on L and find the vector
• If is the angle between and d, then the required distance h is so h =
2)* Find the co-ordinates of F, the foot of the perpendicular from A to L and find AF. F can be found as follows:
• As F lies on L, express the position vector of F as a single vector involving
• Use the fact that to find
• As is perpendicular to L, then so use this to find and hence F
#### Distance between 2 skew lines
• Locate any point, A on and any point B on and find the vector = ba
• Find the common normal, i.e. the vector which is perpendicular to both and . This is given by n = and may be simplified if appropriate
• If is the angle between and n, then the required distance h is so h =
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# Worksheet on Absolute Value of an Integer | Absolute Value Worksheet PDF with Answers
Printable Absolute Value of Integers Worksheets available here are designed to aid students of 6th Grade Math to learn the concept of Absolute Value in detail. You will find Problems on finding Absolute Values of both Positive and Negative Integers, Adding, Subtracting, Multiplying, Dividing the Absolute Value of Real Numbers, and much more in our Absolute Value Worksheets PDF. Begin your practice using these handy worksheets and check with the answers provided to assess your strengths and weaknesses.
Also, Refer:
## Absolute Value of Integers Worksheet
I. Write the absolute value for each of the following
I. 20
II. +12
III. -34
IV. 1
V. 0
VI. +475
VIII. -238
Solution:
I. The absolute value of 20 is 20.
II. The absolute value of +12 is 12.
III. The absolute value of -34 is 34.
IV. The absolute value of 1 is 1.
V. The absolute value of 0 is 0.
VI. The absolute value of +475 is 475.
VII. The absolute value of -238 is 238.
II. Write the absolute values of each of the given numbers
I. |-50|
ii. |-220|
iii. |189|
iv. |+29|
v. |-58|
vi. |-14|
Solution:
I. The absolute value of |-50| is 50.
II. The absolute value of |-220| is 220.
III. The absolute value of |189| is 189.
IV. The absolute value of |+29| is 29.
V. The absolute value of |-58| is 58.
VI. The absolute value of |-14| is 14.
III. Evaluate the following
I. |12|
ii. |-10|
iii. |10 – 12|
iv. |10| + |-12|
v. |10| – |12|
vi. |12 – 10|
Solution:
I. The absolute value of |12| is 12.
II. The absolute value of |-10| is 10.
III. |10 – 12|
= |- 2|
=2
The absolute value of |10 – 12| is 2
IV. |10| + |-12|
=10 + 12
=22
The absolute value of |10| + |-12| is 22.
V. |10| – |12|
=10-12
=-2
The absolute value of |10| – |12| is -2.
VI. |12 – 10|
=|2|=2
The absolute value of |12 – 10| is 2.
IV. Evaluate the following
I. Â |-4| + |+15| + |0|
ii. |13| – |-18| + |+14|
iii. – |+4| – |-3| + |-8|
iv. |-9| – |25| + |-10|
Solution:
I. |-4| + |+15| + |0|
=4 + 15 + 0
=19
II. |13| – |-18| + |+14|
=13 – 18 + 14
=9
III. – |+4| – |-3| + |-8|
=- 4- 3 + 8
=1
IV. |-9| – |25| + |-10|
=9 – 25 +10
=6
V. State whether the statements are true or false
i. The absolute value of -7 is 7.
ii. The absolute value of an integer is always less than the integer.
iii. |+4| = +4
iv. |-8| = -8
v.- |+8| = -8
vi. – |-8| = -8
Solution:
i. True
ii. False
iii. True
iv. False
v. True
Vi. True
VI. Arrange the given numbers in ascending order
i. -|-18|, |14|, |9|, |-95|, |-7|, |-8|, |-63|, |3|
Solution:
First, we have to find the absolute values of the given numbers
-18,14,9,95,7,8,63,3
Now arrange the numbers in ascending order.
-18, 3, 7, 8, 9, 14, 63, 95.
ii. -|-52|, |43|, |18|, |-102|, |-13|, |-38|, |-89|, |14|
Solution:
First, we have to find the absolute values of the given numbers.
-52,43,18,102,13,38,89,14
Now arrange the numbers in ascending order.
-52, 13, 14, 18, 38, 43, 89, 102.
iii. |-12|, |23|,- |-81|, |-16|, |-14|, |-78|, |-81|, |19|
Solution:
First, we have to find the absolute values of the given numbers.
12, 23, -81, 16,14, 78, 81, 19
Now arrange the numbers in ascending order.
-81, 12, 14, 16, 19, 23, 78, 81.
VII. Arrange the given numbers in descending order
i. |-33|,- |-12|, |4|, |-90|, |-5|, |-60|, |-68|, |3|
Solution:
First, we have to find the absolute values of the given numbers.
33, -12, 4, 90, 5, 60, 68, 3
Now arrange the numbers in descending order.
90, 68, 60, 33, 5, 4, 3, -12.
ii. |-73|,- |-66|, |7|, |-99|, |-3|, |-61|, |-88|, |35|
Solution:
First, we have to find the absolute values of the given numbers.
73, -66, 7, 99, 3, 61, 88, 35
Now arrange the numbers in descending order.
99, 88, 73, 61, 35, 7, 3, -66.
iii. |-23|,|-36|, -|5|, |-54|, |-69|, |-73|, |-81|, |-25|
Solution:
First, we have to find the absolute values of the given numbers.
23, 36, -5, 54, 69, 73, 81, 25.
Now arrange the numbers in descending order.
81, 73, 69, 54, 36, 25, 23,-5.
VIII.
i. Graph the absolute value of the number -6.
Solution:
The absolute value of the number |-6| is 6.
So the graph of the number -6 will look like the following.
ii. Graph the absolute value of the number -3.
Solution:
The absolute value of the number |-3| is 3.
So the graph of the number -3 will look like the following.
Scroll to Top
Scroll to Top |
# Basic Money Math Practice Test for 2nd Grade – [Medium]
Money is a fascinating subject, especially for 2nd graders. At this age, children are becoming more aware of the concept of money, its importance, and its various forms. Brighterly aims to simplify the world of math and money for young learners. Today, we dive into the basic money concepts tailored for 2nd graders.
## Understanding Coins and Their Values
Before we get to counting money, it’s essential to recognize the different coins. Here are the primary ones:
• Penny:
• Worth: \$0.01
• Fun Fact: It features Abraham Lincoln on one side!
• Nickel:
• Worth: \$0.05
• Quick Tip: It’s thicker and bigger than a penny.
• Dime:
• Worth: \$0.10
• Remember: Even though it’s the smallest coin, it’s worth more than both the penny and the nickel.
• Quarter:
• Worth: \$0.25
• Cool Fact: There are different designs on the back, showcasing various U.S. states and territories.
Understanding the different coin values is crucial. It sets the foundation for more advanced money math in the future.
## Counting Money
Now that we’re familiar with the various coins let’s move on to counting money. Using physical coins can be a fun activity. Here’s a simple exercise:
2. Move to Nickels: Remember, each nickel is like counting by fives.
3. Dimes are Next: Counting by tens is a breeze!
4. End with Quarters: They’re like counting by twenty-fives.
With a bit of practice, kids will start to combine different coins and count mixed amounts. For example, 2 quarters, 3 dimes, and 4 pennies equal 79 cents!
## Making Purchases and Getting Change
One of the most thrilling experiences for a child is buying something with their money. But this also introduces the idea of receiving change. When they hand over more money than the item’s price, they should get some money back. This concept strengthens their subtraction skills. Let’s say a toy costs \$0.50, and they give a dollar bill. How much should they get back? By deducting the cost from the amount given, they’ll find out they should receive 50 cents back!
## Saving Money
Finally, it’s not just about spending; saving is equally important. Introduce your child to the concept of piggy banks or savings jars. Every time they save, they’re setting aside money for bigger things in the future. It’s an excellent lesson in patience and foresight.
In conclusion, understanding basic money concepts in the 2nd grade is more than just math. It’s about laying the foundation for essential life skills. At Brighterly, our mission is to illuminate young minds, and with lessons like these, we’re confident we’re on the right path!
Basic Money Practice Test for 2nd Grade
Get ready for math lessons with Brighterly! This practice test is designed to do just that, specifically focusing on fundamental money concepts. Here, our young learners will delve into identifying various coins, counting their value, making simulated purchases, and even getting a taste of how to save and budget.
1 / 15
How much is a nickel worth?
2 / 15
If you have three dimes and two pennies, how much money do you have in total?
3 / 15
Which coin is the smallest in size but worth more than a penny and a nickel?
4 / 15
Sarah has two quarters. How much money does she have?
5 / 15
Which of the following combinations will give you 28 cents?
6 / 15
How much more is a quarter worth than a dime?
7 / 15
If you buy a toy for 46 cents and give a dollar, how much change should you get back?
8 / 15
How much is five nickels worth?
9 / 15
Which coin has the smallest value?
10 / 15
Tom has three quarters and four dimes. How much money does he have?
11 / 15
If a candy costs 12 cents and you pay with two dimes, how much change will you receive?
12 / 15
How much money do you have if you combine a quarter, a dime, and two pennies?
13 / 15
Which combination gives 16 cents?
14 / 15
How much is a dime and five pennies worth combined?
15 / 15
If an eraser costs 8 cents and you pay with a dime, what will be your change?
0%
Poor Level
Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence.
Mediocre Level
Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence.
Needs Improvement
Start practicing math regularly to avoid your child`s math scores dropping to C or even D.
High Potential
It's important to continue building math proficiency to make sure your child outperforms peers at school. |
# SAT Math: Guide to Quadratic Equations & Radicals
tl;dr: Quadratic equations and radicals in the SAT Math section include topics such as using the quadratic formula, factoring polynomials, interpreting graphs and functions, and determining roots and square roots. The practice problems below will test your ability to rewrite expressions, add/subtract/multiply polynomial expressions, utilize the distributive property, use the quadratic formula, and apply the principles of quadratic equations & factoring.
These questions (both with-calculator and no-calculator) test on four major content areas: the Heart of Algebra, Problem Solving, Data Analysis, and the Passport to Advanced Math.
### How long is the SAT Math section?
The SAT Math section contains approximately 60 questions, separated into:
• No-Calculator Section
• 15 Multiple Choice
• 5 Grid-Ins
• 25 Minutes
• Calculator Section
• 30 Multiple Choice
• 8 Grid-Ins
• 55 Minutes
• Total Time: 80 Minutes
Quadratic Equations and Radicals fall under the Passport to Advanced Math section. Topics pertaining to “Quadratic Equations & Radicals”:
• Quadratic Equations & Expressions
• Function Interpretation
• Graph Interpretation
• Roots
• Square Roots
• Squaring
What you will have to be able to do:
• Use the quadratic formula
• Factor and multiply polynomial expressions (word problems, etc.)
• Interpret graphs and functions to write quadratic expressions
• Determine the roots & square roots of various functions and equations
### SAT Math Section Rules
Directions for the Non-Calculator Section
• The use of a calculator is not permitted.
• All variables and expressions used represent real numbers unless otherwise indicated.
• Figures provided in this test are drawn to scale unless otherwise indicated.
• All figures lie in a plane unless otherwise indicated.
## SAT Math Quadratics Practice Problems
### Non-calculator practice problems:
1. Which of the following is equivalent to 16s4 - 4t2?
(a) 4(s2 − t )(4s2 + t )
(b) 4(4s2 - t)(s2 - t)
(c) 4(2s2 − t )(2s2 + t )
(d) (8s2 − 2t )(8s2 + 2t )
The correct answer is C.
• Recognize something familiar about this problem? It’s a difference of squares!
• This classic quadratic pattern draws from x2 - y2, which can be factored into (x-y)(x+y). Recognizing these patterns can help you answer questions quickly and help you set a steady pace on the exam.
• How does this question demonstrate an understanding of the difference of squares (more specifically, the difference of perfect squares)?
• Let’s break it down!
⇒ 16s4 - 4t2
⇒ = (4s2)2 - (2t)2 (difference of perfect squares)
⇒ = (4s2 − 2t ) (4s2 + 2t) (an extension of (x+y)(x-y))
⇒ = 2(2s2 − t )(2)(2s2 + t ) (factor out a 2!)
⇒ = 4(2s2 − t ) (2s2 + t ) (Multiply your factors of 2 = 4!)
⇒ This leads us to Option C!
• This question tests your ability to rewrite expressions, and add/subtract/multiply polynomial expressions.
Feeling good? Let’s try another one!
2. (x2 + bx − 2) (x + 3) = x3 + 6x2 + 7x − 6
In the equation above, b is a constant. If the equation is true for all values of x, what is the value of b?
(a) 2
(b) 3
(c) 7
(d) 9
The correct answer is B.
• This question tests your ability to utilize the distributive property. Let’s expand this equation!
• The key to this question is to convert one side of the equation to match the other side in order to compare like terms and determine the value of b!
• Left Hand Side = (x2 + bx − 2)(x + 3)
⇒ = (x3 + bx2 − 2x ) + (3x2 + 3bx − 6)
⇒ = x3 + (3 + b) x2 + (3b − 2)(x − 6)
• Right Hand Side = x3 + 6x2 + 7x − 6
⇒ Now that all like terms have been combined and expanded on the Left Hand Side, we can compare them to the Right Hand Side! It’s visible that (3+b) = 6, therefore b = 3
• This question tests your ability to rewrite expressions, and add/subtract/multiply polynomial expressions.
Time for Question 3! Looking good!
3) What are the solutions to the equation x2 − 3 = x?
(a) (−1 ± √11)/2
(b) (−1 ± _√13)/2
(c) (1 ± √11)/2
(d) (1 ± √13)/2
The correct answer is D.
• This question draws on your ability to use the quadratic formula and completing the square. How so? Let’s find out!
• First, subtract x from each side to convert the equation into standard form
• Standard Form: ax2 + bc + c
• Now, you’re left with x2 - x - 3 = 0
• Comparing this to the standard form above, you can note that a = 1, b = -1, and c = -3!
• You can plug these values into the quadratic formula, and simplify to get Choice D!
• This question tests your ability to utilize radicals, and apply the quadratic formula.
Time for Question 4! How are you feeling?
(4) x − 12 = √(x + 44)
What are all possible solutions to the given equation?
(a) 5
(b) 20
(c) −5 and 20
(d) 5 and 20
The correct answer is B.
• This question draws on your ability to use the radicals and use the quadratic equation. Let’s break it down:
• x − 12 = √(x + 44)
• First, square both sides of the equation!
⇒ = (x - 12)2 = (x + 44)
⇒ = x2 - 24x + 144 = x + 144
⇒ x2 - 25x + 100 = 0
⇒ (x-5)(x-20)
Seems like 5 and 20 both satisfy our requirements! But wait, there’s one more check! We need to plug both 5 and 20 back into the original equation to ensure they work.
After plugging both 5 and 20 in, only 20 works, therefore Choice B is correct!
• This question tests your ability to utilize radicals, and apply the principles of quadratic equations & factoring.
### Closing
Congratulations! You’ve made it to the end of this prep activity 🙌. You learned about “SAT Math - Quadratic Equations & Radicals”. You should have a better understanding of the Math sections for the SAT© , topic highlights, what you will have to be able to do in order to succeed, as well as have seen some practice questions that put the concepts in action. Good luck studying for the SAT Math section 👏!
Need more resources? Check out our complete SAT Math Study Guide w/ Practice Problems.
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## Questions
### Subject:Pre-Calculus
TutorMe
Question:
Solve the triangle ABC if a=2, b=6, and A = 30 degrees.
Murali Krishna Varma N.
Every triangle ABC has six elements: 1) Three angles: <A, <B, and <C 2) The sides opposite to them: a, b, and c. By solving a triangle, we mean finding the unknown elements of the triangle. It is possible only in the following cases: 1) The three sides are given. 2) Two sides and the included angle are given. 3) Two sides and the angle opposite to one of them given. 4) One side and two angles are given. In our problem, two sides 'a' and 'b' and the angle 'A' which is opposite to the side 'a' are given. So, this is the case (3) and in which we use the law of sines. That is, sin(A)/a = sin(B)/b = sin(C)/c. From, sin(A)/a = sin(B)/b, we get sin(B)= b*sin(A)/a. Now, we might have three possibilities: 1) sin(B)>1 2) sin(B)=1 3) 0<sin(B)<1 Note that 0<=sin(B)<=1 for any angle B, and sin(B) is not equal to zero as b>0. Also, sin(B)>0 when B is in the first two quadrants. So, 1) sin(B)>1 implies existence of no such triangle, 2) sin(B)=1 results in a unique right-triangle, right angled at B, and 3) 0<sin(B)<1 results two triangles. If we can find the angle B, we can easily find the values of C and c. m <C = 180-(m<A + m <B). Now, sin(B) = b*sin(A)/a = 6*sin(30)/2 = 6*(1/2)/2 = 3/2 = 1.5>1. This result shows that there is no such triangle.
### Subject:Geometry
TutorMe
Question:
A regular hexagon with a perimeter of 24 inches is inscribed in a circle. How far from the center is each side?
Murali Krishna Varma N.
A regular hexagon has six sides with equal length. Let the length of each side be x inches. Then, its perimeter would be 6x inches. So, we have 6x= 24 so that x= 24/6 = 4 inches. That is, each side of the hexagon is of length 4 inches. Each side of the hexagon subtends an angle of 360/6 = 60 degrees at the center. By choosing one side of the hexagon and joining its endpoints (vertices) with the center, we form an isosceles triangle with vertex angle 60 degrees. Let the altitude of this triangle be h inches. This altitude bisects the vertex angle and the base of the isosceles triangle and hence bisects the triangle into two 30-60-90 special triangles. Since the base is 4 inches, half of it would be 2 inches. So, in a 30-60-90 triangle, the side opposite to 30 degrees is of length 2 inches, and the side opposite to 60 degrees is of length h inches. The ratio of the sides of a 30-60-90 triangle is 1: sqrt(3):2. So, 2/h= 1/sqrt(3) and hence h= 2sqrt(3) inches. That is, the center is located at a distance of 2*sqrt(3) inches from each side of the given regular hexagon.
### Subject:Calculus
TutorMe
Question:
Does the series 1+ (1/2) +(1/3)+(1/4)+......converge?
Murali Krishna Varma N.
This is one of the basic results in this concept that we can use to check the convergence of other series. Here, we use the Monotone Convergence Theorem. It states: The sequence {s(n)} is convergent if it is EITHER 1. Increasing and 2. Bounded above OR 1. Decreasing and 2. Bounded below. Now, let us consider nth partial sums s(n) of the terms of the given series: 1+(1/2)+(1/3)+(1/4)+..... s(1) = 1 s(2) = 1+(1/2) s(4) = 1+ 1/2 + (1/3 + 1/4) = 1+ 1/2 +7/12> 1+ 1/2 + 1/2=1+(1/2)(2) (Note: 2^2 = 4) s(8) = 1+1/2 + (1/3 + 1`/4) + (1/5+1/6+1/7+1/8)> 1+ 1/2 + 1/2 + 1/2 = 1 + (1/2)(3) (Note: 2^3=8) ------- s(2^k) > 1 + (1/2 + 1/2 + .....(k times)) = 1+ (1/2)k From this, we can see that the sequence {s(n)} of partial sums is increasing and not bounded above. Therefore, it cannot be convergent by the Monotone Convergence Theorem and hence it is divergent. Since the sequence of partial sums is divergent, the given series is also divergent.
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### Coterminal Angles and Radian Measure
```Radian Measure and
Coterminal Angles
Take out your
homework from
Friday!!!
Warm-up (1:30 m)
from Friday, describe how you convert
Converting Between Degrees
To convert degrees To convert radians
to radians, multiply to degrees, multiply
by
by
Converting Between and
220
π
5
Picture of Unit Circle with missing degrees
and radian measures. Students fill missing
measures.
180
57.3
π
Another way of measuring angles
Convenient because major measurements of a
circle (circumference, area, etc.) are involve pi
Radians result in easier numbers to use
The Unit Circle – An Introduction
1 Revolution = 360°
2 Revolutions = 720°
Positive angles move
counterclockwise around
the circle
Negative angles move
clockwise around the
circle
Sketching
90°
0°
180°
360°
270°
Trick: Convert the fractions into decimals
and use the leading coefficients of pi
π
2
π
3π
2
2π
5π
Example #1 6
6π
Example #2 4
π
Example #3 4
9π
Example #4 7
7
π
3
7
5π
3
13
17π
9
Experiment
Graph
3
and
2
2
do you notice?
on the axes below. What
Coterminal Angles
co – terminal
with, joint,
or together
ending
Coterminal Angles – angles that end
at the same spot
Coterminal Angles, cont.
Each positive angle has a negative
coterminal angle
Each negative angle has a positive
coterminal angle
Solving for Coterminal Angles
If the angle is
If the angle is less
greater than 2 pi, than 0, add 2 pi
subtract 2 pi
to the given
from the given
angle.
angle.
You may need to add or subtract 2 pi more than
once!!!
Trick: Add or subtract the coefficients of pi
rather than the entire radian measure
Examples: Find a coterminal angle
between 0 and 2 pi
2π
3
29π
6
Your Turn: Find a coterminal angle
between 0 and 2 pi
14π
13
18π
5
9π
4
6π
4
Group Exit Ticket
7
π
17
π
Are
and
coterminal? Why or
6
6
why not?
Exit Ticket, cont.
1.
2.
Multiply:
2 * 18
Rationalize:
2
2
``` |
# 9.3 Double-angle, half-angle, and reduction formulas (Page 3/8)
Page 3 / 8
## Using the power-reducing formulas to prove an identity
Use the power-reducing formulas to prove
${\mathrm{sin}}^{3}\left(2x\right)=\left[\frac{1}{2}\text{\hspace{0.17em}}\mathrm{sin}\left(2x\right)\right]\text{\hspace{0.17em}}\left[1-\mathrm{cos}\left(4x\right)\right]$
We will work on simplifying the left side of the equation:
Use the power-reducing formulas to prove that $\text{\hspace{0.17em}}10\text{\hspace{0.17em}}{\mathrm{cos}}^{4}x=\frac{15}{4}+5\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right)+\frac{5}{4}\text{\hspace{0.17em}}\mathrm{cos}\left(4x\right).$
## Using half-angle formulas to find exact values
The next set of identities is the set of half-angle formulas , which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}\frac{\alpha }{2},$ the half-angle formula for sine is found by simplifying the equation and solving for $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\alpha }{2}\right).\text{\hspace{0.17em}}$ Note that the half-angle formulas are preceded by a $\text{\hspace{0.17em}}±\text{\hspace{0.17em}}$ sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which $\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ terminates.
The half-angle formula for sine is derived as follows:
$\begin{array}{ccc}\hfill {\mathrm{sin}}^{2}\theta & =& \frac{1-\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ \hfill {\mathrm{sin}}^{2}\left(\frac{\alpha }{2}\right)& =& \frac{1-\left(\mathrm{cos}2\cdot \frac{\alpha }{2}\right)}{2}\hfill \\ & =& \frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}\hfill \\ \hfill \mathrm{sin}\left(\frac{\alpha }{2}\right)& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \end{array}$
To derive the half-angle formula for cosine, we have
$\begin{array}{ccc}\hfill {\mathrm{cos}}^{2}\theta & =& \frac{1+\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ \hfill {\mathrm{cos}}^{2}\left(\frac{\alpha }{2}\right)& =& \frac{1+\mathrm{cos}\left(2\cdot \frac{\alpha }{2}\right)}{2}\hfill \\ & =& \frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}\hfill \\ \hfill \mathrm{cos}\left(\frac{\alpha }{2}\right)& =& ±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \end{array}$
For the tangent identity, we have
$\begin{array}{ccc}\hfill {\mathrm{tan}}^{2}\theta & =& \frac{1-\mathrm{cos}\left(2\theta \right)}{1+\mathrm{cos}\left(2\theta \right)}\hfill \\ \hfill {\mathrm{tan}}^{2}\left(\frac{\alpha }{2}\right)& =& \frac{1-\mathrm{cos}\left(2\cdot \frac{\alpha }{2}\right)}{1+\mathrm{cos}\left(2\cdot \frac{\alpha }{2}\right)}\hfill \\ & =& \frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }\hfill \\ \hfill \mathrm{tan}\left(\frac{\alpha }{2}\right)& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }}\hfill \end{array}$
## Half-angle formulas
The half-angle formulas are as follows:
$\mathrm{sin}\left(\frac{\alpha }{2}\right)=±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}$
$\mathrm{cos}\left(\frac{\alpha }{2}\right)=±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}$
$\begin{array}{l}\mathrm{tan}\left(\frac{\alpha }{2}\right)=±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{\mathrm{sin}\text{\hspace{0.17em}}\alpha }\hfill \end{array}$
## Using a half-angle formula to find the exact value of a sine function
Find $\text{\hspace{0.17em}}\mathrm{sin}\left(15°\right)\text{\hspace{0.17em}}$ using a half-angle formula.
Since $\text{\hspace{0.17em}}15°=\frac{30°}{2},$ we use the half-angle formula for sine:
$\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\frac{30°}{2}& =& \sqrt{\frac{1-\mathrm{cos}30°}{2}}\hfill \\ & =& \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\hfill \\ & =& \sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}}\hfill \\ & =& \sqrt{\frac{2-\sqrt{3}}{4}}\hfill \\ & =& \frac{\sqrt{2-\sqrt{3}}}{2}\hfill \end{array}$
Remember that we can check the answer with a graphing calculator.
Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.
1. Draw a triangle to represent the given information.
2. Determine the correct half-angle formula.
3. Substitute values into the formula based on the triangle.
4. Simplify.
## Finding exact values using half-angle identities
Given that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha =\frac{8}{15}$ and $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ lies in quadrant III, find the exact value of the following:
1. $\mathrm{sin}\left(\frac{\alpha }{2}\right)$
2. $\mathrm{cos}\left(\frac{\alpha }{2}\right)$
3. $\mathrm{tan}\left(\frac{\alpha }{2}\right)$
Using the given information, we can draw the triangle shown in [link] . Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =-\frac{8}{17}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\alpha =-\frac{15}{17}.$
1. Before we start, we must remember that if $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ is in quadrant III, then $\text{\hspace{0.17em}}180°<\alpha <270°,$ so $\text{\hspace{0.17em}}\frac{180°}{2}<\frac{\alpha }{2}<\frac{270°}{2}.\text{\hspace{0.17em}}$ This means that the terminal side of $\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ is in quadrant II, since $\text{\hspace{0.17em}}90°<\frac{\alpha }{2}<135°.$
To find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\frac{\alpha }{2},$ we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in [link] and simplify.
$\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ & =& ±\sqrt{\frac{1-\left(-\frac{15}{17}\right)}{2}}\hfill \\ & =& ±\sqrt{\frac{\frac{32}{17}}{2}}\hfill \\ & =& ±\sqrt{\frac{32}{17}\cdot \frac{1}{2}}\hfill \\ & =& ±\sqrt{\frac{16}{17}}\hfill \\ & =& ±\frac{4}{\sqrt{17}}\hfill \\ & =& \frac{4\sqrt{17}}{17}\hfill \end{array}$
We choose the positive value of $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ because the angle terminates in quadrant II and sine is positive in quadrant II.
2. To find $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{\alpha }{2},$ we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in [link] , and simplify.
$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ & =& ±\sqrt{\frac{1+\left(-\frac{15}{17}\right)}{2}}\hfill \\ & =& ±\sqrt{\frac{\frac{2}{17}}{2}}\hfill \\ & =& ±\sqrt{\frac{2}{17}\cdot \frac{1}{2}}\hfill \\ & =& ±\sqrt{\frac{1}{17}}\hfill \\ & =& -\frac{\sqrt{17}}{17}\hfill \end{array}$
We choose the negative value of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ because the angle is in quadrant II because cosine is negative in quadrant II.
3. To find $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\frac{\alpha }{2},$ we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in [link] and simplify.
$\begin{array}{ccc}\hfill \mathrm{tan}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }}\hfill \\ & =& ±\sqrt{\frac{1-\left(-\frac{15}{17}\right)}{1+\left(-\frac{15}{17}\right)}}\hfill \\ & =& ±\sqrt{\frac{\frac{32}{17}}{\frac{2}{17}}}\hfill \\ & =& ±\sqrt{\frac{32}{2}}\hfill \\ & =& -\sqrt{16}\hfill \\ & =& -4\hfill \end{array}$
We choose the negative value of $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ lies in quadrant II, and tangent is negative in quadrant II.
x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
How can I solve for a domain and a codomains in a given function?
ranges
EDWIN
Thank you I mean range sir.
Oliver
proof for set theory
don't you know?
Inkoom
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
hi
Raihany
Hi
Solomon
need help
Raihany
maybe provide us videos
Nasasira
Raihany
Hello
Cromwell
a
Amie
What do you mean by a
Cromwell
nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
Denisse
b=p-4a-2c
Suddhen
b= p - 4a - 2c
Snr
p=2(2a+C)+b
Suraj
b=p-2(2a+c)
Tapiwa
P=4a+b+2C
COLEMAN
b=P-4a-2c
COLEMAN
like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has |
# How big is an acre?
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How big is an acre? 43560 square feet is an acre.
Related articles about How big is an acre
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How many inches in a yard or mile
Let us examine the acre which is a popular term for calculating the size of lands. It could be that you heard someone is selling a specific number of acres of land and you are wondering what is the actual size. Is it big or is it small? It could be that your math papers require you to find out acres to all the units possible like square meters and square feet.
You should keep in mind when dealing with acres that it can be in any shape like circles, square or rectangle. The rule is that it has to be 43 560 square feet no matter what its shapes. A regular shape of an acre is 660 feet by 66 feet. It is a usually misinterpreted information when looking for plots for a new home is that you should have minimum 1 acre of land in order to start construction.
Let us find by calculating:
660 feet X 66 feet = 42 560 square feet
Two feet make square feet.
If you are thinking of converting into other units, then you can first convert into another unit popular for measuring land which is the yard. It is 4840 square yard. There are two basic ways. One of them is through calculation and other is using formulae.
The first method
There are 0.00020661 acres in a square yard which means you have to multiply this with 4840 to get the result.
The result will be 0.9999924 which is approximately equal to 1.
0.00020661 X 4840 = 0.9999924 (approx. 1)
The second method
acres = feet square ÷ 4,840
The formula to convert from acres to ft. square is:
Feet square = acres x 4,840
If you are calculating in miles, then you should know that 1 acre is 0.0015625 square mile. 1 square mile is equal to 540 acres. To find out multiply by the factor or use the line “1 Square mile = 640 acre” Square mile is another popular unit used in both the British and American systems used to measure land. It goes like this, 1 square mile is 640 acres and 2 square miles is 1280 acre, 3 square miles is 1920 acre and so on.
To be more precise, 1 acre is 90 percent of a hundred yard or 91.44 meters long by 53.33 yards (48.76 meters) wide American field for football excluding the end zones. If you take into account the end zones, it will be about 1.32 acres.
1 Acre = 43,560 square feet
1 Acre = 160 square rods
1 Acre = 10 square chains
1 Acre = 160 square rods
1 Acre = 160 perches 1 Acre = 160 poles
1 Acre = .4047 hectare
1 Acre = 4047 square meters
1 Acre = is about 208 3/4 feet square
If you can remember them, you will be able to calculate the size of an acre using any formula.
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# Difference between revisions of "2012 AMC 10B Problems/Problem 13"
## Problem
It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$
## Solution 1
Let $s$ be the speed of the escalator and $c$ be the speed of Clea. Using $d = v t$, the first statement can be translated to the equation $d = 60c$. The second statement can be translated to $d = 24(c+s)$. Since the same distance is being covered in each scenario, we can set the two equations equal and solve for $s$. We find that $s = \dfrac{3c}{2}$. The problem asks for the time it takes her to ride down the escalator when she just stands on it. Since $t = \dfrac{d}{s}$ and $d = 60c$, we have $t = \dfrac{60c}{\dfrac{3c}{2}} = 40$ seconds. Answer choice $\boxed{B}$ is correct.
## Solution 2
Let $s$ be the speed of the escalator and $c$ be the speed of Clea. Then without loss of generality, assume that the length of the escalator be 1. Then $c=\dfrac{1}{60}$ and $c+s=\dfrac{1}{24}$, so $s=\dfrac{1}{24}-\dfrac{1}{60}=\dfrac{1}{40}$. Thus the time it takes for Clea to ride down the operating escalator when she just stands on it is $\dfrac{1}{\dfrac{1}{40}}=\boxed{\textbf{(B)}\ 40}$.
## See Also.
2012 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 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 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# 11.8: Sums of Integers with Different Signs
Difficulty Level: At Grade Created by: CK-12
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Callie continued to work on her sweater throughout the evening. For some reason, once she changed the thread, the stitches did look even.
"What is wrong with this?" She commented in frustration.
Callie's Mom came over to look at her work.
"What's the matter?" her Mom asked.
"These stitches don't look right. I made 14 and then took out 4 before adding 8 more and it still looks funny," Callie said showing her Mom the sweater.
"You have the wrong size needle. If you change to a smaller needle, the stitches will look a lot better," her Mom instructed.
Callie was relieved and went to the sewing box to get a new needle.
Callie's stitches present a situation where you could demonstrate adding positive and negative numbers.
Do you know how to write this expression? How many stitches did Callie sew?
This Concept is all about adding integers with different signs. By the end of the Concept, you will know how to answer these two questions.
### Guidance
You now know how to add integers with the same sign. What about when two integers have different signs? We started working on this when we used a number line for adding earlier in this Concept. But if you don’t have a number line, you can still add integers with different sign. Let’s learn how to do this now.
How do we add integers with different signs?
We can add integers with different signs by ignoring the sign and by finding the difference between the two values. Then the sign of the greater loss or gain becomes the sign of the answer.
8 + -3 = ____
First, ignore the sign and find the difference between 8 and 3. Difference means to subtract. We subtract 8 and 3 and get 5.
8 - 3 = 5
Next, think about losses and gains. The gain is greater than the loss. So our sign is positive.
8 + -3 = 5
We can check our work by using a number line.
We add a negative three to that, so we move three units towards the negative side of the number line.
Let's try another one.
-9 + 4 = ____
First, ignore the signs and find the difference between the two values.
9 - 4 = 5
Next, think about losses and gains. Here the loss is negative nine. That is a big loss. The loss is greater than the gain. A loss of nine is greater than a gain of four, so our sign in negative.
-9 + 4 = -5
Now it’s time for you to try a few on your own. Figure out each sum.
7 + -13 = ____
Solution: -6
-22 + 10 = ____
Solution: -12
#### Example C
-1 + 16 = ____
Solution: 15
Now back to Callie and the sweater. Here is the original problem once again.
Callie continued to work on her sweater throughout the evening. For some reason, once she changed the thread, the stitches did look even.
"What is wrong with this?" She commented in frustration.
Callie's Mom came over to look at her work.
"What's the matter?" her Mom asked.
"These stitches don't look right. I made 14 and then took out 4 before adding 8 more and it still looks funny," Callie said showing her Mom the sweater.
"You have the wrong size needle. If you change to a smaller needle, the stitches will look a lot better," her Mom instructed.
Callie was relieved and went to the sewing box to get a new needle.
Callie's stitches present a situation where you could demonstrate adding positive and negative numbers.
Do you know how to write this expression? How many stitches did Callie sew?
To write this expression, we begin by writing the stitches. A positive value is when Callie added a stitch and a negative value is when she took one out.
Next, we add them together. We add the first two values and then add the third to that sum.
Callie added 18 stitches in all.
### Vocabulary
Here are the vocabulary words in this Concept.
Integer
the set of whole numbers and their opposites. Positive and negative whole numbers are integers.
Absolute Value
the number of units that an integer is from zero. The sign does not make a difference.
### Guided Practice
-2 + 8 = _____
First, ignore the signs and find the difference between the two values.
8 - 2 = 6
Next, think about losses and gains. This problem starts with a loss of 2, that’s the negative two, and then there is a gain of 8. That is a sum of positive 6. Since the gain is greater than the loss, the answer is positive.
### Video Review
Here are videos for review.
### Practice
Directions: Add the following pairs of integers.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Absolute Value The absolute value of a number is the distance the number is from zero. Absolute values are never negative.
Integer The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
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# Integral Of Sin Square 2x
Integral of sin square 2x along with its formula and proof with examples. Also learn how to calculate integration of sin^2(2x) with step by step examples.
Alan Walker-
Published on 2023-04-13
## Introduction to integral of sin^2(2x)
In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration calculates the integrals. This process is defined as finding an antiderivative of a function.
Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to a trigonometric function sine squared. You will also understand how to compute sin^(2x) integral by using different integration techniques.
## What is the integral of sin2(2x)?
The integral of sin2(2x) is an antiderivative of sine function which is equal to x/2–sin4x/8. It is also known as the reverse derivative of sine function which is a trigonometric identity.
The sine function is the ratio of opposite side to the hypotenuse of a triangle which is written as:
Sin = opposite side / hypotenuse
### Integral of sin^2(2x) formula
The formula of integral of sin contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(sin2 (2x))dx. In mathematical form, the integral of sin^2(2x) is:
$∫\sin^2(2x)dx=\frac{x}{2}–\frac{\sin4x}{8}+ c$
Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. The above formula is also used to calculate the integral of sin cube x.
## How to calculate the integral of sin^2(2x)?
The integral of sin^2(2x) is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of sine by using:
1. Integration by parts
2. Substitution method
3. Definite integral
## Integral of sin 2x squared by using integration by parts
The derivative of a function calculates the rate of change, and integration is the process of finding the antiderivative of a function. The integration by parts is a method of solving integral of two functions combined together. Let’s discuss calculating the integral of sin squared x by using by parts integral calculator.
### Proof of integral of sin^2(2x) by using integration by parts
Since we know that the function sine squared x can be written as the product of two functions. Therefore, we can calculate the integral of sin^2(2x) by using integration by parts. For this, suppose that:
$I = \sin 2x.\sin 2x$
Applying the integral we get,
$I=∫\sin 2x.\sin 2xdx$
Since the method of integration by parts is:
$∫[f(x).g(x)]dx = f(x).∫g(x)dx - ∫[f’(x).∫g(x)dx]dx$
Now replacing f(x) and g(x) by sin x, we get,
$I =-\sin 2x\frac{\cos 2x}{2}+∫[\cos 2x.\cos 2x]dx$
To eveluate the integral of sin^4x, f(x) and g(x) can be replace by sin^2x in the above formula.
It can be written as:
$I =-\sin(2x)\frac{\cos(2x)}{2}+∫[\cos^2(2x)]dx$
Now by using a trigonometric identity cos22x = 1+cos4x/2. Therefore, substituting the value of cos2(2x) in the above equation, we get:
$I=-\sin (2x)\frac{\cos (2x)}{2}+∫\left(\frac{1+\cos4x}{2}\right)dx$
Integrating remaining terms,
$I=-\sin(2x)\frac{\cos(2x)}{2}+\frac{x}{2}+\frac{\sin4x}{8}$
Or,
$I=-\frac{\sin4x}{4}+\frac{x}{2}+\frac{\sin4x}{8}$
Or,
$I=\frac{x}{2}–\frac{\sin4x}{8}$
Hence the integral of sin^2(2x) is equal to,
$∫\sin^2(2x)dx=\frac{x}{2}–\frac{\sin4x}{8}$
## Integral of sin^2(2x) by using substitution method
The substitution method calculator involves many trigonometric formulas. We can use these formulas to verify the integrals of different trigonometric functions such as sine, cosine, tangent, etc. Let’s understand how to prove the integral of sin squared by using the substitution method.
### Proof of Integral of sin^2(2x) by using substitution method
To proof the integral of sin2(2x) by using substitution method, suppose that:
$I=∫\sin^2(2x)=∫(1- \cos^2(2x))dx$
Further we can cos2(2x) can be substituted as cos2(2x) = 1+cos4x/2. Then the above equation will become.
$I=x-∫\left(\frac{1+ \cos4x}{2}\right)dx$
Integrating,
$I=x–\frac{x}{2}–\frac{\sin4x}{8}$
Moreover,
$I=\frac{x}{2}–\frac{\sin4x}{8}$
Hence the integration of sin^2(2x) is verified by using substitution method.
## Integral of sin^2(2x) by using definite integral
The definite integral is a type of integral that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral can be written as:
$∫^b_a f(x) dx = F(b) – F(a)$
Let’s understand the verification of the integral of sin^2(2x) by using the definite integral.
### Proof of integral of sin^2(2x) by using definite integral
To compute the integral of sin^2(2x) by using a definite integral, we can use the interval from 0 to 2π or 0 to π. Let’s compute the integral of sin^2(2x) from 0 to 2π.
The definite integral of sin^2(2x) can be written as:
$∫^{2π}_0 \sin^2(2x)dx=\left|\frac{x}{2}-\frac{\sin4x}{8}\right|^{2π}_0$
Substituting the value of limit we get,
$∫^{2π}_0 \sin^2(2x)dx=\left[\frac{2π}{2}-\frac{\sin8π}{8}\right]-\left[0 - \frac{\sin 0}{8}\right]$
$∫^{2π}_0 \sin^2(2x)dx= π – frac{0}{8}$
Therefore, the integral of sin2(2x) from 0 to 2π is
$∫^{2π}_0 \sin^2(2x)dx = π$
Which is the calculation of the definite integral of sin^2(2x). Now to calculate the integral of sinx between the interval 0 to π, we just have to replace 2π by π. Therefore,
$∫^π_0 \sin^2(2x)dx = \left|\frac{x}{2} -\frac{\sin4x}{8}\right|^π_0$
$∫^π_0 \sin^2(2x)dx=\left[\frac{π}{2}-\frac{\sin4π}{8}\right] -\left[0 -\frac{\sin 0}{8}\right]$
$∫^π_0 \sin^2(2x)dx=\frac{π}{2}-\frac{0}{8}$
$∫^π_0 \sin^2(2x)dx=\frac{π}{2}$
Therefore, the integral of sin2 2x from 0 to π is π/2. |
## Elementary Technical Mathematics
Solve each equation for y, then find and graph 3 points for each equation. $x-y=2$ $x-y+y-2=2-y-2$ $x-2=y$ $\underline{\ x\ \ \ \ \ \ \ \ \ \ \ x-2\ \ \ \ \ \ \ \ \ \ \ \ \ y\ \ }$ $-1\ \ \ \ \ \ \ -1-2\ \ \ \ \ \ -3$ $\ \ \ 0\ \ \ \ \ \ \ \ \ 0-2\ \ \ \ \ \ \ \ \ -2$ $\ \ \ 5\ \ \ \ \ \ \ \ \ 5-2\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3$ $x+3y=6$ $x+3y-x=6-x$ $3y\div3=(6-x)\div3$ $y=2-\frac{x}{3}$ $\underline{\ x\ \ \ \ \ \ \ \ \ \ \ \ 2-\frac{x}{3}\ \ \ \ \ \ \ \ \ \ y}$ $3\ \ \ \ \ \ \ \ \ \ \ \ \ 2-1\ \ \ \ \ \ \ \ \ \ \ 1$ $0\ \ \ \ \ \ \ \ \ \ \ \ \ 2-0\ \ \ \ \ \ \ \ \ \ \ 2$ $-6\ \ \ \ \ \ \ \ \ 2-(-2)\ \ \,\ \ \ 4$ The lines intersect where x=3 and y=1. When $x=3, x-2=3-2=1.$ When $x=3, 2-\frac{x}{3}=2-1-1$. |
Lesson Objectives
• Demonstrate an understanding of how to translate phrases into algebraic expressions and equations
• Learn the six-step method used for solving applications of linear equations
• Learn how to solve word problems that involve "sums of unknown quantities"
• Learn how to solve word problems that involve the simple interest formula "I = prt"
• Learn how to solve word problems that involve percentages
## How to Solve Word Problems with Linear Equations
In our Algebra 1 course, we learned the six-step method for solving a word problem that involves a linear equation in one variable. Let's quickly review this procedure and then jump into some sample word problems.
### Six-step method for Solving Word Problems with Linear Equations in One Variable
1. Read the problem and determine what you are asked to find
2. Assign a variable to represent the unknown
• If more than one unknown exists, we express the other unknowns in terms of this variable
3. Write out an equation which describes the given situation
4. Solve the equation
5. State the answer using a nice clear sentence
6. Check the result
• We need to make sure the answer is reasonable. In other words, if asked how many students were on a bus, the answer shouldn't be (-4) as we can't have a negative amount of students on a bus.
Fortunately, pretty much all common Algebra word problems can be broken down into a "type" of problem. The first type of word problem that we will look at involves finding the individual amounts when the sums are known. This problem type is known as "sums of unknown quantities" or "finding unknown quantities".
### Finding Unknown Numerical Quantities - Word Problem
Let's look at an example.
Example 1: Solve each word problem
Jay’s Place is a local restaurant that has only three items on its menu. Their menu consists of lasagna, tacos with chips, or steak and rice. One Sunday, the restaurant sold a total of 80 plates from the menu. They sold twice as many plates of tacos with chips as plates of lasagna. Additionally, the sales of plates with Steak and rice were only one-fifth of the sales of plates of lasagna.
Step 1) After reading the problem, it is clear we need to find the number of plates sold for each menu item: lasagna, tacos with chips, and steak with rice.
Step 2) Here we have three unknowns. When this happens, look for the unknown that is involved in both comparisons. This will make our problem much easier to solve.
Since lasagna is involved in each comparison, let's let x be equal to the number of plates of lasagna sold.
x = plates of lasagna
We are told that the restaurant sold twice the number of plates of tacos with chips as plates of lasagna. Since x has been assigned to represent the number of plates of lasagna sold, we can think of twice this number as 2x. This leads us to the following:
then: 2x = plates of tacos with chips
Lastly, we are told that the sales for plates of steak with rice were one-fifth the number of sales of plates of lasagna. Again, since x has been assigned to represent the number of plates of lasagna sold, we can think of one-fifth this number as (1/5)x. This leads us to the following:
then: (1/5)x = plates of steak with rice
Step 3) To write an equation, think about what we know. So far, we have represented how much of each menu item was sold using the variable x. Additionally, we know the total number of plates sold for the day (80). This means we can sum the individual amounts and set this equal to 80 (total plates sold for the day). $$\frac{1}{5}x + x + 2x = 80$$ Step 4) Solve the equation: $$\frac{1}{5}x + x + 2x = 80$$ Multiply both sides by 5, this will clear the fraction: $$x + 5x + 10x = 400$$ $$16x = 400$$ $$x = 25$$ Step 5) Since x represented the number of plates of lasagna sold, we know the following:
25 plates of lasagna
50 plates of tacos with chips (2 x 25 = 50)
5 plates of steak with rice (1/5 x 25 = 5)
Jay's Place sold 25 plates of lasagna, 50 plates of tacos with chips, and 5 plates of steak with rice.
Step 6) We can read back through the problem to check our answer. First, we can check the total number of plates sold:
25 + 50 + 5 = 80
80 = 80
Additionally, we can check that the number of plates sold of tacos with chips are twice that of plates of lasagna:
50 = 2 • 25
50 = 50
Lastly, we can check that the number of plates sold of steak with rice are one-fifth that of plates of lasagna:
25 • 1/5 = 5
5 = 5
### Solving Simple Interest Word Problems » I = prt
When we work with investment problems in Algebra, we will encounter two types of interest: simple interest and compound interest. The main difference between the two comes from the fact that compound interest will earn interest on interest. This means as the account balance grows, interest is paid on the current account balance, not the original amount invested (known as the principal). When we work with simple interest word problems, interest is only earned on the principal or amount initially invested. In other words, when we have simple interest, we do not earn interest on interest. Although this makes our calculation much simpler, it is not often used in real life.
With simple interest word problems, we will use the simple interest formula:
I = prt
I » simple interest earned
p » principal or amount invested
r » rate (interest rate as a decimal)
t » time (normally in years, but could be months, days, or whatever given time frame)
Let's look at an example.
Example 2: Solve each word problem
After speaking with a retirement professional, James decides to invest in two different accounts. A bond fund, which pays 5% annual simple interest and a savings account which pays 2% annual simple interest. James first decides to invest $29,000 into the bond fund. If his goal is to make 3% annual simple interest, how much additional money should he invest in the savings account? Step 1) After reading the problem, it is clear that we need to find the amount that James needs to invest in the savings account in order to have a combined interest rate of 3%. Step 2) We have one unknown, the amount that is to be invested in the savings account. let x = amount of money in dollars that will be invested in the savings account Step 3) To write an equation, let's think about what we know. James wants an overall interest rate of 3% for the two investments. Currently, we know James will invest$29,000 in the bond fund and earn 5% annual simple interest. If we plug this into the simple interest formula:
I = 29,000(.05)(1)
I = 1450
This amount will be added to the 2% interest obtained from investing an unknown amount (x) in a savings account:
Using our simple interest formula for the savings account yields:
I = x (.02)(1)
I = .02x
If we sum the two, we think about the simple interest earned from the two investments:
I = 1450 + .02x
Since we want I or the simple interest earned to be 3% of the principal invested, we can set up the following equation:
.03(x + 29,000) = .02x + 1450
In other words, 3% multiplied by the principal invested (x + 29,000) will be equal to the interest earned from the savings account (.02x) and the interest earned from the bond fund (1450).
Step 4) Solve the equation:
.03(x + 29,000) = .02x + 1450
.03x + 870 = .02x + 1450
We can clear the decimals by multiplying both side of the equation by 100:
3x + 87,000 = 2x + 145,000
3x - 2x = 145,000 - 87,000
x = 58,000
Step 5) Since x represented the amount that needed to be invested in the savings account, we can state our answer as:
James needs to invest $58,000 dollars in the savings account in order to achieve a simple interest rate of 3% from the two investments. Step 6) We can read back through the problem to check our answer. Think about the simple interest earned from his two investments: I = .02(58,000) + .05(29,000) I = 1160 + 1450 I = 2610 The simple interest earned from the two investments is 2610. If we solve the simple interest formula for r (rate): $$I = prt$$ $$r = \frac{I}{pt}$$ Since time is 1, we can rewrite our equation as: $$r = \frac{I}{p}$$ Since his goal is 3% interest, let's plug everything in and check: $$.03 = \frac{2610}{29,\hspace{-.1em}000 + 58,\hspace{-.1em}000}$$ $$.03 = \frac{2610}{87,\hspace{-.1em}000}$$ .03 = .03 ### Solving Percent Word Problems Another common type of word problem involves percentages. These usually deal with percent increase or percent decrease. Let's look at an example. Example 3: Solve each word problem Heather works as a cashier for a local grocery store. After one busy weekend shift, she had a total of$2725 in total receipts. This amount included the 9% state and local sales taxes. What was the amount of the tax collected?
Step 1) After reading the problem, it is clear that we need to find the amount of taxes that were collected.
Step 2) We have two unknowns, the pre-tax amount of goods and services and the amount of tax collected.
let x = amount of pre-tax goods and services sold
then .09x = amount of tax collected
Step 3) Write an equation, let's think about what we know.
If we were to sum the amount of pre-tax goods and services sold (x) with the amount of the tax collected (.09x), we would get 2725 (total receipts):
x + .09x = 2725
Step 4) Solve the equation:
x + .09x = 2725
We can multiply both sides by 100 and clear the decimal: 100x + 9x = 272,500
109x = 272,500
x = 2500
Step 5) Since x (2500) represents the amount of pre-tax goods and services sold, we have to subtract this away from the total receipts (2725) to get the amount of the tax collected:
2725 - 2500 = 225
We can state our answer as:
Heather collected \$225 in tax during her shift.
Step 6) We can read back through the problem to check our answer. We want to check that the amount of pre-tax goods and services (2500) plus the amount of the tax collected (225) is equal to the total receipts (2725):
2500 + 225 = 2725
2725 = 2725 |
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# Perfect Cube
A perfect cube is a number that can be expressed as the product of an integer multiplied by itself three times. This concept establishes a strong foundation for understanding higher mathematical principles and finds practical applications in various fields.
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## What are Perfect Cubes?
A perfect cube is a number that can be obtained by multiplying an integer (a whole number) by itself three times. In mathematical notation, a perfect cube can be represented as n3 where n is an integer. For example, 23 is a perfect cube because 2×2×2 equals 8. Similarly, 33 is also a perfect cube, yielding the value 27 (3×3×3).
We can find cubes of any number, whether integer or fraction. For example, the cube of 0.6 is:
0.6 × 0.6 × 0.6 = 0.216
How to Find the Perfect Cube?
How to determine if a number is a perfect cube by using prime factorization:
Step 1: Begin by finding the prime factorization of the given number, starting from the smallest prime number (2).
Step 2: Group the prime factors in sets of three, as each set of three factors will represent one cube.
Step 3: Repeat the step for all the group sets of the same three factors.
If any prime factors are left ungrouped, the number is not a perfect cube. The number is a perfect cube if all factors are grouped in three sets.
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Example 1
To prime factorize 216, we need to find its prime factors. Let’s go through the process:
Step 1: Divide by the smallest prime number, which is 2.
216=2×108
Step 2: Divide 108 by 2 again:
108=2×54
Step 3: Divide 54 by 2 once more:
54=2×27
Step 4: Now, let’s divide by the next prime number, which is 3:
27=3×9
Step 5: Finally, divide 9 by 3:
9=3×3
At this point, we’ve reached a prime factorization where all the factors are prime numbers:
216=23×33
So, the prime factorization of 216 is 23×33, which means 216 can be expressed as the product of three 2s and three 3s.
Therefore, 216 is a perfect cube.
Example 2
To prime factorize 200, we need to find its prime factors. Let’s go through the process:
Step 1: Divide by the smallest prime number, which is 2.
200=2×100
Step 2: Divide 100 by 2 again:
100=2×50
Step 3: Divide 50 by 2 once more:
50=2×25
Step 4: Now, let’s divide by another prime number, which is 5:
25=5×5
Step 5: Finally, divide 5 by 5:
5=5×1
The number 200 on prime factorization gives 2 × 2 × 2 × 5 × 5. Here, the prime factor 5 is not in the power of 3. Therefore, 200 is not a perfect cube.
## FAQs
#### What is a perfect cube?
A perfect cube is a number that can be expressed as the product of an integer multiplied by itself three times. In mathematical notation, it is represented as n3, where n is a whole number.
#### How do I determine if a number is a perfect cube?
You can use methods like prime factorization or checking the sum of digits to determine if a number is a perfect cube. If you can find an integer whose cube equals the given number, then the number is a perfect cube.
#### Can a negative number be a perfect cube?
Yes, negative numbers can be perfect cubes. For instance, -27 is a perfect cube because (−3)3=−27.
#### What is the connection between perfect cubes and volume?
The volume of a cube with side length n is n3, the same as the cube of n. So, perfect cubes are related to the concept of three-dimensional volume.
#### How are perfect cubes used in real life?
Perfect cubes have engineering, architecture, and computer graphics applications. They help calculate volumes, surface areas, and measurements in three-dimensional objects.
Gloria Mathew writes on math topics for K-12. A trained writer and communicator, she makes math accessible and understandable to students at all levels. Her ability to explain complex math concepts with easy to understand examples helps students master math. LinkedIn
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## Section3.1Using derivatives to identify extreme values
###### Motivating Questions
• What are the critical numbers of a function $f$ and how are they connected to identifying the most extreme values the function achieves?
• How does the first derivative of a function reveal important information about the behavior of the function, including the function's extreme values?
• How can the second derivative of a function be used to help identify extreme values of the function?
In many different settings, we are interested in knowing where a function achieves its least and greatest values. These can be important in applications — say to identify a point at which maximum profit or minimum cost occurs — or in theory to understand how to characterize the behavior of a function or a family of related functions. Consider the simple and familiar example of a parabolic function such as $s(t) = -16t^2 + 32t + 48$ (shown at left in Figure 3.1.1) that represents the height of an object tossed vertically: its maximum value occurs at the vertex of the parabola and represents the highest value that the object reaches. Moreover, this maximum value identifies an especially important point on the graph, the point at which the curve changes from increasing to decreasing.
More generally, for any function we consider, we can investigate where its lowest and highest points occur in comparison to points nearby or to all possible points on the graph. Given a function $f\text{,}$ we say that $f(c)$ is a global or absolute maximum provided that $f(c) \ge f(x)$ for all $x$ in the domain of $f\text{,}$ and similarly call $f(c)$ a global or absolute minimum whenever $f(c) \le f(x)$ for all $x$ in the domain of $f\text{.}$ For instance, for the function $g$ given at right in Figure 3.1.1, $g$ has a global maximum of $g(c)\text{,}$ but $g$ does not appear to have a global minimum, as the graph of $g$ seems to decrease without bound. We note that the point $(c,g(c))$ marks a fundamental change in the behavior of $g\text{,}$ where $g$ changes from increasing to decreasing; similar things happen at both $(a,g(a))$ and $(b,g(b))\text{,}$ although these points are not global mins or maxes.
