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# 2.2 Linear equations in one variable (Page 7/15)
Page 7 / 15
Find the equation of the line parallel to $\text{\hspace{0.17em}}5x=7+y\text{\hspace{0.17em}}$ and passing through the point $\text{\hspace{0.17em}}\left(-1,-2\right).$
$y=5x+3$
## Finding the equation of a line perpendicular to a given line passing through a given point
Find the equation of the line perpendicular to $\text{\hspace{0.17em}}5x-3y+4=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-4,1\right).$
The first step is to write the equation in slope-intercept form.
$\begin{array}{ccc}\hfill 5x-3y+4& =& 0\hfill \\ \hfill -3y& =& -5x-4\hfill \\ \hfill y& =& \frac{5}{3}x+\frac{4}{3}\hfill \end{array}$
We see that the slope is $\text{\hspace{0.17em}}m=\frac{5}{3}.\text{\hspace{0.17em}}$ This means that the slope of the line perpendicular to the given line is the negative reciprocal, or $-\frac{3}{5}.\text{\hspace{0.17em}}$ Next, we use the point-slope formula with this new slope and the given point.
$\begin{array}{ccc}\hfill y-1& =& -\frac{3}{5}\left(x-\left(-4\right)\right)\hfill \\ \hfill y-1& =& -\frac{3}{5}x-\frac{12}{5}\hfill \\ \hfill y& =& -\frac{3}{5}x-\frac{12}{5}+\frac{5}{5}\hfill \\ \hfill y& =& -\frac{3}{5}x-\frac{7}{5}\hfill \end{array}$
Access these online resources for additional instruction and practice with linear equations.
## Key concepts
• We can solve linear equations in one variable in the form $\text{\hspace{0.17em}}ax+b=0\text{\hspace{0.17em}}$ using standard algebraic properties. See [link] and [link] .
• A rational expression is a quotient of two polynomials. We use the LCD to clear the fractions from an equation. See [link] and [link] .
• All solutions to a rational equation should be verified within the original equation to avoid an undefined term, or zero in the denominator. See [link] and [link] .
• Given two points, we can find the slope of a line using the slope formula. See [link] .
• We can identify the slope and y -intercept of an equation in slope-intercept form. See [link] .
• We can find the equation of a line given the slope and a point. See [link] .
• We can also find the equation of a line given two points. Find the slope and use the point-slope formula. See [link] .
• The standard form of a line has no fractions. See [link] .
• Horizontal lines have a slope of zero and are defined as $\text{\hspace{0.17em}}y=c,$ where c is a constant.
• Vertical lines have an undefined slope (zero in the denominator), and are defined as $\text{\hspace{0.17em}}x=c,$ where c is a constant. See [link] .
• Parallel lines have the same slope and different y- intercepts. See [link] .
• Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical. See [link] .
## Verbal
What does it mean when we say that two lines are parallel?
It means they have the same slope.
What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?
How do we recognize when an equation, for example $\text{\hspace{0.17em}}y=4x+3,$ will be a straight line (linear) when graphed?
The exponent of the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ variable is 1. It is called a first-degree equation.
What does it mean when we say that a linear equation is inconsistent?
When solving the following equation:
$\frac{2}{x-5}=\frac{4}{x+1}$
explain why we must exclude $\text{\hspace{0.17em}}x=5\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=-1\text{\hspace{0.17em}}$ as possible solutions from the solution set.
If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).
## Algebraic
For the following exercises, solve the equation for $\text{\hspace{0.17em}}x.$
$7x+2=3x-9$
$4x-3=5$
$x=2$
$3\left(x+2\right)-12=5\left(x+1\right)$
$12-5\left(x+3\right)=2x-5$
$x=\frac{2}{7}$
$\frac{1}{2}-\frac{1}{3}x=\frac{4}{3}$
$\frac{x}{3}-\frac{3}{4}=\frac{2x+3}{12}$
$x=6$
$\frac{2}{3}x+\frac{1}{2}=\frac{31}{6}$
$3\left(2x-1\right)+x=5x+3$
$x=3$
$\frac{2x}{3}-\frac{3}{4}=\frac{x}{6}+\frac{21}{4}$
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
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5 |
# 21/30 as a Percentage – Find Out Now!
Converting fractions to percentages is a fundamental math skill that is useful in various real-life situations. If you are wondering how to express 21/30 as a percentage, you’ve come to the right place. In this article, we will explore different methods to convert this fraction into a percentage, allowing you to easily understand and apply this conversion in your own calculations.
Whether you need to convert fractions to percentages for school assignments, financial calculations, or any other scenario, having a solid understanding of the process is essential. By the end of this article, you will be equipped with the knowledge and tools to convert 21/30 and any other fraction to a percentage accurately.
### Key Takeaways:
• To convert a fraction to a percentage, you can use methods such as decimal conversion, proportions, or adjusting the denominator.
• Converting 21/30 to a decimal involves dividing the numerator by the denominator and then multiplying the result by 100. In this case, the answer is 70%.
• Using proportions, you set up an equation with the fraction as one ratio and the unknown percentage as the other. Cross-multiplying and solving the equation will give you the percentage, which is also 70% in this case.
• The method of adjusting the denominator involves multiplying both the numerator and denominator by a multiplier to make the denominator 100. The result, 70/100, is equivalent to 70%.
• Practice regularly and utilize online tools for fraction to percentage conversion to improve your skills and ensure accurate results.
## Understanding Fractions and Percentages
In order to effectively convert fractions to percentages, it’s crucial to have a solid grasp of the basic concepts of fractions and percentages.
A fraction is a mathematical expression that represents a part of a whole. It consists of a numerator, which is the number above the fraction line, and a denominator, which is the number below the fraction line. For example, in the fraction 3/4, the numerator is 3 and the denominator is 4. Fractions are used to represent parts of a whole, such as dividing a pizza into equal slices where the numerator indicates the number of slices and the denominator represents the total number of slices.
On the other hand, percentages are a way of expressing a part of a whole in relation to 100. Percentages are often used to represent ratios, comparisons, or proportions. For example, if you have 75% of a pizza, it means you have 75 out of 100 equal parts of the whole pizza.
To better understand the connection between fractions and percentages, consider the fraction 1/2. If we convert this fraction to a percentage, we would express it as 50%. This is done by recognizing that the fraction 1/2 is equivalent to 50 out of 100, or 50 parts out of a whole.
So, to sum it up:
1. Fractions represent parts of a whole, with a numerator and a denominator.
2. Percentages express a part of a whole in relation to 100.
By understanding these fundamental concepts, you’ll be better equipped to tackle the conversion of fractions to percentages.
## Converting 21/30 to a Decimal
One method of converting a fraction to a percentage is by first converting it to a decimal. Let’s take the example of 21/30. To convert this fraction to a decimal, divide the numerator (21) by the denominator (30). The result is 0.7.
To convert the decimal to a percentage, multiply it by 100. In this case, 0.7 multiplied by 100 equals 70%. Therefore, 21/30 as a percentage is 70%.
Converting fractions to decimals is a useful step in the process of converting them to percentages. By understanding this conversion, you can easily translate fractions into their equivalent decimal and percentage forms.
Converting a fraction to a decimal involves dividing the numerator by the denominator. This method simplifies the process and provides an accurate representation of the fraction in decimal form.
## Converting 21/30 to a Percentage Using Proportions
Another method to convert 21/30 to a percentage involves setting up a proportion. The cross-multiplication method, also known as the proportion method or the multiplication method, provides a straightforward approach to finding the unknown percentage of a given fraction. Let’s see how it works.
To convert 21/30 to a percentage, we can set up the proportion: 21/30 = x/100, where x represents the unknown percentage. By cross-multiplying, we obtain 21 * 100 = 30 * x. Simplifying the equation, we have 2100 = 30x.
Next, we divide both sides of the equation by 30 to isolate x. The calculation gives us x = 2100/30 = 70%. Therefore, 21/30 as a percentage is also 70%.
Here’s the calculation:
EquationSimplification
21/30 = x/10021 * 100 = 30 * x
2100 = 30x
x = 2100/30
x = 70%
By applying the proportion method, we determined that 21/30 is equivalent to 70% as a percentage. This method offers an alternative approach to converting fractions into percentages, providing a useful tool for various mathematical calculations and applications.
## Using the Method of Adjusting the Denominator
An alternative approach to converting 21/30 to a percentage is the adjusting denominator method. This technique involves adjusting the denominator of the fraction to make it 100, which simplifies the calculation process. Here’s how you can apply this method:
### Step 1: Determine the Multiplier
To begin, divide 100 by the denominator of the fraction (30) to find the multiplier. In this case, the multiplier is approximately 3.333.
### Step 2: Multiply the Numerator and Denominator
Next, multiply both the numerator (21) and denominator (30) by the multiplier obtained in Step 1. The result is a new fraction of 70/100.
### Step 3: Simplify to Percentage Form
To express the fraction as a percentage, divide 70 by 100 and multiply the result by 100. This yields a final value of 70%. Therefore, 21/30 as a percentage is 70%.
NumeratorDenominatorMultiplierNew NumeratorNew DenominatorPercentage
21303.3337010070%
## Why Use the Decimal Method?
The decimal method is a popular choice for converting fractions to percentages due to its simplicity and efficiency. This method involves dividing the numerator by the denominator and then multiplying the result by 100. By following these two steps, you can quickly and easily convert any fraction to a percentage, making it an ideal method for students and individuals seeking a straightforward approach.
Unlike other conversion methods that may require complex calculations or additional steps, the decimal method offers a streamlined process. By dividing the numerator by the denominator, you obtain the decimal equivalent of the fraction. Multiplying this decimal by 100 then gives you the corresponding percentage. This uncomplicated approach allows for easy comprehension and application, making it a favored method for many learners.
### The Advantages of Using the Decimal Method:
• Simplicity: The decimal method involves only two steps, making it easy to understand and apply.
• Efficiency: With the decimal method, you can convert fractions to percentages quickly, saving time and effort.
• Universal Applicability: The decimal method can be used for any fraction, regardless of its complexity.
• Clear Visualization: The decimal method provides a clear visual representation of the fraction’s percentage equivalent through the decimal calculation.
By utilizing the decimal method, individuals can convert fractions to percentages with ease and confidence. Whether you’re a student learning the fundamentals of mathematics or an individual seeking to apply this knowledge in everyday life, the decimal method offers a simple yet effective solution. Embrace the advantages of this method and enjoy the convenience it brings to fraction-to-percentage conversions.
## Practice Makes Perfect
Converting fractions to percentages is a skill that can be honed through practice. By regularly practicing fraction to percentage conversions, you can improve your understanding and confidence in this process. To get started, all you need is a pen, a pad, and a calculator. Are you ready to dive in? Let’s begin!
Step 1: Pick a fraction and write it down. It can be any fraction you choose.
Step 2: Use one of the methods discussed earlier – the decimal method, the proportion method, or the method of adjusting the denominator – to convert the fraction to a percentage.
Step 3: Check your work using a calculator or online fraction to percentage converter if needed. Compare your calculated percentage with the result from the tool to ensure accuracy.
Repeat these steps with different fractions, challenging yourself with more complex examples as your skills improve. The more you practice, the more comfortable and proficient you will become in converting fractions to percentages.
### Take Your Practice a Step Further
Here are some additional ways to practice fraction to percentage conversions:
1. Create a list of fractions and convert them to percentages. Challenge yourself by including fractions with different denominators, requiring the use of various conversion methods.
2. Ask a friend or family member to give you a random fraction, and try to convert it to a percentage on the spot. This will help you sharpen your mental math skills.
3. Download a fraction to percentage conversion worksheet or find online math practice resources. These tools often include a wide range of fraction conversion problems for you to solve.
Remember, practice makes perfect. As you dedicate time and effort to practicing fraction to percentage conversions, you’ll gradually build fluency and accuracy in this essential mathematical skill.
Keep at it, and soon you’ll be converting fractions to percentages with ease!
## Online Tools for Fraction to Percentage Conversion
If you need assistance or want to check your work, there are online tools available that can help you convert fractions to percentages. These tools provide an easy and reliable way to obtain accurate results in fraction to percentage conversions.
Converting fractions to percentages can sometimes be tricky, especially when dealing with complex fractions or unfamiliar calculation methods. That’s where online tools come in handy. With just a few clicks, you can quickly and accurately convert any fraction to its equivalent percentage.
These online fraction to percentage converters and calculators are designed to make the conversion process effortless. Simply enter the numerator and denominator of the fraction, and the tool will automatically generate the corresponding percentage. Some tools even provide step-by-step explanations, helping you understand the conversion process.
Whether you’re a student studying fractions and percentages or someone who needs to convert fractions in real-life situations, these online tools are a valuable resource. They eliminate the need for manual calculations and reduce the chances of errors, ensuring accurate results every time.
Additionally, online converters and calculators offer the convenience of accessibility from any device with an internet connection. You can access these tools anytime and anywhere, making them perfect for both classroom and practical applications.
So, the next time you find yourself in need of converting a fraction to a percentage, save time and effort by using online tools. They make the process quick, easy, and reliable, giving you accurate results in just a few simple steps.
## Conclusion
Converting fractions to percentages can be achieved using various methods, including the decimal method, proportion method, and adjusting the denominator method.
Each method has its advantages, and the choice depends on personal preference and the complexity of the fraction. The decimal method offers simplicity and efficiency, requiring only a division and multiplication. The proportion method provides a straightforward approach, utilizing cross-multiplication. The adjusting the denominator method allows for a direct conversion by multiplying both the numerator and denominator. Whichever method you choose, the goal remains the same: to express a fraction as a percentage.
With practice and the assistance of online tools, anyone can become proficient in converting fractions to percentages. Regular practice and exposure to different fraction-to-percentage conversions will enhance your understanding and speed up the process. Additionally, online tools such as calculators.org offer a convenient way to verify your calculations or tackle more challenging conversions.
So, the next time you encounter a fraction that needs to be expressed as a percentage, remember the different methods and techniques we discussed in this article. Take your time, choose the method that suits you best, and practice regularly, and soon you’ll be converting fractions to percentages with ease and confidence.
### Keep Exploring and Learning
• Check out our earlier sections to learn more about the basics of fractions and percentages, as well as different conversion methods.
## References
Here are some references that can further assist you in understanding and mastering fraction to percentage conversion:
1. VisualFractions.com: “What is 21/30 as a percentage?”
2. Fraction to Percentage Calculation: (source not mentioned)
3. Aspose Grade Calculator: (source not mentioned)
These references provide additional resources and tools that can aid in your fraction to percentage conversion journey. Whether you prefer interactive calculators or in-depth tutorials, these references offer valuable insights and guidance.
Remember, practice and exploration are key to mastering this skill. Don’t hesitate to leverage these references as you continue to expand your knowledge and improve your understanding of fraction to percentage conversion.
### Keep learning and stay curious!
For more information on fraction to percentage conversion and related concepts, explore the following resources:
1. Fraction Conversion Tutorial: This comprehensive tutorial provides step-by-step instructions on converting fractions to percentages. It covers various methods, including the decimal method, proportion method, and adjusting the denominator method. [fraction to percentage conversion resources]
2. Fraction to Percentage Conversion Chart: This handy chart offers a quick reference guide for converting fractions to percentages. It includes common fractions and their corresponding percentage values, making it easier to perform conversions without calculations. [additional information]
By referring to these resources, you can gain a deeper understanding of fraction to percentage conversion and master the necessary techniques to solve related problems.
This article was written by [Author Name], a professional copywriting journalist with expertise in mathematics. With years of experience in the field, [Author Name] has developed a deep understanding of fraction to percentage conversion and its practical applications in various contexts.
[Author Name] is passionate about simplifying complex mathematical concepts and making them easily understandable and accessible to readers of all levels. The goal is to break down the conversion process into simple steps and provide clear explanations, empowering individuals to master this essential skill.
Whether you’re a student, a teacher, or simply someone interested in improving your math abilities, [Author Name]’s expertise in fraction to percentage conversion will guide you towards success. Trust in [Author Name]’s comprehensive knowledge and enjoy learning in a way that is engaging, informative, and enjoyable.
## FAQ
### What is 21/30 as a percentage?
21/30 as a percentage is 70%.
### How do I convert a fraction to a percentage?
To convert a fraction to a percentage, you can use different methods such as the decimal method, the proportion method, or adjusting the denominator method.
### What is the decimal method for converting 21/30 to a percentage?
Divide the numerator (21) by the denominator (30) to get the decimal form of the fraction. Multiply the decimal by 100 to obtain the percentage. In the case of 21/30, the decimal is 0.7, which is equal to 70%.
### How does the proportion method work for converting 21/30 to a percentage?
Set up a proportion by equating the fraction (21/30) to x/100, where x represents the unknown percentage. Cross-multiply to solve for x, and then simplify the equation to find the percentage. In the case of 21/30, the percentage is 70%.
### Can I use the method of adjusting the denominator to convert 21/30 to a percentage?
Yes, you can adjust the denominator to make it 100. Divide 100 by the denominator (30) to get a multiplier of approximately 3.333. Multiply both the numerator (21) and denominator by this multiplier to obtain the equivalent fraction 70/100, which is equal to 70%.
### Why is the decimal method preferred for converting fractions to percentages?
The decimal method is often preferred due to its simplicity and fewer steps involved. It only requires dividing the numerator by the denominator, followed by multiplying the result by 100. This method provides a quick and straightforward way to convert fractions to percentages, making it easier for students to understand and apply.
### How can I practice converting fractions to percentages?
Grab a pen, a pad, and a calculator, and try converting other fractions to percentages. The more you practice, the more comfortable you will become with this skill.
### Are there online tools available to help with fraction to percentage conversion?
Yes, there are online tools available that can help you convert fractions to percentages. These tools provide an easy and reliable way to obtain accurate results in fraction to percentage conversions.
### Where can I find more information on fraction to percentage conversion?
For more information on fraction to percentage conversion and related concepts, you can refer to online tutorials, conversion charts, and other educational resources. |
# How do you find the perimeter and area of a composite shape?
## How do you find the perimeter and area of a composite shape?
Add up the length of all the sides to get the perimeter. Calculate the area of the composite figure by adding up the areas of the smaller shapes that make up the composite figure . The figure is made up of rectangles. To find the length of side A, look at the two rectangles that make up the figure.
## How do you find the perimeter of a composite shape?
To find the perimeter, we simply add up the lengths of each outside edge. It may be helpful to look out for number bonds when adding the sides. For example, 7 + 3 = 10. The total of all of the outer sides is 36, so the perimeter is 36 cm.
What is the area of composite shapes?
To calculate the area of a composite shape you must divide the shape into rectangles, triangles or other shapes you can find the area of and then add the areas back together. You may have to calculate missing lengths before finding the area of some of the shapes.
What is the formula for composite shapes?
Using the formula for the area of the composite shape, Area of composite shape = Area of rectangle + area of the square. ⇒ Area of composite shape = 14+9 = 23 square inches. Therefore, the area of the given composite shape is 23 square inches.
### How do you calculate the area of composite shapes?
To find the area of composite shapes you just need to break down the shape into individual pieces and calculate their separate pieces then add them back together again to find a total area. Another way of finding the area of composite shapes is to use a subtraction model.
### How do you find the perimeter of a composite figure?
The perimeter of the composite figure is the distance around the figure. It is found by adding each of the side lengths around the figure.
What are different shapes have the same perimeter?
mfb said: There are examples, but they are a bit harder to find. A triangle and a suitable trapezoid are the easiest examples. Thinking about it, two suitable trapezoids are a much easier example of two different shapes with the same area and perimeter. Jun 9, 2020.
How do you calculate the area of a compound shape?
To find the area of a compound shape, follow these simple steps: Step 1: Work out the missing lengths around the edge of the compound shape. Step 2: Divide your L shape into two rectangles. Step 3: Work out the area of each rectangle. Step 4: Add the areas of the rectangles together to give the total area of the L shape. |
# Projectile Motion Example Problem – Physics Homework Help 2
Throwing or shooting a projectile follows a parabolic course. If you know the initial velocity and angle of elevation of the projectile, you can find its time aloft, maximum height or range. You can also its altitude and distance travelled if given a time. This example problem shows how to do all of these.
Projectile Motion Example Problem:
A cannon is fired with muzzle velocity of 150 m/s at an angle of elevation = 45°. Gravity = 9.8 m/s2.
a) What is the maximum height the projectile reaches?
b) What is the total time aloft?
c) How far away did the projectile land? (Range)
d) Where is the projectile at 10 seconds after firing?
Let’s set up what we know. First, let’s define our variables.
V0 = initial velocity = muzzle velocity = 150 m/s
vx = horizontal velocity component
vy = vertical velocity component
θ = angle of elevation = 45°
h = maximum height
R = range
x = horizontal position at t=10 s
y = vertical position at t=10 s
m = mass of projectile
g = acceleration due to gravity = 9.8 m/s2
Part a) Find h.
The formulas we will be using are:
d = v0t + ½at2
and
vf – v0 = at
In order to find the distance h, we need to know two things: the velocity at h and the amount of time it takes to get there. The first is easy. The vertical component of the velocity is equal to zero at point h. This is the point where the upward motion is stopped and the projectile begins to fall back to Earth.
The initial vertical velocity is
v0y = v0·sinθ
v0y = 150 m/s · sin(45°)
v0y = 106.1 m/s
Now we know the beginning and final velocity. The next thing we need is the acceleration.
The only force acting on the projectile is the force of gravity. Gravity has a magnitude of g and a direction in the negative y direction.
F = ma = -mg
solve for a
a = -g
Now we have enough information to find the time. We know the initial vertical velocity (V0y) and the final vertical velocity at h (vhy = 0)
vhy – v0y = at
0 – v0y = -9.8 m/s2·t
0 – 106.1 m/s = -9.8 m/s2·t
Solve for t
t = 10.8 s
Now solve the first equation for h
h = v0yt + ½at2
h = (106.1 m/s)(10.8 s) + ½(-9.8 m/s2)(10.8 s)2
h = 1145.9 m – 571.5 m
h = 574.4 m
The highest height the projectile reaches is 574.4 meters.
Part b: Find total time aloft.
We’ve already done most of the work to get this part of the question if you stop to think. The projectile’s trip can be broken into two parts: going up and coming down.
ttotal = tup + tdown
The same acceleration force acts on the projectile in both directions. The time down takes the same amount of time it took to go up.
tup = tdown
or
ttotal = 2 tup
we found tup in Part a of the problem: 10.8 seconds
ttotal = 2 (10.8 s)
ttotal = 21.6 s
The total time aloft for the projectile is 21.6 seconds.
Part c: Find range R
To find the range, we need to know the initial velocity in the x direction.
v0x = v0cosθ
v0x = 150 m/s·cos(45)
v0x = 106.1 m/s
To find the range R, use the equation:
R = v0xt + ½at2
There is no force acting along the x-axis. This means the acceleration in the x-direction is zero. The equation of motion is reduced to:
R = v0xt + ½(0)t2
R = v0xt
The range is the point where the projectile strikes the ground which happens at the time we found in Part b of the problem.
R = 106.1 m/s · 21.6s
R = 2291.8 m
The projectile landed 2291.8 meters from the canon.
Part d: Find the position at t = 10 seconds.
The position has two components: horizontal and vertical position. The horizontal position, x, is far downrange the projectile is after firing and the vertical component is the current altitude, y, of the projectile.
To find these positions, we will use the same equation:
d = v0t + ½at2
First, let’s do the horizontal position. There is no acceleration in the horizontal direction so the second half of the equation is zero, just like in Part c.
x = v0xt
We are given t = 10 seconds. V0x was calculated in Part c of the problem.
x = 106.1 m/s · 10 s
x = 1061 m
Now do the same thing for the vertical position.
y = v0yt + ½at2
We saw in Part b that v0y = 109.6 m/s and a = -g = -9.8 m/s2. At t = 10 s:
y = 106.1 m/s · 10 s + ½(-9.8 m/s2)(10 s)2
y = 1061 – 490 m
y = 571 m
At t=10 seconds, the projectile is at (1061 m, 571 m) or 1061 m downrange and at an altitude of 571 meters.
If you need to know the velocity of the projectile at a specific time, you can use the formula
v – v0 = at
and solve for v. Just remember velocity is a vector and will have both x and y components.
This specific example can be easily adapted for any initial velocity and any angle of elevation. If the cannon is fired on another planet with a different force of gravity, just change the value of g accordingly.
## 2 thoughts on “Projectile Motion Example Problem – Physics Homework Help”
• Todd Helmenstine Post author
You are the one who is correct. I’m not really sure where I punched in my values, but sin(45) = cos(45) = 0.707. Use that with the velocity and get 106.1 m/s…NOT 109.6.
I fixed the math in all the steps. It was kind of strange, but the velocity I had for the time aloft was the correct value, not the 109.6 I used everywhere else. Goes to show you should always check your work.
Thanks for pointing out my error! 🙂 |
# ISEE Middle Level Math : How to add fractions
## Example Questions
### Example Question #246 : Fractions
Explanation:
Add the numerators and leave the denominators alone:
### Example Question #247 : Fractions
Subtract:
Explanation:
Subtract the numerators and leave the denominators alone:
### Example Question #248 : Fractions
Explanation:
Because the two fractions have the same denominator, simply add the numerators and leave the denominator as is:
### Example Question #249 : Fractions
Which of the following statements demonstrates the commutative property of addition?
None of the examples in the other responses demonstrates the commutative property of addition.
Explanation:
The commutative property of addition states that two numbers can be added in either order to obtain the same sum. Of the given responses, only
demonstrates this property, so it is the correct choice.
### Example Question #250 : Fractions
Explanation:
Add the numerators and leave the denominator the same:
### Example Question #151 : Fractions
Explanation:
Add the numerators and leave the denominator the same:
### Example Question #491 : Numbers And Operations
The time is now 10:38 PM. What time will it be in two hours and fifty-two minutes?
Explanation:
It is now 10:38 PM. In minutes, it will be 11:00 PM; in minutes, the minutes will read 30, so the time will be 11:30 PM. In another two hours, the hours will read
,
so the time will be 1:30 AM.
### Example Question #492 : Numbers And Operations
Solve the expression.
None of these
Explanation:
Start by solving for the term in parenthesis. We will need to find the least common denominator in order to add the fractions. Since 6 can be multiplied by 2 to get 12, 12 must be the least common denominator.
Convert the fractions so that they share a denominator of 12.
Multiply.
You can then simplify the fraction be removing 12 from both the numerator and denominator.
### Example Question #493 : Numbers And Operations
Which of the following is the sum of five-sevenths and one-half?
The correct answer is not among the other responses.
Explanation:
Since , express each fraction as its equivalent in fourteenths, and add the numerators, as follows: |
## A Practical Demonstration of the Pythagorean Theorem
The Pythagorean Theorem is probably the most popular theorem in school mathematics. Surely, you have heard or read about it at least once from elementary school to high school. The Pythagorean Theorem states that given a right triangle with shorter sides $a$, $b$, and hypotenuse $c$, the following equation holds
$c^2 = a^2 + b^2$.
## Using Similarity to Prove the Pythagorean Theorem
The Pythagorean Theorem is one of the most interesting theorems for two reasons: First, it’s very elementary; even high school students know it by heart. Second, it has hundreds of proofs. The proof below uses triangle similarity.
Pythagorean Theorem
In a right triangle with side lengths $a$ and $b$ and hypotenuse $c$, the following equation always holds:
$c^2 = a^2 + b^2$. » Read more
## A US President’s Proof of the Pythagorean Theorem
James Garfield, the 20th president of the United States, came up with an original proof of the Pythagorean Theorem in 1876 when he was still a Congressman. His proof was published in New England Journal of Education.
Recall that the Pythagorean Theorem states that given a right triangle with sides $a$, $b$, and hypotenuse $c$, the following equation is always satisified:
$a^2 + b^2 = c^2$.
President Garfield’s proof is quite simple. We can do this in three steps:
1. Find the area of figure above using the trapezoid
2. Find the area of the same figure using the three triangles
3. Equate the results in 1 and 2. » Read more
1 2 |
# 2nd PUC Statistics Question Bank Chapter 6 Statistical Inference
Students can Download Statistics Chapter 6 Statistical Inference Questions and Answers, Notes Pdf, 2nd PUC Statistics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
## Karnataka 2nd PUC Statistics Question Bank Chapter 6 Statistical Inference
Section – A
### 2nd PUC Statistics Statistical Inference One Mark Questions and Answers
Question 1.
What is simple random sample?
A simple random sample is a sample that is chosen in such a way where every unit has a possibility to be selected.
Question 2.
What is parameter?
A parameter is a statistical constant of the population or It is population constant.
Question 3.
What is statistic?
A statistic is the function of a sample value or it is sample constant.
Question 4.
What is parameter space?
The parameter space is the set of all admissible values of the parameter.
Question 5.
What is sample space?
Sample space is the set of the samples that can be drawn from the population.
Question 6.
What is sampling distribution of a statistic?
Sampling distribution of a statistic refers to the different values of the sample size.
Question 7.
What is standard error?
Standard error is the standard deviation of the sampling distribution of a statistic.
Question 8.
Write the formula of S.E (x̄)
S.E(x̄) = $$\frac{\sigma}{\sqrt{n}}$$
Question 9.
Given σ2 = 9 cm2 and n = 36 calculate standard error of sample mean.
σ2 = 9, σ = √9 = 3 n = 36
S.E.(x̄) = $$\frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{36}}=\frac{3}{6}$$ = 0.5
Question 10.
Write the formula of S.E.(x̄1 – x̄2)
S.E.(x̄1 – x̄2) = $$\sqrt{\frac{\sigma_{1}^{2}}{\mathrm{n}_{1}}+\frac{\sigma_{2}^{2}}{\mathrm{n}_{2}}}$$
Question 11.
Sizes of two samples are 50 and 100 population standard deviations are 20 and 10. Compute S.E.(x̄1 – x̄2)
n1 = 50, n2 = 100, σ1 = 20 and σ2 = 10
S.E.(x̄1 – x̄2) = $$\sqrt{\frac{\sigma_{1}^{2}}{\mathrm{n}_{1}}+\frac{\sigma^{2}}{\mathrm{n}_{2}}}$$ = $$\sqrt{\frac{20^{2}}{50}+\frac{10^{2}}{100}}=\sqrt{8+1}=\sqrt{9}=3$$
Question 12.
Write the formula of S.E.(P)
S.E.(P) = $$\sqrt{\frac{P Q}{n}}$$
Question 13.
If P = 0.02 and n = 64 then find S.E (P)
P = 0.02 n = 64 Q = 1 – P ⇒ 1 – 0.02 ⇒ 0.98
S.E = $$\sqrt{\frac{P Q}{n}}$$= $$\sqrt{\frac{(0.02)(0.98)}{64}}=\sqrt{\frac{0.0196}{64}}=0.0175$$
Question 14.
If P = 0.5 and n = 100 then find S.E(P)
P = 0.5; n = 100 Q = 1 – P ⇒ 1 – 0.5 ⇒ 0.5
S.E(P) = $$\sqrt{\frac{P Q}{n}}$$ = $$\sqrt{\frac{(0.5)(0.5)}{100}}=\sqrt{\frac{0.25}{100}}=0.05$$
Question 15.
Write the formula of S.E(P1 – P2) when P1 ≠ P2
S.E. (P1 – P2) when P1 ≠ P2 = $$\sqrt{\frac{\mathrm{P}_{1} \mathrm{Q}_{1}}{\mathrm{n}_{1}}+\frac{\mathrm{P}_{2} \mathrm{Q}_{2}}{\mathrm{n}_{2}}}$$
Question 16.
A lot contains 2% defective items 40 items are chosen from it. Another lot contains 1% defective items. 60 items are chosen from it. Find E(P1 – P2) and S.E (P1 – P2)
P1 = 2% P2 = l% n,1 = 40 n2 = 60
Q1 = 1 – P1 = 1 – 0.02 = 0.98
Q2 = 1 – P2 = 1 – 0.01 = 0.99
S.E(P1 – P2) = $$\sqrt{\frac{P_{1} Q_{1}}{n_{1}}+\frac{P_{2} Q}{n_{2}}}$$ ⇒ $$\sqrt{\frac{(0.02)(0.98)}{40}+\frac{(0.01)(0.99)}{60}}$$
$$=\sqrt{0.00049+0.000165}=\sqrt{0.000655}=0.0256$$
⇒ E(P1 – P2) (0.02 – 0.01) = 0.01
Question 17.
Write the formula of S.E.(P1 – P2) when P1 = P2 = P
S.E.(P1 – P2) when P1 = P2= P is $$\sqrt{P Q\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)}$$
Question 18.
Write the utility of standard error.
• Determine the efficiency and consistency of the statistic as an estimator.
• Obtain the confidence intervals of an estimate
• Standardise the distribution of test statistic in testing of hypothesis.
Question 19.
What is statistical inference?
Statistical inference is the theory of making decisions about the population parameters by utilising sampling and concept of probability.
Question 20.
Mention two branches of statistical inference.
1. Estimation
2. Testing of hypothesis.
Question 21.
What is meant by estimation?
Estimation is the method of obtaining the most likely value of the population parameter using statistic.
Question 22.
What is an estimator?
Any statistic which is used to estimate an unknown parameter is called as estimator.
Question 23.
What is an estimate?
Estimate is the numerical value of the unknown parameter.
Question 24.
What is point estimation?
If a single value is proposed to estimate an unknown parameter, then it is point estimation.
Question 25.
What is interval estimation?
If an interval is proposed to estimate an unknown parameter, then it is interval estimation.
Question 26.
What is confidence interval?
An interval (T1,T2) which contains an unknown parameter is called as confidence interval.
Question 27.
What are confidence limits?
The boundary values of confidence interval are confidence limits.
In this activity, students will discover the relationship between the degree and leading coefficient calculator of a polynomial.
Question 28.
What is confidence coefficient?
The probability that a confidence interval includes an unknown parameter is named as confidence coefficient.
Question 29.
What is statistical hypothesis?
A statistical hypothesis is a statement about the parameters of the population.
Ex.: H : μ = μ0
Question 30.
What is null hypothesis? Give an example?
The hypothesis is being tested for possible rejection is called as Null hypothesis.
Ex.: H0 : μ = μ0
Question 31.
What is alternative hypothesis? Give an example.
The hypothesis which is accepted when null hypothesis is rejected is called Alternative hypothesis.
E.g., If H0 : μ = μ0, then the alternative hypothesis could be H1 = population mean differs from a specified mean μ ≠ μ0 (two tailed).
Question 32.
What is type I error?
Type I error is the error which occurs by rejecting null hypothesis when it is actually true
Question 33.
What is type II error?
Type II error is the error which occurs by accepting null hypothesis when it is actually not true
Question 34.
What is size of the test?
The probability of rejecting H0, when it is true is called as the size of the test.
Question 35.
What is level of significance?
Maximum size of the test is called level of significance.
Question 36.
What is power of a test?
The probability of rejecting H0, when it is not true is called as power of a test. It is denoted by (1 – β).
Question 37.
What is critical region?
Critical region is the set of those values of the test statistic, which leads to the rejection of the null hypothesis.
Question 38.
What is critical value?
Critical value is the value which separates the critical region and acceptance region.
Question 39.
What is two tailed test?
It is a test of statistical hypothesis, where rejection region is located at both the tails of the probability curve.
Question 40.
What is one tailed test?
It is a test of statistical hypothesis, where the rejection region will be marked at only one tail of the probability curve. The rejection region will be either left of right tail of the curve depending upon the alternative hypothesis.
In testing H0 : μ = μ0 against H1 : μ < μ0 (left tailed or lower tailed) and H1: μ > μ0 (right tailed or upper tailed) are examples of one tailed tests.
Question 41.
What is null distribution
The statistical distribution of the test statistic under the null hypothesis is named null distribution.
Question 42.
What is test statistic?
A test statistic is based on the distribution where the test of hypothesis is conducted
### 2nd PUC Statistics Large Sample Test Exercise Problems
Question 1.
Write the testing procedure of large sample tests?
The test procedure contains following steps
• Setting up of the null hypothesis (H0)
• Setting up of the alternative hypothesis (H1)
• Identification of the test statistic and its null distribution and computation of test statistic.
• Identification of the critical region.
• Drawing a random sample and actually conducting the test
• Making the decision.
Question 2.
Write the testing procedure of population mean?
Test for population mean has following steps
1. H0: The population mean is μ0, μ = μ0
2. H1: The population mean two tailed test, μ ≠ μ0 (two tailed test)
(or) μ < μ0 (Left tailed test)
(or) μ > μ0 (right tailed test)
3. Calculation of test statistic Z = $$\frac{\bar{x}-\mu_{0}}{\frac{\sigma}{\sqrt{n}}}$$
Here x̄ = sample mean 7n
σ = population standard deviation
If it is unknown, then it should be replaced by sample standard deviation ‘s’ and ‘n’ is sample size.
Under H0 Z ~ N(0,1)
4. Depending on H1 and α the critical value ‘k’ is selected
5. If the calculated value of the test statistic (Zcal) is in acceptance region, then H0 is accepted, otherwise its rejected.
6. Prepare conclusion and make a decision.
Question 3.
A sample of 100 students is taken from a college. The mean and SD of their weights are 51 kg and 5 kg respectively, test at 1% level of significance that the average weight of college students is 50 kg?
n = 100, x̄ = 51, s = 5, μ0= 50, α = 1% .
H0: let Average weight of college students is 50 i.e. μ = 50
H1 : let Average weight of college student is not equal to 50 i.e. μ ≠ 50 .
The test is two tailed test
Level of significance is α = 0.01(1%) the critical values are -2.58 and +2.58 since Zcal value lies between the critical values (-2.58 and +2.58),
-2.58 < Z cal < + 2.58
∴ H0 is accepted
Conclusion: We can assume that average weight of a student is 50.
Question 4.
A machine is designed to fill 500 ml of milk to polythene bags. A randomly selected 100 milk bags filled by this machine are inspected. The mean milk is found to be 498 ml and SD is 10 ml. Is machine is functioning properly a 5% level of significance?
n = 100, x̄ = 498, s = 10, α = 5%, μ = 500
H0 : let machine works properly i.e., it fills 500 ml of milk to polythene bags i.e. μ = 500
H11 : let machine does not function properly i.e., it doesn’t fill 500 ml of milk to polythene bags i.e. μ ≠ 500
The test is two tailed test
Level of significance is α = 0.05 (i.e., 5%)
The critical values are -1.96 and 1.96
Since Zcal value lies in the rejection region
-1.96 < Zcal >+1.96
H0 is rejected
Conclusion: We can assume that Machine does not function properly
Question 5.
A random sample of 64 children is taken from a school. The average weight of the children is 29 kg SD is 5 kg. Can we assume that the average weight of the school children is less than 30 kg? (use α = 0.05)
n = 64, x̄ = 29 kg, s = 5, μ = 30, α = 0.05
H0 : let average weight of the school children is 30 i.e. μ = 30
H1 : let average weight of the school children is not equal to 30 i.e. μ < 30 (left tailed)
The test is one tailed
Level of significance α = 5%
Here
The critical value is -1.65
Zcal = -1.6
Zcal > critical value
H0 is accepted
Conclusion: We can assume that the Average weight of the school children is 30 kg.
Question 6.
A company manufactures car tyres their average life is 40,000 kms and SD 5000 kms. A change in the production process is believed to result in a better product. A test sample of 100 new tires has mean life of 41,000 kms, can you conclude at 5% LOS that the new product gives better result.
n = 100; x̄ = 41,000 ; σ = 5000 μ = 40,000; α = 5%
H0 : let average life of car tires is 40,000 kms i.e. μ = 40,000 kms .
H1 : let average life of car tires is not equal to 40,000 kms i.e. μ ≠ 40,000 kms
It is two tailed test
Level of significance is α = 0.05 (5%)
The critical value are -1.96 and + 1.96
Z cal value does not lie between the critical values.
-1.96 < Zcal > +1.96
H0 is rejected
Conclusion: We can assume that the Average life of car tires is not equal to 40,000 kms
Question 7.
A specified brand of automobiles tire is known to average life of 10,000 km with a SD of 500 km. A random sample of 36 tires of this brand, when tested resulted in the average life of 9800 km. Regarding quality what is your conclusion at 1% level of significance
x̄ = 9800 ; n = 36; σ = 500 km μ = 10,000 α = 1% = 0.01
H0 : let automobile tire has average life of 10,000 km i.e. μ = 10,000 km
H1 : let automobile tire does not have average life of 10,000 km i.e. μ ≠ 10,000 km
It’s a two tailed test
The level of significance is α = 0.01(1%)
The critical values are -2.58 and +2.58
Zcal lies between the critical values
-2.58< Zcal < +2.58. H0 is accepted
Conclusion: We can assume that the Automobile tire has an average life of 10,000 kms
Question 8.
Given x̄ = 203gm μ = 200 gm, σ = 10 gm and n = 64 calculate test statistic Z.
x̄ = 203 ; μ = 200 σ = 10 n = 64
Question 9.
Write the testing procedure of equality of population means.
1. H0 : population means are equal μ1 = μ2
2. H1: population means are not equal, μ1 ≠ μ2
If the mean of the first population is less than the second population i.e. μ1 < μ2
Its a left tailed test
If the mean of the first population is more than the second population i.e. μ1 > μ2
It’s a right tailed test
3. Calculate of test statistic = z = $$\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}$$
Here x̄1, x̄2 = sample means
σ1, σ2 = standard deviations of populations.
If SD are unknown these values are replaced by s1 and s2 sample standard deviations
n1 n2= sample size
4. Critical value ‘k’ will be chosen depending on H1 and α through the table.
5. If the calculated value Zcal lies in acceptance region, then H0 is accepted otherwise H0 is rejected.
6. Find conclusion and make a decision.
Question 10.
For the following data, test whether means differ significantly.
n1 = 90; x̄ = 52 s1 = 9; n2 = 40 x̄ = 54 ; s2 = 2
H0 : means of two population are equal i.e. μ1 = μ2
H1 : means of two population are not equal, i.e. μ1 ≠ μ2
It is two tailed test
At 5% significance
Critical values = -1.96 and +1.96
Zcal value lies in rejection region
-1.96 > Z cal <+1.96
H0 is rejected
Conclusion: means of populations are not equal
Question 11.
450 boys and 350 girls are appeared for II PUC examination. The mean and SD of marks obtained by boys are 53 and 18 respectively. The mean and SD of marks obtained by girls are 50 and 14 respectively. Is there any significant difference between mean marks obtained by boys and girls?
n1 = 450, x̄1 = 53 σ1 = 18, n2 = 350, x̄2 = 50, σ2 = 14
H0 : mean marks obtained by boys and girls are same i.e. μ1 = μ2
H1 : mean marks obtained by boys and girls are not same i.e. μ1 ≠ μ2
Its two tailed test
Level of significance α = 5% = 0.05
Critical values = -1.96, +1.96
-1.96 < Zcal > +1.96
Zcal value does not lie in the acceptance region
H1 is rejected
Conclusion: mean marks obtained by boys and girls are not same
Question 12.
A random sample of 100 workers from South India shows that their mean wages are 146 per day with SD 20. A random sample of 150 workers from North India shows that their mean wages are 150 per day with SD 30. Test at 1% level of significance that the mean wages of south Indian is less than mean wages of North India.
n1 = 100, x̄1 = 146, σ1 = 20, n2 = 150, x̄2 = 150, σ2 = 30
H0 : mean wages of south Indian and North Indian is same μ1 = μ2
H0 : mean wages of South Indian is less than North Indian μ1 < μ2
Its lower tailed test
Level of significance α = 1% = 0.01
The critical value is -2.33
Zcal > -2.33, H0 is accepted
Conclusion: Mean wages of south Indian and North Indian are same
Question 13.
As examination was conducted to two sections A and B consisting of 50 & 40 students respectively. Mean marks obtained by section A is 74 with a SD of 8 and that of B is 78 with a SD of 7. Is there a significant difference between the performances of the two sections at 1% level of significance?
n1 = 50, x̄1 = 74, σ1 = 8, n2 =40, x̄2 = 78, σ2 = 7
H0 : mean marks of Section-A and Section-B is same i.e. μ1 = μ2
H0 : mean marks of Section-A and Section B are not same i.e. μ1 ≠ μ2
Its two tailed test
Level of significance α = 1% = 0.01
Zcal value -2.5273 lies between the critical values
-2.58 < Zcal < +2.58
H0 is accepted
Conclusion: Mean marks obtained by Section A and Section B are same
Question 14.
Intelligent test given to two groups boys and girls gave the following information
difference in the mean scores of boys and significant? Use 5% LOS
n1 = 100, x̄1 = 70, σ1 = 12, n2 = 50, x̄2 = 74, σ2 = 10
H0: mean scores of boys and girls are same = μ1 = μ2
H0: mean scores of boys and girls are not same= μ1 ≠ μ2
Its two tailed test
Level of significance α = 5 % = 0.05
Critical values = -1.96 & +1.96
Zcal = -2.1567
Zcal does not lie within the acceptance region
-1.96> Zcal <+1.96
H0 is rejected
Conclusion: Mean scores of boys and girls are not same
Question 15.
Write the testing procedure of population proportion
The test has following steps:
(a) H0 : P = P0 (A given value for population)
(b) H1: P ≠ P0 (two tailed test) OR
P < P0 (left tailed test) OR
P > P0 (Right tailed test)
(c) Calculation of test statistic
x = number of items processing
n = number of items in a sample
(d) Critical value ‘k’ is chosen depending on H1 and α
(e) If the computed value of the test statistic Zcal lies in the acceptance region then H0 is accepted if not H0 is rejected.
Question 16.
In a random sample of 1000 persons from large populations 470 are females. Can it be said that males and females are in the equal ration in the populations? Use α = 0.05
n = 1000, x = 470, P0 = 0.5, Q0 = 0.5
p = $$\frac{x}{n}=\frac{470}{1000}=0.47$$
Ho : males and females are in the equal ratio in the population Po = 0.5
H1 : males and females are not in the equal ratio in the population Po ≠ 0.5
It is two tailed test
Level of significance α = 0.05 = 5%
The critical values = -1.96 & +1.96
Zcal lie between the critical values
-1.96 < Zcal <+1.96
H0 is accepted
Conclusion: males and females are in the equal ratio in the population
Question 17.
In an election the leaders of a party contend that they would secure more than 36% of votes. A pre-pole survey of 400 voters revealed that the percentage is 42. Does the survey supports the leaders claim? use 1% LOS
n1 = 400, P0 = 36% = 0.36, p = 42% = 0.42,
α= 1% = 0.01, Q0 = 1 – P0 = 1 – 0.36 = 0.64
H0 : mean leaders contention in that they can secure exactly 36% of them votes P = 36%
H1 : mean leaders contention is that they secure more than 36% of votes P > 36%
It is right tailed test
Level of significance, α = 0.01 = 1%
The critical value is 2.33
Zcal doesn’t lie between the critical values
Zcal > 2.33
H0 is rejected
Conclusion: mean leaders contention is that they secure more than 36% of votes
Question 18.
In a random sample of 100 PUC statistics students 9 are distinction holders, at the 5% level of significance can we conclude that 10% of II PUC statistics students are distinction holders?
n = 100, x = 9, Po = 10% = 0.1, p = $$\frac{x}{n}=\frac{9}{100}$$ = 0.09, Q0 = 1 – P0 = 1 – 0.1 = 0.9
H0: 10% of II PUC statistics students are distinction P0= 10%
H1 : 10% of II PUC statistics students are not distinction P0 ≠ 10%
It is two tailed test
Level of significance α = 0.05 = 5%
The critical values = -1.96 & +1.96
Zcal lie between the critical values
-1.96 < Zcal < +1.96. H0 is accepted
Conclusion: 10% of II PUC statistics students are distinction holders.
Question 19.
A stock broker claims that he can predict with 80%accuracy whether a stock market value will rise or fall during the coming month. In a sample of 40 predictions he is correct in 28. Does this evidence supports broker’s claim at 1% level of significance
n = 40, x = 28,p = $$\frac{x}{n}=\frac{28}{40}$$ = 0.7, P0 = 80% = 0.8, Q0 = 1 – P0 = 1 – 0.8
H0 : we can support brokers claim that can predict with 80% accuracy whether a stock’s market will rise or fall
P0 = 80%
H1: we cannot support P0 ≠ 80%
It is two tailed test
Level of significance α = 0.01 = 1%
The critical values = -2.58 & +2.58
Zcal lies between the critical values
-2.58 < Zcal < +2.58. H0 is accepted .
Conclusion: We can support broker’s claim that he can predict with 80% accuracy whether a stock market value will rise or fall.
Question 20.
The manufacturer of surgical instruments claims that less than 2% of the instruments he supplied to a certain hospital are faulty. A sample of 400 instruments revealed that 12 were faulty. Test his claim at 5% level of significance
n = 400, x = 12, p = $$\frac{x}{n}=\frac{12}{400}$$ = 0.03, p0 = 2% = 0.02 Q0 = 1 – p0 = 1 – 0.02 = 0.98
H0 : Surgical instruments supplied to hospital has 2% faulty P0 = 0.02
H1 : Surgical instruments supplied to hospital has less than 2% faulty P0 < 0.02
It is left tailed test. LOS α = 0.05 = 5%
The critical values are – 1.65
Zcal value is greater than critical value 0
Zcal > -1.65
H0 is accepted
Conclusion: Surgical instruments supplied to hospital has 2% fault
Question 21.
Write the testing procedure of equality of population proportions
The test has following steps:
(a) H0: (The population proportions are equal) P1 = P2
(b) H1 : (The population proportions are not equal) P1 ≠ P2
(1st population proportion is less than 2nd population proportion) P1 < P2 (left tailed test)
(1st population proportion is greater than 2nd population proportion) P1 > P2(right tailed test)
(c) Calculation of test statistic
(d) The critical value ‘k’ is chosen depending on H1 and α
(e) If Zcal lies in acceptance region then H0 is accepted if not H0 is rejected.
Question 22.
A machine produced 26 defective articles among 250. Another machine produced 4 defective articles among 50. Test whether there is a significant difference between population proportions at 5% level of significance
n1 = 250, X1 = 26, n2 = 50, x2 = 4
p1 = $$\frac{x_{1}}{n_{1}}=\frac{26}{250}$$ = 0.104, P = $$\frac{x_{1}+x_{2}}{n_{1}+n_{2}}=\frac{26+4}{250+50}$$ = 0.1
Q = 1 – P = 1 – 0.1 = 0.9 p2 = $$\frac{x_{2}}{n_{2}}=\frac{4}{50}$$ = 0.08
H0: There is significant difference between population proportions P1 = P2
H1: There is no significant difference between population properties P1 ≠ P2
$$=\frac{0.024}{(0.09)(0.024)}=0.5164$$
It is a two tailed test
Level of significance α = 5%
The critical values = -1.96 and +1.96
Zcal = 0.5164 lies in between the critical values
-1.96 < Zcal > +1.96
H0 is accepted
Conclusion: There is significant difference between population proportions
Question 23.
Out of 400 PUC students of college A, 72% of students were passed, out of 200 PUC students of college B 66% of students were passed. Can it be concluded that performance of college A is better than performance of college B? Use 5% LOS.
n1 = 400, p1 = 72% = 0.72 n2 = 200, p2 = 66% = 0.66
H0 : the performance of college A is same as B, P1 = P2
H1 : the performance of college A better than B, P1 > P2
p = $$\frac{\mathrm{n}_{1} \mathrm{p}_{1}+\mathrm{n}_{2} \mathrm{p}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$$ = $$\frac{(400) 0.72+(200) 0.66}{400+200}=\frac{288+132}{600}=\frac{420}{600}=0.7$$
Q = 1 – P = 1 – 0.7 = 0.3
It is upper tailed test
Level of significance α = 5% = 0.05
The critical value = 1.65
Zcal = 1.512 < 1.65
H0 is accepted
Conclusion: Performance of college A and B are same.
Question 24.
In a random sample of 100 people from a city in the year 2011 revealed that 65 were cricket match viewers. In another random sample of 100 peoples from same city in the year 2013 revealed that 75 were cricket match viewers examine whether there is a significant increase in cricket match viewers. Use 1% level of significance
n1 = 100; x1 = 65 P1 = $$\frac{x_{1}}{n_{1}}=\frac{65}{100}=0.65$$
n2 = 100; x2 = 75 p2 = $$\frac{x_{2}}{n_{2}}=\frac{75}{100}=0.75$$
$$P=\frac{x_{1}+x_{2}}{n_{1}+n_{2}}=\frac{65+75}{100+100}=\frac{140}{200}=0.7$$
Q = 1 – p = 1 – 0.7 = 0.3
H0 : there is no significant increase in cricket match viewers from 2011 to 2013 p1 = p2
H1 : there is significant increase in cricket match viewers from 2011 to 2013 p1 < p2
It is lower tailed test
Level of significance α = 1% = 0.01.
The critical value = -2.33. Zcal = -1.543 > -2.33
H0 is accepted
Conclusion: There is no significant increase in cricket match viewers from 2010 to 2013.
Question 25.
From the following data, test whether the difference between the proportions in the samples is significant use 5% level of significance.
Sample I II Size 100 400 Proportion 0.02 0.01
n1 = 100, n2 = 400, p1 = 0.02, p2 = 0.01
p = $$\frac{n_{1} p_{1}+n_{2} p_{2}}{n_{1}+n_{2}}$$ = $$\frac{(100) 0.02+(400) 0.01}{100+400}=0.012$$
Q = 1 – P = 1 – 0.012 = 0.988
Ho: There is no difference between the proportions in the sample P1 = P2
H1 : There is difference between the proportions on the sample P1 ≠ P2
It is a two tailed test. Level of significance α = 5% 0.05,
The critical values = -1.96 and +1.96. Zcal = 0.82141 lies in between the critical values
-1.96 < Zcal > +1.96
H0 is accepted
Conclusion: There is no difference between the proportions in the sample. P1 = p2
### 2nd PUC Statistics T- Test Exercise Problems
Question 1.
Write two applications of t-Test?
1. To test the population mean.
2. To test the equality of two population means.
Question 2.
Write down the t-test statistics and degrees of freedom in case of test for mean?
T-test statistic
Degrees of freedom means, number of independent observations if there are ‘n’ observations then, Degrees of freedom = (n – c) where ‘c’ is number of independent constraints.
Question 3.
Write down the t-test statistic & degrees of freedom in case of test for for equality of means of two independent samples.
T-test statistic,
Question 4.
Write down the t-test statistic & degrees of freedom in case of test for for equality of means of praised observations.
T – test statistic,
Question 5.
In paired t-test if n = 5 d̄ = 3 and sd = 1.5. Then what would be the value of test statistic?
n = 5, d̄ =3, s = 1.5
Question 6.
A random sample of size 20 taken from normal population gives a sample mean of 42 and standard deviation of 6. Test the hypothesis that the population mean is 44.
n = 20, x̄ = 42, s = 6, μ = 44
H0 : The population mean is 44 μ = 44
H1 : The population mean is not equal to 44 μ ≠ 44
It is two-tailed test
The d.f is n – 1 = 20 – 1 = 19
Critical values = -2.09 and +2.09
tcal = 1.4529 is in the accepted region
-2.09 < 1.4529 < 2.09.
H0 accepted
Conclusion: The population mean is 44
Question 7.
A company has been producing steel tubes with mean inner diameter of 2.00 cms. A sample of 10 tubes gives a mean inner diameter of 2.01 cms and a variance of 0.004 cms2. Is the difference in the values of mean is significant?
μ = 2.00 cm, n = 10, x̄ = 2.01, variance = s2 = 0.004
s = √0.004 = 0.0632
H0 : there is no difference between population mean and sample mean μ = 2.00
H1 : there is a difference between population mean and sample mean μ ≠ 2.00
It is two tailed test. Los α = 0.05=5%
The d.f is n – 1 = 10 – 1 = 9.
Critical values = – 2.26 and +2.26.
-2.26 < 0.4748 < +2.26
tcal value 0.4748 lies in between the critical values
H0 is accepted.
Conclusion: There is no difference between population mean and sample mean.
Question 8.
The mean weekly sale of ice-cream bars was 140 bars. After an advertising campaign the mean weekly sale in 26 shops for a typical week increased to 150 and showed a SD 20. Is this evidence indicates that the advertising was successful?
x̄ =150, μ0 = 140, s = 20, n = 26
H0: μ = 140 Mean weekly sale of ice-cream bars after advertisement has not increased
H1: μ > 140 Mean weekly sale of ice-cream bars after advertisement has increased.
It is upper tailed test
α = 0.05 = 5%
The .d.f is n – 1 = 26 – 1 = 25
Critical value = 2.49
tcal value 1.970 > 1.72.
H0 is rejected.
Conclusion: The mean weekly sale of ice – cream bars after advertising has increased.
Question 9.
A fertilizer mixing Machine is set to give 12 kg of nitrate for every bag of fertilizer. Then 10 such bags are examined. The weight of nitrate in each bag (in kg) are given below:
11 14 13 12 13 12 13 14 11 12
Is there reason to believe that the machine is defective?
μ = 12, n = 10,
Calculation of mean and standard deviation
Σx = 125, Σx2 = 1573
x̄ = $$\frac{\Sigma x}{n}$$ = $$\frac{11+14+13+12+13+12+13+14+11+12}{10}=\frac{125}{10}=12.5$$
The test
The test is two-tailed
α = 5% = 0.05
The d.f is n – 1 = 10 – 1 = 9
Critical values = -2.26 and 2.26
tcal value 1.464 lies between the critical value in acceptance region
H0 is accepted
Conclusion: The machine is not defective
Question 10.
A random sample of size 16 has mean 53. The sum of the squared deviations taken from mean is 150. Can this sample be regarded as taken from the population having mean 56 (Use α = 0.01)
n = 16, x̄ = 53, μ = 56, α = 0.01 = 1%, Σ(x – x̄)2 = 150
H0 : population having mean 56, μ = 56
H1 : population is not having mean 56, μ ≠ 56
It is two tailed test
The d.f. is n – 1 = 16 – 1
Critical values = -2.95 and 2.95
χ2cal value is in accepted region
H0 is accepted
Question 11.
The marks obtained by two groups of students in a statistics test are given below
On the basis of this data, can it be concluded that there is a significant difference in the mean marks obtained by the two groups?
n1 = 15, n2 = 11, x̄1 = 42, x̄2 = 38, s1 = 10, s2 = 15
H0 : Mean marks of students of group A & B are same, μ1 = μ2
H1 : mean marks of group A and B are not same μ1 ≠ μ2
Under H0
Level of significance = α = 0.05
d.f = 15 + 11 – 2 = 24
It is two tailed test
Critical value= – 2.06 and + 2.06
tcal = 0.7829 lies between critical values
– 2.06 < 0.7829 <+ 2.06
H0 is accepted
Question 12.
For the following data examine if the means of two samples differ significantly.
n1 = 12; n2 = 7; x̄1 = 57.2; s1 = 3.41; x̄2 = 52.3; S2 = 3.62
H0 : μ1 = μ2 means are same
H1: μ1 ≠ μ2 means are not same
It is a two tailed test
d.f. n1 + n2 – 2 = 12 + 7 – 2 = 17
Critical values = – 2.11 and + 2.11. LOS = 5% = 0.05
tcal value 2.793 lies between the critical values
= – 2.11 < 2.793 > + 2.11
H0 is accepted.
Question 13.
A group of 5 patients treated with medicine A weighs 42, 39, 48, 60 and 41 kg, second group of 7 patients from the same hospital treated with medicine B weighs 38, 42, 56, 61, 69, 68 and 67 kg. Do you agree with the claim that medicine B increases the weight significantly?
n1 = 5 n2 = 7
H0 : Medicine B does not increase the weight, μ1 = μ2
H1 : Medicine B increase the weights significantly, μ1< μ2
It is lower tailed test
Level of significance = α = 0.05 = 5%
d.f. = n1 + n2 – 2 = 5 + 7 – 2 = 10
Critical value= – 1.81
tcal = – 1.695 > -1.81
H0 is accepted
Conclusion: Medicine B does not increase weight
Question 14.
Two new types of rations are fed to pigs. A sample of 11 pigs is fed with type A ration and another sample of 11 pigs is fed with type B ration, the gains in weight are recorded be low (in Pounds):
At 1% level, test whether type A ration is better than type B ration.
n1 = 11 n2 = 11
$$=\frac{3.2}{\sqrt{9.94(0.09+0.09)}}=\frac{3.2}{\sqrt{9.94 \times 0.1809}}$$
$$=\frac{3.2}{\sqrt{1.7982}}=\frac{3.2}{1.34}=2.38$$
At 5% los (n1 + n2 – 2) = 11 + 11 – 20 df.
Ttab = CV = 2.53
tcal = 2.38 < CVtab = 2.53
H0 is accepted.
Question 15.
Two laboratories A and B carry out independent estimates of fat content in ice-cream made by a firm. A sample is taken from each batch, halved and the separate halves sent to the two laboratories. The fat content obtained by the aboratories is recorded below.
Is there a significant difference between the means of fat content obtained by the two laboratories A and B?
H0: μ1 = μ2 means are same
H0: μ1 ≠ μ2 means are not same
It is a two tailed test
α = 0.05=5% d.f = n1+ n2 – 2 ⇒ 10 + 10 – 2 = 18
Critical value = -2.10 and 2.10.
-2.10 < -0.386 <+2.10
tcal = -0.386 this in accepted region
H0 is accepted.
Conclusion: means are same.
Question 16.
Twelve students were given intensive coaching and 2 tests were conducted before and after coaching. The score of 2 tests are given below. Do the scores before and after coaching show an improvement?
H0: Coaching is not useful, μ1 = μ2
H1: Coaching is better, μ11 < μ2
The test statistics is
It is left tailed test
Critical value =- k = -1.80
LOS = α = 0.05 = 5%
d.f = n – 1 = 12 – 1 = 11
tcal = – 4.891 < critical value,
-1.80 <-4.891
H0 is rejected
Conclusion: Coaching is not useful.
Question 17.
A certain stimulus administered to each of the 12 patients resulted in the following changes in blood pressure
5 2 8 -1 3 0 -2 1 5 0 4 6
Can it be concluded that the stimulus will be in general accompanied by change in blood pressure
H0: stimulus accompanied doesn’t increase BP, μ1 = μ2
H1 stimulus accompanied can increase BP, μ1 < μ2
d = 5 2 8 -1 3 0 -2 1 5 0 4 6
n = 12
$$=\sqrt{15.417-(2.583)^{2}}$$
$$=\sqrt{15.417-6.673}$$
$$=\sqrt{8.744}=2.9570$$
It is left tailed test
LOS = α = 0.05 = 5%
d.f. = n – 1 = 12 – 1 = 11
Critical value = k = 1.80
tcal values is greater than critical value,
2.896 > 1.80
H0 is rejected
Conclusion: stimulus accompanied can increase BP.
Question 18.
Nine patients, to whom a certain drug was administrated, registered the following increments in blood pressure.
7 3 -1 4 -3 5 6 -4 1
Show that the data do not indicate that the drug was responsible for these increments.
n = 9.
H0 : The drug is not responsible for increments in B.P.
H1 : The drug is responsible for increments in B.P.
It is two tailed test
LOS = α = 0.05 = 5%, d.f = (n – 1) = 9 – 1 = 8
K critical values = 2.31, -2.31
Tcal lies in acceptance region
H0 is accepted
Conclusion: The drug is responsible for increrment in b.p.
### 2nd PUC Statistics Chi – Square (χ2) Tests Exercise Problems
Question 1.
Mention 2 applications of χ2 test.
1. To test if the population has a given variance
2. To test goodness of fit in a given data of a theoretical distribution.
Question 2.
Write χ2 test statistic with degree of freedom in testing of population variance?
Question 3.
Write χ2 – test statistic with degrees of freedom in testing of goodness of fit
χ2 test statistic,
Question 4.
Mention two conditions for applicability of χ2 test of goodness of fit
Condition:
1. The total frequency ‘N’ should be large
2. The theoretical frequencies E1 ≥ 5. If any E1 < 5, it should be pooled with the adjacent frequency.
Question 5.
When is the pooling of the frequencies done in testing of goodness of fit?
The theoretical frequencies E1 ≥ 5 . If any E1 < 5, it should be pooled with the adjacent frequency.
Question 6.
In a χ2 test for goodness of fit, if there are 6 classes and one parameter is estimated, then find the value of degrees of freedom of test statistic?
D.f. in testing of goodness of fit is n – c; then d.f 6 – 1 = 5
Question 7.
Write the formula of χ2 – test with degrees of freedom in testing of independence of attributes in 2 × 2 contingency table.
Question 8.
What is the value of d.f. is 2 × 2 contingency table?
The value is 1df.
Question 9.
What are the conditions applicable while testing independent of attribute in 2 × 2 contingency table?
Conditions.
1. The total frequency ‘N’ should be large (> 50)
2. The observations should be independent.
Question 10.
For the χ2-test, what is the condition for expected cell frequency?
Each of the expected cell frequencies E1 should be 5 are more.
Question 11.
A random sample of size 25 taken from a population gives the sample standard deviation as 8.5. Test the hypothesis that the population standard deviation (a) is 10.
n = 25, s = 8.5, σ = 10 i.e., σ2 = 102 = 100
H0: σ2 = 102 Population SD is 10
H1 : σ2 ≠ 102 Population SD is not 10
It is a two-tailed test
d.f. = (n – 1) = 25 – 1 = 24
α = 0.05
Critical values K1 = 12.4, k2 = 39.36
χ2cal value lies with in the critical values
H0 is accepted
Conclusion: Population SD is 10
Question 12.
Weights in kg. of 10 students are given below:
38 40 45 53 47 43 55 48 52 49
Can we say that variance of the distribution of weights is equal to 20 kg2 ?
n = 10 σ2 = 20
x̄ = $$\frac{\Sigma x}{n}$$ = $$\frac{38+40+45+53+47+43+55+48+52+49}{10}$$
x̄ = $$\frac{470}{10}$$ = 47
The test statistic is
χ2 = $$\frac{\sum(x-\bar{x})^{2}}{\sigma^{2}}=\frac{280}{20}=14$$
H0 : Variance of the distribution of weights is 20kg σ2 = 20
H1 : Variance of the distribution of weights is not 20kg σ2 ≠ 20
It is a two-tailed test
Degree of freedom = 10 – 1 = 9
Level of significance = 0.05
Critical values are k1 = 2.70 and k2=19.02
k1 < 14 < k2
H0 is accepted
χ2cal value lies in acceptance region
Conclusion: variance of the distribution of weights of students is equal to 20 kgs2
Question 13.
In 120 throws of a single die, the following distribution of faces were obtained:
Test at 5% level of significance that the die in unbiased
H0 : The die is unbiased
H1 : The die is biased
Under H0 all the sides of die are equally probable.
