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# Optimizing Problem, max/min, rectangle inside a triangle
• Mar 24th 2010, 12:22 PM
ellenberger
Optimizing Problem, max/min, rectangle inside a triangle
Find the area of the largest rectangle that can be inscribed in a right triangle with legs adjacent to the right angle of lengths 5cm and 12cm. The two sides of the rectangle lie along the legs.
• Mar 24th 2010, 04:15 PM
apcalculus
Quote:
Originally Posted by ellenberger
Find the area of the largest rectangle that can be inscribed in a right triangle with legs adjacent to the right angle of lengths 5cm and 12cm. The two sides of the rectangle lie along the legs.
Ellen:
How much progress have you made with this problem?
Let x and y be the sides.
x will be the side of the rectangle along the triangle side of length 5, and y will be the side along the triangle side of length 12.
Identify the right triangle (lower right if side of length 5 is the base) with legs (5-x) and y. This triangle is similar to the big right triangle, so the following ratio is valid:
$\displaystyle \frac{y}{12} = \frac{5-x}{5}$
Solve for y:
$\displaystyle y = 12 - \frac{12x}{5}$ (*)
The objective function is the area of the rectangle:
$\displaystyle A = x y$
Write this area as a single variable function by using substitution (*) above, then apply differential calculus techniques to optimize.
I hope this helps.
Good luck!
• Mar 24th 2010, 06:26 PM
ellenberger
Quote:
Originally Posted by apcalculus
Ellen:
How much progress have you made with this problem?
Let x and y be the sides.
x will be the side of the rectangle along the triangle side of length 5, and y will be the side along the triangle side of length 12.
Identify the right triangle (lower right if side of length 5 is the base) with legs (5-x) and y. This triangle is similar to the big right triangle, so the following ratio is valid:
$\displaystyle \frac{y}{12} = \frac{5-x}{5}$
Solve for y:
$\displaystyle y = 12 - \frac{12x}{5}$ (*)
The objective function is the area of the rectangle:
$\displaystyle A = x y$
Write this area as a single variable function by using substitution (*) above, then apply differential calculus techniques to optimize.
I hope this helps.
Good luck!
|
Mixed Number to Percent Calculator
# Mixed Number to Percent Calculator
Instructions:
• Enter a whole number, numerator, and denominator.
• Choose the conversion type (Decimal or Percent).
• Click "Convert" to perform the conversion.
• View the result and calculation details below.
• Calculations are stored in the calculation history.
• Click "Clear" to reset the inputs and results.
• Click "Copy Result" to copy the result to the clipboard.
Result:
Calculation History:
## Introduction
In the realm of mathematics, the conversion of mixed numbers into percentages plays a vital role in various applications. This process involves converting a mixed number, which consists of an integer and a proper fraction, into a percentage, expressed as a fraction over 100. The “Mixed Number to Percent Calculator” is a valuable tool that simplifies this conversion, making it accessible and efficient for students, professionals, and anyone dealing with numerical data.
## The Concept
The concept behind the Mixed Number to Percent Calculator revolves around the need to express mixed numbers as percentages. A mixed number consists of an integer part and a fractional part. To convert this into a percentage, we need to express the fractional part as a percentage of the whole, taking into account that percentages are based on a denominator of 100.
## Formulae
### Convert the Fractional Part to a Percentage
The first step in converting a mixed number to a percentage is to convert the fractional part into a percentage. This can be done using the following formula:
Percentage = (Fractional Part / 1) * 100
For example, if you have the mixed number 2 1/4, you would first convert the 1/4 to a percentage as follows:
Percentage = (1/4 / 1) * 100 = 25%
Once you have the percentage of the fractional part, you simply need to add it to the integer part of the mixed number to get the final percentage. Continuing with our example:
Final Percentage = 2 + 25% = 2 + 0.25 * 100 = 2 + 25 = 27%
This is the final percentage representation of the mixed number 2 1/4.
## Example Calculations
Let’s go through a few more example calculations to illustrate the conversion process using the Mixed Number to Percent Calculator:
### Example 1: 3 3/5
Step 1: Convert the fractional part to a percentage: Percentage = (3/5 / 1) * 100 = 60%
Step 2: Add the integer part: Final Percentage = 3 + 60% = 3 + 0.60 * 100 = 3 + 60 = 63%
So, 3 3/5 is equivalent to 63% as a percentage.
### Example 2: 5 1/8
Step 1: Convert the fractional part to a percentage: Percentage = (1/8 / 1) * 100 = 12.5%
Step 2: Add the integer part: Final Percentage = 5 + 12.5% = 5 + 0.125 * 100 = 5 + 12.5 = 17.5%
Hence, 5 1/8 is equivalent to 17.5% as a percentage.
## Real-World Use Cases
The Mixed Number to Percent Calculator finds relevance in various real-world scenarios:
### Cooking and Recipes
In culinary arts, recipes require precise measurements. Converting mixed numbers to percentages can help chefs accurately adjust ingredient proportions when scaling recipes up or down. For instance, if a recipe calls for 1 1/2 cups of flour and you need to make half the recipe, you can use the calculator to quickly determine that you need 75% of the original quantity, which is 1.5 cups * 0.75 = 1.125 cups.
Financial analysts frequently deal with data in mixed number formats. For instance, if a company’s profit margin is 3 1/2%, the calculator can assist in converting it into a decimal form for further financial analysis. In this case, the calculator can quickly show that the profit margin is 0.035 when expressed as a decimal.
### Education
In the classroom, teachers and students can use the Mixed Number to Percent Calculator as a learning tool to grasp the concept of converting mixed numbers into percentages. It can serve as a valuable aid in math education and help students build confidence in their math skills.
## Conclusion
The Mixed Number to Percent Calculator simplifies the process of converting mixed numbers into percentages by providing a straightforward method for performing this conversion. The underlying formulae, which involve converting the fractional part into a percentage and adding it to the integer part, make the process accessible to users of all levels of mathematical proficiency. This tool finds practical application in fields such as cooking, finance, and education, where precise numerical conversions are essential.
## References
1. Smith, J. (2009). Mathematical Tools for Everyday Life. Academic Press.
2. Johnson, L. (2015). Fraction Conversions in Practical Mathematics. Educational Research Journal, 20(2), 45-62.
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# In this article, we will explore the process of finding the derivative of a complex function that combines polynomial terms, radical terms, and trigonometric functions. Specifically, we will demonstrate how to find the derivative of the function: f(x) = 7x⁵ - x⁴ + 4x³/² + 8/√x - 4sec(x). Calculus, the mathematical study of rates of change and accumulation, is a critical tool in various scientific and engineering disciplines. One of its core concepts is finding the derivative of a function, which reveals how the function's values change concerning its independent variable.
## Understanding Basic Differentiation Rules
Before we embark on finding the derivative of this multifaceted function, let's review some fundamental rules of differentiation:
Power Rule: The derivative of x^n with respect to x, where n is a constant, is n * x^(n-1).
Constant Rule: The derivative of a constant C (where C is any real number) is 0.
Sum/Difference Rule: The derivative of the sum or difference of two functions is the sum or difference of their derivatives, respectively.
Derivative of √x: The derivative of √x with respect to x is (1/2) * x^(-1/2).
Derivative of sec(x): The derivative of sec(x) with respect to x is sec(x) * tan(x).
### Now, let's proceed to find the derivative of 7x⁵ - x⁴ + 4x³/² + 8/√x - 4sec(x) step by step.
#### Step 1: Differentiate the Polynomial Terms
Let's differentiate the polynomial terms individually:
• Derivative of 7x⁵: Using the power rule, the derivative of 7x⁵ with respect to x is 5 * 7x^(5-1) = 35x⁴.
• Derivative of -x⁴: Similarly, the derivative of -x⁴ is -4x^(4-1) = -4x³.
• Derivative of 4x³/²: The derivative of 4x³/² can be computed using the power rule for fractional exponents: Derivative = (3/2) * 4x^(3/2-1) = 6x^(1/2) = 6√x.
#### Step 2: Differentiate the Radical Term
Now, let's differentiate the radical term, 8/√x:
• Derivative of 8/√x: Applying the power rule for radicals, the derivative is -4x^(-1/2) = -4/√x.
#### Step 3: Differentiate the Trigonometric Term
Lastly, let's differentiate the trigonometric term, -4sec(x):
• Derivative of -4sec(x): The derivative of -4sec(x) with respect to x is -4 * sec(x) * tan(x).
#### Step 4: Combine the Derivatives
Now that we have found the derivatives of all the individual terms, we can combine them to find the derivative of the entire function f(x).
• f'(x) = 35x⁴ - 4x³ + 6√x - 4/√x - 4sec(x)tan(x).
#### Step 5: Simplify the Derivative
To simplify the derivative further, we have:
• f'(x) = 35x⁴ - 4x³ + 6√x - 4/√x - 4sec(x)tan(x).
### Conclusion
Calculus, particularly the process of finding the derivative of a function, is a foundational concept in mathematics that plays a crucial role in various scientific and engineering disciplines. By applying basic differentiation rules and understanding the derivatives of radical and trigonometric functions, we successfully found the derivative of the complex function f(x) = 7x⁵ - x⁴ + 4x³/² + 8/√x - 4sec(x). The derivative, f'(x) = 35x⁴ - 4x³ + 6√x - 4/√x - 4sec(x)tan(x), provides valuable insights into how the original function changes as the independent variable x varies. This process exemplifies the essential role of calculus in analyzing and understanding dynamic relationships between variables in a wide range of applications.
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# 2-D Kinematics Review
Μηχανική
14 Νοε 2013 (πριν από 4 χρόνια και 6 μήνες)
96 εμφανίσεις
2
-
D Kinematics Review
Scalars
vs
Vectors
Scalars
have magnitude only
Distance, speed, time, mass
Vectors
have both magnitude and direction
displacement, velocity, acceleration
R
tail
Inverse Vectors
Inverse vectors have the same length, but opposite
direction.
A
-
A
A
B
R
A
+
B
=
R
-
to
-
tail.
The sum is called the resultant.
The inverse of the sum is called the equilibrant
Unit Vectors
Unit vectors are quantities that specify direction only.
They have a magnitude of exactly one, and typically
point in the x, y, or z directions.
ˆ
points in the x direction
ˆ
points in the y direction
ˆ
points in the z direction
i
j
k
Unit Vectors
z
y
x
i
j
k
Unit Vectors
Instead of using magnitudes and directions, vectors can be
represented by their components combined with their unit
vectors.
Example: displacement of 30 meters in the +x direction added to a
displacement of 60 meters in the
displacement of 40 meters in the +z direction yields a displacement
of:
ˆ
ˆ ˆ
(30 -60 40 ) m
30,-60,40 m
i j k
i
components together, all
the
j
components together, and all the
k
components together.
Sample problem:
Consider two vectors,
A
= 3.00
i
+ 7.50
j
and
B
=
-
5.20
i
+ 2.40
j.
Calculate
C
where
C
=
A
+
B
.
Sample problem:
Consider two vectors,
A
= 3.00
i
+ 7.50
j
and
B
=
-
5.20
i
+ 2.40
j.
Calculate
C
where
C
=
A
+
B
.
Suppose I need to convert unit vectors to
a magnitude and direction?
Given the vector
2 2 2
ˆ
ˆ ˆ
x y z
x y z
r r i r j r k
r r r r
Sample problem:
You move 10 meters north and 6 meters east. You
then climb a 3 meter platform, and move 1 meter west on the
platform. How far are you from your starting point?
Sample problem:
You move 10 meters north and 6 meters east. You
then climb a 3 meter platform, and move 1 meter west on the
platform. How far are you from your starting point?
1 Dimension
2 or 3 Dimensions
x: position
x: displacement
v: velocity
a: acceleration
r
: position
r
: displacement
v
: velocity
a
: acceleration
r
= x
i
+ y
j
+ z
k
r
=
x
i
+
y
j
+
z
k
v
= v
x
i
+ v
y
j
+ v
z
k
a
= a
x
i
+ a
y
j
+ a
z
k
In Unit Vector
Notation
Sample problem:
The position of a particle is given by
r
= (80 + 2t)
i
40
j
-
5t
2
k.
Derive the velocity and acceleration
vectors for this particle. What does motion “look like”?
Sample problem:
The position of a particle is given by
r
= (80 + 2t)
i
40
j
-
5t
2
k.
Derive the velocity and acceleration
vectors for this particle.
Trajectory of Projectile
g
g
g
g
g
This shows the parabolic trajectory of a projectile fired over
level ground.
Acceleration points down at 9.8 m/s
2
for the entire
trajectory.
Trajectory of Projectile
v
x
v
y
v
y
v
x
v
x
v
y
v
x
v
y
v
x
The velocity can be resolved into components all along its path.
Horizontal velocity remains constant; vertical velocity is
accelerated.
Remember…
To work projectile problems…
…resolve the initial velocity into components.
V
o
V
o,y
= V
o
sin
V
o,x
= V
o
cos
Sample problem:
A soccer player kicks a ball at 15 m/s at an angle
of 35
o
above the horizontal over level ground. How far
horizontally
will the ball travel until it strikes the ground?
Sample problem:
A soccer player kicks a ball at 15 m/s at an angle
of 35
o
above the horizontal over level ground. How far will the ball
travel until it strikes the ground?
Sample problem:
A cannon is fired at a 15
o
angle above the
horizontal from the top of a 120 m high cliff. How long will it take
the cannonball to strike the plane below the cliff? How far from the
base of the cliff will it strike?
Sample problem:
A cannon is fired at a 15
o
angle above the
horizontal from the top of a 120 m high cliff. How long will it take
the cannonball to strike the plane below the cliff? How far from the
base of the cliff will it strike?
Uniform Circular Motion
Occurs when an object moves in a circle without
changing speed.
Despite the constant speed, the object’s
velocity
vector
is continually changing; therefore, the object must be
accelerating.
The acceleration vector is pointed toward the center of
the circle in which the object is moving, and is referred
to as
centripetal acceleration
.
Vectors in
Uniform Circular Motion
a
v
a = v
2
/ r
v
a
v
a
v
a
Sample Problem
The Moon revolves around the Earth every 27.3 days. The radius
of the orbit is 382,000,000 m. What is the magnitude and
direction of the acceleration of the Moon relative to Earth?
Sample Problem
The Moon revolves around the Earth every 27.3 days. The
radius of the orbit is 382,000,000 m. What is the magnitude
and direction of the acceleration of the Moon relative to
Earth?
Tangential acceleration
Sometimes the speed of an object in circular motion is not
constant (in other words, it’s not uniform circular motion).
An acceleration component may be tangent to the path, aligned
with the velocity. This is called
tangential acceleration
. It causes
speeding up or slowing down.
The centripetal acceleration component causes the object to
continue to turn as the tangential component causes the speed
to change. The centripetal component is sometimes called the
, since it lies along the radius.
v
Tangential Acceleration
component (a
r
or a
c
)
tangential component (a
T
)
a
If tangential acceleration
exists, either the speed or
This is no longer UCM.
Sample Problem:
Given the figure at right rotating at
acceleration components if
= 30
o
and
a
has a
magnitude of 15.0 m/s
2
. What is the speed of the
particle at the location shown? How is the particle’s
speed changing?
5.00 m
a
v
Sample Problem:
Given the figure at right rotating at
acceleration components if
= 30
o
and
a
has a
magnitude of 15.0 m/s
2
. What is the speed of the
particle? How is it behaving?
5.00 m
a
|
# How do you solve the following system?: -x+4y=8 , x=2y+1
Aug 21, 2017
See a solution process below:
#### Explanation:
Step 1) Because the second equation is already solved for $x$ we can substitute $\left(2 y + 1\right)$ for $x$ in the first equation and solve for $y$:
$- x + 4 y = 8$ becomes:
$- \left(2 y + 1\right) + 4 y = 8$
$- 2 y - 1 + 4 y = 8$
$4 y - 2 y - 1 = 8$
$\left(4 - 2\right) y - 1 = 8$
$2 y - 1 = 8$
$2 y - 1 + \textcolor{red}{1} = 8 + \textcolor{red}{1}$
$2 y - 0 = 9$
$2 y = 9$
$\frac{2 y}{\textcolor{red}{2}} = \frac{9}{\textcolor{red}{2}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} y}{\cancel{\textcolor{red}{2}}} = \frac{9}{2}$
$y = \frac{9}{2}$
Step 2) Substitute $\frac{9}{2}$ for $y$ in the second equation and calculate $x$:
$x = 2 y + 1$ becomes:
$x = \left(2 \cdot \frac{9}{2}\right) + 1$
$x = \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot \frac{9}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}\right) + 1$
$x = 9 + 1$
$x = 10$
The Solution Is:
$x = 10$ and $y = \frac{9}{2}$ or $\left(10 , \frac{9}{2}\right)$
Aug 21, 2017
$\left(x , y\right) \to \left(10 , \frac{9}{2}\right)$
#### Explanation:
$- \textcolor{red}{x} + 4 y = 8 \to \left(1\right)$
$\textcolor{red}{x} = 2 y + 1 \to \left(2\right)$
$\text{substitute "color(red)(x)=2y+1" into } \left(1\right)$
$\Rightarrow - \left(2 y + 1\right) + 4 y = 8$
$\Rightarrow - 2 y - 1 + 4 y = 8$
$\Rightarrow 2 y = 9 \Rightarrow y = \frac{9}{2}$
$\text{substitute this value into } \left(2\right)$
$\Rightarrow x = \left(2 \times \frac{9}{2}\right) + 1 = 9 + 1 = 10$
$\textcolor{b l u e}{\text{As a check}}$
$\text{substitute these values into } \left(1\right)$
$- 10 + \left(4 \times \frac{9}{2}\right) = - 10 + 18 = 8 \leftarrow \text{ True}$
$\Rightarrow \text{point of intersection } = \left(10 , \frac{9}{2}\right)$
Aug 21, 2017
$x = 10 \mathmr{and} y = 4.5$
#### Explanation:
Notice that the $x$-terms are ADDITIVE INVERSES. That means that they will add together to give $0$.
Change the second equation into the same form so we have:
$\textcolor{w h i t e}{\times \times \times} \textcolor{b l u e}{- x} + 4 y = 8 \text{ } \ldots \ldots . A$
$\textcolor{w h i t e}{\times \times \times} \textcolor{b l u e}{+ x} - 2 y = 1 \text{ } \ldots \ldots \ldots B$
$A + B : \text{ } \textcolor{b l u e}{0 x} + 2 y = 9$
$\textcolor{w h i t e}{\times \times \times \times \times x} y = 4.5$
Substitute this value for $y$ into the original equation for $x$
$x = 2 y + 1$
$x = 2 \left(4.5\right) + 1$
$x = 10$
Check in the other equation:$\text{ } - x + 4 y = 8$
$- \left(10\right) + 4 \left(4.5\right)$
$- 10 + 18$
$= 8 \text{ } \leftarrow$ the answer is correct.
|
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# Class 10 RD Sharma Solutions- Chapter 14 Coordinate Geometry – Exercise 14.1
• Last Updated : 10 May, 2021
### Problem 1: On which axis do the following points lie?
(i) P (5, 0)
Solution:
As its ordinate is 0. So, it lies on x-axis.
(ii) Q (0, -2)
Solution:
As its abscissa is 0. So, it lies on y-axis (negative half).
(iii) R (-4, 0)
Solution:
As its ordinate is 0. So, it lies on x-axis (negative half).
(iv) S (0, 5)
Solution:
As its abscissa is 0. So, it lies on y-axis.
### (ii) The centre of the square is at the origin and coordinate axes are parallel to the sides AB and AD respectively.
Solution:
(I)
(II)
(i) Coordinate of the vertices of the square ABCD of side 2a will be – A(0, 0), B(2a, 0), C(2a, 2a) and D(0, 2a)
(ii) Coordinate of the vertices of the square ABCD of side 2a will be – A(a, a), B(-a, a), C(-a, -a) and D(a, -a)
### Problem 3: The base PQ of two equilateral triangles PQR and PQR’ with side 2a lies along y- axis such that the mid-point of PQ is at the origin. Find the coordinates of the vertices R and R’ of the triangles.
Solution:
Here, We have two equilateral triangles PQR and PQR’ with side 2a lying along y-axis.
O is the mid-point of PQ.
Now, in ∆QOR, ∠QOR = 90°
Now, By using Pythagoras theorem –
OR2 + OQ2 = QR2
OR2 = (2a)2 – (a)2
OR2 = 3a2
OR = (√3)a
Thus, the coordinate of vertex R is (√3 a, 0) and coordinate of vertex R’ is (-√3 a, 0)
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Related Articles
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# Sequences and Series
## Sequences and Series
Patterns of numbers separated by commas are called sequences e.g. $2,4,6,8,10{\ldots }$ and $1,3,5,7,9{\ldots }$
A sequence of terms can be defined by a rule for the $n^{\rm th}$ term e.g. $u_{n} = 2n + 3$
This would generate the sequence $u_{1} = 5$, $u_{2} = 7$, $u_{3} = 9$, $u_{4} = 11{\ldots }$, written as $5,7,9,11{\ldots }$
An inductive definition for a sequence is given by stating the first term and a rule showing how to get to the next term from the previous one. For example, the definition $u_{1} = 1, u_{n+1} = 2u_{n} + 3$ would generate this sequence $1,5,13,29{\ldots }$
### Arithmetic Sequences
$a, a+d, a+2d \ldots$
These are sequences which have a common difference e.g. $4,6,8,10,12{\ldots }$
In this case the common difference is 2.
The $n^{\rm th}$ term of an arithmetic sequence $u_{1}, u_{2}, u_{3}{\ldots }$ is given by $u_{n} = a + (n-1) d$ where
\begin{eqnarray*} a & =& {\hbox{first term}} \\ n & =& {\hbox{number of terms}} \\ d & =& {\hbox{common difference}} \\ \end{eqnarray*}
### Arithmetic Series
When the terms of an arithmetic sequence are added together they form an arithmetic series, e.g. $2 + 3 + 4 + 5 + 6 + {\ldots }$
The sum to n terms of an arithmetic series is:
$S_ n = {1\over 2}n(a+\ell ) = {1\over 2} n\left\{ 2a+(n-1)d\right\}$
where
\begin{eqnarray*} a & =& {\hbox{first term}} \\ n & =& {\hbox{number of terms}} \\ d & =& {\hbox{common difference}} \\ \ell & =& {\hbox{last term}} \\ \end{eqnarray*}
### Geometric Sequences
$a, ar, ar^2 \ldots$
These are sequences which have a common ratio e.g. $2,6,18,54,162{\ldots }$
In this case the common ratio is 3.
To find the $n^{th}$ term use the formula
$n^{th} \hbox{term} = ar^{n-1}$
where
\begin{eqnarray*} a & =& \hbox{first term} \\ r & =& \hbox{common ratio} \\ n & =& \hbox{number of terms} \\ \end{eqnarray*}
### Geometric Series
The sum of the first $n$ terms of a geometric series is:
$S_ n = {a(1-r^ n)\over 1-r}$
where
\begin{eqnarray*} a & =& \hbox{first term} \\ r & =& \hbox{common ratio} \\ \end{eqnarray*}
### The sum to infinity of a geometric series
$S_\infty = {a\over 1-r} \quad -1<r<1$
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# How do you differentiate f(x)=ln(cos(e^(x) )) using the chain rule?
Mar 17, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \tan {e}^{x} \cdot {e}^{x}$
#### Explanation:
So, we got three functions here:
$\ln \left(\cos \left({e}^{x}\right)\right)$
$\cos \left({e}^{x}\right)$
and
${e}^{x}$
Let $y = \ln \left(\cos \left({e}^{x}\right)\right)$
differentiating w.r.t. $x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \ln \left(\cos \left({e}^{x}\right)\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left({e}^{x}\right) \cdot \frac{d}{\mathrm{dx}} \cos \left({e}^{x}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\cos} \left({e}^{x}\right) \cdot \sin \left({e}^{x}\right) \cdot \frac{d}{\mathrm{dx}} {e}^{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\cos} \left({e}^{x}\right) \cdot \sin \left({e}^{x}\right) \cdot {e}^{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sin \frac{{e}^{x}}{\cos} \left({e}^{x}\right) \cdot {e}^{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \tan {e}^{x} \cdot {e}^{x}$
This will be the differentiated function.
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# High School: Number and Quantity
### The Complex Number System HSN-CN.C.7
7. Solve quadratic equations with real coefficients that have complex solutions.
When students dealt with parabolas, they only saw the good parabolas. The ones with nice simple roots like -1 and 1. The roots could be found just by looking at the graph or, if they were ugly, by using the quadratic formula.
They probably didn't think too much about those "other" parabolas. The parabolas from the wrong side of the tracks. Yeah, we mean the ones that don't cross the x-axis at all. Your students didn't consider that those parabolas have roots (and feelings), too.
Well, it's time. They're old enough to know the truth.
Those quadratic equations have imaginary roots. Now that your students know that "imaginary" doesn't mean "make believe," we can make it up to these forgotten, mistreated parabolas. Flowers and chocolate just won't cut it; we'll have to work with them.
Let's take a look at one: x2 + 4x + 6 = 0. This one doesn't factor, so we'll have to use the quadratic formula:
For this particular parabola, our values are a = 1, b = 4 and c = 6. When we plug in those numbers, we get
But wait. We can't have the square root of a negative number if we're working with real numbers. Good thing we're working with imaginary numbers, then, isn't it?
That means we can turn that fraction into which simplifies to . So our parabola has 2 perfectly lovely roots. The only problem is that the graph doesn't cross the x-axis because it only contains real numbers, and the roots are imaginary, not real.
Quadratics with imaginary roots aren't always easy to identify until we get to the point in the problem where we have to take the square root. At that point, if the number under the radical is negative, that parabola's roots will be imaginary. (That number under the radical, b2 – 4ac, is called the discriminant. Probably because it discriminates between real and imaginary roots.)
Now, your students should be able to calculate the roots of these other parabolas and tell whether or not a parabola will have real or imaginary roots by looking at its discriminant.
#### Drills
1. Does the parabola with equation x2 + 7x – 3 = 0 have real or imaginary roots?
Correct Answer:
Real, because the discriminant is positive
Answer Explanation:
To find out whether a parabola has real or imaginary roots, all we need to do is calculate the discriminant, or the part of the quadratic formula under the radical (b2 – 4ac). Since a = 1, b = 7, and c = -3, the discriminant is 72 – 4(1)(-3) = 49 + 12 = 61. A positive number under the square root is real, so the answer is (B).
2. Does the parabola with equation x2 – x + 4 = 0 have real or imaginary roots?
Correct Answer:
Imaginary, because the discriminant is negative
Answer Explanation:
The discriminant of a parabola is b2 – 4ac, which is equal to 1 – 4(1)(4) = -15 in this case. Since the discriminant is under the radical in the quadratic formula, a positive discriminant will yield a real number and a negative will yield an imaginary number. Since the discriminant in this case is -15, the roots of the parabola are imaginary.
3. Does the parabola with equation -7x2 = 18 – 2x have real or imaginary roots?
Correct Answer:
Imaginary, because the discriminant is negative
Answer Explanation:
In this case, we need to set our equation to equal zero before we can calculate the discriminant. That means we rearrange the equation so that it takes the form -7x2 + 2x – 18 = 0 and take our a, b, and c values from there. If we use those values to calculate the discriminant, we end up with 22 – 4(-7)(-18) = 4 – 504 = -500. The discriminant is very negative, and the square root of negative numbers is why we have imaginary numbers in the first place. That means our answer is (C).
4. Solve 2x2 + 2x + 5 = 0 for x and express the answer in a + bi form.
Correct Answer:
Answer Explanation:
Since the equation is already set to equal zero, our values are a = 2, b = 2, and c = 5. Plugging that into the quadratic formula, we have , or That reduces to or .
5. Solve x2 + 4x = -6 for x and express the answer in a + bi form.
Correct Answer:
Answer Explanation:
Once we set the equation to equal to zero, a = 1, b = 4, and c = 6. All we need to do is use the quadratic formula and make sure we do the algebra correctly. It will give us , or , which is (B).
6. Solve 3x2 + 6x = -6 for x and express it in a + bi form.
Correct Answer:
-1 ± i
Answer Explanation:
If we rearrange the equation to 3x2 + 6x +6 = 0, we can use the values a = 3, b = 6, and c = 6 in quadratic formula. This means we have or . If we simplify this to its fullest extent, we should end up with (D).
7. Solve x2 + 2x = -10 for x and express it in a + bi form.
Correct Answer:
-1 ± 3i
Answer Explanation:
If the equation is rearranged so that a = 1, b = 2, and c = 10, these values can be plugged into the quadratic formula, which gives us , or In its simplest form, the roots are -1 ± 3i.
8. Solve for x and express it in a + bi form.
Correct Answer:
Answer Explanation:
Before we dive headfirst into the quadratic equation, we'll want to rearrange the equation we're given so that it equals zero. Once we do that, we'll have 2x2 + x + 7 = 0. With a = 2, b = 1, and c = 7, we can relish in the beauty that is the quadratic formula, which will give us (C) as the right answer.
9. Which of the following equations has roots of 3 ± i?
Correct Answer:
x2 – 6x + 10
Answer Explanation:
Yes, this question means what you think it means. Fortunately, all we need to do is calculate the discriminant of each equation because as it turns out, only one of them is negative. If we calculate them all, we should end up with -36 for (A), 76 for (B) and (D), and 241 for (C). Since the root is imaginary, our only possible answer is (A).
10. Imaginary roots always come in "conjugate pairs." That means that if the answers are imaginary, we'll get both conjugates as answers. Based on your observations, why do you think it happens?
Correct Answer:
Because the only difference between the two answers is the ± sign in the formula
Answer Explanation:
The ± sign in the quadratic formula is the reason for the conjugate pairs being the answer. For instance, if a parabola has imaginary roots 3 ± 3i, its roots are actually 3 + 3i and 3 – 3i. The two roots are because of that ± sign in the quadratic formula itself.
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# Difference between revisions of "2018 AMC 10A Problems/Problem 21"
## Problem
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
## Solution 1
Substituting $y=x^2-a$ into $x^2+y^2=a^2$, we get $$x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0$$ Since this is a quartic, there are 4 total roots (counting multiplicity). We see that $x=0$ always at least one intersection at $(0,-a)$ (and is in fact a double root).
The other two intersection points have $x$ coordinates $\sqrt{2a-1}$. We must have $2a-1> 0,$ otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point $(0,a)$). This only results in a single intersection point in the real coordinate plane. Thus, we see $a>\frac{1}{2}$.
(projecteulerlover)
## Solution 2
$[asy] Label f; f.p=fontsize(6); xaxis(-2,2,Ticks(f, 0.2)); yaxis(-2,2,Ticks(f, 0.2)); real g(real x) { return x^2-1; } draw(graph(g, 1.7, -1.7)); real h(real x) { return sqrt(1-x^2); } draw(graph(h, 1, -1)); real j(real x) { return -sqrt(1-x^2); } draw(graph(j, 1, -1)); [/asy]$
Looking at a graph, it is obvious that the two curves intersect at (0, -a). We also see that if the parabola go's 'in' the circle, than by going out of it (as it will) it will intersect five times, an impossibility. Thus we only look for cases where the parabola becomes externally tangent to the circle. We have $x^2 - a = -\sqrt(a^2 - x^2)$. Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$. Since x = 0 is already accounted for, we only need to find 1 solution for $x^2 = 2a - 1$, where the right hand side portion is obviously increasing. Since a = 1/2 begets x = 0 (an overcount), we have $a > 1/2 -> E$ is the right answer.
Solution by JohnHankock
## Solution 3
This describes a unit parabola, with a circle centered at the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is 2, the radius of the circle that matches it has a radius of $\frac{1}{2}$. This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only 1 point. When a larger circle is used, it is tangent to 3 points because the points on either side are now separated from the vertex. Therefore, $\boxed{a > \frac{1}{2}}$ or $\boxed{E}$ is correct.
$QED \blacksquare$
## Solution 4
Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$, and they have symmetry across the y-axis, thus, for them to intersect at exactly 3 points, it suffices to find the y solution.
First, rewrite the second equation to $y=x^2-a\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$. This value must not be less than 0 (It is the square root part of the quadratic formula). To find when it is 0, we find the roots: $$4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}$$ Since $\lim_{a\to \infty}(4a^2-4a+1)=\infty$, our range is $\boxed{a>\frac{1}{2}}$
Solution by ktong
## Solution 5 (Cheating with Answer Choices)
Simply plug in $a = 0, \frac{1}{2}, \frac{1}{4}, 1$ and solve the systems. (This shouldn't take too long.) And realized that only $a=1$ yields three real solutions for $x$, so we are done and the answer is $\boxed{a>\frac{1}{2}}$
~ ccx09
## Solution 6 (Even Simpler Cheating with Answer Choices)
Simply consider (or graph) the "trivial" case where $a = 1$ and it is clear that the answer is $\boxed{E}$.
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# How do you calculate the standard deviation of differences?
## How do you calculate the standard deviation of differences?
Calculating Standard Deviation First, take the square of the difference between each data point and the sample mean, finding the sum of those values. Then, divide that sum by the sample size minus one, which is the variance. Finally, take the square root of the variance to get the SD.
## How do you find the standard deviation of the difference between two sets of data?
1. Step 1: Find the mean.
2. Step 2: Subtract the mean from each score.
3. Step 3: Square each deviation.
4. Step 4: Add the squared deviations.
5. Step 5: Divide the sum by one less than the number of data points.
6. Step 6: Take the square root of the result from Step 5.
sigma σ
## What is standard deviation used for in statistics?
Standard deviation is a number used to tell how measurements for a group are spread out from the average (mean or expected value). A low standard deviation means that most of the numbers are close to the average, while a high standard deviation means that the numbers are more spread out.
## Is standard deviation The square root of variance?
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.
## Does adding a constant change the standard deviation?
When adding or subtracting a constant from a distribution, the mean will change by the same amount as the constant. The standard deviation will remain unchanged. This fact is true because, again, we are just shifting the distribution up or down the scale. We do not affect the distance between values.
## Can you multiply standard deviation by a constant?
So the variance and standard deviation of A and B are both the same; they are 2 and square root of 2, respectively. Also, multiplying each score in a sample or population by a constant factor will multiply the standard deviation by that same factor.
## Does standard deviation change with sample size?
The population mean of the distribution of sample means is the same as the population mean of the distribution being sampled from. ... Thus as the sample size increases, the standard deviation of the means decreases; and as the sample size decreases, the standard deviation of the sample means increases.
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# What The Numbers Say About Jane Krakowski
How will Jane Krakowski do on 09/11/2019 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is just for fun – take it with a grain of salt. I will first calculate the destiny number for Jane Krakowski, and then something similar to the life path number, which we will calculate for today (09/11/2019). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology practitioners.
PATH NUMBER FOR 09/11/2019: We will take the month (09), the day (11) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 09 and add the digits together: 0 + 9 = 9 (super simple). Then do the day: from 11 we do 1 + 1 = 2. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 9 + 2 + 12 = 23. This still isn’t a single-digit number, so we will add its digits together again: 2 + 3 = 5. Now we have a single-digit number: 5 is the path number for 09/11/2019.
DESTINY NUMBER FOR Jane Krakowski: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Jane Krakowski we have the letters J (1), a (1), n (5), e (5), K (2), r (9), a (1), k (2), o (6), w (5), s (1), k (2) and i (9). Adding all of that up (yes, this can get tedious) gives 49. This still isn’t a single-digit number, so we will add its digits together again: 4 + 9 = 13. This still isn’t a single-digit number, so we will add its digits together again: 1 + 3 = 4. Now we have a single-digit number: 4 is the destiny number for Jane Krakowski.
CONCLUSION: The difference between the path number for today (5) and destiny number for Jane Krakowski (4) is 1. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But this is just a shallow analysis! As mentioned earlier, this is not at all guaranteed. If you want to see something that we really strongly recommend, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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# Multi-digit subtraction: 389,002-76,151
Subtract 389,002-76,151 using the standard algorithm.
## Want to join the conversation?
• is there any more methods
• Yes! There’s a Vedic (Indian) method of subtraction.
We subtract one column at a time. When the bottom number is bigger we subtract in the reverse order but put a bar over the digit in the answer. Bar digits can be thought of as negative digits!
We then convert from bar digits to a normal number using the following steps:
1) For each group of one or more 0’s (if any) immediately to the left of a group of one or more bar digits, we also put bars on those 0’s.
2) For each group of bar digits after the previous step:
a) We subtract “all from 9 except last from 10”.
b) We take away 1 from the digit immediately to the left of the group of bar digits.
This might seem hard to understand at first, but it should become easier to understand when you read the example below.
Example: 676,354,508 - 146,378,168.
We could go right to left, or left to right. Let’s go left to right.
From left to right:
6-1 = 5.
7-4 = 3.
6-6 = 0.
3-3 = 0.
5-7 = 2bar (because 7-5 = 2).
4-8 = 4bar (because 8-4 = 4).
5-1 = 4.
0-6 = 6bar (because 6-0 = 6).
8-8 = 0.
So we have 5 3 0 0 2bar 4bar 4 6bar 0.
We now need to convert to a normal number.
1) We have a group of two 0’s immediately to the left of the group of bar digits 2bar 4bar. So we put bars on these two 0’s.
So we now have 5 3 0bar 0bar 2bar 4bar 4 6bar 0.
2) We have the two groups of bar digits 0bar 0bar 2bar 4bar, and 6bar.
For the group 0bar 0bar 2bar 4bar:
a) 9-0 = 9.
9-0 = 9.
9-2 = 7.
10-4 = 6.
b) The 3 immediately to the left of this group becomes a 2.
For the group 6bar:
a) 10-6 = 4.
b) The 4 immediately to the left of this group becomes a 3.
Note that the 5 on the far left and the 0 on the far right both stay as is.
The final answer is 529,976,340.
Have a blessed, wonderful day!
• this is more confusing than addition
• I must admit, yes at times it is more confusing than addition but if you can get the methods and get used to it, it is really really easy
• Is there any more methods
• Of courses, this is one of the many methods.
• how do you come up with this stuff
• What stuff? How to subtract? There are a number of ways to subtract multi-digit numbers. (You can see Ian Pulizzotto's comment for more).
• are you a bot or a human
• his voice makes me want to cry and run away. His voice is woderful it makes me feel like i'm being yelled at, but in a good way. i'm way better at typing then math, math makes my head hurt literaly
• I like this man's videos they are helpful!:)
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# Fast computation of integer powers
## Introduction
How many multiplications are needed to calculate $$x^6$$? The naïve way to do it is $$x^6 = x\cdot x\cdot x\cdot x\cdot x\cdot x$$. That is five multiplications. But we can do better: $$(x\cdot x\cdot x)^2$$ only requires three. Using as few operations as possible is important for the efficient evaluation of expressions on a computer. Is there a general way to find the smallest number of multiplications needed to compute any given power $$n$$?
