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# Question Video: Studying the Collision of Two Spheres Where One Is Moving and the Other Is Resting on the Same Line Mathematics
A sphere of mass 299 g was moving horizontally in a straight line at 51 cm/s. It collided with another sphere of mass 390 g that was at rest. As a result of the impact, the first sphere came to rest. Determine the speed of the second sphere after the impact.
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### Video Transcript
A sphere of mass 299 grams was moving horizontally in a straight line at 51 centimeters per second. It collided with another sphere of mass 390 grams that was at rest. As a result of the impact, the first sphere came to rest. Determine the speed of the second sphere after the impact.
We can call the mass of the first sphere thatβs mentioned, 299 grams, π sub one. And the speed of π sub one before it reaches the second ball, 51 centimeters per second, weβll call π£ sub one. The second sphere, the one thatβs initially at rest, has a mass of 390 grams, which weβll name π sub two. Knowing that the first mass comes to a rest after the collision, we want to calculate the speed of the second sphere after the impact. Weβll call that speed π£ sub two.
Letβs start off by sketching out this collision. In this situation, before the collision, mass one is moving at speed π£ one to encounter mass two, which has a speed of zero. Itβs stationary. Then after the spheres collide, mass one is stationary, not in motion at all, and mass two is moving at a speed π£ sub two. Itβs that speed that we want to solve for. And weβll solve for π£ sub two by recalling that momentum is conserved.
Mathematically, we can write this out as saying that the initial momentum, π sub π, of a system is equal to its final momentum, π sub π. Written another way, we can say that the initial mass of a system multiplied by its initial speed is equal to its final mass times its final speed.
In our given system, we have two masses: π sub one and π sub two. So the initial momentum in the system is equal to π one π£ one plus π two times its initial speed. Since π sub two is initially at rest, the second term is zero. And the initial momentum of our system reduces to π one times π£ one.
When we consider our system after the collision, we see that now π one is not in motion and π two is at a speed π£ sub two. So the term π one times zero is equal to zero. And our final momentum of the system is π two times π£ two.
And by the principle of the conservation of momentum, the initial momentum of our system is equal to the final momentum. That is, π one π£ one is equal to π two π£ two. Since π£ two is what we want to solve for, we can rearrange this equation. And we see that π£ sub two is equal to the ratio of mass one to mass two multiplied by the speed of mass one initially, π£ one.
When we plug in the given values for these three terms and calculate π£ sub two, we find itβs 39.1 centimeters per second. Thatβs the speed of the second ball after the collision. |
# Detecting Infinite Solutions: Unveiling the Mystery Behind Systems of Equations
How to Know if a System of Equations has Infinite Solutions
In the realm of Mathematics education, understanding the nature of solutions in a system of equations is essential. This article aims to shed light on the concept of infinite solutions in a system of equations and equip readers with the tools to identify them. By delving into the principles of linear algebra and exploring key indicators such as coefficients and consistent equations, we will unravel the mystery behind whether a system has infinitely many solutions. Join us as we unlock the secrets behind this fascinating aspect of Mathematics!
## What is a system of equations?
A system of equations is a set of two or more equations that are related to each other and share common variables. It represents a way to describe relationships between different quantities or unknowns in mathematical problems.
Explanation: In Mathematics education, it is essential to understand the concept of a system of equations before determining if it has infinite solutions. A system of equations can be represented using variables and mathematical symbols, such as constants, coefficients, and operators. By solving the system, we can find the values of the variables that satisfy all the equations simultaneously.
## Determining if a system has infinite solutions
To know if a system of equations has infinite solutions, we can use various methods such as graphing, substitution, or elimination. The following criteria help identify infinite solutions:
a) Parallel lines: If the graphs of two linear equations are parallel, they will never intersect, indicating an infinite number of solutions.
b) Identical equations: If all equations in the system are equivalent or identical, there are infinitely many solutions since any value of the variables will satisfy all the equations.
c) Dependent equations: If all equations in the system are dependent, meaning one equation can be derived from another through algebraic manipulation, the system has infinite solutions.
d) Consistent but underdetermined system: In some cases, a system may have fewer equations than variables, leading to infinitely many solutions.
## Solving systems with infinite solutions
When a system of equations has infinite solutions, it means there are multiple sets of values that satisfy all the equations simultaneously. To solve such systems, we typically express the solution in terms of one or more free variables.
Example: For a system with two free variables, we can choose arbitrary values for any one of the variables and then express the other variables in terms of that chosen variable.
Note: It is important to emphasize that infinite solutions do not imply that any arbitrary value can be assigned to the variables. The solutions still need to satisfy all the equations in the system.
## Practical applications of systems with infinite solutions
Systems of equations with infinite solutions have real-world applications in various fields, including engineering, physics, and economics.
a) Overdetermined systems: When dealing with more equations than unknowns, overdetermined systems may have infinitely many solutions due to measurement errors or inconsistent data.
b) Linear dependence: In linear algebra, studying systems with infinitely many solutions helps to understand concepts like linear independence, vector spaces, and rank of matrices.
c) Optimization problems: Systems with infinite solutions can arise in optimization problems where there is more than one solution that optimizes the objective function.
By understanding how to identify and solve systems with infinite solutions, students can develop a deeper understanding of linear equations and their applications in real-world scenarios.
### What are the conditions for a system of equations to have infinite solutions?
A system of equations has infinite solutions if the equations are dependent and represent the same line or plane.
### How can we determine if a system of equations has infinitely many solutions?
A system of equations has infinitely many solutions if and only if the equations are dependent and represent the same line or planes in a higher dimensional space. This can be determined by analyzing the coefficients and constants in the equations, looking for patterns or relationships that indicate dependency. Additionally, graphical methods, such as plotting the equations on a graph, can help visualize if the lines or planes overlap or coincide, indicating infinite solutions.
### What are some strategies or techniques for identifying whether a system of equations has an infinite number of solutions?
One strategy for identifying whether a system of equations has an infinite number of solutions is to determine if the equations are dependent. This can be done by simplifying the equations and checking if one equation can be obtained from the other(s) by a combination of algebraic operations. If the equations are dependent, it means they represent the same line or plane, and therefore have an infinite number of solutions. Another technique is to solve the system using elimination or substitution and check if the resulting equation is always true, indicating that any values of the variables would satisfy the system.
### Are there any specific patterns or characteristics that can help us recognize when a system of equations has infinite solutions?
Yes, there are specific patterns and characteristics that can help us recognize when a system of equations has infinite solutions. One key indicator is when all the equations in the system are dependent, meaning they can be obtained by multiplying one equation by a constant factor or adding/subtracting multiple equations together. Additionally, if the system has fewer equations than unknowns, it is likely to have infinitely many solutions.
### Can you provide examples or real-life applications where systems of equations have infinite solutions?
One example of a real-life application where systems of equations have infinite solutions is in the field of economics. In economics, systems of equations are often used to model supply and demand relationships. In some cases, there may be multiple combinations of prices and quantities that satisfy the given equations, leading to an infinite number of solutions. This can occur when the supply and demand curves intersect at multiple points or when there is a range of equilibrium prices and quantities. The concept of infinite solutions in systems of equations helps economists understand the complexities of market behavior and make more accurate predictions.
In conclusion, understanding whether a system of equations has infinite solutions is a crucial concept in mathematics education. By analyzing the coefficients and constants of the equations and applying various strategies such as substitution or elimination, students can determine if the system has multiple solutions or none at all. This knowledge not only enhances their problem-solving skills but also lays a strong foundation for more complex topics like linear algebra. Therefore, mastering this topic is essential for students aiming to excel in mathematics education. |
# 5.1: Midsegments of a Triangle
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Identify the midsegments of a triangle.
• Use the Midsegment Theorem to solve problems involving side lengths, midsegments, and algebra.
## Review Queue
Find the midpoint between the given points.
1. (-4, 1) and (6, 7)
2. (5, -3) and (11, 5)
3. (0, -2) and (-4, 14)
4. Find the equation of the line between (-2, -3) and (-1, 1).
5. Find the equation of the line that is parallel to the line from #4 through (2, -7).
Know What? A fractal is a repeated design using the same shape (or shapes) of different sizes. Below, is an example of the first few steps of a fractal. What does the next figure look like? How many triangles are in each figure (green and white triangles)? Is there a pattern?
## Defining Midsegment
Midsegment: A line segment that connects two midpoints of adjacent sides of a triangle.
Example 1: Draw the midsegment \begin{align*}\overline{DF}\end{align*} between \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*}. Use appropriate tic marks.
Solution: Find the midpoints of \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*} using your ruler. Label these points \begin{align*}D\end{align*} and \begin{align*}F\end{align*}. Connect them to create the midsegment.
Don’t forget to put the tic marks, indicating that \begin{align*}D\end{align*} and \begin{align*}F\end{align*} are midpoints, \begin{align*}\overline{AD} \cong \overline{DB}\end{align*} and \begin{align*}\overline{BF} \cong \overline{FC}\end{align*}.
Example 2: Find the midpoint of \begin{align*}\overline{AC}\end{align*} from \begin{align*}\triangle ABC\end{align*}. Label it \begin{align*}E\end{align*} and find the other two midsegments of the triangle.
Solution:
For every triangle there are three midsegments.
Let’s transfer what we know about midpoints in the coordinate plane to midsegments in the coordinate plane. We will need to use the midpoint formula, \begin{align*}\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\end{align*}.
Example 3: The vertices of \begin{align*}\triangle LMN\end{align*} are \begin{align*}L(4, 5), M(-2, -7)\end{align*} and \begin{align*}N(-8, 3)\end{align*}. Find the midpoints of all three sides, label them \begin{align*}O, P\end{align*} and \begin{align*}Q\end{align*}. Then, graph the triangle, it’s midpoints and draw in the midsegments.
Solution: Use the midpoint formula 3 times to find all the midpoints.
\begin{align*}L\end{align*} and \begin{align*}M = \left (\frac{4+(-2)}{2}, \frac{5+(-7)}{2}\right)=(1,-1)\end{align*}, point \begin{align*}O\end{align*}
\begin{align*}L\end{align*} and \begin{align*}N = \left(\frac{4+(-8)}{2}, \frac{5+3}{2}\right)=(-2,4)\end{align*}, point \begin{align*}Q\end{align*}
\begin{align*}M\end{align*} and \begin{align*}N = \left(\frac{-2+(-8)}{2}, \frac{-7+3}{2}\right)=(-5,-2)\end{align*}, point \begin{align*}P\end{align*}
The graph would look like the graph to the right. We will use this graph to explore the properties of midsegments.
Example 4: Find the slopes of \begin{align*}\overline{NM}\end{align*} and \begin{align*}\overline{QO}\end{align*}.
Solution: The slope of \begin{align*}\overline{NM}\end{align*} is \begin{align*}\frac{-7-3}{-2-(-8)}=\frac{-10}{6}=-\frac{5}{3}\end{align*}.
The slope of \begin{align*}\overline{QO}\end{align*} is \begin{align*}\frac{-1-4}{1-(-2)}=-\frac{5}{3}\end{align*}.
From this we can conclude that \begin{align*}\overline{NM} \ || \ \overline{QO}\end{align*}. If we were to find the slopes of the other sides and midsegments, we would find \begin{align*}\overline{LM} \ || \ \overline{QP}\end{align*} and \begin{align*}\overline{NL} \ || \ \overline{PO}\end{align*}. This is a property of all midsegments.
Example 5: Find \begin{align*}NM\end{align*} and \begin{align*}QO\end{align*}.
Solution: Now, we need to find the lengths of \begin{align*}\overline{NM}\end{align*} and \begin{align*}\overline{QO}\end{align*}. Use the distance formula.
\begin{align*}NM &= \sqrt{(-7-3)^2+(-2-(-8))^2}=\sqrt{(-10)^2+6^2}=\sqrt{100+36}=\sqrt{136} \approx 11.66\\ QO &= \sqrt{(1-(-2))^2+(-1-4)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34} \approx 5.83\end{align*}
From this we can conclude that \begin{align*}QO\end{align*} is half of \begin{align*}NM\end{align*}. If we were to find the lengths of the other sides and midsegments, we would find that \begin{align*}OP\end{align*} is half of \begin{align*}NL\end{align*} and \begin{align*}QP\end{align*} is half of \begin{align*}LM\end{align*}. This is a property of all midsegments.
## The Midsegment Theorem
The conclusions drawn in Examples 4 and 5 can be generalized into the Midsegment Theorem.
Midsegment Theorem: The midsegment of a triangle is half the length of the side it is parallel to.
Example 6: Mark everything you have learned from the Midsegment Theorem on \begin{align*}\triangle ABC\end{align*} above.
Solution: Let’s draw two different triangles, one for the congruent sides, and one for the parallel lines.
Because the midsegments are half the length of the sides they are parallel to, they are congruent to half of each of those sides (as marked). Also, this means that all four of the triangles in \begin{align*}\triangle ABC\end{align*}, created by the midsegments are congruent by SSS.
As for the parallel midsegments and sides, several congruent angles are formed. In the picture to the right, the pink and teal angles are congruent because they are corresponding or alternate interior angles. Then, the purple angles are congruent by the \begin{align*}3^{rd}\end{align*} Angle Theorem.
To play with the properties of midsegments, go to http://www.mathopenref.com/trianglemidsegment.html.
Example 7: \begin{align*}M, N,\end{align*} and \begin{align*}O\end{align*} are the midpoints of the sides of the triangle.
Find
a) \begin{align*}MN\end{align*}
b) \begin{align*}XY\end{align*}
c) The perimeter of \begin{align*}\triangle XYZ\end{align*}
Solution: Use the Midsegment Theorem.
a) \begin{align*}MN = OZ = 5\end{align*}
b) \begin{align*}XY = 2(ON) = 2 \cdot 4 = 8\end{align*}
c) The perimeter is the sum of the three sides of \begin{align*}\triangle XYZ\end{align*}.
\begin{align*}XY + YZ + XZ = 2 \cdot 4 + 2 \cdot 3 + 2 \cdot 5 = 8 + 6 + 10 = 24\end{align*}
Example 8: Algebra Connection Find the value of \begin{align*}x\end{align*} and \begin{align*}AB\end{align*}.
Solution: First, \begin{align*}AB\end{align*} is half of 34, or 17. To find \begin{align*}x\end{align*}, set \begin{align*}3x-1\end{align*} equal to 17.
\begin{align*}3x - 1 &= 17\\ 3x &= 18\\ x &= 6\end{align*}
Let’s go back to the coordinate plane.
Example 9: If the midpoints of the sides of a triangle are \begin{align*}A(1, 5), B(4, -2)\end{align*}, and \begin{align*}C(-5, 1)\end{align*}, find the vertices of the triangle.
Solution: The easiest way to solve this problem is to graph the midpoints and then apply what we know from the Midpoint Theorem.
Now that the points are plotted, find the slopes between all three.
slope \begin{align*}AB = \frac{5+2}{1-4}=-\frac{7}{3}\end{align*}
slope \begin{align*}BC = \frac{-2-1}{4+5}=\frac{-3}{9}=-\frac{1}{3}\end{align*}
slope \begin{align*}AC = \frac{5-1}{1+5}=\frac{4}{6}=\frac{2}{3}\end{align*}
Using the slope between two of the points and the third, plot the slope triangle on either side of the third point and extend the line. Repeat this process for all three midpoints. For example, use the slope of \begin{align*}AB\end{align*} with point \begin{align*}C\end{align*}.
The green lines in the graph to the left represent the slope triangles of each midsegment. The three dotted lines represent the sides of the triangle. Where they intersect are the vertices of the triangle (the blue points), which are (-8, 8), (10, 2) and (-2, 6).
Know What? Revisited To the left is a picture of the \begin{align*}4^{th}\end{align*} figure in the fractal pattern. The number of triangles in each figure is 1, 4, 13, and 40. The pattern is that each term increase by the next power of 3.
## Review Questions
\begin{align*}R, S, T,\end{align*} and \begin{align*}U\end{align*} are midpoints of the sides of \begin{align*}\triangle XPO\end{align*} and \begin{align*}\triangle YPO\end{align*}.
1. If \begin{align*}OP = 12\end{align*}, find \begin{align*}RS\end{align*} and \begin{align*}TU\end{align*}.
2. If \begin{align*}RS = 8\end{align*}, find \begin{align*}TU\end{align*}.
3. If \begin{align*}RS = 2x\end{align*}, and \begin{align*}OP = 20\end{align*}, find \begin{align*}x\end{align*} and \begin{align*}TU\end{align*}.
4. If \begin{align*}OP = 4x\end{align*} and \begin{align*}RS = 6x - 8\end{align*}, find \begin{align*}x\end{align*}.
5. Is \begin{align*}\triangle XOP \cong \triangle YOP\end{align*}? Why or why not?
For questions 6-13, find the indicated variable(s). You may assume that all line segments within a triangle are midsegments.
1. The sides of \begin{align*}\triangle XYZ\end{align*} are 26, 38, and 42. \begin{align*}\triangle ABC\end{align*} is formed by joining the midpoints of \begin{align*}\triangle XYZ\end{align*}.
1. Find the perimeter of \begin{align*}\triangle ABC\end{align*}.
2. Find the perimeter of \begin{align*}\triangle XYZ\end{align*}.
3. What is the relationship between the perimeter of a triangle and the perimeter of the triangle formed by connecting its midpoints?
Coordinate Geometry Given the vertices of \begin{align*}\triangle ABC\end{align*} below, find the midpoints of each side.
1. \begin{align*}A(5, -2), B(9, 4)\end{align*} and \begin{align*}C(-3, 8)\end{align*}
2. \begin{align*}A(-10, 1), B(4, 11)\end{align*} and \begin{align*}C(0, -7)\end{align*}
3. \begin{align*}A(0, 5), B(4, -1)\end{align*} and \begin{align*}C(-2, -3)\end{align*}
4. \begin{align*}A(2, 4), B(8, -4)\end{align*} and \begin{align*}C(2, -4)\end{align*}
Multi-Step Problem The midpoints of the sides of \begin{align*}\triangle RST\end{align*} are \begin{align*}G(0, -2), H(9, 1)\end{align*}, and \begin{align*}I(6, -5)\end{align*}. Answer the following questions.
1. Find the slope of \begin{align*}GH, HI\end{align*}, and \begin{align*}GI\end{align*}.
2. Plot the three midpoints and connect them to form midsegment triangle, \begin{align*}\triangle GHI\end{align*}.
3. Using the slopes, find the coordinates of the vertices of \begin{align*}\triangle RST\end{align*}.
4. Find \begin{align*}GH\end{align*} using the distance formula. Then, find the length of the sides it is parallel to. What should happen?
More Coordinate Geometry Given the midpoints of the sides of a triangle, find the vertices of the triangle. Refer back to problems 19-21 for help.
1. (-2, 1), (0, -1) and (-2, -3)
2. (1, 4), (4, 1) and (2, 1)
Given the vertices of \begin{align*}\triangle ABC\end{align*}, find:
a) the midpoints of \begin{align*}M, N\end{align*} and \begin{align*}O\end{align*} where \begin{align*}M\end{align*} is the midpoint of \begin{align*}\overline{AB}\end{align*}, \begin{align*}N\end{align*} is the midpoint of \begin{align*}\overline{BC}\end{align*} and \begin{align*}C\end{align*} is the midpoint of \begin{align*}\overline{AC}\end{align*}.
b) Show that midsegments \begin{align*}\overline{MN}, \overline{NO}\end{align*} and \begin{align*}\overline{OM}\end{align*} are parallel to sides \begin{align*}\overline{AC}, \overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*} respectively.
c) Show that midsegments \begin{align*}\overline{MN}, \overline{NO}\end{align*} and \begin{align*}\overline{OM}\end{align*} are half the length of sides \begin{align*}\overline{AC}, \overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*} respectively.
1. \begin{align*}A(-3, 5), B(3, 1)\end{align*} and \begin{align*}C(-5, -5)\end{align*}
2. \begin{align*}A(-2, 2), B(4, 4)\end{align*} and \begin{align*}C(6, 0)\end{align*}
For questions 27-30, \begin{align*}\triangle CAT\end{align*} has vertices \begin{align*}C(x_1,y_1), A(x_2,y_2)\end{align*} and \begin{align*}T(x_3,y_3)\end{align*}.
1. Find the midpoints of sides \begin{align*}\overline{CA}\end{align*} and \begin{align*}\overline{CT}\end{align*}. Label them \begin{align*}L\end{align*} and \begin{align*}M\end{align*} respectively.
2. Find the slopes of \begin{align*}\overline{LM}\end{align*} and \begin{align*}\overline{AT}\end{align*}.
3. Find the lengths of \begin{align*}\overline{LM}\end{align*} and \begin{align*}\overline{AT}\end{align*}.
4. What have you just proven algebraically?
1. \begin{align*}\left ( \frac{-4+6}{2}, \frac{1+7}{2} \right ) = (1, 4)\end{align*}
2. \begin{align*}\left ( \frac{5+11}{2}, \frac{-3+5}{2} \right ) = (8, 1)\end{align*}
3. \begin{align*}\left (\frac{0 - 4}{2}, \frac{-2 + 14}{2} \right ) = (-2, 6)\end{align*}
4. \begin{align*}m=\frac{-3-1}{-2-(-1)} = \frac{-4}{-1}=4\!\\ y=mx+b\!\\ -3=4(-2)+b\!\\ b=5, \ y=4x+5\end{align*}
5. \begin{align*}-7=4(2)+b\!\\ {\;} \ b=-15, \ y=4x-15\end{align*}
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If O is any point in the interior of a rectangle ABCD. Prove that $O{A^2} + O{C^2} = O{B^2} + O{D^2}$. Hence find the length of OD, if the lengths of OA, OB and OC are 3 cm, 4 cm and 5 cm respectively.
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Hint:Analyze the situation with a diagram. Take a random point O inside the square and join it with every vertex. Draw two lines passing through O and parallel to the sides AB and BC respectively. Consider four right angled triangles each to determine the values of OA, OB, OC and OD respectively in terms of some variables. Then verify the result $O{A^2} + O{C^2} = O{B^2} + O{D^2}$.
According to the question, a rectangle ABCD is said to have any interior point O. We have to prove that $O{A^2} + O{C^2} = O{B^2} + O{D^2}$.
Consider the rectangle ABCD shown below with a point O lying inside it.
OA, OB, OC and OD are the lines joining the vertex of the square and point O. We have also drawn EG and FH parallel to the sides of the square and passing through point O.
From this we can conclude that AD, FH and BC are parallel and equal. Similarly AB, EG and DC are also parallel and equal.
So let $AF = OE = DH = a$.
Similarly we will assume some variable for other sides also as shown below:
$\Rightarrow FB = OG = HC = b \\ \Rightarrow AE = OF = BG = c \\ \Rightarrow ED = OH = GC = d \\$
To find the values of OA, OB, OC and OD, we’ll consider right angled triangles.
So in right angled triangle $AOF$, we have:
$\Rightarrow O{A^2} = O{F^2} + A{F^2} = {c^2} + {a^2}{\text{ }}.....{\text{(1)}}$
Similarly in triangle $BOG$, we have:
$\Rightarrow O{B^2} = O{G^2} + B{G^2} = {b^2} + {c^2}{\text{ }}.....{\text{(2)}}$
In triangle $COH$, we have:
$\Rightarrow O{C^2} = O{H^2} + H{C^2} = {d^2} + {b^2}{\text{ }}.....{\text{(3)}}$
And in triangle $DOE$, we have:
$\Rightarrow O{D^2} = O{E^2} + E{D^2} = {a^2} + {d^2}{\text{ }}.....{\text{(4)}}$
Now adding equation (1) and (3), we’ll get:
$\Rightarrow O{A^2} + O{C^2} = {c^2} + {a^2} + {d^2} + {b^2} \\ \Rightarrow O{A^2} + O{C^2} = {a^2} + {b^2} + {c^2} + {d^2}{\text{ }}.....{\text{(5)}} \\$
And adding equation (2) and (4), we’ll get:
$\Rightarrow O{B^2} + O{D^2} = {b^2} + {c^2} + {a^2} + {d^2} \\ \Rightarrow O{B^2} + O{D^2} = {a^2} + {b^2} + {c^2} + {d^2}{\text{ }}.....{\text{(6)}} \\$
On comparing equation (5) and (6), we can say that:
$\Rightarrow O{A^2} + O{C^2} = O{B^2} + O{D^2}$
Hence this is proved.
Further we have to calculate the length of OD such that OA, OB and OC are 3 cm, 4 cm and 5 cm respectively.
So using the same result:
$\Rightarrow O{A^2} + O{C^2} = O{B^2} + O{D^2}$
Putting the values, we’ll get:
$\Rightarrow {3^2} + {5^2} = {4^2} + O{D^2} \\ \Rightarrow 16 + O{D^2} = 9 + 25 = 34 \\ \Rightarrow O{D^2} = 18 \\ \Rightarrow OD = \sqrt {18} = 3\sqrt 2 \\$
Thus the length of OD is $3\sqrt 2$ cm.
Note: Although we have proved the above result for rectangles, this will hold true for squares also. Since we have only used the property of square that it’s opposite sides are parallel and equal and all of its angles are ${90^ \circ }$ and this property is also followed by square, thus the result will be equally valid for squares. |
# How to Calculate Function From Ordered Pairs
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Put strawberries into a blender and a smoothie comes out; put carrots into a blender and chopped carrots come out. A function is the same: it produces one output for each individual input and the same input cannot produce two different outputs. For example, you cannot put strawberries into a blender and get both a smoothie and chopped carrots. This is what mathematicians mean when they write a function such as f(x) = x + 1. Put strawberries (x) into the function, and you get a smoothie (x + 1).
Write the ordered pairs you want to analyze. For example, write, (3,7) and (7,2).
Write the quotient of the difference of the second term of the second pair and the second term of the first pair divided by the difference of the first term of the second pair and the first term of the first pair. Solve using a calculator. For example, (2 - 7)/(7 - 3) = -1.25.
Substitute your answer as the value of the variable m in the equation y = mx + b. For example, write, y = -1.25x + b.
Substitute the first term of the first ordered pair into the same equation in place of the variable x. For example, write, y = (-1.25 x 3) + b.
Substitute the second term of the first ordered pair into the same equation in place of the variable y. For example, write, 7 = (-1.25 x 3) + b.
Simplify your equation by completing the multiplication in the parentheses using a calculator. For example, write, 7 = -3.75 + b.
Simplify your equation again by adding a term to both sides of the equation that will leave the b variable alone on its side of the equation. For example, if you add 3.75 to both sides of the equation, 3.75 and -3.75 on the right side of the equation will cancel, leaving the variable b alone. Write, 7 + 3.75 = -3.75 + 3.75 + b.
Simplify your equation by performing the indicated addition operations. For example, write, 10.75 = b.
Substitute your answer for the variable b in the original equation y = mx + b. For example, write, y = mx + 10.75.
Substitute into the same equation your original value for m. For example, your original value for m was -1.25. Write, y = -1.25x + 10.75. You have calculated a function from the ordered pairs (3,7) and (7,2).
• Paper
• Pencil
• Calculator |
# Rotation of Complex number and Power of i
## Rotation of Complex Number
• Multiplying i is a rotation by 90 degrees counter-clockwise
• Multiplying by -i is a rotation of 90 degrees clockwise
Example
z=1
If we multiply it by i, it becomes
z=i so that it has rotated by the angle 90 degrees
## What is the significance of Complex Numbers? Why they are required?
• Real numbers such as natural number,rational number , irrational number are invented in the history as and when we encounter various mathematical needs. Same happen with the complex numbers.
We had no solution for the problem
x2=-1
Eular was the first mathematicain to introduce solution to this problem,he introduced the symbol i
$i=\sqrt{(-1)}$
So i2=-1
So solution to the problem becomes
x=i or -i
• He called the symbol i as imaginary unit.
• Just like all the other number ,this number was added to our Number vocabulary. This like other numbers is useful in explaining where physical explanation.
• It is very useful in the field Electrical and electronics.
## Power of i
We know that
i2 =-1
i3=-i
i4=1
i5=i
i6=-1
i7=-i
This can be generalized as
For any integer k, i4k = 1, i4k + 1 = i, i4k + 2 = - 1, i4k + 3 = -i
Example
Find the value $(\frac {i}{2})(\frac {-2i}{3}) (\frac {i^3}{4})$
Solution
$(\frac {i}{2})(\frac {-2i}{3}) (\frac {i^3}{4})$
$=\frac {-i^5}{12} = \frac {-i}{5}$
## Identities for Complex Numbers
Here are some the identities for Complex numbers
For all complex numbers z1 and z2
1. (z1 + z2)2 = z12 + z22 + 2z1z2
2. (z1 - z2)2 = z12 + z22 - 2z1z2
3. (z1 + z2)3 = z13 + z23 + 3z1z22+ 3z12z2
4. (z1 - z2)3 = z13 - z23 + 3z1z22 -3z12z2
5. z12 - z22 = (z1 + z2 )(z1 - z2)
6. ## Solved Example
Example 1
Find the non-zero integral Solution for this equation
$|1-i|^x =2^x$
Solution
$|1-i|= \sqrt {2} =2^{1/2}$
So
$2^{x/2}=2^x$
2x-x/2=1
2x/2 =1
or x=0
So there is no non-zero integral solution for this equation |
# What is the additive inverse property?
Definition. The additive inverse of a number is what you add to a number to create the sum of zero. So in other words, the additive inverse of x is another number, y, as long as the sum of x + y equals zero. For example, the additive inverse of the positive number 5 is -5.
Moreover, what is an example of the inverse property of multiplication?
A multiplicative inverse is a reciprocal. A reciprocal is one of a pair of numbers that when multiplied with another number equals the number 1. For example, if we have the number 7, the multiplicative inverse, or reciprocal, would be 1/7 because when you multiply 7 and 1/7 together, you get 1!
What is the inverse of 12?
In other words, a reciprocal is a fraction flipped upside down. Multiplicative inverse means the same thing as reciprocal. For example, the multiplicative inverse (reciprocal) of 12 is and the multiplicative inverse (reciprocal) of is . Note: The product of a number and its multiplicative inverse is 1.
What is the definition of inverse operation?
Mathematically, inverse operations are opposite operations. Addition is the opposite of subtraction; division is the opposite of multiplication, and so on. Inverse operations are used to solve simple algebraic equations to more difficult equations that involve exponents, logarithms, and trigonometry.
## What is the additive inverse of 6?
Mathwords: Additive Inverse of a Number. The opposite of a number. For example, the additive inverse of 12 is –12. The additive inverse of –3 is 3.
## What is the additive property of equality?
Mathwords: Additive Property of Equality. The formal name for the property of equality that allows one to add the same quantity to both sides of an equation. This, along with the multiplicative property of equality, is one of the most commonly used properties for solving equations.
## What is an example of the multiplicative inverse property?
A multiplicative inverse is a reciprocal. What is a reciprocal? A reciprocal is one of a pair of numbers that when multiplied with another number equals the number 1. For example, if we have the number 7, the multiplicative inverse, or reciprocal, would be 1/7 because when you multiply 7 and 1/7 together, you get 1!
## What is the additive inverse of 8?
The additive inverse of 8 is -8. To find the additive inverse of a number, we simply change the sign of the number. In other words, the additive inverse of a given number x is -x. Therefore, to find the additive inverse of 8, we just change the sign of 8 to get -8.
## What is the additive inverse property examples?
Definition. The additive inverse of a number is what you add to a number to create the sum of zero. So in other words, the additive inverse of x is another number, y, as long as the sum of x + y equals zero. For example, the additive inverse of the positive number 5 is -5.
## What is the absolute value of 7?
Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. The absolute value of a number is never negative. The absolute value of 2 + 7 is 5.
## What is an example of inverse property of multiplication?
The purpose of the inverse property of addition is to get a result of zero. The purpose of the inverse property of multiplication is to get a result of 1. We use inverse properties to solve equations. Inverse Property of Addition says that any number added to its opposite will equal zero.
## What is the inverse of subtraction?
Mathematically, inverse operations are opposite operations. Addition is the opposite of subtraction; division is the opposite of multiplication, and so on. Inverse operations are used to solve simple algebraic equations to more difficult equations that involve exponents, logarithms, and trigonometry.
## What is the additive inverse of 2?
additive inverse. See more synonyms on Thesaurus.com noun Mathematics. the number in the set of real numbers that when added to a given number will yield zero: The additive inverse of 2 is −2.
## What is additive inverse of 0?
The Additive Inverse of Zero: Zero is a special number in mathematics because it holds special properties. For this reason, we call 0 the additive identity. Furthermore, if the sum of two numbers is equal to 0, then we say that the two numbers are additive inverses.
## What is the distributive property of addition?
The Distributive Property is easy to remember, if you recall that “multiplication distributes over addition”. Formally, they write this property as “a(b + c) = ab + ac”. In numbers, this means, for example, that 2(3 + 4) = 2×3 + 2×4.
## What is the meaning of closure property?
Closure Property. The closure property means that a set is closed for some mathematical operation. For example, the set of even natural numbers, [2, 4, 6, 8, . . .], is closed with respect to addition because the sum of any two of them is another even natural number, which is also a member of the set.
## What is the definition of the zero property?
One of zero’s unique rules is called the multiplication property. The multiplication property states that the product of any number and zero is zero. It doesn’t matter what the number is, when you multiply it to zero, you get zero as the answer. So: 2 x 0 = 0.
## What is the multiplicative inverse?
In other words, a reciprocal is a fraction flipped upside down. Multiplicative inverse means the same thing as reciprocal. For example, the multiplicative inverse (reciprocal) of 12 is and the multiplicative inverse (reciprocal) of is . Note: The product of a number and its multiplicative inverse is 1. Observe that.
## What is the associative property?
Definition: The associative property states that you can add or multiply regardless of how the numbers are grouped. By ‘grouped’ we mean ‘how you use parenthesis’. In other words, if you are adding or multiplying it does not matter where you put the parenthesis. Add some parenthesis any where you like!.
## What is the definition of zero pair?
a pair of numbers whose sum is zero, e.g. +1, -1. • used to illustrate addition and subtraction problems. with positive and negative integers. EXAMPLES: © Jenny Eather 2014.
## What is the definition of multiplicative inverse?
In mathematics, a multiplicative inverse or reciprocal for a number x, denoted by 1/x or x−1, is a number which when multiplied by x yields the multiplicative identity, 1. The multiplicative inverse of a fraction a/b is b/a. Multiplicative inverses can be defined over many mathematical domains as well as numbers.
## What are the properties of addition?
There are four mathematical properties which involve addition. The properties are the commutative, associative, additive identity and distributive properties. Commutative property: When two numbers are added, the sum is the same regardless of the order of the addends.
## What is the additive inverse of a complex number?
The additive identity in the complex number system is zero (the same as in the real number system). The additive inverse of the complex number a + bi is –(a + bi) = –a – bi.
## What is an integer greater than zero?
An integer is a whole number that can be either greater than 0, called positive, or less than 0, called negative. Zero is neither positive nor negative. Two integers that are the same distance from the origin in opposite directions are called opposites.
Categories FAQ |
## Precalculus (6th Edition) Blitzer
The probability is $\frac{1}{6}$.
We know that there are six outcomes, so $\text{n}\left( \text{S} \right)=\text{ 6}$. There are two outcomes in the green stopping event, so $\text{n}{{\left( \text{E} \right)}_{\text{green}}}=\text{ 2}$. And the probability of stopping on red is given below, $\text{P}{{\left( \text{E} \right)}_{\text{green}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{green}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{2}{6}\text{ }$ And the event of stopping on a number less than 4 can be represented by ${{\left( \text{E} \right)}_{\text{less than 4}}}\text{ }=\text{ }\left\{ 1,2,3 \right\}$ There are three outcomes in this event, so $\text{n}{{\left( \text{E} \right)}_{\text{less than 4}}}=\text{ 3}$. And the probability of stopping on less than $4$ is given below $\text{P}{{\left( \text{E} \right)}_{\text{less than 4}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{less than 4}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{3}{6}\text{ }$ Thus, the probability of stopping on green on first spin and a number less than $4$ on second spin is \begin{align} & \text{P}\left( \text{red on first spin and less than 4 on second spin} \right)=\text{P}\left( \text{red} \right)\times \text{P}\left( \text{less than 4} \right) \\ & =\frac{2}{6}\times \frac{3}{6} \\ & =\frac{6}{36} \\ & =\frac{1}{6} \end{align} |
# Concepts 1.1
```Bell Work
9/16/14
Solve the equation.
1.
4.
3x 7 5
1
2
x 7 9
2.
6 x 4 8
3.
5.
2 7x 9
6.
x
5
3 12
11 13 3 x
Yesterday’s Homework
1. Any questions?
• Make sure the homework is 100% complete.
• Incomplete work will NOT be accepted.
4/10/2015
2 .3A
Solving Multi-Step
Equations (3-Step)
TSWBAT: solve 3-step equations.
Students solve multi-step problems, including
word problems, involving linear equations and
linear inequalities in one variable and provide
justification for each step.
Notes
• Steps for Solving Linear Equations
1. Simplify both sides of the equation.
• Distribute and/or combine like terms.
2. Collect the variables to one side of the equation.
• Collect to the side with the greater variable.
3. Use inverse operations to isolate the variable.
• PEMDAS backwards.
• The Golden Rule (of Mathematics)
What you do to one side,
you must do to the other.
Notes
Solve the equation.
Ex.
2 x 3 24
2 x 6 24
6 6
2 x 18
2
2
x9
Now you try.
Ex.
Distribute first.
4 x 5 8
What operation4 x 20 8
do I do next?
20 20
What operation
do I do next?
4 x 28
4
4
x7
Notes
Solve the equation.
Ex.
3 x 4 27
3 x 12 2 7
12 12
3 x 15
3
3
x 5
Now you try.
Ex.
Distribute first.
2 x 6 18
What operation 2 x 12 1 8
do I do next?
12 12
What operation
2 x 30
do I do next?
2
2
x 15
Notes
Solve the equation.
Ex.
3 x 4 x 35
7 x 35
7
7
x 5
Ex.
Combine like
20
terms first.
20
What operation
do I do next? 5
4
4x 9x
5 x
5
x
Now you try.
Ex.
Ex.
2 x 6 x 28
4 x 28
4
4
x7
32
32
4
8
2 x 2 x
4 x
4
x
Notes
Now you try.
Solve the equation.
Ex.
Ex.
2 x 3 5 x 9 27
7 x 6 27
6 6
7 x 21
7
7
x 3
1like
4 terms
4 x first.
6 7x 2
Combine
What
operation
8
14
3 x do
you
8do next? 8
Whatoperation
6 3 x do
you do next?
3
3
2 x
Summary
To solve a 3-step equation you first have to
combine like terms. After that isolate
distribute and/or ________
the variable term by _______
isolate the variable completely by multiplying or
dividing
________.
Today’s Homework
#1-13 on next slide
& Worksheet 2.3A
Rules for Homework
1. Pencil ONLY.
2. Must show all of your work.
• NO WORK = NO CREDIT
3. Must attempt EVERY problem.
Homework
2.3A
Solve the equation.
1.
2 x 5 16
2.
3 x 2 12
3.
4 x 3 32
4.
14 2 x 5
5.
30 6 x 2
6.
24 6 x 3
7.
7 x 2 x 25
8.
27 4 x 7 x
9.
3 x 5 x 40
10.
3 x 5 x 4 20
11.
3 x 2 5 x 10
12.
12 x 5 x 6
13.
20 3 x 8 7 x
Ticket Out the Door
Complete the Ticket Out the Door without talking!!!!!
Talking = time after the bell!
Put your NAME on the paper.
When finished, turn your paper face DOWN.
Solve the equation.
5 x 2 20
``` |
Procedure: To divide a polynomial (in the numerator) by a monomial (in the denominator)
# Procedure: To divide a polynomial (in the numerator) by a monomial (in the denominator)
## Procedure: To divide a polynomial (in the numerator) by a monomial (in the denominator)
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. There are two types of division problems to consider. For both types, the numerator of the fraction will be a polynomial. If the denominator is a monomial, we will break up the fraction into several simpler fractions. If the denominator has more than one term (not a monomial), we will use long division. Objective A: To divide a polynomial by a monomial Procedure: To divide a polynomial (in the numerator) by a monomial (in the denominator) Note: The sign of each term in the original numerator becomes the sign of its resulting fraction. In step 2, if there are two negatives, the answer is positive; if there is one negative, the answer is negative. Step 1. Rewrite the problem, making separate fractions for each term in the numerator. The denominator of each fraction is the original monomial denominator. Step 2. Simplify each fraction.
2. Your Turn Problem #1 1. Separate into three fractions, one for each term in the numerator. 2. Simplify each fraction.
3. Your Turn Problem #2 1. Separate into three fractions, one for each term in the numerator. 2. Simplify each fraction.
4. Objective B: To divide a polynomial by a polynomial Procedure: To divide a polynomial (in the numerator) by a polynomial in the denominator Step 2. Divide the first term of the dividend by the first term of the divisor. Write the answer in the quotient. Step 3. Multiply answer from Step 2 by the divisor. Write answer directly under the first term of the dividend. Step 4. Subtract (change signs) answer found in Step 3 from the dividend. Add the columns and write the answer directly below, under a horizontal bar. Step 5. Bring down the next term of the dividend under the horizontal bar, next to the difference found in Step 4. Step 6. Repeat steps 2 to 5 using the polynomial under the horizontal bar as the new dividend. Step 7. If no terms are left to bring down, then the polynomial remaining under the horizontal bar (if any) is the remainder. Write the remainder in fractional form.
5. Your Turn Problem #3 1. Rewrite in long division format. 2. Divide first term of dividend by first term of divisor. Write in quotient. 3. Multiply 2x by the divisor. Write below dividend. 4. Write a line below and change the signs. 5. Add the columns and bring down next term. 6. Repeat the steps. Divide first term (from step 5) by the first term of the divisor. Write in quotient. 7. Multiply the divisor by 1. 8. Write a line below and change the signs. 9. Add columns. Note: To check the answer, multiply the divisor and the quotient which should equal the dividend. i.e. Multiply (x+2)(2x+1)
6. Your Turn Problem #4 1. Rewrite in long division format. 2. Divide first term of dividend by first term of divisor. Write in quotient. 3. Multiply x2 by the divisor. Write below dividend. 4. Write a line below and change the signs. 5. Add the columns and bring down next term. 6. Repeat the steps. Divide first term (from step 5) by the first term of the divisor. Write in quotient. 7. Multiply the divisor by –5x. 8. Write a line below and change the signs. 9. Add columns, bring down next term. 10. Divide first term by the divisor; write in quotient. 11. Multiply 4 by divisor. 12. Write a line below and change the signs. To check: multiply divisor and quotient, then add remainder. This will give the dividend. 13. Add columns, this is the remainder.
7. 1. The polynomials must be in descending order. 2. If the dividend has a gap in the exponent sequence, insert the variable(s) with a coefficient of 0. Next Slide Note: There are two conditions which are to be followed before the steps for long division can be performed.
8. Your Turn Problem #5 1. Rewrite in long division format. 2. Divide first term of dividend by first term of divisor. Write in quotient. 3. Multiply x2 by the divisor. Write below dividend. 4. Write a line below and change the signs. 5. Add the columns and bring down next term. 6. Repeat the steps. Divide first term (from step 5) by the first term of the divisor. Write in quotient. 7. Multiply the divisor by 2x. 8. Write a line below and change the signs. 9. Add columns, bring down next term. 10. Divide first term by the divisor; write in quotient. 11. Multiply 4 by divisor. 12. Write a line below and change the signs. 13. Add columns. This is the remainder. The End. B.R. 12-28-06 |
# 2016 AMC 10B Problems/Problem 13
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $1000$ babies were in sets of quadruplets?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 160$
## Solution 1
We can set up a system of equations where $a$ is the sets of twins, $b$ is the sets of triplets, and $c$ is the sets of quadruplets. $$\begin{split} 2a + 3b + 4c & = 1000 \\ b & = 4c \\ a & = 3b \end{split}$$
Solving for $c$ and $a$ in the second and third equations and substituting into the first equation yields $$\begin{split} 2 (3b) + 3b + 4 (0.25b) & = 1000 \\ 6b + 3b + b & = 1000 \\ b & = 100 \end{split}$$
Since we are trying to find the number of babies and $\textbf{NOT}$ the number of sets of quadruplets, the solution is not $c$, but rather $4c$. Therefore, we strategically use the second initial equation to realize that $b$ $=$ $4c$, leaving us with the number of babies born as quadruplets equal to $\boxed{\textbf{(D)}\ 100}$.
## Solution 2
Say there are $12x$ sets of twins, $4x$ sets of triplets, and $x$ sets of quadruplets. That's $12x\cdot2=24x$ twins, $4x\cdot3=12x$ triplets, and $x\cdot4=4x$ quadruplets. A tenth of the babies are quadruplets and that's $\frac{1}{10}(1000)=\boxed{\textbf{(D) }100}$
~savannahsolver |
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# If in a triangle ABC, $\cos A\cos B + \sin A\sin B\sin C = 1$, then find the value of $\dfrac{{a + b}}{c}$.
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Hint: Find the relation between the angles A, B and C from the given equation to find the type of the triangle, then find the ratio involving the sides.
For triangle ABC, we solve to find the relation between the angles A, B and C.
It is given that,
$\cos A\cos B + \sin A\sin B\sin C = 1..........(1)$
We know that the value of sine is less than or equal to 1.
$\sin C \leqslant 1$
We, now, multiply sinA sinB on both sides of the inequality.
$\sin A\sin B\sin C \leqslant \sin A\sin B$
We, now, add the term cosA cosB to both sides of the inequality.
$\cos A\cos B + \sin A\sin B\sin C \leqslant \cos A\cos B + \sin A\sin B$
The left-hand side of the inequality is equal to 1 from equation (1).
$1 \leqslant \cos A\cos B + \sin A\sin B$
The right-hand side is the cosine of difference between the angles A and B.
$1 \leqslant \cos (A - B)$
We know that the value of cosine of any angle is less than or equal to one, then, we have:
$1 \leqslant \cos (A - B) \leqslant 1$
Hence, the value of cos(A – B) is 1.
$\cos (A - B) = 1$
We know that the value of cos0 is equal to one, then the angles A and B are equal.
$A = B..........(2)$
Substituting equation (2) in equation (1), we get:
${\cos ^2}A + {\sin ^2}A\sin C = 1$
We know that the value of ${\cos ^2}A + {\sin ^2}A$ is equal to one, then, we have the value of sinC equal to one.
$\sin C = 1$
$C = 90^\circ$
Hence, triangle ABC is isosceles right angles triangle with right angle at C.
Hence, the sides a and b are equal and c is the hypotenuse.
Using, Pythagoras theorem, we have:
${a^2} + {b^2} = {c^2}$
Since, sides a and b are equal, we have:
$2{a^2} = {c^2}$
Finding ratio of a and c, we have:
$\dfrac{{{a^2}}}{{{c^2}}} = \dfrac{1}{2}$
$\dfrac{a}{c} = \dfrac{1}{{\sqrt 2 }}........(3)$
We need to find the value of $\dfrac{{a + b}}{c}$, we know that a and b are equal, therefore, we have:
$\dfrac{{a + b}}{c} = \dfrac{{2a}}{c}$
Using equation (3), we get:
$\dfrac{{a + b}}{c} = \dfrac{2}{{\sqrt 2 }}$
$\dfrac{{a + b}}{c} = \sqrt 2$
Hence, the value is $\sqrt 2$.
Note: You know that the sum of two sides is always greater than the third side, so the answer should be greater than 1, this fact can be used for cross-checking. |
## Archimedes' Cattle Problem
This is Archimedes' cattle problem, also called the bovinum problem:
The sun god had a herd of cattle consisting of bulls and cows, one part of which was white, a second black, a third spotted, and a fourth brown. Among the bulls, the number of white ones was one half plus one third the number of the black greater than the brown; the number of the black, one quarter plus one fifth the number of the spotted greater than the brown; the number of the spotted, one sixth and one seventh the number of the white greater than the brown. Among the cows, the number of white ones was one third plus one quarter of the total black cattle; the number of the black, one quarter plus one fifth the total of the spotted cattle; the number of spotted, one fifth plus one sixth the total of the brown cattle; the number of the brown, one sixth plus one seventh the total of the white cattle. What was the composition of the herd?
Step #1: The first task is to convert all of that into equations. Using W, X, Y, Z, w, x, y, z for white bulls, black bulls, spotted bulls, brown bulls, white cows, black cows, spotted cows, and brown cows, respectively, we get these equations:
1. W = (5/6)X + Z
2. X = (9/20)Y + Z
3. Y = (13/42)W + Z
4. w = (7/12)(X+x)
5. x = (9/20)(Y+y)
6. y = (11/30)(Z+z)
7. z = (13/42)(W+w)
Solution:
There are seven equations with eight unknowns. So there are many possible solutions. And we want the smallest positive integer solution (positive integer because we don't want negative numbers or fractions of cattle). I will explain my reasoning below. Here is the solution:
• W = 10,366,482
• X = 7,460,514
• Y = 7,358,060
• Z = 4,149,387
• w = 7,206,360
• x = 4,893,246
• y = 3,515,820
• z = 5,439,213
Sometimes further restrictions are set on the numbers, in order to make the herd much larger. Such numbers were far beyond the limits of the Greek or Roman or Egyptian number systems. And that may have been the entire idea behind the problem. We can precisely describe numbers beyond the limits of our number systems.
Reasoning:
Combining the first three equations, we get W = (5/6)((9/20)Y+Z)+Z = (5/6)((9/20)((13/42)W+Z)+Z)+Z = (13/112)W + (3/8)Z + (5/6)Z + Z = (13/112)W + (53/24)Z, so (99/14)W = (53/3)Z, and 297W = 742Z. This is reduced to its smallest values, so W is divisible by 742 and Z is divisible by 297. By equation #3, W is divisible by 42. 2226 (3x742) is the smallest number divisible by both 742 and 42, so W is divisible by 2226. Let's try some W's:
``` W = 2226 4452 6677 8903 ...
X = 1602 3204 4806 6408 ...
Y = 1580 3160 4740 6320 ...
Z = 891 1782 2673 3564 ...```
The second column is twice the first; the third is three times the first, etc. For each of these columns, we can plug in the values of W, X, Y, and Z into equations #4 through #7. So we have four equations with four unknowns, and we should be able to solve for w, x, y, and z. Choosing the first column, we get these four equations:
• w = (7/12)(1602+x)
• x = (9/20)(1580+y)
• y = (11/30)(891+z)
• z = (13/42)(2226+w)
Solving these four equations for z, we get:
• w = 7206360 / 4657
• x = 4893246 / 4657
• y = 3515820 / 4657
• z = 5439213 / 4657
after reducing z to its lowest terms. So for these four number to be positive integers, W, X, Y, and Z must be 4657 times the first column above. W = 2226 x 4657 = 10366482. And the rest of the number follow.
Problems which call for integer solutions are called Diophantine problems. |
### Precalculus and Calculus
```Students Mentoring Students Presents:
Ruben Sanchez
[email protected]
BEFORE WE START:
Do NOT be afraid to ask
questions.
• There are no dumb
questions!
• The only dumb thing to
do is not ask for help
when you are stuck.
Looking at Linear
Equations and Graphs
• There are many different
types of graphs but the
most common ones we
see, look something like
parent function graphs.
• To the right, there are
various examples of
some of the parent
functions that exist.
So what do these
functions mean?
•
All these functions can be used to
express information gathered or it
represents a relationship between
2 variables and then plotted on a
graph, in this case the Cartesian
Coordinate System.
•
Information that can be read off of
graphs include:
•
Distances
•
Velocity
•
Acceleration
•
Money/Interest Earned
Lets look at examples
on how to graph on the
coordinate system.
There are 2 main ways to
graph on a coordinate
plane.
1. Point- Slope form
2. Slope intercept form
We will look at both and
see how they are related.
We will also see how to
convert from point- slope
to slope intercept
Lets say they give us a point of ( 8, 6) and
that we are given a slope of 5.
Once we have our information, all we do is plug our
variables into the equation to the left.
We are going to plug in the coordinates
Into the x and y with sub-script 1.
We just plug in our variables where it asks us to in the formula and that is all .
Once you do that we have our answer, a point-slope formula!
Point Slope Form
In Point slope Form we are given a point on the graph. And the slope of a line. When
we have these 2 main components we are able to plug into an equation to find an
equation for a line.
1)
Point (7,8)
Slope 6
3) Point (2, 8)
Slope 5
2)
Point (3,9)
Slope 2
4) Point (5,2)
Slope 8
5) Point (4,4)
Slope 9
Practice with Point-Slope Form
Practice putting these points/ slope examples into the point slope formula. Only set
up.
Point-Slope
What do we have to do to make it look like this?
Isolate Y! Y wants to be on its own!
So first we distribute the right side to
make things easier and we get
Then we add 6 on the left to leave Y by itself.
Remember the rule, whatever we do to one
side, we do to the other. Afterwards Y is left by
itself and we reach our answer.
Changing Point-Slope to Slope Intercept Form
Once we have our equation in point slope form, simple algebra is used to convert this
equation to slope intercept form, making it possible to graph on graph calculators.
Change from point-slope to slope-intercept
Using and understanding the Pythagorean Theorem
Pythagorean theorem deals with measurement of triangle sides and simple algebra
used to find lengths of sides. NOTE- this method only works when you have a right
triangle, a triangle with at least one angle measuring 90 degrees.
Take the Square root of both sides
Example:
Let us find the length of the hypotenuse for the triangle MLO above. We are given
the side lengths but missing the hypotenuse. All we have to do now is substitute into
the equation a2+b2=c2
Main Street
Commercial
Street
4 Miles
Park Avenue 3 Miles
Real Life Example with 90 Degree Triangles
If Park Ave. is 3 miles long and Commercial Street is 4 miles long, how long is Main
Street? (hint- set up Park Ave. as side A, and Commercial St. as side B)
Trowbridge Drive
6 Miles
Paisano
Drive
Montana Street
4 Miles
Real Life Example with 90 Degree Triangles
If Trowbridge is 6 miles long and Montana Street is 4 miles long, how long is Paisano
Street? (hint- set up Montana as side A, and Trowbridge Drive as side C )
Didn’t see this coming!
Let us say that the top bar (olive) of the bicycle is 3 feet, and the bar that goes down
from the seat to the pedals (violet) is 2 feet. How long is the bar that that connects
the handlebars to the pedals (blue)?
More real life applications with pre-calculus
We found out that right triangles could be used with distances or lengths but what
about other useful math terms or calculations? We can use perimeters and area and
relate it to real life applications. Fore example the length of track, perimeter of a
farm, etc…
circle is 25 Meters
Find the total distance around the track
What we need to do is find the “perimeter” of the track to find out the length of one
lap around the field. How do we do this problem?
Put a fence around this barn
The owner of this barn decides to put up a fence around his bar. Looking at it from a top
view it makes a rectangle measuring 35 feet long and 20 feet wide. If the farmer wants to
have “extra” space and put the fence 5 feet away from the barn on every side,
what will be the amount of fence the farmer will have to buy?
What will be the new total area of the enclosed land including the barn?
What is the Perimeter of the cardboard box?
If each can has a diameter of 2.5 inches, and the box is 4 cans wide and 6 cans long,
what is the perimeter of the cardboard? What is the area of the base of the
cardboard box?
If each side of the parking aisle can
Park 9 cars, how many cars can 8
Aisles hold?
Multiplication practice
How many cars can be parked if each parking aisle can park 9 cars on each side?
What if there are a total of 8 aisles?
How many desks will fit?
How many desks will fit if the desks are 3 feet wide and 4 feet long and the room
measures 16 feet by 21. The teacher wants to leave a gap of 3 feet in between each
row of desks.
How many test tubes do you need?
The doctors office you work for needs to order test tubes but the Doctor doesn’t
know how many test tubes to order for the 15 racks he ordered. Each rack’s
dimensions are 5 by 12 slots. How many test tubes does he need to order to fill each
slot on the racks?
How much does it cost to fill up?
Lets use the example with the man on the bicycle, if you are to pull up in a big V8
truck that requires you to fuel up with premium grade fuel, how much would it cost
to fill up the 24 Gallons the truck has? What if you are in a Prius and only have to fill
up 10 Gallons of Regular grade fuel?
Rates including speed
Nascar engines are built to last and endure very high speeds and accelerations. These
fine tuned engines can reach top speeds of 200+ miles per hour (MPH). How long
(time) would it take to complete 550 miles at a race? (set up as an equation)
Rates with distances
The Tour de France typically comprises 20 professional teams of nine riders each and
covers some 2,200 miles of flat and mountainous country, mainly in France, with
occasional and brief visits to Belgium, Italy, Germany, and Spain. How long would it
take to complete the race at a constant pace of 15 miles per hour?
Proportions
If 250 songs take up 1 gigabyte of space, how many songs can fit in modern day
electronics such as iPod’s/ iPad’s/ iPhone’s/ etc… that can hold 16 gigabytes? How
Rates
The average rate that water comes out from a residential outlet, or hose, is about
200 gallons per hour. At this rate, how long would it take to fill up a pool of 22,000
gallons?
Basic Calculus
Calculus can be used in many different applications. One key application that this
field of math is used in is Physics and Engineering. Others may use calculus to
calculate distances or positions. Whatever you wish to apply it to, we will cover the
basics to understand how to carry out processes in calculus.
Lets practice an example.
Derivatives
In our real life applications, derivatives can be understood as a rate of change. One way
we can think of a rate of change that we may not even notice is when we drive, miles per
hour. This is an example of a rate of change. Derivatives don’t always have a simple chart
that we can refer to but for integrals, we have charts that can help us set up the problem.
Let’s take the derivative of each function
Integrals
Here for integrals, there are some set integral guides that can help us set up the
problem and
In most cases and in textbooks, the constant shown here is is written as a “K” or a “C”
Integrals
Lets practice an example solving integrals.
Let us solve the definite integral
``` |
# Long Division Alternative: The Area or Box Method
Long division is often considered one of the most challenging topics to teach. Luckily, there are strategies that we can teach to make multi-digit division easier to understand and perform.
The Box Method, also referred to as the Area Model, is one of these strategies. It is a mental math based approach that will enhance number sense understanding. Students solve the equation by subtracting multiples until they get down to 0, or as close to 0 as possible.
If you plan on teaching the partial quotients strategy in your classroom (which I highly recommend) the Box Method is a great way to get started. It uses the same steps as partial quotients, but is organized a bit differently.
Let’s learn how to perform the Box Method/Area Model for long division!
Wait! Are you looking for the Area Model for multiplication rather than division? Find it HERE.
Below, I have included both a video tutorial and step-by-step instructions.
## AREA MODEL/BOX METHOD FOR LONG DIVISION: STEP-BY-STEP INSTRUCTIONS
Suppose that we want to solve the equation 324÷2.
Step 1:
First we draw a box. We write the dividend inside the box, and the divisor on the left side.
Step 2:
We want to figure out how many groups of 2 can be made from 324. We will do this in parts to make it easier. We could start by making 100 groups of 2, since we know that we have at least this many groups. So we multiply 100×2 to make 200, and then take that 200 away from 324. Now we have 124 left.
Step 3:
We make another box and carry the 124 over to it. Now let’s take away another easy multiply of 2. How about 50 groups of 2? We know that we can take out another 50 groups of 2 from 124. 50×2=100, so we take 100 from 124. Now we have 24 left.
Step 4:
We make another box and carry the 24 over to it. We know that 12 groups of 2 makes 24, so let’s write a 12 on top and take away 24 from the 24. Now we end up with 0, so we know that we are finished our equation.
Step 5:
Now we add the “parts” from the top of the boxes to find our quotient. 100+50+12=162, so we know that 324÷2=162.
## ONE MORE EXAMPLE (WITH A REMAINDER)
Let’s take a look at one more example. In this example, we will solve 453÷4.
1. First we wrote our dividend inside the box, and our divisor on the left side.
2. We took out 100 groups of 4 first. This made 400. We subtracted 400 from 453 and were left with 53.
3. We carried the 53 to the next box, and then took out another 10 groups of 4 to make 40. We took the 40 away from the 53 and were left with 13.
4. We carried the 13 over to the next box, and then took out 3 groups of 4 to make 12. We took the 12 away from the 13 and were left with 1.
5. We cannot take any more groups of 4 out, so our remainder is 1. To find our final quotient, we add 100+10+3+remainder 1 to make 113 R1.
## The Area Model for Multiplication
The area model is fantastic for multiplication as well! If you’d like to read about how to teach this in a concrete way, here’s a post you may be interested in.
These task cards give students the opportunity to practice the box method/area model for long division in a variety of different ways. Students will calculate quotients, solve division problems, figure out missing dividends and divisors, think about how to efficiently solve an equation using the box method, and more. See the Box Method Task Cards HERE or the Big Bundle of Long Division Task Cards HERE.
THE LONG DIVISION STATION
The Long Division Station is a self-paced, student-centered math station for long division. Students gradually learn a variety of strategies for long division, the box method being one of them. One of the greatest advantages to this Math Station is that is allows you to target every student and their unique abilities so that everyone is appropriately challenged. See The Long Division Station HERE.
OR SEE ALL RESOURCES |
# Find the vector equation of the line passing through (1, 2, -4) and perpendicular to the two lines $\frac{\large x-8}{\large 3} = \frac{\large y+19}{\large -16} = \frac{\large z-10}{\large 7}$ and $\frac{\large x-15}{\large 3} = \frac{\large y-29}{\large 8} = \frac{\large z-5}{\large -5}$
Toolbox:
• Vector equation of a line passing through a given point and parallel to a given vector is $\overrightarrow r=\overrightarrow a+\lambda\overrightarrow b$
• Where $\lambda\in R$
Step 1:
Let the vector equation of the line passing through the point $(1,2,-4)$ be
$\overrightarrow r=\hat i+2\hat j-4\hat k+\lambda(b_1\hat i+b_2\hat j+b_3\hat k)$------(1)
Where $b_1\hat i+b_2\hat j+b_3\hat k$ is the vector parallel to the given line.
Step 2:
Let us consider the lines $\large\frac{x-8}{3}=\large\frac{y+19}{-16}=\large\frac{z-10}{7}$-----(2)
The direction cosines of the lines are $(3,-16,7)$
The vector parallel to that line is $3\hat i-16\hat j+7\hat k$
But it is given the line (1) and (2) are perpendicular to each other.
Therefore $(b_1\hat i+b_2\hat j+b_3\hat k).(3\hat i-16\hat j+7\hat k)=0$
$\Rightarrow 3b_1-16b_2+7b_3=0$-----(3)
Step 3:
Consider the line:
$\large\frac{x-11}{3}=\large\frac{y-29}{8}=\large\frac{x-5}{-5}$----(4)
The vector which is parallel to the line is $5\hat i+3\hat j-5\hat k$
The lines (1) and (4) are perpendicular to each other.