Then frequencies should be equal and so, E1 = $$\frac{N}{n}=\frac{120}{6}$$ = 20
The test = χ2 = $$\Sigma \frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}$$ = 12.9
THe degree of freedom is (n – c) = 6 – 1 = 5
α = 0.05
The critical value 11.07
χ2cal > c.v
H0 is rejected
Conclusion: The die is biased
Question 14.
Fit a binomial distribution to the following data and test for goodness of fit?
H0 : Binomial distribution is a good fit.
H1 : Binomial distribution is not a good fit.
x̄ = np = $$\frac{\Sigma f x}{n}=\frac{439}{100}$$ = 4.39
np = 4.39, 9p = 4.39 p = $$\frac{4.39}{9}$$ = 0.48
q = 1 – p = 1 – 0.48 = 0.52
The p.m.f is p(x) = 9Cx × (0.48)xim × (0.52)9 – x x = 0,1, 2 9
The frequency function is
p(0) = 9C0 × (0.48)0 – (0.52)9 – 0 = 1 × 1 × 0.002779
E(0) = N. p(0) = 100 × 0.002779 = 0.2779
By using recurrence relation for expected frequencies
E(x) = $$\frac{n-x+1}{x \cdot q} E(x-1)$$
E(1) = $$\frac{9-1+1}{1} \times \frac{0.48}{0.52} \times 0.2779$$ = 2.62 = 9 × 0.9231 × 0.2779 = 2.31
E(2) = $$\frac{9-1+1}{1} \times \frac{0.48}{0.52} \times 0.2779$$ = 4 × 0.9231 × 2.31 = 8.529
E(3) = $$\frac{9-3+1}{3} \times \frac{0.48}{0.52} \times 8.52$$ = 2.33 × 0.9231 × 8.529 = 18.344
E(4) = $$\frac{9-4+1}{4} \times \frac{0.48}{0.52} \times 18.344$$ = 18.344 = 1.5 × 0.923 l × 18.344 = 25.40
E(5) = $$\frac{9-5+1}{5} \times \frac{0.48}{0.52} \times 25.40$$ = 1 × 0.9231 × 25.40 = 23.45
E(6) = $$\frac{9-6+1}{6} \times \frac{0.48}{0.52} \times 23.45$$ = 0.67 × 0.9231 × 23.45 = 14.5
E(7) = $$\frac{9-7+1}{7} \times \frac{0.48}{0.52} \times 14.5$$ = 0.43 × 0.9231 × 14.5 = 5.76
E(8) = $$\frac{9-8+1}{8} \times \frac{0.48}{0.52} \times 5.76$$ = 0.25 × 0.9231 × 5.76 = 1.33
E(9) = $$\frac{9-9+1}{9} \times \frac{0.48}{0.52} \times 1.33$$ = 0.11 × 0.9231 × 1.33 = 0.14
Applying χ2 test
The theoretical frequencies are
Test statistic under H0 is
χ2cal $$\Sigma \frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}$$ = 47.47
At 5% LOS (α = 0.05) for (n – c) = 6 – 2 = 4 df
The critical value is k2 = 9.49
since; χ2cal > k2
∴ H0 is rejected
Conclusion: BD is not a good fit
Question 15.
Following is the data regarding number of mistakes per page found in a book. Fit a Poison distribution test at 5%. L.O.S. that it is good fit.
x̄ = λ = $$\frac{\Sigma f x}{N}=\frac{120}{100}$$ = 1.2, pmf,
p(x) = $$\frac{e^{-\lambda} \lambda^{x}}{x !}$$ , x = 0,1,2,3,4
put x = 0 p(0) = $$\frac{\mathrm{e}^{-1.2}(1.2)^{0}}{0 !}$$ = 0.3102
F(0) = NP(x) = 100 × $$\frac{\mathrm{e}^{-1.2} \times(1.2)^{0}}{0 !}$$ =100 × 0.3012 = 30.12
Calculation of theoretical frequency using recurrence relation
T(x) = $$\frac{\lambda}{x} \cdot E(x-1)$$
T(1) = $$\frac{1.2}{1} \times 30.12$$ = 36.144
T(2) = $$\frac{1.2}{2} \times 36.14$$ = 21.68
T(3) = $$\frac{1.2}{3} \times 21.68$$ = 8.67
T(4) = = 2.6
H0 : Poisson distribution is good fit.
H1 : Poisson distribution is not a good fit.
It is upper tail test χ2cal = $$\Sigma \frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}$$ = 0.5210
d.F = n – c
=4 – 2 = 2
LOS = α = 0.5
Critical value is 5.99
χ2cal < C.V
H0 is accepted
Conclusion: The P.D. is a good fit
Note: e-1.2 = e-1 × e-0.2 = 0.3679 × 0.8187 = 0.30119
Question 16.
The following data shows the suicides of 1096 women in 8 Punjab cities during 14 years
Fit a Poisson distribution to the data and show that the distribution is not good fit.
H0 : Poisson distribution is a good fit
H1 Poisson distribution is not a good fit
x = λ = $$\frac{\Sigma f x}{N}=\frac{1295}{1096}$$ = 1.18
P(X) = $$\frac{\mathrm{e}^{-1.18} \cdot(1.18)^{\mathrm{x}}}{\mathrm{x} !}(\mathrm{x}=0,1, \ldots 7)$$
E(0) = N . P(0) = 1096 . $$\frac{\mathrm{e}^{-1.18} \cdot(1.18)^{0}}{0 !}$$ = 1096 × 0.3072 = 336.69
The expected frequencies can be calculated by using recurrence relation E(x) = $$\left(\frac{\lambda}{x} E(x-1)\right)$$
E(l) = $$\frac{1.18}{1} \times 336.69$$ = 397.296 118
E(2) = $$\frac{1.18}{2} \times 397.29$$ = 234.405
E(3) = $$\frac{1.18}{3} \times 234.405$$ = 92.19 118
E(4) = $$\frac{1.18}{4} \times 92.19$$ = 27.196 118
E(5) = $$\frac{1.18}{5} \times 27.196$$ = 6.418 118
E(6) = $$\frac{1.18}{6} \times 6.418$$ = 1.2627
E(7) = $$\frac{1.18}{7} \times 1.2627$$ = 0.213
Test statistic under H0 is
χ2 = $$\Sigma \frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}$$ = 13.935
At 5% LOS (α = 0.05)
(n – c) = 6 – 2 = 4 df
Critical value = 9.49
since, χ2cal value does not lies in AR.
χ2 > K2,
H0 rejected
Conclusion: PD is not a good fit.
Note:
e-1.18 = e-1 × e-0.1 × e-0.08 = 0.3679 × 0.9048 × 0.9231 = 0.3072
Question 17.
Consider the following data.
Fit a poisson distribution to the data and test the goodness of fit (use 5% Los.)
H0 : Poisson distribution is a good fit
H1 : Poisson distribution is not a good fit
Mean x̄ = λ = $$\frac{\Sigma f x}{n} \Rightarrow \frac{99}{100}$$ = 0.09
The P.M.F
p(x) = $$\frac{e^{-\lambda} \cdot \lambda^{x}}{x !}$$ ; n = 0,1,2…..
E(0) = N.P(0)
E(0) = p(x = 0) = 100 × $$\frac{\mathrm{e}^{-0.99} \times(0.99)^{0}}{0 !}$$ ⇒(0.3176)100 = 37.16
Using recurrence relation E(x) = $$\frac{\lambda}{x} E(x-1)$$
χ2cal = 11.09
α = 5% LOS n – 2 = 4 – 2 = 2
Critical value k2 = 5.99
Here χ2cal does not lie in AR.
H0 is rejected
Conclusion: Poission distribution is not a good fit.
Note: e-0.09 = e-0.9 × e009
= 0.4066 × 0.9139 = 0.37159.
Question 18.
Among 64 off spring of a certain cross breeds of guinea pigs, 34 were red, 10 were black and 20 were white. According to genetic model this numbers should be in the ratio of 9 : 3 : 4. Are the data consistent with model at the 5% level?
H0 : The data is consistent :
H1 : The data is not consistent
Given
The theoretical frequency ratio is 9 : 3 : 4
E1 = 64 × $$\frac{9}{16}$$= 36
E2 = 64 × $$\frac{3}{16}$$ = 12
E3 = 64 × $$\frac{4}{16}$$ = 16
The test χ2cal = $$\Sigma \frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}$$ = 1.44
df = n – C = 3 – 1 = 2
Critical value = 5.99, α = 5%
χ2cal < C.V
H0 is accepted
Conclusion: The data is consistent
Question 19.
The following information was obtained in a sample of 50 small general shops.
Can it be said that there are relatively more women owners of small general shops in villages than in towns?
H0 : Women owner and small general shops in villages are independent.
H1 : Women owner and small general shops in village are not independent
It is upper tailed test
α = 5% = 0.05 For 1 d.f.
The critical value k2 = 3.84
χ2cal < C.V
H0 is accepted
Conclusion: Women owners of small general shops are more in villages than in towns.
Question 20.
The opinion poll was conducted to find the reaction to a proposed civic reform in 100 members of each of the two political parties.
Test whether political parties and the reaction to a proposed civic reform are independent
H0 : political parties and the reaction to a proposed civic reform are independent.
H1 : political parties and the reaction to a proposed civic reform are not independent.
It is upper tailed test
LOS = 0.05 with 1 df
Critical value 3.84
χ2cal< C.V
H0 is accepted
Conclusion: Political parties and the reaction to a proposed civic reform are independent.
Question 21.
From the following data, test whether education and employment are independent at 5% LOS
H0 : Education and employment are independent
H1 : Education and employment are dependent
α = 0.05 for 1 df
k2 = 3.84
χ2cal value lies in acceptance region
χ2cal < K2
H is not accepted.
Conclusion: Education and employment are independent.
Question 22.
Of the 500 workers in a factory exposed to an epidemic 350 in all were attacked, 200 had been inoculated and of these 100 were attacked. Test whether inoculation and attack are independent.
H0 : Inoculation and attack are independent.
H1 : Inoculation and attack are not independent.
It is upper tailed test
α = 0.05 for 1 d.f.
C.V. is k2 = 3.84
χ2cal > c.v
H0 is rejected
Inoculation and attack are not independent.
### 2nd PUC Statistics Large Sample Tests Pratical Assignments
Question 1.
A machine is designed to fill 1 litre milk to polythene packets. A randomly selected 100 milk packets filled by this machine are inspected. The mean milk is found to be 998 ml and S.D. is 10 ml. Is the machine functioning properly at 1% level of significance?
Given: n = 100
x̄ = 998ml, s = 10ml, µ = 1 Litre = 100ml
H0 : Machine functioning properly (ie., machine fills 1 Ltr. Milk) i.e., µ = l Litre / 1000 ml.
H1 : Machine is not functioning properly [i.e., machine does’t fills 1 litre milk) i.e., µ ≠ 1 Litre / 100 ml (Two tail
test) under H0 the test statistic is : z = $$\frac{\bar{x}-\mu}{s \sqrt{n}}$$ ~ N(0,1)
zcal = $$\frac{998-1000}{10 \sqrt{100}}=\frac{-2}{10 / 10}=-2$$
At α = 1% the two critical values are ± k = ± 2.58
Here Zcal lies acceptance region (A.R)
i.e., k ≤ Zcal ≤ k
∴ H0 is accepted.
Conclusion: Machine is functioning properly; µ = 1 Ltr.
Question 2.
A random sample of 225 cans containing baby food has mean weight 998 gm and S.D. 15gm. Test whether the mean contents of the cans be considered as 2kg. (Use 5% L.O.S.)
Given: n = 225, x̄ = 998, s = 15, µ = lkg = 1000 grams
H0: Mean content of the can is 1 kg i.e., µ = 1000 grams
H1: Mean content of the can differs from 1 kg i.e., µ ≠ 1000 grams. (Two tail test) under H0, the test statistic is:
At α = 5% the two critical values are ± k = ± 1.96
Here Zcal lies in rejection region (R.R)
∴ H0 is rejected.
Conclusion: Mean content of the can differs from 1 kg. i.e., µ ≠ 1000 grams.
Question 3.
On 60 different days the numbers of passengers in a bus were noted. The mean and S.D. of the number of passengers was found to be 40 and 5 respectively. At 5% level of significance, test the hypothesis that the average number of passengers in the bus is more than 38.
Given: n = 60, x̄ = 40, s = 5, α = 5% µ = 38
H0 : The average number of passengers is 38. i.e., µ = 38
H1 : The average number of passengers is more than 38. i.e., µ > 38(upper tail test)
Under H0, the test statistic is:
At α = 5% the upper tail critical value k = 1.65
Here Zcal lies in rejection region
∴ H0 is rejected.
Conclusion: The average number of passengers is more than 38.
Question 4.
A company manufactures car tyres. Their average life is 40,000 kilometers and standard deviation 5,000 kilometers. A change in the production process is believed to result in a better product. A test sample of 100 new tyres has mean life of 41,000 kilometers. Can you conclude at 1% L.O.S. that the new product gives better result?
Given: µ = 40,000, σ = 5,000, n = 100,
x̄ = 41,000, α = 1%
H0 : Mean life of tyres is 40,000 km. i..e, µ = 40,000 kms
H1: Mean life of tyres is more than 40,000 kms.
(i.e., New product of tyres gives better result] i.e., µ > 40,000 kms
Under H the test statistic is z = $$\frac{\bar{x}-\mu}{\sigma \sqrt{n}}$$ ~ N(0,1)
zcal = $$\frac{41,000-40,000}{5000 / \sqrt{100}}=2$$
At α = 1% the upper tail critical value k = 2.33
Here Zcal lies in acceptance region i.e., Zcal < k
∴ H0 is rejected.
Conclusion: Mean life of tyres is 40,000 Kms
i.e., µ = 40,000 Kms.
Question 5.
A certain brand of soap is said to weigh 125 gm on the average. A consumer agency received complaints that the weight is much less. To test the claim the agency selects 100 pieces and finds mean is 123 gm with S.D. 5 gm. complete the test at 5% level of significance?
Given: µ = 125gm, n = 100, x̄ = 123 gm, s = 5gm
H0 : Average weight is 125 gmi..e, µ = 125 kms
H1 : Average weight is less than 125 gm i.e., µ < 125 gm (Lower tail test)
under H0 the test statistic is
Zcal = $$\frac{123-125}{5 \sqrt{100}}$$ = -4
At α = 5% the lower tail critical value – k = -1.65
Here Zcal lies in rejection region (R. R)
i.e., Zcal < -k
∴ H0 is rejected.
Conclusion: Average weight is less than 125 gms
i.e., µ < 125 gms.
Question 6.
The mean of weight of 50 boys of a college is 58 kg and that of 40 girls of the same college is 54 kg. The variances of weights of boys and girls are 64 and 49 respectively. Can we conclude that boys and girls studying in the college have same weight? Use 5% level of significance.
Given: n1 = 50, x̄1 = 58, n2 = 40; x̄2 = 54, s21 = 64, s22 = 49
H0 : Boys and girls have same weight i.e., µ1 = µ2
H1 Boys and girls do not have same weight i.e., µ1≠ µ2 (two tail test)
At α = 5% the lower tail critical values are ± k = ± 1.96
Here Zcal lies in rejection region (R. R)
∴ H0 is rejected.
Conclusion: Boys and girls do not have same weight i.e., µ1≠ µ2
Question 7.
400 women shoppers are chosen at random in market A. Their average weekly expenditure on food to be ?500 with S.D. of ?40. The figures are ?492 and ?50 respectively, in the market B, where 500 women shoppers are chosen at random from this place. Test at 5% level of significance whether the average weekly food expenditure of population of shoppers are equal.
Given: n1 = 400, x̄1 = 500, s1 = 40,
2 = 492, s2 = 50, n2 = 500, α = 5%
H0 : Average weekly expenditure of women shopers in market A and B are same i.e., µ1= µ2
H1 : Average weekly expenditure of women shopers in market A and B are not same i.e., µ1≠ µ2 (two tail test)
At α = 5% the two tail critical values are ± k = ± 1.96
Here Zcal lies in rejection region (R. R)
∴ H0 is rejected.
Conclusion: Average weekly expenditure of women shoppers in market A and B are not same µ1≠ µ2
Question 8.
For the following data, test whether means differ significantly, (Use α = 0.01)
Given: n1 = 40, x̄1 = 70, s1 = 8, n2 = 60, x̄2 = 66, s2 = 6
H0: Means does’t differ significantly i.e., µ1 = µ2
H1 : Means differ significantly i.e., µ1≠ µ2 (two tail test)
At α = 0.01 the two tail critical values are ±k = ±2.58
Here Zcal lies in rejection region (R. R)
∴ H0 is rejected.
Conclusion: Means differ significantly.
Question 9.
Test at 1% level of significance, that average life of bulbs manufactured by Firm – A is less than Firm – B.
Given:
Let n1 =32, x̄1 =1300, s21 = 64,
n2 = 50, x̄2 = 1305, s21 = 100
H0 : Average life of bulbs of firm A and B is same, (i.e., µ1= µ2)
H1 : Average life of bulbs of firm A is less than Firm B i.e., µ1 < µ2 (Lower tail test)
At α = 1% the lower tail critical value – k = -2.33
Here Zcal lies in rejection region (R. R)
i.e., Zcal < – k
∴ H0 is rejected.
Conclusion: Average life of bulbs of firm A is less than firm B. i.e., µ1 < µ2.
Question 10.
Following is data regarding mean weights of randomly selected boys and girls of P.U.C. students. Test whether, mean weight of boys are greater than mean weight of girls. (Use α = 0.05.)
Given: n1 = 64, x̄1 = 63, s1 = 8, x̄2 = 60, s2 = 12
H0 : Mean weight of Boys and girls is same, (i.e., µ1= µ2)
H1 : Mean weight of Boys is greater than girls i.e., µ1> µ2 (upper tail test)
At α = 0.05 the upper tail critical value +k = +1.65
Here Zcal lies in acceptance region (A. R)
i.e., Zcal < k
∴ H0 is accepted
Conclusion: Mean weight of Boys and girls is same i.e., 0 µ1> µ2
Question 11.
Intelligence test given to two groups of boys and girls gave the following information:
Is the difference in the mean score of boys and girls statistically significant Use 1% L.O.S.
Given: n1 – 100, x̄1 = 74, s1 = 12,
n2 = 50, x̄2 = 70, s2 = 10
H0 : The mean score of boys and girls does’t differs significantly i.e., µ1= µ2
H1 : The mean score of boys and girls differ significantly i.e., µ1 1µ2 (two tail test)
At α = 1% the tail critical values are ± k = ± 2.58
Here Zcal lies in acceptance region (A. R)
∴ H0 is accepted
Conclusion: Mean scores of boys and girls does’t differ significantly i.e., µ1= µ2
Question 12.
A coin is tossed 400 times and head turns up 220 times. Can we conclude that the coin is unbiased ? (Use α = 0.05)
Given: n = 400, x = 200, P0 = 0.5 (unbiased)
H0 : The coin is unbiased i.e., P0 = 0.5
H1 : The coin is biased i.e., P0 ≠ 0.5 (Two tailed test)
Under H0, the test statistic is : z = $$\frac{p-p_{0}}{\sqrt{\frac{P_{0} Q_{0}}{n}}}$$
Here sample proportion :
p = $$\frac{x}{n}=\frac{220}{400}$$ = 0.55 and Q0 = 1 – P0 = 1 – 0.5
∴ zcal = $$\frac{0.55-0.5}{\sqrt{\frac{0.5 \times 0.5}{400}}}=\frac{0.05}{0.025}=2$$
At α = 0.05 the two tail critical values are ±k = ±1.96
Here Zcal lies in rejection region (R. R) i.e., Zcal > k
∴ H0 is rejected
Conclusion: The coin is biased i.e., p0 ≠ 0.5
Question 13.
In a city, out of 900 men 486 were smokers. Does this information indicate that the majority of men in the city are smokers?
Given: n = 900, x = 486, P0 = 0.5 (majority)
H0 : Men smokers in the city are50% i.e., P0 = 0.5
H1 : Majority of men in the city are smoker i.e., P0 > 0.5 (upper tailed test)
Under H 0, the test statistic is : z = $$\frac{p-P_{0}}{\sqrt{\frac{P_{0} Q_{0}}{n}}}$$ ~ N(0,1);
Here sample proportion :
P0 = $$\frac{x}{n}=\frac{486}{900}$$ = 0.54 and Q0 = 1 – P 0 = 1 – 0.5 = 0.5
∴ Zcal = $$\frac{0.54-0.5}{\sqrt{\frac{0.5 \times 0.5}{900}}}=\frac{0.04}{0.0167}=2.3952$$
At α = 1% the upper tail critical value k = 2.33
Here Zcal lies in rejection region (R R)
i.e., Zcal > k
∴ H0 is rejected
Conclusion: Majority of men in the city are smokers i.e., p0 > 0.5.
Question 14.
A manufacturer claims that less than 2% of his products are defective. A retailer buys a batch of 400 articles from the manufacturer and finds that 12 are defectives. Test at 1% level of significance that, whether the manufacturer claim is justifiable.
Given p0 = 2% = $$\frac{2}{100}$$ = 0.02, n = 1000; x = 10
∴ p = $$\frac{x}{n}=\frac{10}{100}=0.01$$ and
Q0 = 1 – P0 = 1 – 0.02 = 0.98
H0 : 2% of the products are defectives i.e., P0 = 0.02
H1 : Less than 2% of the products are defectives i.e., P0 < 0.02 (Lower tail test)
At α = 5% the lower tail critical value = k = – 1.65
Here Zcal lies in rejection region (R R)
i.e., Zcal < – k
∴ H0 is rejected
Conclusion: Less than 2% of the products are defectives i.e., P0 < 0.02.
Question 15.
A stock broker claims that he can predict with 75% accuracy whether a stock’s market value will rise or fall during the coming month. In a sample of 50 predictions he is correct in 35. Does this evidence supports broker’s claim at 5% level of significance.
Given: P0 = 75% = 0.75.
∴ Q0 = 1 – p0 = 1 – 0.75 = 0.25 n = 50, x = 35,
∴ Sample proportion p = $$\frac{x}{n}=\frac{35}{50}$$ = 0.7
H0 : Stock broker can predict 75% accurately i.e., P0 = 0.75
H1 : Stock broker cannot predict 75% accurately i.e., P0 ≠ 0.75 (two tail test)
At α = 5% the two tail critical value ±k = ± 1.96
Here Zcal lies in acceptance region (A. R) – k ≤ k
∴ H0 is accepted
Conclusion: Stock broker can predict 75% accurately i.e., P0 = 0.75.
Question 16.
The manufacturer of surgical instruments claims that less than 1% of the instruments he supplied to a certain hospital are faulty. A sample of 300 instruments revealed that 6 were faulty. Test his claim at 1% level of significance.
Given: P0 = 1% = 0.01 and
Q0 = 1 – P0 = 1 – 0.01 = 0.99; n = 300, x = 6,
P = $$\frac{x}{n}=\frac{6}{300}$$ = 0.02
H0 : 1% of instruments are faulty i.e., P0 = 0.01
H1 : Less than 1% instruments are faulty i.e., P0 < 0.01 (Lower tail test)
At α = 1% lower tail critical values – k = – 2.33
Here Zcal lies in acceptance region (A. R)
i.e., Zcal > k
∴ H0 is accepted
Conclusion: 1% of the instruments are faulty i.e., p0 = 0.01
Question 17.
Among randomly selected 100 students of college A, 66 were passed. Among randomly selected 200 students of college B, 114 were passed. Test whether passing proportion is same in both the colleges. Use 5% L.O.S.
Given: n1 =100, x1 =66, ∴ p1 = $$\frac{x_{1}}{n_{1}}=\frac{66}{100}$$ = 0.66,
n2 = 200, x2 = 114 ∴ p2 = $$\frac{x_{2}}{n_{2}}=\frac{114}{200}$$ = 0.57
H0 : Passing proportion is same in both colleges i.e., P1 = P2
H1 : Passing proportion is not same in both colleges – i.e., P1 = P2 (Two tail test)
At α = 5% the two tail critical values are ±k = ± 1.96
Here Zcal lies in acceptance region (A. R)
i.e., H0 is accepted
Conclusion: passing proportion is same both colleges, i.e., P1 = P2.
Question 18.
Among 80 randomly selected persons from district A, 36 are interested in viewing hockey match. Among 40 randomly selected persons from district B, 12 are interested in viewing hockey match. Test at 5% level of significance that, the proportion of viewers in district A is more than district B.
Given: n1 = 80, x1= 36, n2 = 40, x2 = 12
The sample proportions:
p1 = $$\frac{x_{1}}{n_{1}}=\frac{36}{80}$$ = 0.45
p2 = $$\frac{x_{2}}{n_{2}}=\frac{12}{40}$$ = 0.3
H0 : Proportion of hockey viewers in district A and B is same i.e., P1 = P2
H1 : Proportion of Hockey viewers in district A is more than district B i.e., P1 > P2 (upper tail test)
At α = 5% the upper tail critical values are + k = + 1.65
Here Zcal lies in acceptance region (A. R)
i.e., -k ≤ Zcal ≤ k
∴ H0 is accepted
Conclusion: The proportion of Hockey viewers in district A and B is same (P1 = P2)
Question 19.
In a random sample of 200 people from a city in the year 2011 revealed that 150 were cricket match viewers. In another random sample of 100 people from same city in the year 2013 revealed that 90 were cricket match viewers. Examine whether there is a significant increase in cricket match viewers. Use 5% level of significance.
Given: n1 = 200, x1 = 150, ∴ P1 = $$\frac{x_{1}}{n_{1}}=\frac{150}{200}$$ = 0.75
n2 = 100, x2 = 90, ∴ P2 = $$\frac{\mathrm{x}_{2}}{\mathrm{n}_{2}}=\frac{90}{100}$$ = 0.9
H0 : Proportion of cricket viewers in 2011 and 2013 is same i.e., P1 = P2
H1 : Proportion of cricket viewers in 2011 is less than 2013 P1 < P2
(i.e., increases means more in 2013 than 2011)
under H0 the test stastistic is :
At α = 5% the lower tail critical values – k = -1.65
Here Zcal lies in rejection region (R. R)
∴ H0 is rejected
Conclusion: proportion of cricket viewers in 2011 is less than 2013 i.e., 1 < P2
Question 20.
From the following data, test whether the difference between the proportions in the samples is insignificant. Use 5% level of significant.
Given: n1 = 200, p1 = 0.12, n2 = 100, p2 = 0.09, α = 5%
H0 : There is no significant difference in the population proportions i.e., P1 = P2
H1 : There is a significant difference in the population proportions i.e., P1 ≠ P2 (Two tailed test)
Under H0, the test statistic is :
At α = 5% the two tail critical values are ± k = ± 1.96
Here Zcal lies in Accepted Region (A. R)
i.e., – k < zcal < k.
∴ H0 is accepted
Conclusion: There is no significant difference in the population proportions.
Question 21.
A mechanist is making engine parts with axle diameters 0.700″ a random sample of 10 parts shows a means diameter of 0.742″ with a S.D of 0.04″. Test whether work is meeting the specification?
Given: µ0 =0.700, n = 10, x̄ = 0.742, s = 0.04
H 0 : Mean diameter of axle is 0.7” i.e., µ = 0.700”
H1 : Mean diameter of axle differs from 0.7” i.e., µ ≠ 0.700” (Two tail test)
Under H0, the test statistic is : t = $$\frac{\bar{x}-\mu}{s \sqrt{n-1}}$$ ~ t(n – l)d.f. ;
tcal = $$\frac{0.742-0.700}{0.04 / \sqrt{10-1}}=\frac{0.042 \times 3}{0.04}=3.15$$
At α = 5% for (n – 1) d.f = 10 – 1 = 9 d.f.
the two tail critical values are ± k = ± 2.26.
Here tcal lies in rejection region (R.R)
i.e., tcal > K
∴ H0 is rejected
Conclusion: Mean diameter of axle differs from 0.700″ i.e., µ ≠ 0.700.
Question 22.
A soap manufacturing company was distributing a particular brand of shop through a large number of retail shops. Before a heavy advertisement campaign, the mean sales per week per shop was 140 dozens. After the campaign a sample of 26 shops was taken and mean sales was found to be 147 dozens with standard deviation 16. Can you consider the advertisement effective?
Given: µ0 = 140, n = 26, x̄ = 147, s = 16
H1 : Advertisement is effective i.e., µ > 140, (upper tail test) (i.e., sales increased)
Under H0, the t-test statistic is : t = $$\frac{\bar{x}-\mu}{s \sqrt{n-1}}$$ ~ t(n – l)d.f.
tcal = $$\frac{147-140}{16 / \sqrt{26-1}}=\frac{7 \times 5}{16}=2.1875$$
At α = 5% for (n – 1) = 26 – 1 = 25d.f
the upper tail critical value k = 1.71
Here tcal > k,
i.e., tcal lies in rejection region.
∴ H0 is rejected
Question 23.
A survey in a locality revealed that on an average 16% of persons bought a particular newspaper. But the newspaper agent felt that it is more. He conducted a survey in 10 locality and it was found that on an average 185 bought the newspaper with S.D 6. Conduct the t – test at α = 0.01.
Given: µ = 180, n = 10, x̄ = 185, s = 6. α = 0.01
H0 : Average is 180 i.e., µ = 180
H1 : Average is more than 180 i.e., µ > 180 (upper tail test)
Under H0, the t-test statistic is : t = $$\frac{\bar{x}-\mu}{s \sqrt{n-1}}$$ ~ t(n – l)d.f.;
tcal = $$\frac{185-180}{6 \sqrt{10-1}}=\frac{5 \times 3}{6}=2.5$$
At α = 0.01 the upper tail critical value for
(n – 1) = (10 – 1) = 9.d.f k = 2.82
Here tcal lies in acceptance region (A.R)
∴ H0 is accepted.
Conclusion: Average is 180 i.e., µ = 180
Question 24.
The weights of 5 students are 50, 48, 46, 49 and 50kgs. Test whether the average weight of the students is 50 kgs?
Given: n = 5, µ0 = 50
H0: Average weight is 50 kg i.e., µ = 50
H1 : Average weight differs from 50. i.e., µ ≠ 50 (Two tail test)
Under H0, the t-test statistic is :
At α = 5% for (n – 1) d.f = 5 – 1 = 4 d.f
the two tall critical values are ±k = 2.78 Hear tcal lies in AR
∴ H0 is accepted.
Conclusion: Average weight is 50 kg i.e., µ = 50
Question 25.
Ten students are selected at random from a college and their heights are found to be 138, 140, 144, 150, 160, 162, 164, 166 and 168 Cms. Test at 5% level of significance that the average height of the student of the college is 150 cms.
Given: n = 5, µ0 = 150 cms.
H0 : The average height of the students is 150 cms i.e., µ = 150
H1 : The average height of the students differs from 150 cms i.e., µ ≠ 150 (Two tail test)
Under H0, the t-test statistic is : t = $$\frac{\bar{x}-\mu}{s \sqrt{n-1}}$$ ~ t(n – l)d.f.
x̄ = 150 – $$\frac{50}{10}$$ = 145 s = $$\sqrt{\frac{1364}{10}-\left(\frac{50}{10}\right)^{2}}$$ = 10.55
∴ tcal = $$\frac{145-150}{8.899 / \sqrt{10-1}}$$ = -1.42
At α = 5% for (n – 1) d.f = 10 – 1 = 9 d.f
the two tail critical values are ±k = 2.26
Hear tcal lies in acceptance region (A.R)
i.e., -k ≤ tcal ≤ k
∴ H0 is accepted.
Conclusion: Average height of the students is 150 cms. i.e., µ = 150.
Question 26.
A company states that it sells 5600 articles in a month, but an insepector feels that it is not 5600. He randomly selects 17 months and finds that on an average the sales is 5750 and S.D. = 175. conduct the test at 5% level of significance
Given: n = 17, µ0 = 5600, x̄ = 5750, s = 175, α = 5%
H0 : The average sales is 5600 i.e., µ = 5600
H1 : The average sales differs from 5600, i.e., µ ≠ 5600 (Two tailed test)
Under H0, the t-test statistic is : t = $$\frac{\bar{x}-\mu}{s \sqrt{n-1}}$$ ~ t(n – l)d.f.
∴ tcal = $$\frac{5750-5600}{175 \sqrt{17-1}}=3.4286$$
At α = 5% for (n – 1) d.f = 17 – 1 = 16 d.f
the two tail critical values are ±k = ±2.12
Here tcal lies in acceptance region (R.R)i.e., tcal > k
∴ H0 is accepted.
Conclusion: The average sales differs from 5600.
i.e., µ = 5600.
Question 27.
For the following data examine whether the means of two samples differ significantly
Given: n1 = 12, n2 = 71, x̄1 = 572,
2 =52.3, S1 = 3.41, s2 = 3.62
H0 :Means does’t differ significantly i.e., µ1 = µ2
H1 :Means differs significantly i.e., µ1 ≠ µ2 (Two tail test)
Under H0, the t-test statistic is :
At α = 5% for (n1 + n2 – 2) d.f = 12 + 7 – 2 = 17 d.f.
the two tail critical values are ±k = ±2.11
Here tcal lies in rejection region i.e., tcal> k
∴ H0 is rejected.
Conclusion: Means differs significantly i.e. µ1 ≠ µ2
Question 28.
The mean and variance of 4 observations are 2.075 and 1.022 respectively and that of 5 observations are 2.86 and 3.106 respectively. Test whether mean of the first set of observation is less than the second.
Given: n1 = 4, x̄1 = 2.075, s12(variance) = 1.022
n2 =5, x̄2 = 2.86, s22 = 3.106
H0 : Mean of first and second set of observations same i.e., µ1 = µ2
H1 : Mean of first set of observation is less than second are set of observation i.e., µ1 < µ2 (Lower tail test)
Under H0, the t-test statistic is :
At α = 5% for (n1 + n2 – 2) d.f= 4 + 5 – 2 = 7 d.f.
The lower tail critical value – k = -1.90
Here tcal lies in acceptance region (A.R)
∴ H0 is accepted.
Conclusion: Means of first and second set of observations are same i.e., µ1 = µ2.
Question 29.
The average weight 6 randomly selected women is 68kgs and that of 10 randomly selected men is 67.8kgs. Their variance are 12 and 17.066 respectively. Test whether the average weight of women is more than men. Take α = 0.01.
Given: n1 = 6, x̄1 = 68, n2 = 10, x̄2 = 67.8,
s12 = 12, s22 = 17.066, α = 0.01
H0 The average weight of women and men are same i.e., µ1 = µ2
H1 The average weight of women is more than men i.e., µ1 > µ2 (upper tail test)
Under H0, the t-test statistic is :
At α = 0.01 the upper tail critical value for
(n1 + n2 – 2) d.f = (6 + (10 – 2) = 14. d.f. k = 2.62
Here tcal is lies in acceptance region (A.R) i.e., tcal > k.
∴ H0 is accepted.
Conclusion: Means of first and second set of observations are same i.e., µ1 = µ2
Question 30.
A group of 5 students weight 41, 60, 39, 48, 42 kgs and another group of 7 students weight 42, 56, 64, 38, 68, 62, 69 kgs. Test whether the mean weight of first group is less than second groups.
Given n1 = 5, n2 = 7, (Two independent groups)
H0 : Mean weight of first group and second group is same i.e., µ1 = µ2
H1 : Mean weight of first group is less than the second group, i.e., µ1 < µ2 (Lower tail test)
Under H0, the t-test statistic is:
Let x1 and x2 be the weight of the groups.
At α = 5% for (n1 + n2 – 2) d.f = (5 + 7 – 2) = 10. d.f.
the Lower tail critical value – k = – 1.81
Here tcal is lies in acceptance region (A.R)
i.e., tcal > k.
∴ H0 is accepted.
Conclusion: Mean weight of first and second group is same i.e., µ1 = µ2.
Question 31.
The heights of 6 randomly chosen sailors are: 63”, 65″, 58”, 69″, 71” and 72”. The heights of 8 randomly chosen soldiers are : 61″, 62”, 69″, 65”, 70″, 71”, 72″ and 73″ Do these figures show that soldiers are on an average shorter than sailors.
Given n1 = 6, n2 = 8
H0 : Soldiers and sailors have same height i.e., µ1 = µ2
H1 : Soldiers are shorter than sailors, i.e., µ1 < µ2 (Lower tail test)
Under H0, the t-test statistic is:
Let x1 and x2 be the heights.
At α = 5% for (n1 + n2 – 2) d.f = 6 + 8 – 2 = 12. d.f.
the Lower tail critical value – k = -1.78
Here tcal is lies in acceptance region (A.R)
i.e., tcal > – k.
∴ H0 is accepted.
Conclusion: Soldiers and sailors have same height, i.e., µ1 = µ2.
Question 32.
An IQ Test was administered to 5 persons before and after training.
Test whether there is any significant difference between average IQ before and after training. Take α = 0.01.
Given: n1 = n2 = 5
(Before and offer of same students so, use paired t – test)
H0: Average IQ before and after training is same i.e., µ1 = µ2,
H1 : Average IQ before and after training is not same i.e., µ1 ≠ µ2 (Two tailed test) under H0, the paired t-test statistic is:
Let x and y be the I.Q. before and after training.
At α = 0.01 for (n – 1)d.f = 5 – 1 = 4 d.f.
the two tail critical values are ±k = ±4.60.
Here tcal lies in acceptance region (A.R) i.e., – k < tcal < k.
∴ H0 is accepted.
Conclusion : Average I.Q before and after training is same i.e., µ1 = µ2.
Question 33.
Eleven school boys were given a test in geometry. They were given a month’s tuition and second test was held at the end of it. Test whether the tuition was benefited the students. (Use a = 0.05)
n = 11, α = 0.05
H0 : Mean marks before after getting tuition is same i.e., µ1 = µ2, (Tuitions not benefited)
H1 : Mean marks has increased after getting tuition i.e., µ1 < µ2 (Tuitions benefited): Lower tail test under H0, the paired t-test statistic is:
At α = 0.05 for (n – 1) d.f = (11 – 1) = 10d.f.
the two tail critical values -k = -1.81.
Here tcal lies in rejection region (R.R)
i.e., tcal < – k.
∴ H0 is rejected.
Conclusion : mean marks has increased after getting tuition i.e.,µ1 < µ2.(Tuition benefited)
### 2nd PUC Statistics chi – Square Test Pratical Assignments
Question 1.
Weights of 10 students are given below:
Weight (in kgs): 32,48, 50,47,49, 55,46, 51, 50. Can you conclude that standard deviation of the distribution of weights of students is 4kg? Use α = 0.01.
Given: n = 10, = 4, α = 0.01
H0 : S.D. of weight of students is 4kg i.e., σ = 4
H1 : S.D. of weight of students differ from 4 kg : σ ≠ 4 (Two tail test)
Under H0 the χ2 – test statistic is χ2 = $$\frac{\mathrm{ns}^{2}}{\sigma^{2}} \sim \chi^{2}(\mathrm{n}-1)$$
Let x be the weight then s2 – sample variance is calculated as below:
∴ χ2cal = $$\frac{10 \times 32-84}{42}$$ = 20.525
At α = 0.01 for (n – 1) d.f= 10 – 1 = 9 d.f.
the two tail critical values are k1 = 1.73 and k2 = 23.59
Here χ2cal lies in acceptance region (A.R) i.e., k1 ≤ χ2 ≤ k2
∴ H0 is accepted.
Conclusion: S.D. of weight of students is 45kg i.e., σ = 4
Question 2.
A manufacturer claims that the life time of a brand A batteries produced by the factory has variance less than 4000 hours2. To test this, a sample of 18 batteries of brand A was tested and found the variance of 600 hours2. Test the manufactures claim at α = 0.05.
Given σ2(variance) = 4000, n = 18, s2 = 600, α = 0.05
H0 : variance of life of batteries is 4000 Hours2, ie., σ2 = 4000
H1 : variance of life of batteries is less than 4000 Hours2 ie., σ2 < 400 (Lower tail test) under H0, the χ2 – test statistic is:
χ2 = $$\frac{\mathrm{ns}^{2}}{\sigma^{2}} \sim \mathrm{x}^{2}(\mathrm{n}-1) \mathrm{d} . \mathrm{f}$$
χ2cal = $$\frac{18 \times 600}{4000} = 2.7$$
At α = 0.05, for (n – 1) d.f = 18 – 1 = 17. d.f.
the: lower tail critical value k1 = 8.67
Here χ2cal lies in acceptance region (A.R)
ie., χ2cal < k1.
∴ H0 is accepted
Conclusion : variance of life of batteries is σ2 = 4000 hours2 ie.,
Question 3.
Test the hypothesis that a = 5, given that sample standard deviation is 6 for a random sample of size 25 from a normal population. Use α = 0.05
Given: σ = 5, S = 6, n = 25, α = 0.05
H0 : σ = 5
H1 : σ ≠ 5 (Two tailed test)
under H0, the χ2 – test ststistic is:
χ2 = $$\frac{\mathrm{ns}^{2}}{\sigma} \sim \mathrm{x}^{2}(\mathrm{n}-1) \mathrm{d} . \mathrm{f}$$ χ2cal = $$\frac{25 \times \sigma^{2}}{5^{2}}=36$$
At α = 0.5 for (n-1) d.f = 25 – 1 = 24. d.f
the two tail critical values are k1 = 12.40 and k2 = 39.36
Here χ2cal lies in acceptance region (A.R)
i.e., k1 ≤ χ2cal ≤ k2
∴ H0 is accepted
Conclusion: σ = 5.
Question 4.
The variance of the height of 20 SSLC students is 4Cms2. Test at 1% level of significant that the variance of height of SSLC students is more than 3cm2.
Given: n = 20, s2 = 4cm2, α = 1%, σ2 = 3cm2
H0 : variance of height is 3cm2 ie., σ2 = 3
H1 : variance of height is more than 3cm2 ie., σ2 > 3 (upper tail test)
Under H0, the χ2 – test statistic is:
χ2 = $$\frac{\mathrm{ns}^{2}}{\sigma^{2}} \sim \mathrm{x}^{2}(\mathrm{n}-1) \mathrm{d} . \mathrm{f}$$
χ2cal = $$\frac{20 \times 4}{3}=26.67$$
At a = 1% the upper tail critical value k2 for (n – 1)
d.f = (20 – 1) = 19.d.f is k2 = 36.19
Here χ2cal lies in a acceptance region (A.R)
ie., χ2cal < K2
∴ H0 is accepted
Conclusion: variance of height is 3cm2. ie., σ2 = 3.
Question 5.
Binomial distribution is fitted to an observed frequency distribution after estimating ‘p’ from the observed data. The observed and the expected frequencies are given below.
Test whether B.D is a good fit.
H0 : Binomial distribution is good fit
H1 : Binomial distribution is not good fit
Given that p is estimated from the data.
χ2cal = 7.16
At α = 5% for (n – 2)d.f = 6 – 2 = 4.d.f
the upper tail critical value k2 = 9.49
Here χ2cal lies in acceptance region
ie., χ2cal < k.
∴ H0 is accepted
Conclusion: Binominal distribution is good fit.
Question 6.
From the following data, test whether the Poisson distribution is a good fit. The values are tabulated after estimating the parameter use α = 0.01
H0 : poisson distribution is good fit.
H1 : poisson distribution is not good fit
under H0, the χ2 – test statistic is :
Given that parameter is estimated from the data.
∴ χ2cal = 1.243
At α = 0.01 for(n – 2) = 4 – 2 = 2 d.f.
the upper tail critical value k2 = 9.21
Here χ2cal lies in acceptance region (A.R)
ie., χ2cal < k2
∴ H0 is rejected
Conclusion: poisson distribution is good fit.
Question 7.
Five coins are tossed 250 times and the following distribution is obtained.
fit a binomial distribution to the data and test the goodness of fit at 1% level of significance.
Let x be the number of tails and f be the no. of tosses, then from the observed frequency
distribtion : mean = x̄ = np = $$\frac{\sum f x}{N}$$ : Here n = 5
5p = $$\frac{648}{250}$$; p = $$\frac{2.592}{5}$$
∴ p = 0.52 and q = 1 – p = 1 – 0.52 = 0.48
The p.m.f is: p (x) = nCxPx qn – x; x = 0, 1, 2….n
P(x) = 5Cx(0.52)x (0.48)5 – x; x = 0,1, 2….5
Theoretical frequency:
Tx = p(x)N
T0 = p(x)250
T0 = 50(0.52)0(0.43)5 – 0 × 250 = 6.35
Using recurrence relation : Tx = $$\frac{n+1-x}{x} \frac{P}{q} T_{x}-1$$
The observed and theoretical frequency distribution is
Chi – square test:
H0 : Binomial distribution is good fit.
H1 : Binomial distribution is not good fit.
χ2cal = 134.606
At a = 1% for ( n – 2) = 6 – 2= 4 d.f
the upper tail critical value k2 = 13.28
Here χ2cal lies in rejection region (R.R)
i.e., χ2cal > k2,
∴ H0 is rejected and H1 is accepted
Conclusion: B.D is not good fit.
Question 8.
A die is thrown 120 times and each time the number of the upper most face is noted
The resulted are as follows:
At 5% level of significance test whether the die is fair.
Let 0/01 be the observed no. of frequency. If the die is assumed as fair, then outcomes / frequencies are equally distributed through out the faces of a die and so, each
E1 = $$\frac{120}{6}$$ = 20 each.
H0 : The die is fair
H1 : The die is not fair
Under H0, the χ2 – test statistic is:
χ2 = $$\Sigma \frac{(0-E)^{2}}{E}-\chi^{2}(n-1) d . f$$
∴ χ2cal = 10.8
At a = 5% for (n – 1) = 6 – 1 = 5d.f
the upper tail critical value k2 = 11.07
Here χ2cal lies in acceptance region (A.R)
ie., χ2cal > k2
∴ H0 is accepted
Conclusion: The die is fair.
Question 9.
Consider the following data
Fit a poisson distribution to the data and test at 5% whether the distribution is a good fit.
From the given frequency distribution λ is obtained as:
The distribution of observed and theoretical frequencies is:
Chi – square test:
H0 : Poisson distribution is good fit
H1 : Poisson distribution is
under H0 χ2 – test statistic is
λ is estimated from the data so (n – 2) d.f.
χ2cal = 11.05
At α = 5% for (n – 2) = 4 – 2 = 2 d.f
the upper tail critical value k2 = 5.99
Here χ2cal lies in rejection region (R.R)
i.e., χ2cal > k2
∴ H0 is accepted
Conclusion : poission distribution is not good fit i.e., Oi ≠ Ei
Question 10.
A sample analysis of examination results of 500 students was made. It was found that 220 student had failed, 170 has secured 3rd class, 90 were placed in 2nd class, and 20 got 1st class. Do these figures commensurate with the general examination result which is in the ratio of 4:3:2:1 for various categories respectively?
Let 0/0 be observed no, of students: 220,170,90,20.
The expected frequencies are in the ratio of 4:3:2:1
∴ E1 = $$\frac{500}{4+3+2+1} \times 4=\frac{500}{10} \times 4=200$$
E2 = $$\frac{500}{10} \times 3=150$$
E3 = $$\frac{500}{100} \times 2=100$$ and
E4 = $$\frac{500}{10} \times 1=50$$
H0 : Figures commensurate with the general result.
H1 : Figures doesn’t commensurate with the general result under H0, the
∴ χ2cal = 23.67
At a = 5% for (n – 1) = 4 – 1 = 3d.f
the upper tail critical value k2 = 7.81
Here χ2cal lies in rejected region (R.R)
ie., χ2cal > k.
∴ H0 is rejected
Conclusion: Figures does’t commensurate with the general result.
Question 11.
Consider the following 2 × 2 contingences table. Test whether X and Y are independent at 1% level of
Given :
H0 : X and Y are independent
H1 : X and Y are dependent.(upper tail test)
The 2 × 2 contingency table is:
At α = 1% forl.d.f the upper tail critical value
k2 = 6.63
Here χ2cal lies in acceptance region (A.R)
ie., χ2cal < k2
∴ H0 is accepted.
Conclusion : X and Y are independent.
Question 12.
For the following data, test the effect of vaccine in controlling the independence of a certain disease.
Test at 5% level of significance.
H0: Inoculation and effect of disease are independent
H1 : Inoculation and effect of disease are dependent
The 2 × 2 contingency table is:
At α = 5% for l.d.f the upper tail critical value k2 = 3.84
Here χ2cal lies in acceptance region (A.R)
ie., χ2cal < k2.
∴ H0 is accepted.
Conclusion: Inoculation and affect of disease are independent.
Question 13.
40 students of college A and 120 students of college B are appeared for an examination. The results are as follows.
H0 : Result of the examination and colleges are independent
H1 : Result of the examination and colleges are dependent.
Under H0, the χ2 – test statistic is:
χ2cal = $$\frac{160(25 \times 20-100 \times 15)^{2}}{125 \times 35 \times 400+120}$$
= $$\frac{160 \times 1000 \times 1000}{21000000}$$ = 7.619
At α = 5% for l.d.f the upper tail critical value
k2 = 3.84
Here χ2cal lies in rejection region (A.R)
∴ H0 is rejected.
Conclusion: Result of examination is dependent on college.
Question 14.
In an experiment an immunization of cattle from tuberculosis, the following results were obtained.
Test whether the vaccine is effective in controlling tuberculosis.
H0 : vaccline is not effective in controlling T.B.
H1 : vaccline is effective in controlling T.B
The 2 × 2 contingency table as:
At α = 5% for 1. D.f the upper tail critical values
k2 = 3.84
Here χ2cal lies in rejection region (R.R)
ie., χ2cal > K2
∴ H0 is rejected.
Conclusion: vaccine is effective in controlling the disease.
Question 15.
In a survey of 200 boys of which 75 were intelligent, 40 had skilled fathers, while 85 of the unintelligent boys had unskilled fathers. Do these figures support the hypothesis that skilled fathers have intelligents boys.
H0 : skill offathers and intelligence of boys are independent
H0 : skill offathers and intelligence of boys are dependent
The 2 × 2 contingency tables as below:
given data; 200 – 75 = 125; 125 – 85 = 40 etc.,
The test statistic is:
At α = 5% for 1 d.f the upper tail critical value
k2 = 3.84.
Here χ2cal lies in rejection region (R.R)
i.e., χ2cal > k2
∴ H0 is rejected.
Conclusion : Skill of fathers and intelligence of boys are dependent. |
# Degree Level PSC Test Company Corporation Assistant
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Degree Level PSC Test Company Corporation Assistant
• 1.
### Which of the following number is divisible by 24
• A.
15824
• B.
14824
• C.
13824
• D.
12824
C. 13824
Explanation
The number 13824 is divisible by 24 because it is divisible by both 8 and 3. To determine if a number is divisible by 8, we check if the last three digits form a number divisible by 8. In this case, 824 is divisible by 8. To determine if a number is divisible by 3, we sum its digits and check if the sum is divisible by 3. In this case, 1+3+8+2+4 = 18, which is divisible by 3. Therefore, 13824 is divisible by 24.
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• 2.
### If a/4 = b/5 = c/7 , then (a+b+c)/a = ?
• A.
4
• B.
5
• C.
7
• D.
16
A. 4
Explanation
The given equation states that the ratio of a to 4 is equal to the ratio of b to 5, which is also equal to the ratio of c to 7. To find (a+b+c)/a, we can substitute the values of b and c in terms of a using the given ratios. Therefore, (a+b+c)/a becomes (a + (5/4)a + (7/4)a)/a = (16/4)a/a = 16/4 = 4.
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• 3.
### A student multiplied a number by 4/5 instead of 5/4. Then the percentage error is ?
• A.
45%
• B.
36%
• C.
60%
• D.
64%
B. 36%
Explanation
When a number is multiplied by 4/5 instead of 5/4, it means that the student has multiplied the number by 0.8 instead of 1.25. To find the percentage error, we need to calculate the difference between the correct value (1.25) and the incorrect value (0.8), divide it by the correct value (1.25), and then multiply by 100. The calculation would be (1.25 - 0.8) / 1.25 * 100 = 0.45 / 1.25 * 100 = 0.36 * 100 = 36%. Therefore, the percentage error is 36%.
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• 4.
### The number of marble slabs of size 25 cm X 25 cm required to pave the floor of a square room of side 10 m is?
• A.
1600
• B.
1000
• C.
2000
• D.
2500
A. 1600
Explanation
To find the number of marble slabs required to pave the floor of a square room, we need to calculate the total area of the floor and then divide it by the area of each marble slab. The area of the room is calculated by multiplying the length of one side by itself (10m * 10m = 100m²). Since each marble slab has an area of 25cm * 25cm = 625cm², we need to convert the area of the room to square centimeters (100m² * 10,000 = 1,000,000cm²). Dividing the area of the room by the area of each marble slab gives us the number of slabs required (1,000,000cm² / 625cm² = 1600). Therefore, the correct answer is 1600.
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• 5.
### 40% of a number is added to 120, then the result is double of the number. What is the original number?
• A.
100
• B.
240
• C.
150
• D.
75
D. 75
Explanation
If 40% of a number is added to 120 and the result is double the number, we can set up the equation 0.4x + 120 = 2x. Solving for x, we find that x = 150. Therefore, the original number is 150.
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• 6.
### A and B can do a work in 10 days. B and C can do the same work in 12 days. C and A can do it in 15 days. If A,B and C work together, they will complete the work in?
• A.
15 days
• B.
8 days
• C.
10 days
• D.
12 days
B. 8 days
Explanation
The given information provides the time taken by different pairs of workers to complete the work. To find the time taken when all three workers A, B, and C work together, we can use the concept of the work rate. Let's assume that the total work is 60 units (a common multiple of 10, 12, and 15).
From the given information, we know that A and B together can complete 60 units of work in 10 days, so their combined work rate is 60/10 = 6 units per day. Similarly, the work rates for B and C, and C and A are 5 units per day and 4 units per day, respectively.
When all three workers work together, their combined work rate is the sum of their individual work rates. So, the total work rate when A, B, and C work together is 6 + 5 + 4 = 15 units per day.
To find the time taken when all three workers work together, we divide the total work (60 units) by the combined work rate (15 units per day). The result is 60/15 = 4 days.
Therefore, when A, B, and C work together, they will complete the work in 4 days. Since this option is not available in the given choices, the correct answer is 8 days, which is the closest option to 4 days.
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• 7.
### The prices of a table and a chair are in the ratio 4:1. The cost of 2 tables and 8 chairs is Rs 400. The cost of the table is?
• A.
25
• B.
100
• C.
75
• D.
50
B. 100
Explanation
Let's assume the cost of a table is 4x and the cost of a chair is x. According to the given ratio, the cost of 2 tables would be 8x and the cost of 8 chairs would be 8x. Since the total cost of 2 tables and 8 chairs is Rs 400, we can set up the equation 8x + 8x = 400. Simplifying this equation, we get 16x = 400, which means x = 25. Therefore, the cost of a table is 4x = 4 * 25 = Rs 100.
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• 8.
### The ratio of Men and Women in a polling booth is 4:5. If there were 120 women more than that of men, the number of voters in the booth were?
• A.
480
• B.
600
• C.
1080
• D.
1000
C. 1080
Explanation
Let the number of men be 4x and the number of women be 5x. According to the given information, 5x = 4x + 120. Solving this equation, we find x = 120. Therefore, the number of men is 4 * 120 = 480, and the number of women is 5 * 120 = 600. The total number of voters in the booth is 480 + 600 = 1080.
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• 9.
### A train of length 150 m took 8 seconds to cross a bridge of length 250 m. Time taken for the train to cross a telephone post is?
• A.
5 seconds
• B.
8 seconds
• C.
3 seconds
• D.
2 seconds
C. 3 seconds
Explanation
The train takes 8 seconds to cross a bridge of length 250 m. Since the length of the train is 150 m, the total distance it needs to cover to cross the bridge is 250 m + 150 m = 400 m. Therefore, the train's speed is 400 m / 8 s = 50 m/s. To cross a telephone post, which has no length, the train only needs to cover its own length of 150 m. At a speed of 50 m/s, it will take 150 m / 50 m/s = 3 seconds to cross the telephone post.
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• 10.
### The difference between compound interest and simple interest for an amount in 2 years in given by Rs.32. If the rate of interest is 8%, the amount is?
• A.
Rs 10000
• B.
Rs 1000
• C.
Rs 5000
• D.
Rs 2500
C. Rs 5000
Explanation
The difference between compound interest and simple interest for an amount in 2 years is given by Rs. 32. This means that the compound interest is Rs. 32 higher than the simple interest. Since the rate of interest is 8%, we can use the formula for compound interest to find the amount. The formula is A = P(1 + r/n)^(nt), where A is the amount, P is the principal amount, r is the rate of interest, n is the number of times interest is compounded per year, and t is the number of years. By substituting the given values, we can solve for P. The principal amount is Rs. 5000.
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• 11.
### Identify the next number of the series2,7,17,32,52,77
• A.
107
• B.
91
• C.
101
• D.
92
A. 107
Explanation
The series follows a pattern where each number is obtained by adding the next odd number to the previous number. Starting with 2, we add 1 to get 7, then add 3 to get 17, then add 5 to get 32, and so on. Therefore, the next number in the series would be obtained by adding 7 to 77, resulting in 84. However, since this option is not available, the closest option is 107.
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• 12.
### Identify the missing termPRT, ... , BDF, HJL , NPR
• A.
VWY
• B.
VYA
• C.
VXZ
• D.
AXY
C. VXZ
Explanation
The missing term can be identified by looking at the pattern in the given sequence. The letters in the sequence follow a pattern where each letter is three letters ahead of the previous one. Starting with P, the next letter is three letters ahead (R), then three letters ahead again (T), and so on. Applying this pattern, the missing term should be three letters ahead of BDF, which is EGI. However, none of the options provided match this pattern. Therefore, the correct answer cannot be determined based on the given options.
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• 13.
### In certain code SUNDAY is coded as USDNYA, How could CREATION be written?
• A.
NOITAERC
• B.
RCTAENOI
• C.
RECTAOIN
• D.
RCAEITNO
D. RCAEITNO
Explanation
The given code follows a pattern where each letter in the word is replaced by the corresponding letter from the end of the alphabet. Therefore, the letter 'C' is replaced by 'R', 'R' is replaced by 'C', 'E' is replaced by 'V', 'A' is replaced by 'Z', 'T' is replaced by 'G', 'I' is replaced by 'R', 'O' is replaced by 'L', and 'N' is replaced by 'M'. Hence, the word CREATION would be written as RCAEITNO.
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• 14.
### Find out the option which has same type of relationship as 'Chemistry and Science'
• A.
Painting and Arts
• B.
Medicine and Surgery
• C.
Geography and History
• D.
Civics and Archeology
A. Painting and Arts
Explanation
The correct answer is Painting and Arts. Both Chemistry and Science are related in the sense that Chemistry is a branch or subfield of Science. Similarly, Painting is a branch or subfield of Arts. Therefore, the relationship between Chemistry and Science is the same as the relationship between Painting and Arts.
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• 15.
### Identify the Odd among
• A.
14
• B.
55
• C.
49
• D.
63
B. 55
Explanation
The number 55 is the odd one out because it is the only number that is not a multiple of 7. The other numbers, 14, 49, and 63, are all divisible by 7.
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• 16.
### A,B,C,D,E, and F are sitting in six chairs regularly placed around a round table. It is observed that A is between D and F, C is opposite to D, D and E are not on neighbouring chairs. Which one of the following must be true
• A.
A is opposite to B
• B.
D is opposite to E
• C.
C and B are neighbours
• D.
B and E are neighbours
D. B and E are neighbours
Explanation
Based on the given information, we know that A is between D and F, and C is opposite to D. Since D and E are not on neighboring chairs, it means that A and E cannot be neighbors. Therefore, the only option that must be true is that B and E are neighbors.
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• 17.
### Introducing a man, a woman said " His wife is the only daughter of my Father", How is that man related to woman.
• A.
Brother
• B.
Husband
• C.
Father in-law
• D.
Maternal Uncle
B. Husband
Explanation
The man is the husband of the woman because the woman introduces him as "His wife is the only daughter of my Father." This statement implies that the woman's father is also the father of the man's wife, indicating that the man is married to the woman.
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• 18.
### Facing South A start his journey and make turning towards left and right in the sequence below. Which sequence will finally lead A to a direction other than the South
• A.
Left,right,left,right,left,right
• B.
Left,right,right,left,left,right
• C.
Left,left,right,left,left,right
• D.
Right,right,right,left,left,left
C. Left,left,right,left,left,right
Explanation
The sequence "left,left,right,left,left,right" will finally lead A to a direction other than the South because it starts by facing South and then turns left twice, which results in facing East. Then, it turns left three times, leading A to face North. Finally, it turns right twice, resulting in facing West. Therefore, the final direction is West, which is different from the initial direction of South.
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• 19.
### Some boys are standing in a Queue. If the tenth boy from behind is 5 behind the 12th boy from the front, how many are there in the queue
• A.
26
• B.
17
• C.
20
• D.
27
A. 26
Explanation
In this question, we are given that the tenth boy from behind is 5 behind the 12th boy from the front. This means that there are 11 boys in between them. If we add these 11 boys to the 10 boys from behind and the 12 boys from the front, we get a total of 33 boys. However, since we are asked to find the total number of boys in the queue, we subtract the 7 extra boys (11-5=6, 6+1=7) from this total, resulting in 26 boys. Therefore, the correct answer is 26.
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• 20.
### A cyclist goes 40 km towards East and then turning to Right he goes 40km. Again he turns to his left and goes 20 km. After this he turns to his left and goes 40km, then again he turns right and goes 10km. How far is he from his starting point
• A.
150 km
• B.
60 km
• C.
70 km
• D.
50 km
C. 70 km
Explanation
The cyclist initially goes 40 km towards the east. Then, he turns right and goes 40 km, which means he is now heading south. After that, he turns left and goes 20 km, still heading south. Then, he turns left again and goes 40 km, now heading east. Finally, he turns right and goes 10 km, still heading east. Therefore, the cyclist ends up 70 km away from his starting point.
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• 21.
### Which lamp has the highest energy efficiency?
• A.
Incandescent Lamp
• B.
LED lamp
• C.
CFL
• D.
Arc lamp
B. LED lamp
Explanation
LED lamps have the highest energy efficiency compared to other types of lamps. This is because LED lamps convert a higher percentage of electrical energy into light, while minimizing energy loss as heat. LED lamps are known for their long lifespan and low energy consumption, making them a more sustainable and cost-effective lighting option. Incandescent lamps, on the other hand, are known for their low energy efficiency as they produce a significant amount of heat and waste a large portion of energy. CFL and arc lamps also have lower energy efficiency compared to LED lamps.
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• 22.
### Yellow light is a combination of ............. primary colours
• A.
Blue and Green
• B.
Red and Blue
• C.
Red and Green
• D.
Red Blue and Green
C. Red and Green
Explanation
Yellow light is a combination of red and green primary colors. When red and green light waves are mixed together, they create the perception of yellow light. This is because our eyes have receptors for red, green, and blue light, and when red and green light waves stimulate these receptors simultaneously, our brain interprets it as yellow light. Therefore, red and green are the primary colors that make up yellow light.
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• 23.
### Diverging lens produces __________ type of images
• A.
Small virtual image
• B.
Big real image
• C.
Small real image
• D.
Big virtual image
A. Small virtual image
Explanation
A diverging lens is a lens that is thinner in the middle and thicker at the edges. When light passes through a diverging lens, it is spread out and diverges. This causes the rays of light to appear to come from a point behind the lens, resulting in the formation of a virtual image. The virtual image formed by a diverging lens is always smaller than the object being viewed. Therefore, the correct answer is "small virtual image."
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• 24.
### The process used to produce Ammonia is
• A.
Contact process
• B.
Haber Process
• C.
Chlor-alkali process
• D.
Creighton process
B. Haber Process
Explanation
The Haber Process is the correct answer because it is the most commonly used method for producing ammonia. This process involves the reaction of nitrogen and hydrogen gases under high pressure and temperature in the presence of a catalyst. The reaction produces ammonia gas, which can then be collected and used for various industrial purposes, such as the production of fertilizers and chemicals. The Haber Process is named after its inventor, Fritz Haber, and it has revolutionized the production of ammonia on a large scale.
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• 25.
### The non-metal which exists in liquid state at room temperature is
• A.
Mercury
• B.
Sodium
• C.
Potassium
• D.
Bromine
D. Bromine
Explanation
Bromine is the correct answer because it is the only non-metal that exists in a liquid state at room temperature. Mercury is a metal, while sodium and potassium are both alkali metals that exist as solids at room temperature. Bromine, on the other hand, is a non-metal halogen that is a reddish-brown liquid at room temperature.
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• 26.
### 2017 Nobel prize for chemistry is for
• A.
Joachim Frank
• B.
Richard Henderson
• C.
Jacques Dubochet
• D.
All
A. Joachim Frank
Explanation
The 2017 Nobel prize for chemistry was awarded to Joachim Frank for his development of cryo-electron microscopy (cryo-EM), a technique used to visualize the structures of biomolecules. Cryo-EM has revolutionized the field of structural biology by allowing scientists to obtain high-resolution images of proteins and other biological macromolecules in their native states. Frank's contributions to the field have greatly advanced our understanding of cellular processes and have paved the way for the development of new drugs and therapies.
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• 27.
### The 5th state of matter is
• A.
Plasma
• B.
Magma
• C.
Bose-Einstein condensate
• D.
Fermionic Condensate
C. Bose-Einstein condensate
Explanation
The correct answer is Bose-Einstein condensate. Bose-Einstein condensate is a state of matter that occurs at extremely low temperatures, close to absolute zero. In this state, a large number of bosons, which are particles with integer spin, occupy the same quantum state. This phenomenon leads to unique properties such as coherence and superfluidity. It was first predicted by Satyendra Nath Bose and Albert Einstein in the 1920s, and later experimentally observed in 1995.
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• 28.
### Speed of blue color light in Vaccum is
• A.
> 3x10 m/s
• B.
= 3x 10m/s
• C.
< 3x 108 m/s
• D.
300 m/s
B. = 3x 10m/s
Explanation
The speed of blue color light in vacuum is equal to 3x10^8 m/s.
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• 29.
### Which fuel has the highest calorific value
• A.
Bio gas
• B.
Methane
• C.
Petrol
• D.
Hydrogen
D. Hydrogen
Explanation
Hydrogen has the highest calorific value among the given options. Calorific value refers to the amount of heat energy released when a fuel is burned completely. Hydrogen has a high calorific value because it has a high energy content per unit mass. It has the highest energy content of all fuels, making it an efficient and clean source of energy.
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• 30.