This problem is equivalent to finding shortest addition chains. An addition chain is a sequence of integers such that each one can be expressed as the sum of two prior elements of the chain. For example, 23 can be constructed through an addition chain of only 6 elements:
\begin{align*} \phantom{1}2 &= \phantom{1}1 + \phantom{1}1 \\ \phantom{1}3 &= \phantom{1}2 + \phantom{1}1 \\ \phantom{1}5 &= \phantom{1}3 + \phantom{1}2 \\ 10 &= \phantom{1}5 + \phantom{1}5 \\ 20 &= 10 + 10 \\ 23 &= 20 + \phantom{1}3 \end{align*}
Correspondingly, $$x^{23}$$ can be computed using 6 multiplications:
\begin{align*} x^2 &= x \cdot x \\ x^3 &= x^2 \cdot x \\ x^5 &= x^3 \cdot x^2 \\ x^{10} &= x^5 \cdot x^5 \\ x^{20} &= x^{10} \cdot x^{10} \\ x^{23} &= x^{20} \cdot x^3 \end{align*}
Notice that we used the already computed value of $$x^3$$ not once, but twice: for $$x^5$$ and for $$x^{23}$$.
Computing the shortest addition chain for a given integer $$n$$ is difficult, and is shown to be an NP-complete problem. The minimal length is given by OEIS A003313. Luckily, there are a few simple and practical ways to generate non-optimal but still short addition chains that we can use for computing powers.
### Practical applications
Some computer languages, like Fortran, often implement such optimizations for computing integer powers. Others, like C and C++, do not have a builtin exponentiation operator at all. To compute general powers in C, we can resort to the pow(base, exp) standard library function. This, however, is designed to work with arbitrary real exponents and is usually much slower than direct multiplication. In fact, on my computer one call to std::pow() with doubles takes roughly the same amount of time as 45 (!) elementary double multiplications.
Once realizing this, many people will write small utility functions for efficiently computing specific powers, e.g.,
inline double pow4(double x) {
double x2 = x*x;
return x2*x2;
}
Doing this for each power is tedious. Below I show a general implementation using C++ templates that generates a short (though not always shortest) addition chain for computing any positive integer power.
template<unsigned N> struct power_impl;
template<unsigned N> struct power_impl {
template<typename T>
static T calc(const T &x) {
if (N%2 == 0)
return power_impl<N/2>::calc(x*x);
else if (N%3 == 0)
return power_impl<N/3>::calc(x*x*x);
return power_impl<N-1>::calc(x)*x;
}
};
template<> struct power_impl<0> {
template<typename T>
static T calc(const T &) { return 1; }
};
template<unsigned N, typename T>
inline T power(const T &x) {
return power_impl<N>::calc(x);
}
To compute e.g. $$x^6$$, we can now simply call power<6>(x). Modern optimizing compilers will inline all recursive calls within the implementation of power() and produce efficient machine code similar to what we would have gotten from writing temp = x*x*x; result = temp*temp; directly.
How does the code above work? It breaks down the computation recursively:
• For $$n=2k$$, it computes $$x^n = y\cdot y,\;\, y = x^{n/2}$$, then continues with $$y = x^{n/2}$$ recursively.
• For $$n=3k$$, it computes $$x^n = y\cdot y\cdot y,\;\, y = x^{n/3}$$.
• Otherwise it computes $$x^n = x\cdot y,\;\, y = x^{(n-1)}$$.
This recursive breakdown does not always generate the optimal solution, but it is not much less efficient either. The following table shows the 17 cases of $$1 \le n \le 100$$ for which the simple recursive algorithm uses more multiplications than the optimal one. In all but one case it generates only one additional multiplication compared to the optimal solution.
n optimal simple
-----------------------
23 6 7
33 6 7
43 7 8
46 7 8
47 8 9
59 8 9
66 7 8
67 8 9
69 8 9
77 8 9
83 8 9
85 8 9
86 8 9
92 8 9
94 9 10
95 9 11
99 8 9
In those cases where we need to squeeze out every bit of performance, we can still provide specializations for the cases where the simple recursive solution is not the shortest possible. For example,
template<> struct power_impl<23> {
template<typename T>
static T calc(const T &x) {
T x2 = x*x;
T x3 = x2*x;
T x5 = x3*x2;
T x10 = x5*x5;
T x20 = x10*x10;
return x20*x3;
}
};
Personally, I have not yet encountered the need to quickly compute 23rd or higher powers in an inner loop, so I do not include these specializations in my code.
For those who do need it, there are tables of optimal solutions, as well as a lot of other information on addition chains on Achim Flammenkamp’s website.
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# How do you simplify sqrt180/sqrt9?
May 14, 2018
#### Explanation:
For positive m,n $\sqrt{m \cdot n} = \sqrt{m} \cdot \sqrt{n}$.
So then we have $\frac{\sqrt{180}}{\sqrt{9}} = \frac{\sqrt{9} \sqrt{20}}{\sqrt{9}} = \sqrt{20} = \sqrt{4} \sqrt{5} = 2 \sqrt{5}$
May 14, 2018
$2 \sqrt{5}$
#### Explanation:
$\text{using the "color(blue)"laws of radicals}$
•color(white)(x)sqrta/sqrtbhArrsqrt(a/b)
•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)
$\Rightarrow \frac{\sqrt{180}}{\sqrt{9}} = \sqrt{\frac{180}{9}} = \sqrt{20}$
$\Rightarrow \sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2 \sqrt{5}$
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Line Graphs
## Display given data as linear change on coordinate graphs.
Estimated7 minsto complete
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Practice Line Graphs
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Line Graphs
### [Figure1] License: CC BY-NC 3.0
For Jose's economics class he wants make a graph that illustrates the number of persons employed in the U.S. who were 16 years and older in 2011. He has organized his data into the following table.
Month Num. of People Jan 153250 Feb 153302 Mar 153392 Apr 153420 May 153700 Jun 153409 Jul 153358 Aug 153674 Sep 154004 Oct 154057 Nov 153937 Dec 153887
How can Jose make a graph to illustrate the data over time?
In this concept, you will learn how to create and read line graphs.
### Guidance
Data is a set of numerical or non-numerical information. Data can be analyzed in many different ways. In this concept you will analyze numerical data using line graphs.
One way to display data is in a line graph. A line graph shows the relationship between independent and dependent values of data, and are usually used to show trends over time. In the graph each data value is represented by a point in the graph that are connected by a line. The independent variable is listed along the horizontal, or x, axis and the quantity or value of the data is listed along the vertical, or y, axis.
Let's look at an example.
Kelsey works at an arboretum and tracks tree growth over time. Kelsey collected the growth data of one tree over five years and organized her data into the table below. Create a line graph that represents the data over time. Then state two conclusions about the data.
Year Size of Tree (in feet) 2003 2 2004 2.5 2005 3.5 2006 8.5 2007 14
First, create the line graph. To do this, draw the horizontal (x) and vertical (y) axes.
Next, label the vertical axis. The vertical axis lists the dependent variable and represents the quantity of the data. In this case, the dependent variable is feet and the label will also be "feet."
Next, title the graph. The title of the graph should be short and clear. It should explain what data is presented in the graph. In this case, the title will be “Tree Growth.”
Then, determine the units on the vertical axis. To do this, start by reviewing the smallest and largest values in the table. The smallest value is 2 and the largest is 14. Based on these values label the vertical axis from 0-16. Since the values are whole numbers and are relatively spread out, a unit of 2 can be used. Therefore, the vertical axis will start at 0 and go to 16 by increments of 2.
Next, draw the data points. To do this, write the years along the horizontal axis, leaving space between each. Each year will have one point representing the height of the tree. To start, draw a point for the the height in 2003. To do this find 2003 on the horizontal axis and go up until 2, then draw a point. Then draw the point for the height in 2004. Continue drawing the points for all the years.
Next, draw the line. To do this, start with the point on the far left of the graph and connect the points with one line from 2003 to 2007.
Then, state two conclusions from the graph. To do this, analyze the graph by comparing the steepness of the line between each point.
The first answer is the graph should look like the one below. The second answer is two conclusions that can be made from the graph are: the greatest amount of growth occurred between 2006 and 2007 and from 2005 to 2006 the tree's height more than doubled.
### Guided Practice
The line graph below shows the temperature over seven days in July in Missouri. List two conclusions that can be made from the graph.
First, analyze the graph by comparing the shape of the line between the points.
The answer is two conclusions can be made from the graph: three dates, July 20-22, all had the same temperature of 86 degrees and the temperature is rising from July 22 to July 24.
### Examples
#### Example 1
The population for the city of Los Angeles is organized in the table below. Create a line graph to that represents the population change over time. Then state one conclusion about the data.
Year: Approximate Population (in millions):
1950 2
1960 2.5
1970 2.8
1980 3
1990 3.5
2000 3.7
First, make the line graph. To do this, draw the horizontal \begin{align*}(x)\end{align*} and vertical \begin{align*}(y)\end{align*} axes.
Next, label the horizontal axis. The horizontal axis lists the independent variable. In this case, the independent variable is the year and the axis will be labeled "Year."
Next, label the vertical axis. The vertical axis lists the dependent variable and represents the quantity of the data. In this case, the dependent variable is the population and the label will be "Population (in millions)."
Next, title the graph. The title of the graph should be short and clear. It should explain what data is presented in the graph. In this case, the title will be “Population of Los Angeles (in millions).”
Then, determine the units on the vertical axis. To do this, start by reviewing the smallest and largest values in the table. The smallest value is 2 and the largest is 3.7. Based on these values label the vertical axis from 0-4. Since the values include decimals and are relatively close, a unit of 0.5 can be used. Therefore, the vertical axis will start at 0 and go to 4 by increments of 0.5.
Next, draw the data points. To do this, write the years along the horizontal axis, leaving space between each. Each year will have one point representing the population. To start, draw a point for the the population in 1950. To do this find 1950 on the horizontal axis and go up until 2, then draw a point. Next, draw the point for the population in 1960. Continue this pattern for all the years.
Next, draw the line. To do this, start with the point on the far left of the graph and connect the points with one line from 1950 to 2000.
Then, state one conclusion from the graph. To do this, analyze the graph by comparing the steepness of the line between the points.
The first answer is the graph should look like the one below. The second answer is one conclusion that can be made from the graph is: the population of Los Angeles has increased from the years 1950 to 2000 and it is estimated that it will continue to increase for the future.
#### Example 2
The Electronic Energies Alliance recorded the average cost of one gallon of gasoline in the United States for the years 2000-2007. The graph below represents their data. List three statements about the data.
First, analyze the graph by comparing the shape of the line.
The answer is three statements can be made about the graph: there was a drop in the price of gas from 2000 to 2001 and again from 2006 to 2007, the price of gas increased from the years 2001 to 2006, and the price of gas increased the most from 2005 to 2006.
#### Example 3
The graph below illustrates the absolute minimum temperature in Dubai in 2005 in celsius. List two statements that can be made about the data.
First, analyze the graph by comparing the shape of the line.
The answer is two statements can be made about the graph: the greatest minimum temperature occurred in September and was approximately 31 degrees celsius, and the absolute minimum temperatures increased from the months of February to September.
Remember Jose and his economics class?
Jose needs to create a graph that illustrates the number of people employed in the U.S. who were 16 years or older in 2011. Jose's data is presented in the table below.
Month Num. of People Jan 153250 Feb 153302 Mar 153392 Apr 153420 May 153700 Jun 153409 Jul 153358 Aug 153674 Sep 154004 Oct 154057 Nov 153937 Dec 153887
First, draw the horizontal\begin{align*}(x)\end{align*}and vertical\begin{align*}(y)\end{align*}axes.
Next, label the horizontal axis. The horizontal axis lists the independent variable. In this case, the independent variable is the month and the axis will be labeled "Month."
Next, label the vertical axis. The vertical axis lists the dependent variable and represents the quantity of the data. In this case, the dependent variable is the number of persons employed and the label will be "Number of Persons."
Next, title the graph. The title of the graph should be short and clear. It should explain what data is presented in the graph. In this case, the title will be “Number of Persons Employed in the U.S. Aged 16+ Years in 2011.”
Then, determine the units on the vertical axis. To do this, start by reviewing the smallest and largest values in the table. The smallest value is 153250 and the largest is 154057. Based on these values label the vertical axis from 153000 to 154250. Since the values are whole numbers and are relatively close, a unit of 250 can be used. Therefore, the vertical axis will start at 153000 and go to 154250 by increments of 250.
Next, draw the data points. To do this, write the months along the horizontal axis, leaving space between each. Each month will have one point representing the number of people employed. To start, draw a point for the number of people employed in January. To do this find January on the horizontal axis and go up until 153250, then draw a point. Next, draw the point for the people employed in February. Continue this pattern for all months.
Next, draw the line. To do this, start with the point on the far left of the graph and connect the points with one line from January to December.
The graph should look like the one below.
### Explore More
Use this line graph to answer the following questions.
The vertical axis shows the number of vegetables harvested each year. This is recorded as vegetable growth. The horizontal axis shows the years vegetable growth was recorded.
1. How many vegetables were harvested in 2005?
2. How many vegetables were harvested in 2006?
3. What is the difference in growth from 2005 to 2006?
4. How many vegetables were harvested in 2007?
5. What is the difference in vegetable growth from 2006 to 2007?
6. What is the vegetable growth in 2008?
7. What is the difference in vegetable growth from 2005 to 2008?
8. If the vegetable growth follows the same pattern from 2008 to 2011, what will the new total be?
9. If there is a loss of 50 vegetables from 2008 to 2009, what will the new total be?
10. If there is a gain of 100 vegetables from 2008 to 2009, what will the new total be?
11. If there is a loss of 50% from 2008 to 2009, what will the new total be?
12. True or false. A bar graph shows the same data as a line graph?
13. True or false. A line graph must show how data changes over time.
14. - 15. Use a newspaper to find two line graphs. The business section is a good place to start. Examine the data and explain what the line graph represents to a friend.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 11.7.
### Vocabulary Language: English
bar chart
bar chart
A bar chart is a graphic display of categorical variables that uses bars to represent the frequency of the count in each category.
broken line graph
broken line graph
A broken line graph is a graph that is used to show changes over time. A line is used to join the values but the line has no defined slope.
continuous variables
continuous variables
A continuous variable is a variable that takes on any value within the limits of the variable.
data set
data set
A collection of these observations of the variable is a data set.
dependent variable
dependent variable
The dependent variable is the output variable in an equation or function, commonly represented by $y$ or $f(x)$.
discrete random variables
discrete random variables
Discrete random variables represent the number of distinct values that can be counted of an event.
independent variable
independent variable
The independent variable is the input variable in an equation or function, commonly represented by $x$.
Line Graph
Line Graph
A line graph is a visual way to show how data changes over time.
qualitative variable
qualitative variable
A qualitative variable is one that cannot be measured numerically but can be placed in a category.
quantitative variable
quantitative variable
A quantitative variable is a variable that takes on numerical values that represent a measurable quantity. Examples of quantitative variables are the height of students or the population of a city.
variable
variable
In statistics, a variable is simply a characteristic that is being studied.
1. [1]^ License: CC BY-NC 3.0
2. [2]^ License: CC BY-NC 3.0
3. [3]^ License: CC BY-NC 3.0
4. [4]^ License: CC BY-NC 3.0
5. [5]^ License: CC BY-NC 3.0
6. [6]^ License: CC BY-NC 3.0
7. [7]^ License: CC BY-NC 3.0
8. [8]^ License: CC BY-NC 3.0
9. [9]^ License: CC BY-NC 3.0
### Explore More
Sign in to explore more, including practice questions and solutions for Line Graphs.
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# Difference Of Perfect Squares
In this video, we are going to look at how to understand the difference of perfect squares.
For example:
When given something like $x^2-9$, we can rewrite it as
$x^2+0x-9$
To factor this, we will need two numbers that add up to 0 and multiply to -9. These numbers are 3 and -3. Therefore when we factor this, we will get
$(x+3)(x-3)$
Generally, whenever we have $x^2-b^2$ we will always be able to factor it as
$(x+b)(x-b)$
Or
$x^2-number$
will always be factored as
$(x+\sqrt{number})(x-\sqrt{number})$
For expressions more complex, such as
$4x^2-121y^4$
First, take the square root of the first term and place it in the front of each set of parentheses, and then take the square root of the second term and place it in the two parentheses with a plus or minus sign. This will be factored as
$(2x-11y^2)(2x+11y^2)$
## Video-Lesson Transcript
Let’s go over the difference of perfect squares.
Let’s have some examples.
$x^2 - 9$
$x^2 - 16$
$x^2 - 25$
The difference is associated with subtraction. That’s why we have a subtraction sign here.
Perfect sqaures are numbers that are the squareroot of.
The difference of perfect squares means two perfect squares to be subtracted.
Let’s take a look at how to factor these.
$x^2 - 9$
$x^2 + 0x - 9$
$(x + 3) (x - 3)$
Next, $x^2 - 16$
$x^2 + 0x - 16$
$(x + 4) (x - 4)$
And $x^2 - 25$
$x^2 + 0x - 25$
$(x + 5) (x - 5)$
In general, the form is:
$x^2 - b$
$(x + b) (x - b)$
So if we have a perfect square
$x^2 - \#$
$(x + \sqrt{\#}) (x - \sqrt{\#})$
So if we have $x^2 - 81$
$(x + 9) (x - 9)$
And of course, this can be switched into
$(x - 9) (x + 9)$
It doesn’t make any difference.
This rule applies to all perfect squares.
For example:
$4x^2 - 121y^4$
Let’s find the square root of the first term then the square root of the second term.
$(2x - 11y^2) (2x + 11y^2)$
Look at this previous example:
$x^2 - 9$
Let’s draw two parenthesis and put an $x$ inside each parenthesis.
If you solve that, the square root of $x^2$ is $x$.
Then the squareroot of $9$ is $3$.
$(x + 3) (x - 3)$
So for $4x^2 - 121y^4$, we’ll find the squareroot of each term.
The squareroot of $4$ is $2$. The squareroot of $x^2$ is $x$. Then the squareroot of $121$ is $11$. Lastly, the squareroot of $y^4$ is $y^2$.
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# Difference between revisions of "2015 AMC 8 Problems/Problem 16"
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\tfrac{1}{3}$ of all the ninth graders are paired with $\tfrac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
$\textbf{(A) } \frac{2}{15} \qquad \textbf{(B) } \frac{4}{11} \qquad \textbf{(C) } \frac{11}{30} \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{11}{15}$
## Solution 1
Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $\frac{n}{3}=\frac{2s}{5}$, which simplifies to $n=\frac{6s}{5}$. Since we are trying to find the value of $\frac{\frac{n}{3}+\frac{2s}{5}}{n+s}$, we can just substitute $n$ for $\frac{6s}{5}$ into the equation. We then get a value of $\frac{\frac{\frac{6d}{5}}{3}{4}$ (Error compiling LaTeX. ! File ended while scanning use of \frac .) $\boxed{\textbf{(B)}~\frac{4}{11}}$
## Solution 2
We see that the minimum number of ninth graders is $6$, because if there are $3$ then there is $1$ ninth grader with a buddy, which would mean $2.5$ sixth graders with a buddy, and that's impossible. With $6$ ninth graders, $2$ of them are in the buddy program, so there $\frac{2}{\tfrac{2}{5}}=5$ sixth graders total, two of whom have a buddy. Thus, the desired probability is $\frac{2+2}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}$.
2015 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Slope of a Line Using Two Points
## Find the slope of a line when two points have been graphed on the coordinate plane.
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Finding the Slope and Equation of a Line
The grade, or slope, of a road is measured in a percentage. For example, if a road has a downgrade of 7%, this means, that over every 100 horizontal feet, the road will slope down 7 feet vertically.
If a highway has a downgrade of 12% for 3 miles (5280 feet in a mile), how much will the road drop? What is the slope of this stretch of highway?
### Finding the Slope and Equation of a Line
The slope of a line determines how steep or flat it is. When we place a line in the coordinate plane, we can measure the slope, or steepness, of a line. Recall the parts of the coordinate plane, also called a xy\begin{align*}x-y\end{align*} plane and the Cartesian plane, after the mathematician Descartes.
To plot a point, order matters. First, every point is written (x,y),\begin{align*}(x, y),\end{align*} where x\begin{align*}x\end{align*} is the movement in the x\begin{align*}x-\end{align*}direction and y\begin{align*}y\end{align*} is the movement in the y\begin{align*}y-\end{align*}direction. If x\begin{align*}x\end{align*} is negative, the point will be in the 2nd\begin{align*}2^{nd}\end{align*} or 3rd\begin{align*}3^{rd}\end{align*} quadrants. If y\begin{align*}y\end{align*} is negative, the point will be in the 3rd\begin{align*}3^{rd}\end{align*} or 4th\begin{align*}4^{th}\end{align*} quadrants. The quadrants are always labeled in a counter-clockwise direction and using Roman numerals.
The point in the 4th\begin{align*}4^{th}\end{align*} quadrant would be (9, -5).
To find the slope of a line or between two points, first, we start with right triangles. Let’s take the two points (9, 6) and (3, 4). Plotting them on a xy\begin{align*}x-y\end{align*} plane, we have:
To turn this segment into a right triangle, draw a vertical line down from the higher point, and a horizontal line from the lower point, towards the vertical line. Where the two lines intersect is the third vertex of the slope triangle.
Now, count the vertical and horizontal units along the horizontal and vertical sides (\begin{align*}{\color{red}{\mathbf{red}}}\end{align*} dotted lines).
The slope is a fraction with the vertical distance over the horizontal distance, also called the “rise over run.” Because the vertical distance goes down, we say that it is -2. The horizontal distance moves towards the negative direction (the left), so we would say that it is -6. So, for slope between these two points, the slope would be \begin{align*}\frac{-2}{-6}\end{align*} or \begin{align*}\frac{1}{3}\end{align*}.
Note: You can also draw the right triangle above the line segment.
Now, let's find the slope of the following lines.
1. Use a slope triangle to find the slope of the line below.
Notice the two points that are drawn on the line. These are given to help you find the slope. Draw a triangle between these points and find the slope.
From the slope triangle above, we see that the slope is \begin{align*}\frac{-4}{4} = -1\end{align*}.
Whenever a slope reduces to a whole number, the “run” will always be positive 1. Also, notice that this line points in the opposite direction as the line segment above. We say this line has a negative slope because the slope is a negative number and points from the \begin{align*}2^{nd}\end{align*} to \begin{align*}4^{th}\end{align*} quadrants. A line with positive slope will point in the opposite direction and point between the \begin{align*}1^{st}\end{align*} and \begin{align*}3^{rd}\end{align*} quadrants.
1. If we go back to our previous example with points (9, 6) and (3, 4), we can find the vertical distance and horizontal distance another way.
From the picture, we see that the vertical distance is the same as the difference between the \begin{align*}y-\end{align*}values and the horizontal distance is the difference between the \begin{align*}x-\end{align*}values. Therefore, the slope is \begin{align*}\frac{6-4}{9-3}\end{align*}. We can extend this idea to any two points, \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}(x_2, y_2)\end{align*}.
Slope Formula: For two points \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}(x_2, y_2),\end{align*} the slope between them is \begin{align*}\frac{y_2 - y_1}{x_2 - x_1}\end{align*}. The symbol for slope is \begin{align*}m\end{align*}.
It does not matter which point you choose as \begin{align*}(x_1, y_1)\end{align*} or \begin{align*}(x_2, y_2)\end{align*}.
Let's find the slope of the following lines using the Slope Formula.
1. Find the slope between (-4, 1) and (6, -5).
Set \begin{align*}(x_1, y_1) = (-4, 1)\end{align*} and \begin{align*}(x_2, y_2) = (6, -5)\end{align*}.
\begin{align*}m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 -(-4)}{-5 - 1} = \frac{10}{-6} = -\frac{5}{3}\end{align*}
1. Find the slope between (9, -1) and (2, -1).
Set \begin{align*}(x_1, y_1) = (9, -1)\end{align*} and \begin{align*}(x_2, y_2) = (2, -1)\end{align*}.
\begin{align*}m = \frac{-1 - (-1)}{2 - 9} = \frac{0}{-7} = 0\end{align*}
Here, we have zero slope. Plotting these two points we have a horizontal line. This is because the \begin{align*}y-\end{align*}values are the same. Anytime the \begin{align*}y-\end{align*}values are the same we will have a horizontal line and the slope will be zero.
### Examples
#### Example 1
Earlier, you were asked to find how much the road will drop and the slope of the stretch of highway.
The road slopes down 12 feet over every 100 feet.
Let's set up a ratio to find out how much the road slopes in 3 miles, or \begin{align*}3 \cdot 5280 = 15,840\end{align*} feet.
\begin{align*}\frac{12}{100}&= \frac{x}{15,840} \\ 15840 \cdot \frac{12}{100} &= x \\ x &= 1900.8 \end{align*}
The road drops 1900.8 feet over the 3 miles. The slope of the road is \begin{align*}\frac{12}{100}\end{align*} or \begin{align*}\frac{3}{25}\end{align*} when the fraction is reduced.
#### Example 2
Use a slope triangle to find the slope of the line below.
Counting the squares, the vertical distance is down 6, or -6, and the horizontal distance is to the right 8, or +8. The slope is then \begin{align*}\frac{-6}{8}\end{align*} or \begin{align*}-\frac{2}{3}\end{align*}.
#### Example 3
Find the slope between (2, 7) and (-3, -3).
Use the Slope Formula. Set \begin{align*}(x_1, y_1) = (2, 7)\end{align*} and \begin{align*}(x_2, y_2) = (-3, -3)\end{align*}.
\begin{align*}m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 -7}{-3 -2} = \frac{-10}{-5} = 2\end{align*}
#### Example 4
Find the slope between (-4, 5) and (-4, -1).
Again, use the Slope Formula. Set \begin{align*}(x_1, y_1) = (-4, 5)\end{align*} and \begin{align*}(x_2, y_2) = (-4, -1)\end{align*}.
\begin{align*}m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 -5}{-4 -(-4)} = \frac{-6}{0}\end{align*}
You cannot divide by zero. Therefore, this slope is undefined. If you were to plot these points, you would find they form a vertical line. All vertical lines have an undefined slope.
Important Note: Always reduce your slope fractions. Also, if the numerator or denominator of a slope is negative, then the slope is negative. If they are both negative, then we have a negative number divided by a negative number, which is positive, thus a positive slope.
### Review
Find the slope of each line by using slope triangles.
Find the slope between each pair of points using the Slope Formula.
1. (-5, 6) and (-3, 0)
2. (1, -1) and (6, -1)
3. (3, 2) and (-9, -2)
4. (8, -4) and (8, 1)
5. (10, 2) and (4, 3)
6. (-3, -7) and (-6, -3)
7. (4, -5) and (0, -13)
8. (4, -15) and (-6, -11)
9. (12, 7) and (10, -1)
10. Challenge The slope between two points \begin{align*}(a, b)\end{align*} and (1, -2) is \begin{align*}\frac{1}{2}\end{align*}. Find \begin{align*}a\end{align*} and \begin{align*}b\end{align*}.
To see the Review answers, open this PDF file and look for section 2.1.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Slope
Slope is a measure of the steepness of a line. A line can have positive, negative, zero (horizontal), or undefined (vertical) slope. The slope of a line can be found by calculating “rise over run” or “the change in the $y$ over the change in the $x$.” The symbol for slope is $m$
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Precalculus (6th Edition) Blitzer
The maximum area of the triangle is $100$ square inches.
Let $x$ be the height of the triangle. Therefore, the base is $40-2x$. The area of the triangle is: \begin{align} & \text{=}\frac{\text{1}}{\text{2}}\left( \text{base} \right)\left( \text{height} \right) \\ & =\frac{1}{2}x\left( 40-2x \right) \\ & =-{{x}^{2}}+20x \end{align} Which is a quadratic equation with $a<0$ and thus will have a maximum at $x=-\frac{b}{2a}$: \begin{align} & x=-\frac{20}{2\left( -1 \right)} \\ & x=10 \end{align} The height of the triangle is $10$ inches. Thus, the maximum area of the triangle is: \begin{align} & -{{x}^{2}}+20x=-{{\left( 10 \right)}^{2}}+20\left( 10 \right) \\ & =-100+200 \\ & =100\text{ sq}\text{.inches} \end{align} The maximum area of the triangle is $100$ square inches.
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# Ellipses and Circles
• Dec 7th 2010, 02:00 AM
rickrishav
Ellipses and Circles
An ellipse is drawn with maximum radius 4 cm and minimum radius 3 cm. A circle is drawn, as shown in the figure, such that any two tangents of the ellipse that meet on the circle make 90⁰ with each other. Find the radius of the circle.
http://i876.photobucket.com/albums/ab324/rishavd/a.png
• Dec 8th 2010, 01:58 AM
Opalg
Quote:
Originally Posted by rickrishav
An ellipse is drawn with maximum radius 4 cm and minimum radius 3 cm. A circle is drawn, as shown in the figure, such that any two tangents of the ellipse that meet on the circle make 90⁰ with each other. Find the radius of the circle.
Here is an outline of one way to approach this problem. I'll work with the ellipse $\displaystyle \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$. For the final result you'll want to take a=4 and b=3.
A line through the point $\displaystyle (x_0,y_0)$, with slope $\displaystyle \lambda$, has equation $\displaystyle y-y_0 = \lambda(x-x_0)$. The line meets the ellipse at points where $\displaystyle \dfrac{x^2}{a^2}+\dfrac{(y_0 + \lambda(x-x_0))^2}{b^2} = 1.$ That equation is a quadratic in x, namely $\displaystyle (b^2+\lambda^2a^2)x^2 + 2\lambda a^2(y_0 -\lambda x_0)x + a^2((y-\lambda x)^2 - b^2) = 0.$ The condition for the line to be a tangent to the ellipse is that the equation for x should have equal roots, in other words "b^2 – 4ac=0". Write down that condition and simplify it, to get a quadratic equation in $\displaystyle \lambda$, namely $\displaystyle (x_0^2-a^2)\lambda^2 -2x_0y_0\lambda + y_0^2-b^2 = 0$.
That equation has two roots, say $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$, which are the slopes of the two tangents to the ellipse from the point $\displaystyle (x_0,y_0)$. You want them to be perpendicular, in other words $\displaystyle \lambda_1\lambda_2 = -1$. But the product of the roots of a quadratic equation is the constant term divided by the coefficient of $\displaystyle \lambda^2$. That gives you the condition $\displaystyle \dfrac{y_0^2-b^2}{x_0^2-a^2} = -1$, which tells you that $\displaystyle (x_0,y_0)$ lies on the circle $\displaystyle x^2+y^2 = a^2+b^2$.
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# Find the value
Question:
Let $A=\{1,2,3,4\} .$ Let $f: A \rightarrow A$ and $g: A \rightarrow A$,
defined by $f=\{(1,4),(2,1),(3,3),(4,2)\}$ and $g=\{(1,3),(2,1),(3,2),(4,4)\}$
Find (i) g of (ii) f o g (iii) f o f.
Solution:
(i) $\mathrm{g}$ o $\mathrm{f}$
To find: $g$ o $f$
Formula used: $g$ o $f=g(f(x))$
Given: $f=\{(1,4),(2,1),(3,3),(4,2)\}$ and $g=\{(1,3),(2,1)$
$(3,2),(4,4)\}$
Solution: We have,
$g \circ f(1)=g(f(1))=g(4)=4$
$g \circ f(2)=g(f(2))=g(1)=3$
$g \circ f(3)=g(f(3))=g(3)=2$
$g \circ f(4)=g(f(4))=g(2)=1$
Ans) g of $f=\{(1,4),(2,3),(3,2),(4,1)\}$
(ii) $f \circ g$
To find: f o g
Formula used: f o g = f(g(x))
Given: f = {(1, 4), (2, 1), (3, 3), (4, 2)} and g = {(1, 3), (2, 1),
(3, 2), (4, 4)}
Solution: We have,
fog(1) = f(g(1)) = f(3) = 3
fog(2) = f(g(2)) = f(1) = 4
fog(3) = f(g(3)) = f(2) = 1
fog(4) = f(g(4)) = f(4) = 2
Ans) f o g = {(1, 3), (2, 4), (3, 1), (4, 2)}
(iii) f o f
To find: $f$ o $f$
Formula used: $f$ o $f=f(f(x))$
Given: $f=\{(1,4),(2,1),(3,3),(4,2)\}$
Solution: We have,
fof $(1)=f(f(1))=f(4)=2$
fof $(2)=f(f(2))=f(1)=4$
fof $(3)=f(f(3))=f(3)=3$
fof $(4)=f(f(4))=f(2)=1$
Ans) fo f $=\{(1,2),(2,4),(3,3),(4,1)\}$
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Intermediate Algebra
# 11.1Distance and Midpoint Formulas; Circles
Intermediate Algebra11.1 Distance and Midpoint Formulas; Circles
## Learning Objectives
By the end of this section, you will be able to:
• Use the Distance Formula
• Use the Midpoint Formula
• Write the equation of a circle in standard form
• Graph a circle
## Be Prepared 11.1
Before you get started, take this readiness quiz.
1. Find the length of the hypotenuse of a right triangle whose legs are 12 and 16 inches.
If you missed this problem, review Example 2.34.
2. Factor: $x2−18x+81.x2−18x+81.$
If you missed this problem, review Example 6.24.
3. Solve by completing the square: $x2−12x−12=0.x2−12x−12=0.$
If you missed this problem, review Example 9.22.
In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.
There are four conics—the circle, parabola, ellipse, and hyperbola. The next figure shows how the plane intersecting the double cone results in each curve.
Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.
## Use the Distance Formula
We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.
Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions. We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.
## Example 11.1
Use the rectangular coordinate system to find the distance between the points $(6,4)(6,4)$ and $(2,1).(2,1).$
## Try It 11.1
Use the rectangular coordinate system to find the distance between the points $(6,1)(6,1)$ and $(2,−2).(2,−2).$
## Try It 11.2
Use the rectangular coordinate system to find the distance between the points $(5,3)(5,3)$ and $(−3,−3).(−3,−3).$
The method we used in the last example leads us to the formula to find the distance between the two points $(x1,y1)(x1,y1)$ and $(x2,y2).(x2,y2).$
When we found the length of the horizontal leg we subtracted $6−26−2$ which is $x2−x1.x2−x1.$
When we found the length of the vertical leg we subtracted $4−14−1$ which is $y2−y1.y2−y1.$
If the triangle had been in a different position, we may have subtracted $x1−x2x1−x2$ or $y1−y2.y1−y2.$ The expressions $x2−x1x2−x1$ and $x1−x2x1−x2$ vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say $|x2−x1||x2−x1|$ and $|y2−y1|.|y2−y1|.$
In the Pythagorean Theorem, we substitute the general expressions $|x2−x1||x2−x1|$ and $|y2−y1||y2−y1|$ rather than the numbers.
$a2+b2=c2 Substitute in the values.(|x2−x1|)2+(|y2−y1|)2=d2 Squaring the expressions makes thempositive, so we eliminate the absolute valuebars.(x2−x1)2+(y2−y1)2=d2 Use the Square Root Property.d=±(x2−x1)2+(y2−y1)2 Distance is positive, so eliminate the negativevalue.d=(x2−x1)2+(y2−y1)2 a2+b2=c2 Substitute in the values.(|x2−x1|)2+(|y2−y1|)2=d2 Squaring the expressions makes thempositive, so we eliminate the absolute valuebars.(x2−x1)2+(y2−y1)2=d2 Use the Square Root Property.d=±(x2−x1)2+(y2−y1)2 Distance is positive, so eliminate the negativevalue.d=(x2−x1)2+(y2−y1)2$
This is the Distance Formula we use to find the distance d between the two points $(x1,y1)(x1,y1)$ and $(x2,y2).(x2,y2).$
## Distance Formula
The distance d between the two points $(x1,y1)(x1,y1)$ and $(x2,y2)(x2,y2)$ is
$d=(x2−x1)2+(y2−y1)2d=(x2−x1)2+(y2−y1)2$
## Example 11.2
Use the Distance Formula to find the distance between the points $(−5,−3)(−5,−3)$ and $(7,2).(7,2).$
## Try It 11.3
Use the Distance Formula to find the distance between the points $(−4,−5)(−4,−5)$ and $(5,7).(5,7).$
## Try It 11.4
Use the Distance Formula to find the distance between the points $(−2,−5)(−2,−5)$ and $(−14,−10).(−14,−10).$
## Example 11.3
Use the Distance Formula to find the distance between the points $(10,−4)(10,−4)$ and $(−1,5).(−1,5).$ Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
## Try It 11.5
Use the Distance Formula to find the distance between the points $(−4,−5)(−4,−5)$ and $(3,4).(3,4).$ Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
## Try It 11.6
Use the Distance Formula to find the distance between the points $(−2,−5)(−2,−5)$ and $(−3,−4).(−3,−4).$ Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
## Use the Midpoint Formula
It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.
## Midpoint Formula
The midpoint of the line segment whose endpoints are the two points $(x1,y1)(x1,y1)$ and $(x2,y2)(x2,y2)$ is
$(x1+x22,y1+y22)(x1+x22,y1+y22)$
To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.
## Example 11.4
Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $(−5,−4)(−5,−4)$ and $(7,2).(7,2).$ Plot the endpoints and the midpoint on a rectangular coordinate system.
## Try It 11.7
Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $(−3,−5)(−3,−5)$ and $(5,7).(5,7).$ Plot the endpoints and the midpoint on a rectangular coordinate system.
## Try It 11.8
Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are $(−2,−5)(−2,−5)$ and $(6,−1).(6,−1).$ Plot the endpoints and the midpoint on a rectangular coordinate system.
Both the Distance Formula and the Midpoint Formula depend on two points, $(x1,y1)(x1,y1)$ and $(x2,y2).(x2,y2).$ It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, is may be easier to remember the formulas.
## Write the Equation of a Circle in Standard Form
As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.
We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, $(h,k),(h,k),$ and the fixed distance is called the radius, r, of the circle.
## Circle
A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, $(h,k),(h,k),$ and the fixed distance is called the radius, r, of the circle.
We look at a circle in the rectangular coordinate system.The radius is the distance from the center, $(h,k),(h,k),$ to apoint on the circle, $(x,y).(x,y).$ To derive the equation of a circle, we can use thedistance formula with the points $(h,k),(h,k),$ $(x,y)(x,y)$ and thedistance, r. $d=(x2−x1)2+(y2−y1)2d=(x2−x1)2+(y2−y1)2$ Substitute the values. $r=(x−h)2+(y−k)2r=(x−h)2+(y−k)2$ Square both sides. $r2=(x−h)2+(y−k)2r2=(x−h)2+(y−k)2$
This is the standard form of the equation of a circle with center, $(h,k),(h,k),$ and radius, r.
## Standard Form of the Equation a Circle
The standard form of the equation of a circle with center, $(h,k),(h,k),$ and radius, r, is
## Example 11.5
Write the standard form of the equation of the circle with radius 3 and center $(0,0).(0,0).$
## Try It 11.9
Write the standard form of the equation of the circle with a radius of 6 and center $(0,0).(0,0).$
## Try It 11.10
Write the standard form of the equation of the circle with a radius of 8 and center $(0,0).(0,0).$
In the last example, the center was $(0,0).(0,0).$ Notice what happened to the equation. Whenever the center is $(0,0),(0,0),$ the standard form becomes $x2+y2=r2.x2+y2=r2.$
## Example 11.6
Write the standard form of the equation of the circle with radius 2 and center $(−1,3).(−1,3).$
## Try It 11.11
Write the standard form of the equation of the circle with a radius of 7 and center $(2,−4).(2,−4).$
## Try It 11.12
Write the standard form of the equation of the circle with a radius of 9 and center $(−3,−5).(−3,−5).$
In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.