Therefore $(b_1\hat i+b_2\hat j+b_3\hat k).(3\hat i+8\hat j-5\hat k)=0$
$\Rightarrow 3b_1+8b_2-5b_3=0$-----(5)
Step 4:
On solving equ(3) and equ(5)
$\large\frac{b_1}{\begin{vmatrix}-16 & 7\\8 & -5\end{vmatrix}}=\large\frac{b_2}{\begin{vmatrix}7 & 3\\-5 & 3\end{vmatrix}}=\large\frac{b_1}{\begin{vmatrix}3 & -16\\3& 8\end{vmatrix}}$
$\large\frac{b_1}{80-56}=\large\frac{b_2}{21+15}=\large\frac{b_3}{24+48}$
$\Rightarrow \large\frac{b_1}{24}=\large\frac{b_2}{36}=\large\frac{b_3}{72}$
Step 5:
Dividing throughout by 12
$\Rightarrow \large\frac{b_1}{2}=\large\frac{b_2}{3}=\large\frac{b_3}{6}$
Therefore direction of the vector which is parallel to line (1) is $2\hat i+3\hat j+6\hat k$
Therefore Equation of the line (1) is
$\overrightarrow r=\hat i+2\hat j-4\hat k+\lambda(2\hat i+3\hat j+6\hat k)$ |
# Fractions of Fractions
This is day 1 of multiplication of fraction by a fraction and I can already see this will dramatically increase my blogging! So much to write about (for reflection, excitement and possibly confusion). With the implementation of CCSS this year, this is new in the Investigations curriculum and I am finding some things I love about it already and some things I am struggling with just a bit.
Before this lesson, students have worked in the context of a bike race of “x” number of miles and found a fraction of the race various bikers have completed. Looked like this:
This lesson went very smoothly and I found it was more of a struggle to have them model what was happening on the fraction bar since finding the fraction of the whole number was an action they could do mentally. To some, it seemed like an unnecessary step and to be honest, I wavered between unnecessary and yet completely necessary to make their thinking visual. I knew how important it would be in fraction x fraction, so I made them construct the model of what was happening in the story.
Today we started fraction of a fraction. It incorporates the same visual image of the fraction bar, so I love that continuation from previous lessons. It did lack a context, which at first bothered me but as we continued working, and heard the discussions, I moved past that. Tomorrow, I am actually going to have them come up with a story to go along with a few problems to see if they can contextualize the math they are doing. We started with a fraction of a half and then a fraction of a third, writing the expressions (some equations) as we went:
Of course, you always have the students who fly through the work and finish early as I am walking around and having discussions with the students who need some extra help, so I asked those who finished early to think about the denominator each time. Why is the product’s denominator changing from the denominators of the factors? Did you have an idea what the denominator would be before you used the fraction bar? There thoughts were so interesting:
Absolutely LOVE all of this scratching out, changing her reasoning!
This one brings up the issue of vocabulary….fours instead of fourths, eights instead of eighths. Something I have to bring out in our discussions.
This one I struggle with because of the words double and triple. I know the number itself is doubling and tripling, but I would like to have them expand that is it happening because there is another half to split or two other thirds to split.
I love that this makes the fractions factors and products are just like whole number factors and product.
Again with the “double” word. Is it just me that struggles with this one??
I am thinking this will be one of MANY multiplication and division of fraction posts! I am just amazed at the ease the students work with the fraction bars and I like what Investigations has done thus far with these lessons. One tweak I would like made would be the directions…students are asked to “stripe 1/2 of the shaded portion” and it is becoming a tongue-twister for me 🙂 I keep saying shaded when I mean striped, minor detail but they keep correcting me!
These conversations are so rich and valuable for this understanding that it blows my mind that a teacher could just say “multiply the numerators. multiply the denominators. That is multiplication of fractions.” If I had learned fractions this way, it would have all made SO much more sense!
To be continued…
# A Fraction of our Time in Math Class…
I absolutely love fraction work with my students because there is always something interesting that leaves me pondering the whys and hows of my practice….
Being a K-5 Math Specialist for a couple years offered me the opportunity to really see the trajectory of our fraction work. Now being back in the classroom, I feel I have a much better grasp as to the work the students have previously done within our math program. In third grade, they work tremendously with halves, thirds, and sixths using polygons to represent fractions of a hexagon whole for comparison and addition/subtraction. In fourth grade, students use arrays and known equivalencies to compare and add/subtract fractions with unlike denominators by choosing an appropriate array that works for both fractions (common denominator). In addition, at each grade level, students in need of RTI enrichment, work in Marilyn Burns’ Do The Math Program which utilizes fraction strips to compare and add/subtract fractions. All of this work focuses heavily on the students’ understandings of equivalencies.
Knowing all of this still never prepares you for the power of a new model….time! I have to admit, I am a huge fan of fraction strips and array work, however today I felt the power of clocks in developing equivalencies. I have taught this lesson in previous years and to be completely honest, never really liked it. It felt contrived, like a pizza divided into slices in another form. This year I have realized it was not the context that was lending itself to the “pizza feel,” it was me.
The class began with a discussion of a blank clock face. I asked the class if the minute hand stayed at 12 and the hour hand moved to the 1, what fraction of the clock did it turn? They said 1/12 and we chatted about how we can prove that, divided it up and went from there. Next I asked if the hands were reversed, would that give us a different fraction? Some said no, some said yes and we talked about the equivalency of 5/60.
The student questions that followed took my appreciation of the clock to another level:
“Is this the same as degrees since it is a circle?”
“Could we do the fraction for a whole day (24 hours)?”
“Can we split the minutes in half to do eighths?”
“What fraction does the clock go at the time we go to lunch?”
Holy cow, how many directions could I take this lesson?? I moved forward with having the students work with partners to find all of the fractions they could represent on the clock. Then I asked them to use that model to add 1/3 and 1/4 on the clock. It was interesting to see the students who know how to “find common denominators” by multiplying the numerator and denominator by the same number were challenged to make a proof of their equivalencies on the clock face, while the students who needed the clock as a tool had it as their disposal to see that 1/4 is 3/12 and 1/3 is 4/12. That clock face immediately went from something I saw as just one more pizza, to both a tool and model at the same time in my classroom.
The follow up activity is called Roll Around The Clock (http://tinyurl.com/p8sm7wa). It has fantastic variations to the game and I have student work on the positive/negative scoring system that I will post soon, it was the perfect extension for the students who needed it!
So today, in just a fraction of time, I found a new appreciation for the analog clock and hopefully improved my practice by a fraction!
-Kristin |
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# A bag contains $4$ red, $5$ black and $6$ white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is: red or white.
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Hint: The formula that is needed to find the probability is
$P(R) = \dfrac{{n(R)}}{{n(S)}}$ ,
where $n(R)$ is no. of favorable outcome and $n(S)$ is total no. of events in the sample space.
The probability of two disjoint events $A$ or $B$ is given by
$P(AorB) = P(A) + P(B)$
It is given that the bag contains $4$ red, $5$ black and $6$ white balls.
Then the sample space is $S = \{ R,R,R,R,B,B,B,B,B,W,W,W,W,W,W\}$
Therefore, the total number of balls in the bag is $4 + 5 + 6 = 15$
That is, the total no. of event in the sample $n(S) = 15$
To find: Probability of getting a red or white ball.
Let $R$ be the event of getting a red ball, then the probability of getting a red ball is given by
$P(R) = \dfrac{{n(R)}}{{n(S)}}$ ,
where $n(R)$ is no. of favorable outcome and $n(S)$ is total no. of events in the sample space.
From the sample space we get, $n(R) = 4$
Therefore, $P(R) = \dfrac{4}{15}$
Let $W$ be the event of getting a white ball, then the probability of getting a white ball is given by
$P(W) = \dfrac{{n(W)}}{{n(S)}}$ ,
where $n(W)$ is no. of favorable outcome and $n(S)$ is total no. of events in the sample space.
Again, from the sample space we get, $n(W) = 6$
Therefore, $P(W) = \dfrac{6}{{15}}$
Let $A$ be the event of getting a red or white ball, then the probability of $A$ is given by
$P(A) = P(R) + P(W)$
Therefore, $P(A) = \dfrac{4}{{15}} + \dfrac{6}{{15}}$
Simplifying this we will get,
$\Rightarrow P(A) = \dfrac{{(4 + 6)}}{{15}}$
$\Rightarrow P(A) = \dfrac{{10}}{{15}}$
Thus, the probability of getting a red or white ball is $\dfrac{{10}}{{15}}$
Note: In this problem both the events are disjoints that is event of getting red ball and event of getting white ball are disjoint event (i.e. There is no intersection between these two events) so we used the formula $P(AorB) = P(A) + P(B)$ . If the events are not disjoint events, then we have to use the formula $P(AorB) = P(A) + P(B) - P(AandB)$ where $P(AandB)$ is intersection between the events $A$ and $B$ . |
# University of South Carolina High School Math Contest/1993 Exam/Problem 6
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
After a $p\%$ price reduction, what increase does it take to restore the original price?
$\mathrm{(A) \ }p\% \qquad \mathrm{(B) \ }\frac p{1-p}\% \qquad \mathrm{(C) \ } (100-p)\% \qquad \mathrm{(D) \ } \frac{100p}{100+p}\% \qquad \mathrm{(E) \ } \frac{100p}{100-p}\%$
## Solution
Let the unknown be $x$. Initially, we have something of price $Q$. We reduce the price by $p\%$ to $Q - Q\cdot p\% = Q - Q\frac p{100} = Q\cdot\frac{100 - p}{100}$. We now increase this price by $x\%$ to get $\left(Q\cdot\frac{100 - p}{100}\right) + \left(Q\cdot\frac{100 - p}{100}\right)\cdot x\% = \left(Q\cdot\frac{100 - p}{100}\right)\cdot(1 + x\%) = Q$ We can cancel $Q$ from both sides to get $\frac{100 - p}{100}\cdot\left(1 + x\%\right) = 1$ so $1 + x\% = \frac{100}{100 - p}$ and $x\% = \frac{p}{100 - p}$ and $x = \frac{100p}{100 - p}$, so our answer is $\mathrm{(E) \ }$.
Alternatively, select a particular value for $p$ such that the five answer choices all have different values. For instance, let $p=10$. Thus if $100$ dollars was the original price, after the price reduction, we have $90$ dollars. We need $10$ dollars. Thus, $90(1+x\%)=100 \Longrightarrow x\%=\frac{10}{90}$ and $x = \frac{100 \cdot 10}{90}$. This only matches up with answer $\mathrm{(E) \ }$ when we plug in $p = 10$.
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How trigonometry works
Monday, June 10th, 2013
I’ve never been a very mathy person, and I came to trigonometry particularly late in life—surprisingly so, considering I’m a programmer who has to draw graphics from time to time. (Guess why I started learning it.)
So, for folks like me who can’t read Greek, here’s an introduction to trigonometry.
Trigonometry largely revolves around three basic functions:
• Cosine
• Sine
• Tangent
You know these from the famous mnemonic acronym “SOHCAHTOA”, which is where I’ll start from.
The acronym summarizes the three functions thusly:
• sine = opposite / hypotenuse
• cosine = adjacent / hypotenuse
• tangent = opposite / adjacent
Very buzzwordy, and nonsensical when every time you use them, you pass in an angle. And yet, 100% correct.
The cosine, sine, and tangent functions work by creating an imaginary triangle whose hypotenuse has the given angle, and returning the ratio of two of that triangle’s sides.
Given the angle of 30° (or π × 30180 radians, or τ × 30360 radians):
All three functions create this triangle, and then return the ratio of two of its sides.
Note the proximity of the three sides to the origin.
• The opposite side is the vertical side, literally on the opposite side of the triangle from the origin.
• The adjacent side is the horizontal side, extending from the origin to the opposite side. It’s the adjacent side because it touches (is adjacent to) the origin.
• The hypotenuse is the (usually) diagonal side that extends from one end of the adjacent side (namely, from the origin) to one end of the opposite side (namely, the end that isn’t touching the other end of the adjacent side).
Let’s consider a different case for each function—namely, for each function, the case in which it returns 1.
Cosine
With the hypotenuse at 0°, there basically is no opposite side: The hypotenuse is in exactly the same space as the adjacent side, from the origin to the lines’ ends. Thus, they are equal, so the ratio is 1.
Sine
Definition: opposite / hypotenuse
With the hypotenuse at 90° (or τ/4), there basically is no adjacent side: The hypotenuse is in exactly the same space as the opposite side, from the origin to the lines’ ends. Thus, they are equal, so the ratio is 1.
Cosine and sine: What if we swap them?
Try `sin 0` or `cos τ/4`. What do you get?
Zero, of course. The 0° triangle has effectively no opposite side, so the sine of that (tri)angle is 01, which is zero.
Likewise, the 90° triangle has effectively no adjacent side, so the cosine (adjacent/hypotenuse) of that (tri)angle is 01.
Tangent
You should be able to guess what the triangle for which tangent returns 1 looks like. Go on, take a guess before you scroll down.
A 45° (tri)angle’s adjacent and opposite sides are equal, which is what makes the tangent function return 1.
Cosine and sine: The unit circle
Cosine and sine return the ratio of one side or the other to the hypotenuse.
Accordingly, the length of the hypotenuse affects the result. But, again, these functions take only an angle, so where do you tell it what hypotenuse to use? And why do these functions, on any calculator and in any programming language, return only a single number?
The trigonometric functions are defined in terms of the unit circle, which is a circle with radius 1.
If you look at the diagrams above, you’ll notice that the hypotenuse of the triangle always extends to the perimeter of the circle—that is, it’s always equal to the radius. This is no accident: The hypotenuse of the constructed triangle is the radius of the circle. And since the radius of the unit circle is 1, that means the hypotenuse of the imaginary triangle is 1.
Thus, the fractions that cosine and sine return are adjacent / 1 and opposite / 1. That’s why they return single numbers: the “/ 1” is simplified out.
From this follows the method to compute cosine or sine for an arc with a different radius: Multiply the cosine or sine by the desired radius.
Cosine and sine: What if we use an angle greater than 90°?
What happens if we take the cosine and sine of an angle like, say, 4 radians (about 230°)?
Let’s draw out the triangle:
Geometrically, the origin is 0,0. As long as we’re in the 0–90° range, no problem, because both the x (cosine) and y (sine) values in that quadrant are positive. But now we’re in negative territory.
With the hypotenuse in this quadrant, the adjacent and opposite sides are now negative numbers. `cos π` = `cos τ⁄2` is -1, and `sin (τ×3⁄4)` is likewise -1. For this triangle, they’re similarly negative, though not -1.
(Exercise: What about the other two quadrants? What are the cosine and sine of, say, 110° and 300°?)
Tangent: What if we use an angle greater than 45°?
As we saw above, if we give the tangent function an angle of τ/8, the ratio is 1. What if we go higher?
Well, then the ratio goes higher. Very quickly.
The half-curve at left is the quadrant from 0 to τ/4 (the upper-right quadrant).
The curve in the middle is the two quadrants from τ/4 to τ×34 (the entire left half of the circle).
The half-curve at right is the quadrant from τ×34 to τ (the lower-right quadrant).
In words, the tangent function returns a value from 0 to 1 (inclusive) for any angle that is a multiple of π plus or minus τ4 (45°). 0 counts (it’s 0π), as does π, as does π2 (= τ = 360°), and so on. Likewise 45°, 360-45=315°, 180-45=135°, 180+45=215°, etc.
Outside of those left and right quadrants, the tangent function curves very quickly off the chart—it approaches infinity.
(Programmer note: In some environments, there are both positive and negative values of zero, in which case `tan 0` returns positive zero and `tan π` returns negative zero. Mathematically, there is only one zero and it is neither positive nor negative.)
Tangent is the only one of the three that can barf on its input. Namely, a hypotenuse angle of τ/4 (90°) equates to the opposite (vertical) side being 1 and the adjacent (horizontal) side being 0 (as shown above for the sine function), so `tan τ⁄4` = `1/0`, which is undefined. The same goes for `tan τ3⁄4`, which equates to -10.
The tangent of an angle is its slope, which you can use to reduce an angle down to whether it is more horizontal (-1..+1), more vertical (< -1 or > +1), perfectly horizontal (0), or perfectly vertical (undefined).
As a practical matter, whenever I need to compute a slope ratio, I special-case perfectly vertical angles to result in ∞.
Cosine and sine: Width and height
From the above definitions, the practical use of cosine and sine emerges: They return the width and height of the right triangle whose hypotenuse has that angle.
As described above, these results are typically interpreted in terms of the unit circle (a circle with radius 1), meaning that the hypotenuse of the triangle is 1. Thus, if you’re working with an arc or circle with a different radius, you need to multiply your cosine or sine value by that radius.
A practical problem
For example, let’s say your friend has a 50″ TV, and you’re wondering what its width and height is. Maybe she’s moving, or giving or selling it to you, or both, so one of you is going to need to know whether and where it’ll fit.
The length of the hypotenuse is the radius of the circle; in the unit circle, it’s 1, but we’re dealing with a hypotenuse (diagonal measurement of the screen) whose length is something else. Our radius is 50″.
Next, we need the angle. No need for a protractor; TVs typically have an aspect ratio of either 16:9 (widescreen) or 4:3 (“standard”). The aspect ratio is width / height, which is the inverse of the slope ratio: the ratio that the tangent function gives us (which is opposite / adjacent, or height / width). Dividing 1 by the aspect ratio gives us the slope.
Only problem is now we need to go the opposite direction of tangent: we need to go from the slope ratio to the angle.
No problem! That’s what the `atan` (arctangent) function is for. (Each of the trigonometric functions has an inverse, with the same name but prefixed with “arc” for reasons I have yet to figure out.)
`atan` takes a slope ratio and gives us, in radians (fraction of τ), the angle that corresponds to it.
Let’s assume it’s an HDTV. (I don’t want to think about trying to move an old 50″ rear-projection SDTV.) The aspect ratio is 16/9, so the slope is 9/16 (remember, tangent is opposite over adjacent); `atan 9⁄16` is about 29–30°, or about 0.5 radians.
I promise that my choice of 30° for the first example and subsequently deciding to measure an HDTV as the example use case was merely a coincidence.
So we have our angle, 0.5 radians, and our radius, which is 50″. From this, we compute the width and height of the television:
• Take the cosine and sine of the angle. (Roughly 0.867 and 0.577, respectively, but use your calculator.)
• Multiply each of these by 50 to get the width and height (respectively) in inches. (Roughly 44″ and 29″, respectively, rounding up for interior-decorative pessimism.)
• Add an inch or two to each number to account for the frame around the viewable area of the display.
So the TV needs about 45 by 30 inches of clear space in order to not block anything.
Apple documentation search that works
Sunday, March 6th, 2011
You’ve probably tried searching Apple’s developer documentation like this:
Edit: That’s the filter field, which is not what this post is about. The filter sucks. This isn’t just an easy way to use the filter field; it’s an entirely different solution. Read on.
You’ve probably been searching it like this:
(And yes, I know about site:developer.apple.com. That often isn’t much better than without it. Again, read on.)
There is a better way.
Better than that: A best way.
Setup
First, you must use Google Chrome or OmniWeb.
Go to your list of custom searches. In Chrome, open the Preferences and click on Manage:
In OmniWeb, open the Preferences and click on Shortcuts:
Then add one or both of these searches:
For the Mac
Chrome OmniWeb
Name ADC Mac OS X Library
URL http://developer.apple.com/library/mac/search/?q=%s http://developer.apple.com/library/mac/search/?q=%@
For iOS
Chrome OmniWeb
URL http://developer.apple.com/library/ios/search/?q=%s http://developer.apple.com/library/ios/search/?q=%@
Result
Notice how the results page gives you both guides and references at once, even giving specific-chapter links when relevant. You even get relevant technotes and Q&As. No wild goose chases, no PDF mines, no third-party old backup copies, no having to scroll past six hits of mailing-list threads and Stack Overflow questions. You get the docs, the right docs, and nothing but the docs.
For this specific purpose, you now have something better than Google.
Nearest Neighbor Image Unit
Saturday, February 6th, 2010
I originally wrote this as an application using NSImage (with NSImageInterpolationNone), but decided to rewrite it as an Image Unit. So, here it is.
Ship-It Saturday: IconGrabber 2.0.1
Sunday, January 3rd, 2010
The last time I released a version of IconGrabber was only a week after Valve released Half-Life 2—way back in 2004. That game wasn’t even on my radar then, since I couldn’t run it on my PowerPC-based Mac!
Just over five years later, I’ve played all of the Half-Life 2 games and love them, and IconGrabber returns with some bug fixes and support for the new bigger icon sizes introduced in Tiger and Leopard. Version 2.0.1 is available from the IconGrabber home page.
iPhone app settings
Wednesday, January 7th, 2009
One of the ongoing debates among users of iPhone OS devices is whether an app’s settings belong in the app itself, or the Settings app.
I’m of the opinion that this wouldn’t even be a debate if it weren’t for Apple’s prescription in the iPhone HIG that every iPhone app’s settings should be in the Settings app. Mac apps don’t come with prefpanes for their preferences (with the exception of faceless background apps like Growl). Windows apps don’t, either, that I know of. GNOME and KDE apps don’t pollute Ubuntu’s Control Panel.
The iPhone is the only OS I know of whose developer recommends that app developers put their application settings in the system-wide Settings app.
As we’ve seen several times on every platform, it’s OK to break one of the local Human Interface Guidelines if and only if the violation makes the interface better.
I think this guideline is one that iPhone developers should violate flagrantly.
But there’s a problem. The iPhone doesn’t really have an icon for a Settings button. Most developers seem to use the icon that one of the frameworks apparently provides, but this isn’t the proper use of that icon. The means info, not application settings.
Another choice is the gear icon for Action buttons:
But, again, we have a conflation of functions. The button in question is not an Action button; it is a Settings button. This icon is not a Settings icon. (I suspect that most people who use the Action icon use it because it doesn’t have any particular association with “action”, either, other than Apple’s endorsement of it for that.)
The Iconfactory, wisely, chose differently in Twitterrific. I suspect that this was largely coincidence, as the Mac version of Twitterrific came first and already had a Settings icon; for the iPhone version, the developers simply used the same icon. It works well enough:
But it’s not perfect. A wrench does not say “settings”. (I offer myself as evidence: When I first saw it in the Mac version, I didn’t know it was the Preferences button.) Generally, a wrench means “edit this”, as in the context of a game.
What we need is an icon that says “settings”. Ideally, this icon should either convey the notion of a changeable state value (as the previously-favored light switch [Mac OS X through Tiger] and slider [Mac OS] did), or build on an existing association with the concept of settings.
Let’s go with the latter. I nominate the Settings app’s icon:
Familiar enough, wouldn’t you say?
That’s Apple’s version. Here’s my button-icon (a.k.a. template) version, in the 16-px size:
I tried it out in the iPhone version of Twitterrific on my iPod touch. Before and a mock-up of after:
After I created this icon, I wondered what it would look like in the Mac version of Twitterrific.
Here’s the original:
… And right away we have a problem. These buttons are already framed; my white frame will glare here.
Fortunately, that’s easy to solve. With ten seconds of work, I created a frameless version. Here’s what that looks like:
I think we could all get used to this icon. This wouldn’t have worked at all before Apple changed the icon of System Preferences to match the iPhone Settings app, but now it can.
I don’t think it’s perfect. Perhaps a real icon designer (I’m just a programmer) can refine it. But I think it’s a good first draft. I’m curious to hear your opinions; please post constructive feedback in the comments.
If you want to use this icon, go ahead. Here’s the original Opacity document , from which you can build all the many variations of the icon. (Click on Inspector, then Factories, then find the version you want in the list and click its Build button.)
“Photoshop sucks” updated
Sunday, March 23rd, 2008
Upon inspiration by a comment, I’ve just updated my rant from a couple years ago, “Photoshop sucks”, to include a list of alternatives. Topping the list, of course, is Acorn; also included are Core Image Fun House, Pixelmator, DrawIt, and Iris.
I’m very glad that there are now solutions to the problem that is Photoshop. I dislike bitching about something without a solution to offer; now I have six to offer, so that rant is now complete.
A public thank-you
Monday, March 17th, 2008
This goes out to whichever engineers at Apple fixed the Image Unit template to not suck.
Before Xcode 2.5, that template was useless. Now, it contains everything I need already set up, to the maximum extent possible.
Thank you, Apple engineers.
What to do if Core Image is ignoring your slider attributes
Wednesday, February 27th, 2008
So you’re writing an Image Unit, and it has a couple of numeric parameters. You expect that Core Image Fun House will show a slider for each of them, and it does—but no matter what you do, the slider’s minimum and maximum are both 0. Furthermore, Core Image Fun House doesn’t show your parameters’ real display names; it simply makes them up from the parameters’ KVC keys.
The problem is that you are specifying those attributes in the wrong place in your Description.plist. And yes, I know you’re specifying them where the project template had them—so was I. The template has it wrong.
One of the filter attributes that Core Image recognizes is `CIInputs`. The value for this key is an array of dictionaries; each dictionary represents one parameter to the filter. The template has all the parameter attributes in these dictionaries. That makes sense, but it’s not where Core Image looks for them.
In reality, Core Image only looks for three keys in those dictionaries:
• `CIAttributeName`
• `CIAttributeClass`
• `CIAttributeDefault`
Anything else, it simply ignores.
The correct place to put all those other keys (including `CIAttributeSliderMin`, `CIAttributeSliderMax`, and `CIAttributeDisplayName`) is in another dictionary—one for each parameter. These dictionaries go inside the `CIFilterAttributes` dictionary. In other words, the `CIFilterAttributes` dictionary should contain:
• `CIInputs` => Aforementioned array of (now very small) dictionaries
• `inputFoo` => Dictionary fully describing the `inputFoo` parameter, including slider attributes and display name
• `inputBar` => Dictionary fully describing the `inputBar` parameter, including slider attributes and display name
• `inputBaz` => Dictionary fully describing the `inputBaz` parameter, including slider attributes and display name
Finally, an example:
```<key>CIFilterAttributes</key>
<dict>
⋮
<key>CIInputs</key>
<array>
<dict>
<key>CIAttributeClass</key>
<string>CIImage</string>
<key>CIAttributeName</key>
<string>inputImage</string>
</dict>
<dict>
<key>CIAttributeClass</key>
<string>NSNumber</string>
<key>CIAttributeDefault</key>
<real>1.0</real>
<key>CIAttributeName</key>
<string>inputWhitePoint</string>
</dict>
<dict>
<key>CIAttributeClass</key>
<string>NSNumber</string>
<key>CIAttributeDefault</key>
<real>0.0</real>
<key>CIAttributeName</key>
<string>inputBlackPoint</string>
</dict>
</array>
<key>inputWhitePoint</key>
<dict>
<key>CIAttributeClass</key>
<string>NSNumber</string>
<key>CIAttributeDefault</key>
<real>1.0</real>
<key>CIAttributeDisplayName</key>
<string>White point</string>
<key>CIAttributeIdentity</key>
<real>1.0</real>
<key>CIAttributeMin</key>
<real>0.0</real>
<key>CIAttributeMax</key>
<real>1.0</real>
<key>CIAttributeName</key>
<string>inputWhitePoint</string>
<key>CIAttributeSliderMin</key>
<real>0.0</real>
<key>CIAttributeSliderMax</key>
<real>1.0</real>
<key>CIAttributeType</key>
<string>CIAttributeTypeScalar</string>
</dict>
<key>inputBlackPoint</key>
<dict>
<key>CIAttributeClass</key>
<string>NSNumber</string>
<key>CIAttributeDefault</key>
<real>0.0</real>
<key>CIAttributeDisplayName</key>
<string>Black point</string>
<key>CIAttributeIdentity</key>
<real>0.0</real>
<key>CIAttributeMin</key>
<real>0.0</real>
<key>CIAttributeMax</key>
<real>1.0</real>
<key>CIAttributeName</key>
<string>inputBlackPoint</string>
<key>CIAttributeSliderMin</key>
<real>0.0</real>
<key>CIAttributeSliderMax</key>
<real>1.0</real>
<key>CIAttributeType</key>
<string>CIAttributeTypeScalar</string>
</dict>
</dict>```
You can see how the descriptions under `CIInputs` are as terse as possible; everything besides the absolute necessities is specified in the outer dictionaries.
How to convert an alpha channel to a mask
Monday, February 18th, 2008
Updated 2008-04-17 to clarify the marked-up screenshot of the Color Matrix view. If you’ve seen this post before, check out the before and after.
So, let’s say you want to convert an image to a mask.
This is easy to do with the Color Matrix filter in Core Image.
If you’ve ever looked at the Color Matrix filter out of curiosity, you were probably frightened by its imposing array of 20 text fields:
Don’t worry. The fields are actually very simple, though not explained in the UI:
• The Red, Green, Blue, and Alpha rows each represent a component of an output pixel.
• Each column represents a component of an input pixel.
• Each cell in the component rows is a multiplier.
• Each cell in the “Bias vector” row is an addend.
(The documentation for the Color Matrix filter actually does explain it, but I like my explanation better for not using fancy math terms like “dot product”.)
So with this tool, our task is redefined like so:
How to replace the color channels of an image with the alpha channel, and set the alpha channel to all-100%
1. Set the three color-component vectors (Red Vector, Green Vector, and Blue Vector) to 0, 0, 0, 1. (In other words, multiply every input color component by 0, and the input alpha by 1, and set all three color components to that.)
2. Set the Alpha Vector to 0, 0, 0, 0. (In other words, multiply all input components by 0, and set the output alpha component to that. In other other words, set every output alpha component to 0.)
3. Set the Bias Vector also to 0, 0, 0, 1. (In other words, add 0 to all three color components, and add 1 to the alpha component.)
You can generalize this to the extraction of other channels. Let’s say you want to make a mask of the blue channel:
1. Set the three color-component vectors to 0, 0, 1, 0. (For every output color component, multiply every color component by 0, except for blue. Multiply blue by 1—i.e., don’t change it.)
2. Set the Alpha Vector to 0, 0, 0, 0. (Multiply every alpha component by 0—i.e., set every output alpha component to 0.)
3. Set the Bias Vector to 0, 0, 0, 1. (Add 1 to the alpha component. This step is invariant; you always add to the alpha component.)
To demonstrate this, here’s a red-blue gradient (shown in Acorn to visualize the gradient image’s own transparency):
If we extract the blue channel, as shown above, we get this:
Note how the red parts of the gradient are black, because we extracted the blue channel, and there was little to no blue there.
Likewise, if we extract the red channel, we get this:
In this case, the converse of the blue-channel mask.
(By the way, in case you’re wondering: No, I don’t know what caused the white pixels along the edge. It could be a Lineform bug, or a Core Image bug, or a graphics-card bug. I didn’t keep the original Lineform file for the source image, stupidly, but in case you’d like to test it on your own machine, I re-created it. Here’s a PDF of the replica; you can use image to convert it to PNG. I can confirm that I saw similar results with this image to the results with the image I used for this post.)
You can even mix up the colors of an image. Suppose we want to reverse that gradient:
1. Set the Red Vector to 0, 0, 1, 0. (In other words, replace red with blue.)
2. Set the Blue Vector to 1, 0, 0, 0. (In other words, replace blue with red.)
3. Leave the Alpha and Bias Vectors at the default values. (In other words, we’re leaving the alpha channel unchanged this time.)
So what is this good for?
Well, mainly, so you can create mask images. Several filters require these, such as the Blend with Mask filter in the Stylize category. The Color Matrix filter makes this easy, although you still have to save the mask image somewhere.
It’s even easier in Opacity, where you can create a Color Matrix filter layer, configure it using the Layer Inspector, then hide it by clicking its eye icon. This way, the filter layer won’t show up in the rendered document (or in any of its build products), but you can still use its result as the mask to another filter layer.
Opacity
Wednesday, February 13th, 2008
As you may have read on wootest’s weblog, Like Thought Software released its new image editor, Opacity, today.
Before I go any further, here’s full disclosure: The developer invited me to beta-test the app, and I did. He also gave me a free license for this purpose (the app normally costs \$89 USD). Also, I have some code in the app, because it uses IconFamily, which I contributed a patch to a long time ago.
OK, that’s everything. Now, to borrow from wootest’s disclaimer on the same topic:
Don’t confuse this as simple tit-for-tat back-scratching, though. Had I … had no involvement whatsoever, the application would still have been every bit as brilliant, and I would have come out just as strongly in favor of it.
I love this app.
Opacity is an image editor designed to enable app developers to create multiple-resolution and any-resolution graphics easily. It’s built for that specific purpose, and the Opacity website even says so. This app really is not intended for anything other than user-interface graphics.
Key points:
• It’s mostly vector-based, but it also has primitive raster tools.
• It has non-destructive Core Image filter layers, similar to Photoshop’s adjustment layers. (Contrast with Acorn, which makes you apply each filter permanently. You can’t go back and edit the filter parameters.)
• It has built-in templates for most common icon types.
Opacity has several important features over past editors:
• It has built-in support for multiple resolutions. Every Opacity document has one or more resolutions, and you can add and delete them at will.
• It has a target-based workflow. Each Opacity document is, essentially, a “project” for one image; every target in the document results in one image file in an external format, such as TIFF or IconFamily (.icns). (The application now calls these “factories”, but early betas did, in fact, call them targets, and I prefer that terminology.) You can build each target factory or all targets factories at will, and there’s an option to build all whenever you Save.
• You are not limited to the stock suite of transformations (e.g., Rotate 90°, Scale, Flip Vertical); you can make your own.
• You can create folder layers to group layers (especially filter layers) together, and these folder layers can be nested as deeply as you want.
• When configuring a Core Image filter that accepts an image as a parameter (e.g., Shaded Material, Blend with Mask, or one of the Transition or Composite filters), you can use any layer in the document—even folder layers.
Opacity is not perfect. Some things don’t quite work like you would expect: for example, vector objects do automatically appear in every resolution, but pixels that you draw or paste don’t automatically get mirrored to the other resolutions; instead, Opacity waits for your explicit say-so (the Clone Current Layer’s Pixels to Other Resolutions command). Opacity also still has a couple of major bugs: Flip Horizontal, for example, takes way too long in one document that I created. Personally, I didn’t expect it to go final this early, and I recommend that you wait until at least 1.0.1.
But those are dark linings in a silver cloud. Once all the major bugs are fixed, I believe that this app is how you will create your application’s custom toolbar and button images for the modern resolution-independent world.
How to make a 512-px version of the Network icon
Saturday, February 2nd, 2008
UPDATE 2008-01-02: Ahruman commented that you can just use NSNetwork in IconGrabber. No need to go through all these steps and fake one.
If you’ve ever needed a high-resolution version of the Network icon for anything, you may have noticed that Mac OS X does not ship with one. When you select the Network and Copy it, then create a new document from the clipboard in Preview or Acorn, the largest size available is 128-px.
Fortunately, the .Mac icon is available in 512-px, and you can easily change it into the Network icon.
You will, of course, need Leopard (for no other version of Mac OS X has 512-px icons).
1. Obtain the built-in image NSImageNamedDotMac in either Core Image Fun House or Acorn.
2. Apply a Hue Adjust filter: +5°.
3. Apply a Color Controls filter: Saturation × 0.25.
The easiest way to get the .Mac image is IconGrabber. Enter the name “NSDotMac”, then click Draw, then set the size to 512×512, then save. (Note: On an Intel Mac, you’ll need to build from source, because the pre-built version for PowerPCs doesn’t run on Intel for some reason.)
I do believe we have a record
Monday, September 24th, 2007
```pngout \ %~/Projects/@otherpeoplesprojects/growl/trunk/Core/Resources(0)
> NotifyOSX.growlStyle/Contents/Resources/sidetitle.png
In: NotifyOSX.growlStyle/Contents/Resources/sidetitle.png
In: 29644 bytes
Out: NotifyOSX.growlStyle/Contents/Resources/sidetitle.png
Out: 527 bytes
Chg: -29117 bytes ( 1% of original)```
Report-an-Apple-Bug Friday! 58
Saturday, May 12th, 2007
Slightly late because I had to devise a way to determine whether a GIF file is interlaced. (I settled on GifBuilder, in case you’re curious.) This ties in with the next two bugs; I’ll blog both at once next week.
This bug is NSImageInterlaced documented as working on half of known interlaceable types. It was filed on 2007-05-12 at 00:27 PDT.
Report-an-Apple-Bug Friday! 57
Friday, April 27th, 2007
This bug is NSFrameRectWithWidth uses the current color, not the stroke color. It was filed on 2007-04-27 at 16:17 PDT.
A novel way to reduce the size of a grayscale PNG file
Sunday, April 8th, 2007
Today, I scanned in one of my old drawings: a study of five-pointed stars that I made when I was trying to figure out how to draw a proper star (this was at the time of me working on Keynote Bingo MWSF2007 Edition, and a derivative of the same star is used in TuneTagger).
The odd thing is, after I corrected the image using Preview’s Black Point and Aperture controls (no relation to the photo-management program), the image weighed about two-fifths as much:
du -b Five-pointed\ star\ study* %~/Pictures(0)
3346498 Five-pointed star study.png
(These sizes are after pngout, but even if I re-correct the original image and save it elsewhere, it comes out 1790244 bytes long.)
Go figure.
Why Mac programmers should learn PostScript
Saturday, April 7th, 2007
I’ll follow this up with a tutorial called “PostScript for Cocoa programmers”, but today brings my list of reasons why you should care in the first place.
New utility: exif-confer
Monday, March 26th, 2007
Not too long ago, I was at the bank and decided to take this photo of a couple of magazines sitting next to each other. As you can see, I edited out the bank’s address.
I did this using Lineform. The problem is, Lineform is a vector app, so it doesn’t keep any EXIF data from the original image (most of the time, that would not make sense). In my situation, I did want to keep the EXIF info, but there’s no way to make Lineform do that.
So I wrote a command-line tool to bring EXIF properties over from one image to another image. I call this image exif-confer. Enjoy.
How to make the HP Photosmart M425 work on a Mac
Monday, March 12th, 2007
1. Get out the HP drivers CD.
2. Put it in one of these.
3. Push the button.
Silly me, trying to use a device with the Mac drivers that come with the device. Turns out it works just fine with the built-in Mac OS X drivers, either via PTP (whatever that is), or as a mass-storage device. In fact, Image Capture works the same either way.
With the HP drivers, a program called “HPCamera_PTP” would crash whenever I plugged in the camera, whether I did this in Image Capture or iPhoto. I found that switching the camera to mass-storage mode (“Disk Drive” in the USB Configuration menu) worked around that problem nicely, and Image Capture (and iPhoto) even work transparently in this mode.
Later, I was tinkering with Image Capture in some way (I forget why) and noticed that it has its own PTP driver. This gave me an idea, and having long ago uninstalled the HP uselessware, I switched the camera back to PTP mode (“Digital Camera” in the USB Configuration menu) and plugged it back in. Huzzah! It worked exactly as it did in mass-storage mode.
Kudos to Apple for making it do the Right Thing either way. Antikudos to HP for making non-functional drivers.
I also got a new scanner yesterday, a CanoScan LiDE 600F. Unfortunately, it doesn’t work without drivers. Fortunately, its drivers work. (Both devices let me use Image Capture without touching any of the apps that come with them, which I consider mandatory given the nearly-consistent asstasticity of the UIs of such apps in general.)
What’s the resolution of your screen?
Sunday, February 4th, 2007
A few weeks ago, I installed Adobe Reader to view a particular PDF, and noticed something interesting in its Preferences:
“Wow”, I thought, “I wonder how it knows that.” So I went looking through the Quartz Display Services documentation, and found it.
The function is CGDisplayScreenSize. It returns a struct CGSize containing the number of millimeters in each dimension of the physical size of the screen. Convert to inches and divide the number of pixels by it, and you’ve got DPI.
Not all displays support EDID (which is what the docs for CGDisplayScreenSize say it uses); if yours doesn’t, CGDisplayScreenSize will return CGSizeZero. Watch for this; failure to account for this possibility will lead to division-by-zero errors.
Here’s an app to demonstrate this technique:
ShowAllResolutions will show one of these windows on each display on your computer, and it should update if your display configuration changes (e.g. you change resolution or plug/unplug a display). If CGDisplayScreenSize comes back with CGZeroSize, ShowAllResolutions will state its resolution as 0 dpi both ways.
The practical usage of this is for things like Adobe Reader and Preview (note: Preview doesn’t do this), and their photographic equivalents. If you’re writing an image editor of any kind, you should consider using the screen resolution to correct the magnification factor so that a 8.5×11″ image takes up exactly 8.5″ across (and 11″ down, if possible).
“Ah,”, you say, “but what about Resolution Independence?”.
The theory of Resolution Independence is that in some future version of Mac OS X (possibly Leopard), the OS will automatically set the UI scale factor so that the interface objects will be some fixed number of (meters|inches) in size, rather than some absolute number of pixels. So in my case, it would set the UI scale factor to roughly 98/72, or about 1+⅓.
This is a great idea, but it screws up the Adobe Reader theory of automatic magnification. With its setting that asks you what resolution your display is, it inherently assumes that your virtual display is 72 dpi—that is, that your UI is not scaled. Multiplying by 98/72 is not appropriate when the entire UI has already been multiplied by this same factor; you would essentially be doing the multiplication twice (the OS does it once, and then you do it again).
The solution to that is in the bottom half of that window. While I was working on ShowAllResolutions, I noticed that NSScreen also has a means to ascertain the screen’s resolution: `[[[myScreen deviceDescription] objectForKey:NSDeviceResolution] sizeValue]`. It’s not the same as the Quartz Display Services function, as you can see; it seemingly returns `{ 72, 72 }` constantly.
Except it doesn’t.
In fact, the size that it returns is premultiplied by the UI scale factor; if you set your scale factor to 2 in Quartz Debug and launch ShowAllResolutions, you’ll see that NSScreen now returns `{ 144, 144 }`.
The Resolution-Independent version of Mac OS X will probably use CGDisplayScreenSize to set the scale factor automatically, so that on that version of Mac OS X, NSScreen will probably return `{ 98.52, 98.52 }`, `{ 96.33, 96.33 }`, or `{ 98.52, 96.33 }` for me. At that point, dividing the resolution you derived from CGDisplayScreenSize by the resolution you got from NSScreen will be a no-op, and the PDF view will not be doubly-magnified after all. It will be magnified by 133+⅓% by the UI scale factor, and then magnified again by 100% (CGDisplayScreenSize divided by NSDeviceResolution) by the app.
Obviously, that’s assuming that the app actually uses NSScreen to get the virtual resolution, or corrects for HIGetScaleFactor() itself. Adobe Reader doesn’t do that, unfortunately, so it suffers the double-multiplication problem.
So, the summary:
• To scale your drawing so that its size matches up to real-world measurements, scale by NSDeviceResolution divided by `{ 72.0f, 72.0f `}. For example, in my case, you would scale by `{ 98.52, 96.33 }` / `{ 72.0, 72.0 }` (that is, the x-axis by 98.52/72 and the y-axis by 96.33/72). The correct screen to ask for its resolution is generally `[[self window] screen]` (where self is a kind of NSView).
• You do not need to worry about HIGetScaleFactor most of the time. It is only useful for things like `-[NSStatusBar thickness]`, which return a number of pixels rather than points (which is inconvenient in, say, your status item’s content view).
A Core-Image-less Image Unit
Wednesday, January 17th, 2007
Can you imagine an Image Unit that didn’t actually use Core Image?
I just wrote one.
Well, OK, so I did use CIFilter and CIImage — you can’t get away without those. But I did not use a CIKernel. That’s right: This simple filter does its work without a kernel.
For the uninitiated, a kernel is what QuartzCore compiles to either a pixel shader or a series of vector (AltiVec or SSE) instructions. All Image Units (as far as I know) use one — not only because it’s faster than any other way, but because that’s all you see in the documentation.
But I was curious. Could an Image Unit be written that didn’t use a kernel? I saw nothing to prevent it, and indeed, it does work just fine.
The image unit that I wrote simply scales the image by a multiplier, using AppKit. I call it the AppKit-scaling Image Unit. Feel free to try it out or peek at the source code; my usual BSD license applies.
Obviously, this Image Unit shouldn’t require a Core Image-capable GPU. |
# Thread: Probability of a girl?
1. ## Probability of a girl?
Q: Find the probability of a couple having a baby girl when their fourth child is born, given that the first 3 children are girls.
A: I think the answer is 0.5. Because regardless of the situation the probability will always be 0.5 as each event is independent.
How can I solve this problem in terms of P(B|A)=P(A and B)/P(A)
2. Hello, crazydeo!
Good question!
Q: Find the probability of a couple's fourth child is a girl,
given that their first 3 children are girls.
A: I think the answer is 0.5, because regardless of the situation,
the probability will always be 0.5, as each event is independent.
How can I solve this problem in terms of: . $P(B|A)\:=\:\frac{P(A \wedge B)}{P(A)}$
We want: . $P(\text{4th is girl }|\text{ first 3 are girls}) \;=\;\frac{P(\text{[4th is girl] }\wedge\text{ [first 3 are girls]})}{P\text{(first 3 are girls})}$
The numerator is: . $P(\text{all 4 are girls}) \;=\;\left(\frac{1}{2}\right)^4 \:=\:\frac{1}{16}$
The denominator is: . $P(\text{first 3 are girls}) \:=\:\left(\frac{1}{2}\right)^3 \:=\:\frac{1}{8}$
Therefore: . $P(\text{4th is a girl }|\text{ first 3 are girls}) \;=\;\frac{\frac{1}{16}}{\frac{1}{8}} \;=\;\frac{1}{2}$
3. Just a comment on the problem--
Both crazydeo and Soroban assume that the genders of the children are (statistically) independent. I think this is very likely true, but it is not automatically so. One can imagine biological mechanisms which would result in non-independence.
(According to this web page
Houghton Mifflin Science for Families: Cricket Connections
there was an article in "Chance" which concluded there is no compelling evidence for non-independence.)
,
,
,
,
,
,
,
,
# the possibility of a couple having a daughter as a fourth child is
Click on a term to search for related topics. |
# Question 072cb
May 17, 2017
To find the slope, evaluate $f ' \left(0\right)$:
$f ' \left(0\right) = - 2$
#### Explanation:
The equation for the slope of the tangent line is:
$f ' \left(x\right) = 5 {x}^{4} + 3 {x}^{2} - 2$
We can find local maxima by computing the next derivative, setting that equal to 0, and then solving for the value(s) of x.
Compute the next derivative:
$f ' ' \left(x\right) = 20 {x}^{3} + 6 x$
Set it equal to 0:
20x^3+ 6x = 0
Factor:
$2 x \left(10 {x}^{2} + 3\right) = 0$
$x = 0 \mathmr{and} x = \pm \sqrt{\frac{3}{10}} i$
$x = 0$
Compute the next derivative:
$f ' ' ' \left(x\right) = 60 {x}^{2} + 6$
Evaluate it at $x = 0$:
$f ' ' ' \left(0\right) = 60 {\left(0\right)}^{2} + 6 = 6$
$6 > 0$, therefore, the slope is a local minimum at $x = 0$
To find the slope, evaluate $f ' \left(0\right)$:
$f ' \left(0\right) = - 2$
May 17, 2017
$- 2$
#### Explanation:
The tangent line will be the derivative of the function.
$f ' \left(x\right) = 5 {x}^{4} + 3 {x}^{2} - 2$
To find the minimum value of this line we have to find the critical points. So we find the second derivative
$f ' ' \left(x\right) = 20 {x}^{3} + 6 x$
The minimum will be when this function is equal to zero
$0 = 20 {x}^{3} + 6 x$
$0 = x \left(20 {x}^{2} + 6\right)$
From this we can see that either factor will only be equal to zero when $x = 0$.
So we plug this into the tangent line formula to get the slope at this point.
$f ' \left(0\right) = 5 {\left(0\right)}^{4} + 3 {\left(0\right)}^{2} - 2$
$f ' \left(0\right) = 0 + 0 - 2$
$f ' \left(0\right) = - 2$ |
# FAQ: How To Get S1 Standard Devi?
## How do you find S in statistics?
Standard deviation ( S ) = square root of the variance Standard deviation is the measure of spread most commonly used in statistical practice when the mean is used to calculate central tendency.
## How do I calculate the standard deviation?
1. The standard deviation formula may look confusing, but it will make sense after we break it down.
2. Step 1: Find the mean.
3. Step 2: For each data point, find the square of its distance to the mean.
4. Step 3: Sum the values from Step 2.
5. Step 4: Divide by the number of data points.
6. Step 5: Take the square root.
## What does s1 mean in statistics?
s1 is the standard deviation of sample 1. n1 is the sample size of sample 1. x2 is the mean of sample 2. s2 is the standard deviation of sample 2.
## What is the difference between σ and S?
The distinction between sigma ( σ) and ‘s ‘ as representing the standard deviation of a normal distribution is simply that sigma ( σ ) signifies the idealised population standard deviation derived from an infinite number of measurements, whereas ‘ s ‘ represents the sample standard deviation derived from a finite number of
You might be interested: Question: What Ais A Number That Is Devi Sable By 6 And 9?
## What is S 2 in statistics?
The statistic s² is a measure on a random sample that is used to estimate the variance of the population from which the sample is drawn. Numerically, it is the sum of the squared deviations around the mean of a random sample divided by the sample size minus one.
## What is the easiest way to find standard deviation?
To calculate the standard deviation of those numbers:
1. Work out the Mean (the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result.
3. Then work out the mean of those squared differences.
4. Take the square root of that and we are done!
## What is the symbol for standard deviation?
The symbol ‘σ’ represents the population standard deviation.
## What is the formula for standard deviation for grouped data?
How to calculate grouped data standard deviation? step 1: find the mid-point for each group or range of the frequency table. step 2: calculate the number of samples of a data set by summing up the frequencies.
## What does μ0 mean?
Tests whether the mean of a normally distributed population is different from a specified value. Null Hypothesis (H0): states that the population mean is equal to some value ( μ0 ) Alternative Hypothesis (Ha): states that the mean does not equal/ is greater than/ is less than μ0.
## How do I calculate mean?
The mean is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count.
You might be interested: Often asked: Mukambika Devi What Weapon Does She Carry?
## What is coding method in statistics?
Coding of data refers to the process of transforming collected information or observations to a set of meaningful, cohesive categories. It is a process of summarizing and re-presenting data in order to provide a systematic account of the recorded or observed phenomenon.
## What are the 1-VAR stats?
The variables 1 – Var Stats affects are: is the mean (average) of the elements, as returned by mean( Σx is the sum of the elements, as returned by sum( Σx² is the sum of the squares of the elements.
## How do you do statistics on a calculator?
To calculate mean, median, standard deviation, etc. Press STAT, then choose CALC, then choose 1-Var Stats. Press ENTER, then type the name of the list (for example, if your list is L3 then type 2 nd 3). If your data is in L1 then you do not need to type the name of the list.
## What does SX mean in math?
The symbol Sx stands for sample standard deviation and the symbol σ stands for population standard deviation. If we assume this was sample data, then our final answer would be s =2.71. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
4.3: What's the Weight
Difficulty Level: At Grade Created by: CK-12
What’s the Weight? – Solve for Unknowns
Teacher Notes
Given two scales, each holding one, two, or three blocks of two different types, students have to figure out the weights of the individual blocks. They do this by first identifying the scale that holds only one type of block (one unknown) and solving for the weight of that block. Next, students replace that block with its weight on the other scale, and then solve for the weight of the second block.
These problems provide introduction to the solution of systems of two equations with two unknowns. Each scale represents one equation, and can be represented symbolically. For example, a scale that shows two cubes with a total weight of 4 pounds can be displayed as x+x=4\begin{align*}x + x = 4\end{align*}.
Solutions
What’s the Weight? 1
1. Cube is 5 pounds; sphere is 1 pound
2. Sphere is 8 pounds; cylinder is 2 pounds
What’s the Weight? 2
1. Triangular prism is 6 pounds; cylinder is 9 pounds
2. Sphere is 5 pounds; triangular prism is 8 pounds
What’s the Weight? 3
1. Triangular prism is 7 pounds; cube is 2 pounds
2. Cylinder is 6 pounds; sphere is 1 pound
What’s the Weight? 4
1. Cube is 5 pounds; sphere is 3 pounds
2. Cylinder is 7 pounds; triangular prism is 2 pounds
What’s the Weight?
(Teacher page)
\begin{align*}& \mathbf{Describe:} && \text{Two scales}\\ &&& A: \ \text{2 cubes weigh 40 pounds}\\ &&& B: \ \text{1 cube and 1 cylinder weigh 6 pounds}\\ & \mathbf{My \ Job:} && \text{Figure out the weight of one cube and one cylinder.}\\ & \mathbf{Plan:} && \text{Start with} \ A. \ \text{Figure out the weight of one cube.}\\ &&& \text{In} \ B, \ \text{replace the cube with its weight.}\\ &&& \text{Figure out the weight of the cylinder.}\\ & \mathbf{Solve:} && A: \ \text{cube + cube = 4 pounds, so one cube is half of 4, or 2 pounds.}\\ &&& B: \ \text{Replace cube with 2.}\\ &&& \text{Then 2 + cylinder = 6 pounds.}\\ &&& \text{Cylinder = 6 - 2, or 4 pounds.}\\ & \mathbf{Check:} && \text{Replace each cube with 2 pounds.}\\ &&& \text{Replace the cylinder with 4 pounds.}\\ &&& \text{Check with the scale weights.}\\ &&& A: \ \text{2 + 2 = 4 pounds} \quad B: \ \text{2 + 4 = 6 pounds}\end{align*}
What’s the Weight? 1
Figure out the weight of each block.
What’s the Weight? 2
Figure out the weight of each block.
What’s the Weight? 3
Figure out the weight of each block.
What’s the Weight? 4
Figure out the weight of each block.
Extra for Experts: What’s the Weight? – Solve for Unknowns
Solutions
Extra for Experts: What’s the Weight? 1
1. Cylinder is 3 pounds; sphere is 9 pounds
2. Cube is 1 pound; cylinder is 8 pounds
Extra for Experts: What’s the Weight? 2
1. Triangular prism is 2 pounds; sphere is 6 pounds
2. Cube is 3 pounds; triangular prism is 7 pounds
Extra for Experts: What’s the Weight? 1
Extra for Experts: What’s the Weight? 2
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# How do you evaluate e^( ( 11 pi)/6 i) - e^( ( 17 pi)/12 i) using trigonometric functions?
Oct 30, 2016
${e}^{11 \frac{\pi}{6} i} - {e}^{\frac{17 \pi}{12} i} = \frac{2 \sqrt{3} + \sqrt{6} - \sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6} - 2}{4}$
#### Explanation:
Evaluating the expression is determined by using Euler's theorem
$\textcolor{b l u e}{{e}^{i x} = \cos x + i \sin x}$
e^(11pi/6i)=color(blue)(cos(11pi/6)+isin(11pi/6)
${e}^{11 \frac{\pi}{6} i} = \cos \left(\frac{12 \pi}{6} - \frac{\pi}{6}\right) + i \sin \left(\frac{12 \pi}{6} - \frac{\pi}{6}\right)$
${e}^{11 \frac{\pi}{6} i} = \cos \left(2 \pi - \frac{\pi}{6}\right) + i \sin \left(2 \pi - \frac{\pi}{6}\right)$
${e}^{11 \frac{\pi}{6} i} = \cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)$
Applying the trigonometric identities:
$\textcolor{g r e e n}{\cos \left(- \alpha\right) = \cos \alpha}$
$\textcolor{g r e e n}{\sin \left(- \alpha\right) = - \sin \alpha}$
Let us continue computing ${e}^{\frac{11 \pi}{6}}$
${e}^{11 \frac{\pi}{6} i} = \cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)$
${e}^{11 \frac{\pi}{6} i} = \textcolor{g r e e n}{\cos \left(\frac{\pi}{6}\right)} \textcolor{g r e e n}{- i \sin \left(\frac{\pi}{6}\right)}$
${e}^{11 \frac{\pi}{6} i} = \frac{\sqrt{3}}{2} - i \frac{1}{2}$
Let us compute ${e}^{\frac{17 \pi}{12} i}$
e^((17pi)/12i)=color(blue)(cos((17pi)/12)+isin((17pi)/12)
${e}^{\frac{17 \pi}{12} i} = \cos \left(\frac{12 \pi}{12} + \frac{5 \pi}{12}\right) + i \sin \left(\frac{12 \pi}{12} + \frac{5 \pi}{12}\right)$
${e}^{\frac{17 \pi}{12} i} = \cos \left(\pi + \frac{5 \pi}{12}\right) + i \sin \left(\pi + \frac{5 \pi}{12}\right)$
Applying the trigonometric identities
$\textcolor{b r o w n}{\cos \left(\pi + \alpha\right) = - \cos \alpha}$
$\textcolor{b r o w n}{\sin \left(\pi + \alpha\right) = - \sin \alpha}$
Let us continue computing ${e}^{\frac{17 \pi}{12} i}$
${e}^{\frac{17 \pi}{12} i} = \cos \left(\pi + \frac{5 \pi}{12}\right) + i \sin \left(\pi + \frac{5 \pi}{12}\right)$
e^((17pi)/12i)=color(brown)(-cos((5pi)/12)color(brown)-isin((5pi)/12)
${e}^{\frac{17 \pi}{12} i} = - \frac{\sqrt{6} - \sqrt{2}}{4} - i \frac{\sqrt{2} + \sqrt{6}}{4}$
${e}^{\frac{17 \pi}{12} i} = \frac{- \sqrt{6} + \sqrt{2}}{4} - i \frac{\sqrt{2} + \sqrt{6}}{4}$
hint:To evaluate $\cos \left(\frac{5 \pi}{12}\right) \mathmr{and} \sin \left(\frac{5 \pi}{12}\right)$
$\cos \left(\frac{5 \pi}{12}\right) = \cos \left(\frac{\pi}{6} + \frac{\pi}{4}\right)$
$\sin \left(\frac{5 \pi}{12}\right) = \sin \left(\frac{\pi}{6} + \frac{\pi}{4}\right)$
So,
color(blue)(e^(11pi/6i)-e^((17pi)/12i)
$= \frac{\sqrt{3}}{2} - i \frac{1}{2} - \left(\frac{- \sqrt{6} + \sqrt{2}}{4} - i \frac{\sqrt{2} + \sqrt{6}}{4}\right)$
$= \frac{\sqrt{3}}{2} - i \frac{1}{2} - \left(\frac{- \sqrt{6} + \sqrt{2}}{4}\right) + i \frac{\sqrt{2} + \sqrt{6}}{4}$
$= \frac{\sqrt{3}}{2} - i \frac{1}{2} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6}}{4}$
$= \frac{2 \sqrt{3}}{4} - \frac{2 i}{4} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6}}{4}$
$= \frac{2 \sqrt{3}}{4} + \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6}}{4} - \frac{2 i}{4}$
$= \frac{2 \sqrt{3} + \sqrt{6} - \sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6} - 2}{4}$ |
# How do you solve x-2/3=4/5?
Apr 29, 2016
$x = \frac{22}{15}$
To solve the equation $x$ must be on it's own on one side of the equals sign.
#### Explanation:
First, cancel out the $- \frac{2}{3}$ by adding $\frac{2}{3}$ to both sides.
To get the correct answer you must do everything you have done on one side of the equals sign to the other side of the equation.
This would mean $x = \frac{4}{5} + \frac{2}{3}$
To solve $\frac{4}{5} + \frac{2}{3}$, both fractions must have the same denominator. To get the same denominator, multiply $\frac{4}{5}$ by $3$ and multiply $\frac{2}{3}$ by $5$. This doesn't change the value of the fraction, just how it is written.
This would make:
$x = \frac{12}{15} + \frac{10}{15}$
Then simply add across the fractions.
$x = \frac{22}{15}$ |
# Proving Angle Relationships
## Presentation on theme: "Proving Angle Relationships"— Presentation transcript:
Proving Angle Relationships
Postulate 2.10 – Protractor Postulate Given ray AB and a number r between 0 and 180, there is exactly one ray with endpoint A extending on either side of ray AB, such that the measure of the angle formed is r.
Proving Angle Relationships
Postulate 2.11 – Angle Addition Postulate If R is in the interior of PQS, then mPQR + mRQS = m PQS. If mPQR + mRQS = mPQS, then R is in the interior of PQS.
and is a right angle, find
QUILTING The diagram below shows one square for a particular quilt pattern. If and is a right angle, find Answer: 50 Example 8-1c
Proving Angle Relationships
Theorem 2.3 – Supplement Theorem If two angles form a linear pair, then they are supplementary angles. Theorem 2.4 – Complement Theorem If the noncommon sides of two adjacent angles form a right angle, then the angles are complementary angles.
are complementary angles and . and If find
Proving Angle Relationships
Theorem 2.5 – Angle Congruence Theorem Congruence of angles is reflexive, symmetric, and transitive. Reflexive: 1 1 Symmetric: If 1 2, then 2 1. Transitive: If 1 2 and 2 3, then 1 3.
Proving Angle Relationships
Theorem 2.6 Angles supplementary to the same angle or to congruent angles are congruent. Theorem 2.7 Angles complementary to the same angle or to congruent angles are congruent. Vertical Angles Theorem If two angles are vertical angles, then they are congruent.
In the figure, NYR and RYA form a linear pair, AXY and AXZ form a linear pair, and RYA and AXZ are congruent. Prove that RYN and AXY are congruent. Example 8-3c
2. If two s form a linear pair, then they are suppl. s.
Proof: Statements Reasons 1. Given 2. If two s form a linear pair, then they are suppl. s. 3. Given 4. 1. 2. 3. linear pairs. Example 8-3d
If and are vertical angles and and
find and If and are vertical angles and and Answer: mA = 52; mZ = 52 Example 8-4c
Proving Angle Relationships
Theorem 2.9 Perpendicular lines intersect to form four right angles. Theorem 2.10 All right angles are congruent. Theorem 2.11 Perpendicular lines form congruent adjacent angles.
Proving Angle Relationships
Theorem 2.12 If two angles are congruent and supplementary, then each angle is a right angle. Theorem 2.13 If two congruent angles form a linear pair, then they are right angles. |
Lesson Explainer: Properties of Combinations | Nagwa Lesson Explainer: Properties of Combinations | Nagwa
# Lesson Explainer: Properties of Combinations Mathematics • Third Year of Secondary School
## Join Nagwa Classes
In this explainer, we will learn how to use the properties of combinations to simplify expressions and solve equations.
A combination is a selection of items chosen without repetition from a collection of items in which order does not matter. The key difference between a combination and a permutation is the idea that order does not matter. For a permutation, order matters. Consider counting the number of ways we can assign the role of president and vice president to a group of 5 people: Mona, Amer, Samar, Bassem, and Dalia. If we choose Mona, then Dalia, this would not be the same as Dalia, then Mona since the first choice would be the president and the second the vice president. However, if we just wanted a committee of two people, it would not matter if we chose Mona, then Dalia or Dalia, then Mona. Hence, counting with permutations results in us overcounting the number of possible choices if order does not matter. In fact, we overcount by a factor of exactly . Therefore, we can define the number of combinations from as the number of permutations of divided by .