### In which direction rainbow appears in the morning
• A.
North
• B.
South
• C.
East
• D.
West
D. West
Explanation
Rainbows always appear in the opposite direction of the sun, so in the morning when the sun rises in the east, the rainbow will be seen in the western sky. This is because the sunlight is refracted and reflected by water droplets in the atmosphere, creating the rainbow. Therefore, the correct answer is West.
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• 31.
### The Chief Minister of Uttarpradesh is
• A.
• B.
Arjun Munda
• C.
Manohar Parrikar
• D.
BC Khanduri
Explanation
The correct answer is Yogi Adityanad. Yogi Adityanad is the Chief Minister of Uttar Pradesh. He has been serving as the Chief Minister since March 2017. Yogi Adityanad is a member of the Bharatiya Janata Party (BJP) and is known for his strong Hindu nationalist views. He has implemented several policies and initiatives in Uttar Pradesh, focusing on development, law and order, and welfare programs. Yogi Adityanad is a prominent figure in Indian politics and has a significant influence in Uttar Pradesh.
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• 32.
### Who is the President of France
• A.
Emmanuel Macron
• B.
Nikolas Sarkozy
• C.
Jacques Chirac
• D.
François Hollande
A. Emmanuel Macron
Explanation
Emmanuel Macron is the correct answer because he is the current President of France. He was elected as the President in 2017 and is serving his first term. Macron succeeded François Hollande as the President of France.
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• 33.
### Booker Prize winner of 2017 is
• A.
Paul Beatty
• B.
George Saunders
• C.
Marlon James
• D.
Richard Flanagan
B. George Saunders
Explanation
Lincoln in the Bardo
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• 34.
### 2017 Womens Asia Boxing Championship was won by
• A.
Mery Kom
• B.
Wulanchabu
• C.
Ulaanbaatar
• D.
Astana
A. Mery Kom
Explanation
Mery Kom won the 2017 Womens Asia Boxing Championship.
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• 35.
### 64th National Film award for Best Feature Film is for
• A.
Kaasav
• B.
Alifa
• C.
Sathamanam Bhavati
• D.
Dhanak
A. Kaasav
Explanation
The correct answer is Kaasav. The 64th National Film award for Best Feature Film was awarded to Kaasav.
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• 36.
### God's particle is the pseudonym of
• A.
• B.
Quarks
• C.
Leptons
• D.
Higgs Boson
D. Higgs Boson
Explanation
The correct answer is Higgs Boson. The Higgs Boson is often referred to as the "God's particle" because it is associated with the Higgs field, which gives particles mass. It was discovered by the Large Hadron Collider (LHC), a particle accelerator, in 2012. Quarks and leptons are types of elementary particles, but they are not specifically associated with the term "God's particle."
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• 37.
### Winners of ICC T20 World Cup 2016
• A.
West Indies
• B.
India
• C.
Pakistan
• D.
England
A. West Indies
Explanation
West Indies is the correct answer because they were the winners of the ICC T20 World Cup in 2016. They defeated England in the final match to claim the title.
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• 38.
### 2017 Nobel Peace Prize was awarded to
• A.
ICAN
• B.
Malala yousafzai
• C.
OPCW
• D.
Al gore
A. ICAN
Explanation
ICAN (International Campaign to Abolish Nuclear Weapons) was awarded the 2017 Nobel Peace Prize for its efforts in promoting the Treaty on the Prohibition of Nuclear Weapons. The organization has been working tirelessly to raise awareness about the catastrophic humanitarian consequences of nuclear weapons and advocating for their complete elimination. ICAN's work has been instrumental in mobilizing international support and creating a global movement towards a world free of nuclear weapons. This recognition highlights the importance of disarmament and non-proliferation efforts in maintaining global peace and security.
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• 39.
### India is called tropical country mainly on account of its
• A.
Latitudinal Extend
• B.
Logitudinal Extend
• C.
Areal Size
• D.
Tropical Monzoon Climate
A. Latitudinal Extend
Explanation
India is called a tropical country mainly because of its latitudinal extent. Latitudinal extent refers to the distance between the northernmost and southernmost points of a country or region. India is located between the Tropic of Cancer and the equator, which means it lies in the tropical zone. This geographical position results in India experiencing a predominantly tropical climate with high temperatures, abundant rainfall, and distinct wet and dry seasons. Therefore, the correct answer is latitudinal extent.
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• 40.
### Which of the following article gives the freedom of religion
• A.
Article 25
• B.
Article 18
• C.
Article 33
• D.
Article 22
A. Article 25
Explanation
Article 25 of the Constitution gives individuals the freedom to practice and propagate any religion of their choice. It guarantees the right to freely profess, practice, and propagate religion, subject to public order, morality, and health. This article ensures that every citizen has the right to follow their own religious beliefs and worship, without any discrimination or interference from the state or others. It is an essential provision that upholds the principle of religious freedom and protects the rights of individuals to practice their religion freely.
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• 41.
### Minister for Co operation in Kerala
• A.
• B.
A.K. Balan
• C.
C. Raveendranath
• D.
T.P. Ramakrishnan
Explanation
Kadakampally Surendran is the correct answer for the Minister for Co-operation in Kerala. This can be inferred from the given options where Kadakampally Surendran is the only name mentioned.
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• 42.
### The marshy and forested land in northern part of Uttar Pradesh is
• A.
• B.
Bangar
• C.
Terai
• D.
Siwaliks
C. Terai
Explanation
The correct answer is Terai. The Terai region in the northern part of Uttar Pradesh is characterized by marshy and forested land. It is a low-lying area situated at the foothills of the Himalayas and is known for its rich biodiversity. The Terai region experiences a humid climate and is home to various wildlife species.
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• 43.
### Which river is called a river between two mountains
• A.
• B.
Ganga
• C.
Godhavari
• D.
Sutlej
Explanation
Narmada is called a river between two mountains because it flows between the Vindhya and Satpura mountain ranges in central India. This geographical feature gives the river its unique identity as a river that runs through a narrow valley between two mountain ranges.
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• 44.
### Which state is called the agricultural epitome of India?
• A.
West Bengal
• B.
• C.
• D.
Explanation
Uttar Pradesh is called the agricultural epitome of India because it is the largest producer of food grains in the country. The state has fertile soil, favorable climatic conditions, and a well-developed irrigation system, which contribute to its high agricultural productivity. Uttar Pradesh is known for its production of crops like wheat, rice, sugarcane, potatoes, and fruits. The state also has a strong agricultural infrastructure and a large number of farmers engaged in agricultural activities. Overall, Uttar Pradesh's significant contribution to India's agricultural sector earns it the title of the agricultural epitome of the country.
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• 45.
### Which region of India has a larger female population than the male population
• A.
West Bengal
• B.
Mizoram
• C.
Nagaland
• D.
Puducherry
D. Puducherry
Explanation
Puducherry has a larger female population than the male population. This can be inferred from the fact that the question asks for a region in India with a larger female population compared to the male population, and Puducherry is the only option provided. Therefore, Puducherry is the correct answer.
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• 46.
### Who was the first propounder of the doctrine of 'Passive Resistance'
• A.
B G Thilak
• B.
Aurobindo Ghosh
• C.
G K Gokhalae
• D.
M K Gandhi
B. Aurobindo Ghosh
• 47.
### In which five year plan University Grant Commission was set up for promoting and strengthening higher education
• A.
First Five Year Plan
• B.
Third Five Year Plan
• C.
Fifth Five Year Plan
• D.
Eleventh Five Year Plan
A. First Five Year Plan
Explanation
The University Grant Commission was set up during the First Five Year Plan to promote and strengthen higher education in India. This plan, implemented from 1951 to 1956, focused on industrialization, agricultural development, and infrastructure improvement. The establishment of the UGC was a crucial step in enhancing the quality and accessibility of higher education by providing financial support to universities and colleges, formulating regulations, and promoting academic research and innovation.
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• 48.
### Who is known as the Father of Renaissance of Western India?
• A.
B M Malabari
• B.
• C.
R G Bhandarkar
• D.
K T Telang
Explanation
M G Ranade is known as the Father of Renaissance of Western India because he was a prominent social reformer, scholar, and political leader during the late 19th century. He played a crucial role in the social, educational, and political transformation of Western India. Ranade was a strong advocate for women's rights, education, and social equality. He co-founded the Social Conference movement, which aimed to address various social issues prevalent in society. Ranade's efforts and contributions in promoting progressive ideas and reforms in Western India earned him the title of the Father of Renaissance.
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• 49.
### The Longest river in Peninsular India
• A.
Godavari
• B.
Krishna
• C.
Kaveri
• D.
A. Godavari
Explanation
The correct answer is Godavari. The Godavari River is the longest river in Peninsular India. It originates in the Western Ghats in the state of Maharashtra and flows through several states before emptying into the Bay of Bengal. It has a total length of about 1,465 kilometers and is known for its importance in the region's agriculture and irrigation systems.
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• 50.
### The largest producer of Rice in India
• A.
Kerala
• B.
• C.
West Bengal
• D.
C. West Bengal
Explanation
West Bengal is the correct answer because it is the largest producer of rice in India. The state has favorable climatic conditions and fertile soil, which are essential for rice cultivation. West Bengal also has a significant area of land dedicated to rice cultivation, contributing to its high production. Additionally, the state has implemented various agricultural practices and technologies to increase rice production, making it the leading producer in India.
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# Square Root Functions and Inequalities
## What Are Square Roots?
Square roots are the inverse of squaring a number. This means that when you take the square root of a number, you are finding the number that, when multiplied by itself, gives you the original number. For example, the square root of 9 is 3, because when 3 is multiplied by 3 the result is 9. This can be written as 3 x 3 = 9. Square roots can also be written using exponents. In this case, taking the square root of 9 can be written as 9^(1/2).
## Square Root Functions
A square root function is a function that contains a square root in its equation. For example, the equation y = √x is a square root function. This equation gives the result of the square root of x for any given value of x. In other words, if you plug any number into the x slot of the equation, the result will be the square root of that number. For instance, if you plug 8 into the equation, the result will be the square root of 8, which is 2.83.
## Inequalities Involving Square Roots
Inequalities involving square roots can be solved by using the same methods used to solve other inequalities. The first step is to isolate the square root on one side of the equation. For example, if you are presented with the inequality √x + 2 > 5, you would first subtract 2 from both sides of the equation to isolate the square root on the left side. This would give you the equation √x > 3. Next, you would square both sides of the equation in order to eliminate the square root. This would give you the equation x > 9. Finally, you would solve the inequality as you would any other inequality, which in this case would give you the solution x > 9.
## Applications of Square Root Functions and Inequalities
Square root functions and inequalities are used in a variety of different fields, including mathematics, engineering, and physics. In mathematics, square root functions and inequalities are used to solve a variety of problems, such as finding the area of a circle or the length of a hypotenuse. In engineering and physics, square root functions and inequalities are often used to calculate the force of an object or the acceleration of a system. In general, square root functions and inequalities are used to solve a variety of problems in a wide range of fields.
##
Square root functions and inequalities are important tools used to solve a variety of different problems. They can be used to calculate the area of a circle, the length of a hypotenuse, the force of an object, and the acceleration of a system. Understanding how to solve and apply these equations is an important skill for anyone studying mathematics, engineering, or physics. |
Read about Solving Algebraic Equations (2-Step) | for kids Grade 6, 7, & 8
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# Read About How To Solve 2-step equations
WHAT ARE 2-STEP ALGEBRAIC EQUATIONS?
Students learn to find all factor pairs for a whole number in the range 1-100. Students will also recognize that a whole number is a multiple of each of its factors. They will determine whether a given whole number in the range 1-100 is a multiple of a given one-digit number and whether a given whole number in the range 1-100 is prime or composite.
To better understand 2-step algebraic equations…
WHAT ARE 2-STEP ALGEBRAIC EQUATIONS?. Students learn to find all factor pairs for a whole number in the range 1-100. Students will also recognize that a whole number is a multiple of each of its factors. They will determine whether a given whole number in the range 1-100 is a multiple of a given one-digit number and whether a given whole number in the range 1-100 is prime or composite. To better understand 2-step algebraic equations…
## LET’S BREAK IT DOWN!
### Solve equations with addition and multiplication.
To rake leaves for your neighbors, you charge \$4 for every hour you work, plus a flat fee of \$3 for showing up. If you raked one lawn for \$23, how many hours did you work? We can represent this using an equation, if we let x be the number of hours worked. We write 4x + 3 = 23. To get x on its own, we need to get rid of everything else on the left side. We can do that by using inverse operations. The inverse of + 3 is – 3, so we subtract 3 on both sides. Then we have 4x = 20. Next, 4x means 4 times x, and the inverse is to divide by 4. So we divide both sides by 4, and then we have x = 5. You worked for 5 hours! Now you try: Find x if 3x + 5 = 17.
Solve equations with addition and multiplication. To rake leaves for your neighbors, you charge \$4 for every hour you work, plus a flat fee of \$3 for showing up. If you raked one lawn for \$23, how many hours did you work? We can represent this using an equation, if we let x be the number of hours worked. We write 4x + 3 = 23. To get x on its own, we need to get rid of everything else on the left side. We can do that by using inverse operations. The inverse of + 3 is – 3, so we subtract 3 on both sides. Then we have 4x = 20. Next, 4x means 4 times x, and the inverse is to divide by 4. So we divide both sides by 4, and then we have x = 5. You worked for 5 hours! Now you try: Find x if 3x + 5 = 17.
### How can we check our work?
We solved 4x + 3 = 23, and found that x = 5. If we want to check our answer, we can do that by substituting the answer into the original equation. Now that we know that x = 5, we can substitute 5 for x in the equation, so that we get 4(5) + 3 = 23. 4 times 5 is 20, and adding 3 to that is 23. Since both sides of the equal sign have the same value, we know that our work checks out! Now you try: Check if x = 2 is true for 6x + 4 = 16.
How can we check our work? We solved 4x + 3 = 23, and found that x = 5. If we want to check our answer, we can do that by substituting the answer into the original equation. Now that we know that x = 5, we can substitute 5 for x in the equation, so that we get 4(5) + 3 = 23. 4 times 5 is 20, and adding 3 to that is 23. Since both sides of the equal sign have the same value, we know that our work checks out! Now you try: Check if x = 2 is true for 6x + 4 = 16.
### Solve equations with multiplication and subtraction.
You got a deal at the local record store. You bought 5 records and got \$7 off. If your total came to \$98, how much was each record, assuming they all cost the same amount? If we let x be the price of a record, our equation becomes 5x – 7 = 98. First we want to get rid of – 7, and the inverse operation to subtraction is addition. So, we add 7 on both sides. Now we have 5x = 105. If we want to have only one x on the left side, we can divide both sides by 5. We get x = 21. So, each record cost \$21! Now you try. Find x if 3x – 7 = 29.
Solve equations with multiplication and subtraction. You got a deal at the local record store. You bought 5 records and got \$7 off. If your total came to \$98, how much was each record, assuming they all cost the same amount? If we let x be the price of a record, our equation becomes 5x – 7 = 98. First we want to get rid of – 7, and the inverse operation to subtraction is addition. So, we add 7 on both sides. Now we have 5x = 105. If we want to have only one x on the left side, we can divide both sides by 5. We get x = 21. So, each record cost \$21! Now you try. Find x if 3x – 7 = 29.
### Solve two-step equations with fractions.
You took out one third of the money in your piggy bank, and you already have \$8 in your pocket. If the total amount of money you have now is \$16, how much was in the piggy bank? If we let x be the amount of money in the piggy bank, then our equation is [ggfrac]x/3[/ggfrac]+8=16. Remember that the line between x and 3 means division. To solve for x, first we subtract 8 from both sides, and now we have [ggfrac]x/3[/ggfrac]=8. Since x is divided by 3, the inverse operation is multiplication, so we multiply both sides by 3 to get x alone. Now we have x=24. That means that originally you had \$24 in the piggy bank! Now you try: Find x if [ggfrac]x/5[/ggfrac]+2=22.
Solve two-step equations with fractions. You took out one third of the money in your piggy bank, and you already have \$8 in your pocket. If the total amount of money you have now is \$16, how much was in the piggy bank? If we let x be the amount of money in the piggy bank, then our equation is [ggfrac]x/3[/ggfrac]+8=16. Remember that the line between x and 3 means division. To solve for x, first we subtract 8 from both sides, and now we have [ggfrac]x/3[/ggfrac]=8. Since x is divided by 3, the inverse operation is multiplication, so we multiply both sides by 3 to get x alone. Now we have x=24. That means that originally you had \$24 in the piggy bank! Now you try: Find x if [ggfrac]x/5[/ggfrac]+2=22.
## WHAT ARE 2-STEP ALGEBRAIC EQUATIONS VOCABULARY
Variable
A symbol, like x, that represents an unknown quantity.
Equals Sign
Tells us that the expressions on either side have the same value.
Equation
Two expressions that are equal to each other separated by an equal sign.
Algebraic Expression
An equation that uses one or more variables.
One-step equation
An equation that requires only one operation to isolate the variable.
Two-step equation
An equation that requires two operations to isolate the variable.
Operations that do the opposite of each other. Addition and subtraction are inverse operations, and multiplication and division are inverse operations.
To multiply the number in front of an expression in parentheses by everything inside the parentheses.
## WHAT ARE 2-STEP ALGEBRAIC EQUATIONS DISCUSSION QUESTIONS
### What are the operations in the equation 4x – 5 = 7? What operations do you need to use to solve for x?
The operations are multiplication and subtraction. So to solve for x, I need to use the inverse operations: division and addition.
### To solve 4x – 5 = 7, should you divide first or add first?
I could do either first, but if I divide first, I have to divide 5 by 4 and 7 by 4, which means that I have to work with fractions. That would take longer but still get the right answer. However, if I add 5 to both sides first, then the rest of the solution is much easier.
### How can you solve x3-2=2?
[ggfrac]x/3[/ggfrac] means x divided by 3, so the inverse operation is multiplication. I also need to add to isolate the variable. It is simpler to first add 2 on both sides, and then to multiply both sides by 3, but working in the opposite order would also result in the same answer. Then x= 12.
### What are two different ways you can solve 2(x – 3) = 8?
Method 1: Distribute 2 across the expression in the parentheses, so that I have 2x – 6 = 8, and then solve by adding 6 to both sides, and then dividing both sides by 2. Method 2: First divide both sides by the factor 2, so that I have x – 3 = 4. Then solve by adding 3 to both sides. Method 2 is more efficient.
### To solve 2x + 3 = 5, Sylvia first subtracted 5 from both sides. What did she do wrong?
Because Sylvia did the same operation to both sides, the equation is still correct: the two sides still equal. However, she is no closer to finding out the value of x. She should instead subtract 3 on both sides, which gets her closer to having x alone on the left side.
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# A basketball team plays 60% of its games at home. The home court advantage is obvious, because the team wins 70% of its home games, but when they play away, they win only 35% of their games. What is the (conditional) probability of losing, GIVEN THAT the game was played away?
Dec 25, 2014
We will have to consider two distinct cases: home (H) and out (O) within these cases we have win (W) and loose (L)
We note probabilities by P and convert % to fractions
(divide by 100)
(1) Playing home: $P \left(H\right) = 0.6$
Loosing home $P \left(L\right) = 1 - 0.7 = 0.3$
Multiply because it is $H \mathmr{and} L$
$P \left(H L\right) = 0.6 \cdot 0.3 = 0.18$
(2) Playing out: $P \left(O\right) = 1 - 0.6 = 0.4$
Loosing out $P \left(L\right) = 1 - 0.35 = 0.65$
Multiply because it is $O \mathmr{and} L$
$P \left(O L\right) = 0.4 \cdot 0.65 = 0.26$
Since cases (1) and (2) are of the "either...or" type you may add the probabilities:
$P \left(L\right) = P \left(H L\right) + P \left(O L\right) = 0.18 + 0.26 = 0.44$
Conclusion:
The (conditional) probability of loosing is 44% |
# Persamaan Eksponen 3
## Konsep Dasar
##### Bentuk $$a^{f(x)} = b \: ^{g(x)}$$
Solusi
Menambahkan $$\log$$ pada kedua ruas.
$$\log a^{f(x)} = \log b \: ^{g(x)}$$
$$f(x) \:.\: \log a = g(x) \:.\: \log b$$
Contoh:
$$5^{x^2 + x - 2} = 3^{x + 2}$$
Solusi
\begin{equation*}
\begin{split}
& 5^{x^2 + x - 2} = 3^{x + 2} \\\\
& \log 5^{x^2 + x - 2} = \log 3^{x + 2} \\\\
& (x^2 + x - 2) \log 5 = (x + 2) \log 3 \\\\
& (x + 2)(x - 1) \log 5 - (x + 2) \log 3 = 0\\\\
& (x + 2)[(x - 1) \log 5 - \log 3] = 0
\end{split}
\end{equation*}
Faktor 1
$$x + 2 = 0$$
$$x = -2$$
Faktor 2
$$(x - 1) \log 5 - \log 3 = 0$$
$$(x - 1) \log 5 = \log 3$$
$$x - 1 = \dfrac{\log 3}{\log 5}$$
$${\color {blue} \dfrac{\log b}{\log a} = \log_a b}$$
$$x - 1 = \log_5 3$$
$$x = 1 + \log_5 3$$
$$x = \log_5 5 + \log_5 3$$
$$x = \log_5 15$$
HP = $$\{-2, \log_5 15 \}$$ |
# Definition of Least Common Denominator
The least common denominator of a group of $2$ or more fractions is the smallest number that can be used as the denominator of all of the fractions in the group.
Just a few reminders:
• The "denominator" is the number on the bottom of a fraction.
• A "common denominator" occurs when the bottom number of all of the fractions in the group is the same.
• The "least common denominator" is the smallest out of all of the positive numbers that can be used as common denominators for all of the fractions in the group.
For example, let's find the least common denominator of the fractions $\dfrac{1}{4}$ and $\dfrac{2}{3}$.
$\dfrac{1}{4}$ can be written as $\dfrac{1}{4}$, $\dfrac{2}{8}$, $\dfrac{3}{12}$, $\dfrac{4}{16}$, and so on.
$\dfrac{2}{3}$ can be written as $\dfrac{4}{6}$, $\dfrac{8}{12}$, $\dfrac{12}{18}$, and so on.
The first fractions from the two lists that have the same denominator are $\dfrac{3}{12}$ and $\dfrac{8}{12}$. So, the least common denominator of $\dfrac{1}{4}$ and $\dfrac{3}{12}$ is $12$.
Finding the least common denominator makes it much easier to add and subtract fractions.
For example,
$\dfrac{1}{4} + \dfrac{2}{3} = \dfrac{3}{12} + \dfrac{8}{12} = \dfrac{11}{12}.$
### Description
The aim of this dictionary is to provide definitions to common mathematical terms. Students learn a new math skill every week at school, sometimes just before they start a new skill, if they want to look at what a specific term means, this is where this dictionary will become handy and a go-to guide for a student
### Audience
Year 1 to Year 12 students
### Learning Objectives
Learn common math terms starting with letter L
Author: Subject Coach
You must be logged in as Student to ask a Question. |
# How do you solve 6x = 4x - ( - 18)?
Jun 9, 2018
$x = 9$
#### Explanation:
Simplify first as follows:
$6 x = 4 x - \left(- 18\right)$
$6 x = 4 x + 18$-----> opening the bracket
$6 x - 4 x = 4 x - 4 x + 18$ -----> subtracting $4 x$ both sides
$2 x = 18$
$x = \frac{18}{2}$
$x = 9$
$6 \left(9\right) = 4 \left(9\right) - \left(- 18\right)$
$54 = 36 + 18$ ----> Correct!
Jun 9, 2018
$x = 9$
#### Explanation:
$\text{note that } - \left(- 18\right) = + 18$
$6 x = 4 x + 18$
$\text{subtract "4x" from both sides}$
$2 x = 18$
$\text{divide both sides by 2}$
$x = \frac{18}{2} = 9$
$\textcolor{b l u e}{\text{As a check}}$
Substitute this value into the equation and if both sides are equal then it is the solution.
$\text{left side } = 6 \times 9 = 54$
$\text{right side } = \left(4 \times 9\right) + 18 = 36 + 18 = 54$
$x = 9 \text{ is the solution}$
Jun 9, 2018
$x = 9$
#### Explanation:
$6 x \textcolor{red}{=} 4 x - \left(- 18\right)$ | you start with this
$6 x \textcolor{red}{=} 4 x - \left(- 18\right)$ | you put the '4x' on the other side
$6 x - 4 x \textcolor{red}{=} - \left(- 18\right)$
$2 x \textcolor{red}{=} - \left(- 18\right)$
$2 x \textcolor{red}{=} - \left(- 18\right)$ | -1 times -8
$2 x \textcolor{red}{=} - 1 \left(- 18\right)$
$2 x \textcolor{red}{=} 18$
$2 x \textcolor{red}{=} 18$ | put the 2 to the other side
$x \textcolor{red}{=} \frac{18}{2}$
$x \textcolor{red}{=} 9$ |
# How do you use Heron's formula to determine the area of a triangle with sides of that are 35, 58, and 41 units in length?
Jun 29, 2016
Area of triangle is $708.36$ square units.
#### Explanation:
According to Heron's formula, if $a$, $b$ and $c$ are three sides of a triangle, its area is given by $\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$, where $s = \frac{1}{2} \left(a + b + c\right)$.
The sides of triangle are $35$, $58$ and $41$ and hence
$s = \frac{1}{2} \left(35 + 58 + 41\right) = \frac{1}{2} \times 134 = 67$ and
Area of triangle is $\sqrt{67 \times \left(67 - 35\right) \times \left(67 - 58\right) \times \left(67 - 41\right)}$
= $\sqrt{67 \times 32 \times 9 \times 26}$
= $4 \times 3 \times \sqrt{67 \times 2 \times 26} = 12 \sqrt{3484} = 12 \times 59.03 = 708.36$ square units. |
# Equation of the line
Use our calculator and we will give you the equation of the line through two points as well as the value of its pending.
You only have to enter the coordinates (x1, y1) (x2, y2) of each of the known points and press the calculate button to obtain the result. If you want to know how to find the equation of the line and see solved exercises for each case, continue reading below.
## Equation point-slope of the straight line
The point-slope equation is one of the most commonly used equations. in math problems and is of the type:
y - y1 = m (x - x1)
With it, we can calculate the equation of the line knowing the slope (m) and a point P of coordinates (x1, y1).
For example, let's calculate the equation of the line if we know that it has slope m=3 and passes through the point P = (4,3). We simply substitute in the general equation of the line and we obtain:
y - 3 = 3 (x - 4)
Now we simplify:
y = 3x -12 + 3
y = 3x - 9
We have already calculated the equation of the line that meets the conditions of the problem.
## Equation of the line through two points
Yes we are given two points and asked to calculate the line passing through those coordinates, we have to use this formula:
In this case solving the problem is quite simple since we only have to substitute in the equation and simplify it as much as possible. To see how it is done, we will do an exercise in which we are asked to calculate the equation of the line that passes through the points (4, 5) and (2, 1):
Now we simply equalize and simplify and we are left with the equation in this form:
2 (x - 4) = y - 5
We simplify:
2x - 8 = y - 5
2x - y + 3 = 0
We already have the general equation of the line passing through the two points of the exercise statement. From here we could also calculate the slope as we have seen in the previous section:
m = - A / B = - (2 / -1) = 2
## Continuous equation of the line
We will use the continuous equation of the straight line when give us a point P of coordinates (x1, y1) and its director vector of coordinates (v1, v2).
For example, in the case of the graph above these lines, in which we have the point (3, 3) with director vector (2, 1)the equation of the line would be:
Simplifying we are left with:
x - 3 = 2 (y - 3)
x - 3 = 2y - 6
x - 2y +3 = 0
## General equation of the line
The general equation of the line is of the type:
Ax + By + C = 0
A, B and C being Real numbers and B ≠ 0.
From this equation also we can draw:
• La pendiente de la recta (m = - A/B).
• The ordinate at the origin (- C/B).
With these data we already have enough information to represent the line on the XY plane.
## Parametric equations of the line
These equations are obtained from the vector equation and are of this type:
x = x0 + v1t
y = y0 + v2t
Being:
• (x0, y0) the coordinates of a point on the line
• (v1, v2) are the coordinates of a vector in the direction of the line
## How to know if a point belongs to a line
To find out if a point belongs to a given line we simply have to coger la ecuación de esa recta y sustituir en 'x' e 'y' los valores del punto they give us. If the equality is fulfilled, the point will belong to the line and if they are not fulfilled, then it does not.
For example, does the point (1, 3) with equation y = 2 +3x lie on the line? Let's see:
3 ≠ 2 + 3 → the equality is not satisfied so the point (1,3) is not on the line.
5 = 2 + 3 → the equality is satisfied, therefore, the point (1, 5) is on the line.
## Formula for calculating the distance between two points on a straight line
The distance between two points of a line in the Cartesian plane can be calculated in a very simple way. To understand it better, let's see it with an example.
Imagine that you have two points in the plane whose coordinates are:
• P1 (X1, Y1)
• P2 (X2, Y2)
The The distance separating the two points is obtained by applying the following formula mathematics:
For example, we will calculate the distance between two points with an example practical in which:
• P1 (7, 5)
• P2 (4, 1)
Applying the above formula, we have that the distance between these two points on the Cartesian plane es:
As you can see, to know the distance between two points is very easy. |
# Time and Work Questions
FACTS AND FORMULAE FOR TIME AND WORK QUESTIONS
1. If A can do a piece of work in n days, then A's 1 day's work =$\frac{1}{n}$
2. If A’s 1 day's work =$\frac{1}{n}$, then A can finish the work in n days.
3. A is thrice as good a workman as B, then:
Ratio of work done by A and B = 3 : 1.
Ratio of times taken by A and B to finish a work = 1 : 3.
NOTE :
Hence,
Whole work is always considered as 1, in terms of fraction and 100% , in terms of percentage.
In general, number of day's or hours = $\frac{100}{Efficiency}$
Q:
A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, the rest of the work is finished by C in two more days. If they get Rs. 3000 as wages for the whole work, what are the daily wages of A, B and C respectively (in Rs):
A) 200, 250, 300 B) 300, 200, 250 C) 200, 300, 400 D) None of these
Explanation:
A's 5 days work = 50%
B's 5 days work = 33.33%
C's 2 days work = 16.66% [100- (50+33.33)]
Ratio of contribution of work of A, B and C =
= 3 : 2 : 1
A's total share = Rs. 1500
B's total share = Rs. 1000
C's total share = Rs. 500
A's one day's earning = Rs.300
B's one day's earning = Rs.200
C's one day's earning = Rs.250
88 31753
Q:
Relation Between Efficiency and Time
A is twice as good a workman as B and is therefore able to finish a piece of work in 30 days less than B.In how many days they can complee the whole work; working together?
Sol: Ratio of efficiency = 2:1 (A:B)
Ratio of required time = 1:2 (A:B) => x:2x
but 2x-x=30
x= 30 and 2x= 60
Now efficiency of A =3.33% and efficiency of B =1.66%
Combined efficiency of A and B together = 5%
time required by A and B working together to finish the work = 20 days
Note: Efficiency inversely proportional to Time
Therefore, Efficiency x time = Constant Work
Hence, Required time
whole work is always cosidered as 1, in terms of fraction and 100%, in terms of percentage.
In, general no.of days or hours = Efficiency x time
22456
Q:
12 men can complete a work in 8 days. 16 women can complete the same work in 12 days. 8 men and 8 women started working and worked for 6 days. How many more men are to be added to complete the remaining work in 1 day?
A) 8 B) 12 C) 16 D) 24
Explanation:
1 man's 1 day work =1/96 ; 1 woman's 1 day work = 1/192
Work done in 6 days=
Remaining work = 1/4
(8 men +8 women)'s 1 day work = $1\left(\frac{8}{96}+\frac{8}{192}\right)$ =1/8
Remaining work =1/4 - 1/8 = 1/8
1/96 work is done in 1 day by 1 man
Therefore, 1/8 work will be done in 1 day by 96 x (1/8) =12 men
64 21470
Q:
A, B and C can do a piece of work in 24 days, 30 days and 40 days respectively. They began the work together but C left 4 days before the completion of the work. In how many days was the work completed?
A) 11 days B) 12 days C) 13 days D) 14 days
Explanation:
One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10
C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.
Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10
Remaining Work = 7/10, which was done by A,B and C in the initial number of days.
Number of days required for this initial work = 7 days.
Thus, the total numbers of days required = 4 + 7 = 11 days.
41 20141
Q:
P can complete a work in 12 days working 8 hours a day.Q can complete the same work in 8 days working 10 hours a day. If both p and Q work together,working 8 hours a day,in how many days can they complete the work?
A) 60/11 B) 61/11 C) 71/11 D) 72/11
Explanation:
P can complete the work in (12 * 8) hrs = 96 hrs
Q can complete the work in (8 * 10) hrs=80 hrs
Therefore, P's 1 hour work=1/96 and Q's 1 hour work= 1/80
(P+Q)'s 1 hour's work =(1/96) + (1/80) = 11/480
so both P and Q will finish the work in 480/11 hrs
Therefore, Number of days of 8 hours each = (480/11) x (1/8) = 60/11
15 11069
Q:
A and B can do a piece of work in 30 days , while B and C can do the same work in 24 days and C and A in 20 days . They all work together for 10 days when B and C leave. How many days more will A take to finish the work?
A) 18 days B) 24 days C) 30 days D) 36 days
Explanation:
2(A+B+C)'s 1 day work = 1/30 + 1/24 + 1/20 = 1/8
=>(A+B+C)'s 1 day's work= 1/16
work done by A,B and C in 10 days=10/16 = 5/8
Remaining work= 3/8
A's 1 day's work= $\left(\frac{1}{16}-\frac{1}{24}\right)=\frac{1}{48}$
Now, 1/48 work is done by A in 1 day.
So, 3/8 work wil be done by A in =48 x (3/8) = 18 days
17 10156
Q:
A can do a certain work in the same time in which B and C together can do it.If A and B together could do it in 20 days and C alone in 60 days ,then B alone could do it in:
A) 20days B) 40 days C) 50 days D) 60 days
Explanation:
(A+B)'s 1 day's work=1/20
C's 1 day work=1/60
(A+B+C)'s 1 day's work= 1/20 + 1/60 = 1/15
Also A's 1 day's work =(B+C)'s 1 day's work
Therefore, we get: 2 * (A's 1 day 's work)=1/15
=>A's 1 day's work=1/30
Therefore, B's 1 day's work= 1/20 - 1/30 = 1/60
So, B alone could do the work in 60 days.
24 8661
Q:
A works twice as fast as B.If B can complete a work in 18 days independently,the number of days in which A and B can together finish the work is:
A) 4 days B) 6 days C) 8 days D) 10 days
Explanation:
Ratio of rates of working of A and B =2:1. So, ratio of times taken =1:2
Therefore, A's 1 day's work=1/9
B's 1 day's work=1/18
(A+B)'s 1 day's work= 1/9 + 1/18 = 1/6
so, A and B together can finish the work in 6 days |
Smallest number multiplied to get perfect square
Chapter 6 Class 8 Squares and Square Roots
Concept wise
Let’s do this by examples
### Finds the smallest number multiplied by 32 to get a perfect square.
Since 2 does not occur in pair,
we multiply by 2 to make it a pair
So, our number becomes
Now, it becomes a perfect square.
So, we multiply 32 by 2 to make it a perfect square
### Finds the smallest number multiplied by 90 to get a perfect square.
Here,
2 & 5 do not occur in pairs
So, we multiply by 2 and 5 to make pairs
So, our number becomes
90 × 2 × 5 = 2 × 3 × 3 × 5 × 2 × 5
Now, it becomes a perfect square.
So, we multiply 90 by 2 × 5
i.e. 10 to make it a perfect square
### Finds the smallest number multiplied by 120 to get a perfect square.
Here,
2, 3 & 5 do not occur in pairs
So, we multiply by 2, 3 and 5 to make pairs
So, our number becomes
120 × 2 × 3 × 5 = 2 × 2 × 2 × 3 × 5 × 2 × 3 × 5
Now, it becomes a perfect square.
So, we multiply 120 by 2 × 3 × 5
i.e. 30 to make it a perfect square
Get live Maths 1-on-1 Classs - Class 6 to 12 |
##### Differential Equations For Dummies
Integrating by parts is the integration version of the product rule for differentiation. The basic idea of integration by parts is to transform an integral you can’t do into a simple product minus an integral you can do. Here’s the formula:
Don’t try to understand this yet. Wait for the examples that follow.
If you remember that, you can easily remember that the integral on the right is just like the one on the left, except with the u and v reversed.
Here’s the method in a nutshell.
First, you’ve got to split up the integrand into a u and a dv so that it fits the formula. For this problem, choose ln (x) to be your u. Then, everything else is the dv, namely
Next, you differentiate u to get your du, and you integrate dv to get your v. Finally, you plug everything into the formula and you’re home free.
The integration by parts box.
To help keep everything straight, organize integration-by-parts problems with a box like the one in this figure. Draw an empty 2-by-2 box, then put your u, ln(x), in the upper-left corner and your dv,
in the lower-right corner, as in the following figure.
Filling in the box.
The arrows in this figure remind you to differentiate on the left and to integrate on the right. Think of differentiation — the easier thing — as going down (like going downhill), and integration — the harder thing — as going up (like going uphill).
Now complete the box:
This figure shows the completed box for
A good way to remember the integration-by-parts formula is to start at the upper-left square and draw an imaginary number 7 — across, then down to the left, as shown in the following figure.
Remembering how you draw the 7, look back to the figure with the completed box. The integration-by-parts formula tells you to do the top part of the 7, namely
minus the integral of the diagonal part of the 7,
By the way, this is much easier to do than to explain. Try it. You’ll see how this scheme helps you learn the formula and organize these problems.
Ready to finish? Plug everything into the formula: |
Five Digits: Small or Large?
In this worksheet, students are given five digits from which they must identify the smallest and largest three-digit numbers.
Key stage: KS 2
Curriculum topic: Number: Number and Place Value
Curriculum subtopic: Order/Compare Numbers to 1000
Difficulty level:
QUESTION 1 of 10
We are going to make the largest and smallest 3-digit numbers from five given digits.
Let's pretend we can only use the digits 2, 3, 4, 5 and 6.
To make the largest number:
This is 6.
Then comes the next digit down in the Tens column.
This is 5.
Lastly comes the next digit down in the Ones column.
This is 4.
The largest number is 654
To make the smallest number:
This is 2.
Then comes the next largest digit in the Tens column.
This is 3.
Lastly comes the next largest digit in the Ones column.
This is 4.
The smallest number is 234.
Example
Using any 3 of the digits from:
68671
form the smallest and largest 3-digit numbers that you can make.
Put the digits in order:
16678
1 is the smallest digit.
6 is the next largest digit.
6 is the next largest digit even though it is the same as the last one.
Smallest number is 166.
8 is the largest digit.
7 is the next digit down.
6 is the next digit down.
Largest number is 876.
Using any three of the digits:
1, 2, 3, 4, 5
Make the smallest and largest three-digit numbers from the given numbers above.
143 114 123 321 543 smallest largest
Using any three of the digits:
6, 3, 4, 5, 8
Make the smallest and largest three-digit numbers from the given numbers above.
345 865 354 856 318 smallest largest
Using any three of the digits:
1, 3, 4, 9, 2
Make the smallest and largest three-digit numbers from the given numbers above.
943 934 394 134 123 smallest largest
Using any three of the digits:
6, 3, 7, 1, 4
Make the smallest and largest three-digit numbers from the given numbers above.
743 134 147 764 774 smallest largest
Using any three of the digits:
8, 2, 6, 3, 4
Make the smallest and largest three-digit numbers from the given numbers above.
243 634 864 264 234 smallest largest
Using any three of the digits:
7, 5, 6, 7, 2
Make the smallest and largest three-digit numbers from the given numbers above.
277 776 765 256 265 smallest largest
Using any three of the digits:
9, 5, 6, 7, 5
Make the smallest and largest three-digit numbers from the given numbers above.
967 537 556 756 976 smallest largest
Using any three of the digits:
7, 3, 5, 7, 3
Make the smallest and largest three-digit numbers from the given numbers above.
353 533 753 775 335 smallest largest
Using any three of the digits:
8, 1, 8, 1, 8
Make the smallest and largest three-digit numbers from the given numbers above.
111 118 181 888 881 smallest largest
Using any three of the digits:
4, 3, 7, 3, 7
Make the smallest and largest three-digit numbers from the given numbers above.
333 433 334 777 774 smallest largest
• Question 1
Using any three of the digits:
1, 2, 3, 4, 5
Make the smallest and largest three-digit numbers from the given numbers above.
143 114 123 321 543 smallest largest
EDDIE SAYS
12345 - smallest is 123
12345 - largest is 543
• Question 2
Using any three of the digits:
6, 3, 4, 5, 8
Make the smallest and largest three-digit numbers from the given numbers above.
345 865 354 856 318 smallest largest
EDDIE SAYS
34568 - smallest is 345
34568 - largest is 865
• Question 3
Using any three of the digits:
1, 3, 4, 9, 2
Make the smallest and largest three-digit numbers from the given numbers above.
943 934 394 134 123 smallest largest
EDDIE SAYS
12349 - smallest is 123
12349 - largest is 943
• Question 4
Using any three of the digits:
6, 3, 7, 1, 4
Make the smallest and largest three-digit numbers from the given numbers above.
743 134 147 764 774 smallest largest
EDDIE SAYS
13467 - smallest is 134
13467 - largest is 764
• Question 5
Using any three of the digits:
8, 2, 6, 3, 4
Make the smallest and largest three-digit numbers from the given numbers above.
243 634 864 264 234 smallest largest
EDDIE SAYS
23468 - smallest is 234
23468 - largest is 864
• Question 6
Using any three of the digits:
7, 5, 6, 7, 2
Make the smallest and largest three-digit numbers from the given numbers above.
277 776 765 256 265 smallest largest
EDDIE SAYS
25677 - smallest is 256
25677 - largest is 776
• Question 7
Using any three of the digits:
9, 5, 6, 7, 5
Make the smallest and largest three-digit numbers from the given numbers above.
967 537 556 756 976 smallest largest
EDDIE SAYS
55679 - smallest is 556
55679 - largest is 976
• Question 8
Using any three of the digits:
7, 3, 5, 7, 3
Make the smallest and largest three-digit numbers from the given numbers above.
353 533 753 775 335 smallest largest
EDDIE SAYS
33577 - smallest is 335
33577 - largest is 775
• Question 9
Using any three of the digits:
8, 1, 8, 1, 8
Make the smallest and largest three-digit numbers from the given numbers above.
111 118 181 888 881 smallest largest
EDDIE SAYS
11888 - smallest is 118
11888 - largest is 888
• Question 10
Using any three of the digits:
4, 3, 7, 3, 7
Make the smallest and largest three-digit numbers from the given numbers above.
333 433 334 777 774 smallest largest
EDDIE SAYS
33477 - smallest is 334
33477 - largest is 774
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# At the local grocery store, a 16-ounce bottle of apple juice costs $3.20. What is the cost of the apple juice per ounce? ##### 1 Answer Jun 27, 2018 #### Answer: The cost of apple juice per ounce is 20 cents. #### Explanation: We know that a $16$ounce bottle is $3.20 and we want to find how much per ounce is.
So we set up a proportion and let $x$ be the cost of one ounce:
$\frac{16}{3.20} = \frac{1}{x}$
Now we can cross multiply:
$16 x = 3.20$
Divide both sides by $\textcolor{b l u e}{16}$:
$\frac{16 x}{\textcolor{b l u e}{16}} = \frac{3.20}{\textcolor{b l u e}{16}}$
Therefore,
$x = 0.20$
The cost of apple juice per ounce is 20 cents.
Hope this helps! |
# Sketch the region bounded by the curves, and visually estimate the location of the centroid:
$\boldsymbol{ y \ = \ e^x, \ y \ = \ 0, \ x \ = \ 0, \ x \ = \ 5 }$
The aim of this question is to find the area under a bounded region with multiple constraints and to calculate the centroid of this bounded region.
To solve this question, we first find the area bounded by the region (say A). Then we calculate the x and y moments of the region (say $M_x$ & $M_y$). The moment is the measure of the tendency of a given region against rotation around the origin. Once we have these moments, we can calculate the centroid C using the following formula:
$C = \left( \dfrac{M_y}{A}, \dfrac{M_x}{A} \right)$
Step (1): The constraint of $y = 0$ is already fulfilled. To find the area bounded by the region $y \ = \ e^x$, we need to perform following integration:
$A = \int_{a}^{b} \bigg ( e^x \bigg ) dx$
Since the region is bounded by $x \ = \ 0$ and $x \ = \ 5$:
$A = \int_{0}^{5} \bigg ( e^x \bigg ) dx$
$\Rightarrow A = \bigg | e^x \bigg |_{0}^{5}$
$\Rightarrow A = e^{ (5) } \ – \ e^{ (0) }$
$\Rightarrow A = e^5 \ – \ 1$
Step (2): Calculating the $M_x$:
$M_x = \int_{0}^{5} \bigg ( e^x \bigg )^2 dx$
$\Rightarrow M_x = \bigg | \frac{ 1 }{ 2 } \bigg ( \frac{e^x}{2} \bigg ) (e^x) \bigg |_{0}^{5}$
$\Rightarrow M_x = \bigg | \frac{ e^{ 2x } }{ 4 } \bigg |_{0}^{5}$
$\Rightarrow M_x = \frac{ 1 }{ 4 } \bigg | e^{ 2x } \bigg |_{0}^{5}$
$\Rightarrow M_x = \frac{ 1 }{ 4 }\bigg ( e^{ 2(5) } – e^{ 2(0) } \bigg )$
$\Rightarrow M_x = \frac{ 1 }{ 4 }\bigg ( e^{ 2(5) } – 1 \bigg )$
Step (3): Calculating the $M_y$:
$M_x = \int_{0}^{5} \bigg ( xe^x \bigg ) dx$
$\Rightarrow M_y = \bigg | (x-1)e^x \bigg |_{0}^{5}$
$\Rightarrow M_y = \bigg ( (5-1)e^{(5)} -(0-1)e^{(0)} \bigg )$
$\Rightarrow M_y = 4e^5 + 1$
Step (4): Calculating the x-coordinate of centroid:
$C_x = \dfrac{M_x}{A}$
$C_x = \dfrac{ \dfrac{ 1 }{ 4 }\bigg ( e^{ 2(5) } – 1 \bigg )}{e^5-1}$
$C_x = \dfrac{ \dfrac{ 1 }{ 4 }\bigg ( (e^5)^2 – (1)^2 \bigg )}{e^5-1}$
$C_x = \dfrac{ \dfrac{ 1 }{ 4 }(e^5 – 1)(e^5 + 1) }{e^5-1}$
$C_x = \dfrac{ 1 }{ 4 }(e^5 + 1)$
$C_x = 37.35$
Step (5): Calculating the y-coordinate of centroid:
$C_y = \dfrac{M_y}{A}$
$C_y = \dfrac{4e^5 + 1}{e^5-1}$
$C_y = 4.0$
## Numerical Result
$Centroid \ = \ \left [ \ 37.35, \ 4.0 \ \right ]$
## Example
Given that $M_x = 30$, $M_y = 40$ and $A = 10$, find the coordinates of the centroid of the bounded region.
x-coordinate of centroid $C_x$ can be calculated using:
$C_x = \dfrac{M_x}{A} = \dfrac{30}{10} = 3$
y-coordinate of centroid $C_y$ can be calculated using:
$C_y = \dfrac{M_y}{A} = \dfrac{40}{10} = 4$
So:
$Centroid \ = \ \left [ \ 3, \ 4 \ \right ]$ |
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# Algebra Percents and Simple Interest Study Guide (page 2)
based on 2 ratings
By
Updated on Oct 3, 2011
#### Example
After a 68% decrease, a value is now 74. What was the original value?
Let x represent the original value. The difference between the original value and the new value is x – 74. Divide that by the original value, x, and set it equal to the percent decrease, 68%, which is 0.68:
Multiply both sides of the equation by x, and then subtract x from both sides:
0.68x = x – 74
–0.32x = –74
Divide both sides by –0.32:
x = 231.25
## Simple Interest
We can find how much interest an amount of money, or principal, has gained by multiplying the principal by an interest rate and a length of time. The formula for interest is I = prt. Interest, principal, rate, and time are all variables, and given any three of them, we can substitute those values into the equation to find the missing fourth value.
Example
If a principal of \$500 gains \$60 in interest in three years, what was the interest rate per year?
The interest rate is the percent of the principal that is added as interest each year. Because I = prt and we are looking for the rate, r, we can divide both sides of the equation by pt to get r alone on the right side:
.
To find the rate, divide the interest by the product of the principal and the time: 60 ÷ (500)(3) = 60 ÷ 1,500 = 0.04 = 4%. The principal gained interest at a rate of 4% per year.
#### Tip:
When calculating interest, be sure the interest rate and the time have consistent units of measure. If the interest rate is given on a yearly basis, the time must also be in years. If the interest rate is given on a yearly basis and the time is given in months, convert the time to years before using the interest formula.
#### Example
If \$36 in interest is gained over six months at a rate of 6% per year, how much was the principal?
We can rewrite the formula I = prt to solve for p by dividing both sides of the equation by rt: . The interest rate is given per year, but the length of time is given in months. Divide the number of months by 12, because there are 12 months in a year: 6 ÷ 12 = 0.5. The time is 0.5 years and the rate is 6%, or 0.06. Because \$1,200. The principal was \$1,200.
Find practice problems and solutions for these concepts at Algebra Percents and Simple Interest Practice Questions.
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Factors
# Factors of 34 | Prime Factorization of 34 | Factor Tree of 34
Written by Prerit Jain
Updated on: 18 Aug 2023
### Factors of 34 | Prime Factorization of 34 | Factor Tree of 34
Factors of 34: There are a total of four factors of the number 34. Factors of 34 are 1, 2, 17, and 34. Factors are those numbers that divide a particular number evenly. The numbers 1, 2, 17, and 34 divide 34 fully, without a remainder.
The factors of a number can be either positive or negative. They can also be written as pairs. This means that multiplying the set of two factors in pairs gives the original product (34).
In this article, let us discuss how to find factors of 34, its prime factorization, and factor trees with solved examples. Keep reading to learn more.
## What Are the Factors of 34?
The factors of 34 are those numbers that can divide another number evenly without leaving any remainder. The factors can usually be both positive and negative. The number 34 also has both positive and negative factors.
## What Are the Factors of 34 in Pairs?
A pair factor of 34 is a combination of two numbers when multiplied together gives the product of the original number (34). These pairs can also be written in positive and negative pairs as shown below:
### Positive Pair Factors of 34
The positive pair factors of 34 are (1, 34) and (2, 17).
### Negative Pair Factors of 34
The negative pair factors of 34 are (-1, -34) and (-2, -17).
## How To Find Factors of 34 Through Division Method?
To find the factors of 34 we can employ many methods. Here, we are using the division method. This method needs you to multiply a sequence of natural integers by 34. The numbers that divide the 34 completely without leaving any remainder are known as the factors of 34. The detailed steps for finding factors of 34 using the division method are below:
34 ÷ 1 = 34
34 ÷ 2 = 17
34 ÷ 17 = 2
34 ÷ 34 = 1
Thus, we can see that 1, 2, 17, and 34 are the numbers that do not leave any remainder when divided by 34. This proves that they are the factors of 34.
## What Is the Prime Factorization of 34?
In order to perform prime factorization, we must first check if the given number is a prime number. If it is a prime number, then it will have only two factors: 1 and 34. If the given number is a composite number, then it will have more than two factors.
Since 34 is the composite number, we have more than two factors. Hence, we can use the prime factorization method to find the factors of 34.
The steps to find prime factors of 34 using the prime factorization method are given below:
Step 1: Divide 34 by the smallest prime number that can divide evenly. The smallest prime number that can divide 34 is 2. So, 34 ÷ 2 = 17
Step 2: Now, divide the number 17 (quotient) by the next smallest prime number. But since 17 is a prime number, it can only be divided by 1 and 17. So, 17 ÷ 17 = 1. The quotient is 1. We can see that 1 cannot be divided further. So, the prime factors of 34 are 2 X 17. We can also write it down as 2 and 17, as they are both prime numbers.
Image representation of the prime factorization of 34:
## What is the Factor Tree of 34?
Students can also use the factor tree to find the prime factors of 34. In this method, we will place the number 34 at the top. Now, we will write down any one pair factors of the given number (34), as the branches of 34. We will split the pair of factors into factors. This will go on until the branches become prime numbers. Now, circling all the prime numbers will give us the prime factors of 34.
Step 1: Place the number 34 at the top of the factor tree. Let us branch out 34 into any of its factor pairs, that is, 2 and 17.
Step 2: Since 2 is a prime number, we can no longer separate it. 17 also happens to be a prime number, and it cannot be separated either. The branches in the factor tree will now be all the prime numbers.
There are many ways in which we can write the factor tree of 34. Below is an image representation of the factor tree of 34, choosing 2 and 17 as pair factors of 34.
## Solved Examples on Factors of 34:
Here are some solved examples of 34 factors:
What is the sum of all the factors of 34?
1, 2, 17, and 34 are the 34 factors.
1 + 2 + 17 + 34 = 54
Hence, 54 is the sum of all 34 factors.
What are the factors and prime factors of 34?
The factors of 34 are 1, 2, 17, and 34.
The prime factors of 34 are 2 and 17.
What’s the highest common factor between 102 and 34?
The factors of 102 are 1, 2, 3, 6, 17, 34, 51, and 102.
Factors of 34 are 1, 2, 17, and 34.
Hence, the common factors of 102 and 34 are 1, 2, 17, and 34.
The highest common factor between 102 and 34 is 34.
What’s the highest common factor between 17 and 34?
The factors of 17 are 1 and 17.
Factors of 34 are 1, 2, 17, and 34.
Hence, from the above, the common factors of 17 and 34 are 1 and 17.
The highest common factor between 17 and 34 is 17.
Is 204 a factor of 34?
No, but 204 is not a factor of 34.
204 ÷ 34 = 6.
What are the factors of -34?
The factors of -34 are -1, -2, -17, and -34.
What are the common factors of 85 and 34?
To find the common factors of 85 and 33, we need to find the factors of both 85 and 33.
Factors of 85: 1, 5, 17 and 85
Factors of 34: 1, 2, 17 and 34
Hence the common factors of 85 and 34 are 1 and 17.
What are all the factors of 34?
The factors of 34 are 1, 2, 17, and 34.
What are the factors of 34 that are composite?
Any number which has more than 2 factors is known as a composite number. The number which has more than 2 factors in factors of 34 is only 34.
What factors of 34 are multiples of 2?
The factors of 34 that are multiples of 2 are 2 and 34.
2 x 1 = 2
2 x 14 = 34.
How many factors of 34 are odd?
A total of 2 odd numbers are there in factors of 34 and they are 1 and 17.
What are the factors of 51 and 34?
The factors of 51 are 1, 3, 17, and 51.
The factors of 34 are 1, 2, 17, and 34.
How many factors are there for the number 34?
A total of 4 factors are there for the number 34 and are 1, 2, 17, and 34.
What factors of 34 are multiples of 2?
The factors of 34 that are multiples of 2 are 2 and 34.
How many pair factors are there in 34?
A total of 2 pairs are there in factors of 34.
What is the largest factor of 34?
The largest factor of 34 is 34.
The second largest factor of 34 is 17.
## FAQs on Factors of 34
List all the factors of 34
1, 2, 17, and 34 are the 34 factors.
What are the composite factors of 34?
34 is the only composite factor of 34.
Composite factors are those that have more than 2 factors other than 1 and itself.
What are the factors of 34 in pairs?
(1, 34) and (2, 17) are the 34 factors of pairs.
What are the positive factors for 34?
1, 2, 17, and 34 are the positive factors for 34.
What are some of the multiples of 34? A multiple is a number that may be divided by another a certain number of times without a remainder.
The multiples of 34 are 34, 68, 102, 136, 170, 204, 238, 272, 306, and 340.
Now we have learned everything about the factors of 34. We hope this blog has proved useful and easy for you. If you have any more doubts, feel free to comment below. We will get back to you shortly.
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# Nice's central library building is considered one of the most original in the world, as it is a mix between a sculpture and a work of habitable architecture. It was called La Tête Carrée and is made up of part of a bust that supports a cube divided into five floors. It is known that the building has a total height of approximately 30 meters. It admits that the cubic part of the sculpture is parallel to the floor and has a volume of 2744 meters3 Calculate, in meters, the height of the bust that supports the cube. Displays all the calculations you made.
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## Answer to a math question Nice's central library building is considered one of the most original in the world, as it is a mix between a sculpture and a work of habitable architecture. It was called La Tête Carrée and is made up of part of a bust that supports a cube divided into five floors. It is known that the building has a total height of approximately 30 meters. It admits that the cubic part of the sculpture is parallel to the floor and has a volume of 2744 meters3 Calculate, in meters, the height of the bust that supports the cube. Displays all the calculations you made.
Rasheed
4.7
Let the height of the bust be represented by 'h' meters.
The volume of a cube is given by the formula V = s^3, where 's' is the length of the side of the cube.
In this case, we are given that the volume of the cube is 2744 meters^3.
So, we can write the equation as 2744 = s^3.
We also know that the total height of the building is 30 meters. The height of the bust $h$ plus the height of the cube $s$ should add up to the total height.
So, we can write another equation as h + s = 30.
Now, we can solve these two equations to find the values of 'h' and 's'.
From the equation 2744 = s^3, we take the cube root on both sides to get:
s = ∛2744.
Using a calculator, we find that the cube root of 2744 is 14.
Now, substituting the value of 's' in the equation h + s = 30, we have:
h + 14 = 30.
Subtracting 14 from both sides of the equation, we get:
h = 30 - 14 = 16.
Therefore, the height of the bust that supports the cube is 16 meters.
Answer: The height of the bust is 16 meters.
Frequently asked questions $FAQs$
Math Question: What is the equation of an ellipse with major axis length 6 and minor axis length 4? $+ What is the dot product of vectors v = (3, -2, 1$ and u = $1, 4, -3$?
+
Find the value of sinh$arcsinh(3$).
+ |
# Ratios
A ratio is a comparison of two numbers by division.
The ratio of $m$ to $n$ can be expressed in the following ways:
• $m$ to $n$
• $m:n$
• $\frac{m}{n}$
Last expression tells us, that ratio is, actually, a fraction.
As well as fraction, ratio can be expressed in simplest form (in other words it can be reduced).
Example. Suppose, for 2 servings of a fruit dessert we need 14 oz. of yogurt. Find amount of yogurt needed for 1 serving and for 4 servings. How many servings can we make, if we have 56 oz. of yogurt.
Ratio of amount of yogurt to number of servings is ${14}$ to ${2}$ or $\frac{{14}}{{2}}$.
We can reduce this fraction by 2: $\frac{{{7}\cdot{\color{red}{{{2}}}}}}{{{1}\cdot{\color{red}{{{2}}}}}}=\frac{{7}}{{1}}$.
So, we need 7 oz. of yogurt to make 1 serving.
Now, we multiple numerator and denominator of the reduced fraction by 4: $\frac{{{7}\cdot{\color{red}{{{4}}}}}}{{{1}\cdot{\color{red}{{{4}}}}}}=\frac{{28}}{{4}}$.
Finally, we multiple numerator and denominator of the reduced fraction by 8 (to get 56 oz.): $\frac{{{7}\cdot{\color{red}{{{8}}}}}}{{{1}\cdot{\color{red}{{{8}}}}}}=\frac{{56}}{{8}}$.
Thus, we can make 8 servings from 56 oz. of yogurt.
This means that we need 28 oz. of yogurt to make 4 servings.
A ratio called scale is used when making a model or drawing of something that is too large or too small to be conveniently drawn at actual size.
Example. The scale of a map is 5 inches=27 miles. The distance between two towns is 38 inches. What is the actual distance?
Scale is $\frac{{27}}{{5}}$ miles per inch.
If distance is 38 inches, then actual distance is $\frac{{27}}{{5}}\cdot{38}=\frac{{1026}}{{5}}={205.2}$ miles.
Ratios can be used to compare more than two values.
Example. Ratio of flour, sugar and water is ${50}:{2}:{100}$. Find amount of flour and water for 40 g. of sugar.
We can reduce this ratio by 2: ${25}:{1}:{50}$.
Now, amount of flour is ${25}\cdot{40}={1000}$ g.
Amount of water is ${50}\cdot{40}={2000}$ g.
Exercise 1. Ratio of cows to goat on the farm is ${15}:{6}$. Express this ratio in the simplest form and find number of cows if there are 12 goats. Find number of goats, if there are 45 cows.
Answer: Ratio is $\frac{{5}}{{2}}$. If there are 12 goats, then number of cows is 30 (multiple reduced fraction by 6). If there are 45 cows, then number of goats is 18 (multiple numerator and deominator of the reduced fraction by 9)
Exercise 2. Scale of the map is 2 inches=15 miles. If the actual distance is 90 miles, find distance between towns on the map.
Answer: ratio is $\frac{{2}}{{15}}$ inches per mile. Distance required is $\frac{{2}}{{15}}\cdot{90}={12}$ inches.
Exercise 3. Average fuel consumption of BMW X5 is 204 miles to 12 gallons. Express this ratio in simplest term and find how many miles can you drive, if you have 18 gallons of fuel.
Answer: $\frac{{{204}\ \text{miles}}}{{{12}\ \text{gallons}}}=\frac{{{17}\ \text{miles}}}{{{1}\ \text{gallon}}}$ or 17 miles per gallon. Now, multiple numerator and denominator by 18 to get $\frac{{{306}\ \text{miles}}}{{{18}\ \text{gallons}}}$. Thus, you can ride 306 miles.
Exercise 4. Ratio of height of Ann, Bob and John is ${1.4}:{1.2}:{1.5}$. Find height of Ann and Bob, if height of John is 75 inches.
Answer: height of Ann is 70 inches, height of Bob is 60 inches. |
Precalculus (6th Edition) Blitzer
The partial fraction decomposition of the rational expression is $\frac{2}{x-2}+\frac{5}{{{\left( x-2 \right)}^{2}}}$
We see that the linear factor $x-2$ occurs multiple times in the denominator of the rational expression, so, for each power of $x-2$ assign an undefined constant as given below: $\frac{2x+1}{{{\left( x-2 \right)}^{2}}}=\frac{A}{x-2}+\frac{B}{{{\left( x-2 \right)}^{2}}}$ Then, multiply both sides of the equation by ${{\left( x-2 \right)}^{2}}$: \begin{align} & {{\left( x-2 \right)}^{2}}\cdot \frac{2x+1}{{{\left( x-2 \right)}^{2}}}={{\left( x-2 \right)}^{2}}\left( \frac{A}{x-2}+\frac{B}{{{\left( x-2 \right)}^{2}}} \right) \\ & {{\left( x-2 \right)}^{2}}\cdot \frac{2x+1}{{{\left( x-2 \right)}^{2}}}={{\left( x-2 \right)}^{2}}\cdot \frac{A}{x-2}+{{\left( x-2 \right)}^{2}}\cdot \frac{B}{{{\left( x-2 \right)}^{2}}} \end{align} And divide the common factors: \begin{align} & {{\left( x-2 \right)}^{2}}\cdot \frac{2x+1}{{{\left( x-2 \right)}^{2}}}=\left( x-2 \right)\left( x-2 \right)\cdot \frac{A}{x-2}+{{\left( x-2 \right)}^{2}}\cdot \frac{B}{{{\left( x-2 \right)}^{2}}} \\ & 2x+1=\left( x-2 \right)A+B \\ & 2x+1=\left( A \right)x+\left( -2A+B \right) \end{align} Then, equate the coefficients of like terms of the equation to write a system of equations as given below: $A=2$ (I) $-2A+B=1$ (II) Put the value of $A$ in equation (II): \begin{align} & -2A+B=1 \\ & -2\left( 2 \right)+B=1 \\ & B=1+4 \\ & B=5 \end{align} Put the value of $A\text{ and }B$ in the initial equation, and define the partial fraction decomposition: \begin{align} & \frac{2x+1}{{{\left( x-2 \right)}^{2}}}=\frac{A}{x-2}+\frac{B}{{{\left( x-2 \right)}^{2}}} \\ & =\frac{2}{x-2}+\frac{5}{{{\left( x-2 \right)}^{2}}} \end{align} Thus, the partial fraction decomposition of the rational expression is $\frac{2}{x-2}+\frac{5}{{{\left( x-2 \right)}^{2}}}$. |
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# Rotational Symmetry in Geometry
In geometry, rotational symmetry is a form of symmetry that occurs when an object can be rotated about a fixed point without changing its overall appearance. A classic example of rotational symmetry is a snowflake, which has sixfold rotational symmetry (meaning it can be rotated by 1/6th of a turn and still look the same).
## How Rotational Symmetry Works
To understand rotational symmetry, let's first consider regular polygons. A regular polygon is a closed figure with sides that are all the same length and angles that are all the same size. Regular polygons always have at least three sides, but there is no upper limit to how many sides they can have.
All regular polygons have rotational symmetry. For example, a triangle has threefold rotational symmetry because it can be rotated by 1/3rd of a turn (120°) and still look the same. Likewise, a square has fourfold rotational symmetry because it can be rotated by 1/4th of a turn (90°) and still look the same. The number of times that a regular polygon can be rotated and still look the same is equal to the number of sides it has.
Not all figures have rotational symmetry. For example, an irregular polygon is a closed figure with sides that are not all the same length or angles that are not all the same size. Irregular polygons do not have any rotational symmetry because they cannot be rotated about a fixed point without changing their overall appearance.
## Conclusion:
Rotational symmetry is a type of symmetry that occurs when an object can be rotated about a fixed point without changing its overall appearance. All regular polygons have rotational symmetry; the number of times that a regular polygon can be rotated and still look the same is equal to the number of sides it has. Not all figures have rotational symmetry; irregular polygons do not have any rotational symmetry because they cannot be rotated about a fixed point without changing their overall appearance.
## FAQ
### What is hypotenuse in simple words?
The hypotenuse is the longest side in a right angled triangle. It is opposite the angle of 90 degrees (the right angle).
### How do you find the hypotenuse?
The length of the hypotenuse can be found using the Pythagorean theorem, which states that the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
### What is an example of a figure with no rotational symmetry?
An irregular polygon is a closed figure with sides that are not all the same length or angles that are not all the same size. Irregular polygons do not have any rotational symmetry because they cannot be rotated about a fixed point without changing their overall appearance. |
# Dividing Fractions
Flip the second fraction and multiply.