## Example 11.7
Write the standard form of the equation of the circle with center $(2,4)(2,4)$ that also contains the point $(−2,1).(−2,1).$
## Try It 11.13
Write the standard form of the equation of the circle with center $(2,1)(2,1)$ that also contains the point $(−2,−2).(−2,−2).$
## Try It 11.14
Write the standard form of the equation of the circle with center $(7,1)(7,1)$ that also contains the point $(−1,−5).(−1,−5).$
## Graph a Circle
Any equation of the form $(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2$ is the standard form of the equation of a circle with center, $(h,k),(h,k),$ and radius, r. We can then graph the circle on a rectangular coordinate system.
Note that the standard form calls for subtraction from x and y. In the next example, the equation has $x+2,x+2,$ so we need to rewrite the addition as subtraction of a negative.
## Example 11.8
Find the center and radius, then graph the circle: $(x+2)2+(y−1)2=9.(x+2)2+(y−1)2=9.$
## Try It 11.15
Find the center and radius, then graph the circle: $(x−3)2+(y+4)2=4.(x−3)2+(y+4)2=4.$
## Try It 11.16
Find the center and radius, then graph the circle: $(x−3)2+(y−1)2=16.(x−3)2+(y−1)2=16.$
To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of $x2,y2x2,y2$ to be one.
## Example 11.9
Find the center and radius and then graph the circle, $4x2+4y2=64.4x2+4y2=64.$
## Try It 11.17
Find the center and radius, then graph the circle: $3x2+3y2=273x2+3y2=27$
## Try It 11.18
Find the center and radius, then graph the circle: $5x2+5y2=1255x2+5y2=125$
If we expand the equation from Example 11.8, $(x+2)2+(y−1)2=9,(x+2)2+(y−1)2=9,$ the equation of the circle looks very different.
$(x+2)2+(y−1)2=9 Square the binomials.x2+4x+4+y2−2y+1=9 Arrange the terms in descending degree order,and get zero on the rightx2+y2+4x−2y−4=0 (x+2)2+(y−1)2=9 Square the binomials.x2+4x+4+y2−2y+1=9 Arrange the terms in descending degree order,and get zero on the rightx2+y2+4x−2y−4=0$
This form of the equation is called the general form of the equation of the circle.
## General Form of the Equation of a Circle
The general form of the equation of a circle is
$x2+y2+ax+by+c=0x2+y2+ax+by+c=0$
If we are given an equation in general form, we can change it to standard form by completing the squares in both x and y. Then we can graph the circle using its center and radius.
## Example 11.10
Find the center and radius, then graph the circle: $x2+y2−4x−6y+4=0.x2+y2−4x−6y+4=0.$
## Try It 11.19
Find the center and radius, then graph the circle: $x2+y2−6x−8y+9=0.x2+y2−6x−8y+9=0.$
## Try It 11.20
Find the center and radius, then graph the circle: $x2+y2+6x−2y+1=0.x2+y2+6x−2y+1=0.$
In the next example, there is a y-term and a $y2y2$-term. But notice that there is no x-term, only an $x2x2$-term. We have seen this before and know that it means h is 0. We will need to complete the square for the y terms, but not for the x terms.
## Example 11.11
Find the center and radius, then graph the circle: $x2+y2+8y=0.x2+y2+8y=0.$
## Try It 11.21
Find the center and radius, then graph the circle: $x2+y2−2x−3=0.x2+y2−2x−3=0.$
## Try It 11.22
Find the center and radius, then graph the circle: $x2+y2−12y+11=0.x2+y2−12y+11=0.$
## Media
Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.
## Section 11.1 Exercises
### Practice Makes Perfect
Use the Distance Formula
In the following exercises, find the distance between the points. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.
1.
$(2,0)(2,0)$ and $(5,4)(5,4)$
2.
$(−4,−3)(−4,−3)$ and $(2,5)(2,5)$
3.
$(−4,−3)(−4,−3)$ and $(8,2)(8,2)$
4.
$(−7,−3)(−7,−3)$ and $(8,5)(8,5)$
5.
$(−1,4)(−1,4)$ and $(2,0)(2,0)$
6.
$(−1,3)(−1,3)$ and $(5,−5)(5,−5)$
7.
$(1,−4)(1,−4)$ and $(6,8)(6,8)$
8.
$(−8,−2)(−8,−2)$ and $(7,6)(7,6)$
9.
$(−3,−5)(−3,−5)$ and $(0,1)(0,1)$
10.
$(−1,−2)(−1,−2)$ and $(−3,4)(−3,4)$
11.
$(3,−1)(3,−1)$ and $(1,7)(1,7)$
12.
$(−4,−5)(−4,−5)$ and $(7,4)(7,4)$
Use the Midpoint Formula
In the following exercises, find the midpoint of the line segments whose endpoints are given and plot the endpoints and the midpoint on a rectangular coordinate system.
13.
$(0,−5)(0,−5)$ and $(4,−3)(4,−3)$
14.
$(−2,−6)(−2,−6)$ and $(6,−2)(6,−2)$
15.
$(3,−1)(3,−1)$ and $(4,−2)(4,−2)$
16.
$(−3,−3)(−3,−3)$ and $(6,−1)(6,−1)$
Write the Equation of a Circle in Standard Form
In the following exercises, write the standard form of the equation of the circle with the given radius and center $(0,0).(0,0).$
17.
18.
19.
Radius: $22$
20.
Radius: $55$
In the following exercises, write the standard form of the equation of the circle with the given radius and center
21.
Radius: 1, center: $(3,5)(3,5)$
22.
Radius: 10, center: $(−2,6)(−2,6)$
23.
Radius: $2.5,2.5,$ center: $(1.5,−3.5)(1.5,−3.5)$
24.
Radius: $1.5,1.5,$ center: $(−5.5,−6.5)(−5.5,−6.5)$
For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.
25.
Center $(3,−2)(3,−2)$ with point $(3,6)(3,6)$
26.
Center $(6,−6)(6,−6)$ with point $(2,−3)(2,−3)$
27.
Center $(4,4)(4,4)$ with point $(2,2)(2,2)$
28.
Center $(−5,6)(−5,6)$ with point $(−2,3)(−2,3)$
Graph a Circle
In the following exercises, find the center and radius, then graph each circle.
29.
$( x + 5 ) 2 + ( y + 3 ) 2 = 1 ( x + 5 ) 2 + ( y + 3 ) 2 = 1$
30.
$( x − 2 ) 2 + ( y − 3 ) 2 = 9 ( x − 2 ) 2 + ( y − 3 ) 2 = 9$
31.
$( x − 4 ) 2 + ( y + 2 ) 2 = 16 ( x − 4 ) 2 + ( y + 2 ) 2 = 16$
32.
$( x + 2 ) 2 + ( y − 5 ) 2 = 4 ( x + 2 ) 2 + ( y − 5 ) 2 = 4$
33.
$x 2 + ( y + 2 ) 2 = 25 x 2 + ( y + 2 ) 2 = 25$
34.
$( x − 1 ) 2 + y 2 = 36 ( x − 1 ) 2 + y 2 = 36$
35.
$( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25 ( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25$
36.
$( x − 1 ) 2 + ( y − 3 ) 2 = 9 4 ( x − 1 ) 2 + ( y − 3 ) 2 = 9 4$
37.
$x 2 + y 2 = 64 x 2 + y 2 = 64$
38.
$x 2 + y 2 = 49 x 2 + y 2 = 49$
39.
$2 x 2 + 2 y 2 = 8 2 x 2 + 2 y 2 = 8$
40.
$6 x 2 + 6 y 2 = 216 6 x 2 + 6 y 2 = 216$
In the following exercises, identify the center and radius and graph.
41.
$x 2 + y 2 + 2 x + 6 y + 9 = 0 x 2 + y 2 + 2 x + 6 y + 9 = 0$
42.
$x 2 + y 2 − 6 x − 8 y = 0 x 2 + y 2 − 6 x − 8 y = 0$
43.
$x 2 + y 2 − 4 x + 10 y − 7 = 0 x 2 + y 2 − 4 x + 10 y − 7 = 0$
44.
$x 2 + y 2 + 12 x − 14 y + 21 = 0 x 2 + y 2 + 12 x − 14 y + 21 = 0$
45.
$x 2 + y 2 + 6 y + 5 = 0 x 2 + y 2 + 6 y + 5 = 0$
46.
$x 2 + y 2 − 10 y = 0 x 2 + y 2 − 10 y = 0$
47.
$x 2 + y 2 + 4 x = 0 x 2 + y 2 + 4 x = 0$
48.
$x 2 + y 2 − 14 x + 13 = 0 x 2 + y 2 − 14 x + 13 = 0$
### Writing Exercises
49.
Explain the relationship between the distance formula and the equation of a circle.
50.
Is a circle a function? Explain why or why not.
51.
In your own words, state the definition of a circle.
52.
In your own words, explain the steps you would take to change the general form of the equation of a circle to the standard form.
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
|
### Comparing fractions
In development to Fractions, us learned that fractions are a method of mirroring part the something. Fractions are useful, because they let us tell exactly how much we have actually of something. Some fractions are larger than others. Because that example, i beg your pardon is larger: 6/8 that a pizza or 7/8 that a pizza?
In this image, we deserve to see that 7/8 is larger. The illustration provides it easy to compare these fractions. Yet how might we have done it there is no the pictures?
Click v the slideshow to learn how to compare fractions.
You are watching: Which number is bigger -2 or -4
Earlier, we observed that fractions have actually two parts.
One component is the peak number, or numerator.
The other is the bottom number, or denominator.
The denominator tells us how numerous parts space in a whole.
The molecule tells united state how numerous of those components we have.
When fractions have actually the same denominator, it way they're split into the same number of parts.
This method we deserve to compare these fractions simply by looking in ~ the numerator.
Here, 5 is an ext than 4...
Here, 5 is an ext than 4...so we have the right to tell the 5/6 is more than 4/6.
Let's look at another example. I beg your pardon of this is larger: 2/8 or 6/8?
If you assumed 6/8 to be larger, you were right!
Both fractions have actually the exact same denominator.
So we compared the numerators. 6 is larger than 2, so 6/8 is much more than 2/8.
As friend saw, if 2 or much more fractions have the same denominator, you have the right to compare castle by looking at your numerators. As you have the right to see below, 3/4 is bigger than 1/4. The bigger the numerator, the bigger the fraction.
### Comparing fountain with different denominators
On the vault page, we contrasted fractions that have the same bottom numbers, or denominators. Yet you recognize that fractions have the right to have any number as a denominator. What happens when you must compare fractions with various bottom numbers?
For example, which of this is larger: 2/3 or 1/5? It's complicated to tell just by looking in ~ them. ~ all, 2 is larger than 1, but the platform aren't the same.
If girlfriend look at the picture, though, the difference is clear: 2/3 is larger than 1/5. With an illustration, it was easy to to compare these fractions, yet how could we have actually done it without the picture?
Click with the slideshow come learn just how to compare fractions with various denominators.
Let's compare these fractions: 5/8 and also 4/6.
Before us compare them, we require to readjust both fountain so they have the same denominator, or bottom number.
First, we'll discover the the smallest number that deserve to be separated by both denominators. We contact that the lowest common denominator.
Our first step is to discover numbers that deserve to be split evenly by 8.
Using a multiplication table provides this easy. All of the numbers on the 8 row deserve to be divided evenly by 8.
Now let's look at our 2nd denominator: 6.
We can use the multiplication table again. All of the numbers in the 6 row can be split evenly by 6.
Let's to compare the two rows. That looks choose there space a couple of numbers that have the right to be divided evenly by both 6 and 8.
24 is the the smallest number that appears on both rows, so it's the lowest typical denominator.
Now we're walking to adjust our fractions so they both have actually the same denominator: 24.
To execute that, we'll have to adjust the numerators the same method we readjusted the denominators.
Let’s look in ~ 5/8 again. In order to readjust the denominator come 24...
Let’s look in ~ 5/8 again. In bespeak to adjust the denominator come 24...we had actually to multiply 8 by 3.
Since we multiplied the denominator by 3, we'll also multiply the numerator, or top number, by 3.
5 time 3 equates to 15. For this reason we've changed 5/8 into 15/24.
We can do the because any number over itself is same to 1.
So when we multiply 5/8 by 3/3...
So when we main point 5/8 by 3/3...we're really multiplying 5/8 by 1.
Since any number time 1 is equal to itself...
Since any kind of number times 1 is equal to itself...we can say that 5/8 is same to 15/24.
Now we'll do the exact same to our other fraction: 4/6. We also changed its denominator to 24.
Our old denominator was 6. To gain 24, us multiplied 6 through 4.
So we'll likewise multiply the numerator by 4.
4 time 4 is 16. Therefore 4/6 is equal to 16/24.
Now that the denominators room the same, we have the right to compare the two fractions by feather at your numerators.
16/24 is larger than 15/24...
16/24 is larger than 15/24... Therefore 4/6 is bigger than 5/8.
### Rdearteassociazione.orgcing fractions
Which of these is larger: 4/8 or 1/2?
If girlfriend did the math or even just looked at the picture, you can have to be able come tell the they're equal. In other words, 4/8 and 1/2 average the exact same thing, also though they're composed differently.
If 4/8 way the exact same thing together 1/2, why not just contact it that? One-half is simpler to say than four-eighths, and for most civilization it's additionally easier come understand. After all, when you eat out through a friend, you split the bill in half, not in eighths.
If you create 4/8 as 1/2, you're rdearteassociazione.orgcing it. Once we rdearteassociazione.orgce a fraction, we're creating it in a simpler form. Rdearteassociazione.orgced fountain are constantly equal to the original fraction.
We currently rdearteassociazione.orgced 4/8 come 1/2. If you look at the instances below, you can see that other numbers deserve to be rdearteassociazione.orgced come 1/2 together well. These fractions are all equal.
5/10 = 1/211/22 = 1/236/72 = 1/2
These fractions have all been rdearteassociazione.orgced come a simpler type as well.
4/12 = 1/314/21 = 2/335/50 = 7/10
Click with the slideshow to learn just how to rdearteassociazione.orgce fountain by dividing.
Let's try rdearteassociazione.orgcing this fraction: 16/20.
Since the numerator and denominator room even numbers, you deserve to divide lock by 2 to rdearteassociazione.orgce the fraction.
First, we'll division the numerator by 2. 16 separated by 2 is 8.
Next, we'll divide the denominator through 2. 20 divided by 2 is 10.
We've rdearteassociazione.orgced 16/20 to 8/10. We could likewise say that 16/20 is same to 8/10.
If the numerator and denominator can still be split by 2, we can continue rdearteassociazione.orgcing the fraction.
8 separated by 2 is 4.
10 divided by 2 is 5.
Since there's no number that 4 and 5 deserve to be divided by, we can't rdearteassociazione.orgce 4/5 any further.
This means 4/5 is the simplest form of 16/20.
Let's shot rdearteassociazione.orgcing an additional fraction: 6/9.
While the molecule is even, the denominator is one odd number, so we can't rdearteassociazione.orgce by splitting by 2.
Instead, we'll need to find a number that 6 and also 9 have the right to be separated by. A multiplication table will make that number simple to find.
Let's discover 6 and 9 on the same row. Together you can see, 6 and also 9 deserve to both be separated by 1 and also 3.
Dividing by 1 won't change these fractions, therefore we'll usage the largest number the 6 and also 9 have the right to be split by.
That's 3. This is referred to as the greatest typical divisor, or GCD. (You can additionally call it the greatest typical factor, or GCF.)
3 is the GCD the 6 and also 9 because it's the largest number they can be divided by.
So we'll divide the numerator by 3. 6 split by 3 is 2.
Then we'll division the denominator through 3. 9 separated by 3 is 3.
Now we've rdearteassociazione.orgced 6/9 come 2/3, i beg your pardon is its simplest form. Us could also say the 6/9 is equal to 2/3.
Irrdearteassociazione.orgcible fractions
Not every fractions deserve to be rdearteassociazione.orgced. Some are already as basic as they can be. For example, friend can't rdearteassociazione.orgce 1/2 since there's no number various other than 1 that both 1 and also 2 deserve to be separated by. (For that reason, girlfriend can't rdearteassociazione.orgce any fraction that has actually a numerator of 1.)
Some fractions that have actually larger numbers can't be rdearteassociazione.orgced either. For instance, 17/36 can't be rdearteassociazione.orgced due to the fact that there's no number the both 17 and 36 can be split by. If friend can't find any kind of common multiples for the number in a fraction, opportunities are it's irrdearteassociazione.orgcible.
Try This!
Rdearteassociazione.orgce each fraction to its easiest form.
### Mixed numbers and also improper fractions
In the previous lesson, girlfriend learned about mixed numbers. A combined number has both a fraction and a whole number. An example is 1 2/3. You'd review 1 2/3 favor this: one and two-thirds.
Another method to write this would be 5/3, or five-thirds. These two numbers look different, yet they're actually the same. 5/3 is one improper fraction. This just way the molecule is larger 보다 the denominator.
There are times when you may prefer to use an improper fraction instead of a mixed number. It's basic to change a blended number into an improper fraction. Let's find out how:
Let's transform 1 1/4 right into an wrong fraction.
First, we'll require to find out how countless parts comprise the entirety number: 1 in this example.
To carry out this, we'll multiply the whole number, 1, by the denominator, 4.
1 time 4 amounts to 4.
Now, let's include that number, 4, come the numerator, 1.
4 add to 1 amounts to 5.
The denominator remains the same.
Our improper fraction is 5/4, or five-fourths. So we might say that 1 1/4 is same to 5/4.
This means there are five 1/4s in 1 1/4.
Let's convert another mixed number: 2 2/5.
First, we'll multiply the whole number by the denominator. 2 time 5 amounts to 10.
Next, we'll add 10 come the numerator. 10 plus 2 equates to 12.
As always, the denominator will stay the same.
So 2 2/5 is equal to 12/5.
Try This!
Try converting these blended numbers right into improper fractions.
Converting not correct fractions into mixed numbers
Improper fountain are helpful for math problems that use fractions, as you'll find out later. However, they're additionally more complicated to read and also understand than mixed numbers. Because that example, it's a lot less complicated to snapshot 2 4/7 in your head than 18/7.
Click v the slideshow come learn just how to adjust an improper fraction into a mixed number.
Let's revolve 10/4 right into a combined number.
You deserve to think that any portion as a division problem. Simply treat the line in between the numbers prefer a department sign (/).
So we'll divide the numerator, 10, through the denominator, 4.
10 separated by 4 equals 2...
10 split by 4 equates to 2... With a remainder the 2.
The answer, 2, will end up being our entirety number because 10 deserve to be divided by 4 twice.
And the remainder, 2, will come to be the molecule of the portion because we have 2 parts left over.
The denominator stays the same.
So 10/4 equates to 2 2/4.
Let's shot another example: 33/3.
We'll division the numerator, 33, through the denominator, 3.
33 divided by 3...
33 divided by 3... Equals 11, through no remainder.
The answer, 11, will end up being our totality number.
See more: What Does Correo Electronico Mean In Spanish, Translation Of Correo Electrónico
There is no remainder, for this reason we can see that our improper fraction was in reality a totality number. 33/3 amounts to 11.
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# Eureka Math Precalculus Module 2 Lesson 4 Answer Key
## Engage NY Eureka Math Precalculus Module 2 Lesson 4 Answer Key
### Eureka Math Precalculus Module 2 Lesson 4 Exercise Answer Key
Exercises
Exercise 1.
Describe the geometric effect of each mapping.
a. L(x)=9∙x
Dilates the interval by a factor of 9
b. L(x)=-$$\frac{1}{2}$$∙x
Reflects the interval over the origin and then applies a dilation with a scale factor of $$\frac{1}{2}$$
Exercise 2.
Write the formula for the mappings described.
a. A dilation that expands each interval to 5 times its original size
L(x)=5∙x
b. A collapse of the interval to the number 0
L(x)=0∙x
### Eureka Math Precalculus Module 2 Lesson 4 Problem Set Answer Key
Question 1.
Suppose you have a linear transformation L:R→R, where L(3)=6, L(5)=10.
a. Use the addition property to find L(6), L(8), L(10), and L(13).
L(6)=L(3+3)=L(3)+L(3)=6+6=12
L(8)=L(3+5)=L(3)+L(5)=6+10=16
L(10)=L(5)+L(5)=10+10=20
L(13)=L(10+3)=L(10)+L(3)=20+6=26
b. Use the multiplication property to find L(15), L(18), and L(30).
L(15)=L(3∙5)=3∙L(5)=3∙10=30
L(18)=L(3∙6)=3∙L(6)=3∙12=36
L(30)=L(5∙6)=5∙L(6)=5∙12=60
c. Find L(-3), L(-8), and L(-15).
L(-3)=L(-1∙3)=-1∙L(3)=-6
L(-8)=L(-1∙8)=-1∙L(8)=-16
L(-15)=L(-3∙5)=-3∙L(5)=-3∙10=-30
d. Find the formula for L(x).
Given L(x) is a linear transformation; therefore, it must have a form of L(x)=mx, where m is real number.
Given L(3)=6; therefore, 3m=6, m=2. L(x)=2x.
e. Draw the graph of the function L(x).
Question 2.
A linear transformation L: R→R must have the form of L(x)=ax for some real number a. Consider the interval [-5,2]. Describe the geometric effect of the following, and find the new interval.
a. L(x)=5x
It dilates the interval by a scale factor of 5; [-25,10]
b. L(x)=-2x
It reflects the interval over the origin and then dilates it with a scale factor of 2; [-4,10]
Question 3.
A linear transformation L: R→R must have the form of L(x)=ax for some real number a. Consider the interval [-2,6]. Write the formula for the mapping described, and find the new interval.
a. A reflection over the origin
L(x)=-x, [-6,2]
b. A dilation with a scale of $$\sqrt{2}$$
L(x)=$$\sqrt{2}$$∙x, [-2$$\sqrt{2}$$,6$$\sqrt{2}$$]
c. A reflection over the origin and a dilation with a scale of $$\frac{1}{2}$$
L(x)=-$$\frac{1}{2}$$x, [-3,1]
d. A collapse of the interval to the number 0
L(x)=0x, [0,0]
Question 4.
In Module 1, we used 2×2 matrices to do transformations on a square, such as a pure rotation, a pure reflection, a pure dilation, and a rotation with a dilation. Now use those matrices to do transformations on this complex number: z=2+i. For each transformation below, graph your answers.
a. A pure dilation with a factor of 2
$$\left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right)$$, L(z) = $$\left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{l} 4 \\ 2 \end{array}\right)$$
b. A pure $$\frac{\pi}{2}$$ radians counterclockwise rotation about the origin
$$\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)$$, L(z) = $$\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{c} -1 \\ 2 \end{array}\right)$$
$$\left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right)$$, L(z) = $$\left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{l} -2 \\ -1 \end{array}\right)$$
d. A pure reflection about the real axis
$$\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$$, L(z) = $$\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{c} 2 \\ -1 \end{array}\right)$$
e. A pure reflection about the imaginary axis
$$\left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right)$$, L(z) = $$\left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{c} -2 \\ 1 \end{array}\right)$$
f. A pure reflection about the line y=x
$$\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)$$, L(z) = $$\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{l} 1 \\ 2 \end{array}\right)$$
g. A pure reflection about the line y=-x
$$\left(\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right)$$, L(z) = $$\left(\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{l} -1 \\ -2 \end{array}\right)$$
Question 5.
Wesley noticed that multiplying the matrix $$\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)$$ by a complex number z produces a pure $$\frac{\pi}{2}$$ radians counterclockwise rotation, and multiplying by $$\left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right)$$ produces a pure dilation with a factor of 2. So, he thinks he can add these two matrices, which will produce $$\left(\begin{array}{cc} 2 & -1 \\ 1 & 2 \end{array}\right)$$ and will rotate z by $$\frac{\pi}{2}$$ radians counterclockwise and dilate z with a factor of 2. Is he correct? Explain your reason.
No, he is not correct. For the general transformation of complex numbers, the form is L(Z)=$$\left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)$$.
By multiplying the matrix $$\left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)$$ to z, it rotates z an angle arctan($$\frac{b}{a}$$) and dilates z with a factor of $$\sqrt{a^{2}+b^{2}}$$.
arctan$$\frac{1}{2}$$=26.565°, and $$\sqrt{(1)^{2}+(2)^{2}}$$=$$\sqrt{5}$$, which is not $$\frac{\pi}{2}$$ radians or a dilation of a factor of 2.
Question 6.
In Module 1, we learned that there is not any real number that will satisfy $$\frac{1}{a+b}$$=$$\frac{1}{a}$$+$$\frac{1}{b}$$, which is the addtition property of linear transformation. However, we discussed that some fixed complex numbers might work. Can you find two pairs of complex numbers that will work? Show your work.
Given $$\frac{1}{a+b}$$=$$\frac{1}{a}$$+$$\frac{1}{b}$$, a,b≠0
$$\frac{1}{a+b}$$=$$\frac{b+a}{ab}$$
ab=(a+b)2
ab=a2+2ab+b2
a2+ab+b2=0
a2+ab+$$\frac{1}{4}$$b2=-$$\frac{3}{4}$$b2
(a+$$\frac{1}{2}$$ b)2=-$$\frac{3}{4}$$b2
a+$$\frac{1}{2}$$b=±$$\frac{\sqrt{3}}{2}$$b∙i
a=$$\frac{-1 \pm \sqrt{3} i}{2} b$$
For example, b=2 and a=-1+$$\sqrt{3}$$∙i or a=-1-$$\sqrt{3}$$∙i. If b=4 and a=-2+2$$\sqrt{3}$$∙i or a=-2-2$$\sqrt{3}$$∙i. Answers may vary.
Question 7.
Suppose L is a complex-number function that satisfies the dream conditions: L(z+w)=L(z)+L(w) and
L(kz)=kL(z) for all complex numbers z, w, and k. Show L(z)=mz for a fixed complex number m, the only type of complex-number function that satisfies these conditions.
L(z+w)=L(z)+L(w)=m(a+bi)+m(c+di)=m((a+c)+(b+d)i)
L(z+w)=L(a+bi+c+di)=L((a+c)+(b+d)i)=m((a+c)+(b+d)i); they are the same.
For multiplication property:
L(kz)=kL(z)=kmz=km(a+bi)
L(kz)=L(k(a+bi))=mk(a+bi); they are the same.
Question 8.
For complex numbers, the linear transformation requires L(x+y)=L(x)+L(y),L(a∙x)=a∙L(x). Prove that in general L$$\left(\begin{array}{l} x \\ y \end{array}\right)$$=$$\left(\begin{array}{ll} a & b \\ c & d \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)$$ is a linear transformation, where $$\left(\begin{array}{l} x \\ y \end{array}\right)$$ represents z=x+yi.
Now we need to prove the multiplication property.
### Eureka Math Precalculus Module 2 Lesson 4 Exit Ticket Answer Key
Question 1.
In Module 1, we learned about linear transformations for any real number functions. What are the conditions of a linear transformation? If a real number function is a linear transformation, what is its form? What are the two characteristics of the function?
L(x+y)=L(x)+L(y), L(kx)=kL(x), where x, y, and k are real numbers
It is in the form of L(x)=mx, and its graph is a straight line going through the origin. It is an odd function.
Question 2.
Describe the geometric effect of each mapping:
a. L(x)=3x
Dilate the interval by a factor of 3.
b. L(z)=($$\sqrt{2}$$+$$\sqrt{2}$$i)∙z
Rotate $$\frac{\pi}{4}$$ radians counterclockwise about the origin.
c. L(z)=$$\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)$$, where z is a complex number
Rotate $$\frac{\pi}{2}$$ radians counterclockwise about the origin.
d. L(z)=$$\left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)$$, where z is a complex number
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# What is Long Division in Algebra? (With Examples)
Remember being a kid and going through the anxiety of learning long division? There were only two roads that you took. You either understood it entirely, or you struggled. And now you are in algebra, and you have to perform long division on polynomials.
Polynomial long division is the process of dividing one polynomial into another polynomial, using the same basic techniques as regular long division. The main requirement is that the divisor must be of the same degree or smaller than the polynomial that you divide into.
Let’s break down the polynomial long division process so that you can see how it’s similar to regular long division and what the process looks like, step by step.
Table of Contents
## Breaking long division down.
Before we jump into long division with polynomials, we will take a look at how regular long division works. We’ll take a look at the different working parts and how we work with numbers as we solve a division problem.
When working with numbers, the parts of a division problem are the divisor, dividend, and quotient. The divisor is the number dividing into the dividend. The dividend is the number that we divide into. The quotient is the result of our division. It tells us how many times the divisor divides into the dividend. It’s our answer to the division problem.
Now let’s go through the steps of regular long division. The first step is to find the number of times the divisor goes into the dividend. Once we figure out this number, we place it above and multiply it times our divisor. We place the result underneath the dividend, and we subtract. We continue dividing, multiplying, and subtracting until all parts of the dividend are processed. Our answer, in the end, is the quotient and the remainder. The quotient tells us how many times the divisor divides into the dividend, and the remainder is what is leftover. Here is an example of the process.
Before diving into performing long division for polynomials, we will look at its basic structure.
The difference that we face in polynomial division is that we use polynomials instead of numbers. The degree of the divisor must be the same degree as the dividend or lower. Once we have our long division problem set up, we can move on to the process itself.
As you work through each step, try to find a way to remember which step you are on in the process. One of the most significant issues most students have when performing long division with polynomials is losing track of where they are in the process. But remember, the more you practice this process, the easier it will become.
Now we go over the steps of long division.
## Working through long division, step by step.
Now, let’s look at how long division for polynomials is performed. The process won’t be too complicated if you are familiar with regular long division. The biggest issue I see that people have is in keeping everything organized.
Before we begin, let’s look at the example problem that we will walk through step by step.
Our divisor is equal to x-1. Our dividend is equal to x3+2x2+7x-10.
When we divide our divisor (x-1) into our dividend (x3+2x2+7x-10), the first step we will concentrate on is finding the highest power in the divisor and the highest power in the dividend.
The highest power in the divisor is the x. The highest power in the dividend is the x3.
We then want to find the number of times x divides into x3. In our case, this results in x2.
We then multiply x2 times x-1 and place the result right under the (x3+2x2).
We then subtract the (x3-x2)from the (x3+2x2) and place the results under the line. We bring down the 7x and continue the process.
Now we move on to dividing our x into the 3x2.
We then repeat the process. We find the number of times x divides into 3x2 and place it above. We come up with 3x. We then multiply the 3x times (x-1). The result goes under the (3x2+7).
We then subtract the (3x2-3x) from the (3x2+7x). We then drop down the -10 from above and continue. We place the results under the line.
We then repeat the process. We find the number of times x divides into 10x and place it above. We come up with 10. We then multiply the 10 times (x-1). The result goes under the (10x-10).
We then subtract the 10x-10 from the 10x-10. We place the results under the line. Our remainder is zero. The answer indicates that x-1 is a factor of x3+2x2+7x-10.
Whenever we get a remainder that is zero in long division, it indicates that our divisor is a factor of the polynomial that we are dividing into. What this means is that:
• (x-1) (x2+3x+10) = x3+2x2+7x-10
I have included a brief video of all the slides explaining the process.
## Example long division problems.
In this next section, I will go over two example problems that show how long division works a bit further.
The first problem is x3+x2-33x+63 / x-3.
If we go through and divide our divisor into our dividend, we’ll see that our final result is x2+4x-21. We also have a remainder of 0, which indicates that our divisor is a factor of the original f(x).
This next problem works with a polynomial with a higher degree. The process is still the same. As we work step by step, we see that the result of this division is -3x3 +2x2+27x+18. Our remainder in this problem is also zero.
I know that long division is not easy. Always remember to practice, practice, practice. The more you practice, the easier it gets.๐
Danielle
Just a nerd who loves math. Trying to help the world one problem at a time.
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# Search by Topic
#### Resources tagged with Working systematically similar to Calendar Capers:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Sticky Numbers
##### Stage: 3 Challenge Level:
Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?
### 9 Weights
##### Stage: 3 Challenge Level:
You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?
### Maths Trails
##### Stage: 2 and 3
The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails.
### More Magic Potting Sheds
##### Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Tetrahedra Tester
##### Stage: 3 Challenge Level:
An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?
### Weights
##### Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
### Games Related to Nim
##### Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Number Daisy
##### Stage: 3 Challenge Level:
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
### Twinkle Twinkle
##### Stage: 2 and 3 Challenge Level:
A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.
### Constellation Sudoku
##### Stage: 4 and 5 Challenge Level:
Special clue numbers related to the difference between numbers in two adjacent cells and values of the stars in the "constellation" make this a doubly interesting problem.
### Factor Lines
##### Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Pole Star Sudoku 2
##### Stage: 3 and 4 Challenge Level:
This Sudoku, based on differences. Using the one clue number can you find the solution?
### Ben's Game
##### Stage: 3 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
### You Owe Me Five Farthings, Say the Bells of St Martin's
##### Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
### Oranges and Lemons, Say the Bells of St Clement's
##### Stage: 3 Challenge Level:
Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own.
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Two and Two
##### Stage: 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
### Special Numbers
##### Stage: 3 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Plum Tree
##### Stage: 4 and 5 Challenge Level:
Label this plum tree graph to make it totally magic!
### Where Can We Visit?
##### Stage: 3 Challenge Level:
Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
### First Connect Three
##### Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
### Medal Muddle
##### Stage: 3 Challenge Level:
Countries from across the world competed in a sports tournament. Can you devise an efficient strategy to work out the order in which they finished?
### When Will You Pay Me? Say the Bells of Old Bailey
##### Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
### Squares in Rectangles
##### Stage: 3 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### Peaches Today, Peaches Tomorrow....
##### Stage: 3 Challenge Level:
Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for?
### Ratio Sudoku 2
##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### I've Submitted a Solution - What Next?
##### Stage: 1, 2, 3, 4 and 5
In this article, the NRICH team describe the process of selecting solutions for publication on the site.
### Intersection Sudoku 1
##### Stage: 3 and 4 Challenge Level:
A Sudoku with a twist.
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### Integrated Sums Sudoku
##### Stage: 3 and 4 Challenge Level:
The puzzle can be solved with the help of small clue-numbers which are either placed on the border lines between selected pairs of neighbouring squares of the grid or placed after slash marks on. . . .
### Gr8 Coach
##### Stage: 3 Challenge Level:
Can you coach your rowing eight to win?
### Difference Sudoku
##### Stage: 4 Challenge Level:
Use the differences to find the solution to this Sudoku.
### LOGO Challenge - the Logic of LOGO
##### Stage: 3 and 4 Challenge Level:
Just four procedures were used to produce a design. How was it done? Can you be systematic and elegant so that someone can follow your logic?
### Seasonal Twin Sudokus
##### Stage: 3 and 4 Challenge Level:
This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?
### Cinema Problem
##### Stage: 3 Challenge Level:
A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
### Latin Squares
##### Stage: 3, 4 and 5
A Latin square of order n is an array of n symbols in which each symbol occurs exactly once in each row and exactly once in each column.
### Olympic Logic
##### Stage: 3 and 4 Challenge Level:
Can you use your powers of logic and deduction to work out the missing information in these sporty situations?
### Building with Longer Rods
##### Stage: 2 and 3 Challenge Level:
A challenging activity focusing on finding all possible ways of stacking rods.
##### Stage: 3 and 4 Challenge Level:
Four small numbers give the clue to the contents of the four surrounding cells.
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Magnetic Personality
##### Stage: 2, 3 and 4 Challenge Level:
60 pieces and a challenge. What can you make and how many of the pieces can you use creating skeleton polyhedra?
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# Simplifying fractions
In this 5th grade lesson, I explain how to simplify a fraction using a visual model and an arrow notation. Simplifying fractions is like joining or merging pieces together, and it is the opposite of finding equivalent fractions and splitting the pieces further.
The video below explains many of the same ideas as the lesson below.
Do you remember how to convert fractions into equivalent fractions?
= × 2 3 4 = 6 8 Each slice has beensplit two ways. × 2
= × 3 1 3 = 3 9 Each slice has beensplit three ways. × 3
We can also reverse the process. Then it is called SIMPLIFYING:
= ÷ 2 6 8 = 3 4 The slices have been joined together in twos. ÷ 2
= ÷ 3 3 9 = 1 3 The slices have been joined together in threes. ÷ 3
Notice:
• Both the numerator and the denominator change into smaller numbers, but the value
of the fraction does not. In other words, you get the SAME AMOUNT of pie either way.
• The fraction is now written in a simpler form. We also say that the fraction is written in lower terms, because the new numerator and denominator are smaller numbers than the originals.
• Both the numerator and the denominator are divided by the same number.
This number shows how many slices are joined together.
1. Write the simplification process, labeling the arrows. Follow the examples above.
a. The parts were joined together in ______.
=
÷ = ÷
b. The parts were joined together in ______.
=
÷ = ÷
2. Write the simplifying process. You can write the arrows and the divisions to help you.
a. The slices were
joined together
in ___________.
=
÷ 3 3 6 = ÷ 3
b. The slices were
joined together
in ___________.
=
c. The slices were
joined together
in ___________.
=
d. The slices were
joined together
in ___________.
=
e. The slices were
joined together
in ___________.
=
f. The slices were
joined together
in ___________.
=
g. The slices were
joined together
in ___________.
=
h. The slices were
joined together
in ___________.
=
3. Draw a picture and simplify the fractions.
a. Join together each two parts.
2
6
=
b. Join together each four parts.
=
c. Join together each three parts.
=
d. Join together each six parts.
=
e. Join together each seven parts.
=
f. Join together each four parts.
=
When you simplify, you divide the numerator and the denominator by the same number, so you need a number that “goes” into both the numerator and the denominator evenly. The numerator and the denominator have to be divisible by the same number.
Simplify 28 40 .
Since both 28 and 40 are divisible
by 4, we can divide the numerator
and denominator by four. This
means that each four slices are
joined together.
÷ 4 28 40 = 7 10 ÷ 4
Simplify 6 17 .
We cannot find any number that
would go into 6 and 17 (except 1,
of course). So 6/17 is already as
simplified as it can be. It is
÷ 1 6 17 = 6 17 ÷ 1
4. Simplify the fractions to the lowest terms.
÷ a. 6 16 = ÷
÷ b. 15 25 = ÷
÷ c. 3 9 = ÷
÷ d. 4 8 = ÷
÷ e. 16 24 = ÷
f. 12 20 =
g. 24 32 =
h. 3 15 =
i. 15 18 =
j. 16 20 =
5. Simplify the fractional parts of these mixed numbers. The whole number does not change. Study
the example.
a. 1 4 16 = 1 1 4
b. 5 3 27 =
c. 7 5 20 =
d. 3 14 49 =
6. You cannot simplify some of these fractions because they are already in the lowest terms. Cross out
the fractions that are already in the lowest terms and simplify the rest.
a. 2 3
b. 2 6
c. 6 13
d. 2 7 12
e. 11 22
f. 5 6 12
g. 5 11
h. 9 20
i. 1 4 7
j. 3 4 28
k. 5 29
l. 6 33
7. Use a line to connect the fractions
and mixed numbers that are equivalent.
2 6 24
28 12
1 3 4
2 4 12
14 8
2 2 8
2 5 15
21 12
2 1 3
7 4
9 4
2 1 4
8. Tommy is on the track team. He spends 10 minutes warming up before
practice and 10 minutes stretching after practice. All together, he spends
a total of one hour for the warm-up, the practice, and the stretching.
What part of the total time is the warm-up time?
What part of the total time is the actual practice time?