### Definition: Number of Combination of a Given Size
The number of combinations of size taken from a collection of items is given by
The notation can be read as -- or as choose and is also referred to as the binomial coefficient. Another extremely common notation for is ; however, there are also various other forms of notation commonly used such as , , , and .
This explainer will focus on the key properties of and how we can apply these to simplify expressions and solve equations. We begin by looking at an example where we use the formula to evaluate an expression involving combinations.
### Example 1: Evaluating Combinations
Determine the value of without using a calculator.
Recall that
Substituting and , we have
Similarly, substituting in and , we have
Substituting these into the given expression, we get
Using rules of fractions, we can rewrite this as
Canceling the common factor of , we have
Since , we can simplify this further to get
To solve the previous example, we could have simply used the combinations function on our calculator to evaluate the expression. However, growing in fluency in manipulating the formulae for permutations and combinations will give us the necessary skills we need to tackle more challenging problems.
Let’s consider an example where we find an unknown from an equation that involves a permutation and a combination.
### Example 2: Equality of Combinations and Permutations
If , find the value(s) of .
Recall, from the definition of combinations, that we have
Substituting this into the given equation results in
Cross multiplying by and dividing by , we can rewrite this as
We might be temped to immediately jump to the conclusion that . However, this would only be a partial answer since, recalling the definition of the factorial, we also have that .
Notice that when , we have and when , we have
Hence, the two possible values of are 1 and 0.
In the next example, we will find the expression involving permutations which is equal to a given expression involving combinations.
### Example 3: Relationship between Combinations and Permutations
Which of the following is equal to ?
We begin by noting that . Hence, we can rewrite out the expression:
Since all we are trying to find is an expression involving permutations, we should try to express the combinations in terms of permutations. To do this, we can use the definition that to rewrite our expression as
Canceling , we have which we can also write as
Recalling the property of permutations that , we can we rewrite
Hence,
Therefore, the correct answer is C.
Thus far, we have simply used the definition and formula for to solve problems. Many problems involving combinations can be solved this way. However, oftentimes, we can solve problems in a simpler and more straightforward manor by being familiar with the properties of combinations. One such property is related to the symmetry of combinations.
Notice from the definition of that there is a symmetry about the denominator. If we substitute for in the formula, we find that we get the same expression:
This leads to the general identity for combinations.
### Identity: Symmetry of Combinations
Given positive integers and satisfying , we have
This has some interesting implications for solving equations involving with unknowns in . The next example will demonstrate one such implication.
### Example 4: Symmetry of Combinations
Find the possible values of which satisfy the equation .
Using the rule , we get that
Thus, or .
The last example demonstrated that if then or .
Let’s consider another example that requires symmetry of combinations.
### Example 5: Using the Symmetry of Combinations
If , find .
Using the property that , we can rewrite . Substituting this into the given equation, we find
This implies that or . Since the latter of these is inconsistent, we have that the only solution is .
In the next example, we will determine an unknown constant in combinations when we are given that the expressions involving combinations form an arithmetic sequence.
### Example 6: Solving Combinations Problems
Given that is an arithmetic sequence, find all possible values of .
In an arithmetic sequence, there is a constant difference between consecutive terms. Hence, the difference between the first two and last two terms will be equal and we can write
Rearranging, we get
Using the definition we can rewrite this as
Dividing by the common factor of , we have
We can now multiply through by to get
Using the property of the factorial that , we can rewrite this as
Canceling the common factors in the numerators and denominators, we have
We can now divide through by 6 to get
Expanding the parentheses, we get
By gathering like terms, we arrive at the quadratic
Solving this by factoring or by the quadratic formula yields and .
One of the other key properties of combinations is the recursive relationship:
### Formula: Recursive Relationship in Combinations
where .
To derive this formula, we can use the definition to write the left-hand side as
We would like to express this as a single fraction over the common denominator of . We can do this by multiplying the first term by and the second term by as follows:
Using the properties of factorials that , we can rewrite this as
Expressing this as a single fraction and expanding the parentheses, we have
Simplifying and using the same rule of factorials, we have as required.
We will now turn our attention to one example where we apply this property to simplify an equation.
### Example 7: Pairwise Sums of Combinations
Determine the value of .
This expression looks like it will be extremely laborious to evaluate or difficult to simplify. However, the first insight we gain is through noticing that when in the sum, we have the term . Taking this term out of the summation, we have
At this point, we can apply the recursive relationship, and simplify this to
Now we see that if we do the same thing again and take the last term out of the summation, we have
Hence,
Continuing the same method, we will eventually get to the last term in the sum, , and have the expression
Therefore, the whole expression simplifies to
For the final couple of examples, we will consider the sums of all combinations for a given .
### Example 8: Sums of Combinations
Find the value of .
Using the definition we can rewrite this expression as
Evaluating each term, we have
In the last example, we found that the sum of all combinations for is 32; it is no coincidence that this is equal to . In fact, the general rule is that the sum of all for any given is equal to . We can write this as or more concisely, we have the following identity.
### Identity: Summation of Combinations
For any positive integer , we have
This rule is maybe not so surprising when we consider the recursive relationship for each term:
Since this does not apply for or , we can rewrite the sum as
Since and , we can rewrite this expression as
Regrouping the terms, we have
Therefore, the sum of the is twice the sum of . Moreover, since , we can see that the sum of for a given will be a power of two; in particular, it will be .
Finally, we will consider the alternating sum of combinations.
### Example 9: Alternating Sums of Combinations
Find the value of .
Recall that . Using this, we see that
Evaluating each term gives us
Once again, this is a general rule, that the alternating sums of are zero: or, more concisely,
### Identity: Alternating Sum of Combinations
For any positive integer , we have
An alternative way to represent this is that the sums of odd and even terms are equal. This is not surprising when is odd due to the reflective symmetry: . However, as the previous example showed, this is also true for even .
Let’s recap a few important concepts from the explainer.
### Key Points
• The number of combinations of size taken from a set of size is given by
• Combinations have the following key properties: given positive integers and satisfying ,
• Symmetry property:
• Recursive property:
• Summation:
• Alternating sum:
• Using the definition of and its properties, we can simplify many expressions and solve equations involving combinations.
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# Find the size of two radius at once
I got my exam on Thursday, and just got a few questions left. Anyway I would aprreciate help a lot! Can anyone please help me to solve this task? You can see the picture below. The need is to finde the size of the two radius. I thought about working with cords, like the cord AC is the same size like another one. Still couldn´t really find something usefull.
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The are of $ABM_2C=\frac{1}{2}(12+r_2)(r_1+r_2)$ which is equal to area of $ABC+BCM_2$. – Babak S. Mar 25 '13 at 12:28
The key insight is that $\angle ACB=90°$.
To show this, we draw a line passing through $C$ that is tangent to both the circles $k_1,k_2$ at $C$. This is possible since the two circles are tangent to each other. Let this tangent intersect the line $AB$ at $D$.
Now, we have $DA=DC$ since $DA,DC$ are lines tangent to circle $k_1$ at $A,C$ respectively. Similarly, $DB=DC$. Combining these two equations, we have $DA=DC=DB$; therefore, $D$ is the center of the circle passing through points $A,B,C$, that is, the circumcircle of triangle $ABC$. Furthermore, $AB$ must be the diameter of this circle, and thus $\angle ACB=90°$ (by Thales' Theorem).
Once we have shown that $\angle ACB=90°$, the rest of the problem can be solved by trigonometry. However, for a more elegant approach, you can consider the following.
Let the midpoint of $AC$ be $P_1$ and the midpoint of $CB$ be $P_2$. Then $P_1M_1C$ is similar to $CAB$, and $P_2M_2C$ is similar to $CBA$. (This can be shown by labeling one of the angles - say $\angle CAB=\alpha$ and then finding the size of the rest of the angles). Once you have these relations, you can compare the lengths via similar triangles to obtain the results
$$r_2=M_2C=P_2C\times \frac {BA}{CA}=\frac{1}{2}BC\times\frac {BA}{CA}=5.625$$
and similarly
$$r_1=M_1C=P_1C\times \frac {AB}{CB}=\frac{1}{2}AC\times\frac {AB}{CB}=10$$
-
thank you vincent tjeng! – Sophia Mar 25 '13 at 13:37
@my pleasure. It was a nice problem, especially the fact that there was a right angle! – Vincent Tjeng Mar 25 '13 at 23:57
$AM_1M_2B$ is a right angle trapez from figure we can see that $$\frac{r_1+r_2}{2} AB=\frac {AB+r_1+r_2}{2}r_1=AM_1C+BM_2C+ABC$$ where $$AB=\sqrt{12^2+9^2}=15$$ $$AM_1C=6\sqrt{r_1^2-6^2}$$ $$BM_2C=9/2\sqrt{r_2^2-(9/2)^2}$$ $$ABC=54$$ so we get the system
$$\frac{r_1+r_2}{2} 15=\frac {15+r_1+r_2}{2}r_1$$ $$\frac {15+r_1+r_2}{2}r_1=6\sqrt{r_1^2-6^2}+9/2\sqrt{r_2^2-(9/2)^2}+54$$
$$15(r_1+r_2)=(15+r_1+r_2)r_1$$ $$(15+r_1+r_2)r_1=12\sqrt{r_1^2-6^2}+9\sqrt{r_2^2-(9/2)^2}+108$$
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$AC=12$, $CB=9$ and values of $AB$ can be $\{4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}$. How do you know its only $15$ unless you prove the angle is right angled.? – Inceptio Mar 25 '13 at 12:52
Since $ABC$ is right triangle only $AB=15$ is possible – Adi Dani Mar 25 '13 at 13:16
It isn't proved ! You need to prove it before you consider the length. – Inceptio Mar 25 '13 at 13:17
thank you for help! – Sophia Mar 25 '13 at 13:31
You are welcome! – Adi Dani Mar 25 '13 at 13:33
$\triangle MAC$ and $\triangle MBC$ are isosceles.
Construct a tangent $CD$ which is common at $C$.
Now $\angle DCM= \angle M_2BD=90^0$, which means $BDCM_2$ is a cyclic Quadrilateral.
Similarly, prove $ADCM_1$ is a cyclic quadrilateral.
$$\angle BCM_2=\angle CBM_2=x$$(Isosceles)
$\angle BCM_2=BDM_2=x$(Why? Since it is cyclic, $M_2B$ projects equal angles)
$$\angle M_2DC= \angle M_2BC=x$$
$\angle CDB=2x$
In quadrilateral $ADCM_1$,
$\angle M_1AC=\angle M_1CA =y$(Isosceles)
$\angle M_1DA=M_1CA=y \implies \angle CDA=2y$
$2x+2y=90^0 \implies x+y=90^0$
In $\triangle ACB$,
$\angle CAB=90-y$ and $\angle CBA=90-x$
$\angle BCA= x+y$, But we have $x+y=90^0$. Therefore, $\triangle ACB$ is right angled.
-
yes I saw that wioth isocleles. but even if it´s right angled, how can I get the radius? – Sophia Mar 25 '13 at 12:15
It is clear that $\sqrt{144+81}=15$ – Babak S. Mar 25 '13 at 12:18
@BabakS.: It doesn't work if we don't prove that. – Inceptio Mar 25 '13 at 12:19
okay i try to prove now.. – Sophia Mar 25 '13 at 12:25
i dont really know how to prove the right angle, because there is no angle given.do you have any idea? – Sophia Mar 25 '13 at 12:28 |
# IB grade 9 Math book-Chapter5
## Full text
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Vocabulary direct variation constant of variation partial variation slope rise run rate of change fi rst differences
### Graphs
There are many types of puzzles—logic puzzles, word puzzles, number puzzles, mechanical puzzles, and puzzles based on diagrams. Puzzles may be solved by methods such as guessing, trial and error, and analysing data. The solution to a puzzle may rely on patterns and relations. Being able to represent data in a variety of ways is a useful skill in puzzle solving. In this chapter, you will investigate graphs and data and develop equations that represent relations between two variables. You will also analyse and interpret data so that you can make conclusions and extensions.
## 5
Linear Relations
2 Construct tables of values, graphs, and equations.
2 Identify, through investigation, some properties of linear relations, and apply these properties to determine whether a relation is linear or non-linear.
2 Compare the properties of direct variation and partial variation in applications, and identify the initial value.
2 Determine values of a linear relation by using a table of values and by using the equation of the relation.
2 Determine other representations of a linear relation, given one representation.
2 Describe the effects on a linear graph and make the corresponding changes to the linear equation when the conditions of the situation they represent are varied.
Analytic Geometry
2 Determine, through investigation, various formulas for the slope of a line segment or a line, and use the formulas to determine the slope of a line segment or a line.
2 Determine, through investigation, connections among the
representations of a constant rate of change of a linear relation.
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### Chapter Problem
Toothpick patterns are popular puzzles. How does each pattern relate to the number of toothpicks? In this chapter, you will develop equations to represent relationships like these.
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1. In each part, decide which rational number is not equivalent to the others.
a) , 0.75, , b) 2.5, 2 , , c) 0.5, , ,
2. Express each rational number in decimal form.
a) b) c) d)
3. Express each rational number in lowest terms. a) b) c) d) 30 12 12 48 15 10 ˛˛ 3 9 12 5 35 40 ˛˛ 7 10 2 5 1 2 1 2 1 2 5 2 5 2 1 2 3 4 3 4 3 4
### Rational Numbers
, and 1.5 are equivalent rational numbers.
Often, rational numbers can be simplified, or expressed in lowest terms.
3 2 9 3 6 3 ˛˛ 9 6 11 2, ˛˛ 3 2, 3 2 , 3 2 —2 —1 0 1 2 This dot is at 11— 2 . I could also name this point 3—
2or 1.5.
This dot is at 1—3 4. I could also name this point —7
4or 1.75.
### Ratio and Proportion
The body of a 50-kg woman contains about 25 kg of water. The body of an 80-kg man contains about 48 kg of water. Find the mass of water in the bodies of a 60-kg woman and a 60-kg man. Woman:
Man:
The body of a 60-kg woman contains 30 kg of water, and the body of a 60-kg man contains 36 kg of water.
3 5 60 36 1 2 60 30 Person Ratio of Water to Mass Ratio in Simplest Form Ratio in Fraction Form 50-kg woman 25:50 1:2 —25 50 1 — 2 80-kg man 48:80 3:5 —48 80 3 — 5
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4. Write a ratio to compare each quantity to its total. Express each ratio in simplest form. a) 5 kg of potassium in 20 kg of fertilizer b) 12 g of fat in 96 g of meat
c) 12 L of water in 14 L of juice
d) 40 mL of chlorine in 850 mL of solution
5. Seven out of ten people prefer Fresh toothpaste. How many would prefer Fresh in a group of 120 people?
6. To convert from centimetres to inches, you can use the fact that a 30-cm ruler is just over 12 inches long. A person is 160 cm tall. What is the person’s approximate height, in inches?
7. The table lists the number of days with rain during July in four Canadian cities. For each city, express the number of rainy days as a percent of the 31 days in July. Round to one decimal place.
8. The three key elements in lawn fertilizer are expressed as a ratio of their percents of the total mass. For example, in 20:4:8 fertilizer, 20% is nitrogen, 4% is phosphorus, and 8% is potassium.
Calculate the mass of each element in each bag of fertilizer.
a) 10-kg bag of 20:4:8 fertilizer b) 25-kg bag of 21:7:7 fertilizer c) 50-kg bag of 15:5:3 fertilizer
Location Number of Days With Rain
Toronto, ON 10
Vancouver, BC 7
Charlottetown, PE 12
St. John’s, NL 14
### Percents
Over the track and field season, the height Fred cleared in the high jump
increased from 1.81 m to 1.96 m. He was hoping to have a 20% increase in height. Height increase 1.96 1.81
0.15 Percent increase
100% 8.3%
Fred increased his jump height by about 8.3%.
For a 20% increase, multiply by 20% or 0.2, and then add the result to the original height. 0.20 1.81 0.362
1.81 0.362 2.172
Fred would have to clear about 2.17 m to increase his jump height by 20%.
0.15 1.81
height increase
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### Direct Variation
The distance that a person can jog is related to time. If you are jogging at a constant speed of 100 m/min, how far can you jog in 10 min? in 1 h?
### Investigate
What is the relationship between distance and time?
1. Susan can jog at a steady pace of 150 m/min for the first hour.
a) Create a table showing the distance that Susan jogs in 0 min, 1 min, 2 min, and so on up to 10 min.
b) Identify the independent variable and the dependent variable. Graph this relationship. c) Describe the shape of the graph. Where does it
intersect the vertical axis?
d) Write an equation to find the distance, d, in metres, that Susan jogs in t minutes.
e) Use the equation to determine the distance that Susan can jog in 40 min.
f) Consider the distance Susan jogged in 2 min. What happens to the distance when the time is doubled? What happens to the distance when the time is tripled?
2. Trish’s steady jogging pace is 175 m/min. Repeat step 1 using Trish’s speed.
3. Reflect Describe how to develop an equation for distance when you know the average speed.
䊏 grid paper
Tools
independent and dependent variables in Chapter 2: Relations.
Makin
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The relationship between distance and time is an example of a . For example, the table shows distances travelled in various time periods at a constant speed of 10 m/s. When time is multiplied by a specific number, distance is also multiplied by the same number. Another way to describe this direct variation is to say that distance varies directly with time.
In a direct variation, the ratio of corresponding values of the variables does not change. So, if d is distance and t is time, then k, where k is called the . Multiplying both sides of the equation by t gives d kt.
For the data in the table, 10, or d 10t. The constant of variation is 10.
### Algebraic Direct Variation
The Fredrick family travels 250 km to a relative’s home. The distance,
d, in kilometres, varies directly with the time, t, in hours.
a) Find the equation relating d and t if d 43 when t 0.5. What does the constant of variation represent?
b) Use the equation to determine how long it will take the Fredricks to reach their destination.
Solution
a) Since d varies directly with t, the equation has the form d kt. To find k, substitute the given values into k .
k
86
The constant of variation represents the constant average speed, 86 km/h. The equation relating d to t is d 86t. b) Substitute d 250.
250 86t
t Divide both sides by 86.
2.9 t 250 86 43 0.5 d t d t constant of variation d t direct variation Time (s) Distance (m) 1 10 2 20 3 30 4 40 5 50
䊏 a relationship between two variables in which one variable is a constant multiple of the other
direct variation
䊏 in a direct variation, the ratio of corresponding values of the variables, often represented by k, or the constant multiple by which one variable is multiplied
if d varies directly as t, then the constant of variation, k, is given by
kd
t or d kt
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### Hourly Rate of Pay
Amir works part-time at a local bookstore. He earns \$7.50/h. a) Describe the relationship between his pay, in dollars, and
the time, in hours, he works.
b) Illustrate the relationship graphically and represent it with an equation.
c) One week, Amir works 9 h. Find his pay for that week. Solution
a) To get Amir’s pay, multiply the time worked, in hours, by \$7.50. This means that Amir’s pay, P, in dollars, varies directly with the time, t, in hours worked.
b) Method 1: Pencil and Paper
This direct variation can be modelled by the equation P 7.50t, where k 7.50 is the constant of variation.
Method 2: Use a Graphing Calculator
Use the data from the table in Method 1.
• To clear all lists, press n[MEM] to display the MEMORY menu, select 4:ClrAllLists, and press e.
• To enter the data into the lists, press qand select 1:Edit. Under list L1, enter the values for time worked, in hours. Under list L2, enter the values for pay, in dollars.
• To display the scatter plot, set up Plot1 as shown. Press zand select 9:ZoomStat. Draw the line of best fit.
• Press q, cursor over to display the CALC menu, and then select
4:LinReg(axⴙb). Enter L1, a comma, L2, and another comma. Then, press v, cursor over to display the Y-VARS menu, then select 1:FUNCTION, and then 1:Y1. Press e, and then press g.
Time Worked, t (h) Pay, P (\$)
0 0 2 15 4 30 6 45 8 60 10 75 70 60 50 40 30 20 10 0 2 4 6 Time (h) P a y (\$) 8 1012 80 t P
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• Press yto see the equation representing the relationship between the time, in hours, worked and Amir’s pay, in dollars.
Y1 7.50X
Method 3: Use Fathom
From the Tool shelf, click and drag the Case Table icon into the workspace. Name two attributes Time and Pay. Enter the data from the table in Method 1 into the appropriate cells.
From the Tool shelf, click and drag the New Graph icon into the workspace. Drag the Time attribute to the horizontal axis and the Pay attribute to the vertical axis. You will see a scatter plot of the data. From the Graph menu, select Least-Squares Line.
The equation representing the relationship between the time worked and Amir’s pay will be indicated in the space below the graph.
Pay 7.50Time
c) Interpolate from the graph. Read up from 9 h on the horizontal axis to the line. Then, read across to find that Amir’s pay is about \$68.
You can also use the equation. Substitute t 9 into P 7.50t.
P 7.50(9) 67.50
Amir’s pay for 9 h is \$67.50.
In this case, if I use the graph, I only get an approximate answer, but if I use the equation, I get an exact answer.
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### Key Concepts
Direct variation occurs when the dependent variable varies by the same factor as the independent variable.
䊏 Direct variation can be defined algebraically as k or y kx, where k is the constant of variation.
The graph of a direct variation is a straight line that passes through the origin.
Consider the two equations A 2C 5 and A 2C. Which is an example of a direct variation? Explain.
Consider the graphs of d 2t and d 3t.
a) Describe the similarities.
b) Describe the differences. Explain why these differences occur.
### Practise
For help with questions 1 and 2, see Example 1.
1. Determine the constant of variation for each direct variation. a) The distance travelled by a bus varies directly with time. The
bus travels 240 km in 3 h.
b) The total cost varies directly with the number of books bought. Five books cost \$35.
c) The volume of water varies directly with time. A swimming pool contains 500 L of water after 5 min.
C2 C2 C1 C1 y x y = kx 0 x y 12 10 8 6 4 2 0 1 2 3 4 d t d = 2t 12 10 8 6 4 2 0 1 2 3 4 d t d = 3t
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2. The cost, C, in dollars, of building a concrete sidewalk varies directly with its length, s, in metres.
a) Find an equation relating C and s if a 200-m sidewalk costs \$4500.
b) What does the constant of variation represent?
c) Use the equation to determine the cost of a 700-m sidewalk.
For help with questions 3 to 5, see Example 2.
3. Passent’s pay varies directly with the time, in hours, she works. She earns \$8/h.
a) Choose appropriate letters for variables. Make a table of values showing Passent’s pay for 0 h, 1 h, 2 h, and 3 h.
b) Graph the relationship.
c) Write an equation in the form y kx.
4. The total cost of apples varies directly with the mass, in kilograms, bought. Apples cost \$1.50/kg.
a) Choose appropriate letters for variables. Make a table of values showing the cost of 0 kg, 1 kg, 2 kg, and 3 kg of apples.
b) Graph the relationship.
c) Write an equation in the form y kx. 5. A parking garage charges \$2.75/h for parking.
a) Describe the relationship between the cost of parking and the time, in hours, parked.
b) Illustrate the relationship graphically and represent it with an equation.
c) Use your graph to estimate the cost for 7 h of parking.
d) Use your equation to determine the exact cost for 7 h of parking.
### Connect and Apply
6. The cost of oranges varies directly with the total mass bought. 2 kg of oranges costs \$4.50.
a) Describe the relationship in words.
b) Write an equation relating the cost and the mass of the oranges. What does the constant of variation represent?
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7. To raise money for a local charity, students organized a wake-a-thon where they attempted to stay awake for 24 h. At this event, the amount of money raised varied directly with the time, in hours, a participant stayed awake. Tania raised \$50 by staying awake for 16 h. a) Graph this direct variation for times from 0 h to 16 h, using pencil
and paper or technology.
b) Write an equation relating the money Tania raised and the amount of time, in hours, she stayed awake.
c) How much would she have raised by staying awake for 24 h? 8. At his summer job, Sam’s regular wage is \$9.50/h. For any overtime,
Sam earns 1.5 times his regular wage.
a) Write an equation representing Sam’s regular pay.
b) Write a separate equation representing Sam’s overtime pay. c) Sam gets a raise to \$10/h. How does this change affect the
equations?
9. At a bulk store, 0.5 kg of sugar costs \$1.29.
a) Explain why this relationship is considered a direct variation. b) Graph this relationship, using pencil and paper or technology. c) What would happen to the graph if the price increased to \$1.49
for 0.5 kg?
10. Describe a situation that could be illustrated by each graph.
a) b)
11. A bat uses sound waves to avoid flying into objects. A sound wave travels at 342 m/s. The times for sound waves to reach several objects and return to the bat are shown in the table. Set up an equation to determine the distance from the bat to the object. Then, copy and complete the table.
6 5 4 3 2 1 0 2 4 6 Time (min) Cos t (\$) 8 1012 C t 30 25 20 15 10 5 0 1 2 3 Time (s) Dis tanc e (m) 4 5 6 d t
Object Time (s) Distance (m)
Tree 0.1
House 0.25
Cliff wall 0.04
Selecting Tools
Representing
Reasoning and Proving
Communicating
Connecting Reflecting
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12. The volume of water in a swimming pool varies directly with time. 500 L of water is in the pool after 4 min.
a) Write an equation relating the volume of water and time. What does the constant of variation represent?
b) Graph this relationship using pencil and paper or technology. c) What volume of water is in the swimming pool after 20 min? d) How long will it take to fill a swimming pool that holds
115 000 L of water?
e) Describe the changes to the equation and graph if only 400 L of water is in the pool after 4 min.
13. The freezing point of water varies directly with the salt content of the water. Fresh water (no salt content) freezes at a temperature of 0°C. Ocean water has a salt content of 3.5% and freezes at 2°C.
a) Which is the independent variable? Why?
b) Write an equation relating the freezing point of water and the salt content.
c) At what temperature will water with a salt content of 1% freeze?
d) What is the salt content of water that freezes at 3°C?
### Extend
14. To convert from kilometres to miles, multiply by 0.62. Write an equation to convert miles to kilometres.
15. Determine the set of ordered pairs that lists the diameter and circumference of four different coins: a penny, a nickel, a dime, and a quarter. Does the circumference vary directly with the diameter? Explain.
16. Math ContestFrom a bag of disks numbered 1 through 100, one disk is chosen. What is the probability that the number on the disk contains a 3? Justify your answer.
17. Math ContestThe digits 2, 3, 4, 5, and 6 are used to create five-digit odd numbers, with no digit being repeated in any number. Determine the difference between the greatest and least of these numbers.
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### Partial Variation
Earth is made up of several distinct layers. Beneath the oceans, the outermost layer, or crust, is 5 km to 12 km thick. Below the continents, Earth’s crust is 25 km to 90 km thick. The mantle is about 2750 km thick. The outer core is about 2260 km thick, while the inner core has a thickness of about 1228 km. The deeper layers are hotter and denser because the temperature and pressure inside Earth increase with depth.
Depth (km) Temperature Under Oceans (°C) Temperature Under Continents (°C)
Inner Core Outer Core Crust 870°C 3700°C 4300°C 7200°C Mantle 䊏 grid paper Tools
### Investigate
What is the relationship between temperature and depth? The temperature of Earth’s crust increases by about 145°C for every kilometre below the oceans. The temperature increases by about 21.75°C for every kilometre below the continents.
1. Starting at a temperature of 10°C at the surface of Earth’s crust, make a table showing the depth and temperature of Earth’s crust under the ocean and under the continents, at depths between 0 km and 5 km.
2. a) Plot temperature versus depth for your data under the oceans. b) On the same grid, plot temperature versus depth for your
data under the continents.
3. Compare and contrast the two graphs.
4. How do these graphs differ from those you made for direct variation in Section 5.1?
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5. Consider the equation T 145d 10, where T represents the temperature, in degrees Celsius, under the oceans and d represents the depth, in kilometres.
a) Substitute d 1 and calculate T. Repeat for d 2, 3, 4, 5. Compare the results with those you obtained in step 1. b) Explain why this equation works.
c) Write a similar equation relating the temperature under the continents with depth.
6. Reflect Describe the parts of each equation and how they relate to the data in your table and graph.
The graph illustrates the total cost, C, in dollars, of a taxi fare for a distance, d, in kilometres.
The fixed cost of \$2 represents the initial meter fare. The distance travelled by taxi changes, or is variable, depending on the passenger’s destination. For this reason, the variable cost is \$0.50 times the distance.
C20.5d
fixed cost variable cost
The graph is a straight line, but it does not show a direct variation because the line does not pass through the origin. This type of
relationship is called a . Another way to describe this partial variation is to say that “C varies partially with d.”
In general, if y varies partially with x, the equation is of the form
y mx b, where m and b are constants, and • m represents the constant of variation • b represents the fixed, or initial, value of y
partial variation 5 6 4 3 2 1 0 1 2 3 Distance (km) Taxi Fare Cos t (\$) 4 5 6 C d
䊏 a relationship between two variables in which the dependent variable is the sum of a constant number and a constant multiple of the independent variable
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### Graph a Partial Variation
a) Copy and complete the table of values
given that y varies partially with x. b) Identify the initial value of y and the
constant of variation from the completed table. Write an equation relating y and x in the form y mx b.
c) Graph this relation. d) Describe the graph. Solution
a) As x changes from 0 to 1, y changes from 6 to 9. Therefore, y increases by 3 as x increases by 1.
b) The initial value of y occurs when x 0. The initial value of y is 6. As x increases by 1, y increases by 3. So, the constant of variation is 3.
Use b 6 and m 3 to obtain the equation y 3x 6. c)
d) The graph is a straight line that intersects the y-axis at the point (0, 6). The y-values increase by 3 as the x-values increase by 1.
y 15 18 21 24 27 30 12 9 6 3 0 1 2 3 4 5 6 7 8 x x y 0 6 1 9 2 12 3 15 4 18 7 27 x y 0 6 1 9 2 3 15 4 27
The pattern of increasing the y-values by 3 checks for the other values that were given.
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### School Awards Banquet
A school is planning an awards banquet. The cost of renting the banquet facility and hiring serving staff is \$675. There is an additional cost of \$12 per person for the meal.
a) Identify the fixed cost and the variable cost of this partial variation.
b) Write an equation to represent this relationship.
c) Use your equation to determine the total cost if 500 people attend the banquet.
Solution
a) The fixed cost is \$675. The variable cost is \$12 times the number of people.
b) Let C represent the total cost, in dollars. Let n represent the number of people attending.
Multiply the number of people by 12 and add 675.
C 12n 675
c) Substitute n 500.
C 12(500) 675 6675
The total cost for 500 people is \$6675.
### Key Concepts
A partial variation has an equation of the form y mx b, where b represents the fixed, or initial, value of y and m represents the constant of variation.
The graph of a partial variation is a straight line that does not pass through the origin.
y = mx + b
0
y
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Classify each graph as a direct variation, a partial variation, or neither. Justify your answer.
a) b) c)
The cost to repair a television set is made up of a service charge of \$50, which covers the travel time and gas for the repairperson, plus \$40/h. Describe the steps involved in developing a partial variation equation that relates the cost and the time required to complete the repairs.
The table models a partial variation. Discuss how you can use the table to find m and b in the equation y mx b.
### Practise
1. Identify each relation as a direct variation, a partial variation, or neither. Justify your answer.
a) y 3x b) y 2x 1
c) C 20n 500 d) d 5t
For help with questions 2 and 3, see Example 1.
2. a) Copy and complete the table of values given that y varies partially with x.
b) Identify the initial value of y and the constant of variation from the table.
c) Write an equation relating y and x in the form
y mx b.
d) Graph the relation. e) Describe the graph.
C3 C3 C2 C2 C1 C1 x y 0 5 1 10 2 3 20 4 40 x y 0 10 1 14 2 18 3 22 4 26 0 y x 0 y x 0 y x
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3. a) Copy and complete the table of values given that y varies partially with x.
b) Identify the initial value of y and the constant of variation from the table.
c) Write an equation relating y and x in the form
y mx b.
d) Graph the relation. e) Describe the graph.
For help with questions 4 and 5, see Example 2.
4. A small pizza costs \$7.00 plus \$1.50 per topping.
a) Identify the fixed cost and the variable cost of this partial variation.
b) Determine the equation relating the cost, C, in dollars, and the number of toppings, n.
c) Use the equation to determine the cost of a small pizza with five toppings.
5. A class is planning a field trip to an art gallery. The cost of renting a bus is \$250. There is an additional cost of \$4 per student for the entrance fee.
a) Identify the fixed cost and the variable cost of this partial variation.
b) Write an equation relating the cost, C, in dollars, and the number of students, n.
c) Use your equation to determine the total cost if 25 students attend.
### Connect and Apply
6. A fitness club offers two types of monthly memberships:
• membership A: \$4 per visit
• membership B: a flat fee of \$12 plus \$2 per visit
a) Graph both relations for 0 to 10 visits. b) Classify each relation as a direct variation or
a partial variation.
c) Write an equation relating the cost and the number of visits.
d) Compare the monthly membership costs. When is membership A cheaper than membership B? When is membership B cheaper than membership A? x y 0 —2 1 3 2 3 13 4 33
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7. The table shows the amount a printing company charges for advertising flyers.
a) Identify the fixed cost this company charges for producing the flyers. What do you think this amount might represent? b) Determine the variable cost for producing one flyer. Explain
how you found this.
c) Write an equation representing the price for the flyers. d) What is the cost to produce 1000 flyers?
e) How many flyers can be produced for \$280?
8. Chapter Problem Toothpick patterns are popular puzzles. Here is an example of a toothpick pattern.
a) Write an equation relating the diagram number and the number of toothpicks. Is this a partial variation? Explain.
b) Use your equation to determine the number of toothpicks in Diagram 20.
9. At the surface of a lake, a scuba diver experiences 102.4 kPa (kilopascal) of pressure. As the diver descends, the pressure increases by 101.3 kPa for every 10 m.
a) Write an equation that relates the pressure experienced by a diver and the depth that the diver has descended.
b) Divers must be aware of nitrogen narcosis, which occurs when too much nitrogen dissolves in the blood. Narcosis becomes possible when the diver is exposed to a pressure of about 400 kPa. At what depth does the danger from narcosis begin? 10. Describe a situation that might lead to
this graph.
Diagram 1 Diagram 2 Diagram 3
70 C 60 50 40 30 20 10 0 1 2 3 Time (h) Cos t (\$) 4 5 6 t
Number of Flyers, n Cost, C (\$)
0 100
100 120
200 140
300 160
Scuba divers will ascend slowly to the surface to avoid decompression sickness, or the bends. After deep or long dives, a scuba diver needs to undergo decompression by returning to the surface slowly and in stages.
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11. At 12:05 P.M., a parachutist was 8000 m above the ground. At 12:06 P.M., the parachutist was 7750 m above the ground. At 12:07 P.M., the parachutist was 7500 m above the ground. a) Graph this relation.
b) Find the average rate of descent, in metres per minute. c) Write an equation for this relation.
Achievement Check
12. A theatre company produced the musical Cats. The company had to pay a royalty fee of \$1250 plus \$325 per performance. The same theatre company also presented the musical production of Fame in the same year. For the production of Fame, they had to pay a royalty fee of \$1400 plus \$250 per performance.
a) Write an equation that relates the total royalties and the number of performances for each musical.
b) Graph the two relations on the same grid.
c) When does the company pay the same royalty fee for the two productions?
d) Why do you think the creators of these musicals would set royalties in the form of a partial variation instead of a direct variation?
### Extend
13. In Earth’s atmosphere, the speed of sound can be approximated using partial variation. The speed of sound is approximately 331 m/s at 0°C and
approximately 343 m/s at 20°C.
a) What is the approximate speed of sound at i) 30°C? ii) 30°C?
b) Jenny yells out “Hello” in a canyon when the air temperature is 10°C. It takes 1.4 s to hear her echo. How far away is the wall of the canyon? 14. A battery was recharged, remained fully charged, and
then slowly lost its charge, as shown in the table.
a) Graph the battery’s charge over time.
b) Determine an appropriate set of equations for the charge of the battery.
c) What was the remaining charge after i) 12 h? ii) 26 h? iii) 71 h?
Time (h) 0 5 10 15 20 25 30 35 40 45 50 55 60
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### 5.3
Good skiers enjoy skiing on hills with a greater because they can go faster. Ski runs are rated on a variety of factors, including the slope, or steepness. The steeper the ski run, the more challenging it is.
slope
### Slope
䊏 the vertical distance between two points
䊏 the horizontal distance between two points
run rise 䊏 grid paperTools
### Investigate A
How can you determine the steepness of a hill? The diagrams
represent ski hills.
1. Rank the hills in order of their steepness, from least to greatest. 2. A hill rises 2 m over a horizontal run of 8 m. A second hill rises
4 m over a horizontal run of 10 m. Which is the steeper hill? Explain.
3. Reflect Describe your technique for determining steepness.
### Investigate B
How can you determine the slope of any line segment?
The steepness of a line segment is measured by its slope. The slope is the ratio of the to the and is often represented by the letter m.
m rise run run rise 60 m 60 m 70 m 110 m 100 m 80 m 4 A B rise run y x 3 2 1 0 1 2 3 4 4 rise C D run y x 3 2 1 0 1 2 3 4 䊏 a measure of the steepness of a line 䊏 calculated as ——— run rise slope
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When you are looking at a graph on a Cartesian grid, read from left to right. A line segment rising from left to right has a positive slope. A line segment falling from left to right has a negative slope.
1. Consider the graph of line segment AB.
a) Is the slope positive or negative? Explain how you know. b) Determine the rise and run by counting grid units.
c) Determine the slope of the line segment AB using m . 2. Consider the graph of line segment CD.
a) Is the slope positive or negative? Explain how you know. b) Determine the rise and run by counting of grid units. c) Determine the slope of the line segment CD using m . 3. a) On a piece of grid paper, set up coordinate axes. Plot the points
A(1, 1) and D(5, 1). Join the points to form line segment AD. b) Determine the rise and the run.
c) Describe what happens when you calculate the slope of a horizontal line segment.
4. a) On the same set of axes, plot the point E(1, 5). Join points to form line segment AE.
b) Determine the rise and the run.
c) Do you think it is possible to calculate the slope of a vertical line segment? Justify your answer.
5. Reflect Describe how you can find the slope of any line segment.
### Example 1
The ramp at a loading dock rises 2.50 m over a run of 4.00 m. a) Calculate the slope of the ramp.
b) Explain the meaning of the slope. Solution
a) m
0.625
The slope of the ramp is 0.625. 2.50 4.00 rise run rise run rise run 4.00 m 2.50 m 2.50 m 2.50 m
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### Slope of Line Segments
Calculate the slope of each line
segment, where possible. Describe the direction and how it relates to the slope. a) AB b) CD c) EF d) GH Solution
a) Count the number of grid units to find the rise and the run.
m
The slope of AB is . The direction is up 3 units as you go to the right 4 units.
b) m
The slope of CD is . The direction is down 5 units as you go to the right 4 units. This is why it is negative. ˛˛ 5 4 ˛˛ 5 4 5 4 rise run 3 4 3 4 rise run A B H G D C F E 6 4 2 —2 2 4 6 y x 0 8 A B 6 4 2 —2 2 4 6 8 y x 0 run = 4 rise = 3
The rise tells me if the direction is up or down. The run tells me if the direction is right or left. So, a rise of 3 means go up 3, and a run of 4 means go right 4. C D 6 4 2 —2 2 4 6 8 y x 0 run = 4 rise = —5 A rise of 5 means go down 5, and a run of 4 means go right 4.
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c) EF is a horizontal line segment. This line segment has no rise.
m
0
The slope of EF is 0. EF does not have an up or down direction. It is flat. The rise is 0 for a run of 3.
d) GH is a vertical line segment. This line segment has no run.
m
Since division by zero is undefined, the slope of GH is undefined. EF does not have a left or right direction. The run is 0 for a rise of 8.
### Use the Slope to Find a Point
A line segment has one endpoint, A(4, 7), and slope of . Find the coordinates of another possible endpoint, B.
Solution
Method 1: Draw a Graph Plot the point A(4, 7).
Use the slope to find another point. Another possible endpoint is B(7, 2).
5 3 5 3 8 0 rise run 0 3 rise run 6 4 2 —2 2 4 6 8 y x 0 run = 3 rise = 0 E F 6 4 2 —2 2 4 6 8 y x 0 run = 0 rise = 8 H G Since ——5 3 —5 — 3 rise — run, I will go to the right 3 and down 5.
There is an infinite number of solutions. What if I had used a rise of 5 and a run of 3? or 10 and 6? or 10 and 6? A (4, 7) B (7,2) 6 4 2 2 4 6 8 8 y x 0 right 3 down 5
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Method 2: Use the Coordinates
The run is 3 and the rise is 5. Add these values to the x- and y-coordinates,
respectively, of point A. (4 3, 7 (5)) (7, 2)
Another possible endpoint is B(7, 2).
### Key Concepts
The slope, m, is a measure of the steepness of a line segment. It is calculated as m .
䊏 A line segment rising from left to right has a positive slope.
䊏 A line segment falling from left to right has a negative slope.
Kelly looked at this line segment and concluded that the slope had to be negative because the
coordinates of the points contained negative numbers. Is her reasoning correct? Explain.
A ramp rises 2 m over a run of 5 m. a) How would you change the rise
to make the slope steeper? b) How would you change the run
to make the slope steeper?
C2 C2 C1 C1 rise run 6 4 2 2 4 6 8 y x 0 rise run P (1, —2) Q (4, —9) —4 —6 —8 2 4 6 8 —2 y x 0 2 m 5 m Selecting Tools Representing
Reasoning and Proving
Communicating Connecting Reflecting Problem Solving A (4, 7) B (7,2) 6 4 2 2 4 6 8 8 y x 0
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### Practise
For help with questions 1 to 3, see Example 1.
1. Determine the slope of each object.
a) b)
2. A section of road is built with a vertical rise of 2.5 m over
a horizontal run of 152 m. Find the slope, to the nearest hundredth. 3. To be safe, a wheelchair ramp needs to have a slope no greater
than 0.08. Does a wheelchair ramp with a vertical rise of 1.4 m along a horizontal run of 8 m satisfy the safety regulation?
For help with questions 4 and 5, see Example 2.
4. For each line segment,
• count grid units to find the rise • count grid units to find the run • determine the slope
a) b)
5. Calculate the slope of each line segment, where possible. a) AB b) CD c) EF d) GH e) IJ f) KL C D 6 5 4 3 2 1 1 2 3 4 5 6 y x 0 A B 6 5 4 3 2 1 1 2 3 4 5 6 y x 0 4.4 m 3.2 m 3 m 5 m 6 4 2 —2 —4 —6 2 4 y x 0 —2 —4 J G H C I E F K L A B D
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For help with questions 6 and 7, see Example 3.
6. A line segment has one endpoint of A(3, 1). a) Plot the point A on a grid.
b) Use the slope to locate another possible endpoint B. What are the coordinates of point B?
7. A line segment has one endpoint of A(6, 2) and slope of . Find the coordinates of another possible endpoint B by adding the appropriate values to the coordinates of point A.
### Connect and Apply
8. For safety reasons, a staircase should have a slope of between 0.58 and 0.70. Determine whether each staircase is within the safe range.
a) b)
9. Given a point A(2, 5), find the coordinates of a point B so that the line segment AB has each slope.
a) b) c) 4
d) 3 e) 0 f) undefined
10. A ramp needs to have a slope of .
Determine the length of each vertical brace.
11. Slopes of roads are called grades and are expressed as percents. a) Calculate the grade of a road that rises 21 m over a run of 500 m. b) For a road to have a grade of 3%, how far does it have to rise over
a run of 600 m? 3 5 23 2 3 24 cm 28 cm 5 m 6 m 34 3 2 1 m 1 m 1 m 1 m brace 1 m
Saint John, New Brunswick has the steepest main street in Canada. King Street has an 8% grade.
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12. Roofers call the slope of a roof its pitch. Roofs have different pitch classifications, which indicate how safe they are for roofers to walk on. They are classified as shown in this table.
a) Classify each roof by its pitch. i)
ii)
b) A roof is 10 m wide and has a pitch of . Find the height. 13. Two ramps are being built with the same slope.
The first ramp is twice the height of the second ramp. Does the first ramp have to be twice as long as the second ramp? Explain. 14. A steel beam goes between the tops of two buildings that are 7 m
apart. One building is 41 m tall. The other is 52 m tall. What is the slope of the beam?
15. For safety reasons, an extension ladder should have a slope of between 6.3 and 9.5 when it is placed against a wall. If a ladder reaches 8 m up a wall, what are the maximum and
minimum distances from the foot of the ladder to the wall?
16. The Great Pyramid of Cheops has a height of about 147 m and a base width of about 230 m. How does its slope compare to a standard staircase with slope 0.7? 5 12 9 m 30 m 3 m 16 m Classification Pitch Shallow m ≤—3 12 Medium —3 12 < m ≤ 6 — 12 Steep m > —6 12
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17. In 1967, Montreal hosted Expo 67, an international fair, to celebrate Canada’s 100th birthday. Canada’s pavilion was an upside-down pyramid called Katimavik, which means meeting place in Inuktitut, the language of the Inuit. The base width is about 55 m and the height is about 18 m. Calculate the slope of the sides. Compare the slope of the sides to the slope of the Great Pyramid of Cheops, which you found in question 16.
### Extend
18. A cross-country ski area classifies its courses based on the range of slopes. If the slopes are less than 0.09, the course is classified as easy. For slopes between 0.09 and 0.18, the course is intermediate. For slopes greater than 0.18, the course is difficult. For a ski hill 10 m tall, what range of horizontal runs is appropriate for each classification?
19. A hiking trail has been cut diagonally along the side of a hill, as shown. What is the slope of the trail?
20. A regular hexagon has six sides of equal length. One is drawn on a grid as shown. Determine the slope of the line segment from the centre to the vertex indicated. Explain your reasoning.
12 m 30 m 80 m 4 2 —2 —4 2 4 x y 0 —2 —4
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21. How safe are the stairs around your home? To answer this question, carry out the following investigation.
a) For different sets of stairs around your home, measure the tread width and riser height. Try to get measurements for several sets of stairs. Record your measurements in a table, and compute the slope for each set of stairs. Draw conclusions about which set of stairs is the least safe in your home.
b) Collect data from at least five classmates. Construct a scatter plot of the data, and draw a line of best fit. Analyse your results and write a report on your findings.
22. Math ContestRailroad trains cannot go up tracks with a grade (slope) greater than 7%. To go over hills steeper than this, the railroad company builds switchbacks. How many switchbacks are needed to get to the top of a hill that is 250 m high? Assume that the maximum length of the run is 1 km. Explain your solution.
23. Math ContestThe area of the shaded region is 12 square units. What is the slope of the line through AB?
A B C 4 D E ˛˛ 2 3 2 3 ˛˛ 3 2 3 2 slope 7% slope 7% slope 7% A (—6, 0) B y x riser height tread width
The steepest railroad in the world is in the Blue Mountains of New South Wales, Australia. The maximum gradient is 122%.
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### Investigate
How can you find a rate of change from a graph?
The graph shows the average distance, in metres, that each animal or person can run in 10 s.
1. What do you think is meant by average distance?
2. Visually compare the steepness of each graph. Determine the slope of each graph. Rank the slopes from least to greatest.
Time (s) A v er age Dis tanc e (m) 150 100 50 300 350 250 200 2 4 6 8 10 12 14 16 y x 0 Cheetah (10, 311) Professional Cyclist (10, 165) Alligator (10, 155) Polar Bear (10, 111) Olympic Sprinter (10, 102)
### Slope as a Rate of Change
Asafa Powell of Jamaica set the men’s 100-m world record in Athens, Greece, on June 14, 2005. He ran 100 m in 9.77 s. His average speed can be found by dividing the distance by his time.
Average speed 10.2
Asafa’s average speed was about 10.2 m/s. This means that, on average, he covered a distance of 10.2 m per second of the race. Speed is an example of a , because it is a rate that refers to the change in distance relative to the change in time.
rate of change 100 9.77
䊏 a change in one quantity relative to the change in another quantity
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3. Calculate the speed of each animal or person as . Rank the speeds from least to greatest.
4. Reflect Describe how the rate of change relates to the graph of a relation.
### Speed
Sarah is on the soccer team and runs every morning before school. One day, she ran 5 km in 20 min.
a) Calculate the rate of change of Sarah’s distance from her starting point.
b) Graph Sarah’s distance as it relates to time.
c) Explain the meaning of the rate of change and how it relates to the graph. Solution
a) The rate of change is the
distance travelled over the elapsed time. rate of change
0.25
The rate of change is 0.25 km/min. b)
c) The rate of change is Sarah’s average running speed. It is also the slope of the graph.
6 5 4 3 2 1 10 20 d t 0 Dis tanc e (km) Time (min) 5 20 change in distance change in time distance time
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### Fuel Consumption
The graph shows the volume of gasoline remaining in a car’s tank.
a) Calculate the slope of the graph. b) Interpret the slope as a rate of change. Solution
a) Finding the rise and run by counting grid squares may not always be practical. You can also find the rise and the run from the coordinates of two points on the graph.
m
0.12
b) The rate of change of the volume of gasoline is 0.12 L/km. The car uses an average of 0.12 L of gasoline per kilometre driven. The rate of change is negative because the volume of gasoline in the tank is decreasing.
60 500 565 5000 rise run 60 70 50 40 30 20 10 200 400 600 y x 0 (0, 65) (500, 5) Distance (km) run = 500 — 0 rise = 5 — 65 V olume (L)
I can subtract the y-values to get the rise. I can subtract the x-values in the same order to get the run. I’ll subtract the coordinates of the left point from the coordinates of the right point. 60 70 50 40 30 20 10 200 400 600 y x 0 (0, 65) (500, 5) Distance (km) V olume (L)
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### Key Concepts
Rate of change is the change in one quantity relative to the change in another.
A rate of change requires units, such as kilometres per hour. 䊏 When a relation is graphed, the slope describes the rate of change. 䊏 To find the slope of a line segment
joining two points, subtract the
y-values to get the rise and
subtract the x-values in the same order to get the run.
A car travelled 400 km in 5 h. Ahmed calculated the speed as 80 km/h but, when he graphed the relation, he calculated a slope of 0.0125. What do you think Ahmed did incorrectly?
When Carlos goes to the gym, he likes to lift weights. The gym has produced graphs illustrating the weight a person should lift over a number of visits. Match each graph with the appropriate situation.
a) Begin with a small weight and slowly increase the weight at a constant rate.
b) Begin with a small weight and lift the same weight each visit. c) Start with a large weight and slowly decrease the weight lifted
each visit.
C2 C2 C1 C1 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 y x 0 = 6 — 1 = 5 run = 5 — 2 = 3 rise (6, 5) (1, 2) 0 Number of Visits W eigh t (k g) A 0 Number of Visits W eigh t (k g) B 0 Number of Visits W e igh t (k g) C 0 Number of Visits W e igh t (k g) D Selecting Tools Representing
Reasoning and Proving
Communicating
Connecting Reflecting
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### Practise
For help with questions 1 to 3, see Example 1.
1. The average adult breathes in about 37 L of air every 5 min. What is the rate of change of volume of air?
2. A teenager’s heart pumps an average of 7200 L of blood every 24 h. What is the rate of change of volume of blood?
3. A hummingbird can flap its wings an average of 1800 times every 30 s. What is the rate of change of wing flaps?
For help with questions 4 and 5, see Example 2.
4. The graph shows the height above the ground of a rock climber over time.
a) Calculate the slope of the graph. b) Interpret the slope as a rate of change.
5. The graph shows the relationship between temperature and altitude. a) Calculate the slope of the graph. b) Interpret the slope as a rate of
change.
### Connect and Apply
6. The price of a loaf of bread increased from \$1.45 in 2003 to \$1.78 in 2006. What is the average price increase per year?
7. The graph shows the height of a plant over a 2-month growth period.
Calculate the rate of change per day.
30 35 25 20 15 10 5 2000 1000 0 (200, 25) (2200, 13) T emper a tur e (°C) Altitude (m) y x 12 14 10 8 6 4 2 20 40 60 80 0 (61 ,16) (0, 2) Heigh t (cm) Time (days) 16 h t 30 25 20 15 10 5 4 8 0 (0, 26) (8, 0) Time (s) Heigh t (m) y x
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8. The table shows the approximate number of downloads of a freeware program on the Internet over a 2-month period. a) Graph the data.
b) Calculate the slope and describe it as a rate of change. c) Do you think this is a popular piece of software? Why or
why not?
9. Chapter Problem
a) Plot a graph relating the number of toothpicks to the diagram number.
b) Calculate the slope of the line through these points. c) Interpret the slope as a rate of change.
10. From age 12 to 16, girls grow at an average of 8.5 cm/year, while boys grow at an average of 9.5 cm/year. Helen and John are both 12 years old. Helen is 150 cm tall and John is 146 cm tall. Graph their heights on the same grid. When can they expect their heights to be the same?
11. A fire hose can deliver water at a maximum rate of 500 L/min. a) Plot a graph showing the maximum volume of water that a
fire hose can pour onto a fire in time spans of up to 30 min. b) Suppose two fire hoses are used. How will this affect the slope
of the graph?
12. The table shows the minimum volume of water needed to fight a typical fire in rooms of various sizes.
a) Graph the data in the table. b) Calculate the rate of change.
c) If a fire truck is pumping water at a rate of 200 L/min, how long will it take to put out a fire in a room with a floor area of 140 m2?
Floor Area (m2) Minimum Volume of Water (L)
25 39 50 78 75 117 Date Downloads Sept 3 52 000 Sept 10 70 000 Sept 17 88 000 Sept 24 106 000 Oct 1 124 000 Oct 8 142 000 Oct 15 160 000 Oct 22 178 000 Oct 29 196 000 Nov 5 214 000
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13. A large party balloon is being filled with helium at a constant rate. After 8 s, there is 2.5 L of helium in the balloon.
a) Graph this relation.
b) The balloon will burst if there is more than 10 L of helium in it. How long will it take to fill the balloon with that much helium? Mark this point on your graph.
14. The distance-time graph shows two cars that are travelling at the same time.
a) Which car has the greater speed, and by how much?
b) What does the point of intersection of the two lines represent?
15. The table shows the number of people who have a university degree in Canada.
a) Graph the data with a broken-line graph.
b) When was the rate of change relatively constant?
c) When was the rate of change different? How was it different? 16. A scuba tank holds 2.6 m3of compressed
air. A diver at a shallow depth uses about 0.002 m3per breath and takes about
15 breaths per minute.
a) How much air will the diver use in 1 min?
b) How long will the air in the tank last at this rate?
c) At a depth of 10 m, the diver is breathing compressed air at 0.004 m3per breath. How long will
the air last at this depth?
d) At the maximum depth recommended for sport diving, a diver is breathing air at a rate of 0.01 m3per breath.
How long will the air last at this depth?
Year Number of People With Degrees (millions)
1990 2.3 1995 3.0 2000 3.7 2005 4.7 360 420 480 300 240 180 120 60 0 1 2 Time (h) d t Dis tanc e (km) 4 5 6 7 3 Car A Car B Selecting Tools Representing
Reasoning and Proving
Communicating
Connecting Reflecting
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17. The table shows the number of people in Canada employed in the tourism industry by year.
a) Is the rate of change constant over the 10-year period?
b) Are the rates of change large or small relative to the total number of jobs? Explain.
Achievement Check
18. The fuel efficiency of cars is stated in litres per 100 km. Kim’s car has a fuel efficiency of 8 L/100 km. The fuel tank on Kim’s car holds 32 L. a) Graph the relationship between the amount of gasoline remaining
in the car’s tank and the distance Kim drives. Assume that she started with a full tank.
b) Find the slope of the graph. What does this slope represent? c) Kim’s car uses 25% more gas when she drives in the city. Redraw
the graph. Find and interpret the slope for driving in the city.
### Extend
19. A store is holding a special clearance sale on a \$200 coat. Initially, there is a discount of 5%. Every 2 h, an additional 5% is taken off the latest price.
a) Make a table showing the price over the 16 h the sale is in effect. b) Graph the price over the 16 h of the sale.
c) Explain the shape of the graph.
20. A cell phone company does not have a monthly fee but charges by the minute. The graph shows the monthly cost of phone calls based on the number of minutes talked. Describe this cell phone company’s rate plan.
Year Jobs (thousands)
1995 38.8 1996 37.4 1997 37.5 1998 37.9 1999 39.9 2000 41.8 2001 41.7 2002 42.0 2003 41.6 2004 43.1 120 140 160 180 200 220 240 100 80 60 40 20 0 200 Number of Minutes Mon thly Cos t (\$) 400 600 800 1000 (1000, 220) (200, 60) (100, 35) (0, 0) C m
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### Investigate
How can you use a table of values to determine if a relation is linear or non-linear?
Method 1: Use Pencil and Paper 1. Consider the relation y 3x.
a) Copy and complete the table of values. b) Graph the relation.
c) Classify the relation as linear or non-linear.
2. a) Describe the pattern in the x-values.
b) Add a third column to your table to record the change in y. Calculate each entry by subtracting consecutive values of y. The values in the third column are called .
c) What do you notice about the values in the third column? first differences
### First Differences
Immediately after jumping from an airplane, a skydiver falls toward Earth with increasing speed. How can you tell if the speed is increasing linearly or non-linearly over time?
In Chapter 2: Relations, you learned how to identify linear and non-linear relations through graphing. In this section, you will learn how to use tables of values to identify the type of relation.
x y First Differences 0 0 1 3 3 2 3 4 x y 0 0 1 3 2 3 4 䊏 grid paper Tools 䊏 differences between consecutive y-values in tables of values with evenly spaced x-values
first differences
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x y —4 —10 —1 —5 2 0 5 5 8 10 x y 0 —3 2 —1 4 3 6 9 8 17 x y 0 7 1 3 2 —1 3 —5 4 —9
3. Repeat steps 1 and 2 for each relation using a table of values with the x-values shown. a) y 2x 7
b) y x2
c) y 2x
4. Consider your results for the four relations. Make an observation regarding linear relations and first differences.
5. Use first differences to determine which of these relations are linear and which are non-linear.
a) b) c)
6. Reflect Write a rule for using first differences to determine whether a relation is linear or non-linear.
7. Reflect Describe how you can tell if the equation of a relation represents a linear relation.
Method 2: Use Technology
1. Create a table of values for y 3x using five values for
x: 0, 1, 2, 3, 4, 5.
• To clear all lists, press n[MEM] to display the MEMORY menu, select 4:ClrAllLists, and press e.
• To enter the data into the lists, press qand select 1:Edit. Under list L1, enter the x-values.
• To generate the y-values, scroll over to list L2, then up to the L2 heading. Type the expression for y, substituting the list L1 for x. Press 3 *n [L1]e.
2. Graph the relation. Is it linear?
• Press n[STATPLOT] to display the STAT PLOTS menu. Select 1:Plot1. • Press eto select On.
• For Type:, select line graph.
• Ensure that Xlist: is set to L1 and Ylist: is set to L2.
• Press zand select 9:ZoomStat.
x y 0 1 2 3 4 䊏 graphing calculator Tools
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x y —4 —10 —1 —5 2 0 5 5 8 10 x y 0 —3 2 —1 4 3 6 9 8 17 x y 0 7 1 3 2 —1 3 —5 4 —9
3. a) Describe the pattern in the x-values displayed in list L1.
b) Find the differences between successive
y-values.
• To find the first differences, scroll over and up to the L3 heading.
• Press n[LIST] to display the LIST MATH menu. • From the OPS menu, select 7:∆List(.
• Then, press n[L2] ) e.
c) What do you notice about the values in list L3? 4. Repeat steps 1 to 3 for each relation.
a) y 2x 7
b) y x2
c) y 2x
5. Consider your results in steps 2 and 3 for the four relations. Make an observation about linear relations and first differences.
6. Use first differences to determine which of these relations are linear and which are non-linear.
a) b) c)
7. Reflect Write a rule for using first differences to determine whether a relation is linear or non-linear.
8. Reflect Describe how you can tell if the equation of a relation represents a linear relation.
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x y 0 7 1 10 3 16 6 25 x y 0 7 1 10 2 13 3 16
### Key Concepts
To work with first differences, the values of x (independent variable) must change by a constant amount.
To find first differences, subtract consecutive values of y (dependent variable).
2 0 2 4 2 2 6 4 2 8 6 2
䊏 If the first differences of a relation are constant, the relation is linear. 䊏 If the first differences of a relation are not constant, the relation is
non-linear.
For each table of values, decide whether it is possible to use first differences to determine whether the relation is linear or non-linear. Explain your decision.
a) b)
Jacob’s rate of pay is \$9.50/h. If you made a table of values of Jacob’s earnings, how would his hourly wage relate to the first differences?
### Practise
1. Look at each equation. Predict whether it represents a linear relation or a non-linear relation. Use a graphing calculator to confirm your answers.
a) y 5x 6 b) y 3x 2 c) y 4x2 1 d) y 10x e) y 4x1 f) y 6 C2 C2 C1 C1 x y First Differences 0 0 1 2 2 2 4 2 3 6 2 4 8 2
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x y —5 8 —3 4 —1 0 1 —4 x y —1 1 0 0 1 1 2 4 x y 3 —4 4 —1 5 2 6 5 x y 0 5 1 6 2 8 3 12
2. Copy each table and include a third column to record first differences. Classify each relation as linear or non-linear.
a) b)
c) d)
3. Each table shows the speed of a skydiver before the parachute opens. Without graphing, determine whether the relation is linear or non-linear. a) There is no air resistance. b) There is air resistance.
### Connect and Apply
4. Use first differences to determine which relations are linear and which are non-linear. Write an equation representing each linear relation. Extrapolate the relation to predict the outcome for the seventh step. a) b) Time (s) Speed (m/s) 0 0 1 9.6 2 16.6 3 23.1 4 30.8 5 34.2 Time (s) Speed (m/s) 0 0 1 9.8 2 19.6 3 29.4 4 39.2 5 49.0
Number of Houses Number of Segments
1 2 3 4
Base Side Length Total Number of Tiles
1 2 3 4
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5. Use first differences to determine which relations are linear and which are non-linear. Write an equation representing each linear relation. Extrapolate the relation to predict the outcome for the seventh step.
a)
b)
6. Chapter Problem A pattern is made from toothpicks as shown.
a) Create a table comparing the diagram number to the number of toothpicks.
b) Use first differences to show that the pattern is a linear relation. c) Write an equation for the relation.
d) Extrapolate the relation to predict the outcome for the 10th step. 7. A rectangular piece of cardboard is 16 cm wide. It is dipped in water
and is wet from the bottom up.
a) Create a table comparing the height of the wet cardboard to its area as the height increases from 0 cm to 10 cm.
b) Use first differences to determine whether the relation is linear. c) What is the area of wet cardboard if the height is 50 cm?
Diagram 1 Diagram 2 Diagram 3
height
16 cm
Number of Sides Number of Diagonals
4 5 6 7
Number of Circles Number of Intersection Points
1 2 3 4
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### Extend
8. The triangle’s base is twice its height. The triangle is painted from the bottom up. a) Create a table comparing the height of the
painted portion to its area as the height increases.
b) Use first differences to determine whether the relation is linear. 9. A class conducted an experiment to see how high a ball
would bounce from various heights. The results of one group’s experiment are shown in the table.