For any function $f\text{,}$ we say that $f(c)$ is a local maximum or relative maximum provided that $f(c) \ge f(x)$ for all $x$ near $c\text{,}$ while $f(c)$ is called a local or relative minimum whenever $f(c) \le f(x)$ for all $x$ near $c\text{.}$ Any maximum or minimum may be called an extreme value of $f\text{.}$ For example, in Figure 3.1.1, $g$ has a relative minimum of $g(b)$ at the point $(b,g(b))$ and a relative maximum of $g(a)$ at $(a,g(a))\text{.}$ We have already identified the global maximum of $g$ as $g(c)\text{;}$ this global maximum can also be considered a relative maximum.
We would like to use fundamental calculus ideas to help us identify and classify key function behavior, including the location of relative extremes. Of course, if we are given a graph of a function, it is often straightforward to locate these important behaviors visually. We investigate this situation in the following preview activity.
###### Preview Activity3.1.1
Consider the function $h$ given by the graph in Figure 3.1.2. Use the graph to answer each of the following questions.
1. Identify all of the values of $c$ for which $h(c)$ is a local maximum of $h\text{.}$
2. Identify all of the values of $c$ for which $h(c)$ is a local minimum of $h\text{.}$
3. Does $h$ have a global maximum on the interval $[-3,3]\text{?}$ If so, what is the value of this global maximum?
4. Does $h$ have a global minimum on the interval $[-3,3]\text{?}$ If so, what is its value?
5. Identify all values of $c$ for which $h'(c) = 0\text{.}$
6. Identify all values of $c$ for which $h'(c)$ does not exist.
7. True or false: every relative maximum and minimum of $h$ occurs at a point where $h'(c)$ is either zero or does not exist.
8. True or false: at every point where $h'(c)$ is zero or does not exist, $h$ has a relative maximum or minimum.
### Subsection3.1.1Critical numbers and the first derivative test
If a function has a relative extreme value at a point $(c,f(c))\text{,}$ the function must change its behavior at $c$ regarding whether it is increasing or decreasing before or after the point.
For example, if a continuous function has a relative maximum at $c\text{,}$ such as those pictured in the two leftmost functions in Figure 3.1.3, then it is both necessary and sufficient that the function change from being increasing just before $c$ to decreasing just after $c\text{.}$ In the same way, a continuous function has a relative minimum at $c$ if and only if the function changes from decreasing to increasing at $c\text{.}$ See, for instance, the two functions pictured at right in Figure 3.1.3. There are only two possible ways for these changes in behavior to occur: either $f'(c) = 0$ or $f'(c)$ is undefined.
Because these values of $c$ are so important, we call them critical numbers. More specifically, we say that a function $f$ has a critical number at $x = c$ provided that $c$ is in the domain of $f\text{,}$ and $f'(c) = 0$ or $f'(c)$ is undefined. Critical numbers provide us with the only possible locations where the function $f$ may have relative extremes. Note that not every critical number produces a maximum or minimum; in the middle graph of Figure 3.1.3, the function pictured there has a horizontal tangent line at the noted point, but the function is increasing before and increasing after, so the critical number does not yield a location where the function is greater than every value nearby, nor less than every value nearby.
We also sometimes use the terminology that, when $c$ is a critical number, that $(c,f(c))$ is a critical point . of the function, or that $f(c)$ is a critical value
The first derivative test summarizes how sign changes in the first derivative indicate the presence of a local maximum or minimum for a given function.
###### First Derivative Test
If $p$ is a critical number of a continuous function $f$ that is differentiable near $p$ (except possibly at $x = p$), then $f$ has a relative maximum at $p$ if and only if $f'$ changes sign from positive to negative at $p\text{,}$ and $f$ has a relative minimum at $p$ if and only if $f'$ changes sign from negative to positive at $p\text{.}$
We consider an example to show one way the first derivative test can be used to identify the relative extreme values of a function.
###### Example3.1.4
Let $f$ be a function whose derivative is given by the formula $f'(x) = e^{-2x}(3-x)(x+1)^2\text{.}$ Determine all critical numbers of $f$ and decide whether a relative maximum, relative minimum, or neither occurs at each.
Solution
Since we already have $f'(x)$ written in factored form, it is straightforward to find the critical numbers of $f\text{.}$ Since $f'(x)$ is defined for all values of $x\text{,}$ we need only determine where $f'(x) = 0\text{.}$ From the equation
\begin{equation*} e^{-2x}(3-x)(x+1)^2 = 0 \end{equation*}
and the zero product property, it follows that $x = 3$ and $x = -1$ are critical numbers of $f\text{.}$ (Note particularly that there is no value of $x$ that makes $e^{-2x} = 0\text{.}$)
Next, to apply the first derivative test, we'd like to know the sign of $f'(x)$ at inputs near the critical numbers. Because the critical numbers are the only locations at which $f'$ can change sign, it follows that the sign of the derivative is the same on each of the intervals created by the critical numbers: for instance, the sign of $f'$ must be the same for every $x \lt -1\text{.}$ We create a first derivative sign chart to summarize the sign of $f'$ on the relevant intervals along with the corresponding behavior of $f\text{.}$
The first derivative sign chart in Figure 3.1.5 comes from thinking about the sign of each of the terms in the factored form of $f'(x)$ at one selected point in the interval under consideration. For instance, for $x \lt -1\text{,}$ we could consider $x = -2$ and determine the sign of $e^{-2x}\text{,}$ $(3-x)\text{,}$ and $(x+1)^2$ at the value $x = -2\text{.}$ We note that both $e^{-2x}$ and $(x+1)^2$ are positive regardless of the value of $x\text{,}$ while $(3-x)$ is also positive at $x = -2\text{.}$ Hence, each of the three terms in $f'$ is positive, which we indicate by writing “$+++\text{.}$” Taking the product of three positive terms obviously results in a value that is positive, which we denote by the “$+$” in the interval to the left of $x = -1$ indicating the overall sign of $f'\text{.}$ And, since $f'$ is positive on that interval, we further know that $f$ is increasing, which we summarize by writing “INC” to represent the corresponding behavior of $f\text{.}$ In a similar way, we find that $f'$ is positive and $f$ is increasing on $-1 \lt x \lt 3\text{,}$ and $f'$ is negative and $f$ is decreasing for $x \gt 3\text{.}$
Now, by the first derivative test, to find relative extremes of $f$ we look for critical numbers at which $f'$ changes sign. In this example, $f'$ only changes sign at $x = 3\text{,}$ where $f'$ changes from positive to negative, and thus $f$ has a relative maximum at $x = 3\text{.}$ While $f$ has a critical number at $x = -1\text{,}$ since $f$ is increasing both before and after $x = -1\text{,}$ $f$ has neither a minimum nor a maximum at $x = -1\text{.}$
###### Activity3.1.2
Suppose that $g(x)$ is a function continuous for every value of $x \ne 2$ whose first derivative is $g'(x) = \frac{(x+4)(x-1)^2}{x-2}\text{.}$ Further, assume that it is known that $g$ has a vertical asymptote at $x = 2\text{.}$
1. Determine all critical numbers of $g\text{.}$
2. By developing a carefully labeled first derivative sign chart, decide whether $g$ has as a local maximum, local minimum, or neither at each critical number.
3. Does $g$ have a global maximum? global minimum? Justify your claims.
4. What is the value of $\lim_{x \to \infty} g'(x)\text{?}$ What does the value of this limit tell you about the long-term behavior of $g\text{?}$
5. Sketch a possible graph of $y = g(x)\text{.}$
### Subsection3.1.2The second derivative test
Recall that the second derivative of a function tells us several important things about the behavior of the function itself. For instance, if $f''$ is positive on an interval, then we know that $f'$ is increasing on that interval and, consequently, that $f$ is concave up, which also tells us that throughout the interval the tangent line to $y = f(x)$ lies below the curve at every point. In this situation where we know that $f'(p) = 0\text{,}$ it turns out that the sign of the second derivative determines whether $f$ has a local minimum or local maximum at the critical number $p\text{.}$
In Figure 3.1.6, we see the four possibilities for a function $f$ that has a critical number $p$ at which $f'(p) = 0\text{,}$ provided $f''(p)$ is not zero on an interval including $p$ (except possibly at $p$). On either side of the critical number, $f''$ can be either positive or negative, and hence $f$ can be either concave up or concave down. In the first two graphs, $f$ does not change concavity at $p\text{,}$ and in those situations, $f$ has either a local minimum or local maximum. In particular, if $f'(p) = 0$ and $f''(p) \lt 0\text{,}$ then we know $f$ is concave down at $p$ with a horizontal tangent line, and this guarantees $f$ has a local maximum there. This fact, along with the corresponding statement for when $f''(p)$ is positive, is stated in the second derivative test.
###### Second Derivative Test
If $p$ is a critical number of a continuous function $f$ such that $f'(p) = 0$ and $f''(p) \ne 0\text{,}$ then $f$ has a relative maximum at $p$ if and only if $f''(p) \lt 0\text{,}$ and $f$ has a relative minimum at $p$ if and only if $f''(p) \gt 0\text{.}$
In the event that $f''(p) = 0\text{,}$ the second derivative test is inconclusive. That is, the test doesn't provide us any information. This is because if $f''(p) = 0\text{,}$ it is possible that $f$ has a local minimum, local maximum, or neither. 1 Consider the functions $f(x) = x^4\text{,}$ $g(x) = -x^4\text{,}$ and $h(x) = x^3$ at the critical point $p = 0\text{.}$
Just as a first derivative sign chart reveals all of the increasing and decreasing behavior of a function, we can construct a second derivative sign chart that demonstrates all of the important information involving concavity.
###### Example3.1.7
Let $f(x)$ be a function whose first derivative is $f'(x) = 3x^4 - 9x^2\text{.}$ Construct both first and second derivative sign charts for $f\text{,}$ fully discuss where $f$ is increasing and decreasing and concave up and concave down, identify all relative extreme values, and sketch a possible graph of $f\text{.}$
Solution
Since we know $f'(x) = 3x^4 - 9x^2\text{,}$ we can find the critical numbers of $f$ by solving $3x^4 - 9x^2 = 0\text{.}$ Factoring, we observe that
\begin{equation*} 0 = 3x^2(x^2 - 3) = 3x^2(x+\sqrt{3})(x-\sqrt{3})\text{,} \end{equation*}
so that $x = 0, \pm\sqrt{3}$ are the three critical numbers of $f\text{.}$ It then follows that the first derivative sign chart for $f$ is given in Figure 3.1.8.
Thus, $f$ is increasing on the intervals $(-\infty, -\sqrt{3})$ and $(\sqrt{3}, \infty)\text{,}$ while $f$ is decreasing on $(-\sqrt{3},0)$ and $(0, \sqrt{3})\text{.}$ Note particularly that by the first derivative test, this information tells us that $f$ has a local maximum at $x = -\sqrt{3}$ and a local minimum at $x = \sqrt{3}\text{.}$ While $f$ also has a critical number at $x = 0\text{,}$ neither a maximum nor minimum occurs there since $f'$ does not change sign at $x = 0\text{.}$
Next, we move on to investigate concavity. Differentiating $f'(x) = 3x^4 - 9x^2\text{,}$ we see that $f''(x) = 12x^3 - 18x\text{.}$ Since we are interested in knowing the intervals on which $f''$ is positive and negative, we first find where $f''(x) = 0\text{.}$ Observe that
\begin{equation*} 0 = 12x^3 - 18x = 12x\left(x^2 - \frac{3}{2}\right) = 12x\left(x+\sqrt{\frac{3}{2}}\right)\left(x-\sqrt{\frac{3}{2}}\right)\text{,} \end{equation*}
which implies that $x = 0, \pm\sqrt{\frac{3}{2}}\text{.}$ Building a sign chart for $f''$ in the exact same way we do for $f'\text{,}$ we see the result shown in Figure 3.1.9.
Therefore, $f$ is concave down on the intervals $(-\infty, -\sqrt{\frac{3}{2}})$ and $(0, \sqrt{\frac{3}{2}})\text{,}$ and concave up on $(-\sqrt{\frac{3}{2}},0)$ and $(\sqrt{\frac{3}{2}}, \infty)\text{.}$
Putting all of the above information together, we now see a complete and accurate possible graph of $f$ in Figure 3.1.10.
The point $A = (-\sqrt{3}, f(-\sqrt{3}))$ is a local maximum, as $f$ is increasing prior to $A$ and decreasing after; similarly, the point $E = (\sqrt{3}, f(\sqrt{3})$ is a local minimum. Note, too, that $f$ is concave down at $A$ and concave up at $B\text{,}$ which is consistent both with our second derivative sign chart and the second derivative test. At points $B$ and $D\text{,}$ concavity changes, as we saw in the results of the second derivative sign chart in Figure 3.1.9. Finally, at point $C\text{,}$ $f$ has a critical point with a horizontal tangent line, but neither a maximum nor a minimum occurs there since $f$ is decreasing both before and after $C\text{.}$ It is also the case that concavity changes at $C\text{.}$
While we completely understand where $f$ is increasing and decreasing, where $f$ is concave up and concave down, and where $f$ has relative extremes, we do not know any specific information about the $y$-coordinates of points on the curve. For instance, while we know that $f$ has a local maximum at $x = -\sqrt{3}\text{,}$ we don't know the value of that maximum because we do not know $f(-\sqrt{3})\text{.}$ Any vertical translation of our sketch of $f$ in Figure 3.1.10 would satisfy the given criteria for $f\text{.}$
Points $B\text{,}$ $C\text{,}$ and $D$ in Figure 3.1.10 are locations at which the concavity of $f$ changes. We give a special name to any such point: if $p$ is a value in the domain of a continuous function $f$ at which $f$ changes concavity, then we say that $(p,f(p))$ is an inflection point of $f\text{.}$ Just as we look for locations where $f$ changes from increasing to decreasing at points where $f'(p) = 0$ or $f'(p)$ is undefined, so too we find where $f''(p) = 0$ or $f''(p)$ is undefined to see if there are points of inflection at these locations.
It is important at this point in our study to remind ourselves of the big picture that derivatives help to paint: the sign of the first derivative $f'$ tells us whether the function $f$ is increasing or decreasing, while the sign of the second derivative $f''$ tells us how the function $f$ is increasing or decreasing.
###### Activity3.1.3
Suppose that $g$ is a function whose second derivative, $g''\text{,}$ is given by the following graph.
1. Find the $x$-coordinates of all points of inflection of $g\text{.}$
2. Fully describe the concavity of $g$ by making an appropriate sign chart.
3. Suppose you are given that $g'(-1.67857351) = 0\text{.}$ Is there is a local maximum, local minimum, or neither (for the function $g$) at this critical number of $g\text{,}$ or is it impossible to say? Why?
4. Assuming that $g''(x)$ is a polynomial (and that all important behavior of $g''$ is seen in the graph above), what degree polynomial do you think $g(x)$ is? Why?
As we will see in more detail in the following section, derivatives also help us to understand families of functions that differ only by changing one or more parameters. For instance, we might be interested in understanding the behavior of all functions of the form $f(x) = a(x-h)^2 + k$ where $a\text{,}$ $h\text{,}$ and $k$ are numbers that may vary. In the following activity, we investigate a particular example where the value of a single parameter has considerable impact on how the graph appears.
###### Activity3.1.4
Consider the family of functions given by $h(x) = x^2 + \cos(kx)\text{,}$ where $k$ is an arbitrary positive real number.
1. Use a graphing utility to sketch the graph of $h$ for several different $k$-values, including $k = 1,3,5,10\text{.}$ Plot $h(x) = x^2 + \cos(3x)$ on the axes provided. What is the smallest value of $k$ at which you think you can see (just by looking at the graph) at least one inflection point on the graph of $h\text{?}$
2. Explain why the graph of $h$ has no inflection points if $k \le \sqrt{2}\text{,}$ but infinitely many inflection points if $k \gt \sqrt{2}\text{.}$
3. Explain why, no matter the value of $k\text{,}$ $h$ can only have finitely many critical numbers.
### Subsection3.1.3Summary
• The critical numbers of a continuous function $f$ are the values of $p$ for which $f'(p) = 0$ or $f'(p)$ does not exist. These values are important because they identify horizontal tangent lines or corner points on the graph, which are the only possible locations at which a local maximum or local minimum can occur.
• Given a differentiable function $f\text{,}$ whenever $f'$ is positive, $f$ is increasing; whenever $f'$ is negative, $f$ is decreasing. The first derivative test tells us that at any point where $f$ changes from increasing to decreasing, $f$ has a local maximum, while conversely at any point where $f$ changes from decreasing to increasing $f$ has a local minimum.
• Given a twice differentiable function $f\text{,}$ if we have a horizontal tangent line at $x = p$ and $f''(p)$ is nonzero, then the fact that $f''$ tells us the concavity of $f$ will determine whether $f$ has a maximum or minimum at $x = p\text{.}$ In particular, if $f'(p) = 0$ and $f''(p) \lt 0\text{,}$ then $f$ is concave down at $p$ and $f$ has a local maximum there, while if $f'(p) = 0$ and $f''(p) \gt 0\text{,}$ then $f$ has a local minimum at $p\text{.}$ If $f'(p) = 0$ and $f''(p) = 0\text{,}$ then the second derivative does not tell us whether $f$ has a local extreme at $p$ or not.
### SubsectionExercises
###### 4
This problem concerns a function about which the following information is known:
• $f$ is a differentiable function defined at every real number $x$
• $f(0) = -1/2$
• $y = f'(x)$ has its graph given at center in Figure 3.1.13
1. Construct a first derivative sign chart for $f\text{.}$ Clearly identify all critical numbers of $f\text{,}$ where $f$ is increasing and decreasing, and where $f$ has local extrema.
2. On the right-hand axes, sketch an approximate graph of $y = f''(x)\text{.}$
3. Construct a second derivative sign chart for $f\text{.}$ Clearly identify where $f$ is concave up and concave down, as well as all inflection points.
4. On the left-hand axes, sketch a possible graph of $y = f(x)\text{.}$
###### 5
Suppose that $g$ is a differentiable function and $g'(2) = 0\text{.}$ In addition, suppose that on $1 \lt x\lt 2$ and $2 \lt x \lt 3$ it is known that $g'(x)$ is positive.
1. Does $g$ have a local maximum, local minimum, or neither at $x = 2\text{?}$ Why?
2. Suppose that $g''(x)$ exists for every $x$ such that $1 \lt x \lt 3\text{.}$ Reasoning graphically, describe the behavior of $g''(x)$ for $x$-values near $2\text{.}$
3. Besides being a critical number of $g\text{,}$ what is special about the value $x = 2$ in terms of the behavior of the graph of $g\text{?}$
###### 6
Suppose that $h$ is a differentiable function whose first derivative is given by the graph in Figure 3.1.14.
1. How many real number solutions can the equation $h(x) = 0$ have? Why?
2. If $h(x) = 0$ has two distinct real solutions, what can you say about the signs of the two solutions? Why?
3. Assume that $\lim_{x \to \infty} h'(x) = 3\text{,}$ as appears to be indicated in Figure 3.1.14. How will the graph of $y = h(x)$ appear as $x \to \infty\text{?}$ Why?
4. Describe the concavity of $y = h(x)$ as fully as you can from the provided information.
###### 7
Let $p$ be a function whose second derivative is $p''(x) = (x+1)(x-2)e^{-x}\text{.}$
1. Construct a second derivative sign chart for $p$ and determine all inflection points of $p\text{.}$
2. Suppose you also know that $x = \frac{\sqrt{5}-1}{2}$ is a critical number of $p\text{.}$ Does $p$ have a local minimum, local maximum, or neither at $x = \frac{\sqrt{5}-1}{2}\text{?}$ Why?
3. If the point $(2, \frac{12}{e^2})$ lies on the graph of $y = p(x)$ and $p'(2) = -\frac{5}{e^2}\text{,}$ find the equation of the tangent line to $y = p(x)$ at the point where $x = 2\text{.}$ Does the tangent line lie above the curve, below the curve, or neither at this value? Why? |
# 4.5: Math Functions
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Python has a math module that provides most of the familiar mathematical functions. Before we can use the module, we have to import it:
>>> import math
This statement creates a module object named math. If you print the module object, you get some information about it:
>>> print(math)
<module 'math' (built-in)>
The module object contains the functions and variables defined in the module. To access one of the functions, you have to specify the name of the module and the name of the function, separated by a dot (also known as a period). This format is called dot notation.
>>> ratio = signal_power / noise_power
>>> decibels = 10 * math.log10(ratio)
>>> height = math.sin(radians)
The first example computes the logarithm base 10 of the signal-to-noise ratio. The math module also provides a function called log that computes logarithms base e.
The second example finds the sine of radians. The name of the variable is a hint that sin and the other trigonometric functions (cos, tan, etc.) take arguments in radians. To convert from degrees to radians, divide by 360 and multiply by 2π:
>>> degrees = 45
>>> radians = degrees / 360.0 * 2 * math.pi
0.7071067811865476
The expression math.pi gets the variable pi from the math module. The value of this variable is an approximation of π, accurate to about 15 digits.
>>> math.sqrt(2) / 2.0
0.7071067811865476 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Graphing Polynomials
## Absolute and local extrema, critical values, end behavior
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Practice Graphing Polynomials
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Finding and Defining Parts of a Polynomial Function Graph
The prototype for a roller coaster is represented by the equation $y = x^5 - 8x^3 + 10x + 6$ . What is the maximum height the coaster will reach over the domain [-1, 2].
### Watch This
First watch this video.
Then watch this video.
### Guidance
By now, you should be familiar with the general idea of what a polynomial function graph does. It should cross the $x-$ axis as many times as the degree, unless there are imaginary solutions. It will curve up and down and can have a maximum and a minimum. Let’s define the parts of a polynomial function graph here.
Notice that in both the cubic (third degree, on the left) and the quartic (fourth degree, on the right) functions, there is no vertex. We now have minimums and maximums. If there are more than one minimum or maximum, there will be an absolute maximum/minimum , which is the lowest/highest point of the graph. A local maximum/minimum is a maximum/minimum relative to the points around it. The places where the function crosses the $x-$ axis are still the solutions (also called $x-$ intercepts, roots or zeros). In the quartic function, there is a repeated root at $x = 4$ . A repeated root will touch the $x-$ axis without passing through or it can also have a “jump” in the curve at that point (see Example A). All of these points together (maximums, minimums, $x-$ intercepts, and $y-$ intercept) are called critical values.
Another important thing to note is end behavior. It is exactly what it sounds like; how the “ends” of the graph behaves or points. The cubic function above has ends that point in the opposite direction. We say that from left to right, this function is mostly increasing. The quartic function’s ends point in the same direction, both positive, just like a quadratic function. When considering end behavior, look at the leading coefficient and the degree of the polynomial.
#### Example A
Use a table to graph $y=x^3$ .
Solution: Draw a table and pick at least 5 values for $x$ .
$x$ $x^3$ $y$
-2 $(-2)^3$ -8
-1 $(-1)^3$ -1
0 $0^3$ 0
1 $1^3$ 1
2 $2^3$ 8
Plot the points and connect. This particular function is the parent graph for cubic functions. Recall from quadratic functions, that the parent graph has a leading coefficient of 1, no other $x-$ terms, and no $y-$ intercept. $y=x^4$ and $y=x^5$ are also parent graphs.
#### Example B
Analyze the graph below. Find the critical values, end behavior, and find the domain and range.
Solution: First, find the solutions. They appear to be (-2, 0), (1, 0), and (2, 0). Therefore, this function has a minimum degree of 3. However, look at the $y-$ intercept. The graph slightly bends between the maximum and minimum. This movement in the graph tells us that there are two imaginary solutions (recall that imaginary solutions always come in pairs). Therefore, the function has a degree of 5. Approximate the other critical values:
maximum: (-1.1, 10)
minimum: (1.5, -1.3)
$y-$ intercept: (0, 5)
In general, this function is mostly increasing and the ends go in opposite directions. The domain and range are both all real numbers.
When describing critical values, you may approximate their location. In the next concept, we will use the graphing calculator to find these values exactly.
Sometime it can be tricky to see if a function has imaginary solutions from the graph. Compare the graph in Example B to the cubic function above. Notice that it is smooth between the maximum and minimum. As was pointed out earlier, the graph from Example B bends. Any function with imaginary solutions will have a slightly irregular shape or bend like this one does.
#### Example C
Sketch a graph of a function with roots $-4, -3, \frac{1}{2}$ , and 3, has an absolute maximum at (2, 5), and has negative end behavior. This function does not have any imaginary roots.
Solution: There are several possible answers for this graph because we are only asking for a sketch. You would need more information to get an exact answer. Because this function has negative end behavior and four roots, we know that it will pass through the $x-$ axis four times and face down. The absolute maximum is located between the roots $\frac{1}{2}$ and 3. Plot these five points and connect to form a graph.
Intro Problem Revisit
Use a table to graph $x^5 - 8x^3 + 10x + 6$ .
Draw a table and pick at least 5 values for $x$ . Remember that we are dealing only with x values between and including -1 and 2.
$x$ $y$
-1 3
0 6
0.5 10.03125
1 9
2 -16
Plot the points and connect.
From your graph you can see that the maximum height the roller coaster reaches is just slightly over 10.
### Guided Practice
1. Use a table to graph $f(x)=-(x+2)^2(x-3)$ .
2. Analyze the graph. Find all the critical values, domain, range and describe the end behavior.
3. Draw a graph of the cubic function with solutions of -6 and a repeated root at 1. This function is generally increasing and has a maximum value of 9.
1. This function is in intercept form. Because the factor, $(x + 2)$ is squared, we know it is a repeated root. Therefore, the function should just touch at -2 and not pass through the $x-$ axis. There is also a zero at 3. Because the function is negative, it will be generally decreasing. Think of the slope of the line between the two endpoints. It would be negative. Select several points around the zeros to see the behavior of the graph.
$x$ $y$
-4 14
-2 0
0 12
2 16
3 0
4 -36
2. There are three real zeros at approximately -3.5, 1, and 7. Notice the curve between the zeros 1 and 7. This indicated there are two imaginary zeros, making this at least a fifth-degree polynomial. Think about an imaginary horizontal line at $y = 3$ . This line would touch the graph five times, so there should be five solutions. Next, there is an absolute minimum at (-0.5, -7.5), a local maximum at (2.25, 5), a local minimum at (2.25, 2.25) and an absolute maximum at (5, 6). The $y-$ intercept is at (0, -6). The domain and range are both all real numbers and the end behavior is mostly decreasing.
3. To say the function is “mostly increasing” means that the slope of the line that connects the two ends (arrows) is positive. Then, the function must pass through (-6, 0) and touch, but not pass through (1, 0). From this information, the maximum must occur between the two zeros and the minimum will be the double root.
### Vocabulary
Absolute Maximum/Minimum
The highest/lowest point of a function. When referring to the absolute maximum/minimum value, use the $y-$ value.
Local Maximum/Minimum
The highest/lowest point relative to the points around it. A function can have multiple local maximums or minimums.
Solutions
The $x-$ intercepts. Also called roots or zeros.
Critical Values
The $x-$ intercepts, maximums, minimums, and $y-$ intercept.
End Behavior
How the ends of a graph look. End behavior depends on the degree of the function and the leading coefficient.
Parent Graph
The most basic function of a particular type. It has a leading coefficient of 1, no additional $x-$ terms, and no constant.
### Practice
Use the given $x-$ values to make a table and graph the functions below.
1. $f(x) &= x^3-7x^2+15x-2\\x &= -2, -1, 0, 1, 2, 3, 4$
2. $g(x) &= -2x^4 - 11x^3 - 3x^2+37x+35\\x &= -5, -4, -3, -2, -1, 0, 1, 2$
3. $y &=2x^3+25x^2+100x+125\\x &= -7,-6,-5,-4,-3,-2,-1,0$
Make your own table and graph the following functions.
1. $f(x)=(x+5)(x+2)(x-1)$
2. $y=x^4$
3. $y=x^5$
4. Analyze the graphs of $y=x^2, y=x^3, y=x^4$ , and $y=x^5$ . These are all parent functions. What do you think the graph of $y=x^6$ and $y=x^7$ will look like? What can you say about the end behavior of all even functions? Odd functions? What are the solutions to these functions?
5. Writing How many repeated roots can one function have? Why?
Analyze the graphs of the following functions. Find all critical values, the domain, range, and end behavior.
For questions 13-15, make a sketch of the following real-solution functions.
1. Draw two different graphs of a cubic function with zeros of -1, 1, and 4.5 and a minimum of -4.
2. A fourth-degree polynomial with roots of -3.2, -0.9, 1.2, and 8.7, positive end behavior, and a local minimum of -1.7.
3. A fourth-degree function with solutions of -7, -4, 1, and 2, negative end behavior, and an absolute maximum at $\left(-\frac{11}{2}, \frac{1755}{128}\right)$ .
4. Challenge Find the equation of the function from #15.
### Vocabulary Language: English
Absolute Maximum/Minimum
Absolute Maximum/Minimum
The highest and lowest points of a function are referred to as the absolute maximum and minimum, respectively. When referring to the absolute maximum/minimum value, use the $y-$value.
Critical Values
Critical Values
The critical values are the $x-$intercepts of a quadratic function.
End behavior
End behavior
End behavior is a description of the trend of a function as input values become very large or very small, represented as the 'ends' of a graphed function.
Local Maximum
Local Maximum
A local maximum is the highest point relative to the points around it. A function can have more than one local maximum.
Local Minimum
Local Minimum
A local minimum is the lowest point relative to the points around it. A function can have more than one local minimum.
Parent Graph
Parent Graph
A parent graph is the simplest form of a particular type of graph. All other graphs of this type are usually compared to the parent graph.
solution
solution
A solution to an equation or inequality should result in a true statement when substituted for the variable in the equation or inequality. |
# The Four Fundamental Forces that Enable Aircraft to Fly
In this article, we will identify the four main forces concerned with aircraft in flight. We will discuss the nature of these forces and deduce mathematical and descriptive definitions. If an aircraft is to be considered in straight and level flight then it will not be gaining altitude, rolling, or yawing. It will be maintaining a constant airspeed, altitude, and heading. If we were to consider the forces acting on the aircraft, it would look something similar to figure 1.
Figure 1
## Lift and Drag
Lift and drag are important forces that act upon bodies in a fluid flow, in this case our fluid is air. Firstly, it is important to define these forces, we will then examine the mathematical formulas that describe them. Lift is a force that acts perpendicular to the relative motion of the airflow and acts through the aerodynamic centre of the wing. It is a result of different pressure on two sides of an object, if there is a greater pressure on the underside, lift will be positive and there will exist an upwards force.
Figure 2 demonstrates that the lift is perpendicular to the relative airflow, in straight and level flight. The magnitude of the lift force will depend upon a few key factors and for a wing those factors are the relative velocity between wing and air flow, or air speed, the density of the air, wing area and lift coefficient of the wing. All these values can vary throughout the course of flight. The lift coefficient of an aerofoil depends on several parameters but perhaps the most important is the camber. Lift is always perpendicular to the motion. The position of this force is always at a location called the aerodynamic centre of the wing.
Mathematically we can give the lift force as:
### Example 1
If a Boeing 747 has a wing area of 510 m2 and is cruising at 263 m/s, the density of the air is 1.20 kgm-3, and the lift coefficient is 0.6, what is the lift force generated?
Solution:
The most important thing is to check all units are SI, and in this case they are, so we can input them directly into the lift equation above
L= (1/2) (1.2) (263)2 (510) (0.6)
L=12.7 MN
### Example 2
Now we will give some thought to the drag force experienced by the aircraft. Drag is a mechanical force that is caused by the difference in velocity between the aircraft and the velocity of the fluid it is moving through (air). For drag to be generated, the solid body must be in contact with the fluid. If there is no fluid, there is no drag. Drag always opposes the motion of the aircraft. The equation that governs drag for a surface moving through a fluid is like that of lift. However, it is slightly more complex and various factors can influence drag, such as:
• Body shape
• Velocity of air
• Turbulence of air
• Surface roughness of body
• Angle of attack of the bod
• Air density
The drag formula is presented below:
where cD is the drag coefficient.
One of the sources of drag is the skin friction between the molecules of the air and the solid surface of the aircraft. Because the skin friction is an interaction between a solid and a gas, the magnitude of the skin friction depends on properties of both solid and gas. For the solid, a smooth, waxed surface produces less skin friction than a roughened surface. For the gas, the magnitude depends on the viscosity of the air and the relative magnitude of the viscous forces to the motion of the flow.
We can also think of drag as aerodynamic resistance to the motion of the object through the fluid. This source of drag depends on the shape of the aircraft and is called form drag. As air flows around a body, the local velocity and pressure are changed. Since pressure is a measure of the momentum of the gas molecules and a change in momentum produces a force, a varying pressure distribution will produce a force on the body.
There is an additional drag component caused by the generation of lift. Aerodynamicists have named this component the induced drag. It is also called “drag due to lift” because it only occurs on finite, lifting wings. Induced drag occurs because the distribution of lift is not uniform on a wing, but varies from root to tip
Notice that the area (S) given in the drag equation is given as a reference area. The drag depends directly on the size of the body. Since we are dealing with aerodynamic forces, the dependence can be characterised by some area. But which area do we choose? If we think of drag as being caused by friction between the air and the body, a logical choice would be the total surface area of the body. If we think of drag as being a resistance to the flow, a more logical choice would be the frontal area of the body that is perpendicular to the flow direction. And finally, if we want to compare with the lift coefficient, we should use the same wing area used to derive the lift coefficient. Since the drag coefficient is usually determined experimentally by measuring drag and the area and then performing the division to produce the coefficient, we are free to use any area that can be easily measured. If we choose the wing area, rather than the cross-sectional area, the computed coefficient will have a different value. But the drag is the same, and the coefficients are related by the ratio of the areas. In practice, drag coefficients are reported based on a wide variety of object areas. In the report, the aerodynamicist must specify the area used; when using the data, the reader may have to convert the drag coefficient using the ratio of the areas.
## Thrust and Weight
Thrust is the force which moves an aircraft through the air. Thrust is used to overcome the drag of an aeroplane, and to overcome the weight of a rocket. Thrust is generated by the engines of the aircraft through some kind of propulsion system.
Thrust is a mechanical force, so the propulsion system must be in physical contact with a working fluid to produce thrust. Thrust is generated most often through the reaction of accelerating a mass of gas. Since thrust is a force, it is a vector quantity having both a magnitude and a direction. The engine does work on the gas and accelerates the gas to the rear of the engine; the thrust is generated in the opposite direction from the accelerated gas. The magnitude of the thrust depends on the amount of gas that is accelerated and on the difference in velocity of the gas through the engine.
In an aircraft, the thrust is generated in different ways according to the type of propulsion:
• Turbojet: all the thrust is generated in the form of jet efflux from the rear of the engine. (Now used mostly in military aircraft).
• Turbofan: most of the thrust is generated by a large fan at the front of the engine; a small percentage is generated by jet efflux.
• Turboprop: most of the thrust is generated by the propeller; a small percentage is generated by jet efflux.
• Piston: all the thrust is generated by the propeller.
The Power required to generate thrust depends on a number of factors, but in simple terms it may be said that the power is proportional to the thrust required multiplied by the aircraft speed.
Finally the weight of the aircraft must be considered, it is important to note here that students often confuse weight with mass. The mass of an object is given in kilograms and is a scalar quantity. Weight is a force, measured in newtons, and must be equal to the mass multiplied by the acceleration due to gravity. The direction of the force is always downwards to the centre of the Earth. Just as lift, drag and thrust can all vary throughout the course of a flight so too can the weight. As fuel is burned the mass of the aircraft will decrease significantly.
The thrust can be divided by the weight at specific times to give a useful property called the thrust to weight ratio. This parameter is an important indicator and increases with the aerodynamic performance of an aircraft.
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# Use Exponents to Name Place Value Units
Solutions, examples, and videos to help Grade 5 students learn how to use exponents to name place value units and explain patterns in the placement of the decimal point.
Common Core Standards: 5.NBT.1, 5.NBT.2, 5.MD.1
New York State Common Core Math Grade 5, Module 1, Lesson 3
Grade 5, Module 1, Lesson 3 Worksheets (pdf)
The following figure gives the place value chart using exponents to represent each place value. Scroll down the page for examples and solutions.
Grade 5 Lesson 3 Concept Development
Problem 1:
Problem 2: Write ten to the fifth power as a product of tens.
Problem 3: 10 × 100
Problem 4: 3 × 102
Problem 5: 3.4 × 103
Problem 6: 700 ÷ 102
Problem 7: 7.1 ÷ 102
Lesson 3 Problem Set
1. Write the following in exponential form (e.g., 100 = 102).
a. 10,000 = _____
b. 1000 = _____
c. 10 × 10 = _____
d. 100 × 100 = _____
e. 1,000,000 = ______
f. 1000 × 1000 = _____
2. Write the following in standard form (e.g., 5 × 102= 500).
a. 9 × 103 = _____
b. 39 × 104 = _____
c. 7200 ÷ 102 = _____
e. 4.025 × 103 = _____
f. 40.25 × 104 = _____
g. 725 ÷ 103 = _____
This video shows how exponents work in relation to place value. Charts are used to compare exponential form, standard form, and expanded form. Multiplication problems are solved using place value charts.
1. Complete the patterns.
a. 0.02 0.2 ____ 20 ___ ___
Lesson 3 Exit Ticket
1. Write the following in exponential form and as a multiplication sentence using only 10 as a factor (e.g., 100 = 102 = 10 × 10).
a. 1,000 = ______________ = ______________
b. 100 × 100 = ______________ = ______________
2. Write the following in standard form (e.g., 4 × 102 = 400).
a. 3 × 102 = ______________
c. 800 ÷ 102 = ______________
b. 2.16 × 104= ______________
d. 754.2 ÷ 103 = ______________
Use exponents to name place value units and explain patterns in the placement of the decimal point.
Lesson 3 Homework
1. Write the following in exponential form (e.g., 100 = 102).
a. 1000 = _____
b. 10 × 10 = ____
c. 100,000 = ____
d. 100 × 10 = ____
e. 1,000,000 = ____
f. 10,000 × 10 = ____
2. Write the following in standard form (e.g., 4 × 102 = 400).
a. 4 × 103 =
b. 64 × 104 =
c. 5300 ÷ 102 =
d. 5,300,000 ÷ 103 =
e. 6.072 × 103 =
f. 60.72 × 104 =
g. 948 ÷ 103 =
h. 9.4 ÷ 102 =
3. Complete the patterns.
a. 0.02 0.2 _____ 20 _____ _____
b. 3,400,000 34,000 ____ 3.4 _____
c. _____ 8,570 _____ 85.7 8.57 _____
d. 444 4440 44,400 _____ _____ _____
e. ____ 9.5 950 95,000 _____ _____
4. After a lesson on exponents, Tia went home and said to her mom, “I learned that 104 is the same as 40,000.” She has made a mistake in her thinking. Use words, numbers or a place value chart to help Tia correct her mistake.
5. Solve 247 ÷ 102 and 247 × 102. a. What is different about the two answers? Use words, numbers or pictures to explain how the decimal point shifts.
b. Based on the answers from the pair of expressions above, solve 247 ÷ 103 and 247 × 103.
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. |
# Study Notes for CAT 2021: Cyclicity
Updated : April 9th, 2021
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CAT 2021 aspirants need to strengthen their basics before going out guns blazing on Quantitative Aptitude. To help them, this post provides essential concepts to build your basics in any chapter. Here are some topics that we have covered:
CAT 2021 aspirants need to strengthen their basics before going out guns blazing on Quantitative Aptitude. To help them, this post provides essential concepts to build your basics in any chapter. Here are some topics that we have covered:
Basic Concept of Cyclicity
The concept of cyclicity is used to identify the last digit of the number.
Let’s take an example to understand this:
Example: Find the unit digit of 354.
Solution: Now it’s a very big term and not easy to calculate but we can find the last digit by using the concept of cyclicity. We observe powers of 3
31 = 3
32 = 9
33 = 27
34 = 81
35 = 243
36= 729
So now pay attention to the last digits we can observe that the last digit repeats itself after a cycle of 4 and the cycle is 3, 9, 7, 1 this repetition of numbers after a particular stage is called the cyclicity of numbers.
Therefore, when we need to find the unit digit of any number like 3n we just need to find the number on which the cycle halts. So we divide power n by 4 to check remainder
• If remainder is 1 then the unit digit will be 3
• If remainder is 2 then the unit digit will be 9
• If remainder is 3 then the unit digit will be 7
• If remainder is 0 then the unit digit will be 1
We divided the power by 4 because cycle repeats itself after 4 values. Now the main question was that how much is the last digit of 354 and we know the cycle repeat itself after 4 so we will divide the 54 with 4, so on dividing 54 by 4 the remainder becomes 2.
Now as we discussed above if the remainder is 2 the last digit would be 9. Hence unit digit of 354 is 9.
Example: What will be the unit digit of 34745
Solution: Let’s observe unit digit of 347 x 347 = 9.
The main purpose of the above is that the unit digit of any multiplication depends upon the unit digit of numbers, whatever is the number big or small the unit digit always depends upon the multiplication of the last digit.
So the last digit of 34745 can be found by calculating last digit of 745
We observe unit digit while calculating powers of 7
71 = 7
72 = 49
73 = 343
74 = 2401
75 = 16807
So on dividing 45 with 4, 1 will be the remainder and the last digit would be 7
Application of Concept of Cyclicity
How to calculate unit digit if a number contains power of a power
Example: What will be the last digit of (1223)45
Solution:
To find the last digit of this type of number we will start the question from the base the base is given to be 12.
It means we will see the cyclicity of 2 because the last digit depends upon the unit digit of 12 i.e. 2.
We observe unit digit while calculating powers of 2
21 = 2
22 = 4
23 = 8
24 = 6
25 = 2
Step 1: Now we know that cyclicity of last digit of 12 i.e. 2 is of 4, hence we divide the power of 12 i.e. 2345 with 4.
Step 2: Now let’s calculate the remainder of 2345 when divided by 4 and then we will determine the last digit.
Step 3: The remainder will be 3 because we can write remainder of 23 by 4 as 3 or -1. Hence we can write 2345 as (-1)45 but odd power of -1 will be again -1 and thus 2345 when divided by 4 will give us remainder as -1 or 3.
Hence unit digit of (1223 )45will be same as unit digit of 23 i.e. 8
Example: Find the unit digit of (3225)95
Solution: To find the last digit of this type of number we will start the question from the base the base is given to be 32. It means we will see the cyclicity of 2 because the last digit depends upon the unit digit of 32 i.e. 2.
We observe unit digit while calculating powers of 2
21 = 2
22 = 4
23 = 8
24 = 6
25 = 2
Step 1: Now we know that cyclicity of last digit of 32 i.e. 2 is of 4, hence the divide the power of 12 i.e. 2595 with 4.
Step 2: Now let’s calculate the remainder of 2595 when divided by 4 and then we will determine the last digit.
Step 3: The remainder will be 1 because we can write remainder of 25 by 4 as 1. Hence we can write 2595 as (1) 95 but any power of 1 will be again 1 and thus 2595 when divided by 4 will give us remainder as 1.
Hence unit digit of (3225)95 will be same as unit digit of 21 i.e. 2
Questions: Find the unit digit of 4686
1. a) 4 b) 6 c) 2 d) 3
Solution: We know
41 = 4
42 = 6
43 = 4
44 = 6
Here, we see that powers of four repeat after a cycle of 2 i.e.
Odd power of 4 gives unit digit as 4 and even power of 4 gives unit digit as 6
So unit digit of 4686 is 6.
Question: What will be the unit digit of 651234
1. a) 5 b) 0 b) 2 d) 3
Solution: We know
51 = 5
52 = 25
53 = 125
54 = 625
So, we observe that unit digit of any power of 5 is 5. Hence unit digit of 651234 is 5.
Question: Find the unit digit of 36512
1. a) 4 b) 6 c) 2 d) 3
Solution: We know
61 = 6
62 = 36
63 = 216
So, we observe that unit digit of any power of 6 is 6. Hence unit digit of 36512 is 6.
Question: Find the unit digit of 185693
1. a) 4 b) 6 c) 2 d) 8
Solution: We observe unit digit while calculating powers of 8
Unit digit in 81 = 8
Unit digit in 82 = 4
Unit digit in 83 = 2
Unit digit in 84 = 6
Unit digit in 85 = 8
So on dividing 5693 with 4, 1 will be the remainder and hence the last digit would be 8.
Universal Cyclicity: Since all the numbers start repeating themselves after 4, the universal cyclicity of all numbers is 4.
Example: Calculate 5781029
Solution: Let`s look at the following steps.
Step 1: We divide the exponent of the number by 4. So we divide 1029 by 4 and get the remainder as 1. Step 2: Now the unit`s digit of the number can be calculated by solving 81. Here 8 is the unit`s digit of the number 578 and 1 is the remainder.
Step 3: If the remainder is zero, then the unit digit will be same as the unit digit of N4.
=======================
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# Difference between revisions of "2009 AIME I Problems/Problem 11"
## Problem
Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$. Find the number of such distinct triangles whose area is a positive integer.
## Solution 1
Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).
$\det \left({\matrix {P \above Q}}\right)=\det \left({\matrix {x_1 \above x_2}\matrix {y_1 \above y_2}\right).$ (Error compiling LaTeX. ! Package amsmath Error: Old form `\matrix' should be \begin{matrix}.)
Since the triangle has half the area of the parallelogram, we just need the determinant to be even.
The determinant is
$$(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))$$
Since $2009$ is not even, $((x_1)-(x_2))$ must be even, thus the two $x$'s must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$. There are $25$ even and $25$ odd numbers available for use as coordinates and thus there are $(_{25}C_2)+(_{25}C_2)=\boxed{600}$ such triangles.
## Solution 2
As in the solution above, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$.
If the coordinates of $P$ and $Q$ have nonnegative integer coordinates, $P$ and $Q$ must be lattice points either
• on the nonnegative x-axis
• on the nonnegative y-axis
• in the first quadrant
We can calculate the y-intercept of the line $41x+y=2009$ to be $(0,2009)$ and the x-intercept to be $(49,0)$.
Using the point-to-line distance formula, we can calculate the height of $\triangle OPQ$ from vertex $O$ (the origin) to be:
$\dfrac{|41(0) + 1(0) - 2009|}{\sqrt{41^2 + 1^2}} = \dfrac{2009}{\sqrt{1682}} = \dfrac{2009}{29\sqrt2}$
Let $b$ be the base of the triangle that is part of the line $41x+y=2009$.
The area is calculated as: $\dfrac{1}{2}\times b \times \dfrac{2009}{29\sqrt2} = \dfrac{2009}{58\sqrt2}\times b$
Let the numerical area of the triangle be $k$.
So, $k = \dfrac{2009}{58\sqrt2}\times b$
We know that $k$ is an integer. So, $b = 58\sqrt2 \times z$, where $z$ is also an integer.
We defined the points $P$ and $Q$ as $P=(x_1,y_1)$ and $Q=(x_2,y_2)$.
Changing the y-coordinates to be in terms of x, we get:
$P=(x_1,2009-41x_1)$ and $Q=(x_2,2009-41x_2)$.
The distance between them equals $b$.
Using the distance formula, we get
$PQ = b = \sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\sqrt2 \times |x_2 - x_1| = 58\sqrt2\times z$ $(*)$
WLOG, we can assume that $x_2 > x_1$.
Taking the last two equalities from the $(*)$ string of equalities and putting in our assumption that $x_2>x_1$, we get
$29\sqrt2\times (x_2-x_1) = 58\sqrt2\times z$.
Dividing both sides by $29\sqrt2$, we get
$x_2-x_1 = 2z$
As we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well.
• There are 48 ordered pairs $(x_2,x_1)$ such that their positive difference is 2.
• There are 46 ordered pairs $(x_2,x_1)$ such that their positive difference is 4.
...
• Finally, there are 2 ordered pairs $(x_2,x_1)$ such that their positive difference is 48.
Summing them up, we get that there are $2+4+\dots + 48 = \boxed{600}$ triangles.
2009 AIME I (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions |
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# A cistern, internally measuring $150cm \times 120 \times 110cm$, has $129600c{m^3}$of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being $22.5cm \times 7.5cm \times 6.5cm$.
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Hint: The bricks are porous. So, they will absorb some water. The empty space left in the cistern will be covered by the volume of the brick.
Dimensions of the cistern is given as $150cm \times 120 \times 110cm$.
Volume of the cistern $= 150 \times 120 \times 110 = 1980000c{m^3}$
Volume of water filled in the cistern $= 129600c{m^3}$
Empty space left in the cistern $= 1980000 - 129600 = 1850400c{m^3}$
Dimensions of brick are given as $22.5cm \times 7.5cm \times 6.5cm$.
Volume of each brick $= 22.5 \times 7.5 \times 6.5 = 1096.875c{m^3}$
Let there be $x$ number of bricks that can be placed in the cistern just to avoid overflow. Then,
Total volume of all the bricks $= 1096.875x$
It is also given that each brick absorbs one-seventeenth of its own volume of water. Then we hane:
Total volume of water absorbed by all the bricks $= \dfrac{1}{{17}} \times 1096.875x$
This will increase the empty spaces in the cistern. So we have:
Total empty space left in the cistern$= 1850400 + \dfrac{1}{{17}} \times 1096.875x$
This empty space will be capitalized by the space taken by the bricks themselves. So, we have:
$\Rightarrow 1850400 + \dfrac{1}{{17}} \times 1096.875x = 1096.875x, \\ \Rightarrow 1096.875x - \dfrac{1}{{17}} \times 1096.875x = 1850400, \\ \Rightarrow \dfrac{{16}}{{17}} \times 1096.875x = 1850400, \\ \Rightarrow x = \dfrac{{17 \times 115650}}{{1096.875}}, \\ \Rightarrow x = 1792.41 \\$
Therefore, a maximum of 1792 bricks can be placed in the cistern without overflowing the water.
Note: Above, we are getting the value of $x$as 1792.41. The number of bricks can only be integer.
If we take 1793 as the total number of bricks, this will result in some amount of water overflow because 1793 is greater than 1792.41. That’s why we can take 1792 as the maximum number of bricks. |
# TI-83/84 How To Series
```TI-83/84 How To Series
Topic: Graphing Inequalities and Finding the Feasibility Region
Need to find the feasibility region for a set of linear inequalities? No problem, the TI-83/84 can
The easiest way to learn this is through an example. Take the following scenario:
Rei volunteers to bring origami swans and giraffes to sell at a charity crafts fair. It takes her
three minutes to make a swan and six minutes to make a giraffe. She plans to sell the swans for
\$4 each and the giraffes for \$6 each. If she has only 16 pieces of origami paper and can’t spend
more than one hour folding, how many of each animal should Rei make to maximize the
charity’s profit? (Bennett, Chanan, Bergofsky (2002) “Exploring Algebra with the Geometer’s
Before we even pick up the calculator, we must decipher what the question is telling us and
Steps
1. Decide what is being asked.
How many of swans and giraffes should Rei make to maximize profit?
2. Assign variables to the unknowns. This is a very important step because it let’s the
reader/marker know what is what in the problem without having to read the student’s mind.
Let x = number of swans
Let y = number of giraffes
We have explicitly stated what the variables are and what they represent. As a student you
should get into the practice of assigning and labeling variables.
3. Pull out the constraints of the problem. Constraints are the limitations of the scenario. In
this case there are two:
i. Number of pieces of origami paper
ii. Time to fold paper into either a swan or a giraffe.
4. Assign equations to each constraint.
i. The total number of pieces of paper to make origami swans and/or giraffes is
16. This means Rei can make at most 16 figures. If Rei wants she can make
as few as 1 or as many as 16, but not more than 16. Our job is to figure out
how many of each type she needs to make to maximize profit.
Let’s make an equation:
x + y ≤ 16
We use ≤ because Rei can make at most 16 figures
ii. The total time she has to make these figures is at most 1 hour. In addition we
know it takes Rei 3 minutes to make each swan and 6 minutes to make each
giraffe. If she make 2 swans and 3 giraffes, how many minutes has she spent
making them?
Our equation is set up similar to the question posed.
3 x + 6 y ≤ 60
Again we use ≤ because Rei has at most 1 hour to make
the figures. Notice how we changed the 1 hour into 60
minutes. This is because we must maintain the same units.
Since everything else is measured in minutes, we just
converted the 1 hour.
Both of these equations are known as inequalities because the = has been
replaced with ≤ .
5. The constraints have been set, and we can use these to graph with. However, the TI-83/84
requires the equations in the form of y = mx + b . Performing a little algebra (it is assumed
you know how to do this), the equations become:
i. y ≤ 16 − x
ii. y ≤ 10 − 0.5 x
Inequalities are just like regular line equations except that instead of solutions
being along the line, solutions, in this particular case, are not only along the line
but below the line as well. The inequality determines whether the solutions are on
the line, below the line, above the line, etc. These are known as feasibility
regions. The following table will help clarify.
Inequality Feasibility Region
=
All solutions fall on the line
Graphical example
Inequality Feasibility Region
All solutions fall on and below
≤
the line. The shaded area plus the
line.
≥
All solutions fall on and above
the line. The shaded area plus the
line.
<
All solutions fall below the line
Graphical example
The difference between this graph
and ≤ is the line is dotted to
indicate it is not included in the
solution set.
>
All solutions fall above the line
The difference between this graph
and ≥ is the line is dotted to
indicate it is not included in the
solution set.
6. We are now in a position to put these equations into our calculator.
Enter in your two equations. Notice how we
entered in the equations as y =. We have to tell
the calculator to look at the region below the
line to compensate for ≤ in both cases.
Use your arrow keys to go to the left of Y1.
Using the ENTER key scroll through the
selections until you get this symbol.
Do the same for Y2.
Change your window to accommodate the
graph. Why did we choose these values?
7. We are now ready to interpret the graph. Each inequality has its own feasibility region.
Let’s look at each equation separately.
y ≤ 16 − x
The solution set for this equation is the shaded
(feasibility) region and the line.
y ≤ 10 − 0.5 x
The solution set for this equation is the shaded
(feasibility) region and the line.
Where the two regions overlap is where both solutions are satisfied and a new feasibility region
is created.
New feasibility region created that
satisfies both equations
8. You can use the TRACE function to figure out where the important points are. Hint: you
will always maximize your profit at a corner, never within the feasibility region itself. There
are four corners on this feasibility region.
To find where the two lines intersect, use the intersect function of the TI-83/84 calculator.
This tells us that at this point we should make 12 swans (x) and 4 giraffes (y).
We also see this in the table screen of the
calculator. Just scroll down using your arrow
keys and where Y1 and Y2 are equal, this tells
us where the two lines intersect.
For your information: This is the point where Rei would maximize profit, but that’s another
lesson.
``` |
# x - jpiichspapalgebraii
```Section 1.7
Solving Absolute Value Equations
and Inequalities
Definition of Absolute Value
Absolute value of a number x, written |x|, is the
distance the number is from 0 on a number line.
if x is positive
x,
x 0,
if x 0
x, if x is negative
Interpreting Absolute Value Equations
Equation:
|x| = |x – 0| = k
Meaning:
The distance between x and 0 is k.
k
Graph:
−k
k
0
k
Solutions:
x – 0 = −k
x = −k
or
or
x–0=k
x=k
Equation:
|x − b| = k
Meaning:
The distance between x and b is k.
k
Graph:
b−k
k
b
b+k
Solutions:
x – b = −k
or
x = b − k or
x–b=k
x=b+k
Example 1
Solve |x – 9| = 3. Graph the solutions.
x – 9 = -3
x=6
x–9=3
x = 12
or
or
3
2
3
4
5
6
7
3
8
9 10 11 12 13 14 15
Solving an Absolute Value Equation
Use these steps to solve an absolute value
equation |ax + b| = c where c > 0.
1. Write two equations:
ax + b = -c or ax + b = c
2. Solve each equation.
3. Check each solution in the original
absolute value equation.
Example 2
Solve |4x + 12| = 28.
4x + 12 = -28
or
4x = -40
or
x = -10
or
4x + 12 = 28
4x = 16
x=4
Extraneous Solutions
An extraneous solution is an apparent solution
that must be rejected because it does not satisfy
the original equation.
Example 3
Solve |4x + 10| = 6x.
4x + 10 = -6x
or
10x = -10
or
x = -1
or
4x + 10 = 6x
-2x = -10
x=5
|4(-1) + 10| = 6(-1)
6 = -6
|4(5) + 10| = 6(5)
30 = 30
Absolute Value Inequalities
1.
|ax + b| < c
Equivalent Form: -c < ax + b < c
Graph of Solution:
2.
|ax + b| ≤ c
Equivalent Form: -c ≤ ax + b ≤ c
Graph of Solution:
3.
|ax + b| > c
Equivalent Form: ax + b < -c or ax + b > c
Graph of Solution:
4.
|ax + b| ≤ c
Equivalent Form: ax + b ≤ -c or ax + b ≥ c
Graph of Solution:
Example 4
Solve |3x – 7| ≥ 5.