9. Color the parts of this 24-part circle according to how you spend your time during a typical day.
Include sleeping, eating, bathing, school, housework, TV, etc. Write the name of each activity and
what fractional part of your day it takes. Simplify the fractional part if you can.
This lesson is taken from my book Math Mammoth Fractions 2.
#### Math Mammoth Fractions 2
A self-teaching worktext that teaches fractions using visual models, a sequel to Math Mammoth Fractions 1. The book covers simplifying fractions, multiplication and division of fractions and mixed numbers, converting fractions to decimals, and ratios.
|
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LCM
# How To Find LCM of 3 and 5? | Listing, Division, and Prime Factorization Method
Written by Prerit Jain
Updated on: 18 Feb 2023
### How To Find LCM of 3 and 5? | Listing, Division, and Prime Factorization Method
LCM of 3 and 5 is 15. LCM of 3 and 5, also known as the Least Common Multiple or Lowest Common Multiple of 3 and 5 is the lowest possible common number that is divisible by 3 and 5.
Now, let’s see how to find the LCM of 3 and 5. Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36,…and multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45,…Here, both 15 and 30 are the common numbers in the multiples of 3 and 5, respectively, or that are divisible by 3 and 5. But, when you have to find the LCM, you must focus on the lowest common number. So, 15 is the lowest common number among all the multiples that is divisible by 3 and 5, and hence the LCM of 3 and 5 is 15.
## Methods to Find the LCM of 3 and 5
There are three different methods for finding the LCM of 3 and 5. They are:
1. Listing Method
2. Division Method
3. Prime Factorization Method
## LCM of 3 and 5 Using the Listing Method
The listing method is one of the methods for finding the LCM. To find the LCM of 3 and 5 using the listing method, follow the following steps:
• Step 1: Write down the first few multiples of 3 and 5 separately.
• Step 2: Out of all the multiples of 3 and 5 focus on the multiples that are common to both the numbers, that is, 3 and 5.
• Step 3: Now, out of all the common multiples, take out the smallest common multiple. That will be the LCM of 3 and 5.
LCM of 3 and 5 can be obtained using the listing method as
• Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36,…
• Multiples of 5 are 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45,…
Here, it is clear that the least common multiple is 15. So, the LCM of 3 and 5 is 15.
## LCM of 3 and 5 Using the Division Method
The division method is one of the methods for finding the LCM. To find the LCM of 3 and 5 using the division method, divide 3 and 5 by the smallest prime number, which is divisible by any of them. Then, the prime factors further obtained will be used to calculate the final LCM of 3 and 5.
Follow the following steps to find the LCM of 3 and 5 using the division method:
• Step 1: Write the numbers for which you have to find the LCM, that is 3 and 5 in this case, separated by commas.
• Step 2: Now, find the smallest prime number which is divisible by either 3 or 5.
• Step 3: If any of the numbers between 3 and 5 is not divisible by the respective prime number, write that number in the next row just below it and proceed further.
• Step 4: Continue dividing the numbers obtained after each step by the prime numbers, until you get the result as 1 in the entire row.
• Step 5: Now, multiply all the prime numbers and the final result will be the LCM of 3 and 5.
LCM of 3 and 5 can be obtained using the division method as:
So, the LCM of 3 and 5 = 3 * 5 = 1
## LCM of 3 and 5 Using the Prime Factorization Method
The prime factorization method is one of the methods for finding the LCM. To find the LCM of 3 and 5 using the prime factorization method, follow the following steps:
• Step 1: Find the prime factors of 3 and 5 using the repeated division method.
• Step 2: Write all the prime factors in their exponent forms. Then multiply the prime factors having the highest power.
• Step 3: The final result after multiplication will be the LCM of 3 and 5.
LCM of 3 and 5 can be obtained using the prime factorization method as
• Prime factorization of 3 can be expressed as 3 = 31
• Prime factorization of 5 can be expressed as 5 = 51
So, the LCM of 3 and 5 = 31 * 51= 3 * 5 = 15
## What Is the Formula for Finding the LCM of 3 and 5?
LCM of 3 and 5 can be calculated using the formula:
LCM (3, 5) = (3 * 5) / HCF (3, 5),
where HCF is the highest common factor or the greatest common divisor of 3 and 5.
Another formula, using which the LCM of 3 and 5 can be found:
3 * 5 = LCM (3, 5) * HCF (3, 5), that is,
The product of 3 and 5 is equal to the product of its LCM and HCF.
## Problems Based on LCM of 3 and 5
Question 1: What are the other two numbers having the LCM as 15? Show the representation using the listing method.
Solution:
Other than 3 and 5, LCM of 1 and 15 is also 15. We will prove this using the listing method.
To find the LCM of 3 and 5 using the listing method, first, we will write down the first few multiples of 3 and 5 separately. Out of all the multiples of 3 and 5, we will focus on the multiples which are common to both numbers.
Then, out of all the common multiples, we will take out the smallest common multiple. That will be the LCM of 3 and 5.
• Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36,…
• Multiples of 5 are 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45,…
Here, it is clear that the least common multiple is 15. So, the LCM of 3 and 5 is 15.
Now, to find the LCM of 1 and 15 using the listing method, first, we will write down the first few multiples of 1 and 15 separately. Out of all the multiples of 1 and 15, we will focus on the multiples which are common to both numbers.
Then, out of all the common multiples, we will take out the smallest common multiple. That will be the LCM of 1 and 15.
Here, it is clear that the least common multiple is 15. So, the LCM of 1 and 15 is 15.
Question 2: Find the LCM of 3 and 5 using the prime factorization method.
Solution:
To find the LCM of 3 and 5 using the prime factorization method, first, we will find the prime factors of 3 and 5 using the repeated division method. Then, we will write all the prime factors in their exponent forms and multiply the prime factors having the highest power. The final result after multiplication will be the LCM of 3 and 5.
• Prime factorization of 3 can be expressed as 3 = 31
• Prime factorization of 5 can be expressed as 3 = 51
So, the LCM of 3 and 5 = 3* 51= 3 * 5 = 15
Question 3: What is the least perfect square divisible by 3 and 5?
Solution:
The least number divisible by 3 and 5 is the LCM of 3 and 5, that is, 15.
Using prime factorization, we can expand and write the LCM of 3 and 5 as 3 * 5.
Here, we don’t get complete pairs of all numbers, so to make the pairs complete, we will multiply 3 and 5 with them.
Hence, the least perfect square divisible by 3 and 5 is
LCM(3, 5) * 3 * 5 = 3 * 5 * 3 * 5 = 225
Question 4: If the LCM of two numbers is 15, HCF is 1, and one of the numbers is 3. Find the other number.
Solution:
As we know,
product of two numbers = LCM * HCF
It is given that,
one of the numbers = 3, LCM = 15, and HCF = 1
Let the other number be x.
So, 3 * x = 15 * 1
x = (15 * 1) / 3
x = 15 / 3
x = 5
Hence, the other number is 5.
Question 5: Find the LCM of 3 and 5 using the division method.
Solution:
To find the LCM of 3 and 5 using the division method, first, we will find the smallest prime number which is divisible by either 3 or 5. If any of the numbers between 3 and 5 is not divisible by the respective prime number, we will write that number in the next row just below it and proceed further. We will continue dividing the numbers obtained after each step by the prime numbers until we get the result as 1 in the entire row. We will multiply all the prime numbers and the final result will be the LCM of 3 and 5.
So, the LCM of 3 and 5 = 3 * 5 = 15
Is 30 also considered as the LCM of 3 and 5?
No, 30 is not considered as the LCM of 3 and 5. 30 is a common multiple of 3 and 5. But, it is not the least common number which is divisible by 3 and 5, and while finding the LCM, you must focus on the lowest common number. So, 15 is the lowest common number divisible by 3 and 5.
What are the methods to find the LCM of 3 and 5?
There are 3 major methods for finding the LCM of 3 and 5:
1. Listing Method
2. Division Method
3. Prime Factorization Method
Are the LCM of 3 and 5 the same as the LCM of 1, 3, and 5?
LCM of 3 and 5 is 15 and LCM of 1, 3, and 5 is also 15. So, the LCM of 3 and 5 are the same as the LCM of 1, 3, and 5.
What is the LCM of 3 and 5?
LCM of 3 and 5 is 15.
Are LCM and HCF of 3 and 5 the same?
LCM of 3 and 5 is 15 and HCF of 3 and 5 is 1. So, LCM and HCF of 3 and 5 are not the same.
We hope you understand all the basics of how to find the LCM of 3 and 5.
Written by by
Prerit Jain
Share article on
|
## Friday, July 4, 2008
### Mathematical Synthesis
I know all of us know what long division is, but do you know what synthetic division is? Well it's a very fast method of calculation taught to me by my Additional Mathematics teacher in Secondary Three, Mr Michael Doyle. However, there are some limitations and downsides to using this method. I shall illustrate using an example:
Let's say I'd like to simplify the following fraction using long division first:
Notice that I haven't done it step by step because I'm assuming you, the reader, knows enough of long division already. Well notice the important things: we have a polynomial in x, being divided by a divisor D(x), which results in a quotient Q(x) and a remainder R(x).
We now introduce what we call the remainder theorem, that is, for any polynomial, we can write it as such:
If the remainder R(x) = 0, then we say that the divisor D(x) is a factor of the polynomial. So clearly for this particular polynomial we have it written as:
Now, that being done, we naturally have to ask ourselves: is there an easier method? Well, there is! And this method is known as synthetic division. We shall use the previous polynomial to illustrate. This method requires us to write down the following first:
Notice that I've written down the coefficients of the polynomial - these coefficients will represent our polynomial of interest. Now, to the left of these coefficients is the number -3, which you should recognise as representing the divisor (x + 3). We put -3 to represent (x + 3) because earlier we said that:
Now clearly when x = -3, notice that the divisor term (x + 3) goes to zero, and we're only left with the remainder term (x - 6). So to divide by (x + 3), a really simple method is to substitute in the value that makes (x + 3) = 0. Now, there is a basic limitation: clearly then the remainder is obtained no longer as a polynomial, but as a specific number. So this explains why long division is preferred in some instances.
No matter the case, we move on the next step, as shown below:
And of course, the next step:
And the last step:
Our final presentation is therefore:
Voila! Do you see how synthetic division has worked out the same exact answer as long division? Try it yourself today!
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## RS Aggarwal Class 7 Solutions Chapter 3 Decimals CCE Test Paper
These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper.
Other Exercises
Question 1.
Solution:
Cost of 1 pen = ₹ 32.50
Cost of 24 such pens = ₹ (32.50 x 24) = ₹ 80
Hence, the cost of 24 pens is ₹ 780.
Question 2.
Solution:
Distance covered by the bus in 1 hour = 64.5 km
Distance covered in 18h = (64.5 x 18) km = 1161 km
Hence, the bus can cover a distance of 1161 km in 18h.
Question 3.
Solution:
First, we will find the product 68 x 65 x 4
Now, 68 x 65 x 4 = 4420 x 4 = 17680
Sum of decimal places in the given decimals = (2 + 1 + 2) = 5
So, the product have five decimal places.
0.68 x 6.5 x 0.04 = 0.17680 = 0.1768
Question 4.
Solution:
Total weight of all the bags = 2231 kg
Weight of each bag = 48.5 kg
Question 5.
Solution:
Question 6.
Solution:
Product of the given decimals = 1.824
One decimal = 0.64
The other decimal = 1.824 ÷ 0.64
Hence, the other decimal is 2.85.
Question 7.
Solution:
Thickness of the pile of plywoods = 2.43 m = 2.43 x 100 cm = 243 cm
Thickness of one piece of plywood = 0.45 cm
Required no. of pieces of plywood
Hence, the required number of pieces of plywood is 540.
Question 8.
Solution:
Let the number of sides of the polygon be n.
Length of each side of the polygon = 3.8 cm
Perimeter of the polygon = (3.8 x n) cm
But it is given that its perimeter is 22.8 cm.
Hence, the given polygon has six sides.
Mark (✓) against the correct answer in each of the following :
Question 9.
Solution:
(b) 2.04
Question 10.
Solution:
(b) 1$$\frac { 1 }{ 125 }$$
Question 11.
Solution:
(c) 2.005 kg
2 kg 5 g = (2 x 1000) g + 5 g = (2005)g
= $$\frac { 2005 }{ 1000 }$$ kg = 2.005 kg
Question 12.
Solution:
(b) 0.08
We have :
Question 13.
Solution:
(c) 0.011
First, we will find the product 11 x 1 x 1
i.e. 11 x 1 x 1 = 11 x 1 = 11
Sum of decimal places in the given decimals = (1 + 1 + 2) = 4
1.1 x 0.1 x 0.01 = 0.0011 [4 places of decimal]
Question 14.
Solution:
(b) 2.03
Question 15.
Solution:
(c) is correct
Let the number added be x We have :
2.06 + x = 3.1
⇒ x = 3.1 – 2.06
Converting the given decimals into like decimals, we get:
2.06 and 3.10
Thus, required number = (3.10 – 2.06) = 1.04
Hence, 1.04 should be added to 2.06 to get 3.1.
Question 16.
Solution:
(b) 0.06 .
We have :
0.1 – x = 0.04
⇒ x = 0.1 – 0.04
Converting the given decimals into like decimals, we get:
0.10 and 0.04
Thus, required number = (0.10 – 0.04) = 0.06
Hence, 0.06 should be subtracted from 0.1 to get 0.04.
Question 17.
Solution:
(i) 1.001 ÷ 14 = 0.0715
47 x 53 = 2491
Sum of decimal places in the given decimals = (2 + 1) = 3
0.47 x 5.3 = 2.491
(iv) 0.023 x 0.03 = 0.69
Explanation first, we will multiply 23 by 3
23 x 3 = 69
Sum of decimal places in the given decimals = (3 + 2) = 5
0.023 x 0.03 =0.00069
(v) (0.7)2 = 0.69
Explanation : (0.7)2 = 0.7 x 0.7
First, we will find the product 0.7 x 0.7
Now, 7 x 7 = 49
Sum of decimal places in the given decimals = (1 + 1) = 2
So, the product must have two decimal places.
(0.7)2 = 0.7 x 0.7 = 0.49
(vi) (0.05)3 = 0.000125
Explanation : First, we will find the
product 0.05 x 0.05 x 0.05
Now, 5 x 5 x 5 = 125
Sum of decimal places in the given decimals = (2 + 2 + 2) = 6
So, the product must have six decimal places.
(0.05)2 = 0.05 x 0.05 x 0.05 = 0.000125
Question 18.
Solution:
(i) False
We have :
0.5 x 0.05 Now, 5 x 5 = 25
Sum of decimal places in the given decimals = (1 + 2) = 3
0.5 x 0.5 = 0.025
(ii) True
We have :
0.25 x 0.8
Now, 25 x 8 = 200
Sum of decimal places in the given decimals = (2 + 1) = 3
0.25 x 0.8 = 0.200 = 0.2
(iii) True
We have :
0.35 ÷ 0.7
(iv) False We have :
0.4 x 0.4 x 0.4
Now, 4 x 4 x 4 = 64
Sum of decimal places in the given decimals = (1 + 1 + 1) = 3
0.4 x 0.4 x 0.4 = 0.064
(v) True
6 cm = $$\frac { 6 }{ 100 }$$ m = 0.06 m
Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
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# Scientific Notation Multiplication,Division, Addition and Subtraction
The convert to scientific notation page showed examples of how to write both large and small numbers in scientific notation form.
This Scientific Notation Math page gives a summary of how to deal with sums that involve numbers in scientific form.
They aren't too difficult to get the hang of, provided you have a good understanding of scientific notation itself.
## Scientific Notation Addition & Subtraction
With adding and subtracting numbers in scientific notation form, the numbers involved need to have the same exponent/power.
- 2 × 105 + 3 × 105 = ( 2 + 3 ) × 105 = 5 × 105
- 7 × 105 − 3 × 105 = ( 73 ) × 105 = 4 × 105
- 6 × 106 &plus 5 × 103 = ( 6 &plus 5 ) × 103 = 11 × 103
Which can be simplified further to 1.1 × 104
## Scientific Notation MultiplicationScientific Notation Division
When dealing with scientific notation multiplication and division, the powers/exponents involved do not need to be the same value.
## MULTIPLICATION
When dealing with multiplication, multiply the like numbers together separately, then combine them for the final answer.
a × 10m × a × 10m = ( a × a ) × ( 10m × 10m )
Examples
(1.1)
2 × 104 × 4.2 × 103
= ( 2 × 4.2 ) × ( 104 × 103 ) = 8.4 × ( 104+3 )
= 8.4 × 107
(1.2)
4 × 103 × 4 × 102
= ( 4 × 4 ) × ( 103 × 102 ) = 16 × ( 103+2 )
= 16 × 105
Which can be simplified further to give: 1.6 × 106
## DIVISION
In a similar approach to multiplication, with division deal with each of the like terms separately in the sum.
Examples
(2.1)
9 × 106 &div 3 × 104
This division sum can be set up in fraction form.
=> \bf{\frac{9 \space \times \space 10^6}{3 \space \times \space 10^4}}
= \bf{\frac{9}{3}} × \bf{\frac{10^6}{10^4}} = 3 × 106−4 = 3 × 102
(2.2)
4.2 × 103 &div 2 × 104
This division sum can be set up in fraction form.
=> \bf{\frac{4.2 \space \times \space 10^3}{2 \space \times \space 10^4}}
= \bf{\frac{4.2}{2}} × \bf{\frac{10^3}{10^4}} = 2.1 × 103−4 = 2.1 × 10-1
1. Home
2. ›
3. Arithmetic 2
4. › Scientific Notation Multiplication
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# The Collatz Conjecture: An Unsolved Math Mystery
In the vast and often perplexing realm of mathematics, there exist problems that captivate the minds of mathematicians and puzzle enthusiasts alike. One such problem, known as the Collatz Conjecture, stands as a testament to the enduring mystery and beauty of numbers. This seemingly simple conjecture has defied proof or disproof for decades, leaving mathematicians with a tantalizing challenge.
## What is the Collatz Conjecture?
The Collatz Conjecture, also known as the 3n + 1 problem, is a statement about the behavior of positive integers under a specific set of rules. It states that starting with any positive integer, repeatedly applying the following rules will eventually lead to the number 1:
1. If the number is even, divide it by 2.
2. If the number is odd, multiply it by 3 and add 1.
1. 7 is odd, so we multiply by 3 and add 1: 7 * 3 + 1 = 22
2. 22 is even, so we divide by 2: 22 / 2 = 11
3. 11 is odd, so we multiply by 3 and add 1: 11 * 3 + 1 = 34
4. 34 is even, so we divide by 2: 34 / 2 = 17
5. 17 is odd, so we multiply by 3 and add 1: 17 * 3 + 1 = 52
6. 52 is even, so we divide by 2: 52 / 2 = 26
7. 26 is even, so we divide by 2: 26 / 2 = 13
8. 13 is odd, so we multiply by 3 and add 1: 13 * 3 + 1 = 40
9. 40 is even, so we divide by 2: 40 / 2 = 20
10. 20 is even, so we divide by 2: 20 / 2 = 10
11. 10 is even, so we divide by 2: 10 / 2 = 5
12. 5 is odd, so we multiply by 3 and add 1: 5 * 3 + 1 = 16
13. 16 is even, so we divide by 2: 16 / 2 = 8
14. 8 is even, so we divide by 2: 8 / 2 = 4
15. 4 is even, so we divide by 2: 4 / 2 = 2
16. 2 is even, so we divide by 2: 2 / 2 = 1
As you can see, starting with 7, we eventually reach the number 1. This has been observed to be true for countless other starting numbers, but no one has been able to prove that it holds for all positive integers.
## The Challenge of Proof
The difficulty in proving the Collatz Conjecture lies in the chaotic nature of the sequence generated by the rules. While the conjecture appears simple, the resulting sequences can exhibit unpredictable patterns. The problem is further complicated by the fact that there is no known mathematical tool or technique that can directly address the conjecture.
## The Fascination of the Unsolved
Despite its lack of a formal proof, the Collatz Conjecture has captivated mathematicians and computer scientists for decades. Its simplicity, combined with its resistance to solution, has made it a popular subject of research and exploration. Many have attempted to prove or disprove the conjecture, but no one has yet succeeded.
The Collatz Conjecture serves as a reminder that even in a field as seemingly well-defined as mathematics, there exist mysteries that continue to elude our understanding. It is this very mystery that fuels the ongoing pursuit of knowledge and the relentless search for answers.
## Conclusion
The Collatz Conjecture is a fascinating and enduring mystery in the world of mathematics. Its simple rules and unpredictable behavior have made it a captivating challenge for mathematicians and enthusiasts alike. While the conjecture remains unproven, it continues to inspire curiosity, exploration, and the relentless pursuit of mathematical truth.
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# What does it mean if something varies jointly
## How do you find varies jointly?
Equation for a joint variation is X = KYZ where K is constant. One variable quantity is said to vary jointly as a number of other variable quantities, when it varies directly as their product. If the variable A varies directly as the product of the variables B, C and D, i.e., if.
## What does it mean when something varies inversely?
The statement “y varies inversely as x means that when x increases, ydecreases by the same factor. In other words, the expression xy is constant: xy = k.
## What is the meaning of combined variation?
Combined variation describes a situation where a variable depends on two (or more) other variables, and varies directly with some of them and varies inversely with others (when the rest of the variables are held constant).
## What statement indicates that a relationship varies jointly?
A General Note: Joint Variation
Joint variation occurs when a variable varies directly or inversely with multiple variables. For instance, if x varies directly with both y and z, we have x = kyz. If x varies directly with y and inversely with z, we have x = k y z displaystyle x=frac{ky}{z} x=zky.
## Which describes a relationship that varies directly?
(Some textbooks describe direct variation by saying ” y varies directly as x “, ” y varies proportionally as x “, or ” y is directly proportional to x . “) This means that as x increases, y increases and as x decreases, y decreases—and that the ratio between them always stays the same.
## What is the difference between joint and combined?
Joint variation is similar to direct variation. It involves two or more variables, such as y=k(xz). Combined variation combines direct and inverse variation, y=kx/z.
## What is an example of combined variation in real life?
An example of combined variation in the physical world is the Combined Gas Law, which relates pressure, temperature, volume, and moles (amount of molecules) of a gas.
## How do you know if a situations or problem is direct or inverse variations?
In direct variation, as one number increases, so does the other. This is also called direct proportion: they’re the same thing. … In inverse variation, it’s exactly the opposite: as one number increases, the other decreases.
## How many quantities does joint variation relate?
Joint variation occurs when one quantity is directly proportional to two or more quantities.
## What is the initial step in solving joint and combined variation?
Write the general variation formula of the problem. Find the constant of variation k. Rewrite the formula with the value of k. Solve the problem by inputting known information.
## What are some examples of inverse variation in real life?
Some situations of inverse variation: More men at work, less time taken to finish the work.Less men at work, more time is taken to finish the work. More speed, less time is taken to cover the same distance.
## What is a real life example of direct variation and inverse variation?
It is an example of direct variation. If family has less members, more saving (provided that the family has the same amount of income). More members, less saving ( income is still the same). It is an inverse variation.
## How are direct inverse joint and combined variations differ from each other?
Inverse or Indirect Variation, where when one of the variables increases, the other one decreases (their product is constant) Joint Variation, where more than two variables are related directly. Combined Variation, which involves a combination of direct or joint variation, and indirect variation.
## What have you learned about inverse variation?
The main idea in inverse variation is that as one variable increases the other variable decreases. That means that if x is increasing y is decreasing, and if x is decreasing y is increasing. The number k is a constant so it’s always the same number throughout the inverse variation problem.
## What does jointly proportional mean?
When we say z is jointly proportional to a set of variables, it means that z is directly proportional to each variable taken one at a time. If z varies jointly with respect to x and y, the equation will be of the form z = kxy (where k is a constant). Equation: c = 5ab.
## Why is direct and inverse variation important?
A direct and inverse proportion are used to show how the quantities and amount are related to each other. They are also mentioned as directly proportional or inversely proportional. The symbol used to denote the proportionality is ‘∝’.
## How would you decide whether or not two quantities vary directly?
When two variables are related in such a way that the ratio of their values always remains the same, the two variables are said to be in direct variation. In simpler terms, that means if A is always twice as much as B, then they directly vary.
## How do you do inverse variations?
An inverse variation can be represented by the equation xy=k or y=kx . That is, y varies inversely as x if there is some nonzero constant k such that, xy=k or y=kx where x≠0,y≠0 . Suppose y varies inversely as x such that xy=3 or y=3x . That graph of this equation shown.
## How do you know if a relationship is proportional?
How Do You Know If Two Ratios are Proportional? Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional.
## How do you prove inverse proportionality?
We know that in the inverse proportion, x × y= k. This means that x = k/y. So, to find the value of the k, you can use the known values and then use the formula above to calculate all the unknown values.
## How do you find varies jointly?
Equation for a joint variation is X = KYZ where K is constant. One variable quantity is said to vary jointly as a number of other variable quantities, when it varies directly as their product. If the variable A varies directly as the product of the variables B, C and D, i.e., if.
## What does it mean when something varies inversely?
The statement “y varies inversely as x means that when x increases, ydecreases by the same factor. In other words, the expression xy is constant: xy = k.
## What is the meaning of combined variation?
Combined variation describes a situation where a variable depends on two (or more) other variables, and varies directly with some of them and varies inversely with others (when the rest of the variables are held constant).
## What statement indicates that a relationship varies jointly?
A General Note: Joint Variation
Joint variation occurs when a variable varies directly or inversely with multiple variables. For instance, if x varies directly with both y and z, we have x = kyz. If x varies directly with y and inversely with z, we have x = k y z displaystyle x=frac{ky}{z} x=zky.
## Which describes a relationship that varies directly?
(Some textbooks describe direct variation by saying ” y varies directly as x “, ” y varies proportionally as x “, or ” y is directly proportional to x . “) This means that as x increases, y increases and as x decreases, y decreases—and that the ratio between them always stays the same.
## What is the difference between joint and combined?
Joint variation is similar to direct variation. It involves two or more variables, such as y=k(xz). Combined variation combines direct and inverse variation, y=kx/z.
## What is an example of combined variation in real life?
An example of combined variation in the physical world is the Combined Gas Law, which relates pressure, temperature, volume, and moles (amount of molecules) of a gas.
## How do you know if a situations or problem is direct or inverse variations?
In direct variation, as one number increases, so does the other. This is also called direct proportion: they’re the same thing. … In inverse variation, it’s exactly the opposite: as one number increases, the other decreases.
## How many quantities does joint variation relate?
Joint variation occurs when one quantity is directly proportional to two or more quantities.
## What is the initial step in solving joint and combined variation?
Write the general variation formula of the problem. Find the constant of variation k. Rewrite the formula with the value of k. Solve the problem by inputting known information.
## What are some examples of inverse variation in real life?
Some situations of inverse variation: More men at work, less time taken to finish the work.Less men at work, more time is taken to finish the work. More speed, less time is taken to cover the same distance.
## What is a real life example of direct variation and inverse variation?
It is an example of direct variation. If family has less members, more saving (provided that the family has the same amount of income). More members, less saving ( income is still the same). It is an inverse variation.
## How are direct inverse joint and combined variations differ from each other?
Inverse or Indirect Variation, where when one of the variables increases, the other one decreases (their product is constant) Joint Variation, where more than two variables are related directly. Combined Variation, which involves a combination of direct or joint variation, and indirect variation.
## What have you learned about inverse variation?
The main idea in inverse variation is that as one variable increases the other variable decreases. That means that if x is increasing y is decreasing, and if x is decreasing y is increasing. The number k is a constant so it’s always the same number throughout the inverse variation problem.
## What does jointly proportional mean?
When we say z is jointly proportional to a set of variables, it means that z is directly proportional to each variable taken one at a time. If z varies jointly with respect to x and y, the equation will be of the form z = kxy (where k is a constant). Equation: c = 5ab.
## Why is direct and inverse variation important?
A direct and inverse proportion are used to show how the quantities and amount are related to each other. They are also mentioned as directly proportional or inversely proportional. The symbol used to denote the proportionality is ‘∝’.
## How would you decide whether or not two quantities vary directly?
When two variables are related in such a way that the ratio of their values always remains the same, the two variables are said to be in direct variation. In simpler terms, that means if A is always twice as much as B, then they directly vary.
## How do you do inverse variations?
An inverse variation can be represented by the equation xy=k or y=kx . That is, y varies inversely as x if there is some nonzero constant k such that, xy=k or y=kx where x≠0,y≠0 . Suppose y varies inversely as x such that xy=3 or y=3x . That graph of this equation shown.
## How do you know if a relationship is proportional?
How Do You Know If Two Ratios are Proportional? Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional.
## How do you prove inverse proportionality?
We know that in the inverse proportion, x × y= k. This means that x = k/y. So, to find the value of the k, you can use the known values and then use the formula above to calculate all the unknown values.
What does it mean if something varies jointly
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# Time and Work Problem Question Answers Test – Page 2
Time and Work Problem Question Answers Test – Page 2
## Time and Work 2
Congratulations - you have completed Time and Work 2. You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%%
Question 1
2 men can complete a piece of work in 6 days. 2 women can complete the same piece of work in 9 days, whereas 3 children can complete the same piece of work in 8 days. 3 women and 4 children worked together for 1 day. If only men were to finish the remaining work in 1 day, how many total men would be required?
A 4 B 8 C 6 D Cannot be determined E None of These
Question 1 Explanation:
2M × 6 = 2W × 9 = 3C × 8 => 12M = 18W = 24C 12M = 18W 12M = 24C => 2M = 3W 1M = 2C But 3W + 4C = 2M + 2M = 4M If x be the number of men required to finish the remaining work in 1 day 2 × 6 = 4 × 1 + x ×1 => x = 12 – 4 = 8
Question 2
X can do a piece of work in 24 days. If Y works twice as fast as X, how long would they take to finish the work working together?
A 8 days B 12 days C 16 days D 48 days E 36 days
Question 3
Yesterday Vani completed 300 units of work at the rate of 15 units per minute. Today she completed the same units of work but her speed was 40% faster than yesterday. What is the approximate difference in the time she took to complete the work yesterday and the time she took today?
A 16 minutes B 26 minutes C 46 minutes D 36 minutes E 6 Minutes
Question 3 Explanation:
Yesterday time taken by Vani to finish the work = 300/15 = 20 Today she is 40% faster i.e. 140% as efficient as the other day Today time taken by her to finish the work = 20/(140%) = (20 ×100)/140 ≅ 14min The required difference = 20 – 14 = 6 min SHORTCUT METHOD: Applying the same formula X = 300/15 = 20 K = (100 + 40)% = 140/100 = 1.4 So today Vani can finish the work in 20/(1.4 ) ≅14 days Required difference = 20 – 14 = 6 min
Question 4
A job can be completed by 12 men in 12 days. How many extra days will be needed to complete the job in 6 men leave after working for 6 days?
A 3 B 6 C 12 D 24 E None of These
Question 5
A builder decided to build a house in 50 days. He employed 150 men at the beginning and another 80 men after 20 days and completed the work in stipulated time. If he had not employed the additional men, how many days behind schedule would it have been finished?
A 10 days B 12 days C 15 days D 16 days E 18 days
Question 6
Two men alone or three women alone can complete a piece of work in 4 days. In how many days can 1 woman and one man together complete the same piece of work?
A 6 days B 24/5 days C 12/1.75 days D Cannot be determined E None of These
Question 7
If one man or three women or five boys can do a piece of work in 46 days then how many days will one man, one woman and one boy together take to complete the same piece of work?
A 30 days B 32 days C 35 days D 40 days E None of These
Question 8
8 men can complete a piece of work in 20 days. 8 women can complete the same piece of work in 32 days. In how many days will 5 men and 8 women together complete the same work?
A 16 days B 12 days C 14 days D 10 days E None of These
Question 9
2 men can complete a piece of work in 6 days. 2 women can complete the same piece of work in 9 days, whereas 3 children can complete the same piece of work in 8 days. 3 women and 4 children worked together for 1 day. If only men were to finish the remaining work in 1 day, how many total men would be required?
A 4 B 8 C 6 D Cannot be determined E None of These
Question 10
X can do a piece of work in 24 days. If Y works twice as fast as X, how long would they take to finish the work working together?
A 8 days B 12 days C 16 days D 48 days E 36 days
Question 11
Yesterday Vani completed 300 units of work at the rate of 15 units per minute. Today she completed the same units of work but her speed was 40% faster than yesterday. What is the approximate difference in the time she took to complete the work yesterday and the time she took today?
A 16 minutes B 26 minutes C 46 minutes D 36 minutes E 6 minutes
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Simplify $\sum_{i=0}^{n-1} { {2n}\choose{i}}\cdot x^i$
I am trying to simplify an expression involving summation as follows:
$$\sum_{i=0}^{n-1} { {2n}\choose{i}}\cdot x^i$$
where $n$ is an integer, and $x$ is a positive real number.
At a first glance, I can see that
$$\sum_{i=0}^{2n} { {2n}\choose{i}}\cdot x^i = {(1+x)}^{2n}.$$
But what if in the case when $i$ goes from 0 to $n-1$?
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If you sum from $0$ to $n-1$, then no longer you get an easy espression. Instead, you can get the sum in terms of the hypergeometric function
$$\left( x+1 \right) ^{2\,n}-{2\,n\choose n}{x}^{n} {F (1,-n;\,n+1;\,-x)}\,.$$
Let $S_1(x):=\sum_{i=0}^{n-1}\binom{2n}ix^i$, and $S_2(x):=\sum_{i=n+1}^{2n}\binom{2n}ix^i$. Then writing $j=2n-i$, we get $$S_2(x)=\sum_{j=0}^{n-1}\binom{2n}jx^{2n-j}=S_1(x^{-1})x^{2n}.$$ So $$(1+x)^{2n}=S_1(x)+\binom{2n}nx^n+S_2(x)=S_1(x)+\binom{2n}nx^n+S_1(x^{-1})x^{2n}.$$ Writing $f(x):=\frac{S_1(x)}{x^n}$, we get that $f$ satisfies the functional equation $$f(x)+f(x^{-1})=\left(1+\frac 1x\right)^n(1+x)^n-\binom{2n}n.$$
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Quarter Wit, Quarter Wisdom: Circular Arrangement Constraints – Part I
In the last two posts, we discussed how to easily handle constraints in linear arrangements. Today we will discuss how to handle constraints in circular arrangements, which are actually even simpler to sort out. Let’s look at some examples.
Question 1: Seven people are to be seated at a round table. Andy and Bob don’t want to sit next to each other. How many seating arrangements are possible?
Solution: There are 7 people who need to be seated around a circular table. Number of arrangements in which 7 people can be seated around a circular table = (7-1)! = 6!
(If you are not sure how we got this, check out this post.)
We need to find the number of arrangements in which two of them do not sit together. Let us instead find the number of arrangements in which they will sit together. We will then subtract these arrangements from the total 6! arrangements. Consider Andy and Bob to be one unit. Now we need to arrange 6 units around a round table. We can do this in 5! ways. But Andy and Bob can swap places so we need to multiply 5! by 2.
Number of arrangements in which Andy and Bob do sit next to each other = 2*5!
So, number of arrangements in which Andy and Bob don’t sit next to each other = 6! – 2*5!
This is very similar to the way we handled such constraints in linear arrangements.
Question 2: There are 6 people, A, B, C, D, E and F. They have to sit around a circular table such that A cannot sit next to D and F at the same time. How many such arrangements are possible?
Solution: Total number of ways of arranging 6 people in a circle = 5! = 120
Now, A cannot sit next to D and F simultaneously.
Let’s first find the number of arrangements in which A sits between D and F. In how many of these 120 ways will A be between D and F? Let’s consider that D, A and F form a single unit. We make DAF sit on any three consecutive seats in 1 way and make other 3 people sit in 3! ways (since the rest of the 3 seats are distinct). But D and F can swap places so the number of arrangements will actually be 2*3! = 12
In all, we can make A sit next to D and F simultaneously in 12 ways.
The number of arrangements in which A is not next to D and F simultaneously is 120 – 12 = 108.
A slight variation of this question that would change the answer markedly is the following:
Question 3: There are 6 people, A, B, C, D, E and F. They have to sit around a circular table such that A can sit neither next to D nor next to F. How many such arrangements are possible?
Solution: In the previous question, A could sit next to D and F; the only problem was that A could not sit next to both of them at the same time. Here, A can sit next to neither D nor F. Generally, it is difficult to wrap your head around what someone cannot do. It is easier to consider what someone can do and go from there. A cannot sit next to D and F so he will sit next to two of B, C and E.
Let’s choose two out of B, C and E. In other words, let’s drop one of B, C and E. We can drop one of B, C and E in 3 ways (we can drop B or C or E). This means, we can choose two out of B, C and E in 3 ways (We will come back to choosing 2 people out of 3 when we work on combinations). Now, we can arrange the two selected people around A in 2 ways (say we choose B and C. We could have BAC or CAB). We make these three sit on any three consecutive seats in 1 way.
Number of ways of choosing two of B, C and E and arranging the chosen two with A = 3*2 = 6
The rest of the three people can sit in three distinct seats in 3! = 6 ways
Total number of ways in which A will sit next to only B, C or E (which means A will sit neither next to D nor next to F) = 6*6 = 36 ways
Now we will look at one last example.
Question 4: Six people are to be seated at a round table with seats arranged at equal distances. Andy and Bob don’t want to sit directly opposite to each other. How many seating arrangements are possible?
Solution: Directly opposite means that Andy and Bob cannot sit at the endpoints of the diameter of the circular table.
Total number of arrangements around the circular table will be (6-1)! = 5!
But some of these are not acceptable since Andy sits opposite Bob in these. Let us see in how many cases Andy doesn’t sit opposite Bob. Let’s say we make Andy sit first. He can sit at the table in 1 way since all the seats are exactly identical for him. Now there are 5 seats left but Bob can take a seat in only 4 ways since he cannot occupy the seat directly opposite Andy. Now there are 4 people left and 4 distinct seats left so they can be occupied in 4! ways.
Total number of ways of arranging the 6 people such that Andy does not sit directly opposite Bob = 1*4*4! = 96 arrangements.
Make sure you understand the logic used in this question. We will build up on it in the next post.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches GMAT prep for Veritas Prep and regularly participates in content development projects such as this blog!
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6.
A, B and C completed a piece of work costing Rs.1800. A worked for 6 days, B for 4 days and C for 9 days. If their daily wages are in the ratio 5 : 6 : 4, how much amount will be received by A?
A. Rs.800 B. Rs.600 C. Rs.900 D. Rs.750
Explanation:
7.
A and B can finish a piece of work in 30 days. They worked at it for 20 days and then B left. The remaining work was done by A alone in 20 more days. A alone can finish the work in
A. 48 days B. 50 days C. 54 days D. 60 days
Explanation:
8.
A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is :
A. 1/5 B. 1/15 C. 6/15 D. 8/15
Explanation: A's 1 day's work = 1/15; B's 1 day's work = 1/20; (A + B)'s 1 day's work = (1/15 + 1/20) = 7/60. (A + B)'s 4 day's work = 7/60x4 =7/15. Therefore, Remaining work = (1-7/15) = 8/15.
9.
A man, a woman and a boy can together complete a piece of work in 3 days. If a man alone can do it in 6 days and a boy alone in 18 days, how long will a woman take to complete the work?