Provide two or more pieces of evidence to show whether this relationship is linear or non-linear.
10. The first few figures in a pattern are shown.
a) Copy and complete the table.
b) Use Technology A graphing calculator can be used to compute first differences. Follow the steps below.
• Enter the values from column 1 (Figure Number) in list L1. • Enter the values from column 2 (Number of Circles in Pattern)
in list L2.
• Place the cursor on L3 using the cursor keys.
• Press n[LIST]. From the OPS menu, select 7:∆List(. Press n[L2] )e.
What information is in L3? Use this information to create a non-linear equation for this pattern.
Figure 1 Figure 2 Figure 3 Figure 4
Figure Number Number of Circles in Pattern
1 1 2 3 3 4 5 6 7 8 Drop Height (cm) 50 100 150 200 250 300 Bounce Height (cm) 41 82 125 166 208 254 b h
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### First Differences
You have learned to identify a linear relation from its graph, equation, and table of values. For example, from the graph of a linear relation, you can tell if it is a direct variation or a partial variation and calculate its slope. In addition, you can identify a linear relation from its table of values by calculating first differences.
Consider the distance travelled by a snail over time. Is the graph of this relationship linear? How could you find the slope?
In this section, you will learn how variation, slope, and first differences are connected.
Time, t (min) Height, h (m)
0 —3 3 1 6 5 9 9 12 13
### Investigate
How are variation, slope, and first differences connected? The table shows the height, compared to the ground, of a snail as it crawls up a pipe.
1. Graph the relation. Is this a direct variation or a partial variation? 2. Describe the pattern in the t-values. Use first differences to
confirm that the relation is linear. 3. Calculate the slope.
4. How does the slope relate to the first differences and the pattern in the t-values?
5. What is the initial value of the height? 6. Write an equation of the line.
7. Reflect Describe how first differences, slope, and partial 䊏 grid paper
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The slope of a linear relation remains constant. The first differences also remain constant when the changes in the x-values are constant. The slope, m, of a line can be calculated by dividing the change in
y by the change in x.
m
The equation of a line has the form y mx b, where m represents the slope and b represents the vertical intercept, or the value of the dependent variable where the line intersects the vertical axis.
### Fuel Consumption
The graph shows the relationship between the volume of gasoline remaining in a car’s fuel tank and the distance driven.
a) Calculate the slope and describe its meaning. b) Determine the vertical intercept.
c) Write an equation for this relation.
y2 y1 x2 x1 change in y change in x rise run
This is sometimes abbreviated as—∆y
“delta y over delta x.” The Greek letter delta is the symbol for change in.
6 7 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 (x2, y2) (x1, y1) run x2 — x1 rise y2 — y1 y x 78 91 65 52 39 26 13 0 200 400 d V Distance (km) V olume (L)
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Solution
a) Use the first two points on the line to calculate the slope. Use (x1, y1) (0, 65) and (x2, y2) (100, 52).
m
The rate of change of the volume of fuel in the tank is L/km. The car uses an average of 13 L of gasoline per 100 km driven. This is a negative quantity because the volume of gasoline is decreasing.
b) The vertical intercept is the value of V when d 0. From the graph, V 65 when d 0. Therefore, b 65. c) This is a partial variation, so its equation has the form
V md b. The equation of this relation is V .
### Slope and the Constant of Variation
Make a table of values and graph the relation y 2x 5. Draw a right triangle on your graph to find the slope. Solution m 21 rise run 13 100d 65 13 100 13 100 13 100 5265 1000 y2 y1 x2 x1
Since the slope of a linear relation is constant, I can use any pair of points and the slope will be the same. x y 1 —3 2 —1 3 1 4 3 4 2 —2 0 2 4 run = 1 rise = 2 x y
The slope is the same as the constant of variation in the equation y 2x 5.
Updating...
## References
Related subjects : |
# Euler’s Method for Making Money
Chip Rollinson is a teacher at Buckingham Browne & Nichols School in Cambridge, Massachusetts. Today’s post is a note he sent to the AP Calculus Community bulletin board that I found interesting. I will share it with you with his permission. I made some minor edits. Chip wrote on February 5, 2015:
I had an epiphany today about the relationship between Euler’s Method and compounding growth. I had never made the connection before. I thought it was cool so I decided I needed to share it.
Consider the differential equation $\frac{dA}{dt}=rA$ with the condition A(0) = P. Solving this equation gives $A\left( t \right)=P{{e}^{rt}}$, but let’s ignore this for now.
Let’s look at this differential equation using Euler’s Method.
Leonhard Euler
1707 – 1783
Let’s start with a step size of $\tfrac{1}{12}$ and use (0, P) as the “starting point.”
After one step, you arrive at the point $\displaystyle \left( \tfrac{1}{12},P+\tfrac{r}{12}P \right)=\left( \tfrac{1}{12},P\left( 1+\tfrac{r}{12} \right) \right)$
After 2 steps, you arrive at the point $\displaystyle \left( \tfrac{2}{12},P+\tfrac{2}{12}P+\tfrac{{{r}^{2}}}{144}P \right)=\left( \tfrac{2}{12},P{{\left( 1+\tfrac{r}{12} \right)}^{2}} \right)$
After 3 steps, you arrive at the point $\displaystyle \left( \tfrac{3}{12},P+\tfrac{3}{12}rP+\tfrac{3}{144}{{r}^{2}}P+P\tfrac{3}{1728}{{r}^{3}} \right)=\left( \tfrac{3}{12},P{{\left( 1+\tfrac{r}{12} \right)}^{3}} \right)$
After 4 steps, you arrive at the point $\displaystyle \left( \tfrac{4}{12},P{{\left( 1+\tfrac{r}{12} \right)}^{4}} \right)$
And so on …
After 12 steps, you arrive at the point $\displaystyle \left( 1,P{{\left( 1+\tfrac{r}{12} \right)}^{12}} \right)$
After 120 steps, you arrive at the point $\displaystyle \left( 10,P{{\left( 1+\tfrac{r}{12} \right)}^{120}} \right)$
After 12t steps, you arrive at the point $\displaystyle \left( t,P{{\left( 1+\tfrac{r}{12} \right)}^{12t}} \right)$
Do these y-values look familiar? It’s the amount you’d have if you were compounding monthly with a yearly interest rate of r for a year, 10 years, and t years.
If the step size were $\displaystyle \tfrac{1}{365}$ instead, you would arrive at the points
$\displaystyle \left( 1,P{{\left( 1+\tfrac{r}{365} \right)}^{365}} \right)$, and $\left( 10,P{{\left( 1+\tfrac{r}{365} \right)}^{3650}} \right)$, and $\displaystyle \left( t,P{{\left( 1+\tfrac{r}{365} \right)}^{365t}} \right)$.
These y-values are the amount you’d have if you were compounding daily for a year, 10 years, and t years.
If the step size were $\tfrac{1}{n}$, the point after t years is $\displaystyle \left( t,P{{\left( 1+\tfrac{r}{n} \right)}^{nt}} \right)$
If n went to infinity, you would arrive at the point $\displaystyle \left( t,P{{e}^{rt}} \right)$, the y-value for continuously compounding. This is the solution of the differential equation mentioned above.
I’d never made this connection before, but it makes perfect sense now.
The actual problem that got me thinking about all of this was:
Suppose that you now have $6000, you expect to save an additional$3000 during each year, and all of this is deposited in a bank paying 4% interest compounded continuously.
This generates the differential equation: $\displaystyle \frac{dA}{dt}=0.04t+3000$
This was a fun one to do out. I started with a step size of 1/12 and then made them smaller. I derived the solution from Euler’s Method!
So writes Chip Rollinson. He included the following links to his computations: here and here
Then there were some interesting comments from others.
1. Mark Howell pointed out that if $\displaystyle \frac{dy}{dx}=f\left( x \right)$, that is if the derivative is a function of x only, then Euler’s method is the same as a left Riemann sum approximation.
This would make a good exercise for your students to show given a derivative, a starting point, and a small number of steps. Try $\frac{dy}{dx}={{x}^{3}}$, and use Euler’s Method with 4 steps starting at (0, 0) and then a left-Riemann sum with 4 terms to approximate $\int_{0}^{1}{{{x}^{3}}dx}$, (Answer: $\tfrac{9}{64}$)
2. Dan Teague went further and pointed out that this is really the development of the Fundamental Theorem of Calculus for a particular function. He contributed this development of the concept.
So Thank You Chip for this and the several other comments you’ve contributed to other post on this blog. And thank you Mark and Dan for your contributions.
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## Monday, March 17, 2008
### Pythagoras Growing Post
Part 4: Now that you have seen many Pythagorean problems, create your own word problem.
My word problemo
My mom told me to get some cups from Wal-Mart. Wal-Mart is 12 km east from my house. Suddenly i got a phone call and she told me to get milk and SafeWay which is 10 km south from Wal-Mart. Now I had to go home. When i got home, i asked myself how much i walked from SafeWay to my house i guessed 99 km. How much did i walk from SafeWay to my house? was my normal guess right or wrong?
Part 1:
Describe what a Pythagorean Triple is and use your perfect square chart from 1 squared to 10 squared to find another one other than 3,4,5. (Note: You need a picture, not simply text.)
This is a picture of a Pythagorem Triple. A pythagorem triple says that the legs ( a and b) and the hypotenuse (c) of a right triangle follows the equation, a2 + b2 = c2. so if you add the legs together, you'll get the hypotenuse. This also shows that if you add the small square with the medium sized square, you'll get the biggest square. It uses perfect squares to find out the hypotenuse.
Proof
Perfect Square Chart
Here is the next pythagoream triple. It was quite easy to find out after you look at the perfect sqaure chart.
Part 2
Using embeddable web 2.0, describe how to find the missing side of a Right Triangle using the Pythagorean Theorem, show how to solve both for a missing leg and the hypotenuse (note: you may use numerical examples ie. a=4 b=6 c=?)
Part 3
Explain how to solve a Pythagoras word problem. Use one of the examples we covered in class (Worksheet A, B, LeFrog or Bonus Problem).
I am going to explain how we solved LeFrog. First we had the question. Lefrog stands 1.5 meters from a wall where a fly is 2 meters up. If Lefrog's tongue comes out of his mouth 70 cm from the ground, how far does his tongue have to stretch to eat the fly?
LeFrogs tongue had to stretch 1.98 m to eat the fly.
To figure out this problem, you first look at the question and find out the lengths of the triangle to find the hypotenuse. The lengths of our triangle was 1.5 and 1.3. It was 1.3 on one side because of the calculations i did on the top left corner. We then had to find out the length of LeFrogs tounge. We then used our formula ( a2 + b2 = c2 ). We used this formula, when we got to c, we used our calculators to find the sqaure root of the number. Finally, we found the answer to how far LeFrogs tounge had to stretch to eat the fly.
PART 4 ABOVVE !! |
## transformations of quadratic functions in vertex form
The table of values for a base parabola look like this: The reason this small equation forms a parabola, is because it still has the degree 2, something discussed in the previous lesson. Pre AP PreCalculus 20(Ms. Carignan) P20.7: Chapter 3 – Quadratic Functions Page 8 2. When a quadratic is written in vertex form, the transformations can easily be identified because you can pinpoint the vertex (h, k) as well as the value of a. Given the equation y = 3 (x + 4) 2 + 2, list the transformations of y = x 2. The base parabola has a step pattern of 1,2,5,7 (the step pattern can never be negative). Quadratic functions are second order functions, meaning the highest exponent for a variable is two. Identify the transformations of in each of the given functions: Graph the following quadratic functions. Start studying Quadratic Functions in Vertex Form. Notes: Vertex Form, Families of Graphs, Transformations I. Use finite differences to determine if a function is quadratic. You can represent a vertical (up, down) shift of the graph of $f(x)=x^2$ by adding or subtracting a constant, $k$. ( Log Out / Change ), This entry was posted on Friday, November 12th, 2010 at 6:50 am and tagged with, Lesson 3: Graphing and Solving Vertex Form. Determine the equation for the graph of $f(x)=x^2$ that has been shifted right 2 units. Transformations of Quadratic Functions and the Vertex Form of a Quadratic 4 e. f. Find the maximum or the minimum value of a quadratic function. The vertex form of a parabola contains the vital information about the transformations that a quadratic functions undergoes. In particular, the coefficients of $x$ must be equal. The parent function of a quadratic is f(x) = x². Transformations of quadratic functions in vertex form: Transformations of a quadratic function is a change in position, or shape or the size of the quadratic parent function. All parabolas are the result of various transformations being applied to a base or “mother” parabola. In a quadratic function, the variable is always squared. the x-coordinate of the vertex, the number at the end of the form gives the y-coordinate. For the two sides to be equal, the corresponding coefficients must be equal. We have learned how the constants a, h, and k in the functions, and affect their graphs. The U-shaped graph of a quadratic function is called a parabola. A parent function is the simplest function of a family of functions.The parent function of a quadratic is f(x) = x².Below you can see the graph and table of this function rule. Shifting parabolas. Vertex Form of Parabolas Date_____ Period____ Use the information provided to write the vertex form equation of each parabola. If $|a|>1$, the point associated with a particular $x$-value shifts farther from the $x$–axis, so the graph appears to become narrower, and there is a vertical stretch. Vertex Form and Transformations A. Vertex form is the form of the quadratic equation that will allow us to use transformations to graph. The parent graph of a quadratic function … After having gone through the stuff given above, we hope that the students would have understood, "Vertex Form of a Quadratic Equation".Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. In Section 1.1, you graphed quadratic functions using tables of values. Transformations include reflections, translations (both vertical and horizontal) , expansions, contractions, and rotations. Change ), You are commenting using your Twitter account. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Families of Graphs Families of graphs: a group of graphs that displays one or more characteristics Parent graph: A basic graph that is transformed to create other members in a family of graphs. Review (Answers) To see the Review answers, open this PDF file and look for section 3.9. f (x) = a (x – h)2 + k (a ≠ 0). . Learn vocabulary, terms, and more with flashcards, games, and other study tools. We can now put this together and graph quadratic functions $$f(x)=ax^{2}+bx+c$$ by first putting them into the form $$f(x)=a(x−h)^{2}+k$$ by completing the square. In order to verify this, however, we can find the second differences of the table of values. f(x) = a(x h)2 + k. This is called vertex form. Change ), You are commenting using your Facebook account. Investigating Quadratic Functions in Vertex Form Focus on . Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. From the vertex form, it is easily visible where the maximum or minimum point (the vertex) of the parabola is: The number in brackets gives (trouble spot: up to the sign!) Again, for the equation above, for which the a value is 2, we can determine the step pattern of the parabola, which is 2, 4, 10, 14. Big Idea The Parent Function is the focus of this lesson to identify transformations of every point on the graph by identifying the transformation of the Vertex. Email. can tell you about direction of opening of graph of given quadratic function. If the value of k is 4, then the base parabola is shifted to the point 4 on the y-axis. These transformed functions look similar to the original quadratic parent function. • identifying quadratic functions in vertex form • determining the effect of a, p, and q on the graph of y= a(x-p)2 + q • analysing and graphing quadratic functions using transformations The Bonneville Salt Flats is a large area in Utah, in the United This form is sometimes known as the vertex form or standard form. Showing top 8 worksheets in the category - 2 1 Additional Practice Vertex Form Of A Quadratic Function. You can represent a horizontal (left, right) shift of the graph of $f(x)=x^2$ by adding or subtracting a constant, $h$, to the variable $x$, before squaring. If , direction of opening is upwards and if then direction of opening is downwards. It tells a lot about quadratic function. Start studying Transformations of Quadratic Functions. We must be careful to both add and subtract the number to the SAME side of the function to complete the square. Some of the worksheets displayed are Th, 2 1 transformations of quadratic functions, Section quadratic functions and their graphs, Quadratic functions and equations, Factoring quadratic form, Quadratics in context, Vertex form 1, Unit 2 2 writing and graphing quadratics … (3, 9). Below you can see the graph and table of this function rule. Factored Form y=a(x−s)(x−t) Vertex Form y=a(x−h)2+k convert to standard form, then convert to factored form or solve for zeros and substitute into factored form, “a” will be the same Standard Form y=ax2+bx+c factor, if possible or use quadratic formula to find zeros and substitute into factored form Standard Fo rm Vertex Fo rm Factored rm Intro to parabola transformations. Start studying Quadratic Functions in Vertex Form. Vertex form of Quadratic Functions is . ( Log Out / Now that we know about the base parabola, we can discuss the transformations which the various values in the vertex form of an equation apply. parabola axis Of symmetry Quadratic Functions and Transformations Make sure to state transformations, the vertex and show the new tables of values. In Chapters 2 and 3, you studied linear functions of the form f(x) = mx + b. Intro to parabola transformations. Algebra 2Unit: Quadratic FunctionsLesson 2: Vertex Form of Quadratic FunctionsBest if used with the following power point presentation.This worksheet provides practice in graphing quadratic functions in vertex form and identifying transformations. Honors Algebra 2 Notes: Graphs of Quadratic Functions Transformations/Intro to Vertex Form Name Answer key included.Lesson 1: Graphing quadratic fu transformations for quadratic functions in vertex form. The vertex form is a special form of a quadratic function. A handy guide for students to reference while practicing transformations of quadratic functions (graphing from vertex form). I use this graphic organizer as a way to review the concepts before assessments. About "Vertex Form of a Quadratic Function Worksheet" Worksheet given in this section is much useful to the students who would like to practice problems on vertex form of a quadratic function. $-2ah=b,\text{ so }h=-\dfrac{b}{2a}$. where $\left(h,\text{ }k\right)$ is the vertex. transformations to graph any graph in that family. Also, determine the equation for the graph of $f(x)=x^2$ that has been shifted down 4 units. 2.1 - Transformations of Quadratic Functions Before look at the worksheet, if you would like to know the stuff related to vertex form of a quadratic function, … If the value of k is -4, then the base parabola is shifted to the point -4 on the y-axis. If $h>0$, the graph shifts toward the right and if $h<0$, the graph shifts to the left. A quadratic function is a function that can be written in the form of . The graph below contains three green sliders. To make the shot, $h\left(-7.5\right)$ would need to be about 4 but $h\left(-7.5\right)\approx 1.64$; he doesn’t make it. Also, determine the equation for the graph of $f(x)=x^2$ that has been shifted left 2 units. The standard form of a quadratic function presents the function in the form $f\left(x\right)=a{\left(x-h\right)}^{2}+k$ where $\left(h,\text{ }k\right)$ is the vertex. Explain your reasoning. II. To write an equation in vertex form from a graph, follow these steps: Does the shooter make the basket? ( Log Out / !2 also determines if the parabola is vertically compressed or stretched. Determine the equation for the graph of $f(x)=x^2$ that has been compressed vertically by a factor of $\frac{1}{2}$. The vertex form is a special form of a quadratic function. The general rule which comes into play while looking at the h value in the vertex form of a quadratic relation is: Finally, the k value of the equation translates the base parabola vertically k units. This is the $x$ coordinate of the vertexr and $x=-\dfrac{b}{2a}$ is the axis of symmetry we defined earlier. http://cnx.org/contents/9b08c294-057f-4201-9f48-5d6ad992740d@5.2, Graph vertical and horizontal shifts of quadratic functions, Graph vertical compressions and stretches of quadratic functions, Write the equation of a transformed quadratic function using the vertex form, Identify the vertex and axis of symmetry for a given quadratic function in vertex form. Explain your reasoning. Using the following mapping rules, write the equation, in vertex form, that represents the image of . We can now put this together and graph quadratic functions by first putting them into the form by completing the square. When identifying transformations of functions, this original image is called the parent function. From the vertex form, it is easily visible where the maximum or minimum point (the vertex) of the parabola is: The number in brackets gives (trouble spot: up to the sign!) The standard form of a quadratic function presents the function in the form, $f\left(x\right)=a{\left(x-h\right)}^{2}+k$. You can apply transformations to the graph of y = x 2 to create a new graph with a corresponding new equation. The equation for the graph of $f(x)=x^2$ that has been shifted up 4 units is, The equation for the graph of $f(x)=x^2$ that has been shifted down 4 units is. However, there is a key piece of information to remember when plotting the h value. Graph the following functions using transformations. the x-coordinate of the vertex, the number at the end of the form … It is helpful when analyzing a quadratic equation, and it can also be helpful when creating an equation that fits some data. quadraticfunction, a function of the form Y = ax2 + bx + c. Main Idea: A parabola is symmetrical around its axis ofsymmetry, a line passing through the vertex, A parabola can open upward or downward. With the vertex form of a quadratic relation, determining things like the vertex of the parabola, the axis of symmetry, whether the parabola will open upwards or downwards, and whether the vertex will be maximum or minimum value is very simple, and can done by simply looking at the equation. The standard form and the general form are equivalent methods of describing the same function. Quadratic functions can be written in the form Now check Take a moment to work with a partner to match each quadratic function with its graph. The magnitude of $a$ indicates the stretch of the graph. Google Classroom Facebook Twitter. Vertex Form of a Quadratic Function. We can see this by expanding out the general form and setting it equal to the standard form. Transformations of Quadratic Functions | College Algebra 2.1 Transformations of Quadratic Functions Obj: Describe and write transformations for quadratic functions in vertex form. Quadratic Functions(General Form) Quadratic functions are some of the most important algebraic functions and they need to be thoroughly understood in any modern high school algebra course. A quadratic function is a function that can be written in the form f (x) = a (x - h) 2 + k (a ≠ 0). If the value of h is subtracted from x in the equation, it is plotted on the right (positive) x-axis. Change ), You are commenting using your Google account. a) yx2 2 d) f x x( ) 4 2 2 b) yx 3 4 2 22 e) 1 ( ) 1 1 3 f x x In a quadratic function, the variable is always squared. With the vertex form of a quadratic relation, determining things like the vertex of the parabola, the axis of symmetry, whether the parabola will open upwards or downwards, and whether the vertex will be maximum or minimum value is very simple, and can done by simply looking at the equation. This form is sometimes known as the vertex form or standard form. A coordinate grid has been superimposed over the quadratic path of a basketball in the picture below. The table shows the linear and quadratic parent functions. ID: 1240168 Language: English School subject: Math Grade/level: Grade 10 Age: 13-15 Main content: Quadratic equations Other contents: grap quadratic equations Add to my workbooks (2) Download file pdf Embed in my website or blog Add to Google Classroom They're usually in this form: f(x) = ax 2 + bx + c . The equation for the graph of $f(x)=x^2$ that has been shifted right 2 units is, The equation for the graph of $f(x)=^2$ that has been shifted left 2 units is. Vertex form: y=a (x-h)^2+k. Learn vocabulary, terms, and more with flashcards, games, and other study tools. ( Log Out / Take a moment to work with a partner to match each quadratic function with its graph. Transforming quadratic functions. Vertex Form of Parabolas Date_____ Period____ Use the information provided to write the vertex form equation of each parabola. For example, if we have the equation: y=(x-2)^2, we would do this: As you can see, the real value of h is 2. SWBAT graph quadratic functions in Vertex Form by identifying the Vertex from the equation, and plotting 2 points on each side of the vertex. This is the currently selected item. . For example, if we had the equation: 2(x-3)^2+5, the vertex of the parabola would be (3,5). The equation for the graph of $f(x)=x^2$ that has been compressed vertically by a factor of $\frac{1}{2}$ is, The equation for the graph of $f(x)=x^2$ that has been vertically stretched by a factor of 3 is. Since every other parabola is created by applying transformations to the base parabola, the step pattern of any other parabola can be found by multiplying the a value of the equation by the step pattern of the base parabola. By completing the square graph opens up or down form or standard form and setting equal. 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cistG
2021-08-15
An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What is the probability that all of the balls selected are white? What is the conditional probability that the die landed on 3 if all the balls selected are white?
insonsipthinye
1.Events:
A- ll of the choose balls are white
${E}_{i}$ - result of the die rill is i, $i=1,2,3,4,5,6$
Probabilities:
Since the die is fair:
$P\left({E}_{i}\right)=\frac{1}{6}f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}i=\left\{1,2,3,4,5,6\right\}$
If the die rolls i we choose a combination of i balls, among black and five white balls, therefore $P\left(A\mid {E}_{1}\right)=\frac{\left(\begin{array}{c}5\\ 1\end{array}\right)}{\left(\begin{array}{c}15\\ 1\end{array}\right)}=\frac{5}{15}=\frac{1}{3}$
$P\left(A\mid {E}_{2}\right)=\frac{\left(\begin{array}{c}5\\ 2\end{array}\right)}{\left(\begin{array}{c}15\\ 2\end{array}\right)}=\frac{10}{105}=\frac{2}{21}$
$P\left(A\mid {E}_{3}\right)=\frac{\left(\begin{array}{c}5\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}=\frac{10}{455}=\frac{2}{91}$
$P\left(A\mid {E}_{4}\right)=\frac{\left(\begin{array}{c}5\\ 4\end{array}\right)}{\left(\begin{array}{c}15\\ 4\end{array}\right)}=\frac{1}{273}$
$P\left(A\mid {E}_{5}\right)=\frac{\left(\begin{array}{c}5\\ 5\end{array}\right)}{\left(\begin{array}{c}15\\ 5\end{array}\right)}=\frac{1}{3003}$
$P\left(A\mid {E}_{6}\right)=\frac{\left(\begin{array}{c}5\\ 6\end{array}\right)}{\left(\begin{array}{c}15\\ 6\end{array}\right)}=0$
Calculate: $P\left(A\right),P\left({E}_{3}\mid A\right)$
2.${E}_{1},{E}_{2},{E}_{3},{E}_{4},{E}_{5},{E}_{6}$ are competing hypothesis, that is, mutually exclusive events, union of which is the whole outcome space, so conditioning on the roll of the die: $P\left(A\right)=\sum _{i=1}^{6}P\left(A\mid {E}_{i}\right)P\left({E}_{i}\right)$
Substituting $P\left({E}_{i}\right),P\left(A\mid {E}_{i}\right)$ yields: $P\left(A\right)=\frac{1}{6}\left(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}\right)=\frac{5}{66}$
$P\left({E}_{3}\mid A\right)$ can be calculated from the definition if we note the identity $P\left({}_{}$
Jeffrey Jordon
A-event that all balls drawn are white
${D}_{i}$ - outcome of roll of the die is $i,i=1,2,\dots ,6$
$P\left(A\right)=\sum _{i=1}^{6}P\left(A|{D}_{i}\right)P\left({D}_{i}\right)$
$=\frac{1}{6}\left(P\left(A|{D}_{i}\right)+\dots +P\left(A|{D}_{6}\right)\right)$
$=\frac{1}{6}\left(\frac{5}{15}+\frac{{5}_{{C}_{2}}}{{15}_{{C}_{2}}}+\frac{{5}_{{C}_{3}}}{{15}_{{C}_{3}}}+\frac{{5}_{{C}_{4}}}{{15}_{{C}_{4}}}+\frac{{5}_{{C}_{5}}}{{15}_{{C}_{5}}}+0\right)=\frac{5}{66}$
$P\left({D}_{3}|A\right)=\frac{P\left(A|{D}_{3}\right)P\left({D}_{3}\right)}{P\left(A\right)}$
$=\frac{\frac{{5}_{{C}_{3}}}{{15}_{{C}_{3}}}}{\sum _{i=1}^{5}\frac{{5}_{{C}_{i}}}{{15}_{{C}_{i}}}}=\frac{22}{455}$
Vasquez
Result:
$P\left(A\right)=\frac{22}{455}$
$P\left(C|A\right)=\frac{1}{22}$
Solution:
1. Probability of selecting all white balls:
Let's denote the event of selecting all white balls as $A$, and the event of rolling a die and choosing that number of balls from the urn as $B$. We need to find the probability of event $A$ given that event $B$ has occurred.
The total number of balls in the urn is 15 (5 white + 10 black). The probability of selecting all white balls from the urn depends on the number rolled on the die. Let $n$ be the number rolled on the die.
The probability of selecting all white balls given that $n$ is rolled is the ratio of the number of ways to select all white balls from the urn to the total number of possible selections when rolling $n$ on the die.
The number of ways to select all white balls from the urn is $\left(\genfrac{}{}{0}{}{5}{n}\right)$ (choosing $n$ white balls out of 5), and the total number of possible selections when rolling $n$ is $\left(\genfrac{}{}{0}{}{15}{n}\right)$ (choosing $n$ balls out of 15).
Therefore, the probability of selecting all white balls given that $n$ is rolled can be expressed as:
$P\left(A|B=n\right)=\frac{\left(\genfrac{}{}{0}{}{5}{n}\right)}{\left(\genfrac{}{}{0}{}{15}{n}\right)}$
To find the overall probability of selecting all white balls, we need to consider the probabilities for each possible value of $n$ and weigh them by their respective probabilities.
The die is fair, so each number from 1 to 6 has an equal probability of $\frac{1}{6}$.
Hence, the probability of selecting all white balls can be calculated as:
$P\left(A\right)={\sum }_{n=1}^{6}P\left(A|B=n\right)·P\left(B=n\right)={\sum }_{n=1}^{6}\left(\frac{\left(\genfrac{}{}{0}{}{5}{n}\right)}{\left(\genfrac{}{}{0}{}{15}{n}\right)}\right)·\left(\frac{1}{6}\right)$
2. Conditional probability of the die landing on 3 given that all selected balls are white:
Let's denote the event of the die landing on 3 as $C$. We need to find the conditional probability of event $C$ given event $A$.
Using Bayes' theorem, the conditional probability can be calculated as:
$P\left(C|A\right)=\frac{P\left(A|C\right)·P\left(C\right)}{P\left(A\right)}$
We already have the expression for $P\left(A\right)$ from the previous part. The probability of the die landing on 3 is $\frac{1}{6}$, since it is a fair die. To find $P\left(A|C\right)$, we need to substitute $n=3$ in the expression we derived earlier for $P\left(A|B=n\right)$.
Substituting all the values, the conditional probability can be calculated as:
$P\left(C|A\right)=\frac{\left(\frac{\left(\genfrac{}{}{0}{}{5}{3}\right)}{\left(\genfrac{}{}{0}{}{15}{3}\right)}\right)·\left(\frac{1}{6}\right)}{P\left(A\right)}$
To get the final answer, we substitute the expression for $P\left(A\right)$ and calculate the conditional probability.
Calculating these values yields the following result:
$P\left(A\right)=\frac{22}{455}$
$P\left(C|A\right)=\frac{1}{22}$
Therefore, the probability that all the balls selected are white is $\frac{22}{455}$, and the conditional probability that the die landed on 3 given that all the balls selected are white is $\frac{1}{22}$.
RizerMix
To solve this problem, let's break it down into two parts.
Part 1: Probability of selecting all white balls
We have an urn containing 5 white balls and 10 black balls. A fair die is rolled, and that number of balls is randomly chosen from the urn. We want to find the probability that all of the balls selected are white.
Let's denote the event of selecting all white balls as $A$ and the event of rolling a particular number on the die as $B$. We need to find $P\left(A\right)$, the probability of event $A$ occurring.
The total number of balls in the urn is 15 (5 white + 10 black). Let's assume the die rolls a number $n$ (where $n$ can take values from 1 to 6). The number of ways we can choose $n$ balls from the urn is given by the binomial coefficient $\left(\genfrac{}{}{0}{}{15}{n}\right)$. The number of ways we can choose all white balls (5) from the 5 white balls in the urn is $\left(\genfrac{}{}{0}{}{5}{5}\right)=1$. Therefore, the probability of event $A$ given that the die rolls a number $n$ is:
$P\left(A|B=n\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{n}\right)}$
Since the die is fair, the probability of rolling any particular number $n$ is $\frac{1}{6}$. Therefore, the probability of event $A$ occurring is:
$P\left(A\right)=\frac{1}{6}\left(P\left(A|B=1\right)+P\left(A|B=2\right)+P\left(A|B=3\right)+P\left(A|B=4\right)+P\left(A|B=5\right)+P\left(A|B=6\right)\right)$
Now let's calculate the probabilities for each roll of the die:
$P\left(A|B=1\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{1}\right)}$
$P\left(A|B=2\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{2}\right)}$
$P\left(A|B=3\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{3}\right)}$
$P\left(A|B=4\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{4}\right)}$
$P\left(A|B=5\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{5}\right)}$
$P\left(A|B=6\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{6}\right)}$
Now we can substitute these values into the formula for $P\left(A\right)$:
$P\left(A\right)=\frac{1}{6}\left(\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{1}\right)}+\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{2}\right)}+\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{3}\right)}+\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{4}\right)}+\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{5}\right)}+\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{6}\right)}\right)$
Part 2: Conditional probability of die landing on 3 given all balls selected are white
We want to find the conditional probability of the die landing on 3 given that all the balls selected are white. Let's denote this event as $C$ (die lands on 3)
and event $A$ (all balls selected are white). We need to find $P\left(C|A\right)$.
Using Bayes' theorem, the conditional probability is given by:
$P\left(C|A\right)=\frac{P\left(A|C\right)·P\left(C\right)}{P\left(A\right)}$
We have already calculated $P\left(A\right)$ in Part 1. The probability of the die landing on 3 is $\frac{1}{6}$, since the die is fair. To calculate $P\left(A|C\right)$, we need to find the probability of selecting all white balls given that the die lands on 3.
$P\left(A|C\right)=\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{3}\right)}$
Now we can substitute these values into the formula for $P\left(C|A\right)$:
$P\left(C|A\right)=\frac{\frac{\left(\genfrac{}{}{0}{}{5}{5}\right)}{\left(\genfrac{}{}{0}{}{15}{3}\right)}·\frac{1}{6}}{P\left(A\right)}$
Therefore, the probability that all of the balls selected are white is $P\left(A\right)$, and the conditional probability that the die landed on 3 given all the balls selected are white is $P\left(C|A\right)$.
Do you have a similar question? |
# Construct Parallelogram, Square and Rectangle
In these lessons, we will learn
• how to construct a parallelogram given the lengths of its sides and an angle
• how to construct a parallelogram given the lengths of its diagonals
• how to construct a square given the length of the diagonal.
• how to construct a square given the length of one side.
• how to construct a rectangle
Related Topics: More Geometry Lessons
### Construct A Parallelogram given sides and angle
Example:
Construct a parallelogram ABCD with sides AB = 4 cm and AD = 5 cm and angle A = 60˚.
Solution:
Step 1: Construct a line segment AB = 4 cm. Construct a 60˚ angle at point A.
Step 2: Construct a line segment AD = 5 cm on the other arm of the angle. Then, place the sharp point of the compasses at B and make an arc 5 cm above B.
Step 3: Stretch your compasses to 4 cm, place the sharp end at D and draw an arc to intersect the arc drawn in step 2. Label the intersecting point C. Join C to D and B to C to form the parallelogram ABCD.
How to construct a parallelogram with a given angle and certain side lengths?
Example: Construct a parallelogram with side lengths c and d and with a 45° angle. How to construct a parallelogram, using two lengths and one angle provided?
Step 1: Make a copy of the given angle.
Step 2: Make a copy of the lengths for each side.
Step 3: Use the compass to mark the fourth vertex.
### Construct a Parallelogram given its diagonals
This video shows how to construct a parallelogram given the lengths of its diagonals, using the property that diagonals of a parallelogram bisect each other. An infinite number of parallelograms are possible.
Example: Construct a parallelogram ABCD given the diagonals AC and BD.
Step 1: Bisect AC and BD.
Step 2: Draw a circle with the radius equals to 1/2 of AC.
Step 3: Draw a concentric circle with the radius equals to 1/2 of BD.
Step 4: Mark two points that will form the diameter on the bigger circle.
Step 5: Mark two points that will form the diameter on the smaller circle.
Step 6: Join the four points to form the parallelogram. How to construct a parallelogram using the intersecting diagonals property of a parallelogram?
The diagonals of a parallelogram bisect each other.
### How to construct a square given the length of its diagonal?
The following video shows how to construct a square with a compass and straight edge given the length of a diagonal. How to construct a square using a diagonal?
### How to construct a square given its side?
The following video shows how to construct a square given the length of one side. How to construct a square given the length of one side? Two methods are shown.
### How to construct a rectangle?
How to draw a Rectangle Using a Compass and Straight Edge, and perpendicular bisector constructions. Construct a rectangle with compass and straight edge.
Step 1: Construct two perpendicular segments.
Step 2: Mark off congruent sides.
Step 3: Check for 4 right right angles.
Step 4: Check that the diagonals are equal in lengths.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
### Functions and Transformation of Graphs
#### Functions
y = x2 + 3x -2 is said to be a function of x - an expression of x. This can also be written, in a bit more advanced way as f(x) = x2 + 3x -2 - it is read as you write, function of x, f(x)
So, f(x) = x2 + 3x -2
Now whatever you put in as 'x' on the left, substitutes 'x' on the right.
E.g.
• f(2) = 22 + 3x2 -2 = 4 + 6 - 2 = 8
• f(-2) = (-2)2 + 3x(-2) -2 = 4 - 6 - 2 = -4
• f(0) = 02 + 3x0 -2 = 0 + 0 - 2 = -2
• f(x+1) = (x+1)2 + 3(x+1) -2 = x2 + 2x + 1 + 3x + 3 -2 = x2 + 5x + 2
• f(x-1) = (x-1)2 + 3(x-1) -2 = x2 - 2x + 1 + 3x - 3 -2 = x2 + x - 5
• f(2x) = (2x)2 + 3(2x) -2 = (2x)2 + 6x -2 = 4x2 + 6x - 2
• f(x/2) = (x/2)2 + 3(x/2) -2 = x2/4 + 3x/2 -2 = x2/4 + 3x/2 - 2
There are three major transformations:
1. Translation
2. Reflection
3. Stretching
Let's apply the above transformations to the following curve.
#### Translation
f(x) -----> f(x+1)f(x) -----> f(x-1)
translation in the -ve x-axistranslation in the +ve x-axis
f(x) -----> f(x) + 2f(x) -----> f(x) - 2
translation in the +ve y-axistranslation in the -ve y-axis
You can animate the following curves to see how translation works. Please select an option and then move the sliders.
Choose slider, a, for the translation in the x-axis.
Choose slider, c, for the translation in the y-axis.
#### Reflection
f(x) -----> f(-x)f(x) -----> -f(x)
reflection in the y-axisreflection in the x-axis
#### Stretching
f(x) -----> f(2x)f(x) -----> 2f(x)
stretching parallel to x-axisstretching parallel to y-axis
You can animate the following curves to see how stretching works. Please select an option and then move the sliders.
Choose slider, a, for the stretching parallel to y-axis.
Choose slider, c, for the stretching parallel to x-axis.
Now, let's apply the transformation to some of the well-known curves.
E.g.1
Sketch f(x) = x2 + 2x -3, and find the following:
• line of symmetry
• minimum value
• translations from f(x) = x2
• solutions
First of all, let's use the completing the square method:
Let x2 + 2x -3 = (x+a)2 +b
x2 + 2x -3 = x2 + 2ax + a2+ b
Make the coefficients of x2, x and the constant on either side equal.
2a = 2 => a = 1
a2+ b = -3 => b = -4
f(x) = (x+1)2 -4
Line of symmetry, x = 1
Minimum value = -4
This is a translation of f(x) = x2, 1 in the negative x-axis and 4 in the negative y-axis.
The solutions are, x = -3 and x = 1
E.g.2
This question is based on the curve on the left; find the coordinates of A, B and C after the following transformations. The answers are on the table.
TransformationNew Coordinates
f(x+1)A( -3 , 0 )B( -1 , 3 )C( 3 , -4 )
f(x-1) + 2A( -1 , 2 )B( 1 , 5 )C( 5 , -2 )
f(2x)A( -1 , 0 )B( 0 , 3 )C( 2 , -4 )
f(x/2)A( -4 , 0 )B( 0 , 3 )C( 8 , -4 )
2f(x)A( -2 , 0 )B( 0 , 6 )C( 4 , -8 )
f(-x) + 1A( 2 , 1 )B( 0 , 4 )C( -4 , -3 )
-f(x) + 2A( -2 , 2 )B( 0 , -1 )C( 4 , 6 )
#### Combination of Transformations
Here is an opportunity for you to practise transformations using a sine curve. The curve is in the form of y = a + b sin(cx + d). Change the values of a, b, c and d or to see the corresponding change. This is the best form of mastering the transformations.
Now, in order to complement what you have just learnt, work out the following questions:
1. Use the completing the square method to sketch f(x) = x2 + 3x -4. Then describe the transformations of f(x) = x2 to get this particular shape. Write down the minimum value and the line of symmetry too.
• Show that the reflection in the y-axis of f(x) = x4 - 2x2 + 5 is the same as f(x).
• Write down a function that shows f(x) = -f(-x). Hence, sketch 2f(x) and f(2x) of the same function.
2. If f(x) = x2 + 2x, sketch the following:
• f(-x)
• -f(x) + 2
• f(x+2) - 3
• f(2-x)
• 2f(x) + 1
• f(2x) - 3
3. Sketch f(x) = x2 -x - 6 and hence sketch g(x) = (x+2)(x+3) and show that f(x+1) = g(x).
4. Sketch f(x) = 1 +2 sin(x) and solve f(x) = 1.3 for 0 < x < 360. What is the minimum value and maximum value of the function?<
5. Sketch f(x) = 1 + cos(3x) and hence find the period of the function, maximum value and minimum value.
6. Sketch f(x) = 3 - 2sin(3x) and hence find the period of the function, maximum value and minimum value.
7. Sketch f(x) = 1/x and then transform as follows:
• f(-x) + 2
• -f(x) + 2
• 2f(x) - 3
8. The maximum value of f(x) = q + p cos(x) graph is 7 and the minimum is 1. find p and q.
Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7th edition in print.
### Recommended - GCSE & iGCSE
This is the best book available for the new GCSE(9-1) specification and iGCSE: there are plenty of worked examples; a really good collection of problems for practising; every single topic is adequately covered; the topics are organized in a logical order.
### Recommended for A Level
This is the best book that can be recommended for the new A Level - Edexcel board: it covers every single topic in detail;lots of worked examples; ample problems for practising; beautifully and clearly presented. |
# Knowing Our Numbers – Class 6
IMPORTANT QUESTIONS FOR CBSE CLASS 6 MATHEMATICS
Knowing Our Numbers – Chapter 1
1. Write the number name of 8,88,88,888.
2. Write the number in figures: Thirteen lakh three thousand thirty three.
3. Write the place value and face value of 6 in the number 54, 67,879.
4. Write the expanded form of the number: 89, 60,754.
5. Write the standard form of: 30, 00,000 + 6,000 + 800 + 30 + 1
6. Put commas correctly and write the number name in International System of Numeration: 9876435.
7. Write the number in figures according to International System of Numeration: Five million eight hundred three thousand one hundred forty.
8. Use the digits without repetition to form the greatest 7-digit number:
9,0,4,3,6,8,2.
9. Use the digits with repetition to form the smallest 7 digit number:
1,0,7,9 4, 3.
10. Round off to the nearest ten, hundred and thousand of 87654.
11. Write the Roman numeral for
a) 654
b) 42
12. Write the Hindu-Arabic numerals for
a) CXX
b) XCIV
13. Estimate: 568 + 785 (Round off to nearest hundred)
14. Make the greatest 4 digit number with condition 5 is always at tens place.
15. Make the smallest 4 digit number with condition 6 is always at hundreds place.
1. Eight crore eighty eight lakh eighty eight thousand eight hundred eighty eight.
2. 13, 03,033
3. Place value of 6 is 60,000 and face value is 6 itself.
4. 8 x 10, 00,000 + 9 x 1, 00,000 + 6 x 10,000 +7 x 100 + 5 x 10 + 4 x 1.
5. 30, 06,831
6. 9,876,435
Number name: Nine million eight hundred seventy six thousand four hundred thirty five.
7. 5,803,140
8. 9864320
9. 1003479
10. 87654
Nearest ten = 87650
Nearest hundred = 87700
Nearest thousand = 88000
11. a) 654 = DCLIV
b) 42 = XLII
12. a) CXX = 120
b) XCIV = 94
13. 568 rounds off to 600.
785 rounds off to 800.
Estimated sum = 600 + 800 = 1400.
14. 9857
15. 1602. |
1. Problem
The last digit of a six-digit number is 2. If the 2 is moved to the start of the number, the new six-digit number is only a third of the original number. Find the original number.
2. Originally Posted by not happy jan
The last digit of a six-digit number is 2. If the 2 is moved to the start of the number, the new six-digit number is only a third of the original number. Find the original number.
Digitwise, the number is abcde2
the new number is 2abcde
so 3 times the new number equals the old number
$\displaystyle \begin{array}{lllllll} &a&b&c&d&e&2\\ =&6&3a&3b&3c&3d&3e\\ \end{array}$
3e must have a ones digit of 2. So it could be equal to 2, 12, 22, 32, etc
But we know it is a digit, so it is less than 10, and since 3*9 = 27, it must equal 2, 12, or 22. And since 12 is the only one of those which is a multiple of 3, it must be 12. then 12/3=4. so e =4
$\displaystyle \begin{array}{lllllll} &a&b&c&d&4&2\\ =&6&3a&3b&3c&3d&12\\ \end{array}$
we can't have a ones digit of 12, so we carry the one.
$\displaystyle \begin{array}{lllllll} &a&b&c&d&4&2\\ =&6&3a&3b&3c&3d+1&2\\ \end{array}$
now 3d+1 returns a ones digit of 4. so options are 4, 14, and 24 (because d cannot be greater than 9, so 3d+1 cannot be greater than 28) subtracting one from each of these we get 3d = 3, 13, or 23. and clearly 3 is the only option returning an integer. so 3d=3. thus d = 1
$\displaystyle \begin{array}{lllllll} &a&b&c&1&4&2\\ =&6&3a&3b&3c&4&2\\ \end{array}$
now we know that 3c has a ones digit of 1. so 3c equals 1, 11, or 21 (again, must be less than 27, because c < 9). 21 is the only multiple of 3. so 3c = 21. therefore c = 7
$\displaystyle \begin{array}{lllllll} &a&b&7&1&4&2\\ =&6&3a&3b&21&4&2\\ \end{array}$
21 is not a digit, so we carry the 2.
$\displaystyle \begin{array}{lllllll} &a&b&7&1&4&2\\ =&6&3a&3b+2&1&4&2\\ \end{array}$
now we have that 3b+2 must have a ones digit of 7. so 7, 17, or 27. Subtracting 2 we get that 3b must equal 5, 15, or 25. 15 is the only multiple of 3. so 3b=15. therefore b = 5
$\displaystyle \begin{array}{lllllll} &a&5&7&1&4&2\\ =&6&3a&17&1&4&2\\ \end{array}$
17 is not a digit so carry the one.
$\displaystyle \begin{array}{lllllll} &a&5&7&1&4&2\\ =&6&3a+1&7&1&4&2\\ \end{array}$
Now we get that 3a+1 has a ones digit of 5. so 5, 15, or 25. Subtracting one we get that 3a must be 4, 14, or 24. 24 is the only multiple of 3. so 3a=24. therefore a = 8
$\displaystyle \begin{array}{lllllll} &8&5&7&1&4&2\\ =&6&25&7&1&4&2\\ \end{array}$
25 is not a digit, so carry the 2.
$\displaystyle \begin{array}{lllllll} &8&5&7&1&4&2\\ =&8&5&7&1&4&2\\ \end{array}$
Now looking back at the original, equation we see that this becomes
$\displaystyle \begin{array}{lllllll} &a&b&c&d&e&2\\ =&8&5&7&1&4&2\\ \end{array}$
Therefore the number is 857142, and the new number is 285714.
3. Hello, not happy jan!
Lucky for both of us, I'm familiar with this type of problem . . .
The last digit of a six-digit number is 2. If the 2 is moved to the front,
the new six-digit number is only a third of the original number.
Find the original number.
We have: .$\displaystyle ABCDE2$
Then: .$\displaystyle 2ABCDE \:=\:\frac{1}{3}\times ABCDE2 \quad\Rightarrow\quad 2ABCDE \times 3 \;=\;ABCDE2$
And we have a multiplication problem . . .
. . $\displaystyle \begin{array}{cccccc} ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\ 2 & A & B & C & D & E \\ \times & & & & & 3 \\ \hline A & B & C & D & E & 2 \end{array}$
In column-6, $\displaystyle 3E \text{ ends in 2}\quad\Rightarrow\quad\boxed{ E = 4}\quad\hdots$ ("carry 1" to column-5)
. . $\displaystyle \begin{array}{cccccc} ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\ 2 & A & B & C & D & 4 \\ \times & & & & & 3 \\ \hline A & B & C & D & 4 & 2 \end{array}$
In column-5, $\displaystyle 3D + 1\text{ ends in 4}\quad\Rightarrow\quad\boxed{D = 1}$
. . $\displaystyle \begin{array}{cccccc} ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\ 2 & A & B & C & 1 & 4 \\ \times & & & & & 3 \\ \hline A & B & C & 1 & 4 & 2 \end{array}$
In column-4, $\displaystyle 3C\text{ ends in 1}\quad\Rightarrow\quad\boxed{C = 7}\quad\hdots$ ("carry 2" to column-3)
. . $\displaystyle \begin{array}{cccccc} ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\ 2 & A & B & 7 & 1 & 4 \\ \times & & & & & 3 \\ \hline A & B & 7 & 1 & 4 & 2 \end{array}$
In column-3, $\displaystyle 3B + 2\text{ ends is 7}\quad\Rightarrow\quad\boxed{B = 5}\quad\hdots$ ("carry 1" to column -2)
. . $\displaystyle \begin{array}{cccccc} ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\ 2 & A & 5 & 7 & 1 & 4 \\ \times & & & & & 3 \\ \hline A & 5 & 7 & 1 & 4 & 2 \end{array}$
In column-2, $\displaystyle 3A + 1\text{ ends in 5}\quad\Rightarrow\quad\boxed{A = 8}$
. . $\displaystyle \begin{array}{cccccc} 2 & 8 & 5 & 7 & 1 & 4 \\ \times & & & & & 3 \\ \hline 8 & 5 & 7 & 1 & 4 & 2 \end{array}\qquad\hdots$ta-DAA!
Finally took a look at your solution, angel.white . . .
Excellent job . . . nice explanations!
4. Originally Posted by Soroban
Finally took a look at your solution, angel.white . . .
Excellent job . . . nice explanations!
Thanks ^_^ your way was a bit clearer b/c it didn't rely on unconventional mathematical notations like mine. (I sort of figured it out / made it up as I went). |
HomeEducationThe Ultimate Guide for the Concept of Percentage
# The Ultimate Guide for the Concept of Percentage
When you’re struggling with a concept in Math, it’s important to understand what you’re solving for. With that said, let’s dive into the idea of percentages and how they work.
## What is the Percentage?
The percentage is a mathematical term that refers to a number divided by 100. Percent can be used in percentages and fractions, as well as in decimal form.
Percentage can also be abbreviated as P. Percent is often used to describe how much of a certain item or quantity there is.
For example, if someone has 10 items and wants to divide them into two groups, they would say that the first group has 50% of the total and the second group has 25% of the total.
## How to Calculate a Percentage?
• When it comes to percentages, there are a few things that you need to keep in mind. The first is that percentages are calculated with a whole number as the base and then a fractional part.
For example, if you have an equation with a whole number in the numerator (the top number) and a fractional part in the denominator (the bottom number), then you would solve for the percentage by dividing the numerator by the denominator.
• Another thing to remember is that percentage values always range from 0% to 100%. That means that no matter your percentage value, there will always be one more percent than what’s written on either side of the decimal point.
• Finally, remember that when working with Average Percentage Calculator, you’re always dealing with fractions. That means that if you have an equation like 3/10 or 1/5, you’ll need to use fractions when solving for the percentage.
### What are the Different Types of Percentages?
Percentages are a mathematical tool that allows us to calculate proportions. There are three types of percentages: simple, compound, and percentage-of.
• Simple percentages are just number divided by 100. For example, 50% is equal to 50/100 or half.
• Compound percentages are made up of two simple percentages. For example, 75% comprises 50% and 25%.
When it comes to percentages, there are a few things you need to know.
• The first thing to understand is that percentages are a way of describing how much of something there is.
 For example, if you have 10 items and want to know how many are in total, you would say that 100% of the items are in total.
• The second thing to know about percentages is that they work slightly differently than other measurements.
For example, if you have 10 pieces of candy and want to divide them up into thirds, you would say that each piece is 33%. However, if you wanted to divide the candy into 2 piles – one with 5 pieces and one with 7 – then the percent would be 50% for the pile with 5 pieces and 66% for the pile with 7 pieces.
There are a few other things to remember when working with percentages.
• First of all, remember that they always work in proportion. That means that if one part of the equation changes (like the number of items being dealt with), everything else in the equation will also change.
• Another important thing to remember is that percentages can also be used as a decimal value and an integer value.
• If someone were negotiating a salary contract and wanted their salary to be paid out in monthly installments of \$500, they could say that their salary was equivalent to 5% of their total yearly income.
In business, it’s important to understand the concept of percentages. The percentage is a mathematical term used to describe a fraction representing a whole’s quantity. Average Percentage Calculator is used in many different ways in business, and understanding them can help you make better decisions.
One common use of percentages is in pricing. When you sell something, you may want to charge a certain price per unit (for example, \$10 per shirt), or you may want to charge a percentage of the total cost (for example, 20% of the cost of the shirt). In either case, you’re charging based on how much of the product each customer buys.
### Payouts or Rewards
Percentages can also be used when calculating payouts or rewards. For example, if your employee earns an annual salary of \$50,000 and you offer a performance bonus worth 10% of their annual salary, their bonus would be \$5,000 (\$50,000 x .10 = \$5,000).
When working with percentages, it’s important to keep two things in mind: the number that’s being percentage zed (in this case, \$50,000) and the number that’s being divided (in this case, 10%).
This is known as % notation. To calculate 45%, for example, divide 45 by 100 and then multiply that answer by 100: 45 ÷ 100 = 0.45 x 100 = 45%.
### Why do people prefer to use a Calculator?
There are many reasons why people prefer to use an Average Percentage Calculator. These calculators allow users to calculate a percentage and understand it better easily.
 They can also use the average percentage to compare different calculations. Additionally, these calculators can be helpful when working with money or other numerical data.
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# Pharmaceutical calculations by alligation method
by egpat Posted on 11-06-2017
Alligation is one of the simple and illustrative methods in pharmaceutical calculations for the pharmacy technicians. Dilution is a simple method for preparing a lower concentration from a solution of higher concentration. On the other hand, many of the times we require an intermediate concentration from two different concentrations where allegation plays its role.
The solutions should be mixed at a specific proportion in order to get required concentration.
Solutions can be mixed at a fixed proportion called as alligation ratio.
Here we will see few types of working examples on this method and different steps required for its calculation.
Let’s start our discussion with a simple example.
Working example 1:
How many parts of 10% w/v solution and 4% w/v are to be mixed to produce 8%w/v solution.
Solution:
Step 1: Identify the data given.
Higher concentration=10% w/v
Lower concentration=4% w/v
Required concentration=8% w/v
What is required ?
Alligation ratio=?
Step 2: Convert all the data into similar form and units
Here all the concentration terms are expressed in % w/v terms, hence no need to convert the format.
Step 3 : Apply alligation
Let’s indicate alligation ratio as H:L
Now to apply this method, subtract required concentration from higher concentration
H=8-4=4
Similarly, subtract lower concentration from required concentration
L=10-8=2
Hence alligation ratio is=H : L =4 :2=2:1
Watch here a video explaining alligation method along with examples.
Working example 2:
How many grams of 1.5 sp. gravity ointment should be mixed with 1.2 sp.gravity ointment to produce 15 gm of 1.3 sp.graviy ointment.
Solution:
Now let’s go another step in this pharmaceutical calculation i.e. to find out the weights of ointments to be mixed.
Step 1: Identify the data given.
Higher concentration (sp. gravity)=1.5
Lower concentration (sp. gravity)=1.2
Required concentration (sp. gravity)=1.3
What is required
Alligation ratio=?
Alligation volumes=?
Step 2: Convert all the data into similar form and units
Here again all the concentration terms are expressed in % w/v terms.
Step 3 : Apply alligation
Now applying the method,
H=1.3-1.2=0.1
Similarly,
L=1.5-1.3=0.2
Hence alligation ratio is=H : L =0.1 : 0.2=1:2
Therefore 1 part of 1.5 sp. gravity ointment should be mixed with 2 parts of 1.2 sp.gravity ointment.
Step 4 : Calculate individual weight/volumes to be mixed
The weight of the higher concentration can be calculated by the following formula
Since final weight of ointment is 15 g,the weight of the 1.5 sp gravity ointment required is
Similarly, the weight of lower concentration can be calculated by the following formula.
Hence the weight of 1.2 sp gravity ointment required is
Hence 5g of 1.5 sp gravity ointment should be mixed with 10g of 1.2 sp.gravity ointment to prepare 15g of 1.3 sp.gravtiy ointment.
Working example 3:
Alligations – Soln volumes
How many millilitres of drug suspensions containing 80 mg/5 mL and 30 mg/5ml should be mixed with to prepare 500 mL of a suspension of 10mg /ml
Solution:
Step 1: Identify the data given.
Higher concentration=80 mg/5ml
Lower concentration=30 mg/5ml
Required concentration=10 mg/ml
Final volume=500 ml
What is required
Alligation ratio=?
Alligation volumes=?
Step 2: Convert the two different concentrations into similar form and units.
Here two initial concentrations are given as gm per 5ml (80 mg/5 mL and 30 mg/5ml) whereas final concentration as gm/ml (10mg /ml).
Let’s convert all in equal format.
Higher concentration is 80 mg/5 mL =16mg /ml
Lower concentration is 30 mg/5 mL =6mg /ml
Step 3 : Apply alligation
Now applying the method,
H=10-6=4
Similarly,
L=16-10=6
Hence alligation ratio is=H : L =4 : 6=2:3
Therefore 2 parts of 80mg/5ml suspension should be mixed with 3 parts of 30mg/5ml suspension.
Step 4 : Get the volumes required
The fraction of higher concentration in mixture is= 2 / (3+2)
The fraction of lower concentration in mixture is= 3 / (3+2)
Final volume=500 ml
Hence the part/volume of higher concentration required is
Similarly the part/volume of lower concentration required is
Working example 4:
Alligations – Soln volumes
How many millilitres of a 1:20 v/v solution of a drug is can be prepared from 100 mL of 6% v/v solution
Solution:
Step 1: Identify the data given.
Since the diluting solution is not given it can be considered as water. Hence lower concentration is treated as 0% v/v.
Higher concentration=6% v/v
Lower concentration=0% v/v
Required concentration=1:20 v/v
Higher concentration volume=100 ml
What is required
Alligation ratio=?
Alligation volumes=?
Final volume=?
Step 2: Convert the two different concentrations into similar form and units.
Here we can clearly observe that all concentration terms are not expressed equal. Higher and lower concentrations are expressed in percentage (6% v/v) whereas required concentration is expressed in the ratio (1:20 v/v).
Let’s convert required concentration in to percentage expression.
The term 1:20 indicates 1 parts of solute in 20 parts of solution.
Hence 20 ml of solution contains 1ml of drug.
100 ml of solution contains
Now the concentrations are,
Higher concentration=6% v/v
Lower concentration=0% v/v
Required concentration=5% v/v
Step 3 : Apply alligation
Just as above,
H=5-0=5
Similarly,
L=6-5=1
Hence alligation ratio is=H : L =5 : 1
Therefore 5 parts of 6% v/v solution should be mixed with 1 part of water to produce 5% v/v solution.
Step 4 : Calculate individual volumes to be mixed
The volume of the higher concentration =100 ml
By substituting in the formula,
Hence a total volume of 120 ml of 5 % v/v solution can be prepared from 100 ml of 6% v/v and 20 ml of 0% v/v (water). |
Arithmetic Progression
NCERT Exercise 5.2
Part 6
Question 16: Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution: Solution: Given a3 = 16 and a7 – a5 = 12
a_3 = a + 2d = 16
a_5 = a + 4d
a_7 = a + 6d
As per question;
a + 6d – a – 4d = 12
Or, 2d = 12
Or, d = 6
Substituting the value of d in third term, we get;
a + 2 xx 12 = 16
Or, a + 24 = 16
Or, a = 16 – 24 = - 8
Thus, the AP can be given as follows:
-8, 4, 16, 28, 40, ………..
Question 17: Find the 20th term from the last term of the AP; 3, 8, 13, …………….253.
Solution: a = 3, d = 5
Now, 253 = a + (n + 1) d
⇒ 253 = 3 + (n -1) xx 5
⇒ 253 = 3 + 5n – 5 = – 2
⇒ 5n = 253 + 2 = 255
⇒ n = (255)/(5) = 51
Therefore, 20th term from the last term = 51 – 19 = 32
a_(32) = a + 31d
= 3 + 31 xx 5
= 3 + 155 = 158
Thus, required term is 158
Question 18: The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution: Given, a8 + a4 = 24 and a10 + a6 = 44
a_8 = a + 7d
a_4 = a + 3d
As per question;
a + 7d + a + 3d = 24
Or, 2a + 10d = 24
Or, a + 5d = 12 …………(1)
a_(10) = a + 9d
a + 9d + a + 5d = 44
Or, 2a + 14d = 44
Or, a + 7d = 22 ………….(2)
Subtracting equation (1) from equation (2);
a + 7d – a – 5d = 22 – 12
Or, 2d = 10
Or, d = 5
Substituting the value of d in equation (1), we get;
a + 5 xx 5 = 12
Or, a + 25 = 12
Or, a = - 13
Hence, first three terms of AP: – 13, – 8, – 3
Question 19: Subha Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000?
Solution: Here, a = 5000, d = 200 and an = 7000
We know, a_n = a + (n – 1)d
Or, 7000 = 5000 + (n – 1)200
Or, (n -1)200 = 7000 – 5000
Or, (n – 1)200 = 2000
Or, n – 1 = 10
Or, n = 11
Thus, 1995 + 10 = 2005
Hence, his salary reached at Rs. 7000 in 2005.
Question: 20 – Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs. 20.75, find n.
Solution: Here, a = 5, d = 1.75 and an = 20.75
We know, a_n = a + (n -1)d
Or, 20.75 = 5 + (n – 1)1.75
Or, (n – 1)1.75 = 20.75 – 5
Or, (n – 1)1.75 = 15.75
Or, n – 1 = 15.75/1.75 = 9
Or, n = 10 |
## new experiment 2
Data Handling
The word statistics is used for two different meanings.
In the singular sense, it is used as a science or a subject which deals with the collection,
classification, tabulation, representation and interpretation of the data.
In the plural sense, it is sometimes used for the numerical facts collected in the form of
numbers.
If we have collected information about the heights of class 6 children from ten different
schools of Delhi, then this information in the form of numbers is called statistics.
1. Data : Each number, collected for giving a required information, is called the data.
2. Bar Graph (Column Graph) : Bar graph is the simplest form of presenting a data. It
consists of bars (usually vertical), all of same widths. The heights of these bars are
drawn according to the number they represent.