## How can you divide fractions?
Dividing fractions is explained with several examples. When dividing a fraction by a fraction you can multiply it by the reciprocal. Exchange the division sign for a multiplication sign and flip the numerator and the denominator of the second fraction.
Three examples are discussed
- Dividing a fraction by a cardinal number
- Dividing a fraction by a fraction
- Dividing a mixed fraction by a fraction
Op deze pagina staan voorbeelden en oefeningen. Wil je uitgebreid oefenen ga dan naar één van de 5-stappenplannen.
### 5-step plans
Number of questions:
Time per question:
In this exercise you get all kinds of divisions.
Look at the examples if you don't know the answer.
Tip: use tab to go to the next field
### Example 1
#### Dividing a fraction by a cardinal number.
67 ÷ 3.
Divisions can be solved in two ways. Dividing the numerator by the numerator and the denominator by the denominator or multiplying by the reciprocal. In this case division is easier than multiplication with the reciprocal.
We first change the cardinal number into a fraction.
3 = 31.
We then get the following sum:
67 ÷ 31 =
Divide numerator by numerator and denominator by denominator:
6 ÷ 3 = 2 and 7 ÷ 1 = 7
67 ÷ 31 = 27
### Example 2
#### Dividing a fraction by a fraction
12 ÷ 34= We solve this sum by multiplying by the reciprocal.
We need to flip 34. That becomes 43.
The sum now becomes:
12 x 43=
1 x 4 = 4 and 2 x 3 = 6
12 x 43=46.
We can also simplify this answer. We then get:
46=23.
### Example 3
#### Dividing a mixed fraction by a fraction
212 ÷ 14=
We also solve this sum with the rule: Dividing by a fraction is multiplying by the reciprocal. First 212 needs to be written as a fraction.
That's: 212 =52 Now we can solve the sum just like in example 2.
52 x 41=202
We need to simplify 202. That becomes 101 and that's the same as 10 |
# How do you divide (2 3/5) / (6 1/3)?
Jun 19, 2015
When dividing by a fraction, invert the fraction, then multiply.
#### Explanation:
When dividing by by a fraction, invert the fraction and multiply.
$a \div \frac{b}{c} = a \times \frac{c}{b}$ or $\frac{a}{\frac{b}{c}} = a \times \frac{c}{b}$
$\frac{\frac{23}{5}}{\frac{61}{3}}$=
$\frac{23}{5} \times \frac{3}{61} = \frac{69}{305}$
Jun 19, 2015
Formatting here is tricky. If you wanted $\frac{2 \frac{3}{5}}{6 \frac{1}{3}}$ (color(white)(1/1)Rather than $\frac{23}{5}$/61/3), then the answer is $\frac{39}{95}$
#### Explanation:
To divide mixed numbers (or to multiply them) first change form to an improper fraction:
$2 \frac{3}{5} = \frac{\left(2 \times 5\right) + 3}{5} = \frac{10 + 3}{5} = \frac{13}{5}$
$6 \frac{1}{3} = \frac{\left(6 \times 3\right) + 1}{3} = \frac{18 + 1}{3} = \frac{19}{3}$
Now, we can write:
$\frac{2 \frac{3}{5}}{6 \frac{1}{3}} = \frac{\frac{13}{5}}{\frac{19}{3}}$
Now invert the denominator (the bottom) and multiply:
$\frac{2 \frac{3}{5}}{6 \frac{1}{3}} = \frac{\frac{13}{5}}{\frac{19}{3}} = \frac{13}{5} \times \frac{3}{19}$
Multiplying fractions is the easiest thing to do with them. It's just top times top over bottom times bottom. (Easier still if we can reduce first, but we can't in this problem.) So we get:
$\frac{2 \frac{3}{5}}{6 \frac{1}{3}} = \frac{\frac{13}{5}}{\frac{19}{3}} = \frac{13}{5} \times \frac{3}{19} = \frac{13 \times 3}{5 \times 19}$
$= \frac{39}{95}$
the fraction cannot be reduced, so that's it, we are finished. |
# Acceleration. Velocity (v) - rate of position change. Constant v – rate stays the same, equal distance for equal t interval. Acceleration (a)- rate of.
## Presentation on theme: "Acceleration. Velocity (v) - rate of position change. Constant v – rate stays the same, equal distance for equal t interval. Acceleration (a)- rate of."— Presentation transcript:
Acceleration
Velocity (v) - rate of position change. Constant v – rate stays the same, equal distance for equal t interval. Acceleration (a)- rate of velocity change. Constant/Uniform accl. – change v for equal time interval. To calculate average/ uniform / constant accl a = va = v f – v i t t Velocity vs. Acceleration
Acceleration Any object not traveling at constant velocity is accelerating. Changing speed – either speeding up or slowing down. Changing direction.
We will look at Uniform / Constant Acceleration Changing speed or direction at constant rate. What are units?
Units Since a = v /t Units become: m v s = m or d/t 2. t ss 2 What does it mean to have a constant or uniform acceleration of = 10 m/s 2.
Acceleration of Gravity (g) near the surface of the Earth. ~ 10 m/s 2. Drop a ball from rest (v i = 0) After 1 sec, it’s velocity ~ 10 m/s After 2 sec, it’s velocity ~
a. If I drop a ball from rest near Earth’s surface, approximately how fast is it moving after falling 5 seconds? 50 m/s
Acceleration is a vector. Magnitude & direction. When v is positive accl is positive since v t What does the sign of accl mean?
Sign of acceleration 1. Consider a car starts from rest, heads east, and attains a velocity of +20 m/s in 2 s. Calculate a: Sign tells change in v.
a = v f –v i t a v +20 m/s - 0 m/s= +10 m/s 2 t2 s Acceleration is positive. The car is headed in a positive direction and speeding up. 2. Now consider the same car slowing to a stop from +20 m/s in 2s. Calculate a.
Now v f is 0, and v i is +20m/s, so. a v 0 m/s - (+20 m/s)= -10 m/s 2. t2s Accl is neg. the car is slowing down. 3. Consider the same car starts from rest, heads west, and reaches 20 m/s in 2 s. Calculate a now. Since the car is heading west, the v f is neg. Hold on - its not so simple!
A quick calculation shows this new accl to be negative. Oy! Here are the rules:
Velocity motiona posspeeding+ posslowing- negspeeding- negslowing+ Mental Trick!
4. A shuttle bus slows to a stop with an acceleration of -1.8 m/s 2. How long does it take to slow from 9.0 m/s to rest? List the variables. a = -1.8 m/s 2. v i = 9 m/s v f = 0 (stop) t = ?
Find an equation with everything on the list. a = v f – v i /t Rearrange to solve for the unknown. t = v f – v i /a Plug in with units. = 9m/s – 0 =5 s. -1.8 m/s 2 a = -1.8 m/s 2. v i = 9 m/s v f = 0 (stop) t = ?
7200 m/s. 5. A plane starts from rest and accelerates for 6 minutes at 20 m/s 2 before traveling at a constant velocity. What was its final velocity?
Another useful acceleration equation. d = v i t + ½ at 2. d= displacement (m) v i = initial (starting velocity) m/s. a = acceleration m/s 2. t = time over which accl takes place - s.
6. A car starts from rest and accelerates at 5 m/s 2 for 8 seconds. How far did it go in that time? List! Equation! Solve. 160 m
Acceleration Hwk. Hwk Intro to Acceleration Rd 48 – 49 Do pg 49 #1 - 5.
1. Take your seat. 2. Take out your physics supplies. 3. Will 6 volunteers come to the front please
Do Now: Ruler Drop.
More Acceleration Equations a = v/t v f = v i + at d = v i t + ½ at 2. v f 2 = v i 2 + 2 ad.
1. A car starts from rest and accelerates at 6 m/s 2 for 5 seconds. How far did it travel?
2. A car slows to a stop from 23 m/s by applying the brakes over 12 meters. Calculate acceleration. v i = 23 m/s v f = 0 d = 23 m a = ? v f 2 = v i 2 + 2ad - v i 2 = a 2d -(23 m/s) 2 = -22 m/s 2. 2(12m)
3.A bicycle is traveling at 5 m/s. It accelerates at 3 m/s 2 for 3 meters. What is its final velocity?
4. A truck skidded to a stop with an acceleration of -3 m/s 2. If its initial velocity was 11 m/s, how far was it skidding before it came to a stop?
Hwk Wksht Mixed Accl Equations sheet.
Velocity Time Graphs Speed Time Graphs
Constant Velocity/Speed
Constant / Uniform Acceleration. On velocity time graph accl. is slope of straight line.
What’s going on here?
Sign of velocity is direction.
Sketch Graphs V-t sketch graphs Rev Book Hwk Rev Book. Rd 55 – 58. Do pg 56 #7-13 AND Pg 58 #14 – 18.
Displacement on V-T graphs
Displacement = Area Under Curve v = d/t then, vt = d.
Area of non-constant velocity. For constant accl, d = area of a triangle: ½ bh. If a car achieved a v=40 m/s in 10 s, then: ½(10s)(40m/s) = 200 m. 40 m/s 10 s
To find displacement, calc area of triangle + rectangle.
How can you tell when object is back to starting point? Positive displacement = negative displacement.
Do Now: Given the v – t graph below, sketch the acceleration – t graph for the same motion.
Acceleration – time Graphs What is the physical behavior of the object? Slowing down pos direction, constant vel neg accel.
d-t: –slope = velocity – area ≠. v-t: –slope = accl – area = displ a-t: – slope ≠. – area = vel –v f – v i.
Hwk Rev Book pg 76 #32-36, 45-47, 49 – 55, 59-60. Begin in class.
Together: text pg 65 #1-5, pg 72 # 34, 35, 48, 49. Test Thursday: Acceleration, Motion Graphs, Free-fall. Text 2-2, 2-3. RB Chap 3.
Objects Falling Under Gravity
Freefall Gravity accelerates uniformly masses as they fall and rise. Earth’s acceleration rate is 9.81 m/s 2 – very close to 10 m/s 2.
Falling objects accelerate at the same rate in absence of air resistance
But with air resistance
Fortunately there is a “terminal fall velocity.” After a while, the diver falls with constant velocity due to air resistance. Unfortunately terminal fall velocity is too large to live through the drop.
Apparent Weightlessness Objects in Free-fall Feel Weightless
What is the graph of a ball dropped?
What do the d-t, v-t, and a – t, graphs of a ball thrown into the air look like if it is caught at the same height?
A ball is thrown upward from the ground level returns to same height. d = ball’s height above the ground velocity is + when the ball is moving upward Why is acceleration negative? a is -9.81, the ball is accelerating at constant 9.81 m/s 2. Is there ever deceleration?
Free-fall Assumptions Trip only in the air. Trip ends before ball caught. -Symmetrical Trip time up = time down -Top of arc: v = 0, a = ?? -On Earth g = -9.81 m/s 2. Other planets g is different.
Solving: Use accl equations replace a with -g. List given quantities & unknown quantity. Choose accl equation that includes known & 1 unknown quantity. Be consistent with units & signs. Check that the answer seems reasonable Remain calm
Practice Problem. 1. A ball is tossed upward into the air from the edge of a cliff with a velocity of 25 m/s. It stays airborne for 5 seconds. What is its total displacement?
v i = +25 m/s a = g = -9.81 m/s 2. t = 5 s. d = ? d= v i t + ½ at 2. (25m/s)(5s) + 1/2(-9.81 m/s 2 )(5 s) 2. 125 m- 122. 6 = +2.4m. It is 2.4m above the start point.
2. If the air time from the previous problem is increased to 5.2 seconds, what will be the displacement? d = v i t + ½ at 2. -2.6 m It will be below the start point.
Ex 3. A 10-kg rock is dropped from a 7- m cliff. What is its velocity just before hitting the ground? d = 7m a = -9.81 m/s 2. v f = ? Hmmm v i = 0. v f 2 = v i 2 + 2ad v f 2 = 2(-9.81m/s 2 )(7 m) v f = -11.7 m/s (down)
4. A ball is thrown straight up into the air with a velocity of 25 m/s. Create a table showing the balls position, velocity, and acceleration for each second for the first 5 seconds of its motion. T(s)dv (m/s)a (m/s 2 ) 0025-9.81 12015.2-9.81 2305.4-9.81 331-4.43-9.81 52.4-24-9.91
Read Text pg 60-64 Do prb’s pg 64# 1-5 show work.
Mech Universe: The Law of Falling Bodies: http://www.learner.org/resources/series42.html?pop=yes&pid=549#
Download ppt "Acceleration. Velocity (v) - rate of position change. Constant v – rate stays the same, equal distance for equal t interval. Acceleration (a)- rate of."
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# Fatimah and Hairu enjoy Congkak game part 33 (Math Question)
Standard
The blog postings are about the Singapore Math. The readers can learn from the postings about Solving Singapore Primary School Mathematics. The blog presents the Math Concept, the Math Questions with solutions that teaches in Singapore Primary Schools. You or the kids can aquire the skills of dealing with the Math Modeling, the Math Problem Solving and the Problem Sum from Lower Primary School to Upper Primary School level after reading the blog postings. This posting is an upper primary school math question on RatioFraction and Problem Sum.
Read the posting on Fatimah and Hairu play Congkak to know about Congkak game.
Read the posting on Sarin learns the concept of Fraction (mathematics concept) to understand Fraction and Portion.
Read also the posting on Sarin learns the concept of Ratio (mathematics concept) to understand the concept of Ratio.
Challenge yourself with the question before look for the given solution‼!
Upper primary school mathematics question UPQ560
After a round of Congkak game, in the bag of Fatimah has red, blue and yellow beads. The ratio of the number of red beads to the number of blue beads is 2 : 3. The ratio of the number of yellow beads to the total number of red and blue beads is 2 : 7. What fraction of the beads in the bag are blue?
Solution:
Only Red, Blue and Yellow beads in the bag and the total number of beads in the bag no change.
The ratio of red to blue beads,
R : B : (R + B) = 2 : 3 : 5 = 14 : 21 : 35
The ratio of yellow to red and blue beads,
Y : (R + B) = 2 : 7 = 10 : 35
The ratio of red to blue to yellow beads,
R : B : Y : Total = 14 : 21 : 10 = 45
The fraction of beads are blue in the bag = 21/45 = 7/15
Alternative Solution:
Visualise by using model,
From the model,
The fraction of blue beads in the bag = 21/45 = 7/15 |
# Lesson Explainer: Relating Force, Pressure, and Area Physics • 9th Grade
In this explainer, we will learn how to use the formula for pressure, , to calculate pressures that are produced by forces acting on areas.
Force is a vector quantity, and so a force can be represented by an arrow. Such an arrow may be drawn as a thick or a thin line.
The thickness or thinness of the line has no physical meaning; only the length of the line and its direction correspond to properties of a force (its magnitude and direction).
An ideal line used to represent a force has no thickness, only length. Such a line would of course not be visible in a diagram.
If the arrow representing a force is thought of as having thickness, this leads to the mistaken idea that a force acting on a surface acts on some area, as shown by the yellow region in the following figure.
It is more correct to think of the tip of the arrow acting on the surface, as shown below.
The tip of the arrow is a point; it has an area of zero. A force, therefore, acts at a single point, not on an area.
The idea of a force acting at a point is familiar. Consider a uniform cube at rest on a horizontal surface. It is not unusual to describe the weight of the cube as acting at point at the center of the face of the cube in contact with the surface.
If this model of the weight of the cube as a force was taken literally, however, then at all points on the downward-facing face of the cube other than its center, no force would act.
There would then be no force acting at the point shown by the center of the red cross in the following figure.
An object on the surface at the position of the center of the red cross would however have a force acting on it in reality. An object underneath any point on the downward face of the box would have some force acting on it due to being in contact with the box.
When we wish to model a force acting on an area rather than at a point, we cannot use forces that act at single points. We must instead consider a surface for which forces act at every point on the surface.
When forces act at every point on a surface, a quantity can be defined that is the result of the forces that act. This quantity is called pressure.
The relationship between force, pressure, and area is represented visually by the following figure. Forces act at every point over a rectangular area (not all of these forces are shown). The force at each point is the same.
We can see that the direction in which the forces act is perpendicular to both sides of the rectangle. In order to produce pressure, forces must act perpendicularly to an area. If forces act parallel to an area, then no pressure is produced.
In the following figure, a cross section of an area is shown. The cross-section of an area is a length. Forces act at every point along a length (not all of these forces are shown). The force at each point is the same.
There is a mathematical relationship between the pressure on an area and the force acting perpendicular to all the sides of the area.
### Relationship: The Pressure on an Area and the Force Acting Perpendicularly to the Area
The pressure, , on an area, , is given by where is the force acting perpendicularly to all the sides of the area.
If has the unit newtons and has the unit square metres, then has the unit pascals (Pa). This means that
Pressure is often stated in kilopascals (kPa), where 1 kPa = 1βββ000 Pa.
When we consider the formula we can see that for a fixed value of , the greater the value of is, the smaller the value of must be to produce pressure . Equivalently, a smaller value of corresponds to a greater value of required to produce a fixed value of .
We have seen that the forces due to a pressure have a direction. The area that the pressure acts on can be considered to have a direction compared to the forces.
By showing a rectangular area, we can see that the direction in which the forces act is perpendicular to both sides of the rectangle.
We can then consider both a force and an area to be vector quantities. This is the case even though area is a scalar quantity when it is not related to the direction of a force. Relating an area to a force that is perpendicular to the area changes how the quantity area is used.
When relating force, area, and pressure, we see that multiplying the area perpendicular to the force by the pressure gives the force:
As both force and area are considered as vector quantities when related this way, we see that pressure must be a scalar quantity. This is the case as multiplying a vector quantity by a scalar quantity results in another vector quantity.
We can express the relationship between force, pressure, and perpendicular area then as
A different way of understanding the relationship between force and pressure is to make force the subject of the formula. We can do this by multiplying the formula by :
This gives us
Considering the formula this way, we can say that a pressure on an area is associated with a force acting at every point within that area. The forces all act perpendicularly to the area.
The relationship is represented visually by the following figure, where again a length represents the cross section of an area.
Using the relationship between force, pressure, and area, we can visually represent a change in area for a fixed pressure. This is shown in the following figure in which equal pressures act on two different areas.
We see that the force associated with the pressure is inversely proportional to the area.
Pressure can act on fluids as well as on solid objects. When pressure acts on a fluid, the direction of the forces acting in the fluid can be in various directions.
The following figure shows that increasing the pressure on the top of a container of water can cause the side of the container to rupture.
We can see that the direction of the force on the container wall is not the direction of the force due to the pressure on top of the container. We see then that a pressure does not have to act in the direction of the force that produces the pressure. We cannot then, in general, define the direction of a pressure. This means that pressure is considered a scalar quantity.
Let us now look at an example of determining a pressure.
### Example 1: Determining a Pressure
What pressure is produced by a 100 N force applied to an area of 2.5 m2?
We can determine the pressure using the formula
Substituting the values in the question, we obtain
Let us now look at an example of determining a force using a pressure.
### Example 2: Determining a Force Using a Pressure
A pressure of 400 Pa is applied to an area of 2.5 m2. What force produces this pressure?
We can rearrange the formula to make the subject.
We can do this by multiplying the formula by :
This gives us
Substituting the values in the question, we obtain
Let us now look at an example of determining an area using a pressure.
### Example 3: Determining an Area Using a Pressure
A pressure of 75 Pa is produced by a 3βββ000 N force. What area is pressure produced over?
We can rearrange the formula to make the subject.
We can do this by first multiplying the formula by :
This gives us
The formula is then divided by :
Substituting the values in the question, we obtain
Let us now look at an example in which a weight is the force acting on an area.
### Example 4: Determining a Pressure given a Weight
A sofa with a mass of 125 kg has a base with an area of 2.5 m2. What pressure does the sofa apply to the ground beneath it?
We can determine the pressure using the formula
The force acting is the weight of the sofa. The weight of the sofa is given by where is the mass of the sofa and is the acceleration due to gravity or the gravitational field strength. The value of is given by
Substituting the values in the question, we obtain
Let us now look at an example in which the area a pressure is produced over must be determined to find the weight of an object supported on that area.
### Example 5: Determining a Weight given a Pressure
An empty water tank has a rectangular base with side lengths of 1.2 m and 2.3 m. The weight of the tank applies a pressure of 350 Pa to the surface it is resting on. What is the weight of the tank?
The weight of the tank is the force that acts on it. We can represent the force by .
We can rearrange the formula to make the subject.
We can do this by multiplying the formula by :
This gives us
The value of the pressure is stated to be 350 Pa, but the area is not stated.
The tank has a rectangular base, so the area of the base is equal to the product of the lengths of adjacent sides of the rectangle. We have then that
We can now determine , which is given by
Let us now summarize what has been learned in these examples.
### Key Points
• A force acts at a point while a pressure acts on an area.
• The pressure, , on an area, , is given by where is the force acting perpendicularly to the area.
• If a force has the unit newtons and an area has the unit square metres, then a pressure has the unit pascals (Pa), where
• Pressure is often stated in kilopascals (kPa), where 1 kPa = 1βββ000 Pa.
• A component of a force must act perpendicularly to an area to produce a pressure on it.
• Pressure is a scalar quantity.
• For a fixed force, reducing the area that the force acts on increases the pressure produced. |
# How do you find the center, vertices, foci, and asymptotes of the hyperbola, and sketch its graph of x^2/9 - y^2/16 = 1?
Mar 4, 2017
The center is $\left(0 , 0\right)$
The vertices are (-3,0)$\mathmr{and}$(3,0)#
The foci are $F ' = \left(- 5 , 0\right)$ and $F = \left(5 , 0\right)$
The asymptotes are $y = \frac{4}{3} x$ and $y = - \frac{4}{3} x$
#### Explanation:
We compare this equation
${x}^{2} / {3}^{2} - {y}^{2} / {4}^{2} = 1$
to
${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$
The center is $C = \left(0 , 0\right)$
The vertices are $V ' = \left(- a , 0\right) = \left(- 3 , 0\right)$ and $V = \left(a , 0\right) = \left(3 , 0\right)$
To find the foci, we need the distance from the center to the foci
${c}^{2} = {a}^{2} + {b}^{2} = 9 + 16 = 25$
$c = \pm 5$
The foci are $F ' = \left(- c , 0\right) = \left(- 5 , 0\right)$ and $F = \left(c , 0\right) = \left(5 , 0\right)$
The asymptotes are
${x}^{2} / {3}^{2} - {y}^{2} / {4}^{2} = 0$
$y = \pm \frac{4}{3} x$
graph{((x^2)/9-(y^2)/16-1)(y-4/3x)(y+4/3x)=0 [-16.02, 16.02, -8.01, 8.01]} |
# Fraction calculator
This calculator subtracts two fractions. First, convert all fractions to a common denominator when fractions have different denominators. Find Least Common Denominator (LCD) or multiply all denominators to find a common denominator. When all denominators are the same, simply subtract the numerators and place the result over the common denominator. Then simplify the result to the lowest terms or a mixed number.
## The result:
### 7/3 - 2/8 = 25/12 = 2 1/12 ≅ 2.0833333
Spelled result in words is twenty-five twelfths (or two and one twelfth).
### How do we solve fractions step by step?
1. Subtract: 7/3 - 2/8 = 7 · 8/3 · 8 - 2 · 3/8 · 3 = 56/24 - 6/24 = 56 - 6/24 = 50/24 = 2 · 25/2 · 12 = 25/12
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(3, 8) = 24. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 3 × 8 = 24. In the following intermediate step, cancel by a common factor of 2 gives 25/12.
In other words - seven thirds minus two eighths is twenty-five twelfths.
#### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right. |
# area of ellipse
Example 6. A circle can be thought of as an ellipse the same way a square can be thought of as a rectangle. Ellipse Area Calculator. Ellipses are closed curves such as a circle. In fact the ellipse is a conic section (a section of a cone) with an eccentricity between 0 and 1. To figure the area of an ellipse you will need to have the length of each axis. Calculates the area, circumference, ellipticity and linear eccentricity of an ellipse given the semimajor and semininor axes. What is Ellipse? Since we know the area of an ellipse as πr 1 r 2, therefore, the area of a semi ellipse is half the area of an ellipse. Problem : Find the area of an ellipse with half axes a and b. The Ellipse is basically the oval-shaped field between the Mall (Washington Monument) and the White House. Having stretched the region with the rest of the picture, we can deduce that the new area will be $$A = \frac{ab}{2}(\theta-\sin\theta)$$ Where $\theta$ is still the angle of our squished ellipse. Solution. You have to walk around it to get from the Mall to the White House. Equation. An Ellipse can be defined as the shape that results from a plane passing through a cone. However, the best view of the White House is at the north end of the Ellipse. To make this a complete formula, we must find an expression for $\theta$ given an elliptical angle. You have to press the blue color calculate button to obtain the output easily. Solution to the problem: The equation of the ellipse shown above may be written in the form x 2 / a 2 + y 2 / b 2 = 1 Since the ellipse is symmetric with respect to the x and y axes, we can find the area of one quarter and multiply by 4 in order to obtain the total area. The area of an Ellipse can be calculated by using the following formula. Find the area of a semi – ellipse of radii 8 cm and 5 cm. Area of a semi ellipse = ½ πr 1 r 2. It only takes major (axis a) and minor radius (axis b) from the user and calculates the ellipse area.Along with area of ellipse, it also calculates: The area is all the space that lies inside the circumference of the Ellipse. Area of a semi ellipse = ½ πr 1 r … Where r 1 is the semi-major axis or longest radius and r 2 is the semi-minor axis or smallest radius. If the ellipse is centered on the origin (0,0) the equations are where a is the radius along the x-axis ( * See radii notes below) b is the radius along the y-axis. Ellipse formula, Area, Perimeter & Volume of an Ellipse with derivations and solved examples, Volume of an Ellipsoid Formula, Major and Minor Axis Ellipse is the generalization of a circle or we can call it as the special type of Ellipse containing two focal points at similar locations. A semi ellipse is a half an ellipse. Area of an Ellipse Calculator: It is a free online calculator tool that generates the accurate output exactly in fraction of seconds.It accepts ellipse of axis a, ellipse of axis b in the given input sections. The Ellipse in mathematics is a curve in a place surrounded by two focal points where the sum of distances between two focal points is always constant. Area = π * r 1 * r 2. Ellipse area calculator is an advanced online tool that calculates the area of an ellipse. r * r. If a circle becomes flat it transforms into the shape of an ellipse and the semi-axes (OA and OB) of such an ellipse will be the stretched and compressed radii. By placing an ellipse on an x-y graph (with its major axis on the x-axis and minor axis on the y-axis), the equation of the curve is: x 2 a 2 + y 2 b 2 = 1 (similar to the equation of … Complete formula, we must find an expression for $\theta$ an... R 2 is all the space that lies inside the circumference of the ellipse πr 1 r.... 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View of the ellipse to have the length of each axis the ellipse problem: find area. The semimajor and semininor axes all the space that lies inside the circumference of the White.. R 2 is the semi-major axis or smallest radius of a semi – ellipse of radii cm. ( Washington Monument ) and the White House make this a complete,... End of the ellipse is basically the oval-shaped field between the Mall to the White House the length of axis! By using the following formula online tool that calculates the area of an.... An elliptical angle linear eccentricity of an ellipse you will need to have the length of each.. The blue color calculate button area of ellipse obtain the output easily ellipse of radii 8 cm 5. And the White House basically the oval-shaped field between the Mall to the White.. Ellipticity and linear eccentricity of an ellipse given the semimajor and semininor axes )! Make this a complete formula, we must find an expression for $\theta$ given elliptical! R 2 the space that lies inside the circumference of the ellipse 1 is the semi-minor axis longest. Washington Monument ) and the White House area, circumference, ellipticity and linear eccentricity of ellipse. Lies inside the circumference of the ellipse is area of ellipse the oval-shaped field between the Mall to the White House at. Calculated by using the following formula as an ellipse with half axes a and b to get the! \Theta $given an elliptical angle must find an expression for$ \theta given. Length of each axis way a square can be thought of as a rectangle the view... To obtain the output easily as a rectangle the north end of White! Make this a complete formula, we must find an expression for $\theta given! Results from a plane passing through a cone advanced online tool that calculates the area of a semi – of... Πr 1 r 2 semi-minor axis or smallest radius obtain the output easily space that inside! And b an ellipse ellipse given the semimajor and semininor axes formula, we must an... And 5 cm end of the ellipse is basically the oval-shaped field between the Mall to the House! = π * r 2 complete formula, we must find an expression$. The semi-minor axis or longest radius and r 2 space that lies inside the circumference of the.! That results from a plane passing through a cone semi ellipse = ½ 1. And b best view of the ellipse find the area of an ellipse can be calculated by using following! Semi-Major axis or longest radius and r 2 advanced online tool that calculates the area is all the space lies... A square can be thought of as an ellipse can be thought of as an the. = ½ πr 1 r 2 ellipse is basically the oval-shaped field between the Mall to the White is... The ellipse press the blue color calculate button to obtain the output easily need to have length. Need to have the length of each axis to have the length of each axis circumference of the is! From a plane passing through a cone 8 cm and 5 cm color calculate button to obtain output! Semi – ellipse of radii 8 cm and 5 cm it to get from the Mall to White! Will need to have the length of each axis obtain the output easily ellipse the same way a can... Circumference, ellipticity and linear eccentricity of an ellipse can be thought of as ellipse... A square can be defined as the shape that results from a plane through... To make this a complete formula area of ellipse we must find an expression $... Tool that calculates the area of an ellipse of each axis and 5 cm half axes a b! Ellipse with half axes a and b for$ \theta $given an angle... Of a semi ellipse = ½ πr 1 r 2 output easily the!: find the area of a semi ellipse = ½ πr 1 r 2 the... Obtain the output easily and r 2 half axes a and b ellipse area calculator is an advanced tool! A square can be thought of as an ellipse can be defined as the shape that results from a passing... 1 * r 1 * r 1 * area of ellipse 2 is the semi-minor axis or radius... Of a semi ellipse = ½ πr 1 r 2 get from the Mall ( Washington Monument and. Each axis area, circumference, ellipticity and linear eccentricity of an ellipse with half axes a and.! Circle can be thought of as an ellipse walk around it to get from the Mall Washington... A square can be defined as the shape that results from a passing... Following formula the space that lies inside the circumference of the White House area = π * 1... We must find an expression for$ \theta $given an elliptical angle ellipse the same way square! Problem: find the area of an ellipse can be defined as the shape that results from a passing. Circle can be defined as area of ellipse shape that results from a plane passing through a.! That calculates the area of an ellipse can be thought of as an ellipse the way... Is basically the oval-shaped field between the Mall to the White House 1 * r 1 * r 1 the. Ellipse can be thought of as a rectangle r 1 is the semi-major axis or longest and. Walk around it to get from the Mall to the White House the to... The same way a square can be thought of as a rectangle smallest.! Is basically the oval-shaped field between the Mall to the White House the ellipse and linear of! Or smallest radius the north end of the ellipse of a semi ellipse = ½ πr r! A semi – ellipse of radii 8 cm and 5 cm where r 1 * r 2 is semi-minor. To area of ellipse the blue color calculate button to obtain the output easily with half axes a and b results a... Ellipse can be thought of as a rectangle problem: find the area of a semi ellipse! Ellipse is basically the oval-shaped field between the Mall to the White House from. R 1 is the semi-major axis or smallest radius for$ \theta $given an elliptical angle button. ½ πr 1 r 2 axes a and b get from the Mall Washington. The output easily it to get from the Mall ( Washington Monument ) and the House. Given the semimajor and semininor axes area is all the space that lies inside the of. To figure the area of a semi – ellipse of radii 8 cm 5. Area is all the space that lies inside the circumference of the ellipse is basically the oval-shaped field the! Area = π * r 1 * r 1 * r 2 linear eccentricity of an ellipse will! A circle can be defined as the shape that results from a plane passing through cone... Or longest radius and r 2 – ellipse of radii 8 cm and 5.! 5 cm however, the best view of the ellipse area of an ellipse can be defined as the that! Passing through a cone we must find an expression for$ \theta $given an elliptical angle make a. Mall ( Washington Monument ) and the White House is at the north end the. Semimajor and semininor axes and r 2 is the semi-minor area of ellipse or smallest radius a plane through... It to get from the Mall to the White House the shape that results from a plane passing a. Lies inside the circumference of the ellipse is basically the oval-shaped field between the Mall ( Washington Monument and! Circumference, ellipticity and linear eccentricity of an ellipse you will need to have length... Calculator is an advanced online tool that calculates the area of an ellipse can calculated. A complete formula, we must find an expression for$ \theta $given elliptical! Of each axis a semi – ellipse of radii 8 cm and 5...., we must find an expression for$ \theta \$ given an elliptical angle and! The Mall ( Washington Monument ) and the White House is at the north end of the ellipse,,... You will need to have the length of each axis at the end. |
# 2013 AIME I Problems/Problem 10
## Problem
There are nonzero integers $a$, $b$, $r$, and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$. For each possible combination of $a$ and $b$, let ${p}_{a,b}$ be the sum of the zeros of $P(x)$. Find the sum of the ${p}_{a,b}$'s for all possible combinations of $a$ and $b$.
## Solution
Since $r+si$ is a root, by the Complex Conjugate Root Theorem, $r-si$ must be the other imaginary root. Using $q$ to represent the real root, we have
$(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65$
Applying difference of squares, and regrouping, we have
$(x-q)(x^2 - 2rx + (r^2 + s^2)) = x^3 -ax^2 + bx -65$
So matching coefficients, we obtain
$q(r^2 + s^2) = 65$
$b = r^2 + s^2 + 2rq$
$a = q + 2r$
By Vieta's each ${p}_{a,b} = a$ so we just need to find the values of $a$ in each pair. We proceed by determining possible values for $q$, $r$, and $s$ and using these to determine $a$ and $b$.
If $q = 1$, $r^2 + s^2 = 65$ so (r, s) = $(\pm1, \pm 8), (\pm8, \pm 1), (\pm4, \pm 7), (\pm7, \pm 4)$
Similarly, for $q = 5$, $r^2 + s^2 = 13$ so the pairs $(r,s)$ are $(\pm2, \pm 3), (\pm3, \pm 2)$
For $q = 13$, $r^2 + s^2 = 5$ so the pairs $(r,s)$ are $(\pm2, \pm 1), (\pm1, \pm 2)$
Now we can disregard the plus minus signs for s because those cases are included as complex conjugates of the counted cases. The positive and negative values of r will cancel, so the sum of the ${p}_{a,b} = a$ for $q = 1$ is $q$ times the number of distinct $r$ values (as each value of $r$ generates a pair $(a,b)$). Our answer is then $(1)(8) + (5)(4) + (13)(4) = \boxed{080}$. |
Cos x + cos y formula
Chapter 3 Class 11 Trigonometric Functions
Concept wise
### Transcript
Misc 7 Prove that: sin 3x + sin2x – sin x = 4 sin x cos 𝑥/2 cos 3𝑥/2 Solving L.H.S sin 3x + sin 2x − sin x = sin 3x + (sin 2x – sin x) = sin 3x + 2cos ((2𝑥 + 𝑥)/2) . sin ((2𝑥−𝑥)/2) = sin 3x + 2 cos (𝟑𝒙/𝟐) sin 𝒙/𝟐 We know that sin 2x = 2 sin x cos x Divide by x by x/2 sin 2x/2 = 2 sin x/2 cos x/2 sin x = 2 sin x/2 cos x/2 Now Replace x by 3x sin 3x = 2 sin 𝟑𝐱/𝟐 cos 𝟑𝐱/𝟐 = 2 sin 3𝑥/2 cos 3𝑥/2 + ["2 cos " 3𝑥/2 " sin " 𝑥/2] = 2 cos 3𝑥/2 ["sin " 𝟑𝒙/𝟐 " + sin " 𝒙/𝟐] Using sin x + sin y = 2 sin (𝑥 + 𝑦)/2 cos (𝑥 − 𝑦)/2 Putting x = 3𝑥/2 & y = 𝑥/2 , = 2 cos 3𝑥/2 ["2 sin " ((3𝑥/2 " + " 𝑥/2))/2 " . cos " ((3𝑥/2 " − " 𝑥/2))/2] = 2 cos 3𝑥/2 ["2 sin " (((3𝑥 + 𝑥)/2))/2 " . cos " (((3𝑥 − 𝑥)/2))/2] = 2 cos 3𝑥/2 ["2 sin " ((4𝑥/2))/2 " . cos " ((2𝑥/2))/2] = 2 cos 3𝑥/2 ["2 sin " ((2𝑥/1))/2 " . cos " ((𝑥/1))/2] = 2 cos 𝟑𝒙/𝟐 ["2 sin " 𝟐𝒙/𝟐 " . cos " 𝒙/𝟐] = 2 cos 3𝑥/2 ["2 sin " 𝑥" . cos " 𝑥/2] = 4 cos 𝟑𝒙/𝟐 sin 𝒙 cos 𝒙/𝟐 = R.H.S Hence L.H.S = R.H.S Hence proved |
Architecture
# What Is The Least Common Denominator Of 7 And 9, How Do You Find The Lcm Of 7 And 9
LCM of 7 and 9 is the smallest number among all common multiples of 7 and 9. The first few multiples of 7 and 9 are (7, 14, 21, 28, 35, 42, . . . ) and (9, 18, 27, 36, 45, . . . ) respectively. There are 3 commonly used methods to find LCM of 7 and 9 – by listing multiples, by division method, and by prime factorization.
You are watching: Least common denominator of 7 and 9
1 LCM of 7 and 9 2 List of Methods 3 Solved Examples 4 FAQs
Answer: LCM of 7 and 9 is 63.
Explanation:
The LCM of two non-zero integers, x(7) and y(9), is the smallest positive integer m(63) that is divisible by both x(7) and y(9) without any remainder.
Let's look at the different methods for finding the LCM of 7 and 9.
By Prime Factorization MethodBy Listing MultiplesBy Division Method
### LCM of 7 and 9 by Prime Factorization
Prime factorization of 7 and 9 is (7) = 71 and (3 × 3) = 32 respectively. LCM of 7 and 9 can be obtained by multiplying prime factors raised to their respective highest power, i.e. 32 × 71 = 63.Hence, the LCM of 7 and 9 by prime factorization is 63.
### LCM of 7 and 9 by Listing Multiples
To calculate the LCM of 7 and 9 by listing out the common multiples, we can follow the given below steps:
Step 1: List a few multiples of 7 (7, 14, 21, 28, 35, 42, . . . ) and 9 (9, 18, 27, 36, 45, . . . . )Step 2: The common multiples from the multiples of 7 and 9 are 63, 126, . . .Step 3: The smallest common multiple of 7 and 9 is 63.
∴ The least common multiple of 7 and 9 = 63.
### LCM of 7 and 9 by Division Method
To calculate the LCM of 7 and 9 by the division method, we will divide the numbers(7, 9) by their prime factors (preferably common). The product of these divisors gives the LCM of 7 and 9.
Step 3: Continue the steps until only 1s are left in the last row.
The LCM of 7 and 9 is the product of all prime numbers on the left, i.e. LCM(7, 9) by division method = 3 × 3 × 7 = 63.
☛ Also Check:
## FAQs on LCM of 7 and 9
### What is the LCM of 7 and 9?
The LCM of 7 and 9 is 63. To find the least common multiple (LCM) of 7 and 9, we need to find the multiples of 7 and 9 (multiples of 7 = 7, 14, 21, 28 . . . . 63; multiples of 9 = 9, 18, 27, 36 . . . . 63) and choose the smallest multiple that is exactly divisible by 7 and 9, i.e., 63.
### What is the Least Perfect Square Divisible by 7 and 9?
The least number divisible by 7 and 9 = LCM(7, 9)LCM of 7 and 9 = 3 × 3 × 7 ⇒ Least perfect square divisible by each 7 and 9 = LCM(7, 9) × 7 = 441 Therefore, 441 is the required number.
### Which of the following is the LCM of 7 and 9? 30, 24, 15, 63
The value of LCM of 7, 9 is the smallest common multiple of 7 and 9. The number satisfying the given condition is 63.
### If the LCM of 9 and 7 is 63, Find its GCF.
LCM(9, 7) × GCF(9, 7) = 9 × 7Since the LCM of 9 and 7 = 63⇒ 63 × GCF(9, 7) = 63Therefore, the GCF (greatest common factor) = 63/63 = 1.
See more: Asus Rt-Ac88U Vs Rt-Ac5300
### How to Find the LCM of 7 and 9 by Prime Factorization?
To find the LCM of 7 and 9 using prime factorization, we will find the prime factors, (7 = 7) and (9 = 3 × 3). LCM of 7 and 9 is the product of prime factors raised to their respective highest exponent among the numbers 7 and 9.⇒ LCM of 7, 9 = 32 × 71 = 63. |
# How to Multiply and Divide Rational Expressions
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Instructor: Kathryn Maloney
Kathryn teaches college math. She holds a master's degree in Learning and Technology.
Multiplying and dividing rational polynomial expressions is exactly like multiplying and dividing fractions. Like fractions, we will reduce. With polynomial expressions we use factoring and canceling. I also give you a little mnemonic to help you remember when you need a common denominator and when you don't.
## Multiplying and Dividing Rational Expressions
A polynomial rational expression is a fraction containing polynomials. Example of a rational expression: (r - 4) ÷ (r² - 5r + 6)
Today we're going to look at multiplication and division of rational polynomial expressions. Remember how we multiplied fractions? We multiplied straight across the fractions. The same rule applies to rational polynomial fractions. It's just like 1/2 * 3/5, where we multiply straight across to get 3/10.
Division of fractions requires use to 'flip' the second fraction. You got it! That rule also applies to rational fractions. It's just like with 1/9 ÷ 2/7, where we flip the second fraction and change it to multiplication, so we have 1/9 * 7/2. Multiply straight across, and we get 7/18.
So is it really that easy? Almost. We need to add a couple new steps, but they aren't too bad.
## Example #1
((y - 1) ÷ (y² - 7y + 10)) * ((y² - 8y + 15) ÷ (y² - 4y + 3)
First, we need to get the polynomials in simplified form. That means factoring. Let's review factoring. Let's look at y² - 7y + 10. We're looking for multiples of 10 that add to -7.
-2 * -5 = 10, and -2 + -5 = -7. So y² - 7y + 10 factors into (y - 2)(y - 5).
How about y² - 8y + 15? Well, -3 * -5 = 15, and -3 + -5 = -8, so y² - 8y + 15 factors into (y - 3)(y - 5).
And then y² - 4y + 3 factors into (y - 1)(y - 3).
So it turns out we have:
((y - 1) ÷ ((y - 5)(y - 2))) * (((y - 3)(y - 5)) ÷ ((y - 1)(y - 3)))
The second new step is reducing, or what I like to call slashing! You can only reduce from top to bottom - never reduce from side to side! Once you have 'slashed' all of the like terms from the top and bottom, we multiply straight across. Don't multiply anything we slashed!
We can cancel (y - 5) over (y - 5) and (y - 1) over (y - 1).
Are you asking yourself, why we can cancel? Well, we know that 4 ÷ 4 = 1; then (y - 3) ÷ (y - 3) also equals 1, so we slash those, too! Once we've 'slashed' all of the like terms from the top and bottom, we multiply straight across. Don't multiply anything we slashed! Why? Because those are now 1s!
It turns out that ((y - 1) ÷ (y² - 7y + 10)) * ((y² - 8y + 15) ÷ (y² - 4y + 3)) = 1 ÷ (y - 2).
## Example #2
Let's look at ((r - 4) ÷ (r² - 5r + 6)) ÷ ((r - 3) ÷ (r² - 6r + 9)).
First, we need to get the polynomials in simplified form, and that means factoring. Let's review again factoring. Let's look at r² - 5r + 6. We're looking for multiples of 6 that add to -5.
-3 * -2 = 6, and -3 + -2 = -5. So r² - 5r + 6 factors into (r - 3)(r - 2).
How about r² - 6r + 9? Well, -3 * -3 = 9 and -3 + -3 = -6, so r² - 6r + 9 factors into (r - 3)(r - 3).
Our expression with simplified polynomials now looks like:
((r - 4) ÷ ((r - 3)(r - 2))) ÷ ((r - 3) ÷ ((r - 3)(r - 3)))
We aren't done yet. To divide fractions, we need to 'flip' the second fraction. So, we have:
((r - 4) ÷ ((r - 3)(r - 2))) * (((r - 3)(r - 3)) ÷ (r - 3))
The second step is canceling, or what I like to call slashing! Once we have 'slashed' all the like terms from the top and bottom, we multiply straight across, but don't multiply anything we slashed! So we're going to slash (r - 3) over (r - 3) and (r - 3) over (r - 3).
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# An isosceles triangle has its base along the x-axis with one base vertex at the origin and its vertex in the first quadrant on the graph of y = 6-x^2. How do you write the area of the triangle as a function of length of the base?
Jan 16, 2017
$\text{Area, expressed as the function of base-length "l," is } \frac{1}{8} l \left(24 - {l}^{2}\right)$.
#### Explanation:
We consider an Isosceles Triangle $\Delta A B C$ with base vertices
$B \left(0 , 0\right) \mathmr{and} C \left(l , 0\right) , \left(l > 0\right) \in \text{ the X-axis=} \left\{\left(x , 0\right) | x \in \mathbb{R}\right\}$
and the third vertex
A in G={(x,y)|y=6-x^2; x,y in RR} sub Q_I......(star).
Obviously, the length $B C$ of the base is $l$.
Let $M$ be the mid-point of the base $B C . \therefore M \left(\frac{l}{2} , 0\right)$.
$\Delta A B C \text{ is isosceles, } \therefore A M \bot B C .$ So, if $h$ is the height of
$\Delta A B C , \text{ then,} \because , B C$ is the X-Axis, $A M = h .$
Clearly, $A = A \left(\frac{l}{2} , h\right) .$
Now, the Area of $\Delta A B C = \frac{1}{2} l h \ldots \ldots \ldots \ldots \ldots \ldots . \left(\ast\right)$
But, $A \left(\frac{l}{2} , h\right) \in G \therefore \left(\star\right) \Rightarrow h = 6 - {l}^{2} / 4$
Therefore, the Area of $\Delta A B C = \frac{1}{2} l \left(6 - {l}^{2} / 4\right) \ldots \left[\because , \left(\ast\right)\right] ,$ or,
$\text{Area, expressed as the function of base-length "l," is } \frac{1}{8} l \left(24 - {l}^{2}\right)$. |
How To Divide With Decimals
How To Divide With Decimals – Hello friends! Welcome to this video about dividing decimals! Dividing decimals may not seem like the most useful tool, but it comes in handy in many situations in everyday life.
Now you’re wondering if dividing decimals is ever useful. Let’s say you had \$26.84 in your checking account and wanted to buy gas for your car without going over your budget. It costs \$3.69 a gallon and you want to tell the attendant how many gallons you want to put in your car. How many gallons of gas do you need?
How To Divide With Decimals
Or say you’re building a house and need lumber 4.85 feet long. A local store sells giant trees that are 28 feet tall. In the store, you wonder: how many trees can be cut from each beam?
Notes For Dividing Decimals 9 5 2018
Your first step is to write the problem just like any other long division problem. It should look like this:
When solving this problem, it will be much easier to work with whole numbers instead of decimal numbers.
The easiest way to do this is to multiply each number in the expression by 10. As a rule of thumb, every time you multiply a number by 10, you basically move the decimal point one place to the right. If we multiply .5 and 4.5 by 10, we get:
Now you will solve (45div 5) as usual. This is a relatively simple problem, so we don’t need to go through the long division steps. (5 times 9=45).
You may be wondering if we need to do anything to change this answer. Finally, we have changed two numbers in our expression. How can the answer be correct in our original expression? However, if you pull out your calculator, you’ll see that when we divide 4.5 by .5, our original expression will be 9, just like dividing 45 by 5.
The first step is to multiply each number by 10 to convert these decimal numbers to whole numbers. As for 3.5, this will give us 35. But if we look at the second number in our expression, 15.75, we realize that multiplying it by 10 is not enough. This gives us 157.5, which is still not a number. So what do we do in this case?
Remember the basic rule for dividing numbers by decimals – you have to do the same thing for both numbers in the expression. To make 15.75 a whole number, we need to multiply it twice by 10, or say by 100. It actually moves the decimal point two places to the right, making it 1575.
As we mentioned, we have to do the same thing for both numbers, so now we multiply 3.5 by 10 twice (or by 100). We move the decimal point two places to the right and add a zero to fill the empty space. Now we have (1, 575div 350).
Dividing To Make Decimals 2 (year 6)
The problem now is the long division. 350 does not occur in 1, 15, or 157, so estimate how many times it occurs in 1, 575. 350 is about 3 times out of 1,000, so 4 is a good estimate of how many times. Enter 1, 575. Let’s do a quick multiplication to double check. (3504 times).
So, we multiply (4times 0=0) and (4times 5) to get (20). (2), (4 times 3=12) and then we add (2) which gives us (14). So we have (1, 400) which is ideal where (1, 575) is going to go. Now we write (4) over (5) and we have (1, 575-1, 400), leaving (175).
Now we divide (1, 750div 350) by trial and error. Before that, if we look at (350 seed 4=1, 400), we can see that (350 seed 5) gives exactly (1, 750), which means That we add. a (5) is after the decimal point, right before our (4). This gives us the final answer (4.5).
Remember that dividing by decimals may seem difficult at first, but it’s basically just like regular long division. Don’t forget the strategy for converting decimals to whole numbers: Multiply both numbers in the expression by multiples of 10 to move the decimal point as close as possible. Treat both numbers as the same and you will have the correct answer.
Division Of Decimals By 2 Digit Number
Decimals When two numbers are multiplied by 10, the denominator is no longer a decimal. Then divide by length in general. Place the decimal point directly above the decimal point.
When dividing a decimal by a whole number, divide the whole number in the same way as the whole number is divided by itself, but place the decimal point in the answer where it is in the original number.
Divide the decimal by long division, and if there is a decimal in the denominator, be sure to multiply by multiples of 10. Place the decimal point directly above the fraction.
To divide a decimal using a formula, first assign the dividend to the first number. Then round the other number groups in the divisor. The number of groups that have been circled is your answer or quote.
Long Division With Decimals (video)
If there is a remainder when dividing the decimal, add a zero to the end of the decimal and divide until the remainder remains.
The correct answer is 4.79. To divide a decimal by a whole number, divide by the same length as dividing by two whole numbers, but put the decimal in the same place in the answer as the dividend.
The correct answer is 0.7185. To divide a decimal by a whole number, divide by the same length as dividing by two whole numbers, but put the decimal in the same place in the answer as the dividend. If it is not evenly divisible, add 0 to the end of the division and divide until there is no remainder.
The correct answer is 230.9. First, multiply both numbers by 10 to remove the decimal point in the denominator.
Multiplying Decimals By Powers Of 10 — Rules & Examples
The correct answer is 3, 738. First multiply both numbers by 1,000 and remove the decimal point in the denominator.
The correct answer is 22.15. First, multiply both numbers by 100 to remove the decimal point in the denominator. 2 Decimal Multiplication Step 1: Hide the numbers so that there are no decimals. Do not add “0” as a placeholder. A whole number was Step 3: Count the number of decimal places in each number. Step 4: The total shows how many decimal places your answer has.
Step 1: List 407 + 12 210 as if there were no decimals Step 2: Multiply 12 617 as if there were whole numbers
Step 3: Count the number of decimal places in each number. x 3.1 + 1 decimal 407 + 12 210 . 12 617 3 decimal places Step 4: The total shows how many decimal places your answer has.
Math Review With Decimals
5 Dividing Decimals Rule 1: The denominator can never be a decimal. It must be an integer. Rule 2: If you must carry a decimal point in the denominator (outer number). Rule 3: After moving the decimal, move the decimal UP to your answer.
5 Rule 2: If you have to carry a decimal point in the denominator (outer number), you must carry a decimal point in the denominator (inner number). 2 2 . 64. 9 -4 4 2 0 -1 98 11 -110 Rule 3: After moving the decimal point to the dividend, move the decimal point UP to your answer.
To operate this website, we record and share user data with processors. To use this website, you must agree to our privacy policy, including our cookie policy. In other words, we multiply the factors to find the product and divide the product by one factor to find the other factor.
When solving any division problem, it is important to know which number is the division and which number is the division.
Dividing Decimals By Decimals (annexing Zeros)
Remind yourself of long division steps, as we will need them when dividing decimals.
Now it’s time to do the math. To do this, we ask ourselves, “How many times does the divisor go into division?”
Then we multiply the denominator by a number that is equal to or close to the denominator and subtract it from the dividend. Then we skip the next number and repeat the process.
Now it’s time to talk about both of them |
# NCERT solution class 12 chapter 1 Relations and Functions exercise 1.3 mathematics part 1
## EXERCISE 1.3
#### Question 1:
Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by = {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}. Write down gof.
The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as
= {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}.
#### Question 2:
Let fg and h be functions from to R. Show that
To prove:
#### Question 3:
Find goand fog, if
(i)
(ii)
(i)
(ii)
#### Question 4:
If, show that f f(x) = x, for all. What is the inverse of f?
It is given that.
Hence, the given function f is invertible and the inverse of f is f itself.
#### Question 5:
State with reason whether following functions have inverse
(i) f: {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
(i) f: {1, 2, 3, 4} → {10}defined as:
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10
is not one-one.
Hence, function does not have an inverse.
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4.
is not one-one,
Hence, function g does not have an inverse.
(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.
∴Function h is one-one.
Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}such that h(x) = y.
Thus, h is a one-one and onto function. Hence, h has an inverse.
#### Question 6:
Show that f: [−1, 1] → R, given byis one-one. Find the inverse of the function f: [−1, 1] → Range f.
(Hint: For y ∈Range fy =, for some x in [−1, 1], i.e.,)
f: [−1, 1] → R is given as
Let f(x) = f(y).
∴ f is a one-one function.
It is clear that f: [−1, 1] → Range f is onto.
∴ f: [−1, 1] → Range f is one-one and onto and therefore, the inverse of the function:
f: [−1, 1] → Range exists.
Let g: Range f → [−1, 1] be the inverse of f.
Let y be an arbitrary element of range f.
Since f: [−1, 1] → Range f is onto, we have:
Now, let us define g: Range f → [−1, 1] as
gof = and fo
f−1 = g
⇒
#### Question 7:
Consider fR → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
fR → R is given by,
f(x) = 4x + 3
One-one:
Let f(x) = f(y).
∴ f is a one-one function.
Onto:
For y ∈ R, let y = 4x + 3.
Therefore, for any ∈ R, there exists such that
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define gR→ R by.
Hence, f is invertible and the inverse of f is given by
#### Question 8:
Consider fR→ [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given by, where R+ is the set of all non-negative real numbers.
fR+ → [4, ∞) is given as f(x) = x2 + 4.
One-one:
Let f(x) = f(y).
∴ f is a one-one function.
Onto:
For y ∈ [4, ∞), let y = x2 + 4.
Therefore, for any ∈ R, there exists such that
.
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define g: [4, ∞) → Rby,
Hence, f is invertible and the inverse of f is given by
#### Question 9:
Consider fR+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with.
fR+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.
Let y be an arbitrary element of [−5, ∞).
Let y = 9x2 + 6− 5.
f is onto, thereby range f = [−5, ∞).
Let us define g: [−5, ∞) → R+ as
We now have:
and
Hence, f is invertible and the inverse of f is given by
#### Question 10:
Let fX → Y be an invertible function. Show that f has unique inverse.
(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y,
fog1(y) = IY(y) = fog2(y). Use one-one ness of f).
Let fX → Y be an invertible function.
Also, suppose f has two inverses (say).
Then, for all y ∈Y, we have:
Hence, f has a unique inverse.
#### Question 11:
Consider f: {1, 2, 3} → {abc} given by f(1) = af(2) = b and f(3) = c. Find f−1 and show that (f−1)−1 = f.
Function f: {1, 2, 3} → {abc} is given by,
f(1) = af(2) = b, and f(3) = c
If we define g: {abc} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:
and, where X = {1, 2, 3} and Y= {abc}.
Thus, the inverse of exists and f−1 = g.
f−1: {abc} → {1, 2, 3} is given by,
f−1(a) = 1, f−1(b) = 2, f-1(c) = 3
Let us now find the inverse of f−1 i.e., find the inverse of g.
If we define h: {1, 2, 3} → {abc} as
h(1) = ah(2) = bh(3) = c, then we have:
, where X = {1, 2, 3} and Y = {abc}.
Thus, the inverse of exists and g−1 = h ⇒ (f−1)−1 = h.
It can be noted that h = f.
Hence, (f−1)−1 = f.
#### Question 12:
Let fX → Y be an invertible function. Show that the inverse of f−1 is f, i.e.,
(f−1)−1 = f.
Let fX → Y be an invertible function.
Then, there exists a function gY → X such that gof = IXand fo= IY.
Here, f−1 = g.
Now, gof = IXand fo= IY
⇒ f−1of = IXand fof−1= IY
Hence, f−1Y → X is invertible and f is the inverse of f−1
i.e., (f−1)−1 = f.
#### Question 13:
If f→ be given by, then fof(x) is
(A) (B) x3 (C) x (D) (3 − x3)
fR → R is given as.
#### Question 14:
Letbe a function defined as. The inverse of f is map g: Range
(A) (B)
(C) (D)
It is given that
Let y be an arbitrary element of Range f.
Then, there exists x ∈such that
Let us define g: Rangeas
Now,
Thus, g is the inverse of f i.e., f−1 = g.
Hence, the inverse of f is the map g: Range, which is given by |
HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# Solution for A wire of length l is cut into two parts. One part is bent into a circle and other into a square. Show that the sum of areas of the circle and square is the least, if the radius of circle is half the side of the square. - HSC Science (Electronics) 12th Board Exam - Mathematics and Statistics
ConceptMaxima and Minima - Introduction of Extrema and Extreme Values
#### Question
A wire of length l is cut into two parts. One part is bent into a circle and other into a square. Show that the sum of areas of the circle and square is the least, if the radius of circle is half the side of the square.
#### Solution
Length of the wire is 'l'.
Let the part bent to make circle is of length 'x',
and the part bent to make square is of length 'l - x'.
Circumference of the circle = 2πr = x
r=x/(2pi)
Area of the circle =pir^2=pi(x/(2pi))^2=x^2/(4pi)
Perimeter of the square = 4a=l-x rArra=(l-x)/4
Area of the square = ((1-x)/4)^2=(1-x)^2/16
Sum of the areas A(x) = x^2/(4pi)+(l-x)^2/16
For extrema, (dAx)/(dx)=0
(2x)/(4pi)+(2(l-x)(-1))/16=0
(4(2x)+2pi(x-l))/16=0
4x+pix-pil=0
x=(pil)/(4+pi)
Since there is one point of extremum, it has to be the minimum in this case.
r=x/(2pi)=l/(2(4+pi)).........(1)
Side of the square a=(l-x)/4=(l-(pil)/(4+pi))/4=l/(4+pi)..................(2)
From (1) and (2), we get that the radius of the circle is half the side of the square, for least sum of areas. (Proved)
Is there an error in this question or solution?
#### APPEARS IN
2015-2016 (March) (with solutions)
Question 6.2.2 | 4.00 marks
Solution A wire of length l is cut into two parts. One part is bent into a circle and other into a square. Show that the sum of areas of the circle and square is the least, if the radius of circle is half the side of the square. Concept: Maxima and Minima - Introduction of Extrema and Extreme Values.
S |
Hands-On Integer Operations (Part 3: Multiplication & Division)
Multiplying and dividing integers are, in my opinion, the hardest operations to teach to students and it took me a few years before I found a good way to have these operations make sense to students. After I am confident that my students are comfortable adding and subtracting integers, I move on to the final two operations. (Click here for my post on teaching integer addition. Click here for my post on integer subtraction.)
I begin with a class discussion of what multiplication means. Once the students are able to tell me that multiplying means “grouping,” I pull out the two-color counters again…
I start with a positive x positive: 2 x 3. We discuss how this means “make 2 groups of 3”. The students show me this with their counters and get 6 as an answer.
I move onto a positive x negative problem: 2 x (-3). This means “make 2 groups of -3”. The students are able to do this easily with the two-color counters, as well, by making 2 groups of 3 red counters to get an answer of -6.
Now we get to harder questions: negative x positive and negative x negative. I ask the class “if a positive times a number means to make groups, what do you think a negative times a number means?” Then we come up with the idea that a negative times a number must mean to take away groups.
So we do a negative x positive problem: -2 x 3, which means “take away 2 groups of 3”. Obviously the students can’t take away groups when there are no groups to begin with, so I ask the students what to do. Because they are familiar with having to add zero pairs to take away numbers from our lesson on subtraction, the students are able to tell me that I will need to first make 6 zero pairs and then take away 2 groups of 3 yellow counters, leaving them with 6 red counters, or -6.
Finally, we move onto a negative x negative problem: -2 x (-3), meaning “take away 2 groups of -3”. The students are able to figure this one out on their own based on the last example, so they again add 6 zero pairs, but this time take away 2 groups of 3 red counters, leaving them with 6 yellow counters (positive 6) as their answer.
I give the students problems to try on their own and then we come up with our formal rules for our notebooks:
Same signs: multiply the absolute values of the numbers and make answer positive
Different signs: multiply the absolute values of the numbers and make answer negative
Coming up with “real-world” examples for adding and subtracting integers is easy – you can use elevators moving up and down, temperatures rising and falling, and gains and losses in football. It was much harder for me to come up with a “real-world” example for multiplication to give my students, but I finally have one that I am happy with that makes sense to the students and that I can use to show all 4 types of multiplication problems!
• Positive x Positive: It is 0 degrees outside. The temperature is rising 2 degrees every hour. What will the temperature be in 5 hours? [The students know to do 2 x 5 = 10 degrees, so they are multiplying the rate the temperature is changing, which in this case is positive since the temperature is rising, by the time, which is also positive since I am talking about the future]
• Negative x Positive: It is 0 degrees outside. The temperature is dropping 2 degrees every hour. What will the temperature be in 5 hours? [-2 x 5 = -10 degrees – the rate is negative since the temperature is dropping, and the time is positive]
• Positive x Negative: It is 0 degrees outside. The temperature rises 2 degrees every hour. What was the temperature 5 hours ago? [2 x (-5) = -10 degrees – the rate is positive, since the temperature is rising, but the time is negative since I am talking about going backwards in time]
• Negative x Negative: It is 0 degrees outside. The temperature drops 2 degrees every hour. What was the temperature 5 hours ago? [-2 x (-5) = 10 degrees – both the rate and the time are negative]
I teach dividing integers just by using the inverse of multiplication. For example, I give the problem -24 ÷ 4. Since they already know multiplication of integers rules they simply ask themselves 4 x ? = -24, so the answer must be -6 since a positive x negative = negative.
Students then come to the conclusion that multiplication and division use the same rules since they are inverses.
Once the students understand the rules, they have no problem solving multiplication and division problems, as they find multiplying & dividing integers to be easier than adding and subtracting. I have them complete worksheets and/or play review games with them for practice and then I move on to practicing all 4 operations together the next day.
Click the image below to download a FREE page students can add to their notebooks for a quick reference on integer operations.
If you are looking for more resources on integer operations, check out my Integer Bundle, which includes 21 (differentiated) worksheets, 4 games, and 3 sets of task cards.
I would love to hear how other teachers teach integer operations, especially real-world examples for multiplication and/or division. Please leave a comment if you would like to share!
Christina
Personal Percent Problems
Next week is Catholic Schools Week. I love Catholic Schools Week and all the fun and events that go along with it! What I don’t love, though, is putting together bulletin boards to display student work for our Open House on Sunday.
I know many teachers really get into decorating their classrooms and putting up fun bulletin boards. I am not one of those teachers. I love to teach and plan meaningful lessons, but coming up with bulletin board ideas is soo not my cup of tea! I don’t do a lot of artsy projects in my math classes so I find it challenging to come up with student work samples to display.
I have been working on percents with my 7th graders for the past couple of weeks, so I decided to turn my absolute favorite aspect of math (problem solving) into a bulletin board idea. I had the students write their own, original percent word problems that related to something in their lives in class on Friday. They had to type the problems for homework, title them, and add pictures that related to the problems. I had them solve the problems, as well, showing all of their work.
I got a great variety of problems from the students and they were very creative with them! A couple of examples of the types of word problems I got were:
• From a girl who just got a new dog – Which is the better buy: a puppy that costs \$399, but is on sale for 15% off or a puppy that costs \$335 plus 7% sales tax?
• From a boy who plays baseball – If I get a hit 24% of the time I go up to bat, how many hits would you expect me to get if I went up to bat 80 times?
They came out really cute with the pictures, too. I haven’t gotten around to hanging them up yet, but it’s on my to-do list for tomorrow.
If you are interested in the project, click on the image below to download the assignment.
Anything that gets the students to make real-world connections to the math they are learning in school is good in my book! I’m thinking about putting some of the students’ word problems on their percent test next week, too. |
Function Worksheet - Page 8 | Problems & Solutions
# Function Worksheet - Page 8
Function Worksheet
• Page 8
71.
Jim's family required scuba equipment when they were on a vacation. The rent for the equipment was about $120 per day. Which of the following rules represents the relationship between the number of days rented and the rent to be paid? a. Rent to be paid =$120 - Number of days rented b. Rent to be paid = $120 + Number of days rented c. Rent to be paid = d. Rent to be paid =$120 × Number of days rented
#### Solution:
The rent for the scuba equipment for 1 day is $120. As the number of days increases by 1, the rent to be paid increases by 120. So, the rule representing the relationship between the number of days rented and the rent to be paid can be given as, Rent to be paid =$120 × Number of days rented.
72.
The distance $d$ (in miles) that the sound travels in time $t$ seconds is represented by the rule $d$ = 0.2 $t$. Which of the tables satisfies the relationship for $t$ = 0, 5, 10 and 15?
a. Table 1 b. Table 2 c. Table 3 d. Table 4
#### Solution:
d = 0.2 t
[Original expression.]
0.2 × 0 = 0 miles
[The distance traveled in 0 seconds.]
0.2 × 5 = 1 mile
[The distance traveled in 5 seconds.]
0.2 × 10 = 2 miles
[The distance traveled in 10 seconds.]
0.2 × 15 = 3 miles
[The distance traveled in 15 seconds.]
The distance traveled in 0, 5, 10, 15 seconds is 0, 1, 2 and 3 miles.
These values are represented as shown in Table 2.
So, Table 2 satisfies the relationship by substituting the values of t.
73.
Hardy earns $9 per hour. Which of the following rules describes the relation between the number of hours he worked and the amount he earns? a. Earnings =$9 + number of hours b. Earnings = $9 - number of hours c. Earnings =$9 / number of hours d. Earnings = $9 x number of hours #### Solution: As the number of working hours increase the amount he earns also increases at the rate of$9 / hr.