3x – 7 ≤ -5
3x ≤ 2
2
x
3
-1
0
1
2
The graph the solution.
or
3x – 7 ≥ 5
or
3x ≥ 12
x 4
or
3
4
5
6
7
Example 5
A food manufacturer specifies that every familysize box of cereal should have a net weight of 25
ounces, with a tolerance of 1.2 ounces. Write
and solve an absolute value inequality that
describes the acceptable net weights for the
cereal in a family-size box.
Tolerance is the maximum acceptable deviation
of an item from some ideal or mean
measurement.
Let w = the actual weight of a box of cereal
Verbal model:
|actual weight – ideal weight| ≤ tolerance
|w – 25| ≤ 1.2
-1.2 ≤ w – 25 ≤ 1.2
23.8 ≤ w ≤ 26.2
The net weight for a family-size box of cereal is
between 23.8 ounces and 26.2 ounces, inclusive.
Example 6
You have found that your new winter coat is
comfortable to wear when the outdoor
temperature is between 10°F and 42°F, inclusive.
Write an absolute value inequality for this
temperature range, where t represents the
temperature in degrees Fahrenheit.
1.
Calculate the means of the extreme
temperatures.
42 10
Mean of extremes
26
2
2. Find the tolerance.
tolerance = 42 – 26 = 16
3. Write a verbal model.
|Actual temperature – Mean temperature| ≤
tolerance
|t – 26| ≤ 16
``` |
# Completing the square
Animation depicting the process of completing the square. (Details, animated GIF version)
In elementary algebra, completing the square is a technique for converting a quadratic polynomial of the form
$ax^2 + bx + c\,\!$
to the form
$a(\cdots\cdots)^2 + \mbox{constant}.\,$
In this context, "constant" means not depending on x. The expression inside the parenthesis is of the form (x + constant). Thus
$ax^2 + bx + c\,\!$ is converted to
$a(x + h)^2 + k\,$
for some values of h and k.
Completing the square is used in
In mathematics, completing the square is considered a basic algebraic operation, and is often applied without remark in any computation involving quadratic polynomials. Completing the square is also used to derive the quadratic formula.
## Overview
### Background
There is a simple formula in elementary algebra for computing the square of a binomial:
$(x + p)^2 \,=\, x^2 + 2px + p^2.\,\!$
For example:
\begin{alignat}{2} (x+3)^2 \,&=\, x^2 + 6x + 9 && (p=3)\\[3pt] (x-5)^2 \,&=\, x^2 - 10x + 25\qquad && (p=-5). \end{alignat}
In any perfect square, the number p is always half the coefficient of x, and the constant term is equal to p2.
### Basic example
Consider the following quadratic polynomial:
$x^2 + 10x + 28.\,\!$
This quadratic is not a perfect square, since 28 is not the square of 5:
$(x+5)^2 \,=\, x^2 + 10x + 25.\,\!$
However, it is possible to write the original quadratic as the sum of this square and a constant:
$x^2 + 10x + 28 \,=\, (x+5)^2 + 3.$
This is called completing the square.
### General description
Given any monic quadratic
$x^2 + bx + c,\,\!$
it is possible to form a square that has the same first two terms:
$\left(x+\tfrac{1}{2} b\right)^2 \,=\, x^2 + bx + \tfrac{1}{4}b^2.$
This square differs from the original quadratic only in the value of the constant term. Therefore, we can write
$x^2 + bx + c \,=\, \left(x + \tfrac{1}{2}b\right)^2 + k,$
where k is a constant. This operation is known as completing the square. For example:
\begin{alignat}{1} x^2 + 6x + 11 \,&=\, (x+3)^2 + 2 \\[3pt] x^2 + 14x + 30 \,&=\, (x+7)^2 - 19 \\[3pt] x^2 - 2x + 7 \,&=\, (x-1)^2 + 6. \end{alignat}
### Non-monic case
Given a quadratic polynomial of the form
$ax^2 + bx + c\,\!$
it is possible to factor out the coefficient a, and then complete the square for the resulting monic polynomial.
Example:
\begin{align} 3x^2 + 12x + 27 &= 3(x^2+4x+9)\\ &{}= 3\left((x+2)^2 + 5\right)\\ &{}= 3(x+2)^2 + 15 \end{align}
This allows us to write any quadratic polynomial in the form
$a(x-h)^2 + k.\,\!$
### Formula
The result of completing the square may be written as a formula. For the general case:[1]
$ax^2 + bx + c \;=\; a(x-h)^2 + k,\quad\text{where}\quad h = -\frac{b}{2a} \quad\text{and}\quad k = c - ah^2 = c - \frac{b^2}{4a}.$
Specifically, when a=1:
$x^2 + bx + c \;=\; (x-h)^2 + k,\quad\text{where}\quad h = -\frac{b}{2} \quad\text{and}\quad k = c - \frac{b^2}{4}.$
The matrix case looks very similar:
$x^{\mathrm{T}}Ax + x^{\mathrm{T}}b + c = (x - h)^{\mathrm{T}}A(x - h) + k \quad\text{where}\quad h = -\frac{1}{2}A^{-1}b \quad\text{and}\quad k = c - \frac{1}{4}b^{\mathrm{T}}A^{-1}b$
where $A$ has to be symmetric.
If $A$ is not symmetric the formulae for $h$ and $k$ have to be generalized to:
$h = -(A+A^{\mathrm{T}})^{-1}b \quad\text{and}\quad k = c - h^{\mathrm{T}}A h = c - b^{\mathrm{T}} (A+A^{\mathrm{T}})^{-1} A (A+A^{\mathrm{T}})^{-1}b$.
## Relation to the graph
Graphs of quadratic functions shifted to the right by h = 0, 5, 10, and 15.
Graphs of quadratic functions shifted upward by k = 0, 5, 10, and 15.
Graphs of quadratic functions shifted upward and to the right by 0, 5, 10, and 15.
In analytic geometry, the graph of any quadratic function is a parabola in the xy-plane. Given a quadratic polynomial of the form
$(x-h)^2 + k \quad\text{or}\quad a(x-h)^2 + k$
the numbers h and k may be interpreted as the Cartesian coordinates of the vertex of the parabola. That is, h is the x-coordinate of the axis of symmetry, and k is the minimum value (or maximum value, if a < 0) of the quadratic function.
One way to see this is to note that the graph of the function ƒ(x) = x2 is a parabola whose vertex is at the origin (0, 0). Therefore, the graph of the function ƒ(x − h) = (x − h)2 is a parabola shifted to the right by h whose vertex is at (h, 0), as shown in the top figure. In contrast, the graph of the function ƒ(x) + kx2 + k is a parabola shifted upward by k whose vertex is at (0, k), as shown in the center figure. Combining both horizontal and vertical shifts yields ƒ(x − h) + k = (x − h)2 + k is a parabola shifted to the right by h and upward by k whose vertex is at (hk), as shown in the bottom figure.
## Solving quadratic equations
Completing the square may be used to solve any quadratic equation. For example:
$x^2 + 6x + 5 = 0,\,\!$
The first step is to complete the square:
$(x+3)^2 - 4 = 0.\,\!$
Next we solve for the squared term:
$(x+3)^2 = 4.\,\!$
Then either
$x+3 = -2 \quad\text{or}\quad x+3 = 2,$
and therefore
$x = -5 \quad\text{or}\quad x = -1.$
This can be applied to any quadratic equation. When the x2 has a coefficient other than 1, the first step is to divide out the equation by this coefficient: for an example see the non-monic case below.
### Irrational and complex roots
Unlike methods involving factoring the equation, which is only reliable if the roots are rational, completing the square will find the roots of a quadratic equation even when those roots are irrational or complex. For example, consider the equation
$x^2 - 10x + 18 = 0.\,\!$
Completing the square gives
$(x-5)^2 - 7 = 0,\,\!$
so
$(x-5)^2 = 7.\,\!$
Then either
$x-5 = -\sqrt{7} \quad\text{or}\quad x-5 = \sqrt{7},\,$
so
$x = 5 - \sqrt{7}\quad\text{or}\quad x = 5 + \sqrt{7}. \,$
In terser language:
$x = 5 \pm \sqrt{7}.\,$
Equations with complex roots can be handled in the same way. For example:
$\begin{array}{c} x^2 + 4x + 5 \,=\, 0 \\[6pt] (x+2)^2 + 1 \,=\, 0 \\[6pt] (x+2)^2 \,=\, -1 \\[6pt] x+2 \,=\, \pm i \\[6pt] x \,=\, -2 \pm i. \end{array}$
### Non-monic case
For an equation involving a non-monic quadratic, the first step to solving them is to divide through by the coefficient of x2. For example:
$\begin{array}{c} 2x^2 + 7x + 6 \,=\, 0 \\[6pt] x^2 + \tfrac{7}{2}x + 3 \,=\, 0 \\[6pt] \left(x+\tfrac{7}{4}\right)^2 - \tfrac{1}{16} \,=\, 0 \\[6pt] \left(x+\tfrac{7}{4}\right)^2 \,=\, \tfrac{1}{16} \\[6pt] x+\tfrac{7}{4} = \tfrac{1}{4} \quad\text{or}\quad x+\tfrac{7}{4} = -\tfrac{1}{4} \\[6pt] x = -\tfrac{3}{2} \quad\text{or}\quad x = -2. \end{array}$
## Other applications
### Integration
Completing the square may be used to evaluate any integral of the form
$\int\frac{dx}{ax^2+bx+c}$
using the basic integrals
$\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| +C \quad\text{and}\quad \int\frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right) +C.$
For example, consider the integral
$\int\frac{dx}{x^2 + 6x + 13}.$
Completing the square in the denominator gives:
$\int\frac{dx}{(x+3)^2 + 4} \,=\, \int\frac{dx}{(x+3)^2 + 2^2}.$
This can now be evaluated by using the substitution u = x + 3, which yields
$\int\frac{dx}{(x+3)^2 + 4} \,=\, \frac{1}{2}\arctan\left(\frac{x+3}{2}\right)+C.$
### Complex numbers
Consider the expression
$|z|^2 - b^*z - bz^* + c,\,$
where z and b are complex numbers, z* and b* are the complex conjugates of z and b, respectively, and c is a real number. Using the identity |u|2 = uu* we can rewrite this as
$|z-b|^2 - |b|^2 + c , \,\!$
which is clearly a real quantity. This is because
\begin{align} |z-b|^2 &{}= (z-b)(z-b)^*\\ &{}= (z-b)(z^*-b^*)\\ &{}= zz^* - zb^* - bz^* + bb^*\\ &{}= |z|^2 - zb^* - bz^* + |b|^2 . \end{align}
As another example, the expression
$ax^2 + by^2 + c , \,\!$
where a, b, c, x, and y are real numbers, with a > 0 and b > 0, may be expressed in terms of the square of the absolute value of a complex number. Define
$z = \sqrt{a}\,x + i \sqrt{b} \,y .$
Then
\begin{align} |z|^2 &{}= z z^*\\ &{}= (\sqrt{a}\,x + i \sqrt{b}\,y)(\sqrt{a}\,x - i \sqrt{b}\,y) \\ &{}= ax^2 - i\sqrt{ab}\,xy + i\sqrt{ba}\,yx - i^2by^2 \\ &{}= ax^2 + by^2 , \end{align}
so
$ax^2 + by^2 + c = |z|^2 + c . \,\!$
### Idempotent matrix
A matrix M is idempotent when M 2 = M. Idempotent matrices generalize the idempotent properties of 0 and 1. The completion of the square method of addressing the equation
$a^2 + b^2 = a ,$
shows that the idempotent 2 × 2 matrices are parametrized by a circle in the (a,b)-plane.
The matrix $\begin{pmatrix}a & b \\ b & 1-a \end{pmatrix}$ will be idempotent provided $a^2 + b^2 = a ,$ which, upon completing the square, becomes
$(a - \tfrac{1}{2})^2 + b^2 = \tfrac{1}{4} .$
In the (a,b)-plane, this is the equation of a circle with center (1/2, 0) and radius 1/2.
## Geometric perspective
Consider completing the square for the equation
$x^2 + bx = a.\,$
Since x2 represents the area of a square with side of length x, and bx represents the area of a rectangle with sides b and x, the process of completing the square can be viewed as visual manipulation of rectangles.
Simple attempts to combine the x2 and the bx rectangles into a larger square result in a missing corner. The term (b/2)2 added to each side of the above equation is precisely the area of the missing corner, whence derives the terminology "completing the square". [1]
## A variation on the technique
As conventionally taught, completing the square consists of adding the third term, v 2 to
$u^2 + 2uv\,$
to get a square. There are also cases in which one can add the middle term, either 2uv or −2uv, to
$u^2 + v^2\,$
to get a square.
### Example: the sum of a positive number and its reciprocal
By writing
\begin{align} x + {1 \over x} &{} = \left(x - 2 + {1 \over x}\right) + 2\\ &{}= \left(\sqrt{x} - {1 \over \sqrt{x}}\right)^2 + 2 \end{align}
we show that the sum of a positive number x and its reciprocal is always greater than or equal to 2. The square of a real expression is always greater than or equal to zero, which gives the stated bound; and here we achieve 2 just when x is 1, causing the square to vanish.
### Example: factoring a simple quartic polynomial
Consider the problem of factoring the polynomial
$x^4 + 324 . \,\!$
This is
$(x^2)^2 + (18)^2, \,\!$
so the middle term is 2(x2)(18) = 36x2. Thus we get
\begin{align} x^4 + 324 &{}= (x^4 + 36x^2 + 324 ) - 36x^2 \\ &{}= (x^2 + 18)^2 - (6x)^2 =\text{a difference of two squares} \\ &{}= (x^2 + 18 + 6x)(x^2 + 18 - 6x) \\ &{}= (x^2 + 6x + 18)(x^2 - 6x + 18) \end{align}
(the last line being added merely to follow the convention of decreasing degrees of terms). |
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# Eureka Math Lesson 5 Homework 4.3 Answer Key
## Eureka Math Lesson 5 Homework 4.3 Answer Key Introduction:
Eureka Math is a new math curriculum that is gaining popularity among students and educators alike. It offers a unique approach to teaching mathematical concepts that have been proven to be extremely effective in helping students understand math better. However, like any new curriculum, there are bound to be challenges, especially when it comes to homework and testing. In this blog post, we will delve into Eureka Math Lesson 5 Homework 4.3 and provide you with a complete answer key to help you master the topic.
### Blog Body:
Eureka Math Lesson 5 Homework 4.3 is designed to help students understand division by fractions. The lesson involves dividing fractions by whole numbers, which can be quite challenging, especially for those who are not familiar with the concept. The homework consists of a series of questions that require students to use their understanding of division by fractions to solve mathematical problems.
Question 1: Sarah has 2/3 of a pan of brownies. If she wants to divide the brownies into 4 equal pieces, how much of a pan of brownies will each piece be?
Answer: 2/3 divided by 4 = 2/12 or 1/6.
Question 2: Sam has 1/5 of a pizza left. If he wants to divide the pizza into 10 equal slices, how much of a pizza will each slice be?
Answer: 1/5 divided by 10 = 1/50.
Question 3: If I divide 2/3 of a gallon of milk into 6 equal cups, how much milk will each cup contain?
Answer: 2/3 divided by 6 = 2/18 or 1/9.
These answers may seem quite simple, but they require a good understanding of the basic concepts of division by fractions to solve.
It is important to note that Eureka Math Lesson 5 Homework 4.3 is only one part of the curriculum, and students should also focus on mastering the other lessons to achieve a comprehensive understanding of the subject. Students should also practice similar problems to build their confidence and ensure that they fully understand the concepts.
### Conclusion:
In conclusion, Eureka Math Lesson 5 Homework 4.3 requires a solid understanding of division by fractions and can be quite challenging for some students. However, by mastering the basic concepts and practicing similar problems, students can easily solve the homework problems. We hope that this blog post has helped you understand the topic better and provided you with the confidence to approach Eureka Math Lesson 5 Homework 4.3 with ease. Remember, practice makes perfect!
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Age of Exploration Worksheet Pdf Answers: A Comprehensive Resource for Learning about the Era of Discovery. |
# RonnieChen
+1
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2
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### Tau and Totient Functions
RonnieChen May 6, 2023
0
37
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### Cube Roots In The Denominators Of Fractions
RonnieChen Apr 27, 2023
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May 14, 2023
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Using combination, the number of ways of placing 8 counters in the square is 12 ways and the number of ways of placing 12 counters in the square is 8 ways
What is the number of ways of placing 8 counters in the square
a) To place 8 counters in the 4 x 4 grid, with exactly two counters in each row, we can first choose the two columns in the first row that will have counters. Using combination, there are {4 C 2} = 6 ways to do this. Then, we choose the two columns in the second row that will have counters, but we need to make sure that these columns are not the same as the ones chosen in the first row. There are only two columns left to choose from, so there are only two ways to do this. Similarly, we choose the two columns in the third row that will have counters, making sure that they are different from the columns chosen in the first two rows. There is only one column left to choose from, so there is only one way to do this. Finally, the columns for the fourth row are uniquely determined. Therefore, the total number of ways to place 8 counters in the grid with exactly two counters in each row is:
6 * 2 = 12
(b) To place 12 counters in the 4 x 4 grid, with exactly three counters in each column, we can start by choosing the three rows that will have counters in the first column. Using combination, there are {4 C3} = 4 ways to do this. Then, we choose the three rows that will have counters in the second column, but we need to make sure that these rows are not the same as the ones chosen for the first column. There are only two rows left to choose from, so there are only two ways to do this. Similarly, we choose the three rows that will have counters in the third column, making sure that they are different from the rows chosen for the first two columns. There is only one row left to choose from, so there is only one way to do this. Finally, the rows for the fourth column are uniquely determined. Therefore, the total number of ways to place 12 counters in the grid with exactly three counters in each column is:
4 * 2 = 8
Thus, there are 12 ways to place 8 counters in the grid with exactly two counters in each row, and 8 ways to place 12 counters in the grid with exactly three counters in each column.
Apr 2, 2023
#1
+49
+1
Using combination, the number of ways of placing 8 counters in the square is 12 ways and the number of ways of placing 12 counters in the square is 8 ways
What is the number of ways of placing 8 counters in the square
a) To place 8 counters in the 4 x 4 grid, with exactly two counters in each row, we can first choose the two columns in the first row that will have counters. Using combination, there are {4 C 2} = 6 ways to do this. Then, we choose the two columns in the second row that will have counters, but we need to make sure that these columns are not the same as the ones chosen in the first row. There are only two columns left to choose from, so there are only two ways to do this. Similarly, we choose the two columns in the third row that will have counters, making sure that they are different from the columns chosen in the first two rows. There is only one column left to choose from, so there is only one way to do this. Finally, the columns for the fourth row are uniquely determined. Therefore, the total number of ways to place 8 counters in the grid with exactly two counters in each row is:
6 * 2 = 12
(b) To place 12 counters in the 4 x 4 grid, with exactly three counters in each column, we can start by choosing the three rows that will have counters in the first column. Using combination, there are {4 C3} = 4 ways to do this. Then, we choose the three rows that will have counters in the second column, but we need to make sure that these rows are not the same as the ones chosen for the first column. There are only two rows left to choose from, so there are only two ways to do this. Similarly, we choose the three rows that will have counters in the third column, making sure that they are different from the rows chosen for the first two columns. There is only one row left to choose from, so there is only one way to do this. Finally, the rows for the fourth column are uniquely determined. Therefore, the total number of ways to place 12 counters in the grid with exactly three counters in each column is:
4 * 2 = 8
Thus, there are 12 ways to place 8 counters in the grid with exactly two counters in each row, and 8 ways to place 12 counters in the grid with exactly three counters in each column.
Mar 31, 2023
#1
+49
+1
Mar 26, 2023
#1
+49
+1
Using combination, the number of ways of placing 8 counters in the square is 12 ways and the number of ways of placing 12 counters in the square is 8 ways
What is the number of ways of placing 8 counters in the square
a) To place 8 counters in the 4 x 4 grid, with exactly two counters in each row, we can first choose the two columns in the first row that will have counters. Using combination, there are {4 C 2} = 6 ways to do this. Then, we choose the two columns in the second row that will have counters, but we need to make sure that these columns are not the same as the ones chosen in the first row. There are only two columns left to choose from, so there are only two ways to do this. Similarly, we choose the two columns in the third row that will have counters, making sure that they are different from the columns chosen in the first two rows. There is only one column left to choose from, so there is only one way to do this. Finally, the columns for the fourth row are uniquely determined. Therefore, the total number of ways to place 8 counters in the grid with exactly two counters in each row is:
6 * 2 = 12
(b) To place 12 counters in the 4 x 4 grid, with exactly three counters in each column, we can start by choosing the three rows that will have counters in the first column. Using combination, there are {4 C3} = 4 ways to do this. Then, we choose the three rows that will have counters in the second column, but we need to make sure that these rows are not the same as the ones chosen for the first column. There are only two rows left to choose from, so there are only two ways to do this. Similarly, we choose the three rows that will have counters in the third column, making sure that they are different from the rows chosen for the first two columns. There is only one row left to choose from, so there is only one way to do this. Finally, the rows for the fourth column are uniquely determined. Therefore, the total number of ways to place 12 counters in the grid with exactly three counters in each column is:
4 * 2 = 8
Thus, there are 12 ways to place 8 counters in the grid with exactly two counters in each row, and 8 ways to place 12 counters in the grid with exactly three counters in each column.
Mar 26, 2023 |
# 5.1 Quadratic functions (Page 8/15)
Page 8 / 15
$f\left(x\right)=-2{\left(x+3\right)}^{2}-6$
$f\left(x\right)={x}^{2}+6x+4$
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left[-5,\infty \right).$
$f\left(x\right)=2{x}^{2}-4x+2$
$k\left(x\right)=3{x}^{2}-6x-9$
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left[-12,\infty \right).$
For the following exercises, use the vertex $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ and a point on the graph $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ to find the general form of the equation of the quadratic function.
$\left(h,k\right)=\left(2,0\right),\left(x,y\right)=\left(4,4\right)$
$f\left(x\right)={x}^{2}-4x+4$
$\left(h,k\right)=\left(-2,-1\right),\left(x,y\right)=\left(-4,3\right)$
$\left(h,k\right)=\left(0,1\right),\left(x,y\right)=\left(2,5\right)$
$f\left(x\right)={x}^{2}+1$
$\left(h,k\right)=\left(2,3\right),\left(x,y\right)=\left(5,12\right)$
$\left(h,k\right)=\left(-5,3\right),\left(x,y\right)=\left(2,9\right)$
$f\left(x\right)=\frac{6}{49}{x}^{2}+\frac{60}{49}x+\frac{297}{49}$
$\left(h,k\right)=\left(3,2\right),\left(x,y\right)=\left(10,1\right)$
$\left(h,k\right)=\left(0,1\right),\left(x,y\right)=\left(1,0\right)$
$f\left(x\right)=-{x}^{2}+1$
$\left(h,k\right)=\left(1,0\right),\left(x,y\right)=\left(0,1\right)$
## Graphical
For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts.
$f\left(x\right)={x}^{2}-2x$
Vertex Axis of symmetry is $\text{\hspace{0.17em}}x=1.\text{\hspace{0.17em}}$ Intercepts are
$f\left(x\right)={x}^{2}-6x-1$
$f\left(x\right)={x}^{2}-5x-6$
Vertex $\text{\hspace{0.17em}}\left(\frac{5}{2},\frac{-49}{4}\right),\text{\hspace{0.17em}}$ Axis of symmetry is $\text{\hspace{0.17em}}\left(0,-6\right),\left(-1,0\right),\left(6,0\right).$
$f\left(x\right)={x}^{2}-7x+3$
$f\left(x\right)=-2{x}^{2}+5x-8$
Vertex Axis of symmetry is $\text{\hspace{0.17em}}x=\frac{5}{4}.\text{\hspace{0.17em}}$ Intercepts are
$f\left(x\right)=4{x}^{2}-12x-3$
For the following exercises, write the equation for the graphed quadratic function.
$f\left(x\right)={x}^{2}-4x+1$
$f\left(x\right)=-2{x}^{2}+8x-1$
$f\left(x\right)=\frac{1}{2}{x}^{2}-3x+\frac{7}{2}$
## Numeric
For the following exercises, use the table of values that represent points on the graph of a quadratic function. By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function.
$x$ –2 –1 0 1 2 $y$ 5 2 1 2 5
$f\left(x\right)={x}^{2}+1$
$x$ –2 –1 0 1 2 $y$ 1 0 1 4 9
$x$ –2 –1 0 1 2 $y$ –2 1 2 1 –2
$f\left(x\right)=2-{x}^{2}$
$x$ –2 –1 0 1 2 $y$ –8 –3 0 1 0
$x$ –2 –1 0 1 2 $y$ 8 2 0 2 8
$f\left(x\right)=2{x}^{2}$
## Technology
For the following exercises, use a calculator to find the answer.
Graph on the same set of axes the functions
What appears to be the effect of changing the coefficient?
Graph on the same set of axes $\text{\hspace{0.17em}}f\left(x\right)={x}^{2},f\left(x\right)={x}^{2}+2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(x\right)={x}^{2},f\left(x\right)={x}^{2}+5\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}-3.\text{\hspace{0.17em}}$ What appears to be the effect of adding a constant?
The graph is shifted up or down (a vertical shift).
Graph on the same set of axes
What appears to be the effect of adding or subtracting those numbers?
The path of an object projected at a 45 degree angle with initial velocity of 80 feet per second is given by the function $\text{\hspace{0.17em}}h\left(x\right)=\frac{-32}{{\left(80\right)}^{2}}{x}^{2}+x\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is the horizontal distance traveled and $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ is the height in feet. Use the TRACE feature of your calculator to determine the height of the object when it has traveled 100 feet away horizontally.
50 feet
A suspension bridge can be modeled by the quadratic function $\text{\hspace{0.17em}}h\left(x\right)=.0001{x}^{2}\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}-2000\le x\le 2000\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}|x|\text{\hspace{0.17em}}$ is the number of feet from the center and $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ is height in feet. Use the TRACE feature of your calculator to estimate how far from the center does the bridge have a height of 100 feet.
## Extensions
For the following exercises, use the vertex of the graph of the quadratic function and the direction the graph opens to find the domain and range of the function.
Vertex $\text{\hspace{0.17em}}\left(1,-2\right),\text{\hspace{0.17em}}$ opens up.
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left[-2,\infty \right).$
Vertex $\text{\hspace{0.17em}}\left(-1,2\right)\text{\hspace{0.17em}}$ opens down.
Vertex $\text{\hspace{0.17em}}\left(-5,11\right),\text{\hspace{0.17em}}$ opens down.
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left(-\infty ,11\right].$
Vertex $\text{\hspace{0.17em}}\left(-100,100\right),\text{\hspace{0.17em}}$ opens up.
For the following exercises, write the equation of the quadratic function that contains the given point and has the same shape as the given function.
Contains $\text{\hspace{0.17em}}\left(1,1\right)\text{\hspace{0.17em}}$ and has shape of $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
$f\left(x\right)=2{x}^{2}-1$
Contains $\text{\hspace{0.17em}}\left(-1,4\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
Contains $\text{\hspace{0.17em}}\left(2,3\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=3{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
$f\left(x\right)=3{x}^{2}-9$
Contains $\text{\hspace{0.17em}}\left(1,-3\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=-{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
Contains $\text{\hspace{0.17em}}\left(4,3\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=5{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
$f\left(x\right)=5{x}^{2}-77$
Contains $\text{\hspace{0.17em}}\left(1,-6\right)\text{\hspace{0.17em}}$ has the shape of $\text{\hspace{0.17em}}f\left(x\right)=3{x}^{2}.\text{\hspace{0.17em}}$ Vertex has x-coordinate of $\text{\hspace{0.17em}}-1.$
## Real-world applications
Find the dimensions of the rectangular corral producing the greatest enclosed area given 200 feet of fencing.
50 feet by 50 feet. Maximize $\text{\hspace{0.17em}}f\left(x\right)=-{x}^{2}+100x.$
Find the dimensions of the rectangular corral split into 2 pens of the same size producing the greatest possible enclosed area given 300 feet of fencing.
Find the dimensions of the rectangular corral producing the greatest enclosed area split into 3 pens of the same size given 500 feet of fencing.
125 feet by 62.5 feet. Maximize $\text{\hspace{0.17em}}f\left(x\right)=-2{x}^{2}+250x.$
Among all of the pairs of numbers whose sum is 6, find the pair with the largest product. What is the product?
Among all of the pairs of numbers whose difference is 12, find the pair with the smallest product. What is the product?
$6\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}-6;\text{\hspace{0.17em}}$ product is –36; maximize $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}+12x.$
Suppose that the price per unit in dollars of a cell phone production is modeled by $\text{\hspace{0.17em}}p=\text{}45-0.0125x,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in thousands of phones produced, and the revenue represented by thousands of dollars is $\text{\hspace{0.17em}}R=x\cdot p.\text{\hspace{0.17em}}$ Find the production level that will maximize revenue.
A rocket is launched in the air. Its height, in meters above sea level, as a function of time, in seconds, is given by $\text{\hspace{0.17em}}h\left(t\right)=-4.9{t}^{2}+229t+234.\text{\hspace{0.17em}}$ Find the maximum height the rocket attains.
2909.56 meters
A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, is given by $\text{\hspace{0.17em}}h\left(t\right)=-4.9{t}^{2}+24t+8.\text{\hspace{0.17em}}$ How long does it take to reach maximum height?
A soccer stadium holds 62,000 spectators. With a ticket price of $11, the average attendance has been 26,000. When the price dropped to$9, the average attendance rose to 31,000. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue?
\$10.70
A farmer finds that if she plants 75 trees per acre, each tree will yield 20 bushels of fruit. She estimates that for each additional tree planted per acre, the yield of each tree will decrease by 3 bushels. How many trees should she plant per acre to maximize her harvest?
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
yeah
Morosi
prime number?
Morosi
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1 |
# The tangent at a point A of a circle with centre O intersects the diameter PQ of the circle (when extended) at the point B. If ∠BAP = 125°, then ∠AQP is equal to:
Free Practice With Testbook Mock Tests
## Options:
1. 45°
2. 55°
3. 60°
4. 50°
### Correct Answer: Option 2 (Solution Below)
This question was previously asked in
SSC CHSL Previous Paper 107 (Held On: 20 Oct 2020 Shift 3)
## Solution:
Given:
A circle with center O and diameter PQ extended outside the circle to point B
∠BAP = 125°
Concept Used:
Angle formed between radius and tangent of a circle is always 90°
Angles opposite to equal sides are equal.
Sum of two opposite interior angles of a triangle is equal to the opposite exterior angle.
Sum of all angles of triangle is 180°
Calculation:
OA ⊥ BA (Angle between Radius and Tangent)
∠OAP = ∠BAP - ∠OAB
⇒ ∠OAP = 125° - 90° = 35°
Since,
OA = OP = OQ (Radius of circle)
So,
∠OAP = ∠OPA = 35°
∠AOQ = ∠OAP + ∠OPA
⇒ ∠AOQ = 35° + 35° = 70°
In Δ AOQ,
∠AOQ = 70°
∠OAQ = ∠OQA (since OA = OQ)
∠AOQ + ∠OAQ + ∠OQA = 180°
⇒ 70° + ∠OAQ + ∠OQA = 180°
⇒ ∠OAQ + ∠OQA = 180° - 70° = 110°
⇒ 2∠OQA = 110°/2 = 55°
Since ∠OQA is the same angle as ∠AQP
∠AQP = 55°
The value of AQP = 55° |
# Slope. Direct Variation What you’ll learn To write and graph an equation of a direct variation. Direct variation and Constant of variation for a direct.
## Presentation on theme: "Slope. Direct Variation What you’ll learn To write and graph an equation of a direct variation. Direct variation and Constant of variation for a direct."— Presentation transcript:
Slope. Direct Variation What you’ll learn To write and graph an equation of a direct variation. Direct variation and Constant of variation for a direct variation To find the slope using a graph without the formula.
Problem 1: Finding the slope. If you travel at 50 mph for one hour, then you would have traveled 50 miles. If you travel for 2 hours at that speed, you would have traveled 100 miles. 3 hours would be 150 miles, etc. To graph these you could use a table Time (h)DistanceOrdered pair 150(1,50) 2100(2,100) 3150(3,150) 4200(4,200) Let’s be y= the distance And x=the time(hours) The graphs of the ordered pairs(time,distance) lie on a line. The relationship between time and distance is linear. When data are linear; the rate is constant.
0 1 2 3 4 5 6 300 50 100 150 200 250 Notice that the rate is just ratio of the rise over the run between two points. This rate of change is called slope
There is a worksheet for finding slope from the the graph only. KUTA and you have to do it now
Hours Worked (x) 1234 Dollars Earned (y) 10203040 If John earns \$10.00 an hour, the above table shows the relationship between hours worked and dollars earned. From this table, we can see that dollars earned is equal to hours worked multiplied by 10. The equation y=10x describes this relationship. In this relationship, the number of dollars earned varies directly with the number of hours worked. Problem 2: Direct Variation
direct variation A direct variation is a relationship that can be represented by a function in the Form y=kx, where k‡0. the constant of variation for a direct variation k is the coefficient of x. By dividing each side by x you can see the ratio is constant
The money collected and the hours worked The distance traveled and the speed of the car The circumference of a circle and its diameter The area of a circle and its radius M=kh d=kt C=kd Examples of direct variation
Problem 1 : Does the equation represent a direct variation? Identifying a direct variation a) b) First:Solve the equation for y The equation has the Form Y=kx, so the equation is a direct variation. Its constant of variation is You cannot write the equation in the form y=kx. It is not a direct variation
Your turn Does 4x+5y=0 represent a direct variation, if so Find the constant of variation. Answer:
Graphing a direct variation Weight on Mars y varies directly with weight on Earth x. The weights of the Science instruments onboard the Phoenix Mars Lander on Earth and mars are shown. Weight on Mars 50 lbs and the weight on Earth 130lbs a)What is an equation that relates weight, in pounds on Earth x and weight on Mars y?
What is the graph? x0.38xy 00.38(0)0 500.38(50)19 1000.38(100)38 1500.38(150)57 Ordered pairs (x,y) (0,0) (50,19) (100,38) (150,57) 50 100 150 60 40 20
Your turn Weight on the moon y varies directly with the weight on Earth x. A person who weights 100lb on Earth weights 16.6 lb on the moon. What is an equation that relates weight on Earth x and weight on the moon y? what is the graph of this equation. 2 4 6 0.4 0.2 Answer:
Take a note: the graph of a direct variation equation y=kx is a line with the following properties. The line passes through (0,0) The slope of the line is k When k›0 when k‹0 x x y y
Writing a Direct Variation From a Table For the data in the table, does y vary directly with x? if it does, write an equation for the direct variation. xy 46 812 1015 xy -23.2 12.4 41.6 a) b) Findfor each ordered pair Y=1.5x So y does not vary directly
Your turn For the data in the table, does y vary directly with x? if it does, write an equation for the direct variation. xy -32.25 1-0.75 4-3 xy 35.4 712.6 1221.6 xy -69 1-1.5 8-12 A B C Answers
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# EP-Program
## Dr.Wattana Toutip - Department of Mathematics Khon Kaen University
2010 :Wattana Toutip
wattou@kku.ac.th
http://home.kku.ac.th/wattou
## 3 The binomial theorem
3.1 Positive integer powers
Pascals Triangle is the triangle of numbers shown below, in which each term is obtained
by adding the two terms above it.
1
n0
1
1
n 1
1
2
1
n2
1
3
3
1
n3
1
1
1
4
5
6
10
15
4
10
20
n4
n5
n6
1
5
15
1
6
The binomial coefficient nCr is the r th term of the n th row of Pascals Triangle,
(counting from 0 instead of from 1 ).
5
4
So for example :
C0 1;
C2 6 ;
C6 1
## These coefficients are also given by the formula:
n!
n
Cr
r !(n r )!
n
The binomial coefficients are sometimes written as
r
The binomial theorem state that:
(a b)n a n nCr a n1b nCr a nr br
bn .
## So for example (a b)5 a5 5a 4b 10a 3b 2 10a 2b3 5ab 4 b5 .
The binomial theorem can be used to find approximations for expressions of the form
(1 x)n , where x is small. Continue taking terms until they are so small that they do
not affect the answer to the required degree of accuracy.
3.1.1 Examples
1. Expand ( x 2 y )4 .
Solution
Use the binomial theorem, with the fourth row of Pascals Triangle.
Put a x and b 2 y .
( x 2 y) 4 x 4 4 x3 (2 y) 6 x 2 (2 y) 2 4 x(2 y)3 (2 y) 4
(1 2 y) 4 x 4 8 x3 y 24 x 2 y 2 32 xy 3 16 y 4 .
1. Use the binomial theorem to find an approximation for (0.997)7 ,taking the first three
term of expansion of (1 x)7 .
Solution
The first three terms of (1 x)7 are : 1 7C1 x 7C2 x2 .
Put x 0.003 to obtain 1 7 0.003 21 0.0032
(0.977)7 0.979189 .
3.1.2 Exercises
1. Expand the following using the binomial theorem , and simplify as far as possible.
(a) (2 b)5
(b) (1 3b)4
(c) (2 x 3 y ) 4
(d) (1 2 z )6
## (f) (1 0.5 x)7
2. In the following expansions, write down the appropriate terms. Simplify your answer as
far as possible .
(a) (1 x)8 x 3 tern
(b) (1 y )9 y 5 term
(c) (2 x)10 x 8 term
## (d) (1 2 x)8 x 3 term
(e) (3 2 y )5 y 3 term
## (h) (2a 3b)7 a5b 2 term
1
(i) ( x )8 x 4 term
x
2
(j) ( y )7 y 3 term
y
1
1
(k) ( x 2 )7 x8 term
(l) ( y 2 )9 constant term.
y
x
3. Find the greatest coefficient of a power of x in the following expansion :
(a) (1 x)17
(b) (1 2 x)7
(c) (1 3 x)8
(d) (2 3x)9
## 4. Expand and simplify (1 3)6 (1 3)6 ,without using a calculator.
5. Expand (1 x)12 up to the term in x 3 .Hence find approximations for :
(a) 1.0212
(b) 0.9612
6. Use the binomial expansion up to the x 3 term to find approximations for :
(a) 1.018
(b) 0.957
(c) 2.019
(d) 2.965
7. The first two terms of the expansion of (2 ax)n are 1024 15360x .Find a and n .
b
8. The constant term in the expansion of (ax )10 is 8064 .Find a in terms of b .
x
2
9. There is no x term the expansion of (1 ax)6 (2 bx)7 .Find the ratio of a to b .
## 10. Use the expansion of (1 y )10 to find 1 x x
2 10
, up to the term in x 2 .
## 11. Expand the following, up to the term indicated:
(a) (1 x x 2 )7 x 2 term
(b) (1 x 2 )8 x 6 term
(c) (1 2 x 3x 2 )6 x 2 term
(d) (2 x 2 x 2 )7 x 2 term
## 3.2.Binomial theorem for other powers
If n th is not a positive integer, the binomial theorem is still true , with certain restrictions
.Provided that 1 x 1 :
## n(n 1) x3 n(n 1)(n 2) x3
2!
3!
In general, this series continues to infinity .So if only the first few terms are take then the
result is only an approximation.
3.2.1 Examples
(1 x)n 1 nx
## 1. Find an expansion for 1 x , up to the term in x 2 .Hence find an approximation for 17
.
Solution
Write 1 x as (1 x)0.5 .Putting n 0.5 in the binomial expansion:
0.5(0.5 1) 2
x
2!
1 0.5 x 0.125 x 2
(1 x)0.5 1 0.5 x
## Write 17 as (16 1)0.5 4(1 0.0625)0.5
Put x 0.0625 in the expansion above :
## 17 4[1 (0.5)(0.0625) (0.125)(0.0625)2 ]
4.1230 .
1 3x
1. Expand
up to the terms in x 2 . For which values of x is the expansion valid?
1 2x
Solution
1 3x
Write
as (1 3x)(1 2 x) 1
1 2x
1x 2 x(2 x) 2
(1 2 x) 1 1 1x(2 x)
2!
2
1 2x 4x
1 3x
(1 3x)(1 2 x 4 x 2 ) .
1 2x
Expand this ignoring the term in x 3 :
1 3x
1 5 x 10 x 2
1 2x
The expansion is valid if 1 2 x 1 .
## The expansion is valid for 0.5 x 0.5
3.2.2 Exercises
1. Expand the following up to the term x 2 .
1
(b) (1 3 x)
(a) (1 x) 4
(c) (1 x )
3
2
(d) (1 2 x)
4
3
1
1 2x
1
3
(g)
(h)
2
1 x
1 x2
2. For which values of x are the expansions in Question 1 valid?
3. Use the expansion up to x 2 to find approximations for:
(e) (1 x)2
(f)
1
2
(a) 1.01
(c) 0.98
(b) 1.05
1
2
(d) 0.96
## 4. Without using a calculator or table , use the binomial expansion up to x 3 to find an
approximation for
1.01 0.99 .
1
2
## 5. Using the binomial expansion up to x of (1 x) , and writing 50 as 49 1 , find an
3
approximation for 50 .
6. Use the method of Question 5 to find approximations for :
(a) 101
(b) 98
1
3
(c) 28
(d) 63
2
7. Expand the following up to term in x :
1
3
1 x2
(b)
1 2x
1 x
(a)
1 x
1 2 x x2
(c)
1 3x
(d)
1 x
1 x2
## (1 x) (1 x) up to the tern in x 3 is of the form a bx 2
.Find a and b .
9.
1 ax bx 2
has no tern in either x or x 2 .Find a and b .
(1 x) 2
## 10. Find the expansion of
1
(1 x) , up to the x 2 tern. By putting x ,show that
8
800
.
253
3.3.Examination questions
1. Find the first three terms in the expansion in ascending powers of x , of
(i) (1 3x)5
(ii) (2 x) 4
approximately
7 is
## Hence find the coefficient of x 2 in the expansion of (1 3x)5 (2 x) 4 .
2. (i) Write down the expansion of (1 x)5 .Deduce that the first three terms in the expansion
of (1 3x)(1 x)5 are 1 8 x 25 x 2 and find the term in x 3 .
1
by taking the first three terms
100
of the expansion. Show that the inclusion of t he term in x 3 does not affect the
approximation when working to 4 decimal places.
3. Obtain the expansion in ascending powers of x of (1 x)8 .Hence calculate the value of
## (0.95)8 correct to three decimal places.
When a ball is dropped on to level ground it always rebounds to a height of 0.95 times the
height from, which it fell. If it is dropped from a height of 10 metres, calculate, correct to
three significant figures, the height to which it rebounds after eight bounces.
4. (a) Evaluate the tern which is independent of x in the expansion of
1
( x )8
x
(b) Expand (1 x)2 (1 2 x) in ascending powers of x as far the term in x 3 . For which
values of x is this expansion valid?
1 px
5. The first three terms in ascending power of
and (1 x)10 are identical. Find the
1 qx
values of p and q , assuming that x is sufficiently small for both expansions to be valid.
1
## Us e this result to that the tenth root of 33 is approximately
651 2
.
649
1
in partial fractions .Hence or otherwise, given that x is
(1 x)(2 x)
sufficiently small for powers of x above the second to be neglected, show that
1
1
(4 6 x 7 x 2 )
(1 x)(2 x) 8
(b) Show that the first three non-zero terns in the series expansion of
6. (a) Express
1 n
1
n
in ascending powers of
1
are :
n
2
1 11
1
n 2 n
4
1
and find the term in .
n
By giving n a suitable value, use the first four non-zero terms of the series to find a value for
1
(0.9)10
## giving your answer to five decimal places.
1
4
7. Use the binomial series to expand (1 x ) in ascending powers of x as far as the term in
x3 .
1
in the expression and in the series, calculate the value of 51/4 correct to
81
five decimal places, explaining why your working ensures this degree of accuracy.
Common errors
1. Positive integer powers
(a) When writing out the expansion of (a b) n do nor forget the binomial coefficients.
By putting x
(b) When expanding something like (a 2b)5 , do not forget to take powers of 2 as of b .
For example , the third term is 10a3 (2b)2 40a3b 2 20a3b 2 .
(c) Similarly, be careful of negative numbers. If you are expanding (a 2b)5 , then the
terms will be alternately positive and negative.
2. Other powers
If n is not a positive integer, then the binomial theorem only holds for expansions of
the form (1 x) n , where 1 x 1 .Do not try to expand
5 as (1 4)1/2 ,using x 4 .
1/2
Similarly, the first term inside the bracket must be 1 .Do not write
1 1
5 as 4
2 2
## and attempt to expand .
1
5 is to write it as (4 1)1/2 2(1 )1/2 . This can now be
4
expand using the Binomial theorem.
## The correct way to find
Solution (Exercise)
3.1.2
1.
(a) 32 80b 80b2 40b3 10b4 b5
(b) 1 12b 54b2 108b3 82b4
(c) 16 x 4 96 x3 y 216 x 2 y 2 216 xy 3 81y 4
(d) 1 12 z 60 z 2 160 z 3 240 z 4 192 z 5 64 z 6
(e) 128a 7 448a 6b 672a5b2 560a 4b3 280a3b4 84a 2b5 14ab6 b7
1
5
7
1 7
(f) 128 244 x 168 x 2 70 x3 17 x 4 2 x5 x 6
x
2
8
32
128
2.
(a) 56x 3
(b) 126 y 5
(c) 180x8
(d) 448x 3
(e) 720 y 3
(f) 11250x 2
(g) 4368x5 y11
(h) 6048a5b2
(i) 28x 4
(j) 672 y 3
(k) 21x8
(l) 84
3.
(a) 24310
(b) 672
(c) 20412
(d) 489888
4. 416
5. 1 12 x 66 x 2 220 x3
(a) 1.26816
(b) 0.61152
6.
(a) 1.083
(b) 0.698
(c) 535.5
(d) 227.2
7. a 3, n 10
8.
9.
2
b
224 : 5
10. 1 10 x 55 x 2
11.
(a) 1 7 x 28 x 2
(b) 1 8 x 2 28 x 4 56 x 6
(c) 1 12 x 78 x 2
(d) 128 488 x 1568 x 2
(e) 1 8 x 28 x 2
(f) 1 9 x 30 x 2
(g) 1 8 x 29 x 2 64 x3
(h) 1 11x 56x 2 175x3
3.2.2
1.
(a) 1
x 3x 2
4 32
3x 9 x 2
2
8
3x 3x 2
(c) 1
2
8
(b) 1
8x 8x2
3
9
1 2 x 3x 2
1 2 x 4 x2
1 x2
3 3x 2
(d) 1
(e)
(f)
(g)
(h)
2. a, c, e, g , h, 1 x 1
3.
(a) 1.00498
(b) 1.02468
(c) 0.98995
(d) 0.9798
4. 1.999975
5. 7.07106
6.
(a) 10.0499
(b) 9.8995
(c) 3.0366
(d) 3.97906
7.
(a) 1 2 x 2 x 2
(b) 1 2 x 5 x 2
(c) 1 x 2 x 2
(d) 1 x x 2
1
8. a 2, b
4
9. a 2, b 1
x x2
10. 1
3 9
1
1
b, x
3
3
1
1
d, f , x
2
2
x x2
11. 1
2 8
===========================================================
References:
Solomon, R.C. (1997), A Level: Mathematics (4th Edition) , Great Britain, Hillman
Printers(Frome) Ltd. |
## Algebra 2 (1st Edition)
$\left\{\begin{array}{l} a_{1}=15,\\ a_{n}=a_{n-1}-4 \end{array}\right.$
$a_{1}=15$ We obtain $a_{2}=11$ from $a_{1}=15$ by adding $(-4)$ to it.$\quad a_{2}=a_{1}-4$ We obtain $a_{3}=7$ from $a_{2}=11$ by adding $(-4)$ to it.$\quad a_{3}=a_{2}-4$ We obtain $a_{4}=3$ from $a_{3}=7$ by adding $(-4)$ to it.$\quad a_{4}=a_{3}-4$ We obtain $a_{5}=-1$ from $a_{4}=3$ by adding $(-4)$ to it.$\quad a_{5}=a_{4}-4$ We obtain $a_{6}=-5$ from $a_{4}=3-1$ by adding $(-4)$ to it.$\quad a_{5}=a_{4}-4$ $...$ We obtain $a_{n}$ from $a_{n-1}$ by adding $(-4)$ to it.$\quad a_{n}=a_{n-1}-4$ A recursive rule for the sequence: $\left\{\begin{array}{l} a_{1}=15,\\ a_{n}=a_{n-1}-4 \end{array}\right.$ |
# NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.4
Go back to 'Continuity and Differentiability'
## Chapter 5 Ex.5.4 Question 1
Differentiating the following wrt x: $$\frac{{{e^x}}}{{\sin x}}$$
### Solution
Let $$y = \frac{{{e^x}}}{{\sin x}}$$
By using the quotient rule, we get
\begin{align}\frac{{dy}}{{dx}} &= \frac{{\sin x\frac{d}{{dx}}\left( {{e^x}} \right) - {e^x}\frac{d}{{dx}}\left( {\sin x} \right)}}{{{{\sin }^2}x}}\\& = \frac{{\sin x.\left( {{e^x}} \right) - {e^x}.\left( {\cos x} \right)}}{{{{\sin }^2}x}}\\ &= \frac{{{e^x}\left( {\sin x - \cos x} \right)}}{{{{\sin }^2}x}}\end{align}
## Chapter 5 Ex.5.4 Question 2
Differentiating the following $${e^{{{\sin }^{ - 1}}x}}$$
### Solution
Let $$y = {e^{{{\sin }^{ - 1}}x}}$$
By using the quotient rule, we get
\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{e^{{{\sin }^{ - 1}}x}}} \right)\\ &= {e^{{{\sin }^{ - 1}}x}}.\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right)\\ &= {e^{{{\sin }^{ - 1}}x}}.\frac{1}{{\sqrt {1 - {x^2}} }}\\ &= \frac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}\\ &= \frac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }},x \in \left( { - 1,1} \right)\end{align}
## Chapter 5 Ex.5.4 Question 3
Differentiating the following wrt $$x:{e^{{x^3}}}$$
### Solution
Let $$y = {e^{{x^3}}}$$
By using the quotient rule, we get
\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left( {{e^{{x^3}}}} \right)\\ &= {e^{{x^3}}}.\frac{d}{{dx}}\left( {{x^3}} \right)\\& = {e^{{x^3}}}.3{x^2}\\ &= 3{x^2}{e^{{x^3}}}\end{align}
## Chapter 5 Ex.5.4 Question 4
Differentiate the following wrt $$x:\sin \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)$$
### Solution
Let $$y = \sin \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)$$
By using the chain rule, we get
\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left[ {\sin \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)} \right]\\ &= \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right).\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\\& = \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right).\frac{1}{{1 + {{\left( {{e^{ - x}}} \right)}^2}}}.\frac{d}{{dx}}\left( {{e^{ - x}}} \right)\\& = \frac{{\cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)}}{{1 + {e^{ - 2x}}}}.{e^{ - x}}.\frac{d}{{dx}}\left( { - x} \right)\\ &= \frac{{{e^{ - x}}\cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)}}{{1 + {e^{ - 2x}}}} \times \left( { - 1} \right)\\ &= \frac{{ - {e^{ - x}}\cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)}}{{1 + {e^{ - 2x}}}}\end{align}
## Chapter 5 Ex.5.4 Question 5
Differentiate the following wrt $$x:\log \left( {\cos {e^x}} \right)$$
### Solution
Let $$y = \log \left( {\cos {e^x}} \right)$$
By using the chain rule, we get
\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left[ {\log \left( {\cos {e^x}} \right)} \right]\\&= \frac{1}{{\cos {e^x}}}.\frac{d}{{dx}}\left( {\cos {e^x}} \right)\\ &= \frac{1}{{\cos {e^x}}}.\left( { - \sin {e^x}} \right).\frac{d}{{dx}}\left( {{e^x}} \right)\\ &= \frac{{ - \sin {e^x}}}{{\cos {e^x}}}.{e^x}\\& = - {e^x}\tan {e^x},{e^x} \ne \left( {2n + 1} \right)\frac{\pi }{2},n \in {\bf{N}}\end{align}
## Chapter 5 Ex.5.4 Question 6
Differentiate the following wrt $$x:{e^x} + {e^{{x^2}}} + \ldots + {e^{{x^5}}}$$
### Solution
$$\frac{d}{{dx}}\left( {{e^x} + {e^{{x^2}}} + \ldots + {e^{{x^5}}}} \right)$$
Differentiating wrt x, we get
\begin{align}\frac{d}{{dx}}\left( {{e^x} + {e^{{x^2}}} + \ldots + {e^{{x^5}}}} \right) &= \frac{d}{{dx}}\left( {{e^x}} \right) + \frac{d}{{dx}}\left( {{e^{{x^2}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^3}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^4}}}} \right) + \frac{d}{{dx}}\left( {{e^{{x^5}}}} \right)\\ &= {e^x} + \left[ {{e^{{x^2}}} \times \frac{d}{{dx}}\left( {{x^2}} \right)} \right] + \left[ {{e^{{x^3}}} \times \frac{d}{{dx}}\left( {{x^3}} \right)} \right] \\& \qquad + \left[ {{e^{{x^4}}} \times \frac{d}{{dx}}\left( {{x^4}} \right)} \right] + \left[ {{e^{{x^5}}} \times \frac{d}{{dx}}\left( {{x^5}} \right)} \right]\\ &= {e^x} + \left( {{e^{{x^2}}} \times 2x} \right) + \left( {{e^{{x^3}}} \times 3{x^2}} \right) + \left( {{e^{{x^4}}} \times 4{x^3}} \right) + \left( {{e^{{x^5}}} \times 5{x^4}} \right)\\& = {e^x} + 2x{e^{{x^2}}} + 3{x^2}{e^{{x^3}}} + 4{x^3}{e^{{x^4}}} + 5{x^4}{e^{{x^5}}}\end{align}
## Chapter 5 Ex.5.4 Question 7
Differentiating the following wrt $$x:\sqrt {{e^{\sqrt x }}} ,x > 0$$
### Solution
Let $$y = \sqrt {{e^{\sqrt x }}}$$
Then, $${y^2} = {e^{\sqrt x }}$$
Differentiating wrt x, we get
$${y^2} = {e^{\sqrt x }}$$
\begin{align}&\frac{d}{{dx}}\left( {{y^2}} \right) = \frac{d}{{dx}}\left( {{e^{\sqrt x }}} \right)\\& \Rightarrow \;2y\frac{{dy}}{{dx}} = {e^{\sqrt x }}\frac{d}{{dx}}\left( {\sqrt x } \right)\\& \Rightarrow \;2y\frac{{dy}}{{dx}} = {e^{\sqrt x }}\frac{1}{2}.\frac{1}{{\sqrt x }}\\ &\Rightarrow \;\frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4y\sqrt x }}\\& \Rightarrow \;\frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4\sqrt {{e^{\sqrt x }}} \sqrt x }}\\& \Rightarrow \;\frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{4\sqrt {x{e^{\sqrt x }}} }},x > 0\end{align}
## Chapter 5 Ex.5.4 Question 8
Differentiating the following wrt $$x:\log \left( {\log x} \right),x > 1$$
### Solution
Let $$y = \log \left( {\log x} \right)$$
By using the chain rule, we get
\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left[ {\log \left( {\log x} \right)} \right]\\ &= \frac{1}{{\log x}}.\frac{d}{{dx}}\left( {\log x} \right)\\ &= \frac{1}{{\log x}}.\frac{1}{x}\\ &= \frac{1}{{x\log x}},x > 1\end{align}
## Chapter 5 Ex.5.4 Question 9
Differentiating the following wrt $$x:\frac{{\cos x}}{{\log x}},x > 0$$
### Solution
Let $$y = \frac{{\cos x}}{{\log x}}$$
By using the quotient rule, we get
\begin{align}\frac{{dy}}{{dx}}& = \frac{{\frac{d}{{dx}}\left( {\cos x} \right).\log x - \cos x.\frac{d}{{dx}}\left( {\log x} \right)}}{{{{\left( {\log x} \right)}^2}}}\\& = \frac{{ - \sin x\log x - \cos x.\frac{1}{x}}}{{{{\left( {\log x} \right)}^2}}}\\& = - \left[ {\frac{{x\log x.\sin x + \cos x}}{{x{{\left( {\log x} \right)}^2}}}} \right],x > 0\end{align}
## Chapter 5 Ex.5.4 Question 10
Differentiate the following wrt $$x:\cos \left( {\log x + {e^x}} \right),x > 0$$
### Solution
Let $$y = \cos \left( {\log x + {e^x}} \right)$$
By using the chain rule, we get
\begin{align}\frac{{dy}}{{dx}} &= - \sin \left[ {\log x + {e^x}} \right].\frac{d}{{dx}}\left( {\log x + {e^x}} \right)\\ &= - \sin \left( {\log x + {e^x}} \right).\left[ {\frac{d}{{dx}}\left( {\log x} \right) + \frac{d}{{dx}}\left( {{e^x}} \right)} \right]\\& = - \sin \left( {\log x + {e^x}} \right).\left( {\frac{1}{x} + {e^x}} \right)\\ &= - \left( {\frac{1}{x} + {e^x}} \right)\sin \left( {\log x + {e^x}} \right),x > 0\end{align}
Related Sections
Related Sections |
# MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers
Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Principle of Mathematical Induction Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 4 Principle of Mathematical Induction Objective Questions.