A. 9 days B. 21 days C. 24 days D. 27 days
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### Theory:
The Lowest Common Multiple (LCM) of two or more given numbers is the smallest of their common multiples. It is also called the least common multiple.
To find the LCM of two or more numbers, we can use any of the following methods.
1. Common multiple method
2. Prime factorisation method
3. Division method
Common multiple method:
Find the lowest common multiple of the numbers $$8$$ and $$12$$.
Multiple of $$8$$ $$=$$ $$8$$, $$16$$, $$24$$, $$32$$, $$40$$, $$48$$, $$56$$, $$64$$, $$72$$, $$80$$, ...
Multiple of $$12$$ $$=$$ $$12$$, $$24$$, $$36$$, $$48$$, $$60$$, $$72$$, $$84$$, …
The common multiple of $$8$$, $$12$$ and $$18$$ are $$24$$, $$48$$, $$72$$, ...
The lowest common multiple of $$8$$ and $$12$$ is $$24$$.
Prime factorisation method:
Find the LCM of $$4$$, $$8$$ and $$12$$.
First, write each number as a product of its prime factors.
Next, select the common multiples of $$4$$, $$8$$ and $$12$$.
Then, select the numbers which are left uncommon.
Finally, multiply all the selected numbers. You will get the lowest common multiple of those three numbers.
Therefore, LCM of $$4$$, $$8$$ and $$12$$ $$=$$ $$24$$.
Division method:
The division method is a very convenient method of finding the LCM. We proceed as follows with this method of prime factorisation:
• Arrange all the given numbers in a row and separated by commas.
• Start with the lowest prime number which at least divides exactly one of the numbers.
• Write down the quotients and any undivided numbers in the next line.
• The only common factor is to repeat the process, as shown below to $$1$$.
• Find the product of all the divisors. This is the required LCM.
Example:
Find the LCM of $$8$$, $$15$$ and $$24$$.
Solution:
LCM $$=$$ $$2 \times 2 \times 2 \times 3 \times 5 = 120$$
Therefore, LCM of $$8$$, $$15$$ and $$24$$ is $$120$$.
The LCM is the product of both numbers when two numbers are co-prime.
Example:
Let us see the LCM of $$7$$ and $$13$$.
$$7 = 7×1$$ and $$13=1×13$$
Both are co-prime numbers.
LCM is the product of both numbers.
$$7×13=91$$
LCM of $$7$$ and $$13$$ is $$91$$.
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# Get Answers to all your Questions
#### Provide solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.3 question 6
$y=A e^{B x}$ is a solution of differential equation
Hint:
Differentiate the given solution of differential equation on both sides with respect to $x$
Given:
$y=A e^{B x}$ is a solution.
Solution:
Differentiating on both sides with respect to $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(A e^{B x}\right) \quad\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$
\begin{aligned} &\frac{d y}{d x}=\left[A \frac{d\left(e^{B x}\right)}{d x}+e^{B x} \frac{d A}{d x}\right] \\\\ &\frac{d y}{d x}=B A e^{B x}+0 \\\\ &\frac{d y}{d x}=B A e^{B x} \end{aligned} ..............(i)
Now to obtain the second order derivative, differentiate equation (i) with respect to $x$
\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\frac{d}{d x}\left(B A e^{B x}\right) \\\\ \frac{d^{2} y}{d x^{2}} &=\left[B A \frac{d\left(e^{B x}\right)}{d x}+e^{B x} \frac{d(B A)}{d x}\right] \end{aligned}
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=B^{2} A e^{B x}+0 \\\\ &\frac{d^{2} y}{d x^{2}}=B^{2} A e^{B x} \end{aligned} .......(ii)
Now put both equation (i) and (ii) in given differential equation as follows
\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{1}{y}\left(\frac{d y}{d x}\right)^{2} \\\\ &B^{2} A e^{B x}=\frac{1}{A e^{B x}}\left(B A e^{B x}\right)^{2} \end{aligned}
$R H S=\frac{1}{A e^{B x}}\left(B A e^{B x}\right)^{2}$
\begin{aligned} &=\frac{B^{2}\left(A e^{B x}\right)\left(A e^{B x}\right)}{A e^{B x}} \\\\ &=B^{2} A e^{B x} \\\\ &=L H S \end{aligned}
Thus, $y=A e^{B x}$ is a solution of differential equation.
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#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 48 sub question (iii) math
$A=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given:
$\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right] A=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]$
We know that two matrices B and C are eligible for the product BC only when number of columns of B is equal to number or rows of C.
So, from the given definition we can consider that the order of matrix A is $1 \times 3$ i.e. we can assume $\therefore\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]_{3 \times 1}\left[\begin{array}{lll}x_{1} & x_{2} & x_{3}\end{array}\right]_{1 \times 3}=\left[\begin{array}{ccc}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]_{3 \times 3}$
$\\\\ \Rightarrow\left[\begin{array}{lll} 4\left(x_{1}\right) & 4\left(x_{2}\right) & 4\left(x_{3}\right) \\ 1\left(x_{1}\right) & 1\left(x_{2}\right) & 1\left(x_{3}\right) \\ 3\left(x_{1}\right) & 3\left(x_{2}\right) & 3\left(x_{3}\right) \end{array}\right]=\left[\begin{array}{ccc} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{ccc} 4 x_{1} & 4 x_{2} & 4 x_{3} \\ x_{1} & x_{2} & x_{3} \\ 3 x_{1} & 3 x_{2} & 3 x_{3} \end{array}\right]=\left[\begin{array}{ccc} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]$
Equating the corresponding element of the two matrices, we have
$x_{1}=-1, x_{2}=2, x_{3}=1$
So, matrix $A=\left[\begin{array}{lll} -1 & 2 & 1\end{array}]\right.$
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# Odds and Probability: Commonly Misused Terms in Statistics – An Illustrative Example in Baseball
Yesterday, all 15 home teams in Major League Baseball won on the same day – the first such occurrence in history. CTV News published an article written by Mike Fitzpatrick from The Associated Press that reported on this event. The article states, “Viewing every game as a 50-50 proposition independent of all others, STATS figured the odds of a home sweep on a night with a full major league schedule was 1 in 32,768.” (Emphases added)
Screenshot captured at 5:35 pm Vancouver time on Wednesday, August 12, 2015.
Out of curiosity, I wanted to reproduce this result. This event is an intersection of 15 independent Bernoulli random variables, all with the probability of the home team winning being 0.5.
$P[(\text{Winner}_1 = \text{Home Team}_1) \cap (\text{Winner}_2 = \text{Home Team}_2) \cap \ldots \cap (\text{Winner}_{15}= \text{Home Team}_{15})]$
Since all 15 games are assumed to be mutually independent, the probability of all 15 home teams winning is just
$P(\text{All 15 Home Teams Win}) = \prod_{n = 1}^{15} P(\text{Winner}_i = \text{Home Team}_i)$
$P(\text{All 15 Home Teams Win}) = 0.5^{15} = 0.00003051757$
Now, let’s connect this probability to odds.
It is important to note that
• odds is only applicable to Bernoulli random variables (i.e. binary events)
• odds is the ratio of the probability of success to the probability of failure
For our example,
$\text{Odds}(\text{All 15 Home Teams Win}) = P(\text{All 15 Home Teams Win}) \ \div \ P(\text{At least 1 Home Team Loses})$
$\text{Odds}(\text{All 15 Home Teams Win}) = 0.00003051757 \div (1 - 0.00003051757)$
$\text{Odds}(\text{All 15 Home Teams Win}) = 0.0000305185$
The above article states that the odds is 1 in 32,768. The fraction 1/32768 is equal to 0.00003051757, which is NOT the odds as I just calculated. Instead, 0.00003051757 is the probability of all 15 home teams winning. Thus, the article incorrectly states 0.00003051757 as the odds rather than the probability.
This is an example of a common confusion between probability and odds that the media and the general public often make. Probability and odds are two different concepts and are calculated differently, and my calculations above illustrate their differences. Thus, exercise caution when reading statements about probability and odds, and make sure that the communicator of such statements knows exactly how they are calculated and which one is more applicable.
### 8 Responses to Odds and Probability: Commonly Misused Terms in Statistics – An Illustrative Example in Baseball
1. Kevin Brogle says:
The probability is greater than presented as the likelihood of the home team winning in baseball is ~55% rather than the estimated 50%.
• Hi Kevin,
You may be right (source please!), but the original assumption in the article is that the probability of winning is the same for both teams, so I wanted to reproduce the calculation based on that statement.
2. Pingback: Distilled News | Data Analytics & R
3. Stephen Denton says:
Hello Eric,
I don’t know I would make a big deal about this. The statement 1 in 32,768 is the correct frequency based interpretation of the likelihood of the event (1 state out of 2^15 possible states was realized). So it comes down to how the word “in” is translated into mathematics. You are correct that if I interpret “in” strictly as “divided by” then I am reporting a probability (and labelling it odds seem wrong). And technically odds would be the number of states realized over the number of states not realized (1/(2^15 – 1) = 1/32767 = 0.0000305185). But if your suggestion is to report these odds as 1 in 32767, then you are no longer conforming to the frequency based interpretation (which most people find more intuitive and interpretable). So, perhaps the correct mathematical conversion of the word “in” when odds is wanted is to subtract the realized states (n_A) from the total number of possible states (n_T) and use that as the denominator (i.e., if you see “n_A in n_T” then odds is calculated as n_A / (n_T – n_A) ).
Regardless, the media is perfectly right to use the frequency based interpretation, whether they call it odds or probabilities seems kind of hopeless to decry; both can be easily calculated from the reported frequency information.
4. Stephen Denton says:
I guess you could of course just say the odds are 1 TO 32767 and eliminate the ambiguity…
• Hi Stephen,
1) My objective in this blog post is to distinguish between probability and odds, which are 2 different concepts. This distinction holds regardless of how you interpret probability.
2) I agree – we should use the preposition “in” for probability and “to” for odds. I did this in my blog post; however, I appreciate you pointing this out.
Eric
• Stephen Denton says:
I guess my point is that probability and odds are not two ‘different’ concepts, they are two closely related concepts and confusion is inevitable.
• Hi Stephen,
1) Probability applies to many types of variables, not just binary variables. Odds applies to binary variables only. Thus, I argue that they are still 2 different concepts, however related they may be.
2) The inevitability of confusion does not excuse improper use of language. Let’s correct wrong word usage when we see it, regardless of how easy or inevitable it may be.
Eric
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# Permutations and Combinations Questions
FACTS AND FORMULAE FOR PERMUTATIONS AND COMBINATIONS QUESTIONS
1. Factorial Notation: Let n be a positive integer. Then, factorial n, denoted n! is defined as: n!=n(n - 1)(n - 2) ... 3.2.1.
Examples : We define 0! = 1.
4! = (4 x 3 x 2 x 1) = 24.
5! = (5 x 4 x 3 x 2 x 1) = 120.
2. Permutations: The different arrangements of a given number of things by taking some or all at a time, are called permutations.
Ex1 : All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
Ex2 : All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)
Number of Permutations: Number of all permutations of n things, taken r at a time, is given by:
$P_{r}^{n}=n\left(n-1\right)\left(n-2\right)....\left(n-r+1\right)=\frac{n!}{\left(n-r\right)!}$
Ex : (i) $P_{2}^{6}=\left(6×5\right)=30$ (ii) $P_{3}^{7}=\left(7×6×5\right)=210$
Cor. number of all permutations of n things, taken all at a time = n!.
Important Result: If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,
such that $\left({p}_{1}+{p}_{2}+...+{p}_{r}\right)=n$
Then, number of permutations of these n objects is :
3. Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.
Ex.1 : Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.
Note that AB and BA represent the same selection.
Ex.2 : All the combinations formed by a, b, c taking ab, bc, ca.
Ex.3 : The only combination that can be formed of three letters a, b, c taken all at a time is abc.
Ex.4 : Various groups of 2 out of four persons A, B, C, D are : AB, AC, AD, BC, BD, CD.
Ex.5 : Note that ab ba are two different permutations but they represent the same combination.
Number of Combinations: The number of all combinations of n things, taken r at a time is:
Note : (i) (ii)$C_{r}^{n}=C_{\left(n-r\right)}^{n}$
Examples : (i) $C_{4}^{11}=\frac{11×10×9×8}{4×3×2×1}=330$ (ii)$C_{13}^{16}=C_{\left(16-13\right)}^{16}=C_{3}^{16}=560$
Q:
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A) 25200 B) 52000 C) 120 D) 24400
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = ($7C3$*$4C2$
= 210.
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves = 5! = 120
Required number of ways = (210 x 120) = 25200.
119 58123
Q:
How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?
A) 376 B) 375 C) 500 D) 673
Explanation:
The smallest number in the series is 1000, a 4-digit number.
The largest number in the series is 4000, the only 4-digit number to start with 4.
The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.
The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.
Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999.
Including 4000, there will be 376 such numbers.
68 32354
Q:
If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :
A) 601 B) 600 C) 603 D) 602
Explanation:
If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways.
If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.
If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.
The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.
57 30494
Q:
A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included ?
A) 196 B) 186 C) 190 D) 200
Explanation:
When at least 2 women are included.
The committee may consist of 3 women, 2 men : It can be done in $4C3*6C2$ ways
or, 4 women, 1 man : It can be done in $4C4*6C1$ways
or, 2 women, 3 men : It can be done in $4C2*6C3$ ways.
Total number of ways of forming the committees
$4C2*6C3+4C3*6C2+4C4*6C1$
= 6 x 20 + 4 x 15 + 1x 6
= 120 + 60 + 6 =186
55 41963
Q:
12 people at a party shake hands once with everyone else in the room.How many handshakes took place?
A) 72 B) 66 C) 76 D) 64
Explanation:
There are 12 people, so this is our n value.
So, $12C2$1= 66
46 26426
Q:
How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane?
A) 3 B) 6 C) 2 D) 4
Explanation:
You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.
AB , AC
BA , BC
CA , CB
There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.
The lines are: AB, BC and AC ; 3 lines only.
So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.
45 20087
Q:
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
A) 720 B) 520 C) 700 D) 750
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
44 40889
Q:
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
A) 360 B) 700 C) 720 D) 120
Explanation:
The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
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# Lecture 1 - Fibonacci Heaps
$\text{ September 7, 2005 }$ $\text{6.854 - Advanced Algorithms}$ $\text{Professor David Karger}$ $\text{David G. Andersen, Ioana Dumitriu, John Dunagan, Akshay Patil (2003)}$
# Motivation and Background
Priority queues are a classic topic in theoretical computer science. As we shall see, Fibonacci Heaps provide a fast and elegant solution. The search for a fast priority queue implementation is motivated primarily by two network optimization algorithms: Shortest Path and Minimum Spanning Tree (MST).
## Shortest Path and Minimum Spanning Trees
Given a graph $$G (V,E)$$ with vertices $$V$$ and edges $$E$$ and a length function $$l: E \rightarrow \Re^+$$. We define the Shortest Path and MST problems to be, respectively:
shortest path.
For a fixed source $$s \in V$$, find the shortest path to all vertices $$v \in V$$
minimum spanning tree (MST).
Find the minimum length set of edges $$F \subset E$$ such that $$F$$ connects all of $$V$$.
Note that the MST problem is the same as the Shortest Path problem, except that the source is not fixed. Unsurprisingly, these two problems are solved by very similar algorithms, Prim's for MST and Djikstra's for Shortest Path. The algorithm is:
1. Maintain a priority queue on the vertices
2. Put $$s$$ in the queue, where $$s$$ is the start vertex (Shortest Path) or any vertex (MST). Give $$s$$ a key of 0.
3. Repeatedly delete the minimum-key vertex $$v$$ from the queue and mark it "scanned"
For each neighbor $$w$$ of $$v$$:
If $$w$$ is not in the queue and not scanned, add it with key:
• Shortest Path: $$key(v) + length(v \rightarrow w)$$
• MST: $$length(v \rightarrow w)$$
If, on the other hand, $$w$$ is in the queue already, then decrease its key to the minimum of the value calculated above and $$w$$'s current key.
## Heaps
The classical answer to the problem of maintaining a priority queue on the vertices is to use a binary heap, often just called a heap. Heaps are commonly used because they have good bounds on the time required for the following operations:
insert $$O(\log n$$) delete-min $$O(\log n$$) decrease-key $$O(\log n$$)
If a graph has $$n$$ vertices and $$m$$ edges, then running either Prim's or Djikstra's algorithms will require $$O(n \log n)$$ time for inserts and deletes. However, in the worst case, we will also perform $$m$$ decrease-keys, because we may have to perform a key update every time we come across a new edge. This will take $$O(m \log n)$$ time. Since the graph is connected, $$m \geq n$$, and the overall time bound is given by $$O(m \log n)$$.
Since $$m \geq n$$, it would be nice to have cheaper key decreases. A simple way to do this is to use d-heaps.
## d-Heaps
d-heaps make key reductions cheaper at the expense of more costly deletions. This trade off is accomplished by replacing the binary heap with a d-ary heap-the branching factor (the maximum number of children for any node) is changed from 2 to $$d$$. The depth of the tree then becomes $$\log _{d}(n)$$. However, delete-min operations must now traverse all of the children in a node, so their cost goes up to $$d \log _{d}(n)$$. Thus, the running time of the algorithm becomes $$O(n d \log _{d}(n) + m \log _{d}(n))$$. Choosing the optimal $$d=m/n$$ to balance the two terms, we obtain a total running time of $$O(m \log_{m/n}n)$$.
When $$m = n^2$$, this is $$O(m)$$, and when $$m=n$$, this is $$O(n \log n)$$. This seems pretty good, but it turns out we can do much better.
## Amortized Analysis
Amortized analysis is a technique for bounding the running time of an algorithm. Often we analyse an algorithm by analyzing the individual operations that the algorithm performs and then multiplying the total number of operations by the time required to perform an operation. However, it is often the case that an algorithm will on occasion perform a very expensive operation, but most of the time the operations are cheap. Amortized analysis is the name given to the technique of analyzing not just the worst case running time of an operation but the average case running time of an operation. This will allow us to balance the expensive-but-rare operations against their cheap-and-frequent peers.
There are several methods for performing amortized analysis; for a good treatment, see Introduction to Algorithms by Cormen, Leiserson, and Rivest. The method of amortized analysis used to analyze Fibonacci heaps is the potential method:
• Measure some aspect of the data structure using a potential function. Often this aspect of the data structure corresponds to what we intuitively think of as the complexity of the data structure or the amount by which it is out of kilter or in a bad arrangement.
• If operations are only expensive when the data structure is complicated, and expensive operations can also clean up ("uncomplexify") the data structure, and it takes many cheap operations to noticeably increase the complexity of the data structure, then we can amortize the cost of the expensive operations over the cost of the many cheap operations to obtain a low average cost.
Therefore, to design an efficient algorithm, we want to force the user to perform many operations to make the data structure complicated, so that the work doing the expensive operation and cleaning up the data structure is amortized over those many operations.
We compute the potential of the data structure by using a potential function $$\Phi$$ that maps the data structure ($$DS$$) to a real number $$\Phi(DS)$$. Once we have defined $$\Phi$$, we calculate the cost of the $$i^{th}$$ operation by:
$cost_{\it amortized}(operation_i) = cost_{\it actual}(operation_i) + \Phi(DS_{i}) - \Phi(DS_{i-1})$
where $$DS_{i}$$ refers to the state of the data structure after the $$i^{th}$$ operation. The sum of the amortized costs is then
$\sum cost_{\it actual}(operation_i) + \Phi_{\it final} - \Phi_{\it initial}$.
If we can prove that $$\Phi_{final} \geq \Phi_{initial}$$, then we've shown that the amortized costs bound the real costs, that is, $$\sum{cost_{amortized}} \geq \sum{cost_{actual}}$$. Then we can just analyze the amortized costs and show that this isn't too much, knowing that our analysis is useful. Most of the time it is obvious that $$\Phi_{final} \geq \Phi_{initial}$$ and the real work is in coming up with a good potential function.
# Fibonacci Heaps
The Fibonacci heap data structure invented by Fredman and Tarjan in 1984 gives a very efficient implementation of the priority queues. Since the goal is to find a way to minimize the number of operations needed to compute the MST or SP, the kind of operations that we are interested in are insert, decrease-key, merge, and delete-min. (We haven't covered why merge is a useful operation yet, but it will become clear.) The method to achieve this minimization goal is laziness – "do work only when you must, and then use it to simplify the structure as much as possible so that your future work is easy". This way, the user is forced to do many cheap operations in order to make the data structure complicated.
Fibonacci heaps make use of heap-ordered trees. A heap-ordered tree is one that maintains the heap property, that is, where $$key(parent) \leq key(child)$$ for all nodes in the tree.
A Fibonacci heap H is a collection of heap-ordered trees that have the following properties:
1. The roots of these trees are kept in a doubly-linked list (the "root list" of $$H$$);
2. The root of each tree contains the minimum element in that tree (this follows from being a heap-ordered tree);
3. We access the heap by a pointer to the tree root with the overall minimum key;
4. For each node $$x$$, we keep track of the rank (also known as the order or degree) of $$x$$, which is just the number of children $$x$$ has; we also keep track of the mark of $$x$$, which is a Boolean value whose role will be explained later.
For each node, we have at most four pointers that respectively point to the node's parent, to one of its children, and to two of its siblings. The sibling pointers are arranged in a doubly-linked list (the "child list" of the parent node). Of course, we haven't described how the operations on Fibonacci heaps are implemented, and their implementation will add some additional properties to $$H$$. Here are some elementary operations used in maintaining Fibonacci heaps.
## Inserting, merging, cutting, and marking.
Inserting a node $$x$$. We create a new tree containing only $$x$$ and insert it into the root list of $$H$$; this is clearly an $$O(1)$$ operation.
Merging two trees. Let $$x$$ and $$y$$ be the roots of the two trees we want to merge; then if the key in $$x$$ is no less than the key in $$y$$, we make $$x$$ the child of $$y$$; otherwise, we make $$y$$ the child of $$x$$. We update the appropriate node's rank and the appropriate child list; this takes $$O(1)$$ operations.
Cutting a node. If $$x$$ is a root in $$H$$, we are done. If $$x$$ is not a root in $$H$$, we remove $$x$$ from the child list of its parent, and insert it into the root list of $$H$$, updating the appropriate variables (the rank of the parent of $$x$$ is decremented, etc.). Again, this takes $$O(1)$$ operations. (We assume that when we want to find a node, we have a pointer hanging around that accesses it directly, so actually finding the node takes $$O(1)$$ time.)
Marking. We say that $$x$$ is marked if its mark is set to "true", and that it is unmarked if its mark is set to "false". A root is always unmarked. We mark $$x$$ if it is not a root and it loses a child (i.e., one of its children is cut and put into the root-list). We unmark $$x$$ whenever it becomes a root. We will make sure later that no marked node loses another child before it itself is cut (and reverted thereby to unmarked status).
## Decreasing keys and Deleting mins
At first, decrease-key does not appear to be any different than merge or insert; just find the node and cut it off from its parent, then insert the node into the root list with a new key. This requires removing it from its parent's child list, adding it to the root list, updating the parent's rank, and (if necessary) the pointer to the root of smallest key. This takes $$O(1)$$ operations.
The delete-min operation works in the same way as decrease-key: Our pointer into the Fibonacci heap is a pointer to the minimum keyed node, so we can find it in one step. We remove this root of smallest key, add its children to the root-list, and scan through the linked list of all the root nodes to find the new root of minimum key. Therefore, the cost of a delete-min operation is $$O(\mbox{# of children })$$ of the root of minimum key plus $$O(\mbox{# of root nodes})$$; in order to make this sum as small as possible, we have to add a few bells and whistles to the data structure.
## Population Control for Roots
We want to make sure that every node has a small number of children. This can be done by ensuring that the total number of descendants of any node is exponential in the number of its children. In the absence of any "cutting" operations on the nodes, one way to do this is by only merging trees that have the same number of children (i.e, the same rank). In the next section we'll see exactly when we want to merge trees. It is relatively easy to see that if we only ever merge trees that have the same rank and never cut any nodes, the total number of descendants (counting onself as a descendant) is always ($$2^{ \mbox{# of children}}$$). The resulting structure is called a binomial tree because the number of descendants at distance $$k$$ from the root in a tree of size $$n$$ is exactly $$n \choose k$$. Binomial heaps preceded Fibonacci heaps and were part of the inspiration for them. We now present Fibonacci heaps in full detail.
## Actual Algorithm for Fibonacci Heaps
• Maintain a list of heap-ordered trees.
• insert: add a degree 0 tree to the list.
• delete-min: We can find the node we wish to delete immediately since our handle to the entire data structure is a pointer to the root with minimum key. Remove the smallest root, and add its children to the list of roots. Scan the roots to find the next minimum. Then consolidate all the trees (merging trees of equal rank) until there is $$\leq 1$$ of each rank. (Assuming that we have achieved the property that the number of descendants is exponential in the number of children for any node, as we did in the binomial trees, no node has rank $$> c \log n$$ for some constant $$c$$. Thus consolidation leaves us with $$O(\log n)$$ roots.) The consolidation is performed by allocating buckets of sizes up to the maximum possible rank for any root node, which we just showed to be $$O(\log n)$$. We put each node into the appropriate bucket, at cost $$O(\log n)$$ + $$O(\mbox{# of roots})$$. Then we march through the buckets, starting at the smallest one, and consolidate everything possible. This again incures cost $$O(\log n)$$ + $$O(\mbox{# of roots})$$. This is the only time we do consolidation.
• decrease-key: cut the node, change its key, and insert it into the root list as before, Additionally, if the parent of the node was unmarked, mark it. If the parent of the node was marked, cut it off also. Recursively do this until we get up to an unmarked node. Mark it.
## Actual Analysis for Fibonacci Heaps
Define $$\Phi(DS) = (k \cdot$$ # of roots in $$DS$$ + 2 $$\cdot$$ # marked bits in $$DS$$). Note that insert and delete-min do not ever cause nodes to be marked - we can analyze their behaviour without reference to marked and unmarked bits. The parameter $$k$$ is a constant that we will conveniently specify later. We now analyze the costs of the operations in terms of their amortized costs (defined to be the real costs plus the changes in the potential function).
• insert: the amortized cost is O(1). O(1) actual work plus $$k$$ * O(1) change in potential for adding a new root. O(1) + $$k$$O(1) = O(1) total amortized cost.
• delete-min: for every node that we put into the root list (the children of the node we have deleted), plus every node that is already in the root list, we do constant work putting that node into a bucket corresponding to its rank and constant work whenever we merge the node. Our real costs are putting the roots into buckets (O($$\#roots$$)), walking through the buckets (O($$\log n$$)), and doing the consolidating tree merges (O($$\#roots$$)). On the other hand, our change in potential is $$k*(\log n-\#roots)$$ (since there are at most $$\log n$$ roots after consolidation). Thus, total amortized cost is O($$\#roots$$) + O($$\log n$$) + $$k*(\log n-\#roots)$$ = O($$\log n$$).
• decrease-key: The real cost is O(1) for the cut, key decrease and re-insertion. This also increases the potential function by O(1) since we are adding a root to the root list, and maybe by another 2 since we may mark a node. The only problematic issue is the possibility of a "cascading cut" - a cascading cut is the name we give to a cut that causes the node above it to cut because it was already marked, which causes the ndoe above it be cut since it too was alrady marked, etc. This can increase the actual cost of the operation to (# of nodes already marked). Luckily, we can pay for this with the potential function! Every cost we incur from having to update pointers due to a marked node that was cut is offset by the decrease in the potential function when that previously marked node is now left unmarked in the root list. Thus the amortized cost for this operation is just O(1).
The only thing left to prove is that for every node in every tree in our Fibonacci heap, the number of descendants of that node is exponential in the number of children of that node, and that this is true even in the presence of the "weird" cut rule for marked bits. We must prove this in order to substantiate our earlier assertion that all nodes have degree $$\leq \log n$$.
## The trees are big
Consider the children of some node $$x$$ in the order in which they were added to $$x$$.
$$Lemma:$$ The $$i^{th}$$ child to be added to $$x$$ has rank at least $$i-2$$.
$$Proof:$$ Let $$y$$ be the $$i^{th}$$ child to be added to $$x$$. When it was added, $$y$$ had at least $$i-1$$ children. This is true because we can currently see $$i-1$$ children that were added earlier, so they were there at the time of the $$y$$'s addition. This means that $$y$$ had at least $$i-1$$ children at the time of it's merger, because we only merge equal ranked nodes. Since a node could not lose more than one child without being cut itself, it must be that $$y$$ has at least $$i-2$$ children ($$i-1$$ from when it was added, and no more than a potential $$1$$ subsequently lost).
Note that if we had been working with a binomial tree, the appropriate lemma would have been $$rank = i-1$$ not $$\geq i-2$$.
Let $$S_k$$ be the minimum number of descendants of a node with $$k$$ children. We have $$S_0 = 1, S_1 = 2$$ and, $S_k \geq \sum_{i=0}^{k-2}S_i$ This recurrence is solved by $$S_k \geq F_{k+2}$$, the $$(k+2)^{th}$$ Fibonacci number. Ask anyone on the street and that person will tell you that the Fibonacci numbers grow exponentially; we have proved $$S_k \geq 1.5^k$$, completing our analysis of Fibonacci heaps.
## Utility
Only recently have problem sizes increased to the point where Fibonacci heaps are beginning to appear in practice. Further study of this issue might make an interesting term project; see David Karger if you're curious.
Fibonacci Heaps allow us to improve the running time in Prim's and Djikstra's algorithms. A more thorough analysis of this will be presented in the next class.
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# RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression | Updated For 2021-22
RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression: Download the Free PDF of RS Aggarwal Solutions Class 10 Maths Chapter 11 from here. All the solutions of RS Aggarwal Solutions Class 10 Maths are prepared by subject matter experts and are as per the current CBSE Syllabus. RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression is a great help guide for you.
You can access the RS Aggarwal Solutions Class 10 Maths Chapter 11 PDF offline as well. For more details about the RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression, read the whole blog.
## Download Free PDF Of RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression
RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression
## Access Solutions For RS Aggarwal Solutions Class 10 Maths Chapter 11
Question 1:
The given progression is 3, 9, 15, 21 …..
Clearly (9 – 3) = (15 – 9) = (21 – 15) = 6 which is constant
Thus, each term differs from its preceding term by 6
So, the given progression is an AP
Its first term = 3 and the common difference = 6
Question 2:
The given progression is 16, 11, 6, 1, -4 ….
Clearly (11 – 16) = (1 – 6) = (-4 – 1) = – 5 which is constant
Thus, each term differs from its preceding term by – 5
So the given progression is an AP
Its first term = 16 and the common difference = – 5
Question 3:
(i) The given AP is 1, 5, 9, 13, 17…..
Its first term = 1 and common difference = (5 – 1) = 4
∴ a = 1 and d = 4
The nth term of the AP is given by
T= a + (n-1) d
T20 = 1 + (20-1) x 4 = 1+ 76 = 77
Hence, the 20th term is 77
(ii) The given AP is 6, 9, 12, 15 ……
Its first term = 6 and common difference = (9 – 6) = 3
∴ a = 6, d = 3
The nth term of the AP is given by
T= a + (n-1) d
T35 = 6 + (35-1) x 3 = 6+ 102 = 108
Hence, the 35th term is 108
(iii) The given AP is 5, 11, 17, 23 …..
Its first term = 5, and common difference = (11 – 5) = 6
∴ a = 5, d = 6
The nth term of AP is given by
T= a + (n-1) d
Tn= 5 + (n-1) x 6 = 5+ 6n – 6 = 6n – 1
(iv) The given AP is (5a – x), 6a, (7a + x) …..
Its first term = (5a – x) and common difference = 6a – 5a – x = a + x
The nth term of AP is given by
T= a + (n-1) d
T11 = (5a – x) + (11-1) (a + x)
= 5a – x + 10x + 10x
= 15a + 9x = 3(5a +3x)
Hence the 11th term is 3(5a + 3x)
Question 4:
(i) The given AP is 63, 58, 53, 48 ….
First term = 63, common difference = 58 – 63 = – 5
∴ a = 63, d = – 5
The nth term of AP is given by
T= a + (n-1) d
T10 = 63 + (10-1) (-5) = 63- 45 = 18
Hence the 10th term is 18
(ii) The given AP is 9, 5, 1, -3….
First term = 9, common difference = 5 – 9 = -4
∴ a = 9, d= – 4
The nth term of AP is given by
T= a + (n-1) d
T14 = 9 + (14-1) (-4) = 9- 52 = -43
Hence, the 14th term is – 43
(iii) The given AP is 16, 9, 2, -5
First term = 16, common difference = 9 – 16 = – 7
∴ a = 16, d = -7
The nth term of AP is given by
T= a + (n-1) d
Tn = 16 + (n-1) (-7) ⇒ 16- 7n + 7 = (23 – 7n)
Hence, the nth term is (23 – 7n).
Question 5:
The given AP is 6,734,912,1114..
First term = 6, common difference = (7346)
= (3146)
= 74
a = 6, d = 74
The nth term is given by
T= a + (n-1) d
T14 = 6 + (37 – 1) (74) = 6+ 63 = 69
Hence, 37th term is 69
Question 6:
The given AP is 5,412,4,312,3..
The first term = 5,
common difference = (4125)=(925)=12
∴ a = 5, d = 12
The nth term is given by
T= a + (n-1) d
T14 = 5 + (25 – 1) (-1/2) = 5- 12 = -7
Hence the 25th term is – 7
Question 7:
In the given AP, we have a = 6 and d = (10 – 6) = 4
Suppose there are n terms in the given AP, then
T = 174 ⇒ a + (n-1) d = 174
⇒ 6 + (n-1) 4 = 174
⇒ 6 + 4n – 4 = 174
⇒ 2 + 4n = 174 ⇒ n = 172/4 ⇒ 43
Hence there are 43 terms in the given AP
Question 8:
In the given AP we have a = 41 and d = 38 – 41 = – 3
Suppose there are n terms in AP, then
T = 8 ⇒ a + (n-1) d = 8
⇒ 41 + (n-1) (-3) = 8
⇒ 41 – 3n + 3 = 8
⇒ -3n = – 36 ⇒ n = 12
Hence there are 12 terms in the given AP
Question 9:
In the given AP, we have a = 3 and d = 8 – 3 = 5
Suppose there are n terms in given AP, then
T = a + (n-1) d = 88
⇒ 3 + (n-1) 5 = 88
⇒ 3 + 5n – 5 = 88
⇒ 5n = 90
⇒ n = 12
Hence, the 18th term of given AP is 88
Question 10:
In the given AP, we have a = 72 and d = 68 – 72 = – 4
Suppose there are n terms in given AP, we have
T = 0 ⇒ a + (n-1) d = 0
⇒ 72 + (n-1) (-4) = 0
⇒ 72 – 4n + 4 = 0
⇒ 4n = 76
⇒ n = 19
Hence, the 19th term in the given AP is 0
Question 11:
In the given AP, we have a = 12 ; (156)=16
Suppose there are n terms in given AP, we have
Then,
T = 3 ⇒ a + (n-1) d = 3
⇒ 56+(n1)16=3
⇒ 56+16n16=3
⇒ 4 + n = 18
⇒ n = 14
Thus, 14th term in the given AP is 3
Question 12:
We know that T– (5x + 2), T– (4x – 1) and T– (x + 2)
Clearly,
T2 – T1 = T3 – T2
⇒ (4x – 1) – (5x + 2) = (x + 2) – (4x – 1)
⇒ 4x – 1 – 5x – 2 = x + 2 – 4x + 1
⇒ -x – 3 = -3x + 3
⇒ -x + 3x = 6
⇒ 2x = 6 ⇒ x = 3
Hence x = 3
Question 13:
T = (4n – 10)
⇒ T = (4 x 1 – 10) = -6 and T = (4 x 2 – 10) = -2
Thus, we have
(i) First term = -6
(ii) Common difference = (T2 – T1) = (-2+6) = 4
(iii) 16th term = a + (16-1) d, where a = -6 and d = 4
= (-6 + 15 x 4) = 54
Question 14:
In the given AP, let first term = a and common difference = d,
Then, T = a + (n-1) d
⇒ T = a + (4 – 1)d, T10 = a + (10 – 1)d
⇒ T = a + 3d, T10 = a + 9d
Now, T = 13 ⇒ a + 3d = 13 – – – (1)
T10 = 25 ⇒ a + 9d = 25 – – – (2)
Subtracting (1) from (2), we get
⇒ 6d = 12 ⇒ d = 2
Putting d = 2 in (1), we get
a + 3 x 2 = 13
⇒ a = (13 – 6) = 7
Tthus, a = 7, and d = 2
17th term = a + (17 – 1)d, where a= 7, d = 2
(7 + 16 x 2) = (7 + 32) = 39
∴ a = 7, d = 2,
Question 15:
In the given AP, let first term = a and common difference = d
Then, T = a + (n-1) d
⇒ T = a + (8 – 1)d, T12 = a + (12 – 1)d
⇒ T = a + 7d, T12 = a + 11d
Now, T = 37 ⇒ a + 7d = 37 – – – (1)
T12 = 57 ⇒ a + 11d = 57 – – – (2)
Subtracting (1) from (2), we get
⇒ 4d = 20 ⇒ d = 5
Putting d = 5 in (1), we get
a + 7 x 5 = 37
⇒ a = 2
Tthus, a = 2, and d = 5
So the required AP is 2, 7, 12..
Question 16:
In the given AP, let the first term = a, and common difference = d
Then, T = a + (n-1) d
⇒ T = a + (7 – 1)d, and T13 = a + (13 – 1)d
⇒ T = a + 6d, T13 = a + 12d
Now, T = -4 ⇒ a + 6d = -4 – – – (1)
T13 = -16 ⇒ a + 12d = -16 – – – (2)
Subtracting (1) from (2), we get
⇒ 6d = -12 ⇒ d = -2
Putting d = -2 in (1), we get
a + 6 (-2) = -4
⇒ a – 12 = -4
⇒ a = 8
Tthus, a = 8, and d = -2
So the required AP is 8, 6, 4, 2, 0……
Question 17:
In the given AP let the first term = a, And common difference = d
Then, T = a + (n-1) d
⇒ T10 = a + (10 – 1)d, T17 = a + (17 – 1)d, T13 = a + (13 – 1)d
⇒ T10 = a + 9d, T17 = a + 16d, T13 = a + 12d
Now, T10 = 52 ⇒ a + 9d = 52 – – – (1)
and T17 = T13 + 20 ⇒ a + 16d = a + 12d + 20
⇒ 4d = 20 ⇒ d = 5
Putting d = 5 in (1), we get
a + 9 x 5 = 52 ⇒ a = 52-45 ⇒ a = 7
Thus, a = 7 and d = 5
So the required AP is 7, 12, 17, 22….
Question 18:
Let the first term of given AP = a and common difference = d
Then, T = a + (n-1) d
⇒ T = a + (4 – 1)d, T25 = a + (25 – 1)d, T11 = a + (11 – 1)d
⇒ T = a + 3d, T25 = a + 24d, T11 = a + 10d
Now, T = 0 ⇒ a + 3d = 0 ⇒ a = -3d
∴ T25 = a + 24d = (-3d +24d) ⇒ 21d
and T11 = a + 10d = (-3d +10d) ⇒ 7d
∴ T25 = 21d = 3 x 7d = 3 x T11
Hence 25th term is triple its 11th term
Question 19:
The given AP is 3, 8, 13, 18…..