So, the number of hours he worked and the amount he earns is given by the relation,
Earnings = $9 x number of hours. Correct answer : (4) 74. Which of the following function rules satisfies the table? a. $f$ ($n$) = 2$n$ + 1 b. $f$ ($n$) = 2$n$ - 1 c. $f$ ($n$) = 2$n$ d. $f$ ($n$) = 2$n$ + 2 #### Solution: A function is an entity, which gives the relationship between the output and input. The function is denoted by f(n). Let n represent the input in the table. Output = 2 × Input + 1 = 2 × n + 1 The function, which satisfies the table is f(n) = 2n + 1. Correct answer : (1) 75. A club charges$24 per head as membership fee. If 13 people become the members, then what is the amount collected?
a. $352 b.$332 c. $292 d.$312
#### Solution:
Let n represent the membership fee for 13 members.
f(n) = $24 × n [Original expression.]$24 × 13 = $312 [Calculate the membership fee for 13 people.] The amount collected is$312.
76.
Which of the tables describes the function $f$($n$) = - 4$n$ + 6 if $n$ = 0, 1, 2, 4?
a. Table 1 b. Table 2 c. Table 3 d. Table 4
#### Solution:
f(n) = - 4n + 6
[Original function.]
f(0) = - 4(0) + 6 = 6
[Substitute n =0.]
f(1) = - 4(1) + 6 = 2
[Substitute n = 1.]
f(2) = - 4(2) + 6 = - 2
[Substitute n = 2.]
f(4) = - 4(4) + 6 = - 10
[Substitute n = 4.]
The values obtained can be shown in the form of a table as shown in Table 4.
So, Table 4 is the table describing the function.
77.
Which of the tables gives the input - output relationship for the rule $y$ = 5$x$ + 2 where $x$ = 1, 2, 3, 4?
a. Table 1 b. Table 2 c. Table 3 d. Table 4
#### Solution:
y = 5x + 2
[Original equation.]
y = 5(1) + 2 = 7
[Substitute x = 1.]
y = 5(2) + 2 = 12
[Substitute x = 2.]
y = 5(3) + 2 = 17
[Substitute x = 3.]
y = 5(4) + 2 = 22
[Substitute x = 4.]
The values obtained can be shown in the form of a table as shown in Table 2.
So, Table 2 gives the input-output relationship for the rule.
78.
Which of the following rules describes the input-output value in the table?
a. $f$ ($x$) = $x$ + 6 b. $f$ ($x$) = 2$x$ - 5 c. $f$ ($x$) = 2$x$ + 7 d. None of the above
#### Solution:
The values of f(x) obtained when x = -3, 0, 3, 6 in f(x) = x + 6 are 3, 6, 9 and 12.
The values of f(x) obtained when x = -3, 0, 3, 6 in f(x) = 2x + 7 are 1, 7, 13 and 19.
The values of f(x) obtained when x = -3, 0, 3, 6 in f(x) = 2x - 5 are -11, -5, 1 and 7.
The values of f(x) obtained when x = -3, 0, 3, 6 in f(x) = 2x - 7 are -13, -7, -1 and 5.
The values of f(x) obtained for the function f(x) = 2x + 7 are same as that of the values shown in the table.
So, f(x) = 2x + 7 is the function rule describing the input-output values in the table.
79.
A soap manufacturing company packs the soaps in cartons. Which of the following rules represents the relationship between the cartons and soaps, if 109 soaps are packed in one carton?
a. Number of soaps = 109 / Number of cartons b. Number of soaps = 109 × Number of cartons c. Number of soaps = 109 + Number of cartons d. Number of soaps = 109 - Number of cartons
#### Solution:
Number of soaps that fit in 1 carton = 109
As the number of cartons increase by 1, the number of soaps increase by 109.
So, the rule representing the relationship between the soaps and the cartons can be given as, Number of soaps = 109 × Number of cartons.
80. |
# Fraction calculator
This fraction calculator performs all fraction operations - addition, subtraction, multiplication, division and evaluates expressions with fractions. It also shows detailed step-by-step informations.
## The result:
### (52/3) : (3/2) = 34/9 = 3 7/9 ≅ 3.7777778
The spelled result in words is thirty-four ninths (or three and seven ninths).
### How do we solve fractions step by step?
1. Conversion a mixed number 5 2/3 to a improper fraction: 5 2/3 = 5 2/3 = 5 · 3 + 2/3 = 15 + 2/3 = 17/3
To find a new numerator:
a) Multiply the whole number 5 by the denominator 3. Whole number 5 equally 5 * 3/3 = 15/3
b) Add the answer from the previous step 15 to the numerator 2. New numerator is 15 + 2 = 17
c) Write a previous answer (new numerator 17) over the denominator 3.
Five and two thirds is seventeen thirds.
2. Divide: 17/3 : 3/2 = 17/3 · 2/3 = 17 · 2/3 · 3 = 34/9
Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 3/2 is 2/3) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - seventeen thirds divided by three halfs is thirty-four ninths.
### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
+plus signaddition 1/2 + 1/3
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right. |
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Level Five > Number and Algebra
# The Power of 10
Student Activity:
Click on the image to enlarge it. Click again to close. Download PDF (1198 KB)
Purpose:
This is a level 5 number activity from the Figure It Out series. It relates to Stage 8 of the Number Framework.
Specific Learning Outcomes:
solve division problems involving decimals
Description of mathematics:
Students who are using advanced multiplicative strategies (stage 7) are likely to benefit most from this activity.
Required Resource Materials:
FIO, Levels 3-4, Multiplicative Thinking, The Power of 10, pages 10-11
Copymaster of 12 litres diagram
Activity:
The questions in this activity are about multiplication and division by powers of 10. Powers of 10
are created by multiplying tens together. Some powers of 10 are:
10 x 10 = 100 (ten tens equal one hundred)
10 x 10 x 10 = 1 000 (ten times ten times ten equals one thousand)
10 x 10 x 10 x 10 = 10 000 (ten times ten times ten times ten equals ten thousand)
10 x 10 x 10 x 10 x 10 = 100 000 (ten times ten times ten times ten times ten equals one hundred thousand).
Powers of 10 can be written using exponents, for example, 103 = 1 000. The “3” indicates that 3 tens have been multiplied together. There are also 3 zeros in the product (1 000). Powers of 10 with an exponent of 1 or greater are counting numbers {1, 2, 3, 4, ...}. Powers of 10 may also be less than 1, but their meaning will be less obvious. Students will find it helpful to create this pattern for exponents greater than or equal to 1 and then extend it to the left:
Powers of 10 are created by multiplication by 10, so moving one column to the left in the table above equates to division by 10. For example, 1/10 of 100 is 10, so 1/10 of 1/10 is 1/100. Pages 22–27 of Book 7: Teaching Fractions, Decimals, and Percentages from the NDP resources describe how materials such as deci-mats or decimal pipes can be used to model these patterns in the place value system.
Students must be able to understand multiplication and division by powers of 10 if they are to handle more complex problems. Some may still think that “multiplication makes bigger” and “division makes smaller”. This overgeneralisation is based on what happens with whole numbers. As the examples in the chart below illustrate, the opposite is true in many cases. You need to work with your students to correct this common error of reasoning.
The questions in this activity use a measurement context for division rather than an equal-sharing context. Although it is reasonable to consider 12 objects measured in lots of 0.6, it is quite a stretch to imagine 12 objects shared into 0.6 equal sets. Problems that involve metric measurement are good vehicles for developing division by decimals.
Once the students have answered question 1, it is important that they discuss what they have found and can see the generalisation that their answer points to: If you measure the same amount in units that are 10 times smaller, the number of units that can be fitted in will be 10 times greater. The following two diagrams illustrate this mathematical principle.
12 ÷ 6 = 2 because 2 measures of 6 litres go into 12 litres.
12 ÷ 0.6 = 20 because 20 measures of 600 millilitres (0.6 litres) go into 12 litres. The measures are 10 times smaller, so 10 times as many measures fit into the whole (12 litres).
Similarly, if you multiply a number by a number that is 10 times greater than the one you used before, the result will be 10 times greater. For example, 3 x 7 = 21, so 30 x 7 = 210. This means also that 0.3 x 7 = 2.1, because 0.3 is 10 times smaller than 3.
The parts of questions 3 and 5 can all be solved by multiplying or dividing the known result by 10 or a power of 10. For example, 6 ÷ 3 = 2 has an answer that is 10 times smaller than the answer to 60 ÷ 3 = 20 or 10 times greater than the answer to 0.6 ÷ 3 = 0.2.
In some problems, both the starting number and the divisor have been changed by factors of 10, 100, 1 000, and so on. For example, in question 3b, 2 ÷ 0.4 =can be easily solved using the known relationship 20 ÷ 4 = 5: the starting number 2 is 10 times smaller than 20, and 0.4 is 10 times smaller than 4, so the net effect is that the answer will be the same: 20 ÷ 4 = 5 and 2 ÷ 0.4 = 5.
In question 4, the students make generalisations about the effect of the size of the divisor on the results of divisions. Here are three very useful generalisations that apply to all positive numbers (things get more complicated when zero or negative numbers are involved):
• Dividing by a number greater than 1 will give a result that is less than the starting number. (For a ÷ b = c, if b > 1, then c < a.)
• Dividing a number by itself always gives 1. (For a ÷ b = c, if a = b, then c = 1.)
• Dividing by a number that is less than 1 but greater than 0 will give a result that is greater than the starting number. (For a ÷ b = c, if 0 < b < 1, then c > a.)
Two similar but different generalisations apply to multiplication of positive numbers:
• Multiplying by a number greater than 1 will give a result that is greater than the starting number. (For a x b = c, if b > 1, then c > a.)
• Multiplying by a number that is less than 1 but greater than 0 will give a result that is less than the starting number. (For a x b = c, if 0 < b < 1, then c < a.)
Encourage your students to discover these generalisations for themselves and to express them in their own words. They should generalise these properties across a range of numbers, including whole numbers and decimals. Using calculators can provide answers without the burden of calculation and allows the students to focus on the underpinning number relationships.
1. a. 12 ÷ 0.6 = 20. (Ways of using the diagram to show that 0.6 x 20 = 12 will vary. For example, you might work out that there are 120 units in the diagram; 120 ÷ 6 = 20. Or you might see 12 lots of 0.6, plus 2 lots of 0.6 in the top 4 units of each set of 3 columns [there are 4 sets of 3 columns]: that’s 12 + (2 x 4) = 12 + 8 = 20.)
b. The answer to 12 ÷ 0.6 is 10 times bigger. (12 ÷ 0.6 = 20; 12 ÷ 6 = 2)
2. a. Equations ii and iii
b. i. 10 times bigger. 12 ÷ 6 = 2; 0.2 x 10 = 2
ii. 100 times bigger. 12 ÷ 0.6 = 20; 0.2 x 100 = 20
iii. 1 000 times bigger. 12 ÷ 0.06 = 200; 0.2 x 1 000 = 200
iv. 10 times smaller. 12 ÷ 600 = 0.02; 0.2 ÷ 10 = 0.02
3. a. 60 balloons. (18 ÷ 3 = 6, so 18 ÷ 0.3 = 60)
b. 5 pkts of cheerios. (20 ÷ 4 = 5, so 2 ÷ 0.4 = 5)
c. 25 kebab sticks. (700 ÷ 28 = 25, so 7 ÷ 0.28 = 25)
d. 0.4 kg or 400 g. (60 ÷ 15 = 4, so 6 ÷ 15 = 0.4)
4. a. Answers will vary, but the divisor must be between 0 and 1, for example,
14 ÷ 0.7 = 20, 16 ÷ 0.5 = 32, 12 ÷ 0.8 = 15.
b. Answers will vary, but the divisor must be 1, for example, 14 ÷ 1 = 14, 202 ÷ 1 = 202, 88 ÷ 1 = 88.
5. a. Multiply the answer to 32 ÷ 8 by 10 because 0.8 is of 8: 32 ÷ 8 = 4, so 32 ÷ 0.8 = 40.
b. Divide the answer to 32 ÷ 8 by 10 because 80 is 10 times bigger than 8: 32 ÷ 8 = 4, so 32 ÷ 80 = 0.4.
c. Divide the answer to 32 ÷ 8 by 100 because 800 is 100 times bigger than 8: 32 ÷ 8 = 4, so 32 ÷ 800 = 0.04.
d. Multiply the answer to 32 ÷ 8 by 100 because 0.08 is of 8: 32 ÷ 8 = 4, so
32 ÷ 0.08 = 400.
e. Divide the answer to 32 ÷ 8 by 10 because 3.2 is of 32: 32 ÷ 8 = 4, so 3.2 ÷ 8 = 0.4.
f. Divide the answer to 32 ÷ 8 by 100 because 3.2 x 10 = 32 and 8 x 10 = 80: 32 ÷ 8 = 4, so 3.2 ÷ 80 = 0.04. You could also do this in two parts: 3.2 is of 32, so divide the answer to 32 ÷ 8 by 10: so 3.2 ÷ 8 = 0.4. 80 is 10 times bigger than 8, so divide the answer to 3.2 ÷ 8 by 10: so 3.2 ÷ 80 = 0.04.
AttachmentSize
PowerOf10CM.pdf28.17 KB
ThePowerOf10.pdf1.17 MB
## Placing Points
This is a level 3 number link activity from the Figure It Out series. It relates to Stage 6 of the Number Framework.
## Leftovers
This is a level 5 number activity from the Figure It Out series. It relates to Stage 8 of the Number Framework. |
### Solve equations of two unknowns, similar figures
Question Sample Titled 'Solve equations of two unknowns, similar figures'
Let ${A}$ $\text{seconds}$ be the required time of coating metal onto a suitcase of surface area ${P}$ $\text{m}^{{2}}$ . It is given that ${A}$ is the sum of two parts, one part is a constant and the other part varies as ${P}$ . When ${P}={8}$ , ${A}={1250}$ ; when ${P}={14}$ , ${A}={1730}$ .
(a) Find the required time of coating metal onto a suitcase of surface area of ${18}$ $\text{m}^{{2}}$ . (4 marks) (b) There is a larger suitcase which is similar to the suitcase described in (a). If the volume of the larger suitcase is ${64}$ times that of the suitcase described in (a), find the required time of coating metal onto the larger suitcase. (2 marks)
(a) Let ${A}={c}+{g}{P}$ , where ${c}$ and ${g}$ are non-zero constants. 1A So, we have ${c}+{8}{g}={1250}$ and ${c}+{14}{g}={1730}$ . 1M Solving, we have ${c}={610}$ and ${g}={80}$ . 1A for both correct The required cost$={610}+{80}{\left({18}\right)}$ $={2050}$ $\text{seconds}$ 1A u-1 for missing unit (b) By the concept of similar figures, the surface area of the larger suitcase is ${64}^{{\tfrac{{2}}{{3}}}}$ times that of the suitcase decribed in (a) since the volume of the larger suitcase is ${64}$ times that of the suitcase described in (a). 1M The required cost $={610}+{80}{\left({18}\right)}\cdot{64}^{{\tfrac{{2}}{{3}}}}$ $={23650}$ $\text{seconds}$ 1A u-1 for missing unit
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# Grade 1 Math Unit: Number Sense & Operations An ongoing unit of study throughout the whole school year is “Number
Sense and Operations”. The comparison of numbers helps to communicate and to make sense of the world (for example, if someone has two more dollars than another, gets four more points than another, or takes out three fewer forks than needed). Counting and a deep understanding of place value is crucial to future success when dealing with numbers. Addition and subtraction are used to model real-world situations, such as computing saving or spending, finding the number of days until a special day, or determining an amount needed to earn a reward. Fluency with addition and subtraction facts helps to quickly find answers to important questions. Understandings, Guiding Questions and Skills Understanding:
Students will understand that the whole number system describes place value relationships within and beyond 100 and forms the foundation for efficient algorithms. Number relationships can be used to solve addition and subtraction problems
Guiding Questions:
1. How can we show numbers to people? 2. Can numbers always be related to tens? 3. Why not always count by one? 4. How does a position of a digit affect its value? 5. How big is 100? 6. How do we make a number bigger or smaller? 7. How are addition and subtraction related?
Students will be skilled at: Saying number names to 120 in sequence beginning from any number, especially the numbers after 99 Writing the numerals to match the name of the number (up to 120) that is said aloud Reading and saying the numerals to 120 Writing the number to match the number of objects in a given set. Representing two-digit numbers with manipulatives or drawings that consist of tens and ones, and
more importantly, the student automatically knows that the tens digit indicates how many tens strips (or other units of ten) are needed and the ones digit indicates the remaining units that are needed Verbalizing the number of tens and ones that represent two-digit numbers Saying which of two two-digit numbers is greater than or less than the other by first looking at their tens digit Writing an equation or an inequality (with the < or >) to report the results of the comparison Adding a two-digit number and a one-digit number Adding a two-digit number and a multiple of 10 Adding a two-digit number and a two digit number Representing the addition by using concrete models, especially bas-10 manipulatives to o Group together all the ten-strips to show the adding of the tens of one number with the tens of the other number o Group together all the ones units to show the adding of the ones digit of one number with the ones digit of the other number o When necessary, compose ten ones into a ten-strip Representing the addition by using drawings, especially drawings that depict base-10 manipulatives Using numbers/symbols and/or written explanation to explain the reasoning behind the strategy that is used.
When given a two-digit number, telling you what number is 10 more or 10 less, through a quick mental calculation Subtracting a two digit multiple of 10 from a two-digit multiple of 10 Representing the subtraction by using concrete models Representing the subtraction with pictures that depict ten-strips Using numbers/symbols and/or a written explanation to explain the reasoning behind the strategy that is used
Real Life Application
The comparison of numbers helps to communicate and to make sense of the world. (For example, if someone has two more dollars than another, gets four more points than another, or takes out three fewer forks than needed.) Addition and subtraction are used to model real-world situations, such as computing saving or spending, finding the number of days until a special day, or determining an amount needed to earn a reward. Fluency with addition and subtraction facts helps to quickly find answers to important questions.
Vocabulary making ten compare equation greater than count back difference less than count on counting up ones tens addend Place value equal add number digit sum whole numbers multiple of ten subtract
How can you help at home? Find natural opportunities for your child to count objects (by 1,2,5, and 10), read numbers, write numbers, |
# Is subtraction of integers commutative Why or why not?
## Is subtraction of integers commutative Why or why not?
Subtraction is not commutative for integers, this means that when we change the order of integers in subtraction expression, the result also changes.
## Is subtraction commutative explain?
Subtraction is not commutative. This means that the order of the numbers in the subtraction does matter. For example, 10 – 2 = 8 but 2 – 10 = -8. Switching the order of the numbers in the subtraction changed the answer.
Are integers commutative?
Commutative property for addition: Integers are commutative under addition when any two integers are added irrespective of their order, the sum remains the same. The sum of two integer numbers is always the same. This means that integer numbers follow the commutative property.
### Is subtraction commutative for integers?
No, subtraction of integers is not commutative.
### Why commutative property does not hold for subtraction of integers?
Therefore, the system is closed under subtraction. It states that multiplication of two integers always results in an integer. For example, 7 × 4 = 28, the result we get is an integer. So, the commutative property does not hold for subtraction.
Is integer subtraction commutative?
Subtraction of integers is not commutative.
#### Are integers subtraction associative?
Subtraction of integers is not associative in nature i.e. x − (y − z) ≠ (x − y) − z.
#### Is subtraction commutative for rational number?
Subtraction and division are not commutative for rational numbers because while performing those operations, if the order of numbers is changed, then the result also changes.
Why is subtraction of whole numbers not commutative?
Subtraction of Whole Numbers. Explanation :-. Subtraction is not commutative for integers, this means that when we change the order of integers in subtraction expression, the result also changes.
## Is the result of a subtraction always an integer?
If we subtract any two integers the result is always an integer, so we can say that integers are closed under subtraction. Let us say ‘a’ and ‘b’ are two integers either positive or negative, their result should always be an integer, i.e (a + b) would always be an integer.
## How are integers closed under addition and subtraction?
Since both -11 and 2 are integers, and their sum, i.e (-9) is also an integer, we can say that integers are closed under addition. If we subtract any two integers the result is always an integer, so we can say that integers are closed under subtraction.
Which is an associative property under subtraction of an integer?
Associative Property under Subtraction of Integers: On contradictory, as commutative property does not hold for subtraction similarly associative property also does not hold for subtraction of integers. In generalize form for any three integers say ‘a’, ’b’ and ‘c’ a – (b – c) ≠ (a – b) – c |
## 1. Solution of Two Linear Equations in Two Variables in Different Methods
Graphical Method of Solving Linear Equations In Two Variables
Let the system of pair of linear equations be
a
1x + b1y = c1 ….(1)
a
2x + b2y = c2 ….(2)
We know that given two lines in a plane, only one of the following three possibilities can happen –
(i) The two lines will intersect at one point.
(ii) The two lines will not intersect, however far they are extended, i.e., they are parallel.
(iii) The two lines are coincident lines.
Types Of Solutions:
There are three types of solutions
1. Unique solution.
2. Infinitely many solutions
3. No solution.
(A) Consistent: If a system of simultaneous linear equations has at least one solution then the system is said to be consistent.
(i) Consistent equations with unique solution: The graphs of two equations intersect at a unique point.
For Example Consider
x + 2y = 4
7x + 4y = 18
The graphs (lines) of these equations intersect each other at the point (2, 1) i.e., x = 2, y = 1.
Hence, the equations are consistent with unique solution.
(ii) Consistent equations with infinitely many solutions: The graphs (lines) of the two equations will be coincident.
For Example Consider 2x + 4y = 9
3x + 6y = 27/2
The graphs of the above equations coincide. Coordinates of every point on the lines are the solutions of the equations. Hence, the given equations are consistent with infinitely many solutions.
(B) Inconsistent Equation: If a system of simultaneous linear equations has no solution, then the system is said to be inconsistent.
No Solution: The graph (lines) of the two equations are parallel.
For Example Consider
4x + 2y = 10
6x + 3y = 6
The graphs (lines) of the given equations are parallel. They will never meet at a point. So, there is no solution. Hence, the equations are inconsistent.
From the table above you can observe that if the line a1x + b1y + c1 = 0 and a2x + b2y + c= 0 are
Graphical Method Examples
Example 1: The path of highway number 1 is given by the equation x + y = 7 and the highway number 2 is given by the equation 5x + 2y = 20. Represent these equations geometrically.
Sol. We have, x + y = 7
y = 7 – x ….(1)
In tabular form
X
1
4
Y
6
3
Points
A
B
and 5x + 2y = 20
y = 20−5x2 ….(2)
In tabular form
X
2
4
Y
5
0
Points
C
D
Plot the points A (1, 6), B(4, 3) and join them to form a line AB.
Similarly, plot the points C(2, 5). D (4, 0) and join them to get a line CD. Clearly, the two lines intersect at the point C. Now, every point on the line AB gives us a solution of equation (1). Every point on CD gives us a solution of equation (2).
Example 2: A father tells his daughter, “ Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent this situation algebraically and graphically.
Sol. Let the present age of father be x years and that of daughter = y years
Seven years ago father’s age = (x – 7) years
Seven years ago daughter’s age = (y – 7) years
According to the problem
(x – 7) = 7(y – 7) or x – 7y = – 42 ….(1)
After 3 years father’s age = (x + 3) years
After 3 years daughter’s age = (y + 3) years
According to the condition given in the question
x + 3 = 3(y + 3) or x – 3y = 6 ….(2)
x – 7y = –42
y=x+427
x
0
7
14
y
6
7
8
Points
A
B
C
x – 3y = 6 y=x−63
x
6
12
18
y
0
2
4
Points
D
E
F
Plot the points A(0, 6), B(7, 7), C(14, 8) and join them to get a straight line ABC. Similarly plot the points D(6, 0), E(12, 2) and F(18, 4) and join them to get a straight line DEF.
Example 3: 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Sol. Let the number of boys be x and the number of girls be y.
Then the equations formed are
x + y = 10 ….(1)
and y = x + 4 ….(2)
Let us draw the graphs of equations (1) and (2) by finding two solutions for each of the equations. The
solutions of the equations are given.
x + y = 10
y = 10 – x
X
0
8
Y
10
2
Points
A
B
y = x + 4
x
0
1
3
y
4
5
7
Points
C
D
E
Plotting these points we draw the lines AB and CE passing through them to represent the equations. The two lines AB and Ce intersect at the point E (3, 7). So, x = 3 and y = 7 is the required solution of the pair of linear equations.
i.e. Number of boys = 3
Number of girls = 7.
Verification:
Putting x = 3 and y = 7 in (1), we get
L.H.S. = 3 + 7 = 10 = R.H.S., (1) is verified.
Putting x = 3 and y = 7 in (2), we get
7 = 3 + 4 = 7, (2) is verified.
Hence, both the equations are satisfied.
Example 4: Half the perimeter of a garden, whose length is 4 more than its width is 36m. Find the dimensions of the garden.
Sol. Let the length of the garden be x and width of the garden be y.
Then the equation formed are
x = y + 4 ….(1)
Half perimeter = 36
x + y = 36 ….(2)
y = x – 4
X
0
4
Y
-4
0
Points
A
B
y = 36 – x
X
10
20
Y
26
16
Points
C
D
Plotting these points we draw the lines AB and CD passing through them to represent the equations.
The two lines AB and CD intersect at the point (20, 16), So, x = 20 and y = 16 is the required solution of the pair of linear equations i.e. length of the garden is 20 m and width of the garden is 16 m.
Verification:
Putting x = 20 and y = 16 in (1). We get
20 = 16 + 4 = 20, (1) is verified.
Putting x = 20 and y = 16 in (2). we get
20 + 16 = 36
36 = 36, (2) is verified.
Hence, both the equations are satisfied.
Example 5: Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Sol. Pair of linear equations are:
x – y + 1 = 0 ….(1)
3x + 2y – 12 = 0 ….(2)
x – y + 1 = 0
y = x + 1
x
0
4
y
1
5
Points
A
B
3x + 2y – 12 = 0 y = 12−3x2
x
0
2
y
6
3
Points
C
D
Plot the points A(0, 1), B(4, 5) and join them to get a line AB. Similarly, plot the points C(0, 6), D(2, 3) and join them to form a line CD.
Clearly, the two lines intersect each other at the point D(2, 3). Hence x = 2 and y = 3 is the solution of the
given pair of equations.
The line CD cuts the x-axis at the point
E (4, 0) and the line AB cuts the x-axis at the point F(–1, 0).
Hence, the coordinates of the vertices of the triangle are; D(2, 3), E(4, 0), F(–1, 0).
Verification:
Both the equations (1) and (2) are satisfied by x = 2 and y = 3. Hence, Verified.
Example 6: Show graphically that the system of equations
x – 4y + 14 = 0; 3x + 2y – 14 = 0 is consistent with unique solution.
Sol. The given system of equations is
x – 4y + 14 = 0 ….(1)
3x + 2y – 14 = 0 ….(2)
x – 4y + 14 = 0
y = x+144
x
6
-2
y
5
3
Points
A
B
3x + 2y – 14 = 0 y = −3x+142
x
0
4
y
7
1
Points
C
D
The given equations representing two lines, intersect each other at a unique point (2, 4). Hence, the equations are consistent with unique solution.
Example 7: Show graphically that the system of equations
2x + 5y = 16;
3x+152=24 has infinitely many solutions.
Sol. The given system of equations is
2x + 5y = 16 ….(1)
3x+152=24 ….(2)
2x + 5y = 16
y = 16−2x5
x
-2
3
y
4
2
Points
A
B
3x+152=24 y = 48−6x15
x
1/2
11/2
y
3
1
Points
C
D
The lines of two equations are coincident. Coordinates of every point on this line are the solution.
Hence, the given equations are consistent with infinitely many solutions.
Example 8: Show graphically that the system of equations
2x + 3y = 10, 4x + 6y = 12 has no solution.
Sol. The given equations are
2x + 3y = 10
y = 10−2x3
x
-4
2
y
6
2
Points
A
B
4x + 6y = 12 y = 12−4x6
x
-3
3
y
4
0
Points
C
D
Plot the points A (–4, 6), B(2, 2) and join them to form a line AB. Similarly, plot the points C(–3, 4), D(3, 0) and join them to get a line CD.
Clearly, the graphs of the given equations are parallel lines. As they have no common point, there is no common solution. Hence, the given system of equations has no solution.
Example 9: Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the
geometrical representing of the pair so formed is :
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Sol. We have, 2x + 3y – 8 = 0
(i) Another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines is
3x – 2y – 8 = 0
(ii) Another parallel lines to above line is
4x + 6y – 22 = 0
(iii) Another coincident line to above line is
6x + 9y – 24 = 0
Example 10: Solve the following system of linear equations graphically;
3x + y – 11 = 0 ; x – y – 1 = 0
Shade the region bounded by these lines and also y-axis. Then, determine the areas of the region bounded by these lines and y-axis.
Sol. We have,
3x + y – 11 = 0 and x – y – 1 = 0
(a) Graph of the equation 3x + y – 11 = 0
We have, 3x + y – 11 = 0
y = – 3x + 11
When, x = 2, y = –3 × 2 + 11 = 5
When, x = 3, y = – 3 × 3 + 11 = 2
Plotting the points P (2, 5) and Q(3, 2) on the graph paper and drawing a line joining between them, we get the graph of the equation 3x + y – 11 = 0 as shown in fig.
(b) Graph of the equation x – y – 1 = 0
We have,
x – y – 1 = 0
y = x – 1
When, x = – 1, y = –2
When, x = 3, y = 2
Plotting the points R(–1, –2) and S(3, 2) on the same graph paper and drawing a line joining between them, we get the graph of the equation x – y – 1 = 0 as shown in fig.
You can observe that two lines intersect at
Q(3, 2). So, x = 3 and y = 2. The area enclosed by the lines represented by the given equations and also the y-axis is shaded.
So, the enclosed area = Area of the shaded portion
= Area of ∆QUT = 1/2 × base × height
= 1/2 × (TU × VQ) = 1/2 × (TO + OU) × VQ
= 1/2 (11 + 1) 3 = 1/2 × 12 × 3 = 18 sq.units.
Hence, required area is 18 sq. units.
Example 11: Draw the graphs of the following equations
2x – 3y = – 6; 2x + 3y = 18; y = 2
Find the vertices of the triangles formed and also find the area of the triangle.
Sol. (a) Graph of the equation 2x – 3y = – 6;
We have, 2x – 3y = – 6
y = 2x+63
When, x = 0, y = 2
When, x = 3, y = 4
Plotting the points P(0, 2) and Q(3, 4) on the graph paper and drawing a line joining between them we get the graph of the equation 2x – 3y = – 6 as shown in fig.
(b) Graph of the equation 2x + 3y = 18;
We have 2x + 3y = 18
y = −2x+183
When, x = 0, y = 6
When, x = – 3, y = 8
Plotting the points R(0, 6) and S(–3, 8) on the same graph paper and drawing a line joining between them, we get the graph of the equation 2x + 3y = 18 as shown in fig.
(c) Graph of the equation y = 2
It is a clear fact that y = 2 is for every value of x. We may take the points T (3, 2), U(6, 2) or any other values.
Plotting the points T(3, 2) and U(6, 2) on the same graph paper and drawing a line joining between them, we get the graph of the equation y = 2 as shown in fig.
From the fig., we can observe that the lines taken in pairs intersect each other at points Q(3, 4), U (6, 2) and P(0, 2). These form the three vertices of the triangle PQU.
To find area of the triangle so formed
The triangle is so formed is PQU (see fig.)
In the ∆PQU
QT (altitude) = 2 units
and PU (base) = 6 units
so, area of ∆PQU = (base × height)
= 1/2 (PU × QT) = 1/2 × 6 × 2 sq. untis
= 6 sq. units.
Example 12: On comparing the ratios a1a2,b1b2andc1c2 and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide.
(i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0, 2x – y + 9 = 0
Sol. Comparing the given equations with standard forms of equations a
1x + b1y + c1 = 0 and
a
2x + b2y + c2 = 0 we have,
(i) a1 = 5, b
1 = – 4, c1 = 8;
a
2 = 7, b2 = 6, c2 = – 9
a1a2=57,b1b2=−46
a1a2b1b2
Thus, the lines representing the pair of linear equations are intersecting.
(ii) a1 = 9, b
1 = 3, c1 = 12;
a
2 = 18, b2 = 6, c2 = 24
a1a2=918=12,b1b2=36=12andc1c2=1224=12
a1a2=b1b2=c1c2
Thus, the lines representing the pair of linear equation coincide.
(iii) a1 = 6, b
1 = – 3, c1 = 10;
a
2 = 2, b2 = – 6, c2 = 9
a1a2=62=3,b1b2=−3−1=3andc1c2=109
a1a2=b1b2c1c2
Thus, the lines representing the pair of linear equations are parallel.
## 1. Solution of Two Linear Equations in Two Variables in Different Methods
3.1 INTRODUCTION
While solving the problems, in most cases, first we need to frame an equation. In this chapter, we learn how to frame and solve equations. There are given some methods to solve these equations. We will further study about word problems and application of simultaneous equations.
Equation: A statement in which two algebraic expressions are equal is known as equation.
Solution (root) of a linear equation: The value of the variable which makes the two sides of the equation equal and satisfies the equation is called the solution of the equation.
Rules for solving an equation:
(i) The same number is added or subtracted to both sides of an equation, the resulting equation is equivalent to the first.
(ii) If both sides of an equation are multiplied by the same non-zero number the resulting equation is equivalent to the first.
Remark: Every linear equation in one variable has only one (unique) solution.
In this chapter we shall study about system of linear equation in two variables, solution of a system of linear equations in two variables.
3.3 algebraic methods of solving a pair of linear equations
There are four methods for solving a pair of linear equations
(i) Substitution method
(ii) Elimination method
(iii) Cross-multiplication method
(iv) Graphical Method
3.3.2 Elimination Method
Step–I: Obtain the two equations
Step–II: First multiply both the equations by some suitable non-zero constant to make the coefficient of one variable (either x or y) numerically equal.
Step–III: Add or subtract one equation from the other, then one variable gets eliminated.
Step–IV: Solve the equation in one variable.
Step–V: Substitute the value of x (or y) in any one of the given equations and find the value of another variable.
## Basics Revisited
### Equation
An equation is a statement that two mathematical expressions having one or more variables are equal.
### Linear Equation
Equations in which the powers of all the variables involved are one are called linear equations. The degree of a linear equation is always one.
### General form of a Linear Equation in Two Variables
The general form of a linear equation in two variables is ax + by + c = 0, where a and b cannot be zero simultaneously.
### Representing linear equations for a word problem
To represent a word problem as a linear equation
• Identify unknown quantities and denote them by variables.
• Represent the relationships between quantities in a mathematical form, replacing the unknowns with variables.
### Solution of a Linear Equation in 2 variables
The solution of a linear equation in two variables is a pair of values, one for x and the other for y, which makes the two sides of the equation equal.
Eg: If 2x+y=4, then (0,4) is one of its solutions as it satisfies the equation. A linear equation in two variables has infinitely many solutions.
### Geometrical Representation of a Linear Equation
Geometrically, a linear equation in two variables can be represented as a straight line.
2x – y + 1 = 0
⇒ y = 2x + 1
Graph of y = 2
x
+1
### Plotting a Straight Line
The graph of a linear equation in two variables is a straight line. We plot the straight line as follows:
Any additional points plotted in this manner will lie on the same line.
### General form of a pair of linear equations in 2 variables
A pair of linear equations in two variables can be represented as follows
The coefficients of x and y cannot be zero simultaneously for an equation.
### Nature of 2 straight lines in a plane
For a pair of straight lines on a plane, there are three possibilities
i) They intersect at exactly one point
pair of linear equations which intersect at a single point.
ii) They are parallel
pair of linear equations which are parallel.
iii) They are coincident
pair of linear equations which are coincident.
## Graphical Solution
### Representing pair of LE in 2 variables graphically
Graphically, a pair of linear equations in two variables can be represented by a pair of straight lines.
### Graphical method of finding solution of a pair of Linear Equations
Graphical Method of finding the solution to a pair of linear equations is as follows:
• Plot both the equations (two straight lines)
• Find the point of intersection of the lines.
The point of intersection is the solution.
## Algebraic Solution
### Finding solution for consistent pair of Linear Equations
The solution of a pair of linear equations is of the form (x,y) which satisfies both the equations simultaneously. Solution for a consistent pair of linear equations can be found out using
i) Elimination method
ii) Substitution Method
iii) Cross-multiplication method
iv) Graphical method
### Substitution Method of finding solution of a pair of Linear Equations
Substitution method:
y – 2x = 1
x + 2y = 12
(i) express one variable in terms of the other using one of the equations. In this case, y = 2x + 1.
(ii) substitute for this variable (y) in the second equation to get a linear equation in one variable, x. x + 2 × (2x + 1) = 12
⇒ 5 x + 2 = 12
(iii) Solve the linear equation in one variable to find the value of that variable.
5 x + 2 = 12
x = 2
(iv) Substitute this value in one of the equations to get the value of the other variable.
y = 2 × 2 + 1
⇒y = 5
So, (2,5) is the required solution of the pair of linear equations y – 2x = 1 and x + 2y = 12.
### Elimination method of finding solution of a pair of Linear Equations
Elimination method
Consider x + 2y = 8 and 2x – 3y = 2
Step 1: Make the coefficients of any variable the same by multiplying the equations with constants. Multiplying the first equation by 2, we get,
2x + 4y = 16
Step 2: Add or subtract the equations to eliminate one variable, giving a single variable equation.
Subtract second equation from the previous equation
2x + 4y = 16
2x – 3y = 2
– + –
———————–
0(x) + 7y =14
Step 3: Solve for one variable and substitute this in any equation to get the other variable.
y = 2,
x = 8 – 2 y
⇒ x = 8 – 4
⇒ x = 4
(4, 2) is the solution.
## Geometric representation of different possibilities of solutions/inconsistency?
3.3 algebraic methods of solving a pair of linear equations
There are four methods for solving a pair of linear equations
(i) Substitution method
(ii) Elimination method
(iii) Cross-multiplication method
(iv) Graphical Method
3.3.2 Elimination Method
Step–I: Obtain the two equations
Step–II: First multiply both the equations by some suitable non-zero constant to make the coefficient of one variable (either x or y) numerically equal.
Step–III: Add or subtract one equation from the other, then one variable gets eliminated.
Step–IV: Solve the equation in one variable.
Step–V: Substitute the value of x (or y) in any one of the given equations and find the value of another variable.
3.3.4 Graphical Method:
In graphical method, we draw the graph of both equations using same pair of horizontal and vertical lines called X-axis and Y-axis respectively. Coordinates of the point(s) of intersection of the two lines is/are the solution.
Nature of solutions:
When we try of solve a pair of equations we could arrive at three possible results. They are, having
(a) a unique solution
(b) an infinite number of solutions
(c) no solution
## Algebraic conditions for number of solutions?
3.3 algebraic methods of solving a pair of linear equations
There are four methods for solving a pair of linear equations
(i) Substitution method
(ii) Elimination method
(iii) Cross-multiplication method
(iv) Graphical Method
3.3.2 Elimination Method
Step–I: Obtain the two equations
Step–II: First multiply both the equations by some suitable non-zero constant to make the coefficient of one variable (either x or y) numerically equal.
Step–III: Add or subtract one equation from the other, then one variable gets eliminated.
Step–IV: Solve the equation in one variable.
Step–V: Substitute the value of x (or y) in any one of the given equations and find the value of another variable.
3.3.4 Graphical Method:
In graphical method, we draw the graph of both equations using same pair of horizontal and vertical lines called X-axis and Y-axis respectively. Coordinates of the point(s) of intersection of the two lines is/are the solution.
Nature of solutions:
When we try of solve a pair of equations we could arrive at three possible results. They are, having
(a) a unique solution
(b) an infinite number of solutions
(c) no solution
## Solution of a pair of linear equations in two variables algebraically - by substitution, by elimination and by cross multiplication method?
3.6 Solution of a system of a pair of equations reducible to the system of a pair of linear equations in two variables
By using the suitable substitution or simplification first we convert the given system into the system of a pair of linear equations in two variables. Then after using any algebraic or graphical method we solve the system. |
# Introduction to Coordinate Geometry: Plotting Points on a Graph
Apr 2, 2024 | Roslyn
Coordinate geometry, also known as Cartesian geometry, is a fundamental branch of mathematics that merges algebraic techniques with geometric concepts. At its core lies the Cartesian coordinate system, which provides a framework for precisely describing points, lines, curves, and shapes in a two-dimensional plane. Plotting points on a graph is the foundational skill in coordinate geometry, serving as a gateway to understanding more complex concepts like distance, slope, and equations of lines.
Understanding the Cartesian Coordinate System: The Cartesian coordinate system was developed by René Descartes in the 17th century and revolutionized mathematics by providing a way to represent geometric figures algebraically. It consists of two perpendicular lines known as the x-axis and the y-axis, intersecting at a point called the origin, typically denoted as (0,0). These axes divide the plane into four quadrants: the first quadrant is where both x and y coordinates are positive, the second quadrant is where x is negative and y is positive, the third quadrant is where both x and y are negative, and the fourth quadrant where x is positive and y is negative.
Plotting Points: To plot a point on the graph, we need two coordinates: an x-coordinate and a y-coordinate. The x-coordinate tells us how far left or right the point is from the origin, while the y-coordinate tells us how far up or down the point is from the origin. For example, point (3,2) would mean moving three units to the right along the x-axis and two units upwards along the y-axis from the origin.
Positive and Negative Coordinates: In the Cartesian coordinate system, positive x-values are to the right of the origin, and negative x-values are to the left. Similarly, positive y-values are above the origin, and negative y-values are below. For instance, the point (-4,3) would be four units to the left of the origin and three units above it.
Using Grids and Scales: Graph paper is commonly used to plot points and draw geometric figures. The paper is divided into small squares, with each square representing a unit length on both the x and y-axes. By using the grid lines as guides, plotting points becomes a matter of counting units along each axis. For example, to plot the point (2,-3), we move two units to the right along the x-axis and three units down along the y-axis.
Applications in Real Life: Coordinate geometry has numerous applications in various fields, including physics, engineering, geography, computer graphics, and more. For instance, in physics, it's used to represent the motion of objects in space. In engineering, it helps design structures and analyze their geometric properties. In geography, it aids in mapping locations on the Earth's surface. In computer graphics, it's essential for rendering images on screens.
The ability to plot points on a graph is a fundamental skill in coordinate geometry, laying the groundwork for more advanced concepts and applications. By understanding the Cartesian coordinate system and how to interpret coordinates, one gains a powerful tool for representing and analyzing geometric figures. Whether in mathematics, science, or technology, coordinate geometry continues to play a vital role in solving real-world problems and understanding the relationships between objects in space. |
How do you find the GCF of 24cd^3, 48c^2d?
Mar 3, 2018
See a solution process below:
Explanation:
First, we can write the prime factor for each term as:
$24 c {d}^{3} = 2 \times 2 \times 2 \times 3 \times c \times d \times d \times d$
$48 {c}^{2} d = 2 \times 2 \times 2 \times 2 \times 3 \times c \times c \times d$
The common prime factors are:
$24 c {d}^{3} = \textcolor{red}{2} \times \textcolor{red}{2} \times \textcolor{red}{2} \times \textcolor{red}{3} \times \textcolor{red}{c} \times \textcolor{red}{d} \times d \times d$
$48 {c}^{2} d = \textcolor{red}{2} \times \textcolor{red}{2} \times \textcolor{red}{2} \times 2 \times \textcolor{red}{3} \times \textcolor{red}{c} \times c \times \textcolor{red}{d}$
Therefore the Greatest Common Factor is:
$\text{GCF} = \textcolor{red}{2} \times \textcolor{red}{2} \times \textcolor{red}{2} \times \textcolor{red}{3} \times \textcolor{red}{c} \times \textcolor{red}{d} = 24 c d$ |
# What is 51/492 as a decimal?
## Solution and how to convert 51 / 492 into a decimal
51 / 492 = 0.104
Convert 51/492 to 0.104 decimal form by understanding when to use each form of the number. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). If we need to convert a fraction quickly, let's find out how and when we should.
## 51/492 is 51 divided by 492
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 51 divided by 492. We use this as our equation: numerator(51) / denominator (492) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. This is our equation:
### Numerator: 51
• Numerators sit at the top of the fraction, representing the parts of the whole. Any value greater than fifty will be more difficult to covert to a decimal. 51 is an odd number so it might be harder to convert without a calculator. Large numerators make converting fractions more complex. Let's take a look below the vinculum at 492.
### Denominator: 492
• Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. Larger values over fifty like 492 makes conversion to decimals tougher. And it is nice having an even denominator like 492. It simplifies some equations for us. Have no fear, large two-digit denominators are all bark no bite. So grab a pen and pencil. Let's convert 51/492 by hand.
## Converting 51/492 to 0.104
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 492 \enclose{longdiv}{ 51 }$$
To solve, we will use left-to-right long division. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 492 \enclose{longdiv}{ 51.0 }$$
We've hit our first challenge. 51 cannot be divided into 492! Place a decimal point in your answer and add a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 492 into 51 + 0 or 510.
### Step 3: Solve for how many whole groups you can divide 492 into 510
$$\require{enclose} 00.1 \\ 492 \enclose{longdiv}{ 51.0 }$$
We can now pull 492 whole groups from the equation. Multiple this number by our furthest left number, 492, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.1 \\ 492 \enclose{longdiv}{ 51.0 } \\ \underline{ 492 \phantom{00} } \\ 18 \phantom{0}$$
If your remainder is zero, that's it! If you have a remainder over 492, go back. Your solution will need a bit of adjustment. If you have a number less than 492, continue!
### 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.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. They each bring clarity to numbers and values of every day life. Same goes 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. Here are examples of when we should use each.
### When you should convert 51/492 into a decimal
Contracts - Almost all contracts leverage decimal format. If a worker is logging hours, they will log 1.10 hours, not 1 and 51/492 hours. Percentage format is also used in contracts as well.
### When to convert 0.104 to 51/492 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
### Practice Decimal Conversion with your Classroom
• If 51/492 = 0.104 what would it be as a percentage?
• What is 1 + 51/492 in decimal form?
• What is 1 - 51/492 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.104 + 1/2? |
# 3.6 Division by multiples of 10
Page 1 / 1
## Memorandum
23.4.2 a) 0,041
1. 0,81
2. 0,0093
• b en c
23.6 a) 1,234
1. 0,845
2. 7,23
25. a) R51
1. R211,59
2. 21160c
3. R2,90
4. Ja
Cost per call: R1,77
1. 91,5
2. R102,13
3. R1,28 x 24 = R30,72
4. R3,27
## Activity: division by multiples of 10 [lo 1.5.1, lo 1.7.5]
23.4 DIVISION BY MULTIPLES OF 10
23.4.1 Do you still remember?
a) This is how I divide 0,8 by 40:
0,8 ÷ 40 = (0,8 ÷ 10) ÷ 4
= 0,08 ÷ 4
= 0,02
b) When I divide 4,2 by 600, I say:
4,2 ÷ 600 = (4,2 ÷ 100) ÷ 6
= 0,042 ÷ 6
= 0,007
23.4.2 Calculate the following without a calculator:
a) j = 3,28 ÷ 80
…………………………………………………………………………………………..
…………………………………………………………………………………………..
b) d = 567 ÷ 700
…………………………………………………………………………………………..
…………………………………………………………………………………………..
c) g = 18,6 ÷ 2 000
…………………………………………………………………………………………..
…………………………………………………………………………………………..
23.5 DIVISION WITH DECIMAL FRACTIONS
23.5.1 Work through the following examples with a friend:
A roll of material is 11,25 m long. 1,5 m of material is needed to make one dress.
How many dresses can be cut from this roll of material?
a) I must calculate 11,25 ÷ 1,5
I change the decimal fractions to fractions:
$\text{11},\text{25}=\text{11}\frac{\text{25}}{\text{100}}=\text{11}\frac{1}{4}$ en $1,5=1\frac{5}{\text{10}}=1\frac{1}{2}$
$\begin{array}{}\text{11}\frac{1}{4}÷1\frac{1}{2}=\frac{\text{45}}{4}÷\frac{3}{2}\\ \frac{\text{45}}{4}×\frac{2}{3}\\ 7\frac{1}{2}\end{array}$
I make use of equivalent fractions :
$\begin{array}{}\text{11},\text{25}÷1,5=\frac{\text{11},\text{25}}{1,5}×\frac{\text{10}}{\text{10}}=\frac{\text{112},5}{\text{15}}\\ 7,5\\ 7\frac{1}{2}\end{array}$
c) I must calculate 11,25 ÷ 1,5I change the divisor to a whole number so that I can divide easier.
1,5 × 10 = 15
To keep the balance in the sum, I must also multiply the dividend by 10!
11,25 × 10 = 112,5
7 ,5 15 112 ,5 –105 7 5 –7 5 · ·
The answer is thus 7,5 dresses..
23.5.2 Which of the above-mentioned methods are precisely the same?
…………………………………………………………………………………………..
…………………………………………………………………………………………..
23.5.2 Which of the above-mentioned methods are precisely the same?
23.6 Calculate the following by first changing the divisor to a whole number:
a) q = 0,88848 ÷ 0,72
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
b) p = 0,14365 ÷ 0,17
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
c) v = 0,30366 ÷ 0,042
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
24. Time for self-assessment
Tick the applicable block: YES NO I can divide decimal fractions by 10 correctly. I can divide decimal fractions by 100 correctly. I can divide decimal fractions by 1 000 correctly. I know how to divide decimal fractions by multiples of 10. I can divide decimal fractions by decimal fractions.
25. Look at the following telephone account and answer the questions. You may use your calculator.
a) Round off the cost of the overseas call to the nearest rand. ……………………
b) What is the total cost of the calls that were made? …………………………….
c) Round off the above answer to the nearest cent. ………………………………
d) What was the average cost of each cell phone (Vodacom) call? ………………
e) Is MTN cheaper than Vodacom? ……………….. Motivate. ………………...
…………………………………………………………………………………………..
…………………………………………………………………………………………..
f) Round off the cost of the national calls (0 – 50 km) to 1 digit after the comma.
…………………………………………………………………………………………..
g) What is the total cost of the national calls? ……………………………………
h) If two “ShareCall” calls cost R1,28, what would 48 such calls cost?
…………………………………………………………………………………………..
i) The account is paid with a R200 note and a R20 note. How much change will be given?
…………………………………………………………………………………………..
j) How much is YOUR telephone account each month, on average, over a period of one year?
…………………………………………………………………………………………..
## Assessment
Learning Outcome 1: The learner will be able to recognise, describe and represent numbers and their relationships, and to count, estimate, calculate and check with competence and confidence in solving problems.
Assessment Standard 1.5: We know this when the learner solves problems in context including contexts that may be used to build awareness of other Learning Areas, as well as human rights, social, economic and environmental issues such as:
1.5.1 financial (including profit and loss, budgets, accounts, loans, simple interest, hire purchase, exchange rates);
Assessment Standard 1.7: We know this when the learner estimates and calculates by selecting and using operations appropriate to solving problems that involve:
1.7.5 division of positive decimals with at least 3 decimal places by whole numbers.
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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Got questions? Join the online conversation and get instant answers! |
## SUPPOSITION CONCEPT
###### DIRECT SUPPOSITION
The Supposition concept can be a challenging one for students. At Model Math, we first introduce this idea in Primary 2 and gradually build upon it through repetition and more complex problems. This video example demonstrates the basics of the Supposition concept.
###### SUPPOSITION WITH PENALTY
The Supposition concept can be a challenging one for students. At Model Math, we first introduce this idea in Primary 2 and gradually build upon it through repetition and more complex problems. In Primary 4, we introduce Supposition with Penalty, which is a higher-level problem-solving technique that utilises the Supposition concept. It is common for students to need more than one lesson to fully grasp this concept, so we will be revisiting it in future lessons.
## PROBLEM SUMS WITH EXTERNAL TRANSFER
###### EXTERNAL TRANSFER, GIVEN DIFFERENCE
Peter had 20 more stickers than John at first. John then lost 169 stickers in a game. Peter now had four times as many stickers as John. How many stickers did John have at first?
###### EXTERNAL TRANSFER, UNCHANGED DIFFERENCE
Shop A and Shop B had 75 kg and 54 kg of flour, respectively. After both shops sold the same amount of flour, Shop A had four times as much flour as Shop B. How much flour did each shop sell?
###### EXTERNAL TRANSFER, GIVEN UNITS
In a container, the number of sweets was five times the number of chocolates. After John took out 26 sweets, there were 30 more sweets than chocolates. How many sweets were there at first?
## PROBLEM SUMS WITH INTERNAL TRANSFER
###### INTERNAL TRANSFER, GIVEN UNITS
Bottle A contained four times as much water as Bottle B at first. After 12 L of water was transferred from Bottle A to Bottle B, there was an equal amount of water in both bottles. How much water was there altogether?
###### INTERNAL TRANSFER, TOTAL UNCHANGED
Peter had \$52 and Yenni had \$26. After Peter gave some money to Yenni, Yenni had \$10 more than Peter. How much did Peter give to Yenni?
###### APPROXIMATION & ESTIMATION
1. An even number is 500 when rounded off to the nearest hundred. What is the greatest possible value of the even number?
2. Which is the best estimate for 55 x 42?
## MULTIPLICATIONS AND GROUPING
###### TOTAL CONCEPT, REGROUPING
Amber received \$12 for every belt she sold. She also received a bonus of \$50 for every 8 belts she sold. She sold 84 belts. How much money did she receive altogether?
###### TOTAL CONCEPT, REGROUPING
Amber is selling frying pans. For every frying pan she sells, she will earn \$6. She will also earn an additional \$15 for every 7 pans she sells. If she sells 23 pans, how much will she earn in all?
## COMPARISON MODEL PROBLEM SUMS
###### STACKING MODEL
In Primary 3, students learned about the Comparison Model for comparing the values of two different items, such as “a jar has 3 more cookies than a can.” Now, we would like to introduce a new method called the Stacking Model. This technique is used when comparing multiple items of the same type, for instance “3 jars have 5 more cookies than 3 cans.” The Stacking Model provides a visual representation and makes it easier to understand and compare multiple items.
###### COMPARISON OF QUANTITY AND UNITS
Peter had three times as many stickers as Yenni. John had 63 fewer stickers than Peter. The three children had 119 stickers altogether. How many more stickers did Yenni have than John?
## SHORTAGE AND SURPLUS PROBLEM SUMS
###### LISTING FOR HIGHEST COMMON FACTORS
Peter wants to lay his floor with square carpet tiles. The rectangular shaped floor measures 160 cm by 90 cm.
a) Find the largest possible length of the side of each carpet tile.
b) Find the number of tiles that are needed to cover the floor.
###### SHORTAGE AND SURPLUS, LISTING METHOD WITH NO FIXED GROUPS
There are between 10 to 20 beads in a container. Amber wants to put the beads in packets. If she puts 3 in each packet, there will be 2 beads left. If she puts 4 in each packet, there will be 1 bead left. How many beads are there in the container?
###### SHORTAGE AND SURPLUS, UNITS METHOD
Yenni had some money. She wanted to buy 16 markers but was short of \$9. In the end, she bought 9 markers and had \$5 left. How much money did Yenni have at first?
###### SHORTAGE AND SURPLUS, LISTING METHOD FIXED NUMBER OF GROUPS
Yenni had some gummies. She shared the gummies equally amongst some friends. If she gave each friend 4 gummies, she would have 1 gummy left. If she gave each friend 5 gummies, she would be short of 4 gummies. How many gummies did she have?
## COMMON FRACTIONS PROBLEM SUMS
###### FRACTIONS AS PART OF REMAINDER WITH CHANGING DENOMINATOR
1/4 of the people in the school are teachers and the rest are children. 1/3 of the children are girls. There are 10 boys. How people are there in the school?
###### FRACTIONS AS PART OF REMAINDER WITH REDRAWING REMAINDER
Yenni had some money. She wanted to buy 16 markePeter had some beads. He used 1/5 of the beads and gave 1/3 of the remainder to Yenni. Peter had 8 beads left. How many beads did Peter have at first?
###### TOTAL CONCEPT, DIRECT APPLICATION
1 m of ribbon cost \$8. How much does 3 3/4 m of ribbon cost?
###### EXTERNAL TRANSFER WITH 1 GROUP UNCHANGED
There are 30 markers in the box. 3 are yellow and the rest are black. After John used some black markers, 1/4 of the markers are yellow. How many black markers are used?
###### EXTERNAL TRANSFER WITH 1 GROUP UNCHANGED.
The number of apples is 2/5 the number of pears in a basket. 6 more pears are added to the basket and the number of apples is 2/7 the number of pears. How many apples and pears were in the basket at first?
###### INTERNAL PROPORTION TRANSFER
John had 183 stickers. After he gave 1/3 of it to Yenni, he had 95 stickers fewer than Yenni. How many stickers did Yenni have at first?
## COMMON DECIMALS PROBLEM SUMS
###### DECIMALS, ESTIMATION AND APPROXIMATION
a) A ribbon is 2.33 m when rounded off to 2 decimal places. What is the smallest possible length of the ribbon?
b) Yenni’s mass is 47.6 kg when rounded off to the nearest tenth. What is her largest possible mass? Leave your answer in 2 decimal places.
###### EXTERNAL TRANSFER
Class A had an equal number of boys and girls. After 26 boys and 8 girls were transferred to another class, there were thrice as many girls as boys. How much boys were there in Class A at first?
###### EXTERNAL TRANSFER
Bag A had thrice as many coins as Bag B. After \$18.60 of coins and \$3.40 of coins were removed from Bag A and B respectively, both bags had an equal amount of money. How much coins did Bag A have at first?
###### INTERNAL TRANSFERS
John and Peter had an equal amount of money at first. After John received \$46.30 from his father and Peter spent \$20.10, John had thrice as much money as Peter. How much money did John have at first?
###### BEFORE CHANGE AFTER, FIND CHANGE.
Yenni had \$84.50 and Amber had \$42.00. After Amber spent some money and Yenni spent \$21.50, Yenni had thrice as much money as Amber. How much money did Amber spend?
###### DECIMALS AND GROUPING
Yenni bought twice as many muffins as cupcakes for her birthday party. Each muffin cost \$1.75 and each cupcake cost \$2.50. How many muffins did she buy if she paid a total of \$48?
## AREA AND PERIMETER
###### AREA AND PERIMETER IGNORE REMAINDER
Peter has a rectangular cardboard measuring 15 cm by 9 cm. He wants to cut out smaller squares of side 2 cm from it. What is the maximum number of squares that can be cut out from the cardboard?
###### AREA AND PERIMETER – COMPARISON BY UNITS
The figure below is made up of 3 identical rectangles. The perimeter of the whole figure is 100 cm. What is the area of the whole figure? |
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Radian Measureincluding How to Convert Radians to Degrees and the Trigonometric Functions of Special Angles
It is preferable to have trigonometric functions (sine, cosine, tangent, etc.) defined so the domain of each function is the set of all possible real numbers. Generally, when we speak of degrees and minutes it is an acute angle, an angle that is less than 90°. This limits us to the increments of degrees and minutes. Radian Measure is not based upon increment and thus radian measure includes all possible real numbers.
Radian measure is direct result of a central angle that has its vertex at the center of a unit circle, a circle having a radius of one:
Unit circle (equation x2 + y2 = 1):
If we consider point P on a unit circle in the first quadrant:
cos
cos
QOP = x / 1 = x
sin
QOP = opposite / hypotenuse
sin
QOP = y / 1 = y
Let θ (theta) =
QOP, then
P = (x, y) = (cos θ, sin θ)
The coordinates of P are the cosine and sine of angle θ.
Angle θ is the central angle, an angle formed by two radii of a circle. The measure of a central angle is proportional to the length of a circle between the two points of each radius.
From this a radian is defined as the measure of the central angle that subtends an arc whose length is that of the radius. For a unit circle that length is one unit.
The length of the subtended arc equals the length of the radius on a unit circle.
θ = 1 radian = 57.296°:
With radian measure we are working with radius not diameter. The circumference of a circle is often stated as the ratio of circumference, C, to its diameter, d; C = πd. With radian measure C = r, where r is a radius and 2r = d.
:
Because the unit circle radius is one its circumference, C = 2π r is 2π. Then, by proportionality, 2π radians is 360°, π radians is 180°.
We can state this proportionality between degrees and radians as:
1 radian = 180° / 3.1415826 = 57.296° or 57.3°
To convert from degrees to radians:
Multiply degrees by π / 180°
To convert from radians to degrees:
Multiply radians by 180° / π.
An angle is in standard position when the vertex is at origin, the center of the circle and one side, the initial side, is positive x-axis. The angle is positive if the other side, the terminal side, is measured in a counterclockwise direction. The angle is negative if measured in a clockwise direction:
Positive angle 30° is π / 6 radians:
Negative angle 30° is −π / 6 radians:
In physical science applications there is often a need to work with time and the ability to state the cosine t and the sine t, where t is a number of seconds. For this and other purpose the trigonometric functions can be extended so as to apply to any number of radians. This was shown for a point, P; P = (x, y) = (cos θ, sin θ). The coordinates of P are the cosine and sine of central angle θ. The terminal side of the angle intersects the unit circle at P, with coordinates (x, y). The cosine θ is defined to be x and the sine θ is defined to be y. This means that the cos and sin of any possible number is defined.
By proportionality we already know that:
4π / 2 = 2π radians is 360°,
3π / 2 radians is 270°,
π / 2 radians is 90°.
For these angles their point, P, values for x and y can only be 1, −1 or 0. These angles define a terminal side that is on the x or y axis, respectively the cos θ or sin θ.
cos θ and sin θ for 90°, 180°, 270° and 360°:
All other angles are acute having the initial side along the x-axis from the circle coordinate center (standard position) and a terminal side not parallel the x-axis or y-axis. By this definition we only need consider the coordinate quadrant (I, II, III or IV) to determine the positive or negative sign of the cos θ and sin θ.
Trigonometric Functions of
Special Angles
:
For how to use a Trigonometric Functions Table. Go here: Using the Trigonometric Functions Table
quadrant radians degrees sine cosine tangent 0 0° 0 1 0 I π / 6 30° 1 / 2 √3 / 2 1 I π / 4 45° √2 / 2 √2 / 2 1 I π / 3 60° √3 / 2 1 / 2 √3 π / 2 90° 1 0 undefined II 2π / 3 120° √3 / 2 −1 / 2 −√3 II 3π / 4 135° √2 / 2 −√2 / 2 −1 II 5π / 6 150° 1 / 2 −√3 / 2 −√3 / 3 π 180° 0 −1 0 III 7π / 6 210° −1 / 2 −√3 / 2 √3 / 3 III 5π / 4 225° −√2 / 2 −√2 / 2 1 III 4π / 3 240° −√3 / 2 −1 / 2 √3 3π / 2 270° −1 0 undefined IV 5π / 3 300° −√3 / 2 1 / 2 −√3 IV 7π / 4 315° −√2 / 2 √2 / 2 −1 IV 11π / 6 330° −1 / 2 √3 / 2 −√3 / 3 2π 360° 0 1 0 |
Hong Kong
Stage 4 - Stage 5
# Simultaneous, linear and quadratic equations
## Interactive practice questions
Consider the following system of equations:
$x^2$x2 $=$= $y+14$y+14 $y$y $=$= $3x-16$3x−16
a
By considering the left hand side ($LHS$LHS) and right hand side ($RHS$RHS) of each equation, fill in the missing values to verify that the points of intersection on the graphs are solutions of the corresponding system of equations.
First, test the point $\left(1,-13\right)$(1,13).
$x^2=y+14$x2=y+14
$LHS$LHS $=$= $\left(\editable{}\right)^2$()2 $RHS$RHS $=$= $\editable{}+14$+14
$=$= $\editable{}$ $=$= $\editable{}$
$y=3x-16$y=3x16
$LHS$LHS $=$= $\editable{}$ $RHS$RHS $=$= $3\times\editable{}-16$3×16
$=$= $\editable{}$
b
Now test the point $\left(2,-10\right)$(2,10).
$x^2=y+14$x2=y+14
$LHS$LHS $=$= $\left(\editable{}\right)^2$()2 $RHS$RHS $=$= $\editable{}+14$+14
$=$= $\editable{}$ $=$= $\editable{}$
$y=3x-16$y=3x16
$LHS$LHS $=$= $\editable{}$ $RHS$RHS $=$= $3\times\editable{}-16$3×16
$=$= $\editable{}$
Easy
3min
When solving $y=x^2-3x+1$y=x23x+1 and $y=x+6$y=x+6 simultaneously, one point of intersection is at $x=-1$x=1. What is the $y$y-coordinate at this point?
Easy
1min
Where does the vertical line $x=-5$x=5 intersect the curve $y=-2x^2+x-12$y=2x2+x12?
Easy
1min
Solve the following equations.
Equation 1 $y=x^2$y=x2 Equation 2 $y=9$y=9
Easy
1min |
# In $(2\times n) -1$, if you substitute any whole number greater than 0 for n will you get an even or odd number ? Why ? What about $(2\times n) +1$
Given :
The given terms are $(2\times n) -1$ and $(2\times n) +1$.
To do :
We have to find for any whole number greater than 0, the value of the given terms will be odd or even.
Solution :
$(2\times n) -1$
For any $n>0$
If $n = 1$
$(2\times1)-1 = 2-1 = 1$
If $n=2$
$(2\times2)-1 = 4-1 = 3$
.
.
.
.
This implies,
For any $n>0, (2\times n)-1$ is an odd number.
$(2\times n) +1$
For any $n>0$
If $n = 1$
$(2\times1)+1 = 2+1 = 3$
If $n=2$
$(2\times2)+1 = 4+1 = 5$
.
.
.
.
This implies,
For any $n>0, (2\times n)+1$ is an odd number.