3. Pie Graph : When the given data is represented by the sectors of a circle, the
resulting diagram (graph) obtained is called a pie-graph or a pie-chart.
EXERCISE 33 (A)
Question 1.
Marks scored by 30 students of class VI are as given below :
38, 46, 33, 45, 63, 53, 40, 85, 52, 75, 60, 73, 62, 22, 69, 43, 45, 33, 47, 41, 29, 43, 37,
49, 83, 44, 55, 22, 35 and 45. State:
(i) the highest marks scored.
(ii) the lowest marks scored.
(iii) the range of marks.
Solution:
(i) Highest marks scored = 85 .
(ii) Lowest marks scored = 22
(iii) Range of marks = 85
– 22 = 63
Question 2.
For the following raw data, form a discrete frequency distribution :
30,32,32, 28,34,34,32,30,30,32,32,34,30,32,32. 28,32,30, 28,30,32,32,30,28 and 30.
Solution:
Question 3.
Define :
(i) data
(ii) frequency of an observation.
Solution:
(i) Data : The word data means information in the form of numerical figures.
(ii) Frequency of an observation : The number of times a particular observation
occurs is called its frequency.
Question 4.
Rearrange the following raw data in descending order :
5.3, 5.2, 5.1, 5.7, 5.6, 6.0, 5.5, 5.9, 5.8, 6.1, 5.5, 5.8, 5.7, 5.9 and 5.4. Then write the :
(i) highest value
(ii) lowest value
(iii) range of values
Solution:
Writing these numbers in descending order we get:
6.1, 6.0, 5.9, 5.9, 5.8, 5.8, 5.7, 5.7, 5.6, 5.5, 5.5, 5.4, 5.3, 5.2, 5.1
(i) Highest value = 6.1
(ii) Lowest value = 5.1
(iii) Range of values = Highest value
– lowest value = 6.1 -5.1 = 1.0
Question 5.
Represent the following data in the form of a frequency distribution :
52, 56, 72, 68, 52, 68, 52, 68, 52, 60, 56, 72, 56, 60, 64, 56, 48, 48, 64 and 64.
Solution:
Question 6.
In a study of number of accidents per day, the observations for 30 days were
obtained as follows :
Solution:
The required frequency table will be as shown below :
Question 7.
The following data represents the w
eekly wages (in ₹) of 15 workers in a factory :
900, 850, 800, 850, 800, 750, 950, 900, 950, 800, 750, 900, 750, 800 and 850.
Prepare a frequency distribution table. Now find,
(i) how many workers are getting less than ₹850 per week?
(ii) how many workers
are getting more than ₹800 per week?
Solution:
(i)
Workers getting less than ₹850 per week
No. of workers getting ₹750 = 3 workers
No. of workers getting ₹800 = 4 works
∴ Workers getting less than ₹ 850 = 4 + 3 = 7 workers
(ii)
Workers are getting more than ₹800 per week
No. of workers getting ₹850 = 3
No. of workers getting ₹900 = 3
No. of workers getting ₹950 = 2
∴Workers getting more than ₹800 = 3 + 3 + 2 = 8 workers
Question 8.
Using the data, given below, construct a frequency distribution table : 9, 17, 12,
20, 9, 18, 25, 17, 19, 9, 12, 9, 12, 18, 17, 19, 20, 25, 9 and 12. Now answer the
following :
(i) How many numbers are less than 19?
(ii) How many numbers are more than 20?
(iii) Which of the numbers, given above, is occuring most frequently?
Solution:
The required frequency table will be as shown below :
(i) There are 14 numbers are less than 19.
(ii) There are 2 numbers more than 20.
(iii) 9 is occuring most frequently i.e. 5 times.
Using the following data, construct a frequency distribution table : 46, 44, 42, 54,
52, 60, 50, 58, 56, 62, 50, 56, 54, 58 and 48.
(i) What is the range of the numbers?
(ii) How many numbers are greater than 50?
(iii) How many numbers are between 40 and 50?
Solution:
(i) Range of numbers = Highest number
– Lowest number = 62 – 42 = 20
(ii) 9 numbers are greater than 50
(iii) 6 numbers are between 40 and 50 Ans.
EXERCISE 33 (B)
Question 1.
The sale of vehicles, in a particular city, during the first six months of the year
2016 is shown below :
vehicles sold
Draw a pictograph to represent the above data.
Question 2.
Solution:
(i) Cars sold by dealer A = 6 x 50 = 300
Cars sold by dealer D = 4 x 5 = 200 ,
∴ A sold more cars than dealer D by = 300 – 200 = 100
∴A has sold 100 more cars than dealer D.
(ii) No. of cars = 23
Scale = 50 cars
∴Total no. of cars = 23 x 50 = 1150 cars Ans.
Question 3.
The following pictograph shows the number of watches manufactured by a
factory, in a particular weeks.
Find
(i) on which day were the least number of watches manufactured ?
(ii) total number of watches manufactured in the whole week ?
Solution:
(i) On Friday least no. of watches manufactured by = 100 x 5 = 500 watches
(ii) Total no. of watches manufactured in the whole week = 100 x 42.5 = 4250 watches
Question 4.
The number or animals in five villages are as follows :
Prepare a pictograph of these animals using one symbol to represent 20 animals.
Question 5.
The following pictograph shows different subject books which are kept in a
school library.
Taking symbol of one book = 50 books, find :
(i) how many History books are there in the library ?
(ii) how many Science books are there in the library ?
(iii) which books are maximum in number ?
(i) There are 50 x 4 = 200 History books in the library.
(ii) There are 50 x 5.5 = 275 Science books in the library.
(iii) English books are maximum in number = 500 x 9 = 450 books.
EXERCISE 33 (C)
Question 1.
The following table gives the number of students in class VI in a school during
Represent the above data by a bar graph.
Solution:
Question 2.
The attendance of a particular class for the six days of a week are as given below
Draw a suitable graph.
Solution:
Question 3.
The total number of students present in class VI B, for the six days in a week
were as given below. Draw a suitable bar graph.
Question 4.
The following table shows the population of a particular city at different years :
Represent the above information with the help of a suitable bar graph.
Solution:
In a survey of 300 families of a colony, the number of children in each family was
recorded and the data has been represented by the bar graph, given below :
(i) How many families have 2 children each ?
(ii) How many families have no child ?
(iii) What percentage of families have 4 children ?
Solution:
(i) 60 families have 2 children each.
(ii) Zero
(iii) The percentage of families having 4 children =
x 100 = 20%
Question 6.
Use the data, given in the following table, to draw’ a bar graph
Out of A, B, C, D, E and F
(i) Which has the maximum value.
(ii) Which is greater A + D or B + E.
Solution:
(i) D has the maximum value of 350
(ii) A + D = 250 + 350 = 600
B + E = 300 + 275 = 575
Hence A + D is greater.
Question 7.
The bar graph drawn below shows the number of tickets sold during a fair by 6
students A, B, C, D, E and F.
Using the Bar graph, answer the following questions :
(i) Who sold the least number of tickets?
(ii) Who sold the maximum number of tickets ?
(iii) How many tickets were sold by A, B and C taken together ?
(iv) How many tickets were sold by D, E and F taken together ?
(v) What is the average number of tickets sold per student ?
Solution:
Question 8.
The following bar graph shows the number of children, in various classes, in a
school in Delhi.
Using the given bar graph, find :
(i) the number of children in each class.
(ii) the total number of children from Class 6 to Class 8.
(iii) how many more children there are in Class 5 compared to Class 6 ?
(iv) the total number of children from Class 1 to Class 8.
(v) the average number of children in a class.
Solution:
(i) In, Class 1 = 100, Class 2 = 90, Class 3 = 100, Class 4 = 80,
Class 5 = 120, Class 6 = 90, Class 7 = 70, Class 8 = 50
(ii) Class 6 = 90, Class 7 = 70, Class 8 = 50, Total number = 210
(iii) Number of student in class 5 = 120, Number of student in class 6 = 90
More children is class 5 = (120
– 90) = 30
(iv) Total number of children class 1 to 8 = 100 + 90+ 100+ 80 + 120 +90 + 70 + 50 =
700
Question 9.
The column graph, given above , shows the number of patients, examined by Dr.
V.K. Bansal, on different days of a particular week.
Use the graph to answer the following:
(i) On which day were the maximum number of patients examined ?
(ii) On which day were the least number of patients examined ?
(iii) On which days were equal number or patients examined ?
(iv) What is the total number of patients examined in the week ?
Solution:
(i) Tuesday were the maximum number of patients examined.
(ii) Friday were the least number of patients examined.
(iii) Sunday and Thursday were equal number of patient examined.
(iv) Total number of patients examined in the week .
= 50 + 40 + 70 + 60 + 50 + 30 = 300
Question 10.
A student spends his pocket money on various items, as given below :
Books : Rs. 380, Postage : Rs. 30, Cosmetics : Rs. 240, Stationary : Rs. 220 and
Entertainment : Rs. 120.
Draw a bar graph to represent his expenses.
Solution:
Amount spent on
Books = Rs. 380
Postage = Rs. 30
Cosmetics = Rs. 240
Stationary = Rs. 220
Entertainment = Rs. 120 |
## Key Concept
Case 1 :
If the denominator is in the form of √b (where b is a rational number), then we have to multiply both the numerator and denominator by the same √b to rationalize the denominator.
Case 2 :
If the denominator is in the form of a ± √b or a ± c√b (where b is a rational number), then we have to multiply both the numerator and denominator by its conjugate.
a + √b and a - √b are conjugate to each other
a + c√b and a - c√b are conjugate to each other
Case 3 :
If the denominator is in the form of ± √b (where a and b are rational numbers), then we have to multiply both the numerator and denominator by its conjugate.
a + √b and a - √b are conjugate to each other
## Examples
Example 1 :
Simplify :
18 / √6
Solution :
Simplifying the above radical expression is nothing but rationalizing the denominator.
So, rationalize the denominator.
Here, the denominator is √6.
In the given fraction, multiply both numerator and denominator by √6.
18 / √6 = (18√6) / (√6 ⋅ √6)
18 / √6 = 18√6 / 6
18 / √6 = 3√6
Example 2 :
Simplify :
1 / (2 + √5)
Solution :
Simplifying the above radical expression is nothing but rationalizing the denominator.
So, rationalize the denominator.
Here, the denominator is 2 + √5.
In the given fraction, multiply both numerator and denominator by the conjugate of 2 + √5. That is 2 - √5.
Example 3 :
Simplify :
(6 + √5) / (6 - √5)
Solution :
Simplifying the above radical expression is nothing but rationalizing the denominator.
So, rationalize the denominator.
Here, the denominator is 6 - √5.
In the given fraction, multiply both numerator and denominator by the conjugate of 6 - √5. That is 6 + √5.
(6 + √5) / (6 - √5) = [(6+√5)(6+√5)] / [(6-√5)(6+√5)]
(6 + √5) / (6 - √5) = [(6+√5)(6+√5)] / [(6-√5)(6+√5)]
(6 + √5) / (6 - √5) = (6 + √5)2 / [6- (√5)2]
(6 + √5) / (6 - √5) = [62 + 2(6)(√5) + (√5)2] / (36 - 5)
(6 + √5) / (6 - √5) = [36 + 12√5 + 5] / 31
(6 + √5) / (6 - √5) = (41 + 12√5) / 31
Example 4 :
Find the values of 'x' and 'y' :
(2 + √3)/(2 - √3) = x + y√3
Solution :
(2 + √3)/(2 - √3) = x + y√3
On the left side of the above equation, multiply both numerator and denominator by the conjugate of 2 - √3. That is 2 + √3.
[(2+√3)(2+√5)] / [(2-√3)(2+√3)] = x + y√3
(2 + √3)2 / [2- (√3)2] = x + y√3
[22 + 2(2)(√3) + (√3)2] / (4 - 3) = x + y√3
[4 + 4√3 + 3] / 1 = x + y√3
7 + 4√3 = x + y√3
Therefore,
x = 7
y = 4
Example 5 :
Simplify :
√(12x2) / √(30x)
Solution :
√(12x2) / √(30x) = √(12x2/30x)
√(12x2) / √(30x) = √(2x/5)
√(12x2) / √(30x) = √(2x) / √5
On the right side, multiply both numerator and denominator by √5.
√(12x2) / √(30x) = √(2x) ⋅ 5 / √5 ⋅ √5
√(12x2) / √(30x) = √(2x ⋅ 5) / 5
√(12x2) / √(30x) = √10x / 5
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Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Binomial Theorem Introduction:
#### What is Factorial (!):
The continuous multiplication of positive integers from 1 to n is called factorial. It is denoted by the exclamation symbol (!). The factorial "n" would be written as n! where "n" is an integer.
#### Expansion of n!:
The expansion of factorial "n" will be $$n! = n \ (n - 1) \ (n - 2) \ (n - 3).....$$
Note(1): The value of 0! is always 1 (0! = 1).
Note(2): The value of 1! is always 1 (1! = 1).
Example(1): Find the value of 5!?
Solution: $$5! = 5 (5 - 1) (5 - 2) (5 - 3) (5 - 4) (5 - 5)$$ $$5! = 5 \times 4 \times 3 \times 2 \times 1$$ $$5! = 120$$
Solution: $$5! = 5 (5 - 1) (5 - 2) (5 - 3) \\ (5 - 4) (5 - 5)$$ $$5! = 5 \times 4 \times 3 \times 2 \times 1$$ $$5! = 120$$
Example(2): Find the value of 12!?
Solution: $$12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$ $$12! = 479,001,600$$
Solution: $$12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \\ \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$ $$12! = 479,001,600$$
#### What is Binomial:
The binomial is a polynomial expression with two terms adding together such as $$(x + y)$$, $$(x - y)$$, $$(2x - y)$$, $$(x - 2y)$$, $$(2x + y)$$, etc. Here each binomial has two terms either positive or negative.
#### What is Exponent:
An exponent is a number that says how many times any term to be multiplied by itself.
Example: In $$5^2$$, the number 2 is an exponent. Here the term 5 has to multiply by itself two times.$$5^2 = 5 \times 5 = 25$$
Note(1): The value of exponent zero (0) is always 1. For example $$8^0 = 1$$, $$(a + b)^0 = 1$$, etc.
Note(2): The value of exponent one (1) is always the term that needs to be multiplied by 1 time. For example $$8^1 = 8$$, $$(a + b)^1 = (a + b)$$, etc.
#### What is Binomial Theorem:
The binomial theorem is an expansion of the power of a binomial. It makes it easy to solve any mathematical binomial expression.
The below expansion came from the multiplication of $$(a + b)$$, "n" number of times. If we multiply $$(a + b)$$, "n" number of times, it will consume a lot of time and effort. The binomial theorem makes it short and effortless to solve any binomial expression.
#### Binomial theorem formula:
For any positive integer "n", the binomial expansion formula would be $$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{(n - k)} b^k$$
The value of "k" in the above formula starts from "0" and goes till "n".
#### Binomial theorem Expansion:
We can get the binomial expansion by expanding the above formula as given below.
$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$
$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \\ \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$
Example(1): By using the binomial theorem, expand the binomial $$(x + 5)^4 ?$$
Solution: Given $$a = x$$, $$b = 5$$, and $$n = 4$$.
by putting these values into the binomial theorem expression till the fourth term as $$n = 4$$.
$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$
$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \\ \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$
$$(x + 5)^4 = \binom{4}{0} x^4 + \binom{4}{1} x^{(4 - 1)} (5) + \binom{4}{2} x^{(4 - 2)} (5)^2 + \binom{4}{3} x^{(4 - 3)} (5)^3 + \binom{4}{4} x^{(4 - 4)} (5)^4$$
$$(x + 5)^4 = \binom{4}{0} x^4 + \binom{4}{1} x^{(4 - 1)} (5) + \\ \binom{4}{2} x^{(4 - 2)} (5)^2 + \binom{4}{3} x^{(4 - 3)} (5)^3 + \\ \binom{4}{4} x^{(4 - 4)} (5)^4$$
$$= \frac{4!}{(4 - 0)! \ 0!} \ x^4 + \frac{4!}{(4 - 1)! \ 1!} \ x^3 (5) + \frac{4!}{(4 - 2)! \ 2!} \ x^2 (5)^2 + \frac{4!}{(4 - 3)! \ 3!} \ x^1 (5)^3 + \frac{4!}{(4 - 4)! \ 4!} \ x^0 (5)^4$$
$$= \frac{4!}{(4 - 0)! \ 0!} \ x^4 + \frac{4!}{(4 - 1)! \ 1!} \ x^3 (5) + \\ \frac{4!}{(4 - 2)! \ 2!} \ x^2 (5)^2 + \frac{4!}{(4 - 3)! \ 3!} \ x^1 (5)^3 + \\ \frac{4!}{(4 - 4)! \ 4!} \ x^0 (5)^4$$
$$= \frac{4!}{4! \ 0!} \ x^4 + \frac{4!}{3! \ 1!} \ x^3 (5) + \frac{4!}{2! \ 2!} \ x^2 (5)^2 + \frac{4!}{1! \ 3!} \ x^1 (5)^3 + \frac{4!}{0! \ 4!} \ x^0 (5)^4$$
$$= \frac{4!}{4! \ 0!} \ x^4 + \frac{4!}{3! \ 1!} \ x^3 (5) + \\ \frac{4!}{2! \ 2!} \ x^2 (5)^2 + \frac{4!}{1! \ 3!} \ x^1 (5)^3 + \\ \frac{4!}{0! \ 4!} \ x^0 (5)^4$$
$$= 1 \ x^4 + \frac{4 \times 3!}{3! \ 1!} \ x^3 (5) + \frac{4 \times 3 \times 2!}{2! \ 2!} \ x^2 (25) + \frac{4 \times 3!}{1! \ 3!} \ x (125) + \frac{4!}{1 \ 4!} \ (1) \ (625)$$
$$= 1 \ x^4 + \frac{4 \times 3!}{3! \ 1!} \ x^3 (5) + \frac{4 \times 3 \times 2!}{2! \ 2!} \ x^2 (25) + \\ \frac{4 \times 3!}{1! \ 3!} \ x (125) + \frac{4!}{1 \ 4!} \ (1) \ (625)$$
$$(x + 5)^4 = x^4 + 20 x^3 + 150 x^2 + 500 x + 625$$
$$(x + 5)^4 = x^4 + 20 x^3 + \\ 150 x^2 + 500 x + 625$$
Example(2): By using the binomial theorem, expand the binomial $$(2x - y)^3 ?$$
Solution: Given $$a = 2x$$, $$b = y$$, and $$n = 3$$.
by putting these values into the binomial theorem expression till the third term as $$n = 3$$.
$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$
$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \\ \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$
$$(2x - y)^3 = \binom{3}{0} (2x)^3 + \binom{3}{1} (2x)^{(3 - 1)} (-y) + \binom{3}{2} (2x)^{(3 - 2)} (-y)^2 + \binom{3}{3} (2x)^{(3 - 3)} (-y)^3$$
$$(2x - y)^3 = \binom{3}{0} (2x)^3 + \\ \binom{3}{1} (2x)^{(3 - 1)} (-y) + \\ \binom{3}{2} (2x)^{(3 - 2)} (-y)^2 + \\ \binom{3}{3} (2x)^{(3 - 3)} (-y)^3$$
$$= \frac{3!}{(3 - 0)! \ 0!} \ (2x)^3 + \frac{3!}{(3 - 1)! \ 1!} \ (2x)^2 (-y) + \frac{3!}{(3 - 2)! \ 2!} \ (2x)^1 (-y)^2 + \frac{3!}{(3 - 3)! \ 3!} \ (2x)^0 (-y)^3$$
$$= \frac{3!}{(3 - 0)! \ 0!} \ (2x)^3 + \frac{3!}{(3 - 1)! \ 1!} \ (2x)^2 (-y) + \\ \frac{3!}{(3 - 2)! \ 2!} \ (2x)^1 (-y)^2 + \\ \frac{3!}{(3 - 3)! \ 3!} \ (2x)^0 (-y)^3$$
$$= \frac{3!}{3! \ 0!} \ 8 \ (x)^3 + \frac{3!}{2! \ 1!} \ 4 \ (x)^2 (-y) + \frac{3!}{1! \ 2!} \ 2x \ (-y)^2 + \frac{3!}{0! \ 3!} \ (1) \ (-y)^3$$
$$= \frac{3!}{3! \ 0!} \ 8 \ (x)^3 + \frac{3!}{2! \ 1!} \ 4 \ (x)^2 (-y) + \\ \frac{3!}{1! \ 2!} \ 2x \ (-y)^2 + \\ \frac{3!}{0! \ 3!} \ (1) \ (-y)^3$$
$$= 8 \ x^3 + \frac{3 \times 2!}{2! \ 1!} \ 4 \ (x)^2 (-y) + \frac{3 \times 2!}{1! \ 2!} \ 2x \ (-y)^2 + (-y)^3$$
$$= 8 \ x^3 + \frac{3 \times 2!}{2! \ 1!} \ 4 \ (x)^2 (-y) + \\ \frac{3 \times 2!}{1! \ 2!} \ 2x \ (-y)^2 + (-y)^3$$
$$(2x - y)^3 = 8x^3 - 12 x^2 y + 6 x y^2 - y^3$$
$$(2x - y)^3 = 8x^3 - 12 x^2 y + \\ 6 x y^2 - y^3$$ |
Trig Comic strip part 2
Updated: 4/8/2021
This storyboard was created with StoryboardThat.com
Storyboard Text
• 18°
• The missiles are set up about 200 meters away from the watchtower, facing 18° upwards. How tall is the watchtower
• 200 meters
• 18°
• This a regular trig problem since we have both one angle and one side length. Since we are trying to find the opposite side length, we will using tangent or tan.
• the equation we will be using is Tan(18°)=x/200. First we will cross multiply so that the equation is now 200 times tan(18°) which is 64.9
• So the watchtower is about 65 meters tall
• 200 meters
• 65 meters
• Sir!
• Well done Sergeant, but your not done yet. There one more, it a bit more strenuous.
• 20°
• 50°
• The watchtower is sitting atop a wall that is 130 meters high. At 20° the missiles can hit the wall, and at 50° they can hit the tower. How tall is the tower
• 130 meters
• When we calculate that we get 357.17So the missiles are 357.2 meters away from the cliff. But we are not done yet
• 20°
• 50°
• Okay, the first triangle we have an angle and side length. We have the opposite angle and we are trying to find the adjacent angle. Meaning we using Tan.
• the equation we will be using is Tan(20°)=130/x. First we will cross multiply so that the equation is x=130/tan(20)
• 357.2 meters
• 130 meters
• When we calculate that we get 425.69. the the walls and watchtower combine height is about 426 meters. Now we subtract 130 from 426 which is 296. So the Watchtower is 296 meters tall
• 20°
• 50°
• Now we can do the second triangle. Since we have one angle and an adjacent side length we are using tan.
• the equation we will be using is Tan(50°)=x/357.2. First we will cross multiply so that the equation is x(tan(50) over tan(50) equal to 357.2 over tan(50)
• 357.2 meters
• 130 meters
• 296 meters |
50 in Words
# 50 in Words
Edited By Team Careers360 | Updated on Feb 10, 2023 04:46 PM IST
## How to write 50 in words?
In words, 50 is represented as "Fifty" The number 50 is used to signify a count or a value that is equivalent to the word "Fifty" in mathematics. For example, 50 Rupees would be written as 50 Rupees only or Fifty Rupees only on a check to indicate the value in words. When counting objects, we use the number fifty to begin a sentence. If you spent Rs. 50 on a new pen, for example, you may say, "I spent Rs. 50 on a new pen." You will discover an interesting way to translate the number 50 into words in this article.
## How to write 50 in words?
We typically use the English alphabet to represent numerals in words. Thus, 50 can be read as "Fifty" in English. We should create a place value chart that shows the place values for each of the two digits in the given number 50 since it has two digits. The Indian numbering system defines the order of place values for the digits in a number.
## The number 50 place value chart
The following observations for the place values of these constituent 2 digits of 50.
• The unit’s place digit in the number "50" is "0".
• The ten’s place digit in the number "50" is "5".
The place value of each digit can be observed as follows.
Ten’s place value Unit’s or One’s place value 5\times {{10}^{1}} 0\times {{10}^{0}}
Mathematically you can conclude the following
\begin{align} & 5\times {{10}^{1}}+0\times {{10}^{0}} \\ & =50 \\ \end{align}
And, thus 50 consists of 5 ten’s and 0 one’s, and hence, it is written as “fifty” in words.
## Is 50 an even number?
Even numbers are numbers that can be divided by two exactly without leaving a remainder. Examples of even numbers include 2, 6, 10, 20, and 50.
Similarly, on dividing 50 by two it results in the answer without remainder. It is clear that it is an even number as a result.
## Is 50 a prime number?
A whole number higher than 1 whose only factors are 1 and itself is known to be a prime number. Therefore, the number 50 is not a prime number because 50 can be divided by 1, 2, 5, 10, 25, and 50. A number must have exactly two factors in order to be considered as a prime number.
## Is 50 a cardinal number?
The cardinal numbers are the ones that are used to count anything. The counting numbers that begin at 1 and grow successively are known as cardinal numbers; they are not fractions.
You may have noticed that whenever you get a natural or counting number, you refer to it as a cardinal number. The number 50 obviously falls within the category of a cardinal number because it is a specific counting number.
## Can you write 50 as an ordinal number?
Ordinal numbers are those that represent something or someone's exact location in a place. If the number of things or people is given as a list, ordinal numbers are used to identify a set of the items or people. The adjective terms 1st, 2nd, 3rd, 4th, 5th, 6th, and so on are used to indicate something or someone's place in the order.
Thus, you can write 50th which indicates an ordinal number.
## What are the digits we use in mathematics?
Digits are used in mathematics to represent numbers as a single sign. As a result, we use the digits 0, 1, 2, 3, 4, 5, 6, 7 and 9 to perform mathematical operations and represent groups of numbers in our daily lives.
## How are numbers important in our daily life?
Our lives depend heavily on numbers. There are numbers involved in practically everything we do in mathematics. Whether we like it or not, since the day of our birth, numbers have been in charge of our lives. Numerous numbers have an impact on our life, either directly or indirectly.
Here are some examples of how we use numbers in real-world situations:
• Making a daily budget that accounts for costs like food, travel, and other expenses.
• Calculating the cost of the goods being offered in a mall.
• How long you spent at work or in class.
• Making a mobile phone call to a friend or a family member.
You can find out the detailed number in words article list below:-
20000 in Words 350000 in Words 1 to 100 in Words 65000 in Words 18000 in Words 28000 in Words 150 in Words 80 in Words 49000 in Words 400 in Words 11800 in Words 13500 in Words 12 in Words 71000 in Words 55000 in Words 150000 in Words 24000 in Words 59000 in Words 900 in Words 27000 in Words 12000 in Words 14400 in Words 12400 in Words 300000 in Words 17 in Words 12800 in Words 75 in Words 33000 in Words 21000 in Words 1180 in Words 41000 in Words 36000 in Words 1999 in Words 28 in Words 57000 in Words 4500 in Words 85000 in Words 95000 in Words 1200 in Words 14500 in Words 83000 in Words 48000 in Words 49 in Words 500000 in Words 60 in Words 70 in Words 10500 in Words 1100 in Words 25 in Words 27000 in Words 31000 in Words 34000 in Words 400000 in Words 7500 in Words 800 in Words 15500 in Words 16500 in Words 19500 in Words 250000 in Words 43000 in Words 44 in Words 45 in Words 47000 in Words 63000 in Words 6500 in Words 9500 in Words 99 in Words 10000000 in Words 1400 in Words 140000 in Words 200 in Words 21 in Words 4 in Words 41 in Words 52000 in Words 110000 in Words 125000 in Words 130000 in Words 2 in Words 23 in Words 24 in Words 33 in Words 35 in Words 450000 in Words 48 in Words 51000 in Words 54000 in Words 58000 in Words 5900 in Words 7 in Words 700 in Words 95 in Words 96000 in Words 12600 in Words 1900 in Words 20500 in Words 2400 in Words 26 in Words 2800 in Words 31 in Words 42 in Words 47 in Words 56000 in Words 600000 in Words 66000 in Words 68000 in Words 84000 in Words 88000 in Words 98 in Words 99999 in Words 11200 in Words 1200000 in Words 21500 in Words 27 in Words 29 in Words 32 in Words 32500 in Words 3300 in Words 350 in Words 3540 in Words 3600 in Words 37500 in Words 43 in Words 46 in Words 61000 in Words 67000 in Words 76000 in Words 800000 in Words 9 in Words 99000 in Words 96 in Words 10 in Words 1050 in Words 1111 in Words 120 in Words 1250 in Words 13200 in Words 180000 in Words 2000000 in Words 2300 in Words 24500 in Words 25500 in Words 27500 in Words 29500 in Words 26500 in Words 31500 in Words 37 in Words 38500 in Words 450 in Words 5000000 in Words 700000 in Words 72 in Words 77000 in Words 81000 in Words 85 in Words 86000 in Words 88 in Words 92000 in Words 93000 in Words 94 in Words 11100 in Words 11300 in Words 12200 in Words 123 in Words 13600 in Words 13900 in Words 15400 in Words 175000 in Words 1770 in Words 23600 in Words 29500 in Words 30500 in Words 33500 in Words 3700 in Words 3800 in Words 4200 in Words 42500 in Words 69000 in Words 7080 in Words 79000 in Words 82000 in Words 900000 in Words 97000 in Words 100000000 in Words 1000000000 in Words 3100 in Words 4700 in Words 4900 in Words 10700 in Words 1 to 20 in Words
1. Is the number 50 a perfect square?
No, 50 is not a perfect square. A perfect square is a number whose square root is a whole number and since the square root of 50 is $5\sqrt{2}$ , which is not a whole number. Therefore, 50 is not a perfect square.
2. What is the prime factorisation of 50?
The solution for the prime factorization of 50 is
$50={{2}^{1}}\times {{5}^{2}}$
where 2 and 5 are prime numbers.
3. What is the successor and predecessor of 50?
Successor of 50 is 51 and predecessor of 50 is 49.
4. What is an odd number in mathematics?
Odd numbers are those that cannot be evenly divided by two. It cannot be evenly split into two different numbers. An odd number will leave a remainder when divided by two. 1, 3, 5, 7, and other odd numbers are examples.
5. What are composite numbers?
Composite numbers are those that can be produced by multiplying the two lowest positive integers and have at least one factor other than the number "1" and themselves. Always more than two components are involved in these numbers.
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# Class 11 NCERT Solutions – Chapter 8 Binomial Theorem – Exercise 8.2
• Last Updated : 01 Apr, 2021
### Question 1. Find the coefficient of x5 in (x+3)8
Solution:
The (r+1)th term of (x+3)8 is given by Tr+1 = 8Cr(x)8-r(3)r (eq1).
Therefore for x5 we need to get 8-r =5 (Because we need to find x5. Therefore, power ox must be equal to 5)
So we get r=3.
Now, put r=3 in eq1. We get,
Coefficient of x5 = 8C3(x)5(3)3 = 8!*33/(4!*4!) = 1512
Coefficient of x5 is 1512.
### Question 2. Find the coefficient of a5b7 in (a-2b)12.
Solution:
The (r+1)th term of (a-2b)12 is given by Tr+1 = 12Cr(a)12-r(-2b)r × (eq1)
In the question it is given that exponent of b is 7. Therefore, r should be equal to 7.
By putting r=7 in eq1, we get
Coefficient of a5b7 = 12C7(-2)7 = -101376
### Question 3. Write the general term in the expansion of (x2−y)6.
Solution:
General term of the equation (a+b)n is given as Tr+1 = nCr(a)n-r(b)r.
In this question a= x2 and b=-y. After putting the value of a, b and n, we get the general term as
Tr+1 = 6Cr(x2)(12-r)(-y)6 = (-1)r 6Cr(x)(12-2r)(y)r.
### Question 4. Write the general term in the expansion of (x2-yx)12.
Solution:
General term of the equation (a+b)n is given as Tr+1 = nCr(a)n-r(b)r.
In this question a= x2 and b=-yx. After putting the value of a, b and n, we get the general term as
Tr+1 = 12Cr(x2)(12-r)(-yx)r = (-1)r 12Cr(x)(24-2r)(y)r(x)r = (-1)r 12Cr(x)(24-r)(y)
### Question 5. Find the 4th term in the expansion of (x-2y)12
Solution:
General term in the expansion of (a+b)is written as Tr+1 = nCr(a)n-r(b)r
In the question we are given that a = x, b = -2y and n=12.
To get the 4th term, we need to put r = 3 (Because r+1=4, therefore r=3).
Therefore, T4 = 12C3(x)12-3(-2y)3 = −1760x9y3
### Question 6. Find the 13th term in the expansion of (9x-1/3√x)18
Solution:
General term in the expansion of (a+b)n is written as Tr+1 = nCr(a)n-r(b)r
In this question a = 9x, b= -1/3√x and n=18.
To get the 13th term, we need to put r=12 (Because r+1=13, therefore r=12).
Therefore, T13 = 18C12(9x)18-12(-1/3√x)12 = 18564
### Question 7. Find the middle terms in the expansion of (3−x3/6)7.
Solution:
In the expansion of (a+b)n, if n is odd, then there are two middle terms, namely, ((n+1)/2)th and ((n+1)/2+1)th term.
Therefore, middle terms in the expansion of (3−x3/6)7are 4th term and 5th term.
T4 = T3+1 = 7C3(3)7-3(−x3/6)3 = (-1)37!34x9/4!.3!63=-105x9/8
T5 = T4+1 = 7C4(3)7-4(−x3/6)4= (-1)47!33x12/4!3!64 = 35x12/48
Thus, the middle terms are -105x9/8 and 35x12/48.
### Question 8. Find the middle terms in the expansion of (x/3+9y)10
Solution:
In the expansion of (a+b)n, if n is even, then the middle term is (n/2+1)th term.
Therefore, the middle term is 6th term.
T6 = T5+1 = 10C5(x/3)5(9y)5= (10!x595y5)/(5!5!35) = 61236x5y5
Thus, the middle term is 61236x5y5
### Question 9. In the expansion of (1+a)m+n, prove that coefficients of am and an are equal.
Solution:
Let us assume that am occurs in the (r+1)th term, we obtain
Tr+1 = m+nCr(1)m+n-r(a)r = m+nCrar
Comparing the indices of a in am and in Tr+1, we obtain r=m
Therefore, the coefficient of am is m+nCm = (m+n)!/m!n! …(1)
Let us assume that an occurs in the (k+1)th, we obtain
Tk+1 = m+nCk(1)m+n-k(a)k = m+nCkak
Comparing the indices of a in an and Tk+1, we obtain k-n
Therefore, the coefficient of an is m+nCn = (m+n)!/m!n! ….(2)
Thus, from (1) and (2), it can be obtained that the coefficients of am and an are equal.
### Question 10. The coefficients of the (r-1)th, rth and (r+1)th terms in the expansion of (x+1)n are in the ratio of 1:3:5. Find n and r.
Solution:
(r-1)th term in the expansion of (x+1)n is Tr-1 = nCr-2(x)n-(r-2)(1)r-2 = nCr-2(x)n-r+2
rth term in the expansion of (x+1)n is Tr = nCr-1(x)n-(r-1)(1)r-1 = nCr-1(x)n-r+1
(r+1)th term in the expansion of (x+1)n is Tr+1= nCr(x)n-r(1)r = nCr(x)n-r
Therefore the coefficients of (r-1)th , rth and (r+1)th in the expansion of (x+1)n are nCr-2, nCr-1 and nCr respectively.
Since these coefficients are in the ratio of 1:3:5, we obtain
nCr-2/nCr-1 = 1/3 and nCr-1/nCr = 3/5
Solving these two equations we get n-4r+5=0 and 3n-8r+3=0.
After solving these two equations, we get n=7 and r=3
Thus n=7 and r=3.
### Question 11. Prove that the coefficient of x^n in the expansion of (1+x)2n is twice the coefficient of xn in the expansion of (1+x)2n-1.
Solution:
In the expansion of (1+x)2n, Tn+1 = 2nCn(1)2n-n(x)n = 2nCnxn
Therefore, coefficient of xn in the expansion of (1+x)2n is 2nCn
2nCn = (2n)!/(n!)2 ….(1)
Similarly, coefficient of xn in the expansion of (1+x)2n-1 is 2n-1Cn
2n-1Cn = (2n)!/2.(n!)2 …(2)
From (1) and (2), 2.2n-1Cn = 2nCn
Hence, it is proved that the coefficient of xn in the expansion of (1+x)2n is twice the coefficient of xn in the expansion of (1+x)2n-1
### Question 12. Find a positive value of m for which the coefficient of x2 in the expansion of (1+x)m is 6.
Solution:
General term Tr+1 = mCr(1)m-r(x)r = mCr(x)r
Comparing the indices of x in x2 and Tr+1, we get r=2
Therefore, mC2 = 6
= 6
m!/(m-2)!=12
m(m-1) = 12=> m2-m-12 = 0
(m-4)(m+3) = 0
m cannot be negative. Therefore, m=4
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# Slope Intercept Form Parallel
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Slope Intercept Form Parallel – One of the numerous forms that are used to represent a linear equation one of the most commonly seen is the slope intercept form. The formula of the slope-intercept to solve a line equation as long as you have the straight line’s slope , and the yintercept, which is the coordinate of the point’s y-axis where the y-axis meets the line. Read more about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three main forms of linear equations: the standard, slope-intercept, and point-slope. Although they may not yield identical results when utilized in conjunction, you can obtain the information line produced more efficiently through this slope-intercept form. It is a form that, as the name suggests, this form makes use of an inclined line, in which you can determine the “steepness” of the line is a reflection of its worth.
This formula can be utilized to discover the slope of a straight line. It is also known as y-intercept, or x-intercept, where you can apply different available formulas. The line equation of this formula is y = mx + b. The slope of the straight line is indicated through “m”, while its intersection with the y is symbolized by “b”. Each point of the straight line can be represented using an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” need to remain variables.
## An Example of Applied Slope Intercept Form in Problems
The real-world in the real world, the slope intercept form is frequently used to depict how an object or problem evolves over it’s course. The value given by the vertical axis demonstrates how the equation addresses the magnitude of changes in the value provided through the horizontal axis (typically time).
An easy example of the use of this formula is to find out how the population grows in a specific area as the years go by. Using the assumption that the population in the area grows each year by a fixed amount, the amount of the horizontal line increases by one point for every passing year, and the point worth of the vertical scale will rise in proportion to the population growth by the set amount.
Also, you can note the starting point of a question. The starting value occurs at the y value in the yintercept. The Y-intercept represents the point where x is zero. In the case of a problem above the starting point would be at the time the population reading starts or when the time tracking starts, as well as the changes that follow.
The y-intercept, then, is the location that the population begins to be tracked in the research. Let’s assume that the researcher begins to do the calculation or the measurement in the year 1995. The year 1995 would become”the “base” year, and the x 0 points will occur in 1995. So, it is possible to say that the population of 1995 represents the “y”-intercept.
Linear equation problems that utilize straight-line formulas are nearly always solved in this manner. The starting value is represented by the yintercept and the change rate is represented in the form of the slope. The most significant issue with this form typically lies in the horizontal variable interpretation in particular when the variable is attributed to one particular year (or any kind in any kind of measurement). The most important thing to do is to make sure you know the meaning of the variables. |
## HOW MANY SQUARES ARE THERE IN A CHESSBOARD
Question
The chessboard is a rectangular grid of 64 squares. A diagonal square on the chessboard is then equal to two orthogonal squares. Hence, there are 32 white and 32 black squares. There are 32 white and 32 black squares on the chessboard.
## The chessboard is a rectangular grid of 64 squares.
The chessboard is a rectangular grid of 64 squares. Each square on the board has a unique coordinate, from 1 to 64.
## A diagonal square on the chessboard is then equal to two orthogonal squares.
A diagonal square on the chessboard is then equal to two orthogonal squares. Since these squares are also called oblique or long diagonals, we can write:
d2 = 2 * (1/2) = 1/4
## Hence, there are 32 white and 32 black squares.
Thus, there are 32 white and 32 black squares.
The chessboard is a rectangular grid of 64 squares. A diagonal square on the chessboard is then equal to two orthogonal squares (e.g., if you take any black square and cut it diagonally, you get two smaller orthogonal black squares). So if we multiply by 2 we get
32 x 2 = 64
## There are 32 white and 32 black squares on the chessboard.
There are 32 white and 32 black squares on the chessboard. The diagonal square is equal to two orthogonal squares, so there are 32 white and 32 black squares in total.
## Takeaway:
The chessboard is a rectangular grid of 64 squares. A diagonal square on the chessboard is then equal to two orthogonal squares, but we don’t need to know that for this problem. So all we need to do is count how many white and black squares there are on the board, which we can easily do using any basic counting method. There are 32 white and 32 black squares on a standard chessboard!
The chessboard is a rectangular grid of 64 squares. A diagonal square on the chessboard is then equal to two orthogonal squares. Hence, there are 32 white and 32 black squares on the chessboard.
1. # HOW MANY SQUARES ARE THERE IN A CHESSBOARD
## Introduction
We all know that a chessboard has 64 squares, but do you know how we arrived at that number? In this blog post, we’ll take a look at the math behind the chessboard and explore some of the patterns that emerge from its structure. Whether you’re a chess enthusiast or just interested in numbers and patterns, this post is for you!
## The mathematics of a chessboard
A chessboard consists of 64 squares arranged in an 8×8 grid. The math behind a chessboard is actually quite simple – each row contains eight squares, and there are eight rows. So, the total number of squares on a chessboard is 8×8, or 64.
Interestingly, the same math can be applied to determine the number of squares on any size grid. For example, a 4×4 grid would have 16 squares (4×4), and a 10×10 grid would have 100 squares (10×10). So, if you ever need to know how many squares are in a grid, just multiply the number of rows by the number of columns!
## Why the number of squares on a chessboard is important
The number of squares on a chessboard is important for several reasons. First, it allows the game to be played on a board of any size. Second, it provides a consistent playing surface for all players. Third, it eliminates the need for players to count squares during the game. Finally, it makes the game more visually appealing.
## How to use the number of squares on a chessboard to your advantage
In chess, the number of squares on the board is a major factor in determining the outcome of the game. The more squares there are, the more possible moves and combinations there are.
Chess experts have long used the number of squares on a chessboard to their advantage. In fact, many grandmasters (the highest-ranking chess players in the world) use what is called the “64 Square Principle” to help them plan their moves and strategies.
The 64 Square Principle is based on the fact that there are 64 squares on a chessboard. By understanding how many squares there are, and how they are arranged, grandmasters can better visualize the possibilities for move combinations and strategic planning.
While the 64 Square Principle is most often used by grandmasters, it can be applied to any level of chess playing. If you understand how many squares are on a chessboard, and how they are arranged, you can use this knowledge to your advantage when planning your own moves and strategies.
## Conclusion
A chessboard has 64 squares.
2. Chess is one of the oldest games in the world, and it’s still going strong today. But did you know that it’s also one of the most entertaining? This game has been played for centuries, but there are still some things about it that we don’t understand. One question that has puzzled many people: how many squares are there on a chessboard? Well today we’re going to solve this conundrum once and for all!
## There are 8 rows of 8 squares each.
There are 8 rows of 8 squares each. The first four rows are filled with black pieces and the last four rows are filled with white pieces. The first square is surrounded by white squares and has a black square at its center.
## The first square is surrounded by white squares and has a black square at its center.
The first square is surrounded by white squares and has a black square at its center.
The first square is the corner square of the chessboard, so it has two white neighbors and one black neighbor (or vice versa).
## The third square is surrounded by white squares and has a black square at its center.
The third square is surrounded by white squares and has a black square at its center. The fourth square, in turn, contains the third, with a white square on one side and two black ones on the other. The fifth square contains all four previous squares: one black in its center and three surrounding it (two white and one more with another black).
## The fourth square is surrounded by white squares and has a black square at its center.
The fourth square is surrounded by white squares and has a black square at its center.
It’s important to note that the fourth square can’t be any other color than black, because when you get to the fifth square, if it were another color than black then there would be two different-colored squares next to each other (which is impossible).
## The fifth square is surrounded by white squares and has a black square at its center.
The fifth square is surrounded by white squares and has a black square at its center. It is the middle square of the second row, which makes it unique among all other squares on the chessboard.
## The seventh square is surrounded by white squares and has a black square at its center.
The seventh square is surrounded by white squares and has a black square at its center. This is the only square that does not contain an opposing color, making it unique among all chessboards. |
# Conjugate Hyperbola
What is conjugate hyperbola?
If the transverse axis and conjugate axis of any hyperbola be respectively the conjugate axis and transverse axis of another hyperbola then the hyperbolas are called the conjugate hyperbola to each other.
The conjugate hyperbola of the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 is - $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1
The transverse axes of the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 is along x-axis and its length = 2a.
The conjugate axes of the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 is along y-axis and its length = 2b.
Therefore, the hyperbola conjugate to $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 will have its transverse and conjugate axes along y and x-axes respectively while the length of transverse and conjugate axes will be 2b and 2a respective.
Therefore, the equation of the hyperbola conjugate to $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 is - $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1
Thus, the hyperbolas $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 and - $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 are conjugate to each other.
The eccentricity of the conjugate hyperbola is given by a$$^{2}$$ = b$$^{2}$$(e$$^{2}$$ - 1).
Now we will come across various results related to the hyperbola $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 ……………. (i) and its conjugate - $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 ………………. (ii).
1. The co-ordinates of the centre of both the hyperbola (i) and its conjugate hyperbola (ii) are (0, 0).
2. The co-ordinates of the vertices of the hyperbola (i) are (-a, 0) and (a, 0) and its conjugate hyperbola (ii) are (0, -b) and (0, b).
3. The co-ordinates of the foci of the hyperbola (i) are (-ae, 0) and (ae, 0) and its conjugate hyperbola (ii) are (0, be) and (0, -be).
4. The length of the transverse axis of the hyperbola (i) is 2a and its conjugate hyperbola (ii) is 2b.
5. The length of the conjugate axis of the hyperbola (i) is 2b and its conjugate hyperbola (ii) is 2a.
6. The eccentricity of the hyperbola (i) is e = $$\sqrt{\frac{a^{2} + b^{2}}{a^{2}}}$$ or, b$$^{2}$$ = a$$^{2}$$(e$$^{2}$$ - 1) and its conjugate hyperbola (ii) is e = $$\sqrt{\frac{b^{2} + a^{2}}{b^{2}}}$$ or, a$$^{2}$$ = b$$^{2}$$(e$$^{2}$$ - 1).
7. The length of the latusrectum of the hyperbola (i) is $$\frac{2b^{2}}{a}$$ and its conjugate hyperbola (ii) is $$\frac{2a^{2}}{b}$$.
8. The equation of the transverse axis of the hyperbola (i) is y = 0 and its conjugate hyperbola (ii) is x = 0.
9. The equation of the conjugate axis of the hyperbola (i) is x = 0 and its conjugate hyperbola (ii) is y = 0.
The Hyperbola
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# Problem 9: The Graph of a Derivative
## 2011-2012
In the below graph, two functions are pictured, f(x) and its derivative, but I can’t seem to tell which is which.
According to those graphs, which is greater:
### Solution:
You might have noticed that the red function has an even degree whereas the blue has an odd degree. Why? An even-degreed function’s ends will either both point up or both point down (here they both point up at the edges of the graph), whereas a function with an odd degree has ends that go in opposite directions (like the blue graph which goes down to the left but up to the right). Even though this is interesting, it is not enough to answer the question.
You may also say that the red graph is definitely a lesser degree because it has a fewer number of x-intercepts (4) than the blue graph (5), but this, too, is not sufficient information to answer the question correctly. It is true that the blue graph is f(x) and the red is f ‘(x), but for different reasons.
Here are two acceptable justifications for identifying the blue graph as f(x):
• Whenever the blue graph has a relative max or min (hilltop or valley bottom, respectively), the red graph has an x-intercept; remember that the derivative of a function is either 0 or undefined when the original function has a relative extrema point
• Whenever the blue graph is increasing, the red graph is positive (i.e., above the x-axis), and when the blue graph is decreasing, the red is negative
Now that we’ve got that straight, let’s get down to answering the question. You have no idea what the exact function values are, but it turns out that you don’t really need to know them! Since the red graph is decreasing at x = –1/4, then its derivative must be negative there (for the same reason that the second bullet above is true).
What about f (–1)? Because the red graph represents f ‘(x), it’s easy to see that the red graph is just above the x-axis when x = –1. Therefore, f ‘(–1) > 0.
Back to the original question, which asks you to compare two values. You now know that one of those values is positive and the other is negative, so the positive value must be greater.
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# The variance of 15 observations is 6.
Question:
The variance of 15 observations is 6. If each observation is increased by 8, find the variance of the resulting observations.
Solution:
Let the observations are $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}_{4}, \ldots, \mathrm{x}_{15}$
and Let mean $=\overline{\mathrm{X}}$
Given: Variance $=6$ and $n=15$
We know that,
Variance, $\sigma^{2}=\frac{1}{\mathrm{n}} \sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}$
Putting the given values, we get
$6=\frac{1}{15} \sum\left(x_{i}-\bar{x}\right)^{2}$
$\Rightarrow 6 \times 15=\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}$
$\Rightarrow 90=\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}$
or $\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}=90$ .........(i)
It is given that each observation is increased by 8, we get new observations
Let the new observation be $\mathrm{y}_{1}, \mathrm{y}_{2}, \mathrm{y}_{3}, \ldots, \mathrm{y}_{15}$
where $y_{i}=x_{i}+8 \ldots$ (ii) or $x_{i}=y_{i}-8 \ldots$ (iii)
Now, we find the variance of new observations
i. e. New Variance $=\frac{1}{\mathrm{n}} \sum\left(\mathrm{y}_{\mathrm{i}}-\overline{\mathrm{y}}\right)^{2}$
Now, we calculate the value of $\overline{\mathrm{Y}}$
We know that,
Mean $=\frac{\text { Sum of observations }}{\text { Total number of observations }}$
$\Rightarrow \overline{\mathrm{y}}=\frac{\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{y}_{\mathrm{i}}}{\mathrm{n}}$
$\Rightarrow \bar{y}=\frac{\sum_{1=1}^{15} x_{1}+8}{15}$ [from eq. (ii)]
$\Rightarrow \overline{\mathrm{y}}=\left(\frac{1}{15}\right)\left\{\sum_{\mathrm{i}=1}^{15}\left(\mathrm{x}_{\mathrm{i}}+8\right)\right\}$
$\Rightarrow \overline{\mathrm{y}}=\frac{1}{15}\left[\sum_{\mathrm{i}=1}^{15} \mathrm{x}_{\mathrm{i}}+8 \sum_{\mathrm{i}=1}^{15} 1\right]$
$\Rightarrow \bar{y}=\frac{1}{15} \sum_{i=1}^{15} \mathrm{x}_{\mathrm{i}}+8 \times \frac{15}{15}$
$\Rightarrow \overline{\mathrm{y}}=\overline{\mathrm{x}}+8$
$\Rightarrow \overline{\mathrm{x}}=\overline{\mathrm{y}}-8$ …(iv)
Putting the value of eq. (iii) and (iv) in eq. (i), we get
$\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}=90$
$\sum\left(\mathrm{y}_{\mathrm{i}}-8-(\overline{\mathrm{y}}-8)\right)^{2}=90$
$\Rightarrow \sum\left(y_{i}-8-\bar{y}+8\right)^{2}=90$
$\Rightarrow \sum\left(y_{i}-\bar{y}\right)^{2}=90$ so
New Variance $=\frac{1}{n} \sum\left(y_{i}-\bar{y}\right)^{2}$
$=\frac{1}{15} \times 90$
$=6$ |
what is 1/2 of 5/6 as a fraction
# what is 1/2 of 5/6 as a fraction
These fractions are 1/2, 2/4, 3/6 representing the parts taken from the total number of parts.Mixed fractions can be written either as a whole part plus a proper fraction or entirely as an improper fraction. Percent is another way of writing fractions with a denominator of 100. . Percent refers to a value compared to a whole expressed as 100, rather than.It is the denominator of a fraction equal to a percent fraction. Decimal or Decimal Number. Fraction Tutorial Part I: The Concept of a fraction. A fraction can be thought of as part of a whole. Below are several pictures that attempt to illustrate the concept of the fraction , and respectively. A visual representation of fractions as slices of a pizza pie. Fractions quiz. 6. If you change 2/4 to 4/8 have you simplified the fraction? A) Yes B) No.4. How would you write six quarters as an improper fraction? We arrange the numbers in numerical order to get the list 1, 2, 3, 5, 6, 8, 9, 10. The two middle numbers are 5 and 6. The median is the average of 5 and 6. We add 5 and 6 and then divide the resulting sum by 2.possible outcomes. 28.
a. A pint is what fraction of a quart? Time to look at the numbers in between 0 - 1! Theyre called fractions. Here is a tutorial that shows how numbers can be broken. Now you know what is a A fraction is an equal part of a whole. When you cut an apple.1 5. of the pie. Exercise 1: Anything that is divided into equal parts can be referred to in terms of fractions. Look at the following chart. |
# 2.3 Differentiabilityap Calculus
(1) Find the derivatives of the following functions using first principle.
All students take calculus All sin tan cos rule. Trigonometric ratios of some negative angles. Trigonometric ratios of 90 degree minus theta. Trigonometric ratios of 90 degree plus theta. Trigonometric ratios of 180 degree plus theta. Trigonometric ratios of 180 degree minus theta. Trigonometric ratios of 180 degree plus theta. Prepare your calculus students for the topics they need to know to succeed in Calculus 2 and AP Calculus BC. Students tend to forget the PreCalculus topics they didn't use in Calculus 1 or AB. This review can be used for summer review or between semesters for Calculus BC. Since the first number is negative and the second number is positive and f(x) is a continuous function on the interval 2, 3, by the Intermediate Value Theorem, f(x) must have a solution between 2 and 3. Example 2: Show that the function has only one real solution. Live: Biden on the verge of reaching 270 electoral votes. Anne Hathaway apologizes to disability community. Did the 49ers swing a local California election?
(i) f(x) = 6Solution
(ii) f(x) = -4x + 7Solution
(iii) f(x) = -x2 + 2 Solution
(2) Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?
(i) f(x) = x - 1 Solution
(ii) f(x) = √(1 - x2) Solution
(3) Determine whether the following function is differentiable at the indicated values.
(i) f(x) = x x at x = 0 Solution
(ii) f(x) = x2 - 1 at x = 1 Solution
(iii) f(x) = x + x - 1 at x = 0, 1 Solution
(iv) f(x) = sin x at x = 0 Solution
(4) Show that the following functions are not differentiable at the indicated value of x.
(i)
Solution
(5) The graph of f is shown below. State with reasons that x values (the numbers), at which f is not differentiable.
Solution
(6) If f(x) = x + 100 + x2, test whether f'(-100) exists.
(7) Examine the differentiability of functions in R by drawing the diagrams.
(i) sin x Solution
(ii) cos x Solution
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Example 1:
Show that the function has a solution between 2 and 3.
Solution
:
Plugging in 2 and 3 into f(x), we see that f(2) = ln(2) 1 -0.307 and f(3) = ln(3) 1 0.099. Since the first number is negative and the second number is positive and f(x) is a continuous function on the interval [2, 3], by the Intermediate Value Theorem, f(x) must have a solution between 2 and 3.
Example 2:
Show that the function has only one real solution.
Solution:
First we use the Intermediate Value Theorem to show that there is at least one solution. We can use the theorem since f(x) is a continuous function everywhere. Notice that f(1) = -1 and f(2) = 45. That means that somewhere between 1 and 2, f(x) = 0.
Well, we have shown that there is at least one solution to the equation. Now we have to show that it is the only solution. To do that, we shall show that it is not possible for f(x) to have a second solution.
We do this by taking the derivative. . Notice that . That means that f(x) is a strictly increasing function. A strictly increasing function will only have one x-intercept (solution), and thus f(x) has only one real solution.
Problems For You To Solve
Multiple Choices:
1. In what situation f'(x) might NOT fail to exist？
A. cuspe
B. conner
C. jump
D. continuous
E. IDK
2. Let f be a continuous function on the closed interval [-3.6]. If f (-3) =- 1 and f (6) = 3, then the Intermediate Value Theorem guarantees that
A. f(0)=0
B. for at least one c between -3 and 6
C. for all x between -3 and 6
D. f(c)=1 for at least one c between -3 and 6
E. f(c)=0 for at least one c between -1 and 3
3. The function f is continuous on the closed interval [0, 2] and has values that are given in the table above. The equation must have at least two solutions in the interval [0, 2] if k=
A. 0
B. 1
C. 2
D. 3
E. 4
4. Let f be a function that is differentiable on the open interval (1, 10). If f(2) = -5, f(5) = 5, and f(9) = -5, which of the following must be true?
I. f has at least 2 zeros.
II. The graph of f has at least one horizontal tangent.
III. For some c, 2 < c < 5 , f(c)=3.
A. None
B. I only
C. I and II only
D. I and III only
E. I, II and III
5. The function f is continuous for and differentiable for -2 < x < 1. If f(-2) =-5 and f(l) = 4, which of the following statements could be false?
A. There exists c, where -2 < c < 1, such that f (c) = 0.
B. There exists c, where -2 < c < 1, such that f '(c) = 0.
C. There exists c, where -2 < c < 1, such that f(c) =3.
D. There exists c, where -2 < c < 1, such that f '(c) = 3.
6. Let f(x) = x x and g(x) =sin x. Assertion : gof is differentiable at x = 0 and its derivative is continuous at that point Reason : gof is twice differentiable at x = 0.
### 2.3 Differentiabilityap Calculus Test
C. Both assertion and reason are true but reason is not the correct explanation of assertion
7. Function f(x) =Ixl + x-1 is not differentiable at
B. x=0,1
D. x=1,2
### 2.3 Differentiabilityap Calculus Notes
8. If f(x) = x , then f'(0) =
B. X
D. None of these
9.
10.
11. Let f be a differentiable odd function defined on R. (That is f(-x) = -f(x) for all x in R.) Let a be a positive number. How many solution(s) could the equation af'(x) = f(a) have?
A. 0
C. 2
12. Let f be a quadratic function defined on the interval [a,b] with 0 < a < b. Which one of the followings is the value of c in (a,b) such that f(b)-f(a) = f'(c)(b-a) ?
A. (a+b)/2
13. The Mean Value Theorem is applied to the function f(x) = x3 + qx2 + 5x - 6 on the interval [0,2]. Suppose that the number c determined by the theorem is equal to 2. Which one of the followings is the value of q ?
D. -4
14. Which of the following functions satisfy the conditions of the Mean value Theorem on their domains ?
(I) f(x) = x¾ for all x in [-1,1]. (II) g(x) = x-1 for all x in [-1,1]. (III) h(x) = x/(1-x2) for all x in (-1,1). (IV) k(x) = 1- x 3 for all x in [-1,½].
B. I and IV
D. II and IV
15. Let g(x) be a differentiable function defined on R and f(x) = g(x)sinx. How many real solutions does the equation g(x)cosx + g'(x)sinx = 0 have ?
Free Response: |
# How do you find the solution of a difference equation?
## How do you find the solution of a difference equation?
Steps
1. Substitute y = uv, and.
2. Factor the parts involving v.
3. Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)
4. Solve using separation of variables to find u.
5. Substitute u back into the equation we got at step 2.
6. Solve that to find v.
## What is order difference equation?
Order of a differential equation is the order of the highest derivative (also known as differential coefficient) present in the equation. Example (i): d3xdx3+3xdydx=ey. In this equation, the order of the highest derivative is 3 hence, this is a third order differential equation.
What is first order difference equation?
A first-order differential equation is defined by an equation: dy/dx =f (x,y) of two variables x and y with its function f(x,y) defined on a region in the xy-plane. It has only the first derivative dy/dx so that the equation is of the first order and no higher-order derivatives exist.
### What is a difference equation?
A difference equation is any equation that contains a difference of a variable. The classification within the difference equations depends on the following factors. Order of the equation. The order of the equation is the highest order of difference contained in the equation.
Why does a second-order differential equation have two solutions?
5 Answers. second order linear differential equation needs two linearly independent solutions so that it has a solution for any initial condition, say, y(0)=a,y′(0)=b for arbitrary a,b. from a mechanical point of view the position and the velocity can be prescribed independently.
## What is the difference between first and second-order differential equations?
As for a first-order difference equation, we can find a solution of a second-order difference equation by successive calculation. The only difference is that for a second-order equation we need the values of x for two values of t, rather than one, to get the process started.
## What is the difference between difference equation and differential equation?
The main difference between them, as others have pointed out, is that difference equations take discrete steps of finite size, whereas differential equations are about continuous flows, where the individual steps are so small that they cannot be distinguished, because the differentials are so small that , whereas is …
What is meant by difference equation?
difference equation, mathematical equality involving the differences between successive values of a function of a discrete variable.
### What is difference equation with example?
General Differential Equations. Consider the equation y′=3×2, which is an example of a differential equation because it includes a derivative. There is a relationship between the variables x and y:y is an unknown function of x. Furthermore, the left-hand side of the equation is the derivative of y.
### What is the solution of second order equation?
– Take any equation with second order differential equation – Let us assume dy/dx as an variable r – Substitute the variable r in the given equation – It will form a binomial equation – Solve the equation and find its factors – Find the value of y
What is the second order differential equation?
Second-Order Linear Equations. The order of a differential equation is the order of the highest derivative appearing in the equation. Thus, a second‐order differential equation is one that involves the second derivative of the unknown function but no higher derivatives.
## What is second order in math?
In mathematical logic, second-order arithmetic is a collection of axiomatic systems that formalize the natural numbers and their subsets. It is an alternative to axiomatic set theory as a foundation for much, but not all, of mathematics. It was introduced by David Hilbert and Paul Bernays in their book Grundlagen der Mathematik .
## What is second order?
Definitions for SECOND-ORDER SECOND-ORDER. Here are all the possible meanings and translations of the word SECOND-ORDER. As an adjective it is the re-make, retooling or retrofit of a previous model; be it an idea, machinery, or a physical structure. A 1947 Ford with a newer motor is undergoing second-order upgrades. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Division of Polynomials
## Using long division to divide polynomials
Estimated20 minsto complete
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Progress
Practice Division of Polynomials
Progress
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Division of a Polynomial by a Monomial
Can you divide the polynomial by the monomial? How does this relate to factoring?
### Watch This
James Sousa: Dividing Polynomials by Monomials
### Guidance
Recall that a monomial is an algebraic expression that has only one term. So, for example, , 8, –2, or are all monomials because they have only one term. The term can be a number, a variable, or a combination of a number and a variable. A polynomial is an algebraic expression that has more than one term.
When dividing polynomials by monomials, it is often easiest to separately divide each term in the polynomial by the monomial. When simplifying each mini-division problem, don't forget to use exponent rules for the variables. For example,
.