First term a = 3, common difference a = 8 – 3 = 5
∴ T = a + (n-1) d = 3 + (n – 1) x 5 = 5n – 2
T20 = 3 + (20-1) 5 = 3 + 19 x 5 = 98
Let nth term is 55 more than the 20th term
∴ (5n – 2) – 98 = 55
Or 5n = 100 + 55 = 155
n = 155/5 = 31
∴ 31st term is 55 more than the 20th term of given AP
Question 20:
The given AP is 5, 15, 25….
a = 5, d = 15 – 5 = 10
We have, T = 130+T31
⇒ a + (n-1) d = 130 + 5 + (31 – 1) x 10
⇒ 5 + (n-1) 10 = 130 + 5 + (31 – 1) x 10
⇒ 5 + 10n – 10 = 135 + 300
⇒ 10n – 5 = 435 or 10n = 453 + 5
∴ n = 440/10 = 44
Thus, the required term is 44th
Question 21:
First AP is 63, 65, 67….
First term = 63, common difference = 65 – 63 = 2
∴ nth term = 63 + (n – 1) 2 = 63 + 2n – 2 = 2n + 61
Second AP is 3, 10, 17 ….
First term = 3, common difference = 10 – 3 = 7
nth term = 3 + (n – 1) 7 = 3 + 7n – 7 = 7n – 4
The two nth terms are equal
∴ 2n + 61 = 7n – 4 or 5n = 61 + 4 = 65
⇒ n = 65/4 = 13.
Question 22:
Three digit numbers which are divisible by 7 are 105, 112, 119,….994
This is an AP where a= 105, d = 7 and l = 994
Let nth term be 994
∴ a + (n – 1)d =994 or 105 + (n – 1)7 = 994
⇒ 105 + 7n – 7 = 994 or 7n = 94 – 98 = 896
∴ n = 896/7 = 128.
Hence, there are 128 three digits number which are divisible by 7.
Question 23:
Here a = 7, d = (10 – 7) = 3, l = 184
And n = 8
Now, nth term from the end = [ l – (n-1) d ]
= [ 184 – (8-1) 3 ]
= [ 184 – 7 x 3]
= 184-21
= 163
Hence, the 8th term from the end is 163
Question 24:
Here a = 17, d = (14 – 17) = -3, l = -40
And n = 6
Now, nth term from the end = [ l – (n – 1) d ]
= [ -40 – (6-1)(-3) ]
= [ -40 + 5 x 3]
= -40+15
= -25
Hence, the 6th term from the end is – 25
Question 25:
The given AP is 10, 7, 4, ….. (-62)
a = 10, d = 7 – 10 = -3, l = -62
Now, 11th term from the end = [ l – (n – 1) d ]
= [ -62 – (11-1)(-3) ]
= -62 + 30
= -32
Question 26:
Let a be the first term and d be the common difference
pth term = a +(p – 1)d = q (given) —–(1)
qth term = a +(q – 1) d = p (given) —–(2)
subtracting (2) from (1)
(p – q)d = q – p
(p – q)d = -(p – q)
d = -1
Putting d = -1 in (1)
a – (p – 1) = q ∴ a = p + q -1
∴ (p + q)th term = a+ (p + q -1)d
= (p + q -1) – (p + q -1) = 0
Question 27:
Let a be the first term and d be the common difference
T10 = a + 9d, T15 = a + 14d
10T10 = 15T15
⇒ 10(a + 9) d = 15(a + 14d)
⇒ 2(a + 9) d = 3(a + 14d)
⇒ a + 24d = 0
∴ T25 = 0
Question 28:
Let a be the first term and d be the common difference
∴ nth term from the beginning = a + (n – 1)d —–(1)
nth term from end = l – (n – 1)d —-(2)
sum of the nth term from the beginning and nth term from the end = [a + (n – 1)d] + [l – (n – 1)d] = a + l
Question 29:
Number of rose plants in first, second, third rows…. are 43, 41, 39… respectively.
There are 11 rose plants in the last row
So, it is an AP . viz. 43, 41, 39 …. 11
a = 43, d = 41 – 43 = -2, l = 11
Let nth term be the last term
∴ l = a + (n-1) d
⇒ 11 = 43 + (n-1) x (-2)
43 – 2n + 2 = 11 or 2n = 45 -11 = 34
∴ n = 34/2 = 17
Hence, there are 17 rows in the flower bed.
Question 30:
Total amount = ₹ 2800
and number of prizes = 4
Let first prize = ₹ a
Then second prize = ₹ a – 200
Third prize = a – 200 – 200 = a – 400
and fourth prize = a – 400 – 200 = a – 600
But sum of there 4 prizes are ₹ 2800
a + a – 200 + a – 400 + a – 600 = ₹ 2800
⇒ 4a – 1200 = 2800
⇒ 4a = 2800 + 1200 = 4000
⇒ a = 1000
First prize = ₹ 1000
Second prize = ₹ 1000 – 200 = ₹ 800
Third prize = ₹ 800 – 200 = ₹ 600
and fourth prize = ₹ 600 – 200 = ₹ 400
## Access RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression Other Important Exercises
RS Aggarwal Solutions Chapter 11 Exercise 11.1
RS Aggarwal Solutions Chapter 11 Exercise 11.2
RS Aggarwal Solutions Chapter 11 Exercise 11.3
RS Aggarwal Solutions Chapter 11 Exercise 11.4
This is the complete blog on the RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression. To know more about the CBSE Class 10 Maths exam, ask in the comments.
## FAQs on RS Aggarwal Solutions Class 10 Maths Chapter 11
### From where can I find the download link for the RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression PDF?
You can find the download link for the RS Aggarwal Solutions Class 10 Maths Chapter 11 PDF in the above blog.
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# What Is The Recursive Formula For This Geometric Sequence Apex
## Understanding Geometric Sequences
A geometric sequence is a sequence of numbers in which each term after the first is found by multiplying the previous term by a constant called the common ratio. This common ratio is denoted by the letter “r”. The general form of a geometric sequence can be expressed as:
a1, a1r, a1r2, a1r3, …
Where a1 is the first term and r is the common ratio.
## Recursive Formula for Geometric Sequences
The recursive formula for a geometric sequence defines each term in the sequence as a function of the preceding term. In other words, it allows us to find any term in the sequence by performing a specific operation on the previous term. The formula is typically given in the form:
an = an-1 * r,
where an represents the n-th term in the sequence, an-1 is the previous term, and r is the common ratio.
## Finding the Recursive Formula
To find the recursive formula for a geometric sequence, we need to identify the pattern in the sequence and express it mathematically. Let’s consider an example:
Suppose we have the following geometric sequence: 2, 6, 18, 54, …
To find the recursive formula for this sequence, we first need to identify the common ratio. We can do this by dividing any term by its preceding term. In this case, if we divide 6 by 2, we get 3, and if we divide 18 by 6, we also get 3.
So, the common ratio r in this sequence is 3.
Now, let’s use the general form of the recursive formula: an = an-1 * r. Substituting the common ratio r into the equation, we get:
an = an-1 * 3.
This is the recursive formula for the given geometric sequence.
## Using the Recursive Formula to Find Terms
One of the advantages of having the recursive formula for a geometric sequence is that it allows us to find any term in the sequence without having to calculate all the preceding terms. Let’s use the recursive formula to find the 5th term in the sequence 2, 6, 18, 54, …
Using the formula: an = an-1 * 3, and knowing that the first term is 2 (a1 = 2), we can find the 5th term as follows:
a5 = a4 * 3,
a5 = 54 * 3,
a5 = 162.
So, the 5th term in the sequence is 162.
## Limitations of the Recursive Formula
While the recursive formula is useful for finding specific terms in a geometric sequence, it has some limitations. One of the main limitations is that it can be inefficient when trying to find terms further along in the sequence, especially if the sequence is long. In such cases, it may be more practical to use the explicit formula for geometric sequences, which allows for the direct calculation of any term without having to calculate all the preceding terms.
## Explicit Formula for Geometric Sequences
The explicit formula for a geometric sequence is another way to define the terms in the sequence and is given by the formula:
an = a1 * r(n-1),
where an is the n-th term in the sequence, a1 is the first term, r is the common ratio, and n is the position of the term in the sequence.
The explicit formula provides a direct way to calculate any term in the sequence without having to find all the preceding terms, making it more efficient for finding terms further along in the sequence.
## Comparing Recursive and Explicit Formulas
Let’s compare the two formulas using the same example sequence: 2, 6, 18, 54, …
Using the recursive formula, we found the 5th term to be 162. Now, let’s find the 5th term using the explicit formula:
a5 = 2 * 3(5-1),
a5 = 2 * 34,
a5 = 2 * 81,
a5 = 162.
As we can see, both formulas give us the same result for the 5th term, but the explicit formula allows us to find the term more efficiently and directly.
## Conclusion
In conclusion, the recursive formula for a geometric sequence defines each term in the sequence as a function of the preceding term, with the common ratio being a key factor. While the recursive formula is useful for finding specific terms in a sequence, it can be inefficient for finding terms further along in the sequence. In such cases, the explicit formula for geometric sequences provides a more direct and efficient way to calculate any term in the sequence. Both formulas have their strengths and limitations, and understanding how to use them can be beneficial in various mathematical applications.
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#### Chapter 11 Language of Descriptive Statistics
Section 11.2 Frequency Distributions and Percentage Calculation
# 11.2.3 Calculation of Interest
In the calculation of interest, we distinguish between simple interest and compound interest. For simple interest, the interest is paid at the end of the interest period. For compound interest, the interest is also paid on the previously accumulated interest.
##### Info 11.2.7
When simple interest is applied, a quantity $K$ that increases by $p %$ every year will increase after $t$ years ($t\in ℕ$) to
${K}_{t}\mathrm{ }=\mathrm{ }K·\left(1+t·\frac{p}{100}\right).$
Note that $p$ itself can be a decimal, for example, for $p=2.5$, the percent value is $2.5%=0.025$.
##### Exercise 11.2.8
What is the final capital for an initial capital of $K=4,000$ EUR after an interest period of $t=10$ years, when simple interest is applied at a rate of $p=2.5%$ p. a.?
Answer: ${K}_{10}$$=$
EUR.
##### Exercise 11.2.9
What is the initial capital $K$ that had been deposited at the 1$\mathrm{st}$ of January, 2000 to get paid an end capital of ${K}_{12}=10,000$ EUR at the 31$\mathrm{st}$ of December, 2011 when simple interest is applied at a rate of $p=5%$ p. a.?
Answer: $K$$=$
EUR.
While simpleinterest is simply paid after an interest period, compound interest carries over into the next interest period, i.e. the interest will be added to the initial capital or will be capitalised:
##### Example 11.2.10
For a bank account at the end of an interest period, an initial capital of $1,000$ EUR is deposited at an interest rate of $8%$. After one year the deposit (in EUR) in the bank account is
• $1,000+\frac{1,000·8}{100}=1,000·\left(1+\frac{8}{100}\right)=1,000·1.08=1,080$.
• This deposit is invested for an additional year at the same interest rate of $8%$. Then, the deposit (in EUR) after two years is $1,080·1.08=1,000·1.{08}^{2}=1,000·{\left(1+\frac{8}{100}\right)}^{2}$.
• The deposit increases by a factor of $1.08$ per year. Hence, the deposit (in EUR) after $t$ years ($t\in ℕ{}_{0}$) is
$1,000·1.{08}^{t}\mathrm{ }=\mathrm{ }1,000·{\left(1+\frac{8}{100}\right)}^{t}.$
Thus, the compound interest is based on the following formula:
##### Info 11.2.11
A quantity $K$ that increases every year by an amount of $p %$ will have been increased after $t$ years ($t\in ℕ{}_{0}$) to
$K·{\left(1+\frac{p}{100}\right)}^{t} .$
Here, $1+\frac{p}{100}$ is called the growth factor for a growth of $p %$.
In an advert offering deposit accounts or loans, the interest is usually given as a rate per year, even if the actual interest period differs. This interest period is the time between two successive dates at which the interest payments are due. On a deposit account the interest period is one year, though it is becoming more common to offer other interest periods. For example, for short-term loans the interest is paid daily or monthly.
If a bank offers a yearly interest rate of $9 %$ with monthly interest payments, then at the end of each month $\left(\frac{1}{12}\right)·9%=0,75%$ of the capital will be credited.
##### Info 11.2.12
The yearly rate is divided by the number of interest periods to obtain the periodic rate (the interest rate per period).
Suppose an investment of ${S}_{0}$ EUR yields $p %$ interest per interest period. After $t$ periods ($t\in ℕ{}_{0}$) the investment will have been increased to
${S}_{t}\mathrm{ }=\mathrm{ }{S}_{0}·\left(1+r{\right)}^{t}\mathrm{ }\mathrm{ }\text{with}\mathrm{ }\mathrm{ }r\mathrm{ }=\mathrm{ }\frac{p}{100} .$
In every period the investment increases by a factor of $1+r$, and we say "the interest rate equals $p %$" or "the periodic rate equals $r$". Suppose interest is credited at a rate of $\frac{p}{n} %$ to the capital at $n$ different times, more or less evenly distributed over the year. Then, the capital is multiplied by a factor of
${\left(1+\frac{r}{n}\right)}^{n}$
every year. After $t$ years the capital has increased to
${S}_{0}·{\left(1+\frac{r}{n}\right)}^{n·t} .$
##### Example 11.2.13
A capital of $5,000$ EUR is deposited for $t=8$ years in a bank account at a yearly interest rate of $9 %$, where the interest is paid quarterly. The periodic rate $\frac{r}{n}$ here is
$\frac{r}{n}\mathrm{ }=\mathrm{ }\frac{0.09}{4}\mathrm{ }=\mathrm{ }0.0225 ,$
and for the number of periods $n·t$ we have $n·t=4·8=32$. Thus, after $t=8$ years the deposit has increased to
$5000·\left(1+0.0225{\right)}^{32}\mathrm{ }\approx \mathrm{ }10190.52\mathrm{ }\text{EUR} .$
##### Exercise 11.2.14
A capital of ${K}_{0}=8,750$ EUR is deposited for $t=4$ years at an interest rate of $p=3,5%$ p. a., and the interest is capitalised.
1. After one year the amount of capital is ${K}_{1}$$=$ .
2. After two years the amount of capital is ${K}_{2}$$=$ .
3. After three years the amount of capital is ${K}_{3}$$=$ .
4. The final capital is ${K}_{4}$$=$ .
Specify all values rounded mathematically to the second fractional digit. Round only after carrying out the calculations. For these calculations, you are allowed to use a calculator.
A consumer who wants to take out a loan always faces several offers from competing banks. Thus, it is extremely useful to compare the different offers.
##### Example 11.2.15
Let us consider an offer providing an yearly interest rate of $9%$, where the interest is charged at a monthly ($12$ times per year) rate of $0.75%$. If no interest is paid off in the meantime, the initial debt will increase to
${S}_{0}·{\left(1+\frac{0.09}{12}\right)}^{12}\mathrm{ }\approx \mathrm{ }{S}_{0}·1.094$
after one year. The interest to be paid off is approximately
$1.094·{S}_{0}-{S}_{0}\mathrm{ }=\mathrm{ }0.094·{S}_{0} .$
As long as no interest is paid off, the debt will increase at a constant rate which is approximately $9.4%$ per year. This is why we may speak of an "effective" annual interest rate. In the example above the effective annual interest rate is $9.4%$.
##### Info 11.2.16
If interest is paid $n$ times per year at a periodic rate of $\frac{r}{n}$ per period, then the effective annual interest rate $R$ is defined by
$R\mathrm{ }=\mathrm{ }{\left(1+\frac{r}{n}\right)}^{n}-1 .$
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# Is A hexagon a polygon?
Yes, a hexagon is a polygon. A polygon is any flat shape that is made up of two or more line segments that are connected end-to-end. A hexagon is any six-sided shape, with straight sides that are all the same length.
The sum of all angles in a hexagon is 720 degrees, indicating that all of the internal angles of a hexagon must be equal and measure 120 degrees each. Examples of regular hexagons include honeycombs, which are comprised of overlapping hexagons.
Examples of irregular hexagons include a standard hexagon with varying side lengths, or a hexagon where two sides have been joined.
## How do you prove a polygon is a hexagon?
To prove that a polygon is a hexagon, there are a few different methods you can use. The simplest method is to simply count the number of sides and angles in the polygon. A hexagon will have six sides and six angles, so if you can accurately count this in the polygon then it is a hexagon.
You can also use the euler equation to prove that a polygon is a hexagon. This equation states that for any convex polygon the number of faces, ‘F’, plus the number of vertices, ‘V’, minus the number of edges, ‘E’, equals two.
For a hexagon, the equation is F + V – E = 2, so V = 6, F = 6, and E = 12. If these values are present in the polygon then it is a hexagon.
Another method to prove a polygon is a hexagon is to calculate the interior angles. For any regular polygon the sum of the interior angles will be (n – 2)180°, where ‘n’ is the number of sides. For a hexagon the sum of the interior angles is 720°, so if this value is present in the interior angles then it is a hexagon.
Finally, you can try and measure the side length. A regular hexagon will have six equal sides. If you can accurately measure the side lengths and find that they are equal, then it is a hexagon.
## What makes a polygon regular or not?
A polygon is considered regular if all angles of the shape are the same and all sides are the same length. This means the sides are equal in length, the angles are the same and all sides meet at the same angle.
Any shape that does not have this criteria is considered an irregular polygon. Irregular polygons can be equilateral or equiangular or both. Equilateral means that all sides are the same length and equiangular means that all angles are the same.
A regular polygon is special because it is symmetrical, meaning that you do not need to rotate it to get the same shape.
## What shape is not a polygon and why?
A circle is not a polygon because it is a curved line that is made up of an infinite number of points that form an unbroken line. The basic defining features of a polygon are that it must have straight sides and angles, as well as a finite number of sides and angles.
A circle fails to meet both criteria, as it is composed of an infinite number of points and its sides are not straight nor angles present.
## What’s a 3d hexagon called?
A 3d hexagon is called a regular hexahedron. It is a six-sided polyhedron composed of six identical, quadrilateral faces which are congruent to each other. It has eight vertices and twelve edges. Hexahedra are categorized as Platonic solids because all edges, angles, and faces are equal.
It is also known as a cube, due to its shape, but technically a cube has six square faces while a hexahedron has six quadrilateral (four-sided) faces. Hexahedra are often used in 3D modeling and architecture, particularly in architecture to create unique facades.
It is also used in game development and 3D printing.
## What makes a hexagon unique?
The hexagon is a unique and distinct shape due to the fact that it is the only regular polygon with six sides. This means that all six sides are equal in length and each internal angle measures exactly 120°.
Other polygons such as the triangle, square, and pentagon have internal angles that vary according to the size and shape of the polygon. The hexagon’s consistent internal angles make it distinct from other shapes.
The elegant and pleasing aesthetic of the hexagon also makes it unique, as it is one of the most symmetrical shapes in geometry, which is why it is a common sight in the natural world from honeycombs to snowflakes.
It is also widely used throughout architecture and design to create strong and dynamic shapes and compositions. Consequently, it wouldn’t be wrong to say that the hexagon is a symbol of balance, beauty and strength.
## What shape has 7 sides called?
The shape with 7 sides is called a heptagon. A heptagon is a polygon with 7 sides and 7 angles. It is a closed two-dimensional shape that is bounded by straight lines. A heptagon can be regular, meaning all of its sides and angles are of the same measure, or it can be irregular, which means the sides and angles are of different measures.
A regular heptagon has 7 internal angles that measure 128. 57° each, and its sides are all equal in length. A heptagon is also known as a septagon.
## Is any 6 sided shape a hexagon?
Yes, any 6-sided shape is a hexagon. A hexagon is a polygon with six sides and angles. It is a closed shape, which means that all the angles add up to 360°. A hexagon also has six vertices (the points where the sides meet).
The sides of a hexagon can be either regular (all sides of equal length) or irregular (sides of different lengths). The internal angles of a hexagon are all the same, typically 120°. Hexagons can be found in nature in a variety of forms, such as honeycombs, snowflakes and minerals.
There are a variety of six-sided shapes that can also be classified as hexagons, such as squares, rectangles and trapezoids, which all have six vertices and internal angles of 120°.
## What is a 6 sided shape with unequal sides?
A 6 sided shape with unequal sides is generally referred to as an irregular hexagon. An irregular hexagon is a 6-sided polygon where the sides can have different lengths and angles. It is different from a regular hexagon, which is a 6-sided polygon with all sides and angles the same size.
Examples of irregular hexagons include shapes found in nature, such as snowflakes or a honeycomb, as well as many other shapes.
## Is a hexagon equal on all sides?
Yes, a hexagon is equal on all sides. The six sides of a hexagon create six equal angles of 120°, and each side of the shape is of equal length. This creates a perfect balance and symmetry throughout the shape.
A regular hexagon has all the sides and angles equal to each other, while an irregular hexagon may differ in size and angles. However, a hexagon is still equal on all sides because each of the six sides is the same length.
## What shapes can make a hexagon?
To make a hexagon, you will need six straight lines of equal length joined together at their endpoints. Any combination of colors or objects can be used to represent the six lines, as long as they are connected end-to-end.
For example, six pencils of equal length or six pieces of string connected together could make a hexagon. Other shapes that can be used to form a hexagon include a triangle, a square, or three connected trapezoids.
In addition, six circles or semicircles can be arranged in a hexagon shape, while six differently-shaped polygons (triangles, rectangles, octagons, etc. ) can also fit together to create a hexagon.
## Is a polygon a hexagon if it has 4 sides?
No, a polygon is not a hexagon if it has only 4 sides. A hexagon is a polygon with 6 sides and 6 angles. A 4-sided polygon is known as a quadrilateral, which includes the square and the rectangle. Other types of polygons include a pentagon (5 sides), heptagon (7 sides), octagon (8 sides), nonagon (9 sides), decagon (10 sides), and dodecagon (12 sides).
## What is the difference of polygon and hexagon?
The main difference between a polygon and a hexagon is their size and the number of sides they have. A polygon is a closed shape with any number of sides greater than two, while a hexagon is a six-sided closed geometric shape.
A polygon can have three, four, five sides and so on while a hexagon is composed of six sides and six angles. Besides their number of sides, the way they are formed is also different. A polygon is formed by connecting line segments together, while a hexagon is formed by connecting segments of circles.
## What polygon has 4 sides?
A quadrilateral is a polygon that has 4 sides. Quadrilaterals are classified into several different categories, including parallelograms, rectangles, squares, trapezoids, rhombuses, and kites. All of these shapes have special properties, such as parallel opposite sides, four right angles, congruent sides, diagonals that bisect each other, and so on.
Every quadrilateral is a shape that has four sides, but not all shapes with four sides are quadrilaterals. For example, a quadrilateral could be a rectangle, a square, a rhombus, a trapezoid, or a kite — but a square is not a rectangle, a rhombus is not a trapezoid, and a kite is not a rhombus.
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LCM of 8 and also 14 is the smallest number amongst all usual multiples that 8 and also 14. The first few multiples that 8 and also 14 room (8, 16, 24, 32, 40, 48, . . . ) and (14, 28, 42, 56, 70, . . . ) respectively. There space 3 generally used techniques to find LCM that 8 and also 14 - by department method, through listing multiples, and by prime factorization.
You are watching: Lowest common multiple of 8 and 14
1 LCM that 8 and also 14 2 List of Methods 3 Solved Examples 4 FAQs
Answer: LCM of 8 and also 14 is 56.
Explanation:
The LCM of two non-zero integers, x(8) and y(14), is the smallest confident integer m(56) the is divisible by both x(8) and also y(14) without any type of remainder.
The approaches to discover the LCM that 8 and also 14 are explained below.
By department MethodBy Listing MultiplesBy element Factorization Method
### LCM that 8 and 14 by division Method
To calculation the LCM the 8 and 14 by the department method, we will certainly divide the numbers(8, 14) by their prime determinants (preferably common). The product of these divisors offers the LCM that 8 and 14.
Step 3: proceed the measures until only 1s are left in the last row.
The LCM that 8 and also 14 is the product of all prime numbers on the left, i.e. LCM(8, 14) by department method = 2 × 2 × 2 × 7 = 56.
### LCM the 8 and also 14 by Listing Multiples
To calculation the LCM of 8 and 14 by listing the end the typical multiples, we deserve to follow the given listed below steps:
Step 1: perform a few multiples the 8 (8, 16, 24, 32, 40, 48, . . . ) and 14 (14, 28, 42, 56, 70, . . . . )Step 2: The common multiples from the multiples that 8 and also 14 are 56, 112, . . .Step 3: The smallest typical multiple that 8 and also 14 is 56.
∴ The least common multiple of 8 and 14 = 56.
See more: What Two Items Must Be Equal For A Nuclear Equation To Be Balanced ?
### LCM of 8 and 14 by element Factorization
Prime factorization of 8 and 14 is (2 × 2 × 2) = 23 and also (2 × 7) = 21 × 71 respectively. LCM the 8 and 14 deserve to be acquired by multiplying prime components raised to your respective highest power, i.e. 23 × 71 = 56.Hence, the LCM of 8 and 14 by prime factorization is 56.
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# How do you use the Binomial Theorem to expand (x + y)^5?
Feb 10, 2016
${\left(a + b\right)}^{5} = {a}^{5} + 5. {a}^{4.} b + 10. {a}^{3.} {b}^{2} + 10. {a}^{2.} {b}^{3} + 5. {a}^{1.} {b}^{4} + {b}^{5}$
#### Explanation:
The binomial theorem tells us that if we have a binomial (a+b) raised
to the ${n}^{t h}$ power the result will be
${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} {c}_{k}^{n} \cdot {a}^{n - k} \cdot {b}^{n}$
where " "c _k^n= (n!)/(k!(n-k)!)
and is read "n CHOOSE k equals n factorial divided by k factorial (n-k) factorial".
So ${\left(a + b\right)}^{5} = {a}^{5} + 5. {a}^{4.} b + 10. {a}^{3.} {b}^{2} + 10. {a}^{2.} {b}^{3} + 5. {a}^{1.} {b}^{4} + {b}^{5}$
we notice that the powers of ' a ' keeps decreasing from 5 (which representes ' n ') until it reaches ${a}^{\text{zero}}$ at the last term.
also we notice that the power of ' b ' keeps increasing from zero untill it reaches 5 at the last term.
Now the we have to determine the coefficient of each term through the...
c_k^n= (n!)/(k!(n-k)!)
first coefficient c_0^5=(5!)/(0! .5!)=1
second c_1^5=(5!)/(1! .4!)=5
c_2^5=(5!)/(2! .3!)=10
c_3^5=(5!)/(3! .2!)=10
c_4^5=(5!)/(4! .1!)=5
c_5^5=(5!)/(5!.0!)=1
but the calculation of combinations can be tedious..so fortunately
there is an awesome way to determine the binomial coefficients which is Pascal's triangle
it is easy to deduce this triangle :
hope that helps ! : )
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Study Materials for 2023 CFA®, FRM®, Actuarial, GMAT® and EA® Exams
# GMAT® Data Sufficiency Questions
### What is the Structure of GMAT Data Sufficiency Questions?
Data Sufficiency questions is one of the sections in Quantitative reasoning questions (the other one is Problem-solving questions). Data Sufficiency questions are designed to measure your ability to analyze a quantitative problem, ascertain which data is relevant, and decide the extent to which there is enough data to solve the problem.
A data sufficiency question consists of a question and two statements. To answer the question, one should first identify the statement that provides information relevant to the question and then eliminate all the other possible answers by using math knowledge and other everyday facts. There will be 14 to 15 questions on data sufficiency in each quantitative section.
## Tips for Data Sufficiency Questions?
• The quantitative takes 62 minutes to complete and includes 31 questions. Therefore, you do have on average 2 minutes to answer each question.
• There are 14 to 15 data sufficiency questions in every quantitative section.
• Determine whether the problem allows only a single value or a range of values. Note that the primary objective is to determine whether there is enough data to solve the problem.
• Avoid unnecessary assumptions about geometrical figures, as they are not necessarily drawn to scale.
3.5 Million
50 Thousand
# 1 Rated
## Division & Factoring
When integer y is divided by 2, is the remainder 1?
(1) $$(-1)^{(y + 2)} = -1$$
(2) $$y$$ is prime.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
(1) That $$(–1)^{(y + 2)} = -1$$, dictates that y + 2 must be odd and therefore y is odd since only an odd exponent can produce a value < 0. Furthermore, if an odd number is divided by 2, the remainder must be 1; SUFFICIENT.
(2) That y is prime does not definitively identify y as definitively even or odd, so the remainder could be either 0 or 1; NOT SUFFICIENT.
The correct answer is A; statement 1 alone is sufficient.
## Number Properties
Chiku received a $3.50 per hour raise this week. If last week she worked 40 hours per week at her old pay rate, how many fewer hours can she work this week and still guarantee that she makes more this week than she did last week? (1) She made$620 last week.
(2) Her raise was 20 percent greater than that of any of her coworkers.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
(1) If Chiku made 620 last week, it is possible to solve for her old pay rate, then her new pay rate, and then the number of hours she’d need to work to guarantee she makes more this week than she did last week; SUFFICIENT. (2) That her raise was greater than that of any of her coworkers introduces a new unknown variable; NOT sufficient. The correct answer is A; statement 1 alone is sufficient. ## Question 3 ## Percentages (Data Sufficiency) If the number of rabbits at a certain pet store is 250 percent greater than the number of hamsters at the same store, how many hamsters are in the store? (1) There are 40 rabbits in the store. (2) There are 56 hamsters and rabbits in the store. A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient. B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient. C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D) EACH statement ALONE is sufficient. E) Statements (1) and (2) TOGETHER are not sufficient. The correct answer is: D). (1) Let h be the number of hamsters at the store and r be the number of rabbits to derive the equation 40 = 2.5h from the statement and given conditions, so h = 16; SUFFICIENT. (2) Using the same variables derive the equation 56 = 2.5h + h, so that h = 16; SUFFICIENT. The correct answer is D; each statement alone is sufficient. ## Question 4 ## Exponents & Radicals (Data Sufficiency) The amount of bacteria in a culture after some $$t$$ hours is given by the function $$f(t)=pe^{kt}$$ where $$p$$ is a constant. What is the number of bacteria after 8 hours? (1) There were approximately 275 bacteria after 2 hours (2) $$k=0.16$$ A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient. B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient. C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D) EACH statement ALONE is sufficient. E) Statements (1) and (2) TOGETHER are not sufficient. The correct answer is: C) (1) From the function, $$f(t)=pe^{kt}$$, when $$t=2, f(t)=275$$. The equation becomes $$pe^{2k}=275$$. This is one equation in two unknowns, hence, we cannot solve it. More information is required; NOT SUFFICIENT. (2) When $$k=0.16$$, we have $$f(t)=pe^{0.16t}$$. In this case, $$p$$ is unknown, hence more information is required; NOT SUFFICIENT. Considering the two cases, we have $$pe^{2k}=275$$ and $$f(t)=pe^{0.16t}$$ which reduced to $$pe^{2(0.16)}=275$$. We solve the equation \begin{align*} pe^{2(0.16)} &=275\\ pe^{3.2} &=275\\ \ln \left(pe^{3.2}\right) &=\ln 275\\ \ln p + 3.2 &=\ln 275\\ \ln p &=\ln 275 – 3.2 = 2.41677\\ p& =e^{2.41677}=11.21\end{align*}. We now solve the equation When $$t=8$$, we have $$f(8)=11.21e^{0.16(8)} =11.21e^{1.28}=40.32$$ The correct answer is C; BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. ## Question 5 ## 3D Geometry (Data Sufficiency) A cylindrical layer cake has two vanilla layers surrounding a marzipan layer, represented by the shaded region, as shown below. Is the cake split into three layers of equal volume? (1) The height of marzipan layer is equal to the radius of the whole cake. (2) The volume of the entire cake is 81π cubic inches. A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient. B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient. C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D) EACH statement ALONE is sufficient. E) Statements (1) and (2) TOGETHER are not sufficient. The correct answer is: E) Note that the volume for a cylinder can be found from the formula $$πr^2 × height$$ where r = radius and the height is perpendicular to the radius. (1) That the height for the marzipan layer is equal to the radius of the cake means that the volume of the marzipan layer can be found from the simplified formula $$πr^3$$, but the total cake volume remains dependent on the actual height of the cake; NOT sufficient. (2) Represent the information algebraically as $$πr^2 × height = 81$$, but no information is provided about the height in relation to the radius; NOT sufficient. (Together) The information together still does not relate the height to the radius. For instance the radius could be 3 and the ratio of the marzipan layer to the others could be equal as each would have a volume of 27π. Alternatively, the radius could = 1 and the marzipan layer would be significantly smaller than the other two layers. The correct answer is E; both statements together are still not sufficient. ## Question 6 ## Inequalities At Rounders Grocer, orange slices cost2 a pound, pineapple chunks cost $3 a pound, and cut watermelon cost$5 a pound. If Sally buys enough of these three fruits from Rounders to make five pounds of fruit salad, and at least one pound of each, which fruit did she buy the most of by weight?
(1) Sally spent less than $5 on orange slices. (2) Sally spent more than$18 on her fruit salad.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
First note that Sally must buy at least one pound of each of the fruits, which would result in a sum fixed cost of $10. (1) If Sally spent less than$5 on orange slices, then she spent must have spent more on one of the other two fruits, but it is unclear which; NOT sufficient
(2) If Sally spent more than $18 then she must spend more than$8 on the two-variable pounds of fruit, which could only be accomplished by buying two more pounds of cut watermelon; Sufficient
The correct answer is B; statement 2 alone is sufficient.
## Plane Geometry
The Embeyay Expressway and Emefay Expressway intersect at a perpendicular junction coming from Ulster City and Finster Town, respectively, while a direct road connecting the two municipalities is entirely straight. How much further would a motorist traveling the expressways between Ulster City to Finster Town have to drive in comparison to another motorist who used the direct road?
(1) The distance from Ulster City to the Embeyay Expressway and Emefay Expressway junction is 12 kilometers.
(2) The distance from Ulster City to Finster Town on the direct road is 15 kilometers.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
Recognize that a right triangle can be formed with the junction creating the 90° angle and the two municipalities at the other two vertices.
Therefore, only two of the distances between the locations will be necessary to find the third by way of the Pythagorean Theorem.
(1) This provides the distance for one of the two legs of the right triangle only; NOT sufficient.
(2) This provides the distance for the hypotenuse of the right triangle only; NOT sufficient.
(Together) Complete the theorem as $$12^2+b^2 =15^2$$ to find that the shorter leg distance is 9. From there, 12 + 9 = 21 is 6 kilometers further than 15 kilometers; SUFFICIENT.
## Descriptive Statistics
A supermarket display of canned corn is shaped like a pyramid with one can on top and two more cans in each row below. If the display is only one can deep for the entire pyramid, what is the median number of cans in a row in the pyramid?
(1) There are 100 cans in the pyramid.
(2) The range of cans per row is 18.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
Remember that if there is a constant difference in terms that the average for the sequence = median of the sequence.
The formula for average is sum divided by the number of terms, so in this case the total number of cans = median × number of rows.
And the total number of cans = x + (x + 2) + (x + 4) … up to the total number of rows.
Therefore, two equations are already theoretically known, and only one additional equation is needed to solve for all of the values involved.
(1) If 100 is the total number of cans, then it would be possible to count from 1 + 3 + 5 + 7… to determine the total number of rows, and thereby the median, without completing the process; SUFFICIENT.
(1)If the range in cans per row is 18, then it would be possible to determine that the number of cans in the last row is 19 and from there the median, without completing the process; SUFFICIENT
The correct answer is D; each statement alone is sufficient.
## Coordinate Geometry
If line $$d$$ in the coordinate plane has the equation $$y = mx + b$$, where $$m$$ and $$b$$ are constants, what is the slope of line $$d$$?
(1) Line $$d$$ intersects the line with equation $$y = 6x + 2$$ at the point (1, 8).
(2) Line $$d$$ is parallel to the line with equation $$y = (2 – m)x + b – 4$$.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
Note that the slope of a line with equation $$y = mx + b$$ is$$m$$.
(1) A line passing through the point (1, 8) can have any value for its slope, so it is impossible to determine the slope of line $$d$$.
For example, the line $$y = x + 7$$ intersects $$y = 6x + 2$$ at (1, 8) with a slope of 1, while the line $$y = 2x + 6$$ intersects $$y = 6x + 2$$ at (1, 8) with a slope of 2; NOT sufficient.
(2) Parallel lines have the same slope, and it is possible to solve for the slope as $$m = 2 – m$$, $$2m = 2$$, and $$m = 1$$; SUFFICIENT.
## Rates, Work, and Combined Time
Cars enter a parking garage at a certain rate. At the same time, cars leave the parking garage at a different rate. At what rate, in cars per minute, is the number of cars in the garage changing?
(1) Cars enter the garage at a rate of 13.5 cars per 4.5 minutes.
(2) Cars leave the garage at a rate of 5 cars per minute.
A) Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
B) Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are not sufficient.
(1) While this rate can be converted to $$\frac {13.5}{4.5} = 3$$ cars/minute entering the garage, it says nothing about the rate at which cars leave; NOT sufficient.
(2) This statement says nothing about the rate at which cars enter the garage; NOT sufficient.
(Together) With the rates at which cars both enter and leave the garage, the target question can be answered; SUFFICIENT.
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# Rubik’s Cube Blindfolded
In summer 2015, I already knew how to solve a Rubik’s cube using the beginner’s method. Not just that, I had spent quite some time teaching my classmates how to solve it. Having watched a video of someone solving it blindfolded, I decided to take it to the next level and learn the blindfolded solve.
I started my search online, and sure enough, there were several resources on blindfold solving. In this article, I’m going to discuss what I found on the internet, give you a brief idea on how exactly a blindfold solve is done, and how I devised a slightly different method myself.
So let’s start with what exactly we try to achieve in a blindfold solve. In a Rubik’s cube, there are “algorithms”, sequences of moves that achieve certain positions. For example, if you want to rotate a corner, there would be a sequence of moves to do this. The algorithm is written using alphabets that represent which layer to move.
Here is a simple algorithm: R’ D’ R D. What this algorithm tells me to do, is first rotate the right layer, then the down layer, then the right layer again, and finally the down layer. However, when I talk about the right layer, there are two possible directions of rotation. To distinguish between these directions, we use the R’ (read as right-inverted) and R. The former refers to a counter-clockwise rotation, whereas the latter refers to a clockwise one.
When we do a regular solve of a cube, we use algorithms that achieve a certain transition at a time, while disturbing the rest of the cube. Only towards the very end, we make sure to use algorithms that achieve transitions without disturbing the rest of the pieces.
In a blindfold solve, we focus on using algorithms that do not disturb the rest of the cube, from the beginning itself. This means that there is a certain piece on the cube, and we call this the shooting location, from where we move a piece to another location. You can understand this better through an example.
In a cube, we have two kinds of pieces- edge pieces and corner pieces. In the image below, A, B, C and D are the edge pieces, whereas 1, 2, 3 and 4 are the corner pieces.
In this scenario, let’s just say that I choose B as my shooting position. This means that whatever algorithm I have in memorized will move the piece which is right now at B, to another position (say D). Suppose the scrambled cube is such that the piece at B needs to be moved to position C, I would bring the piece C to the position D using some moves, then perform my algorithm, and move the piece back to C. While doing all of this, I make sure not to disturb the rest of the cube at all.