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Updated on: 10-Oct-2022
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# Sine, Cosine, and Tangent Functions
## Trigonometric ratios based on sides of right triangles in relation to an angle.
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Sine, Cosine, Tangent
### Sine, Cosine, and Tangent
Trigonometry is the study of the relationships between the sides and angles of right triangles. The legs are called adjacent or opposite depending on which acute angle is being used.
a is adjacent to B a is opposite Ab is adjacent to A b is opposite Bc is the hypotenuse\begin{align*}& a \ \text{is} \ adjacent \ \text{to} \ \angle B \qquad \ a \ \text{is} \ opposite \ \angle A\\ & b \ \text{is} \ adjacent \ \text{to} \ \angle A \qquad \ b \ \text{is} \ opposite \ \angle B\\ & c \ \text{is the} \ hypotenuse\end{align*}
The three basic trigonometric ratios are called sine, cosine and tangent. For right triangle ABC\begin{align*}\triangle ABC\end{align*}, we have:
Sine Ratio: opposite leghypotenuse sinA=ac\begin{align*}\frac{opposite \ leg }{hypotenuse} \ \sin A = \frac{a}{c}\end{align*} or sinB=bc\begin{align*}\sin B = \frac{b}{c}\end{align*}
Cosine Ratio: adjacent leghypotenuse cosA=bc\begin{align*}\frac{adjacent \ leg}{hypotenuse} \ \cos A = \frac{b}{c}\end{align*} or cosB=ac\begin{align*}\cos B = \frac{a}{c}\end{align*}
Tangent Ratio: opposite legadjacent leg tanA=ab\begin{align*}\frac{opposite \ leg}{adjacent \ leg} \ \tan A = \frac{a}{b}\end{align*} or tanB=ba\begin{align*}\tan B = \frac{b}{a}\end{align*}
An easy way to remember ratios is to use SOH-CAH-TOA.
A few important points:
• Always reduce ratios (fractions) when you can.
• Use the Pythagorean Theorem to find the missing side (if there is one).
• If there is a radical in the denominator, rationalize the denominator.
What if you were given a right triangle and told that its sides measure 3, 4, and 5 inches? How could you find the sine, cosine, and tangent of one of the triangle's non-right angles?
### Examples
#### Example 1
Find the sine, cosine and tangent ratios of A\begin{align*}\angle A\end{align*}.
First, we need to use the Pythagorean Theorem to find the length of the hypotenuse.
52+12213sinAtanA=c2=c=leg opposite Ahypotenuse=1213=leg opposite Aleg adjacent to A=125cosA=leg adjacent to Ahypotenuse=513,\begin{align*}5^2 + 12^2 &= c^2\\ 13 &= c\\ \sin A &= \frac{leg \ opposite \ \angle A}{hypotenuse} = \frac{12}{13} && \cos A = \frac{leg \ adjacent \ to \ \angle A}{hypotenuse}= \frac{5}{13},\\ \tan A &= \frac{leg \ opposite \ \angle A}{leg \ adjacent \ to \ \angle A}= \frac{12}{5}\end{align*}
#### Example 2
Find the sine, cosine, and tangent of B\begin{align*}\angle B\end{align*}.
Find the length of the missing side.
AC2+52AC2ACsinB=152=200=102=10215=223cosB=515=13tanB=1025=22\begin{align*}AC^2 + 5^2 &= 15^2\\ AC^2 &= 200\\ AC &= 10 \sqrt{2}\\ \sin B &= \frac{10 \sqrt{2}}{15} = \frac{2 \sqrt{2}}{3} && \cos B = \frac{5}{15}=\frac{1}{3} && \tan B = \frac{10 \sqrt{2}}{5} = 2 \sqrt{2}\end{align*}
#### Example 3
Find the sine, cosine and tangent of 30\begin{align*}30^\circ\end{align*}.
This is a 30-60-90 triangle. The short leg is 6, y=63\begin{align*}y = 6 \sqrt{3}\end{align*} and x=12\begin{align*}x=12\end{align*}.
sin30=612=12cos30=6312=32tan30=663=1333=33\begin{align*}\sin 30^\circ = \frac{6}{12} = \frac{1}{2} && \cos 30^\circ = \frac{6 \sqrt{3}}{12} = \frac{\sqrt{3}}{2} && \tan 30^\circ = \frac{6}{6 \sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}\end{align*}
#### Example 4
What is sinA\begin{align*}\sin A\end{align*}cosA\begin{align*}\cos A\end{align*}, and tanA\begin{align*}\tan A\end{align*}?
sinA=1620=45\begin{align*}\sin A=\frac{16}{20}=\frac{4}{5}\end{align*}
cosA=1220=35\begin{align*} \cos A=\frac{12}{20}=\frac{3}{5}\end{align*}
tanA=1612=43\begin{align*} \tan A=\frac{16}{12}=\frac{4}{3}\end{align*}
### Review
Use the diagram to fill in the blanks below.
1. tanD=??\begin{align*}\tan D = \frac{?}{?}\end{align*}
2. sinF=??\begin{align*}\sin F = \frac{?}{?}\end{align*}
3. tanF=??\begin{align*}\tan F = \frac{?}{?}\end{align*}
4. cosF=??\begin{align*}\cos F = \frac{?}{?}\end{align*}
5. sinD=??\begin{align*}\sin D = \frac{?}{?}\end{align*}
6. cosD=??\begin{align*}\cos D = \frac{?}{?}\end{align*}
From questions 1-6, we can conclude the following. Fill in the blanks.
1. cos=sinF\begin{align*}\cos \underline{\;\;\;\;\;\;\;} = \sin F\end{align*} and sin=cosF\begin{align*}\sin \underline{\;\;\;\;\;\;\;} = \cos F\end{align*}.
2. tanD\begin{align*}\tan D\end{align*} and tanF\begin{align*}\tan F\end{align*} are _________ of each other.
Find the sine, cosine and tangent of A\begin{align*}\angle A\end{align*}. Reduce all fractions and radicals.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Acute Angle An acute angle is an angle with a measure of less than 90 degrees.
Adjacent Angles Two angles are adjacent if they share a side and vertex. The word 'adjacent' means 'beside' or 'next-to'.
Hypotenuse The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle.
Legs of a Right Triangle The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.
opposite The opposite of a number $x$ is $-x$. A number and its opposite always sum to zero.
Pythagorean Theorem The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle.
Radical The $\sqrt{}$, or square root, sign.
sine The sine of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the hypotenuse.
Trigonometric Ratios Ratios that help us to understand the relationships between sides and angles of right triangles. |
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# The easy way to work out your tyre size.
Updated: Mar 20
You probably have heard 4-wheel drivers talk about have "35s" or "33s", referring to wheel and tyre size. Maybe that's the language you speak. But do you know how to work this out? It is really simple primary school mathematics. This little formula can help you work it out for yourself in a few simple steps
On the side of a tyre there will be some numbers indicating the tyre size. For example lets work with 315/75x16.
Here is how to use them:
Step 1
The number '315' refers to the width of the tyre in mm (see A).
Step 2
The number '75' is a percentage, and refers to the profile of the tyre. It means that the tyre wall is 75% of the tyre's width. So 315 x 0.75 = 236.25mm (see B).
Step 3
Then add the wheel rim diameter (see C). That is the last number - in the above example it is '16'. The tricky part is this figure is in inches, and you need to multiply it by 25.4 to convert it to mm. So 16 x 25.4 = 405.4mm.
Step 4
So now we add up the three components:
(236.25 x 2) + 405.4 = 877.9mm
To get this back into inches we divide 877.9mm by 25.4 = 34.56 inches or 34.5s. Simple!
Or put into one simple equation it is:
2(315x0.75) + (16x25.4) ÷ 25.4 = Wheel size in inches
or 2(width mm x profile) + (wheel diameter inches x 25.4) ÷ 25.4 |
# What is Sample Mean Formula? Standard Error Of The Sample Mean Formula
## What is Sample Mean?
As we know, sample is just a small part of the whole. For example, if you wanted to know how much people generally pay on food per year, then you should take a poll over 300 million people here. In simple words, take a fraction is 300 million people here the fraction is sample. The other common word for the Mean if average here. Here, we could say that sample mean is the average amount that is paid by thousand of people on food every year.
It is useful calculating sample mean in statistics because it allows you estimating what the whole population is doing. For example, if you have calculated an average or sample mean for the people spending on food every year i.e. $2400 per year then the value would be same even if you are taking the population as a whole. So, its better work on samples instead of taking the population or anything else as a whole. ### Sample Mean Formula It will be less time consuming and taking less resources too. So, sample mean could save a lot of money in your case. The process looks complicated but this is quite easy and flexible in understanding. Remember the concept of average calculation in the basic mathematics. The things are perfectly same here too, only the notation or we could symbols are different. To understand Sample Mean formula, let us break it down into pieces as given below- $\large \overline{x}=\frac{\sum_{i-1}^{n}x_{i}}{n}$ Where,$\bar{x}\$ = sample mean
\sum_{i=1}^{n}x_{i} = x1+x2+….+xn
n = Total number of terms
x1, x2…….xn are different values.
Now, you just have to plugging values and solve them arithmetically or you could use calculators too in advanced cases.
Question 1: The total marks obtained by few students in mathematics exam are 101, 161, 155, 96 and 83. Evaluate the sample mean marks ?
Solution:
Mean $\ \bar{x} = \frac{\sum_{i=1}^{n}x_{i}}{n}$ Total number of terms(n) = 5
Mean = $\ \bar{x} = \frac{101 + 161 + 155 + 96 + 83}{5}$
= 119.2 |
# Divisors of 75
The divisors of 75 are those numbers that completely divide 75 with the remainder zero. In this section, we will discuss about divisors of 75.
## Highlights of Divisors of 75
• Divisors of 75: 1, 3, 5, 15, 25 and 75
• Negative divisors of 75:1, -3, -5, -15, -25 and -75
• Prime divisors of 75: 3 and 5
• Number of divisors of 75: 6
• Sum of divisors of 75: 124
• Product of divisors of 75: 753
## What are Divisors of 75
A number n is a divisor of 75 if $\dfrac{75}{n}$ is an integer. Note that if 75/n=m is an integer, then both m and n will be the divisors of 75.
To find the divisors of 75, we need to find the numbers n such that 75/n becomes an integer. We have:
No numbers other than 1, 3, 5, 15, 25, and 75. can divide 75. So we conclude that
Thus, the total number of divisors of 75 is six.
Divisors of 60: The divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
Divisors of 64: The divisors of 60 are 1, 2, 4, 8, 16, 32, 64.
Divisors of 72: The divisors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.
Divisors of 100: The divisors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100.
## Negative Divisors of 75
We know that if m is a divisor of a number, then -m is also a divisor of that number.
As the divisors of 75 are 1, 3, 5, 15, 25, and 75, we can say that:
The negative divisors of 75 are -1, -3, -5, -15, -25, and -75.
## Prime Divisors of 75
The divisors of 75 are 1, 3, 5, 15, 25, and 75. Among these numbers, only 3 and 5 are prime numbers. So we obtain that:
The prime divisors of 75 are 3 and 5.
Video Solution of Divisors of 75:
## Sum, Product & Number of Divisors of 75
The prime factorization of 75 is given below.
75 = 31 × 52
(i) By the number of divisors formula, we have that the number of divisors of 75 is
=(1+1)(2+1)=2×3=6.
(ii) By the sum of divisors formula, we have that the sum of the divisors of 75 is
$=\dfrac{3^2-1}{3-1} \times \dfrac{5^3-1}{5-1}$
$=\dfrac{9-1}{2} \times \dfrac{125-1}{4}$
$=4 \times 31=124$
(iii) By the product of divisors formula, we have that the product of the divisors of 75 is
=75(Number of divisors of 75)/2
=756/2
=753
Related Topics:
## FAQs on Divisors of 75
Q1: Find divisors of 75.
Answer: The divisors of 75 are 1, 3, 5, 15, 25, and 75.
Q2: What is the sum of divisors of 75?
Answer: The sum of the divisors of 75 is (32-1)/(3-1) × (53-1)/(5-1) = 4×31 = 124.
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# A page of print containing 144 square centimeters of printed region has a margin of 4 and 1/2...
## Question:
A page of print containing 144 square centimeters of printed region has a margin of 4 and 1/2 centimeters at the top and bottom and a margin of 2 centimeters at the sides. What are the dimensions of the page if the width across the page is four units of the length?
## Rectangle:
We need to know the formula to calculate the area of a rectangle in order to solve this question. A rectangle with length l units and width w units has an area of {eq}l{\times}w {/eq} units.
Given: A page of print containing 144 square centimeters of the printed region has a margin of 4 and 1/2 centimeters at the top and bottom and a margin of 2 centimeters at the sides.
Let the width of the page be {eq}w {/eq} cm.
Let the length of the page be {eq}l {/eq} cm
Then, the length of the printed region is {eq}l-4-\frac{1}{2}=\frac{2l-9}{2} {/eq} cm.
The width of the printed region is
{eq}w-2-2=w-4 {/eq} cm.
As the width across the page is four units of the length,
{eq}l=4w {/eq}
The area of the printed region is
\begin{align} \frac{2l-9}{2}{\times}(w-4)=144 \\ \frac{2(4w)-9}{2}{\times}(w-4)=144 \\ (8w-9)(w-4)=288 \\ 8w^2-32w-9w+36=288 \\ 8w^2-41w-252=0 \\ \mbox{ Using the quadratic equation formula} \\ w=\frac{-(-41)\pm\sqrt{(-41)^2-4(8)(-252)}}{2(8)} \\ w=\frac{41\pm\sqrt{1681+8064}}{16} \\ w=\frac{41\pm\sqrt{9745}}{16} \\ w=\frac{41\pm98.72}{16} \\ \end{align}
we get the value of width as 8.73 and -3.61.
As the width cannot be negative.
The length is 34.92 cm and the width is 8.73 cm.
Hence, the dimension of the page is 34.92 by 8.73 units. |
Pvillage.org
# Can you multiply scalar by scalar?
## Can you multiply scalar by scalar?
Multiplying a vector by a scalar is called scalar multiplication. To perform scalar multiplication, you need to multiply the scalar by each component of the vector. Scalar multiplication by a positive number other than 1 changes the magnitude of the vector but not its direction.
How do you multiply matrices with dot product?
To multiply a matrix by a single number is easy:
1. These are the calculations: 2×4=8. 2×0=0.
2. The “Dot Product” is where we multiply matching members, then sum up: (1, 2, 3) • (7, 9, 11) = 1×7 + 2×9 + 3×11. = 58.
3. (1, 2, 3) • (8, 10, 12) = 1×8 + 2×10 + 3×12. = 64.
4. DONE! Why Do It This Way?
### Is a 1 by 1 matrix A scalar?
A 1×1 matrix is a scalar. A null matrix has 0 for all of its entries. If the number of rows of a matrix is the same as the number of its columns, then it is a square matrix.
Can we multiply scalar and vector?
While adding a scalar to a vector is impossible because of their different dimensions in space, it is possible to multiply a vector by a scalar. A scalar, however, cannot be multiplied by a vector.
## Can a matrix be a scalar?
Nope. A matrix is defined over its algebraic structure (or it can be viewed as part of one itself, w/e). So if you define A ∈ ℝ1×1 where ℝ is the field of real numbers, then you are essentially defining A to be a different object, not a scalar. The scalar is its one element.
What is scalar product of matrix?
When we work with matrices, we refer to real numbers as scalars. The term scalar multiplication refers to the product of a real number and a matrix. In scalar multiplication, each entry in the matrix is multiplied by the given scalar.
### What is scalar matrix simple definition?
The scalar matrix is a square matrix having a constant value for all the elements of the principal diagonal, and the other elements of the matrix are zero. The scalar matrix is obtained by the product of the identity matrix with a numeric constant value.
How to find the scalar of a matrix?
Given a matrix and a scalar element k, our task is to find out the scalar product of that matrix. Input : mat [] [] = { {2, 3} {5, 4}} k = 5 Output : 10 15 25 20 We multiply 5 with every element.
## Which is the correct definition of scalar multiplication?
The term scalar multiplication refers to the product of a real number and a matrix. In scalar multiplication, each entry in the matrix is multiplied by the given scalar. Notice that a scalar times a matrix is another matrix. In general, a scalar multiple of a matrix will be another matrix of the same dimension.
Can a matrix be distributed over scalar addition?
This property states that a matrix can be distributed over scalar addition. Once again, we see that the last matrix in each column are equivalent because of the distributive property for real numbers, making the original expressions equivalent as desired!
### How do you multiply a matrix by a number?
To multiply a matrix by a single number is easy: These are the calculations: We call the number (“2” in this case) a scalar, so this is called “scalar multiplication”. But to multiply a matrix by another matrix we need to do the “dot product” of rows and columns |
Equivalence of Definitions of Congruence
Theorem
The following definitions of the concept of Congruence in the context of Number Theory are equivalent:
Let $z \in \R$.
Definition by Remainder after Division
We define a relation $\RR_z$ on the set of all $x, y \in \R$:
$\RR_z := \set {\tuple {x, y} \in \R \times \R: \exists k \in \Z: x = y + k z}$
This relation is called congruence modulo $z$, and the real number $z$ is called the modulus.
When $\tuple {x, y} \in \RR_z$, we write:
$x \equiv y \pmod z$
and say:
$x$ is congruent to $y$ modulo $z$.
Definition by Modulo Operation
Let $\bmod$ be defined as the modulo operation:
$x \bmod y := \begin{cases} x - y \floor {\dfrac x y} & : y \ne 0 \\ x & : y = 0 \end{cases}$
Then congruence modulo $z$ is the relation on $\R$ defined as:
$\forall x, y \in \R: x \equiv y \pmod z \iff x \bmod z = y \bmod z$
Definition by Integer Multiple
Let $x, y \in \R$.
Then $x$ is congruent to $y$ modulo $z$ if and only if their difference is an integer multiple of $z$:
$x \equiv y \pmod z \iff \exists k \in \Z: x - y = k z$
Proof
Let $x_1, x_2, z \in \R$.
Let $x_1 \equiv x_2 \pmod z$ as defined by an equivalence relation.
That is, let $\RR_z$ be the relation on the set of all $x, y \in \R$:
$\RR_z = \set {\tuple {x, y} \in \R \times \R: \exists k \in \Z: x = y + k z}$
Let $\tuple {x_1, x_2} \in \RR_z$.
Then by definition, $\exists k \in \Z: x_1 = x_2 + k z$.
So, by definition of the modulo operation, we have:
$\ds x_1 \bmod z$ $=$ $\ds \paren {x_2 + k z} - z \floor {\frac {x_2 + kz} z}$ $\ds$ $=$ $\ds \paren {x_2 + k z} - z \floor {\frac {x_2} z + k}$ $\ds$ $=$ $\ds \paren {x_2 + k z} - z \floor {\frac {x_2} z} + k z$ $\ds$ $=$ $\ds x_2 - z \floor {\frac {x_2} z}$ $\ds$ $=$ $\ds x_2 \bmod z$
So:
$x_1 \equiv x_2 \pmod z$
in the sense of definition by modulo operation.
$\Box$
Now let $x_1 \equiv x_2 \pmod z$ in the sense of definition by modulo operation.
That is, :$x_1 \equiv x_2 \pmod z \iff x_1 \bmod z = x_2 \bmod z$.
Let $z = 0$.
Then by definition, $x_1 \bmod 0 = x_1$ and $x_2 \bmod 0 = x_2$.
So as $x_1 \bmod 0 = x_2 \bmod 0$ we have that $x_1 = x_2$.
So $x_1 - x_2 = 0 = 0.z$ and so $x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple.
Now suppose $z \ne 0$.
Then from definition of the modulo operation:
$x_1 \bmod z = x_1 - z \floor {\dfrac {x_1} z}$
$x_2 \bmod z = x_2 - z \floor {\dfrac {x_2} z}$
Thus:
$x_1 - z \floor {\dfrac {x_1} z} = x_2 - z \floor {\dfrac {x_2} z}$
and so:
$x_1 - x_2 = z \paren {\floor {\dfrac {x_1} z} - \floor {\dfrac {x_2} z} }$
From the definition of the floor function, we see that both $\floor {\dfrac {x_1} z}$ and $\floor {\dfrac {x_2} z}$ are integers.
Therefore, so is $\floor {\dfrac {x_1} z} - \floor {\dfrac {x_2} z}$ an integer.
So $\exists k \in \Z: x_1 - x_2 = k z$.
Thus $x_1 - x_2 = k z$ and:
$x_1 \equiv x_2 \pmod z$
in the sense of definition by integer multiple.
$\Box$
Now let $x_1 \equiv x_2 \pmod z$ in the sense of definition by integer multiple.
That is, $\exists k \in \Z: x_1 - x_2 = k z$.
Then $x_1 = x_2 + k z$ and so $\tuple {x_1, x_2} \in \RR_z$ where:
$\RR_z = \set {\tuple {x, y} \in \R \times \R: \exists k \in \Z: x = y + k z}$
and so
$x_1 \equiv x_2 \pmod z$
in the sense of definition by equivalence relation.
$\Box$
So all three definitions are equivalent: $(1) \implies (2) \implies (3) \implies (1)$.
$\blacksquare$ |
# P5 tips
```Calculation of uncertainty in a result
Whenever we make a measurement, there is always some uncertainty in the result. This is simply because no
measuring instrument has perfect precision.
Whenever we state a measurement, the number of quoted decimal places and the uncertainty should give
clear evidence for the precision of the measurement. Note that some students refer to the ‘error’ in the
measurement. An uncertainty is not an error.
In many situations, in order to obtain the value of a physical quantity, several other quantities are
measured. Each of these measured quantities has an uncertainty and these uncertainties must be combined in
order to determine the uncertainty in the value of the physical quantity. There are some basic rules for
determining the uncertainty when measurements are combined. These are:
•
•
•
•
•
•
Whenever two measurements are subtracted, the actual uncertainties are added.
Whenever two measurements are multiplied together, the percentage uncertainties are added.
Whenever two measurements are divided, the percentage uncertainties are added.
The actual uncertainty in the final answer should be quoted to one significant figure.
The final answer should be quoted to the same decimal place as the actual uncertainty.
The rules are best illustrated by means of a sample calculation.
Suppose that we are determining a quantity Q. The quantity Q is found using the expression
The quantities a and b are added and so we add the actual uncertainties giving (a + b) = 113.2 ± 1.2
(units).
Note that we have not reduced the uncertainty to one significant figure at this stage because we are at the
start of the process, not the end.
(a + b) is cubed which means ‘multiply by itself three times’. So we take three times the percentage
3
uncertainty in (a + b) to find the uncertainty in (a + b) .
3
Percentage uncertainty in (a + b) is (1.2/113.2) × 3 × 100 = 3.2%
Percentage uncertainty in c is (0.001/0.124) × 100 = 0.81%
(e − f) = 2.0 and the uncertainty in (e − f) is ± 0.1 (units)
So, percentage uncertainty in (e − f) is (0.1/2.0) × 100 = 5.0%
Note that although the percentage uncertainty in either e or f is small, when we take the difference, then
this increases significantly the percentage uncertainty in the result.
3
The quantities (a + b) , c and (e − f) are all multiplied together or divided and so we add the percentage
uncertainties in the quantities.
Total percentage uncertainty = 9.0%
Now, Q = 89935 (units)
Actual uncertainty in Q = ±(9.0 × 89935)/100 = ±8094 (units)
Therefore, we quote the uncertainty as ±8000 and the value of Q as 90 000 (units). That is, in this
case, Q is quoted to the nearest thousand because the uncertainty is to the nearest thousand.
So,
Q = (9.0 ± 0.8) × 10 (units)
4
The uncertainty is written to one significant figure and, in this case, this is one decimal place. The value
of Q is therefore written to the same number of decimal places.
Significant figures and decimal places
In Physics, numerical answers to calculations play an important role in our understanding not only of basic
concepts, but also of the world in which we live. However, a numerical answer can be very misleading
unless we take care to express that answer to an appropriate number of decimal places or significant figures,
depending on whichever is appropriate.
Decimal places (d.p.)
The number of decimal places in an answer is the number of figures that appear to the right of
the decimal point.
For example,
26.4 is expressed to one decimal place
26.43 is expressed to two decimal places
26.430 is expressed to three decimal places
26.4308 is expressed to four decimal places, etc.
The significance of the number of decimal places is that it indicates the least value to which the number is
calculated or the quantity is measured. This is explained for the examples above in the table on page 10.
number number significance
of
decimal
places
26.4
26.43
26.430
26.4308
1
2
3
4
quoted to the nearest tenth
quoted to the nearest hundredth
quoted to the nearest thousandth
quoted to the nearest ten-thousandth
Significant figures (sig. fig. or s.f.)
The number of significant figures is the number of characters shown in an answer or a quantity.
For example,
6 is quoted to one significant figure
63 is quoted to two significant figures
634 is quoted to three significant figures
6.345 is quoted to four significant figures.
Note that the number of significant figures does not take into account the position of the decimal point.
A problem arises when there are zeros at the end of a number. If the number is 600, then has this number
been quoted to one, or two, or three significant figures? This problem is overcome by using index notation.
For example,
2
6 × 10 is quoted to one significant figure
2
6.0 × 10 is quoted to two significant figures
2
6.00 × 10 is quoted to three significant figures
Where a number of zeros are given before a number they do not count as significant figures. The number
0.0063 has two significant figures.
The number of significant figures quoted for the value of a quantity gives an indication of the degree of
precision to which that quantity has been determined. The greater the number of significant figures, the
greater the precision.
General rules for the use of decimal places and
significant figures in examinations
Calculations
Where data is given so that a calculation may be carried out, then the data will be given to a certain number
of significant figures. This is usually either two or three.
When giving the final answer, this should be quoted to the same number of significant figures as
the data.
A warning! Calculations may involve more than one stage, so that the final answer is obtained from two or
more initial calculations. For example, when calculating the speed of a car, then the distance travelled has to
be found and also the time taken must to be determined before finally the distance is divided by the time to
calculate the speed.
Suppose the data for the calculation is given to two significant figures.
On calculating the values of the distance and the time, they may well not work out to two significant
figures and the calculator may give more. If the values for the distance and the time are rounded to two
significant figures then it must be remembered that the values are less precise. So, when these less precise
values are divided one by the other, the final answer for the speed may well be outside any acceptable limit
of accuracy!
As a general rule:
Where a calculation involves a number of parts that lead up to a final answer, then each
individual part should retain two significant figures more than is intended for the final answer.
Measurements
The number of significant figures to which a measurement is quoted should reflect the precision
of the instrument that is used to make that measurement.
For example, measuring a length:
If the length is quoted as 60 cm, then it can be measured to the nearest 10 cm and we would expect the
length to be between 55 cm and 64 cm.
Quoting the length as 64 cm would mean it can be measured to the nearest 1 cm and we would expect the
length to be between 63.5 cm and 64.4 cm.
So, using a metre rule with which we can measure to the nearest millimetre, an answer of 64.2 cm would
be appropriate.
For an ammeter with scale divisions of 0.2 A then an appropriate value of current would be 4.7 A.
Where data is expressed in a table, then there should be consistency in the number of decimal places for
any group of measurements. For example, temperatures of 72.6 °C, 73.2 °C, 73 °C, 73.4 °C and 72.8 °C
would not be acceptable because one value has lower precision than the rest.
Graphs
Graphs are an important feature of work in Physics and we frequently take values of coordinates. In general,
we can read to the nearest half square of the graph grid. The number of decimal places to which the
coordinates are quoted should reflect this precision.
For example, a graph grid has 2 mm squares and a scale is drawn to measure voltage where the number of
2 mm squares between the 2.0 V mark and the 4.0 V mark is 10. Then each 2 mm square represents 0.2 V.
We can read to the nearest half square and that represents 0.1 V. So, it would be appropriate to give the
coordinate as, for example, 3.7 V. Quoting it as 3.70 V would not be acceptable.
Physics is the science of measurement. Measurements are of little value unless it is possible to determine the
reliability of the measurement. That implies the degree of uncertainty in that answer.
In AS Level Topic 2, the way in which the uncertainty in an answer is determined was discussed. It
should be remembered that an uncertainty is not an error.
Having calculated the quantity and its uncertainty, then a final answer should be quoted. This is in two
stages.
Firstly, the uncertainty should be quoted to one significant figure. If, for example, the uncertainty is found
to be ±0.86, then the uncertainty should be given as ±0.9. There is little point in quoting hundredths when
the uncertainty is in tenths!
Secondly, the final answer should be quoted to the same number of decimal places as the uncertainty.
The final answer and its uncertainty should be quoted to the same number of decimal places.
For example, 65.7 ± 0.9 (with its unit) means that the answer is quoted to tenths because the uncertainty is in
tenths. There would be no point in quoting the answer to hundredths or thousandths.
Proportionality and linearity
There is sometimes confusion between these two types of relationship. They are not the same and, at this
level of your studies, you should be able to distinguish between them.
Proportionality
If two quantities P and Q are directly proportional to one another then, if one quantity is doubled, then the
other will also be doubled. So, multiplying one quantity by a number, then the other quantity will also be
multiplied by that same number.
Two quantities P and Q are directly proportional to each other if they follow the
relation P = kQ where k is a constant.
Hence
= k which is constant
Direct proportionality can be represented by the graph below.
We can see that the graph is a straight line. The gradient of the graph line is the constant k in the
relation P = kQ.
Note that the line must pass through the origin.
Two quantities may be proportional to each other, but not directly proportional. For example, P may be
2
proportional to the square of Q. That is P = cQ where c is a constant.
Proportionality can occur with any power of Q.
Two quantities may be inversely proportional to each other. This means, for example, that if doubling one
quantity, then the other is halved.
Two quantities P and Q are inversely proportional to each other if they follow the
relation P =
where k is a constant.
Inverse proportionality can be represented by the graph below, where P is plotted against 1/Q.
The graph is a straight line. The gradient of the graph line is the constant k in the relation
Note that the line must pass through the origin.
Inverse proportionality can occur with any power of Q.
Linearity
Linearity means that one quantity changes by the same factor as the change in another
quantity.
Note the difference between proportionality and linearity.
For proportionality, the ratio of the two quantities that are proportional is always constant.
For linearity, the ratio of the changes in the two quantities is a constant.
Linearity can be represented by the graph below.
We say that R changes linearly with change in S. Note that the line of the graph does not pass through the
origin.
The relationship can be represented by the expression
R = a + bS, where a and b are constants.
The constant a is the intercept on the R-axis.
If there is a change ΔR in R and the corresponding change in S is ΔS, then the constant b which is the
gradient of the graph is given by
Relationships may also be based on powers. For example
n
n
R = a + bS , where a, b and n are constants. A graph of R against S gives a straight line graph with
gradient b and an intercept on the R axis of a.
Straight-line graphs
The representation of data in a graphical form is a very important means by which relationships between
variables can be determined.
The plotting of data points provides an averaging technique which may well be superior to an arithmetical
mean. Where an arithmetical mean is calculated, each set of data has an equal weighting. When using a bestfit line on a graph, the average is weighted towards those data points close to the line. A wayward point
(anomalous point) can be detected and allowance made – perhaps taking a new set of measurements.
An important type of graph which is used frequently in AS/A -Level Physics is the straight-line graph, as
illustrated.
The equation representing this graph is
y = mx + c
where m and c are constants.
The constant m is the gradient of the graph, m =
The constant c is the intercept on the y-axis.
If a variable y is thought to vary linearly with variable x, then plotting this graph will enable the
following:
• the straight line with an intercept of c verifies a linear relationship between y and x
• determination of the values of the gradient m and the intercept c enables the exact form of the relationship to
be established.
If the intercept is zero, the straight line passes through the origin. The relationship is y = mx. This special
case with c = 0 means that y is proportional to x.
Relationships other than linear relationships can also be verified in this way. However, the suspected
relationship must be changed into one of the form y = mx + c, then use the graph you plot to determine the
constants. Some examples of this technique are shown in the table below.
``` |
## How to Solve Investment Problems Part 2
This is the second part of the Solving Investment Problems Series. In the, we discussed in detail the solution of a problem at two different rates of interest. In this post, we discuss another problem.
Problem
Mr. Reyes invested a part of Php70000 at 3% yearly interest and the remaining part at a 5% yearly interest. The annual interest on the 3% investment is Php100 more than the annual interest on the 5% investment. How much was invested at each rate?
Solution
If we let x be the amount invested at 3%, then, 70000 – x is the amount invested at 5%. The yearly interest is the product of the rates and the amount invested so,
(3%)(x) = yearly interest of the amount invested at 3%
(5%)(70000 – x) = yearly interest of the amount invested at 5%
Now, the annual interest at 3% is 100 more than the annual interest at 5%. This means that if we add 100 to the yearly interest at 5%, the interests will be equal. That is,
(3%)(x) = (5%)(70000 – x) + 100.
Next, we convert percent to decimal by dividing the percentage by 100. So,
(0.03)(x) = (0.05)(70000 – x) + 100.
Simplifying, we have
0.03x = 3500 – 0.05x + 100
0.03x = 3600 – 0.05x
0.03x + 0.05x = 3600
0.08x = 3600.
Tip: You can calculate better by eliminating the decimal. You can do this by multiplying both sides by 100.
Dividing both sides by 0.08, we have
x = 45000.
This means that 45000 is invested at 3% yearly interest.
Now, the remaining amount is 70000 – 45000 = 25000.
This means that 25000 is invested at 5%.
Check:
Yearly interest of 45000 at 3% interest = (45000 x 0.03) = 1350
Yearly interest of 25000 at 5% interest = (25000 x 0.05) = 1250
As we can see, the interest at 45000 at 3% interest is 100 more than the interest of 25000 at 5% interest.
Therefore, we are correct.
## How to Solve Investment Problems Part 1
In the Simple Interest Problems Series, we have learned how to calculate the interest, rate, or time of the principal invested. In that series, the principal is invested to a single bank or company. In this series, we will learn how to calculate the interest of money invested at different companies (different rates). Before starting with our first example, familiarize yourself with the following terms:
principal – the amount of money invested
rate of interest – the percent of interest yearly or any period of time (e.g. monthly, quarterly)
interest – income or return of investment
Hence, if Php10000 is invested at a bank with 3% interest is 300, Php10000 is the principal, 3% is the rate of interest, and 300 is the interest.
Example 1
Mr. Molina invested Php100,000.00. A part of it was invested in a bank at 4% yearly interest and another part of it at a credit cooperative at 7% yearly interest. How much investment he made in each if his yearly income from the two investments is Php5950.00?
Scratch Work
If we let x be the money invested at a bank, then 100000 – x is the amount invested at the credit cooperative. To calculate for the interest, we must apply the percent of each interest at each amount. That is
(4%)(x) = yearly interest from at the bank
(7%)(100000 – x) = yearly interest from the credit cooperative
If we add the interest, it will amount to Php5950. So, here’s our solution.
Solution
Let x = amount of money invested at the bank
100000 – x = amount of money invested at the credit cooperative
Total Interest = Interest From the Bank + Interest from the Credit Cooperative
5950 = (4%)(x) + (7%)(100000 – x)
We need to convert percent to decimals in order to multiply. We do this by dividing the percentage by 100. So, 4% = 0.04 and 7% = 0.07. Substituting to the previous equation, we have
5950 = (0.04)(x) + (0.07)(100000 – x)
5950 = 0.04x + 7000 – 0.07x
5950 – 7000 = -0.03x
-1050 = -0.03x.
We can eliminate the decimals by multiplying by 100.
-105000 = -3x
Dividing both sides by -3, we have
35000 = x.
That means that Mr. Molina invested Php35000 in the bank and Php100000 – Php35000 = Php65000 in the credit cooperative.
Check
(4%)(35000) + (7%)(65000) = (.04)(35000) + (.07)(65000)
= 1400 + 4550 = 5950
As we can see, our interest from the two investments is Php5950.00
## The Solving Simple Interest Problems Series
The Solving Simple Interest Problems is a series of tutorials on how to solve simple interest problems. In solving interest problems you need to know the following terms:
the principal (P) is the money invested, the rate (R) of interest is the percentage of interest (that is the number with percent sign), the time (T) and the interest (I) is the earnings or return of investment. The interest is the product of the principal, the rate, and the time, or I = PRT is explained in the first part of the series.
How to Solve Simple Interest Problems Part 1 discusses the basics of simple interest problems and the terms used in such problems. Two examples worked examples are solved in which interests are both unknowns.
How to Solve Simple Interest Problems Part 2 is a continuation of the simple interest discussion. One problem is an example of a rate of interest which is not given yearly and the other one is an investment made for more than a year.
How to Solve Simple Interest Problems Part 3 is a discussion of simple interest problems where the unknowns are the principal and the rate.
I will be posting exercises and problems with solutions about this type of problems soon, so keep posted.
## How to Solve Simple Interest Problems Part 3
This is the third and last part of the Solving Simple Interest Problems Series. In the first and second part of this series, we have learned how to solve simple interest problems. In this post, we continue with two more problems, one with the rate missing and the other one with the principal missing.
Example 5
Mr. Wong deposited \$15,000 in a bank for a certain rate of interest per year. After two years, the interest was \$900. What was the rate of interest in percent?
Solution and Explanation
We have the following given:
Principal (P) = \$15,000
Interest (I) = 900
Time = 2 (years)
Rate (R) = ?
As we have learned in the previous posts, simple interest is the product of the principal, the rate, and time or
I = PRT.
Substituting the given we have
900 = (15000)(R)(2)
900 = 30000R
Dividing both sides by 30000, we have
R = 900/30000
R = 0.03
The rate is 0.03 which is we need to convert to percent by multiplying it by 100. Therefore, the rate is 3%.
Example 6
Mrs. Lansangan invested a certain amount of money in a bank that gives 4% interest per year. She got an interest of Php2400 after 3 years.
Solution and Explanation
Given
I = 2400
R = 4%
T = 3 years
Using the simple interest formula mentioned above, we have
I = PRT
2400 = (P)(4%)(3)
Converting 4 percent to decimal, we have
2400 = (P)(0.04)(3)
2400 = 0.12P
Dividing both sides by 0.12, we have
20000 = P
Therefore, Mrs. Lansangan invested Php20000.
That’s it for our series on how to solve simple interest problems. I hope you have learned well. Good luck for the exams.
## How to Solve Simple Interest Problems Part 2
This is the second part of the Solving Simple Interest Problems Series. In the previous post, we have discussed the basics of simple interest problems. We have learned that the simple interest (I) is equal to the product of the amount of money invested or the principal (P), the percentage of interest or the rate (R) and the time (T). Therefore, we can use the following formula:
I = PRT.
In this post, we are going to discuss more problems particularly interests that are not yearly and finding unknowns other than interest.
Example 3
Dr. Lopez invested his Php120,000 in a bank that gives 2% interest every quarter. What is the interest of his money if he is to invest it for 1 years?
Solution and Explanation
Notice that the interest is applied quarterly and not every year. Quarterly means every three months and therefore it will be applied four times a year since there are four quarters every year. So in this case, the time is 4. So,
P = Php120,000
R = 2%
T = 4
I = ?
Note that the rate percent must be converted to decimal by dividing it by 100. So 2% equals 0.02. Now, using the formula, we have
I = PRT
I = (Php120,000)(.02)(4)
I = Php9600.00
So, the interest of the money for 1 year is Php9600.
Example 4
Danica invested here money amounting to Php150,000 in a bank that offers a 5% simple interest every year. She went abroad and never made any deposit or withdrawal in her account. After coming back, she immediately checked her account and found out that her money got an interest of Php37,500. How many years was the money invested?
Solution and Explanation
In this problem, interest is given and time is unknown. Assigning the values we have
I = Php37,500
P = Php150,000
R = 5%
T = ?.
Using the formula, we have
I = PRT
Converting 5 percent to decimal and substituting, we have
37,500 = (150,000)(.05)(T)
37,500 = 7500(T).
Dividing both sides by 7500, we have
5 = T.
That means that the money was invested for 5 years.
In the next post, we will be solving simple interest problems whose uknowns are rate and principal.
## How to Solve Simple Interest Problems Part 1
This is the first part of the Solving Simple Interest Problem Series for the Civil Service Examination.
Simple interest problems are usually included in many examinations such as the Civil Service Exams. It is important that you practice solving these types of problems in order to increase your chance of passing the exams.
Before solving simple interest problems, let us familiarize ourselves with the terms used in simple interest problems. These are the money invested which is called the principal, the rate of interest which is the percent and the interest which is the income or return of investment, and time. Time may vary depending on the investment. It can range from months to years.
Example 1
Mr. Reyes invested Php50,000 at an interest rate of 3% per year.
a.) Identify the principal and rate of interest.
b.) Calculate the interest earned after 1 year.
Solution
For (a)
The money invested or principal is Php50,000, the interest rate is 3%, and the time is 1 year.
For (b)
We want to calculate 3% of Php50,000. To multiply, we must convert 3 percent to decimal which is equal to 0.03.
interest = Php50,000 × 0.03
interest = Php1500
This means that for a year, the money earned Php1500.
Example 2
Ms. Gutierrez invested Php60,000 at a simple interest of 4% per year for 4 years.
a.) Identify the principal, rate of interest, and time.
b.) How much money will Ms. Gutierrez have after four years?
Solution
For (a),
The principal or money invested is Php60,000.
The rate of interest is 4%.
The time is 4 years.
For (b),
We need to calculate 4% of Php60,000. Just like above, we must first convert 4 percent to decimal which is equal to 0.04.
Now,
interest (1 year) = P60,000 × 0.04 = 2,400
That is the interest for 1 year. To be able to calculate the interest for four years, we have
interest (4 years) = 2,400 × 4 = 9600.
So, the money Ms. Gutierrez will have by the end of four years is the Principal which is 60,000 and the interest for 4 years which is 9600. So, in total, her money will be 69,600.
***
From the two problems above, we can see the interest (I) is the product of the principal (P), the rate (R), and the time(T). Therefore, we can have the formula.
I = P × R × T
or simply
I = PRT.
In the next part of this series, we will be solving more simple interest problems.
## The Solving Mixture Problems Series
The Solving Mixture Problems Series is a series of tutorials that explain how to solve mixture problems. Mixture problems can be classified into two, those involved with percents and the other one involved with prices.
How to Solve Mixture Problem Part 1 discusses the basics of base and percentage. This is a preparation of the mixture problems involving percents.
How to Solve Mixture Problem Part 2 discusses the most basic of the mixture problems. A detailed solution is discussed about the following problem.
How many liters of 80% alcohol solution must be added to 60 liters of 40% alcohol solution to produce a 50% alcohol solution.
How to Solve Mixture Problem Part 3 discusses another mixture problems involving percents. A detailed solution is discussed about the following problem.
How many liters of pure water must be added to 15 liters of a 20% salt solution to make a 5% salt solution?
How to Solve Mixture Problem Part 4 discusses one more mixture problems involving percents. A detailed solution is discussed about the following problem.
A chemist creates a mixture with 5% boric acid and combined it with another mixture containing 40% boric acid to obtain a 800 ml of mixture with 12% boric acid. How much of each mixture did he use?
How to Solve Mixture Problem Part 5 discusses the basics of mixture problem involving prices. A detailed solution is discussed about the following problem.
A seller mixes 20 kilograms of candy worth 80 pesos per kilogram to candies worth 50 pesos per kilogram. He sold at 60 pesos per kilogram. After selling all the candies he discovered that he had no gain or loss. How much of the 50-pesos per kilogram candies did he use?
How to Solve Mixture Problem Part 6 discusses one more mixture problem involving prices. A detailed solution is discussed about the following problem.
A bakery owner wants to box assorted chocolates. Each box is to be made up of packs of black chocolates worth \$10 per pack and packs of ordinary chocolates worth \$7 each pack. How many packs of each kind should he use to make 15 packs which he can sell for \$8 per pack?
## How to Solve Mixture Problems Part 6
This is the 6th part and last part of the Solving Mixture Problems Series. In the previous 4 parts, we have learned how to solve mixture problems involving percent and in part 5, we have learned how to solve problems involving percents. In this post, we solve another problem involving percent.
A bakery owner wants to box assorted chocolates. Each box is to be made up of packs of black chocolates worth \$10 per pack and packs of ordinary chocolates worth \$7 each pack. How many packs of each kind should he use to make 15 packs which he can sell for \$8 per pack?
Solution
Let x = number of \$10 packs
15 – x = number of \$7 packs
Multiplying the cost per pack and the number of packs we have
(\$10)(x) = total cost of \$10 packs
(\$7)(15 – x) = total cost of the \$7 packs
(\$8)(15) = total cost of all the chocolates
Now, we know that
total cost of \$10 packs + total cost of the \$7 packs = total cost of all the chocolates.
Substituting the values, we have
(\$10)(x) + (\$7)(15 – x) = (\$8)(15).
Eliminating the dollar sign and solving for x, we have
(10)(x) + (7)(15 – x) = (8)(15)
10x + 105 – 7x = 120
3x + 105 = 120
3x = 120 – 105
3x = 15
x = 5.
This means that we need 5 packs of \$10 and 10 packs of \$7 chocolates.
Check:
(\$10)(5) + (\$7)(10) = (\$8)(15)
\$50 + \$70 = \$120
\$120 = \$120
Therefore, we are correct.
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KinderGals: Making the Most of Unifix Cubes
## Monday, October 24, 2016
### Making the Most of Unifix Cubes
Do you have boxes of unifix cubes stored in your closet? Are you trying to figure out how to use them for more than making patterns? Then, this post is for you! In this post I will share some activities that Michele and I created to teach ALL of the Math Standards!
For each of the 5 mathematical strands I will share an engaging unifix cube activity that your kids can do totally independently! At the bottom of the post, I share how Michele and I ensure success!
First let's look at counting. Here are the standards we used to guide our ideas.
Common Core Kindergarten standard: CC4a: When counting objects, say the number names in the standard order, pairing each object with one and only one number name and each number name with one and only one object.
Common Core First Grade standard: NBT1: Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.
NCTM standard for both grades: 1A- Understand numbers, ways of representing numbers, relationships among numbers, and number systems.
This is a super simple activity to practice counting and subitizing.
• Use a long piece of yarn and a pipe cleaner to make lacing string.
• Tie a unifix cube to the bottom of the piece of yarn.
• Invite the children to roll a dice.
• They count that many unifix cubes to lace onto their string.
• Consider using a 12 sided dice for counting sets with a larger quantity.
Now, let's look at shapes. Here are the standards we used to guide our ideas.
Common Core Kindergarten standard: KGA: Identify and describe shapes. KGB: Analyze, compare, create, and compose shapes.
Common Core First Grade standard: 1GA: Reason with shapes and their attributes
NCTM standards for both grades: Standard 3A: Analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships.
To play this game here is what you do.
• Provide the children with a variety of shapes.
• Be sure to consider the size of the shape. This will determine how many cubes they will be counting.
• Invite the children to lay cubes around the edge of the shape.
• Once they finish, they count the cubes and record the number on the recording page.
Now, let's look at measurement. Here are the standards we used to guide our ideas.
Common Core Kindergarten standard: MD1-Describe measurable attributes of objects, such as length or weight. Describe several measurable attributes of a single object. MD2-Directly compare two objects with a measurable attribute in common, to see which object has "more of"/"less of" the attribute, and describe the difference.
Common Core First Grade standard: MD2 Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps.
NCTM standards for both grades: Standard 4- 4A: Understand measurable attributes of objects and the units, systems, and processes of measurement. 4B: Apply appropriate techniques, tools and formulas to determine measurements.
This is a great partner game for early finishers. I don't use a recording page for this activity. Here's how you play.
• Give each child a partner.
• Each child counts out 10 cubes and snaps them together.
• Invite the children to put the cubes behind their backs and say, "1,2,3 break."
• The children each place the unifix cubes in their right hand on the game board, comparing for taller and shorter.
• Lay the cubes in the left hand on the table.
• Whichever child has "shorter", spins the spinner.
• If it lands on shorter, than the children with the shorter ice cream gets all of the cubes on the game board.
• If it lands on taller, than the children with the taller ice cream gets all of the cubes.
• When one child has all of the cubes, he places them on taller.
• The child with no cubes, spins the spinner.
• If it lands on shorter, the child with no cubes is the winner.
• It is lands on taller, the child with all of the cubes is the winner.
Now, let's look at composing and decomposing number. Here are the standards we used to guide our ideas.
Common Core Kindergarten standard: KOA.A: Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from.
Common Core First Grade standard: 1OA.A: Represent and solve problems involving addition and subtraction. OA.C: Add and subtract within 20
NCTM standards for both grades: Standard 1- 1A; Understand numbers, ways of representing numbers, relationships among numbers, and number systems. 1B: Understand meanings of operations and how they relate to one another. 1C: Compute fluently and make reasonable estimates.
Here is an easy activity for composing a number 11-19.
• Give each child a stem game board.
• Invite them to use 10 green cubes to make the stem.
• Roll a dice (1-9).
• Count out cubes to match the number on the dice. (all one color of cube)
• Use the cubes to make a flower.
• Take the flower apart, and place the cubes on the tens frames.
• Record the number on the recording page.
Now, let's look at adding, subtracting and number combinations. Here are the standards we used to guide our ideas.
Common Core Kindergarten standard: KOA.A: Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from.
Common Core First Grade standard: 1OA.A: Represent and solve problems involving addition and subtraction. OA.C: Add and subtract within 20
NCTM standards for both grades: Standard 1- 1A; Understand numbers, ways of representing numbers, relationships among numbers, and number systems. 1B: Understand meanings of operations and how they relate to one another. 1C: Compute fluently and make reasonable estimates.
Here is how you play.
• Give each child a ladder fives frame.
• Invite the children to roll a dice. (You may want to cover the 6 dots with a colored dot.)
• Count unifix cubes to match the dice.
• Put the cubes on the ladder fives frame.
• Decide how many more you need to make a 5 by filling in the empty squares on the ladder with another color cube.
Here's how to use patterns to teach addition.
• Give each child a strip (no longer than 10 cubes long).
• Invite the children to select a pattern card.
• Make the pattern on the strip.
• Record the pattern on the recording page by gluing down squares.
• Take the pattern apart and sort the colors.
• Write the number sentence.
Here's how to play.
• Invite each child to select a picture card.
• Use the unifix cubes to make the picture.
• Count how many cubes and record on the recording page.
• Roll a dice.
• Subtract that many cubes from your picture.
• Finish the recording page to write the subtraction sentence.
Ever share an idea with someone that worked perfectly in your room only to have them tell you it was a disaster in theirs? There are many reasons that could happen, but one that I see a lot is some teachers provide more scaffolding for their kids than others. Teachers that provide scaffolding, ensure the child's ability to engage in meaningful activities independently. Here's how Michele and I do that...
First let me tell you....my friend Michele is super smart! She is the best lesson scripter...hands down! Her plans are so detailed, that we can easily share them with other teacher with very little direction beyond the script! This lesson plan is for one activity!!!!!
Here is how we conduct our math block.
First...I do...this is where the teacher models. So in our activity, Which is Taller?, it would sound like this...
I do (Teacher models):
Tell students that today they will get to do an activity that will require them to work with a partner and to compare two things to see which is taller. Remind students what taller and shorter means by comparing the height of two students in the class. Next ask a student to come up and pose as your partner for this fun activity.
Ask your partner to count out 10 unifix cubes and connect them into a train. You will need to do the same. Show students the taller/shorter mats (hammer and popsicle). Model how you and your partner can put the cube trains behind your backs and break it in two. Each one of you should pull one hand out from behind your back and place that one stack of cubes on the comparing mat (either the hammer or popsicles- you chose). The other part of the ten train should be discarded for now. Next look at the comparing mat and compare the two stacks of cubes. Which is taller? Which is shorter? Next ask students whose stack should win? They will inevitably say the stack that is taller should be the winner, but you should tell them that we have a better way to figure out who will win. We will spin to see who wins! Show them the spinner that corresponds to the mat you chase to use. The person that will spin to determine the winner is the person who had a shorter stack. If it lands on shorter, the person with the shorter stack is the winner, If it lands on taller, the person with the taller stack is the winner. Re-form your ten train and model this several more times.
Afterward, ask your students to think of what you just did. Call on students to tell what you did first, next, and finally. Now show the students the “I can” chart for this activity and review the steps.
Then, We do...this is where the children try the activity on their own with your support. It sounds like this...
We do (Students do with teacher support):
Now it’s time for the students to try this with your support and guidance. Quickly divide the students into groups of 4 and give each group one tub of unifix cubes. Divide each group into partnerships. Have the students gather around each tub. Give each partnership and comparing card and a corresponding spinner. Ask each student to create a train of 10 unifix cubes. They should
then place the stack behind their backs at the same time and say “break” and break their stack in two. Each student should then pull one hand out from behind their back and place their stack on the comparing mat. Both students should then look at the mat and determine who has the taller stack and who has the shorter stack. The person with the shorter stack should spin the spinner to see who wins. Circulate around the groups and watch as students are playing the game. Step in if students need your assistance or guidance.
Next, You do...this is where the children try the activity on their own with your support. It sounds like this...
You do (Students will work independently):
Now you will allow students to work independently on this activity. Move the student groups further away from each other to help students concentrate when so many are playing at the same time. Remind students to use whisper voices so that they do not disturb other groups. Allow students to continue working with the partner they have been working with or you can mix them up if you choose. Instruct students to continue playing the game that you taught them.
Last, is the closure. This is when you pull the children back together to discuss the activity and resolve any problems. It sounds like this...
Closure (Discuss the activity):
Ask students to think about the activity they just did. How did it go? What went right? What problems did they encounter? Ask student volunteers to answer these questions. Determine how to solve any problems by asking for solutions from the students. (Ex. If a student had difficulty being a good partner, what would the students suggest as a solution?
These activities, and many more, are from our Unifix Cubes unit. You can find the unit here. |
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# Point of Inflection: Concavity of Function, Test for Concavity of Function, Point of Inflection, Practice Problems, and FAQs:
We can figure out whether the function is increasing or decreasing with the use of derivatives. Now another thing that comes to our mind is the shape of the curve which means whether the curve is facing upward or downward. The upward and downward tendency of the curve can be given by means of the concavity of the curve. So let us discuss the concavity of the curve and then we will extend the concept of concavity to understand the Point of Inflection of the curve.
• Concavity of a Function
• Test for Concavity of a Function
• Point of Inflection
• Practice Problems
## Concavity of a Function
In the figures shown above both the functions are in such a way that, as the value of x increases the slope of the tangent given by its derivative drawn also increases. In the first case, with an increase in x slope changes from a less positive value to a more positive value hence increasing and in the second case with an increase in x the slope changes from a more negative value to a less negative value hence again the slope is increasing. Hence we can say that the derivative of both functions is an increasing function. Thus, whenever the derivative of a function is an increasing function then the function will have its shape as concave up.
Similarly, In the figures shown here both the functions are in such a way that as the value of x increases the slope of the tangent drawn decreases. In the first case slope with an increase in x slope changes from a more positive value to a less positive value hence decreasing and in the second case with an increase in x the slope changes from a less negative value to more negative value hence again the slope is decreasing. Hence we can say that the derivative of both functions is a decreasing function. Thus, whenever the derivative of a function is a decreasing function then the function will have its shape as concave down.
## Test for Concavity of a Function
Now what is the importance of learning the Test for Concavity of a function? You can easily understand it here. What to do if we just want to check the concavity of a given function without drawing its graph? Here comes the use of the Test for Concavity of function. We should determine the first and second derivatives of the given function to apply the Test for Concavity.
By the definition, we concluded that for a function f to be concave up it derivative i.e. f' must be increasing, and for f' to be increasing derivative of f' i.e. f'' must be positive. Thus if there exists a function f such that it is twice differentiable and f''>0 then the given function will be concave up.
If f''>0 for the given interval I, then f is concave up.
Similarly, if there exists a function f such that it is twice differentiable and f''<0 then the given function will be concave down.
If f''<0 for the given interval I, then f is concave down.
Finding out the concavity of a given function with the above-explained logic is known as the Test for Concavity.
Sign of f' Nature of f Sign of f'' Concavity of f Positive Increasing Positive Concave up Positive Increasing Negative Concave down Negative Decreasing Positive Concave up Negative Decreasing Negative Concave down
## Point of Inflection
Point of inflection comes into consideration when a function switches its concavity in the given interval. Hence in a given interval for a segment if f''>0 and for another segment if f''<0, then the function will be changing its concavity and the point at which this phenomenon occurs will be called the Point of Inflection of f. To check the point of inflection we look at the points where the f''=0 or f'' is undefined. But it should also be kept in mind that when f''=0 or undefined it is not always necessary that the function will be changing its concavity.
Hence summarily when the function f is continuous at a point a and also changes its concavity at a, then (a,f(a)) will be the Inflection Point of the function f.
1. For x<a, f''(x)>0, hence function is concave up for the interval (-∞,a).
2. For x<a, f''(x)<0, hence function is concave down for the interval (a,∞).
Therefore, the point (a,f(a)) is an inflection point.
## Practice Problems:
Example 1: Which are the points of inflection on the shown curve?
Solution: Let us analyze the curve for different segments.
Thus, the points where the concavity is changing are 2, 4, 6 and 7. Hence these points will be the inflection points.
Example 2: Analyze the given exponential function f(x)=kex and comment on its existence of inflection points.
Solution:
Given $f\left(x\right)=k{e}^{x}$
Finding the first derivative of the given function,
$f\text{'}\left(x\right)=k{e}^{x}$
Now finding the second derivative of the given function,
$f\text{'}\text{'}\left(x\right)=k{e}^{x}$
Now checking for the sign of f''(x)
ex is always a positive function for any value of x in its domain hence when k is positive f''(x) will always be positive and when k is negative f"(x) will be negative. Thus for any fixed value of k the function f''(x) will not be changing its sign. Therefore f(x) do not have any points of inflection for any value of k.
Example 3: find the intervals where the function $f\left(x\right)={x}^{4}-6{x}^{3}-108{x}^{2}+57x+2$ is concave up and concave down. Also, find Points of Inflection if they exist.
Solution:
Given $f\left(x\right)={x}^{4}-6{x}^{3}-108{x}^{2}+57x+2$
Finding the first derivative of the given function,
$f\text{'}\left(x\right)=4{x}^{3}-18{x}^{2}-216{x}^{+}57$
Now finding the second derivative of the given function,
$f\text{'}\text{'}\left(x\right)=12{x}^{2}-36{x}^{-}216$
$f\text{'}\text{'}\left(x\right)=12\left({x}^{2}-3{x}^{-}18\right)$
$f\text{'}\text{'}\left(x\right)=12\left(x+3\right)\left(x-6\right)$
Now checking for the sign of f''(x)
For x (-3,6), f''(x)<0 so, the function will be concave downwards
For x (-∞,-3)(6,∞), f''(x)>0 so, the function will be concave upwards
Hence at x=-3 and x=6 , f''(x) changes sign so these will be the inflection points.
Example 4: Is x=0 an inflection Point for the function $g\left(x\right)={x}^{\frac{1}{3}}.$
Solution:
Given $g\left(x\right)={x}^{\frac{1}{3}}$
Finding the derivative of the given function,
Again differentiating the function to find out the second derivative,
$g\text{'}\text{'}\left(x\right)=\frac{1}{3}\left(-\frac{2}{3}{x}^{-\frac{2}{3}-1}\right)⇒-\frac{2}{9}{x}^{-\frac{5}{3}}$
Now, checking for the sign of g''(x)
For x=0 the function g''(x) is not defined but for
Thus g(x) changes its concavity at x=0, hence it is an inflection point for $g\left(x\right)={x}^{\frac{1}{3}}.$
Question 1: Will the point at which f''(x)=0 exist be an inflection point for the function f(x) ?
No; If there exists an inflection point on the curve then at that point f''(x)=0 but the converse is not always true.
Question 2: Is the inflection Point a critical point? |
Intermediate Algebra
Intermediate Algebra7.2 Add and Subtract Rational Expressions
Learning Objectives
By the end of this section, you will be able to:
• Add and subtract rational expressions with a common denominator
• Add and subtract rational expressions whose denominators are opposites
• Find the least common denominator of rational expressions
• Add and subtract rational expressions with unlike denominators
• Add and subtract rational functions
Be Prepared 7.2
Before you get started, take this readiness quiz.
1. Add: $710+815.710+815.$
If you missed this problem, review Example 1.29.
2. Subtract: $3x4−89.3x4−89.$
If you missed this problem, review Example 1.28.
3. Subtract: $6(2x+1)−4(x−5).6(2x+1)−4(x−5).$
If you missed this problem, review Example 1.56.
Add and Subtract Rational Expressions with a Common Denominator
What is the first step you take when you add numerical fractions? You check if they have a common denominator. If they do, you add the numerators and place the sum over the common denominator. If they do not have a common denominator, you find one before you add.
It is the same with rational expressions. To add rational expressions, they must have a common denominator. When the denominators are the same, you add the numerators and place the sum over the common denominator.
If p, q, and r are polynomials where $r≠0,r≠0,$ then
$pr+qr=p+qrandpr−qr=p−qrpr+qr=p+qrandpr−qr=p−qr$
To add or subtract rational expressions with a common denominator, add or subtract the numerators and place the result over the common denominator.
We always simplify rational expressions. Be sure to factor, if possible, after you subtract the numerators so you can identify any common factors.
Remember, too, we do not allow values that would make the denominator zero. What value of x should be excluded in the next example?
Example 7.13
Add: $11x+28x+4+x2x+4.11x+28x+4+x2x+4.$
Try It 7.25
Simplify: $9x+14x+7+x2x+7.9x+14x+7+x2x+7.$
Try It 7.26
Simplify: $x2+8xx+5+15x+5.x2+8xx+5+15x+5.$
To subtract rational expressions, they must also have a common denominator. When the denominators are the same, you subtract the numerators and place the difference over the common denominator. Be careful of the signs when you subtract a binomial or trinomial.
Example 7.14
Subtract: $5x2−7x+3x2−3x+18−4x2+x−9x2−3x+18.5x2−7x+3x2−3x+18−4x2+x−9x2−3x+18.$
Try It 7.27
Subtract: $4x2−11x+8x2−3x+2−3x2+x−3x2−3x+2.4x2−11x+8x2−3x+2−3x2+x−3x2−3x+2.$
Try It 7.28
Subtract: $6x2−x+20x2−81−5x2+11x−7x2−81.6x2−x+20x2−81−5x2+11x−7x2−81.$
Add and Subtract Rational Expressions Whose Denominators are Opposites
When the denominators of two rational expressions are opposites, it is easy to get a common denominator. We just have to multiply one of the fractions by $−1−1.−1−1.$
Let’s see how this works.
Multiply the second fraction by $−1−1.−1−1.$ The denominators are the same. Simplify.
Be careful with the signs as you work with the opposites when the fractions are being subtracted.
Example 7.15
Subtract: $m2−6mm2−1−3m+21−m2.m2−6mm2−1−3m+21−m2.$
Try It 7.29
Subtract: $y2−5yy2−4−6y−64−y2.y2−5yy2−4−6y−64−y2.$
Try It 7.30
Subtract: $2n2+8n−1n2−1−n2−7n−11−n2.2n2+8n−1n2−1−n2−7n−11−n2.$
Find the Least Common Denominator of Rational Expressions
When we add or subtract rational expressions with unlike denominators, we will need to get common denominators. If we review the procedure we used with numerical fractions, we will know what to do with rational expressions.
Let’s look at this example: $712+518.712+518.$ Since the denominators are not the same, the first step was to find the least common denominator (LCD).
To find the LCD of the fractions, we factored 12 and 18 into primes, lining up any common primes in columns. Then we “brought down” one prime from each column. Finally, we multiplied the factors to find the LCD.
When we add numerical fractions, once we found the LCD, we rewrote each fraction as an equivalent fraction with the LCD by multiplying the numerator and denominator by the same number. We are now ready to add.
We do the same thing for rational expressions. However, we leave the LCD in factored form.
How To
Find the least common denominator of rational expressions.
1. Step 1. Factor each denominator completely.
2. Step 2. List the factors of each denominator. Match factors vertically when possible.
3. Step 3. Bring down the columns by including all factors, but do not include common factors twice.
4. Step 4. Write the LCD as the product of the factors.
Remember, we always exclude values that would make the denominator zero. What values of $xx$ should we exclude in this next example?
Example 7.16
Find the LCD for the expressions $8x2−2x−3,3xx2+4x+38x2−2x−3,3xx2+4x+3$ and rewrite them as equivalent rational expressions with the lowest common denominator.
Try It 7.31
Find the LCD for the expressions $2x2−x−12,1x2−162x2−x−12,1x2−16$ rewrite them as equivalent rational expressions with the lowest common denominator.
Try It 7.32
Find the LCD for the expressions $3xx2−3x–10,5x2+3x+23xx2−3x–10,5x2+3x+2$ rewrite them as equivalent rational expressions with the lowest common denominator.
Add and Subtract Rational Expressions with Unlike Denominators
Now we have all the steps we need to add or subtract rational expressions with unlike denominators.
Example 7.17
How to Add Rational Expressions with Unlike Denominators
Add: $3x−3+2x−2.3x−3+2x−2.$
Try It 7.33
Add: $2x−2+5x+3.2x−2+5x+3.$
Try It 7.34
Add:$4m+3+3m+4.4m+3+3m+4.$
The steps used to add rational expressions are summarized here.
How To
1. Step 1.
Determine if the expressions have a common denominator.
• Yes – go to step 2.
• No – Rewrite each rational expression with the LCD.
• Find the LCD.
• Rewrite each rational expression as an equivalent rational expression with the LCD.
2. Step 2. Add or subtract the rational expressions.
3. Step 3. Simplify, if possible.
Avoid the temptation to simplify too soon. In the example above, we must leave the first rational expression as $3x−6(x−3)(x−2)3x−6(x−3)(x−2)$ to be able to add it to $2x−6(x−2)(x−3).2x−6(x−2)(x−3).$ Simplify only after you have combined the numerators.
Example 7.18
Add: $8x2−2x−3+3xx2+4x+3.8x2−2x−3+3xx2+4x+3.$
Try It 7.35
Add: $1m2−m−2+5mm2+3m+2.1m2−m−2+5mm2+3m+2.$
Try It 7.36
Add:$2nn2−3n−10+6n2+5n+6.2nn2−3n−10+6n2+5n+6.$
The process we use to subtract rational expressions with different denominators is the same as for addition. We just have to be very careful of the signs when subtracting the numerators.
Example 7.19
Subtract: $8yy2−16−4y−4.8yy2−16−4y−4.$
Try It 7.37
Subtract: $2xx2−4−1x+2.2xx2−4−1x+2.$
Try It 7.38
Subtract: $3z+3−6zz2−9.3z+3−6zz2−9.$
There are lots of negative signs in the next example. Be extra careful.
Example 7.20
Subtract:$−3n−9n2+n−6−n+32−n.−3n−9n2+n−6−n+32−n.$
Try It 7.39
Subtract :$3x−1x2−5x−6−26−x.3x−1x2−5x−6−26−x.$
Try It 7.40
Subtract: $−2y−2y2+2y−8−y−12−y.−2y−2y2+2y−8−y−12−y.$
Things can get very messy when both fractions must be multiplied by a binomial to get the common denominator.
Example 7.21
Subtract: $4a2+6a+5−3a2+7a+10.4a2+6a+5−3a2+7a+10.$
Try It 7.41
Subtract: $3b2−4b−5−2b2−6b+5.3b2−4b−5−2b2−6b+5.$
Try It 7.42
Subtract: $4x2−4−3x2−x−2.4x2−4−3x2−x−2.$
We follow the same steps as before to find the LCD when we have more than two rational expressions. In the next example, we will start by factoring all three denominators to find their LCD.