Remember that a fraction is just a division problem!
#### Example A
What is ?
Solution: This is the same as . Divide each term of the polynomial numerator by the monomial denominator and simplify.
Therefore, .
#### Example B
What is ?
Solution: Divide each term of the polynomial numerator by the monomial denominator and simplify. Remember to use exponent rules when dividing the variables.
Therefore, .
#### Example C
What is ?
Solution: This is the same as . Divide each term of the polynomial numerator by the monomial denominator and simplify. Remember to use exponent rules when dividing the variables.
Therefore, .
#### Concept Problem Revisited
Can you divide the polynomial by the monomial? How does this relate to factoring?
This process is the same as factoring out a from the expression .
Therefore, .
### Vocabulary
Divisor
A divisor is the expression in the denominator of a fraction.
Monomial
A monomial is an algebraic expression that has only one term. , 8, –2, or are all monomials because they have only one term.
Polynomial
A polynomial is an algebraic expression that has more than one term.
### Guided Practice
Complete the following division problems.
1.
2.
3.
1.
2.
3.
### Practice
Complete the following division problems.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 7.12.
### Vocabulary Language: English
Denominator
Denominator
The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
Dividend
Dividend
In a division problem, the dividend is the number or expression that is being divided.
divisor
divisor
In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Polynomial long division
Polynomial long division
Polynomial long division is the standard method of long division, applied to the division of polynomials.
Rational Expression
Rational Expression
A rational expression is a fraction with polynomials in the numerator and the denominator.
Rational Root Theorem
Rational Root Theorem
The rational root theorem states that for a polynomial, $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_n, a_{n-1}, \cdots a_0$ are integers, the rational roots can be determined from the factors of $a_n$ and $a_0$. More specifically, if $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$, then all the rational factors will have the form $\pm \frac{p}{q}$.
Remainder Theorem
Remainder Theorem
The remainder theorem states that if $f(k) = r$, then $r$ is the remainder when dividing $f(x)$ by $(x - k)$.
Synthetic Division
Synthetic Division
Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.
### Explore More
Sign in to explore more, including practice questions and solutions for Division of Polynomials. |
# How To Remember Customary System Measurement Conversions - A Mnemonic Device For Cups, Pints, Quarts, Gallons
## Remembering Customary Unit Measurements
In the United States, we measure things using customary units. The volumetric measures are cups, pints, quarts, and gallons. Remembering these measurements and their conversions can be difficult. People of all ages struggle with knowing how many cups, pint, quarts, and gallons make up everyday things. Many school children have problems remember the amounts and how to convert between measurements when they first learn measurements in grade school. Adults may struggle with these conversions while cooking. How many cups can your 2 quart pot hold? How many pints of water do you need to drink each day?
There is an easy way to remember the measurements. This method can also be used to do the measurement conversions.
## Conversions Using The Square Method
To convert between cups and gallons you can draw a square and divide it in half multiple times. When you are done you will have a graphic representation of cups, pints, quarts, and gallons. If you need to convert measurements you can simply draw the squares and count the boxes.
## 1. Cups
To start off, get out a piece of paper. Draw one square on it. This square is one cup. A cup holds 8 ounces.
## 2. Pints
Take your square and draw a line from the top to the bottom dividing it in half. Then draw a line from left to right. You will now have two rectangles. Each rectangle has two squares in it. The rectangles are pints. The squares are cups. Each pint has two cups.
## 3. Quarts
The big box that now has four squares represents a quart.
Each quart has two pints or four cups.
## 4. Gallons
We are going to take the big box with four squares and divide it in half a few more times. First, divide the left column in half by drawing a line up and down through it. Next, do the same thing to the right column. After that, draw a line dividing the top row in half left to right. Then, do the bottom row the same way. You should now have sixteen squares.
Each of the little squares is one cup. There are sixteen cups in a gallon.
Each of the big squares in one quart (notice how it sounds like quarter, or quarter gallon). There are four quarts in a gallon.
You can also see that 8 pints are in one gallon.
Notice now each big square has four little squares? There are four cups in a quart.
## Popular
28
23
• ### Converting within the Metric System using the Metric Staircase
8
0 of 8192 characters used
• Author
Lena Welch 3 years ago from USA
I am sorry. i don't right off. I learned this one from my mentor teacher when I was doing fieldwork.
• kayla 3 years ago
ok first off i am a new culinary student and are teacher has asked up to remember the cups quarts and pints and oz s and teaspoons and tablespoons. This came in very handy for me to remember this thank you so much. if you know of any sites now that can help me with the teaspoon or tablespoon part be greatly appreciated
• Author
Lena Welch 5 years ago from USA
You're welcome!
• caltex 5 years ago
Never thought of it this way. This sure is very helpful. Thanks for sharing!
• Author
Lena Welch 5 years ago from USA
Not showing up? odd.... You are approved. Wonder if the lag monster got it. I can see them. If they don't show up maybe I will have to ask support that question. Let me know.
• kjforce 5 years ago from Florida
Hey..Lwelch..where did my comment go ?
• Author
Lena Welch 5 years ago from USA
You are welcome! I learned it while doing fieldwork at a school last year. If only I had been taught this trick earlier! I never could remember the differences. That is a great idea for a kitchen tool.
Thanks for stopping by!
• kjforce 5 years ago from Florida
Lwelch..WOW !! what a geat and handy idea for in the kitchen....as a matter of fact you gave me an idea for holiday gifts..I will make small wooden plaques with this info written on them, they can hang in a kitchen......they also can be done as a small poster for wall hanging at home or in the classroom...thank you so much for sharing.... |
# How to Calculate the Sum of Interior Angles
A polygon is any closed figure with sides made from straight lines. At each vertex of a polygon, there is both an interior and exterior angle, corresponding to the angles on the inside and outside of the closed figure. Understanding the relationships that govern these angles is useful in various geometrical problems. In particular, it is helpful to know how to calculate the sum of interior angles in a polygon. This can be done using a simple formula, or by dividing the polygon into triangles.
Method 1
Method 1 of 2:
### Using the Formula
1. 1
Set up the formula for finding the sum of the interior angles. The formula is ${\displaystyle sum=(n-2)\times 180}$, where ${\displaystyle sum}$ is the sum of the interior angles of the polygon, and ${\displaystyle n}$ equals the number of sides in the polygon.[1] [2]
• The value 180 comes from how many degrees are in a triangle. The other part of the formula, ${\displaystyle n-2}$ is a way to determine how many triangles the polygon can be divided into. So, essentially the formula is calculating the degrees inside the triangles that make up the polygon.[3]
• This method will work whether you are working with a regular or irregular polygon. Regular and irregular polygons with the same number of sides will always have the same sum of interior angles, the difference only being that in a regular polygon, all interior angles have the same measurement.[4] In an irregular polygon, some of the angles will be smaller, some of the angles will be larger, but they will still add up to the same number of degrees that are in the regular shape.
2. 2
Count the number of sides in your polygon. Remember that a polygon must have at least three straight sides.[5]
• For example, if you want to know the sum of the interior angles of a hexagon, you would count 6 sides.
3. 3
Plug the value of into the formula. Remember, ${\displaystyle n}$ is the number of sides in your polygon.[6]
• For example, if you are working with a hexagon, ${\displaystyle n=6}$, since a hexagon has 6 sides. So, your formula should look like this:
${\displaystyle sum=(6-2)\times 180}$
4. 4
Solve for . To do this, subtract 2 from the number of sides, and multiply the difference by 180. This will give you, in degrees, the sum of the interior angles in your polygon.[7]
• For example, to find out the sum of the interior angles of a hexagon, you would calculate:
${\displaystyle sum=(6-2)\times 180}$
${\displaystyle sum=(4)\times 180}$
${\displaystyle sum=(4)\times 180=720}$
So, the sum of the interior angles of a hexagon is 720 degrees.
Method 2
Method 2 of 2:
### Drawing Triangles
1. 1
Draw the polygon whose angles you need to sum. The polygon can have any number of sides and can be regular or irregular.
• For example, you might want to find the sum of the interior angles of a hexagon, so you would draw a six-sided shape.
2. 2
Choose one vertex. Label this vertex A.
• A vertex is a point where two sides of a polygon meet.
3. 3
Draw a straight line from Point A to each other vertex in the polygon. The lines should not cross. You should create a number of triangles.
• You do not have to draw lines to the adjacent vertices, since they are already connected by a side.
• For example, for a hexagon you should draw three lines, dividing the shape into 4 triangles.
4. 4
Multiply the number of triangles you created by 180. Since there are 180 degrees in a triangle, by multiplying the number of triangles in your polygon by 180, you can find the sum of the interior angles of your polygon.[8]
• For example, since you divided your hexagon into 4 triangles, you would calculate ${\displaystyle 4\times 180=720}$ to find a total of 720 degrees in the interior of your polygon.
## Community Q&A
Search
• Question
How do I find a single interior angle?
Work out what all the interior adds up to, then divide by however many sides the shape has.
• Question
How do I calculate the number of sides of a polygon if the sum of the interior angles is 1080?
Donagan
Divide that sum by 180°, then add 2. In this example, 1080° / 180° = 6. 6 + 2 = 8. The polygon has 8 sides.
• Question
If two equilateral triangles are placed together to form a rhombus, how do I calculate the value of each interior angle of this rhombus, and how do I find the sum?
Donagan
In the rhombus you describe, the two smaller interior angles would each be 60°, and the two larger interior angles would each be 120°. You wouldn't have to calculate the angles. Simple inspection of the rhombus and the two triangles would show what the angles are, given that equilateral triangles have three 60° angles. The sum is 60° + 60° + 120° + 120°.
200 characters left
## Tips
• Check your work on a piece of paper using a protractor to sum the interior angles manually. When doing this, be careful while drawing the polygon's sides as they should be linear.
⧼thumbs_response⧽
## Things You'll Need
• Pencil
• Paper
• Protractor (optional)
• Pen
• Eraser
• Ruler
## References
Co-authored by:
This article was co-authored by David Jia. David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. This article has been viewed 377,600 times.
Co-authors: 17
Updated: October 3, 2022
Views: 377,600
Categories: Geometry
Article SummaryX
To calculate the sum of interior angles, start by counting the number of sides in your polygon. Next, plug this number into the formula for the "n" value. Then, solve for "n" by subtracting 2 from the number of sides and multiplying the difference by 180. This will give you, in degrees, the sum of the interior angles in your polygon! To learn how to calculate the sum of interior angles by drawing triangles, read on!
Thanks to all authors for creating a page that has been read 377,600 times. |
# Lesson 12Fractional Lengths
Let’s solve problems about fractional lengths.
### Learning Targets:
• I can use division and multiplication to solve problems involving fractional lengths.
## 12.1Number Talk: Multiplication Strategies
Find the product mentally.
## 12.2How Many Would It Take? (Part 1)
1. Jada was using square stickers with a side length of inch to decorate the spine of a photo album. The spine is inches long. If she laid the stickers side by side without gaps or overlaps, how many stickers did she use to cover the length of the spine?
2. How many -inch binder clips, laid side by side, make a length of inches?
3. It takes exactly 26 paper clips laid end to end to make a length of inches.
1. Estimate the length of each paper clip.
2. Calculate the length of each paper clip. Show your reasoning.
### Are you ready for more?
Lin has a work of art that is inches by inches. She wants to frame it with large paper clips laid end to end.
1. If each paper clip is inch long, how many paper clips would she need? Show your reasoning and be sure to think about potential gaps and overlaps. Consider making a sketch that shows how the paper clips could be arranged.
2. How many paper clips are needed if the paper clips are spaced inch apart? Describe the arrangement of the paper clips at the corners of the frame.
## 12.3How Many Times as Tall or as Far?
1. A second-grade student is 4 feet tall. Her teacher is feet tall.
1. How many times as tall as the student is the teacher?
1. What fraction of the teacher’s height is the student’s height?
2. Find each quotient. Show your reasoning and check your answer.
3. Write a division expression that can help answer each of the following questions. Then answer the question. If you get stuck, draw a diagram.
1. A runner ran miles on Monday and miles on Tuesday. How many times her Monday’s distance was her Tuesday’s distance?
2. A cyclist planned to ride miles but only managed to travel miles. What fraction of his planned trip did he travel?
## 12.4Comparing Paper Rolls
The photo shows a situation that involves fractions.
1. Use the photo to help you complete the following statements. Explain or show your reasoning for the second statement.
1. The length of the long paper roll is about ______ times the length of the short paper roll.
2. The length of the short paper roll is about ______ times the length of the long paper roll.
2. If the length of the long paper roll is inches, what is the length of each short paper roll?
Use the information you have about the paper rolls to write a multiplication equation or a division equation for the question. Note that .
3. Answer the question. If you get stuck, draw a diagram.
## Lesson 12 Summary
Division can help us solve comparison problems in which we find out how many times as large or as small one number is compared to another. Here is an example.
A student is playing two songs for a music recital. The first song is minutes long. The second song is minutes long.
We can ask two different comparison questions and write different multiplication and division equations to represent each question.
• How many times as long as the first song is the second song?
Let’s use the algorithm we learned to calculate the quotient:
This means the second song is times as long as the first song.
• What fraction of the second song is the first song?
Let’s calculate the quotient:
The first song is as long as the second song.
## Lesson 12 Practice Problems
1. One inch is around centimeters.
1. How many centimeters long is 3 inches? Show your reasoning.
1. What fraction of an inch is 1 centimeter? Show your reasoning.
1. What question can be answered by finding ?
2. A zookeeper is feet tall. A young giraffe in his care is feet tall.
1. How many times as tall as the zookeeper is the giraffe?
1. What fraction of the giraffe’s height is the zookeeper’s height?
3. A rectangular bathroom floor is covered with square tiles that are feet by feet. The length of the bathroom floor is feet and the width is feet.
1. How many tiles does it take to cover the length of the floor?
1. How many tiles does it take to cover the width of the floor?
4. The Food and Drug Administration (FDA) recommends a certain amount of nutrient intake per day called the “daily value.” Food labels usually show percentages of the daily values for several different nutrients—calcium, iron, vitamins, etc.
In cup of oatmeal, there is of the recommended daily value of iron. What fraction of the daily recommended value of iron is in 1 cup of oatmeal?
Write a multiplication equation and a division equation to represent the question, and then answer the question. Show your reasoning.
5. What fraction of is ? Draw a tape diagram to represent and answer the question. Use graph paper if needed.
6. Noah says, “There are groups of in 2.” Do you agree with his statement? Draw a tape diagram to show your reasoning. Use graph paper, if needed. |
## How Many More?
Pre-K-2
1
Students write subtraction problems, model them with sets of fish-shaped crackers, and communicate their findings in words and pictures. They record differences in words and in symbols. The additive identity is reviewed in the context of comparing equal sets.
Call seven students to the front of the room, and then roll a number cube to decide how many more will come to form a second group. Next, ask each group to form a line so that the lines are parallel and the last person in each line is standing against the board. Then tell each student to hold hands with the student across from him or her. Have a volunteer say how many more students were in the larger group, and then record the subtraction equation where all the students can see it.
Continue the lesson by reading a book that features fish, such as Swimmy, by Leo Lionni. To help the students become more familiar with the set meaning for comparison subtraction, tell the students you are going to compare sets of crackers. Show a plate of fish-shaped crackers. Then roll a number cube and ask how many there will be in a plate with that many more fish-shaped crackers. Make a second plate with that many crackers to verify the students' responses. Then encourage them to write the subtraction equation that would be used to compare the two sets.
Repeat this procedure several times. Then, if necessary, review the terms "addend," "compare," and "difference." Ask what the addends and difference would be if one plate has four crackers and the other has six? [4, 6, 2] Ask what the addends and difference would be if both plates had seven fish-shaped crackers [7, 7, 0]. Prompt the students to create other such entries.
Next, ask the students to watch as you solve a subtraction problem in which two sets are compared. For example, you could say that Jen’s plate has five crackers and Sally’s plate has three crackers. Then create two sets where everyone can see them, surrounding one set with red yarn and the other with blue yarn. Ask the students to imagine the yarn loops represent plates. (If you prefer, you can use red and blue plates for the demonstration and for the student activity.)
Then ask questions such as the following:
• What comparing questions can we ask and answer about the plates?
• How many more crackers are on (student's name) plate?
• How many fewer crackers are on (another student's name) plate?
Now give each student two lengths of yarn and some crackers. Have the students pose comparison situations, model them, and answer similar questions.
When the students are ready, present a subtraction story problem in which a set of three and a set of four are compared. Demonstrate how to make a horizontal bar graph that will allow the students to compare the data. Next, guide the students through the solution of another problem, this one showing the comparison of a plate of two crackers with a plate of three crackers. (For example, Meg had three fish and Pete had two fish. How many more fish did Meg have?) Then ask the students to record the sets in a bar graph.
When they are ready, call on the students to share their problems and the graph. You may wish to suggest that they record a comparison in pictures, as a bar graph, and in an equation for their portfolios.
• Number cubes
• Fish-shaped crackers in resealable bags
• Paper plates
• Yarn in two colors
• Paper and crayons
• Graph paper
• Book of your choice about fish
Assessment Option
You may find it helpful to add to your recordings on the Class Notes recording sheet you began earlier in this unit. This data may be helpful as you plan strategies for regrouping students.
Extension
Move on to the next lesson, Hopping Backwards to Solve Problems.
Questions for Students
1. What do we find when we compare two sets?
[The difference.]
2. Make two plates of crackers. Can you show how to compare the two plates?
[Student responses may vary, but they should be able to identify the major concepts covered in today's lesson.]
3. If we compare a set of five and a set of eight, what will the difference be? How would you verify that?
[3; 8 - 5 = 3; students may use a bar graph, subtract, draw two plates, etc.]
4. What would be the smallest difference we could get between two plates if one plate has four fish-shaped crackers? How will I get that difference?
[0; Put four crackers on each pate.]
5. Suppose you had a plate of five fish-shaped crackers. How many crackers would you put on a second plate so that there would be a difference of 4 with that plate? Is there another way?
[There are two ways--a plate of one and a plate of nine.]
6. Look at one of the bar charts. How could we act it out with lines of students? With sets of crackers?
Teacher Reflection
• Which students were able to pose appropriate comparison problems?
• Which students were able to model problems with objects? Which could record the comparison with an equation?
• Which students could identify addends and differences?
• Can most of the students justify the difference when one addend is 0? Can they justify a difference of 0?
• Which students met all the objectives of this lesson? What extension activities are appropriate for these students?
• Which students did not meet all the objectives of this lesson? What caused them particular difficulty?
### Counting Back
Pre-K-2
Students count back to compare plates of fish-shaped crackers, and then they record the comparison in vertical and horizontal format. They apply their skills of reasoning and problem solving during this lesson in several ways. [Because students have associated the word "more" with addition, the comparative approach to subtraction is typically more challenging for the students to understand.]
### Hopping Backward to Solve Problems
Pre-K-2
In this lesson, students determine differences using the number line to compare lengths. Because this meaning is based on linear measurement, it is a distinctly different representation from the meanings presented in Lessons One and Two. At the end of the lesson, the students use reasoning and problem solving to predict differences and to answer puzzles involving subtraction.
### Balancing Equations
Pre-K-2
This lesson encourages the students to explore another meaning for operations of subtraction, the balance. This meaning leads naturally into recording with equations. The students will imitate the action of a pan balance and record the modeled subtraction facts in equation form.
### Fact Family Fun
Pre-K-2
In this lesson, the relation of addition to subtraction is explored with fish-shaped crackers. The students search for related addition and subtraction facts for a given number and also investigate fact families when one addend or the difference is 0.
### Wrapping Up the Unit
Pre-K-2
During this final lesson in the unit, the students use the mathematical knowledge and skills developed in the previous lessons as they visit five stations to review comparative subtraction.
### Learning Objectives
Students will:
• Find differences by comparing sets.
• Review the terms "addend" and "difference."
• Explore the effects of subtracting 0 and subtracting all.
• Record differences in pictures, words, and symbols.
### NCTM Standards and Expectations
• Count with understanding and recognize "how many" in sets of objects.
• Understand the effects of adding and subtracting whole numbers.
• Understand various meanings of addition and subtraction of whole numbers and the relationship between the two operations.
• Develop and use strategies for whole-number computations, with a focus on addition and subtraction.
• Develop fluency with basic number combinations for addition and subtraction.
• Use a variety of methods and tools to compute, including objects, mental computation, estimation, paper and pencil, and calculators.
### Common Core State Standards – Mathematics
-Kindergarten, Algebraic Thinking
• CCSS.Math.Content.K.OA.A.1
Represent addition and subtraction with objects, fingers, mental images, drawings1, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.
-Kindergarten, Algebraic Thinking
• CCSS.Math.Content.K.OA.A.2
Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.
-Kindergarten, Algebraic Thinking
• CCSS.Math.Content.K.OA.A.5
Fluently add and subtract within 5.
Grade 1, Algebraic Thinking
• CCSS.Math.Content.1.OA.B.4
Understand subtraction as an unknown-addend problem. For example, subtract 10 - 8 by finding the number that makes 10 when added to 8.
Grade 1, Algebraic Thinking
• CCSS.Math.Content.1.OA.C.6
Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 - 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
Grade 2, Algebraic Thinking
• CCSS.Math.Content.2.OA.B.2
Fluently add and subtract within 20 using mental strategies. By end of Grade 2, know from memory all sums of two one-digit numbers.
Grade 2, Number & Operations
• CCSS.Math.Content.2.NBT.B.7
Add and subtract within 1000, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method. Understand that in adding or subtracting three-digit numbers, one adds or subtracts hundreds and hundreds, tens and tens, ones and ones; and sometimes it is necessary to compose or decompose tens or hundreds.
### Common Core State Standards – Practice
• CCSS.Math.Practice.MP4
Model with mathematics.
• CCSS.Math.Practice.MP5
Use appropriate tools strategically.
• CCSS.Math.Practice.MP6
Attend to precision. |
# Example: Universal Variables#
An Earth-centered trajectory has initial velocity of 10 km/s, initial radius of 10,000 km, and initial true anomaly of 30°. Determine the true anomaly 1 hr later, using the universal anomaly.
## Solution#
The general solution procedure is as follows.
1. Determine the type of orbit
2. Determine which eccentric anomaly is appropriate, depending on the type of orbit
3. Determine the value of $$\chi$$ at the later time
4. Determine the value of the eccentric anomaly from $$\chi$$
5. Determine $$\nu$$ from the eccentric anomaly
To start, we need to identify the type of orbit. In the universal formulation, we know that if the semimajor axis is positive, the orbit is elliptical; if the semimajor axis is negative, the orbit is a hyperbola. We can find the semimajor axis from the energy equation:
$a = \left(\frac{2}{r_0} - \frac{v_0^2}{\mu}\right)^{-1}$
import numpy as np
from scipy.optimize import newton
import matplotlib.pyplot as plt
r_0 = 10_000 # km
v_0 = 10 # km/s
mu = 3.986004418E5 # km**3/s**2
a = 1 / (2 / r_0 - v_0**2 / mu)
print(round(a, 3), "km")
-19654.94 km
Since the semimajor axis is negative, we know the orbit is a hyperbola. Therefore, we can solve for the eccentricity using the orbit equation for a hyperbola, in terms of the semimajor axis:
$r_0 = a\frac{e^2 - 1}{1 + e\cos\nu_0}$
Solving this equation for $$e$$, we find:
$0 = e^2 - \frac{r_0}{a}\cos\nu_0 e - \left(1 + \frac{r_0}{a}\right)$
This equation is quadratic in $$e$$, so we can use the quadratic formula to solve it. Notice that the signs of the second and third term are negative. In addition, we need to take the absolute value of the semimajor axis, because the orbit formula was developed assuming that $$a$$ was positive.
e = (r_0/np.abs(a) * np.cos(theta_0) + np.sqrt((-r_0 / np.abs(a) * np.cos(theta_0))**2 + 4 * (1 + r_0/np.abs(a)))) / 2
print(round(e, 3))
1.468
As expected, the eccentricity is larger than 1 for a hyperbola.
Next, to find the universal anomaly at $$t_0$$ + 1 hr, we need the initial radial velocity. From Ch. 2, we know that:
$v_r = \frac{\mu}{h} e \sin\nu$
The only unknown in this equation is $$h$$, since we are interested in the initial radial velocity, that is, when $$\nu = \nu_0$$. For a hyperbola, we can find the orbital angular momentum from the hyperbolic excess velocity:
$h = \frac{\mu}{v_{\infty}} \sqrt{e^2 - 1}$
and the hyperbolic excess velocity in terms of the semimajor axis:
$v_{\infty} = \sqrt{\frac{\mu}{a}}$
Note again that this formula was derived under the assumption that $$a$$ is positive, so we need to take the absolute value.
v_infty = np.sqrt(mu / np.abs(a))
h = mu / v_infty * np.sqrt(e**2 - 1)
v_r0 = mu / h * e * np.sin(theta_0)
print(round(v_r0, 3), "km/s")
3.075 km/s
Now we have enough information to find the universal anomaly from Kepler’s equation:
$f(\chi) = 0 = \frac{r_0 v_{r,0}}{\sqrt{\mu}}\chi^2 C(z) + \left(1 - \alpha r_0\right) \chi^3 S(z)+ r_0 \chi - \sqrt{\mu}\Delta t$
where $$z = \alpha\chi^2$$. The derivative of this function is:
$f'(\chi) = \frac{r_0v_{r,0}}{\sqrt{\mu}} \chi\left(1 - z S(z)\right) + \left(1 - \alpha r_0\right) \chi^2 C(z) + r_0$
The Stumpff functions $$C(z) = c_2(z)$$ and $$S(z) = c_3(z)$$ are:
$\begin{split}C(z) = \begin{cases}\displaystyle \frac{1 - \cos\sqrt{z}}{z} & \left(z > 0\right)\\ \displaystyle\frac{\cosh\sqrt{-z} - 1}{-z} & \left(z < 0\right) \\ \displaystyle\frac{1}{2} & \left(z = 0\right)\end{cases}\end{split}$
and
$\begin{split}S(z) =\begin{cases}\displaystyle \frac{\sqrt{z} - \sin\sqrt{z}}{\left(\sqrt{z}\right)^3} & \left(z > 0\right)\\ \displaystyle \frac{\sinh\sqrt{-z} - \sqrt{-z}}{\left(\sqrt{-z}\right)^3} & \left(z < 0\right) \\ \displaystyle \frac{1}{6} & \left(z = 0\right)\end{cases}\end{split}$
def stumpff_2(z):
"""Solve the Stumpff function C(z) = c2(z). The input z should be
a scalar value.
"""
if z > 0:
return (1 - np.cos(np.sqrt(z))) / z
elif z < 0:
return (np.cosh(np.sqrt(-z)) - 1) / (-z)
else:
return 1/2
def stumpff_3(z):
"""Solve the Stumpff function S(z) = c3(z). The input z should be
a scalar value.
"""
if z > 0:
return (np.sqrt(z) - np.sin(np.sqrt(z))) / np.sqrt(z)**3
elif z < 0:
return (np.sinh(np.sqrt(-z)) - np.sqrt(-z)) / np.sqrt(-z)**3
else:
return 1/6
def universal_kepler(chi, r_0, v_r0, alpha, delta_t, mu):
"""Solve the universal Kepler equation in terms of the universal anomaly chi.
This function is intended to be used with an iterative solution algorithm,
such as Newton's algorithm.
"""
z = alpha * chi**2
first_term = r_0 * v_r0 / np.sqrt(mu) * chi**2 * stumpff_2(z)
second_term = (1 - alpha * r_0) * chi**3 * stumpff_3(z)
third_term = r_0 * chi
fourth_term = np.sqrt(mu) * delta_t
return first_term + second_term + third_term - fourth_term
def d_universal_d_chi(chi, r_0, v_r0, alpha, delta_t, mu):
"""The derivative of the universal Kepler equation in terms of the universal anomaly."""
z = alpha * chi**2
first_term = r_0 * v_r0 / np.sqrt(mu) * chi * (1 - z * stumpff_3(z))
second_term = (1 - alpha * r_0) * chi**2 * stumpff_2(z)
third_term = r_0
return first_term + second_term + third_term
Finally, we need to define the rest of the values for this function. By definition,
$\alpha = \frac{1}{a}$
and the initial guess for $$\chi$$ is given by:
$\chi_{i=0} = \sqrt{\mu} \left\lvert\alpha\right\rvert \Delta t$
delta_t = 1 * 3600
alpha = 1 / a
chi_0 = np.sqrt(mu) * np.abs(alpha) * delta_t
chi = newton(
func=universal_kepler,
fprime=d_universal_d_chi,
x0=chi_0,
args=(r_0, v_r0, alpha, delta_t, mu)
)
print(round(chi, 3))
128.511
With $$\chi$$ determined, we need to relate it back to the eccentric anomaly to determine the true anomaly. The appropriate eccentric anomaly is $$F$$, for hyperbolic trajectories. The relationship between $$\chi$$ and $$F$$ is:
$\chi = \sqrt{-a} \left(F - F_0\right)$
where $$F_0$$ is the eccentric anomaly determined at the initial true anomaly:
$F_0 = 2 \tanh^{-1}\left(\sqrt{\frac{e - 1}{e + 1}}\tan\frac{\nu_0}{2}\right)$
F_0 = 2. * np.arctanh(np.sqrt((e - 1) / (e + 1)) * np.tan(theta_0 / 2))
print(round(F_0, 3))
0.234
Then, we can solve for $$F$$ and relate that back to $$\nu$$:
$F = \frac{\chi}{\sqrt{-a}} + F_0$
and
$\nu = 2 \tan^{-1} \left(\sqrt{\frac{e + 1}{e - 1}}\tanh\frac{F}{2}\right)$
F = chi / np.sqrt(-a) + F_0
print(round(F, 3))
theta = 2 * np.arctan(np.sqrt((e + 1) / (e - 1)) * np.tanh(F / 2))
print(round(theta, 3), f"{np.degrees(theta):.3F}°")
1.151
1.746 100.040° |
Students can download Maths Chapter 3 Algebra Ex 3.10 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.
## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.10
Question 1.
Draw the graph for the following:
(i) y = 2x
Solution:
When x = -2, y = 2 (-2) = -4
When x = 0, y = 2 (0) = 0
When x = 2, y = 2 (2) = 4
When x = 3, y = 2 (3) = 6
Plot the points (-2, -4) (0, 0) (2, 4) and (3, 6) in the graph sheet we get a straight line.
(ii) y = 4x – 1
Solution:
When x = – 1; y = 4 (-1) -1 ⇒ y = -5
When x = 0; y = 4 (0) – 1 = 0 – 1 ⇒ y = -1
When x = 2; y = 4 (2) -1 = 8 – 1 ⇒ y = l
Plot the points (-1, -5) (0, -1) and (2, 7) in the graph sheet we get a straight line. At the time of printing change the direction.
(iii) y = ($$\frac{3}{2}$$)x + 3
Solution:
When x = -2;
y = $$\frac{3}{2}$$(-2) + 3
y = -3 + 3 = 0
when x = 0;
y = $$\frac{3}{2}$$(0) + 3
y = 3
when x = 2;
y = $$\frac{3}{2}$$(2) + 3
y = 3 + 3
= 6
Plot the points (-2, 0) (0, 3) and (2, 6) in the graph sheet we get a straight line.
(iv) 3x + 2y = 14
Solution:
y = $$\frac{-3x+14}{2}$$
y = – $$\frac{3}{2}$$x + 7
when x = -2
y = –$$\frac{3}{2}$$(-2) + 7 = 10
when x = 0
y = –$$\frac{3}{2}$$(0) + 7 = 7
when x = 2
y = –$$\frac{3}{2}$$(2) + 7 = 4
Plot the points (-2, 10) (0, 7) and (2, 4) in the graph sheet we get a straight line.
Question 2.
Solve graphically (i) x + y = 7, x – y = 3
Solution:
x + y = 7
y = 7 – x
Plot the points (-2, 9), (0, 7) and (3, 4) in the graph sheet
x – y = 3
-y = -x + 3
y = x – 3
Plot the points (-2, -5), (0, -3) and (4, 1) in the same graph sheet.
The point of intersection is (5, 2) of lines (1) and (2).
The solution set is (5,2).
(ii) 3x + 2y = 4; 9x + 6y – 12 = 0
Solution:
2y = -3x + 4
y = $$\frac{-3x+4}{2}$$
= $$\frac{-3}{2}$$x + 2
Plot the points (-2, 5), (0, 2) and (2, -1) in the graph sheet
9x + 6y= 12 (÷3)
3x + 2y = 4
2y = $$\frac{-3x+4}{2}$$
= $$\frac{-3}{2}$$x + 2
Plot the points (-2, 5), (0, 2) and (2, -1) the same graph sheet
Here both the equations are identical, but in different form.
Their solution is same.
This equations have an infinite number of solution.
(iii) $$\frac{x}{2}$$ + $$\frac{y}{4}$$ = 1: $$\frac{x}{2}$$ + $$\frac{y}{4}$$ = 2
Solution:
$$\frac{x}{2}$$ + $$\frac{y}{4}$$ = 1
multiply by 4
2x + y = 4
y = -2x + 4
Plot the points (-3, 10), (-1, 6), (0, 4) and (2, 0) in the graph sheet
$$\frac{x}{2}$$ + $$\frac{y}{4}$$ = 2
multiply by 4
2x + y = 8
y = -2x + 8
Plot the points (-2, 12), (-1, 10), (0, 8) and (2, 4) in the same graph sheet.
The given two lines are parallel.
∴ They do not intersect a point.
∴ There is no solution.
(iv) x – y = 0; y + 3 = 0
Solution:
x – y = 0
-y = -x
y = x
Plot the points (-2, -2), (0, 0), (1, 1) and (3, 3) in the same graph sheet.
y + 3 = 0
y = -3
Plot the points (-2, -3), (0, -3), (1, -3) and (2, -3) in the same graph sheet.
The two lines l1 and l2 intersect at (-3, -3). The solution set is (-3, -3).
(v) y = 2x + 1; 3x – 6 = 0
Solution:
y = 2x + 1
Plot the points (-3, -5), (-1, -1), (0, 1) and (2, 5) in the graph sheet
3x – 6 = 0
y = -3x + 6
Plot the points (-2, 12), (-1, 9), (0, 6) and (2, 0) in the same graph sheet
The two lines l1 and l2 intersect at (1, 3).
∴ The solution set is (1, 3).
(vi) x = -3; y = 3
Solution:
x = -3
Plot the points (-3, -3), (-3, -2), (-3, 2) and (-3, 3) in the graph sheet
y = 3
Plot the points (-3, 3), (-1, 3), (0, 3) and (2, 3) in the same graph sheet
The two lines l1 and l2 intersect at (-3, 3)
∴ The solution set is (-3, 3)
Question 3.
Two cars are 100 miles apart. If they drive towards each other they will meet in 1 hour. If they drive in the same direction they will meet in 2 hours. Find their speed by using graphical method.
Solution:
Let the speed of the two cars be “x” and “y”.
By the given first condition
x+ y = 100 → (1)
(They travel in opposite direction)
By the given second condition.
\frac{100}{x-y}= 2 [time taken in 2 hours in the same direction]
2x – 2y = 100
x – y = 50 → (2)
x + y = 100
y = 100 – x
Plot the points (30, 70), (50, 50), (60, 40) and (70,30) in the graph sheet
x – y = 50
-y = -x + 50
y = x – 50
Plot the points (40, -10), (50, 0), (60, 10) and (70, 20) in the same graph sheet
The two cars intersect at (75, 25)
The speed of the first car 75 km/hr
The speed of the second car 25 km/hr |
Home | Math | 10 Examples of Approximation
10 Examples of Approximation
October 18, 2023
written by Rida Mirza
Approximation is a fundamental mathematical concept used to simplify complex calculations and provide reasonable estimates of values.
In this article, we will discuss ten examples of approximation in mathematics.
Examples of Approximation
These are 10 examples of approximation.
1: Rounding Numbers
Rounding numbers is one of the most common forms of approximation.
For example, rounding 3.14159 to 3.14 simplifies calculations while providing a reasonably close value for π (pi).
2: Estimating Square Roots
Estimating square roots is useful when dealing with non-perfect squares.
For example, √7 is approximately 2.65, which is a close approximation.
3: Calculating Percentage Increases
When calculating percentage increases or decreases, approximations are often used.
For example, estimating a 15% increase of \$90 as \$15 simplifies the calculation to \$90 + \$15 = \$105.
4: Trigonometric Functions
In trigonometry, approximations are common when dealing with angles.
For example, sin(30°) is approximately 0.5, providing a close estimate for simple calculations.
5: Taylor Series Approximations
Taylor series are used to approximate functions.
For example, e^x, sin(x), and cos(x, providing a series of terms that closely approach the actual values.
6: Numerical Integration
In calculus, numerical methods like the trapezoidal rule or Simpson’s rule are used to approximate the definite integral of a function.
7: Linear Approximations
Linear approximations use tangent lines to approximate functions near a specific point.
For example, the linear approximation of √x near x = 4 simplifies to 2 + (x – 4)/4.
8: Exponential Growth Models
In finance and biology, exponential growth models, like the compound interest formula, approximate how values grow over time.
9: Finite Difference Approximations
Finite difference methods are used to approximate derivatives and solve differential equations numerically.
10: Monte Carlo Simulations
Monte Carlo simulations involve using random sampling to approximate complex mathematical models, frequently used in risk analysis, finance, and statistical physics.
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## 5th grade math standards and assessment anchors
Standard: CC.2.1.5.B.1 Apply place‐value concepts to show an understanding of operations and rounding as they pertain to whole numbers and decimals.
ASSESSMENT ANCHOR: M05.A-T.1 Understand the place-value system.
• DESCRIPTOR ELIGIBLE CONTENT: M05.A-T.1.1 Demonstrate understanding of place-value of whole numbers and decimals, and compare quantities or magnitudes of numbers.
• M05.A-T.1.1.1 Demonstrate an understanding that in a multi-digit number, a digit in one place represents 1/10 of what it represents in the place to its left. Example: Recognize that in the number 770, the 7 in the tens place is 1/10 the 7 in the hundreds place.
• M05.A-T.1.1.2 Explain patterns in the number of zeros of the product when multiplying a number by powers of 10 and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10. Example 1: 4 × 102 = 400 Example 2: 0.05 ÷ 103 = 0.00005
• M05.A-T.1.1.3 Read and write decimals to thousandths using base-ten numerals, word form, and expanded form. Example: 347.392 = 300 + 40 + 7 + 0.3 + 0.09 + 0.002 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (0.1) + 9 × (0.01) + 2 × (0.001)
• M05.A-T.1.1.4 Compare two decimals to thousandths based on meanings of the digits in each place using >, =, and < symbols.
• M05.A-T.1.1.5 Round decimals to any place (limit rounding to ones, tenths, hundredths, or thousandths place).
Standard: CC.2.1.5.B.2 Extend an understanding of operations with whole numbers to perform operations including decimals.
ASSESSMENT ANCHOR: M05.A-T.2 Perform operations with multi-digit whole numbers and with decimals to hundredths.
• DESCRIPTOR ELIGIBLE CONTENT: M05.A-T.2.1 Use whole numbers and decimals to compute accurately (straight computation or word problems).
• M05.A-T.2.1.1 Multiply multi-digit whole numbers (not to exceed three-digit by three-digit).
• M05.A-T.2.1.2 Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors.
• M05.A-T.2.1.3 Add, subtract, multiply, and divide decimals to hundredths (no divisors with decimals).
Standard: CC.2.1.5.C.1 Use the understanding of equivalency to add and subtract fractions
• ASSESSMENT ANCHOR ASSESSMENT ANCHOR: M05.A-F.1 Use equivalent fractions as a strategy to add and subtract fractions.
• DESCRIPTOR ELIGIBLE CONTENT : M05.A-F.1.1 Solve addition and subtraction problems involving fractions (straight computation or word problems).
• M05.A-F.1.1.1 Add and subtract fractions (including mixed numbers) with unlike denominators. (May include multiple methods and representations.) Example: 2/3 + 5/4 = 8/12 + 15/12 = 23/12
Standards: CC.2.1.5.C.2 Apply and extend previous understandings of multiplication and division to multiply and divide fractions
• ASSESSMENT ANCHOR: M05.A-F.2 Apply and extend previous understandings of multiplication and division to multiply and divide fractions.
• DESCRIPTOR ELIGIBLE CONTENT : M05.A-F.2.1 Solve multiplication and division problems involving fractions and whole numbers (straight computation or word problems).
• M05.A-F.2.1.1 Solve word problems involving division of whole numbers leading to answers in the form of fractions (including mixed numbers).
• M05.A-F.2.1.2 Multiply a fraction (including mixed numbers) by a fraction.
• M05.A-F.2.1.3 Demonstrate an understanding of multiplication as scaling (resizing). Example 1: Comparing the size of a product to the size of one factor on the basis of the size of the other factor without performing the indicated multiplication. Example 2: Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number.
• M05.A-F.2.1.4 Divide unit fractions by whole numbers and whole numbers by unit fractions.
CC.2.2.5.A.1 Interpret and evaluate numerical expressions using order of operations.
• ASSESSMENT ANCHOR M05.B-O.1 Write and interpret numerical expressions.
• DESCRIPTOR ELIGIBLE CONTENT : M05.B-O.1.1 Analyze and complete calculations by applying the order of operations.
• M05.B-O.1.1.1 Use multiple grouping symbols (parentheses, brackets, or braces) in numerical expressions and evaluate expressions containing these symbols.
• M05.B-O.1.1.2 Write simple expressions that model calculations with numbers and interpret numerical expressions without evaluating them. Example 1: Express the calculation “add 8 and 7, then multiply by 2” as 2 × (8 + 7). Example 2: Recognize that 3 × (18,932 + 921) is three times as large as 18,932 + 921 without having to calculate the indicated sum or product.
Standard: CC.2.2.5.A.4 Analyze patterns and relationships using two rules.
• ASSESSMENT ANCHOR : M05.B-O.2 Analyze patterns and relationships.
• DESCRIPTOR ELIGIBLE CONTENT : M05.B-O.2.1 Create, extend, and analyze patterns.
• M05.B-O.2.1.1 Generate two numerical patterns using two given rules. Example: Given the rule “add 3” and the starting number 0 and given the rule “add 6” and the starting number 0, generate terms in the resulting sequences.
• M05.B-O.2.1.2 Identify apparent relationships between corresponding terms of two patterns with the same starting numbers that follow different rules. Example: Given two patterns in which the first pattern follows the rule “add 8” and the second pattern follows the rule “add 2,” observe that the terms in the first pattern are 4 times the size of the terms in the second pattern.
CC.2.3.5.A.1 Graph points in the first quadrant on the coordinate plane and interpret these points when solving real world and mathematical problems
• ASSESSMENT ANCHOR : M05.C-G.1 Graph points on the coordinate plane to solve real-world and mathematical problems.
• DESCRIPTOR ELIGIBLE CONTENT : M05.C-G.1.1 Identify parts of a coordinate grid and describe or interpret points given an ordered pair.
• M05.C-G.1.1.1 Identify parts of the coordinate plane (x-axis, y-axis, and the origin) and the ordered pair (x-coordinate and y-coordinate). Limit the coordinate plane to quadrant I.
• M05.C-G.1.1.2 Represent real-world and mathematical problems by plotting points in quadrant I of the coordinate plane and interpret coordinate values of points in the context of the situation.
Standard: CC.2.3.5.A.2 Classify two‐dimensional figures into categories based on an understanding of their properties.
• ASSESSMENT ANCHOR : M05.C-G.2 Classify two-dimensional figures into categories based on their properties.
• DESCRIPTOR ELIGIBLE CONTENT: M05.C-G.2.1 Use basic properties to classify two-dimensional figures.
• M05.C-G.2.1.1 Classify two-dimensional figures in a hierarchy based on properties. Example 1: All polygons have at least three sides, and pentagons are polygons, so all pentagons have at least three sides. Example 2: A rectangle is a parallelogram, which is a quadrilateral, which is a polygon; so, a rectangle can be classified as a parallelogram, as a quadrilateral, and as a polygon.
Standard: CC.2.4.5.A.1 Solve problems using conversions within a given measurement system.
• ASSESSMENT ANCHOR: M05.D-M.1 Convert like measurement units within a given measurement system.
• DESCRIPTOR ELIGIBLE CONTENT : M05.D-M.1.1 Solve problems using simple conversions (may include multistep, real-world problems).
• M05.D-M.1.1.1 Convert between different-sized measurement units within a given measurement system. A table of equivalencies will be provided. Example: Convert 5 cm to meters.
Standards: CC.2.4.5.A.2 Represent and interpret data using appropriate scale.
CC.2.4.5.A.4 Solve problems involving computation of fractions using information provided in a line plot.
• ASSESSMENT ANCHOR: M05.D-M.2 Represent and interpret data.
• DESCRIPTOR ELIGIBLE CONTENT: M05.D-M.2.1 Organize, display, and answer questions based on data.
• M05.D-M.2.1.1 Solve problems involving computation of fractions by using information presented in line plots.
• M05.D-M.2.1.2 Display and interpret data shown in tallies, tables, charts, pictographs, bar graphs, and line graphs, and use a title, appropriate scale, and labels. A grid will be provided to display data on bar graphs or line graphs.
CC.2.4.5.A.5 Apply concepts of volume to solve problems and relate volume to multiplication and to addition.
• M05.D-M.3.1.2 Find volumes of solid figures composed of two non-overlapping right rectangular prisms.
• M05.D-M.3.1.1 Apply the formulas V = l × w × h and V = B × h for rectangular prisms to find volumes of right rectangular prisms with whole-number edge lengths in the context of solving real-world and mathematical problems. Formulas will be provided.
DESCRIPTOR ELIGIBLE CONTENT: M05.D-M.3.1 Use, describe, and develop procedures to solve problems involving volume.
ASSESSMENT ANCHOR: M05.D-M.3 Geometric measurement: understand concepts of volume and relate volume to multiplication and to addition. |
# Mixed numbers and fractions: how to convert from one to the other
This is a fifth grade lesson about fractions and mixed numbers. First, this lesson has some review exercises about mixed numbers. Then, we learn how to change mixed numbers into fractions - both the concept and the shortcut. Lastly, we convert improper fractions into mixed numbers by thinking of them as DIVISIONS.
### Mixed numbers as pictures
1. Write the mixed numbers that these pictures illustrate.
a. b. c. d.
2. Draw pictures of “pies” that illustrate these mixed numbers.
a. 4 2 3
b. 2 3 5
c. 3 2 6
d. 4 7 8
e. 6 8 10
3. Write the mixed number that is illustrated by each number line.
a.
b.
4. Write the fractions and mixed numbers that the arrows indicate.
a. b. c. d.
5. Mark the fractions on the number lines.
a. 10 6 , 17 6 , 12 6 , 5 6 , 14 6
b. 9 8 , 22 8 , 13 8 , 24 8 , 11 8
6. a. Mark 1 3 5 on the number line. b. Write the mixed number that is 4 5 to its right: _______
c. Mark 2 1 5 on the number line. d. Write the mixed number that is 3 5 to its left: _______
Changing mixed numbers to fractions To write 3 3 4 as a fraction, count how many fourths there are:
Each pie has four fourths, so the three complete pies have 3 × 4 = 12 fourths. Additionally, the incomplete pie has three fourths. The total is 15 fourths or 15/4.
Shortcut:
Numerator: 3 × 4 + 3 = 15
Denominator: 4
= 15 4
Multiply the whole number times the denominator, then add the numerator. The result gives you the number of fourths, or the numerator, for the fraction. The denominator will remain the same.
7. Write as mixed numbers and as fractions.
a. 1 2 5 = 5
b. =
c. =
d. =
e. =
f. =
8. May changed 5 9 13 into a fraction, and explained how the shortcut works. Fill in.
There are ____ whole pies, and each pie has _____ slices. So ____ × ____
tells us the number of slices in the whole pies. Then the fractional part 9/13 means that we add _____ slices to that. In total we get ____ slices, each one a 13th part. So the fraction is .
9. Write as fractions. Think of the shortcut.
a. 7 1 2
b. 6 2 3
c. 8 3 9
d. 6 6 10
e. 2 5 11
f. 8 1 12
g. 2 5 16
h. 4 7 8
### Changing fractions to mixed numbers
To write a fraction, such as 58 7 , as a mixed number, you need to figure out:
• How many whole “pies” there are, and
• How many slices are left over.
In the case of 58 7 , each whole “pie” will have 7 sevenths. (How do you know?) So we ask:
• How many 7s are there in 58? (Those make the whole pies!)
• After the 7s are gone, how many are left over?
That is solved by the division 58 ÷ 7! That division tells you how many 7s there are in 58.
Now, 58 ÷ 7 = 8 R2. So you get 8 whole pies, with 2 slices or 2 sevenths left over.
To write that as a fraction, we get 58 7 = 8 2 7 .
Example: 45 4 is the same as 45 ÷ 4, and 45 ÷ 4 = 11 R1. So, we get 11 whole pies and
1 fourth-part or slice left over. Writing that as a mixed number, 45 4 = 11 1 4 .
The Shortcut: Think of the fraction bar as a division symbol, then DIVIDE. The quotient tells
you the whole number part, and the remainder tells you the numerator of the fractional part.
10. Rewrite the “division problems with remainders” as problems of “changing fractions to mixed
numbers.”
a. 47 ÷ 4 = 11 R3
47 4 = 11 3 4
b. 35 ÷ 8 = 4 R3
=
c. 19 ÷ 2 = ___ R ___
=
d. 35 ÷ 6 = ___ R ___
=
e. 72 ÷ 10 = ___ R ___
=
f. 22 ÷ 7 = ___ R ___
=
11. Write these fractions as mixed numbers (or as whole numbers, if you can).
a. 62 8
b. 16 3
c. 27 5
d. 32 9
e. 7 2
f. 25 4
g. 50 6
h. 32 5
i. 24 11
j. 39 3
k. 57 8
l. 87 9
This lesson is taken from Maria Miller's book Math Mammoth Fractions 1, and posted at www.HomeschoolMath.net with permission from the author. Copyright © Maria Miller.
#### Math Mammoth Fractions 1
A self-teaching worktext for 5th grade that teaches fractions and their operations with visual models. The book covers fractions, mixed numbers, adding and subtracting like fractions, adding and subtracting mixed numbers, adding and subtracting unlike fractions, and comparing fractions. |
FutureStarr
11 8 As a Percent OR"
## 11 8 As a Percent
via GIPHY
8, as a single digit, is 11 as a percent, meaning this percentage is 86. Or, 3,333 as an integer, which is the base value divided by 100 and multiplied by 11.
### Percent
via GIPHY
A mixed number is a whole number plus a fraction. You can convert fraction part of the mixed number to a decimal and then multiply by 100 to get the percent value. Alternatively you can convert mixed number to an improper fraction, and then convert it to a decimal by dividing numerator by denominator. Finally, multiply the decimal by 100 to find the percent value.
The term percent is a ratio or a number that is expressed as a fraction of 100. It is denoted using the percentage sign %. To understand the concept of how the percent represents the fraction of 100, here is an example. 35% can be written in fraction as 35/100. In class, 50% of the students were male, which means out of every 100 students, 50 were male. (Source: byjus.com)
### Convert
via GIPHY
You often need to find out what percent of something in your daily life. To understand how to convert percent to fraction, consider an example. If a school has 865 students out of which 389 are female, then what percent of students are female. To solve this you need to divide both the numbers. It is 389 out of 865 or $$\small \frac{389}{865} = 0.44971$$. To simplify this you need to convert it into percent, which is $$\small 44.9711 \%$$. Now, let’s discuss how to convert percent and fraction in detail.
Percent refers to the fractions of a whole and can be remembered easily than the fraction. It is how much of a whole thing contains. For example, $$\small 50 \%$$ can be written as $$\small \frac{1}{2}$$, and $$\small 25 \%$$ means $$\small \frac{1}{4}$$. In the same way you can convert vice versa: fraction to percent conversion. (Source: byjus.com)
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# A person can swim in still water at 5m/s. He moves in river of velocity 3m/s, First down the stream next same distance up the steam the ratio of times takes are(A) 1:1(B) 1:2(C) 1:4(D) 4:1
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Hint: This could be simply solved by applying the basic formula of speed. The average speed is the distance per time ratio.
Formula used: Here, we will use the basic formula of speed, distance and time:
${{\text{V}}_{{\text{avg}}}}{\text{ = }}\dfrac{{\text{D}}}{{\text{T}}}$
Here, ${{\text{V}}_{{\text{avg}}}}$is the mean speed of the car
${\text{D}}$is the total distance travel
$T$is the travel time
We will start by considering the total distance to be${\text{d}}$,
For the downstream over the distance:
${{\text{v}}_{\text{m}}}{\text{ = 5m/s and }}{{\text{v}}_{\text{r}}}{\text{ = 3m/s}}$
Similarly, for the upstream over the distance:
${{\text{v}}_p}{\text{ = 5m/s}}$
${v_r} = 3m/s$
And also, we find the respective time:
${t_1} = \dfrac{d}{{5 + 3}} = \dfrac{d}{8}$
${t_2} = \dfrac{d}{{5 - 3}} = \dfrac{d}{2}$
Now, the ratios between the time:
$\dfrac{{{t_1}}}{{{t_2}}} = \dfrac{{\dfrac{d}{8}}}{{\dfrac{d}{2}}} = \dfrac{1}{4}$
Thus, we need to select the correct option.
The correct option is C.
Additional Information: Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance.
The first scientist to measure speed as distance over time was Galileo.
Note: It should always be kept in mind that there is a difference between speed and velocity. Just as distance and displacement have distinctly different meanings, same is the situation between speed and velocity. Velocity is a vector quantity that refers to the rate at which an object changes its position whereas speed is a scalar quantity that refers to how fast an object is moving. Velocity gives us a sense of direction whereas speed does not give any sense of direction. If the velocity is constant, which creates a horizontal line
For example: If the person is travelling at a constant speed of 3 miles per hour, we can find the distance travelled by multiplying the speed by the amount of time they are walking. So, the person travelled 6 miles in 2 hours. |
# Angle between two lines Calculator
The calculator for finding the angle between two lines requires the input of the slope of the two lines. The formula for finding the angle between two lines is given by:
$\theta = \tan^{-1} \left( \left| \frac{m_2-m_1}{1+m_1m_2} \right| \right)$
where $m_1$ and $m_2$ are the slopes of the two lines.
In order to use this formula, you first need to find the slopes of the two lines, which can be done using the formula:
$m = \frac{y_2 – y_1}{x_2 – x_1}$
where $(x_1,y_1)$ and $(x_2,y_2)$ are two points on the line. Once you have found the slopes, you can input them into the angle between two lines formula to get the result.
## The angle between two lines with example
The angle between two lines can be calculated using the formula:
θ = tan⁻¹(m₂ – m₁ / 1 + m₁m₂)
Where m₁ and m₂ are the slopes of the two lines.
Let’s take an example to understand this better:
Find the angle between the lines y = 3x – 4 and y = 2x + 1.
Solution:
To find the angle between the lines, we need to calculate the slopes of both lines.
Slope of the first line (m₁) = 3 Slope of the second line (m₂) = 2
Now, we can substitute these values in the formula:
θ = tan⁻¹(2 – 3 / 1 + (3 x 2)) = tan⁻¹(-1 / 7) ≈ -8.13°
Therefore, the angle between the lines y = 3x – 4 and y = 2x + 1 is approximately -8.13 degrees. Note that the angle is negative because the lines are moving in opposite directions. |
# Check if $(\mathbb Z_7, \odot)$ is an abelian group, issue in finding inverse element
Take $\mathbb Z_7$ and the operation $\odot$ defined on it as follows $\forall a,b \in \mathbb Z_7$:
\begin{aligned} a \odot b=a+b+3\end{aligned}
Check if $(\mathbb Z_7, \odot)$ is a group and in particular if it is an abelian group.
Associativity
It can be easily proved that $\odot$ is associative:
\begin{aligned} (a \odot b) \odot c = a \odot (b \odot c)\end{aligned} \begin{aligned}(a + b + 3) \odot c = a \odot (b+c+3)\end{aligned} \begin{aligned} (a + b + 3) + c +3 = a + (b+c+3)+3 \end{aligned} \begin{aligned} a + b + c +6 = a +b+c+6 \end{aligned}
Commutativity
$\odot$ is commutative:
\begin{aligned} a \odot b = b \odot a \end{aligned} \begin{aligned} a + b + 3 = b +a+3 \end{aligned}
Identity element
There is an identity element for $(\odot, \mathbb Z_7)$
\begin{aligned} a \odot \mathbb 1_{\mathbb Z_7} = a \end{aligned} \begin{aligned} a + e + 3 = a \end{aligned} \begin{aligned} e + 3 = 0 \end{aligned} \begin{aligned} e = 4 \end{aligned}
and similarly if we repeat the calculation for $\mathbb 1_{\mathbb Z_7} \odot a$, still $e=4$
What I am struggling with is the inverse element, how to check if there is one. I know that it should be:
\begin{aligned} a \odot a^{-1} = e \\ a + a^{-1} + 3 = e \\ a + a^{-1} + 3 = 4\end{aligned}
Is it acceptable to state $a^{-1} = a+1$ (and similarly for $a^{-1} \odot a = e$)? If so, have I succeeded in proving $(\odot, \mathbb Z_{7})$ to be an abelian group?
• $a^{-1}=1-a=1+6a$ that is $a^{-1}=(1+6a)$ mod 7 isnt it ? – pritam Jul 8 '12 at 9:11
• No, it's not acceptable to state that $a+1$ is the inverse of $a$, since it's not true. For example, take $a=1$. Then $a\odot (a+1)=1\odot 2=1+2+3=6\neq 4$. – Chris Eagle Jul 8 '12 at 9:11
• Note you did not prove the alleged group has an identity. What you did prove is that if it does have an identity, then the identity is 4. It's unclear if you have the same conceptual mistake for your proofs of the other properties. – user14972 Jul 8 '12 at 10:33
• @Hurkyl what about the associativity, commutativity and identity element? If my calculation is right and all of those apply along with the presence of the inverse elements isn't that enough to say that $(\mathbb Z_7, \odot)$ is an abelian group? If not what am I missing? – haunted85 Jul 8 '12 at 10:40
• @haunted85: Your proof outline is fine: the individual steps are true (once you've corrected your typo for inverses) and the conclusion holds. However, the way you did the "prove it has an identity" step is backwards, and I suspect you did the other two steps backwards as well (but I can't know for sure without knowing your actual thoughts about what you were doing). – user14972 Jul 8 '12 at 10:46
The inverse is $a^{-1}=1-a$, as $a\odot (1-a)=a+(1-a)+3=4=e$.
You have $\rm\:a + a^{-1} = 1,\:$ so $\rm\: a^{-1} = 1 - a.\:$ Note that all of your equations should be connected by arrows going both ways, i.e. $\iff\!\!,\:$ since you need to prove both necessity and sufficiency.
Here the group structure arises simply from renaming (or labeling) the elements of the additive group $\,\Bbb Z/7\,$ via the "label" bijection $\rm\:\ell\, n := n-3,\:$ i.e. by naming or labeling each natural mod $7\,$ by the natural congruent to $\rm\,n\!-\!3.\,$ To perform an operation on labels, we first unlabel the operands by applying $\,\ell^{-1}n\, =\, n\!+\!3,\,$ then perform the normal operation, then label the result, i.e.
$$\rm a \oplus b\ :=\ \ell\,(\ell^{-1}a\, +\, \ell^{-1}b)\ =\, -3 + ((a\!+\!3) + (b\!+\!3))\ =\ a+b+3$$
$$\rm \ominus\, a\ :=\ \ell(-\,\ell^{-1}a)\ =\ -3+ (-(a\!+\!3))\ =\ -6-a\ =\ 1-a\quad$$
Thus for $\mu = \ell^{-1}\,$ we have $\rm\ \mu(a \oplus b)\, =\, \mu\,a + \mu\, b,\$ and $\rm\ \mu\ominus a\, =\, -\mu a\$ so $\mu$ is a bijective group homomorphism, hence an isomorphism. In more technical language one says that one has transported the group structure along the bijection $\mu$ (or $\ell).$
For example the equation $\,5+ 6 = 4\,$ transports to $\,\ell\, 5 \oplus \ell\, 6 = \ell\, 4,\,$ i.e. $\it\, 2 \oplus 3 = 1,\,$ and the equation $\,-(5)\, =\, 2\,$ transports to $\,\ominus\,\ell\,5\, =\, \ell\, 2,\,$ i.e. $\it\,\ominus\,2 = 6.\,$ Transporting the entire addition table yields
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline \color{#C00}\oplus &\it\color{#C00}0 &\it\color{#C00}1 &\it\color{#C00}2 &\it\color{#C00}3 &\it\color{#C00}4 &\it\color{#C00}5 &\it\color{#C00}6 \\ \hline \it\color{#C00} 0 &\it 3 &\it 4 &\it 5 &\it 6 &\it 0 &\it 1 &\it 2 \\ \hline \it\color{#C00} 1 &\it 4 &\it 5 &\it 6 &\it 0 &\it 1\, &\it 2 &\it 3 \\ \hline \it\color{#C00} 2 &\it 5 &\it 6 &\it 0 &\it 1\, &\it 2 &\it 3 &\it 4 \\ \hline \it\color{#C00} 3 &\it 6 &\it 0 &\it 1\, &\it 2 &\it 3 &\it 4 &\it 5 \\ \hline \it\color{#C00} 4 &\it 0 &\it 1\, &\it 2 &\it 3 &\it 4 &\it 5 &\it 6 \\ \hline \it\color{#C00} 5 &\it 1\, &\it 2 &\it 3 &\it 4 &\it 5 &\it 6 &\it 0 \\ \hline \it\color{#C00} 6 &\it 2 &\it 3 &\it 4 &\it 5 &\it 6 &\it 0 &\it 1\, \\ \hline \end{array} \ \ \begin{array}{c} \xrightarrow[\large \ \it N\ \to\,\rm N+3\ ]{\large \rm unlabel\,\ \mu} \\ \\ \\ \xleftarrow[\large \ \it N-3\ \leftarrow\, \rm N\ ]{\large \rm label\,\ \ell} \end{array}\ \ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \color{#C00}+ &\color{#C00} 3 &\color{#C00} 4 &\color{#C00} 5 &\color{#C00} 6 &\color{#C00} 0 &\color{#C00} 1 &\color{#C00} 2 \\ \hline \color{#C00}3 & 6 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \color{#C00}4 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \color{#C00}5 & 1 & 2 & 3 & 4 & 5 & 6 & 0 \\ \hline \color{#C00}6 & 2 & 3 & 4 & 5 & 6 & 0 & 1 \\ \hline \color{#C00}0 & 3 & 4 & 5 & 6 & 0 & 1 & 2 \\ \hline \color{#C00}1 & 4 & 5 & 6 & 0 & 1 & 2 & 3 \\ \hline \color{#C00}2 & 5 & 6 & 0 & 1 & 2 & 3 & 4 \\ \hline \end{array}$$
Note that the addition table on the right is the table for the operation of addition mod $7$, except the rows and colums have been reordered (shifted by $3$). Thus the two addition tables are essentially the same, i.e. they differ only in the names chosen for the elements. This is the sense of isomorphism that is captured by the notion of isomorphic groups, i.e. the two groups have exactly the same operation tables after a (renaming) bijection is applied to the elements. The notion of isomorphism is defined so that the algebraic structure is determined completely by the operation tables, i.e. the only properties of the elements that we care about algebraically are how the elements relate to each other under the operations. Any other (internal) structure the elements may possess (names, set-theoretic representation, etc), play no role algebraically.