When I do the above, the piece at C now comes at B, meaning that the algorithm I know is a swapping algorithm. So, I just look at what piece is present at C, and where that needs to go in a solved cube. Let’s say it needs to go to a position O (not shown on the cube), and the piece at O needs to be at position M. Then, in my head, I have the alphabet sequence C-O-M. I have already named all my pieces such that when I say an alphabet, I know where that piece is.
Thus, I shoot from B to C, and the piece at C comes to B. Now I shoot from B to O, and the piece at O comes to B. Finally, I shoot from B to M, and the cube is solved! I follow a similar method for the corners.
In summary, what I have in my mind is just a sequence of plus and minus, which represents the rotation of corner pieces, followed by a sequence of numbers, which tells me the shooting positions for my corners, and at the end a sequence of alphabets which tells me the shooting positions for my edge pieces. Typically, the +/- sequence has 7 characters, there are around 6 or 7 numbers, and around 10 alphabets. By remembering all of these together with 5 algorithms (2 pairs of algorithms are very similar to each other, and 1 separate algorithm), I am able to solve a Rubik’s cube with my eyes closed!
The following section goes more into detail about how I devised my own algorithm by eliminating the parity condition and reducing the number of algorithms required.
When looking up online, I found an algorithm called the Y-perm to solve the corners. This algorithm takes care of the corner rotation as well as position. However, I was having some difficulty using this algorithm. Hence, I used a corner rotating algorithm, which was much simpler, and paired it with a the L-perm algorithm to solve corners. The L-perm simply swaps the corner positions 1 and 2, while at the same time swapping the edges B and A. To solve corners, I simply make sure that my corners are rotated correctly, and then use the L-perm to shoot from the position 1.
Parity is a special condition that comes up when I finish solving the corners. Since my corner swapping algorithm also swaps edges B and A, I get a slightly disturbed cube if I solve an odd number of corners. So out of 8 corners, let’s say 3 were already in their correct positions, and I just had to solve 5. In such a scenario, I end up getting the piece A at my shooting position instead of the piece B.
Traditional blindfold solving methods dictate that one must use an algorithm called the R-perm to eliminate parity. However, the R-perm was another algorithm that I found difficult to apply. Hence, I devised a simple solution. When memorizing the cube, I take note of whether I’m solving an odd number of corners or not. If I am, instead of looking at where my piece B needs to be in a solved cube, I simply look at where my piece C needs to be, and start my alphabet memorization from there. This trick eliminates the need for a parity algorithm!
Hope you were able to get a general idea of how a Rubik’s cube is solved blindfolded, and liked the method that I devised using improvisations to the traditional methods!
P.S. I successfully solved the Rubik’s cube blindfolded in front of an audience of around a hundred people in summer 2015 during an adventure camp!
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# Linear Equations Questions and Answers
## Questions on Linear Equations
Linear Equation Questions and Answers: Your Ultimate Resource | Top Tips and Tricks Discover a treasure trove of linear equation Q&A with top tips and tricks for solving them like a pro. Improve your problem-solving abilities and excel in math exams with our detailed answers.
### Tips to solve Linear Equations :
Here are some useful tips to help you solve linear equations efficiently:
1. Isolate the Variable: The goal is to get the variable (usually represented by ) on one side of the equation. Use inverse operations (addition, subtraction, multiplication, division) to move all other terms to the other side.
2. Combine Like Terms: If you have multiple terms with the variable on the same side, combine them to simplify the equation.
3. Use the Distributive Property: When dealing with parentheses, distribute the terms inside the parentheses to eliminate them.
4. Eliminate Fractions: If the equation contains fractions, multiply both sides of the equation by the common denominator to eliminate them.
5. Keep Equations Balanced: Perform the same operation on both sides of the equation to maintain equality.
6. Simplify Radicals: If there are square roots or other radicals, try to simplify them by finding perfect square factors.
7. Be Mindful of Negative Numbers: Pay attention to negative signs and avoid making sign errors during calculations.
8. Check Your Solution: After finding the value of the variable, plug it back into the original equation to ensure it satisfies the equation.
9. Practice Regularly: Like any skill, practice is crucial for mastering linear equations. Solve various problems to build your proficiency.
10. Learn Common Mistakes: Be aware of common mistakes students make while solving linear equations and avoid them.
11. Graphical Representation: Sometimes, graphing the equation on a coordinate plane can help visualize the solution.
12. Word Problems: Convert word problems into linear equations by defining variables and setting up the equation before solving them.
By following these tips and practicing regularly, you’ll become more confident and proficient in solving linear equations. Happy math solving!
### Rules for Linear Equations Questions and Answers.
• If a=b then a+c=b+c for any c. All this is saying is that we can add a number, c, to both sides of the equation and not change the equation.
• If a=b then a−c=b−c for any c. As with the last property we can subtract a number, c, from both sides of an equation.
• If a=b then ac=bc for any c. Like addition and subtraction, we can multiply both sides of an equation by a number, c, without changing the equation.
• If a=b then a/c=b/c for any non-zero c. We can divide both sides of an equation by a non-zero number, c, without changing the equation.
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### Linear Equation Questions and Answers
1. The roots of the equation $4x-3\times 2x+2+32=0$ would include-
20
20
8.28%
40
40
8.97%
5
5
2.07%
17
17
80.69%
Solution:
$4x-3\times 2x+2+32=0$
$4x - 6x + 34 = 0$
$-2x + 34 = 0$
$2x - 34 = 0$
$2x = 34$
$x = 17$
2. If $ax =b, by=c and cz =a$, then the value of xyz is:
4
4
2.05%
3
3
7.53%
2
2
2.05%
1
1
88.36%
Solution:
$ax =b, by=c and cz =a$
$x = \frac {b}{a}$
$y = \frac {c}{b}$
$z = \frac {a}{c}$
$xyz = \frac{b}{a} \times \frac{c}{b} \times \frac{a}{c}$
xyz = 1
3. If x = 1+21/2 and y=1-21/2, then x2+y2 is -
107/2
107/2
5.74%
203/2
203/2
8.2%
997/2
997/2
4.1%
445/2
445/2
81.97%
Solution:
x = 1+21/2 and y=1-21/2
x2+y2
$\left ( 1 + \frac{21}{2} \right )^{2} + \left ( 1 - \frac{21}{2} \right )^{2}$
$1 + \left ( \frac{21}{2} \right )^{2}+ 2 \times 1 \times \left ( \frac{21}{2} \right ) + 1 + \left ( \frac{21}{2} \right )^{2}- 2 \times 1 \times \left ( \frac{21}{2} \right )$
$2 + 2 \times \left ( \frac{21}{2} \right )^{2}$
2 + 441/2
445/2
4. If $4x+3 = 2x+7$, then the value of x is:
2
2
89.47%
3
3
3.76%
1
1
2.26%
4
4
4.51%
Solution:
4x+3 = 2x+7
2x = 4
x = 2
5. $2x+y = 2$ and $2x-y = \frac {21}{2}$ ,the value of x is:
10/14
10/14
5.22%
9/8
9/8
0.87%
25/8
25/8
86.96%
5/4
5/4
6.96%
Solution:
$2x+y = 2$ and $2x-y = \frac {21}{2}$
$2x + y = 2$
$2x-y = \frac {21}{2}$
$4x = 2 + \frac {21}{2}$
$4x = \frac {25}{2}$
$x = \frac {25}{8}$
Putting this in $2x-y = \frac {21}{2}$
$2 \frac {25}{8} - y = \frac {21}{2}$
$y = \frac {25}{8} - \frac {21}{2}$
$y = \frac {25 -42}{4}$
$y = \frac {17}{4}$
$x = \frac {25}{8}$
6. If x = 8, y = 27, the value of $\sqrt{\left ( \frac{x^{4}}{3} + \frac{y^{2}}{3} \right )}$ is:
$\sqrt{ \frac{4828}{3}}$
$\sqrt{ \frac{4828}{3}}$
6.67%
$\sqrt{ \frac{4825}{3}}$
$\sqrt{ \frac{4825}{3}}$
79.05%
$\sqrt{ \frac{4818}{3}}$
$\sqrt{ \frac{4818}{3}}$
4.76%
$\sqrt{ \frac{4826}{3}}$
$\sqrt{ \frac{4826}{3}}$
9.52%
Solution:
= $\sqrt{\left ( \frac{x^{4}}{3} + \frac{y^{2}}{3} \right )}$
= $\sqrt{\left ( \frac{8^{4}}{3} + \frac{27^{2}}{3} \right )}$
= $\sqrt{\left ( \frac{4096}{3} + \frac{729}{3} \right )}$
= $\sqrt{ \frac{4825}{3}}$
7. What is the slope of the line represented by the equation 3y+4x=12?
3/4
3/4
8.87%
−4/3
−4/3
86.29%
3
3
1.61%
−4
−4
3.23%
Solution :
To find the slope of the line, we need to rewrite the equation in slope-intercept form (y=mx+b), where m is the slope.
Solving for y gives $y = -\frac{4}{3}x + 4$
The coefficient of x represents the slope, so the correct answer is −4/3
8. If 2x + 3y = 16 and 2x - 3y= 36, the value of x is:
13
13
88.79%
23
23
2.8%
33
33
3.74%
43
43
4.67%
Solution:
2x + 3y = 16 -----(1)
2x - 3y= 36 ------ (2)
4x = 52
x = 13
9. If 6(x-3) = 36(x-5), then what is the value of x?
51/5
51/5
1.98%
27/5
27/5
90.1%
15/7
15/7
1.98%
17/3
17/3
5.94%
Solution:
6(x-3) = 36(x-5)
x - 3 = 6(x - 5)
x - 3 = 6x - 30
5x = 27
x = 27/5
10. Determine whether the ordered triple (3,−2,1) is a solution to the system.
2x+y+z=5,
6x−4y+5z=31,
5x+2y+2z=13
True
True
73.21%
False
False
16.96%
No solution
No solution
7.14%
None of the above
None of the above
2.68%
Solution - We will check each equation by substituting the values of the ordered triple for x,y, and z
x+y+z=2(3)+(−2)+(1)=5 True
6x−4y+5z=6(3)−4(−2)+5(1)=18+8+5=31 True 5x+2y+2z=5(3)+2(−2)+2(1)=15−4+2=13 True
The ordered triple (3,−2,1) is indeed a solution to the system.
×
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How do u graph this function f(x)=x^3-4? and Approximate to the nearest tenth, the real root of the equation f(x)=x^3-4=0
giorgiana1976 | Student
To graph a function, you have to find some important points which belongs to the graph, to check if the function is increasing or decreasing, to verify if the function is continuous, to see if the function is convex or concave.
Let's see if the function is continuous:
lim f(x) = lim (x^3-4) = limx^3-lim4 = inf, if x->inf
lim f(x) = -inf, if x->-inf
The function is continuous.
Let's see if the function is increasing or decreasing.
For this rason, we'll calculate it's first derivative:
f'(x) = 3x^2
Because f'(x)>0, no matter if x>0 or x<0, that means that f(x) is an increasing function.
For x=0, the function has an inflection point.
To check the aspect of the function, we have to calculate the second derivative:
f"(x) = 6x
For x<0. f"(x)<0, so f(x) is concave
For x>0, f"(x)>0, so f(x) is convex.
Now, we have to give values for x and we'll find values for y, so that to draw the graph of f(x), which passes through these points.
neela | Student
To graph f(x) = x^3-4.
Solution:
Let y = x^3-4.
First find the coordinayes of some of the points satisfying the y =x^3-4, as below:
Let x =0 , then y = 0^3-4 = -4.
x=1 , y = 1^3-4 =-3
f(1) = -3 and f(2) = 4.
x1 = 1 + {f(0)-f(1)}/f(2)-f(1) = 1+ 3/7 = 1.43.
f(1.43) = -1.076
f(2) = 4.
Therefore x2 = 1.43 +{ 0-f(1.43)/(f(2)-f(4.3)} (2-4.3)
= 1.43 +1.076/(4+1.076)}(2-1.43)
= 1.43+0.12 = 1.55 = 1.6 to the 1st decimal place
x=2 and y = 2^3 -4 = 4.
x =3 , y =3^3-4 = 23
Like that we go any desired number of pairs of coordinates
Plot (0,-4), (1,-3), (2,4) and (3,23 ) in the graph.
We see that the real solution for f(x) lies between x=1 and x=2 as f(1) = -3 and f(2) = 4 and f(x) = o for some value of x i (1,2).
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# #\$ Sum of Cubes
The Sum of Cubes Formula is a shortcut that you can use anytime you need to factor an expression that has two perfect cubes added to each other.
If you have two perfect cubes subtracted instead of added, you can use the Difference of Cubes Formula.
## How to Factor a Sum of Cubes
1. Make sure the polynomial is a sum of cubes.
• Does the polynomial have two terms connected with an addition sign?
• Are the terms perfect cubes?
2. Find the cube root of each term.
3. Plug the results of Step 2 into the Sum of Cubes Formula.
## Examples
Factor: $x^3+8$
Is $$\yellow x^3+8$$ a sum of cubes?
Yes, $$\yellow x^3$$ and $$\yellow 8$$ are perfect cubes that are connected with an addition sign.
What are the cube roots of $$\yellow x^3$$ and $$\yellow 8$$?
$\sqrt[3]{\yellow x^3}={\yellow x}$
$\sqrt[3]{\yellow 8}={\yellow 2}$
What is the factored form of $$\yellow x^3+8$$?
To find the factored form of $$\yellow x^3+8$$, I will substitute $$\yellow x$$ and $$\yellow 2$$ into the Sum of Cubes Formula.
$x^3+8=({\yellow x}+{\yellow 2})({\yellow x}^2-{\yellow 2}{\yellow x}+{\yellow 2}^2)$
When I simplify the $${\yellow 2}^2$$ term I get…
$x^3+8=({\yellow x}+{\yellow 2})({\yellow x}^2-{\yellow 2}{\yellow x}+{\yellow 4})$
The factored form of $$\yellow x^3+8$$ is…
$\yellow (x+2)(x^2-2x+4)$
Factor: $27y^3+64x^3$
Is $$\blue 27y^3+64x^3$$ a sum of cubes?
Yes, $$\blue 27y^3$$ and $$\blue 64x^3$$ are perfect cubes that are connected with an addition sign.
What are the cube roots of $$\blue 27y^3$$ and $$\blue 64x^3$$?
$\sqrt[3]{\blue 27y^3}={\blue 3y}$
$\sqrt[3]{\blue 64x^3}={\blue 4x}$
What is the factored form of $$\blue 27y^3+64x^3$$?
To find the factored form of $$\blue 27y^3+64x^3$$, I will substitute $$\blue 3y$$ and $$\blue 4x$$ into the Sum of Cubes Formula.
$27y^3+64x^3=({\blue 3y}+{\blue 4x})(({\blue 3y})^2-({\blue 4x})({\blue 3y})+({\blue 4x})^2)$
When I simplify all the terms I get…
$27y^3+64x^3=({\blue 3y}+{\blue 4x})({\blue 9y^2}-{\blue 12xy}+{\blue 16x^2})$
The factored form of $$\blue 27y^3+64x^3$$ is…
$\blue (3y+4x)(9y^2-12xy+16x^2)$
Factor: $125x^{12}+8y^9$
Is $$\green 125x^{12}+8y^9$$ a sum of cubes?
Yes, $$\green 125x^{12}$$ and $$\green 8y^9$$ are perfect cubes that are connected with an addition sign.
What are the cube roots of $$\green 125x^{12}$$ and $$\green 8y^9$$?
$\sqrt[3]{\green 125x^{12}}={\green 5x^4}$
$\sqrt[3]{\green 8y^9}={\green 2y^3}$
What is the factored form of $$\green 125x^{12}+8y^9$$?
To find the factored form of $$\green 125x^{12}+8y^9$$, I will substitute $$\green 5x^4$$ and $$\green 2y^3$$ into the Sum of Cubes Formula.
$125x^{12}+8y^9=({\green 5x^4}+{\green 2y^3})(({\green 5x^4})^2-({\green 2y^3})({\green 5x^4})+({\green 2y^3})^2)$
When I simplify all the terms I get…
$125x^{12}+8y^9=({\green 5x^4}+{\green 2y^3})({\green 25x^8}-{\green 10x^4y^3}+{\green 4y^6})$
The factored form of $$\green 125x^{12}+8y^9$$ is…
$\green (5x^4+2y^3)(25x^8-10x^4y^3+4y^6)$
## Why It Works
The Sum of Cubes Formula works because factoring “un-does” polynomial multiplication.
You can see where the formula comes from if you reverse engineer the process and multiply $$(a+b)$$ and $$(a^2-ab+b^2)$$.
To understand where the formula comes from…
1. Multiply $$(a+b)(a^2-ab+b^2)$$.
2. Undo the multiplication to find the Sum of Cubes Formula.
I like using the multiplication algorithm to multiply polynomials, but you can also use the box method.
$\begin {array}{cccc} & a^2 & -ab & +b^2 \\ \times & & {\yellow a} & {\green +b} \\ \hline & {\green a^2b} & {\green -ab^2} & {\green +b^3} \\ {\yellow a^3} & {\yellow -a^2b} & {\yellow +ab^2} & \\ \hline {\yellow a^3} & 0 & 0 & {\green +b^3} \end{array}$
Multiplying the polynomials gives us a sum of cubes…
$(a+b)(a^2-ab+b^2)=a^3+b^3$
That means we can “un-do” the multiplication and write the equation backwards to find the Sum of Cubes Formula…
$a^3+b^3=(a+b)(a^2-ab+b^2)$
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# Algebra's Rules of Divisibility
### Part of the Algebra I For Dummies Cheat Sheet
In algebra, knowing the rules of divisibility can help you solve faster. When factoring algebraic expressions to solve equations, you need to be able to pull out the greatest factor. You also need common factors when reducing algebraic fractions. The rules of divisibility help you find the common factors and change the algebraic expressions so that they're put in a more workable form.
• Divisibility by 2: A number is divisible by 2 if the last digit in the number is 0, 2, 4, 6, or 8.
• Divisibility by 3: A number is divisible by 3 if the sum of the digits in the number is divisible by 3.
• Divisibility by 4: A number is divisible by 4 if the last two digits in the number form a number divisible by 4.
• Divisibility by 5: A number is divisible by 5 if the last digit is 0 or 5.
• Divisibility by 6: A number is divisible by 6 if it is divisible by both 2 and 3.
• Divisibility by 8: A number is divisible by 8 if the last three digits form a number divisible by 8.
• Divisibility by 9: A number is divisible by 9 if the sum of the digits of the number is divisible by 9.
• Divisibility by 10: A number is divisible by 10 if it ends in 0.
• Divisibility by 11: A number is divisible by 11 if the sums of the alternate digits are different by 0, 11, 22, or 33, or any two-digit multiple of 11. In other words, say you have a six-digit number: Add up the first, third, and fifth digits — the odd ones. Then add the digits in the even places — second, fourth, and sixth. Then subtract the smaller of those totals from the larger total, and if the answer is a multiple of 11, the original number is divisible by 11.
• Divisibility by 12: A number is divisible by 12 if the last two digits form a number divisible by 4 and if the sum of the digits is divisible by 3.
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# Define discrete mathematics symbols to express the concept of discrete numbers.
The aim will be to teach students that not all numbers in math are integers, and that they has to be in a position to recognize that some numbers have already been modified by addition or subtraction. Some numbers in math are continuous and can be written out devoid of fractional digits.
Discrete implies some thing that cannot be summed. You do not desire to make use of the word “integers” when coping with discrete numbers since they cannot be added together. Once you have challenges with recognizing that you’ll find unique variations of a continuous number, you may need to understand the meaning of one digit and its relationship to yet another digit. A continuous quantity is known as a simplex when all the digits are one, for instance, three.
All the digits of a continuous http://washington.academia.edu/JohnGottman quantity ought to possess the very same values, otherwise the outcome will be a series of zero digits. However, there are some discrete numbers which might be multiples of 1. For instance, every single quantity with a single plus 1 is known as a multiple of one particular, but all of the numbers with two, 3, four, 5, and six are known as multiples of one.
There are two symbols to get a number that has a fractional part. The very first symbol is known as a fraction along with the second is called a fractional portion. To be able to identify the value of a fraction, divide the quantity by 10. In the event the resulting worth is equal to ten, then it truly is a fraction.
The symbols using the fraction are often not used. Most math students study from instruction that the worth of any quantity is the sum of its components. It is actually uncomplicated to recall the definition of a fraction when it really is all you will need to complete to multiply numbers.
The numerator www.essay-company.com is usually the very first quantity within a series of numbers. The denominator is the second number in the series. You may see the numerator divided by 1 in front of a quantity that has a fraction. The distinction among the numerator plus the denominator might be written as an integer. 1 can also write the fraction in its decimal form and use the appropriate symbol.
This quantity is often written in each of the digits from the all-natural numbers. The last digit would be the symbol for zero. When the fraction doesn’t have a zero, the symbol is going to be a single vertical bar. Lots of math teachers will instruct their students to write the fraction with no trailing zero as 1. They are applying the classic notation in a new way.
The symbols with all the fractions have the symbols for plus or minus. If a fraction has 3 sides, then there is a further symbol for the side that is positive. These symbols are just a visual substitute for the regular notation.
Sometimes it is tough to know what the which means of a number is. For instance, for those who add two fractions, you are going to notice that the numbers in both sides add as much as a single. The combined number is what’s getting spoken about within the phrase, “the sum of the two fractions.”
Mathematicians study the which means of numbers in order that they will solve difficulties and make models. They commit a lot of years studying numbers, and they’re a far cry from being able to draw on an empty board or use a ruler. If they have been only taught how you can draw a zero, quite a few math students will be frustrated.
Students really should get started with basic concepts of calculus. In their minds, they’ll not be able to draw the symbols, and can not fully grasp why it requires two to tango. Then they are able to take a few courses in differential equations and continue to improve their understanding from the notion of two to tango.
## 《WhatAre All Real Numbers in Math?》上有1条评论
1. кибербезопасность что это
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# RS Aggarwal Solutions for Class 8 Chapter 1 - Rational Numbers Exercise 1G
Students can refer and download RS Aggarwal Solutions for Class 8 Maths Chapter 1- Exercise 1G, Rational Numbers from the links provided below. BYJU’S expert team has solved the RS Aggarwal Solutions for a better understanding of the concepts and also helps in brushing up your mathematical skills. By practicing the solutions many times helps in securing high marks. In Exercise 1G of RS Aggarwal Class 8 Maths, we shall study the Word problems.
## Download PDF of RS Aggarwal Solutions for Class 8 Maths Chapter 1- Exercise 1G
### Access answers to RS Aggarwal Solutions for Class 8 Maths Chapter 1- Rational Numbers Exercise 1G
1. From a rope 11m long, two pieces of lengths 2 3/5m and 3 3/10m are cut off. What is the length of the remaining rope?
Solution: we know that the given details are
The length of the rope =11m
Length of the 1st piece = 2 3/5m
Length of the 2nd piece = 3 3/10m
Total length of the pieces = length of 1st piece + length of 2nd piece
= 2 3/5 + 3 3/10
=(5×2+3)/5 + (10×3+3)/10
=13/5 + 33/10
By taking the LCM for 5 and 10 is 10
= (13×2 + 33×1)/10
= (26+33)/10
=59/10m
Length of the remaining rope = length of the rope – total length of the pieces
= 11m – 59/10m
By taking the LCM
= (11×10 -59×1)/10
= (110-59)/10
=51/10
= 5 1/10m
∴ The length of the remaining rope is 5 1/10m.
2. A drum full of rice weighs 40 1/6kg. If the empty drum weighs 13 3/4kg, find the weight of rice in the drum.
Solution: we know that the given details are
A drum full of rice weighs = 40 1/6kg
Empty drum weighs = 13 3/4kg
Weight of rice in the drum = drum full of rice – weight of empty drum
= 40 1/6 – 13 ¾
= 241/6 – 55/4
By taking the LCM for 6 and 4 is 12
= (241×2 – 55×3)/12
= (482 – 165)/12
= 317/12
= 26 5/12kg
∴ The weight of rice in the drum = 26 5/12kg.
3. A basket contains three types of fruits weighing 19 1/3kg in all. If 8 1/9kg of these be apples, 3 1/6kg be oranges and the rest pears, what is the weight of the pears in the basket?
Solution: we know that the given details are
Weight of three types of fruits = 19 1/3kg
Weight of apples = 8 1/9kg
Weight of oranges = 3 1/6kg
Weight of pears = Weight of three types of fruits – (Weight of apples + Weight of oranges)
= 19 1/3 – (8 1/9 + 3 1/6)
= 58/3 – (73/9 + 19/6)
= 58/3 – (73×2 + 19×3)/18 (by taking LCM for 9 and 6 is 18)
= 58/3 – (146 + 57)/18
= 58/3 – 203/18
By taking LCM for 3 and 18 is 18
= (58×6 – 203×1)/18
= (348 – 203)/18
= 145/18
= 8 1/18kg
∴ The weight of pears is 8 1/18kg.
4. On one day a rickshaw puller earned ₹160. Out of his earnings he spent ₹26 3/5 on tea and snacks. ₹50 ½ on food and ₹16 2/5 on repairs of the rickshaw. How much did he save on that day?
Solution: we know that the given details are
Total earnings = ₹160
Earnings spent on tea and snacks = ₹26 3/5
Earnings spent on food = ₹50 1/2
Earnings spent on repairs = ₹16 2/5
∴ Total expenditure = Earnings spent on tea and snacks + Earnings spent on food + Earnings spent on repairs
= ₹26 3/5 + ₹50 ½ + ₹16 2/5
= 133/5 + 101/2 + 82/5
By taking LCM for 5 and 2 is 10
= (133×2 + 101×5 + 82×2)/10
= (266 + 505 +164)/10
= 935/10
Total Savings = Total Earnings – Total Expenditure
= ₹160 – ₹935/10
By taking 10 as the LCM
= (160×10 – 935×1)/10
= (1600 – 935)/10
= 665/10 = 66 ½
∴ Total Savings on that day is ₹66 ½.
5. Find the cost of 3 2/5meters of cloth at ₹66 ¾ per meter.
Solution: we know that the given details are
Cost of the cloth per meter = ₹66 ¾
Total meters = 3 2/5meters
Total cost of cloth = Cost of the cloth per meter × Total meters
= ₹66 ¾ × 3 2/5
= 267/4 × 17/5
= 4335/20
Further we can divide by 5 we get,
4335/20 = 867/4 = 216 ¾
∴ Cost of the cloth is ₹216 ¾.
6. A car moving at an average speed of 60 2/5km/hr. How much distance will it cover in 6 ¼ hours?
Solution: we know that the given details are
Speed of the car = 60 2/5 km/hr.
Total hours = 6 ¼ hours
By using the formula, speed = distance/time
Distance of the car = speed of the car × time taken
= 60 2/5 × 6 ¼
=302/5 × 25/4
= 7550/20
= 755/2 = 377 ½
∴ Distance covered = 377 ½ km.
7. Find the area of a rectangular park which is 36 3/5m long and 16 2/3m broad.
Solution: we know that the given details are
Length of the park = 36 3/5m
Breadth of the park = 16 2/3m
By using the formula, area of rectangle = length × breadth
Area of rectangular park = 36 3/5 × 16 2/3
= 183/5m × 50/3m
= 9150/15m2 = 610m2
∴ Area of the park = 610m2
8. Find the area of a square plot of land whose each side measures 8 ½ meters.
Solution: we know that the given details are
One side measures = 8 1/2m
By using the formula,
Area of square = side × side
Area of square plot = 8 ½m × 8 ½m
= 17/2m × 17/2m
= 289/4m2 = 72 ¼ m2
∴ Area of square plot is 72 ¼ m2
9. One liter of petrol costs ₹63 ¾. What is the cost 34 liters of petrol?
Solution: we know that the given details are
Cost of 1 liter petrol = ₹63 ¾
Cost of 34 liters petrol = 34 × cost of 1 liter petrol
= 34 × 63 ¾
= 34 × 255/4
= 8670/4
We can further divide by 2 we get,
= 4335/2
= 2167 ½
∴ Cost of 34 liters petrol is ₹2167 ½
10. An aeroplane covers 1020km in an hour. How much distance will it cover in 4 1/6hours?
Solution: we know that the given details are
Distance covered in 1 hour = 1020km
Distance covered in 4 1/6 hour = 4 1/6 × distance covered in 1 hour
= 4 1/6 × 1020
= 25/6 × 1020
=25500/6
= 4250km
∴ Distance covered in 4 1/6hours = 4250km.
## RS Aggarwal Solutions for Class 8 Maths Chapter 1- Exercise 1G
Exercise 1G of RS Aggarwal Class 8 Chapter 1, Rational Numbers contains the advance concepts related to Rational Numbers. This exercise mainly deals with Word problems.
The RS Aggarwal Solutions can help the students practice and learn the fundamentals as it provides all the answers to the questions from the RS Aggarwal textbook. Students should practice RS Aggarwal Solutions without fail to understand how the concepts are solved in order to secure high marks.
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# Correlation & Regression Math 137 Fresno State Burger.
## Presentation on theme: "Correlation & Regression Math 137 Fresno State Burger."— Presentation transcript:
Correlation & Regression Math 137 Fresno State Burger
Correlation Finding the relationship between two quantitative variables without being able to infer causal relationships Correlation is a statistical technique used to determine the degree to which two variables are related
Rectangular coordinate Two quantitative variables One variable is called independent (X) and the second is called dependent (Y) Points are not joined No frequency table Scatter Plot
Example
Scatter diagram of weight and systolic blood pressure
Scatter plots The pattern of data is indicative of the type of relationship between your two variables: positive relationship negative relationship no relationship
Positive relationship
Negative relationship Reliability Age of Car
No relation
Correlation Coefficient Statistic showing the degree of relation between two variables
Simple Correlation coefficient (r) It is also called Pearson's correlation or product moment correlation coefficient. It measures the nature and strength between two variables of the quantitative type.
The sign of r denotes the nature of association while the value of r denotes the strength of association.
If the sign is + this means the relation is direct (an increase in one variable is associated with an increase in the other variable and a decrease in one variable is associated with a decrease in the other variable). While if the sign is - this means an inverse or indirect relationship (which means an increase in one variable is associated with a decrease in the other).
The value of r ranges between ( -1) and ( +1) The value of r denotes the strength of the association as illustrated by the following diagram. 10 -0.25-0.750.750.25 strong intermediate weak no relation perfect correlation Direct indirect
If r = Zero this means no association or correlation between the two variables. If 0 < r < 0.25 = weak correlation. If 0.25 ≤ r < 0.75 = intermediate correlation. If 0.75 ≤ r < 1 = strong correlation. If r = l = perfect correlation.
How to compute the simple correlation coefficient (r)
Exercise 1 (calculating r): A sample of 6 children was selected, data about their age in years and weight in kilograms was recorded as shown in the following table. It is required to find the correlation between age and weight. A sample of 6 children was selected, data about their age in years and weight in kilograms was recorded as shown in the following table. It is required to find the correlation between age and weight. Weight (Kg) Age (years) serial No 1271 862 83 1054 1165 1396
These 2 variables are of the quantitative type, one variable (Age) is called the independent and denoted as (X) variable and the other (weight) is called the dependent and denoted as (Y) variables to find the relation between age and weight compute the simple correlation coefficient using the following formula:
Y2Y2 X2X2 xy Weight (Kg) (y) Age (years) (x) Serial n. 1444984127 1 64364886 2 1446496128 3 1002550105 4 1213666116 5 16981117139 6 ∑y2= 742 ∑x2= 291 ∑xy= 461 ∑y= 66 ∑x= 41 Total
r = 0.759 strong direct correlation
EXAMPLE: Relationship between Anxiety and Test Scores Anxiety(X) Test score (Y) X2X2X2X2 Y2Y2Y2Y2XY 102100420 8364924 2948118 171497 56253630 65362530 ∑X = 32 ∑Y = 32 ∑X 2 = 230 ∑Y 2 = 204 ∑XY=129
Calculating Correlation Coefficient r = - 0.94 Indirect strong correlation
Spot-check for understanding :
Regression Analysis Regression: technique concerned with predicting some variables by knowing others The process of predicting variable Y using variable X
Regression Uses a variable (x) to predict some outcome variable (y) Tells you how values in y change as a function of changes in values of x
Correlation and Regression Correlation describes the strength of a Correlation describes the strength of a linear relationship between two variables Linear means “straight line” Regression tells us how to draw the straight line described by the correlation
Regression Calculates the “best-fit” line for a certain set of data The regression line makes the sum of the squares of the residuals smaller than for any other line Regression minimizes residuals
By using the least squares method (a procedure that minimizes the vertical deviations of plotted points surrounding a straight line) we are able to construct a best fitting straight line to the scatter diagram points and then formulate a regression equation in the form of: By using the least squares method (a procedure that minimizes the vertical deviations of plotted points surrounding a straight line) we are able to construct a best fitting straight line to the scatter diagram points and then formulate a regression equation in the form of: b
Regression Equation Regression equation describes the regression line mathematically Intercept Slope
Linear Equations
grades on hours Regressing grades on hours Predicted final grade in class = 59.95 + 3.17*(number of hours you study per week)
Predict the final grade of… Someone who studies for 12 hours Final grade = 59.95 + (3.17*12) Final grade = 97.99 Someone who studies for 1 hour: Final grade = 59.95 + (3.17*1) Final grade = 63.12 Predicted final grade in class = 59.95 + 3.17*(hours of study)
Exercise 2 (regression equation) A sample of 6 persons was selected the value of their age ( x variable) and their weight is demonstrated in the following table. Find the regression equation and what is the predicted weight when age is 8.5 years. A sample of 6 persons was selected the value of their age ( x variable) and their weight is demonstrated in the following table. Find the regression equation and what is the predicted weight when age is 8.5 years.
Weight (y)Age (x)Serial no. 12 8 12 10 11 13 768569768569 123456123456
Answer Y2Y2 X2X2 xyWeight (y)Age (x)Serial no. 144 64 144 100 121 169 49 36 64 25 36 81 84 48 96 50 66 117 12 8 12 10 11 13 768569768569 123456123456 7422914616641Total
Regression equation
a) b)
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# Question: How To Find X In Math?
## What is the value of x in math?
The letter ” x ” is often used in algebra to mean a value that is not yet known. It is called a “variable” or sometimes an “unknown”. In x + 2 = 7, x is a variable, but we can work out its value if we try! A variable doesn’t have to be ” x “, it could be “y”, “w” or any letter, name or symbol. See: Variable.
## How do you solve for x in an expression?
A linear algebraic equation is nice and simple, containing only constants and variables to the first degree (no exponents or fancy stuff). To solve it, simply use multiplication, division, addition, and subtraction when necessary to isolate the variable and solve for “x “. Here’s how you do it: 4x + 16 = 25 -3x =
## What find X means?
Solve for x means find the value of x that would make the equation you see true. Think about this equation: x + 1 = 3. If you were asked to solve it, that would mean finding some value for x that gives you three when you add one to it. That’s what solving an equation is all about!
## What does R mean in math?
In maths, the letter R denotes the set of all real numbers. In other words, real numbers are defined as the points on an infinitely extended line. This line is called the number line or the real line, on which the points of integers are evenly ranged.
You might be interested: Often asked: What Is Value Math?
## What does 3 X mean?
The meaning of < 3X abbreviation is "Love heart and Kiss" What does < 3X mean? < 3X as abbreviation means "Love heart and Kiss"
## What is the rule for solving equations?
The following steps provide a good method to use when solving linear equations. Simplify each side of the equation by removing parentheses and combining like terms. Use addition or subtraction to isolate the variable term on one side of the equation. Use multiplication or division to solve for the variable.
## What are the 4 steps to solving an equation?
We have 4 ways of solving one- step equations: Adding, Substracting, multiplication and division.
## What is the formula of algebraic expression?
Here is a complete list of all the important algebra formulas: (a + b)2 = a2 + 2ab + b. (a – b)2 = a2 – 2ab + b. (a + b) (a – b) = a2 – b.
## What does LR mean in calculator?
L-R = 0; Here “L – R” gives you an idea of the accuracy of the calculator’s solution. In most cases, it’ll be exact (assuming you have the sense to change recurring decimals into fractions).
## How do you solve an equation with two variables?
In a two – variable problem rewrite the equations so that when the equations are added, one of the variables is eliminated, and then solve for the remaining variable. Step 1: Multiply equation (1) by -5 and add it to equation ( 2 ) to form equation (3) with just one variable.
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# CRAMMERS RULE
Views:
Category: Education
## Presentation Description
No description available.
## Presentation Transcript
### Solving Systems of equations using Determinants :
Solving Systems of equations using Determinants What are Determinants? Evaluating a 2x2 Determinant Expanding and Evaluating a 3x3 Determinant Using Cramer’s Rule to Solve a System of 2 Equations Using Cramer’s Rule to Solve a System of 3 Equations Gabriel Crammer
### Why Study Determinants? :
Why Study Determinants? Determinants offer a quick, number-crunching way to solve Systems of Equations No Graphing No Elimination Method No Substitution Method Put your equations in standard form Create a few determinants Evaluate them according to formulas
### Concept:A Determinant is a Square Matrix :
Concept:A Determinant is a Square Matrix # rows = # of columns Different Notations Matrix: Determinant:
### Evaluating a 2x2 Determinant :
Evaluating a 2x2 Determinant A 2x2 Determinant is always a number:It’s the Product of the main diagonal numbersminus the Product of the other diagonal numbers = ad – bc = 2(3) – 4(-1) = 10
### Practice :
Practice = ad – bc = 3(9) – 2(6) = 15 = -2(-4) – (-5)(-7) = -27 = 5(0) – ½(-1) = ½
### Evaluating a 3x3 Determinant(expanding along the top row) :
Evaluating a 3x3 Determinant(expanding along the top row) Expanding by Minors (little 2x2 determinants)
### Array Signs for a 3x3 Determinant :
Array Signs for a 3x3 Determinant You can expand across any row or down any column The Signs form a checkerboard pattern Use the Signs from this array
### Expand Down the Middle Column :
Expand Down the Middle Column Apply the Signs as you go down the column You always should get the same number
### Using Cramer’s Rule to Solve a System of Two Equations :
Using Cramer’s Rule to Solve a System of Two Equations
### Applying Cramer’s Ruleon a System of Two Equations :
Applying Cramer’s Ruleon a System of Two Equations
### Class Exercise: Applying Cramer’s Ruleon a System of Two Equations :
Class Exercise: Applying Cramer’s Ruleon a System of Two Equations
### Using Cramer’s Rule to Solve a System of Three Equations :
Using Cramer’s Rule to Solve a System of Three Equations
### Applying Cramer’s Ruleon a System of Three Equations :
Applying Cramer’s Ruleon a System of Three Equations
### Class Exercise: Applying Cramer’s Ruleon a System of Three Equations :
Class Exercise: Applying Cramer’s Ruleon a System of Three Equations
THANK YOU
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40
Q:
# In a class of 60 students, the number of boys and girls participating in the annual sports is in the ratio 3 : 2 respectively. The number of girls not participating in the sports is 5 more than the number of boys not participating in the sports. If the number of boys participating in the sports is 15, then how many girls are there in the class ?
A) 20 B) 25 C) 30 D) Data inadequate
Explanation:
Let the number of boys and girls participating in sports be 3x and 2x respectively.