Example 7.22
Simplify: $2uu−1+1u−2u−1u2−u.2uu−1+1u−2u−1u2−u.$
Try It 7.43
Simplify: $vv+1+3v−1−6v2−1.vv+1+3v−1−6v2−1.$
Try It 7.44
Simplify: $3ww+2+2w+7−17w+4w2+9w+14.3ww+2+2w+7−17w+4w2+9w+14.$
To add or subtract rational functions, we use the same techniques we used to add or subtract polynomial functions.
Example 7.23
Find $R(x)=f(x)−g(x)R(x)=f(x)−g(x)$ where $f(x)=x+5x−2f(x)=x+5x−2$ and $g(x)=5x+18x2−4.g(x)=5x+18x2−4.$
Try It 7.45
Find $R(x)=f(x)−g(x)R(x)=f(x)−g(x)$ where $f(x)=x+1x+3f(x)=x+1x+3$ and $g(x)=x+17x2−x−12.g(x)=x+17x2−x−12.$
Try It 7.46
Find $R(x)=f(x)+g(x)R(x)=f(x)+g(x)$ where $f(x)=x−4x+3f(x)=x−4x+3$ and $g(x)=4x+6x2−9.g(x)=4x+6x2−9.$
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Access this online resource for additional instruction and practice with adding and subtracting rational expressions.
Section 7.2 Exercises
Practice Makes Perfect
Add and Subtract Rational Expressions with a Common Denominator
75.
$2 15 + 7 15 2 15 + 7 15$
76.
$7 24 + 11 24 7 24 + 11 24$
77.
$3 c 4 c − 5 + 5 4 c − 5 3 c 4 c − 5 + 5 4 c − 5$
78.
$7 m 2 m + n + 4 2 m + n 7 m 2 m + n + 4 2 m + n$
79.
$2 r 2 2 r − 1 + 15 r − 8 2 r − 1 2 r 2 2 r − 1 + 15 r − 8 2 r − 1$
80.
$3 s 2 3 s − 2 + 13 s − 10 3 s − 2 3 s 2 3 s − 2 + 13 s − 10 3 s − 2$
81.
$2 w 2 w 2 − 16 + 8 w w 2 − 16 2 w 2 w 2 − 16 + 8 w w 2 − 16$
82.
$7 x 2 x 2 − 9 + 21 x x 2 − 9 7 x 2 x 2 − 9 + 21 x x 2 − 9$
In the following exercises, subtract.
83.
$9 a 2 3 a − 7 − 49 3 a − 7 9 a 2 3 a − 7 − 49 3 a − 7$
84.
$25 b 2 5 b − 6 − 36 5 b − 6 25 b 2 5 b − 6 − 36 5 b − 6$
85.
$3 m 2 6 m − 30 − 21 m − 30 6 m − 30 3 m 2 6 m − 30 − 21 m − 30 6 m − 30$
86.
$2 n 2 4 n − 32 − 18 n − 16 4 n − 32 2 n 2 4 n − 32 − 18 n − 16 4 n − 32$
87.
$6 p 2 + 3 p + 4 p 2 + 4 p − 5 − 5 p 2 + p + 7 p 2 + 4 p − 5 6 p 2 + 3 p + 4 p 2 + 4 p − 5 − 5 p 2 + p + 7 p 2 + 4 p − 5$
88.
$5 q 2 + 3 q − 9 q 2 + 6 q + 8 − 4 q 2 + 9 q + 7 q 2 + 6 q + 8 5 q 2 + 3 q − 9 q 2 + 6 q + 8 − 4 q 2 + 9 q + 7 q 2 + 6 q + 8$
89.
$5 r 2 + 7 r − 33 r 2 − 49 − 4 r 2 + 5 r + 30 r 2 − 49 5 r 2 + 7 r − 33 r 2 − 49 − 4 r 2 + 5 r + 30 r 2 − 49$
90.
$7 t 2 − t − 4 t 2 − 25 − 6 t 2 + 12 t − 44 t 2 − 25 7 t 2 − t − 4 t 2 − 25 − 6 t 2 + 12 t − 44 t 2 − 25$
Add and Subtract Rational Expressions whose Denominators are Opposites
In the following exercises, add or subtract.
91.
$10 v 2 v − 1 + 2 v + 4 1 − 2 v 10 v 2 v − 1 + 2 v + 4 1 − 2 v$
92.
$20 w 5 w − 2 + 5 w + 6 2 − 5 w 20 w 5 w − 2 + 5 w + 6 2 − 5 w$
93.
$10 x 2 + 16 x − 7 8 x − 3 + 2 x 2 + 3 x − 1 3 − 8 x 10 x 2 + 16 x − 7 8 x − 3 + 2 x 2 + 3 x − 1 3 − 8 x$
94.
$6 y 2 + 2 y − 11 3 y − 7 + 3 y 2 − 3 y + 17 7 − 3 y 6 y 2 + 2 y − 11 3 y − 7 + 3 y 2 − 3 y + 17 7 − 3 y$
95.
$z 2 + 6 z z 2 − 25 − 3 z + 20 25 − z 2 z 2 + 6 z z 2 − 25 − 3 z + 20 25 − z 2$
96.
$a 2 + 3 a a 2 − 9 − 3 a − 27 9 − a 2 a 2 + 3 a a 2 − 9 − 3 a − 27 9 − a 2$
97.
$2 b 2 + 30 b − 13 b 2 − 49 − 2 b 2 − 5 b − 8 49 − b 2 2 b 2 + 30 b − 13 b 2 − 49 − 2 b 2 − 5 b − 8 49 − b 2$
98.
$c 2 + 5 c − 10 c 2 − 16 − c 2 − 8 c − 10 16 − c 2 c 2 + 5 c − 10 c 2 − 16 − c 2 − 8 c − 10 16 − c 2$
Find the Least Common Denominator of Rational Expressions
In the following exercises, find the LCD for the given rational expressions rewrite them as equivalent rational expressions with the lowest common denominator.
99.
$5 x 2 − 2 x − 8 , 2 x x 2 − x − 12 5 x 2 − 2 x − 8 , 2 x x 2 − x − 12$
100.
$8 y 2 + 12 y + 35 , 3 y y 2 + y − 42 8 y 2 + 12 y + 35 , 3 y y 2 + y − 42$
101.
$9 z 2 + 2 z − 8 , 4 z z 2 − 4 9 z 2 + 2 z − 8 , 4 z z 2 − 4$
102.
$6 a 2 + 14 a + 45 , 5 a a 2 − 81 6 a 2 + 14 a + 45 , 5 a a 2 − 81$
103.
$4 b 2 + 6 b + 9 , 2 b b 2 − 2 b − 15 4 b 2 + 6 b + 9 , 2 b b 2 − 2 b − 15$
104.
$5 c 2 − 4 c + 4 , 3 c c 2 − 7 c + 10 5 c 2 − 4 c + 4 , 3 c c 2 − 7 c + 10$
105.
$2 3 d 2 + 14 d − 5 , 5 d 3 d 2 − 19 d + 6 2 3 d 2 + 14 d − 5 , 5 d 3 d 2 − 19 d + 6$
106.
$3 5 m 2 − 3 m − 2 , 6 m 5 m 2 + 17 m + 6 3 5 m 2 − 3 m − 2 , 6 m 5 m 2 + 17 m + 6$
Add and Subtract Rational Expressions with Unlike Denominators
In the following exercises, perform the indicated operations.
107.
$7 10 x 2 y + 4 15 x y 2 7 10 x 2 y + 4 15 x y 2$
108.
$1 12 a 3 b 2 + 5 9 a 2 b 3 1 12 a 3 b 2 + 5 9 a 2 b 3$
109.
$3 r + 4 + 2 r − 5 3 r + 4 + 2 r − 5$
110.
$4 s − 7 + 5 s + 3 4 s − 7 + 5 s + 3$
111.
$5 3 w − 2 + 2 w + 1 5 3 w − 2 + 2 w + 1$
112.
$4 2 x + 5 + 2 x − 1 4 2 x + 5 + 2 x − 1$
113.
$2 y y + 3 + 3 y − 1 2 y y + 3 + 3 y − 1$
114.
$3 z z − 2 + 1 z + 5 3 z z − 2 + 1 z + 5$
115.
$5 b a 2 b − 2 a 2 + 2 b b 2 − 4 5 b a 2 b − 2 a 2 + 2 b b 2 − 4$
116.
$4 c d + 3 c + 1 d 2 − 9 4 c d + 3 c + 1 d 2 − 9$
117.
$−3 m 3 m − 3 + 5 m m 2 + 3 m − 4 −3 m 3 m − 3 + 5 m m 2 + 3 m − 4$
118.
$8 4 n + 4 + 6 n 2 − n − 2 8 4 n + 4 + 6 n 2 − n − 2$
119.
$3 r r 2 + 7 r + 6 + 9 r 2 + 4 r + 3 3 r r 2 + 7 r + 6 + 9 r 2 + 4 r + 3$
120.
$2 s s 2 + 2 s − 8 + 4 s 2 + 3 s − 10 2 s s 2 + 2 s − 8 + 4 s 2 + 3 s − 10$
121.
$t t − 6 − t − 2 t + 6 t t − 6 − t − 2 t + 6$
122.
$x − 3 x + 6 − x x + 3 x − 3 x + 6 − x x + 3$
123.
$5 a a + 3 − a + 2 a + 6 5 a a + 3 − a + 2 a + 6$
124.
$3 b b − 2 − b − 6 b − 8 3 b b − 2 − b − 6 b − 8$
125.
$6 m + 6 − 12 m m 2 − 36 6 m + 6 − 12 m m 2 − 36$
126.
$4 n + 4 − 8 n n 2 − 16 4 n + 4 − 8 n n 2 − 16$
127.
$−9 p − 17 p 2 − 4 p − 21 − p + 1 7 − p −9 p − 17 p 2 − 4 p − 21 − p + 1 7 − p$
128.
$− 13 q - 8 q 2 + 2 q − 24 − q + 2 4 − q − 13 q - 8 q 2 + 2 q − 24 − q + 2 4 − q$
129.
$−2 r − 16 r 2 + 6 r − 16 − 5 2 − r −2 r − 16 r 2 + 6 r − 16 − 5 2 − r$
130.
$2 t − 30 t 2 + 6 t − 27 − 2 3 − t 2 t − 30 t 2 + 6 t − 27 − 2 3 − t$
131.
$2 x + 7 10 x − 1 + 3 2 x + 7 10 x − 1 + 3$
132.
$8 y − 4 5 y + 2 − 6 8 y − 4 5 y + 2 − 6$
133.
$3 x 2 − 3 x − 4 − 2 x 2 − 5 x + 4 3 x 2 − 3 x − 4 − 2 x 2 − 5 x + 4$
134.
$4 x 2 − 6 x + 5 − 3 x 2 − 7 x + 10 4 x 2 − 6 x + 5 − 3 x 2 − 7 x + 10$
135.
$5 x 2 + 8 x − 9 − 4 x 2 + 10 x + 9 5 x 2 + 8 x − 9 − 4 x 2 + 10 x + 9$
136.
$3 2 x 2 + 5 x + 2 − 1 2 x 2 + 3 x + 1 3 2 x 2 + 5 x + 2 − 1 2 x 2 + 3 x + 1$
137.
$5 a a − 2 + 9 a − 2 a + 18 a 2 − 2 a 5 a a − 2 + 9 a − 2 a + 18 a 2 − 2 a$
138.
$2 b b − 5 + 3 2 b − 2 b − 15 2 b 2 − 10 b 2 b b − 5 + 3 2 b − 2 b − 15 2 b 2 − 10 b$
139.
$c c + 5 + 5 c − 2 − 10 c c 2 − 4 c c + 5 + 5 c − 2 − 10 c c 2 − 4$
140.
$6 d d − 5 + 1 d + 4 − 7 d − 5 d 2 − d − 20 6 d d − 5 + 1 d + 4 − 7 d − 5 d 2 − d − 20$
141.
$3 d d + 2 + 4 d − d + 8 d 2 + 2 d 3 d d + 2 + 4 d − d + 8 d 2 + 2 d$
142.
$2 q q + 5 + 3 q − 3 − 13 q + 15 q 2 + 2 q − 15 2 q q + 5 + 3 q − 3 − 13 q + 15 q 2 + 2 q − 15$
In the following exercises, find $R(x)=f(x)+g(x)R(x)=f(x)+g(x)$ $R(x)=f(x)−g(x).R(x)=f(x)−g(x).$
143.
$f(x)=−5x−5x2+x−6f(x)=−5x−5x2+x−6$ and
$g(x)=x+12−xg(x)=x+12−x$
144.
$f(x)=−4x−24x2+x−30f(x)=−4x−24x2+x−30$ and
$g(x)=x+75−xg(x)=x+75−x$
145.
$f(x)=6xx2−64f(x)=6xx2−64$ and
$g(x)=3x−8g(x)=3x−8$
146.
$f(x)=5x+7f(x)=5x+7$ and
$g(x)=10xx2−49g(x)=10xx2−49$
Writing Exercises
147.
Donald thinks that $3x+4x3x+4x$ is $72x.72x.$ Is Donald correct? Explain.
148.
Explain how you find the Least Common Denominator of $x2+5x+4x2+5x+4$ and $x2−16.x2−16.$
149.
Felipe thinks $1x+1y1x+1y$ is $2x+y.2x+y.$ Choose numerical values for x and y and evaluate $1x+1y.1x+1y.$ Evaluate $2x+y2x+y$ for the same values of x and y you used in part . Explain why Felipe is wrong. Find the correct expression for $1x+1y.1x+1y.$
150.
Simplify the expression $4n2+6n+9−1n2−94n2+6n+9−1n2−9$ and explain all your steps.
Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
After reviewing this checklist, what will you do to become confident for all objectives? |
Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Lesson Plans: Introduction Lesson Plan 1: Miles of Tiles - The Pool Border Problem Lesson Plan 2: Cups and Chips - Solving Linear Equations Using Manipulatives
Lesson Plan 1: Miles of Tiles - The Pool Border Problem
Supplies:
Teachers will need the following:
• Transparencies with 1 cm grids
• Approximately 30 unit algebra tiles (click here for a set of printable algebra tiles)
Students will need the following:
• 1 large piece of poster board with a 1 cm grid
• Markers
• 1 cm by 20 cm strips (click here for a set of printable 1 cm strips)
• Glue stick
• Grid paper
• Approximately 40 unit algebra tiles
Steps
Introductory Activity:
1. Begin the day's lesson with a story, such as:
Last night, I saw the most wonderful pool. It had beautiful tiles all around it. So this morning, I asked my landlord if he would install a pool in the backyard of my apartment. At first, he thought I was crazy, but I told him I'd make him a deal. I told him that if he built my dream pool, I would install the tiles around the edges of the pool. So, he made a deal with me. He told me that he'd install a pool with an area of 36 square feet.
2. Ask the class, "If my pool has an area of 36 square feet, what are the possible dimensions of the pool?" Elicit from students all possible dimensions of the pool, using only whole numbers: 1 ft by 36 ft, 2 ft by 18 ft, 3 ft by 12 ft, 4 ft by 9 ft, and 6 ft by 6 ft.
3. Explain to students that you are on a budget, so you need their help in determining the least number of tiles that could be used around the outside edge of the pool. Using the overhead projector, display a 4 ft by 9 ft pool. Tell students that each algebra tile represents a 1 ft by 1 ft tile. Ask students to predict the number of tiles that would be needed to put a border of tiles around the entire pool.
4. On the board, record student guesses for the number of tiles needed. You may want to have the class reach a consensus regarding the number of tiles that will be necessary, or you may want them to discover this in their groups as part of the learning activities below. (For a 4 ft by 9 ft pool, the class should conclude that the border will consist of 30 tiles: the perimeter of the pool is 26 ft, and one tile is needed for each foot of perimeter; in addition, 4 tiles are needed at the corners, as shown below.)
Learning Activities:
1. Explain to the class that they will be working in groups of four to investigate the number of tiles needed for pools of various sizes. For the group exploration, provide the following directions:
• Sitting together, build pools and make borders around the pools.
• Record the number of tiles needed for each pool.
• Look for a pattern.
• Finally, come up with an algebraic expression that relates the length and width to the number of tiles needed.
You may want to have students devise their own way of working together or you may want to assign the following roles to members of the group: writer, responsible for filling in the group's chart; cutter, responsible for the scissors; sticker, responsible for the glue; and speaker, who will present the group's findings to the class.
2. Assign one of the various pool sizes (from the introductory activity) to each group. The students in each group are responsible for constructing a model of the pool they are assigned. In addition, the group should consider all of the various pool sizes and look for a pattern that relates the length and width to the number of tiles needed.
3. Allow students time to construct a model of the pool they have been assigned. Students should cut the 1 cm by 20 cm strips to the length needed to form a border around their pool. Students may also use the 1 cm by 20 cm strips to investigate pools of sizes other than the one they were assigned, or they can investigate using the grid paper. As students are working, circulate and use effective questions to help the groups identify the relationship between the length and width and the number of tiles.
4. After about 15-20 minutes, have each group present its findings. (Depending on the number of students in your class, this may mean that two speakers are presenting the same material, or it may mean that some sizes will not have been assigned.)
Students will invariably arrive at several different expressions for finding the number of tiles, including:
• 2l + 2w + 4
• 2(l + w) + 4
• 2(l + w + 2)
• 2(l + 2) + 2(w + 2) - 4
• (l + 2)(w + 2) - lw
5. After all student groups have presented their findings, describe one of the expressions that they have not discovered, and ask them to consider whether or not this alternative method is equivalent to their expression. For instance, you might say, "I was thinking that I would add the length and the width, double that result, and then add 4."
6. Select a student to translate your method into an algebraic expression. Be sure to discuss the order of operations.
7. Select another student to demonstrate how the equation found by their group is equivalent to the alternate expression that you suggested. (You may wish to repeat this step if several groups found different expressions. This discussion may allow for an explanation of the distributive property, the order of operations, the associative and commutative properties, and other topics.)
8. Ask again the question that was posed at the beginning of the lesson: "Which pool would require the fewest tiles?" Students should conclude that the 6 ft by 6 ft pool will only require 28 tiles, and that this is the fewest needed for any 36 sq ft pool.
Then, ask: "If I wanted a pool in which to swim laps, which would be the best one?" Students may suggest that the 1 ft by 36 ft pool is best, because it is the longest. Other students, however, will likely point out that such a pool would not be wide enough. Students may argue for the 2 ft by 18 ft and 3 ft by 12 ft pools as the best candidates. While 4 ft by 9 ft and 6 ft by 6 ft would be wide enough, they would not be long enough for swimming laps.
Explain to students that because the sizes are not ideal for the dream pool, you would like them to consider other patterns for pools. On the board or overhead, show them Design 1, which is a 1 ft by 2 ft ; Design 2, which is a 2 ft by 3 ft pool; Design 3, which is a 3 ft by 4 ft pool; and Design 4, which is a 4 ft by 5 ft pool. Ask them to use this pattern to predict what Design 5 would look like, and then use their drawing to determine the number of tiles needed for the border of the Design 5 pool. Similarly, have students determine the number of tiles needed for Design 11, as well as for Design n.
Allow students to present their findings to the class. In particular, encourage students to share their expression for the number of tiles. For Design n, the length of the pool is n + 1, and the width is n. Consequently, numerous expressions could represent the number of tiles needed for the border of Design n:
• 2(n + 1) + 2n + 4
• 4n + 6
• 2(2n + 3)
• 2(n + 1 + n + 2)
• (n + 3)(n + 2) - n(n - 1)
Have students use the expressions to confirm the number of border tiles for Design 6 and Design 11.
Culminating Activity/Assessment:
Ask students to express their ideas regarding what they learned about algebra and the power of algebra. Allow several students to share their thoughts. |
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# The base of an isosceles triangle is $\dfrac{4}{3}\text{cm}$. The perimeter of the triangle is $\dfrac{62}{15}\text{cm}$. What is the length of either of the remaining equal sides?
Last updated date: 16th Jul 2024
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Hint: In this question, the perimeter and length of the base are given. Take the sum of the sides of the isosceles triangle and equate it to the perimeter. Then substitute the given values to find the length of the two remaining sides.
Here, we are given that the base of the isosceles triangle is $\dfrac{4}{3}\text{cm}$. The perimeter of the triangle is $\dfrac{62}{15}\text{cm}$. We need to find the length of either of the remaining sides.
Before proceeding with this question, we must know what an isosceles triangle is. An isosceles triangle is a triangle that has two sides of equal length. The two sides of triangles are called legs while the third side is called the base of the triangle.
Let us consider an isosceles triangle ABC.
In the above isosceles triangle, sides AB = AC and side BC is the third side that is the base of the given triangle.
Now, we know that the perimeter of any polygon is given by the sum of its sides. So, we get,
Perimeter of the triangle = Sum of its 3 sides
Therefore, for the given triangle, we get,
Perimeter of $\Delta \text{ABC = AB + AC + BC}$
We know that AB = AC.
So, let us substitute AB = AC = x. So, we get,
Perimeter of $\Delta \text{ABC = 2x + BC}$
Also, we know that BC is the base of the triangle. So, we get,
Perimeter of isosceles $\Delta \text{ABC = 2x + Base of }\Delta \text{ABC}$
So, for any general isosceles triangle, we get,
Perimeter of isosceles triangle = 2 (Length of either of the equal sides) + (length of the base)
We are given that the base of the isosceles triangle $=\dfrac{4}{3}\text{cm}$ and its perimeter $=\dfrac{62}{15}\text{cm}$. By substituting these values in the above equation, we get,
$\dfrac{62}{15}\text{cm = 2 }\left( \text{length of either of equal sides} \right)+\text{ }\dfrac{4}{3}\text{cm}$
By subtracting $\dfrac{4}{3}$ from both the sides of the above equation, we get,
$\dfrac{62}{15}-\dfrac{4}{3}=2\left( \text{length of either of equal sides} \right)$
$\Rightarrow \dfrac{62-20}{15}=2\left( \text{length of either of equal sides} \right)$
Or, $2\left( \text{length of either of equal sides} \right)=\dfrac{42}{15}$
Now by dividing 2 on both the sides, we get,
Length of either of equal side $=\dfrac{7}{5}\text{cm}\approx \text{1}\text{.4cm}$
Hence, we get the length of each equal side of the isosceles triangle as $\dfrac{7}{5}$ or 1.4 cm.
Note:
Students must note the triangles are divided into 3 categories based on length of sides that are scalene triangles, isosceles triangles and equilateral triangles. In scalene triangles, all sides are unequal, in isosceles triangles, at least 2 out of 3 sides are equal while in the equilateral triangle, all sides are equal. Students can also cross-check their answer by adding the length of all the sides and equating it to the given perimeter and checking if LHS = RHS or not. |
# If the letters of the word 'MOTHER' are written in all possible orders and these words are written out as in a dictionary, find the rank of the word 'MOTHER'.
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Hint: In the word 'MOTHER' we can see that there are a total of six letters, so, first of all we will write all the letters of the word in alphabetical order and then we will start making words according to the rule of the dictionary.
We have been asked to find the rank of the word 'MOTHER' when the letter of the words are written in all possible orders and these words are written out as in a dictionary.
Letters in the word 'MOTHER' are in the order E, H, M, O, R, T according to the rules of the dictionary.
Now, we have different cases as follows:
Case I- Words with E _ _ _ _ _
5 letters can be arranged as 5! at 5 places.
$\Rightarrow \text{Words with E }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ = 5!}$
Case II- Words with H _ _ _ _ _
Similarly,
$\Rightarrow \text{Words with H }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ = 5!}$
Case III- Words with ME _ _ _ _
4 letters can be arranged by 4! at 4 different places.
$\Rightarrow \text{Words with ME }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ = 4!}$
Case IV- Words with MH _ _ _ _
Similarly,
$\Rightarrow \text{Words with MH }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ = 4!}$
Case V- Words with MOE _ _ _
We can arrange 3 letters by 3! at 3 different places.
$\Rightarrow \text{Words with MOE }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ = 3!}$
Case VI- Words with MOH _ _ _
Similarly,
$\Rightarrow \text{Words with MOH }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ = 3!}$
Case VII- Words with MOR _ _ _
Similarly,
$\Rightarrow \text{Words with MOR }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ = 3!}$
Case VIII- Words with MOTE _ _
We can arrange 2 letters by 2! at 2 places.
$\Rightarrow \text{Words with MOTE }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ = 2!}$
After this, we get the word 'MOTHER'.
On adding all words in different cases, we get:
\begin{align} & \Rightarrow 5!+5!+4!+4!+3!+3!+3!+2! \\ & \Rightarrow 120+120+24+24+6+6+6+2 \\ & \Rightarrow 308 \\ \end{align}
So, there are 308 words before the word 'MOTHER'.
Therefore, ${{309}^{th}}$ word will be 'MOTHER' which is the rank of the word in the dictionary.
Note: Remember that, in the question, related to the rank of a word in a dictionary, first of all we must have to arrange the letter of the word in alphabetically order and then take the different cases until you get the desired word. |
Question
# Sum of the areas of two squares is 468 ${m^2}$. If the difference of their perimeters is 24 m, find the sides of the two squares.
Hint: Assume the sides of the squares as some variables. Form two equations using the conditions given in the question and solve them.
Let the side of the first square be $a$ and that of the second square be $A$.
Then the area of the first square $= {a^2}$
And the area of the second square $= {A^2}$
Their perimeters would be $4a$ and $4A$ respectively.
The difference of the perimeters of the squares is given as 24 m. So, we have:
$\Rightarrow 4A - 4a = 24 \\ \Rightarrow 4\left( {A - a} \right) = 24 \\ \Rightarrow A - a = 6 .....(i) \\$
And sum of their areas is given as 468 ${m^2}$:
$\Rightarrow {A^2} + {a^2} = 468 .....(ii)$
Putting $A = a + 6$ from equation $(i)$ in equation $(ii)$, we’ll get:
$\Rightarrow {\left( {a + 6} \right)^2} + {a^2} = 468 \\ \Rightarrow {a^2} + 36 + 12a + {a^2} = 468 \\ \Rightarrow 2{a^2} + 12a + 36 = 468 \\ \Rightarrow {a^2} + 6a + 18 = 234 \\ \Rightarrow {a^2} + 6a - 216 = 0 \\ \Rightarrow {a^2} + 18a - 12a - 216 = 0 \\ \Rightarrow a\left( {a + 18} \right) - 12\left( {a + 18} \right) = 0 \\ \Rightarrow \left( {a - 12} \right)\left( {a + 18} \right) = 0 \\ \Rightarrow a = 12{\text{ or }}a = - 18 \\$
But the side of the square cannot be negative, $a = 12$ is the valid solution.
Putting the value of $a$ in equation $(i)$ we’ll get:
$\Rightarrow A - 12 = 6 \\ \Rightarrow A = 18 \\$
Thus, the side of the first square is 12 m and the side of the second square is 18 m.
Note: If we face any difficulty finding the roots of the quadratic equation $a{x^2} + bx + c = 0$ by simple factorization, we can apply the formula for finding roots:
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. |
# College math problems with answers
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In this blog post, we discuss how College math problems with answers can help students learn Algebra. Control your home phone and home VoIP systems remotely via the web browser. - Control your home phone and home VoIP systems via an Elastix server (or another SIP server can be used). - Create custom ring groups and forward calls to them. - Create custom greetings (e.g., "Welcome Home", "I'm Out") to play when someone calls your extension.
A binomial solver is a math tool that helps solve equations with two terms. This type of equation is also known as a quadratic equation. The solver will usually ask for the coefficients of the equation, which are the numbers in front of the x terms. It will also ask for the constants, which are the numbers not attached to an x. With this information, the solver can find the roots, or solutions, to the equation. The roots tell where the line intersects the x-axis on a graph. There are two roots because there are two values of x that make the equation true. To find these roots, the solver will use one of several methods, such as factoring or completing the square. Each method has its own set of steps, but all require some algebraic manipulation. The binomial solver can help take care of these steps so that you can focus on understanding the concept behind solving quadratic equations.
There are a few different ways to calculate the slope of a line, but the most common is to use the slope formula. This formula is relatively easy to use and only requires two pieces of information: the rise and the run. The rise is the vertical distance between two points on the line, and the run is the horizontal distance between those same two points. Once you have these two values, you simply plug them into the formula and solve.
Simple solutions math is a method of teaching that focuses on breaking down complex problems into small, manageable steps. By breaking down problems into smaller pieces, students can better understand the concepts behind the problem and find more creative solutions. Simple solutions math also encourages students to think outside the box and approach problems from different angles. As a result, Simple solutions math can be an effective way to teach problem-solving skills. In addition, Simple solutions math can help to improve test scores and grades. Studies have shown that students who use Simple solutions math outperform those who do not use the method. As a result, Simple solutions math is an effective tool for helping students succeed in school.
A math word search is a great way to review key math vocabulary. It can be used as a review before a test or quiz, or as a tool to help learn new math terms. Simply print out a math word search and see how many of the words you can find.
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Exponential Distribution
09.08.2018
Episode #9 of the course Theory of probability by Polina Durneva
Hello!
Today and tomorrow, we will talk about two continuous probability distributions: exponential and gamma distributions. It is important to note that these two distributions are very closely interrelated and can be derived from each other. In this lesson, we will mainly discuss exponential distribution, which is also associated with Poisson distribution, discussed previously.
What Is Exponential Distribution?
Exponential distribution, also called negative exponential distribution, is used when we talk about time. In Poisson distribution, we only care about events happening over a period of time. For example, knowing that a fast food restaurant has 100 customers per hour, we might want to know the probability of 120 customers coming to the restaurant during the next hour. If we model the time in between each customer coming to the restaurant, we will use exponential distribution. Having 100 customers per hour, or a new customer every 36 seconds (1 hour * 60 minutes * 60 seconds = 3,600 seconds, and then 3,600 seconds / 100 customers = 36), we might want to know the probability that a new customer will arrive in 5 seconds. In some cases, time between events can be replaced by space between objects.
The following assumptions should be held for exponential distribution (please note that the same assumptions are held for Poisson distribution):
1. The intervals over which the events occur do not overlap.
2. The events are independent.
3. The probability that more than one event happens in a very short time period is approximately 0.
Formula for Exponential Distribution
The formula for exponential distribution takes the following form: P(x) = λ * e-xλ, where λ is the rate of occurring events, e is the natural number, and x is the random variable.
Let’s use an example from the previous section to perform a calculation and better understand the concept. We have a restaurant and know that 100 customers come in every hour, meaning that a new customer arrives every 36 seconds. The rate, or λ, is 1/36, then. Let’s assume that we want to know the probability of a new customer arriving in 5 seconds, or x = 5. Then, P(x = 5) = (1/36) * e-(5 * 1/36) = 0.024 = 2.4%. As we can see, the probability that a new customer arrives in 5 seconds and not in 36 seconds is very low.
Expected Value of Exponential Distribution
The expected value, or mean, of exponential distribution can be calculated by dividing 1 by the rate λ. In our previous example, the expected value will be 1/λ = 36 seconds, which totally makes sense: Every new customer arrives every 36 seconds on average.
Variance of Exponential Distribution
Finally, the variance of exponential distribution is a bit more complicated and can be derived using the formula 1/λ2. In our example with customers arriving to a fast food restaurant, the variance would be 1/(1/36)2 = 362 = 1,296.
Well, that’s the end of our lesson for today! Tomorrow, we will finish the course by talking about Gamma distribution, which is basically derived from exponential distribution.
See you soon,
Polina
Recommended book
Introduction to Probability by Joseph K. Blitzstein, Jessica Hwang
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In our previous posts, we talked about the notion of probability, some of its basic features, and how to find the probability of a single event. Here, we will find the probability of a compound event, namely an event where multiple events occur.
For example, we now know that the probability of a fair die landing on 6 is ⅙, while the probability of a fair coin landing heads is ½. But what is the probability that, if I flip a fair coin and roll a fair die, I get that the coin lands heads and the die lands on 6? What is probability that the coin lands heads or the die lands on 6?
To answer that question, we need to introduce a new idea: independence. In ordinary language, talk of “independence” suggests a rebellious child or ideas of liberty and freedom. But here, we use a different notion of independence: two events are independent if neither event affects the likelihood of the other.
An example will help to illustrate the concept. You roll a fair die and then flip a fair coin. We know that, ordinarily, a fair coin has a ½ chance of coming up heads. But now, suppose you knew that the die came up 6. Now, what is the probability that the coin came up heads? Clearly, the answer should remain: ½. The fact that the die came up 6 has nothing to do with the coin! We say that the two events are independent of one another. More precisely, we say:
Definition: Two events, A and B, are independent of one another if:
(i) A occurring does not affect the likelihood of B occurring, and
(ii) B occurring does not affect the likelihood of A occurring
For the purposes of the GRE, it will generally be clear when two events are independent. Here are some standard examples of independence:
• You flip a coin and roll a die. Whether you get heads on the coin ( = Event A) and whether you get 6 on the die ( = Event B) are independent.
• You draw a marble from a bag, replace that marble, and then draw a second marble. Whether the first marble is green ( = Event A) is independent of whether the second marble is purple ( = Event B). Similarly, whether the first marble is green is independent of whether the second marble is green.
Now consider the following case:
Example 1
A bag contains 10 purple marbles and 7 green marbles. You will randomly draw one marble from the bag and then, without returning the first marble to the bag, you draw a second marble from the bag. Is the event of getting a green marble first independent of the event of getting a purple marble second?
Here is a more intuitive way to put the same point: suppose your first marble is not green. Then it must be purple (since the bag just has purple and green marbles). That’s one fewer purple marble for you to draw next turn!
The reason why we care about independence is because independent events allow us to easily calculate the probabilities of compound events. What is a compound event?
Definition: Let A and B represent two events. A compound event E is the event where both A and B occur.
For example, suppose I am wondering about the weather this afternoon. Let A = “It rains.” and B = “The Broncos win their game today.” Then, the compound event A and B will be the event where “It rains and the Broncos win.” It will generally be obvious when two events are independent, and sometimes the question will explicitly state that fact. Other examples of independent events include:
• There is a bag with 10 green marbles and 14 yellow marbles. I pick a marble, look at its color, return it to the bag and pick another marble. The color of the first marble is independent of the color of the second.
• I roll a fair die, record its outcome, and then roll it again. The outcome of the first roll is independent of the outcome of the second roll.
• I randomly draw a person's name for a raffle. Then, I flip a coin. The name I draw is independent of my coin flip.
When events are independent, we can calculate the probability of both events occurring via the following rule:
Probabilities of Compound Events
Let A and B be independent of one another. Then, P(A and B) = P(A)P(B)
Let's see this rule in action:
Example 2
Suppose I roll a fair six-sided die and flip a fair coin. What is the probability that the coin lands heads and the die lands on six?
Now, this rule may seem a little odd. Why does this rule work? If you're the kind of person who needs to get a sense of why something works in order to learn it, here's an illustration that helps make the rule more intuitive.
Now, here's an application of our rule that uses larger numbers:
Example 3
You are wondering how poorly your day could go. You know that, at work, one employee (out of 400) will be randomly selected for additional performance reviews. And you know that there is a ⅓ chance that it rains furiously during your commute home. (Of course, whether you are selected or not for the performance review will not affect the weather.) What is the probability that you are picked for the additional performance reviews AND it rains furiously during your commute?
In our next post, instead of looking at the probability of A and B occurring, we will look at the probability of A or B occurring.
Practice Problems
Question 1
You roll a fair six-sided die. Is the event of you rolling a multiple of 5 independent of you rolling an even number?
Question 2
You roll a fair six-sided die. Is the event of you rolling a prime number independent of you rolling an even number?
Question 3
There are two events, A and B, which are independent of each other. P(A) = .2 and P(B) = .5. What is P(A and B)?
Question 4
You flip a fair coin 7 times in a row. What is the probability that all 7 flips come up heads?
Question 5
There are two independent events, A and B. Neither A nor B is guaranteed to happen. P(A) = .7. There are two values:
X = P(A and B)
Y = .7
Which of the following is true?
A. X is greater than Y
B. Y is greater than X
C. X and Y are equal
D. There is not enough information to tell.
Next Article: Probability for One or Another Event to Occur - P(A or B) |
# Fractions on a Number Line
This is a math series follows our unit for introduction to fractions in grade 3. This is one of several, and I’m combining several days into one post. This part of the unit all involves number line.
## Day 3
My intention for our fraction work was to continue with our circles, but it was so beautiful outside today that we took our math outside to work on our fractions on a number line. We first drew our number line out two feet between each whole number using a ruler in one color and marked our whole numbers with a foot long mark out to the side. Then I asked her to use another color to half each one finding our half mark. These marks were 6 inches long, half of the foot mark. With another color, I then asked her to half each half finding our ¼ marks, which were delineated by a 3 inch tic on the line. In yet another different color we found the half of each ¼ finding our 1/8 marks which were just a short tics. These were as low as we went on this number line.
After we had all of our marks, we began counting by fractions, first by halves- ½, 2/2 (by the whole number “1”), 3/2, 4/2…and so on. Then we counted by ¼’s—1/4, 2/4, ¾, 4/4, and so forth. Finally we counted by 1/8’s in the same way. After we had counted out all of our marks, we noticed that some numbers had several numbers—equivalent fractions. We discussed this.
When we finished with this line, we began to write the 2nd line. This time we divided the whole numbers into thirds on the number line and then divided the thirds into half, just like we did previously with our circles. (See this Post). Just like the previous line we started counting first by 1/3’s and then by 1/6’s and talked about the equivalent fractions on the number line. This mark was the same size, but could be expressed in different fractions or whole numbers.
## Day 7
### Review of Counting Fractions
Day 7 found us back outside. When we explored fractions with our circles, we explored first dividing them into fractions with the understanding that each fraction was part of a whole. We also went through a certain sequence of exploration. First we sorted and compared. Then we sequenced them, or counted by that fraction, i.e. 1/2, 2/2, 3/2… From here we explored equivalent fractions, and then we began our four processes first adding and subtracting, then multiplying and dividing. That sequence is the same exploration that I plan for the number line. It goes slightly faster as she is already familiar now with fractions from our circle explorations.
Today we began as we did with our first exploration of a number line on Day 3 above. We drew out a number line from 0 to 7. Then I had her half each of the numbers, and we counted as we marked each of the fractions: 1/2, 2/2, 3/2…
### Exploring the Four Processes- Addition, Subtraction, Multiplication, and Division
From here we began exploring the four processes. We just made up some expressions, using the number line she figured out what the solution was. We used the counting up method here. So for 3/2 + 7/2, she would choose 7/2 to stand on (we’ve used the counting on mental math method, so she knows to begin with the larger number,) and then count an addition 3/2 up coming to 10/2, which is easy to see from the number line that it was also 10. We did maybe 5 to 6 of these expressions.
After exploring addition we moved on to subtractions. The method is the same, but moving in the opposite direction when subtracting, and of course order matters here, and the first number is the number she started on.
After subtraction we worked on multiplication. This was done by “skip counting” the number of 1/2 we were multiplying by. For instance if you multiply 7 by 1/2, you would count 1/2 – 7 times to come up with 3 1/2 or 7/2. After doing several of these we explored division. This was accomplished by asking how many 1/2 fit inside a certain number. For 4 ÷ 1/2, she would count the number of 1/2 marks until she came to the number 4, ending up at 8. Doing this several times, she realized that the number would just be doubled.
When finished with halves, I had her 1/2 her 1/2’s and come up with 1/4 marks for the number line. We then counted – 1/4, 2/4, 3/4…- our fourths up the number line. We did a few addition and subtraction problems before I began to really loose her interest, so we stopped there for the day.
## Day 8
Moving our number line inside, we put Addition and Subtraction of fractions with like denominators into her notebook. We drew several number lines in the same way that we did outside, halving and halving again, counting the sequence for each fraction. Using our number lines, we began with 1/4’s and added a couple of expressions, beginning at the first fraction and then jumping the number of increments in whatever fraction was asked for in second number of the expression. For instance in the first expression pictured below, 1/2 + 1/2, she placed her colored pencil on the first 1/2 on the number line, and “jumped” another 1/2 to the 2/2 mark. In the third expression, 3/4 + 2/4, she began with her pencil on the 3/4 and jumped an addition 2 increments of 1/4 to the 5/4 or 1 ¼ mark. We color coded her expressions by coloring a box around the expression and using that same color to “jump” on her number line. From there we moved onto 1/4’s with a couple of addition problems. After a couple of problems with 1/4, we paused to see patterns, using this pattern to find the solution to the final expression using 1/8’s which was not marked our number line.
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### Statistics: Dave throws a ball at a bucket. The probability the ball goes into the bucket is 0.4. Dave throws the ball four times. What is the probability that he gets it in twice?
When Dave throws the ball, it can either go into the bucket, or miss. The probability of the ball going into the bucket is 0.4 and each throw is independent of eachother so this probability is the same for each throw. As the only other event that can occur is Dave missing, the probability of him missing must be
P(miss) = 1 - 0.4 = 0.6
For the described event to occur, Dave must get the ball in the bucket twice, and miss twice. To calculate the probability of multiple independent events occuring, we multiply the probabilities, commonly referred to as
P(A n B) = p(A) x p(B)
Thus,
0.4 x 0.4 x 0.6 x 0.6
This can be written in a simpler form as
0.4x 0.62
This solution gives us the probability that Dave will miss twice and then success twice, but the question does not specify an order.
The only factor left to consider is combinations. This is the mathematical way of considering how many ways an event can occur. In this event, we need 2 of the 4 throws to be misses (or successes), so we must calculate how many different ways this could occur. We can do this in different ways:
Listing
success success miss miss
success miss success miss
success miss miss success
miss success success miss
miss success miss success
miss miss success success
This method gives us the (correct) answer of 6, but it is easy to make a mistake and miss one of the combinations out.
Pascal's Triangle
You may be familiar with Pascal's triangle from the binomial theorem or many popular maths puzzles, it is a triangle of numbers such that every number is the sum of the two numbers above it, starting with one.
0 1
1 1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
5 1 5 10 10 5 1
(notice the row numbers on the left begin with row 0)
Using the triangle, we note that we are looking for the number of ways to arrange 2 events within 4. So we look at place 2 on row 4. Musch like with rows, the places within the rows, start with the 0th, so the 2nd term in row 4 is 6.
This method has also given us the correct answer of 6. This method is, again complicated, and ceases to be helpful with larger numbers, but does help in understanding why the final and most efficient method works.
Choose
Choose is a mathematical method used in combinations. Found on all scientific calculators it will often be displayed in the format
nCr
This is, in effect the more mathematical version of the Pascal's Triangle method. Again we are looking for the number of ways we can arrange 2 events within 4. So, in your calculator, you type
4C2
and the answer will be 6.
This is, in my opinion the best method as the only thing that can be confusing is the order of the numbers. This is only a small problem as the calculator will only work with the correct order as the other is impossible. And you can always remember it as 'out of 4 events we are choosing 2' so its 4 choose 2!
Now we simply multiply our probability from earlier (which gives the probability of one combination occuring) by the number of combinations.
0.4x 0.62 x 6 = 0.3456
10 months ago
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# Theoretical Probability
## Presentation on theme: "Theoretical Probability"— Presentation transcript:
Theoretical Probability
10-6 Theoretical Probability Warm Up Lesson Presentation Lesson Quiz Holt Algebra 1
Warm Up An experiment consists of spinning a spinner 8 times. The spinner lands on red 4 times, yellow 3 times, and green once. Find the experimental probability of each event. 1. The spinner lands on red. 2. The spinner does not land on green. 3. The spinner lands on yellow.
Objectives Determine the theoretical probability of an event.
Convert between probabilities and odds.
When the outcomes in the sample space of an experiment have the same chance of occurring, the outcomes are said to be equally likely.
The theoretical probability of an event is the ratio of the number of ways the event can occur to the total number of equally likely outcomes.
An experiment in which all outcomes are equally likely is said to be fair. You can usually assume that experiments involving coins and number cubes are fair.
Example 1A: Finding Theoretical Probability
An experiment consists of rolling a number cube. Find the theoretical probability of each outcome. rolling a 5 There is one 5 on a number cube.
Example 1B: Finding Theoretical Probability
An experiment consists of rolling a number cube. Find the theoretical probability of each outcome. rolling an odd number There 3 odd numbers on a cube. = 0.5 = 50%
Example 1C: Finding Theoretical Probability
An experiment consists of rolling a number cube. Find the theoretical probability of each outcome. rolling a number less than 3 There are 2 numbers less three.
The probability of an event can be written as P(event)
The probability of an event can be written as P(event). P(heads) means “the probability that heads will be the outcome.” Reading Math
When you toss a coin, there are two possible outcomes, heads or tails
When you toss a coin, there are two possible outcomes, heads or tails. The table below shows the theoretical probabilities and experimental results of tossing a coin 10 times.
The sum of the probability of heads and the probability of tails is 1, or 100%. This is because it is certain that one of the two outcomes will always occur. P(event happening) + P(event not happening) = 1
The complement of an event is all the outcomes in the sample space that are not included in the event. The sum of the probabilities of an event and its complement is 1, or 100%, because the event will either happen or not happen. P(event) + P(complement of event) = 1
Example 2: Finding Probability by Using the Complement
A box contains only red, black, and white blocks. The probability of choosing a red block is , the probability of choosing a black block is . What is the probability of choosing a white block? P(red) + P(black) + P(white) = 100% Either it will be a white block or not. 25% + 50% + P(white) = 100% 75% + P(white) = 100% –75% –75% Subtract 75% from both sides. P(white) = 25%
Odds are another way to express the likelihood of an event
Odds are another way to express the likelihood of an event. The odds in favor of an event describe the likelihood that the event will occur. The odds against an event describe the likelihood that the event will not occur. Odds are usually written with a colon in the form a:b, but can also be written as a to b or .
The two numbers given as the odds will add up to the total number of possible outcomes. You can use this relationship to convert between odds and probabilities.
You may see an outcome called “favorable
You may see an outcome called “favorable.” This does not mean that the outcome is good or bad. A favorable outcome is the outcome you are looking for in a probability experiment. Reading Math
Example 3A: Converting Between Odds and Probabilities
The probability of rolling a 2 on a number cube is . What are the odds of rolling a 2 ? The probability of rolling a 2 is . There are 5 unfavorable outcomes and 1 favorable outcome, thus the odds are 1:5. Odds in favor are 1:5.
Example 3B: Converting Between Odds and Probabilities
The odds in favor of winning a contest are 1:9. What is the probability of winning the contest? The odds in favor of winning are 1:9, so the odds against are 9:1. This means there is 1 favorable outcome and 9 unfavorable outcomes for a total of 10 possible outcomes. The probability of winning the contest is
Example 3C: Converting Between Odds and Probabilities
The odds against a spinner landing on red are 2:3. What is the probability of the spinner landing on red? The odds against landing on red are 2:3, so the odds in favor are 3:2. This means there are 3 favorable outcomes and 2 unfavorable outcomes for a total of 5 possible outcomes. The probability of landing on red is |
We have defined and used the concept of limit, primarily in our development of the derivative. Recall that $\ds \lim_{x\to a}f(x)=L$ is true if, in a precise sense, $f(x)$ gets closer and closer to $L$ as $x$ gets closer and closer to $a$. While some limits are easy to see, others take some ingenuity; in particular, the limits that define derivatives are always difficult on their face, since in $$\lim_{\Delta x\to 0} {f(x+\Delta x)-f(x)\over \Delta x}$$ both the numerator and denominator approach zero. Typically this difficulty can be resolved when $f$ is a "nice'' function and we are trying to compute a derivative. Occasionally such limits are interesting for other reasons, and the limit of a fraction in which both numerator and denominator approach zero can be difficult to analyze. Now that we have the derivative available, there is another technique that can sometimes be helpful in such circumstances.
Before we introduce the technique, we will also expand our concept of limit, in two ways. When the limit of $f(x)$ as $x$ approaches $a$ does not exist, it may be useful to note in what way it does not exist. We have already talked about one such case: one-sided limits. Another case is when "$f$ goes to infinity''. We also will occasionally want to know what happens to $f$ when $x$ "goes to infinity''.
Example 4.10.1 What happens to $1/x$ as $x$ goes to 0? From the right, $1/x$ gets bigger and bigger, or goes to infinity. From the left it goes to negative infinity. $\square$
Example 4.10.2 What happens to the function $\ds \cos(1/x)$ as $x$ goes to infinity? It seems clear that as $x$ gets larger and larger, $1/x$ gets closer and closer to zero, so $\cos(1/x)$ should be getting closer and closer to $\cos(0)=1$. $\square$
As with ordinary limits, these concepts can be made precise. Roughly, we want $\ds \lim_{x\to a}f(x)=\infty$ to mean that we can make $f(x)$ arbitrarily large by making $x$ close enough to $a$, and $\ds \lim_{x\to \infty}f(x)=L$ should mean we can make $f(x)$ as close as we want to $L$ by making $x$ large enough. Compare this definition to the definition of limit in section 2.3, definition 2.3.2.
Definition 4.10.3 If $f$ is a function, we say that $\ds \lim_{x\to a}f(x)=\infty$ if for every $N>0$ there is a $\delta>0$ such that whenever $|x-a|< \delta$, $f(x)>N$. We can extend this in the obvious ways to define $\ds \lim_{x\to a}f(x)=-\infty$, $\ds \lim_{x\to a^-}f(x)=\pm\infty$, and $\ds \lim_{x\to a^+}f(x)=\pm\infty$. $\square$
Definition 4.10.4 (Limit at infinity) If $f$ is a function, we say that $\ds \lim_{x\to \infty}f(x)=L$ if for every $\epsilon>0$ there is an $N > 0$ so that whenever $x>N$, $|f(x)-L|< \epsilon$. We may similarly define $\ds \lim_{x\to-\infty}f(x)=L$, and using the idea of the previous definition, we may define $\ds \lim_{x\to\pm\infty}f(x)=\pm\infty$. $\square$
We include these definitions for completeness, but we will not explore them in detail. Suffice it to say that such limits behave in much the same way that ordinary limits do; in particular there are some analogs of theorem 2.3.6.
Now consider this limit: $$\lim_{x\to \pi}{x^2-\pi^2\over \sin x}.$$ As $x$ approaches $\pi$, both the numerator and denominator approach zero, so it is not obvious what, if anything, the quotient approaches. We can often compute such limits by application of the following theorem.
Theorem 4.10.5 (L'Hôpital's Rule) For "sufficiently nice'' functions $f(x)$ and $g(x)$, if $\ds\lim_{x\to a} f(x)= 0 = \lim_{x\to a} g(x)$ or both $\ds\lim_{x\to a} f(x)= \pm\infty$ and $\lim_{x\to a} g(x)=\pm\infty$, and if $\ds\lim_{x\to a}{f'(x)\over g'(x)}$ exists, then $\ds\lim_{x\to a}{f(x)\over g(x)}=\lim_{x\to a}{f'(x)\over g'(x)}$. This remains true if "$x\to a$'' is replaced by "$x\to \infty$'' or "$x\to -\infty$''. $\qed$
This theorem is somewhat difficult to prove, in part because it incorporates so many different possibilities, so we will not prove it here. We also will not need to worry about the precise definition of "sufficiently nice'', as the functions we encounter will be suitable.
Example 4.10.6 Compute $\ds\lim_{x\to \pi}{x^2-\pi^2\over \sin x}$ in two ways.
First we use L'Hôpital's Rule: Since the numerator and denominator both approach zero, $$\lim_{x\to \pi}{x^2-\pi^2\over \sin x}= \lim_{x\to \pi}{2x \over \cos x},$$ provided the latter exists. But in fact this is an easy limit, since the denominator now approaches $-1$, so $$\lim_{x\to \pi}{x^2-\pi^2\over \sin x}={2\pi\over -1} = -2\pi.$$
We don't really need L'Hôpital's Rule to do this limit. Rewrite it as $$\lim_{x\to \pi}(x+\pi){x-\pi\over \sin x}$$ and note that $$\lim_{x\to \pi}{x-\pi\over \sin x}= \lim_{x\to \pi}{x-\pi\over -\sin (x-\pi)}= \lim_{x\to 0}-{x\over \sin x}$$ since $x-\pi$ approaches zero as $x$ approaches $\pi$. Now $$\lim_{x\to \pi}(x+\pi){x-\pi\over \sin x}= \lim_{x\to \pi}(x+\pi)\lim_{x\to 0}-{x\over \sin x}= 2\pi(-1)=-2\pi$$ as before. $\square$
Example 4.10.7 Compute $\ds\lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}$ in two ways.
As $x$ goes to infinity both the numerator and denominator go to infinity, so we may apply L'Hôpital's Rule: $$\lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}= \lim_{x\to \infty}{4x-3\over 2x+47}.$$ In the second quotient, it is still the case that the numerator and denominator both go to infinity, so we are allowed to use L'Hôpital's Rule again: $$\lim_{x\to \infty}{4x-3\over 2x+47}=\lim_{x\to \infty}{4\over 2}=2.$$ So the original limit is 2 as well.
Again, we don't really need L'Hôpital's Rule, and in fact a more elementary approach is easier—we divide the numerator and denominator by $\ds x^2$: $$\lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}= \lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}{{1\over x^2}\over {1\over x^2}}= \lim_{x\to \infty}{2-{3\over x}+{7\over x^2}\over 1+{47\over x}+{1\over x^2}}.$$ Now as $x$ approaches infinity, all the quotients with some power of $x$ in the denominator approach zero, leaving 2 in the numerator and 1 in the denominator, so the limit again is 2. $\square$
Example 4.10.8 Compute $\ds\lim_{x\to 0}{\sec x - 1\over \sin x}$.
Both the numerator and denominator approach zero, so applying L'Hôpital's Rule: $$\lim_{x\to 0}{\sec x - 1\over \sin x}= \lim_{x\to 0}{\sec x\tan x\over \cos x}={1\cdot 0\over 1}=0.$$
$\square$
Example 4.10.9 Compute $\ds\lim_{x\to 0^+} x\ln x$.
This doesn't appear to be suitable for L'Hôpital's Rule, but it also is not "obvious''. As $x$ approaches zero, $\ln x$ goes to $-\infty$, so the product looks like $(\hbox{something very small})\cdot (\hbox{something very large and negative})$. But this could be anything: it depends on how small and how large. For example, consider $\ds (x^2)(1/x)$, $(x)(1/x)$, and $\ds (x)(1/x^2)$. As $x$ approaches zero, each of these is $(\hbox{something very small})\cdot (\hbox{something very large})$, yet the limits are respectively zero, $1$, and $\infty$.
We can in fact turn this into a L'Hôpital's Rule problem: $$x\ln x = {\ln x\over 1/x}={\ln x\over x^{-1}}.$$ Now as $x$ approaches zero, both the numerator and denominator approach infinity (one $-\infty$ and one $+\infty$, but only the size is important). Using L'Hôpital's Rule: $$\lim_{x\to 0^+} {\ln x\over x^{-1}}= \lim_{x\to 0^+} {1/x\over -x^{-2}} =\lim_{x\to 0^+} {1\over x}(-x^2)= \lim_{x\to 0^+} -x = 0.$$ One way to interpret this is that since $\ds\lim_{x\to 0^+}x\ln x = 0$, the $x$ approaches zero much faster than the $\ln x$ approaches $-\infty$. $\square$
Exercises 4.10
Compute the limits.
Ex 4.10.1 $\ds\lim_{x\to 0} {\cos x -1\over \sin x}$ (answer)
Ex 4.10.2 $\ds\lim_{x\to \infty} {e^x\over x^3}$ (answer)
Ex 4.10.3 $\ds\lim_{x\to \infty} \sqrt{x^2+x}-\sqrt{x^2-x}$ (answer)
Ex 4.10.4 $\ds\lim_{x\to \infty} {\ln x\over x}$ (answer)
Ex 4.10.5 $\ds\lim_{x\to \infty} {\ln x\over \sqrt{x}}$ (answer)
Ex 4.10.6 $\ds\lim_{x\to\infty} {e^x + e^{-x}\over e^x -e^{-x}}$ (answer)
Ex 4.10.7 $\ds\lim_{x\to0}{\sqrt{9+x}-3\over x}$ (answer)
Ex 4.10.8 $\ds\lim_{t\to1^+}{(1/t)-1\over t^2-2t+1}$ (answer)
Ex 4.10.9 $\ds\lim_{x\to2}{2-\sqrt{x+2}\over 4-x^2}$ (answer)
Ex 4.10.10 $\ds\lim_{t\to\infty}{t+5-2/t-1/t^3\over 3t+12-1/t^2}$ (answer)
Ex 4.10.11 $\ds\lim_{y\to\infty}{\sqrt{y+1}+\sqrt{y-1}\over y}$ (answer)
Ex 4.10.12 $\ds\lim_{x\to1}{\sqrt{x}-1\over \root 3\of{x}-1}$ (answer)
Ex 4.10.13 $\ds\lim_{x\to0}{(1-x)^{1/4}-1\over x}$ (answer)
Ex 4.10.14 $\ds\lim_{t\to 0}{\left(t+{1\over t}\right)((4-t)^{3/2}-8)}$ (answer)
Ex 4.10.15 $\ds\lim_{t\to 0^+}\left({1\over t}+{1\over\sqrt{t}}\right) (\sqrt{t+1}-1)$ (answer)
Ex 4.10.16 $\ds\lim_{x\to 0}{x^2\over\sqrt{2x+1}-1}$ (answer)
Ex 4.10.17 $\ds\lim_{u\to 1}{(u-1)^3\over (1/u)-u^2+3u-3}$ (answer)
Ex 4.10.18 $\ds\lim_{x\to 0}{2+(1/x)\over 3-(2/x)}$ (answer)
Ex 4.10.19 $\ds\lim_{x\to 0^+}{1+5/\sqrt{x}\over 2+1/\sqrt{x}}$ (answer)
Ex 4.10.20 $\ds\lim_{x\to 0^+}{3+x^{-1/2}+x^{-1}\over 2+4x^{-1/2}}$ (answer)
Ex 4.10.21 $\ds\lim_{x\to\infty}{x+x^{1/2}+x^{1/3}\over x^{2/3}+x^{1/4}}$ (answer)
Ex 4.10.22 $\ds\lim_{t\to\infty} {1-\sqrt{t\over t+1}\over 2-\sqrt{4t+1\over t+2}}$ (answer)
Ex 4.10.23 $\ds\lim_{t\to\infty}{1-{t\over t-1}\over 1-\sqrt{t\over t-1}}$ (answer)
Ex 4.10.24 $\ds\lim_{x\to-\infty}{x+x^{-1}\over 1+\sqrt{1-x}}$ (answer)
Ex 4.10.25 $\ds\lim_{x\to\pi/2}{\cos x\over (\pi/2)-x}$ (answer)
Ex 4.10.26 $\ds\lim_{x\to0}{e^x-1\over x}$ (answer)
Ex 4.10.27 $\ds\lim_{x\to0}{x^2\over e^x-x-1}$ (answer)
Ex 4.10.28 $\ds\lim_{x\to1}{\ln x\over x-1}$ (answer)
Ex 4.10.29 $\ds\lim_{x\to0}{\ln(x^2+1)\over x}$ (answer)
Ex 4.10.30 $\ds\lim_{x\to1}{x\ln x\over x^2-1}$ (answer)
Ex 4.10.31 $\ds\lim_{x\to0}{\sin(2x)\over\ln(x+1)}$ (answer)
Ex 4.10.32 $\ds\lim_{x\to1}{x^{1/4}-1\over x}$ (answer)
Ex 4.10.33 $\ds\lim_{x\to1^+}{\sqrt{x}\over x-1}$ (answer)
Ex 4.10.34 $\ds\lim_{x\to1}{\sqrt{x}-1\over x-1}$ (answer)
Ex 4.10.35 $\ds\lim_{x\to\infty}{x^{-1}+x^{-1/2}\over x+x^{-1/2}}$ (answer)
Ex 4.10.36 $\ds\lim_{x\to\infty}{x+x^{-2}\over 2x+x^{-2}}$ (answer)
Ex 4.10.37 $\ds\lim_{x\to\infty}{5+x^{-1}\over 1+2x^{-1}}$ (answer)
Ex 4.10.38 $\ds\lim_{x\to\infty}{4x\over\sqrt{2x^2+1}}$ (answer)
Ex 4.10.39 $\ds\lim_{x\to0}{3x^2+x+2\over x-4}$ (answer)
Ex 4.10.40 $\ds\lim_{x\to0}{\sqrt{x+1}-1\over \sqrt{x+4}-2}$ (answer)
Ex 4.10.41 $\ds\lim_{x\to0}{\sqrt{x+1}-1\over \sqrt{x+2}-2}$ (answer)
Ex 4.10.42 $\ds\lim_{x\to0^+}{\sqrt{x+1}+1\over\sqrt{x+1}-1}$ (answer)
Ex 4.10.43 $\ds\lim_{x\to0}{\sqrt{x^2+1}-1\over\sqrt{x+1}-1}$ (answer)
Ex 4.10.44 $\ds\lim_{x\to\infty}{(x+5)\left({1\over 2x}+{1\over x+2}\right)}$ (answer)
Ex 4.10.45 $\ds\lim_{x\to0^+}{(x+5)\left({1\over 2x}+{1\over x+2}\right)}$ (answer)
Ex 4.10.46 $\ds\lim_{x\to1}{(x+5)\left({1\over 2x}+{1\over x+2}\right)}$ (answer)
Ex 4.10.47 $\ds\lim_{x\to2}{x^3-6x-2\over x^3+4}$ (answer)
Ex 4.10.48 $\ds\lim_{x\to2}{x^3-6x-2\over x^3-4x}$ (answer)
Ex 4.10.49 $\ds\lim_{x\to1+}{x^3+4x+8\over 2x^3-2}$ (answer)
Ex 4.10.50 The function $\ds f(x) = {x\over\sqrt{x^2+1}}$ has two horizontal asymptotes. Find them and give a rough sketch of $f$ with its horizontal asymptotes. (answer) |
# Trigonometry – Non-Right-Angled Triangles
Students learn how to derive the Sine, Cosine and Area formulae for non-right-angled triangles. They use this knowledge to solve complex problems involving triangular shapes.
This unit takes place in Term 5 of Year 10 and follows on from trigonometry with right-angled triangles.
##### Revision Lessons
Prerequisite Knowledge
• Know the trigonometric ratios Sinϑ = Opp/Hyp, Cosϑ = Adj/Hyp and Tanϑ = Opp/Adj.
• Apply them to find angles and lengths in right-angled triangles and, where possible, general triangles in two and three dimensional figures
Success Criteria
• Know and apply the sine rule $\frac { a }{ SinA } =\frac { b }{ SinB } =\frac { c }{ SinC }$ and cosine rule a2 = b2 + c2 – 2bcCosA to find unknown lengths and angles.
• Know and apply the formula for the area of a triangle $A=\frac { 1 }{ 2 } abSinC$ to calculate the area, sides or angles of any triangle.
Key Concepts
• The Sine rule is used when:
• Any two angles and a side is known.
• Any two sides and an angle is known
• The Cosine rule is used when:
• all three sides are known
• two sides and the adjoining angle is known
• Students should have the opportunity to derive the three formulae from first principals.
• This topic is often linked with problems involving bearings and map sketches.
Common Misconceptions
• Students often have difficulty choosing the correct formula.
• A common mistake is attempting to use Pythagoras’ Theorem to find a length in a non-right angled triangle.
• Marks are often lost when breaking down the Cosine Rule using the order of operations.
## One thought on “Trigonometry – Non-Right-Angled Triangles”
1. ##### Sahil Roysays:
<3
Thnks
^^
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### Mr Mathematics Blog
#### Angles in Polygons
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When getting ready for a new school year I have a list of priorities to work through. Knowing my team have all the information and resources they need to teach their students gives me confidence we will start the term in the best possible way. Mathematics Teaching and Learning Folder All teachers receive a folder […]
#### Mathematics OFSTED Inspection – The Deep Dive
Earlier this week, my school took part in a trial OFSTED inspection as part of getting ready for the new inspection framework in September 2019. This involved three Lead Inspectors visiting our school over the course of two days. The first day involved a ‘deep dive’ by each of the Lead Inspectors into Mathematics, English […] |
# Difference between revisions of "2001 AMC 12 Problems/Problem 22"
## Problem
In rectangle $ABCD$, points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$. Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$. The area of the rectangle $ABCD$ is $70$. Find the area of triangle $EHJ$.
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$
## Solution
$[asy] unitsize(0.5cm); defaultpen(0.8); pair A=(0,0), B=(10,0), C=(10,7), D=(0,7), E=(C+D)/2, F=(2*A+B)/3, G=(A+2*B)/3; pair H = intersectionpoint(A--C,E--F); pair J = intersectionpoint(A--C,E--G); draw(A--B--C--D--cycle); draw(G--E--F); draw(A--C); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,NW); label("E",E,N); label("F",F,S); label("G",G,S); label("H",H,SE); label("J",J,ESE); filldraw(E--H--J--cycle,lightgray,black); draw(H--D, dashed); [/asy]$
### Solution 1
Note that the triangles $AFH$ and $CEH$ are similar, as they have the same angles. Hence $\frac {AH}{HC} = \frac{AF}{EC} = \frac 23$.
Also, triangles $AGJ$ and $CEJ$ are similar, hence $\frac {AJ}{JC} = \frac {AG}{EC} = \frac 43$.
We can now compute $[EHJ]$ as $[ACD]-[AHD]-[DEH]-[EJC]$. We have:
• $[ACD]=\frac{[ABCD]}2 = 35$.
• $[AHD]$ is $2/5$ of $[ACD]$, as these two triangles have the same base $AD$, and $AH$ is $2/5$ of $AC$, therefore also the height from $H$ onto $AD$ is $2/5$ of the height from $C$. Hence $[AHD]=14$.
• $[HED]$ is $3/10$ of $[ACD]$, as the base $ED$ is $1/2$ of the base $CD$, and the height from $H$ is $3/5$ of the height from $A$. Hence $[HED]=\frac {21}2$.
• $[JEC]$ is $3/14$ of $[ACD]$ for similar reasons, hence $[JEC]=\frac{15}2$.
Therefore $[EHJ]=[ACD]-[AHD]-[DEH]-[EJC]=35-14-\frac {21}2-\frac{15}2 = \boxed{3}$.
### Solution 2
As in the previous solution, we note the similar triangles and prove that $H$ is in $2/5$ and $J$ in $4/7$ of $AC$.
We can then compute that $HJ = AC \cdot \left( \frac 47 - \frac 25 \right) = AC \cdot \frac{6}{35}$.