## Principle of Mathematical Induction Class 11 MCQs Questions with Answers
Students are advised to solve the Principle of Mathematical Induction Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Principle of Mathematical Induction Class 11 with answers will boost your confidence thereby helping you score well in the exam.
Explore numerous MCQ Questions of Principle of Mathematical Induction Class 11 with answers provided with detailed solutions by looking below.
Question 1.
For all n∈N, 3n5 + 5n³ + 7n is divisible by
(a) 5
(b) 15
(c) 10
(d) 3
Answer
Answer: (b) 15
Given number = 3n5 + 5n² + 7n
Let n = 1, 2, 3, 4, ……..
3n5 + 5n³ + 7n = 3 × 1² + 5 × 1³ + 7 × 1 = 3 + 5 + 7 = 15
3n5 + 5n³ + 7n = 3 × 25 + 5 × 2³ + 7 × 2 = 3 × 32 + 5 × 8 + 7 × 2 = 96 + 40 + 14 = 150 = 15 × 10
3n5 + 5n³ + 7n = 3 × 35 + 5 × 3³ + 7 × 3 = 3 × 243 + 5 × 27 + 7 × 3 = 729 + 135 + 21 = 885 = 15 × 59
Since, all these numbers are divisible by 15 for n = 1, 2, 3, …..
So, the given number is divisible by 15
Question 2.
{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} =
(a) 1/(n + 1) for all n ∈ N.
(b) 1/(n + 1) for all n ∈ R
(c) n/(n + 1) for all n ∈ N.
(d) n/(n + 1) for all n ∈ R
Answer
Answer: (a) 1/(n + 1) for all n ∈ N.
Let the given statement be P(n). Then,
P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).
When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.
Therefore LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)
Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}]
= [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}]
= [1/(k + 1)] ∙ [(k + 1)/(k + 2)]
= 1/(k + 2)
Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Question 3.
For all n ∈ N, 32n + 7 is divisible by
(a) non of these
(b) 3
(c) 11
(d) 8
Answer
Answer: (d) 8
Given number = 32n + 7
Let n = 1, 2, 3, 4, ……..
32n + 7 = 3² + 7 = 9 + 7 = 16
32n + 7 = 34 + 7 = 81 + 7 = 88
32n + 7 = 36 + 7 = 729 + 7 = 736
Since, all these numbers are divisible by 8 for n = 1, 2, 3, …..
So, the given number is divisible by 8
Question 4.
The sum of the series 1 + 2 + 3 + 4 + 5 + ………..n is
(a) n(n + 1)
(b) (n + 1)/2
(c) n/2
(d) n(n + 1)/2
Answer
Answer: (d) n(n + 1)/2
Given, series is series 1 + 2 + 3 + 4 + 5 + ………..n
Sum = n(n + 1)/2
Question 5.
The sum of the series 1² + 2² + 3² + ……….. n² is
(a) n(n + 1) (2n + 1)
(b) n(n + 1) (2n + 1)/2
(c) n(n + 1) (2n + 1)/3
(d) n(n + 1) (2n + 1)/6
Answer
Answer: (d) n(n + 1) (2n + 1)/6
Given, series is 1² + 2² + 3² + ……….. n²
Sum = n(n + 1)(2n + 1)/6
Question 6.
For all positive integers n, the number n(n² − 1) is divisible by:
(a) 36
(b) 24
(c) 6
(d) 16
Answer
Answer: (c) 6
Given,
number = n(n² − 1)
Let n = 1, 2, 3, 4….
n(n² – 1) = 1(1 – 1) = 0
n(n² – 1) = 2(4 – 1) = 2 × 3 = 6
n(n² – 1) = 3(9 – 1) = 3 × 8 = 24
n(n² – 1) = 4(16 – 1) = 4 × 15 = 60
Since all these numbers are divisible by 6 for n = 1, 2, 3,……..
So, the given number is divisible 6
Question 7.
If n is an odd positive integer, then aⁿ + bⁿ is divisible by :
(a) a² + b²
(b) a + b
(c) a – b
(d) none of these
Answer
Answer: (b) a + b
Given number = aⁿ + bⁿ
Let n = 1, 3, 5, ……..
aⁿ + bⁿ = a + b
aⁿ + bⁿ = a³ + b³ = (a + b) × (a² + b² + ab) and so on.
Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
So, the given number is divisible by (a + b)
Question 8.
n(n + 1) (n + 5) is a multiple of ____ for all n ∈ N
(a) 2
(b) 3
(c) 5
(d) 7
Answer
Answer: (b) 3
Let P(n): n(n + 1)(n + 5) is a multiple of 3.
For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3.
So, the given statement is true for n = 1, i.e. P(1) is true.
Let P(k) be true. Then,
P(k): k(k + 1)(k + 5) is a multiple of 3
⇒ K(k + 1) (k + 5) = 3m for some natural number m, …… (i)
Now, (k + 1) (k + 2) (k + 6) = (k + 1) (k + 2)k + 6(k + 1) (k + 2)
= k(k + 1) (k + 2) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5 – 3) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5) – 3k(k + 1) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5) + 3(k + 1) (k +4) [on simplification]
= 3m + 3(k + 1 ) (k + 4) [using (i)]
= 3[m + (k + 1) (k + 4)], which is a multiple of 3
⇒ P(k + 1) (k + 1 ) (k + 2) (k + 6) is a multiple of 3
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Question 9.
For any natural number n, 7ⁿ – 2ⁿ is divisible by
(a) 3
(b) 4
(c) 5
(d) 7
Answer
Answer: (c) 5
Given, 7ⁿ – 2ⁿ
Let n = 1
7ⁿ – 2ⁿ = 71 – 21 = 7 – 2 = 5
which is divisible by 5
Let n = 2
7ⁿ – 2ⁿ = 72 – 22 = 49 – 4 = 45
which is divisible by 5
Let n = 3
7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5
Question 10.
The sum of the series 1³ + 2³ + 3³ + ………..n³ is
(a) {(n + 1)/2}²
(b) {n/2}²
(c) n(n + 1)/2
(d) {n(n + 1)/2}²
Answer
Answer: (d) {n(n + 1)/2}²
Given, series is 1³ + 2³ + 3³ + ……….. n³
Sum = {n(n + 1)/2}²
Question 11.
(1² + 2² + …… + n²) _____ for all values of n ∈ N
(a) = n³/3
(b) < n³/3
(c) > n³/3
(d) None of these
Answer
Answer: (c) > n³/3
Let P(n): (1² + 2² + ….. + n²) > n³/3.
When = 1, LHS = 1² = 1 and RHS = 1³/3 = 1/3.
Since 1 > 1/3, it follows that P(1) is true.
Let P(k) be true. Then,
P(k): (1² + 2² + ….. + k² ) > k³/3 …. (i)
Now,
1² + 2² + ….. + k²
+ (k + 1)²
= {1² + 2² + ….. + k² + (k + 1)²
> k³/3 + (k + 1)³ [using (i)]
= 1/3 ∙ (k³ + 3 + (k + 1)²) = 1/3 ∙ {k² + 3k² + 6k + 3}
= 1/3[k³ + 1 + 3k(k + 1) + (3k + 2)]
= 1/3 ∙ [(k + 1)³ + (3k + 2)]
> 1/3(k + 1)³
P(k + 1):
1² + 2² + ….. + k² + (k + 1)²
> 1/3 ∙ (k + 1)³
P(k + 1) is true, whenever P(k) is true.
Thus P(1) is true and P(k + 1) is true whenever p(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Question 12.
{1/(3 ∙ 5)} + {1/(5 ∙ 7)} + {1/(7 ∙ 9)} + ……. + 1/{(2n + 1) (2n + 3)} =
(a) n/(2n + 3)
(b) n/{2(2n + 3)}
(c) n/{3(2n + 3)}
(d) n/{4(2n + 3)}
Answer
Answer: (c) n/{3(2n + 3)}
Let the given statement be P(n). Then,
P(n): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + ……. + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3).
Putting n = 1 in the given statement, we get
and LHS = 1/(3 ∙ 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15.
LHS = RHS
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} ….. (i)
Now, 1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}2(k + 1) + 3
= {1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]} + 1/{(2k + 3)(2k + 5)}
= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [using (i)]
= {k(2k + 5) + 3}/{3(2k + 3)(2k + 5)}
= (2k² + 5k + 3)/[3(2k + 3)(2k + 5)]
= {(k + 1)(2k + 3)}/{3(2k + 3)(2k + 5)}
= (k + 1)/{3(2k + 5)}
= (k + 1)/[3{2(k + 1) + 3}]
= P(k + 1): 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}{2(k + 1) + 3}]
= (k + 1)/{3{2(k + 1) + 3}]
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for n ∈ N.
Question 13.
If n is an odd positive integer, then aⁿ + bⁿ is divisible by :
(a) a² + b²
(b) a + b
(c) a – b
(d) none of these
Answer
Answer: (b) a + b
Given number = aⁿ + bⁿ
Let n = 1, 3, 5, ……..
aⁿ + bⁿ = a + b
aⁿ + bⁿ = a³ + b³ = (a + b) × (a² + b² + ab) and so on.
Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
So, the given number is divisible by (a + b)
Question 14.
(2 ∙ 7N + 3 ∙ 5N – 5) is divisible by ……….. for all N ∈ N
(a) 6
(b) 12
(c) 18
(d) 24
Answer
Answer: (d) 24
Let P(n): (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) is divisible by 24.
For n = 1, the given expression becomes (2 ∙ 71 + 3 ∙ 51 – 5) = 24, which is clearly divisible by 24.
So, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) is divisible by 24.
⇒ (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) = 24m, for m = N
Now, (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5)
= (2 ∙ 7k ∙ 7 + 3 ∙ 5k ∙ 5 – 5)
= 7(2 ∙ 7k + 3 ∙ 5k – 5) = 6 ∙ 5k + 30
= (7 × 24m) – 6(5k – 5)
= (24 × 7m) – 6 × 4p, where (5k – 5) = 5(5k-1 – 1) = 4p
[Since (5k-1 – 1) is divisible by (5 – 1)]
= 24 × (7m – p)
= 24r, where r = (7m – p) ∈ N
⇒ P (k + 1): (2 ∙ 7k + 13 ∙ 5k + 1 – 5) is divisible by 24.
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Question 15.
For all n∈N, 52n − 1 is divisible by
(a) 26
(b) 24
(c) 11
(d) 25
Answer
Answer: (b) 24
Given number = 52n − 1
Let n = 1, 2, 3, 4, ……..
52n − 1 = 5² − 1 = 25 – 1 = 24
52n − 1 = 54 – 1 = 625 – 1 = 624 = 24 × 26
52n − 1 = 56 – 1 = 15625 – 1 = 15624 = 651 × 24
Since, all these numbers are divisible by 24 for n = 1, 2, 3, …..
So, the given number is divisible by 24
Question 16.
1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) =
(a) n(n + 1)(n + 2)
(b) {n(n + 1)(n + 2)}/2
(c) {n(n + 1)(n + 2)}/3
(d) {n(n + 1)(n + 2)}/4
Answer
Answer: (c) {n(n + 1)(n + 2)}/3
Let the given statement be P(n). Then,
P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) = (1/3){n(n + 1) (n + 2)}
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + k(k + 1) = (1/3){k(k + 1) (k + 2)}.
Now, 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +…+ k(k + 1) + (k + 1) (k + 2)
= (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ……. + k(k + 1)) + (k + 1) (k + 2)
= (1/3) k(k + 1) (k + 2) + (k + 1)(k + 2) [using (i)]
= (1/3) [k(k + 1) (k + 2) + 3(k + 1)(k + 2)
= (1/3){(k + 1) (k + 2)(k + 3)}
⇒ P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +……+ (k + 1) (k + 2)
= (1/3){k + 1 )(k + 2) (k +3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all values of ∈ N.
Question 17.
1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)} =
(a) {n(n + 3)}/{4(n + 1)(n + 2)}
(b) (n + 3)/{4(n + 1)(n + 2)}
(c) n/{4(n + 1)(n + 2)}
(d) None of these
Answer
Answer: (a) {n(n + 3)}/{4(n + 1)(n + 2)}
Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)}
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.
Therefore LHS = RHS.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……… + 1/{k(k + 1) (k + 2)} = {k(k + 3)}/{4(k + 1) (k + 2)}. …….(i)
Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1) (k + 2)} + 1/{(k + 1) (k + 2) (k + 3)}
= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1) (k + 2}] + 1/{(k + 1)(k + 2) (k + 3)}
= [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)]
= {k(k + 3)² + 4}/{4(k + 1)(k + 2) (k + 3)}
= (k³ + 6k² + 9k + 4)/{4(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 1) (k + 4)}/{4 (k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 4)}/{4(k + 2) (k + 3)
⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 2)}/{4(k + 2) (k + 3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Question 18.
For any natural number n, 7ⁿ – 2ⁿ is divisible by
(a) 3
(b) 4
(c) 5
(d) 7
Answer
Answer: (c) 5
Given, 7ⁿ – 2ⁿ
Let n = 1
7ⁿ – 2ⁿ = 71 – 21 = 7 – 2 = 5
which is divisible by 5
Let n = 2
7ⁿ – 2ⁿ = 7² – 2² = 49 – 4 = 45
which is divisible by 5
Let n = 3
7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5
Question 19.
The sum of n terms of the series 1² + 3² + 5² +……… is
(a) n(4n² – 1)/3
(b) n²(2n² + 1)/6
(c) none of these.
(d) n²(n² + 1)/3
Answer
Answer: (a) n(4n² – 1)/3
Let S = 1² + 3² + 5² +………(2n – 1)²
⇒ S = {1² + 2² + 3² + 4² ………(2n – 1)² + (2n)²} – {2² + 4² + 6² +………+ (2n)²}
⇒ S = {2n × (2n + 1) × (4n + 1)}/6 – {4n × (n + 1) × (2n + 1)}/6
⇒ S = n(4n² – 1)/3
Question 20.
For all n ∈ N, 3n5 + 5n³ + 7n is divisible by:
(a) 5
(b) 15
(c) 10
(d) 3
Answer
Answer: (b) 15
Given number = 3n5 + 5n³ + 7n
Let n = 1, 2, 3, 4, ……..
3n5 + 5n³ + 7n = 3 × 1² + 5 × 1³ + 7 × 1 = 3 + 5 + 7 = 15
3n5 + 5n³ + 7n = 3 × 25 + 5 × 2³ + 7 × 2 = 3 × 32 + 5 × 8 + 7 × 2 = 96 + 40 + 14 = 150 = 15 × 10
3n5 + 5n³ + 7n = 3 × 35 + 5 × 3³ + 7 × 3 = 3 × 243 + 5 × 27 + 7 × 3 = 729 + 135 + 21 = 885 = 15 × 59
Since, all these numbers are divisible by 15 for n = 1, 2, 3, …..
So, the given number is divisible by 15
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# Bihar Board 12th Maths Important Questions Long Answer Type Part 6
BSEB Bihar Board 12th Maths Important Questions Long Answer Type Part 6 are the best resource for students which helps in revision.
## Bihar Board 12th Maths Important Questions Long Answer Type Part 6
Vector Algebra
Question 1.
For any three vectors a, b and c
$$(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$$
Solution:
Let the vectors $$(\vec{a}$$, $$(\vec{b}$$ and $$(\vec{c}$$ be represented by $$\overrightarrow{P Q}, \overrightarrow{Q R}$$ and $$\overrightarrow{RS}$$ respectively
The associative property of vector addition enables us to write the sum of three vectors $$\vec{a}, \vec{b}, \vec{c}$$ as $$\vec{a}+\vec{b}+\vec{c}$$ without using brackets.
Three Dimensional Geometry
Question 1.
Find the angle .between the pair of tines given by s
and $$\vec{r}=\dot{3} \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}-2 \hat{k})$$ and $$\vec{r}=5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}-6 \hat{k})$$
Solution:
Here $$\overrightarrow{b_{1}}=\hat{i}-2 \hat{j}+2 \hat{k}$$ and $$\overrightarrow{b_{2}}=3 \hat{i}+2 \hat{j}+6 \hat{k}$$
The angle θ between the two lines is given by
Hence θ = cos-1($$\frac{19}{21}$$)
Question 2.
Find the vector equation of the plane passing through the intersection of the planes $$\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6 \text { d } \vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5$$ and the point (1,1,1).
Solution:
Here $$\overrightarrow{n_{1}}=\hat{i}+\hat{j}+\hat{k}$$ and $$\overrightarrow{n_{2}}=2 \hat{i}+3 \hat{j}+4 \hat{k}$$
and d1 = 6 and d2 = -5
Hence using the relation $$\vec{r} \cdot\left(\vec{n}_{1}+\lambda \vec{n}_{2}\right)=d_{1}+\lambda d_{2}$$
We get $$\vec{r} \cdot[\hat{i}+\hat{j}+\hat{k}+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})]=6-5 \lambda$$
or, $$\vec{r} \cdot[(1+2 \lambda) \hat{i}+(1+3 \lambda) \hat{j}+(1+4 \lambda) \hat{k}]=6-5 \lambda$$………………(i)
where, λ is some real number.
Takipg $$\vec{r} \cdot x \hat{i}+y \hat{j}+2 \hat{\hat{k}}$$,we get
(x$$\hat{i}$$ + y$$\hat{j}$$ + z$$\hat{k}$$)-[(1+2^)$$\hat{i}$$ +(1 + 3A.)$$\hat{j}$$ + (l + 4A.)$$\hat{k}$$] = 6 – 5λ
or, (1 + 2λ)x + (1 + 3λ)y + (1 + 4λ)z = 6 – 5λ
or, (x + y + z – 6) + λ(2x + 3y + 4z + 5) = 0 ………………………(ii)
Given that the plane passes through the point (1,1, 1) it must satisfy (2).
i.e. (1+ 1+ 1-6) + λ(2+3+4+5) = 0
or λ = $$\frac{3}{14}$$
Putting the value of λ in (1), we get
which is the required vector equation of the plane.
Question 3.
Find the shortest distance between the l1 and l2 whose vector equation are
$$\vec{r}=\hat{i}+\hat{j}+\lambda(2 \hat{i}+\hat{j}+\hat{k})$$
and $$\overrightarrow{\boldsymbol{r}}=\mathbf{2} \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})$$
Solution:
Comparing (i) and (ii) with $$\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$$ and $$\vec{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$$ respectively.
We get $$\overrightarrow{a_{1}}$$ = $$\hat{i}$$ + $$\hat{j}$$ $$\overrightarrow{b_{1}}$$ = 2$$\hat{i}$$ – $$\hat{j}$$ + $$\hat{k}$$
$$\overrightarrow{a_{2}}$$ =2$$\hat{i}$$ + $$\hat{j}$$ – $$\hat{k}$$ and & $$\overrightarrow{b_{2}}$$ = 3$$\hat{i}$$ – 5$$\hat{j}$$ + 2$$\hat{k}$$
and $$\overrightarrow{b_{2}} \times \overrightarrow{b_{2}}=(2 \hat{i}-\hat{j}+\hat{k}) \times(3 \hat{i}-5 \hat{j}+2 \hat{k})$$
Therefore, $$\overrightarrow{a_{2}}-\overrightarrow{a_{1}}=\hat{i}-\hat{k}$$
$$\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{array}\right|$$ = 3$$\hat{i}$$ – $$\hat{j}$$ – 7$$\hat{k}$$
$$\left|\overrightarrow{b_{1}} \times \vec{b}_{2}\right|=\sqrt{9+1+49}=\sqrt{59}$$
Hence the shortest distance between the given lines is given by
Question 4.
Find the vector and the cartesian equations of the Hoe through the point (5, 2, -4) and which Is parallel to the vectors $$3 \hat{i}+2 \hat{j}-8 \hat{k}$$
Solution:
We have
$$\overrightarrow{a}$$ = 5$$\hat{i}$$ + 2$$\hat{j}$$ – 4$$\hat{k}$$ and $$\overrightarrow{b}$$ = 3$$\hat{i}$$ +2$$\hat{j}$$ – 8$$\hat{k}$$
Therefore, the vector of the line is ‘
$$\overrightarrow{r}$$ = 5$$\hat{i}$$+2$$\hat{j}$$-4$$\hat{k}$$ + λ(3$$\hat{i}$$+2$$\hat{j}$$-8$$\hat{k}$$)
Now, $$\overrightarrow{r}$$ is the position vector of any point p (x,y, z) on the line.
Therefore x$$\hat{i}$$ + y$$\hat{j}$$ + 2$$\hat{k}$$ = 5$$\hat{i}$$ + 2$$\hat{j}$$ -4$$\hat{k}$$ + 2(3$$\hat{i}$$ + 2 $$\hat{j}$$– 8$$\hat{k}$$)
= (5 + 3λ)$$\hat{i}$$ + (2 + 2λ)$$\hat{j}$$ + (-4 – 8λ)$$\hat{k}$$
Eliminating λ, we get
$$\frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8}$$
which is the equation of the line is cartesian form.
Question 5.
Find the angle between the two planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7 using vector method.
Solution:
The angle between two planes is the angle between their normals. From the equation of the planes the normal vectors are
$$\vec{N}_{1}$$ = 2$$\hat{i}$$ + $$\hat{j}$$ – 2$$\hat{k}$$ and $$\vec{N}_{2}$$ = 3$$\hat{i}$$ – 6$$\hat{j}$$ – 2$$\hat{k}$$
Hence θ = cos-1( $$\frac{4}{21}$$ )
Question 6.
Find the angle between the line
$$\frac{x^{\circ}+1}{2}=\frac{y}{3}=\frac{z-3}{6}$$ and the plane 10x + 2y – 112 = 3
Solution:
Let θ be the angle between the line and the normal to the plane .
Converting the given equations into vector form we have
$$\vec{r}$$ = (-$$\hat{i}$$ + 3$$\hat{k}$$ + (2$$\hat{i}$$ + 3$$\hat{j}$$ – 6$$\hat{k}$$ )
and $$\vec{r}$$ = (10$$\hat{i}$$ + 2$$\hat{j}$$ + (2$$\hat{i}$$ – 11$$\hat{k}$$ ) = 3
Here $$\vec{b}$$ = (2$$\hat{i}$$ + 3$$\hat{j}$$ – 6$$\hat{k}$$ and $$\vec{n} 10 \hat{i}+2 \hat{j}-11 \hat{k}$$
im
Question 7.
Find the equation of the plane that contains the point (1, -1,2) and is perpendicular to each of tfie planes 2x + 3y – 2z – 5 and x + 2y-3z = 8.
Solution:
The equation of the plane containing the given point is
A(x-1) + B(y+1) + C(z-2) = 0 ….(i)
Applying the condition of perpendicularly to the plane given in eqn. (i) with the planes
2x+3y -2z = 5 and x + 2y – 3z = 6 we have
2A + 3B-2C= 10 and A + 2B – 3C = 0
Solving these equations. We find A =-5C and B = 4C.
Hence the required equation is
– 5C (x – 1) + 4C (y+ 1) + C(z – 2) = 0
i.e. 5x – 4y – z = 7
Question 8.
Find the coordinates of the point where the line through the point A (3,4,1) and B (5,1,6) crosses the xy-plane.
Solution:
The vector equation of the line through the points A and B is
$$\vec{r}=3 \hat{i}+4 \hat{j}+\hat{k}+\lambda[(5-3) \hat{i}+(1-4) \hat{j}+(6-1) \hat{k}]$$
i.e. $$\overrightarrow{\boldsymbol{r}}=3 \hat{i}+4 \hat{j}+\hat{k}+\lambda(2 \hat{i}-3 \hat{j}+5 \hat{k})$$ ….(i)
Let P be the point where the line AB crosses the xy-plane. Then the position vector of the point P is of the form x$$\hat{i}$$ + y $$\hat{i}$$
This point must satisfy the eqn. (i)
i.e. x$$\hat{i}$$ + y$$\hat{j}$$ = (3 + 2λ)$$\hat{i}$$ + (4 – 3λ)$$\hat{j}$$ + (1 + 5λ)$$\hat{k}$$
Equating the like coefficient of $$\hat{i}$$ + $$\hat{j}$$ and $$\hat{k}$$.
We have
x = 3 + 2λ
y = 4 – 3λ
0 = 1 + 5λ
Solving the above equations, we get
x = $$\frac{13}{5}$$ and y = $$\frac{23}{5}$$
(13 23 ‘t
Hence the coordinates of the required point are ($$\frac{13}{5}$$, $$\frac{23}{5}$$, 0)
Question 9.
Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin.
Solution:
Since the direction ratio of the normal to the plane are 2, -3,4 the direction cosines of it are
im
Hence dividing the equation 2x – 3y + 4z – 6 = 0.
2x – 3y + 4s = 6 throughout by $$\sqrt{29}$$ we get
$$\frac{2}{\sqrt{29}} x+\frac{-3}{\sqrt{29}} y+\frac{4}{\sqrt{29}} z=\frac{6}{\sqrt{29}}$$
This is of the form fct +my + nz-d. Where d is the distance of the plane from the origin. So the distance of the plane from the orign is $$\frac{6}{\sqrt{29}}$$
Question 10.
A line makes angle α, β, γ and δ with the digonals of a cube, pro e that
cos2 α + cos2 β + cos2 γ + cos2 δ = $$\frac{4}{3}$$
Solution:
A cube is a rectangular parallelopiped having equal length, breadth and height.
Let OADBFEGC be the cube with each side of length a units.
The four diagonals are OE, AF, BG and CD.
The direction cosines of the diagonal OE which is the line joining two point O and E are
Similarly, the direction cosines of AF, BG and CD are.
Let, m, n be the direction cosines of the given line which makes angles α, β, γ, δ with OE, AF, BG, CD respectively.
Then cos α = $$\frac{1}{\sqrt{3}}$$(l + m + n)
cos β = $$\frac{1}{\sqrt{3}}$$(-l + m + n)
cos γ = $$\frac{1}{\sqrt{3}}$$(l – m + n)
c0s δ = $$\frac{1}{\sqrt{3}}$$(l + m-n)
cos2 α + cos2 β + cos2 γ + cos2 δ
= $$\frac{1}{3}$$ [(l + m +n)2 +(-l + m + n)2 + (l – m + n)2 +(l + m – n)2]
= $$\frac{1}{3}$$[4(l2 + m2 + n2)] = $$\frac{4}{3}$$(asl2 + m2 + n2 = 1)
Linear Programming
Question 1.
Minimize
Z = -3x + 4y Subject to constrants ’ x +2y ≤ 8,3x + 2y ≤ 12 ,x ≥ 0,y ≥ 0.
Solution:
x + 2y = 8 ⇒ $$\frac{x}{8}+\frac{y}{4}$$ = 1
3x + 2y = 12 ⇒ $$\frac{x}{4}+\frac{y}{6}$$ = 1
∴ the feasible region is
OCPBO
The coordinates of. P can be determined by solving x + 2y = 8, 3x +2y =12.
∴ coordinates of P = (2, 3)
Z at O (0,0) = – 3x + 4y = 0
Z at C (4,0) = – 3 x 4 = -12
Z at P (2, 3) = – 3 x 2 + 4 x 3 = 6
ZatB (0,4) = 4 x 4= 16
Hence the minimum value of Zis -12 at the point C (4,0).
Question 2.
Solve the following linear programming problem graphically :
Maximize Z = 4x + y…………. (i)
( Subject to the constraints
x + y ≤ 50 ….(ii)
3x + y ≤ 90 ……….(iii)
x ≥ 0, y ≥ 0 …….(iv)
Solution:
The shaded region in fig. is the feasible region determined by the system of constraints (ii) to (iv). We observe that the feasible region OABC is bounded. So we now use comer point method to determine the maximum value of Z. The coordinates of the comer points O, A, B and C are (0,0), (30,0), (20, 30) and respectively.
Now we evaluate Z at each corner point
table
Hence, maximum value of Z is 120 at the point (30, 0)
Question 3.
A Dietician has to develop a special diet using two foods p and q Each packet (containing 30 g) of food P contains 12 unit of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units c f chol e s te r [‘& 3 units of’ – min A, The d let requires at least 240 units of calcium, atleast 460 units of iron and atriiost 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A In the diet 2 What is the minimum amount of viiamiiin A ?
Solution:
Let x and y be the number of packets of food P and Q respectively. Obviously x ≥ 0, y ≥ 0. Mathematical formulation of the given problem is as follows:
Minimise
Z = 6.r+ 3v (vitamin A)
Subject to the constraints
12x+ 3y ≥ 240 (constraint on calcium),
4x + y ≥ 80
4x + 20v ≥ 460 (constraint on iron), x + 5v >115
6x + 4y ≥ 3 w (constraint on cholesterol),
3* + 2y ≤ 150 x ≥ 0, y ≥ 0
Let us graph the inequalities (i) to (iv).
The feasible region (shaded) determined by the constraints (i) to (iv) is show in figure, and note that is bounded.
The coordinates of the comer points L, M and N are (2,7,2), (15,20) and (40, 15) respectively. Let us evaluate Z at these points.
table
From the table, we find that Z is minimum at the point (15,20). Hence, the amount of vitamin A under the constraints given in the problem will be minimum, if 15 packets of food P and 20 Packets of food Q are used in the special diet. The minimum amount of vitamin A will be 150 units.
Probability
Question 1.
A die is thrown three times, Events A and B are fined as below:
A : 4 on the third throw
B : 6 on the first and 5 on the Second throw
Find the probability of A given that B has already occurred,
Solution:
The sample space has 216 outcomes.
B ={(6, 5, 1), (6. 5, 2). (6.5. 3), (6,5,4), (6,5,5), (6,5, 6)} and A ∩ B = {(6,5,4)}
Question 2.
An unbased die is thrown twice. Let the event A be odd number on the first throw and B the event odd number on the second throw. Check the independence of the events A and B.
Solution:
If all the 36 elementary events of the experiment are considered to be equally likely, we have
P(A) = $$\frac{18}{36}=\frac{1}{2}$$
P(B) = $$\frac{18}{36}=\frac{1}{2}$$
Also P (A ∩ B) = P (odd number on both throws)
= $$\frac{9}{36}=\frac{1}{4}$$
Now P(A)P(B) = $$\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$$
Clearly P (A n B) = P (A) x P (B)
Thus A and B are independent events.
Question 3.
An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. ‘What is the probability that both drawn balls are black.
Solution:
Let E and F denote respectively the events that first and second ball drawn are black.
We have to find P (E ∩ F) or P (EF).
Now P (E) = P (black ball in first draw) = $$\frac{10}{15}$$
Also given that first ball drawn is black, i.e. event E has occured. Now therefore 9 black balls and five white balls left in the urn. Therefore the probability that the second ball drawn is black, given that the ball in first draw is black, is nothing the conditional probability of F given that E has occured.
i.e. p(F\E) = $$\frac{9}{14}$$
By multiplication rule of probability, we have P(E∩F) = P(E)P(F|E)
= $$\frac{10}{15} \times \frac{9}{14}=\frac{3}{7}$$
Question 4.
Ten cards numbered 1 to: 10 are placed in a box, raised up thoroughly and then one card is drawn randomly. If It is know n that the number OK the drawn card is more than 3. What is the probability that It is an even number ?
Solution:
Let A be the event the number on the card drawn is even and B be the event the number on the card drawn is greater than 3.
We have to find P (A | B).
Now, the sample space of the experiment is δ ={ 1,2, 3,4, 5,6, 7, 8,9, 10}.
A = {2,4, 6, 8,10}, B = {4,5,6, 7, 8,9,10}
Also A∩B = {4,6, 8, 10}
Also P(A) = $$\frac{5}{10}$$, P(B) = $$\frac{7}{10}$$
and P(A∩B) = $$\frac{4}{10}$$
Then P(A/B) = $$\frac{P(A \cap B)}{P(B)}$$
= $$\frac{\frac{4}{10}}{\frac{7}{10}}=\frac{4}{10} \times \frac{10}{7}=\frac{4}{7}$$
Question 5.
Prove that if E and F are independent events, then so are the events E and F.
Solution:
Since E and F are independent, we have
P(E∩F) = P.E. P (F)
From the even diagram in fig. It is clear that E∩F and E∩F’ are mutually exclusive events and also .
E = (E ∩ F) ∪ (E ∩ F)
Therefore P (E) = P(E∩F) +(E∩F)
or, P (E∩F) = P(E) – P (E∩F)
= P (E) – P (E) P (F) (by (1))
= P(E) (1 – P(F))
= P (E) . P (F)
Hence, E and F are indepent.
Question 6.
Three coins are tossed simultaneously. Consider the event E three heads or three, tails*. F^’at least two heads’ and G ’at most two heads’ of the pairs (E, F), (E,G) and (F, G) which are independent ? Which
are dependent ?
Solution:
The same space of the experiment is given by
S= {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT)
Clearly E = {HHH, TTT)
F = {HHH, HHT, HTH, THH)
and G = {HHT, HTH, THH, HTT, THT, TTH, TTT)
Also E ∩ F = {HHH},
E∩G = {TTT},
F∩G = [HHT, HTH, THH},
Therefore P(E) = $$\frac{2}{8}=\frac{1}{4}$$
P(F) = $$\frac{4}{8}=\frac{1}{2}$$
P(G) = $$\frac{7}{8}$$
and P(E∩F) = $$\frac{1}{8}$$
P(E∩G) = $$\frac{1}{8}$$
P(E∩C) = $$\frac{3}{8}$$
Also P(E)P(F) = $$\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}$$
P(E)P(G) = $$\frac{1}{4} \times \frac{7}{8}=\frac{7}{32}$$
and P(F)P(G) = $$\frac{1}{2} \times \frac{7}{8}=\frac{7}{16}$$
Thus P(E∩F) = P (E)P(F)
P (E∩G) ≠ P(E).P(G)
and P (F ∩G) ≠ P(F) .P (G)
Hence, the events (F and F) are independent, and the events (E and G) and (F and G) are dependent.
Question 7.
A person has under taken a construction job. The probabilities are 0 65 that there will be strike 0 80 that construction job will be completed on time if there is no strike, and 0 32 that construction job w ill be completed on time if there is a strike. Determine the probability that the construction job will be completed on time.
Solution:
Let A be the event that the construction job will be completed on time and B be the event that there will be 9 strike. We have to find P (A).
We have
P (B) = 0.65,
P (no strike) = P(B)= 1 – P(B)
= 1 – 0.65 = 0.35
P (A | B) = 0.32
P(A IB1) = 0.80
Since events B and B’ form a partition of the sample spaces, There fore, by theorem on total probability, we have
P(A) = P(B)P(A|B)+P(B,)P(A|B’)
= 0.65 x 0.32 + 0.35 x 0.8
= 0.208 + 0.28
= 0.488
Thus, the probability that the construction job will be completed in time is 0.488.
We shall now state and prove the Baye s theorem.
Question 8.
Tew eggs are drawn successively with replacement from a lot conjoining 10% defflective eggs. Find the probability that there is at least one deflective eggs.
Solution:
Let X denote the number of defective eggs in the 10 eggs drawn. Since the drawing is done with replacement the trials Bemoullitrials clearly,
X has the binomial distribution with n = 10 and P = $$\frac{10}{100}=\frac{1}{10}$$
Therefore q = 1 – p = $$\frac{9}{10}$$
Now e (at least one defective egg) = P (X ≥ 1) = 1 – P(X = 0)
= 1 – 10c0 ( $$\frac{9}{10}$$ )10
= 1 – $$\frac{9^{10}}{10^{10}}$$ |
# NCERT Class 10 Maths Arithmetic Progressions
In the introduction part of the chapter 5, the concept of Arithmetic Progressions is explained - patterns in which succeeding terms are obtained by adding a fixed number to the preceding terms. Thus, an arithmetic progression is defined as a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. Further, it is noted that the fixed number is called as Common Difference and it can be positive, negative or zero. Furthermore, the general form of an AP is discussed and infinite and finite forms of AP are explained. Thereafter, for the second exercise of the chapter, the formula for finding the nth term of an AP is derived. Finally, the formula for finding the sum of the first n terms of an AP is derived.
## Exercise 5.4
Download FREE PDF of Chapter-5 Arithmetic Progressions
## Chapter 5 Ex.5.1 Question 1
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each $$\rm{}km,$$ when the fare is $$\rm{Rs.} \,15$$ for the first km and $$\rm{Rs.} \,8$$ for each of additional $$\rm{}km.$$
(ii) The amount of air present in a cylinder when a vacuum pump removes \begin{align}\frac{1}{4}\end{align} of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every meter of digging, when it costs $$\rm{Rs}\,150$$ for the first meter and rises by $$\rm{Rs.}\, 50$$ for each subsequent meter.
(iv) The amount of money in the account every year, when $$\rm{Rs.} \,10000$$ is deposited at compound interest at $$8\%$$ per annum.
### Solution
(i)
What is Known?
Charges for the first km and additional $$\rm{km.}$$
What is Unknown?
Whether it is an arithmetic progression or not ?
Reasoning:
An arithmetic progression is a sequence of number in which each term is obtained by adding a fixed number to the preceding term except the first term.
General form of an arithmetic progression is \begin{align}a, \,(a+d),\,( a+2d),\, (a+3d),\end{align} Where
$$a$$ is the first term and $$d$$ is the common difference.
Steps:
Taxi fare for $$1 \,\rm{km}$$ \begin{align}= {\rm{Rs.}}\,15\left( {{a_1}} \right)\end{align}
Taxi fare for $$2 \,\rm{km}$$ \begin{align}= 15+8 = {\rm{Rs}}. 23\left( {{a_2}} \right)\end{align}
Taxi fare for $$3\, \rm{km}$$
\begin{align}= 15+8+8 = \,{\rm{Rs.}} \,31\left( {{a_3}} \right)\end{align}
And so on.
\begin{align}\left( {{a_2}} \right) - \left( {{a_1}} \right) &= \rm{Rs}(23 -15) = \rm{Rs.} \,8\\\left( {{a_3}} \right) - \left( {{a_2}} \right)&= \rm{Rs}(31 - 23) = \rm{Rs. 8}\end{align}
Every time the difference is same.
So, this forms an AP with first term $$15$$ and the difference is $$8.$$
(ii)
What is Known?
The amount of air present.
What is Unknown?
Whether it is an arithmetic progression or not?
Reasoning:
An arithmetic progression is a sequence of number in which each term is obtained by adding a fixed number to the preceding term except the first term.
General form of an arithmetic progression is \begin{align}a, \,(a+d),\,( a+2d),\, (a+3d),\end{align} Where
$$a$$ is the first term and $$d$$ is the common difference.
Steps:
Let the amount of air in the cylinder be $$x.$$
So \begin{align}{a_1= x}\end{align}
After First time removal,
\begin{align}{a_2}&= x - \frac{x}{4}\\{a_2}&= \frac{{3x}}{4}\end{align}
After Second time removal,
\begin{align}{\text{}}\, {a_3}&= \frac{3}{4} - \frac{1}{4}\left( {\frac{{3x}}{4}} \right)\\&= \frac{{3x}}{4} - \frac{{3x}}{{16}}\\&= \frac{{12x - 3x}}{{16}}\\&= \frac{{9x}}{{16}}\end{align}
\begin{align}{a_3} = \frac{{9x}}{{16}}\end{align}
After third time removal,
\begin{align}a_4&= \frac{{9x}}{{16}} - \frac{1}{4}\left( {\frac{{9x}}{{16}}} \right)\\ &= \frac{{9x}}{{16}} - \frac{{9x}}{{64}} \\&= \frac{{36x - 9x}}{{64}} \\&= \frac{{27x}}{{64}}\end{align}
\begin{align}{a_4} &= \frac{27x}{64}\\{a_2} - {a_1} &= \frac{{3x}}{4} - x \\&= \frac{{3x - 4x}}{4} \\&= \frac{{ - x}}{4}\\{a_3} - {a_2} &= \frac{9}{{16}}x - \frac{3}{4}x \\&= \frac{{9x - 12x}}{{16}} \\&= \frac{{ - 3x}}{{16}}\{a_3} - {a_2}) &\ne ({a_2} - {a_1})\end{align} This is not forming an AP. iii) What is Known? Cost of digging well for every meter and subsequent meter. What is unknown? Whether it is an arithmetic progression or not. Reasoning: An arithmetic progression is a sequence of number in which each term is obtained by adding a fixed number to the preceding term except the first term. General form of an arithmetic progression is \(\begin{align}a, \,(a+d),\,( a+2d),\, (a+3d),\end{align} Where
$$a$$ is the first term and $$d$$ is the common difference.
Steps:
Cost of digging the well after $$1$$ meter
\begin{align}= {\rm{Rs.}} \,150\left( {{a_1}} \right)\end{align}
Cost of digging the well after $$2$$ meters
\begin{align}= 150 + 50 = {\rm{Rs.}}\, 200\,\left( {{a_2}} \right)\end{align}
Cost of digging the well after $$3$$ meters
$$=\rm{ Rs.}\, 150+50+50={Rs.\,}250$$ \begin{align}\left( {{a_3}} \right)\end{align}
\begin{align}\left( {{a_2} - {a_1}} \right) &= 200 - 150 = {\rm{Rs}}.50\\\left( {{a_3} - {a_2}} \right) &= 250 - 200 = {\rm{Rs}}.50\\\left( {{a_2} - {a_1}} \right) &= \left( {{a_3} - {a_2}} \right)\end{align}
So, this list of numbers from an AP with first term as $$\rm{Rs.} \,150$$ and common difference is $$\rm{Rs.}\, 50.$$
iv)
What is Known?
Deposited amount and rating interest.
What is Unknown?
Whether it is an arithmetic progression or not.
Reasoning:
An arithmetic progression is a sequence of number in which each term is obtained by adding a fixed number to the preceding term except the first term.
General form of an arithmetic progression is \begin{align}a, \,(a+d),\,( a+2d),\, (a+3d),\end{align} Where
$$a$$ is the first term and $$d$$ is the common difference.
Steps:
Amount present when the amount is $$P$$ and the interest is $$r \%$$ after $$n$$ years is
\begin{align}\mathrm{A}&=\left[P\left(1+\frac{r}{100}\right)\right] \\ \mathrm{P}&=10,000 \\ \mathrm{r}&=8 \%\end{align}
For first year \begin{align}( a_1) = 10000\, \left( 1 + \frac{8}{{100}}\right)\end{align}
For second year
\begin{align}\left( {{a_2}} \right) = 10000{{\left( {1 + \frac{8}{{100}}} \right)}^2}\\\end{align}
For third year \begin{align}\left( {{a_3}} \right) = 10000{{\left( {1 + \frac{8}{{100}}} \right)}^3}\end{align}
For fourth year \begin{align}\left( {{a_4}} \right) = 10000{{\left( {1 + \frac{8}{{100}}} \right)}^4}\end{align}
And so on
\begin{align}&\left( {{a_2}\! - \!{a_1}} \right) \\ \!&=\! 10000{\left( {1 \!+\! \frac{8}{{100}}} \right)^2}\! -\! 10000\left( {1 \!+\! \frac{8}{{100}}} \right)\\ &\!=\! 10000\left( {1 \!+\! \frac{8}{{100}}} \right)\left[ {1 \!+\! \frac{8}{{100}} \!-\! 1} \right]\\ &= 10000\left( {1 + \frac{8}{{100}}} \right)\left( {\frac{8}{{100}}} \right)\\ &\left( {{a_3} - {a_2}} \right) \\&=\! 10000{\left( {1 \!+\! \frac{8}{{100}}} \right)^3}\! \!-\! 10000{\left(\! {1 \!+ \!\frac{8}{{100}}} \!\right)^2}\\ &= 10000{\left( {1 + \frac{8}{{100}}} \right)^2}\left[ {1 + \frac{8}{{100}} - 1} \right]\\ &= 10000{\left( {1 + \frac{8}{{100}}} \right)^2}\left( {\frac{8}{{100}}} \right)\\&\left( {{a_3} - {a_2}} \right) \ne \left( {{a_2} - {a_1}} \right)\end{align}
The amount will not form an AP.
## Chapter 5 Ex.5.1 Question 2
Write first four terms of AP, When the first term $$a$$ and the common difference $$d$$ are given as follows:
(i) \begin{align} a = 10,\,d = 10\end{align}
(ii) \begin{align}a= - 2\quad , d = 0\end{align}
(iii) \begin{align}a= 4,\, d= - 3\end{align}
(iv) \begin{align}a = - 1,\,d = \frac{1}{2}\end{align}
(v) \begin{align}a = - 1.25,\,d = - 0.25\end{align}
### Solution
Reasoning:
General form of an arithmetic progression is \begin{align}{a, (a+d), (a+2d), (a+3d).}\end{align} Where $$a$$ is the first term and $$d$$ is the difference.
(i) \begin{align} a = 10,\,d = 10\end{align}
What is Known?
\begin{align} a= 10\,\, {\rm{and}}\,\, d = 10\end{align}
What is Unknown?
First four terms of the AP.
Steps:
First term \begin{align}{a = 10}\end{align}
Second term \begin{align}{a+d = 10 + 10 = 20}\end{align}
Third term \begin{align}{a + 2d = 10 + 20 = 30}\end{align}
Fourth term \begin{align}{a + 3d = 10 + 30 = 40}\end{align}
The first four terms of AP are $$10, 20, 30,$$ and $$40.$$
(ii) \begin{align}a= - 2,\quad d = 0\end{align}
What is Known?
\begin{align}a= - 2\,\, {\rm{and}}\,\, d = 0\end{align}
What is Unknown?
First four terms of the AP.
Steps:
First term \begin{align}{a =} -2\end{align}
Second term \begin{align}= a+ {d} = -2+0 = -2 \end{align}
Third term \begin{align}= a + {2d} = -2 +0 = -2 \end{align}
Fourth term \begin{align}= a + {3d} = -2 + 0 = -2 \end{align}
The first four terms of AP are $$-2,-2,-2$$ and $$-2.$$
(iii) \begin{align}a= 4,\, d= - 3\end{align}
What is Known?
\begin{align} {a = 4}\,\, {\rm{and}}\,\, {d =} -3 \end{align}
What is Unknown?
First four terms of the AP.
Steps:
First term \begin{align} {= a = 4} \end{align}
Second term \begin{align} { a+d = 4+ (-3) = 1} \end{align}
Third term \begin{align} { a+2d} = 4 – 6= -2\end{align}
Fourth term \begin{align} { a+3d} = 4\; – 9 = -5\end{align}
The first four terms of AP are \begin{align} 4, 1, -2, -5\end{align}.
(iv) \begin{align}a = - 1,\,d = \frac{1}{2}\end{align}
What is Known?
\begin{align}a = -1\,\, {\rm{and}}\,\, {d} =\frac{1}{2}\end{align}
What is Unknown?
First four terms of the AP.
Steps:
First term = $$a = - 1$$
Second term\begin{align}= {{a}} + {{d}} = - 1 + \frac{1}{2} = - \frac{1}{2}\end{align}
Third term \begin{align}= {{a}} + 2{{d}} = - 1 + 1 = 0\end{align}
Fourth term \begin{align}= {{a}} + 3{{d}} = - 1 + \frac{3}{2} = \frac{1}{2}\end{align}
The first four terms of AP are \begin{align} - 1, - \frac{1}{2},0,\frac{1}{2}\end{align} .
(v) \begin{align}a = - 1.25,\,d = - 0.25\end{align}
What is Known?
\begin{align}{a}= -1.25\,\, {\rm{and}}\,\, \quad {d} = - 0.25\end{align}
What is Unknown?
First four terms of the AP.
Steps:
First term\begin{align} = a = - 1.25\end{align}
Second term
\begin{align}&= {{a}} + {{d}}\\ &= - 1.25 + ( - 0.25)\\ &= - 1.25 - 0.25\\ &= - 1.5\\ \end{align}
Third term
\begin{align}&= a + 2d\\ &= - 1.25 + 2 \times ( - 0.25)\\ &= - 1.25 - 0.50\\ &= - 1.75\end{align}
Fourth term
\begin{align}&= a + 3{{d}}\\ &= - 1.25 + 3 \times ( - 0.25)\\ &= - 1.25 - 0.75\\ &= - 2.00\end{align}
The first four terms of AP are
\begin{align}-1.25, -1.5, -1.75, \rm{and} - 2.00\end{align}.
## Chapter 5 Ex.5.1 Question 3
For the following APs, write the first term and the common difference:
i) \begin{align}3,1, - 1, - 3 \ldots \ldots \end{align}
ii) \begin{align}- 5, - 1,3,7 \ldots \ldots \end{align}
iii) \begin{align}\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3} \ldots \ldots.\quad \quad \end{align}
iv) \begin{align}0.6,1.7,2.8,3.9 \ldots \end{align}
### Solution
Reasoning:
General form of an arithmetic progression is \begin{align}{a, (a+d), (a+2d), (a+3d)…}\end{align} where $$a$$ is the first term and $$d$$ is the difference.
i) What is known?
$$3,1,-1,-3$$ are in AP.
What is Unknown?
First term and the common difference of the AP.
Steps:
The AP is \begin{align} 3, 1 ,-1, -3.\end{align}
First term \begin{align}{a = 3}.\end{align}
Common difference
\begin{align} & = {a_2} - {a_1}\\&= 1 – 3\\&= -2\end{align}
First term is $$3$$ and the common difference is $$-2.$$
ii) What is known?
\begin{align}-5, -1, 3, 7\end{align} are in AP
What is Unknown?
First term and the common difference of the AP.
Step:
The AP is \begin{align}-5, -1, 3, 7.\end{align}
First term\begin{align} a = - 5\end{align}
Common difference
\begin{align} & = {a_2} - {a_1}\\&= -1 – (-5)\\&= -1 + 5\\& = 4\end{align}
First term $$-5$$ and the common difference is $$4.$$
iii) What is known?
\begin{align}\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3}\end{align} are in AP.
What is Unknown?
First term and the common difference of the AP.
Steps:
The AP is \begin{align}\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3}\end{align}
First term \begin{align} a = \frac{1}{3}\end{align}
Common difference
\begin{align} &= {a_2} - {a_1}\\&= \frac{5}{3} - \frac{1}{3}\\&= \frac{{5 - 1}}{3}\\&= \frac{4}{3}\end{align}
First term is \begin{align}\frac{1}{3}\end{align} and the common difference is \begin{align}\frac{4}{3}\end{align}
iv) What is known?
\begin{align}{\rm{0}}{\rm{.6,}}\,\,{\rm{1}}{\rm{.7,}}\,\,{\rm{2}}{\rm{.8,}}\,\,{\rm{3}} {\rm{.9}}\end{align} are in AP.
What is Unknown?
First term and the common difference of the AP.
Steps:
The AP is \begin{align} {\rm{0}}{\rm{.6,}}\,\,{\rm{1}}{\rm{.7,}}\,\,{\rm{2}}{\rm{.8,}}\,\,{\rm{3}}{\rm{.9}}\end{align}
First term \begin{align} a = 0.6 \end{align}
Common difference
\begin{align} &= {a_2} - {a_1}\\&= 1.7 – 0.6\\&= 1.1\end{align}
First term $$= 0.6$$ and the common difference $$= 1.1$$
## Chapter 5 Ex.5.1 Question 4
Which of the followings are APs? If they form an AP, Find the common difference d and write three more terms.
i) \begin{align}2,4,16\dots\dots\end{align}
ii)\begin{align}2,\frac{5}{2},3,\frac{7}{2}\dots\dots\end{align}
iii) \begin{align}- 1.2, - 3.2, - 5.2, - 7.2\dots\dots\end{align}
iv)\begin{align}{\kern 1pt} \,\,\, - 10, - 6, - 2,2\dots\dots\end{align}
v)\begin{align}3,3 + \sqrt 2 ,\,3 + 2\sqrt 2 ,3 + 3\sqrt 2 \dots\dots\end{align}
vi) \begin{align}0.2,0.22,0.222,0.2222\dots\dots\end{align}
vii)\begin{align}0, - 4, - 8, - 12\dots\dots\end{align}
viii)\begin{align}- \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}\dots\dots\end{align}
ix) \begin{align}1,3,9,27\dots\dots\end{align}
x)\begin{align}a,2a,3a.4a,\dots\dots\end{align}
xi) \begin{align}a,{a^2},{a^3},{a^4}\end{align}
xii) \begin{align}\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {132} \end{align}
xiii)\begin{align}\sqrt 3 ,\sqrt 6 ,\sqrt 9 ,\sqrt {12} \end{align}
xiv)\begin{align}{1^2},{3^2},{5^2},{7^2}\end{align}
xv)\begin{align}{1^2},{3^2},{5^2},{7^3}\end{align}
### Solution
Reasoning:
General form of an arithmetic progression is \begin{align} a, (a+d), (a+2d), (a+3d)…\end{align} where $$a$$ is the first term and $$d$$ is the common difference.
i) What is Known?
\begin{align} 2, 4, 8, 16…..\end{align}
What is Unknown?
Common difference and next three more terms of AP if it is an AP.
Steps:
The given numbers are $$2,\,4,\,8,\,16$$
First term $$a =2$$
Common difference
$$d=a_{2}-a_{1}=4-2=2$$
Common difference
$$d=a_{3}-a_{2}=8-4=4$$
\begin{align}({a_{3}-a_{2}) \neq (a_{2}-a_{1})}\end{align}
\begin{align}2, 4, 8, 16\end{align} are not in AP, because the common difference is not equal.
ii) What is Known?
\begin{align}2,\,\frac{5}{2},\,3,\,\frac{7}{2}\end{align}
What is Unknown?
Common difference and next three more terms of AP if it is an AP.
Steps:
The given numbers are \begin{align}2,\,\frac{5}{2},\,3,\,\frac{7}{2}\end{align}
First term a $$= 2$$
Common difference
\begin{align}& d =a_{2}-a_{1}=\frac{5}{2}-2=\frac{5-4}{2}=\frac{1}{2}\end{align}
Common difference
\begin{align}& d =a_{3}-a_{2}=3-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}\end{align}
Since
\begin{align}({a_3} - {a_2}) = ({a_2} - {a_1}).\end{align}
\begin{align}2,\,\frac{5}{2},\,3,\,\frac{7}{2}\end{align} forms an AP and common difference is\begin{align}\frac{1}{2}\end{align} .
The next three terms are:
Fifth term
\begin{align} &=a+4 d \\ &=2+4 \times \frac{1}{2} \\ &=2+2 \\ &=4 \end{align}
Sixth term
\begin{align} &=a+5 d \\ &=2+5 \times \frac{1}{2} \\ &=2+\frac{5}{2} \\ &=\frac{4+5}{2} \\ &=\frac{9}{2} \end{align}
Seventh term
\begin{align} &=a+6 d \\ &=2+6 \times \frac{1}{2} \\ &=5 \end{align}
\begin{align}2,\,\frac{5}{2},\,3,\,\frac{7}{2}\end{align} forms an AP and the common difference is \begin{align}\frac{1}{2}\end{align}. The next three terms are \begin{align}4,\,\frac{9}{2}\,,5.\end{align}
iii) What is Known?
\begin{align} - 1.2,\, - 3.2,\, - 5.2,\, - 7.2 \ldots \ldots .\end{align}
What is Unknown?
Whether it forms an AP. If it is find the common difference and the next three terms of AP.
Steps:
The given numbers are
\begin{align} - 1.2,\, - 3.2,\, - 5.2,\, - 7.2 \ldots \ldots .\end{align}
First term \begin{align} a = -1.2\end{align}
Common difference
\begin{align} d &={a_2} - {a_1}\\& = - 3.2 - ( - 1.2)\\& = - 3.2 + 1.2 = - 2\end{align}
Common difference
\begin{align} &={a_3} - {a_2} = - 5.2 - ( - 3.2)\\ &= - 5.2 + 3.2 = - 2 \end{align}
Since
\begin{align}({a_3} - {a_2}) = ({a_2} - {a_1}) . \end{align}
It forms an AP.