Similarly we can transport the group structure along any permutation $\,\ell\,$ of $\,\Bbb Z/7$, and we can label or index any finite group by natural numbers (e.g. which might be addresses in computer memory, where (un)label operations amounts to memory (de)references).
• Irrelevant, but you write good answers. Just appreciating. Thanks. – stranger Feb 9 '17 at 15:43 |
# Median
Median addresses the center incentive for any gathering. It is the place where a portion of the data is more and around 50% of the data is less. Median assists with addressing a huge number of data points with a solitary data point. The median is the most simple applied mathematics to work out. For estimation of the median, the data must be organized in ascending order, and afterward, the middlemost data point addresses the median of the data.
Further, the estimation of the median relies upon the number of data points. For an odd number of data, the median is the middlemost data, and for a much number of data, the median is the normal of the two center values.
## What is the Median?
Median is one of the three measures of central tendency. While portraying a bunch of data, the central position of the data set is recognized. This is what we call the median of central tendency. The three most normal measures of central tendency are mean, median, and mode.
## Definition Of Median
The value of the center most perception got subsequent to orchestrating the data in ascending order is known as the median of the data. Numerous an occurrence, it is hard to think about the total data for portrayal, and here median is helpful. Among the statistical rundown measurements, the median is a simple measurement to work out. The Median is likewise called the Place Average, as the data placed in the sequence is taken as the median.
## Examples Of Median
We should consider a guide to sorting out what is median for a given arrangement of data.
Stage 1: Consider the data: 4, 4, 6, 3, and 2. We should arrange this data in ascending order: 2, 3, 4, 4, 6.
Stage 2: Count the number of values. There are 5 values.
Stage 3: Look for the center’s middle value. The center’s middle value is the median. Along these lines, median = 4.
## Formula Of Median
Utilizing the median formula, the center worth of the organized arrangement of numbers can be determined. For finding this measure of central tendency, composing the parts of the gathering in expanding order is fundamental. The median formula fluctuates in view of the number of perceptions and whether they are odd or even. The accompanying arrangement of formulas would help in tracking down the median of the given data.
## Median Formula For Ungrouped Data
The following advances are useful while applying the median formula for ungrouped data.
Stage 1: Arrange the data in ascending or descending order.
Stage 2: Secondly, count the all-out of a given number of observations ‘n’.
Stage 3: Check if the number of observations ‘n’ is even or odd.
## Median Formula when ‘n’ is Odd
The median formula of a given arrangement of numbers, say having ‘n’ odd number of observations, can be used as:
Median = [(n + 1)/2]th term.
Median Formula when ‘n’ is Even:
The median formula of a given arrangement of numbers say having ‘n’ even number of observations can be used as:
Median = [(n/2)th term + ((n/2) + 1)th term]/2.
## Median Formula For Grouped Data
When the data is continuous and in the type of frequency distribution, the median is determined through the following sequence of steps.
Stage 1: Find the total number of observations(n).
Stage 2: Define the class size(h), and partition the data into different classes.
Stage 3: Calculate the combined frequency of each class.
Stage 4: Identify the class wherein the median falls.
Stage 5: Find the lower limit of the median class(l), and the aggregate frequency of the class preceding the median class (c).
Now, utilize the following formula to find the median value.
## How can we find the median?
We utilize a median formula to find the median worth of given data. For a bunch of ungrouped data, we can follow the beneath-given steps to find the median value.
Stage 1: Sort the given data in increasing order.
Stage 2: Count the number of observations.
Stage 3: If the number of observations is odd utilize median.
Formula: Median = [(n + 1)/2]th term
Stage 4: If the number of observations is even utilize median.
Formula: Median = [(n/2)th term + (n/2 + 1)th term]/2
For a bunch of grouped data, we can follow the following steps to find the median:
When the data is continuous and in the type of frequency distribution, the median is determined through the following sequence of steps.
Stage 1: Find the total number of observations(n).
Stage 2: Define the class size(h), and partition the data into different classes.
Stage 3: Calculate the total frequency of each class.
Stage 4: Identify the class where the median falls.
Stage 5: Find the lower limit of the median class(l), and the aggregate frequency(c).
Stage 6: Apply the formula for the median for gathered data:
Median Of 2 Numbers:
In an ordered series, the median is the number that is mid-way between the range limits. It isn’t generally identical to the mean. How about we understand how to find the median. For a bunch of two values, the median will be equivalent to the mean, or number arithmetic average. The median is the worth at the mid-point of the dataset, not the mid-point of the values. The mean is the number arithmetic average
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## What is the prize money of International Math Olympiad?
The accompanying awards will be given to the winners of...
## What is the Maths Olympiad Exam format?
The International Maths Olympiad (IMO) is a stage given to... |
How to figure out the value of a mathematical foot June 19, 2021 June 19, 2021
Maths math is often used in the context of business calculations, but the way mathematicians calculate value can also be used in other fields.
The value of one foot is often expressed in dollars or other currency.
If we multiply 1 Canadian by 100, that would equal 1 Canadian Dollar.
1 Canadian is a unit of currency, but 1 Canadian foot is a metric foot.
The units of measurement are different, but we use the same formula to calculate the difference between two feet.
We can also use the units of measure in a similar way, but when we use inches to measure a foot, we usually don’t want to multiply by 100.
To convert one metric foot to a foot or vice versa, we divide it by 100 and then add it to the metric foot we have in our possession.
The math behind measuring distance using meters and feet This is where the metric system comes into play.
We use the following formula to determine the distance between two points: (m/feet) / (m).
This formula gives us a distance, or m, or in this case a metric value.
When we multiply a metric by a distance to calculate a metric distance, we can add it back to the first measurement.
We are using the units that we have already seen, but this formula tells us how far we are from the first point.
So if we measure a distance of 100 metres from a point we are 10 metres from, then the distance is 100 meters.
That means that we are 1.5 metres from the last point.
In this example, that means that the distance of a metric point is 100 metres.
In the same example, if we add this distance to the distance to measure from the beginning, we would now measure from 0 metres to the end.
We have already measured distance, so we can continue.
Let’s add the metric value to the measurement to get the final value.
In order to do this, we multiply the measurement by 100 meters, and then divide by the value that we had in our hands.
That value is the distance from the starting point to the ending point.
The equation we used to determine how far from the start we are is: m × 100 metres / (100m × 0.5) × 100 meters = m / 100 metres This is a very simple formula.
Now, to add this metric value into the measurement we have, we need to subtract the measurement from 100 meters to get back to 0 metres.
That is, we subtract 100 metres and then multiply by the metric.
We then multiply the result by 0.75 to get our final value: m / (0.75 × 0) / 100m = m – 0.25.
We will now multiply this by the units in our pocket.
This is done by adding one cent to the unit of measure we have and then subtracting one cent.
The final result is m / 0.125 = 0.05 metres.
We don’t know how many metres we have left to add to the previous measurement.
The calculation continues until we reach the end of the meter.
To measure distance, multiply the value by 1 metre, and subtract the unit we have from 0.01 metres.
If the distance exceeds the number of metres we need, we add one cent and subtract one cent, and so on until we have reached the value we want.
Calculating distance using millimetres This is another method that uses a metre to measure distance.
A metre is a measure of the distance in metres.
A millimetre is a measurement of the number that is a millimetron in diameter.
For the example in the last section, a metre is one metre, or about 1.8 metres.
When using a metre, we measure the distance along the surface of the water using a meter.
This means that our measurement of depth is equal to one metre.
The number of millimetrees we use in our calculation is 1 millimetrey.
We would multiply the number by 0, and multiply that number by the millimetere to get a final value that is the number one millimetree.
We also multiply that value by 100 to get 1 cent.
We multiply the answer to this question by 100 metres to get that value of 1 cent that is equal as 1 millimeter.
The formula that calculates this is: 1.9 / 0 (m × 100 / 0) × 0 (1 millimetrem) = 1.0 metres We use this formula to get an answer that is 1.2 metres.
The distance is 1 metre from the surface to the bottom.
So the answer that we got is 1 meter.
The calculator for calculating the distance can be found at the bottom of this page. |
Subtraction(KS2, Year 4)
The Lesson
Subtraction is the taking away of one number (or object) from another to make a difference. For example, 5 apples take away 3 apples makes 2 apples:
In numbers:
Subtraction is denoted by the minus sign, -.
How to Subtract
The method used to subtract numbers will differ depending on the difficulty of the subtraction. Subtracting short numbers from each other is easier than subtracting long numbers together.
How to Subtract Short Numbers From Each Other
It is easy to subtract short numbers. Children learn to put a certain number of fingers up, then put some of them down to see how many are left. They may do the same with other objects. A number line can be used to subtract. For example, 5 - 3 can be performed on a number line:
How to Subtract Long Numbers From Each Other
It is more difficult to subtract long numbers from each other. For example:
This sum is made easier when we notice numbers are made of hundreds, tens and units (i.e. the place value of the digits in the number).
This allows the numbers to be broken down:
Each part - the hundreds (in blue), the tens (in yellow) and the units (in red) - can then be subtracted from each other:
The solution to 256 - 124 is 132. This method of breaking numbers down into hundreds, tens and units is the basis for long subtraction, which offers a more systematic way of doing subtraction.
How to Do Long Subtraction
Long subtraction involves writing each number in columns and subtracting a column at a time. As each column represents the hundreds, tens and units of the numbers, long subtraction implicitly breaks the subtraction down into subtracting the hundreds, tens and units, but without you having to think about it:
1
Write the numbers you wish to subtract, one underneath the other. Ensure they are aligned with each other so the units of one number is directly underneath the units of the other:
1
Subtract the numbers in the units column.
6 - 4 = 2
1
Subtract the numbers in the tens column.
5 - 2 = 3
1
Subtract the numbers in the hundreds column.
2 - 1 = 1
The solution to 256 - 124 is 132
Lesson Slides
The slider below shows another example of how to do long subtraction: Open the slider in a new tab
Parts of Subtraction
• The number you take away is the subtrahend.
• The result of subtracting the numbers is the difference.
Line Up the Columns! When doing long subtraction, ensure you align the numbers in the correct columns, so the units are all in the units column etc. For example, the following would be incorrect. The units of 11 (1) is underneath the tens of 325 (2):
Borrowing
When subtracting numbers in a column, sometimes the number you are taking away from is smaller than the number being taken away:
5 is less than 7, we should get -2. But that doesn't make any sense here. To do long subtraction correctly, we need to borrow a 1 from the top number in the column to the left.
• Take 1 away from the 3, the number to the left of the 5, leaving 2:
• Place this borrowed 1, and write it in front of the 5:
• The 5 has become a 15, which is bigger than the 7. Do the subtraction:
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# 4/6 Simplified, Simplify 4/6 to Lowest Terms
4/6 simplified as a fraction is equal to 2/3. Here we simplify 4/6 in its simplest form, that is, reduce 4/6 to its lowest terms, and it is equal to 2/3.
There are two ways to simplify 4/6.
1. First method: Write the prime factorisations of the numerator 4 and the denominator 6. Then cancel the common numbers appeared in the prime factorisations of 4 and 6 to reduce 4/6 to its lowest terms.
2. Second method: Divide both the numerator and the denominator of 4/6 by the greatest common divisor of 4 and 6. The resultant fraction will be the simplified form of 4/6.
Let us now simplify the fraction 4/6 in its simplified form which is 2/3.
## Simplify 4/6 to its Simplest Form
To simplify 4/6 in its simplest form, we write down the prime factorisations of both the numbers 4 and 6.
4 = 2 × 2
6 = 2 × 3.
Therefore,
$\dfrac{4}{6} = \dfrac{2 \times 2}{2 \times 3}$
⇒ $\dfrac{4}{6} = \dfrac{\cancel{2} \times 2}{\cancel{2} \times 3}$, here we cancel the common number 2.
⇒ $\dfrac{4}{6} = \dfrac{2}{3}$
So 4/6 simplified in its simplest form is equal to 2/3.
## How to Simplify the Fraction 4/6
Let us follow below steps in order to simplify 4/6 in its simplest form.
At first, divide both the numerator 4 and the denominator 6 by their greatest common divisor (GCD).
Let us now find the GCD of 4 and 6.
• The divisors of 4 are 1, 2, and 4.
• The divisors of 6 are 1, 2, 3, and 6.
See that 2 is the highest number that divides both 4 and 6. Therefore, the greatest common divisor of 4 and 6 is 2. That is,
GCD (4, 6) = 2.
Divide the top and the bottom of 4/6 by GCD (4, 6) = 2 in order to simplify the fraction 4/6. So we get that
$\dfrac{4}{6} = \dfrac{4 \div 2}{6 \div 2} = \dfrac{2}{3}$.
So 4/6 reduced to lowest terms is equal to 2/3.
Video solution on simplifying 4/6:
More Simplifying Fractions:
## FAQs
Q1: Reduce 4/6 to lowest terms.
Answer: The fraction 4/6 when reduced to lowest terms is equal to 2/3.
Q2: How simplify 4/6 in its simplest form?
Answer: To simplify 4/6, divide the top 4 and the bottom 6 by their greatest common divisor (GCD). Note that GCD(4, 6) = 2. Now 4/6 = (4 ÷ 2)/(6 ÷ 2) = 2/3. So 2/3 is the simplest reduced form of the fraction 4/6.
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Riemann integral exercises with detailed answers are offered on this page. These kinds of integrals of bounded function in bounded intervals, almost everywhere continuous. Thus, all continuous or monotone functions are Riemann integrable.
## Lower integral, upper integral, and proper integral
In this section, we give a concise definition of Riemann integrals. To this end, we assume that a real function is defined and bounded on the interval $[a,b]$, with $a,b\in\mathbb{R}$ are such that $a<b$.
For a subdivision $\sigma:=\{x_0,\cdots.x_n\}$ of the interval $[a,b]$, i.e. $a=x_0<x_1<\cdots<x_n=b$, we define the lower sum of $f$ on $[a,b]$, by $$L_n(f,\sigma)=\sum_{i=1}^n m_i(f,\sigma)(x_i-x_{i-1}),$$ where $m_i(f,\sigma)$ is the infinimum of $f$ on the interval $[x_{i-1},x_i]$.
By definition, the lower integral of f over the interval $[a,b]$ is $$\underline{\int^b_a}f(x)ds=\sup_{\sigma} L_n(f,\sigma)=\lim_{n\to\infty}U_n(f,\sigma).$$ Similarly, we define the upper sum of $f$ over the interval $[a,b]$ by $$U_n(f,\sigma)=\sum_{i=1}^n M_i(f,\sigma)(x_i-x_{i-1}),$$ where $M_i(f,\sigma)$ is the supremum of $f$ on the interval $[x_{i-1},x_i]$.
The upper integral of f over the interval $[a,b]$ is $$\overline{\int^b_a}f(x)ds=\inf_{\sigma} U_n(f,\sigma)=\lim_{n\to\infty}U_n(f,\sigma).$$ Definition: A bounded function $f:[a,b]\to\mathbb{R}$ is said to be Riemann integrable on $[a,b]$ if its lower and upper integrals coincide. In this case, the integral of $f$ over $[a,b]$ is given by $$\int^b_a f(x)dx=\overline{\int^b_a}f(x)ds=\underline{\int^b_a}f(x)ds.$$ There exists another equivalent condition for Riemann integrals. Let’s give a summary of it. We take $c_\in [x_{i-1},x_i]$ for $i=1,\cdots,n$ and de define the Riemann sum by $$S_n(f,\sigma):=\sum_{i=1}^n f(c_i)(x_{i}-x_{i-1}).$$ Observe that $L_n(f,\sigma)\le S_n(f,\sigma)\le U_n(f,\sigma)$. Thus if $f$ is Riemann integrable then $S_n(f,\sigma)$ has a limit as $n\to\infty,$ and in this case we have $$\int^b_a f(x)dx=\lim_{n\to\infty} \sum_{i=1}^n f(c_i)(x_{i}-x_{i-1}).$$
A particular cas: Let $f:[0,1]\to \mathbb{R}$ be Riemann integral and take the uniform subdivision $\sigma:=\{\frac{i}{n}: i=0,\cdots,n\}$, and $c_i=\frac{i}{n}$, Then $$\int^1_0 f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right).$$
## Riemann integral exercises
We start our selection of Riemann integral exercises by providing an example of a non-integrable bounded function in the sense of Riemann
Exercise: Prove that the following function $$g(x)=\cos^2(x)1_{\mathbb{Q}}(x),\quad x\in [0,\pi/2],$$ is not Riemann integrable.
Exercise: Let $f:[0,1]\to \mathbb{R}$ be a continuous function. For any $n\in\mathbb{N},$ we set \begin{align*} u_n=n\int^1_0 t^n f(t)dt. \end{align*}
1. Let $\varepsilon>0$. Prove that there exists $a\in (0,1)$ such that $|f(t)|\le \frac{\varepsilon}{2}$ whenever $t\in [a,1]$.
2. We denote $M=\sup{|f(t)|:t\in [0,1]}$, a such $M\in \mathbb{R}^+$ exists because $f$ is continuous on the compact $[0,1]$. Prove that \begin{align*} |u_n|\le M a^n+\frac{\varepsilon}{2},\qquad \forall n\in\mathbb{N}. \end{align*}
3. Deduce that $\lim_{n\to+\infty}u_n=0$.
4. Deduce that, in general, we have \begin{align*} \lim_{n\to+\infty}u_n=f(1). \end{align*}
Solution: The first 3 question: By definition of left continuity at point $1,$ for all $\varepsilon>0,$ there exists $0 < \delta < 1$ such that $t\in [1-\delta,1]$ we have $|f(t)-f(1)|=|f(t)| \le \frac{\varepsilon}{2}$. It suffices to put $a=1-\delta\in (0,1)$.
We can write \begin{align*} |u_n|&\le n \int^1_0 t^n |f(t)|dt\cr &\le n \int^a_0 t^n |f(t)|dt+n \int^1_a t^n |f(t)|dt\cr & \le n M \int^a_0 t^n dt+ n \frac{\varepsilon}{2} \int^1_a t^n dt\cr & \le n M \left[\frac{t^{n+1}}{n+1}\right]^a_0+ n \frac{\varepsilon}{2} \left[\frac{t^{n+1}}{n+1}\right]^1_a \cr &\le M \frac{n}{n+1} a^{n+1}+\frac{\varepsilon}{2} \frac{n}{n+1} (1-a^{n+1}). \end{align*} As $a\in (0,1),$ then $a^{n+1}\le a^n$ and $0 < 1-a^{n+1} < 1$. On the other hand, $\frac{n}{n+1}le 1$. This implies that \begin{align*} |u_n|\le M a^n+\frac{\varepsilon}{2},\qquad \forall n\in\mathbb{N}. \end{align*}
Since $a\in (0,1)$ then the geometric sequence $(a^n)_n$ satisfies $a^n\to 0$ as $n\to+\infty$. Then there exists $N\in \mathbb{N}$ such that for any $n\ge N,$ we have $0 < a^n < \frac{\varepsilon}{2M}$. Hence for $n\ge N,$ we have \begin{align*} |u_n|\le M \frac{\varepsilon}{2M}+\frac{\varepsilon}{2}=\varepsilon. \end{align*} Thus $u_n\to 0$ as $n\to +\infty$.
4) In this question we work with a general continuous function $f:[0,1]\to \mathbb{R}$. We set $g(t)=f(t)-f(1)$. The function $g$ is continuous on $[0,1]$ and satisfies $g(1)=0$. According to the first question we have \begin{align*} \lim_{n\to+\infty} n\int^1_0 t^n g(t)dt=0. \end{align*} In addition, observe that \begin{align*} u_n&=n\int^1_0 t^n g(t)dt+n f(1) \int^1_0 t^n dt \cr &= n\int^1_0 t^n g(t)dt+\frac{n}{n+1}f(1) \end{align*} Hence $u_n\to f(1)$ as $n\to +\infty$.
Exercise: Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function and let $\omega\in\mathbb{R}^\ast$. Define the following function \begin{align*} \varphi(x)=\frac{1}{\omega}\int^x_0 \sin\left(\omega (x-t)\right)f(t),dt,\qquad x\in\mathbb{R}. \end{align*}
• Prove that there exist two functions $\Phi,\Psi:\mathbb{R}\to \mathbb{R}$ such that for any $x\in\mathbb{R},$ \begin{align*} \varphi(x)=\frac{\sin(\omega x)}{\omega} \Phi(x)- \frac{\cos(\omega x)}{\omega}\Psi(x). \end{align*}
• Show that $\varphi$ is twice differentiable on $\mathbb{R}$ and that \begin{align*} \varphi”+\omega^2 \varphi=f. \end{align*}
Solution: 1) Let $x\in\mathbb{R}$ and $t\in\mathbb{R}$. We know from trigonometric formulas that \begin{align*} \sin(\omega x-\omega t)=\sin(\omega x)\cos(\omega t)-\cos(\omega x)\sin(\omega t). \end{align*} Then \begin{align*} \varphi(x)&= \frac{\sin(\omega x)}{\omega} \int^x_0 \cos(\omega t)f(t)dt-\frac{\cos(\omega x)}{\omega} \int^x_0 \sin(omega t)f(t)dt. \end{align*} Thus it suffices to select \begin{align*} \Phi(x)= \int^x_0 \cos(\omega t)f(t)dt\quad\text{and} \quad \Psi(x)=\int^x_0 \sin(\omega t)f(t)dt. \end{align*}
2) The functions $\Phi$ and $\Psi$ are primitives of continuous functions. Thus $\Phi$ and $\Psi$ are $C^1$ functions on $\mathbb{R}$ and that \begin{align*} \Phi'(x)=\cos(\omega x)f(x),\qquad \Psi(x)=\sin(\omega x)f(x) \end{align*} for any $x\in \mathbb{R}$. This proves that the function $varphi$ is differential on $\mathbb{R}$ as the product and sum of differentiable functions. Moreover, \begin{align*} \omega\varphi'(x)&= \omega \cos(\omega x) \Phi(x)+\sin(\omega x) \cos(\omega x)f(x)\cr & \hspace{2cm}+\omega\sin(\omega x)\Psi(x)-\cos(\omega x)\sin(\omega x)f(x)\cr &= \omega \cos(\omega x) \Phi(x)+\omega\sin(\omega x)\Psi(x). \end{align*} As $\omega\neq 0,$ then \begin{align*} \varphi'(x)=\cos(\omega x) \Phi(x)+\sin(\omega x)\Psi(x) \end{align*} This last formula show that $\varphi’$ is differentiable as product and sum of differentiable functions. Hence $\varphi$ is twice differentiable and \begin{align*} \varphi”(x)&=-\omega\sin(\omega x) \Phi(x)+\cos^2(\omega x)f(x)\cr & \hspace{2.5cm}+\omega\cos(\omega x)\Psi(x)+\sin^2(\omega x)f(x)\cr &= -\omega (\sin(\omega x) \Phi(x)-\cos(\omega x)\Psi(x))+f(x)\cr &= -\omega^2 \varphi(x)+f(x) \end{align*} for any $x\in\mathbb{R}$. This proves that \begin{align*} \varphi”+\omega^2 \varphi=f. \end{align*}
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### Which Solids Can We Make?
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Interior angles can help us to work out which polygons will tessellate. Can we use similar ideas to predict which polygons combine to create semi-regular solids?
### First Forward Into Logo 4: Circles
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Learn how to draw circles using Logo. Wait a minute! Are they really circles? If not what are they?
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This is the second in a twelve part introduction to Logo for beginners. In this part you learn to draw polygons.
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How can you make an angle of 60 degrees by folding a sheet of paper twice?
### First Forward Into Logo 5: Pen Up, Pen Down
##### Age 7 to 16 Challenge Level:
Learn about Pen Up and Pen Down in Logo
### Making Maths: Clinometer
##### Age 11 to 14 Challenge Level:
You can use a clinometer to measure the height of tall things that you can't possibly reach to the top of, Make a clinometer and use it to help you estimate the heights of tall objects.
### Making Maths: Equilateral Triangle Folding
##### Age 7 to 14 Challenge Level:
Make an equilateral triangle by folding paper and use it to make patterns of your own.
### Factors and Multiples Game
##### Age 7 to 16 Challenge Level:
A game in which players take it in turns to choose a number. Can you block your opponent?
### More Marbles
##### Age 11 to 14 Challenge Level:
I start with a red, a blue, a green and a yellow marble. I can trade any of my marbles for three others, one of each colour. Can I end up with exactly two marbles of each colour?
### Cubic Conundrum
##### Age 7 to 16 Challenge Level:
Which of the following cubes can be made from these nets?
### Sea Defences
##### Age 7 to 14 Challenge Level:
These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together?
### Cunning Card Trick
##### Age 11 to 14 Challenge Level:
Delight your friends with this cunning trick! Can you explain how it works?
### Whirling Fibonacci Squares
##### Age 11 to 16
Draw whirling squares and see how Fibonacci sequences and golden rectangles are connected.
### Making Maths: Snake Pits
##### Age 5 to 14 Challenge Level:
A game to make and play based on the number line.
### Making Maths: Double-sided Magic Square
##### Age 7 to 14 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
### Tower of Hanoi
##### Age 11 to 14 Challenge Level:
The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.
### Tangram Pictures
##### Age 5 to 14 Challenge Level:
Use the tangram pieces to make our pictures, or to design some of your own!
### Nine Colours
##### Age 11 to 16 Challenge Level:
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
### Constructing Triangles
##### Age 11 to 14 Challenge Level:
Generate three random numbers to determine the side lengths of a triangle. What triangles can you draw?
### Rolling Triangle
##### Age 11 to 14 Challenge Level:
The triangle ABC is equilateral. The arc AB has centre C, the arc BC has centre A and the arc CA has centre B. Explain how and why this shape can roll along between two parallel tracks.
### Marbles
##### Age 11 to 14 Challenge Level:
I start with a red, a green and a blue marble. I can trade any of my marbles for two others, one of each colour. Can I end up with five more blue marbles than red after a number of such trades?
### Triangles to Tetrahedra
##### Age 11 to 14 Challenge Level:
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
### First Forward Into Logo 9: Stars
##### Age 11 to 18 Challenge Level:
Turn through bigger angles and draw stars with Logo.
### Celtic Knotwork Patterns
##### Age 7 to 14
This article for pupils gives an introduction to Celtic knotwork patterns and a feel for how you can draw them.
### Making Rectangles, Making Squares
##### Age 11 to 14 Challenge Level:
How many differently shaped rectangles can you build using these equilateral and isosceles triangles? Can you make a square?
##### Age 11 to 18
Logo helps us to understand gradients of lines and why Muggles Magic is not magic but mathematics. See the problem Muggles magic.
##### Age 7 to 14 Challenge Level:
What shape and size of drinks mat is best for flipping and catching?
### Sociable Cards
##### Age 11 to 14 Challenge Level:
Move your counters through this snake of cards and see how far you can go. Are you surprised by where you end up?
### Factors and Multiples Game for Two
##### Age 7 to 14 Challenge Level:
Factors and Multiples game for an adult and child. How can you make sure you win this game?
### Notes on a Triangle
##### Age 11 to 14 Challenge Level:
Can you describe what happens in this film?
### Amazing Card Trick
##### Age 11 to 14 Challenge Level:
How is it possible to predict the card?
### First Forward Into Logo 7: Angles of Polygons
##### Age 11 to 18 Challenge Level:
More Logo for beginners. Learn to calculate exterior angles and draw regular polygons using procedures and variables.
### Fractions Jigsaw
##### Age 11 to 14 Challenge Level:
A jigsaw where pieces only go together if the fractions are equivalent.
### Back to the Practical?
##### Age 7 to 14
In this article for teachers, Bernard uses some problems to suggest that once a numerical pattern has been spotted from a practical starting point, going back to the practical can help explain. . . .
### Making Maths: Walking Through a Playing Card?
##### Age 7 to 14 Challenge Level:
It might seem impossible but it is possible. How can you cut a playing card to make a hole big enough to walk through?
### The Best Card Trick?
##### Age 11 to 16 Challenge Level:
Time for a little mathemagic! Choose any five cards from a pack and show four of them to your partner. How can they work out the fifth?
### Gym Bag
##### Age 11 to 16 Challenge Level:
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### Factors and Multiples Puzzle
##### Age 11 to 14 Challenge Level:
Using your knowledge of the properties of numbers, can you fill all the squares on the board?
### Making Maths: Archimedes' Spiral
##### Age 7 to 14 Challenge Level:
Make a spiral mobile.
### Attractive Rotations
##### Age 11 to 14 Challenge Level:
Here is a chance to create some attractive images by rotating shapes through multiples of 90 degrees, or 30 degrees, or 72 degrees or...
### Drawing Celtic Knots
##### Age 11 to 14 Challenge Level:
Here is a chance to create some Celtic knots and explore the mathematics behind them.
### Making Maths: Celtic Knot Tiles
##### Age 7 to 16 Challenge Level:
Make some celtic knot patterns using tiling techniques
### Turning the Place Over
##### Age 11 to 18 Challenge Level:
As part of Liverpool08 European Capital of Culture there were a huge number of events and displays. One of the art installations was called "Turning the Place Over". Can you find our how it works?
### Modular Origami Polyhedra
##### Age 7 to 16 Challenge Level:
These models have appeared around the Centre for Mathematical Sciences. Perhaps you would like to try to make some similar models of your own.
### Well Balanced
##### Age 5 to 18 Challenge Level:
Exploring balance and centres of mass can be great fun. The resulting structures can seem impossible. Here are some images to encourage you to experiment with non-breakable objects of your own.
### Making Maths: Make a Pendulum
##### Age 7 to 14 Challenge Level:
Galileo, a famous inventor who lived about 400 years ago, came up with an idea similar to this for making a time measuring instrument. Can you turn your pendulum into an accurate minute timer?
### First Forward Into Logo 6: Variables and Procedures
##### Age 11 to 18 Challenge Level:
Learn to write procedures and build them into Logo programs. Learn to use variables.
### First Forward Into Logo 11: Sequences
##### Age 11 to 18 Challenge Level:
This part introduces the use of Logo for number work. Learn how to use Logo to generate sequences of numbers. |
# ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 8 Ratio and Proportion Ex 8.1
## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 8 Ratio and Proportion Ex 8.1
Question 1.
Express the following ratios in simplest form:
(i) 20 : 40
(ii) 40 : 20
(iii) 81 : 108
(iv) 98 : 63
Solution:
Question 2.
Fill in the missing numbers in the following equivalent ratios:
Solution:
Question 3.
Find the ratio of each of the following in simplest form :
(i) 2.1 m to 1.2 m
(ii) 91 cm to 1.04m
(iii) 3.5 kg to 250gm
(iv) 60 paise to 4 rupees
(v) 1 minute to 15 seconds
(vi) 15 mm to 2 cm
Solution:
Question 4.
The length and the breadth of a rectangular park are 125 m and 60 m respectively. What is the ratio of the length to the breadth of the park?
Solution:
Question 5.
The population of village is 4800. If the numbers of females is 2160, find the ratio of males to that of females.
Solution:
Question 6.
In a class, there are 30 boys and 25 girls. Find the ratio of the numbers of
(i) boys to that of girls.
(ii) girls to that of total number of students.
(iii) boys to that of total numbers of students.
Solution:
Question 7.
In a year, Reena earns ₹ 1,50,000 and saves ₹ 50,000. Find the ratio of
(i) money she earns to the money she saves.
(ii) money that she saves to the money she spends.
Solution:
Question 8.
The monthly expenses of a student have increased from ₹350 to ₹500. Find the ratio of
(i) increase in expenses and original expenses.
(ii) original expenses to increased expenses.
(iii) increased expenses to increased in expenses.
Solution:
Question 9.
Mr Mahajan and his wife are both school teachers and earn ₹20900 and ₹ 18700 per month respectively. Find the ratio of
(i) Mr Mahajan’s income to his wife’s income.
(ii) Mrs Mahajan’s income to the total income of both.
Solution:
Question 10.
Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of
(a) Number of students liking football to number of students liking tennis.
(b) Number of students liking cricket to total number of students.
Solution:
Question 11.
Divide ₹ 560 between Ramu and Munni in the ratio 3 : 2.
Solution:
Question 12.
Two people invested ₹ 15000 and ₹25000 respectively to start a business. They decided to share the profits in the ratio of their investments. If their profit is ₹ 12000, how much does each get?
Solution:
Question 13.
The ratio of Ankur’s money to Roma’s money is 9 : 11. if Ankur has ₹540, how much money does Roma have?
Solution:
Question 14.
The ratio of weights of tin and zinc in on alloy is 2 : 5. How much zinc is there in 31.5g of alloy?
Solution: |
# Multiples of 4
The Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, … i.e. M4 = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, …}. Multiples of 4 are the series of numbers obtained by multiplying 4 with the natural numbers.
4 × 1 = 4
4 × 2 = 8
4 × 3 = 12
4 × 4 = 16
4 × 5 = 20
4 × 6 = 24
4 × 7 = 28
4 × 8 = 32
4 × 9 = 36
4 × 10 = 40
And so on.
From this series, we observe that the multiples of 4 can be derived by multiplying 4 by natural numbers in ascending order. These multiples form an arithmetic sequence with a common difference of 4.
## Properties of Multiples of 4
Multiples of 12 possess several interesting properties that make them unique. Here are a few notable ones:
1. Divisibility by 4: Every multiple of 4 is divisible by 4 without leaving a remainder. This property applies universally to all multiples of 4, as they can be expressed as 4 multiplied by an integer.
2. Even Numbers: Multiples of 4 are always even numbers. An even number is any integer that is divisible by 2. Since 4 is divisible by 2, all its multiples will also be divisible by 2, making them even.
3. Increasing by 4: Each subsequent multiple of 4 is obtained by adding 4 to the previous multiple. For instance, starting from 4, we add 4 to get 8, then 12, and so on. This pattern continues indefinitely.
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## Patterns and Observations:
1. Digit Sum: If we observe the digit sums of multiples of 4, we notice a pattern. The digit sums alternate between 4 and 8. For example, the digit sum of 4 is 4, while the digit sum of 8 is 8. Similarly, the digit sum of 12 is 3 (1 + 2 = 3), and the digit sum of 16 is 7 (1 + 6 = 7). This pattern of alternating digit sums repeats indefinitely.
2. Repeating Last Two Digits: Another interesting pattern emerges when we look at the last two digits of multiples of 4. The last two digits of the multiples follow a cyclical pattern: 04, 08, 12, 16, 20, 24, and so on. This cyclic nature is a result of the fact that each multiple is formed by adding 4 to the previous multiple.
## Applications of Multiples of 4
The study of multiples of 4 finds applications in various fields, including mathematics, science, and everyday life:
1. LCMOne of the applications of multiples of 4 is in finding the lowest common multiple (LCM) of two or more numbers. The LCM is the lowest multiple that two or more numbers have in common. For example, to find the LCM of 4 and 6, we need to find the multiples of both numbers and identify the lowest multiple they have in common. The multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 etc. The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, … etc. The lowest multiple that they have in common is 12. Therefore, the LCM of 4 and 6 is 12.
2. Timekeeping: The concept of multiples of 4 is utilized in our conventional system of timekeeping. Hours, minutes, and seconds are divided into multiples of 4, enabling us to track and measure time efficiently.
3. Computing and Programming: In computer programming, multiples of 4 are often used for memory allocation and addressing. Memory addresses are usually aligned on multiples of 4 for optimal performance.
4. Music and Rhythm: Multiples of 4 play a crucial role in music and rhythm. Musical beats and measures often follow patterns based on multiples of 4, creating the foundation for various musical compositions and genres.
## Conclusion:
The Multiples of 4 are the numbers obtained by multiplying 4 with natural numbers. The multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 etc. The multiples of 4 have various properties, such as divisibility by 4, even numbers, increasing by 4, digit sum, repeating last two digits, etc. The multiples of 4 have several applications in mathematics, such as finding the LCM, timekeeping, computing and programming, and music and rhythm. |
# Lateral Area of Cylinder: Formula, Example and More!
Cylinders are common three-dimensional shapes with different usages in everyday life, from our soup cans to industrial pipes. Furthermore, understanding their surface area is important for different calculations. Moreover, the lateral surface area is a specific part of a cylinder’s total surface area. Read on to learn more in detail about the Lateral Area of a Cylinder, its Formula, example and more.
## Formula of the Lateral Area of a Cylinder
Additionally, the Formula for the Lateral surface area of a Cylinder is:
Lateral Surface Area = 2πrh
To make it simpler, in this formula:
• π (pi) is a mathematical constant roughly equal to 3.14.
• r is the radius of the cylinder’s base (the distance from the centre of the base to the edge).
• h is the height of the cylinder (the vertical distance between the two circular bases).
This formula calculates the area of a rectangle that would perfectly wrap around the curved surface of the cylinder, hence like peeling a label off a Coca Cola can.
## Where is the Lateral Surface Area of the Cylinder?
The Lateral surface area of a Cylinder is the area of the curved surface that runs along the sides of the cylinder. Moreover, it does not include the area of the two circular bases at the top and bottom of the cylinder.
In addition, the Lateral surface area is an important measurement in many applications, like:
• Calculating the amount of material needed to cover the curved surface of a cylinder (e.g. for a storage tank, pipe, or drum).
• Deciding the heat transfer or insulation requirements for the curved surface.
• Calculating the surface area for painting, coating, or decorating the curved surface.
Also Read: Conic Sections
## Example of the Lateral Area of a Cylinder
Let us assume that you have a cylindrical water tank with a base radius of 5 meters and a height of 3 meters. Moreover, you want to paint the curved side of the tank. Thus, to find the amount of paint needed, you will calculate the lateral surface area.
• Radius (r) = 5 meters
• Height (h) = 3 meters
Lateral Surface Area = 2πrh = 2 ✕ 3.14 ✕ 5 ✕ 3 = 94.2 (square meters)
Therefore, you will need approximately 94.2 square meters of paint to cover the curved side of the water tank.
I hope this helps! Did you like learning about the Lateral Area of Cylinder? Keep reading our blogs to learn more about the basic concepts of Maths! |
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# Class 12 RD Sharma Solutions- Chapter 11 Differentiation – Exercise 11.4 | Set 1
• Last Updated : 13 Jan, 2021
### Question 1. xy = c2
Solution:
We have xy=c2
Differentiating both sides with respect to x.
d(xy)/dx = d(c2)/dx
By product rule,
y+x*dy/dx=0
dy/dx=-y/x
### Question 2. y3 -3xy2=x3+3x2y
Solution:
We have
y3-3xy2=x3+3x2y
Differentiating both sides with respect to x,
d(y3-3xy2)/dx=d(x3+3x2y)/dx
By product rule,
=> 3y2dy/dx-3y2-6xydy/dx=3x2+3x2dy/dx+6xy
=>3y2dy/dx-6xydy/dx-3x2dy/dx=3x2+3y2+6xy
=>dy/dx(3y2-3x2-6xy)=3x2+3y2+6xy
=>3dy/dx(y2-x2-2xy)=3(x2+y2+2xy)
=> dy/dx={3(x+y)2}/{3(y2-x2-2xy)
dy/dx=(x+y)2/(y2-x2-2xy)
### Question 3. x2/3+y2/3=a2/3
Solution:
We have,
x2/3+y2/3=a2/3
Differentiating both sides with respect to x,
d(x2/3)/dx +d(y2/3)/dx=d(a2/3)/dx
=> 2/3x1/3 +(2/3y1/3)dy/dx=0
=>1/x1/3 +(1/y1/3)dy/dx =0
=> dy/dx=-y1/3/x1/3
dy/dx=-y1/3/x1/3
### Question 4. 4x+3y=log(4x-3y)
Solution:
We have,
4x+3y= log(4x-3y)
Differentiating both sides with respect to x,
d(4x+3y)/dx=d(log(4x-3y))/dx
=>4+3dy/dx=(1/(4x-3y))(4-3dy/dx)
=>3dy/dx+3dy/dx(1/4x-3y)=4/(4x-3y)-4
=>(3dy/dx)(1+1/(4x-3y))=(4-16x+12y)/(4x-3y)
=>(3dy/dx)((4x-3y+1)/(4x-3y))=(4-16x+12y)/(4x-3y)
=>(3dy/dx)(4x-3y+1)=4-16x+12y
=>(3dy/4dx)(4x-3y+1)=3y-4x+1
=>dy/dx=(4/3)((3y-4x+1)/(4x-3y+1))
dy/dx=4(3y-4x+1)/3(4x-3y+1)
### Question 5. (x2/a2)+ (y2/b2)=1
Solution:
We have,
(x2/a2)+(y2/b2)=1
Differentiating both sides with respect to x,
d(x2/a2)/dx +d(y2/b2)/dx =d(1)/dx
=>(2x/a2)+(2y/b2)(dy/dx)=0
=>(y/b2)(dy/dx)=-x/a2
=>dy/dx = -xb2/ya2
dy/dx =-xb2/ya2.
### Question 6. x5+y5=5xy
Solution:
We have,
x5+y5 =5xy
Differentiating both sides with respect to x,
d(x5)/dx +d(y5)/dx=d(5xy)/ dx
=> 5x4 + 5y4dy/dx=5y+ (5x)dy/dx
=>y4(dy/dx)-x(dy/dx)=y-x4
=>dy/dx(y4-x)=y-x4
=>dy/dx=(y-x4)/(y4-x)
dy/dx=(y-x4)/(y4-x)
### Question 7. (x+y)2=2axy
Solution:
We have,
(x+y)2=2axy
Differentiating with respect to x,
d(x+y)2/dx=d(2axy)/dx
=>2(x+y)(1+dy/dx)=2ax(dy/dx) +2ay
=>(x+y)+(x+y)dy/dx =ax(dy/dx)+ay
=>(x+y)dy/dx-ax(dy/dx)=ay-x-y()
=>(dy/dx)(x+y-ax)=ay-x-y
=>dy/dx=(ay-x-y)/(x+y-ax)
dy/dx=(ay-x-y)/(x+y-ax)
### Question 8. (x2+y2)2=xy
Solution:
We have,
(x2+y2)2=xy
Differentiating both sides with respect to x,
d(x2+y2)2/dx=d(xy)/dx
=>2(x2+y2)(2x+2y(dy/dx))=y+x(dy/dx)
=>4x(x2+y2)+4y(x2+y2)(dy/dx)=y+x(dy/dx)
=>4y(x2+y2)(dy/dx)-x(dy/dx)=y-4x(x2+y2)
=>(dy/dx)(4y(x2+y2)-x)=y-4x(x2+y2)
=>dy/dx=(y-4x(x2+y2))/(4y(x2+y2)-x)
dy/dx=(y-4x(x2+y2))/(4y(x2+y2)-x)
### Question 9. tan-1(x2+y2)=a
Solution:
We have,
tan-1(x2+y2)=a
Differentiating both sides with respect to x ,
d(tan-1(x2+y2))/dx=da/dx
=>(1/(x2+y2))(2x+2y(dy/dx))=0
=>x+y(dy/dx)=0
=> dy/dx=-x/y
dy/dx=-x/y
### Question 10. ex-y=log(x/y)
Solution:
We have,
ex-y=log(x/y)
=>ex-y=log x -log y
Differentiating both sides with respect to x,
d(ex-y)/dx=d(log x- log y)/dx
=>ex-y(1-dy/dx)=1/x-(1/y)(dy/dx)
=>ex-y -ex-y(dy/dx)=1/x -(1/y)(dy/dx)
=>(1/y)(dy/dx) – ex-y(dy/dx)=1/x-ex-y
=> dy/dx((1/y)-ex-y)=(1-xex-y)/x
=> (dy/dx)(1-yex-y)/y=(1-xex-y)/x
=>dy/dx=y(1-xex-y)/x(1-yex-y)
dy/dx=y(1-xex-y)/x(1-yex-y)
### Question 11. sin(xy)+ cos(x+y)=1
Solution:
We have,
sin(xy)+ cos(x+y)=1
Differentiating both sides with respect to x,
d(sin(xy))/dx + d(cos(x+y))/dx=d1/dx
=>cos(xy)(y+xdy/dx) +(-sin(x+y)(1+dy/dx)= 0
=>cos(xy)(y+xdy/dx) = (sin(x+y)(1+dy/dx)
=>ycos(xy)+x*cos(xy)*(dy/dx)= sin(x+y) + sin(x+y)* (dy/dx)
=>x*cos(xy)*(dy/dx) – sin(x+y)* (dy/dx) = sin(x+y) – ycos(xy)
=>(dy/dx)((x*cos(xy))-sin(x+y))= sin(x+y) – ycos(xy)
=>dy/dx =(sin(x+y)-ycos(xy))/((x*cos(xy))-sin(x+y))
dy/dx=(sin(x+y)-ycos(xy))/((x*cos(xy))-sin(x+y))
### Question 12. (1-x2)1/2+(1-y2)1/2=a(x-y)
Solution:
We have,
(1-x2)1/2+(1-y2)1/2=a(x-y)
Let x=sin A and y= sin B
So the expression becomes,
cosA + cosB=a(sinA-sinB)
=>a=(cosA+cosB)/(sinA-sinB)
=>a=(2(cos((A+B)/2))*(cos((A-B)/2)))/(2cos((A+B)/2)*sin((A-B)/2)))
=> a =(cos(A-B)/2)/(sin(A-B)/2)
=> a=cot((A-B)/2)
=>cot-1a=((A-B)/2)
=>2cot-1a=((A-B)/2)
Differentiating both sides with respect to x,
d(2cot-1a)/dx=d(A-B)/dx
=>0=d(sin-1x)/dx -d(sin-1y)/dx
=> 0 = 1/((1-x2)1/2) -(1/(1-y2)1/2)*(dy/dx)
=>(1/(1-y2)1/2)*dy/dx=1/((1-x2)1/2)
=>dy/dx=((1-y2)1/2)/(1-x2)1/2
dy/dx=((1-y2)1/2)/(1-x2)1/2
### Question 13. y(1-x2)1/2+x(1-y2)1/2=1
Solution:
We have,
y(1-x2)1/2+x(1-y2)1/2=1
Let, x=sin A and y=sin B
So, the expression becomes,
(sin B)*(cos A)+(sin A)*(cos B) =1
=> sin(A+B) =1
=> sin-1(1) =A+B
=>A+B =22/(7*2)
=>sin-1x +sin-1y=22/14
Differentiating both sides with respect to x,
d(sin-1x)/dx +d(sin-1 y)/dx=d(22/14)/dx
=>1/((1-x2)1/2)+ (1/((1-y2)1/2))(dy/dx)=0
=>dy/dx=-((1-y2)1/2)/((1-x2)1/2)
dy/dx=-((1-y2)1/2)/((1-x2)1/2)
### Question 14. If xy=1, prove that dy/dx +y2=0
Solution:
We have,
xy=1
Differentiating both sides with respect to x,
d(xy)/dx =d1/dx
=>x(dy/dx)+y=0
=>dy/dx =-y/x
Also x=1/y
so, dy/dx=-y(y)
=>dy/dx+y2=0
Hence, proved.
### Question 15. If xy2=1, prove that 2(dy/dx)+y3=0
Solution:
We have,
xy2=1
Differentiating with respect to x,
d(xy2)/dx=d1/dx
=>2xy(dy/dx)+y2 =0
=>dy/dx=-y2/2xy
=>dy/dx =-y/2x
Also x=1/y2
So, dy/dx=-y(y2)/2
=>2dy/dx=-y3
2dy/d+y3=0
Hence, proved.
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Now, 20 = 2 * 10 = 4 * 5 = 1 * 20
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A number written as a product of prime factors is said to be in the prime factor form.
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4. Which is greater, 0.13 or 0.031? Show the comparison on the number line.
IM Commentary
This task involves using number lines to compare decimal numbers. The numbers selected in this task are purposefully chosen to target student misconceptions. For example, when asked to compare 0.13 and 0.031, students who are not yet proficient in decimal applications of our base ten system may rush to say that 0.031 is greater than 0.13 because 31 is greater than 13. This task also challenges students to think carefully about the value of each digit and how that affects the number's placement on the number line. Students are often presented with number lines that allow them to plot each number on a tick mark. In the last part of this task, students will grapple with placing a number between tick marks. It is important for students to be able to visualize the approximate location of numbers on the number line. A good follow-up task for this one would be 5.NBT Placing Thousandths on the Number Line. Notice that the language in the solutions underscores that the greater number will always be farther right on the number line. This reasoning should be heavily emphasized in discussion as it will transfer across comparisons of all rational numbers.
Solution
1. $0.1 \gt 0.01$ because 0.1 is to the right of 0.01 on the number line.
2. $0.2 \gt 0.03$ because 0.2 is to the right of 0.03 on the number line.
3. $0.21 \gt 0.12$ because 0.21 is to the right of 0.12 on the number line.
4. $0.13 \gt 0.031$ because 0.13 is to the right of 0.031 on the number line. |
# What is 41/7 as a decimal?
## Solution and how to convert 41 / 7 into a decimal
41 / 7 = 5.857
Fraction conversions explained:
• 41 divided by 7
• Numerator: 41
• Denominator: 7
• Decimal: 5.857
• Percentage: 5.857%
Convert 41/7 to 5.857 decimal form by understanding when to use each form of the number. Both are used to handle numbers less than one or between whole numbers, known as integers. The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. If we need to convert a fraction quickly, let's find out how and when we should.
41 / 7 as a percentage 41 / 7 as a fraction 41 / 7 as a decimal
5.857% - Convert percentages 41 / 7 41 / 7 = 5.857
## 41/7 is 41 divided by 7
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 and Denominators. This creates an equation. We use this as our equation: numerator(41) / denominator (7) 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: 41
• 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. 41 is an odd number so it might be harder to convert without a calculator. Values closer to one-hundred make converting to fractions more complex. So how does our denominator stack up?
### Denominator: 7
• Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. Any value less than ten like 7 makes your equation a bit simpler. But 7 is an odd number. Having an odd denominator like 7 could sometimes be more difficult. Overall, a small denominator like 7 could make our equation a bit simpler. So grab a pen and pencil. Let's convert 41/7 by hand.
## How to convert 41/7 to 5.857
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 7 \enclose{longdiv}{ 41 }$$
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: Solve for how many whole groups you can divide 7 into 41
$$\require{enclose} 00.5 \\ 7 \enclose{longdiv}{ 41.0 }$$
We can now pull 35 whole groups from the equation. Multiple this number by our furthest left number, 7, (remember, left-to-right long division) to get our first number to our conversion.
### Step 3: Subtract the remainder
$$\require{enclose} 00.5 \\ 7 \enclose{longdiv}{ 41.0 } \\ \underline{ 35 \phantom{00} } \\ 375 \phantom{0}$$
If your remainder is zero, that's it! If there is a remainder, extend 7 again and pull down the zero
### Step 4: Repeat step 3 until you have no remainder
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. This is also true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But they all represent how numbers show us value in the real world. Here are just a few ways we use 41/7, 5.857 or 585% in our daily world:
### When you should convert 41/7 into a decimal
Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 41/7 MPH. The radar will read: 90.585 MPH. This simplifies the value.
### When to convert 5.857 to 41/7 as a fraction
Cooking: When scrolling through pintress to find the perfect chocolate cookie recipe. The chef will not tell you to use .86 cups of chocolate chips. That brings confusion to the standard cooking measurement. It’s much clearer to say 42/50 cups of chocolate chips. And to take it even further, no one would use 42/50 cups. You’d see a more common fraction like ¾ or ?, usually in split by quarters or halves.
### Practice Decimal Conversion with your Classroom
• If 41/7 = 5.857 what would it be as a percentage?
• What is 1 + 41/7 in decimal form?
• What is 1 - 41/7 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 5.857 + 1/2?
### Convert more fractions to decimals
From 41 Numerator From 7 Denominator What is 41/8 as a decimal? What is 42/7 as a decimal? What is 41/9 as a decimal? What is 43/7 as a decimal? What is 41/10 as a decimal? What is 44/7 as a decimal? What is 41/11 as a decimal? What is 45/7 as a decimal? What is 41/12 as a decimal? What is 46/7 as a decimal? What is 41/13 as a decimal? What is 47/7 as a decimal? What is 41/14 as a decimal? What is 48/7 as a decimal? What is 41/15 as a decimal? What is 49/7 as a decimal? What is 41/16 as a decimal? What is 50/7 as a decimal? What is 41/17 as a decimal? What is 51/7 as a decimal? What is 41/18 as a decimal? What is 52/7 as a decimal? What is 41/19 as a decimal? What is 53/7 as a decimal? What is 41/20 as a decimal? What is 54/7 as a decimal? What is 41/21 as a decimal? What is 55/7 as a decimal? What is 41/22 as a decimal? What is 56/7 as a decimal? What is 41/23 as a decimal? What is 57/7 as a decimal? What is 41/24 as a decimal? What is 58/7 as a decimal? What is 41/25 as a decimal? What is 59/7 as a decimal? What is 41/26 as a decimal? What is 60/7 as a decimal? What is 41/27 as a decimal? What is 61/7 as a decimal?
### Convert similar fractions to percentages
From 41 Numerator From 7 Denominator 42/7 as a percentage 41/8 as a percentage 43/7 as a percentage 41/9 as a percentage 44/7 as a percentage 41/10 as a percentage 45/7 as a percentage 41/11 as a percentage 46/7 as a percentage 41/12 as a percentage 47/7 as a percentage 41/13 as a percentage 48/7 as a percentage 41/14 as a percentage 49/7 as a percentage 41/15 as a percentage 50/7 as a percentage 41/16 as a percentage 51/7 as a percentage 41/17 as a percentage |
### Chapter 8 review
```CHAPTER 8 REVIEW
Algebra 8
PROBLEM #1
Factor the monomial completely.
-33x y
2 3
−1 ∙ 3 ∙ 11 ∙ ∙ ∙ ∙ ∙
PROBLEM #2
Find the GCF.
22b, 33c
GCF: 11
PROBLEM #3
Find the GCF.
21xy, 28x y, 42xy
2
2
GCF: 7
PROBLEM #4
A landscape architect is designing a stone path 36 inches
wide and 120 inches long. What is the maximum size
square stone that can be used so that none of the
stones have to be cut?
GCF: 12
The maximum size square is 12in x 12in
PROBLEM #5
Factor.
12x + 24y
12( + 2)
PROBLEM #6
Factor.
14x y - 21xy + 35xy
2
2
7(2 − 3 + 5)
PROBLEM #7
Factor.
a - 4ac + ab - 4bc
2
( + )( − 4)
PROBLEM #8
Factor.
2x - 3xz - 2xy + 3yz
2
( − )(2 − 3)
PROBLEM #9
Solve.
x(3x - 6) = 0
Set each factor = 0 then solve.
: {0, 2}
PROBLEM #10
Solve.
x = 3x
2
Bring 3x to the other side, factor, then set
each factor = 0 to solve for x.
: {0, 3}
PROBLEM #11
Factor.
x - 8x +15
2
Leading coefficient is 1 (a=1) so break
down the constant.
( − 3)( − 5)
PROBLEM #12
Factor.
x - 5x - 6
2
Leading coefficient is 1 (a=1) so break
down the constant.
( + 1)( − 6)
PROBLEM #13
Solve.
x - 5x - 50 = 0
2
Leading coefficient is 1 (a=1) so break
down the constant.
+ 5 − 10 = 0
: {−5, 10}
PROBLEM #14
Solve.
x +12x + 32 = 0
2
Leading coefficient is 1 (a=1) so break
down the constant.
+4 +8 =0
: {−8, −4}
PROBLEM #15
Factor the trinomial, if possible. If the trinomial cannot
be factored, write prime.
12x + 22x -14
2
Factor out GCF.
Leading coefficient is not 1 (a>1) so make a
MAMA chart.
2 2 − 1 3 + 7
PROBLEM #16
Factor the trinomial, if possible. If the trinomial cannot
be factored, write prime.
2a +13a - 24
2
Leading coefficient is not 1 (a>1) so make a
MAMA chart.
+ 8 2 − 3
PROBLEM #17
Solve each equation. Check your solutions.
6x - 7x - 5 = 0
2
Leading coefficient is not 1 (a>1) so make a
MAMA chart.
3 − 5 2 + 1 = 0
1 5
: {− , }
2 3
PROBLEM #18
Solve each equation. Check your solutions.
40x + 2x = 24
2
Bring over 24 to the other side.
Factor out GCF
Leading coefficient is not 1 (a>1) so make a
MAMA chart.
2 5 + 4 4 − 3
PROBLEM #19
Factor each polynomial.
3x - 3
2
Factor out GCF.
3( 2 − 1)
Now you have difference of two perfect
squares so break down further.
3( + 1)( − 1)
PROBLEM #20
Factor each polynomial.
16a - 21b
2
2
Prime
PROBLEM #21
Solve by factoring. Check your solutions.
9x - 25 = 0
2
3 + 5 3 − 5 = 0
5 5
: {− , }
3 3
PROBLEM #22
Solve by factoring. Check your solutions.
x - 4 =12
2
+4 −4 =0
: {−4, 4}
PROBLEM #23
Factor each polynomial, if possible. If the polynomial
cannot be factored write prime.
x + 5x + 25
2
Prime
PROBLEM #24
Factor each polynomial, if possible. If the polynomial
cannot be factored write prime.
4 - 28a + 49a
2
Perfect square trinomial
2 − 7 2
PROBLEM #25
Factor each polynomial, if possible. If the polynomial
cannot be factored write prime.
x -16x
4
2
Factor out GCF.
2 ( 2 − 4)
Then you have difference of two perfect
squares, so factor more.
2 ( + 2)( − 2)
PROBLEM #26
Solve each equation. Check your solutions.
4y = 64
2
Bring 64 to the other side.
Factor out GCF.
4 2 − 16 = 0
Then you have difference of two perfect squares, so
factor more.
4 −4 +4 =0
Set each factor = 0 and solve for y.
: {−4, 4}
PROBLEM #27
Solve each equation. Check your solutions.
(x - 9) =144
2
FOIL left side and bring 144 to the other
side.
You have a trinomial where a=1, so break
down c.
+ 3 − 21 = 0
Then, set each factor =0 and solve for x.
: {−3, 21}
PROBLEM #28
A sidewalk of equal width is built around a square yard.
What is the width of the sidewalk? |
# Do you square an uncertainty?
## Do you square an uncertainty?
If you are raising an uncertain number to a power n, (squaring it, or taking the square root, for example), then the fractional uncertainty in the resulting number has a fractional uncertainty n times the fractional uncertainty in the original number. Thus if you are calculating a number y = ½ g t2 , where t = 2.36 ± .
## What happens to error when you square?
So squaring a number (raising it to the power of 2) doubles its relative SE, and taking the square root of a number (raising it to the power of ½) cuts the relative SE in half.
How do you find the uncertainty of a set of data?
To summarize the instructions above, simply square the value of each uncertainty source. Next, add them all together to calculate the sum (i.e. the sum of squares). Then, calculate the square-root of the summed value (i.e. the root sum of squares). The result will be your combined standard uncertainty.
How do you find the uncertainty of an area?
The percentage uncertainty in the area of the square tile is calculated by multiplying the percentage uncertainty in the length by 2. The total percentage uncertainty is calculated by adding together the percentage uncertainties for each measurement.
### How do you find the uncertainty of a single measurement?
In general, the uncertainty in a single measurement from a single instrument is half the least count of the instrument.
### How do you find the uncertainty of a square-root?
If you are taking a square-root, you are raising to the one-half power, the relative uncertainty is one half of the number you are taking the square root of.
What is the uncertainty in the area?
The percentage uncertainty in the area of the square tile is calculated by multiplying the percentage uncertainty in the length by 2. The total percentage uncertainty is calculated by adding together the percentage uncertainties for each measurement. the shape of a cube by determining the density of the material.
How do you calculate uncertainty in experimental data?
The most straightforward way to find the uncertainty in the final result of an experiment is worst case error analysis, a method in which uncertainties are estimated from the difference between the largest and smallest possible values that can be calculated from the data.
## Which refers to the uncertainty of available data?
Scientific uncertainty is a quantitative measurement of variability in the data. In other words, uncertainty in science refers to the idea that all data have a range of expected values as opposed to a precise point value. This uncertainty can be categorized in two ways: accuracy and precision.
## How do you find the relative uncertainty of an area?
The relative uncertainty or relative error formula is used to calculate the uncertainty of a measurement compared to the size of the measurement. It is calculated as: relative uncertainty = absolute error / measured value.
How to calculate the uncertainty of a value?
The relative uncertainty gives the uncertainty as a percentage of the original value. Work this out with: Relative uncertainty = (absolute uncertainty ÷ best estimate) × 100%. So in the example above: Relative uncertainty = (0.2 cm ÷ 3.4 cm) × 100% = 5.9%. The value can therefore be quoted as 3.4 cm ± 5.9%.
When to use the negative sign for uncertainty?
When the power is not an integer, you must use this technique of multiplying the percentage uncertainty in a quantity by the power to which it is raised. If the power is negative, discard the negative sign for uncertainty calculations only.
### How do I propagate uncertainties through a calculation?
Rules for combining uncertainties during the step-by-step method of propagating uncertainty The rules below tell you how to combine the uncertainties in each step of the calculation. Rule #1 – Addition and/or Subtraction of numbers with uncertainty Add the absolute uncertainties. Rule #2 – Multiplication and/or Division of numbers with uncertainty
### How many significant figures can you quote for uncertainty?
Significant Figures: Generally, absolute uncertainties are only quoted to one significant figure, apart from occasionally when the first figure is 1. Because of the meaning of an uncertainty, it doesn’t make sense to quote your estimate to more precision than your uncertainty. |
# Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Ex 3.3
Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 3 Linear Regression Ex 3.3 Questions and Answers.
## Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Ex 3.3
Question 1.
From the two regression equations find r, $$\bar{x}$$ and $$\bar{y}$$.
4y = 9x + 15 and 25x = 4y + 17
Solution:
Given 4y = 9x + 15 and 25x = 4y + 17
Since byx and bxy are positive.
∴ r = $$\frac{3}{5}$$ = 0.6
($$\bar{x}$$, $$\bar{y}$$) is the point of intersection of the regression lines
9x – 4y = -15 …….(i)
25x – 4y = 17 ……….(ii)
-16x = -32
x = 2
∴ $$\bar{x}$$ = 2
Substituting x = 2 in equation (i)
9(2) – 4y = -15
18 + 15 = 4y
33 = 4y
y = 33/4 = 8.25
∴ $$\bar{y}$$ = 8.25
Question 2.