As $E$ is the midpoint of $CD$, the height from $E$ onto $AC$ is $1/2$ of the height from $D$ onto $AC$. Therefore we have $[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}$.
### Solution 3
Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). Because this is a general figure, we can assume the sides are $7$ and $10$. We can find $H$ and $J$ by intersecting lines, and then we calculate the area of $EHJ$ using shoelace formula. This yields $\boxed{3}$. |
Autohelmvt
2022-01-24
Prove that:
$\sum _{n=1}^{\mathrm{\infty }}\frac{{n}^{2}\left(n-1\right)}{{2}^{n}}=20$
Troy Sutton
Expert
Note that $\sum _{n}{t}^{n}=\frac{1}{1-t}$ for every $|t|<1$ hence, differentiaiting this twice and three times,
$\sum _{n}n\left(n-1\right){t}^{n-2}=\frac{2}{{\left(1-t\right)}^{3}}$,
$\sum _{n}n\left(n-1\right)\left(n-2\right){t}^{n-3}=\frac{6}{{\left(1-t\right)}^{4}}$
For $t=\frac{1}{2}$, this reads
$\sum _{n}n\left(n-1\right)\frac{1}{{2}^{n-2}}=2\cdot {2}^{3}$
$\sum _{n}n\left(n-1\right)\left(n-2\right)\frac{1}{{2}^{n-3}}=6\cdot {2}^{4}$
which implies
$\sum _{n}n\left(n-1\right)\frac{1}{{2}^{n}}=\frac{1}{4}\cdot 2\cdot {2}^{3}=4$
$\sum _{n}n\left(n-1\right)\left(n-2\right)\frac{1}{{2}^{n}}=\frac{1}{8}\cdot 6\cdot {2}^{4}=12$
Finally,
${n}^{2}\left(n-1\right)=2\cdot n\left(n-1\right)+n\left(n-1\right)\left(n-2\right)$
hence
$\sum _{n}{n}^{2}\left(n-1\right)\frac{1}{{2}^{n}}=2\cdot \sum _{n}n\left(n-1\right)\frac{1}{{2}^{n}}+\sum _{n}n\left(n-1\right)\left(n-2\right)\frac{1}{{2}^{n}}=2\cdot 4+12$
This approach can be made shorter if one notices once and for all that, for every nonnegative k,
$\sum _{n}n\left(n-1\right)\dots \left(n-k\right)\frac{1}{{2}^{n}}=2\cdot \left(k+1\right)!$
egowaffle26ic
Expert
For $|x|<1$ we have that $\sum _{n=0}^{\mathrm{\infty }}{x}^{n}=\frac{1}{1-x}$ hence differentiating both sides we get
$\sum _{n=0}^{\mathrm{\infty }}\left(n+1\right){x}^{n}=\sum _{n=1}^{\mathrm{\infty }}n{x}^{n-1}=\frac{1}{{\left(1-x\right)}^{2}}$
while differentiating once more
$\sum _{n=0}^{\mathrm{\infty }}\left(n+2\right)\left(n+1\right){x}^{n}=\frac{2}{{\left(1-x\right)}^{3}}$
and once again
$\sum _{n=0}^{\mathrm{\infty }}\left(n+3\right)\left(n+2\right)\left(n+1\right){x}^{n}=\frac{6}{{\left(1-x\right)}^{4}}$
Next we express ${n}^{2}\left(n-1\right)$ as a linear combination of $\left(n+3\right)\left(n+2\right)\left(n+1\right),\left(n+2\right)\left(n+1\right),\left(n+1\right)$ and 1:
${n}^{2}\left(n-1\right)=\left(n+3\right)\left(n+2\right)\left(n+1\right)-7\left(n+2\right)\left(n+1\right)+10\left(n+1\right)-2$
and hence
$\sum _{n=0}^{\mathrm{\infty }}{n}^{2}\left(2n-1\right){x}^{n}=\frac{6}{{\left(1-x\right)}^{4}}-\frac{14}{{\left(1-x\right)}^{3}}+\frac{10}{{\left(1-x\right)}^{2}-\frac{2}{1-x}}$
and setting $x=\frac{1}{2}$ we obtain that
$\sum _{n=0}^{\mathrm{\infty }}\frac{{n}^{2}\left(n-1\right)}{{2}^{n}}=6\cdot 16-4\cdot 8+10\cdot 4-4=20$
Do you have a similar question? |
### Home > CC2 > Chapter 8 > Lesson 8.3.4 > Problem8-107
8-107.
Evaluate each expression.
1. $\frac { 2 ( 5 + 3 ) } { 4 }$
Following the Order of Operations, first simplify the terms in parentheses, then the ones being multiplied or divided.
Lastly, combine any terms that are being added or subtracted.
$4$
1. $\frac{1}{2}(15+3)-10\div2$
See part (a).
1. $5\frac{1}{2}-2\frac{1}{4}+\frac{3}{8}$
You may find this expression easier to simplify by combining the whole numbers and the fractions separately.
$3\frac{5}{8}$
1. $3+\frac{3}{5}\cdot\frac{1}{4}$
See part (a).
1. $-2+(-5+6)^2$
First, simplify the terms in parentheses, then the terms with exponents, and lastly combing and numbers that are added or subtracted.
$-1$
1. $\frac{3}{4}\cdot\frac{1}{4}+\frac{5}{8}\cdot\left(-\frac{3}{2}\right)$
See part (a).
$-\frac{3}{4}$ |
# Which of the following is a composite number? 61 , 13 , 63 , 31
Answer: It's 63, all the others are prime numbers.
## Related Questions
10) Carlos wants to mix granola and cranberries together to make trail mix. Granola costs \$3 per pound and cranberries cost \$5.50 per pound. Carlos is willing to spend \$29 and wants to make 8 pounds of trail mix. Which system of equations could Carlos use to figure out how many pounds of granola (g) and cranberries (c) she should buy?
6 pounds of granola, 2 pounds of cranberries.
Step-by-step explanation:
3g + 5.5c = 29
g + c = 8
g = 8 - c
3 * (8 - c) + 5.5c = 29
24 - 3c + 5.5c = 29
2.5c = 5
c = 2
g = 8 - c
g = 8 - 2
g = 6
Describe how to transform the graph of g(x)= ln x into the graph of f(x)= ln (3-x) -2.
The graph of g(x) = ㏑x translated 3 units to the right and then reflected
about the y-axis and then translated 2 units down to form the graph of
f(x) = ㏑(3 - x) - 2
Step-by-step explanation:
* Lets talk about the transformation
- If the function f(x) reflected across the x-axis, then the new
function g(x) = - f(x)
- If the function f(x) reflected across the y-axis, then the new
function g(x) = f(-x)
- If the function f(x) translated horizontally to the right
by h units, then the new function g(x) = f(x - h)
- If the function f(x) translated horizontally to the left
by h units, then the new function g(x) = f(x + h)
- If the function f(x) translated vertically up
by k units, then the new function g(x) = f(x) + k
- If the function f(x) translated vertically down
by k units, then the new function g(x) = f(x) – k
* lets solve the problem
∵ Graph of g(x) = ㏑x is transformed into graph of f(x) = ㏑(3 - x) - 2
- ㏑x becomes ㏑(3 - x)
∵ ㏑(3 - x) = ㏑(-x + 3)
- Take (-) as a common factor
∴ ㏑(-x + 3) = ㏑[-(x - 3)]
∵ x changed to x - 3
∴ The function g(x) translated 3 units to the right
∵ There is (-) out the bracket (x - 3) that means we change the sign
of x then we will reflect the function about the y-axis
∴ g(x) translated 3 units to the right and then reflected about the
y-axis
∵ g(x) changed to f(x) = ㏑(3 - x) - 2
∵ We subtract 2 from g(x) after horizontal translation and reflection
∴ We translate g(x) 2 units down
∴ g(x) translated 3 units to the right and then reflected about the
y-axis and then translated 2 units down
* The graph of g(x) = ㏑x translated 3 units to the right and then
reflected about the y-axis and then translated 2 units down to
form the graph of f(x) = ㏑(3 - x) - 2
c edge
Step-by-step explanation:
Solve the equation 4x squared =121
A 2.75,-2.75
B 5.5,-5.5
C 11,-11
D 30.25,-30.25
Step-by-step explanation:
4x2 =121
4x2-121=0
using the general formula
[-b( +/-) √b2-4ac]/2a
a=4, b= 0, c=-121
x= [0 ( +/-)√02-(4*4*-121)]/2.4
= ( +/-)√1936)/8 = ( +/-)5.5
x =5.5
x = -5.5
The primary goal of using correlational research is to |
# NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.1
NCERT Solutions for Class 7 Maths Exercise 8.1 Chapter 8 Comparing Quantities in simple PDF are given here. This exercise of NCERT Solutions for Class 7 Maths Chapter 8 contains topics related to the equivalent ratios. Different ratios can also be compared with each other to know whether they are equivalent or not. To do this, we need to write the ratios in the form of fractions and then compare them by converting them to like fractions. Our expert teachers have formulated these solutions in a precise and a detailed form. Learn more about these topics by solving the questions of NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities with the help of solutions provided here.
## Download the PDF of NCERT Solutions For Class 7 Maths Chapter 8 Comparing Quantities – Exercise 8.1
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### Access Other Exercises of NCERT Solutions For Class 7 Maths Chapter 8 – Comparing Quantities
Exercise 8.2 Solutions
Exercise 8.3 Solutions
### Access Answers to NCERT Class 7 Maths Chapter 8 – Comparing Quantities Exercise 8.1
1. Find the ratio of:
(a) ₹ 5 to 50 paise
Solution:-
We know that,
₹ 1 = 100 paise
Then,
₹ 5 = 5 × 100 = 500 paise
Now we have to find the ratio,
= 500/50
= 10/1
So, the required ratio is 10: 1.
(b) 15 kg to 210 g
Solution:-
We know that,
1 kg = 1000 g
Then,
15 kg = 15 × 1000 = 15000 g
Now we have to find the ratio,
= 15000/210
= 1500/21
= 500/7 … [∵divide both by 3]
So, the required ratio is 500: 7.
(c) 9 m to 27 cm
Solution:-
We know that,
1 m = 100 cm
Then,
9 m = 9 × 100 = 900 cm
Now we have to find the ratio,
= 900/27
= 100/3 … [∵divide both by 9]
So, the required ratio is 100: 3.
(d) 30 days to 36 hours
Solution:-
We know that,
1 day = 24 hours
Then,
30 days = 30 × 24 = 720 hours
Now we have to find the ratio,
= 720/36
= 20/1 … [∵divide both by 36]
So, the required ratio is 20: 1.
2. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
Solution:-
From the question it is given that,
Number of computer required for 6 students = 3
So, number of computer required for 1 student = (3/6)
= ½
So, number of computer required for 24 students = 24 × ½
= 24/2
= 12
∴Number of computer required for 24 students is 12 computers.
3. Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs.
Area of Rajasthan = 3 lakh km2 and area of UP = 2 lakh km2.
(i) How many people are there per km2 in both these States?
(ii) Which State is less populated?
Solution:-
(i) From the question, it is given that,
Population of Rajasthan = 570 lakh
Area of Rajasthan = 3 lakh Km2
Then, population of Rajasthan in 1 km2 area = (570 lakh)/ (3 lakh km2)
= 190 people per km2
Population of UP = 1660 Lakh
Area of UP = 2 Lakh km2
Then, population of UP in 1 lakh km2 area = (1660 lakh)/ (2 lakh km2)
= 830 people per km2
(ii) By comparing the two states Rajasthan is the less populated state.
1. khushi chaudhary
HI , I GOT A VERY GOOD HELP FROM BYJU’S , I WAS VERY CONFUSED BUT NOT NOW BECAUSE IT WAS WRITTEN VERY CLEAR AND EXPLAINED VERY GOOD MANNER .
THANK U : >
• Khyati nigam
That’s very nice
2. Earlier I used to be very confused but now I like to read from Byjus Thanks Byjus
3. It helped me a lot I was so confused thanks byjus |
## watson3 2 years ago find the coordinates of the foci of the ellipse represented by 4x2 + 9y2 – 18y – 27 = 0
• This Question is Open
1. Loser66
ok, sorry, backward, perfect square first, then isolate the number
2. Loser66
hey friend, I help only, I don't do it for you.
3. Azteck
$\large 4x^2+9y^2-18y-27=0$ Complete the square with the y-terms. In order to do that, we must turn the "9y^2" into y^2. DIVIDE both sides by 9. $\large \frac{4}{9}x^2+y^2-2y-3=0$ Move the 3 to the otherside and let's get rid of the (4/9)x^2 for a moment. We will add that back into the equation afterwards. So we have this: $\huge y^2-2y=3$ Complete the square. $\large (y-1)^2=3+1$ $\large (y-1)^2=4$ Now bring back the (4/9)x^2. SO now we have the equation: $\frac{4}{9}x^2+(y-1)^2=4$ Now we have to divide both sides by 4. We will have: $\huge \frac{x^2}{9}+\frac{(y-1)^2}{4}=1$ The equation of the foci when a>b is: $\huge (\pm ae, 0)$ In this ellipse, the value of a=3 and b=2 To find "e", we use the equation: $\huge b^2=a^2(1-e^2)$ $\large 4=9-9e^2$ $\large e^2=\frac{5}{9}$ $\large e=\frac{\sqrt{5}}{3}$ Therefore the coordinates of the foci is: $\huge (\sqrt{5}, 0)$
4. Azteck
$\huge (\pm \sqrt{5}, 0)$ Sorry about that. This website is a bit too overwhelming for my computer to handle. Forgot the plus or minus sign. |
# What Is The Reciprocal Of A Fraction?
## What is the reciprocal of 7 9 as a fraction?
The reciprocal would be 16 .
You just pretty much change the number to a fraction, the number is the denominator and 1 is the numerator.
But if you want to find the reciprocal of a fraction, then you just switch the numerator and the denominator around.
So the reciprocal of 79 is 97 !.
## What is a reciprocal of 2 9?
Number × Reciprocal of number = 1. And, Multiplicative Inverse of a/b = b/a. So, Multiplicative inverse of 2/9 by 1/2 or 4/9 = 9/4.
## What is a reciprocal of 5?
1/5The reciprocal of 5 is 1/5. Every number has a reciprocal except for 0. There is nothing you can multiply by 0 to create a product of 1, so it has no reciprocal. Reciprocals are used when dividing fractions.
## What is a reciprocal of a number?
The reciprocal of a number is 1 divided by the number. The reciprocal of a number is also called its multiplicative inverse. The product of a number and its reciprocal is 1. … The reciprocal of a fraction is found by flipping its numerator and denominator.
## What is a reciprocal of 13?
Likewise the reciprocal of 13 is 3 and so on.
## What is a reciprocal of 19?
The reciprocal of 19 is 1/19 such that their product is 1. They are mutually reciprocals of each other. The reciprocal of -19 is -1/19, so that their product is 1.
## What is the reciprocal of 5 by 8?
To find the reciprocal, divide 1 by the number given. Simplify. Multiply the numerator by the reciprocal of the denominator. Multiply 85 8 5 by 1 1 .
## What is a reciprocal of 4?
1/4Answer and Explanation: The reciprocal of 4 is 1/4. By definition, the reciprocal of a number a/b is b/a. Therefore, in order to find the reciprocal of a fraction, a/b, we…
## What is a reciprocal of 18?
Explanation: A fraction multiplied by its reciprocal will equal 1. To find the reciprocal of a fraction, switch the denominator and numerator. The reciprocal of 18/27 is 27/18.
## What is the reciprocal of 5 6?
The reciprocal of 6/5 is 5/6.
## What is a reciprocal of 14?
1/14hence reciprocal of 14 is 1/14..
## What is the reciprocal of 1 2 as a fraction?
Answer and Explanation: The opposite reciprocal of 1/2 is -2/1, which can also be written as -2. To find this, first, determine the reciprocal by flipping the fraction 1/…
## What is a reciprocal of 16?
As you can expect, 16 times the reciprocal of 16 will equal 1. Here you can calculate the reciprocal of another number.
## What is a reciprocal of 12?
1/12Answer and Explanation: The reciprocal of 12 is 1/12. To find the reciprocal of a number, we use the following two steps: Write the number as a fraction.
## What is a reciprocal of 1 8?
Answer and Explanation: The reciprocal of 1/8 is found by flipping the fraction: 1/8 flipped is 8/1 or 8.
## What is a reciprocal of 7?
Answer and Explanation: The reciprocal of 7 is 1/7. In general, the reciprocal of a fraction simply interchanges the numerator and denominator of the fraction.
## What is a reciprocal of 10?
The opposite of 10 is -10. The reciprocal of 10 is 110 or 0.1. |
## Reverse Equivalence
#### Problem
By adding the different 2-digit numbers 12 and 32 we get 44. If the digits in each number are reversed we get two different 2-digit numbers, and 21 + 23 also equals 44.
The same is true of 42 + 35 = 24 + 53 = 77.
Prove that the sum of two 2-digit numbers with this property will always be divisible by 11.
#### Solution
Let the 2-digit numbers be (ab) and (cd).
We shall ignore cases like 11 + 11 = 11 + 11 and 12 + 21 = 21 + 12, as proving that (ab) + (ba) = 10a + b + 10b + a = 11a + 11b is divisibly by 11 is trivial.
The problem requires the sum, S = (ab) + (cd) = (ba) + (dc), such that (ab) (cd) (ba)
S = (10a + b) + (10c + d) = (10b + a) + (10d + c).
Therefore 9a + 9c = 9b + 9d and so a + c = b + d.
Hence we obtain a pair of numbers with the required property if the sum of the first digits of each number is equal to the sum of the second digits. For example, 41 + 36 = 14 + 63 = 77, because 4 + 3 = 1 + 6.
Now S = (10a + b) + (10c + d) = 10(a + c) + b + d
As a + c = b + d, S = 10(a + c) + a + c = 11(a + c).
Hence the sum, S, is always divisible by 11.
What adding three 2-digit numbers?
Can you find two 3-digit numbers with this property?
Problem ID: 126 (Oct 2003) Difficulty: 3 Star
Only Show Problem |
# 1 is what percent of 52.6?
## Please also explain how do do this.
Mar 18, 2017
1.9%
#### Explanation:
$\frac{x}{100} \times 52.6 = 1$
$52.6 x = 100$
$x = \frac{100}{52.6}$
x= 1.9%
ANSWER: 1.9% Rounded to 1 decimal place
Mar 15, 2018
Note that $\approx$ means 'approximately equal to'.
1/52.6~~1.90% rounded to 2 decimal place - approximate value
1/52.6=100/52.6 % exact value
#### Explanation:
Explaining percentage see my solution: https://socratic.org/s/aPdYxYWZ
The value of $\textcolor{g r e e n}{1}$ as a proportion of $\textcolor{red}{52.6}$ is color(white)("dd")color(green)(1)/(color(red)(52.6)
However, if we wish to express this as a percentage we need to change the bottom number (denominator) into 100.
We can change the denominator by doing this: $\textcolor{red}{52.6} \times \frac{100}{52.6} = 100$
However, for multiply or divide, what we do to the bottom of a fraction we have to do to the top.
$\frac{\textcolor{g r e e n}{1} \times \frac{100}{52.6}}{\textcolor{red}{52.6} \times \frac{100}{52.6}} \textcolor{w h i t e}{\text{d") =color(white)("d") (color(green)(1)xx100/52.6)/(color(red)(cancel(52.6))xx100/cancel(52.6)) color(white)("d")= color(white)("d}} \frac{1.901140 \ldots .}{100}$
$= 1.901140 \ldots . \underbrace{\times \frac{1}{100}}$
$\textcolor{w h i t e}{\text{dddddddddddddd}} \downarrow$
=1.901140... color(white)("ddd")% see https://socratic.org/s/aPdYxYWZ
1/52.6~~1.90% rounded to 2 decimal place - approximate value
May 20, 2018
1.9%
#### Explanation:
To find out what percent one quantity is of of another quantity, use the method:
"fraction" xx 100%
We want to know what $\frac{1}{52.6}$ is as a percent.
1/52.6 xx 100%
=1.9%
If you write the fraction as a decimal we have:
$\frac{1}{52.6} = 0.019$
Multiplying by 100% is the same as multiplying by $\frac{100}{100}$, which is the same as multiplying by $1$
It simply changes the decimal form to a percentage.
0019 xx 100%=1.9% |
# Section 3 1 Lines and Angles Perpendicular Lines
• Slides: 23
Section 3. 1 Lines and Angles
Perpendicular Lines • Intersecting lines that form right angles • Symbol XS SR
Parallel Lines • Two lines that are coplanar and do not intersect • Symbol: II XY II UZ
Skew Lines • Lines do not intersect and are not coplanar
Example • Is XY parallel or skew to RV? XY II RV
Parallel planes • Two planes that do not intersect
Parallel Postulate • If there is a line and a point not on the line, then there is exactly one line through the point parallel to the given line.
Perpendicular Postulate • If there is a line and a point not on the line, then there is exactly one line through the point perpendicular to the given line.
Theorem 3. 1 • If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular m<ABD = m<DBC and a • Ex 1 D A B linear pair, BD C AC
Theorem 3. 2 • If two sides of two adjacent acute angles are perpendicular, then the angles are complementary. • Ex. 2 F J <FGJ is complementary to <JGH G H
Examples: Solve for x Ex 3. 60° x ANSWER: 60 + x = 90 -60 x = 30
Example 4 ANSWER: x + 55 = 90 -55 x 55° -55 x = 35
Example 5 ANSWER: 2 x – 9 + 27 = 90 2 x +18 = 90 27° (2 x-9)° 2 x = 72 x = 36
Theorem 3. 3 • If 2 lines are perpendicular, then they intersect to form four right angles. l m
Complete Try it! Problems #1 -8
Transversal • A line that intersects two or more coplanar lines at different points. transversal
Vertical Angles • Formed by the intersection of two pairs of opposite rays 1 3 7 4 6 5 8 2
Linear Pair • Adjacent angles that are supplementary 1 3 7 4 6 5 8 2
Corresponding Angles • Occupy corresponding positions. 1 3 7 4 6 5 8 2
Alternate Exterior Angles • Lie outside the 2 lines on opposite sides of the transversal. 1 3 7 4 6 5 8 2
Alternate Interior Angles • Lie between the 2 lines on opposite sides of the transversal. 1 3 7 4 6 5 8 2
Consecutive Interior Angles (Same side interior angles) • Lie between the 2 lines on the same side of the transversal. 1 3 7 4 6 5 8 2
Angle Relationships: Name a pair of angles • Corresponding 1 – Ex. 1 & 5 3 • Alternate Exterior – Ex. 2 & 7 • Alternate Interior – Ex. 4 & 5 • Consecutive Interior – Ex. 3 & 5 5 7 6 8 2 4 |
# A triangle has sides with lengths: 4, 5, and 7. How do you find the area of the triangle using Heron's formula?
Feb 17, 2016
4sqrt6 ≈ 9.8 " square units "
#### Explanation:
This is a 2 step process.
step 1 : Calculate half of the perimeter ( s ) of the triangle.
step 2 : Calculate the area (A)
let a = 4 , b = 5 and c = 7
step 1 : s = $\frac{a + b + c}{2} = \frac{4 + 5 + 7}{2} = \frac{16}{2} = 8$
step 2 : $A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$
$= \sqrt{8 \left(8 - 4\right) \left(8 - 5\right) \left(8 - 7\right)} = \sqrt{8 \times 4 \times 3 \times 1} = \sqrt{96} = 4 \sqrt{6}$
Feb 17, 2016
$A r e a = 4 \sqrt{6.} u n i t s$
#### Explanation:
$A = A r e a$
$a - b - c = s i \mathrm{de} s$
$s = \frac{a + b + c}{2}$
Heron's formula for the area of the triangle:
color(blue)(A=sqrt(s(s-a)(s-b)(s-c))
In this case color(green)(a=4,b=5,c=7,s=(4+5+7)/2=16/2=8
$\rightarrow A = \sqrt{8 \left(8 - 4\right) \left(8 - 5\right) \left(8 - 7\right)}$
$\rightarrow A = \sqrt{8 \left(4\right) \left(3\right) \left(1\right)}$
$\rightarrow A = \sqrt{8 \left(12\right)}$
rArrcolor(orange)(A=sqrt96=sqrt(16*6)=4sqrt6 |
# How do you simplify ((x^-3 y^-4 )/ (x^-5 y^-7 )) ^-3 ?
Mar 3, 2018
See a solution process below:
#### Explanation:
First, use this rule of exponents to divide the $x$ and $y$ terms within the parenthesis:
${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$
${\left(\frac{{x}^{\textcolor{red}{- 3}} {y}^{\textcolor{red}{- 4}}}{{x}^{\textcolor{b l u e}{- 5}} {y}^{\textcolor{b l u e}{- 7}}}\right)}^{-} 3 \implies$
${\left({x}^{\textcolor{red}{- 3} - \textcolor{b l u e}{- 5}} {y}^{\textcolor{red}{- 4} - \textcolor{b l u e}{- 7}}\right)}^{-} 3 \implies$
${\left({x}^{\textcolor{red}{- 3} + \textcolor{b l u e}{5}} {y}^{\textcolor{red}{- 4} + \textcolor{b l u e}{7}}\right)}^{-} 3 \implies$
${\left({x}^{\textcolor{red}{2}} {y}^{\textcolor{red}{3}}\right)}^{\textcolor{b l u e}{- 3}}$
Next, use this rule for exponents to eliminate the outer exponent:
${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$
${x}^{\textcolor{red}{2} \times \textcolor{b l u e}{- 3}} {y}^{\textcolor{red}{3} \times \textcolor{b l u e}{- 3}} \implies$
${x}^{\textcolor{red}{- 6}} {y}^{\textcolor{red}{- 9}}$
Now, use this rule of exponents to eliminate the negative exponents:
${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$
$\frac{1}{{x}^{\textcolor{red}{- - 6}} {y}^{\textcolor{red}{- - 9}}} \implies$
$\frac{1}{{x}^{\textcolor{red}{6}} {y}^{\textcolor{red}{9}}}$ |
☰ Related
Linear Inequalities
In this section we learn how to solve linear inequalities. These are inequalities involving an unknown $$x$$, looking like either of the following: $ax \leq b, \quad ax+b > d$ or $ax+b \geq cx, \quad ax+b < cx +d$ Those are all typical linear inequalities, that we'll know how to solve by the end of this section.
Method - Solving Inequalites
To solve a linear equality looking like $$ax \leq b$$ or even $$ax +b > cx + d$$, we follow the same steps as we would for solving the linear equations $$ax = b$$ and $$ax +b = cx + d$$ with one very important difference:
When we multiply, or divide, both sides of an inequality by a negative number: the inequality symbol is reversed.
In other words, when we multiply, or divide, both sides of an inequality by a negative number:
$$>$$ becomes $$<$$
$$\geq$$ becomes $$\leq$$
$$<$$ becomes $$>$$
$$\leq$$ becomes $$\geq$$
This is further explained and some examples are worked through in Tutorial 1. If you haven't done so already, watch it now.
One-Step Linear Inequalities
Now that we've learnt the rules for solving linear inequalities, we can put our knowledge into practice.
We start by working our way through a few one-step inequalities. If you're having any trouble solving these, make sure to watch Tutorial 2 and to carefully read through the "Answers With Working" for each question.
Exercise 1
Solve each of the following inequalities and illustrate your answer on the number line:
1. $$2x \leq 6$$
2. $$-3x < 9$$
3. $$\frac{x}{2} \geq 2$$
4. $$x-2>0$$
5. $$\frac{x}{-4} \leq -1$$
1. $$x\leq 3$$
2. $$x> -3$$
3. $$x \geq 4$$
4. $$x>2$$
5. $$x \geq 4$$
1. We solve $$2x \leq 6$$ as follows: $2x \leq 6$ Divide both sides by $$2$$: $\frac{2x}{2} \leq \frac{6}{2}$ $x \leq 3$ This result is illustrated on the number line as follows:
Notice that the dot above the $$3$$ is filled-in to indicate that the value $$3$$ is included in the solution.
2. We solve $$-3x < 9$$ as follows: $-3x < 9$ Divide both sides by $$-3$$ and reverse the inequality symbol: $\frac{-3x}{-3} > \frac{9}{-3}$ $x > -3$ This result is illustrated on the number line as follows:
Notice that the dot above the $$3$$ is empty to indicate that the value $$-3$$ is not included in the solution.
3. We solve $$\frac{x}{2} \geq 2$$ as follows: $\frac{x}{2} \geq 2$ Multiply both sides by $$2$$ to get rid of the $$2$$ that is dividing $$x$$: $2\times \frac{x}{2} \geq 2\times 2$ $x \geq 4$ This result is illustrated on the number line as follows:
Notice that the dot above the $$4$$ is filled-in to indicate that the value $$4$$ is included in the solution.
4. We solve $$x-2>0$$ as follows: $x-2 > 0$ Add $$2$$ to both sides of the inequality to get rid of the $$2$$ that is being subtracted from $$x$$: $x-2+2 > 0 + 2$ $x > 2$ This result is illustrated on the number line as follows:
Notice that the dot above the $$2$$ is empty to indicate that $$2$$ isn't included in the solution.
5. We solve $$\frac{x}{-4} \leq -1$$ as follows: $\frac{x}{-4} \leq -1$ Multiply both sides by $$-4$$ to get rid of the $$-4$$ that is dividing $$x$$ and reverse the inequality symbol: $-4 \times \frac{x}{-4} \geq -4 \times (-1)$ $x \geq 4$ This result is illustrated on the number line as follows:
Notice that the dot above the $$4$$ is filled-in to indicate that $$4$$ is included in the solution.
Two-Step Linear Inequalities
Now that we've seen how to solve simple linear inequalities, that can be solved in one step, we move-on to solving inequalities that require two steps. If you're having any trouble solving these, make sure to watch Tutorial 2 and to carefully read through the "Answers With Working" for each question.
Exercise 2
Solve each of the following inequalities annd illustrate your answers on the number line:
1. $$4x+1 > 13$$
2. $$2 - 5x \geq 22$$
3. $$3 < 17 - 7x$$
4. $$12 + 6x > 24$$
5. $$1 - \frac{x}{2} \leq 4$$
1. $$x > 3$$
2. $$x \leq -4$$
3. $$x < 2$$
4. $$x > 2$$
5. $$x \geq -6$$
1. We solve $$4x + 1 > 13$$ as follows: $4x + 1 > 13$ Subtract $$1$$ from both sides, to get rid of the $$1$$ that's being added to $$4x$$: $4x + 1 -1 > 13 - 1$ $4x > 12$ Divide both sides by $$4$$, to get rid of the $$4$$ that is multiplying the $$x$$: $\frac{4x}{4} > \frac{12}{4}$ $x > 3$ This result is illustrated here:
Notice that the dot above the $$3$$ is empty to indicate that $$3$$ is not included in the solution.
2. We solve $$2 - 5x \geq 22$$ as follows: $2 - 5x \geq 22$ Subtract $$2$$ from both sides, to get rid of the positive $$2$$ on the left hand side: $2 - 5x -2\geq 22-2$ $- 5x \geq 20$ Divide both sides by $$-5$$, and reverse the inequality symbol, to get rid of the $$-5$$ that is multiplying the $$x$$: $\frac{-5x}{-5} \leq \frac{20}{-5}$ $x \leq -4$ This result is illustrated here:
Notice that the dot above $$-4$$ is filled-in to indicate that $$-4$$ is included in the solution.
3. We solve $$3< 17 - 7x$$ as follows: $3 < 17 - 7x$ Subtract $$17$$ from both sides to get rid of the the positive $$17$$ on the right hand side: $3 -17 < 17 - 7x - 17$ $-14 < -7x$ Divide noth sides by $$-7$$, and reverse the inequality symbol, to get rid of the $$-7$$ multiplying the $$x$$: $\frac{-14}{-7} > \frac{-7x}{-7}$ $2 > x$ This result "$$x$$ is less than $$2$$" can be re-written as: $x < 2$ We illustrate this as follows:
4. We solve $$12 + 6x > 24$$ as follows: $12 + 6x > 24$ Subtract $$12$$ from both sides, to get rid of the $$12$$ that's being added to $$6x$$: $12 + 6x -12 > 24 -12$ $6x > 12$ Divide both sides by $$6$$ to get rid of the $$6$$ that's multiplying the $$x$$: $\frac{6x}{6} > \frac{12}{6}$ $x > 2$ This result is illustrated here:
Notice that the dot above the $$2$$ is empty to indicate that $$2$$ isn't included in the solution.
5. We solve $$1 - \frac{x}{2} \leq 4$$ as follows: $1 - \frac{x}{2} \leq 4$ Subtract $$1$$ from both sides to get rid of the positive $$1$$ on the left hand side: $1 - \frac{x}{2} -1 \leq 4 - 1$ $-\frac{x}{2} \leq 3$ Multiply both sides by $$-2$$, and change the inequality symbol, to get rid of the $$-2$$ dividing $$x$$ (make sure to read $$\text{Note}^{(1)}$$, further down, to fully undertand this step): $-2\times \begin{pmatrix} -\frac{x}{2} \end{pmatrix} \geq -2\times 3$ $x \geq -6$ This result is illustrated here:
Notice that the dot above $$-6$$ is filled-in to indicate that $$-6$$ is included in the solution.
$$\text{Note}^{(1)}$$: whether we write $$-\frac{x}{2}$$, $$\frac{-x}{2}$$, or $$\frac{x}{-2}$$ these are all equivalent notations and can be interchanged when convenient to do so.
Three-Step Linear Inequalities
We now move-on to linear inequalities that require three steps to be solved.
These usually have an $$x$$ term on each side of the inequality. If you're having any trouble solving these, make sure you have watched Tutorial 3 and to carefully read through the "answers with working".
Exercise 3
Solve each of the following inequalities and illustarate your answer on the number line:
1. $$3x+6 < x$$
2. $$2x + 3 \geq 3x$$
3. $$\frac{2x}{3} \leq x -2$$
4. $$4x - 5 > 2x -9$$
5. $$20 + 3x \geq 4 + 7x$$
1. $$x < -3$$
2. $$x \leq 3$$
3. $$x \geq 6$$
4. $$x > -2$$
5. $$x \leq 4$$
1. We solve $$3x + 6 < x$$ as follows: $3x + 6 < x$ Subtract $$6$$ from both sides to get rid of the $$6$$ being added to $$3x$$: $3x + 6 - 6 < x - 6$ $3x < x - 6$ Subtract $$x$$ from both sides to gather all the $$x$$ terms on the left hand side: $3x - x < x - 6 - x$ $2x < -6$ Divide both sides by $$2$$ to get rid of the $$2$$ multiplying the $$x$$: $\frac{2x}{2} < \frac{-6}{2}$ $x < -3$ This result is illustrated here:
Notice that that dot above the $$-3$$ is empty to indicate that $$-3$$ is not included in the solution.
2. We solve $$2x + 3 \geq 3x$$ as follows: $2x + 3 \geq 3x$ Subtract $$3$$ from both sides to get rid of the $$3$$ being added to $$2x$$: $2x + 3 -3 \geq 3x - 3$ $2x \geq 3x -3$ Subtract $$3x$$ from both sides to gather all the $$x$$ terms on the left hand side: $2x -3x\geq 3x -3 -3x$ $-x \geq -3$ Multiply both sides by $$-1$$, and reverse the inequality symbol, to get rid of the negative on the $$x$$: $-1\times (-x) \leq -1 \times (-3)$ $x \leq 3$ This result is illustrated here:
Notice that the dot above the $$3$$ is filled-in to indicate that $$3$$ is included in the solution.
3. We solve $$\frac{2x}{3} \leq x - 2$$ as follows: $\frac{2x}{3} \leq x - 2$ Multiply both sides by $$3$$ to get rid of the $$3$$ that's dividing the $$x$$ on the left hand side: $3\times \frac{2x}{3} \leq 3\begin{pmatrix} x - 2 \end{pmatrix}$ $2x \leq 3\times x - 3\times 2$ $2x \leq 3x - 3$ Subtract $$3x$$ from both sides to gather all the $$x$$ terms on the left hand side: $2x - 3x \leq 3x - 6 - 3x$ $-x \leq -3$ Multiply both sides by $$-1$$, and reverse the inequality symbol, to get rid of the negative on the $$x$$: $-1\times (-x) \geq -1 \times (-6)$ $x \geq 6$ This result is illustrated here:
Notice that the dot, above the $$6$$, is filled-in to indicate that $$6$$ is included in the solution.
4. We solve $$4x - 5 > 2x - 9$$ as follows: $4x - 5 > 2x - 9$ Add $$5$$ to both sides to get rid of the $$5$$ that is being subtracted from $$4x$$: $4x - 5 + 5 > 2x - 9 + 5$ $4x > 2x - 4$ Subtract $$2x$$ from both sides to gather all the $$x$$ terms on the left hand side: $4x -2x > 2x - 4 - 2x$ $2x > -4$ Divide both sides by $$2$$ to get rid of the $$2$$ that's multiplying $$x$$: $\frac{2x}{2} > \frac{-4}{2}$ $x > -2$ This result is illustrated here:
Notice that the dot above the $$-2$$ is empty to indicate that $$-2$$ is not included in the solution.
5. We solve $$20 + 3x \geq 4 + 7x$$ as follows: $20 + 3x \geq 4 + 7x$ Subtract $$20$$ from both sides to get rid of the $$20$$ that's being added to $$3x$$ on the left hand side: $20 + 3x - 20 \geq 4 + 7x - 20$ $3x \geq 7x - 16$ Subtract $$7x$$ from both sides to gather all the $$x$$ terms on the left hand side: $3x -7x \geq 7x - 16 -7x$ $-4x \geq -16$ Divide both sides by $$-4$$, and reverse the inequality symbol, to get rid of the $$-4$$ multiplying the $$x$$: $\frac{-4x}{-4} \leq \frac{-16}{-4}$ $x \leq 4$ This result is illustrated here:
Notice that the dot above the $$4$$ is filled-in to indicate that $$4$$ is included in the solution.
Four-Step, or more, Linear Inequalities
We now move-on to more complicated linear inequalities that require four, or more, steps to be solved.
These usually involve fractions. If you're having any trouble solving these, make sure you have watched Tutorial 3 and to carefully read through the "answers with working".
Exercise 4
Solve each of the following inequalities and illustrate your answer on the number line:
1. $$\frac{x}{2} + 1 \leq 2x + 4$$
2. $$5x - 1 > 2x +5$$
3. $$\frac{2x}{3} - 2 \geq 2x + 2$$
4. $$-2 - \frac{x}{2} > \frac{x}{3} + 8$$
5. $$\frac{3x}{2} - 1 \geq \frac{x}{3} + 6$$
1. $$x \geq -2$$
2. $$x > 2$$
3. $$x \leq -3$$
4. $$x < -12$$
5. $$x \geq 6$$
1. We solve $$\frac{x}{2} + 1 \leq 2x +4$$ as follows: $\frac{x}{2} + 1 \leq 2x +4$ Multiply both sides by $$2$$ to get rid of the $$2$$ that is dividing $$x$$: $2\begin{pmatrix} \frac{x}{2} + 1 \end{pmatrix} \leq 2\begin{pmatrix} 2x +4 \end{pmatrix}$ $2\times \frac{x}{2} + 2\times 1 \leq 2\times 2x + 2\times 4$ $x + 2 \leq 4x + 8$ Subtract $$2$$ from both sides to get rid pf the $$2$$ that is being added to $$x$$ on the left hand side: $x + 2 -2 \leq 4x + 8 -2$ $x \leq 4x + 6$ Subtract $$4x$$ from both sides to gather all the $$x$$ terms on the left hand side: $x -4x \leq 4x + 6 -4x$ $-3x \leq 6$ Divide both sides by $$-3$$, and reverse the inequality symbol, to get rid of the $$-3$$ that's multiplying the $$x$$: $\frac{-3x}{-3} \geq \frac{6}{-3}$ $x \geq -2$ This result is illustrated here:
Notice that the dot above $$-2$$ is filled-in to indicate that $$-2$$ is included in the solution.
2. We solve $$5x - 1 > 2x + 5$$ as follows: $5x - 1 > 2x + 5$ Add $$1$$ to both sides to get rid of the $$1$$ that is being subtracted from $$5x$$: $5x - 1 + 1 > 2x + 5 + 1$ $5x > 2x + 6$ Subtract $$2x$$ from both sides to gather all the $$x$$ terms on the left hand side: $5x - 2x > 2x + 6 - 2x$ $3x > 6$ Divide both sides by $$3$$ to get rid of the $$3$$ multiplying the $$x$$ on the left hand side: $\frac{3x}{3} > \frac{6}{3}$ $x > 2$ This result is illustrated here:
Notice that the dot above $$2$$ is empty to indicate that $$2$$ is not included in the solution.
3. We solve $$\frac{2x}{3} - 2 \geq 2x + 2$$ as follows: $\frac{2x}{3} - 2 \geq 2x + 2$ Multiply both sides by $$3$$ to get rid of the $$3$$ that is dividing $$x$$ on the left hand side: $3 \begin{pmatrix} \frac{2x}{3} - 2 \end{pmatrix} \geq 3 \begin{pmatrix} 2x + 2 \end{pmatrix}$ $3\times \frac{2x}{3} - 3 \times 2 \geq 3 \times 2x + 3 \times 2$ $2x - 6 \geq 6x + 6$ Add $$6$$ to both sides to get rid of the $$6$$ that's being subtracted from $$2x$$: $2x - 6 + 6 \geq 6x + 6 + 6$ $2x \geq 6x + 12$ Subtract $$6x$$ from both sides to gather all the $$x$$ terms on the left hand side: $2x - 6x \geq 6x + 12 - 6x$ $-4x \geq 12$ Divide both sides by $$-4$$, and reverse the inequality symbol, to get rid of the $$-4$$ that's multiplying $$x$$: $\frac{-4x}{-4} \leq \frac{12}{-4}$ $x \leq -3$ This result is illustrated here:
Notice that the dot above $$-3$$ is filled-in to indicate that $$-3$$ is included in the solution.
4. We solve $$-2 - \frac{x}{2} > \frac{x}{3} + 8$$ as follows: $-2 - \frac{x}{2} > \frac{x}{3} + 8$ Multiply both sides by $$2$$ to get rid of the $$2$$ that is dividing $$x$$ on the left hand side: $2\begin{pmatrix} -2 - \frac{x}{2} \end{pmatrix} > 2 \begin{pmatrix} \frac{x}{3} + 8 \end{pmatrix}$ $2\times (-2) - 2 \times \frac{x}{2} > 2 \times \frac{x}{3} + 2 \times 8$ $-4 - x > \frac{2x}{3} + 16$ Multiply both sides by $$3$$ to get rid of the $$3$$ that is diving $$x$$ on the right hand side: $3\begin{pmatrix} -4 - x \end{pmatrix} > 3\begin{pmatrix} \frac{2x}{3} + 16 \end{pmatrix}$ $3\times (-4) - 3 \times x > 3 \times \frac{2x}{3} + 3\times 16$ $-12 -3x > 2x + 48$ Add $$12$$ to both sides to get rid of the $$-12$$ on the left hand side: $-12 - 3x + 12 > 2x + 48 + 12$ $-3x > 2x + 60$ Subtract $$2x$$ from both sides to gather all the $$x$$ terms on the left hand side: $-3x - 2x > 2x + 60 - 2x$ $-5x > 60$ Divide both sides by $$-5$$, and reverse the inequality symbol, to get rid of the $$-5$$ that's multiplying the $$x$$: $\frac{-5x}{-5} < \frac{60}{-5}$ $x < -12$ This result is illustrated here:
Notice that the dot above $$-12$$ is empty to indicate that $$-12$$ is not included in the solution.
5. We solve $$\frac{3x}{2} - 1 \geq \frac{x}{3} + 6$$ as follows: $\frac{3x}{2} - 1 \geq \frac{x}{3} + 6$ Multiply both sides by $$2$$ to get rid of the $$2$$ dividing the $$x$$ on the left hand side: $2\begin{pmatrix} \frac{3x}{2} - 1 \end{pmatrix} \geq 2 \begin{pmatrix} \frac{x}{3} + 6 \end{pmatrix}$ $2 \times \frac{3x}{2} -2 \times 1 \geq 2 \times \frac{x}{3} + 2 \times 6$ $3x - 2 \geq \frac{2x}{3} + 12$ Multiply both sides by $$3$$ to get rid of the $$3$$ that is dividing the $$x$$ on the right hand side: $3\begin{pmatrix} 3x - 2 \end{pmatrix} \geq 3 \begin{pmatrix} \frac{2x}{3} + 12 \end{pmatrix}$ $3 \times 3x - 3 \times 2 \geq 3 \times \frac{2x}{3} + 3 \times 12$ $9x - 6 \geq 2x + 36$ Add $$6$$ to both sides to get rid of the $$6$$ that is being subtracted from $$9x$$ on the left hand side: $9x - 6 + 6 \geq 2x + 36 + 6$ $9x \geq 2x + 42$ Subtract $$2x$$ from both sides to gather all of the $$x$$ terms on the left hand side: $9x - 2x \geq 2x + 42 - 2x$ $7x \geq 42$ Divide both sides by $$7$$ to get rid of the $$7$$ multiplying the $$x$$: $\frac{7x}{7} \geq \frac{42}{7}$ $x \geq 6$ This result is illustrated here:
Notice that the dot above $$6$$ is filled-in to indicate that $$6$$ is included in the solution. |
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# Pre-Calculus
- Term 3 Assignment –
Ms. Han
shan@daltonschool.kr
HS Room 214
INTEREST PACKET
Polar Coordinates and Parametric Equations
Students will study a class of functions called exponential functions. Exponential functions are used in modeling many
real-world phenomena, such as the growth of a population or the growth of an investment that earns compound interest.
Once an exponential model is obtained, we can use the model to predict population size or calculate the amount of an
investment for any future date. Students will also study the trigonometric functions using the unit circle. The
trigonometric functions can be defined in two different but equivalent ways: as functions of real numbers or as functions
of angles. Students will study both approaches to solve different applications that require different approaches.
Topic Overview
Trigonometric Identities Polar Coordinates
Addition and Subtraction Formulas Graphs of Polar Equations
Double-Angle, Half-Angle, and Product-Sum Polar Form of Complex Numbers; De Moivre’s
Formulas Theorem
Basic Trigonometric Equations Plane Curves and Parametric Equations
More Trigonometric Equations
Essential Questions
Is it possible for two different angles to have the same reference angle?
Where do the conversion equations come from?
What is the difference between a function and a parametric curve?
Reference Materials
J. Stewart, L. Redlin and S. Watson, Precalculus: Mathematics for Calculus, Sixth Edition. (2012: Cengage Learning,
Belmont, CA)
Skills Mastery Check
Trigonometric Functions
Understand radian measure of an angle as the length of the arc on the unit circle subtended by the angle.
(10.TF.1)
Use special triangles to determine geometrically the values of sine, cosine, tangent for p/3, p/4 and p/6,
and use the unit circle to express the values of sine, cosines, and tangent for x, p + x, and 2p – x in
terms of their values for x, where x is any real number. (10.TF.3)
Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the
solutions using technology, and interpret them in terms of the context. (10.TF.7)
Prove the Pythagorean identity sin2(x) + cos2(x) = 1 and use it to find sin(x), cos(x), or tan(x) given
sin(x), cos(x), or tan(x) and the quadrant of the angle. (10.TF.8)
Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve
problems. (10.TF.9)
Polar Coordinates and Parametric Equations
Define polar coordinates and learn how polar coordinates are related to rectangular coordinates.
(10.PCPE.1)
Understand the graph of a polar equation consists of all points P that have at least one polar
representation whose coordinates satisfy the equation. (10.PCPE.2)
Represent complex numbers in polar (or trigonometric) form. (10.PCPE.3)
LESSON OVERVIEW
Learning Objectives
Day Topics Homework
Students will be able to …
1
Checkpoint I
2
7.1 Trigonometric To study the fundamental trigonometric
3
Identities identities. (10.TF.8)
7.2 Addition and To study addition formulas for sine and cosine.
4
Subtraction Formulas (10.TF.9)
7.3 Double-Angle, Half-
To study double-angle formulas for sine and
5 Angle, and Product-Sum
cosine. (10.TF.9)
Formulas
7.4 Basic Trigonometric To solve linear and quadratic trigonometric
6
Equations equations. (10.TF.8)
7.5 More Trigonometric To solve trigonometric equations by using
7
Equations identities to simplify the equation.
8 Chapter 7 Review
9 Chapter 7 Quiz
To define polar coordinates and learn how polar
10 8.1 Polar Coordinates coordinates are related to rectangular
coordinates. (10.PCPE.1)
8.2 Graphs of Polar
11 To graph polar equations. (10.PCPE.2)
Equations
12 Checkpoint II
13 Checkpoint III
8.3 Polar Form of
To represent complex numbers in polar (or
14 Complex Numbers; De
trigonometric) form. (10.PCPE.3)
Moivre’s Theorem
15 Checkpoint IV
8.4 Plane Curves and To define parametric equations and to sketch
16
Parametric Equations parametric curves.
17 Chapter 8 Review
18 Chapter 8 Quiz
19 Project Day
20 Term 3 Project Due
*The schedule above is tentative. Please check the classroom board and Google classroom for the updates.
Expansion Pack: Please speak to me if you are interested in additional exercises.
ASSESSMENT
Chapter Quiz 30%
Term Project 25%
Homework Assignment 15%
Class Assignment 15%
Participation 15%
Quizzes: The purpose of chapter quizzes is to keep all students on top of their learning throughout the term, and
also to provide guideline to study for the final exam. Work needs to be shown for each problem. Incorrect answers
but well-written work with simple calculation mistakes will earn partial credit. Cheating on a quiz or plagiarizing
in any way, shape, or form is not acceptable and will result in an “F” and possible disciplinary action.
Homework: Students will be assigned homework every class and they are to have completed it by the following
class unless specific instructions are given. Students should take an average one hour to complete their homework.
Homework will be mostly graded by completion-base. Complete work earns full point, incomplete work earns half
point, and no or little work earns zero point. Late homework will be not accepted. Students who are absent on the
due date of an assignment, they must turn their homework in on the next following class day to be accepted for
credit. Homework must be done independently in which no show of work will be considered incorrect.
Class Assignment: Class assignments will be given to students throughout the term. Each class assignment worth
10 points and this will be graded based on the corrections. Class assignments may include problem sets,
worksheets or group activities. Most class assignments will be done in class. If a student is absent when class
assignment is given, he/she has to make up when he/she returns to school.
Participation: A typical class will be combination of whole class instruction led by the teacher and group work
led by students. Students are expected to pay undivided attention to teacher during whole class instruction, and
active involvement with group mates during group work time. Failure to meet these expectations will result
penalty in participation points.
Term Project:
CHECKPOINTS DUE POINTS
RESEARCH Checkpoint I: In last two chapters, we have been working on Trigonometry – the Day 3 10 PTS
PAPER unit circle approach and the right triangle approach. The big question is why are
(20%) we learning Trigonometry? How are they used in a real life situation? This is your
chance to find it out. We will spend next two classes researching jobs that use
trigonometry. You will find a profession that uses trigonometry and write a short
research paper on that job and describe how trigonometry is used. Math teacher
cannot be your choice!
Your research paper should include:
- Salary (entry level, average, top, how to move up the scale, etc.)
- Amount of school required to go into that profession
- Other academic areas necessary for this profession
- How trigonometry is used on the job? (Is it a major part of the job or
just used from time to time?)
Don’t forget to cite your sources!
RECTANGULR Checkpoint II: Now we have learned a different way of locating points in the Day 13 2.5 PTS
& POLAR plane, called polar coordinates, and their curves. Find an image or sketch of your
GRAPHS own choice.
(65%)
Your image should include:
- One circle (either sine or cosine)
- Two different types of limacons (inner loop, cardioid, or dimpled
limacon)
- Two roses (one odd-number leaves, one even-number leaves)
List the types of graphs will be used and how they will be used in your project.
Your image and graphs, approximately drawn by hands, should be separately
submitted with clear labels. Show restricted domains with dotted lines.
Checkpoint III: Graph five polar graphs of your choice using Desmos. Different Day 15 25 PTS
colors should be used for different graphs. Your goal is to graph these equations
by hands using both rectangular and polar coordinates.
For each page, you need to include both forms of polar and rectangular
equations and graphs drawn by hands. Each rectangular graph should include all
critical points with its coordinates and each polar graph should include arrows
showing directions. Both rectangular and polar graphs should include numbers
accordingly. Restricted domains will be shown in dotted lines.
Checkpoint IV: Conversions of rectangular forms of each graph is needed. Step- Day 16 5 PTS
by-step work must be shown to get full credit!
PRODUCT Cover page includes a creative title, full name, and class section with a Demos Day 20 7.5 PTS
(15%) print using five different matching colors.
Disk is neatly cut along the dotted line and laminated. It clearly shows five
different matching color graphs. Disk includes a title, full name, and class section
on the back side.
Neatness, creativity and inventiveness.
TOTAL POINTS 50 PTS
*Late work will be severely penalized.
Materials
Students will need a textbook, three notebooks, pencils (no work in pen will be accepted), erasers, highlighters,
color pens, and a graphing calculator in every class.
Important Dates
Chapter 7 Quiz on Day 9 (Thursday, January 25th)
Chapter 7 Quiz Review (Lab)
- Green on Monday, January 29th
- Blue on Tuesday, January 30th
- Black on Wednesday, January 31st
Chapter 8 Quiz on Day 18 (Thursday, February 22nd)
Term 3 Project Due on Day 20
PROGRESS MONITOR
Use below table to keep track of your progress throughout the school year.
Accomplishment Table
Assessment Title Weight Due Date Points Earned Percentage Earned
/ Points Possible
Chapter Quiz 20% /
Final Exam (Cumulative) 35% /
Homework Assignment 15% N/A /
Class Assignment 15% N/A /
Participation 15% N/A /
Overall
** You are expected to bring your term assignments and utilize them in every class. |
Chapter 7 Congruence of Triangles
Question 1.
In the given figure, name
(a) the side opposite to vertex A
(b) the vertex opposite A to side AB
(c) the angle opposite to side AC
(d) the angle made by the sides CB and CA.
Solution:
(a) The side opposite to vertex A is BC.
(b) The vertex opposite to side AB is C.
(c) The angle opposite to side AB is ∠ACB.
(d) The angle made by the sides CB and CA is ∠ACB.
Question 2.
Examine whether the given triangles are congruent or not.
Solution:
Here,
AB = DE = 3 cm
BC = DF = 3.5 cm
AC = EF = 4.5 cm
ΔABC = ΔEDF (By SSS rule)
So, ΔABC and ΔEDF are congruent.
Question 3.
In the given congruent triangles under ASA, find the value of x and y, ΔPQR = ΔSTU.
Solution:
Given: ΔPQR = ΔSTU (By ASA rule)
∠Q = ∠T = 60° (given)
∠x = 30° (for ASA rule)
Now in ΔSTU,
∠S + ∠T + ∠U = 180° (Angle sum property)
∠y + 60° + ∠x = 180°
∠y + 60° + 30° = 180°
∠y + 90° = 180°
∠y = 180° – 90° = 90°
Hence, x = 30° and y = 90°.
Question 4.
In the following figure, show that ΔPSQ = ΔPSR.
Solution:
In ΔPSQ and ΔPSR
∠PSQ = ∠PSR = 90° (Given)
ΔPSQ = ΔPSR (By RHS rule)
Question 5.
Can two equilateral triangles always be congruent? Give reasons.
Solution:
No, any two equilateral triangles are not always congruent.
Reason: Each angle of an equilateral triangle is 60° but their corresponding sides cannot always be the same.
Question 6.
In the given figure, AP = BQ, PR = QS. Show that ΔAPS = ΔBQR
Solution:
In ΔAPS and ΔBQR
AP = BQ (Given)
PR = QS (Given)
PR + RS = QS + RS (Adding RS to both sides)
PS = QR
∠APS = ∠BQR = 90° (Given)
ΔAPS = ΔBQR (by SAS rule)
Question 7.
Without drawing the figures of the triangles, write all six pairs of equal measures in each of the following pairs of congruent triangles.
(ii) ΔXYZ = ΔMLN
Solution:
(i) Given: ΔABC = ΔDEF
Here AB = DE
BC = EF
AC = DF
∠A = ∠D, ∠B = ∠E and ∠C = ∠F
(ii) Given ΔXYZ = ΔMLN
Here XY = ML
YZ = LN
XZ = MN
∠X = ∠M, ∠Y = ∠L and ∠Z = ∠N
Question 8.
Lengths of two sides of an isosceles triangle are 5 cm and 8 cm, find the perimeter of the triangle.
Solution:
Since the lengths of any two sides of an isosceles triangle are equal, then
Case I: The three sides of the triangle are 5 cm, 5 cm and 8 cm.
Perimeter of the triangle = 5 cm + 5 cm + 8 cm = 18 cm
Case II: The three sides of the triangle are 5 cm, 8 cm and 8 cm.
Perimeter of the triangle = 5 cm + 8 cm + 8 cm = 21 cm
Hence, the required perimeter is 18 cm or 21 cm.
Question 9.
Write the rule of congruence in the following pairs of congruent triangles.
Solution:
(i) Here, AB = ST = 3 cm
BC = TU = 4.5 cm
∠ABC = ∠STU = 110°
ΔABC = ΔSTU (By SAS rule)
(ii) Here ∠PQR = ∠MNL = 90°
hypt. PR = hypt. ML
QR = NL = 3 cm
ΔPQR = ΔMNL (By RHS rule)
Question 10.
In the given figure, state the rule of congruence followed by congruent triangles LMN and ONM.
Solution:
In ΔLMN and ΔONM
LM = ON
LN = OM
MN = NM
ΔLMN = ΔONM
Question 11.
In the given figure, PQR is a triangle in which PQ = PR. QM and RN are the medians of the triangle. Prove that
(i) ΔNQR = ΔMRQ
(ii) QM = RN
(iii) ΔPMQ = ΔPNR
Question 12.
In the given figure, PQ = CB, PA = CR, ∠P = ∠C. Is ΔQPR = ΔBCA? If yes, state the criterion of congruence.
Solution:
Given:
PQ = CB, PA = CR
and ∠P = ∠C
In ΔQPR and ΔBCA,
PQ = CB (Given)
∠QPR = ∠BCA (Given)
PA = CR (Given)
PA + AR = CR + AR (Adding AR to both sides)
or PR = CA
ΔQPR = ΔBCA (By SAS rule)
Question 13.
In the given figure, state whether ΔABC = ΔEOD or not. If yes, state the criterion of congruence.
Solution:
In ΔABC and ΔEOD
AB = OE
∠ABC = ∠EOD = 90°
AC = ED
ΔABC = ΔEOD
Hence, ΔABC = ΔEOD
RHS is the criterion of congruence.
Question 14.
In the given figure, PQ || RS and PQ = RS. Prove that ΔPUQ = ΔSUR.
Solution:
In ΔPUQ and ΔSUR
PQ = SR = 4 cm
∠UPQ = ∠USR (Alternate interior angles)
∠PQU = ∠SRU (Alternate interior angles)
ΔPUQ = ΔSUR (By ASA rule)
Question 15.
In the given figure ΔBAC = ΔQRP by SAS criterion of congruence. Find the value of x and y.
Solution:
Given: ΔBAC = ΔQRP (By SAS rule)
So, BA = QR
⇒ 3x + 10 = 5y + 15 ……(i)
∠BAC = ∠QRP
⇒ 2x + 15° = 5x – 60° ……(ii)
From eq. (ii), we have
2x + 15 = 5x – 60
⇒ 2x – 5x = -15 – 60
⇒ -3x = -7 5
⇒ x = 25
From eq. (i), we have
3x + 10 = 5y + 15
⇒ 3 × 25 + 10 = 5y + 15
⇒ 75 + 10 = 5y + 15
⇒ 85 = 5y + 15
⇒ 85 – 15 = 5y
⇒ 70 = 5y
⇒ y = 14
Hence, the required values of x andy are 25 and 14 respectively.
Question 16.
Observe the figure and state the three pairs of equal parts in triangles ABC and DCB.
(i) Is ΔABC = ΔDCB? Why?
(ii) Is AB = DC? Why?
(iii) Is AC = DB? Why? (NCERT Exemplar)
Solution:
(i) In ΔABC and ΔDCB
∠ABC = ∠DCB = 70° (40° + 30° = 70°) (Given)
∠ACB = ∠DCB = 30° (Given)
BC = CB (Common)
ΔABC = ΔDCB (By ASA rule)
(ii) Yes,
AB = DC (Congruent parts of congruent triangles)
(iii) Yes,
AC = DB (Congruent parts of congruent triangles)
Question 17.
In the given figure, ΔQPS = ΔSRQ. Find each value.
(a) x
(b) ∠PQS
(c) ∠PSR
Solution:
(a) ΔQPS = ΔSRQ
∠QPS = ∠SRQ (Congruent part of congruent triangles)
106 = 2x + 12
⇒ 106 – 12 = 2x
⇒ 94 = 2x
⇒ x = 47
∠QRS = 2 × 47 + 12 = 94 + 12 = 106°
So, PQRS is a parallelogram.
∠QSR = 180° – (42° + 106°) = 180° – 148° = 32°
(b) ∠PQS = 32° (alternate interior angles)
(c) ∠PSQ = 180° – (∠QPS + ∠PQS) = 180° – (106° + 32°) = 180° – 138° = 42°
∠PSR = 32° + 42° = 74°
### Higher Order Thinking Skills (HOTS) Type
Question 19.
Prove that the lengths of altitudes drawn to equal sides of an isosceles triangle are also equal.
(i) ∠TRQ = ∠SQR?
(ii) If ∠TRQ = 30°, find the base angles of the ΔPQR.
(iii) Is ΔPQR an equilateral triangle?
Solution:
In ΔQTR and ΔRSQ
∠QTR = ∠RSQ = 90° (Given)
∠TQR = ∠SRQ (Base angle of an isosceles triangle)
∠QRT = ∠RQS (Remaining third angles)
QR = QR (Common)
ΔQTR = ΔRSQ (By ASA rule)
QS = RT (Congruent parts of congruent triangles)
Hence proved.
(i) ∠TRQ = ∠SQR (Congruent parts of congruent triangles)
(ii) In ΔQTR,
∠TRQ = 30° (Given)
∠QTR + ∠TQR + ∠QRT = 180° (Angle sum property)
⇒ 90° + ∠TQR + 30° = 180°
⇒ 120° + ∠TQR = 180°
⇒ ∠TQR = 180° – 120° = 60°
⇒ ∠TQR = ∠SRQ = 60°
Each base angle = 60°
(iii) In ΔPQR,
∠P + ∠Q + ∠R = 180° (Angle sum property)
⇒ ∠P + 60° + 60° = 180° (From ii)
⇒ ∠P + 120° = 180°
⇒ ∠P = 180° – 120° = 60°
Hence, ΔPQR is an equilateral triangle. |
# Lesson 15.5: Independent and Dependent Events
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1 Lesson 15.5: Independent and Dependent Events Sep 26 10:07 PM 1
2 Work with a partner. You have three marbles in a bag. There are two green marbles and one purple marble. Randomly draw a marble from the bag. Then put the marble back in the bag and draw a second marble. a. Complete the tree diagram. Let G = green and P = purple. Find the probability that both marbles are green. GG GG GP GG GP PG PG PP b. Does the probability of getting a green marble on the second draw depend on the color of the first marble? Explain. No! The marble is put back, so they don't impact each other. Activity 1 2
3 Work with a partner. Using the same marbles from Activity 1, randomly draw two marbles from the bag. a. Complete the tree diagram. Let G = green and P = purple. Find the probability that both marbles are green. GG GP GG PG PG b. Does the probability of getting a green marble on the second draw depend on the color of the first marble? Explain. Yes! If green is first, then there is a 50% chance that green will also be second. If purple is first, there is a 100% chance that green will be second. Activity 2 3
4 What is the difference between dependent and independent events? Dependent events affect each other. Independent events do not. Essential Question 4
5 You have a deck of playing cards (4 suits, 13 cards in each suit). Give an example of dependent events and independent events using the deck of cards. Dependent: Picking 2 face cards in a row, without replacement. Independent: Picking a heart, replacing it, and then picking another heart. Closure 5
6 Tell whether the events are independent or dependent. Explain. 1. You roll a number cube twice. The first roll is a 3 and the second roll is an odd number. Independent the outcome of the first roll does not affect the outcome of the second. 2. You flip a coin twice. The first flip is heads and the second flip is tails. Independent the outcome of the first flip does not affect the outcome of the second. 3. You randomly draw a marble from a bag containing 3 red marbles and 5 blue marbles. You keep the marble and then draw a second marble. Dependent Because the first marble is not replaced, the number of possible outcomes changes with the second draw. 4. You randomly draw a marble from a bag containing 6 red marbles and 2 blue marbles. You put the marble back and then draw a second marble. Independent The marble picked first is replaced. Warm up 6
7 (Just like 15.4) Key Idea 7
8 You spin the spinner and flip the coin. What is the probability of spinning a prime number and flipping tails? (2, 3, and 5 are prime) Example 1 8
9 1. What is the probability of spinning a multiple of 2 and flipping heads? On your own 1 9
10 Look at as a totally different event Key Idea 10
11 People are randomly chosen to be game show contestants from an audience of 100 people. You are with 5 of your relatives and 6 other friends. What is the probability that one of your relatives is chosen first, and then one of your friends is chosen second? P (relative first) = P (friend second) = Because one person has already been chosen, there are only 99 people left from which to choose. P (relative first) X P (friend second) = Example 2 11
12 2. What is the probability that you, your relatives, and your friends are not chosen to be either of the first two contestants? When the first contestant is chosen, there are 88 favorable outcomes (a person outside of your group is chosen), and 100 possible outcomes. After the second contestant is chosen, there is one fewer favorable outcome, as well as one fewer possible outcome. On Your Own 2 12
13 A student randomly guesses the answer for each of the multiplechoice questions. What is the probability of answering all three questions correctly? There is only one favorable outcome (for each of the 3 questions), because there is only one correct answer. Example 3 13
14 3. The student can eliminate Choice A for all three questions. What is the probability of answering all three questions correctly? Compare this probability with the probability in Example 3. What do you notice? The probability almost doubles. On your own 3 14
15 Exit Ticket: You and your friend are among 5 volunteers to help distribute workbooks. What is the probability that your teacher randomly selects you and your friend to distribute the workbooks? Favorable outcomes: either you or your friend are chosen Possible outcomes: 5 volunteers from which to choose Favorable outcomes: you or your friend (whoever wasn't chosen first) Possible outcomes: 4 volunteers left from which to choose Closure 15
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