The fifth term
\begin{align} &=a+4d \\ &=-1.2+4(-2) \\ &=-1.2-8=-9.2 \end{align}
The sixth term
\begin{align} &=a+5d \\ &=-1.2+5(-2) \\ &=-1.2-10 \\ &=-11.2 \end{align}
The seventh term
\begin{align} &=a+6d \\ &=-1.2+6(-2) \\ &=-1.2-12 \\ &=-13.2 \end{align}
\begin{align} - 1.2,\, - 3.2,\, - 5.2,\, - 7.2\end{align} forms an AP with common difference -2. The next three terms of AP are \begin{align}-9.2, -11.2, -13.2. \end{align}
iv) What is Known?
\begin{align} -10, -6, -2, 2 \end{align}
What is Unknown?
Whether it forms an AP. If it is find the common difference and the next three terms of AP.
Steps:
The given numbers are \begin{align} -10, -6, -2, 2 \end{align}
First term a $$= -10$$
Common difference
\begin{align} d &=a_{2}-a_{1} \\ &=-6-(-10) \\ &=-6-10 \\ &=4 \end{align}
Common difference
\begin{align} d &=a_{3}-a_{2} \\ &=-2-(-6) \\ &=-2+6 \\ &=4 \end{align}
Since
$$(a_{3}-a_{2})=(a_{2}-a_{1})$$
Fifth Term: $$a+4 d=-10+16=6$$
Sixth Term: $$a+5 d=-10+20=10$$
Seventh Term: $$a+6 d=-10+24=14$$
\begin{align}-10, -6 ,-2, 2 \end{align} forms an AP with common difference $$4$$ and next terms are $$6, 10, 14.$$
v) What is Known?
\begin{align}3,\,\,3 + \sqrt 2 \,\,,3 + 2\sqrt 2 \,\,,3 + 3\sqrt 2 \ldots \ldots \end{align}
What is Unknown?
Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.
Steps:
The list of numbers is
\begin{align}3,\,\,3 + \sqrt 2 \,\,,3 + 2\sqrt 2 \,\,,3 + 3\sqrt 2 \ldots \ldots \end{align}
Common difference
\begin{align} d &=a_{2}-a_{1} \\ &=3+\sqrt{2}-3 \\ &=\sqrt{2} \end{align}
Common difference
\begin{align} d &=a_{3}-a_{2} \\ &=3+2 \sqrt{2}-(3+\sqrt{2}) \\ &=3+2 \sqrt{2}-3-\sqrt{2} \\ &=\sqrt{2} \end{align}
Since
\begin{align} a_{3}-a_{2}=a_{2}-a_{1}\end{align}
So \begin{align}3,\,\,3 + \sqrt 2 \,\,,3 + 2\sqrt 2 \,\,,3 + 3\sqrt 2 \ldots \ldots \end{align} forms an AP with common difference 4.
Next three terms are
\begin{align} \text { Fifth term } &=a+4 d \\ &=3+4 \sqrt{2} \\ \text { Sixth term } &=a+5 d \\ &=3+5 \sqrt{2} \\ \text { Seventh term } &=a+6 d \\ &=3+6 \sqrt{2} \end{align}
It is an AP with common difference \begin{align}\sqrt 2 \end{align} and Next three terms are
\begin{align}3 + 4\sqrt 2 ,\,\,3 + 5\sqrt 2 \,\,,3 + 6\sqrt 2 \end{align} ,.
vi) What is Known?
\begin{align}0.2,0.22,0.222,0.2222.....\end{align}
What is Unknown?
Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.
Steps:
The list of numbers are
\begin{align}0.2,0.22,0.222,0.2222.....\end{align}
Common difference
\begin{align} d &=a_{2}-a_{1} \\ &=0.22-0.2 \\ &=0.20 \end{align}
Common difference
\begin{align} d &=a_{3}-a_{2} \\ &=0.222-0.220 \\ &=0.002\\ (a_{3}-a_{2} )&\neq (a_{2}-a_{1})\end{align}
The given list of numbers does not form an AP.
vii) What is Known?
\begin{align} 0,-4,-8,-12……\end{align}
What is Unknown?
Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.
Steps:
The list of numbers is
\begin{align} 0,-4,-8,-12……\end{align}
Common difference
\begin{align}d = {a_2} - {a_1}= - 4 – 0 = -4\end{align}
Common difference
\begin{align}d = {a_3} - {a_2}= -8 – (-4) = -8 + 4 = -4\end{align}
Since\begin{align}({a_3} - {a_2}) = ({a_2} - {a_1})\end{align} . It forms an AP.
\begin{align} \text { Fifth term } &=a+4 d \\ &=0+4(-4) \\ &=-16 \\ \text { Sixth term } &=a+5 d \\ &=0+5(-4) \\ &=-20 \\ \text { Seventh term } &=a+6 d \\ &=0+6(-4) \\ &=-24 \end{align}
The given numbers form an AP with difference $$-4.$$ The next three terms are \begin{align} -16, -20, -24. \end{align}
viii) What is Known?
\begin{align} - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}.......\end{align}
What is Unknown?
Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.
Steps:
The given list of numbers is
\begin{align} - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}.......\end{align}
Common difference
\begin{align} d &= {a_2} - {a_1}\\&= - \frac{1}{2} - \left( { - \frac{1}{2}} \right)\\ &= - \frac{1}{2} + \frac{1}{2}\\&= 0\end{align}
Common difference
\begin{align} d &= {a_3} - {a_2}\\& = - \frac{1}{2} - \left( { - \frac{1}{2}} \right)\\ &= - \frac{1}{2} + \frac{1}{2}\\&= 0\end{align}
Since \begin{align}({a_3} - {a_2}) = ({a_2} - {a_1})\end{align} .The list of numbers forms an AP.
\begin{align}{\text{The fifth term }} &= a + 4d\\& = - \frac{1}{2} + 4(0)\\ &= - \frac{1}{2}\\{\text{ The sixth term }} &= a + 5d\\ &= - \frac{1}{2} + 5(0)\\ &= - \frac{1}{2}\\{\text{The seventh term}} &= {a} + 6d\\ &= - \frac{1}{2} + 6(0)\\ &= - \frac{1}{2}\end{align}
The given list of numbers form an AP with common difference $$d = 0.$$ Next three terms are \begin{align} - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}.\end{align}
ix) What is Known?
\begin{align}1, 3 ,9 ,27.\end{align}
What is Unknown?
Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.
Steps:
The list of numbers are \begin{align}1, 3 ,9 ,27.\end{align}
Common difference \begin{align} d = {a_2} - {a_1}= 3 – 1 = 2\end{align}
Common difference \begin{align} d = {a_3} - {a_2}= 9 -3 =6\end{align}
Since\begin{align} ({a_2} - {a_1}) \ne ({a_3} - {a_2})\end{align}
The given list of numbers does not form an AP.
x) What is Known?
\begin{align}{a, 2a, 3a, 4a}\end{align}
What is Unknown?
Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.
Steps:
The list of numbers is \begin{align}{a, 2a, 3a, 4a}.\end{align}
Common difference
\begin{align} d &=a_{2}-a_{1} \\ &=2a-a\\ &=a \end{align}
Common difference
\begin{align} d &=a_{3}-a_{2} \\ &=3a-2a \\ &=a \end{align}
since\begin{align}{a_3} - {a_2} = {a_2} - {a_1}, {a, 2a, 3a, 4a}\end{align}
forms an AP.
The fifth term $$=a+4d=a+4a=5a$$
The sixth term $$=a+5d=a+5a=6a$$
The seventh term $$=a+6d=a+6a=7a$$
The given list of numbers form an AP with common difference $$d =a$$ The next three terms are $$5a, 6a, 7a.$$
xi) What is Known?
\begin{align}a,{a^2},{a^3},{a^4}\end{align}
What is Unknown?
Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.
Steps:
The list of numbers is \begin{align}a,{a^2},{a^3},{a^4}\end{align}
Common difference
\begin{align} d &=a_{2}-a_{1} \\ &=a^{2}-a \\ &=a(a-1) \end{align}
Common difference
\begin{align} d &= {a_3} - {a_2}\\ &= {a^3} - {a^2}\\ &= {a^2}\left( {a - 1} \right)\end{align}
Since
\begin{align}({a_2} - {a_1}) \ne ({a_3} - {a_2})\end{align}
The given list of numbers does not form an AP.
xii) What is Known?
\begin{align}\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {32} \end{align}
What is Unknown?
Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.
Steps:
The list of numbers is \begin{align}\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {32} \end{align}
Common difference
\begin{align}d &={a_2} - {a_1}\\&= \sqrt 8 - \sqrt 2 \\ &= \sqrt {4 \times 2} - \sqrt 2 \\ &= 2\sqrt 2 - \sqrt 2 \\ &= \sqrt 2 \end{align}
Common difference
\begin{align}d &= {a_3} - {a_2}\\ &= \sqrt {18} - \sqrt 8 \\ &= \sqrt {9 \times 2} - \sqrt {4 \times 2} \\ &= 3\sqrt 2 - 2\sqrt 2 \\& = \sqrt 2 \end{align}
since
\begin{align}({a_2} - {a_1}) = ({a_3} - {a_2})\end{align} The given numbers form an AP.
\begin{align}{\text{The fifth term }} &= a + 4d\\ &= \sqrt 2 + 4\sqrt 2 \\ &= 5\sqrt 2 \\ &= \sqrt {25 \times 2} \\ &= \sqrt {50} \\{\text{The sixth term}}& = a + 5d\\ &= \sqrt 2 + 5\sqrt 2 \\ &= 6\sqrt 2 \\ &= \sqrt {36 \times 2} \\ &= \sqrt {72} \\{\text{ The seventh term }} &= {\rm{a}} + 6d\\ &= \sqrt 2 + 6\sqrt 2 \\ &= 7\sqrt 2 \\ &= \sqrt {49 \times 2} \\ &= \sqrt {98} \end{align}
The list of numbers forms an AP with common difference\begin{align}\sqrt 2 \end{align}. Next three terms are \begin{align}\sqrt {50} ,\sqrt {72} ,\sqrt {98} \end{align}
xiii ) What is Known?
\begin{align}\sqrt 3 ,\,\sqrt 6 \,,\sqrt 9 ,\,\sqrt {12} \end{align}
What is Unknown?
Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.
Steps:
The list of numbers is
\begin{align}\sqrt 3 ,\,\sqrt 6 \,,\sqrt 9 ,\,\sqrt {12} \end{align}
Common difference
\begin{align}d &= {a_2} - {a_1} = \sqrt 6 - \sqrt 3 \\ &= \sqrt {3 \times 2} - \sqrt 3 \\ &= \sqrt 3 \left( {\sqrt 2 - 1} \right)\end{align}
Common difference
\begin{align} d&= {a_3} - {a_2} = \sqrt 9 - \sqrt 6 \\ &= \sqrt {3 \times 3} - \sqrt {3 \times 2} \\ &= \sqrt 3 (\sqrt 3 - \sqrt 2 )\end{align}
since$$({a_2} - {a_1}) \ne ({a_3} - {a_2})$$
The given list of numbers does not form an AP.
xiv) What is Known?
\begin{align}{1^2},\,{3^2}\,,{5^2}\,,{7^2}\end{align}
What is Unknown?
Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.
Steps:
The list of numbers is
\begin{align}{1^2},\,{3^2}\,,{5^2}\,,{7^2}\end{align}
Common difference
\begin{align} d= {a_2} - {a_1}= 25 – 1 = 24\end{align}
Common difference
$$d ={a_3} - {a_2}= 49 – 25 = 24$$
Since $$({a_2}- {a_1}) = ({a_3} - {a_2})$$ they form an AP
\begin{align} \text { The fifth term } &=a+4 d \\ &=1+4 \times 24 \\ &=1+96 \\ &=97 \end{align}
\begin{align} \text { The sixth term } &=a+5 d \\ &=1+5 \times 24 \\ &=1+120 \\ &=121 \end{align}
\begin{align} \text { The seventh term } &=a+6 d \\ &=1+6 \times 24 \\ &=1+144 \\ &=145 \end{align}
The list of numbers form an AP with common difference $$24.$$ The next three terms are $$97, 121,$$ and $$145.$$
xv)What is Known?
\begin{align}{1^2},\,{3^2},\,{5^2},\;{7^3}\end{align}
What is Unknown?
Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.
Steps:
The list of numbers
\begin{align}{1^2},\,{3^2},\,{5^2},\;{7^3}\end{align}
Common difference
\begin{align} d = {a_2} - {a_1}= 9 – 1 = 8\end{align}
Common difference
\begin{align}d = {a_3} - {a_2}= 25 - 9 = 16\end{align}
Since
\begin{align}({a_2} - {a_1}) \ne ({a_3} - {a_2})\end{align}
The given list of numbers does not form an AP.
Arithmetic Progressions | NCERT Solutions |
# Differential Equation: Application of D.E: Population Growth
### Differential Equation:
Differential Equation: Application of D.E.: Population Growth
A bacterial population B is known to have a rate of growth proportional to (B + 25). Between noon and 2PM the population increases to 3000 and between 2PM and 3PM the population is increased by 1000 in culture. (a) Find an expression for the bacterial population B as a function of time. (b) What is the initial bacterial population in the culture? (c) What is the total bacterial population in the culture at 4:15PM?
### $\dfrac{dB}{dt} = k(B + 25)$
$\dfrac{dB}{dt} = k(B + 25)$
$\dfrac{dB}{B + 25} = k \, dt$
$\displaystyle \int \dfrac{dB}{B + 25} = k \int dt$
$\ln (B + 25) = kt + C$
At 2:00PM, t = 2 and B = 3000
$\ln 3025 = 2k + C$ ← eq. (1)
At 3:00PM, t = 3 and B = 4000
$\ln 4025 = 3k + C$ ← eq. (2)
From eq. (1) and eq. (2)
$k = 0.2856$
$C = 7.4434$
Hence,
$\ln (B + 25) = 0.2856t + 7.4434$ answer for (a)
At noon, t = 0
$\ln (B + 25) = 7.4434$
$B = 1683$ answer for (b)
At 4:15PM, t = 4.25
$\ln (B + 25) = 0.2856(4.25) + 7.4434$
$B = 5726$ answer for (c)
### Another solution (By
Another solution (By Calculator - CASIO fx-991ES PLUS):
MODE 3 5
X Y
2 3000 + 25
3 4000 + 25
AC
$B + 25 = 0\hat{y}$
$B + 25 = 1708$
$B = 1683$ answer for (b)
$B + 25 = 4.25\hat{y}$
$B + 25 = 5751$
$B = 5726$ answer for (c)
Note:
$\hat{y}$ = SHIFT 1 5 5 |
# 5 QUESTIONS ON MARGINAL, AVERAGE, FIXED, VARIABLE AND TOTAL COST
Being the diligent student that you are, your mind would have already been ingrained with the components of economic costs.
In this blog post, I am going to be teaching you how to solve each one of them by going through these examples
Question 1
Given that accompanying cost schedule of a firm producing oranges
Find the total cost, average total cost, average fixed cost of producing 4, 9, 13, 16 oranges
Before I continue, note that all figure in this first question is in dollars.
Ok, let's solve for the total cost(TC) of producing each 4,9,13,16
Recalled that
$$TC=TVC+TFC$$
When producing:
4 oranges
$TC=20+25$
$=45$
9 oranges
$TC=20+50$
$=70$
13 oranges
$TC=20+$75
$=95$
At
16 oranges
$TC=20+100$
$=120$
Now, let's solve for the average total cost(ATC)
Recall that
ATC=$\frac{TC}{OUTPUT}$
When producing:
4 oranges
ATC=$\frac{45}{4}=1.25$
9 oranges
ATC=$\frac{70}{9}=7.8$
13 oranges
ATC=$\frac{95}{13}=7.3$
16 oranges
ATC=$\frac{120}{16}=7.5$
Let's solve for the average fixed cost(AFC)
Recalled that:
$$AFC=\frac{\text{fixed cost}}{\text{output}}$$
When producing:
4 oranges
AFC=$\frac{20}{4}$
=$5$
9 oranges
AFC=$\frac{20}{9}$
=$2.2$
13 oranges
AFC=$\frac{20}{13}$
=$1.5$
16 oranges
AFC=$\frac{20}{16}$
=$1.25$
Question 2
A firm's average variable cost(AVC) is €360, its average total cost(ATC) is $\$400$and its total fixed cost(TFC) is$\$4000$, find its output?
Recalled that,
ATC=AVC+AFC
$\$400=\$360+AFC$
collect like term
$\$400-\$360=AFC$
$\$40=AFCAFC=\$40$
Now, let's solve for the output
Recalled:
$$AFC=\frac{\text{fixed costs}}{\text{output}}$$
€40= $\frac{\$4000}{\text{output}}$Cross multiply$\$40\times \text{output}=\$4000$Divide both side by €40$\frac{\$40\times \text{output}}{40}=\frac{\$4000}{\$40}$
$\text{output}=\$100$Question 3 A firm marginal cost is $$\60$$, its average total cost is $$\100$$, and its output is 1600 units. find the total cost of producing the 1601 units. Answer Recalled that $$ATC=\frac{TC}{output}$$$\$100=\frac{TC}{1600}$
Cross multiply
$TC=\$100\times 1600TC=\$160000$
To find the total cost of the 1601 unit
We add MC(cost of producing one more unit) to the total cost
So
TC of 1601 units=$\$160000+\$60$
$=\$160060$Related post Question 4 A firm's average variable cost and the average total cost are$\$225$ and $\$240$respectively, and its output is 150 units. Find its total fixed cost. Answer First of all, Let's solve for the average fixed cost(AFC) Recalled:$ATC=AFC+AVC$Ok, let make AFC the subject of the formula Collect like term$AFC=ATC-AVCAFC=\$240-\$225=\$15$
Also, remember that
AFC=$\frac{TFC}{output}$
€15=$\frac{TFC}{150}$
cross multiply
TFC=$\$15\times 150=\$750$
Question 5
A firm's average total cost is $\$160$, its fixed cost is$\$2000$, and its output is 200 units, find its average variable cost.
Recalled that
ATC=AVC+AFC
But we need to find the value of AFC,
$AFC=\frac{\text{fixed cost}}{\text{output}}$
$AFC=\frac{\$2000}{200}=\$10$
Now, let's solve for AVC
$\$160=AVC+\$10$
collect like terms
$\$160-\$10=AVC$
$\$150=AVCAVC=\$150$
I hope with this, you can now calculate any question on the components of economic costs. Please, if you have any questions, suggestions, feel free to tell me in the comment box.
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# Divide 16 into two parts such that twice the squares of the larger part exceeds the square of the smaller part by 164.
Question:
Divide 16 into two parts such that twice the squares of the larger part exceeds the square of the smaller part by 164.
Solution:
Let the larger and smaller parts be $x$ and $y$, respectively.
According to the question:
$x+y=16 \ldots$ (i)
$2 x^{2}=y^{2}+164 \ldots$ (ii)
From (i), we get:
$x=16-y \ldots$ (iii)
From (ii) and (iii), we get:
$2(16-y)^{2}=y^{2}+164$
$\Rightarrow 2\left(256-32 y+y^{2}\right)=y^{2}+164$
$\Rightarrow 512-64 y+2 y^{2}=y^{2}+164$
$\Rightarrow y^{2}-64 y+348=0$
$\Rightarrow y^{2}-(58+6) y+348=0$
$\Rightarrow y^{2}-58 y-6 y+348=0$
$\Rightarrow y(y-58)-6(y-58)=0$
$\Rightarrow(y-58)(y-6)=0$
$\Rightarrow y-58=0$ or $y-6=0$
$\Rightarrow y=6 \quad(\because y<16)$
Putting the value of y in equation (iii), we get:
$x=16-6=10$
Hence, the two natural numbers are 6 and $10 .$ |
# How do you solve |x| <3?
Sep 4, 2016
You have to use the definition of $| x |$:
$| x | = x$, if $x \ge 0$
$| x | = - x$, if $x < 0$
#### Explanation:
So we first consider $x \ge 0$. In this case $| x | = x$, and the inequality becomes $x < 3$. Hence, all $x$ such that $x \ge 0$ and $x < 3$ satisfy the inequality. That is all $x$, $0 \le x < 3$
Now consider $x < 0$; is this case $| x | = - x$, and the inequality becomes $- x < 3$. This is the same as $- 3 < x$, so all the $x$ such that $- 3 < x < 0$ satisfy the inequality.
Putting both together, the solution are all $x$ such that $- 3 < x < 3$ |
#### 8.1Functions as Data
It’s interesting to consider how expressive the little programming we’ve learned so far can be. To illustrate this, we’ll work through a few exercises of interesting concepts we can express using just functions as values. We’ll write two quite different things, then show how they converge nicely.
##### 8.1.1A Little Calculus
If you’ve studied the differential calculus, you’ve come across curious sytactic statements such as this:
\begin{equation*}{\frac{d}{dx}} x^2 = 2x\end{equation*}
Let’s unpack what this means: the $$d/dx$$, the $$x^2$$, and the $$2x$$.
First, let’s take on the two expressions; we’ll discuss one, and the discussion will cover the other as well. The correct response to “what does $$x^2$$ mean?” is, of course, an error: it doesn’t mean anything, because $$x$$ is an unbound identifier.
So what is it intended to mean? The intent, clearly, is to represent the function that squares its input, just as $$2x$$ is meant to be the function that doubles its input. We have nicer ways of writing those:
fun sq(x :: Number) -> Number: x * x end
fun dbl(x :: Number) -> Number: 2 * x end
and what we’re really trying to say is that the $$d/dx$$ (whatever that is) of sq is dbl.We’re assuming functions of arity one in the variable that is changing.
So now let’s unpack $$d/dx$$, starting with its type. As the above example illustrates, $$d/dx$$ is really a function from functions to functions. That is, we can write its type as follows:
d-dx :: ((Number -> Number) -> (Number -> Number))
(This type might explain why your calculus course never explained this operation this way—though it’s not clear that obscuring its true meaning is any better for your understanding.)
Let us now implement d-dx. We’ll implement numerical differentiation, though in principle we could also implement symbolic differentiation—using rules you learned, e.g., given a polynomial, multiply by the exponent and reduce the exponent by one—with a representation of expressions (a problem that will be covered in more detail in a future release).
In general, numeric differentiation of a function at a point yields the value of the derivative at that point. We have a handy formula for it: the derivative of $$f$$ at $$x$$ is
\begin{equation*}\frac{f(x + \epsilon) - f(x)}{\epsilon}\end{equation*}
as $$\epsilon$$ goes to zero in the limit. For now we’ll give the infinitesimal a small but fixed value, and later [Combining Forces: Streams of Derivatives] see how we can improve on this.
epsilon = 0.00001
We can now translate the above formula into a function:
d-dx-at :: (Number -> Number), Number -> Number
fun d-dx-at(f, x):
(f(x + epsilon) - f(x)) / epsilon
end
And sure enough, we can check and make sure it works as expected:
check:
d-dx-at(sq, 10) is-roughly dbl(10)
end
Confession: We chose the value of epsilon so that the default tolerance is-roughly works for this example.
However, there is something unsatisfying about this. The function we’ve written clearly does not have the type we described earlier! What we wanted was an operation that takes just a function, and represents the platonic notion of differentiation; but we’ve been forced, by the nature of numeric differentiation, to describe the derivative at a point. We might instead like to write something like this:
fun d-dx(f):
(f(x + epsilon) - f(x)) / epsilon
end
Do Now!
What’s the problem with the above definition?
If you didn’t notice, Pyret will soon tell you: x isn’t bound. Indeed, what is x? It’s the point at which we’re trying to compute the numeric derivative. That is, d-dx needs to return not a number but a function (as the type indicates) that will consume this x:“Lambdas are relegated to relative obscurity until Java makes them popular by not having them.”—James Iry, A Brief, Incomplete, and Mostly Wrong History of Programming Languages
fun d-dx(f):
lam(x):
(f(x + epsilon) - f(x)) / epsilon
end
end
If we want to be a little more explicit we can annotate the inner function:
fun d-dx(f):
lam(x :: Number) -> Number:
(f(x + epsilon) - f(x)) / epsilon
end
end
This is a special case of a concept useful in many programming contexts, which we explore in more detail elsewhere: Staging.
Sure enough, this definition now works. We can, for instance, test it as follows (note the use of num-floor to avoid numeric precision issues from making our tests appear to fail):
d-dx-sq = d-dx(sq)
check:
ins = [list: 0, 1, 10, 100]
for map(n from ins):
num-floor(d-dx-sq(n))
end
is
for map(n from ins):
num-floor(dbl(n))
end
end
Now we can return to the original example that launched this investigation: what the sloppy and mysterious notation of math is really trying to say is,
d-dx(lam(x): x * x end) = lam(x): 2 * x end
or, in the notation of A Notation for Functions,
\begin{equation*}{\frac{d}{dx}} [x \rightarrow x^2] = [x \rightarrow 2x]\end{equation*}
Pity math textbooks for not wanting to tell us the truth!
##### 8.1.2A Helpful Shorthand for Anonymous Functions
Pyret offers a shorter syntax for writing anonymous functions. Though, stylistically, we generally avoid it so that our programs don’t become a jumble of special characters, sometimes it’s particularly convenient, as we will see below. This syntax is
{(a): b}
where a is zero or more arguments and b is the body. For instance, we can write lam(x): x * x end as
{(x): x * x}
where we can see the benefit of brevity. In particular, note that there is no need for end, because the braces take the place of showing where the expression begins and ends. Similarly, we could have written d-dx as
fun d-dx-short(f):
{(x): (f(x + epsilon) - f(x)) / epsilon}
end
but many readers would say this makes the function harder to read, because the prominent lam makes clear that d-dx returns an (anonymous) function, whereas this syntax obscures it. Therefore, we will usually only use this shorthand syntax for “one-liners”.
##### 8.1.3Streams From Functions
People typically think of a function as serving one purpose: to parameterize an expression. While that is both true and the most common use of a function, it does not justify having a function of no arguments, because that clearly parameterizes over nothing at all. Yet functions of no argument also have a use, because functions actually serve two purposes: to parameterize, and to suspend evaluation of the body until the function is applied. In fact, these two uses are orthogonal, in that one can employ one feature without the other. Below, we will focus on delay without abstraction (the other shows up in other computer science settings).
Let’s consider the humble list. A list can be only finitely long. However, there are many lists (or sequences) in nature that have no natural upper bound: from mathematical objects (the sequence of natural numbers) to natural ones (the sequence of hits to a Web site). Rather than try to squeeze these unbounded lists into bounded ones, let’s look at how we might represent and program over these unbounded lists.
First, let’s write a program to compute the sequence of natural numbers:
fun nats-from(n):
end
Do Now!
Does this program have a problem?
While this represents our intent, it doesn’t work: running it—e.g., nats-from(0)creates an infinite loop evaluating nats-from for every subsequent natural number. In other words, we want to write something very like the above, but that doesn’t recur until we want it to, i.e., on demand. In other words, we want the rest of the list to be lazy.
This is where our insight into functions comes in. A function, as we have just noted, delays evaluation of its body until it is applied. Therefore, a function would, in principle, defer the invocation of nats-from(n + 1) until it’s needed.
Except, this creates a type problem: the second argument to link needs to be a list, and cannot be a function. Indeed, because it must be a list, and every value that has been constructed must be finite, every list is finite and eventually terminates in empty. Therefore, we need a new data structure to represent the links in these lazy lists (also known as streams):
<stream-type-def> ::=
data Stream<T>:
| lz-link(h :: T, t :: ( -> Stream<T>))
end
where the annotation ( -> Stream<T>) means a function from no arguments (hence the lack of anything before ->), also known as a thunk. Note that the way we have defined streams they must be infinite, since we have provided no way to terminate them.
Let’s construct the simplest example we can, a stream of constant values:
ones = lz-link(1, lam(): ones end)
Pyret will actually complain about this definition. Note that the list equivalent of this also will not work:
ones = link(1, ones)
because ones is not defined at the point of definition, so when Pyret evaluates link(1, ones), it complains that ones is not defined. However, it is being overly conservative with our former definition: the use of ones is “under a lam”, and hence won’t be needed until after the definition of ones is done, at which point ones will be defined. We can indicate this to Pyret by using the keyword rec:
rec ones = lz-link(1, lam(): ones end)
Note that in Pyret, every fun implicitly has a rec beneath it, which is why we can create recursive functions with aplomb.
Exercise
Earlier we said that we can’t write
ones = link(1, ones)
What if we tried to write
rec ones = link(1, ones)
instead? Does this work and, if so, what value is ones bound to? If it doesn’t work, does it fail to work for the same reason as the definition without the rec?
Henceforth, we will use the shorthand [A Helpful Shorthand for Anonymous Functions] instead. Therefore, we can rewrite the above definition as:
rec ones = lz-link(1, {(): ones})
Notice that {(): …} defines an anonymous function of no arguments. You can’t leave out the ()! If you do, Pyret will get confused about what your program means.
Because functions are automatically recursive, when we write a function to create a stream, we don’t need to use rec. Consider this example:
fun nats-from(n :: Number):
end
with which we can define the natural numbers:
nats = nats-from(0)
Note that the definition of nats is not recursive itself—the recursion is inside nats-fromso we don’t need to use rec to define nats.
Do Now!
Earlier, we said that every list is finite and hence eventually terminates. How does this remark apply to streams, such as the definition of ones or nats above?
The description of ones is still a finite one; it simply represents the potential for an infinite number of values. Note that:
1. A similar reasoning doesn’t apply to lists because the rest of the list has already been constructed; in contrast, placing a function there creates the potential for a potentially unbounded amount of computation to still be forthcoming.
2. That said, even with streams, in any given computation, we will create only a finite prefix of the stream. However, we don’t have to prematurely decide how many; each client and use is welcome to extract less or more, as needed.
Now we’ve created multiple streams, but we still don’t have an easy way to “see” one. First we’ll define the traditional list-like selectors. Getting the first element works exactly as with lists:
fun lz-first<T>(s :: Stream<T>) -> T: s.h end
In contrast, when trying to access the rest of the stream, all we get out of the data structure is a thunk. To access the actual rest, we need to force the thunk, which of course means applying it to no arguments:
fun lz-rest<T>(s :: Stream<T>) -> Stream<T>: s.t() end
This is useful for examining individual values of the stream. It is also useful to extract a finite prefix of it (of a given size) as a (regular) list, which would be especially handy for testing. Let’s write that function:
fun take<T>(n :: Number, s :: Stream<T>) -> List<T>:
if n == 0:
empty
else:
end
end
If you pay close attention, you’ll find that this body is not defined by cases over the structure of the (stream) inputinstead, it’s defined by the cases of the definition of a natural number (zero or a successor). We’ll return to this below (<lz-map2-def>).
Now that we have this, we can use it for testing. Note that usually we use our data to test our functions; here, we’re using this function to test our data:
check:
take(10, ones) is map(lam(_): 1 end, range(0, 10))
take(10, nats) is range(0, 10)
take(10, nats-from(1)) is map((_ + 1), range(0, 10))
end
The notation (_ + 1) defines a Pyret function of one argument that adds 1 to the given argument.
Let’s define one more function: the equivalent of map over streams. For reasons that will soon become obvious, we’ll define a version that takes two lists and applies the first argument to them pointwise:
<lz-map2-def> ::=
fun lz-map2<A, B, C>(
f :: (A, B -> C),
s1 :: Stream<A>,
s2 :: Stream<B>) -> Stream<C>:
f(lz-first(s1), lz-first(s2)),
{(): lz-map2(f, lz-rest(s1), lz-rest(s2))})
end
Now we can see our earlier remark about the structure of the function driven home especially clearly. Whereas a traditional map over lists would have two cases, here we have only one case because the data definition (<stream-type-def>) has only one case! What is the consequence of this? In a traditional map, one case looks like the above, but the other case corresponds to the empty input, for which it produces the same output. Here, because the stream never terminates, mapping over it doesn’t either, and the structure of the function reflects this.This raises a much subtler problem: if the function’s body doesn’t have base- and inductive-cases, how can we perform an inductive proof over it? The short answer is we can’t: we must instead use coinduction.
Why did we define lz-map2 instead of lz-map? Because it enables us to write the following:
rec fibs =
{(): lz-map2({(a :: Number, b :: Number): a + b},
fibs,
lz-rest(fibs))})})
from which, of course, we can extract as many Fibonacci numbers as we want!
check:
take(10, fibs) is [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
end
Exercise
Define the equivalent of map and filter for streams.
Streams and, more generally, infinite data structures that unfold on demand are extremely valuable in programming. Consider, for instance, the possible moves in a game. In some games, this can be infinite; even if it is finite, for interesting games the combinatorics mean that the tree is too large to feasibly store in memory. Therefore, the programmer of the computer’s intelligence must unfold the game tree on demand. Programming it by using the encoding we have described above means the program describes the entire tree, lazily, and the tree unfolds automatically on demand, relieving the programmer of the burden of implementing such a strategy.
In some languages, such as Haskell, lazy evaluation is built in by default. In such a language, there is no need to use thunks. However, lazy evaluation places other burdens on the language, which you can learn about in a programming-languages class.
##### 8.1.4Combining Forces: Streams of Derivatives
When we defined d-dx, we set epsilon to an arbitrary, high value. We could instead think of epsilon as itself a stream that produces successively finer values; then, for instance, when the difference in the value of the derivative becomes small enough, we can decide we have a sufficient approximation to the derivative.
The first step is, therefore, to make epsilon some kind of parameter rather than a global constant. That leaves open what kind of parameter it should be (number or stream?) as well as when it should be supplied.
It makes most sense to consume this parameter after we have decided what function we want to differentiate and at what value we want its derivative; after all, the stream of epsilon values may depend on both. Thus, we get:
fun d-dx(f :: (Number -> Number)) ->
(Number -> (Number -> Number)):
lam(x :: Number) -> (Number -> Number):
lam(epsilon :: Number) -> Number:
(f(x + epsilon) - f(x)) / epsilon
end
end
end
with which we can return to our square example:
d-dx-square = d-dx(square)
Note that at this point we have simply redefined d-dx without any reference to streams: we have merely made a constant into a parameter.
Now let’s define the stream of negative powers of ten:
tenths = block:
fun by-ten(d):
new-denom = d / 10
end
by-ten(1)
end
so that
check:
take(3, tenths) is [list: 1/10, 1/100, 1/1000]
end
For concreteness, let’s pick an abscissa at which to compute the numeric derivative of squaresay 10:
d-dx-square-at-10 = d-dx-square(10)
Recall, from the types, that this is now a function of type (Number -> Number): given a value for epsilon, it computes the derivative using that value. We know, analytically, that the value of this derivative should be 20. We can now (lazily) map tenths to provide increasingly better approximations for epsilon and see what happens:
lz-map(d-dx-square-at-10, tenths)
Sure enough, the values we obtain are 20.1, 20.01, 20.001, and so on: progressively better numerical approximations to 20.
Exercise
Extend the above program to take a tolerance, and draw as many values from the epsilon stream as necessary until the difference between successive approximations of the derivative fall within this tolerance. |
# How to Calculate Square Footage - Step by Step
| Last update: 28 July 2023
Whether you're shopping for carpet, or planning a landscaping or home improvement project, one essential concept is calculating square footage. Join us as we discuss the steps involved in the process.
How to calculate the square footage of a room
Measure the length and width of the room in feet. Next, multiply the length and width together to get the square footage. Example: For a room measuring 10 feet wide by 14 feet long, your calculation is 10 × 14 = 140 square feet.
Having to work out a square footage calculation might give you panicky flashbacks of being late for school! But don't worry, our guide will show you to calculate square footage accurately. We also explain the formula for square footage, so you'll be able to ace any calculations that come up in your future projects, and we look at square feet calculations for irregular-shaped areas. And, as an extra-helpful bonus, we include a square footage calculator to help you check your calculations. Let's begin with step 1...
## 1 How to measure your room/space
To calculate the square feet of a rectangular area, you first need to measure two dimensions: the length and the width (note that we discuss irregular shaped areas further down).
• First, measure the longest side to find the length. Take a tape measure and fix one end of it to one end of the longest side. Note the measurement.
• Next, measure the width - the shortest side of the area to be calculated. Repeat the process above and note down that measurement.
## 2 Square footage formula
For step 2, you're going to use the square footage formula:
Length × Width = Area (in square feet)
Note that the same formula applies if you measure in metric. In this case, the area will be expressed in square metres.
Insert your measurements into the formula. You're going to multiply the length measurement (in feet) by the width measurement. This gives you the area of the space in square feet.
You can also use this formula to calculate smaller areas when the area will be expressed in square inches and larger areas in square yards.
As an example, suppose you want to know how much flooring you might need to renovate a room that's 15 feet long × 10 feet wide. Multiply these dimensions:
15 ft × 10 ft = 150 sq. ft
## 3 Calculate irregular shapes
Sometimes, you might need to calculate a space that's an addition to the main area or a room that isn't evenly shaped. In these situations, think of the space as separate areas. You may find it helpful to sketch out a quick floor-plan to divide the space into more regular shapes.
• Measure the length and width of each separate area and calculate the area of each in square feet.
• Add your calculations to give you the total area in square feet.
### Let's look at an example
Suppose you have a rectangular living room with a narrower dining area attached. Think of the space as being made up of a larger rectangle and a smaller rectangle.
First measure the length and width of the larger section labelled A, then do the same for the smaller section labelled B.
A: 14ft × 10 ft = 140 sq. ft
B: 6 ft × 5 ft = 30 sq. ft.
Now, add both values to calculate the square footage of the room:
140 sq.ft. + 30 sq.ft = 170 sq. ft
So you'll need 170 sq. ft. of flooring materials for this room.
### Triangular and circular shapes
It may be that the area you're measuring isn't square or rectangular, but triangular or circular. If this is the case we have a calculator for square footage that includes multiple room and area shapes. We also have instructions for calculating the square footage of a triangle and a circle. These include details of the measurements you will need to take.
### Measuring the area of a property
If you're selling a property yourself, you can measure the floor space of an entire home in the same way. Take the measurements of each room in turn, calculate the square footage of each, and then add all the values together to give the total square footage.
## 4 Use the square ft calculator
Check your calculations using our square footage calculator below.
ft in
ft in
ft in
ft in
ft in
Calculation results will appear here.
Once you've calculated the square footage of the area, and know how much material you need to order for your project, adjust it to allow for waste. It's best to order 10% more than you think you'll need rather than the exact square footage.
Mistakes can happen, and if you only have the exact amount you need, you might not be able to source matching materials. If you find that you don't need the extra materials, suppliers are usually happy to refund unused tins of paint or unopened packs of tiles. And if you have space, it can be handy to store the surplus materials for any touch-up repairs that are needed in the future.
If you need to calculate the cost of materials for your project, simply multiply your total square feet figure by the price per square foot of the materials you're going to be using.
Once you've learned how to calculate the square footage of an area, you have a skill that will assist you in all your future renovation, landscaping and construction projects. And if time's short, you can use our handy online square footage calculator. |
Various Methods to Rationalize the Numerator
Rationalization in mathematics is a process by which we eliminate the radicals (like square roots) or imaginary numbers from the denominator or numerator of an expression. Rationalizing the denominator is a more commonly known practice. Rationalizing the numerator is equally important and useful for simplifying certain types of expressions and making them easier to work with.
Below, we will explore various methods to rationalize the numerator and provide examples to illustrate these concepts.
Understanding Rationalization
Before diving into the methods, let’s understand rationalization.
In mathematics, irrational numbers, represented by radicals or complex terms, often complicate calculations and expressions, especially when dealing with fractions. Rationalization aims to resolve this complexity by transforming the expression into a more manageable form, typically a rational number or an expression without radicals.
The process of rationalization usually involves multiplying the numerator and denominator of a fraction by carefully chosen expressions, known as conjugates, which effectively eliminate the radicals or complex terms from the expression while preserving its value.
Rationalization is a fundamental technique used in various branches of mathematics, including algebra, calculus, and trigonometry. It allows mathematicians to manipulate and analyze expressions more efficiently, leading to deeper insights into mathematical concepts and facilitating problem-solving processes.
4 Methods to Rationalize the Numerator
1. Rationalizing a Monomial Numerator
A monomial numerator consists of only a single term, typically a single root. Rationalizing this type of numerator can be accomplished by multiplying the numerator and denominator by the same root that exists in the numerator.
Note: This multiplication is simply to help understand the concept of rationalization, as in real-world scenarios, it is common to further rationalize the denominator.
2. Rationalizing a Binomial Numerator with One Radical
A binomial numerator with one radical contains two terms, one of which includes a root. Multiplying the numerator and denominator by the conjugate of the numerator will successfully rationalize this type of numerator.
3. Rationalizing a Binomial Numerator with Two Radicals
When faced with two radicals in a binomial numerator, the approach slightly changes but still employs conjugates. For a numerator like √a + √b, multiply the fraction by its conjugate √a – √b. This harnesses not only the difference of squares but also the property that the product of radicals negates the radical component: (√a + √b)(√a – √b) = a – b, which rids the numerator of radicals.
4. Rationalizing Numerators with Cube Roots or Higher Roots
When dealing with cube roots or higher roots in the numerator, the process remains similar.
To rationalize the numerator, we multiply both numerator and denominator by the appropriate power of the radical to eliminate it. For cube roots, we multiply by the square of the cube root.
These methods of rationalizing the numerator are essential tools for simplifying expressions and solving equations in algebra and calculus. By understanding and applying these techniques, mathematicians can manipulate complex expressions with ease.
Conclusion
Rationalizing the numerator involves multiplying expressions by suitable forms of 1 to eliminate radicals or complex terms. Whether dealing with monomial or binomial numerators, with one or two radicals, or even with cube roots or higher roots, the fundamental principle remains consistent. Mastery of these techniques is fundamental for success in various mathematical disciplines. |
# 2016 AMC 8 Problems/Problem 18
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
In an All-Area track meet, $216$ sprinters enter a $100-$meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?
$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$
## Solution
### Solution 1
From any $n-$th race, only $\frac{1}{6}$ will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal. Starting with the first race: $$\frac{216}{6}=36$$ $$\frac{36}{6}=6$$ $$\frac{6}{6}=1$$ Adding all of the numbers in the second column yields $\boxed{\textbf{(C)}\ 43}$
### Solution 2
Every race eliminates $5$ players. The winner is decided when there is only $1$ runner left. You can construct the equation: $216$ - $5x$ = $1$. Thus, $215$ players have to be eliminated. Therefore, we need $\frac{215}{5}$ games to decide the winner, or $\boxed{\textbf{(C)}\ 43}$
### Solution 3 (Cheap Solution using Answer Choices)
Since $216$ is a power of $6,$ the answer will be in the form of $6x+1.$ We can see that this is odd and the only option is $\boxed{\textbf{(C)}\ 43}$ ~sanaops9
## Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
~savannahsolver |
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Mathematics Grade 1
Educational Links
Strand: OPERATIONS AND ALGEBRAIC THINKING (1.OA)
Represent and solve problems involving addition and subtraction within 20 (Standards 1–2, 5–6). Understand and apply properties of operations and the relationship between addition and subtraction (Standards 3–4). Work with addition and subtraction equations (Standards 7–8).
• \$20 Dot Map
This problem helps students to practice adding three numbers whose sum are 20 or less. It is an open-ended problem with many solutions. This problem would work well either as a whole group or students could work on it as individuals or pairs.
• 20 Tickets
The purpose of the task is for students to add and subtract within 20 and represent complex addition problems with an equation to increase their understanding of and flexibility with the equals sign.
• At the Park
This task includes three different problem types using the "Add To" context with a discrete quantity.
• Boys and Girls, Variation 1
This task represents the Put Together/Take Apart contexts for addition and subtraction. Students may use either addition or subtraction to solve these types of word problems, with addition related to the action of putting together and subtraction related to the action of taking apart.
• Boys and Girls, Variation 2
This task represents the Put Together/Take Apart with both addends unknown context for addition and subtraction.
• Cave Game Subtraction
The purpose of this task is for students to practice creating and memorizing subtraction equations, while focusing on missing addends.
• Daisies in Vases
Given a scenario about daisies and vases, students must try to find as many ways as they can to put the daisies in the vases with the most in the large vase and the least in the smallest vase. They then must explain how they know they have found all the possibilities.
• Domino Addition
The purpose of this task is to help students understand the commutative property of addition. Because the total number of dots is the same regardless of how a domino is oriented, the domino reinforces the idea that the addends can be written in any order.
• Doubles?
Commentary on How 6 + 6 should be 12
• Equality Number Sentences
The purpose of this instructional task is for students to help students understand the meaning of the equal sign and to use it appropriately.
• Fact Families
The purpose of this task is for students to identify and write sets of related addition and subtraction equations; these are often known as "fact families" because the equations are related by the same underlying relationship between the numbers. This task reinforces the commutative property of addition and using the relationship between addition and subtraction.
• Fact Families with Pictures
The purpose of this task is for students to reinforce students' understanding of "fact families." Fact families reinforce the commutative property of addition and using the relationship between addition and subtraction. Working with fact families is a common activity, although often times students are asked only to work with symbols. Also, the scaffolding for these tasks often only supports students in writing four of the eight possible facts in a family; this task purposefully scaffolds them to write all eight. In addition, this task includes a picture to anchor each fact-family; students can graduate from here to a symbols-only version of this task
• Field Day Scarcity
The purpose of this task is for students to relate addition and subtraction problems to money in a context that introduces the concept of scarcity.
• Find the Missing Number
This task asks students to solve addition and subtraction equations with different structures so that they are able to see the connections between addition and subtraction more easily.
• Finding a Chair
This task represents compare contexts for addition and subtraction. The problem explicitly describes one-to-one correspondences without using comparison language.
• Georgia Standards of Excellence Mathematics
GeorgiaStandards.Org (GSO) is a free, public website providing information and resources necessary to help meet the educational needs of students. The goal of this web site is to provide information that will enhance and support teaching and learning of Georgia standards.
• Grade 1 Math Module 1: Sums and Differences to 10 (EngageNY)
In this first module of Grade 1, students make significant progress towards fluency with addition and subtraction of numbers to 10 as they are presented with opportunities intended to advance them from counting all to counting on which leads many students then to decomposing and composing addends and total amounts.
• Grade 1 Math Module 2: Intro to Place Value Through Addition and Subtraction Within 20 (EngageNY)
Module 2 serves as a bridge from students' prior work with problem solving within 10 to work within 100 as students begin to solve addition and subtraction problems involving teen numbers. Students go beyond the Level 2 strategies of counting on and counting back as they learn Level 3 strategies informally called "make ten" or "take from ten."
• Grade 1 Math Module 3: Ordering and Comparing Length Measurements as Numbers (EngageNY)
Module 3 begins by extending students' kindergarten experiences with direct length comparison to indirect comparison whereby the length of one object is used to compare the lengths of two other objects. Longer than and shorter than are taken to a new level of precision by introducing the idea of a length unit. Students then explore the usefulness of measuring with similar units. The module closes with students representing and interpreting data.
• Grade 1 Math Module 4: Place Value, Comparison, Addition and Subtraction to 40 (EngageNY)
Module 4 builds upon Module 2's work with place value within 20, now focusing on the role of place value in the addition and subtraction of numbers to 40. Students study, organize, and manipulate numbers within 40. They compare quantities and begin using the symbols for greater than (>) and less than (<). Addition and subtraction of tens is another focus of this module as is the use of familiar strategies to add two-digit and single-digit numbers within 40. Near the end of the module, the focus moves to new ways to represent larger quantities and adding like place value units as students add two-digit numbers.
• Grade 1 Math Module 6: Place Value, Comparison, Addition and Subtraction to 100 (EngageNY)
In this final module of the Grade 1 curriculum, students bring together their learning from Module 1 through Module 5 to learn the most challenging Grade 1 standards and celebrate their progress. As the module opens, students grapple with comparative word problem types. Next, they extend their understanding of and skill with tens and ones to numbers to 100. Students also extend their learning from Module 4 to the numbers to 100 to add and subtract. At the start of the second half of Module 6, students are introduced to nickels and quarters, having already used pennies and dimes in the context of their work with numbers to 40 in Module 4. Students use their knowledge of tens and ones to explore decompositions of the values of coins. The module concludes with fun fluency festivities to celebrate a year's worth of learning.
• Grade 1 Unit 3: Operations and Algebraic Thinking (Georgia Standards)
In this unit, students will explore, understand, and apply the commutative and associative properties as strategies for solving addition problems. Share, discuss, and compare strategies as a class. Connect counting on to solving subtraction problems. For the problem 15 7 = ? they think about the number they have to count on from 7 to get to 15. Work with sums and differences less than or equal to 20 using the numbers 0 to 20. Identify and then apply a pattern or structure in mathematics. For example, pose a string of addition and subtraction problems involving the same three numbers chosen from the numbers 0 to 20, such as 4 + 13 = 17 and 13 + 4 = 17. Analyze number patterns and create conjectures or guesses.
• Growing Bean Plants
This task adds some rigor to this common activity, by collecting actual growth data, providing practice for students in measuring and recording length measurements.
• IXL Game: Addition: Word problems
This game helps first graders understand addition using word problems. This is just one of many online games that supports the Utah Math core. Note: The IXL site requires subscription for unlimited use.
• Kiri's Mathematics Match Game - 1st Grade
In this activity students play a game in small groups that requires them to find sums or differences.
• Link-Cube Addition
The purpose of this task is for students to identify and represent related addition and subtraction equations with objects and equations.
• Making a ten
This task is designed to help students visualize where the 10's are on a single digit addition table and explain why this is so. This knowledge can then be used to help them learn the addition table.
• Maria's Marbles
This task includes problem types that represent the Compare contexts for addition and subtraction. The student must solve word problems involving marbles.
• Measuring Blocks
This task has students work in pairs. Given two or more blocks of different lengths and paper clips to use to measure them, the pair measures the blocks and tell how many paper clips long it is. They then figure out how long two different blocks would be when laid end to end and explain how they solved the problem.
• Operations and Algebraic Thinking (1.OA) - First Grade Core Guide
The Utah State Board of Education (USBE) and educators around the state of Utah developed these guides for First Grade Mathematics - Operations and Algebraic Thinking (1.OA)
• Peyton's Books
This task was designed to help students to make sense of problems and persevere to solve them, as well as understanding the relationship between addition and subtraction. Students will solve the take from, change unknown problem, and through a teacher-facilitated discussion, understand that the problem can be solved using addition or subtraction.
• School Supplies
In solving a word problem about school supplies, this task could be used for either instructional or assessment purposes, depending on where students are in their understanding of addition.
• Sharing Markers
This task includes the three different problem types using the Take From context: result unknown, change unknown, and start unknown.
• The Pet Snake
Students are given 3 word problems about a pet snake. The task uses a continuous quantity for these examples of addition and subtraction.
• The Very Hungry Caterpillar
In this math lesson the teacher reads the book to the class and asks, "How many things do you think the caterpillar ate in this story?" The students take a minute to share their estimate with a partner. Next, the teacher reads The Very Hungry Caterpillar again. After each page, the teacher pauses so that the students can add counters or unifix cubes to the ten-frame to represent the number of things the caterpillar ate, and then write an equation on the dry-erase board connecting addition to the number of counters used.
• Using lengths to represent equality
In this task students work in pairs and use Cuisenaire rods or paper strips cut to whole centimeter lengths. One student puts a few rods (or strips) end-to-end. The other student matches that length with a different combination of rods (or strips). When two different ways of making the same length are found, the students write a number sentence reflecting the equality.
• Valid Equalities?
The purpose of this task is to help broaden and deepen students understanding of the equals sign and equality. In this task, students must attend to the meaning of the equal sign by determining whether or not the left-hand expression and the right hand expression are equal. This task helps students attend to precision.
http://www.uen.org - in partnership with Utah State Board of Education (USBE) and Utah System of Higher Education (USHE). Send questions or comments to USBE Specialist - Shannon Olson and see the Mathematics - Elementary website. For general questions about Utah's Core Standards contact the Director - Jennifer Throndsen .
These materials have been produced by and for the teachers of the State of Utah. Copies of these materials may be freely reproduced for teacher and classroom use. When distributing these materials, credit should be given to Utah State Board of Education. These materials may not be published, in whole or part, or in any other format, without the written permission of the Utah State Board of Education, 250 East 500 South, PO Box 144200, Salt Lake City, Utah 84114-4200. |
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# Mathematics Class 7 – Chapter 4 – Simple Equation – Exercise – 4.2 – NCERT Exercise Solution
1. Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
Solution:
Adding 1 to both the side of given equation,
we get,
x – 1 + 1 = 0 + 1
x = 1
(b) x + 1 = 0
Solution:
Subtracting 1 to both the side of given equation,
we get,
x + 1 – 1 = 0 – 1
x = – 1
(c) x – 1 = 5
Solution:
Adding 1 to both the side of given equation,
we get,
x – 1 + 1 = 5 + 1
x = 6
(d) x + 6 = 2
Solution:
Substracting 6 to both the side of given equation,
we get,
x + 6 – 6 = 2 – 6
x = – 4
(e) y – 4 = – 7
Solution:
Adding 4 to both the side of given equation,
we get,
y – 4 + 4 = – 7 + 4
y = – 3
(f) y – 4 = 4
Solution:
Adding 4 to both the side of given equation,
we get,
y – 4 + 4 = 4 + 4
y = 8
(g) y + 4 = 4
Solution:
Substracting 4 to both the side of given equation,
we get,
= y + 4 – 4 = 4 – 4
= y = 0
(h) y + 4 = – 4
Solution:
Substracting 4 to both the side of given equation,
we get,
= y + 4 – 4 = – 4 – 4
= y = – 8
2. Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
Solution:
Dividing both sides of the equation by 3,
we get,
3l/3 = 42/3
l = 14
(b) b/2 = 6
Solution:
Multiplying both sides of the equation by 2,
we get,
b/2 × 2= 6 × 2
b = 12
(c) p/7 = 4
Solution:
Multiplying both sides of the equation by 7,
we get,
p/7 × 7= 4 × 7
p = 28
(d) 4x = 25
Solution:
Dividing both sides of the equation by 4,
we get,
4x/4 = 25/4
x = 25/4
(e) 8y = 36
Solution:
Dividing both sides of the equation by 8,
we get,
8y/8 = 36/8
x = 9/2
(f) (z/3) = (5/4)
Solution:
Multiplying both sides of the equation by 3,
we get,
(z/3) × 3 = (5/4) × 3
x = 15/4
(g) (a/5) = (7/15)
Solution:
Multiplying both sides of the equation by 5,
we get,
(a/5) × 5 = (7/15) × 5
a = 7/3
(h) 20t = – 10
Solution:
Dividing both sides of the equation by 20,
we get,
20t/20 = -10/20
x = – ½
3. Give the steps you will use to separate the variable and then solve the equation:
(a) 3n – 2 = 46
Solution:
In step 1 we add 2 to the both sides of the equation,
we get,
3n – 2 + 2 = 46 + 2
= 3n = 48
In step 2 we divide both sides of the equation by 3,
we get,
3n/3 = 48/3
n = 16
(b) 5m + 7 = 17
Solution:
In step 1 we subtract 7 to the both sides of the equation,
we get,
5m + 7 – 7 = 17 – 7
5m = 10
In step 2 we divide both sides of the equation by 5,
we get,
5m/5 = 10/5
m = 2
(c) 20p/3 = 40
Solution:
In step 1 we multiply both sides of the equation by 3,
we get,
(20p/3) × 3 = 40 × 3
20p = 120
In step 2 we divide both sides of the equation by 20,
we get,
20p/20 = 120/20
p = 6
(d) 3p/10 = 6
Solution:
In step 1 we multiply both sides of the equation by 10,
we get,
(3p/10) × 10 = 6 × 10
3p = 60
In step 2 we divide both sides of the equation by 3,
Then, we get,
3p/3 = 60/3
p = 20
4. Solve the following equations:
(a) 10p = 100
Solution:
Dividing both sides of the equation by 10,
we get,
10p/10 = 100/10
p = 10
(b) 10p + 10 = 100
Solution:
Subtracting 10 to the both sides of the equation,
we get,
10p + 10 – 10 = 100 – 10
10p = 90
Now,
Dividing both sides of the equation by 10,
we get,
10p/10 = 90/10
p = 9
(c) p/4 = 5
Solution:
Multiplying both sides of the equation by 4,
we get,
p/4 × 4 = 5 × 4
p = 20
(d) – p/3 = 5
Solution:
Multiplying both sides of the equation by – 3,
we get,
– p/3 × (- 3) = 5 × (- 3)
p = – 15
(e) 3p/4 = 6
Solution:
Multiplying both sides of the equation by 4,
we get,
(3p/4) × (4) = 6 × 4
3p = 24
Now,
Dividing both sides of the equation by 3,
we get,
3p/3 = 24/3
p = 8
(f) 3s = – 9
Solution:
Dividing both sides of the equation by 3,
we get,
3s/3 = -9/3
s = -3
(g) 3s + 12 = 0
Solution:
Substracting 12 to the both sides of the equation,
we get,
3s + 12 – 12 = 0 – 12
3s = -12
Now,
Dividing both sides of the equation by 3,
we get,
3s/3 = -12/3
s = – 4
(h) 3s = 0
Solution:
Dividing both sides of the equation by 3,
we get,
3s/3 = 0/3
s = 0
(i) 2q = 6
Solution:
Dividing both sides of the equation by 2,
we get,
2q/2 = 6/2
q = 3
(j) 2q – 6 = 0
Solution:
Adding 6 to the both sides of the equation,
we get,
2q – 6 + 6 = 0 + 6
2q = 6
Now,
Dividing both sides of the equation by 2,
we get,
2q/2 = 6/2
q = 3
(k) 2q + 6 = 0
Solution:
Subtracting 6 to the both sides of the equation,
we get,
2q + 6 – 6 = 0 – 6
2q = – 6
Now,
Dividing both sides of the equation by 2,
we get,
2q/2 = – 6/2
q = – 3
(l) 2q + 6 = 12
Solution:
Subtracting 6 to the both sides of the equation,
we get,
2q + 6 – 6 = 12 – 6
2q = 6
Now,
Dividing both sides of the equation by 2,
we get,
2q/2 = 6/2
q = 3
👍👍👍
error: |
# Help please? Having trouble with this.
A race car makes two and half laps around a circular track in 3.0 min. What is the car's average angular speed?