In a partially destroyed laboratory record of an analysis of regression data, the following data are legible:
Variance of X = 9
Regression equations:
8x – 10y + 66 = 0 And 40x – 18y = 214.
Find on the basis of the above information
(i) The mean values of X and Y.
(ii) Correlation coefficient between X and Y.
(iii) Standard deviation of Y.
Solution:
Given, $$\sigma_{x}{ }^{2}=9, \sigma_{x}=3$$
(i) ($$\bar{x}$$, $$\bar{y}$$) is the point of intersection of the regression lines
40x – 50y = -330 …….(i)
40x – 50y = +214 ………(ii)
-32y = -544
y = 17
∴ $$\bar{y}$$ = 17
8x – 10(17) + 66 = 0
8x = 104
x = 13
∴ $$\bar{x}$$ = 13
Question 3.
For 50 students of a class, the regression equation of marks in statistics (X) on the marks in Accountancy (Y) is 3y – 5x + 180 = 0. The mean marks in accountancy is 44 and the variance of marks in statistics $$\left(\frac{9}{16}\right)^{t h}$$ of the variance of marks in accountancy. Find the mean in statistics and the correlation coefficient between marks in two subjects.
Solution:
Given, n = 50, $$\bar{y}$$ = 44
$$\sigma_{x}^{2}=\frac{9}{16} \sigma_{y}^{2}$$
∴ $$\frac{\sigma_{x}}{\sigma_{x}}=\frac{3}{4}$$
Since ($$\bar{x}$$, $$\bar{y}$$) is the point intersection of the regression line.
∴ ($$\bar{x}$$, $$\bar{y}$$) satisfies the regression equation.
3$$\bar{y}$$ – 5$$\bar{x}$$ + 180 = 0
3(44) – 5$$\bar{x}$$ + 180 = 0
∴ 5$$\bar{x}$$ = 132 + 180
$$\bar{x}$$ = $$\frac{312}{5}$$ = 62.4
∴ Mean marks in statistics is 62.4
Regression equation of X on Y is 3y – 5x + 180 = 0
∴ 5x = 3y + 180
Question 4.
For bivariate data, the regression coefficient of Y on X is 0.4 and the regression coefficient of X on Y is 0.9. Find the value of the variance of Y if the variance of X is 9.
Solution:
Question 5.
The equation of two regression lines are 2x + 3y – 6 = 0 and 3x + 2y – 12 = 0
Find (i) Correlation coefficient (ii) $$\frac{\sigma_{x}}{\sigma_{y}}$$
Solution:
Question 6.
For a bivariate data $$\bar{x}$$ = 53, $$\bar{y}$$ = 28, byx =-1.5 and bxy = -0.2. Estimate Y when X = 50.
Solution:
Regression equation of Y on X is
(Y – $$\bar{y}$$) = byx (X – $$\bar{x}$$)
(Y – 28) = -1.5(50 – 53)
Y – 28 = -1.5(-3)
Y – 28 = 4.5
Y = 32.5
Question 7.
The equation of two regression lines are x – 4y = 5 and 16y – x = 64. Find means of X and Y. Also, find the correlation coefficient between X and Y.
Solution:
Since ($$\bar{x}$$, $$\bar{y}$$) is the point of intersection of the regression lines.
x – 4y = 5 …..(i)
-x + 16y = 64 …….(ii)
12y = 69
y = 5.75
Substituting y = 5.75 in equation (i)
x – 4(5.75) = 5
x – 23 = 5
x = 28
∴ $$\bar{x}$$ = 28, $$\bar{y}$$ = 5.75
x – 4y = 5
x = 4y + 5
∴ bxy = 4
16y – x = 64
16y = x + 64
y = $$\frac{1}{16}$$x + 4
byx = $$\frac{1}{16}$$
byx . bxy = $$\frac{1}{16}$$ × 4 = $$\frac{1}{4}$$ ∈ [0, 1]
∴ Our assumption is correct
∴ r2 = byx . bxy
r2 = $$\frac{1}{4}$$
r = ±$$\frac{1}{2}$$
Since byx and bxy are positive,
∴ r = $$\frac{1}{2}$$ = 0.5
Question 8.
In partially destroyed record, the following data are available variance of X = 25. Regression equation of Y on X is 5y – x = 22 and Regression equation of X on Y is 64x – 45y = 22 Find
(i) Mean values of X and Y.
(ii) Standard deviation of Y.
(iii) Coefficient of correlation between X and Y.
Solution:
Given $$\sigma_{x}^{2}$$ = 25, ∴ σx = 5
(i) Since ($$\bar{x}$$, $$\bar{y}$$) is the point of intersection of regression lines
-x + 5y = 22 …….(i)
64x – 45y = 22 ………..(ii)
equation (i) becomes
-9x + 45y = 198
64y – 45y = 22
55x = 220
x = 4
Substituting x = 4 in equation (i)
-4 + 5y = 22
5y = 26
∴ y = 5.2
∴ $$\bar{x}$$ = 4, $$\bar{y}$$ = 5.2
Regression equation of X on Y is
64x – 45y – 22
64x = 45y + 22
x = $$\frac{45}{64} y+\frac{22}{64}$$
bxy = $$\frac{45}{64}$$
(ii) Regression equation of Y on X is
5y – x = 22
5y = x + 22
Question 9.
If the two regression lines for a bivariate data are 2x = y + 15 (x on y) and 4y – 3x + 25 (y on x) find
(i) $$\bar{x}$$
(ii) $$\bar{y}$$
(iii) byx
(iv) bxy
(v) r [Given √0.375 = 0.61]
Solution:
Since ($$\bar{x}$$, $$\bar{y}$$) is the point of intersection of the regression line
2x = y + 15
4y = 3x + 25
2x – y = 15 …….(i)
3x – 4y = -25 ……..(ii)
Multiplying equation (i) by 4
8x – 4y = 60
3x – 4y = -25
on Subtracting,
5x = 85
∴ x = 17
Substituting x in equation (i)
2(17) – y = 15
34 – 15 = y
∴ y = 15
Since byx and bxy are positive, ∴ r = 0.61
Question 10.
The two regression equation are 5x – 6y + 90 = 0 and 15x – 8y – 130 = 0. Find $$\bar{x}$$, $$\bar{y}$$, r.
Solution:
Since ($$\bar{x}$$, $$\bar{y}$$) is the point of intersection of the regression lines
5x – 6y + 90 = 0 ……(i)
15x – 8y – 130 = 0
15x – 18y + 270 = 0
15x – 8y – 130 = 0
on subtracting,
-10y + 400 = 0
y = 40
Substituting y = 40 in equation (i)
5x – 6(40) + 90 = 0
5x = 150
x = 30
∴ $$\bar{x}$$ = 30, $$\bar{y}$$ = 40
Since byx and bxy are positive
∴ r = $$\frac{2}{3}$$
Question 11.
Two lines of regression are 10x + 3y – 62 = 0 and 6x + 5y – 50 = 0 Identify the regression equation equation of x on y. Hence find $$\bar{x}$$, $$\bar{y}$$, and r.
Solution:
∴ Our assumption is correct.
∴ Regression equation of X on Y is 10x + 3y – 62 = 0
r2 = byx . bxy
r2 = $$\frac{9}{25}$$
r = ±$$\frac{3}{5}$$
Since, byx and bxy are negative, r = –$$\frac{3}{5}$$ = -0.6
Also ($$\bar{x}$$, $$\bar{y}$$) is the point of intersection of the regression lines
50x + 15y = 310
18x + 15y = 150
on subtracting
32x = 160
x = 5
Substituting x = 5 in 10x + 3y = 62
10(5) + 3y = 62
3y = 12
∴ y = 4
∴ $$\bar{x}$$ = 5, $$\bar{y}$$ = 4
Question 12.
For certain X and Y series, which are correlated the two lines of regression are 10y = 3x + 170 and 5x + 70 = 6y. Find the correlation coefficient between them. Find the mean values of X and Y.
Solution:
Since byx and bxy are positive,
r = $$\frac{3}{5}$$ = 0.6
Since, ($$\bar{x}$$, $$\bar{y}$$) is the point of intersection of the regression lines
3x – 10y = -170 …….(i)
5x – 6y = -70 ………(ii)
9x – 30y = -510
25x – 30y = -350
on subtracting
-16x = -160
x = 10
Substituting x = 10 in equation (i)
3(10) – 10y = -170
30 + 170 = 10y
200 = 10y
y = 20
∴ $$\bar{x}$$ = 10, $$\bar{y}$$ = 20
Question 13.
Regression equation of two series are 2x – y – 15 = 0 and 4y + 25 = 0 and 3x- 4y + 25 = 0. Find $$\bar{x}$$, $$\bar{y}$$ and regression coefficients, Also find coefficients of correlation. [Given √0.375 = 0.61]
Solution:
Since ($$\bar{x}$$, $$\bar{y}$$) is the point of intersection of the regression line
2x = y + 15
4y = 3x + 25
2x – y = 15 ……(i)
3x – 4y = -15 ……..(ii)
Multiply equation (i) by 4
8x – 4y = 60
3x – 4y = -25
on subtracting,
5x = 85
x = 17
Substituting x in equation (i)
2(17) – y = 15
34 – 15 = y
y = 15
∴ $$\bar{x}$$ = 17, $$\bar{y}$$ = 19
∴ Our assumption is correct
r2 = bxy . byx
r2 = $$\frac{3}{8}$$ = 0.375
r = ±√o.375 = ±0.61
Since, byx and bxy are positive, ∴ r = 0.61
Question 14.
The two regression lines between height (X) in includes and weight (Y) in kgs of girls are 4y – 15x + 500 = 0 and 20x – 3y – 900 = 0. Find the mean height and weight of the group. Also, estimate the weight of a girl whose height is 70 inches.
Solution:
Since ($$\bar{x}$$, $$\bar{y}$$) is the point intersection of the regression lines
15x – 4y = 500 ……(i)
20x – 3y = 900 …….(ii)
60x – 16y – 2000
60x – 9y = 2700
on subtracting,
-7y = -700
y = 100
Substituting y = 100 in equation (i)
15x – 4(100) = 500
15x = 900
x = 60
∴ Our assumption is correct
∴ Regression equation of Y on X is
Y = $$\frac{15}{4}$$x – 125
When x = 70
Y = $$\frac{15}{4}$$ × 70 = -125
= 262.5 – 125
= 137.5 kg |
# 4th Grade Mental Math on Factors and Multiples
In 4th grade mental math on factors and multiples students can practice different questions on prime numbers, properties of prime numbers, factors, properties of factors, even numbers, odd numbers, prime numbers, composite numbers, tests for divisibility, prime factorization, common factor, hcf by long division method, common multiples, lowest common multiple and lcm by long division method.
1. Fill in the blanks:
(i) The smallest multiple of a number is __________.
(ii) A prime number has exactly __________ factors.
(iii) Which even number is prime? __________
(iv) Which number has exactly 1 factor? __________
(v) The multiple of an even number is always __________.
(vi) The multiple of an __________ number may be odd or even.
(vii) The product of two numbers is also their __________.
(viii) All multiples of 10 end with __________.
(ix) A number is a multiple of 5 if it has either __________ in ones place.
(x) Every number is a multiple of __________.
(xi) __________ is a factor of every number.
(xii) The greatest factor of a number is the number __________.
(i) 1
(ii) 2
(iii) 2
(iv) 1
(v) even
(vi) odd
(vii) LCM
(viii) 0
(ix) 0 or 5
(x) Itself
(xi) 1
(xii) Itself
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# Mental Subtraction Strategies
This mental math lesson explains 6 different strategies you can use for mentally subtracting 2-digit numbers, meant for 2nd or 3rd grade. It contains the instruction, examples, lots of exercises, and word problems for students to complete.
Strategy 1: Subtract in two parts
53 − 8 = 53 − 3 − 5 = 50 − 5 = 45
72 − 6 = 72 − 2 − 4 = 70 − 4 = 66
Subtract 8 in two parts: first 3, then 5.
Subtract 6 in two parts: first 2, then 4.
In other words, first subtract to the previous whole ten, then the rest.
1. Subtract the elevated number in parts first subtract to the previous whole ten; then the rest.
− 5 / \ a. 51 − 1 − 4 = ______
− 7 / \ b. 62 − ____ − ____ = ______
− 4 / \ c. 33 − ____ − ____ = ______
− 5 / \ d. 92 − ____ − ____ = ______
− 6 / \ e. 75 − ____ − ____ = ______
− 7 / \ f. 63 − ____ − ____ = ______
− 7 / \ g. 35 − ____ − ____ = ______
− 5 / \ h. 74 − ____ − ____ = ______
2. First subtract the balls that are not in the ten-groups.
a. 51 − 7 = _______ 51 − 5 = _______ 51 − 3 = _______ 51 − 6 = _______
b. 42 − 4 = _______ 42 − 5 = _______42 − 3 = _______ 42 − 6 = _______
Strategy 2: Use known subtraction facts Since 14 − 6 = 8, we know that the answer to 74 − 6 will end in 8, but it will be in the sixties sixty-something. So it is 68. Since 15 − 8 = 7, we know that the answer to 55 − 8 will end in 7, but it will be in the forties forty-something. So it is 47.
3. Subtract. The first problem in each box is a “helping problem” for the others.
a. 14 − 9 = _______ 24 − 9 = _______ 44 − 9 = _______ b. 17 − 8 = _______ 27 − 8 = _______ 37 − 8 = _______ c. 12 − 9 = _______ 52 − 9 = _______ 32 − 9 = _______ d. 15 − 9 = _______ 65 − 9 = _______ 45 − 9 = _______ e. 13 − 8 = _______ 33 − 8 = _______ 93 − 8 = _______ f. 16 − 8 = _______ 86 − 8 = _______ 36 − 8 = _______
4. a. Amy has \$32. She bought a comic book for \$7.
How much does she have now?
b. Peter had \$29. A toy train he wants costs \$39.
Mom paid him \$5 for working. How much more
does Peter now need to buy the train?
c. A flower shop has 55 roses. Eight of them are white,
and the rest are red. How many are red?
Strategy 3: Subtract in parts: tens and ones
Break the number being subtracted into its tens and ones. Subtract in parts.
75 − 21 = 75 − 20 − 1 = 55 − 1 = 54
First subtract 20, then 1.
87 − 46 = 87 − 40 − 6 = 47 − 6 = 41
First subtract 40, then 6.
5. Subtract in parts: Break the second number into its tens and ones.
a. 89 − 26 89 − 20 − 6
= __________
b. 56 − 35 56 − −
= __________
c. 75 − 51 75 − −
= __________
d. 69 − 19
e. 67 − 36
f. 64 − 33
Strategy 4: Add. You can “add backwards”. This works well if the two numbers are close to each other. Instead of subtracting, think how much you need to add to the number being subtracted (the subtrahend) in order to get the number you are subtracting from (the minuend). 71 − 67 = ??Think: 67 + ____ = 71 558 − 556 = ??Think: 556 + ____ = 558
6. Subtract.
a. 78 − 75 = _______ 61 − 58 = _______ b. 112 − 108 = _______ 692 − 688 = _______ c. 505 − 499 = _______ 1000 − 994 = _______
7. You had \$50. You purchased two bouquets of roses for \$13 each.
How much do you have left after the purchase?
8. What if you bought three bouquets of roses for \$13 each with your \$50?
How much would you have left after the purchase?
9. Fifteen children were playing on the playground. Seven of them left.
Then, ten more children came. How many are playing on the playground now?
10. A lion chased an antelope for 400 feet, then another 200 feet, and
lastly 200 feet more. Then the lion pounced on the antelope.
What was the total number of feet that the lion chased the antelope?
Strategy 5: Add up to find the difference.
To find the difference, start at the smaller number, and add up until you get to the bigger number. You can first complete the next ten, then add whole tens, then add ones again.
84 − 37 = ?
37 + 3 = 40 40 + 40 = 80 80 + 4 = 84
I added 3, 40, and 4 or a total of 47.
84 − 37 = 47.
92 − 35 = ?
35 + 5 = 40 40 + 50 = 90 90 + 2 = 92
I added a total of 57.
92 − 35 = 57.
11. Add up to find the differences, or use some other strategy such as subtracting in parts.
a.
65 − 26
= _____
+
+
+
26
30
60
65
b.
83 − 35
= _____
+
+
+
35
40
80
83
c. 56 − 28 = _______ 55 − 24 = _______ d. 72 − 18 = _______ 82 − 46 = _______ e. 54 − 37 = _______ 91 − 57 = _______ f. 74 − 55 = _______ 63 − 34 = _______
Strategy 6: Subtract an easy number that is close, and then correct the answer. 74 − 39 = ? First subtract 74 − 40 = 34, since 40 is close to 39 and an easy number to subtract. But, you subtracted one too many. Therefore, add 1 to the answer: 34 + 1 = 35. 81 − 57 = ? First subtract 81 − 60 = 21, since 60 isclose to 57 and an easy number to subtract. But, you subtracted three too many. Therefore, add 3 to the answer: 21 + 3 = 24.
12. Subtract mentally.
a. 34 − 18 = ______ 42 − 29 = ______ b. 65 − 27 = ______ 55 − 38 = ______ c.97 − 49 = ______ 62 − 19 = ______ d.65 − 29 = ______ 83 − 38 = ______
13. A subtraction challenge!
a. 101 − _____ = 92 b. ________ − 60 = 7 c. ________ − 132 = 40
14. Another challenge!
− 30
− 4
− 5
− 20
− 9
− 10
100
______ _____ _____ _____ _____ _____
Solve the mystery numbers and !
Hint: guess and check. Then improve your guess.
a. + = 30 − = 14 = _______ = _______ b. 99 − − − = 36 = _______ c. − 7 = − − = 4 = _______ = _______
This lesson is taken from Maria Miller's book Math Mammoth Add & Subtract 3, and posted at www.HomeschoolMath.net with permission from the author. Copyright © Maria Miller.
#### Math Mammoth Add & Subtract 3
A self-teaching worktext for 2nd-3rd grade that covers mental addition & subtraction with three-digit numbers, regrouping with three-digit numbers (carrying and borrowing), rounding, estimating, and word problems. |
# Precalculus: Functions
### Types of Functions
In this section, we'll briefly cover a few of the most relevant and important classifications of functions.
#### Even and Odd Functions
Every function can either be classified as an even function, an odd function, or neither. Even functions have the characteristic that f (x) = f (- x) . They are symmetrical with respect to the y-axis. A line segment joining the points f (x) and f (- x) will be perfectly horizontal. Odd functions have the characteristic that f (x) = - f (- x) . They are symmetrical with respect to the origin. A line segment joining the points f (x) and - f (- x) always contains the origin. Many functions are neither even nor odd.
Some of the most common even functions are y = k , where k is a constant, y = x 2 , and y = cos(x) . Some of the most common odd functions are y = x 3 and y = sin(x) . Some functions that are neither even nor odd include y = x - 2 , y = , and y = sin(x) + 1 .
Figure %: The function on the left is even; the function on the right is odd. Note the different types of symmetry.
#### Other Types of Functions
Among the types of functions that we'll study extensively are polynomial, logarithmic, exponential, and trigonometric functions. Before we study those, we'll take a look at some more general types of functions.
The inverse of a function is the relation in which the roles of the independent anddependent variable are reversed. Let f (x) = 2x . The inverse of f , f -1 (not to be confused with a negative exponent), equals . It is written like this: f -1(x) = . The inverse of a function can be found by switching the places of x and y in the formula of the function. The inverse of any function is a relation. Whether the inverse is a function depends on the original function f . If f is a one-to-one function, then its inverse is also a function. A one-to-one function is a function for which each element of the range corresponds to exactly one element of the domain. Therefore if a function is not a one-to- one function, its inverse is not a function. The horizontal line test shows us that if a horizontal line can be placed in a graph such that it intersects the graph of a function more than once, that function is not one-to-one, and its inverse is therefore not a function.
Inverse functions are important in solving equations. Sometimes the solution y to a function is known, but the input for that solution x is not known. In situations like these, the inverse of the function can be used to find x . We'll see more inverse functions later.
A piecewise function is a function which is defined by different rules depending on the value of the independent variable. The following piecewise function is graphed below: y = for x≤ 0 , y = x for 0 < x < 2 , and y = 2 for x≥2 .
Figure %: A piecewise function is governed by different rules depending on the value of the independent variable.
A periodic function is a function with the following characteristic: f (x) = f (x + c) , for all values of x , where c is some constant. This means that the values of f repeat themselves in a regular cycle. The most common periodic functions are the trigonometric functions, but others exist as well.
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# Linear equation
From Wikipedia, the free encyclopedia - View original article
Jump to: navigation, search
Graph sample of linear equations.
A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.
Linear equations can have one or more variables. Linear equations occur abundantly in most subareas of mathematics and especially in applied mathematics. While they arise quite naturally when modeling many phenomena, they are particularly useful since many non-linear equations may be reduced to linear equations by assuming that quantities of interest vary to only a small extent from some "background" state. Linear equations do not include exponents.
This article considers the case of a single equation for which one searches the real solutions. All its content applies for complex solutions and, more generally for linear equations with coefficients and solutions in any field.
## One variable
A linear equation in one unknown x may always be rewritten
$ax=b.$
If a ≠ 0, there is a unique solution
$x=\frac{b}{a}.$
If a = 0, then either the equation does not have any solution, if b ≠ 0 (it is inconsistent), or every number is a solution, if b is also zero.
## Two variables
A common form of a linear equation in the two variables x and y is
$y = mx + b,\,$
where m and b designate constants (parameters). The origin of the name "linear" comes from the fact that the set of solutions of such an equation forms a straight line in the plane. In this particular equation, the constant m determines the slope or gradient of that line, and the constant term b determines the point at which the line crosses the y-axis, otherwise known as the y-intercept.
Since terms of linear equations cannot contain products of distinct or equal variables, nor any power (other than 1) or other function of a variable, equations involving terms such as xy, x2, y1/3, and sin(x) are nonlinear.
### Forms for two-dimensional linear equations
Linear equations can be rewritten using the laws of elementary algebra into several different forms. These equations are often referred to as the "equations of the straight line." In what follows, x, y, t, and θ are variables; other letters represent constants (fixed numbers).
#### General (or standard) form
In the general (or standard[1]) form the linear equation is written as:
$Ax + By = C, \,$
where A and B are not both equal to zero. The equation is usually written so that A ≥ 0, by convention. The graph of the equation is a straight line, and every straight line can be represented by an equation in the above form. If A is nonzero, then the x-intercept, that is, the x-coordinate of the point where the graph crosses the x-axis (where, y is zero), is C/A. If B is nonzero, then the y-intercept, that is the y-coordinate of the point where the graph crosses the y-axis (where x is zero), is C/B, and the slope of the line is −A/B. The general form is sometimes written as:
$ax + by + c = 0, \,$
where a and b are not both equal to zero. The two versions can be converted from one to the other by moving the constant term to the other side of the equal sign.
#### Slope–intercept form
$y = mx + b,\,$
where m is the slope of the line and b is the y intercept, which is the y coordinate of the location where line crosses the y axis. This can be seen by letting x = 0, which immediately gives y = b. It may be helpful to think about this in terms of y = b + mx; where the line passes through the point (0, b) and extends to the left and right at a slope of m. Vertical lines, having undefined slope, cannot be represented by this form.
#### Point–slope form
$y - y_1 = m( x - x_1 ),\,$
where m is the slope of the line and (x1,y1) is any point on the line.
The point-slope form expresses the fact that the difference in the y coordinate between two points on a line (that is, y − y1) is proportional to the difference in the x coordinate (that is, x − x1). The proportionality constant is m (the slope of the line).
#### Two-point form
$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1),\,$
where (x1y1) and (x2y2) are two points on the line with x2x1. This is equivalent to the point-slope form above, where the slope is explicitly given as (y2 − y1)/(x2 − x1).
Multiplying both sides of this equation by (x2 − x1) yields a form of the line generally referred to as the symmetric form:
$(x_2 - x_1)(y - y_1)=(y_2 - y_1)(x - x_1).\,$
Expanding the products and regrouping the terms leads to the general form:
$x\,(y_2-y_1) - y\,(x_2-x_1)= x_1y_2 - x_2y_1$
Using a determinant, one gets a determinant form, easy to remember:
$\begin{vmatrix} x&y&1\\ x_1&y_1&1\\ x_2&y_2&1 \end{vmatrix} =0\,.$
#### Intercept form
$\frac{x}{a} + \frac{y}{b} = 1,\,$
where a and b must be nonzero. The graph of the equation has x-intercept a and y-intercept b. The intercept form is in standard form with A/C = 1/a and B/C = 1/b. Lines that pass through the origin or which are horizontal or vertical violate the nonzero condition on a or b and cannot be represented in this form.
#### Matrix form
Using the order of the standard form
$Ax + By = C,\,$
one can rewrite the equation in matrix form:
$\begin{pmatrix} A&B \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}C\end{pmatrix}.$
Further, this representation extends to systems of linear equations.
$A_1x + B_1y = C_1,\,$
$A_2x + B_2y = C_2,\,$
becomes:
$\begin{pmatrix} A_1&B_1\\ A_2 & B_2 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} C_1\\ C_2 \end{pmatrix}.$
Since this extends easily to higher dimensions, it is a common representation in linear algebra, and in computer programming. There are named methods for solving system of linear equations, like Gauss-Jordan which can be expressed as matrix elementary row operations.
#### Parametric form
$x = T t + U\,$
and
$y = V t + W.\,$
Two simultaneous equations in terms of a variable parameter t, with slope m = V / T, x-intercept (VU - WT) / V and y-intercept (WT - VU) / T. This can also be related to the two-point form, where T = p - h, U = h, V = q - k, and W = k:
$x = (p - h) t + h\,$
and
$y = (q - k)t + k.\,$
In this case t varies from 0 at point (h,k) to 1 at point (p,q), with values of t between 0 and 1 providing interpolation and other values of t providing extrapolation.
#### 2D vector determinant form
The equation of a line can also be written as the determinant of two vectors. If $P_1$ and $P_2$ are unique points on the line, then $P$ will also be a point on the line if the following is true:
$\det( \overrightarrow{P_1 P} , \overrightarrow{P_1 P_2} ) = 0.$
One way to understand this formula is to use the fact that the determinant of two vectors on the plane will give the area of the parallelogram they form. Therefore, if the determinant equals zero then the parallelogram has no area, and that will happen when two vectors are on the same line.
To expand on this we can say that $P_1 = (x_1 ,\, y_1)$, $P_2 = (x_2 ,\, y_2)$ and $P = (x ,\, y)$. Thus $\overrightarrow{P_1 P} = (x-x_1 ,\, y-y_1)$ and $\overrightarrow{P_1 P_2} = (x_2-x_1 ,\, y_2-y_1)$, then the above equation becomes:
$\det \begin{pmatrix}x-x_1&y-y_1\\x_2-x_1&y_2-y_1\end{pmatrix} = 0.$
Thus,
$( x - x_1 )( y_2 - y_1 ) - ( y - y_1 )( x_2 - x_1 )=0.$
Ergo,
$( x - x_1 )( y_2 - y_1 ) = ( y - y_1 )( x_2 - x_1 ).$
Then dividing both side by $( x_2 - x_1 )$ would result in the “Two-point form” shown above, but leaving it here allows the equation to still be valid when $x_1 = x_2$.
#### Special cases
$y = b\,$
Horizontal Line y = b
This is a special case of the standard form where A = 0 and B = 1, or of the slope-intercept form where the slope m = 0. The graph is a horizontal line with y-intercept equal to b. There is no x-intercept, unless b = 0, in which case the graph of the line is the x-axis, and so every real number is an x-intercept.
$x = a\,$
Vertical Line x = a
This is a special case of the standard form where A = 1 and B = 0. The graph is a vertical line with x-intercept equal to a. The slope is undefined. There is no y-intercept, unless a = 0, in which case the graph of the line is the y-axis, and so every real number is a y-intercept. This is the only type of line which is not the graph of a function (it obviously fails the vertical line test).
### Connection with linear functions
A linear equation, written in the form y = f(x) whose graph crosses the origin (x,y) = (0,0), that is, whose y-intercept is 0, has the following properties:
$f ( x_1 + x_2 ) = f ( x_1) + f ( x_2 )\$
and
$f ( a x ) = a f ( x ),\,$
where a is any scalar. A function which satisfies these properties is called a linear function (or linear operator, or more generally a linear map). However, linear equations that have non-zero y-intercepts, when written in this manner, produce functions which will have neither property above and hence are not linear functions in this sense. They are known as affine functions.
## More than two variables
A linear equation can involve more than two variables. Every linear equation in n unknowns may be rewritten
$a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b,$
where, a1, a2, ..., an represent numbers, called the coefficients, x1, x2, ..., xn are the unknowns, and b is called the constant term. When dealing with three or fewer variables, it is common to use x, y and z instead of x1, x2 and x3.
If all the coefficients are zero, then either b ≠ 0 and the equation does not have any solution, or b = 0 and every set of values for the unknowns is a solution.
If at least one coefficient is nonzero, a permutation of the subscripts allows to suppose a1 ≠ 0, and rewrite the equation
$x_1 = b -\frac{a_2}{a_1}x_2- \cdots - \frac{a_n}{a_1}x_n.$
In other words, if ai ≠ 0, one may choose arbitrary values for all the unknowns except xi, and express xi in term of these values.
If n = 3 the set of the solutions is a plane in a three-dimensional space. More generally, the set of the solutions is an (n – 1)-dimensional hyperplane in a n-dimensional Euclidean space (or affine space if the coefficients are complex numbers or belong to any field).
## References
• Barnett, R.A.; Ziegler, M.R.; Byleen, K.E. (2008), College Mathematics for Business, Economics, Life Sciences and the Social Sciences (11th ed.), Upper Saddle River, N.J.: Pearson, ISBN 0-13-157225-3 |
# Derivative Rules
In calculus, a derivative can be thought of as an instantaneous rate of change; that is, how much a quantity is changing at a given point. Let’s take a closer look at how we can differentiate a function easily by the use of some helpful rules.
## Understanding the Derivative
Differentiation is a method to compute the rate at which a dependent variable y changes with respect to the change in the independent variable x. This rate of change is called the derivative of y with respect to x. There are many different notations to denote “take the derivative of.” The relationship between y and x is usually denoted by f(x) and its derivative is usually denoted as f‘(x) or y’ or dy/dx. The definition of the derivative is given by:
which is a lengthy procedure used to evaluate the derivative of a function. Thankfully, easier methods have been developed to help evaluate derivatives more quickly.
### The Power Rule
If n is any real number, then d (xn) = nxn-1
dx
Also, remember that the derivatives of a constant is zero: (c) = 0.
dx
Note that the sum and difference rule states:
(Just simply apply the power rule to each term in the function separately).
Example: Find the derivative of
Solution:
First, rewrite the function so that all variables of x have an exponent in the numerator:
Now, apply the power rule to the function:
### The Product Rule
We already know how to find the derivative of a sum or difference of functions, but what about the product of two functions? The product of two functions is when two functions are being multiplied together. If f and g are both differentiable, then the product rule states:
Example: Find the derivative of h(x) = (3x + 1)(8x4 +5x).
Solution:
Using the above formula, let f(x) = (3x+1) and let g(x) = (8x4 + 5x).
Now, find
Lastly, apply the product rule using the above formula:
### The Quotient Rule
Now, let's take a look at the quotient of functions (when two functions are being divided by each other). If f and g are both differentiable, then the quotient rule states:
Example: Find the derivative of
Solution:
Using the above formule, let f(x) = (3x-4) and let g(x) = (2x2-1).
Now, find
Lastly, apply the quotient rule using the above formula:
### The Chain Rule
If f and g are both differentiable and F is the composite function defined by F(x) = f(g(x)), then F is differentiable and the chain rule can be applied to F ', which is given by F '(x) = f '(g(x))g '(x)
Example: Find the derivative of F(x) = (3x4 + 2x2 -7)5.
Solution:
Using the above formula, let g(x) = (3x4 + 2x- 7) and let f(x) = (x)5.
Now, find f '(x) and '(x): f '(x) = 5x4 and '(x) = 12x3 + 4x - 0
So, if f '(x) = 5x4, then the composite function f '(g(x)) = 5(3x4 + 2x2 - 7)4
Lastly, apply the chain rule using the above formula: '(x) = f '(g(x))g '(x)
'(x) = 5(3x4 + 2x2 -7)4 (12x3 + 4x) |
# Subtracting 2-Digit Numbers
In subtracting 2-digit numbers we will subtract or minus a one-digit number from another two-digit number and a two-digit number from another two-digit number. To find the difference between the two numbers we need to ‘ones from ones’ and ‘tens from tens’.
Let us subtract 2 from 39
= 37 ones= 3 tens + 7 ones
Subtract ones and write in the ones place.9 ones – 2 ones = 7 onesSubtract tens and write in the tens place.3 tens – 0 tens = 3 tens
39 – 2 = 37
Let us subtract 34 from 65.
= 31 ones = 3 tens + 1 one
Subtract ones and write in the ones place.5 ones – 4 ones = 1 onesSubtract tens and write in the tens place.6 tens – 3 tens = 3 tens
65 – 34 = 31
How to subtract two digit numbers?
The steps of the examples on subtracting 2-digit numbers i.e. ‘ones from ones’ and ‘tens from tens’:
1. Subtract 41 from 63.
(i) First arrange the numbers vertically so that the tens' place digits and ones' place digits are lined up which means in simple one number should be written above the other number. Draw a line under the bottom number.
(ii) Subtract the digits in the ones place. Subtract (3 - 1 = 2). Place 2 in the ones column as shown.
(iii) Subtract the digits in the tens' place. Subtract (6 - 4 = 2). Place 2 in the tens column as shown.
(iv) The difference of 63 – 41 is 22
2. Subtract 52 from 86.
(i) First arrange the numbers vertically so that the tens' place digits and ones' place digits are lined up which means in simple one number should be written above the other number. Draw a line under the bottom number.
(ii) Subtract the digits in the ones place. Subtract (6 - 2 = 4). Place 4 in the ones column as shown.
(iii) Subtract the digits in the tens' place. Subtract (8 - 5 = 3). Place 3 in the tens column as shown.
(iv) The difference of 86 - 52 is 34
3. Subtract 42 from 78
Solution:
(i) Given numbers are written in column form. The greater number is written first, then the smaller number is written.(ii) First, the digits of one’s place are taken. 2 is subtracted from 8. 8 ones – 2 ones = 6 ones. This 6 is written in one’s column. (iii) Now digits of tens place are written 4 is subtracted from 7. 7 tens – 4 tens = 3 tens. This 3 is written in tens column.
Hence, 78 - 42 = 36
4. Let us subtract 24 from 39.
Let us learn how to subtract 2-digit numbers using an abacus.
1. Subtract 26 from 49.
Step I: Look at the ones rod.
9 – 6 = 3
Step II: Look at the tens rod.Take away 2 beads.4 – 2 = 2We get the number 23.Thus, 49 – 26 = 23
2. Subtract 31 from 73.
Step I: Look at the ones rod.
3 – 1 = 2
Step II: Look at the tens rod.Take away 3 beads.7 – 3 = 4We get the number 42.Thus, 73 – 31 = 42
Note: To do subtraction using an abacus, first subtract the beads o the ones rod and then subtract the beads on the tens rod.
Subtraction without Borrowing:
1. Let us subtract 7 from 29.
Step I: Arrange the two numbers under tens and ones.
Step II: First subtract the ones.
Step III: Subtract the tens.
Thus, 29 – 7 = 22
2. Subtract 23 from 38.
Step I: Arrange the two numbers under tens and ones.
Step II: First subtract the ones.
Step III: Subtract the tens.
Thus, 38 – 23 = 15
In subtraction there are three numbers as:
95 the number from which smaller number is subtracted is called minuend
- 23 the number to be subtracted is called the subtrahend.
72 the number which is the subtracted result is called the difference.
Here, in subtracting 2-digit numbers we get 95 is minuend, 23 is subtrahend and 72 is difference.
Minuend is always greater than subtrahend.
## More Examples on Subtraction of Two Digit Numbers
1. Subtract 23 from 36.
T O
3 6 = 3 tens + 6 ones = 30 + 6
2 3 = 2 tens + 3 ones = 20 + 3
- - - - -
1 3 = 1 ten + 3 ones = 10 + 3 = 13
Method:
Step I: Subtract the ones,
6 ones - 3 ones = 3 ones.
Write 3 under the ones column.
Step II: Subtract the tens,
3 tens - 2 tens = 1 ten.
Write 1 under the tens column.
Therefore, 36 - 23 = 13
We can do the same problem in a short way
Short way:
T O
3 6
- 2 3
1 3 (Steps remain the same as above)
2. Subtract 23 from 36.
T O
7 8
- 3 2
4 6
Method:
Step I: Subtract the ones,
8 ones - 2 ones = 6 ones.
Write 6 under the ones column.
Step II: Subtract the tens,
7 tens - 3 tens = 4 tens.
Write 4 under the tens column.
Therefore, 78 - 32 = 46
Worksheet on Subtracting 2-Digit Numbers:
1. Subtract the following:
(i) 60 - 10 = _____
(ii) 58 - 13 = _____
(iii) 88 - 33 = _____
(iv) 62 - 41 = _____
(v) 43 - 21 = _____
(vi) 57 - 26 = _____
(vii) 72 - 61 = _____
(viii) 47 - 11 = _____
(ix) 97 - 45 = _____
(x) 38 - 26 = _____
(xi) 71 - 60 = _____
(xii) 25 - 14 = _____
(xiii) 63 - 32 = _____
(xiv) 59 - 20 = _____
(xv) 37 - 12 = _____
(xvi) 89 - 69 = _____
1. (i) 50
(ii) 45
(iii) 55
(iv) 21
(v) 22
(vi) 31
(vii) 11
(viii) 36
(ix) 52
(x) 12
(xi) 11
(xii) 11
(xiii) 31
(xiv) 39
(xv) 25
(xvi) 20
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# Essential Question: What are the two types of probability?
## Presentation on theme: "Essential Question: What are the two types of probability?"— Presentation transcript:
Essential Question: What are the two types of probability?
Exact probability of a real event can never be known Probabilities are estimated in two ways: experimentally and theoretically Experimental probability is done by running experiments and calculating the results Theoretical probability is done by making assumptions on the results
Example 1: Experimental Estimate of Probability Throw a dart at a dartboard Red: 43, Yellow: 86, Blue: 71 Write a probability distribution for the experiment Outcomeredyellowblue Probability
Probability Simulation In order for experimental probability to be useful, a large number of simulations need to be run Computer simulations, done via random number generators, prove useful Example 2: Probability Solution Flip a coin 3 times, and count the number of heads (In-class simulation)
Theoretical Estimates of Probability Example 3: Rolling a number cube An experiment consists of rolling a number cube. Assume all outcomes are equally likely. a)Write the probability distribution for the experiment. b)Find the probability of the event that an even number is rolled. Outcome123456 Probability1616 1616 1616 1616 1616 1616 P(2, 4, or 6) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2
Example 4: Theoretical Probability Find the theoretical probability for the experiment you ran in Example 2 (flipping 3 coins) 0 heads: only 1 possible outcome (TTT) 0.5 x 0.5 x 0.5 = 0.125 1 head: 3 possible outcomes (HTT, THT, TTH) 3 x 0.125 = 0.375 2 heads: 3 possible outcomes (HHT, HTH, THH) 3 x 0.125 = 0.375 3 heads: only 1 possible outcome (HHH) 0.5 x 0.5 x 0.5 = 0.125
Homework: Page 882, 1-17 (ALL) Well run the experiment for numbers 8-11 as a class
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Example: Choosing 3 letters of the alphabet Two important factors in determining probability: 1. Are selections chosen with replacement? Is it possible to come up with the outcome AAA or not? 2. Is order important? Is there a difference between EAT and ATE?
With replacementWithout replacement Order mattersxnxn Permutation (nPr) Order unimportant(wont be discussed)Combination (nCr) Permutation: Combination:
Choosing 3 letters of the alphabet 1. With replacement, order important 26 3 = 17,576 2. Without replacement, order important 3. Without replacement, order matters
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Example 8: Pick-6 Lottery 54 numbered balls are used; 6 are randomly chosen. To win anything, at least 3 numbers must match. Whats the probability of matching all 6? 5 of 6? 4 of 6? 3 of 6? Answer: Order doesnt matter, and numbers arent replaced 54 C 6 = 25,827,165
Example 8: Pick-6 Lottery (Answer) Answer: Order doesnt matter, and numbers arent replaced Total combinations: 54 C 6 = 25,827,165 P(jackpot) = 1/25,827,165 P(5 correct) = ( 6 C 5 48 C 1 )/25,827,165=288/25,827,165 P(4 correct) = ( 6 C 4 48 C 2 )/25,827,165=16,920/25,827,165 P(3 correct) = ( 6 C 3 48 C 3 )/25,827,165=345,920/25,827,165 P(win anything) = 363,129/25,827,165 0.014
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# Trigonometrical Ratios of 45°
How to find the trigonometrical Ratios of 45°?
Let a rotating line $$\overrightarrow{OX}$$ rotates about O in the anti-clockwise sense and starting from the initial position $$\overrightarrow{OX}$$ traces out ∠AOB = 45°.
Take a point P on $$\overrightarrow{OY}$$ and draw $$\overline{PQ}$$ perpendicular to $$\overrightarrow{OX}$$.
Now, ∠OPQ = 180° - ∠POQ - ∠PQO
= 180° - 45° - 90°
= 45°.
Therefore, in the △OPQ we have, ∠QOP = ∠OPQ.
Therefore, PQ = OQ = a (say).
Now,
OP2 = OQ2 + PQ2
OP2 = a2 + a2
OP2 = 2a2
Therefore, $$\overline{OP}$$ = √2 a (Since, $$\overline{OP}$$ is positive)
Therefore, from the right-angled △OPQ we get,
sin 45° = $$\frac{\overline{PQ}}{\overline{OP}} = \frac{a}{\sqrt{2} a} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
cos 45° = $$\frac{\overline{OQ}}{\overline{OP}} = \frac{a}{\sqrt{2} a} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
And tan 45° = $$\frac{\overline{PQ}}{\overline{OQ}} = \frac{a}{a} = 1$$.
Clearly, csc 45° = $$\frac{1}{sin 45°}$$ = √2,
sec 45° = $$\frac{1}{cos 45°}$$ = √2
And cot 45° = $$\frac{1}{tan 45°}$$ = 1
Trigonometrical Ratios of 45° are commonly called standard angles and the trigonometrical ratios of these angles are frequently used to solve particular angles.
Trigonometric Functions
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# Limit And Continuity In Calculus
The limit of a function is an interesting but a little bit complex concept where, it may be possible that we try to find the value in the neighborhood of the Point of a function because the value of the function does not exist at a point. On the other hand, continuity of a function is closely related to the concept of limits. Now, talk about the definition of Limit, in many cases the values of the given function ‘f’ for the values of ‘x’ near a point ‘c’ and this value is not equal to function f (c) value or that value lies near no number at all.
Now, next concept to study is how we represent the limit, let a function 'f' is said to tends to a and the limit as 'x' tends to 'a', if x=a then it shows the values is just larger than or just smaller than x=a, f(x) has to move more closely to the value of limit and mathematically it can be written as,
Lim x→ a f(x) = 1, which is equal to
|f(x)-l| < e, x: 0 < |x-a| <d
Where, e and d are positive Numbers.
There are two types of limit that is Right and Left hand limits, in the Right Hand Limit
Lim x→ a+f(x) = 1, the value of a, is positive for function f(x).For example if the value of x tends to 1 than,
Lim x→ 1 (x) = 1,
And in Left Hand Limit the value of a, is negative for function f(x).The Mathematical Expression can be written as,
Lim x→ a-f(x) = 1,
For example if x tends to the value -1 than
Lim x→ 1-(x) = 0,
Now talk about infinity, infinity means something which keeps increasing and passes all limits this is called positive infinity. On the other side, something that continuously decrease and passes all limits is called negative infinity. The symbol of infinity is ∞. Some points given below which are necessary to read.
Infinity cannot be plotted on the paper.
∞+∞=∞
∞-∞ is indeterminate.
∞x∞=∞
0x∞=is indeterminate.
∞/∞,∞0 , are all indeterminate.
Now the definition of continuity, the any function that is f(x) is called continuous on an interval a if
Lim x→ a f(x) = f(a),
Otherwise, the function f(x) is discontinuous at a. Note that the continuity of f(x) at a means two things,
Lim x→ a f(x) the function exist
And this limit is f(a).
Lim x→ a f(x) = f(a)
This property is known as continuity for all Real Numbers a. The f(x) is said to be continuous from the left if the value of a is negative that is
Lim x→a- f(x) = f(a),
And the f(x) is said to be continuous from the right if the value of a is positive that is
Lim x→a+f(x) = f(a)
This shows the continuity and limits.
There are some points of continuity given below, if f(x) and g(x) are continuous at a, Then
(1) f(x) + g(x) is continuous.
(2) f(x) g(x) is continuous at a
(3) f(x)/g(x) is continuous at a g(a)≠0.
## Continuity of function in Calculus
A function is said to be continuous if function f ( x ) at Point ( c , f ( c ) ) if the conditions listed below are satisfied-
1) Function f ( c ) must exist.
2) The lim x → c f ( x ) must also exist.
3) And lim x → c f ( x ) = f ( c ).
The meaning of above conditions function is that there should be no missing point or gaps for f (x...Read More
## Continuity of function of one variable
The continuity of a Functions can be determined by the fact that their graphs are continuous in nature i.e. the graph of the function is continuous. Continuous graphs are the type of the graphs in which there is no break in the graph and which are drawn without lifting the pencil from the paper. In the term of continuous Functions, it is also tr...Read More
## Continuity of function of two or more variable
A function f(x) is said to be continuous at x= a if limx→a- f(x) = limx→a+ f(x)=f(a).
i.e. if Left Hand Limit = Right hand limit = value of the function at 'a'.
i.e. limx→a f(x) = f(a),
If, f(x) is not continuous at x = a, we say that f(x) is discontinuous at x = a.
For a function to be continuous at any Point x=a,...Read More
## Discontinuity of function in Calculus
When we deal with Calculus we have to deal with Continuous Calculus and Discontinuous Calculus. In Continuous Calculus, we focus on continuous Functions and in discontinuous Calculus we deal with discontinuous Functions. Discontinuous functions are those functions which are not continuous means a function whose values does not vary continuously w...Read More
## Types of Discontinuity in Calculus functions
Continuous Functions are those Functions which have no breaks i.e. when we draw them on a graph it seems to be a smooth curve or graph from one end to the other end without any cut or break between it. To understand Discontinuity we need to learn different types of discontinuous functions. |
# What is a prime number factor?
## What is a prime number factor?
A prime factor is a natural number, other than 1, whose only factors are 1 and itself. The first few prime numbers are actually 2, 3, 5, 7, 11, and so on. Now we can also use what’s called prime factorization for numbers which actually consist of using factor trees.
How do you find factors of prime factors?
How to Find Number of Factors?
1. Find its prime factorization, i.e. express it as the product of primes.
2. Write the prime factorization in the exponent form.
3. Add 1 to each of the exponents.
4. Multiply all the resultant numbers.
5. This product would give the number of factors of the given number.
What are the prime factors 2431?
The prime factors of 2431 are 11, 13, 17.
### What is the prime factor of 4?
2 ×2
We know that the number 4 is an even composite number and it can be further factored as the product of 2 and 2. Hence, 4 can be written as 2 ×2. Therefore, the prime factorization of 4 is 2 ×2 or 22.
What is a prime factor of 6?
2 × 3
So, the prime factors of 6 are written as 2 × 3, where 2 and 3 are prime numbers.
How do you find a factor of a number?
The formula for the total number of factors for a given number is given by; Total Number of Factors for N = (a+1) (b+1) (c+1)
#### What is the factor formula?
Factoring Formula 1: (a + b)2 = a2 + 2ab + b.
What is the LCM of 404?
Answer: LCM of 404 and 96 is 9696. Explanation: The LCM of two non-zero integers, x(404) and y(96), is the smallest positive integer m(9696) that is divisible by both x(404) and y(96) without any remainder.
What is the prime factor of 7?
1 and 7
The factors of 7 are 1 and 7. As the number 7 is a prime number, the factors of 7 are one and the number itself.
## What is a factor of 12?
factor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly. The other factors of 12 are 1, 2, 4, and 12.
What is 144×3 in prime factorization form?
144 is a square number. The prime factorization of 144 is 2 × 2 × 2 × 2 × 3 × 3….Factors of 144 in Pairs.
Factors of 144 in Pairs Notation
12 × 12 (12,12)
How do you factor 144?
Factor Pairs: 144 = 1 x 144, 3 x 48, 2 x 72, 4 x 36, 8 x 18, 6 x 24, 9 x 16, 12 x 12. Factors of 144: 1, 4, 2, 3, 6, 8, 16, 9, 12, 18, 24, 36, 48, 72, 144.
### What is the prime factor of 54?
2 × 3 × 3 × 3
Prime factorisation of 54 is 2 × 3 × 3 × 3. Therefore, the highest prime factor of 54 is 3.
What is the prime factorization for 47?
47 × 1
Step 3: As we know 1 and 47 are the prime numbers that have only two factors, i.e., one and the number itself and cannot further factorize it. Step 4: The factorisation of 47 = 47 × 1. Step 5: Therefore, the factorization of 47 is written as 47 = 47 × 1.
How do you find the prime factors of a number?
We cover two methods of prime factorization: find primes by trial division, and use primes to create a prime factors tree. Say you want to find the prime factors of 100 using trial division. Start by testing each integer to see if and how often it divides 100 and the subsequent quotients evenly.
#### How do you factor the first 5000 prime numbers?
For the first 5000 prime numbers, this calculator indicates the index of the prime number. The nth prime number is denoted as Prime [n], so Prime [1] = 2, Prime [2] = 3, Prime [3] = 5, and so on. The limit on the input number to factor is less than 10,000,000,000,000 (less than 10 trillion or a maximum of 13 digits).
What is the maximum number of digits in prime factorization?
The limit on the input number to factor is less than 10,000,000,000,000 (less than 10 trillion or a maximum of 13 digits). What is Prime Factorization? Prime factorization or integer factorization of a number is breaking a number down into the set of prime numbers which multiply together to result in the original number. |
# Chap 1 First-Order Differential Equations
## Presentation on theme: "Chap 1 First-Order Differential Equations"— Presentation transcript:
Chap 1 First-Order Differential Equations
Outline Basic Concepts Separable Differential Equations
substitution Methods Exact Differential Equations Integrating Factors Linear Differential Equations Bernoulli Equations
Basic Concepts Differentiation
Basic Concepts Differentiation
Basic Concepts Integration
Basic Concepts Integration
Basic Concepts Integration
Basic Concepts ODE vs. PDE
Dependent Variables vs. Independent Variables Order Linear vs. Nonlinear Solutions
Basic Concepts Ordinary Differential Equations
An unknown function (dependent variable) y of one independent variable x
Basic Concepts Partial Differential Equations
An unknown function (dependent variable) z of two or more independent variables (e.g. x and y)
Basic Concepts The order of a differential equation is the order of the highest derivative that appears in the equation. Order 2 Order 1 Order 2
Basic Concept The first-order differential equation contain only y’ and may contain y and given function of x. A solution of a given first-order differential equation (*) on some open interval a<x<b is a function y=h(x) that has a derivative y’=h(x) and satisfies (*) for all x in that interval. (*) or
Basic Concept Example : Verify the solution
Basic Concepts Explicit Solution Implicit Solution
Basic Concept General solution vs. Particular solution
arbitrary constant c Particular solution choose a specific c
Basic Concept Singular solutions
Def : A differential equation may sometimes have an additional solution that cannot be obtained from the general solution and is then called a singular solution. Example The general solution : y=cx-c2 A singular solution : y=x2/4
Basic Concepts General Solution
Particular Solution for y(0)=2 (initial condition)
Basic Concept Def: A differential equation together with an initial condition is called an initial value problem
Separable Differential Equations
Def: A first-order differential equation of the form is called a separable differential equation
Separable Differential Equations
Example : Sol:
Separable Differential Equations
Example : Sol:
Separable Differential Equations
Example : Sol:
Separable Differential Equations
Example : Sol:
Separable Differential Equations
Substitution Method: A differential equation of the form can be transformed into a separable differential equation
Separable Differential Equations
Substitution Method:
Separable Differential Equations
Example : Sol:
Separable Differential Equations
Exercise 1
Exact Differential Equations
Def: A first-order differential equation of the form is said to be exact if
Exact Differential Equations
Proof:
Exact Differential Equations
Example : Sol:
Exact Differential Equations
Sol:
Exact Differential Equations
Sol:
Exact Differential Equations
Example
Non-Exactness Example :
Integrating Factor Def: A first-order differential equation of the form is not exact, but it will be exact if multiplied by F(x, y) then F(x,y) is called an integrating factor of this equation
Exact Differential Equations
How to find integrating factor Golden Rule
Exact Differential Equations
Example : Sol:
Exact Differential Equations
Sol:
Exact Differential Equations
Example :
Exact Differential Equations
Exercise 2
Linear Differential Equations
Def: A first-order differential equation is said to be linear if it can be written If r(x) = 0, this equation is said to be homogeneous
Linear Differential Equations
How to solve first-order linear homogeneous ODE ? Sol:
Linear Differential Equations
Example : Sol:
Linear Differential Equations
How to solve first-order linear nonhomogeneous ODE ? Sol:
Linear Differential Equations
Sol:
Linear Differential Equations
Example : Sol:
Linear Differential Equations
Example :
Bernoulli, Jocob Bernoulli, Jocob
Linear Differential Equations
Def: Bernoulli equations If a = 0, Bernoulli Eq. => First Order Linear Eq. If a <> 0, let u = y1-a
Linear Differential Equations
Example : Sol:
Linear Differential Equations
Exercise 3
Summary
Orthogonal Trajectories of Curves
Angle of intersection of two curves is defined to be the angle between the tangents of the curves at the point of intersection How to use differential equations for finding curves that intersect given curves at right angles ?
How to find Orthogonal Trajectories
1st Step: find a differential equation for a given cure 2nd Step: the differential equation of the orthogonal trajectories to be found 3rd step: solve the differential equation as above ( in 2nd step)
Orthogonal Trajectories of Curves
Example: given a curve y=cx2, where c is arbitrary. Find their orthogonal trajectories. Sol:
Existance and Uniqueness of Solution
An initial value problem may have no solutions, precisely one solution, or more than one solution. Example No solutions Precisely one solutions More than one solutions
Existence and uniqueness theorems
Problem of existence Under what conditions does an initial value problem have at least one solution ? Existence theorem, see page 53 Problem of uniqueness Under what conditions does that the problem have at most one solution ? Uniqueness theorem, see page54 |
+0
# Socks
0
1
4
+295
My sock drawer contains two red socks, two white socks, and two blue socks. For three days in a row, I pull two socks out of my sock drawer at random. (I do not replace the socks in the drawer before drawing on the next day.) What is the probability that for each of the three days my socks are unmatched?
Jun 19, 2024
#2
+506
-1
The probability of drawing unmatched socks on any given day depends on the number of socks remaining and the number of matching pairs left. Let's analyze this step-by-step:
Day 1:
Total socks: 6 (2 red, 2 white, 2 blue)
Favorable outcomes (unmatched socks): You can draw any one sock and then a sock from a different pair. There are 4 choices for the first sock (any of the 6 colors) and then 4 remaining socks (excluding the one already drawn). So, there are 4 * 4 = 16 favorable outcomes.
Total possible outcomes: You can draw any two socks from the 6 available. There are 6 choices for the first sock and then 5 remaining socks, resulting in 6 * 5 = 30 total possible outcomes.
Probability on Day 1:
(Favorable outcomes on Day 1) / (Total possible outcomes on Day 1) = 16 / 30 = 8/15
Day 2:
Total socks remaining: 4 (after drawing 2 on Day 1, not replaced)
Favorable outcomes (unmatched socks): Similar to Day 1, you can draw any sock and then one from a different remaining pair. There are 3 choices for the first sock and then 2 remaining socks (excluding the one drawn), resulting in 3 * 2 = 6 favorable outcomes.
Total possible outcomes: There are 4 choices for the first sock and then 3 remaining, resulting in 4 * 3 = 12 total possible outcomes.
Probability on Day 2 (given unmatched socks on Day 1):
We only consider the scenario where you drew unmatched socks on Day 1 because the prompt asks for the probability of this happening for all three days.
So, we only consider the drawers where we have 4 socks remaining (unmatched).
(Favorable outcomes on Day 2) / (Total possible outcomes on Day 2) = 6 / 12 = 1/2
Day 3:
Total socks remaining: 2 (after drawing 2 on Day 2, not replaced)
Favorable outcomes (unmatched socks): There's only one way to draw unmatched socks at this point - you must draw the two remaining socks, which are inherently unmatched.
Total possible outcomes: There are 2 choices for the first sock and then 1 remaining, resulting in 2 * 1 = 2 total possible outcomes.
Probability on Day 3 (given unmatched socks on Days 1 & 2):
Similar to Day 2, we only consider the scenario where you drew unmatched socks on both previous days.
(Favorable outcomes on Day 3) / (Total possible outcomes on Day 3) = 1 / 2
Overall Probability:
The prompt asks for the probability of getting unmatched socks for all three days. To get this probability, we need to multiply the probabilities of getting unmatched socks on each day (considering the condition that you drew unmatched socks on the previous day).
Overall Probability = (Probability on Day 1) * (Probability on Day 2 | Day 1) * (Probability on Day 3 | Day 1 & 2) = 8/15 * 1/2 * 1/2 = 8 / 60
Simplifying the fraction:
We can divide both the numerator and denominator by 4:
Overall Probability = 2 / 15
Therefore, the probability of drawing unmatched socks for all three days is 2/15.
Jun 19, 2024
#3
+42
+1
I just wanted to note that booboo44's solution may be incorrect. First off, there are 6 choose 2 = 15 possible outcomes, not 30. This is because cases will be counted twice. In booboo44's solution, AB and BA are treated as different cases, when they are two equal cases. The favorable cases should also be 12, not 16. There may be more mistakes later on in the solution (but I won't be covering them). If all calculations are done right, the answer should be 8/15. (I created a solution but apparently it's still being moderated)
This is for educational purposes only. I do not want to start a fight with anyone.
Feel free to tell me if I did anything wrong! :D
Jun 19, 2024
edited by Tottenham10 Jun 19, 2024 |
# Hexadecimal to Decimal Conversion of Number System
Hexadecimal is a 16 based number where decimal is 10 based. We can convert a number from hexadecimal to decimal using some rules.
We have to know two rules to convert a number from hexadecimal to decimal where one is for fraction number and other is for integer number.
In this guide we will discuss about these rules to convert hexadecimal to decimal.
## Hexadecimal to decimal number conversion
### Example : Conversion of 13D1A16 to equivalent decimal number.
Solution :
(13D1A)2
= 1 × 164 + 3 × 163 + 13 × 162 + 1 × 161 + 10 × 160
= 65536 + 12288 + 3328 + 16 + 10
= 81178
Here are also another rules to convert this number from hexadecimal to decimal. Let’s see this.
After converting, we have got the result in decimal is 81178. You can convert any hexadecimal integer number using this rules.
### Example: Convert 0.2A516 to equivalent decimal number.
Solution :
(0.2A5)2
= 2 × 16-1 + 10 × 16-2 + 5 × 16-3
= 0.125 + 0.0391 + 0.0012
= 0.1653
Note: We have taken 4 number after radix point here.
We have got the decimal equivalent of the given hexadecimal number is 0.1653 and you can convert any hexadecimal fraction number using this method.
Let’s see the following example to convert a number from hexadecimal to decimal.
Here, we will convert the integer and fraction part separately. Hope this example will help you to understand the topics.
### Example : Conversion of 1351A.2A516 to equivalent decimal number
Solution :
Lets see the conversion of its integer part first.
(1351A)16
= 1 × 164 + 3 × 163 + 5 × 162 + 1 × 161 + 10 × 160
= 65536 + 12288 + 1280 + 16 + 10
= 79130
Here is also another method to convert an integer hexadecimal to decimal number. See this bellow.
As we know the method for converting the integer number, now we will see the method to convert its fraction part from hexadecimal to decimal number. See the method bellow.
(0.2A5)2
= 2 × 16-1 + 10 × 16-2 + 5 × 16-3
= 0.125 + 0.0391 + 0.0012
= 0.1653
Note: We have taken everything after 4 digits of radix point. You can take more if needed.
After converting both the integer and fraction part, we have got the total equivalence of this hexadecimal number is (79130.1653)10 .
However, you can convert any hexadecimal number to its equivalent decimal number by the same way we have followed here.
If you have any doubt still now, then don’t forget to tell us in the comment section bellow. We will try to update our article according to your needs.
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# How do you do addition facts to 20?
## How do you do addition facts to 20?
The addition facts to 20 are simply the sums from 0+0 up to 10+10. They’re the building blocks of arithmetic, and usually the first math facts that children master.
### What are basic addition facts?
The basic facts of addition are those equations in which two single-digit numbers are combined by addition to give a sum Hence they range from 0+0=0 to 9+9=18. For each basic addition fact there is a related basic subtraction fact, for example, 18-9=9.
What is the basic facts in maths?
Number facts are simple addition, subtraction, multiplication and division facts which are sometimes referred to as number bonds. Children learn these basic facts between the ages of 4 and 10.
What are the addition and subtraction facts to 20?
Recall addition and subtraction facts to 20. This is a level 2 number activity from the Figure It Out series. It relates to Stage 5 of the Number Framework.A PDF of the student activity is included. This is a level 2 number activity from the Figure It Out series.
## How to write addition and subtraction worksheets?
Incorporate the addition and subtraction fact family worksheets comprising sorting the number sets, find the missing members in the triangles, circles, number bonds and bar models; writing the four addition and subtraction facts in the house models, dominoes, picture models and more.
### How to add and subtract 2 digit numbers?
Add or subtract 2-digit numbers. All that is expected from you is to add or subtract the 2-digit numbers arranged in a column and horizontal format in here. 2-digit Addition and Subtraction – Column 1. 2-digit Addition and Subtraction – Column 2. 2-digit Addition and Subtraction – Horizontal 1
Why is addition and subtraction important for second graders?
It is also more important for second graders to be fluent in addition and subtraction to 100 than to be fluent in multiplication, per the common core state standards. Why? Math fact fluency is important for many reasons, but most importantly, it takes the frustration out of mathematics for many students. |
# How do you solve (x-2)/(2x+5)<0 using a sign chart?
Dec 27, 2016
The answer is x in ] -5/2, 2[
#### Explanation:
Let $f \left(x\right) = \frac{x - 2}{2 x + 5}$
The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- \frac{5}{2}\right\}$
Now, we can do the sign chart
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- \frac{5}{2}$$\textcolor{w h i t e}{a a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$2 x + 5$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$∣∣$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x - 2$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$∣∣$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a}$∣∣$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a}$$+$
Therefore,
$f \left(x\right) < 0$, when x in ] -5/2, 2[ |
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