## To find the car's average angular speed, we need to know the total angle covered by the car and the time taken.
Step 1: Find the total angle covered by the car.
The car completes two and a half laps around the circular track. Since one lap is equivalent to a full circle, the angle covered by the car is 2.5 times the angle of a full circle, which is 2.5 x 360 degrees.
Total angle covered = 2.5 x 360 = 900 degrees.
Step 2: Convert the time to seconds.
The given time is 3.0 minutes. To calculate the average angular speed, we need the time in seconds since angular speed is measured in units of radians per second.
Time in seconds = 3.0 minutes x 60 = 180 seconds.
Step 3: Calculate the average angular speed.
Angular speed (ω) is given by the formula:
ω = (Total angle covered) / (Time taken)
Substituting the values we found earlier:
ω = 900 degrees / 180 seconds.
However, we need to convert the angle to radians since the angular speed is usually measured in radians.
ω = (900 x π/180) radians / 180 seconds.
Step 4: Simplify the expression.
ω = (5π/2) radians / 180 seconds.
Therefore, the car's average angular speed is (5π/2) radians per 180 seconds.
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## To find the car's average angular speed, we need to calculate the number of radians the car rotates through in a given time. The formula for angular speed (ω) is:
ω = θ/t
Where:
ω - angular speed
θ - angle rotated
t - time interval
In this case, the car makes two and a half laps around the circular track. Since one lap is equal to 2π radians, two and a half laps would be:
2.5 * 2π = 5π radians
The time taken by the car is 3.0 minutes. We need to convert this into seconds, as angular speed is typically measured in radians per second:
3.0 minutes * 60 seconds/minute = 180 seconds
Now we can use the formula to find the angular speed:
ω = θ/t
ω = (5π radians) / (180 seconds)
Calculating this expression will give us the car's average angular speed.
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## one lap (circle) is 2π radians
5/2 * 2π radians in 3.0 min (180 sec) |
Mathematics Solutions Solutions for Class 8 Math Chapter 1 Rational And Irrational Numbers are provided here with simple step-by-step explanations. These solutions for Rational And Irrational Numbers are extremely popular among Class 8 students for Math Rational And Irrational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.
#### Question 1:
If marked price = ₹1700, selling price = ₹1540 then find the discount.
Marked price = ₹ 1,700
Selling price = ₹ 1,540
∴ Discount = Marked price − Selling price = ₹ 1,700 − ₹ 1,540 = ₹ 160
#### Question 2:
If marked price = ₹990 and percentage of discount is 10, then find the selling price.
10% discount is given on the marked price. Therefore, if marked price is ₹ 100, then selling price is ₹ 90.
Let the selling price be ₹ when the marked price is ₹ 990.
$\therefore \frac{x}{990}=\frac{90}{100}$
$⇒x=\frac{90}{100}×990$ = ₹ 891
Thus, the selling price is ₹ 891.
#### Question 3:
If selling price = ₹900. Discount is 20%, then find the marked price.
20% discount is given on the marked price. Therefore, if marked price is ₹ 100, then selling price is ₹ 80.
Let the marked price be ₹ when the selling price is ₹ 900.
$\therefore \frac{900}{x}=\frac{80}{100}$
$⇒x=\frac{100}{80}×900$ = ₹ 1,125
Thus, the marked price is ₹ 1,125.
#### Question 4:
The marked price of the fan is 3000 rupees. Shopkeeper gave 12% discount on it. Find the total discount and selling price of the fan.
12% discount is given on the marked price. Therefore, if marked price is ₹ 100, then selling price is ₹ 88.
Let the selling price be ₹ when the marked price is ₹ 3,000.
$\therefore \frac{x}{3000}=\frac{88}{100}$
$⇒x=\frac{88}{100}×3000$ = ₹ 2,640
So, the selling price of the fan is ₹ 2,640.
∴ Discount = Marked price − Selling price = ₹ 3,000 − ₹ 2,640 = ₹ 360
Thus, the discount on fan is ₹ 360.
#### Question 5:
The marked price of a mixer is 2300 rupees. A customer purchased it for Rs. 1955. Find percentage of discount offered to the customer.
Marked price of the mixer = ₹ 2,300
Selling price of the mixer = ₹ 1,955
∴ Discount = Marked price − Selling price = ₹ 2,300 − ₹ 1,955 = ₹ 345
On marked price of ₹ 2,300, the discount is ₹ 345.
Let the discount be ₹ x.
Now, if the marked price is ₹ 100, then the discount is ₹ x.
∴ $\frac{x}{100}=\frac{345}{2300}$
⇒ $x=\frac{345}{2300}×100$ = 15
Thus, the percentage of discount offered to the customer is 15%.
#### Question 6:
A shopkeeper gives 11% discount on a television set, hence the cost price of it is Rs. 22,250. Then find the marked price of the television set.
Let the marked price of the television set be Rs x.
Discount percent on the television set = 11%
If the marked price of the television set was Rs 100, then the customer would have paid (Rs 100 − Rs 11) Rs 89 for the television set.
It is given that cost price of the television set to the customer is Rs 22,250.
∴ $\frac{89}{100}=\frac{22250}{x}$
⇒ $x=22250×\frac{100}{89}$ = Rs 25,000
Thus, the marked price of the television set is Rs 25,000.
#### Question 7:
After offering discount of 10% on marked price, a customer gets total discount of 17 rupees. To find the cost price for the customer, fill in the following boxes with appropriate numbers and complete the activity.
Suppose, marked price of the item = 100 rupees
Therefore, for customer that item costs = 90 rupees
Hence, when the discount is then the selling price is rupees.
Suppose when the discount is rupees, the selling price is x rupees.
∴ the customer will get the item for 153 rupees.
Suppose, marked price of the item = 100 rupees
Therefore, for customer that item costs = 90 rupees
Hence, when the discount is $\overline{)10}$ then the selling price is $\overline{)90}$ rupees.
Suppose when the discount is $\overline{)17}$ rupees, the selling price is x rupees.
∴ the customer will get the item for 153 rupees.
#### Question 8:
A shopkeeper decides to sell a certain item at a certain price. He tags the price on the item by increasing the decided price by 25%. While selling the item, he offers 20% discount. Find how many more or less percent he gets on the decided price.
Let the decided price of the item be ₹ 100.
∴ Marked price of the item = ₹ 100 + 25% of ₹ 100 = ₹ 100 + ₹ 25 = ₹ 125
Discount on the item = 20% of ₹ 125 = $\frac{20}{100}×125$ = ₹ 25
∴ Selling price of the item = Marked price of the item − Discount on the item = ₹ 125 − ₹ 25 = ₹ 100
Since the decided price of the item is same as the selling price of the item so the shopkeeper neither makes any profit nor any loss on selling the item.
Thus, the shopkeeper gets 0% on the decided price.
#### Question 1:
John sold books worth rupees 4500 for a publisher. For this he received 15% commission. Complete the following activity to find the total commission John obtained.
Selling price of books =
Rate of commission =
Commission obtained =
∴ Commission = rupees.
Selling price of books = $\overline{)4500}$ rupees
Rate of commission = $\overline{)15%}$
Commission obtained = $\frac{\overline{)15}}{\overline{)100}}×\overline{)4500}$ = Rs 675
∴ Commission = $\overline{)675}$ rupees
#### Question 2:
Rafique sold flowers worth ₹ 15,000 by giving 4% commission to the agent. Find the commission he paid. Find the amount received by Rafique.
Selling price of the flowers = ₹ 15,000
Commission rate = 4%
Commission paid to the agent = 4% of ₹ 15,000 = $\frac{4}{100}×15000$ = ₹ 600
So, the commission paid by Rafique is ₹ 600.
∴ Amount received by the Rafique
= Selling price of the flowers − Commission paid to the agent
= ₹ 15,000 − ₹ 600
= ₹ 14,400
Thus, the amount received by Rafique is ₹ 14,400.
#### Question 3:
A farmer sold foodgrains for 9200 rupees through an agent. The rate of commission was 2%. How much amount did the agent get?
Selling price of the foodgrains = ₹ 9,200
Commission rate = 2%
∴ Commission paid to the agent = 2% of ₹ 9,200 = $\frac{2}{100}×9200$ = ₹ 184
Thus, the commission received by the agent is ₹ 184.
#### Question 4:
Umatai purchased following items from a Khadi-Bhandar.
(i) 3 sarees for 560 rupees each.
(ii) 6 bottles of honey for 90 rupees each.
On the purchase, she received a rebate of 12% . How much total amount did Umatai pay?
Cost of each saree = Rs 560
∴ Cost of 3 sarees = 3 × 560 = Rs 1,680
Cost of each bottle of honey = Rs 90
∴ Cost of 6 bottles of honey = 6 × 90 = Rs 540
Total cost of the items purchased = Rs 1,680 + Rs 540 = Rs 2,220
Total rebate received = 12% of Rs 2,220 = $\frac{12}{100}×2220$ = Rs 266.40
∴ Total amount paid by Umatai
= Total cost of the items purchased − Total rebate received
= Rs 2,220 − Rs 266.40
= Rs 1,953.60
Thus, the total amount paid by Umatai is Rs 1,953.60.
#### Question 5:
Use the given information and fill in the boxes with suitable numbers. Smt. Deepanjali purchased a house for ₹7,50,000 from Smt. Leelaben through an agent. Agent has charged 2% brokerage from both of them.
(1) Smt. Deepanjali paid = ₹ brokerage for purchasing the house.
(2) Smt. Leelaben paid brokerage of ₹
(3) Total brokerage received by the agent is ₹
(4) The cost of house Smt. Deepanjali paid is ₹
(5) The selling price of house for Smt. Leelaben is ₹ .
(1) Smt. Deepanjali paid $\overline{)750000}×\frac{\overline{)2}}{\overline{)100}}$ = ₹ $\overline{)15,000}$ brokerage for purchasing the house.
(2) Smt. Leelaben paid brokerage of ₹ $\overline{)15,000}$.
(3) Total brokerage received by the agent is ₹ 15,000 + ₹ 15,000 = ₹ $\overline{)30,000}$.
(4) The cost of house Smt. Deepanjali paid is ₹ 7,50,000 + ₹ 15,000 = ₹ $\overline{)7,65,000}$.
(5) The selling price of house for Smt. Leelaben is ₹ 7,50,000 − ₹ 15,000 = ₹ $\overline{)7,35,000}$.
View NCERT Solutions for all chapters of Class 8 |
# Bank Exam Success: Strategies for Boat and Stream Problems
#### ByGrace
Feb 9, 2024
Bank exams are competitive, and candidates must be well-prepared to tackle various mathematical problems. One common problem frequently appearing in these exams is the “Boat and Stream” problem. These issues assess a candidate’s ability to solve complex mathematical equations involving relative speeds and distances travelled by boats and streams. This article will explore essential strategies to excel in these problems to increase your chances of success in bank exams.
## Ace the Basics
In these problems, you are typically given information about the speed of a vessel in still water and the speed of a stream (river or current). The ship can either move upstream (against the current) or downstream (with the present). You aim to find the vessel’s speed and calculate its time to cover a certain distance.
## Visualise the Situation
To effectively solve these questions, start by visualising the scenario. Imagine a vessel travelling on a river with a current. Try to picture the vessel moving against the current and with the current. This visualisation will help you grasp the relationship between its speed, stream speed, and its combined effect on the vessel’s speed.
## Use the Relative Speed Concept
One key concept in such problems is relative speed. When a vessel moves against the current, its effective speed is reduced because it works against the stream’s flow. Conversely, when moving with the current, its effective speed increases. To calculate these adequate speeds, use the formula:
Speed against the current = boat’s speed – Stream’s speed
Speed with the current = Boat’s speed + stream’s speed
Understanding this concept is essential for solving problems where the vessel’s speed is affected by the stream’s flow.
## Calculate Time and Distance
Once you’ve established the vessel’s relative speed when moving against or with the current, you can then compute the amount of time required to traverse a particular distance. The formula for time is:
Time = Distance / Speed
When solving these problems, remember that time and distance are inversely proportional. If you know the time it takes to travel a certain distance, you can calculate its speed. Similarly, if you have the speed and the time taken, you can find the distance travelled.
## Work with Equations
In many Bank exams, these problems are presented as equations. To solve them, write down the equations based on the information provided and then solve for the unknown variables. For example, if you are given the vessel’s speed in still water and the stream’s speed, you can set up equations for its speed against and with the current, respectively:
Boat’s speed against current = boat’s speed – Stream’s speed
Boat’s speed with current = Boat’s speed + stream’s speed
Utilising these equations, you can ascertain the vessel’s velocity, subsequently enabling you to determine both the time taken and the distance covered.
## Practice Regularly
Like any mathematical skill, mastering these problems requires practice. The greater the number of problems you solve, the more proficient you’ll become in handling the diverse challenges and intricacies that may arise in bank exams. Practice helps you apply the strategies and boosts your confidence in tackling these problems during the exam.
## Work on Speed and Accuracy
Bank exams are time-bound, so it’s essential to be both accurate and efficient in your calculations. Practice mental math techniques to speed up your calculations. Additionally, pay attention to units (e.g., kilometres per hour) and ensure consistency in your units throughout the problem-solving process.
## Conclusion
Such problems revolving around boats and streams are common in bank exams, and mastering them is crucial for success. You can significantly enhance your problem-solving skills by understanding the basics, using the relative speed concept, calculating time and distance, working with equations, practising regularly, improving speed and accuracy, and identifying patterns and shortcuts. |
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# 3.5: Solve Uniform Motion Applications
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##### Learning Objectives
By the end of this section, you will be able to:
• Solve uniform motion applications
##### Note
Before you get started, take this readiness quiz.
1. Find the distance traveled by a car going 70 miles per hour for 3 hours.
If you missed this problem, review Exercise 2.6.1.
2. Solve $$x+1.2(x−10)=98$$.
If you missed this problem, review Exercise 2.4.7.
3. Convert 90 minutes to hours.
If you missed this problem, review Exercise 1.11.1.
## Solve Uniform Motion Applications
When planning a road trip, it often helps to know how long it will take to reach the destination or how far to travel each day. We would use the distance, rate, and time formula, D=rt, which we have already seen.
In this section, we will use this formula in situations that require a little more algebra to solve than the ones we saw earlier. Generally, we will be looking at comparing two scenarios, such as two vehicles traveling at different rates or in opposite directions. When the speed of each vehicle is constant, we call applications like this uniform motion problems.
Our problem-solving strategies will still apply here, but we will add to the first step. The first step will include drawing a diagram that shows what is happening in the example. Drawing the diagram helps us understand what is happening so that we will write an appropriate equation. Then we will make a table to organize the information, like we did for the money applications.
The steps are listed here for easy reference:
##### USE A PROBLEM-SOLVING STRATEGY IN DISTANCE, RATE, AND TIME APPLICATIONS.
1. Read the problem. Make sure all the words and ideas are understood.
• Draw a diagram to illustrate what it happening.
• Create a table to organize the information.
• Label the columns rate, time, distance.
• List the two scenarios.
• Write in the information you know.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
• Complete the chart.
• Use variable expressions to represent that quantity in each row.
• Multiply the rate times the time to get the distance.
4. Translate into an equation.
• Restate the problem in one sentence with all the important information.
• Then, translate the sentence into an equation.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
##### Exercise $$\PageIndex{1}$$
An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in 4 hours and the local train takes 5 hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains.
Step 1. Read the problem. Make sure all the words and ideas are understood.
Draw a diagram to illustrate what it happening. Shown below is a sketch of what is happening in the example.
Create a table to organize the information. Label the columns “Rate,” “Time,” and “Distance.” List the two scenarios. Write in the information you know.
Step 2. Identify what we are looking for.
We are asked to find the speed of both trains. Notice that the distance formula uses the word “rate,” but it is more common to use “speed” when we talk about vehicles in everyday English.
Step 3. Name what we are looking for. Choose a variable to represent that quantity.
Complete the chart Use variable expressions to represent that quantity in each row. We are looking for the speed of the trains. Let’s let r represent the speed of the local train. Since the speed of the express train is 12 mph faster, we represent that as r+12.
\begin{aligned} r &=\text { speed of the local train } \\ r+12 &=\text { speed of the express train } \end{aligned}
Fill in the speeds into the chart.
Multiply the rate times the time to get the distance.
Step 4. Translate into an equation.
Restate the problem in one sentence with all the important information. Then, translate the sentence into an equation.
• The equation to model this situation will come from the relation between the distances. Look at the diagram we drew above. How is the distance traveled by the express train related to the distance traveled by the local train?
• Since both trains leave from Pittsburgh and travel to Washington, D.C. they travel the same distance. So we write:
Step 5. Solve the equation using good algebra techniques.
Now solve this equation. So the speed of the local train is 48 mph. Find the speed of the express train. he speed of the express train is 60 mph.
Step 6. Check the answer in the problem and make sure it makes sense. $\begin{array}{ll}{\text { express train }} & {60 \mathrm{mph}(4 \text { hours })=240 \mathrm{miles}} \\ {\text { local train }} & {48 \mathrm{mph}(5 \text { hours })=240 \mathrm{miles} \checkmark \end{array}$
Step 7. Answer the question with a complete sentence.
The speed of the local train is 48 mph and the speed of the express train is 60 mph.
##### Exercise $$\PageIndex{2}$$
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
Wayne 21 mph, Dennis 28 mph
##### Exercise $$\PageIndex{3}$$
Jeromy can drive from his house in Cleveland to his college in Chicago in 4.5 hours. It takes his mother 6 hours to make the same drive. Jeromy drives 20 miles per hour faster than his mother. Find Jeromy’s speed and his mother’s speed.
Jeromy 80 mph, mother 60 mph
In Exercise $$\PageIndex{4}$$, the last example, we had two trains traveling the same distance. The diagram and the chart helped us write the equation we solved. Let’s see how this works in another case.
##### Exercise $$\PageIndex{4}$$
Christopher and his parents live 115 miles apart. They met at a restaurant between their homes to celebrate his mother’s birthday. Christopher drove 1.5 hours while his parents drove 1 hour to get to the restaurant. Christopher’s average speed was 10 miles per hour faster than his parents’ average speed. What were the average speeds of Christopher and of his parents as they drove to the restaurant?
Step 1. Read the problem. Make sure all the words and ideas are understood.
Draw a diagram to illustrate what it happening. Below shows a sketch of what is happening in the example.
Create a table to organize the information.
Label the columns rate, time, distance.
List the two scenarios.
Write in the information you know.
Step 2. Identify what we are looking for.
We are asked to find the average speeds of Christopher and his parents.
Step 3. Name what we are looking for. Choose a variable to represent that quantity.
Complete the chart.
Use variable expressions to represent that quantity in each row.
We are looking for their average speeds. Let’s let r represent the average speed of the parents. Since the Christopher’s speed is 10 mph faster, we represent that as r+10.
Fill in the speeds into the chart.
Multiply the rate times the time to get the distance.
Step 4. Translate into an equation.
Restate the problem in one sentence with all the important information. Then, translate the sentence into an equation. Again, we need to identify a relationship between the distances in order to write an equation. Look at the diagram we created above and notice the relationship between the distance Christopher traveled and the distance his parents traveled.
The distance Christopher travelled plus the distance his parents travel must add up to 115 miles. So we write:
Step 5. Solve the equation using good algebra techniques.
$$\begin{array} {cc} {} &{1.5(r + 10) + r = 115} \\ {} &{1.5r + 15 + r = 115} \\ {\text{Now solve this equation.}} &{2.5r + 15 = 115} \\{} &{2.5r = 100} \\{} &{r = 40} \\ {} &{\text{so the parents' speed was 40 mph.}} \\ {} &{r + 10} \\ {\text{Christopher's speed is r + 10}} &{40 + 10} \\ {} &{50} \\ {} &{\text{Christopher's speed was 50 mph.}} \\ {} &{} \end{array}$$
Step 6. Check the answer in the problem and make sure it makes sense.
$$\begin{array}{llll} {\text{Christopher drove}} &{50\text{ mph (1.5 hours)}} &{=} &{75\text{ miles}}\\ {\text{His parents drove}} &{40\text{ mph (1 hour)}} &{=} &{\underline{40 \text{ miles}}}\\ {} &{} &{} &{115\text{ miles}} \end{array}$$
$$\begin{array}{ll} {\textbf{Step 7. Answer}\text{ the question with a complete sentence.}} &{} \\{} &{\text{Christopher's speed was 50 mph.}}\\ {} &{\text{His parents' speed was 40 mph.}} \end{array}$$
##### Exercise $$\PageIndex{5}$$
Carina is driving from her home in Anaheim to Berkeley on the same day her brother is driving from Berkeley to Anaheim, so they decide to meet for lunch along the way in Buttonwillow. The distance from Anaheim to Berkeley is 410 miles. It takes Carina 3 hours to get to Buttonwillow, while her brother drives 4 hours to get there. The average speed Carina’s brother drove was 15 miles per hour faster than Carina’s average speed. Find Carina’s and her brother’s average speeds.
Carina 50 mph, brother 65 mph
##### Exercise $$\PageIndex{6}$$
Ashley goes to college in Minneapolis, 234 miles from her home in Sioux Falls. She wants her parents to bring her more winter clothes, so they decide to meet at a restaurant on the road between Minneapolis and Sioux Falls. Ashley and her parents both drove 2 hours to the restaurant. Ashley’s average speed was seven miles per hour faster than her parents’ average speed. Find Ashley’s and her parents’ average speed.
parents 55 mph, Ashley 62 mph
As you read the next example, think about the relationship of the distances traveled. Which of the previous two examples is more similar to this situation?
##### Exercise $$\PageIndex{7}$$
Two truck drivers leave a rest area on the interstate at the same time. One truck travels east and the other one travels west. The truck traveling west travels at 70 mph and the truck traveling east has an average speed of 60 mph. How long will they travel before they are 325 miles apart?
Step 1. Read the problem. Make sure all the words and ideas are understood.
Draw a diagram to illustrate what it happening.
Create a table to organize the information.
Step 2. Identify what we are looking for.
We are asked to find the amount of time the trucks will travel until they are 325 miles apart.
Step 3. Name what we are looking for. Choose a variable to represent that quantity.
We are looking for the time traveled. Both trucks will travel the same amount of time. Let’s call the time t. Since their speeds are different, they will travel different distances. Complete the chart.
Step 4. Translate into an equation.
We need to find a relation between the distances in order to write an equation. Looking at the diagram, what is the relationship between the distance each of the trucks will travel? The distance traveled by the truck going west plus the distance traveled by the truck going east must add up to 325 miles. So we write:
Step 5. Solve the equation using good algebra techniques.
$\begin{array} {lrll} {\text{Now solve this equation. }} & {70 t+60 t} &{=} &{325} \\ {} &{130 t} &{=} &{325} \\ {} &{t} &{=} &{2.5} \end{array}$
Step 6. Check the answer in the problem and make sure it makes sense.
$$\begin{array}{llll} {\text{Truck going West}} &{70\text{ mph (2.5 hours)}} &{=} &{175\text{ miles}}\\ {\text{Truck going East}} &{60\text{ mph (2.5 hour)}} &{=} &{\underline{150 \text{ miles}}}\\ {} &{} &{} &{325\text{ miles}} \end{array}$$
$$\begin{array}{ll} \\{\textbf{Step 7. Answer}\text{ the question with a complete sentence.}} &{\text{It will take the truck 2.5 hours to be 325 miles apart.}} \end{array}$$
##### Exercise $$\PageIndex{8}$$
Pierre and Monique leave their home in Portland at the same time. Pierre drives north on the turnpike at a speed of 75 miles per hour while Monique drives south at a speed of 68 miles per hour. How long will it take them to be 429 miles apart?
3 hours
##### Exercise $$\PageIndex{9}$$
Thanh and Nhat leave their office in Sacramento at the same time. Thanh drives north on I-5 at a speed of 72 miles per hour. Nhat drives south on I-5 at a speed of 76 miles per hour. How long will it take them to be 330 miles apart?
2.2 hours
##### MATCHING UNITS IN PROBLEMS
It is important to make sure the units match when we use the distance rate and time formula. For instance, if the rate is in miles per hour, then the time must be in hours.
##### Exercise $$\PageIndex{10}$$
When Katie Mae walks to school, it takes her 30 minutes. If she rides her bike, it takes her 15 minutes. Her speed is three miles per hour faster when she rides her bike than when she walks. What are her walking speed and her speed riding her bike?
First, we draw a diagram that represents the situation to help us see what is happening.
We are asked to find her speed walking and riding her bike. Let’s call her walking speed r. Since her biking speed is three miles per hour faster, we will call that speed r+3. We write the speeds in the chart.
The speed is in miles per hour, so we need to express the times in hours, too, in order for the units to be the same. Remember, one hour is 60 minutes. So:
$\begin{array}{l}{30 \text { minutes is } \frac{30}{60} \text { or } \frac{1}{2} \text { hour }} \\ {15 \text { minutes is } \frac{15}{60} \text { or } \frac{1}{4} \text { hour }}\end{array}$
Next, we multiply rate times time to fill in the distance column.
The equation will come from the fact that the distance from Katie Mae’s home to her school is the same whether she is walking or riding her bike.
So we say:
Translate into an equation. Solve this equation. Clear the fractions by multiplying by the LCD of all the fractions in the equation. Simplify. 6 mph (Katie Mae's biking speed) Let's check if this works. Walk 3 mph (0.5 hour) = 1.5 miles Bike 6 mph (0.25 hour) = 1.5 miles Yes, either way Katie Mae travels 1.5 miles to school. Katie Mae’s walking speed is 3 mph. Her speed riding her bike is 6 mph.
##### Exercise $$\PageIndex{11}$$
Suzy takes 50 minutes to hike uphill from the parking lot to the lookout tower. It takes her 30 minutes to hike back down to the parking lot. Her speed going downhill is 1.2 miles per hour faster than her speed going uphill. Find Suzy’s uphill and downhill speeds.
uphill 1.8 mph, downhill three mph
##### Exercise $$\PageIndex{12}$$
Llewyn takes 45 minutes to drive his boat upstream from the dock to his favorite fishing spot. It takes him 30 minutes to drive the boat back downstream to the dock. The boat’s speed going downstream is four miles per hour faster than its speed going upstream. Find the boat’s upstream and downstream speeds.
upstream 8 mph, downstream 12 mph
In the distance, rate, and time formula, time represents the actual amount of elapsed time (in hours, minutes, etc.). If a problem gives us starting and ending times as clock times, we must find the elapsed time in order to use the formula.
##### Exercise $$\PageIndex{13}$$
Hamilton loves to travel to Las Vegas, 255 miles from his home in Orange County. On his last trip, he left his house at 2:00 pm. The first part of his trip was on congested city freeways. At 4:00 pm, the traffic cleared and he was able to drive through the desert at a speed 1.75 times as fast as when he drove in the congested area. He arrived in Las Vegas at 6:30 pm. How fast was he driving during each part of his trip?
A diagram will help us model this trip.
Next, we create a table to organize the information.
We know the total distance is 255 miles. We are looking for the rate of speed for each part of the trip. The rate in the desert is 1.75 times the rate in the city. If we let r= the rate in the city, then the rate in the desert is 1.75r.
The times here are given as clock times. Hamilton started from home at 2:00 pm and entered the desert at 4:30 pm. So he spent two hours driving the congested freeways in the city. Then he drove faster from 4:00 pm until 6:30 pm in the desert. So he drove 2.5 hours in the desert.
Now, we multiply the rates by the times.
By looking at the diagram below, we can see that the sum of the distance driven in the city and the distance driven in the desert is 255 miles.
Translate into an equation. Solve this equation. Check. Hamilton drove 40 mph in the city and 70 mph in the desert.
##### Exercise $$\PageIndex{14}$$
Cruz is training to compete in a triathlon. He left his house at 6:00 and ran until 7:30. Then he rode his bike until 9:45. He covered a total distance of 51 miles. His speed when biking was 1.6 times his speed when running. Find Cruz’s biking and running speeds.
biking 16 mph, running 10 mph
##### Exercise $$\PageIndex{15}$$
Phuong left home on his bicycle at 10:00. He rode on the flat street until 11:15, then rode uphill until 11:45. He rode a total of 31 miles. His speed riding uphill was 0.6 times his speed on the flat street. Find his speed biking uphill and on the flat street.
uphill 12 mph, flat street 20 mph
## Key Concepts
• Distance, Rate, and Time
• D = rt where D = distance, r = rate, t = time
• Problem-Solving Strategy—Distance, Rate, and Time Applications
1. Read the problem. Make sure all the words and ideas are understood.
Draw a diagram to illustrate what it happening.
Create a table to organize the information: Label the columns rate, time, distance. List the two scenarios. Write in the information you know.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
Complete the chart.
Use variable expressions to represent that quantity in each row.
Multiply the rate times the time to get the distance.
4. Translate into an equation.
Restate the problem in one sentence with all the important information.
Then, translate the sentence into an equation.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
3.5: Solve Uniform Motion Applications is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. |
# t-test for the mean using a TI83 or TI84 calculator (p-value method)
Do people tend to spend more than 2 hours on a computer every day? Can you say that the mean age of a college freshman in your state is not 18 years old? These are the types of questions that can be answered using collected data and a t-test for the mean. In this guide, you will see how you can use a TI83 or TI84 calculator to perform this test using the p-value method.
We will use an example to see how this process works. For this example, assume that the requirements for a hypothesis test for the mean are met (randomly selected sample, independent observations, large population size).
## Example: performing a t-test on the calculator
Suppose that a marketing firm believes that people who are planning to purchase a new TV spend more than 7 days researching their purchase. They conduct a survey of 32 people who had recently purchased a TV and found that the mean time spent researching the purchase was 7.8 days with a standard deviation of 3.9 days. At a significance level of 0.05, does this survey provide evidence to support the firm’s belief?
### Step 1: Write the null $(H_0)$ and alternative $(H_a)$ hypotheses
The alternative hypothesis is a statement about what you are testing. Here, you are testing the firm’s belief that the mean time spent is more than 7 days. In hypothesis testing, we are trying to understand a population value using a sample, so your hypothesis should be in terms of the population parameter. In this case, that is the population mean, $\mu$.
$\text{mean time spent is more than 7 days} \rightarrow H_a: \mu > 7$
The null hypothesis is the equality* statement using the same value:
$H_0: \mu = 7$
Putting these together, the null and alternative hypotheses are:
$H_0: \mu = 7\\ H_a: \mu > 7$
(*In some books, they use the statement that is the opposite of $H_a$ for the null hypothesis. Here, that would be $H_0: \mu \leq 7$. Make sure you use the form preferred in the class you are taking!)
### Step 2: Calculate the p-value using your calculator and the correct test
A t-test is used here since we have a big enough sample, and the population standard deviation $(\sigma)$ is unknown. (We only have the standard deviation from the sample: $s = 3.9$.) Note that if we knew the population standard deviation, we would use a z-test instead.
1. Press [STAT] then go the the TESTS menu.
2. Select “2. T-test”. Make sure that you highlight Stats and press [ENTER] if your screen looks different from this.
3. Enter the values and select the correct tail for the test.
4. Highlight Calculate and press [ENTER].
### Step 3: Compare the p-value to the significance level alpha $(\alpha)$ and make your decision
From the last line of the calculator, $\text{p-value} \approx 0.1274$. Further, the last part of the problem stated “At a significance level of 0.05, does this survey provide evidence to support the firm’s belief?”. Therefore, $\alpha = 0.05$.
To make the decision, use the decision rule:
In this problem:
### Step 4: Interpret your decision in terms of the problem
If you fail to reject $H_0$, then you are saying there is not enough evidence for $H_a$. Remember for this problem:
$H_0: \mu = 7\\ H_a: \mu > 7$
So, we are saying that there is not enough evidence that the population mean is greater than 7. In context, we are saying:
This sample does not provide evidence that the mean time spent researching a new TV purchase is more than 7 days.
Although our sample mean was in fact larger than 7, it wasn’t quite enough to suggest that this is true for the entire population. Remember, in hypothesis testing, that is what we are trying to determine – is the sample enough to say that the hypothesis holds for the entire population? |
# What is the Endpoint Formula?
Nov 12, 2015
Let's say you had one midpoint given. If you had neither endpoint given nor another midpoint given, then there are an infinite number of endpoints possible and your point is arbitrarily placed (because you only have one point available).
So, to find an endpoint, you need one endpoint and a designated midpoint.
Suppose you have midpoint $M \left(5 , 7\right)$ and the leftmost endpoint $A \left(1 , 2\right)$. That means you have:
${x}_{1} = 1$
${y}_{1} = 2$
So what are $5$ and $7$? The formula for finding the midpoint of a line segment is based on averaging both coordinates in each dimension, assuming 2D cartesian:
$\left(\frac{{x}_{1} + {x}_{\textcolor{red}{2}}}{\textcolor{red}{2}} , \frac{{y}_{1} + {y}_{\textcolor{red}{2}}}{\textcolor{red}{2}}\right)$
where an average is defined as:
$\frac{{a}_{1} + {a}_{2} + {a}_{3} + \ldots + {a}_{\textcolor{red}{N}}}{\textcolor{red}{N}}$
Therefore, you can plug in what you know here to find $B \left({x}_{2} , {y}_{2}\right)$.
$M \left(5 , 7\right) = \left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$
$5 = \frac{{x}_{1} + {x}_{2}}{2} \implies 10 = 1 + {x}_{2}$
$\textcolor{g r e e n}{{x}_{2} = 9}$
$7 = \frac{{y}_{1} + {y}_{2}}{2} \implies 14 = 2 + {y}_{2}$
$\textcolor{g r e e n}{{y}_{2} = 12}$
Therefore, your line segment passes through $A \left(1 , 2\right)$, $M \left(5 , 7\right)$, and $B \left(9 , 12\right)$, and your rightmost endpoint is $\textcolor{b l u e}{B \left(9 , 12\right)}$. |
Rs Aggarwal 2019 2020 Solutions for Class 9 Math Chapter 18 Mean, Median And Mode Of Ungrouped Data are provided here with simple step-by-step explanations. These solutions for Mean, Median And Mode Of Ungrouped Data are extremely popular among Class 9 students for Math Mean, Median And Mode Of Ungrouped Data 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 9 Math Chapter 18 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 9 Math are prepared by experts and are 100% accurate.
#### Question 1:
Find the mean of:
(i) the first eight natural numbers
(ii) the first ten odd numbers
(iii) the first seven multiples of 5
(iv) all the factors of 20
(v) all prime numbers between 50 and 80.
#### Answer:
We know:
(i) The first eight natural numbers are 1, 2, 3, 4, 5, 6, 7 and 8.
Mean of these numbers:
(ii) The first ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19.
Mean of these numbers:
$\frac{1+3+5+7+9+11+13+15+17+19}{10}\phantom{\rule{0ex}{0ex}}=\frac{100}{10}\phantom{\rule{0ex}{0ex}}=10$
(iii) The first seven multiples of 5 are 5, 10, 15, 20, 25, 30 and 35.
Mean of these numbers:
$\frac{5+10+15+20+25+30+35}{7}\phantom{\rule{0ex}{0ex}}=\frac{140}{7}\phantom{\rule{0ex}{0ex}}=20$
(iv) The factors of 20 are 1, 2, 4, 5, 10 and 20.
Mean of these numbers:
(v) The prime numbers between 50 and 80 are 53, 59, 61, 67, 71, 73 and 79.
Mean of these numbers:
$\frac{53+59+61+67+71+73+79}{7}\phantom{\rule{0ex}{0ex}}=\frac{463}{7}\phantom{\rule{0ex}{0ex}}=66.14$
#### Question 2:
The number of children in 10 families of a locality are
2, 4, 3, 4, 2, 0, 3, 5, 1, 6.
Find the mean number of children per family.
#### Answer:
Numbers of children in 10 families = 2, 4, 3, 4, 2, 0, 3, 5, 1 and 6.
Thus, we have:
$=\frac{30}{10}\phantom{\rule{0ex}{0ex}}=3$
#### Question 3:
The following are the number of books issued in a school library during a week:
105, 216, 322, 167, 273, 405 and 346.
Find the average number of books issued per day.
#### Answer:
Numbers of books issued in the school library: 105, 216, 322, 167, 273, 405 and 346
Thus, we have:
#### Question 4:
The daily minimum temperature recorded (in degree F) at a place during a week was as under:
Monday Tuesday Wednesday Thursday Friday Saturday 35.5 30.8 27.3 32.1 23.8 29.9
Find the mean temperature.
#### Answer:
Daily minimum temperatures = 35.5, 30.8, 27.3, 32.1, 23.8 and 29.9
Thus, we have:
#### Question 5:
If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 13, find the value of x and hence find the mean of the last three observations.
#### Answer:
We know that,
The first five observations are x, x + 2, x + 4, x + 6 and x + 8.
Mean of these numbers = $\frac{x+\left(x+2\right)+\left(x+4\right)+\left(x+6\right)+\left(x+8\right)}{5}$
$⇒13=\frac{5x+20}{5}\phantom{\rule{0ex}{0ex}}⇒13×5=5x+20\phantom{\rule{0ex}{0ex}}⇒65=5x+20\phantom{\rule{0ex}{0ex}}⇒5x=65-20\phantom{\rule{0ex}{0ex}}⇒5x=45\phantom{\rule{0ex}{0ex}}⇒x=\frac{45}{5}\phantom{\rule{0ex}{0ex}}⇒x=9\phantom{\rule{0ex}{0ex}}$
Hence, the value of x is 9.
Now, the last three observations are 13, 15 and 17.
Mean of these observations = $\frac{13+15+17}{3}$
= $\frac{45}{3}$
= 15
Hence, the mean of the last three observations is 15.
#### Question 6:
The mean weight of 6 boys in a group is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. Find the weight of the sixth boy.
#### Answer:
The individual weights of five boys are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg.
Now,
Let the weight of the sixth boy be x kg.
We know:
Also,
Given mean = 48 kg
Thus, we have:
Therefore, the sixth boy weighs 53 kg.
#### Question 7:
The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. Find the correct mean.
#### Answer:
Let the marks scored by 50 students be x1, x2,...x50.
Mean = 39
We know:
Thus, we have:
Also, a score of 43 was misread as 23.
#### Question 8:
The mean of 24 numbers is 35. If 3 is added to each number, what will be the new mean?
#### Answer:
Let the numbers be x1, x2,...x24.
We know:
Thus, we have:
After addition, the new numbers become (x1+3), (x2+3),...(x24+3).
New mean:
#### Question 9:
The mean of 20 numbers is 43. If 6 is subtracted from each of the numbers, what will be the new mean?
#### Answer:
Let the numbers be x1, x2,...x20.
We know:
Thus, we have:
Numbers after subtraction: (x1$-$6), (x2$-$6),...(x20$-$6)
∴ New Mean = $\frac{\left({x}_{1}-6\right)+\left({x}_{2}-6\right)+........+\left({x}_{20}-6\right)}{20}$
#### Question 10:
The mean of 15 numbers is 27. If each number is multiplied by 4, what will be the mean of the new numbers?
#### Answer:
Let the numbers be x1, x2,...x15
We know:
Thus, we have:
After multiplication, the numbers become 4x1, 4x2,...4x15
∴ New Mean = $\frac{4{x}_{1}+4{x}_{2}+......+4{x}_{15}}{15}$
#### Question 11:
The mean of 12 numbers is 40. If each number is divided by 8, what will be the mean of the new numbers?
#### Answer:
Let the numbers be x1, x2,...x12.
We know:
Thus, we have:
After division, the numbers become:
#### Question 12:
The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of the new set of 20 numbers.
#### Answer:
Let the numbers be x1, x2,...x20.
We know:
Thus, we have:
New numbers are:
(x1 + 3), (x2 + 3),...(x10 + 3), x11,...x20
New Mean:
#### Question 13:
The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number.
#### Answer:
Let the numbers be x1, x2,..., x6.
Mean = 23
We know:
Thus, we have:
${x}_{1}+{x}_{2}.......{x}_{6}=138$.....................(i)
If one number, say, x6, is excluded, then we have:
${x}_{1}+{x}_{2}......+{x}_{5}=100....................\left(\mathrm{ii}\right)$
Using (i) and (ii), we get:
Thus, the excluded number is 38
#### Question 14:
The average height of 30 boys was calculated to be 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean. Find the correct mean.
#### Answer:
We know that,
Mean of height of 30 boys = $\frac{\sum _{i=1}^{30}{x}_{i}}{30}$
It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean.
Hence, the correct mean is 151.
#### Question 15:
The mean weight of a class of 34 students is 46.5 kg. If the weight of the teacher is included, the mean is rises by 500 g. Find the weight of the teacher.
#### Answer:
Mean weight of 34 students = 46.5 kg
Sum of the weights of 34 students = kg
Increase in the mean weight when the weight of the teacher is included = 500 g = 0.5 kg
∴ New mean weight = (46.5 + 0.5) kg = 47 kg
Now,
Let the weight of the teacher be x kg.
Thus, we have:
Therefore, the weight of the teacher is 64 kg.
#### Question 16:
The mean weight of a class of 36 students is 41 kg. If one of the students leaves the class then the mean is decreased by 200 g. Find the weight of the student who left.
#### Answer:
Mean weight of 36 students = 41 kg
Sum of the weights of 36 students =
Decrease in the mean when one of the students left the class = 200 g = 0.2 kg
Mean weight of 35 students = (41 $-$ 0.2) kg = 40.8 kg
Now,
Let the weight of the student who left the class be x kg.
Hence, the weight of the student who left the class is 48 kg.
#### Question 17:
The average weight of a class of 39 students is 40 kg. When a new student is admitted to the class, the average decreases by 200 g. Find the weight of the new student.
#### Answer:
Average weight of 39 students = 40 kg
Sum of the weights of 39 students =
Decrease in the average when new student is admitted in the class = 200 g = 0.2 kg
∴ New average weight = (40 $-$ 0.2) kg = 39.8 kg
Now,
Let the weight of the new student be x kg.
Thus, we have:
Therefore, the weight of the new student is 32 kg.
#### Question 18:
The average weight of 10 oarsmen in a boat is increased by 1.5 kg when one of the crew who weighs 58 kg is replaced by a new man. Find the weight of the new man.
#### Answer:
Let the average weight of 10 oarsmen be x kg.
Sum of the weights of 10 oarsmen = 10x kg
∴ New average weight = (x + 1.5) kg
Now, we have:
#### Question 19:
The mean of 8 numbers is 35. If a number is excluded then the mean is reduced by 3. Find the excluded number.
#### Answer:
Mean of 8 numbers = 35
Sum of 8 numbers = $35×8=280$
Let the excluded number be x.
Now,
New mean = 35 $-$ 3 = 32
Thus, we have:
Therefore, the excluded number is 56.
#### Question 20:
The mean of 150 items was found to be 60. Later on, it was discovered that the values of two items were misread as 52 and 8 instead of 152 and 88 respectively. Find the correct men.
#### Answer:
Mean of 150 items = 60
Sum of 150 items = $\left(150×60\right)=9000$
New sum = [9000 $-$ (52 + 8) + (152 + 88)] = 9180
Correct mean =
Therefore, the correct mean is 61.2.
#### Question 21:
The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62, find the 16th result.
#### Answer:
Mean of 31 results = 60
Sum of 31 results = $31×60=1860$
Mean of the first 16 results = 58
Sum of the first 16 results = $58×16=928$
Mean of the last 16 results = 62
Sum of the last 16 results = $62×16=992$
Value of the 16th result = (Sum of the first 16 results + Sum of the last 16 results) $-$ Sum of 31 results
= (928 + 992) $-$ 1860
= 1920 $-$ 1860
= 60
#### Question 22:
The mean of 11 numbers is 42. If the mean of the first 6 numbers is 37 and that of the last 6 numbers is 46, find the 6th number.
#### Answer:
Mean of 11 numbers = 42
Sum of 11 numbers = 42$×$11 = 462
Mean of the first 6 numbers = 37
Sum of the first 6 numbers = 37$×$6 = 222
Mean of the last 6 numbers = 46
Sum of the last 6 numbers = 46$×$6 = 276
∴ 6th number = [(Sum of the first 6 numbers + Sum of the last 6 numbers) $-$ Sum of 11 numbers]
= [(222 + 276) $-$ 462]
= [498 $-$ 462]
= 36
Hence, the 6th number is 36.
#### Question 23:
The mean weight of 25 students of a class is 52 kg. If the mean weight of the first 13 students of the class is 48 kg and that of the last 13 students is 55 kg, find the weight of the 13th student.
#### Answer:
Mean weight of 25 students = 52 kg
Sum of the weights of 25 students = (52$×$25) kg = 1300 kg
Mean weight of the first 13 students = 48 kg
Sum of the weights of the first 13 students = (48$×$13) kg = 624 kg
Mean weight of the last 13 students = 55 kg
Sum of the weights of the last 13 students = (55$×$13) kg =715 kg
Weight of the 13th student = (Sum of the weights of the first 13 students + Sum of the weights of the last 13 students) $-$ Sum of the weights of 25 students
= [(624+715)$-$1300] kg
= 39 kg
Therefore, the weight of the 13th student is 39 kg.
#### Question 24:
The mean score of 25 observations is 80 and the mean score of another 55 observations is 60. Determine the mean score of the whole set of observations.
#### Answer:
Mean score of 25 observations = 80
Sum of the scores of 25 observations = 80$×$25 = 2000
Mean score of another 55 observations = 60
Sum of the scores of another 55 observations = 60$×$55 = 3300
$=\frac{2000+3300}{80}\phantom{\rule{0ex}{0ex}}=\frac{5300}{80}\phantom{\rule{0ex}{0ex}}=66.25$
Therefore, the mean score of the whole set of observations is 66.25.
#### Question 25:
Arun scored 36 marks in English, 44 marks in Hindi, 75 marks in mathematics and x marks in science. If he has secured an average of 50 marks, find the value of x.
#### Answer:
Marks scored by Arun in English = 36
Marks scored by Arun in Hindi = 44
Marks scored by Arun in mathematics = 75
Marks scored by Arun in science = x
Average marks = 50
Thus, we have:
$⇒155+x=200\phantom{\rule{0ex}{0ex}}⇒x=200-155=45$
∴ Marks scored by Arun in science = 45
#### Question 26:
A ship sails out to an island at the rate of 15 km/h and sails back to the starting point at 10 km/h. Find the average sailing speed for the whole journey.
#### Answer:
Let the distance from the starting point to the island be x km.
Speed of the ship sailing out to the island = 15 km/h
Speed of the ship sailing back to the starting point = 10 km/h
We know:
Therefore, the average speed of the ship in the whole journey was 12 km/h.
#### Question 27:
There are 50 students in a class, of which 40 are boys. The average weight of the class is 44 kg an that of the girls is 40 kg. Find the average weight of the boys.
#### Answer:
Total students in the class = 50
Number of boys = 40
∴ Number of girls = (50 $-$ 40) = 10
Average weight of students in the class = 44 kg
Average weight of girls in the class = 40 kg
Sum of the weights of girls in the class = (40 $×$10) kg = 400 kg
Thus, we have:
#### Question 28:
The aggregate monthly expenditure of a family was â‚ą 18720 during the first 3 months, â‚ą 20340 during the next 4 months and â‚ą 21708 during the last 5 months of a year. If the total savings during the year be â‚ą 35340 find the average monthly income of the family.
#### Answer:
The aggregate yearly expenditure of a family = Rs (18720 × 3) + Rs (20340 × 4) + Rs (21708 × 5)
= Rs (56160 + 81360 + 108540)
= Rs 246060
The total savings during the year = Rs 35340
The average yearly income = Rs 246060 + Rs 35340
= Rs 281400
∴ The average monthly income of the family =
Hence, The average monthly income of the family is Rs 23450.
#### Question 29:
The average weekly payment to 75 workers in a factory is â‚ą 5680. The mean weekly payment to 25 of them is â‚ą 5400 and that of 30 others is â‚ą 5700. Find the mean weekly payment of the remaining workers.
#### Answer:
Total salary of 75 workers = â‚ą 5680 × 75
= â‚ą 426000
Total salary of 25 workers = â‚ą 5400 × 25
= â‚ą 135000
Total salary of 30 workers = â‚ą 5700 × 30
= â‚ą 171000
No. of remaining workers = 75 − (25 +30)
= 20
Total salary of 20 workers = â‚ą (426000 − 135000 − 171000)
= â‚ą 120000
∴ The mean weekly payment of the 20 workers = $\frac{120000}{20}$
= â‚ą 6000
Hence, the mean weekly payment of the remaining workers is â‚ą 6000.
#### Question 30:
The mean marks (out of 100) of boys and girls in an examination are 70 and 73 respectively. If the mean marks of all the students in that examination is 71, find the ratio of the number of boys to the number of girls.
#### Answer:
Let the number of girls be x and the number of boys be y.
Mean marks of boys =
Mean marks of girls =
Mean marks of all the students =
Hence, the ratio of the number of boys to the number of girls is 2:1.
#### Question 31:
The average monthly salary of 20 workers in an office is â‚ą 45900. If the manager's salary is added, the average salary becomes â‚ą 49200 per month. What's manager's monthly salary?
#### Answer:
Average monthly salary of 20 workers = Rs 45900
Sum of the monthly salaries of 20 workers =
By adding the manager's monthly salary, we get:
Average salary = Rs 49200
Now,
Let the manager's monthly salary be Rs x.
Thus, we have:
Therefore, the manager's monthly salary is Rs 115200.
#### Question 1:
Obtain the mean of the following distribution:
Variable (xi) 4 6 8 10 12 Frequency (fi) 4 8 14 11 3
#### Answer:
We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
For the following data:
Variable (xi) 4 6 8 10 12 Frequency (fi) 4 8 14 11 3
Mean = $\frac{\left(4×4\right)+\left(6×8\right)+\left(8×14\right)+\left(10×11\right)+\left(12×3\right)}{4+8+14+11+3}$
= $\frac{16+48+112+110+36}{40}$
= $\frac{322}{40}$
= 8.05
Hence, the mean of the following distribution is 8.05 .
#### Question 2:
The following table shows the weights of 12 workers in a factory:
Weight (in kg) 60 63 66 69 72 No. of workers 4 3 2 2 1
Find the mean weight of the workers.
#### Answer:
We will make the following table:
Weight (xi) No. of Workers (fi) (fi)(xi) 60 4 240 63 3 189 66 2 132 69 2 138 72 1 72 12 771
Thus, we have:
=
#### Question 3:
The measurements (in mm) of the diameters of the heads of 50 screws are given below:
Diameter (in mm) (xi) 34 37 40 43 46 Number of screws (fi) 5 10 17 12 6
Calculate the mean diameter of the heads of the screws.
#### Answer:
We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
For the following data:
Diameter (in mm) (xi) 34 37 40 43 46 Number of screws (fi) 5 10 17 12 6
Mean = $\frac{\left(34×5\right)+\left(37×10\right)+\left(40×17\right)+\left(43×12\right)+\left(46×6\right)}{5+10+17+12+6}$
= $\frac{170+370+680+516+276}{50}$
= $\frac{2012}{50}$
= 40.24
Hence, the mean diameter of the heads of the screws is 40.24 .
#### Question 4:
The following data give the number of boys of a particular age in a class of 40 students.
Age (in years) 15 16 17 18 19 20 Frequency (fi) 3 8 9 11 6 3
Calculate the mean age of the students.
#### Answer:
We will make the following table:
Age (xi) Frequency (fi) (fi)(xi) 15 3 45 16 8 128 17 9 153 18 11 198 19 6 114 20 3 60 40 $\sum {f}_{i}{x}_{i}=$698
Thus, we have:
#### Question 5:
Find the mean of the following frequency distribution:
Variable (xi) 10 30 50 70 89 Frequency (fi) 7 8 10 15 10
#### Answer:
We will make the following table:
Variable (xi) Frequency (fi) (fi)(xi) 10 7 70 30 8 240 50 10 500 70 15 1050 89 10 890 50 2750
Thus, we have:
#### Question 6:
Find the mean of daily wages of 40 workers in a factory as per data given below:
Daily wages (in â‚ą) (xi) 250 300 350 400 450 Number of workers (fi) 8 11 6 10 5
#### Answer:
We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
For the following data:
Daily wages (in â‚ą) (xi) 250 300 350 400 450 Number of workers (fi) 8 11 6 10 5
Mean = $\frac{\left(250×8\right)+\left(300×11\right)+\left(350×6\right)+\left(400×10\right)+\left(450×5\right)}{8+11+6+10+5}$
= $\frac{2000+3300+2100+4000+2250}{40}$
= $\frac{13650}{40}$
= 341.25
Hence, the mean of daily wages of 40 workers in a factory is 341.25 .
#### Question 7:
If the mean of the following data is 20.2, find the value of p.
Variable (xi) 10 15 20 25 30 Frequency (fi) 6 8 p 10 6
#### Answer:
We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
For the following data:
Variable (xi) 10 15 20 25 30 Frequency (fi) 6 8 p 10 6
Mean = $\frac{\left(10×6\right)+\left(15×8\right)+\left(20×p\right)+\left(25×10\right)+\left(30×6\right)}{6+8+p+10+6}$
$⇒20.2=\frac{60+120+20p+250+180}{30+p}\phantom{\rule{0ex}{0ex}}⇒20.2\left(30+p\right)=610+20p\phantom{\rule{0ex}{0ex}}⇒606+20.2p=610+20p\phantom{\rule{0ex}{0ex}}⇒20.2p-20p=610-606\phantom{\rule{0ex}{0ex}}⇒0.2p=4\phantom{\rule{0ex}{0ex}}⇒p=\frac{4}{0.2}\phantom{\rule{0ex}{0ex}}⇒p=\frac{40}{2}\phantom{\rule{0ex}{0ex}}⇒p=20$
Hence, the value of p is 20.
#### Question 8:
If the mean of the following frequency distribution is 8, find the value of p.
x 3 5 7 9 11 13 f 6 8 15 p 8 4
#### Answer:
We will make the following table:
(xi) (fi) (fi)(xi) 3 6 18 5 8 40 7 15 105 9 p 9p 11 8 88 13 4 52 41 + p 303 + 9p
We know:
Given:
Mean = 8
Thus, we have:
#### Question 9:
Find the missing frequency p for the following frequency distribution whose mean is 28.25.
x 15 20 25 30 35 40 f 8 7 p 14 15 6
#### Answer:
We will prepare the following table:
(xi) (fi) (fi)(xi) 15 8 120 20 7 140 25 p 25p 30 14 420 35 15 525 40 6 240 50 + p 1445 + 25p
Thus, we have:
#### Question 10:
Find the value of p for the following frequency distribution whose mean is 16.6
x 8 12 15 p 20 25 30 f 12 16 20 24 16 8 4
#### Answer:
We will make the following table:
(xi) (fi) (fi)(xi) 8 12 96 12 16 192 15 20 300 p 24 24p 20 16 320 25 8 200 30 4 120 100 1228 + 24p
Thus, we have:
$⇒24p=432\phantom{\rule{0ex}{0ex}}⇒p=18$
#### Question 11:
Find the missing frequencies in the following frequency distribution whose mean is 34.
x 10 20 30 40 50 60 Total f 4 f1 8 f2 3 4 35
#### Answer:
We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
For the following data:
x 10 20 30 40 50 60 Total f 4 f1 8 f2 3 4 35
Mean = $\frac{\left(10×4\right)+\left(20×{f}_{1}\right)+\left(30×8\right)+\left(40×{f}_{2}\right)+\left(50×3\right)+\left(60×4\right)}{35}$
Also, 4 + f1 + 8 + f2 + 3 + 4 = 35
⇒ 19 + f1f2 = 35
f1f2 = 35 − 19
f1f2 = 16
⇒ 26 − 2f2f 2 = 16 (from (1))
⇒ 26 − f2 = 16
⇒ 26 − 16 = f2
⇒ f2 = 10
Putting the value of f2 in (1), we get
f1 = 26 − 2(10) = 6
Hence, the value of f1 and f2 is 6 and 10, respectively.
#### Question 12:
Find the missing frequencies in the following frequency distribution, whose mean is 50.
x 10 30 50 70 90 Total f 17 f1 32 f2 19 120
#### Answer:
We will prepare the following table:
(xi) (fi) (fi)(xi) 10 17 170 30 f1 30f1 50 32 1600 70 f2 70f2 90 19 1710 120 3480 + 30f1 + 70f2
Thus, we have:
Also,
G
iven:
17
+ f1 + 32 + f2 + 19 = 120
68 + f1 + f2 = 120
f1 + f2 = 52
or, f2 = 52 $-$ f1
...(ii)
By putting the value of f2
in (i), we get:
2520
= 30f1 + 70(52 $-$ f1)
2520 = 30f1 + 3640 $-$ 70f1
40f1 = 1120
f1 = 28
Substituting the value in (ii), we get:
f2 = 52 $-$ f1 = 52 $-$ 28 = 24
#### Question 13:
Find the value of p, when the mean of the following distribution is 20.
x 15 17 19 20 + p 23 f 2 3 4 5p 6
#### Answer:
We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
For the following data:
x 15 17 19 20 + p 23 f 2 3 4 5p 6
Mean = $\frac{\left(15×2\right)+\left(17×3\right)+\left(19×4\right)+\left(\left(20+p\right)×5p\right)+\left(23×6\right)}{2+3+4+5p+6}$
$⇒20=\frac{30+51+76+100p+5{p}^{2}+138}{15+5p}\phantom{\rule{0ex}{0ex}}⇒20\left(15+5p\right)=5{p}^{2}+100p+295\phantom{\rule{0ex}{0ex}}⇒300+100p=5{p}^{2}+100p+295\phantom{\rule{0ex}{0ex}}⇒5{p}^{2}=300-295\phantom{\rule{0ex}{0ex}}⇒5{p}^{2}=5\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=\frac{5}{5}\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=1\phantom{\rule{0ex}{0ex}}⇒p=±1$
Hence, the value of p is ±1.
#### Question 14:
The mean of the following distribution is 50.
x 10 30 50 70 90 f 17 5a + 3 32 7a – 11 19
Find the value of a and hence the frequencies of 30 and 70.
#### Answer:
We know that,
Mean = $\frac{\sum _{}{x}_{i}{f}_{i}}{\sum _{}{f}_{i}}$
For the following data:
x 10 30 50 70 90 f 17 5a + 3 32 7a – 11 19
Mean = $\frac{\left(10×17\right)+\left(30×\left(5a+3\right)\right)+\left(50×32\right)+\left(70×\left(7a-11\right)\right)+\left(90×19\right)}{17+5a+3+32+7a-11+19}$
$⇒50=\frac{170+150a+90+1600+490a-770+1710}{60+12a}\phantom{\rule{0ex}{0ex}}⇒50\left(60+12a\right)=2800+640a\phantom{\rule{0ex}{0ex}}⇒3000+600a=2800+640a\phantom{\rule{0ex}{0ex}}⇒640a-600a=3000-2800\phantom{\rule{0ex}{0ex}}⇒40a=200\phantom{\rule{0ex}{0ex}}⇒a=\frac{200}{40}\phantom{\rule{0ex}{0ex}}⇒a=5$
Hence, the value of a is 5.
Also, the frequency of 30 is 28 and the frequency of 70 is 24.
#### Question 1:
Find the median of
(i) 2, 10, 9, 9, 5, 2, 3, 7, 11
(ii) 15, 6, 16, 8, 22, 21, 9, 18, 25
(iii) 20, 13, 18, 25, 6, 15, 21, 9, 16, 8, 22
(iv) 7, 4, 2, 5, 1, 4, 0, 10, 3, 8, 5, 9, 2
#### Answer:
(i) Arranging the numbers in ascending order, we get:
2, 2, 3, 5, 7, 9, 9, 10, 11
Here, n is 9, which is an odd number.
If n is an odd number, we have:
Now,
(ii) Arranging the numbers in ascending order, we get:
6, 8, 9, 15, 16, 18, 21, 22, 25
Here, n is 9, which is an odd number.
If n is an odd number, we have:
Now,
(iii) Arranging the numbers in ascending order, we get:
6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here, n is 11, which is an odd number.
If n is an odd number, we have:
Now,
(iv) Arranging the numbers in ascending order, we get:
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here, n is 13, which is an odd number.
If n is an odd number, we have:
Now,
#### Question 2:
Find the median of
(i) 17, 19, 32, 10, 22, 21, 9, 35
(ii) 72, 63, 29, 51, 35, 60, 55, 91, 85, 82
(iii) 10, 75, 3, 15, 9, 47, 12, 48, 4, 81, 17, 27
#### Answer:
(i) Arranging the numbers in ascending order, we get:
9, 10, 17, 19, 21, 22, 32, 35
Here, n is 8, which is an even number.
If n is an even number, we have:
Now,
(ii) Arranging the numbers in ascending order, we get:
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here, n is 10, which is an even number.
If n is an even number, we have:
Now,
(iii) Arranging the numbers in ascending order, we get:
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Here, n is 12, which is an even number.
If n is an even number, we have:
Now,
#### Question 3:
The marks of 15 students in an examination are:
25, 19, 17, 24, 23, 29, 31, 40, 19, 20, 22, 26, 17, 35, 21.
Find the median score.
#### Answer:
Arranging the marks of 15 students in ascending order, we get:
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
Here, n is 15, which is an odd number.
We know:
Thus, we have:
#### Question 4:
The heights (in cm) of 9 students of a class are 148, 144, 152, 155, 160, 147, 150, 149, 145.
Find the median height.
#### Answer:
Arranging the given data in ascending order:
144, 145, 147, 148, 149, 150, 152, 155, 160
Number of terms = 9 (odd)
Hence, the median height is 149.
#### Question 5:
The weights (in kg) of 8 children are:
13.4, 10.6, 12.7, 17.2, 14.3, 15, 16.5, 9.8.
Find the median weight.
#### Answer:
Arranging the weights (in kg) in ascending order, we have:
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here, n is 8, which is an even number.
Thus, we have:
Hence, the median weight is 13.85 kg.
#### Question 6:
The ages (in years) of 10 teachers in a school are:
32, 44, 53, 47, 37, 54, 34, 36, 40, 50.
Find the median age.
#### Answer:
Arranging the ages (in years) in ascending order, we have:
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, n is 10, which is an even number.
Thus, we have:
#### Question 7:
If 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41 are ten observation in an ascending order with median 24, find the value of x.
#### Answer:
10, 13, 15, 18, x+1, x+3, 30, 32, 35 and 41 are arranged in ascending order.
Median = 24
We have to find the value of x.
Here, n is 10, which is an even number.
Thus, we have:
#### Question 8:
The following observations are arranged in ascending order:
26, 29, 42, 53, x, x + 2, 70, 75, 82, 93.
If the median is 65, find the value of x.
#### Answer:
Arranging the given data in ascending order:
26, 29, 42, 53, x, x + 2, 70, 75, 82, 93
Number of terms = 10 (even)
Hence, the value of x is 64.
#### Question 9:
The numbers 50, 42, 35, (2x + 10), (2x – 8), 12, 11, 8 have been written in a descending order. If their median is 25, find the value of x.
#### Answer:
Arranging the given data in ascending order:
8, 11, 12, (2x – 8), (2x + 10), 35, 42, 50
Number of terms = 8 (even)
Hence, the value of x is 12.
#### Question 10:
Find the median of the data
46, 41, 77, 58, 35, 64, 87, 92, 33, 55, 90.
In the above data, if 41 and 55 are replaced by 61 and 75 respectively, what will be the new median?
#### Answer:
Arranging the given data in ascending order:
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92
Number of terms = 11 (odd)
Hence, the median of the data is 58.
Now, In the above data, if 41 and 55 are replaced by 61 and 75 respectively.
Then, new data in ascending order is:
33, 35, 46, 58, 61, 64, 75, 77, 87, 90, 92
Number of terms = 11 (odd)
Hence, the new median of the data is 64.
#### Question 1:
Find the mode of the following items.
0, 6, 5, 1, 6, 4, 3, 0, 2, 6, 5, 6
#### Answer:
On arranging the items in ascending order, we get:
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
Clearly, 6 occurs maximum number of times.
∴ Mode = 6
#### Question 2:
Determine the mode of the following values of a variable.
23, 15, 25, 40, 27, 25, 22, 25, 20
#### Answer:
On arranging the values in ascending order, we get:
15, 20, 22, 23, 25, 25, 25, 27, 40
Clearly, 25 occurs maximum number of times.
∴ Mode = 25
#### Question 3:
Calculate the mode of the following sizes of shoes sold by a shop on a particular day.
5, 9, 8, 6, 9, 4, 3, 9, 1, 6, 3, 9, 7, 1, 2, 5, 9
#### Answer:
On arranging the shoe sizes in ascending order, we get:
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
Clearly, 9 occurs maximum number of times.
∴ Mode = 9
#### Question 4:
A cricket player scored the following runs in 12 one-day matches:
50, 30, 9, 32, 60, 50, 28, 50, 19, 50, 27, 35.
Find his modal score.
#### Answer:
On arranging the runs in ascending order, we get:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
Clearly, 50 occurs maximum number of times.
∴ Modal score = 50
#### Question 5:
If the mean of the data 3, 21, 25, 17, (x + 3), 19, (x – 4) is 18, find the value of x. Using this value of x, find the mode of the data.
#### Answer:
We know that,
The given data is 3, 21, 25, 17, (x + 3), 19, (x – 4).
Mean of the given data = $\frac{3+21+25+17+\left(x+3\right)+19+\left(x-4\right)}{7}$
$⇒18\left(7\right)=84+2x\phantom{\rule{0ex}{0ex}}⇒126-84=2x\phantom{\rule{0ex}{0ex}}⇒2x=42\phantom{\rule{0ex}{0ex}}⇒x=21$
Hence, the value of x is 21.
Now, the given data is 3, 21, 25, 17, 24, 19, 17
Arranging this data in ascending order:
3, 17, 17, 19, 21, 24, 25
Here, 17 occurs maximum number of times.
∴ Mode = 17
Hence, the mode of the data is 17.
#### Question 6:
The numbers 52, 53, 54, 54, (2x + 1), 55, 55, 56, 57 have been arranged in an ascending order and their median is 55. Find the value of x and hence find the mode of the given data.
#### Answer:
Arranging the given data in ascending order:
52, 53, 54, 54, (2x + 1), 55, 55, 56, 57
Number of terms = 9 (odd)
Hence, the value of x is 27.
Arranging the given data in ascending order:
52, 53, 54, 54, 55, 55, 55, 56, 57
Here, 55 occurs maximum number of times.
∴ Mode = 55
Hence, the mode of the data is 55.
#### Question 7:
For what value of x is the mode of the data 24, 15, 40, 23, 27, 26, 22, 25, 20, x + 3 found 25? Using this value of x, find the median.
#### Answer:
Given: Mode = 25
∴ 25 occurs maximum number of times.
Arranging the given data in ascending order:
15, 20, 22, 23, 24, x + 3, 25, 26, 27, 40
x + 3 = 25
x = 25 − 3
x = 22
Hence, the value of x is 22.
Arranging the given data in ascending order:
15, 20, 22, 23, 24, 25, 25, 26, 27, 40
Number of terms = 10 (even)
Hence, the median is 24.5 .
#### Question 8:
The numbers 42, 43, 44, 44, (2x + 3), 45, 45, 46, 47 have been arranged in an ascending order and their median is 45. Find the value of x. Hence, find the mode of the above data.
#### Answer:
Arranging the given data in ascending order:
42, 43, 44, 44, (2x + 3), 45, 45, 46, 47
Number of terms = 9 (odd)
Hence, the value of x is 21.
Arranging the given data in ascending order:
42, 43, 44, 44, 45, 45, 45, 46, 47
Here, 45 occurs maximum number of times.
∴ Mode = 45
Hence, the mode of the data is 45.
#### Question 1:
If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is
(a) 5
(b) 6
(c) 7
(d) 8
#### Answer:
(c) 7
Mean of 5 observations = 11
We know:
#### Question 2:
If the mean of x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is
(a) $10\frac{1}{3}$
(b) $10\frac{2}{3}$
(c) $11\frac{1}{3}$
(d) $11\frac{2}{3}$
#### Answer:
(c) 11$\frac{1}{3}$
Mean of 5 observations = 9
We know:
#### Question 3:
If $\overline{)x}$ is the mean of then
(a) −1
(b) 0
(c) 1
(d) n − 1
(b) 0
#### Question 4:
If each observation of the data is increased by 8, then their mean
(a) remains the same
(b) is decreased by 8
(c) is increased by 5
(d) becomes 8 times the original mean
#### Answer:
(b) is decreased by 8
1, x2,...xn.
Now the new numbers after decreasing every number by 8 : (x1−8) , (x2−8)...,(xn−8)
Hence, mean is decreased by 8.
#### Question 5:
The mean weight of six boys in a groups is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. The weight of the 6th boy is
(a) 52 kg
(b) 52.8 kg
(c) 53 kg
(d) 47 kg
#### Answer:
(c) 53 kg
Mean weight of six boys = 48 kg
Let the weight of the 6th boy be x kg.
#### Question 6:
The mean of the marks scored by 50 students was found to be 39. Latter on it was discovered that a core of 43 was misread as 23. The correct mean is
(a) 38.6
(b) 39.4
(c) 39.8
(d) 39.2
#### Answer:
(b) 39.4
Mean of the marks scored by 50 students = 39
Sum of the marks scored by 50 students = $\left(39×50\right)=1950$
Correct sum = (1950 + 43 $-$ 23) = 1970
$\therefore \mathrm{Mean}=\frac{1970}{50}=39.4\phantom{\rule{0ex}{0ex}}$
#### Question 7:
The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. The correct mean is
(a) 64.86
(b) 65.31
(c) 64.91
(d) 64.61
#### Answer:
(c) 64.91
Mean of 100 items = 64
Sum of 100 items = $64×100=6400$
Correct sum = (6400 + 36 + 90 $-$ 26 $-$ 9) = 6491
#### Question 8:
The mean of 100 observations is 50. If one of the observations 50 is replaced by 150, the resulting mean will be
(a) 50.5
(b) 51
(c) 51.5
(d) 52
#### Answer:
(b) 51
Mean of 100 observations = 50
Sum of 100 observations = $100×50=5000$
It is given that one of the observations, 50, is replaced by 150.
∴ New sum = (5000 $-$ 50 + 150) = 5100
And,
#### Question 9:
Let $\overline{)x}$ be the mean of and $\overline{)y}$ be the mean of .
If $\overline{)z}$ is the mean of then
(a)
(b)
(c)
(d)
#### Answer:
(b)
$\overline{z}=\frac{\left({x}_{1}+{x}_{2}+...+{x}_{n}\right)+\left({y}_{1}+{y}_{2}+...+{y}_{n}\right)}{2n}$
#### Question 10:
If $\overline{)x}$ is the mean of then for , the mean of is
(a) $\left(a+\frac{1}{a}\right)\overline{)x}$
(b) $\left(a+\frac{1}{a}\right)\frac{\overline{)x}}{2}$
(c) $\left(a+\frac{1}{a}\right)\frac{\overline{)x}}{n}$
(d) $\frac{\left(a+\frac{1}{a}\right)\overline{)x}}{2n}$
#### Answer:
(b) $\left(a+\frac{1}{a}\right)\frac{\overline{)x}}{2}$
#### Question 11:
If are the means of n groups with number of observations respectively, then the mean $\overline{)x}$ of all the groups taken together is
(a) $\sum _{i=1}^{n}{n}_{i}{\overline{)x}}_{i}$
(b) $\sum _{\frac{i=1}{{n}^{2}}}^{n}{n}_{i}{\overline{)x}}_{i}$
(c) $\frac{\sum _{i=1}^{n}{n}_{i}{\overline{)x}}_{i}}{\sum _{i=1}^{n}{n}_{i}}$
(d) $\frac{\sum _{i=1}^{n}{n}_{i}{\overline{)x}}_{i}}{2n}$
#### Question 12:
The mean of the following data is 8.
x 3 5 7 9 11 13 y 6 8 15 p 8 4
The value of p is
(a) 23
(b) 24
(c) 25
(d) 21
#### Answer:
(c) 25
x y x$×$y 3 6 18 5 8 40 7 15 105 9 p 9p 11 8 88 13 4 52 Total 41 + p 303 + 9p
$\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{Mean}=\frac{303+9p}{41+p}\phantom{\rule{0ex}{0ex}}\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\mathrm{Mean}=8\phantom{\rule{0ex}{0ex}}\therefore \frac{303+9p}{41+p}=8\phantom{\rule{0ex}{0ex}}⇒303+9p=328+8p\phantom{\rule{0ex}{0ex}}⇒p=25$
#### Question 13:
The runs scored by 11 members of a cricket team are
15, 34, 56, 27, 43, 29, 31, 13, 50, 20, 0
The median score is
(a) 27
(b) 29
(c) 31
(d) 20
#### Answer:
(b) 29
Arranging the weight of 10 students in ascending order, we have:
0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56
Here, n is 11, which is an odd number.
Thus, we have:
#### Question 14:
The weight of 10 students (in kgs) are
55, 40, 35, 52, 60, 38, 36, 45, 31, 44
The median weight is
(a) 40 kg
(b) 41 kg
(c) 42 kg
(d) 44 kg
#### Answer:
(c) 42 kg
Arranging the numbers in ascending order, we have:
31, 35, 36, 38, 40, 44, 45, 52, 55, 60
Here, n is 10, which is an even number.
Thus, we have:
#### Question 15:
The median of the numbers 4, 4, 5, 7, 6, 7, 7, 12, 3 is
(a) 4
(b) 5
(c) 6
(d) 7
#### Answer:
(c) 6
We will arrange the given data in ascending order as:
3, 4, 4, 5, 6, 7, 7, 7, 12
Here, n is 9, which is an odd number.
Thus, we have:
#### Question 16:
The median of the numbers 84, 78, 54, 56, 68, 22, 34, 45, 39, 54 is
(a) 45
(b) 49.5
(c) 54
(d) 56
#### Answer:
(c) 54
We will arrange the data in ascending order as:
22, 34, 39, 45, 54, 54, 56, 68, 78, 84
Here, n is 10, which is an even number.
Thus, we have:
#### Question 17:
Mode of the data 15, 17, 15, 19, 14, 18, 15, 14, 16, 15, 14, 20, 19, 14, 15 is
(a) 14
(b) 15
(c) 16
(d) 17
#### Answer:
(b) 15
Here, 14 occurs 4 times, 15 occurs 5 times, 16 occurs 1 time, 18 occurs 1 time, 19 occurs 1 time and 20 occurs 1 time. Therefore, the mode, which is the most occurring item, is 15.
#### Question 18:
The median of the data arranged in ascending order 8, 9, 12, 18, (x + 2), (x + 4), 30, 31, 34, 39 is 24. The value of x is
(a) 22
(b) 21
(c) 20
(d) 24
#### Answer:
(b) 21
The given data is in ascending order.
Here, n is 10, which is an even number.
Thus, we have:
View NCERT Solutions for all chapters of Class 9 |
# What is 67/163 as a decimal?
## Solution and how to convert 67 / 163 into a decimal
67 / 163 = 0.411
To convert 67/163 into 0.411, a student must understand why and how. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. Now, let's solve for how we convert 67/163 into a decimal.
## 67/163 is 67 divided by 163
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! To solve the equation, we must divide the numerator (67) by the denominator (163). This is how we look at our fraction as an equation:
### Numerator: 67
• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. Overall, 67 is a big number which means you'll have a significant number of parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Large numerators make converting fractions more complex. Let's take a look below the vinculum at 163.
### Denominator: 163
• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. 163 is a large number which means you should probably use a calculator. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Have no fear, large two-digit denominators are all bark no bite. Next, let's go over how to convert a 67/163 to 0.411.
## Converting 67/163 to 0.411
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 163 \enclose{longdiv}{ 67 }$$
Use long division to solve step one. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 163 \enclose{longdiv}{ 67.0 }$$
Uh oh. 163 cannot be divided into 67. So we will have to extend our division problem. Add a decimal point to 67, your numerator, and add an additional zero. This doesn't add any issues to our denominator but now we can divide 163 into 670.
### Step 3: Solve for how many whole groups you can divide 163 into 670
$$\require{enclose} 00.4 \\ 163 \enclose{longdiv}{ 67.0 }$$
We can now pull 652 whole groups from the equation. Multiply this number by 163, the denominator to get the first part of your answer!
### Step 4: Subtract the remainder
$$\require{enclose} 00.4 \\ 163 \enclose{longdiv}{ 67.0 } \\ \underline{ 652 \phantom{00} } \\ 18 \phantom{0}$$
If your remainder is zero, that's it! If you still have numbers left over, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But they all represent how numbers show us value in the real world. Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 67/163 into a decimal
Dining - We don't give a tip of 67/163 of the bill (technically we do, but that sounds weird doesn't it?). We give a 41% tip or 0.411 of the entire bill.
### When to convert 0.411 to 67/163 as a fraction
Meal Prep - Body builders need to count macro calories. One of the ways of doing this is measuring every piece of food consumed. This is through halves and quarters in order to keep it consistent.
### Practice Decimal Conversion with your Classroom
• If 67/163 = 0.411 what would it be as a percentage?
• What is 1 + 67/163 in decimal form?
• What is 1 - 67/163 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.411 + 1/2? |
## Lesson θ': The Euclidean Proposition
We see how a proof is laid out, and study Book I, Proposition 1 of the Elements as our example.
The Elements consists of definitions, postulates, axioms, and propositions or demonstrations. The latter are of two types, problems and theorems, and are set out in a standard format, which is the same for both. The parts of a demonstration are as follows:
1. Statement of the Theorem or Problem
2. The Figure
3. Application to the Figure
4. Constructions
5. Proof
6. Final statement of what has been proved
At the end of every proposition there is a short conventional phrase saying that what was proposed to be done has now been done. In Latin, a theorem ends with quod erat demonstrandum, "which was to be shown", Q.E.D., and a problem ends with quod erat faciendum, "which was to be done", Q.E.F. The corresponding phrases in Greek are shown in the box below.
The first word is the intensified relative pronoun, neuter singular accusative. By itself, o/( means "that", but adding the intensifier -περ makes it say "precisely that" or something similar. The case is accusative because the pronoun is the object of the infinitive at the end of the phrase.
The second word comes from the verb "to bind" and is used in the third person singular to imply "to be necessary". The form used is the imperfect tense, like the Latin erat. The imperfect describes an action continuing or repeated in the past. It is the typical past tense of "to be". The state of being necessary is certainly continuous, and began at some indefinite time, so the imperfect is the only tense that would do here. To make the imperfect, we augment (and the accent springs ahead) and add the proper personal endings. Here, the third person singular ending is ε (instead of ει as is used in the present tense). Finally, contraction gives us the word.
The third word corresponds to demonstrandum. In Latin, this means precisely "to be demonstrated", and agrees with quod, but in Greek we use an infinitive instead, introduced by the verb implying the necessity. The verb is an -μι verb whose six principal parts are given. Only the present tense is conjugated with different endings than we are used to, and you should be able to form the future and aorist active without trouble. The fourth principal part is the perfect active, the fifth the perfect middle, and the sixth the aorist passive. The stem of the verb ends in -κ which combines with a following σ to give ξ. You should not be stumped by this sort of change any longer. Where it appears, the -νυ- is inserted for euphony. There is, in fact, a present tense using the -ω endings we are familiar with that was an alternative to the -μι forms.
The infinitive is aorist, not future. A future active infinitive would end in -σειν, not -σαι, as this one does. Although the proving is in the future relative to the necessity, in Greek this does not call for the future. The action of proof is a single action, which will be carried out once and then will be finished. This demands the aorist, whenever the deed happens. The same thing happens for the Q.E.F., but you already know how to conjugate poie/w, taking care to contract the final ε of the stem with the vowels of the endings. The method of building the aorist active infinitive is shown in the box. The circumflex is characteristic of the aorist active infinitive.
A casual look at QED might conclude that the pronoun is nominative, the verb is present, and the infinitive is future. Analysis reveals all of these assumptions are incorrect, and that the truth is much more interesting.
Let us now study the very first proposition in the Elements, Book 1, Proposition 1, which is shown in the box. This is a problem, and is necessary for showing that the postulates imply that a line of given length can be moved to any point, which is Proposition 2, and then that a distance equal to any given line can be cut off on any other given line, which is Proposition 3. This proves rigorously that it is permissible to use dividers to transfer a distance. This mechanical procedure is subject to error, but Euclid proves that, in principle, it can be done exactly.
The first paragraph is the statement of the problem. A literal translation is: onto the given straight terminated (perfect) a triangle equilateral to be constructed (aorist). In better English, to construct an equilateral triangle on a given finite straight line. The word for terminated is a perfect passive participle (compare with pe/raj, end). The reduplication is a giveaway of the perfect tense. Perfect is used here because once the line is terminated, the effect persists ever after. The infinitive is an aorist passive infinitive of the verb suni/sthmi, of which the sign is sqai, as sai is of the active, as we saw above.
We see the figure, and the text says: Let there be the given straight terminated (perfect) the AB. It is necessary indeed upon the AB straight a triangle equilateral to be constructed (aorist). That is, Let AB be the given finite line. It is required to construct an equilateral triangle upon it. The same participles and infinitives are used.
Now we proceed to the construction. The text is: At the center A to interval AB let a circle be drawn (perfect) the BΓΔ, and likewise at center B to interval BA let a circle be drawn (perfect) the AΓE, and from the Γ point, under which cut each other the circles, to the A, B points let be joined (aorist) straights the ΓA, ΓB. That is, Draw a circle with center A and radius AB, and another with center B and radius BA. Join the point of intersection of the circles Γ and points A, B.
In the Greek text, no reference is made to previous propositions, but in modern translations it is the custom to note the justification for each step. For example, Postulate 3 says that a circle can be drawn from any center with any radius, and Postulate 1 that a straight line can be drawn joining any two points. This the justification for the construction we have just made. Of course Euclid would know this, and any student who made an unjustified move would be brought up quickly.
The construction done, we can proceed to the proof. The text says: and since the A point center is of the ΓΔB circle, equal is the AΓ to the AB; likewise, since the B point center is of the ΓAE circle, equal is the BΓ to the BA. it was shown (aorist) also the ΓA to the AB equal; each therefore of the ΓA, ΓB to the AB is equal. The to the same equal also to each other is equal; and the ΓA therefore to the ΓB is equal; the three therefore the ΓA, AB, BΓ equal to each other are. In English, this is: Since A is the center of circle GDB, AG = AB; likewise, since B is the center of circle GAE, BG = BA. But GA was shown equal to AB, so each of GA, GB are equal to AB. Things equal to the same things are equal to each other, so GA = GB. Therefore, the three lines GA, AB, BG are equal to each other. Note that a singular verb is used with plural neuters, but plural masculine and feminine always call for a plural verb. This is a peculiarity of Greek.
It is only necessary now to draw the conclusions. The text is: Equilateral therefore is the ABΓ triangle. And it has been constructed (perfect) upon the given straight terminated (perfect) the AB. Which was to be done. In English, Triangle ABG is equilateral, and constructed upon the given line AB, which was to be done.
The tenses of the participles and imperatives have been given; note that they are either aorist or perfect, and that the distinction is not one of time, but of the kind of action involved. Once a line is terminated it is terminated forever (perfect), but when something is proved, the action is complete (aorist). It is good that we do not have to decide which to use, but merely to interpret what is in front of us. The vocabulary box above shows the perfect passive participle for "finite". This participle can be recognized by the reduplication of the stem, and the accent on the penult, so it can be distinguished from the present passive participle with the accent on the antepenult. The box also gives the principal parts of the verbs for draw and stand, with the endings for the third person singular instead of the usual first person. We see these endings more, and should be better acquainted with them. Note that the sixth principal part, the aorist passive, is the one we frequently use both for participles and imperatives. An example of the aorist imperative is given.
### Exercises
1. If you already know some Greek, it will be easy to read the proposition once you have grasped the peculiarities of the language used, with all its third person imperatives and perfect participles. If you are learning, the text will repay word-for-word study, and any time you spend will not be wasted. This demonstration looks very much like all of Euclid's demonstrations, and most of the words are very commonly used indeed. Take each word, and find out why it has the ending it does, and how it is used. Write out the proposition yourself, and read it aloud. The literal translation into English is, of course rather ridiculous, but the reason is that English does not have the cases and tenses. The language is really quite elegant and precise, and you will come to appreciate it.
2. Translate the proposition into good English, and justify all statements, which here can be only by reference to the definitions, postulates and axioms. |
Åk 6–9
English/Soomaali
2.1 From Figures to Expressions
Here below you see a few figures. When we talk about sequences, we usually call every figure for element or term.
In every term, the number of stars is increased. How many stars do you think there will be in term 4 and term 5?
Most likely, you see that there should be 8 stars in term 4 and 10 stars in term 5.
Instead of drawing all 100 terms, we can find an expression for calculating the number of stars.
For every term, the number of stars increases by 2.
We then write up the following table:
We indicate the number of stars in the term “n” for f(n).
The term’s placement in the sequence is called n.
The number of stars in term "n" are f(n) = 2 · n
In the hundredth term, it would be then f(100) = 2 · 100 = 200 stars.
This can also be shown with a graph in a coordinate system. We being by making a value table.
You see now clearly that the graph of the relation creates a straight line. Because of this, the graph is said to be a linear function.
Here below, you see the prices for two mobile phone subscriptions in the form of a table and as a graph in a coordinate system. Foto: Fredrik Enander
We can see in the table that company A that the price is proportional to the number of minutes you talk for if:
• the number of minutes doubles, then the cost becomes twice as much.
1 min = 1 kr
2 min = 2 kr
• the number of minutes becomes half as large, then the costs is half the size.
4 min = 4 kr
2 min = 2 kr
We can see that in company B that the price is not proportional to the number of minutes you talk if:
• the number of minutes doubles, then the cost is not twice as large.
1 min = 1.50 kr
2 min = 2 kr
• the number of minutes becomes half as large, then the cost does not become half as large.
4 min = 3 kr
2 min = 2 kr
We can see this now when both company’s prices are in a coordinate system. |
How to calculate arithmetic sequence? Explained with examples
The arithmetic sequence is a well-known method used to evaluate the sequence and sum of the sequence of the terms. This technique is frequently used in various branches of mathematics. The letters & symbols are used in this technique to represent numbers in the form of an equation.
The collection of things or objects that are well determined is also referred to as arithmetic sequence that is used in set theory for various purposes. In this lesson, we will go through the basics of the arithmetic sequence along with the definition, expression, and examples.
What is the arithmetic sequence?
The arithmetic sequence is a wide concept that is used in algebra and set theory and is defined as a sequence of numbers or integers in which the difference among each consecutive number is the same. The difference between the numbers is known as the common difference.
The common difference is the difference between two consecutive terms in a sequence that gives always the same result. The sequence can be created or the given sequence can be summed up with the help of the expression of the arithmetic sequence.
Such as,
Sequence of numbers = 12, 25, 38, 51, 64, 77, 90, …
The above sequence of numbers is an arithmetic sequence as the difference between each consecutive number is the same that is 13. You can also create a sequence of the numbers if the first term and the common difference are given. Such as,
If the first term is 12 and the common difference is 3 then
12, 15, 18, 21, 24, 27, 30, …
Order of sequence
There are two forms of the order of sequence.
• Increasing sequence
• Decreasing sequence
The order of the arithmetic sequence is depending on the nature of the constant difference of the given sequence. Let us briefly describe the orders of the arithmetic sequence.
1. Increasing sequence
The sequence should be increasing if the values or numbers go from least to greatest such as forming an ascending order. If the constant difference is positive, then it always makes an increasing sequence as the values should go from the least to the greatest.
For example, if the initial number of the sequence is 12 and the common difference is 7 then the sequence that is obtained is:
19, 26, 33, 40, 47, 54, 61, 68, 75, …
The numbers of the sequence make an ascending order as the constant difference is positive then we can say that the given sequence is increasing.
1. Decreasing sequence
The sequence should be decreasing if the values or numbers go from greatest to least such as forms a descending order. If the constant difference is negative, then it always makes a decreasing sequence as the values should go from the greatest to the least.
For example, if the initial number of the sequence is 29 and the common difference is -7 then the sequence that is obtained is:
29, 22, 15, 8, 1, -6, -13, -20, …
The numbers of the sequence make a descending order as the constant difference is negative then we can say that the given sequence is decreasing.
Expression of an arithmetic sequence
There are three different expressions that are used in arithmetic sequences for the calculation of various terms. These terms are
1. For nth term
2. For the sum of the sequence
3. For finding the common difference
The expressions for calculating the above terms are very useful and helpful for the calculation of the problems of the arithmetic sequence manually.
An arithmetic sequence calculator can be used to solve the problems of arithmetic sequence according to the formulas to avoid lengthy calculations.
1. Expression for finding the nth term of the sequence
The nth term of the sequence has a general expression such as:
nth term of the sequence= pn = p1 + (n – 1) * d
where
• pn is the nth term of the sequence
• p1 is the first term of the sequence
• n is the total number of terms
• d is the constant difference.
1. Expression for finding the sum of the sequence
The sum of the sequence has a general expression such as:
Sum of the sequence = s = n/2 * (2p1 + (n – 1) * d)
Where
• f1 is the starting term of the sequence
• n is the total number of terms
• d is a common difference.
1. Expression for finding the common difference
Here is the general expression for calculating the common difference of the sequence.
Common difference = d = pn – pn-1
How to calculate the arithmetic sequence?
The expressions of the arithmetic sequence are used to calculate the problems of the arithmetic sequence. Let us take a few examples to learn how to calculate the arithmetic sequence.
Example 1: For finding the nth term
Evaluate the 15th term of the sequence by taking the values from the given sequence,
2, 9, 16, 23, 30, 37, 44, 51, …
Solution
Step 1: First of all, evaluate the common difference and take out the initial term from the given sequence of numbers.
2, 9, 16, 23, 30, 37, 44, 51, …
Initial term = p1 = 2
Second term = p2 = 9
Common difference = d = p2 – p
Common difference = d = 9 – 2
Common difference = d = 7
We have to calculate the 25th term of the sequence so n = 25
Step 2: Now take the general expression for finding the nth term of the sequence.
nth term of the sequence= pn = p1 + (n – 1) * d
Step 3: Put the values of the constant difference and the first term into the sequence to calculate the nth term of the sequence.
15th term of the sequence= a15 = 2 + (15 – 1) * 7
15th term of the sequence= a15 = 2 + (14) * 7
15th term of the sequence= a15 = 2 + 98
15th term of the sequence= a15 = 100
Example 2: For finding the sum of the sequence
Evaluate the sum of the first 9 terms of the sequence by taking the values from the given sequence,
7, 16, 25, 34, 43, 52, 61, 70, 79, …
Solution
Step 1: First of all, evaluate the common difference and take out the initial term from the given sequence of numbers.
7, 16, 25, 34, 43, 52, 61, 70, 79, …
Initial term = p1 = 7
Second term = p2 = 16
Common difference = d = p2 – p
Common difference = d = 16 – 7
Common difference = d = 9
We have to calculate the sum of the first 30 terms of the sequence so n = 30
Step 2: Now take the general expression for finding the sum of the sequence.
Sum of the sequence = s = n/2 * (2p1 + (n – 1) * d)
Step 3: Put the values of the constant difference and the first term into the sequence to calculate the sum of the sequence.
Sum of the first 9 terms = s = 9/2 * (2(7) + (9 – 1) * 9)
Sum of the first 9 terms = s = 9/2 * (14 + (9 – 1) * 9)
Sum of the first 9 terms = s = 9/2 * (14 + 8 * 9)
Sum of the first 9 terms = s = 9/2 * (14 + 72)
Sum of the first 9 terms = s = 9/2 * 86
Sum of the first 9 terms = s = 4.5 * 86
Sum of the first 9 terms = s = 387
Final words
Now you can solve any problem for finding the nth term and sum of the sequence easily just by following the above post. We have discussed each and every basic of finding the arithmetic sequence along with solved examples.
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Introduction to the Euler Identity
Odin Noble
In mathematics there are some numbers that deserve special attention. The Euler Identity connects five of the most important numbers in mathematics, conventionally named e, pi, i, 0, and 1. The Euler Identity evaluated at pi gives e^(i*pi) - 1 = 0. For those who are already vaguely familiar with these numbers, this is quite disorienting at first glance, but a quick proof might remedy the vertigo. First we need to review the numbers e, pi, and i one by one.
The number e is the most challenging to describe. Roughly, it is the choice of base of an exponential function that makes the function equal to its derivative. Suppose f is a smooth, continuous function of x on the interval (a,b). There are 2 related functions called the derivative of f and the integral of f.
Figure 1. The derivative and integral
Recall from calculus that at any point, x, in the domain of f we can derive the slope of the line tangent to the graph f at that point. We call this quantity the “derivative” of f and denote it by
(d/dx)(f(x)), or f'(x)
This function can be obtained through the evaluation of the limit of a sequence of differences. Similarly the area under the graph of f between the two values of x, a and b, can be found by evaluating a limit of a sequence of sums. We call this quantity the “integral” of f and denote it by
f(x) dx
Fortunately there are easily remembered formulas for the integral and the derivative. The integration formulas are more or less antidifferentiation formulas; the only change is the addition of a constant k. Recall that if the function is a polynomial p_n*x^n
(d/dx)[p_n*x^n] = n*p_n*x^(n-1)
and
[n*p_n*x^(n-1)] dx = p_n*x^n + k .
For the trigonometric functions we have
(d/dx)[sin(x)] = cos(x)
and
(d/dx)[cos(x)] = -sin(x)
and
[sin(x)] dx = -cos(x) + k
and
[cos(x)] dx = sin(x) + k .
The exponential function f(x) = p^x has base p. Its derivative has the form
(d/dx)[p^x] = k*p^x
for some constant k. The constant k will be larger if p is larger. For example,
if p = 2 then k_2 = 0.693147...
if p = 3 then k_3 = 1.098612...
This invites speculation, does there exist a base, which we will call e, for which k_e = 1, giving (d/dx)[e^x] = e^x ? There does, e = 2.7182818459... . Figure 1. shows a plot of y = e^x.
Figure 2. Three exponential functions
A base p exponential function, p^x, has an inverse function called the base p logarithm, log_p(y). Inverse means that
log_p(y) = x if and only if y = p^x .
Notice that if you use either one of these equations for x or y inside the other, then you arrive at the identities
p^(log_p(y)) = y , and log_p(p^x) = x .
The logarithm and the exponential do and undo each other like plus and minus. There is of course a base e logarithm. It is denoted ln(*). Suppose we take
y = ln(x) ,
exponentiate both sides and rewrite it as
x = e^y .
Using the chain rule to differentiate with respect to x gives
(e^y)*(dy/dx) = 1
which implies
(dy/dx) = 1/e^y = 1/x ,
or since y = ln(x)
(dln(x)/dx) = 1/x .
Integrating both sides reveals that
ln(x) = [1/x] dx + k .
Remember this, it is an important stepping stone to Eulers Identity.
Perhaps the most famous number ever is pi. For any circle the ratio of the diameter to the circumference is ~= 3.1415... denoted by the Greek letter pi. This means that it takes pi diameters, or 2*pi radii, called 'radians', to wrap around the outside of the circle. (see Figure 3.) The angle, phi, subtended by the center, and two points on the circles diameter is conveniently characterized by the ratio of the circumference enclosed by the two points to the total circumference. We express this ratio as "radians of circumference enclosed by the angle", or simply "radians". For a given angle, this ratio is the same on any circle regardless of how big or small. We call this property similarity. Similarity is found in other geometric figures, in particular in triangles.
Figure 3. Radian angle measure & similarity
Recall from trigonometry that if we take one of the two circumferential points to lie on the x axis and label the other as the coordinate (x,y), then we can construct the right triangle with (0,0), (x,y), and the point (x,0) lying on the x-axis immediately below (x,y). (see Figure 4. ) The lengths of the sides of the triangle are given by x, y, and r = sqrt(x^2 + y^2). We can scale the size of the triangle by repeating the construction with the same angle on a circle with a greater or smaller radius, but the two triangles are 'similar'. That means that the ratio of any two given sides will be the same in both. These ratios uniquely characterize the triangle and thus the angle phi.
Figure 4. Polar Coordinates & similarity(Note:I couldn't find a more accurate image on the internet. here 'x' represents phi, the angle)
We define the ratios
sin(phi) = y/r , cos(phi) = x/r, and tan(phi) = y/x.
These ratios are the same for any right triangle having angle phi. A transformation from rectangular coordinates to a system of polar coordinates easily follows from these definitions. Our transformation is given by
(x,y) = (r*cos(phi), r*sin(phi))
(r,phi) = (sqrt(x*x + y*y), arctan(y/x))
for some angle theta and radius r. It is often useful to choose r = 1; this set of points is called the "unit circle".
The imaginary number, i, is also important, and deeply connected with a discussion of the unit circle. We denote the square root of negative one by the letter i. A complex number is of the form z = a + i*b, where a and b are real numbers. It is possible that a or b equal zero. That is, purely real or purely imaginary numbers are still complex numbers, just as the x-axis and y-axis are still contained in the plane. Taking this notion a brief step further we could interpret the real and imaginary components of z as rectangular coordinates. Conventionally x is chosen as the real axis and y as the imaginary axis. From the earlier discussion we know that z also has a polar representation for some (r,phi). Pick a z on the unit circle, making r = 1. Then
z = cos(phi)+i*sin(phi)
Differentiate with respect to the angle phi
dz/dphi = -sin(phi)+i*cos(phi) = i(cos(phi)+i*sin(phi)) or
dz/dphi = i*z .
Interchanging z and dphi gives
dz/z = i*dphi .
Integrating both sides gives
ln(z) = i*phi + k ,
and exponentiating both sides gives the Euler Identity
e^ln(z) = e^i*phi or z = e^i*phi = cos(phi) + i*sin(phi) .
It is counterintuitive to accept, however it is easy to evaluate the Euler Identity. Try evaluating it with phi = pi.
e^i*pi = cos(pi) + i*sin(pi) = -1 + 0 or e^i*pi + 1 = 0 .
And there in a single relation we have e, i, pi, 0, and 1, the cornerstones of our number systems! Fortunately this is as close to mysticism that mathematics ever gets. |
## Arithmetic Reasoning Flash Card Set 378226
Cards 10 Topics Adding & Subtracting Radicals, Defining Exponents, Least Common Multiple, PEMDAS, Rational Numbers, Sequence, Simplifying Radicals
#### Study Guide
To add or subtract radicals, the degree and radicand must be the same. For example, $$2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$$ but $$2\sqrt{2} + 2\sqrt{3}$$ cannot be added because they have different radicands.
###### Defining Exponents
An exponent (cbe) consists of coefficient (c) and a base (b) raised to a power (e). The exponent indicates the number of times that the base is multiplied by itself. A base with an exponent of 1 equals the base (b1 = b) and a base with an exponent of 0 equals 1 ( (b0 = 1).
###### Least Common Multiple
The least common multiple (LCM) is the smallest positive integer that is a multiple of two or more integers.
###### PEMDAS
Arithmetic operations must be performed in the following specific order:
1. Parentheses
2. Exponents
3. Multiplication and Division (from L to R)
4. Addition and Subtraction (from L to R)
The acronym PEMDAS can help remind you of the order.
###### Rational Numbers
A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form $${a \over b}$$ where a is the numerator and b is the denominator. An improper fraction ($${5 \over 3}$$) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number ($$1 {2 \over 3}$$) which has a whole number part and a fractional part.
###### Sequence
A sequence is a group of ordered numbers. An arithmetic sequence is a sequence in which each successive number is equal to the number before it plus some constant number.
The radicand of a simplified radical has no perfect square factors. A perfect square is the product of a number multiplied by itself (squared). To simplify a radical, factor out the perfect squares by recognizing that $$\sqrt{a^2} = a$$. For example, $$\sqrt{64} = \sqrt{16 \times 4} = \sqrt{4^2 \times 2^2} = 4 \times 2 = 8$$. |
# Thread: Dimensions of Original Cardboard
1. ## Dimensions of Original Cardboard
A rectangular piece of cardboard is to be formed into an uncovered box. The piece of cardboard is 2 centimeters longer than it is wide. A square that measures 3 centimeters on a side is cut from each corner. When the
sides are turned up to form the box, its volume is 765 cubic centimeters. Find the dimensions, in centimeters, of the original piece of cardboard.
2. Hello, magentarita!
This one takes a few diagrams ... or a good imagination.
A rectangular piece of cardboard is to be formed into an uncovered box.
The piece of cardboard is 2 cm longer than it is wide.
A square that measures 3 cm on a side is cut from each corner.
When the sides are turned up to form the box, its volume is 765 cm³.
Find the dimensions, in centimeters, of the original piece of cardboard.
Let $\displaystyle W$ = width of cardboard.
Then $\displaystyle W+2$ = length of cardboard.
3-cm squares are removed from each corner.
Code:
: - - - W+2 - - - :
- *-------------------* -
: |///: :///| 3
: | - * - - - - - * - | -
: | | | | :
: | | | | :
W | | | | W-6
: | | | | :
: | | | | :
: | - * - - - - - * - | -
: |///: :///| 3
- *-------------------* -
: 3 : W-4 : 3 :
The sides are turned up to form an open-top box.
Code:
*-----------*
/| /|
/ | / |3
*-----------* |
| | *
3| | /
| |/ W-6
*-----------*
W-4
We are told that the volume is 765 cm³.
So we have: .$\displaystyle (W-4)(W-6)3 \:=\:765$
Now solve for $\displaystyle W.$
3. ## great............
Originally Posted by Soroban
Hello, magentarita!
This one takes a few diagrams ... or a good imagination.
Let $\displaystyle W$ = width of cardboard.
Then $\displaystyle W+2$ = length of cardboard.
3-cm squares are removed from each corner.
Code:
: - - - W+2 - - - :
- *-------------------* -
: |///: :///| 3
: | - * - - - - - * - | -
: | | | | :
: | | | | :
W | | | | W-6
: | | | | :
: | | | | :
: | - * - - - - - * - | -
: |///: :///| 3
- *-------------------* -
: 3 : W-4 : 3 :
The sides are turned up to form an open-top box.
Code:
*-----------*
/| /|
/ | / |3
*-----------* |
| | *
3| | /
| |/ W-6
*-----------*
W-4
We are told that the volume is 765 cm³.
So we have: .$\displaystyle (W-4)(W-6)3 \:=\:765$
Now solve for $\displaystyle W.$
Great pictures and set up. Tell me, how do you make those diagrams using a regular keyboard? Yes, I can now find w.
4. Hello, magentarita!
Tell me, how do you make those diagrams using a regular keyboard?
I invented my own system a few years ago.
First, I get into Code mode ... type [ code ] (without the spaces).
Then I make a a few rows of "hyphen-space-hyphen-space..."
Code:
. . . - - - - - - - - - -
. . . - - - - - - - - - -
. . . - - - - - - - - - -
Then I use COPY/PASTE to make more rows.
Then I carefully plot out my points.
I count very carefully, because the characters get out-of-line.
It looks like this on my screen . . .
. . . * - - - - - * - - - - -
. . . | - - - -*- | - - - - -
. . . | - - * - - | - - - - -
. . . | -*- - - - | - - - - -
. . . * - - - - - * - - - -
. . . - - - - - - - - - -
But it looks like this in code mode.
Code:
. . . * - - - - - * - - - - -
. . . | - - - -*- | - - - - -
. . . | - - * - - | - - - - -
. . . | -*- - - - | - - - - -
. . . * - - - - - * - - - -
. . . - - - - - - - - - -
If I add letters and numerals , it really gets out-of-line
. . A * - - - - - * B - - - -
. . . | - - - -*- | - - - - -
. . 5 | - - * - - | - - - - -
. . . | -*- - - - | - - - - -
. . D * - - - - - * C - - -
. . . - - - 8 - - - - - -
But it looks okay in code mode.
Code:
. . A * - - - - - * B - - - -
. . . | - - - -*- | - - - - -
. . 5 | - - * - - | - - - - -
. . . | -*- - - - | - - - - -
. . D * - - - - - * C - - -
. . . - - - 8 - - - - - -
Then I remove the hyphens from the right end.
(Use Delete or Backspace.)
Code:
. . A * - - - - - * B
. . . | - - - -*- |
. . 5 | - - * - - |
. . . | -*- - - - |
. . D * - - - - - * C
. . . - - - 8
Then I replace the other hyphens with spaces.
Code:
A * - - - - - * B
| * |
5 | * |
| * |
D * - - - - - * C
8
Yes, it's a lot of extra work, but I enjoy it.
5. ## great but...
Originally Posted by Soroban
Hello, magentarita!
I invented my own system a few years ago.
First, I get into Code mode ... type [ code ] (without the spaces).
Then I make a a few rows of "hyphen-space-hyphen-space..."
Code:
. . . - - - - - - - - - -
. . . - - - - - - - - - -
. . . - - - - - - - - - -
Then I use COPY/PASTE to make more rows.
Then I carefully plot out my points.
I count very carefully, because the characters get out-of-line.
It looks like this on my screen . . .
. . . * - - - - - * - - - - -
. . . | - - - -*- | - - - - -
. . . | - - * - - | - - - - -
. . . | -*- - - - | - - - - -
. . . * - - - - - * - - - -
. . . - - - - - - - - - -
But it looks like this in code mode.
Code:
. . . * - - - - - * - - - - -
. . . | - - - -*- | - - - - -
. . . | - - * - - | - - - - -
. . . | -*- - - - | - - - - -
. . . * - - - - - * - - - -
. . . - - - - - - - - - -
If I add letters and numerals , it really gets out-of-line
. . A * - - - - - * B - - - -
. . . | - - - -*- | - - - - -
. . 5 | - - * - - | - - - - -
. . . | -*- - - - | - - - - -
. . D * - - - - - * C - - -
. . . - - - 8 - - - - - -
But it looks okay in code mode.
Code:
. . A * - - - - - * B - - - -
. . . | - - - -*- | - - - - -
. . 5 | - - * - - | - - - - -
. . . | -*- - - - | - - - - -
. . D * - - - - - * C - - -
. . . - - - 8 - - - - - -
Then I remove the hyphens from the right end.
(Use Delete or Backspace.)
Code:
. . A * - - - - - * B
. . . | - - - -*- |
. . 5 | - - * - - |
. . . | -*- - - - |
. . D * - - - - - * C
. . . - - - 8
Then I replace the other hyphens with spaces.
Code:
A * - - - - - * B
| * |
5 | * |
| * |
D * - - - - - * C
8
Yes, it's a lot of extra work, but I enjoy it.
Great but there's got to be an easier way. |
?> Rotation Worksheet | Problems & Solutions
# Rotation Worksheet
Rotation Worksheet
• Page 1
1.
Identify the incorrect statement.
I. A rotation is a transformation that turns a figure about a fixed point.
II. A rotation is an isometry.
III. A rotation does not change orientation.
IV. A figure has rotational symmetry, if the image after a rotation of 90° or less exactly fits on the original figure.
a. III b. IV c. II d. I
#### Solution:
According to the definition of rotation, it is a transformation that turns the figure about a fixed point.
Rotaion is an isometry.
A rotation does not change orientation.
A figure has rotational symmetry if the image after a rotation of 180° or less exactly fits on the original figure.
[Definition of rotational symmetry.]
So, statement IV is incorrect.
2.
Rotate the given figure, 210° about the point P.
a. Figure 3 b. Figure 1 c. Figure 2 d. Figure 4
#### Solution:
A type of transformation or movement that results when a geometric figure is turned about a fixed point is called rotation.
Draw a line from A to P.
Use protractor to measure 210° angle with side AP in counter clockwise direction.
Figure 2 makes an angles of 210° with the line AP.
Therefore Figure 2 is the correct answer.
3.
Recognise the figure which has rotational symmetry with an angle of rotation as 120°.
a. Figure 2 b. Figure 4 c. Figure 1 d. Figure 3
#### Solution:
A figure has rotational symmetry if there is a rotation of 180° or less that maps the figure onto itself.
The angle of rotation is 360°n, where n is the number of times the figure repeats.
There are 8 spokes in figure 1. So the angle of rotation is 360 / 8 = 45°.
The angle of rotation for figure 2 is 360 / 4 = 90°.
The angle of rotation for figure 3 is 360 / 3 = 120°.
The angle of rotation for figure 4 is 360 / 5 = 72°.
So, the figure with an angle of rotation of 120° is Figure 3 only.
4.
Identify the center of rotation of the following figure.
a. Figure 3 b. Figure 4 c. Figure 1 d. Figure 2
#### Solution:
Center of rotation is the point around which the figure is turned.
For the given figure, the center of rotation is the center of the figure.
The center of rotation is shown in figure 1.
5.
Identify the figure which has no rotational symmetry.
a. Figure 3 b. Figure 4 c. Figure 2 d. Figure 1
#### Solution:
If you can rotate (or turn) a figure around a center point by 180° or less and the figure appears unchanged, then the figure has rotation symmetry.
Figures 1, 2 and 3 has rotational symmetry as the image maps itself if the figure is turned by 180° or less.
Figure 4 does not have rotational symmetry as it cannot map itself when it is turned to 180° or less.
6.
Rotate the given figure, 180° about P.
a. Figure 3 b. Figure 4 c. Figure 2 d. Figure 1
#### Solution:
A type of transformation or movement, that results when a geometric figure is turned about a fixed point is called rotation.
Use protractor to measure 180° angle with side PQ in counter clockwise direction.
Figure 2 makes an angles of 180° with the actual figure.
7.
Mark and Dennis are playing ferris wheel which is rotating counter clockwise. If they are at the point marked 2, then what is the angle of rotation to reach the point marked 4?
a. 100° b. 90° c. 75° d. 120°
#### Solution:
Place protractor on the line extending from center of the wheel to the point marked 2.
Measure the angle between point marked 2 and point marked 4 extended from the center of the wheel.
The angle is 120°.
8.
Select the rotational image of the given figure after a rotation of 300° in counter clockwise.
a. Figure 4 b. Figure 1 c. Figure 3 d. Figure 2
#### Solution:
Place protractor on the line extending from centre to A.
Identify the figure that is turned 300°.
The figure which is rotated 300° in couter clockwise is Figure 3.
9.
Which of the following figures is not the rotational image?
a. Figure 4 b. Figure 3 c. Figure 1 d. Figure 2
#### Solution:
Figures 1, 2 and 3 are the images obtained by rotating the actual figure.
Figure 4 is not the rotational image.
[Rotation does not change orientation.]
10.
Identify the figure which has rotational symmetry.
a. Figure 2 b. Figure 3 c. Figure 4 d. Figure 1
#### Solution:
A type of transformation or movement, that results when a geometric figure is turned about a fixed point is called rotation.
If you can rotate (or turn) a figure around a center point by 180° or less and the figure appears unchanged, then the figure has rotation symmetry.
Figure 3 has rotational symmetry. |
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Multiplication Chart
# Multiplication Chart
Using a multiplication chart is a great way to learn multiplication of numbers. It's a handy reference that with continued use helps a child memorize their multiplication facts. Quiz your child regularly with our multiplication chart to help them succeed at school.
## How to Use The Multiplication Chart
Step One – Choose a number from the left hand column of the chart (directly under the X)
Step Two – Choose the number you want to multiply your first number with from the row across the top of the chart (directly next to the X)
Step Three – Go along the row from the left hand column number, then go directly down the column from the top row number until both the column and the row meet at one square. That is your answer.
For example: if we chose the number 6 on the left hand column (it's in blue) and the number 7 on the top row (it's in green), we would go across the row from the 6 on the side while going down the column from the number 7 at the top. The box they meet at has the number 42 in it, which tells you that 6 times 7 equals 42.
Throughout this article we have also included some extremely cute times tables to make it a bit more fun. Please enjoy!
## Multiplication Table 1-12
This is the full multiplication table for 1-12. It's a great tool for quick reference.
We have found that it's easier to learn your multiplication tables using smaller individual multiplication tables focused on one number at a time; you can find those below.
## Single Number Times Tables
Learn a single number at a time by reciting the times tables!
One's Times Table
1 x 1 = 1
1 x 2 = 2
1 x 3 = 3
1 x 4 = 4
1 x 5 = 5
1 x 6 = 6
1 x 7 = 7
1 x 8 = 8
1 x 9 = 9
1 x 10 = 10
1 x 11 = 11
1 x 12 = 12
Two's Times Table
2 x 1 = 2
2 x 2 = 4
2 x 3 = 6
2 x 4 = 8
2 x 5 = 10
2 x 6 = 12
2 x 7 = 14
2 x 8 = 16
2 x 9 = 18
2 x 10 = 20
2 x 11 = 22
2 x 12 = 24
Three's Times Table
3 x 1 = 3
3 x 2 = 6
3 x 3 = 9
3 x 4 = 12
3 x 5 = 15
3 x 6 = 18
3 x 7 = 21
3 x 8 = 24
3 x 9 = 27
3 x 10 = 30
3 x 11 = 33
3 x 12 = 36
Four's Times Table
4 x 1 = 4
4 x 2 = 8
4 x 3 = 12
4 x 4 = 16
4 x 5 = 20
4 x 6 = 24
4 x 7 = 28
4 x 8 = 32
4 x 9 = 36
4 x 10 = 40
4 x 11 = 44
4 x 12 = 48
Five's Times Table
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 4 = 20
5 x 5 = 25
5 x 6 = 30
5 x 7 = 35
5 x 8 = 40
5 x 9 = 45
5 x 10 = 50
5 x 11 = 55
5 x 12 = 60
Six's Times Table
6 x 1 = 6
6 x 2 = 12
6 x 3 = 18
6 x 4 = 24
6 x 5 = 30
6 x 6 = 36
6 x 7 = 42
6 x 8 = 48
6 x 9 = 54
6 x 10 = 60
6 x 11 = 66
6 x 12 = 72
Seven's Times Table
7 x 1 = 7
7 x 2 = 14
7 x 3 = 21
7 x 4 = 28
7 x 5 = 35
7 x 6 = 42
7 x 7 = 49
7 x 8 = 56
7 x 9 = 63
7 x 10 = 70
7 x 11 = 77
7 x 12 = 84
Eight's Times Table
8 x 1 = 8
8 x 2 = 16
8 x 3 = 24
8 x 4 = 32
8 x 5 = 40
8 x 6 = 48
8 x 7 = 56
8 x 8 = 64
8 x 9 = 72
8 x 10 = 80
8 x 11 = 88
8 x 12 = 96
Nine's Times Table
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
9 x 6 = 54
9 x 7 = 63
9 x 8 = 72
9 x 9 = 81
9 x 10 = 90
9 x 11 = 99
9 x 12 = 108
Ten's Times Table
10 x 1 = 10
10 x 2 = 20
10 x 3 = 30
10 x 4 = 40
10 x 5 = 50
10 x 6 = 60
10 x 7 = 70
10 x 8 = 80
10 x 9 = 90
10 x 10 = 100
10 x 11 = 110
10 x 12 = 120
Eleven's Times Table
11 x 1 = 11
11 x 2 = 22
11 x 3 = 33
11 x 4 = 44
11 x 5 = 55
11 x 6 = 66
11 x 7 = 77
11 x 8 = 88
11 x 9 = 99
11 x 10 = 110
11 x 11 = 121
11 x 12 = 132
Twelve's Times Table
12 x 1 = 12
12 x 2 = 24
12 x 3 = 36
12 x 4 = 48
12 x 5 = 60
12 x 6 = 72
12 x 7 = 84
12 x 8 = 96
12 x 9 = 108
12 x 10 = 120
12 x 11 = 132
12 x 12 = 144
## How to Help a Child Struggling with Multiplication
Mastering multiplication is essential to your child's success in school. Their ability to perform higher-level mathematical functions is dependent on their ability to multiply. If your child is struggling with multiplication, check out these tips to help them out.
First, make sure you're practicing their multiplication tables frequently enough. Definitely give them a break here and there, but quiz them frequently enough for the concept to fully take hold. If you don't quiz them frequently enough, it'll take longer for them to properly grasp this subject.
If quizzing them frequently isn't helping, it may be time to get them a tutor. A tutor can help them master multiplication, along with any other topic that they're struggling with. Tutors are experts in their subjects. They can also help build your child's confidence, so that they're able to face their math tests without anxiety.
Tutors are available both in-person and online. Online tutors have become increasingly popular, due to their convenience and accessibility. However, they tend to be more expensive than in-person tutors. Ultimately, pick the tutoring method that works best for your child and your budget. Explore your options before settling on one. If you're having a hard time getting started, talk to your child's teacher and ask for recommendations. |
# Multiplication facts
December 12, 2019
When students struggle with multiplication facts, solving complicated problems (like for example finding common denominators to add fractions) becomes hard. When they use most of their working memory on simple calculations they have little mental space left for understanding new concepts.
Students start learning the multiplication facts in grade 3 although multiplication as repeated addition first appears in grade 2. Skip counting by 2 and counting in 5s and 10s are the first steps. It is important that the students master the multiplication tables by grade 4 and be ready to dive into math topics like multi-digit multiplication, equivalent fractions, and division. Of course, before starting with the multiplication facts it is necessary that the students first fully learn and understand addition and subtraction facts.
Parents and teachers often rush into drills to help children memorize the multiplication tables as quickly as possible. There is nothing wrong with memorization if it comes after learning and understanding. Automaticity that comes from endless repetition will not help in the long run. There is no need to rush into memorizing all the multiplication tables. Allow students to take their time and practice solving problems until they start memorizing without even realizing it. A great way to do that is by “playing” with the multiplication table.
Here are some steps you can follow.
### 1. Check students’ understanding
Make sure the children understand the concept of multiplication well. Provide visuals to make facts concrete. Use hands-on manipulatives and ask them to draw pictures and form arrays by arranging a set of objects into rows and columns.
Present the multiplication array table in this form and have them solve many problems by counting and adding.
### 2. Practice solving problems
Once they clearly understand present the multiplication table with numbers. Have the students solve problems with the help of the table. As they solve the problems they will start remembering some of the answers. At the beginning ask them to color the area so that they remember that the same rule applies here as with the table above.
### 3. Find Patterns on the table
Why do some numbers repeat on the table?
Find all the ways to make 12.
Find all the ways to make 15
Count by 3, 4 ,5, 6,….
Look at each row and column and find patterns. Why do the numbers follow that pattern?
You will be surprised by the number of patterns that students will find. These patterns will help them understand and remember.
### 4. Complete the missing numbers on the table.
This is a great way to show students that they can use “easier” facts as stepping stones to the harder facts.
What is 4×6? if 3×6 is 18 then I can add one more 6.
### 5. Square numbers
Talk about these special numbers on the table. What makes them special? Color the area to see the squares. Students always remember the square numbers and they use them as stepping stones.
### 5. Play games
Have the students play games using the multiplication table.
Games like BLOCK IT! help the students get familiar with the table, find patterns and learn the multiplication facts.
Give groups or pairs a table with some missing numbers and see which team will fill in the numbers first. Some friendly competition always motivates students.
### 6. Practice each table on its own
Once students have had a lot of practice with the table you can start learning each table separately starting with the table of 2, 5, 10 which are the easier ones to remember. Make this a game that the child will enjoy and will want to repeat. This is a good time to ask the students to study each table and find patterns (if they haven’t found them already). Understanding these patterns, or “tricks” as students like to call them, will help them figure out the answer fast instead of trying to remember it. Eventually, fluency will come naturally.
Some examples
In the table of 3 the digits of the products add up to 3,6,9 following this pattern.
In the table of 5 the products alternate between products ending with 0 and products ending with 5. Why does that happen? Even numbers multiplied with 5 give half as many tens. Odd numbers multiplied by 5 give a product with 5 as the last digit. 7×5= (6×5)+5= 30+5=35
Students can relate the table of 4 to the table of 2. When you multiply a number x4 you can double it twice. All the products are even and it is like counting by 2 except starting with 4 and skipping one every time.
The table of 9 seems to be the most difficult to remember but is actually one of the easiest. Teach strategies that will help students understand.
We all know that there are huge advantages to having fluency with multiplication facts. However, I think we need to be very careful about how we guide students towards learning and understanding them. Multiplication facts are a very interesting chapter in the math curriculum that deserves to be given time and attention. Time for students to discover, solve and learn. Promoting memorization without meaning could cause problems and push students to dislike math.
Here are some worksheets to practice the multiplication facts.
Here are some free to print (and digital) games to play with students to practice multiplication tables.
Block it! A great game for all students. The game board is the times table and the answers are right there! now with a digital form.
The great escape. Practice factors and multiples.
Maze Escape. Single-player challenge. Practice one table at a time.
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• Level: GCSE
• Subject: Maths
• Word count: 2991
Borders Investigation Maths Coursework
Extracts from this document...
Introduction
Mohammed Sharif 11S Page No: Math’s Coursework
Borders Investigation
Introduction
Below is the starting point of my sequence of cross shapes
Aim: To investigate the sequence of squares in a pattern needed to make any cross shapes built in this way and then extend your investigation to 3 dimensional.
In this investigation I have been asked to find out how many squares would be needed to make up a certain pattern according to its sequence, also to derive algebraic formulae from the sequence each expressing one property in terms of another. I plan to solve my investigation by using a wide variety of mathematical tools such as the nth term and also using formulae which includes the linear and quadratic sequence, formulae will be checked and hopefully proven by different mathematical tools.
My sequence starts with a single white square, and then it’s surrounded by black squares (borders) which will form the next shape. To get the second or third shape the second shape becomes white with a border of black squares around it so in each new cross shape, the previous cross shape can be seen as the area of white squares in the centre. I chose to start from one square because I thought it was creative and will look interesting as it builds up.
Middle
2+1= 3 then I did
3-5 = -2 which is b in the equation
So therefore the equation is complete and will now look like:
tn = 2- 2n + 1
Testing the rule
As you can see above I have now got a complete formula which I will check to see if it works
Sequence two
1. 2(22) -4 +1
2. 2(4) -4 +1
3. 8 – 4 +1
= 5 Correct
As you can see the formula is working I am going to try it on some more sequences to be 100% sure that it is fully working and has no errors
Sequence 3
1. 2(32) -6 +1
2. 2(9) -6 +1
3. 18 – 6 +1
= 13 Correct
Sequence 6
1. 2(62) -12 +1
2. 2(36) -12 +1
3. 72 -12 +1
= 61 Correct
The formula that I found seems to be working fine without any errors as proven above. I am now going to use this formula to find the number of squares (tn) in a higher sequence:
Sequence 11
1. 2(112) -22 +1
2. 2(121) -22 +1
3. 242-22 +1
= 221 Correct
To check if I am correct or incorrect I will start from sequence 6 as the 1st difference for every pattern is 4 and the 2nd difference is constant. I can see that +4 is added to each pattern in the 1st difference so to do this I will add the total number of squares with the 1st difference for sequence 6 through its previous number and then I will add the 2nd difference which is contact +4 the same applies to all other sequences, the table below should explain everything
Sequence Total no of Squares 7 61+20+4 85 8 85+24+4 113 9 113+28+4 145 10 145+32+4 181 11 181+36+4 221
Conclusion
When n = 2 a(2)3+b(2)2+c(2)+d =7
8a + 4b + 2c + d =7 (2)
When n = 3 a(3)3+b(3)2+c(3)+d =25
27a + 9b + 3c + d = 25 (3)
When n = 4 a(4)3+b(4)2+c(4)+d =63
64a + 16b + 4c + d = 63 (4)
Simultaneous Equations
For the first I am going to show the full working out, so it’s easier to understand
(2)-(1) 8a + 4b + 2c + d=7
- a + b + c + d=1
= 7a+3b+c =6 (5)
(3)-(2) 19a+5b+c =18 (6)
(6)-(5) 12a+2b =12 (7)
(4)-(3) 37a+7b+c =38 (8)
(8)-(6) 18a+2b =20 (9)
(9)-(7) 6a =8
6a =8
÷6 ÷6
We can’t do it this way as it comes in decimals therefore we have to do it in fractions
8÷2
6÷2 = 4/3 therefore a= 4/3
Substituting a= 4/3 into equation (9) to find b
Which is
18a+2b=20
18x4/3+2b= 20
=72/3+2b= 20
=24+2b=20
24-24+2b=20-24
2b= -4
To leave b on it’s on I have to divide both sides by two to and then that will help me to find b
So that’s
2b/2= b -4/2= -2 therefore b= -2
To find c substitute a = 4/3 and b = -2 into equation (8)
37a+7b+c=38
37x4/3+7(-2)+c=38
=148/3-14+c=38
148 14
3 1
X3
=
148 42 = 106/3
3 3
106/3 + c = 38
106/3 - 106/3 + c = 38 - 106/3
38-(106÷3)= 8/3 so therefore c = 8/3
To find d substitute a = 4/3, b = -2 and c = 8/3 into equation (1)
a+b+c+d=1
4/3-2+8/3+d=1
4-2+d=1
2+d=1
2-2+d=1-2 which is -1 therfore d = -1
Now I have my complete formula= 4/3n3 - 2n2 + 8/3n – 1
Formula 4/3n3 - 2n2 + 8/3n – 1 checking
n 2 = 7
4/3(2)3 – 2(2)2 + 8/3(2) – 1=7
4/3(8) – 2(4) + 16/3 – 1 = 7
32/3 -8 + 16/3 -1 = 7
32/3+16/3= 16
16-8= 8-1= 7 CORRECT
n 5 = 129
4/3(5)3 – 2(5)2 + 8/3(5) – 1=129
4/3(125) – 2(25) + 40/3 – 1 = 129
500/3 -50 + 40/3 -1 = 129
500/3+40/3= 180
180-50= 130-1= 129 CORRECT
My formula is correct, and therefore it should work on all the sequences as proven above.
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Inequality - A compare of 2 values or expressions.For example, 10x Equation - A statement heralding the equality of 2 expressions.For example, 4x = 8 is an equation conversely, 10x > 20 is an inequality.
## Operating v Inequalities: multiplying & Dividing
Performing multiplication or department with one inequality is nearly identical to multiply or separating parts of traditional equations (with one exception, covered below).
Consider the complying with examples:
### The Exception: an adverse Numbers
There is one very important exemption to the ascendancy that multiply or dividing an inequality is the exact same as multiplying or dividing an equation.
Whenever you main point or division an inequality by a an unfavorable number, you have to flip the inequality sign.
In the following example, notice how the sign once the inequality is split by -2
In the complying with example, notification how the sign becomes a > sign as soon as the inequality is divided by -2
-2x + 15 3-2x + 15 - 15 3 - 15-2x -12x > 6
Warning: Caution as soon as Multiplying or separating Variables
One really important implicit of this dominance is: friend cannot division by one unknown (i.e., a variable) unless you are sure the its sign since you execute not recognize whether you need to flip the sign of the inequality. There room plenty that instances wherein you will understand the authorize of a variable and as a result, you have the right to multiply or divide and also know for certain whether you should flip the inequality sign. However, friend must constantly ask yourself whether you know for certain the authorize of the variable prior to dividing or multiplying when managing an inequality.
If 2x5y 10y, what is the range of potential values for x?You cannot division by y or 5y because you do not understand whether y is an unfavorable or positive and, together such, you do not know whether to upper and lower reversal the inequality.
## Multiple Inequalities
Just together it is possible to deal with two simultaneous equations, so that is possible to resolve two inequalities (or three, or four, etc.). In fixing multiple coincided inequalities using multiplication or division, the many important component is to deal with each inequality separately and also then combine them.
If 2x 150, what is the selection of feasible values for x?1.) fix each inequality alone.2x x -5x x > 2 15x 150x 102.) integrate each inequality and find the overlap (i.e., the areas where every inequality is satisfied--this area is the solution).x x > 2x 10The area of overlap--i.e., the equipment to the collection of inequalities--is where x 2
For many students, the above set of inequalities can best be interpreted graphically. The solution to the set of inequalities is the overlapping graphical area.
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# How do you Find the n-th term of the infinite sequence 1,-2/3,4/9,-8/27,…?
Jul 31, 2014
When looking at sequences, your mind should always be in a pattern-recognizing mode. Sometimes sequences seem daunting, but when you look at them in pieces, they're usually easier to tackle.
With this sequence, we can see that for each term, the numerator doubles, and the denominator triples. Also, it's important to note that each successive term flips sign. (positive to negative to positive)
To tackle the sign flips, we know we'll have to put in our formula a term raising $- 1$ to a power involving $n$. Since the sign flips happen every odd term in the sequence, the term in our formula will need to look something like this:
${\left(- 1\right)}^{n - 1}$
When $n$ is 1, this thing will equate out to ${\left(- 1\right)}^{0}$, which is just $1$. When $n$ is 2, it'll be $- 1$, and so on. In our general formula, we'll multiply this bit against everything else, to cause the sign flips.
Since each successive numerator is double the previous one, we can use ${2}^{n - 1}$ in the formula's numerator. At $n = 1$, this will be $1$, at $n = 2$, it'll be $2$, at $n = 3$, it'll be $4$, and so on.
For the denominator, we'll use ${3}^{n - 1}$ This results in $1 , 3 , 9 ,$ etc..
Now, we simply piece them together:
${x}_{n} = {\left(- 1\right)}^{n - 1} \cdot {2}^{n - 1} / {3}^{n - 1}$ |
## Calculus: Early Transcendentals 8th Edition
$f(\frac{\pi}{6})=\frac{3\sqrt 3}{2}$ is the absolute maximum value. $f(\frac{\pi}{2})=0$ is the absolute minimum value.
The Closed Interval Method 1) Function $f$ must be continuous on the closed interval $[a, b]$ 2) Find the values of $f$ at the critical points in $(a,b)$ and the values of $f$ at the endpoints. 3) Compare the values. - The largest is the absolute maximum value. - The smallest is the absolute minimum value. $$f(t)=2\cos t+\sin 2t,\hspace{1.5cm}[0, \pi/2]$$ First, $f(t)$ is continuous on the interval $[0,\pi/2]$, so the Closed Interval Method can be applied here. 1) Find the critical points of $f$ $$f'(t)=-2\sin t+2\cos 2t$$ $$f'(t)=-2\sin t+2(1-2\sin^2 t)$$ $$f'(t)=-2\sin t+2-4\sin^2 t$$ $$f'(t)=-2(2\sin^2 t+\sin t-1)$$ $$f'(t)=-2(\sin t+1)(2\sin t-1)$$ For $$f'(t)=0$$ $$\sin t=-1\hspace{.5cm}or\hspace{.5cm}\sin t=\frac{1}{2}$$ - For $\sin t=-1$, there is no value of $t$ in the interval $(0,\pi/2)$ that can satisfy the equation. - For $\sin t=\frac{1}{2}$, where t is in the interval $(0,\pi/2)$, $t=\frac{\pi}{6}.$ $t=\frac{\pi}{6}$ would be evaluated next. 2) Calculate the values of $f$ at critical points and endpoints. - Critical points: $t=\frac{\pi}{6}$ $$f(\frac{\pi}{6})=2\cos\frac{\pi}{6}+\sin\frac{2\pi}{6}=\frac{2\sqrt 3}{2}+\frac{\sqrt 3}{2}=\frac{3\sqrt 3}{2}$$ - End points: $t=0$ and $t=\frac{\pi}{2}$ $$f(0)=2\cos0+\sin(2\times0)=2\times1+0=2$$ $$f(\frac{\pi}{2})=2\cos\frac{\pi}{2}+\sin(2\times\frac{\pi}{2})=2\times0+\sin\pi=0+0=0$$ 3) Comparison Among the results: - $f(\frac{\pi}{6})=\frac{3\sqrt 3}{2}$ is the largest, so $f(\frac{\pi}{6})=\frac{3\sqrt 3}{2}$ is the absolute maximum value. - $f(\frac{\pi}{2})=0$ is the smallest, so $f(\frac{\pi}{2})=0$ is the absolute minimum value. |
# Find the maximum value of f(x) = x^3 - 3x if x^4 + 36 < 13x^2. How do I do this? What are the steps.
ishpiro | College Teacher | (Level 1) Educator
Posted on
The given function f(x) is a polynomial function, so it is defined for all x. However, in this case, the domain is restricted by the given inequality
`x^4+36<13x^2` .
Start by finding thw intervals of x for which f(x) is defined by solving this inequality. It can be solved by factoring:
`x^4-13x^2 + 36<0`
`(x^2 -9)(x^2-4)<0`
`(x-3)(x+3)(x-2)(x+2)<0`
The left side of this inequality changes sign when x = -3, -2, 2 and 3. Since for any x greater than 3, for example, x = 4, the left side is positive,
( (4-3)(4+3)(4-2)(4+2) = 84 > 0 )
then it will be negative when x is between 2 and 3, then positive again when x is between -2 and 2, then negative when x is between -3 and 2, and positive when x is less than -3. So, the solutions of the inequality above are intervals
(-3,-2), (2, 3). Notice that the ends of the intervals, x = 3, 2, -2, -3, are not included in the domain, so the function cannot be maximum there.
To determine where the given function will have a maximum, we need to find its derivative:
The derivative of f(x) is
`f'(x) = 3x^2- 3`
It will be zero when `3x^2-3=0` . This quadratic equation can be solved by factoring:
`3(x^2-1)=3(x-1)(x+1)=0`
The roots of this equation are x = 1 and x = -1. So the function f(x) would be maximum at one of these points, but these points do not lie inside the domain.
We can see that the derivative is positive when x is greater than 1 and less than -1. This means that the function is increasing on these intervals. Notice also that these intervals include the domain of the function, intervals
(2, 3) and (-3, -2).
Since the function is only increasing on these intervals, and the ends of the intervals are not included in the domain, the given function cannot attain a maximum value on this domain.
If the ends were included, then the function would be maximum at x =3:
`f(3) = 3^3-3*3=18` . |
# Solve equations with unknown coefficients with Matlab
We have so far been working with numbers, totally ignoring one of the most rewarding ability Matlab put to our use: the possibility to work with symbolic expressions.
In many college algebraic courses, it is taught how to use and simplify equations with symbolic expressions, where one of the major task is to be learn how to express one symbol with respect to others. Here we will attempt to use Matlab to solve some problems we lately worked on using real numbers.
## Example 1
Let’s consider the following equation
We all know that this is second order polynomial equation and we know how to solve it. Let’s try using Matlab to solve this very equation as it is, assuming we don’t know what the value of the coefficients are.
The code
```syms a b c x
f = a*x^2 + b*x + c
solve(f)```
Which returns
Let’s ask Matlab to give us less difficulties reading the answer
The code
`pretty(ans)`
Which returns
Which we all remember from basic algebra.
If you would like to solve the equation with respect to a, you can state it like this
`solve(f,a)`
Which returns
## Example 2
Let’s now use the following equation
Just like the second order polynomial equation in example 1, this will be like it
The code
```syms a b c d x;
f = a*x^3 + b*x^2 + c*x + d;
l = solve(f);
pretty(l)```
Which returns
## Example 3
This is also a nice method you can make use of, to help you remember a formula. Let’s consider the following Matrix
Let’s find the determinant of A
The code
```syms A11 A12 A13 A21 A22 A23 A31 A32 A33
A=[A11 A12 A13 ; A21 A22 A23 ; A31 A32 A33]
l = det(A)```
Which returns
## Example 4
Let’s do some Matrix operation with the following matrices: Addition and Multiplication.
the code
```syms B11 B12 B21 B22 C11 C12 C21 C22;
B = [B11 B12 ; B21 B22];
C = [C11 C12 ; C21 C22];
Add = B + C;
Mul = B*C;
Add
Mul```
Which returns
## Example 5
Let’s end this session with solving the following equation.
The code
```syms a b x
g = a^x + b;
l = solve(g);
pretty(l)```
Which returns |
# How do you find the absolute value of two numbers?
## How do you find the absolute value of two numbers?
The most common way to represent the absolute value of a number or expression is to surround it with the absolute value symbol: two vertical straight lines.
1. |6| = 6 means “the absolute value of 6 is 6.”
2. |–6| = 6 means “the absolute value of –6 is 6.”
3. |–2 – x| means “the absolute value of the expression –2 minus x.”
What is the absolute value?
The absolute value (or modulus) | x | of a real number x is the non-negative value of x without regard to its sign. For example, the absolute value of 5 is 5, and the absolute value of −5 is also 5. The absolute value of a number may be thought of as its distance from zero along real number line.
### What’s the absolute value of 1?
Of course, 1 is the absolute value of both 1 and –1, but it’s also the absolute value of both i and –i since they’re both one unit away from 0 on the imaginary axis. The unit circle is the circle of radius 1 centered at 0. It include all complex numbers of absolute value 1, so it has the equation |z| = 1.
What is the absolute value of -| 14?
The absolute value is the distance between a number and zero. The distance between −14 and 0 is 14 .
#### Why is absolute value used in the formula for distance?
The standard deviation formula may look confusing,but it will make sense after we break it down.
• Find the mean.
• For each data point,find the square of its distance to the mean.
• Sum the values from Step 2.
• Divide by the number of data points.
• Take the square root.
• What does absolute value mean?
The absolute value is always a positive number except for zero, as zero is neither positive or negative. Absolute value refers to the distance of a number from zero, regardless of direction. The distance is always positive, as absolute value of a number cannot be negative.
## How do you find the absolute the absolute value?
Problem:|( − 4 ∗ 5)+3 − 2|{\\displaystyle|(-4*5)+3-2|}
• Simplify inside parenthesis:|( − 20)+3 − 2|{\\displaystyle|(-20)+3-2|} |
Area of a part of a circle | Mathematics Form 2
# KCSE ONLINE
## Esoma Online Revision Resources
### Secondary
Butterfly
Chameleon
#### Area of a part of a circle - Mathematics Form 2
Lesson objective
By the end of the lesson you should be able to:
Find the area of a common region between two circles.
INTRODUCTION
Trigonometric ratio:
Given a right angled triangle, ABC.
OppositeHypotenus A
B C
Diagram 9
The longest side is called the Hypotenuse (AC) (Blink AC as it is read)
Side BC is Adjacent to angle ? and side AB is opposite to angle ?. (Blink AB and BC as they read)
Trigonometric ratios are stated with reference to angle ?
Sin of ? =
Cosine of ?=
Tangent of ? =
The ratios are abbreviated as Cos ?, Sino ? and Tan ?.
Area of sector
The following is a diagram showing a sector.
The arc subtend an angle of ?o at the centre of circle of radius r
The area of sector AOB is
Note that the area of a sector is a fraction of the area of a circle which is r2. The size of the fraction is determined by the angle of the sector eg;
1. If the angle is ?; the fraction is
2. If the angle is 60o, the fraction is
Example 1
Find the area of the following sector.
Solution:
Area of the sector =
Area of the segment.
A sector is made up of a triangle (OAB) and a segment (the shaded part).
The area of the segment:
1. Find the area of the sector.
2. Find the area of triangle OAB.
3. Subtract the area of the triangle OAB from the area of the sector.
Area of the sector =
Area of the triangle = x r2 x sin?
Area of the segment =
= r2sin?.
Example 2
Find the area of the shaded part in the figure below. (? r = 3.142).
Diagram 13
Solution:
The area of the sector is
The area of the triangle OPQ is
x 5 x 5 x sin60 = x 25cm x 0.866
=10.83cm2
Area of the segment (shaded part) is
13.09cm2 10.83cm2
=2.26cm2.
Example 3
Diagram 14
Find the area of the segment in the following diagram given that XY is 12cm and radius is 10 cm.
Solution:
Diagram 15
Is perpendicular OT bisector of the chord XY?
XT=6cm and TY=6cm
Diagram 15
Using the right angled triangle OTY, find the height OT by Pythagoras theory.
Diagram 16
Height (H) = DY2 TY2
H = 102 62
H = 100-36 = 64
H=8cm.
Then:
Area of the triangle XOY
= x base x height
Base = 12cm.
Height = 8cm
Area = x 12cm x 8cm =48cm2
To find the area of the sector, determine the value of ? which the sector angle.
Notice that OT bisects the angle ?.
Using the right angled TOY which is
We use the trigonometric ratios,
Therefore cosine
To find the value of , find the cosine invade of 0.8, Cos-1 0.8 =
Read off the value from the cosine tables
Therefore the sector angle
? = 36.87 x 2 = 73.74
The area of the sector is
The area of the segment then = Area of the sector Area of triangle OXY
=64.36cm2 48cm2
=16.36cm2
Area of a common regions, between two circles.
When two circles interact they form a common region.
Diagram 18
(i) the following are two circles with centre A and B.
Diagram 19
(ii) When the two circles intersect they form a common region.
Diagram 20
1. When the circles are separated, we note that the common region is made of the two segments.
(Animate the two circles by letting them intersect to show the common region and then separate them to show the two segments).
The common area that is shaded is calculated by adding the areas of the two segments.
EXAMPLE 1
Find the area of the common shaded in diagram below where the circles have an equal reading of 7cm and is
PAQ=PBQ=600
Solution
Separate the circles to show the segments.
Diagram 22
To find the area of the common region
Find the area for the segment in the circle centre A
Area of the sector =
=25.67cm2
Area of the triangle APQ
= x 7 x 7x sin60
=21.22cm2
Area of the segment = 25.67-21.22
=4.45cm2
Note that the two segments are equal.
Therefore the area of the common region is
4.45cm x 2 = 8.9cm2
EXAMPLE 2
Find the area of the shaded region in the following diagram.
Solution
1. To find the area of the sector in the circle centre P draw sector PUV.
Diagram 24
Find the area of triangle PUV
To get find the sin inverse 0.5= 30o.
Therefore PM = 10 Cos 30
=10 x 0.866 = 8.66cm
Area of the triangle PVU = x 10 x8.66
=43.3 cm2
To find the area of the sector PUV, find the sector angle.
Sector angle is 2 x 30 =60o
= 37cm2
Area of the segment
= 52.37 43.3
= 9.07 cm2
1. To find the area of the section in circle centre Q. Draw sector
Find the area of the triangle
Sin-1 =0.7143=45.590
= 45.59o
To find QM, use
Cos 45.59 =
QM = 4.899cm
Area of triangle UVQ
= x 10 x 4.899
=24.5cm2
To find the area of sector QUV, find the sector angle.
Sector angle = 45.59 x 2
=91.81o
=
=39cm2
Area of the segment
=39 24.5
=14.49
=14.5 cm2
Area of the common region therefore is the sum of the, area of the segment in circle centre P = 9.07 cm
And
Area of the segment in circle centre
Q = 14.5cm
Which is =9.07 + 14.5
=23.57cm2
(ii) A sector: This an area bounded by two radii and an arc.
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# Algebra 2
```Algebra 2
End Behavior Practice Worksheet
Name _____________________________
Date ________________ Period ______
Decide whether the function is a polynomial function.
If so, state the type of function (constant, quadratic, cubic or quartic), the degree of the polynomial
and the number of total solutions.
If not, circle the reason the function is not a polynomial
1.
f(x) = 6x3 + 2x – 1
2.
f(x) = -3x2 + 4x – 1
3.
g(x) = 3x2 – 4x + 2
4.
h(x) = 5x4 + 3x – 2
5.
h(x) = 3x-2 – 5x3 + 2x4 – 1
6.
g(x) = 4
DO NOT USE A CALCULATOR! Use what you know about end behaviors of polynomial functions to sketch a
possible graph of the functions below, you may use Clouds to fill in the middle of your graph since we are only
concerned with the ends For each write the END BEHAVIOR IN CORRECT NOTATION:
7.
f(x) = 2x5 – 3x3 + 4x – 5
8.
End Behavior:
9.
h(x) = -4x6 – 5x2 + 2x – 1
End Behavior:
g(x) = 7x2 – 3x + 5
End Behavior:
10.
h(x) = -4x3 + 5x – 2
End Behavior:
11.
f(x) = 4x8 – 9x7 + 5x2 – 4
12.
End Behavior:
13.
f(x) = -5x2 – 9x + 20
End Behavior:
g(x) = -3x5 + 4x4 – 5x2 + 1
14.
End Behavior:
h(x) = 2x + 1
End Behavior:
Use what you know about end behaviors of polynomial functions to match the polynomial with its graph. DO
NOT USE A CALCULATOR.
15.
f(x) = 2x4 + 5x - 1
16.
h(x) = -3x4 + 5x - 7
17.
g(x) = 4x5 – 4x2 + 1
18.
f(x) = -2x3 + 3x2 – 1
a)
b)
c)
d)
FACTOR COMPLETELY!
19. 4x2 – 16
20. 9x2 – 64
21. 100 – x2
22. 49y2 – 121
23. x5 – 4x3
24. 81x4 - 16
```
Algebra
20 Cards
Polynomials
21 Cards
Linear algebra
50 Cards
Ring theory
15 Cards
Category theory
14 Cards |
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### Transcript
• 1. 3 8 Adding and Subtracting Fractions By Graeme Henchel http://hench-maths.wikispaces.com
• 2. Adding Fractions with common denominators 3 4 7 + = 8 8 8
• 3. Adding Fractions with different denominators Problem: You can’t add fractions with different denominators without getting them ready first. They will be ready to add when they have common denominators Solution: Turn fractions into equivalent fractions with a common denominator that is find the Lowest Common Multiple (LCM) of the two denominators
• 4. Finding the Lowest Common Denominator • The lowest common multiple of two numbers is the lowest number in BOTH lists of multiples 1 1 + 2 3 Multiples of 2 are 2, 4, 6, 8, 10…… Multiples of 3 are 3, 6, 9, 12, ……… What is the lowest common multiple?
• 5. Finding the Lowest Common Denominator • The lowest common multiple of two numbers is the lowest number they will BOTH divide into 1 1 + 2 3 2 divides into 2, 4, 6, 8….. 3 divides into 3, 6, 9…. What is the lowest number 2 and 3 both divide into?
• 6. 1 1 + 2 3 You can’t add fractions with different denominators + The Lowest Common Multiple of 2 and 3 is 6 so turn all fractions into sixths 1 3 1 2 3 2 5 × + × = + = 2 3 3 2 6 6 6 Special form of 1
• 7. 1 2 + 2 5 Lowest common denominator is 10 so make all fractions tenths 5 4 9 + = 10 10 10
• 8. 1 1 + 3 4 Turn both fractions into twelfths 4 3 7 + = 12 12 12
• 9. ? ? 3 3 2 7 2 9 14 23 × + × = + = + = =1 3 7 3 7 21 21 21 21 21 21 It is 3/3 It is 7/7 So I multiply So I multiply 3/7 by 3/3 2/3 by 7/7 Finally the fractions are READY to add. I just have to add the numerators What is special form 9+14=23 What special form What the lowest number ofBOTH change 7 3 into? 1 of 1 will change will 3 and 7 divide to 21. 21. Hmmmm? to Hmmmm? It is 21. So that is my Hmmmmm?????? common denominator Now 3x3=9 and 2x7=14 Now I know the new numerators
• 10. Adding Mixed Numbers • Separate the fraction and the whole number sections, add them separately and recombine at the end 1 1 2 2 2 +5 2 +5 1 1 3 3 + = = 7+ = 5 6 7 + 5 6 + |
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# TEXTBOOK SOLUTIONS FOR Mathematical Reasoning Writing and Proof 2nd Edition
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Chapter: Problem:
Preview Activity (An Indexed Family of Sets) We often use subscripts to identify sets. For example, in Preview Activity, instead of using A, B, C, and D as the names of the sets, we could have used A1,A2, A3, and A4,. When we do this, we are using the subscript as an indentifying tag, or index, for each set. We can also use this idea to specify an infinite family of sets. For example, for each natural number n, we define
An = {n, n + 1, n + 2,n + 3, n + 4}.
So if we have a family of sets A = {A1, A2, A3, A4}, we use the notation to mean the same thing as
Determine and .
We can see that with the use of subscripts, we do not even have to define the family of sets A.We can work with the infinite family of sets and use the subscripts to indicate which sets to use in a union or an intersection.
Activity
Preview Activity (The Union and Intersection of Four Sets) In Section 4.3, we discussed various properties of set operations. We will now focus on the associative properties for set union and set intersection. Notice that the definition of “set union” tells us how to form the union of two sets. It is the associative law that allows us to discuss the union of three sets. Using the associate law, if A, B,and Care subsets of some universal set, then we can define AU BU C to be (AU B)U C or AU (BU C). That is,
AU BU C = (AU B)U C = AU (BU C).
For this activity, the universal set is ℕ and we will use the following four sets:
A = {1, 2, 3, 4, 5}
B = {2, 3, 4, 5, 6}
C = {3, 4, 5, 6, 7}
D= {4, 5, 6, 7, 8}.
Use the roster method to specify the sets AU BU C, B U C U D, ABC, and BCD.
Preview Activity (The Union and Intersection of Four Sets) In Section 4.3, we discussed various properties of set operations. We will now focus on the associative properties for set union and set intersection. Notice that the definition of “set union” tells us how to form the union of two sets. It is the associative law that allows us to discuss the union of three sets. Using the associate law, if A, B,and Care subsets of some universal set, then we can define AU BU C to be (AU B)U C or AU (BU C). That is,
AU BU C = (AU B)U C = AU (BU C).
For this activity, the universal set is ℕ and we will use the following four sets:
A = {1, 2, 3, 4, 5}
B = {2, 3, 4, 5, 6}
C = {3, 4, 5, 6, 7}
D= {4, 5, 6, 7, 8}.
Use the roster method to specify each of the following sets. In each case, be sure to follow the order specified by the parentheses.
(a) (AU BU C) U D
(b) AU (B U C U D)
(c) AU (B U C) U D
(d) (A U B) U (C U D)
(e) (ABC) ⋂ D
(f) A ⋂ (BCD)
(g) A ⋂ (BC) ⋂ D
(h) (AB) ⋂ (CD)
Preview Activity (The Union and Intersection of Four Sets) In Section 4.3, we discussed various properties of set operations. We will now focus on the associative properties for set union and set intersection. Notice that the definition of “set union” tells us how to form the union of two sets. It is the associative law that allows us to discuss the union of three sets. Using the associate law, if A, B,and Care subsets of some universal set, then we can define AU BU C to be (AU B)U C or AU (BU C). That is,
AU BU C = (AU B)U C = AU (BU C).
For this activity, the universal set is ℕ and we will use the following four sets:
A = {1, 2, 3, 4, 5}
B = {2, 3, 4, 5, 6}
C = {3, 4, 5, 6, 7}
D= {4, 5, 6, 7, 8}.
Based on the work in Part (2), does the placement of the parentheses matter when determining the union (or intersection) of these four sets? Does this make it possible to define AU BU CU Dand AB⋂ C ⋂ D?
SAMPLE SOLUTION
Chapter: Problem:
We don't have the solution to this problem yet.
Get help from a Chegg subject expert. |
Haven
2021-09-22
The reduced row echelon form of a system of linear equations is given. Write the system of equations corresponding to the given matrix. Use x, y;x,y; or x, y, z;x,y,z; or ${x}_{1},{x}_{2},{x}_{3},{x}_{4}$ as variables. Determine whether the system is consistent or inconsistent. If it is consistent, give the solution.
$\left[\begin{array}{cccccc}1& 0& 0& 0& |& 1\\ 0& 1& 0& 0& |& 2\\ 0& 0& 1& 2& |& 3\end{array}\right]$
dieseisB
The goal of the task is to write a system of equations that corresponds to a given matrix. And then determine whether the system is consistent or not and if so give a solution.
The system that corresponds to a given matrix is:
$\left\{\begin{array}{l}1\cdot {x}_{1}+0\cdot {x}_{2}+0\cdot {x}_{3}+0\cdot {x}_{4}=1\\ 0\cdot {x}_{1}+1\cdot {x}_{2}+0\cdot {x}_{3}+0\cdot {x}_{4}=2\\ 0\cdot {x}_{1}+0\cdot {x}_{2}+1\cdot {x}_{3}+2\cdot {x}_{4}=3\end{array}$
$\left\{\begin{array}{l}{x}_{1}=1\\ {x}_{2}=2\\ {x}_{3}+2{x}_{4}=3\end{array}$
The system is consistent.
${x}_{1}=1,{x}_{2}=2,{x}_{3}+2{x}_{4}=3,{x}_{3}=3-2{x}_{4}$
${x}_{4}$ is real number
Do you have a similar question? |
# How to Use Newton's Method to Find Roots of Equations
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Lesson Transcript
Instructor: Erin Ryan
Finding the roots of equations usually requires the use of a calculator. However, in this lesson you'll use Newton's Method to find the root of any equation, even when you can't solve for it explicitly.
## Newton's Method
Remember that Newton's Method is a way to find the roots of an equation. For example, if y = f(x), it helps you find a value of x that y = 0. Newton's Method, in particular, uses an iterative method. That is, you make some guess and you use it to find another guess - a better guess. The method is just a form of linearization (estimation). You estimate x sub (n + 1) - that's your next guess - is going to be equal to your current guess x sub n - y sub n / f '(x sub n). That is, your next guess is equal to your current guess minus the current value of y divided by the derivative.
In this example, we guess x sub 0, find the value of y at x sub 0 and the value of the derivative at x sub 0. We use that information to find a new estimate: x sub 1. Again, we find y at x sub 1, we find the derivative at x sub 1, and we use that information to find another guess, x sub 2. And, we'll continue this until our new value of x gives us y = 0.
In practice, you're going to start with some initial guess, x sub 0. You're going to find the derivative,f '(x) of your equation, and then you're going to use Newton's equation to estimate x sub 1. You're going to estimate x sub 2 (using x sub 1), and so on until your x values converge. They're going to converge somewhere where y = 0. This usually works (we're not going to get into the complex cases where Newton's Method doesn't work in this course).
So, how do I use Newton's Method? I generally make a table to keep track of all my variables. Here's my table:
x y y'
I've got x in one column, y in one column, and the derivative, y ', in another. I make my first guess, which would be the first row of this table. I'm going to make a guess for x, and I'm going to plug it in. Then, I'll find what y and y ' are for that value of x. I'm going to use all that information in the Newton equation to find the next row (particularly, the next x). I'm going to continue from there.
## Solving the Equation
Let's do this. We have the equation f(x) = xˆ2 + 3x - 4. When I graph this, I know that it makes a parabola. I already think there might be two roots, and I can probably solve for them by hand; but for this case, I'm going to use Newton's Method.
I'm going to use x = 0 for my first guess. I'll use my Newton equation, x sub n+1 = x sub n - y sub n / f ' (x sub n). I can also write this as x sub n - f(x sub n) / f '(x sub n), and all I've done is say that f(x sub n) is the same as y sub n. Let's plug in our f(x) and our f '(x) into that equation.
First, what is f '(x)? Let's take the derivative of f(x) and I get 2x + 3. I used the power rule to differentiate xˆ2 as 2x. Then I've got the derivative of 3x, which is just 3.
Let's plug in f(x) and f '(x) into our Newton equation. I get x sub n+1 = x sub n - (x sub nˆ2 + 3x sub n - 4) / 2x sub n + 3. Now I've got an equation. I just need my table, and away we go.
## Graphing the Equation
So, here's my table:
x y y'
I'm going to use an initial guess, x sub 0 = 0. That goes on the first line, 0 for x. When x = 0, f(x) (or y) is -4 and y ' = 3. Let's put those into my table as well:
x y y'
0 -4 3
Let's use Newton's equation to find x sub 1. If my initial guess x sub 0 = 0, I plug in 0 into the equation. I get x sub 1 (my next guess) is 4/3. Let's put that into the table. Let's say that's about 1.3. If x is 1.3, I can find y by plugging 1.3 into my f(x), and I get 1.8. y ' (by plugging 1.3 into this equation) is about 5.66.
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# Section 6.1 Question 1
## What is a frequency distribution?
A local bank is interested in tracking the time it takes a customer transaction to be completed. On Monday, they randomly select 20 customers and measure the amount of time it takes each customer to make a deposit. Each data value is rounded to the nearest minute.
Data like this is easy to record. As banks compete for customers, they strive to lower the amount of time it takes a customer to make a deposit.
With only 20 numbers, it might be possible to examine the data and get some useful information about it. As more and more data is collected, the numbers become lost in the sheer amount of information. It is helpful to represent the data differently. A frequency distribution table allows us to see the different data values and how frequently they occur. The data in the table above are discrete data. This means they only take on a few values. In this case, the data only takes on the values 1, 2, 3, or 4. The frequency distribution table is constructed by counting the number of times each data value occurs. These frequencies are then listed with the corresponding data value in a table.
For instance, the data value 1 occurs 5 times in the table. The data value 2 occurs 6 times, 3 occurs 7 times, and 4 occurs 2 times. List these values in a table.
The total is often included to insure that all data values have been included in the count. Looking at the table, we see that the time of 3 minutes occurs most frequently. In fact, almost all deposits take 3 minutes of less.
### Example 1 Find the Frequency Distribution Table
The bank develops a new training program to help make each teller more efficient. After all tellers have completed the training program, the bank measures the amount of time it takes 40 randomly selected customers to make a deposit. These times are recorded in the table below.
Construct a frequency distribution table for the data.
Solution The data in this table takes on the values 1, 2, 3, 4, and 5. We will need rows in the frequency distribution table for each value. Then count the number of times each value occurs and record the frequency.
The frequencies add to 40 so we have included all of the times in the table. Without this total, we would not know if we miscounted the data values.
Let’s compare the two frequency distribution tables we have created. Add another row to the first table for a deposit taking 5 minutes, but with a frequency of zero.
Since the totals for each table are different, it is harder to make a direct comparison of the tables. This is remedied by computing the relative frequency of each data. The relative frequency is computed by dividing each frequency by the corresponding total number of measurements.
To help us develop a framework for working with discrete data, suppose there are k different data values with k corresponding frequencies. In the case of the bank data, there are five different data values (1, 2, 3, 4, 5). We refer to any individual data value as xi where i can take on the values one through five. The frequencies are referenced as fi. For instance, we would call the first data value, 1, using x1. The corresponding frequency (either 5 or 10 depending on whether we are using the data before or after the training program) is f1. Using this notation, the sum of the frequencies for the first set of data is
For larger numbers of data values, writing a sum in terms of each individual frequency is cumbersome. Sigma notation is used to symbolize a sum. In this case, we would write
The Greek letter sigma (∑ ) is used to indicate a sum of some type. The individual terms of the sum are fi where i takes on the values 1 through 5. This notation may seem a little excessive for only five data values, but it comes in handy when you have 10 or more data values.
The relative frequency for the first data before the training program is
where n is the total number of measurements. Doing a similar calculation for the other frequencies before the training program gives us the table below.
The relative frequency is highest for a 3 minute deposit with 35% of the deposits taking this time. Ten percent of the deposits take 4 minutes. These percentages enable us to compare the data to other sets of data that may have a different number of measurements.
Suppose a dataset contains k different values x1, x2, …, xk. Each data value xi occurs fi times in the data set. The relative frequency of the data value xi is
where n is the total number of measurements in the dataset.
Since the sum of the frequencies is the total number of data measurements, the sum of the relative frequencies is 1.
### Example 2 Find the Relative Frequency Table
Make a relative frequency table for the time it takes to make a deposit after the training program.
Solution The data values are the deposit times. To create the relative frequencies, divide each frequency by the sum of the frequencies.
The relative frequency table allows us to see how frequently each data value occurs. For this purpose, the relative frequency may be written as a percentage instead of a decimal. This is done by multiplying the decimal by 100. This means that the totals in each column will be 100 instead of 1.
From the relative frequencies, we see that deposits taking 3 or 4 minutes prior to the training program are reduced after the training program. This results in an increase in deposits taking 2 minutes. In fact, deposits taking 2 minutes or less increased from 55% of deposits (25 + 35) to 82.5% (25 + 57.5) of deposits.
The time it takes to make a deposit is an example of quantitative data. This means that the data value is a number. In this case, the data values were times in minutes. A relative frequency table may also be computed using qualitative data. Qualitative data describes some attribute of an occurrence.
Suppose a customer satisfaction survey is administered to the customers exiting the bank. The survey asks the customer to rate their satisfaction with their visit as very satisfied, satisfied, neutral, unsatisfied, or very unsatisfied. Based on this survey, the following frequencies are calculated.
The sum of the frequencies is higher since the survey also encompasses customers who utilized services other than making a deposit. Even though the data are not numbers, we can still use the frequencies to compute relative frequencies.
### Example 3 Find a Relative Frequency Table
An investment firm administers a survey to its clients to determine their risk tolerance. Based on this survey, the firm puts each of their 500 clients into one of five categories. The table below reflects these results.
Make a relative frequency table for this data.
Solution For each level of risk tolerance, divide the frequency by the total number of clients. This give the following relative frequency table.
A variable that only takes on a finite number of values is called a discrete variable. In each of the previous examples, the number of possible data values for the variable was small. This meant the frequency tables only needed a few rows. If a variable takes on many values, we need additional rows corresponding to each value the variable may take.
Continuous variables may take on any value over an interval. Examples of this type of variable are weight and length. If we were to make a frequency distribution for a continuous variable, the table could potentially have a huge number of rows since every data value might be different.
Discrete variables with many different data values or continuous variables may be displayed in a meaningful way using a grouped frequency distribution. In this type of frequency distribution, each data value is sorted in a category or class. This allows us to calculate the frequency which data values fall into the class.
Let’s look at a concrete example.
The table above corresponds the market capitalization (in millions of dollars) of 86 companies in the energy sector of the New York Stock Exchange on July 7, 2012. The market capitalization of a company is the value of all shares for the company.
To create a grouped frequency distribution, we start by sorting the data from smallest to largest.
This sorted list, called a data array, shows the smallest market capitalization in the upper left. The market capitalization increases as you move down and to the right in the table.
Now we must define the classes to which each of these data should be placed in. The classes must be defined so that each data value falls into only one of the classes. For these companies, the market capitalizations all fall from 250 through 45,180. This means that the values cover a range of values 45,180 – 250 or 44930 wide. In general, we use between 5 and 15 classes. If possible, we also use classes that are of equal width to make it easier to compare relative frequencies.
Let’s try 8 classes for this data. Each class must have a width of about 44930/3 . This is not a very nice number so we round up to the next convenient value like 6000.
The width of a class is
We could start at 250 with classes that are 6000 wide. However, if we start at 0 and use eight classes that are 6000 wide we will include all of the data values.
The classes above do not overlap and insure that each data value will fall into one of the classes. The second number in each class is chosen carefully to insure the classes contain all data values. In this case, each value is rounded to the nearest integer when calculating the frequency. If all of the data had been written to the hundredths place, we would also want to write the classes to the nearest hundredth. Now we’ll scan the table and calculate the frequency for each class.
We have color coded the appropriate numbers in the table to make them easier to count. Strategies like this help to insure that all values are accounted for and put into the correct class.
Once we have the frequencies, we can calculate the relative frequencies.
Due to rounding, the sum of the relative frequencies exceed 1 by a small amount. This table also includes the cumulative and cumulative relative frequency. The cumulative frequency of a class is the number of data values that are less than the upper boundary of the class. For instance, the cumulative frequency for the market capitalization from 12,000 to 17,999 is the number of companies whose market capitalization is 17,999 or less. This frequency is simply the sum of the frequencies for the class from 12,000 to 17,999 and all of the frequencies for the classes below it or 51 +16 + 10 = 77. The cumulative relative frequency is found by dividing the cumulative frequency by the total. For the class from 12,000 to 17,999, this gives us 77/85 ≈ 0.895. This tells us that approximately 89.5% of the companies have a market capitalization of 17,999 million dollars or less.
### Example 4 Grouped Data
The manager at a bank monitors the amount of time a customer spends in line waiting for a teller. They record this value, in seconds, for 25 customers on a Friday afternoon.
a. Use this data to make a table that includes a column for the relative frequency and the cumulative relative frequency. Use 6 classes to group the data.
Solution Start by putting the data in order.
The class width is
We will round up to a width of 7 and sort the data into the classes below. These classes include all times in the data and do not overlap.
In this table the relative frequencies are calculated by dividing the frequencies by 25. The cumulative frequencies are determined by finding the wait times that are below the upper limit in the class.
b. What proportion of the wait times are at least 14 seconds, but less than 21 seconds?
Solution The relative frequency for the class from 14 seconds to under 21 seconds is the proportion of wait times in the class 14 to under 21:
This means 24 percent of the wait times are 14 minutes to under 21 minutes.
c. What proportion of the wait times are less than 21 seconds?
Solution The cumulative relative frequency for the class from 14 to under 21 minutes is the proportion of wait times less than 21 minutes:
This means 80 percent of the wait times are less than 21 minutes. |
# Position of a Term in a Geometric Progression
We will learn how to find the position of a term in a Geometric Progression.
On finding the position of a given term in a given Geometric Progression
We need to use the formula of nth or general term of a Geometric Progression tn = ar$$^{n - 1}$$.
1. Is 6144 a term of the Geometric Progression {3, 6, 12, 24, 48, 96, .............}?
Solution:
The given Geometric Progression is {3, 6, 12, 24, 48, 96, .............}
The first terms of the given Geometric Progression (a) = 3
The common ratio of the given Geometric Progression (r) = $$\frac{6}{3}$$ = 2
Let nth term of the given Geometric Progression is 6144.
Then,
⇒ t$$_{n}$$ = 6144
⇒ a r$$^{n - 1}$$ = 6144
⇒ 3 (2)$$^{n - 1}$$ = 6144
⇒ (2)$$^{n - 1}$$ = 2048
⇒ (2)$$^{n - 1}$$ = 2$$^{11}$$
⇒ n - 1 = 11
⇒ n = 11 + 1
⇒ n = 12
Therefore, 6144 is the 12th term of the given Geometric Progression.
2. Which term of the Geometric Progression 2, 1, ½, ¼, ............. is $$\frac{1}{128}$$?
Solution:
The given Geometric Progression is 2, 1, ½, ¼, .............
The first terms of the given Geometric Progression (a) = 2
The common ratio of the given Geometric Progression (r) = ½
Let nth term of the given Geometric Progression is $$\frac{1}{128}$$.
Then,
t$$_{n}$$ = $$\frac{1}{128}$$
⇒ a r$$^{n - 1}$$ = $$\frac{1}{128}$$
⇒ 2 (½)$$^{n - 1}$$ = $$\frac{1}{128}$$
⇒ (½)$$^{n - 1}$$ = (½)$$^{7}$$
⇒ n - 2 = 7
⇒ n = 7 + 2
⇒ n = 9
Therefore, $$\frac{1}{128}$$ is the 9th term of the given Geometric Progression.
3. Which term of the Geometric Progression 7, 21, 63, 189, 567, ............. is 5103?
Solution:
The given Geometric Progression is 7, 21, 63, 189, 567, .............
The first terms of the given Geometric Progression (a) = 7
The common ratio of the given Geometric Progression (r) = $$\frac{21}{7}$$ = 3
Let nth term of the given Geometric Progression is 5103.
Then,
t$$_{n}$$ = 5103
⇒ a r$$^{n - 1}$$ = 5103
⇒ 7 (3)$$^{n - 1}$$ = 5103
⇒ (3)$$^{n - 1}$$ = 729
⇒ (3)$$^{n - 1}$$ = 3$$^{6}$$
⇒ n - 1 = 6
⇒ n = 6 + 1
⇒ n = 7
Therefore, 5103 is the 7th term of the given Geometric Progression.
Geometric Progression
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# How do you find the limit lim_(x->0^-)|x|/x ?
Aug 2, 2014
When dealing with one-sided limits that involve the absolute value of something, the key is to remember that the absolute value function is really a piece-wise function in disguise. It can be broken down into this:
$| x | =$
$x$, when $x \ge 0$
-$x$, when $x < 0$
You can see that no matter what value of $x$ is chosen, it will always return a non-negative number, which is the main use of the absolute value function. This means that to evaluate this one-sided limit, we must figure out which version of this function is appropriate for our question.
Because our limit is approaching $0$ from the negative side, we must use the version of $| x |$ that is $< 0$, which is $- x$. Rewriting our original problem, we have:
${\lim}_{x \to {0}^{-}} \frac{- x}{x}$
Now that the absolute value is gone, we can divide the $x$ term and now have:
${\lim}_{x \to {0}^{-}} - 1$
One of the properties of limits is that the limit of a constant is always that constant. If you imagine a constant on a graph, it would be a horizontal line stretching infinitely in both directions, since it stays at the same $y$-value regardless of what the $x$-value does. Since limits deal with finding what value a function "approaches" as it reaches a certain point, the limit of a horizontal line will always be a point along that line, no matter what x-value is chosen. Because of this, we now know:
${\lim}_{x \to {0}^{-}} - 1 = - 1$, Giving us our final answer. |
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