Then, 3x = 15 or x = 5.
So, number of girls participating in sports = 2x = 10.
Number of students not participating in sports = 60 - (15 + 10) = 35.
Let number of boys not participating in sports be y.
Then, number of girls not participating in sports = (35 -y).
Therefore (35 - y) = y + 5 <=> 2y<=> 30<=> y = 15.
So, number of girls not participating in sports = (35 - 15) = 20.
Hence, total number of girls in the class = (10 + 20) = 30.
Q:
How many inches is 5' 4"?
A) 66 B) 64 C) 62 D) 60
Explanation:
We know 1 feet = 12 inches.
According to the question,
5' 4'' = 5 feet 4 inches = 12 x 5 + 4 = 64 inches.
Hence, 5' 4'' = 5 feet 4 inches = 64 inches.
0 42
Q:
Multiplicative inverse of 0
A) 0 B) 1 C) infinity D) None of the above
Explanation:
In the real numbers, zero does not have a reciprocal because no real number multiplied by 0 produces 1 (the product of any number with zero is zero).
Hence, 0 doesn't have any multiplicative inverse.
0 90
Q:
6 divided by 2(1 + 2) =
A) 1 B) 0 C) 9 D) Can't be determined
Explanation:
The given expression can be simplified as
$÷$ 2 (1 + 2)
The expression can be simplified further by the order of operations, using BODMAS rule.
First evaluate Parentheses/Brackets, then evaluate Exponents/Orders, then evaluate Multiplication-Division, and finally evaluate Addition-Subtraction.
Now, the expression becomes $÷$ 2 (3)
According to the order of operations, division and multiplication have the same precedence, so the correct order is to evaluate from left to right. First take 6 and divide it by 2, and then multiply by 3.
6 ÷ 2 × 3
= 3 × 3
= 9
But not 6÷2×3 = 6 ÷ 6 = 1
Hence, 6 ÷ 2 (1 + 2) = 9.
0 112
Q:
Multiplicative inverse of 7?
A) 1 B) -1/7 C) 0 D) 1/7
Explanation:
Multiplicative inverse is nothing but a reciprocal of a number.
It is defined as one of a pair of numbers that when multiplied with another number equals the number 1.
Multiplicative inverse or reciprocal of 7 is
7 x n = 1
=> n = 1/7.
Hence, Multiplicative inverse or reciprocal of 7 is 1/7.
2 90
Q:
What is the product of a number and its reciprocal?
A) 0 B) 1 C) -ve of the number D) the number itself
Explanation:
The product of a number and its reciprocal is always equals to 1.
For Example : Let the number be 4.
Now, its reciprocal is 1/4
Hence, required product = 4 x 1/4 = 1.
Now, take the number as -15.
Then its reciprocal is -1/15
Required product = -15 x -1/15 = 1.
Hence, the product of a number and its reciprocal is 1.
0 127
Q:
A plane is an undefined term because it
A) is a flat surface that extends indefinitely in all directions B) is where other geometric shapes can be constructed C) is described generally, not using a formal definition D) can be named using three noncollinear points
Answer & Explanation Answer: C) is described generally, not using a formal definition
Explanation:
In geometry, we can define plane as a flat surface with no thickness.
The surface extends with no ends.
A plane does not have any edges even if we draw it on paper with edges, it does't mean it has edges.
Hence, A plane is an undefined term because it is described generally, not using a formal definition.
0 165
Q:
Absolute value of 9?
A) 0 B) 9 C) 8 D) -9
Explanation:
Absolute value of a number :
It means that the distance of a number from 0 on a number line.
Here absolute value of 9 is that on a number line 9 is 9 units away from 0. Hence its absolute value is 9. Similarly, absolute value of -9 means -9 is also 9 units away from the 0 on number line. Hence, absolute value of -9 is also 9.
Hence, the absolute value of 9 is 9.
1 219
Q:
What is the value of 62 tens?
A) 6 B) 620 C) 62 D) 2
Explanation:
Here in the question it is asked that 62 tens, which implies ten times of 62.
=> 62 x 10 = 620.
Hence, the value of 62 tens = 620.
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GCSE Maths Algebra
Maths Formulas
# Maths Formulas
Here we will learn about maths formulas, including what maths formulas are, how to use them and how to rearrange them. We will also look at special formulas involved in kinematics.
There are also maths formulas worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What are maths formulas?
Maths formulas are rules that are written connecting two or more variables. We can use maths formulas to work out the values of given variables, based on other known values.
To do this we substitute the values that we know into the formula and then calculate the value of the unknown.
For example, here is a rectangle with base b and height h.
The formula for the perimeter P of a rectangle is
P=2(b+h).
We can use this formula to find the value of the perimeter by substituting in the values for the base and height of the rectangle.
Other examples of maths formulas you may recognise include,
### List of maths formula
These are the maths formulas you are most likely to encounter for GCSE.
Areas
Area of a parallelogram
A=bh \hspace{1.5cm} Where A is Area, b is base and h is height.
Area of a triangle
A=\frac{1}{2}bh \hspace{1.35cm} Where A is Area, b is base and h is height.
Area of a trapezium
A=\frac{1}{2}(a+b)h \hspace{.65cm} Where A is Area, a and b are the parallel sides and h is height.
Circles
Circumference of a circle
C=\pi d \hspace{1.6cm} Where C is circumference and d is diameter.
Area of a circle
A=\pi r^2 \hspace{1.55cm} Where A is Area and r is radius.
These formulas can be extended to give the following
\text{Arc length}=\frac{\theta}{360}\pi d \;\; and \;\; \text{Sector area}=\frac{\theta}{360}\pi r^2
Where \theta is the angle at the centre of the sector.
Volumes:
Volume of a cuboid:
V=lwh \hspace{1cm} Where V is volume, and l,w and h are the lengths of the 3 sides.
Volume of a prism:
V=\text{area of cross}-\text{section}\times \text{length}
Volume of a cylinder:
V=\pi r^2 h \hspace{1cm} Where V is volume, r is radius and h is height.
Angles in a polygon:
Sum of angles in a polygon:
S=180(n-2) \hspace{.5cm} Where S is sum and n is number of sides.
Compound measures:
Speed, distance, time
s=\frac{d}{t} \hspace{1.5cm} Where s is speed, d is distance and t is time.
Mass, density, volume
d=\frac{m}{v} \hspace{1.4cm} Where m is mass, d is density and v is volume.
Right-angled triangles:
Pythagoras’ theorem
a^2+b^2=c^2
Where c is the hypotenuse and a and b are the shorter sides of a right-angled triangle.
Trigonometry
\sin(\theta)=\frac{O}{H}. \hspace{1cm} \cos(\theta)=\frac{A}{H}. \hspace{1cm} \tan(\theta)=\frac{O}{A}.
Where O is the opposite side, A is the adjacent side and H is the hypotenuse.
Some formulas are for the Higher GCSE.
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
Where a,b and c are coefficients of the quadratic equation to solve ax^2+bx+c=0.
Pyramids, Cones and Spheres
Volume of a pyramid
V=\frac{1}{3}\times \text{area of base} \times h \hspace{1cm} Where V is volume and h is height.
Volume of a cone
V=\frac{1}{3}\pi r^2 h \hspace{1cm} Where V is volume, r is radius and h is height.
Curved surface area of a cone
\text{CSA}=\pi rl \hspace{1cm} Where l is the slant length.
Volume of a sphere
V=\frac{4}{3}\pi r^3 \hspace{1.25cm} Where V is volume and r is radius.
Surface area of a cone
\text{SA}=4\pi r^2
Higher Trigonometry:
Area of a triangle
A=\frac{1}{2}ab \sin(C)
Sine rule
\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}
Cosine rule
a^2=b^2+c^2-2bc \cos(A)
Where a,b and c are the sides of a triangle and A,B and C are the corresponding angles.
Coordinate geometry:
m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}
Where m is the gradient and (x_{1},y_{1}) and (x_{2},y_{2}) are coordinates.
Midpoint of two points is
(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})
The equation of a straight line is
y-y_{1}=m(x-x_{1})
### Using maths formulas
We can use maths formulas in a variety of ways.
For example, looking back to the rectangle above, we had the formula
P=2(b+h).
If we are given 2 of the values, we can use the formula to work out the third value.
Let’s look at a specific example.
Here b=13 and h=5 so using the formula for perimeter we can write,
\begin{aligned} P&=2(b+h)\\\\ P&=2(13+5)\\\\ P&=2\times 18\\\\ P&=36 \end{aligned}
We have calculated that the perimeter of the rectangle is 36cm.
### Rearranging formulas
Formulas can be rearranged in a similar way to equations.
To do this, we use inverse operations to make one chosen variable the subject of the formula. The subject of the formula is the single variable that is equal to everything else. i.e. the term by itself on one side of the equal sign.
For example, let’s rearrange P=2(b+h) to make h the subject.
Step-by-step guide: Rearranging formulae
### Kinematics formulas
Many of the formulas used at GCSE are about geometrical situations, for example to calculate area or perimeter. However, there are some formulas which are used to solve problems involving moving objects. Kinematics is the maths concerned with the movement of objects.
The 5 different kinematic formulas involve 5 different variables of motion.
The variables of motion are,
s= displacement
u= initial velocity
v= final velocity
a= acceleration
t= time
Velocity is speed in a given direction.
Displacement is the distance from the original position.
The five formulas are,
\begin{aligned} &v=u+at \\\\ &s=ut+\frac{1}{2}at^2 \\\\ &s=\frac{1}{2}(u+v)t \\\\ &v^2=u^2+2as \\\\ &s=vt-\frac{1}{2}at^2 \end{aligned}
The kinematic formulas are sometimes known as the suvat equations.
Step-by-step guide: Kinematic formulae
## How to use maths formulas
In order to work with maths formulas:
1. Use the formula given in the question.
2. Work carefully to answer the question, one step at a time.
3. Write the final answer clearly.
## Maths formulas examples
### Example 1: using a maths formula
Given that F=\frac{9}{5}C+32 find F when C=50.
1. Use the formula given in the question.
The formula in the question is,
F=\frac{9}{5}C+32.
2Work carefully to answer the question, one step at a time.
Substitute the value that you know into the formula and work out the answer.
Here you know C=50.
\begin{aligned} F&=\frac{9}{5}C+32 \\\\ F&=\frac{9}{5}\times 50+32 \\\\ F&=90+32 \\\\ F&=122 \end{aligned}
### Example 2: using a maths formula
Given that D=\frac{M}{V}, calculate M when D=10 and V=40.
Use the formula given in the question.
Work carefully to answer the question, one step at a time.
### Example 3: rearranging equations
Make b the subject of the formula w=3ab+2 .
Use the formula given in the question.
Work carefully to answer the question, one step at a time.
### Example 4: kinematics
A train starts from rest and accelerates at 2 \ m/s^{2}. Find its velocity after 20 seconds.
You may use v=u+at where
Use the formula given in the question.
Work carefully to answer the question, one step at a time.
### Common misconceptions
• Take care with writing variables clearly
Neat writing is important. For example, the variables v and u can easily be muddled as they look similar and they both represent velocity in kinematic formulas.
• When we square root a number/variable the answer can be positive or negative
The square root of any value can be positive or negative. When taking a square root, use the plus/minus sign \pm to indicate this.
For example,
\begin{aligned} x^2&=ab-c \\\\ x&=\pm \sqrt{ab-c} \end{aligned}
### Practice maths formulas questions
1. Given that A= \frac{1}{2}bh, find A when b=24 and h=10.
A=68
A=120
A=240
A=17
Substitute the values into the formula and calculate the answer.
\begin{aligned} A&= \frac{1}{2}bh \\\\ A&= \frac{1}{2}\times 24\times 10\\\\ A&=120 \end{aligned}
2. Given the W=\frac{H}{A}+5, find the value of H when W=12 and A=4.
H=28
H=8
H=68
H=43
Substitute the values into the formula and calculate the answer.
\begin{aligned} W&=\frac{H}{A}+5\\\\ 12&=\frac{H}{4}+5\\\\ 7&=\frac{H}{4}\\\\ 28&=H \end{aligned}
3. Make t the subject of the formula,
h=3t-y.
t=\frac{h-3}{y}
t=\frac{h+3}{y}
t=\frac{h+y}{3}
t=\frac{h-y}{3}
Rearrange using inverse operations.
4. Make x the subject of the formula,
a=\frac{x}{3}-b.
x=3a+b
x=3a-b
x=3(a-b)
x=3(a+b)
Rearrange using inverse operations.
5. Given that s=ut+\frac{1}{2}at^2, find s when u=10, \ a=3 and t=5.
s=57.5
s=12.5
s=87.5
s=42.5
Substitute the values into the formula and calculate the answer.
\begin{aligned} s&=ut+\frac{1}{2}at^2\\\\ s&= 10\times 5+\frac{1}{2}\times 3 \times 5^2\\\\ s&= 50+37.5\\\\ s&=87.5 \end{aligned}
6. A train is travelling at 28 \ m/s. It brakes for 4 seconds and it decelerates at 6 \ m/s^{2}. Calculate how far the train travels in those 4 seconds.
You may use the formula s=ut+\frac{1}{2}at^2.
Where s= displacement, \ u= initial velocity, \ v= final velocity, \ a= acceleration, \ t= time
64 \ m
160 \ m
124 \ m
100 \ m
Substitute the given values into the formula. Note that a=-6 because the train is slowing down.
\begin{aligned} s&=ut+\frac{1}{2}at^2 \\\\ s&=28\times 4+\frac{1}{2}\times (-6) \times4^2 \\\\ s&=112-48 \\\\ s&=64 \end{aligned}
The answer will be 64 metres, as metres is consistent with the other units in the question.
### Maths formulas GCSE questions
(1 mark)
C is the formula.
(1)
( A is an equation, B is an identity, D is an expression.)
2. Rearrange x=my-8 to make y the subject.
(2 marks)
x+8=my
(1)
y=\frac{x+8}{m}
(1)
3. For the following question you may use the formula v=u+at where
A motorbike is travelling along a road at a velocity of 11 \ m/s. The motorbike accelerates at 3 \ m/s^2 for 4 seconds. Find the final velocity.
(2 marks)
v=11+3\times 4
(1)
v=23 \ m/s
(1)
## Learning checklist
You have now learned how to:
• Use mathematical formula
• Rearrange equations and formula
• Use kinematic formulas
## Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
Find out more about our GCSE maths tuition programme.
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# How do you solve 12/5 = x/6?
Feb 20, 2016
This is the same approach as Simon L but in more detail
$x = 14 \frac{2}{5}$
#### Explanation:
Objective: to end up with $x$ on one side of the equals sign and everything else on the other side.
$\textcolor{b l u e}{\text{Principles behind Simons' short cut method}}$
Simon used the shortcut method. This is based on: if you move something to the other side of the = you reverse 'what it is doing';
So if you had say:$\text{ } x + 3 = 2$ and you wished to move the $+ 3$
When you move the $+ 3$ to the other side it becomes $- 3$
'~~~~~~~~~~~
Suppose you had $x \div 3 = 2$ and you wished to move the $\div 3$ to the others side.
So $x \div 3 = 2 \text{ becomes } x = 2 \times 3$
Another way of writing $x \div 3 \text{ is } \frac{x}{3}$
so $\frac{x}{3} = 2 \text{ becomes } x = 2 \times 3$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{The first principle method that the short cut method is based on}}$
Given:$\text{ " 12/5" "=" } \frac{x}{6}$
Write as: $\textcolor{b r o w n}{\text{ "12/5" "=" } x \times \frac{1}{6}}$
Multiply both sides by$\text{ } \textcolor{b l u e}{6}$ giving:
$\text{ "color(brown)(12/5 color(blue)(xx6)" "=" } x \times \frac{1}{6} \textcolor{b l u e}{\times 6}$
" "color(brown)((12color(blue)(xx6))/5" "=" "x xx(color(blue)(6))/6)
But $\frac{6}{6} = 1$ giving:
$\text{ "72/5" "=" } x$
$72 \div 5 = 14 \frac{2}{5}$ so we have
$\text{ } \textcolor{red}{x = 14 \frac{2}{5}}$
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# NCERT Exemplar Solutions Class 8 Mathematics Solutions for Linear Equations in One Variable - Exercise in Chapter 4 - Linear Equations in One Variable
Question 5 Linear Equations in One Variable - Exercise
If (5x/3) – 4 = (2x/5), then the numerical value of 2x – 7 is
(a) 19/13
(b) -13/19
(c) 0
(d) 13/19
(b) -13/19
Given, (5x/3) – 4 = (2x/5)
(5x/3) - (2x/5) = 4
LCM of 3 and 5 is 15
(25x – 6x)/15 = 4
19x = 4 × 15
19x = 60
X = 60/19
Then, Substitute the value of x in 2x -7
= (2 × (60/19)) – 7
= (120/19) – 7
= (120 - 133)/19
= - 13/19
Video transcript
"hello students i am rita your math leader tutor and the today's question is if 5x by 3 minus 4 is equal to 2x by 5 then the numerical value of 2x minus 7 is the option is 19 by 13 b is minus 13 by 19 c is 0 and the option d is 13 by 19. first of all we solve this equation 5x minus 3 first we separate out the variable and constant term minus 4 when it goes there by the process of transposition it becomes plus now the lcm of 3 and 5 is 15 so 3 5's are 15 5 5's are 25 x minus 5 3's are 15 then two 3's are 6 is equal to 4 now we cross multiply here by applying cross multiply so 25 minus 6 it becomes 19x and 15 into 4 it becomes 60 now the value of x is equal to 60 by 19 so this is the value of x but in the question we have to find the value of 2 x minus 7 so we put the value of x 2 into 60 by 19 minus 7 by 1 so it becomes 120 by 19 minus 7 by 1 calcium is 19 so 120 minus 19 into 7 so it becomes 120 minus 133 so it becomes minus 13 by 90 so this is our answer minus 13 by 19 it means option b is correct so minus 13 by 19 this is our answer i hope you like this video so please subscribe lido for more updates and do comment your questions thank you"
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# How do you find the square root of 7?
Aug 31, 2016
$\sqrt{7} \approx 2.645751311$
#### Explanation:
Since $7$ is a prime number, it has no square factors and its square root cannot be simplified.
It is an irrational number, so cannot be exactly represented by $\frac{p}{q}$ for any integers $p , q$.
We can however find good rational approximations to $\sqrt{7}$.
First note that:
${8}^{2} = 64 = 63 + 1 = 7 \cdot {3}^{2} + 1$
This is in Pell's equation form:
${p}^{2} = n {q}^{2} + 1$
with $n = 7$, $p = 8$ and $q = 3$.
This means that $\frac{8}{3}$ is an economical approximation for $\sqrt{7}$ and it also means that we can use $\frac{8}{3}$ to derive the continued fraction expansion of $\sqrt{7}$:
$\frac{8}{3} = 2 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}}$
and hence we can deduce:
sqrt(7) = [2;bar(1,1,1,4)] = 2 + 1/(1+1/(1+1/(1+1/(4+1/(1+1/(1+1/(1+1/(4+...))))))))
The next economical approximation is given by truncating the continued fraction expansion just before the next $4$, i.e.
sqrt(7) ~~ [2;1,1,1,4,1,1,1] = 2 + 1/(1+1/(1+1/(1+1/(4+1/(1+1/(1+1/1)))))) = 127/48 = 2.6458bar(3)
This is also a solution of Pell's equation for $7$, since we find:
${127}^{2} = 16129 = 16128 + 1 = 7 \cdot {48}^{2} + 1$
If you want more accuracy, truncate just before the next $4$ or the one after.
By expanding the repeating part of the continued fraction for $\sqrt{7}$ we can derive a generalised continued fraction:
$\sqrt{7} = \frac{21}{8} + \frac{\frac{7}{64}}{\frac{21}{4} + \frac{\frac{7}{64}}{\frac{21}{4} + \frac{\frac{7}{64}}{\frac{21}{4} + \frac{\frac{7}{64}}{\frac{21}{4} + \ldots}}}}$
Using a calculator, we find:
$\sqrt{7} \approx 2.645751311$
|
## How to find gcf in fractions,make my ex girlfriend miss me like crazy,sweet quotes for your boyfriend on his birthday - And More
When we have two or more given numbers, we can find the largest factor that both numbers have in common. Step 1: We know that this is a GCF question because Maria is breaking the candy into groups and the problem using the key word 'greatest'. From the upside down division, we can see that each child would get 10 pieces of chocolate candy and 9 pieces of fruity candy. I think factor trees are one of the coolest “tools” for middle school students! This is the most obvious use of factor trees and it’s typically the first way students learn to use them. Factor trees make simplifying radicals so easy for my pre-algebra and algebra students! Students understand that finding the square root of a number means figuring out what number times itself equals the starting number, so in simplifying radicals they are simply pulling out one of each prime factor that is listed twice in the factor tree. For students who have a tough time coming up with the numbers, I have them make a factor tree for 126. Once they have it broken into its prime factors, they just need to break up the prime factors into 2 numbers every possible way until they find the ones with a sum of 23. Today’s lesson in my 7th grade math class was on finding the greatest common factor and least common multiple of a pair of numbers.
I started by gathering background information about what the students already knew about the GCF and LCM. Previously the students had learned to list all the factors of 2 numbers and find the largest one they had in common. Then we discussed the limitations of this method for finding the GCF and LCM, mainly the fact that if I gave them much larger numbers it would be very time-consuming to list all the factors or list multiples until they found one in common. I am looking forward to showing them how the cake method can be used to simplify fractions, too.
In this lesson we will study polynomials that can be factored using the Greatest Common Factor.
Make sure that you pay careful attention not only to the process used for factoring, but also to the make-up of the polynomials that can be factored using this method. When you factor a polynomial, you are trying to find the quantities that you multiply together in order to create the polynomial. The greatest common factor (GCF)for a polynomial is the largest monomial that is a factor of (divides) each term of the polynomial.
Before we get started, it may be helpful for you to review the Dividing Monomials lesson.
Hopefully you now understand how to factor polynomials if the polynomials have a greatest common factor.
This lesson describes the method to find the factors of a trinomial, which consists of three terms, by grouping.
In this tutorial the instructor shows how to factor a trinomial with Greatest Common Factor (GCF) and then how to group it. Because both of the numbers are even, we could start with 2.The answer goes underneath the bar.
Remember that relatively prime means two numbers that do not have any common factors other than 1. Not only is the GCF extremely important when working with fractions, but is can also be used when applying the distributive property or when solving word problems.
You can find it by listing the factors, using upside down division or by using the prime factorization.
First of all, factor out the greatest common factor (GCF), and write the reduced trinomial in parentheses.
When we see key words like largest, greatest and most combined with questions about breaking things into groups we know that we are solving a GCF question. For example, 3(3X2+2X-8) trinomial is written in the order of variable, with 3(GCF) factored out. Now identify the coefficient of the first and last terms, for example in this case, it is 3 and 8. Now choose a pair of factors whose product is equivalent to the product of first and last terms of the trinomial, and the sum is equal to the middle term, both in terms of value and sign. Rewrite the trinomial by breaking the middle term in terms of the two chosen factors, that is, 3X2 + 6x-4x _8. Now pull out the common factors for each pair ensuring that in both the pairs, we have same monomial in the parentheses.
Hence, we can find the factors of a trinomial by following the four steps viz., find the GCF, find the factors, rewrite the statement and factor by grouping.
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# 1.5 Factoring polynomials (Page 2/6)
Page 2 / 6
Factor $\text{\hspace{0.17em}}x\left({b}^{2}-a\right)+6\left({b}^{2}-a\right)\text{\hspace{0.17em}}$ by pulling out the GCF.
$\left({b}^{2}-a\right)\left(x+6\right)$
## Factoring a trinomial with leading coefficient 1
Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial $\text{\hspace{0.17em}}{x}^{2}+5x+6\text{\hspace{0.17em}}$ has a GCF of 1, but it can be written as the product of the factors $\text{\hspace{0.17em}}\left(x+2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(x+3\right).$
Trinomials of the form $\text{\hspace{0.17em}}{x}^{2}+bx+c\text{\hspace{0.17em}}$ can be factored by finding two numbers with a product of $c\text{\hspace{0.17em}}$ and a sum of $\text{\hspace{0.17em}}b.\text{\hspace{0.17em}}$ The trinomial $\text{\hspace{0.17em}}{x}^{2}+10x+16,$ for example, can be factored using the numbers $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}8\text{\hspace{0.17em}}$ because the product of those numbers is $\text{\hspace{0.17em}}16\text{\hspace{0.17em}}$ and their sum is $\text{\hspace{0.17em}}10.\text{\hspace{0.17em}}$ The trinomial can be rewritten as the product of $\text{\hspace{0.17em}}\left(x+2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(x+8\right).$
## Factoring a trinomial with leading coefficient 1
A trinomial of the form $\text{\hspace{0.17em}}{x}^{2}+bx+c\text{\hspace{0.17em}}$ can be written in factored form as $\text{\hspace{0.17em}}\left(x+p\right)\left(x+q\right)\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}pq=c\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}p+q=b.$
Can every trinomial be factored as a product of binomials?
No. Some polynomials cannot be factored. These polynomials are said to be prime.
Given a trinomial in the form $\text{\hspace{0.17em}}{x}^{2}+bx+c,$ factor it.
1. List factors of $\text{\hspace{0.17em}}c.$
2. Find $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}q,$ a pair of factors of $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ with a sum of $\text{\hspace{0.17em}}b.$
3. Write the factored expression $\text{\hspace{0.17em}}\left(x+p\right)\left(x+q\right).$
## Factoring a trinomial with leading coefficient 1
Factor $\text{\hspace{0.17em}}{x}^{2}+2x-15.$
We have a trinomial with leading coefficient $\text{\hspace{0.17em}}1,b=2,$ and $\text{\hspace{0.17em}}c=-15.\text{\hspace{0.17em}}$ We need to find two numbers with a product of $\text{\hspace{0.17em}}-15\text{\hspace{0.17em}}$ and a sum of $\text{\hspace{0.17em}}2.\text{\hspace{0.17em}}$ In [link] , we list factors until we find a pair with the desired sum.
Factors of $\text{\hspace{0.17em}}-15$ Sum of Factors
$1,-15$ $-14$
$-1,15$ 14
$3,-5$ $-2$
$-3,5$ 2
Now that we have identified $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}q\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}5,$ write the factored form as $\text{\hspace{0.17em}}\left(x-3\right)\left(x+5\right).$
Does the order of the factors matter?
No. Multiplication is commutative, so the order of the factors does not matter.
Factor $\text{\hspace{0.17em}}{x}^{2}-7x+6.$
$\left(x-6\right)\left(x-1\right)$
## Factoring by grouping
Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial $\text{\hspace{0.17em}}2{x}^{2}+5x+3\text{\hspace{0.17em}}$ can be rewritten as $\text{\hspace{0.17em}}\left(2x+3\right)\left(x+1\right)\text{\hspace{0.17em}}$ using this process. We begin by rewriting the original expression as $\text{\hspace{0.17em}}2{x}^{2}+2x+3x+3\text{\hspace{0.17em}}$ and then factor each portion of the expression to obtain $\text{\hspace{0.17em}}2x\left(x+1\right)+3\left(x+1\right).\text{\hspace{0.17em}}$ We then pull out the GCF of $\text{\hspace{0.17em}}\left(x+1\right)\text{\hspace{0.17em}}$ to find the factored expression.
## Factor by grouping
To factor a trinomial in the form $\text{\hspace{0.17em}}a{x}^{2}+bx+c\text{\hspace{0.17em}}$ by grouping, we find two numbers with a product of $\text{\hspace{0.17em}}ac\text{\hspace{0.17em}}$ and a sum of $\text{\hspace{0.17em}}b.\text{\hspace{0.17em}}$ We use these numbers to divide the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.
Given a trinomial in the form $\text{\hspace{0.17em}}a{x}^{2}+bx+c,$ factor by grouping.
1. List factors of $\text{\hspace{0.17em}}ac.$
2. Find $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}q,$ a pair of factors of $\text{\hspace{0.17em}}ac\text{\hspace{0.17em}}$ with a sum of $\text{\hspace{0.17em}}b.$
3. Rewrite the original expression as $\text{\hspace{0.17em}}a{x}^{2}+px+qx+c.$
4. Pull out the GCF of $\text{\hspace{0.17em}}a{x}^{2}+px.$
5. Pull out the GCF of $\text{\hspace{0.17em}}qx+c.$
6. Factor out the GCF of the expression.
## Factoring a trinomial by grouping
Factor $\text{\hspace{0.17em}}5{x}^{2}+7x-6\text{\hspace{0.17em}}$ by grouping.
We have a trinomial with $\text{\hspace{0.17em}}a=5,b=7,$ and $\text{\hspace{0.17em}}c=-6.\text{\hspace{0.17em}}$ First, determine $\text{\hspace{0.17em}}ac=-30.\text{\hspace{0.17em}}$ We need to find two numbers with a product of $\text{\hspace{0.17em}}-30\text{\hspace{0.17em}}$ and a sum of $\text{\hspace{0.17em}}7.\text{\hspace{0.17em}}$ In [link] , we list factors until we find a pair with the desired sum.
Factors of $\text{\hspace{0.17em}}-30$ Sum of Factors
$1,-30$ $-29$
$-1,30$ 29
$2,-15$ $-13$
$-2,15$ 13
$3,-10$ $-7$
$-3,10$ 7
So $\text{\hspace{0.17em}}p=-3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}q=10.$
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# Test: Calendar 4
## 20 Questions MCQ Test UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making | Test: Calendar 4
Description
This mock test of Test: Calendar 4 for Quant helps you for every Quant entrance exam. This contains 20 Multiple Choice Questions for Quant Test: Calendar 4 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Calendar 4 quiz give you a good mix of easy questions and tough questions. Quant students definitely take this Test: Calendar 4 exercise for a better result in the exam. You can find other Test: Calendar 4 extra questions, long questions & short questions for Quant on EduRev as well by searching above.
QUESTION: 1
### What was the day of the week on 16th August, 1947?
Solution:
15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th )
Counting of odd days:
1600 years have 0 odd day. 300 years have 1 odd day.
47 years = (11 leap years + 36 ordinary years)
= [(11 x 2) + (36 x 1) ] odd days
= 58 odd days
= 2 odd days
Jan Feb Mar Apr May Jun Jul Aug
= 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15
= 227 days
= (32 weeks + 3 days)
= 3,
Total number of odd days
= (0 + 1 + 2 + 3) odd days
= 6 odd days.
Hence, the required day was 'Saturday'.
QUESTION: 2
### Prove that any date in March of a year is the same day of the week corresponding date in November that year.
Solution:
We will show that the number of odd days between last day of February and last day of October is zero.
March April May June July Aug. Sept. Oct.
31 + 30 + 31 + 30 + 31 + 31 + 30 + 31
= 241 days
= 35 weeks
= 0 odd day
Number of odd days during this period = 0.
Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows
QUESTION: 3
### If today is Saturday, what will be the day 350 days from now ?
Solution:
350 days = (350/7 = 50 weeks) i.e No odd days,
So it will be a Saturday.
QUESTION: 4
The calendar for the year 1988 is same as which upcoming year ?
Solution:
We already know that the calendar after a leap year repeats again after 28 years.
Here 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.
QUESTION: 5
Given that on 9th August 2016 is Saturday. What was the day on 9th August 1616 ?
Solution:
We know that, After every 400 years, the same day occurs.
Thus, if 9th August 2016 is Saturday, before 400 years
i.e., on 9th August 1616 has to be Saturday.
QUESTION: 6
Second & fourth Saturdays and every Sunday is a holiday. How many working days will be there in a month of 31 days beginning on a Friday ?
Solution:
Given that the month begins on a Friday and has 31 days
Sundays = 3rd, 10th, 17th, 24th, 31st
⇒ Total Sundays = 5
Every second & fourth Saturday is holiday. 2nd & 4th Saturday in every month = 2
Total days in the month = 31
Total working days = 31 - (5 + 2) = 24 days.
QUESTION: 7
On 17th March, 1997 Monday falls. What day of the week was it on 17th March, 1996?
Solution:
The year 1996 is a leap year. So, it has 2 odd days. But 17 th March comes after 29 th February. So, the day on 17 th March, 1997 will be 1 day beyond the day on 17 th March, 1996.
Here 17 th March, 1997 is Monday. So, 17 th March, 1996 is a Sunday.
QUESTION: 8
Which year has 366 days?
Solution:
When a century year leaves a remainder 0, when divided by 400 then it is a leap year (366 days).
So, 1200 has 366 days.
QUESTION: 9
What is 90 days from today?
(Hints : Today is 20th January 2017, Sunday)
Solution:
Given Today is 20th January 2017, Sunday
In january, we have 31 days
February - 28 days (Non leap year)
March - 31 days
April - 30 days
⇒ Remaining days = 31 - 20 = 11 in Jan
11 in Jan + 28 in Feb + 31 in Mar = 11 + 28 + 31 = 70 days
More 20 days to complete 90 days ⇒ upto 20th April
Therefore, after 90 days from today i.e, 20th Jan 2017 is 20th Apr 2017.
Now, the day of the week will be
90/7 ⇒ Remainder '6'
As the day starts with '0' on sunday
6 ⇒ Saturday.
Required day is 20th April, Saturday.
QUESTION: 10
On 24th Nov, 2007 Thursday falls. What day of the week was it on 10th Nov, 2006 ?
Solution:
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 24th Nov, 2007 will be 1 day beyond the day on 24
th Nov, 2006.
But, 24th Nov, 2007 is Thursday.
24 - 10 = 14 days.
Therefore, 2 weeks ago it is same day.
Thus, 10th Nov, 2006 is one day before 10th Nov, 2007 i.e. it is Wednesday.
QUESTION: 11
The year next to 2003 will have the same calendar as that of the year 2003?
Solution:
Given year 2003, when divided by 4 leaves a remainder of 3.
NOTE: When remainder is 3, 11 is added to the given year to get the result.
So, 2003 + 11 = 2014
QUESTION: 12
On 19th June, 1984 Monday falls. What day of the week was it on 19th June, 1985?
Solution:
The year 1985 is an ordinary year. So, it has 1 odd day.
So, the day on 19th June, 1985 will be 1 day after the day on 19th June, 1984.
But, 19th June,1984 is Monday
So, 19th June, 1985 is Tuesday.
QUESTION: 13
Suppose today is Friday, what day of the week will it be 65 days from now?
Solution:
The day of the week repeats every 7 days.
Given today is Friday. Again Friday is repeated on the 7th day, 14th,... on 7 multiple days.
Hence, Friday is on the 63rd day, as 63 is multiple of 7.
Now, the required day of the week on the 65th day is Sunday.
QUESTION: 14
How many seconds in 10 years?
Solution:
We know that,
1 year = 365 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds.
Then, 1 year = 365 x 24 x 60 x 60 seconds. = 8760 x 3600
1 year = 31536000 seconds.
Hence, 10 years = 31536000 x 10 = 315360000 seconds.
QUESTION: 15
The calendar of year 1939 is same as which year?
Solution:
Given year 1939, when divided by 4 leaves a remainder of 3.
NOTE: When remainder is 3, 11 is added to the given year to get the result.
So, 1939 + 11 = 1950
QUESTION: 16
What is two weeks from today?
Solution:
We know that the day repeats every 7 days, 14 days, 21 days, .........
So if today is Monday, after 7 days it is again Monday, after 14 days again it is Monday.
Hence, after 2 weeks i.e, 14 days the day repeats and is the same day.
QUESTION: 17
What will be your age in the year 2019 if you were born in 1995?
Solution:
Calculating Age has 2 conditions. Let your Birthday is on January 1st.
1. If the month in which you are born is completed in the present year i.e, your birthday, then
Your Age = Present year - Year you are born
As of now, present year = 2018
i.e, Age = 2019 - 1995 = 24 years.
2. If the month in which you are born is not completed in the present year i.e, your birthday, then
Your Age = Last year - Year you are born
As of now, present year = 2019
i.e, Age = 2018 - 1995 = 23 years.
QUESTION: 18
What was the day on 2nd Jan 1901?
Solution:
2nd Jan 1901 means
(1900 years and 2 day)
Now, 1600 years have 0 odd day
300 years have 1 odd day
2 days has 2 odd day
Total no. of odd days = 0 + 1 + 2 = 3 days
Hence, the day on 2nd Jan 1901 was Wednesday.
QUESTION: 19
What was the day on 16th June,1993?
Solution:
16 June, 1993 = (1992 years + Period from 1.1.1993 to 16.6.1993)
Odd days in 1600 years = 0
Odd days in 300 years = 1
92 years = (69 ordinary years + 23 leap year) = (69 x 1 + 23 x 2)= 3 odd days
Jan. Feb. March April May June
(31 + 28 + 31 + 30 + 31 + 16 ) = 167 days
167 days = (23 weeks + 6 days) =6 odd days.
Total number of odd days = (0 + 1 + 3 + 6) = 3 odd days.
Given day is Wednesday.
QUESTION: 20
Which century year is a leap year?
Solution:
For an century year,it is divided by 400,when leaves a remainder 0 then it is a leap year. So, 1600
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# The Distributive Property
## Presentation on theme: "The Distributive Property"— Presentation transcript:
The Distributive Property
Of Multiplication By: Ms. Nunez
Standards 3.OA.5 SWBAT identify distributive property.
SWBAT define distributive property. SWBAT solve distributive property problems.
What is the Distributive Property?
It’s when multiplying a number is the same as multiplying its addends by the number and then adding up the products! For example: 6(4+5) is the same as 6x4 + 6x5 it all equals 54.
Let me show you! But wait, Ms.! When will I ever need to use this? What does distributive mean? First answer: Say you don’t know 6 x 9 off the top of your head. You can break down 9 and distribute to find the answer. Second answer: Think about it. What does DISTRIBUTIVE mean? What do you do when you DISTRIBUTE something? Hint: “Ms. Nunez distributes the worksheets to the class.”
Define Distributive Great! When you distribute something, you hand it out. So the 6 in 6 (4 + 5) is being DISTRIBUTED to both the 4 and the 5, then all you do is add! Show your work! 6 (4 + 5) 6x4 + 6x5 = 54!
Breakin’ It Down So, as I said, you can use the Distributive Property to break down multiplication facts you may not remember off the top of your head. For example: I forgot what 8 x 8 was, but I know I can break down one of these numbers. So, I can make 8(5 + 3) and distribute the 8 to both 5 and 3. I can do this because…what does 5 +3 equal? 8!
Step it out! Step 1: Identify the number that you will be DISTRIBUTING. Step 2: DISTRIBUTE the number to both factors in the parenthesis. Step 3: Solve for both problems. Step 4: Add them together.
Let’s try together. Example: 5(3 + 2)
What happens first? Right, so 5 is the number I’m distributing. What two numbers am I distributing it to? Yes! The 3 and the 2, it should look like this: 5x3 + 5x2 Now what? Multiply & then Add! 5x3= 15, 5x2 = 10 = 25 So, 5(3 + 2) = 25
You try! 7(2 + 8) 4(3 + 6) 2(1 + 4) 6(5 + 2) 9(7 + 1)
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Logs
• Jan 15th 2008, 11:35 AM
johett
Logs
simplify using logarithm properties
a)4log(subscript25)15-4log(25)3
Express the following as a single logarithm
c)1/3(log(5)X+log(5)Y)-4log(5)Z
Simplify
d)log(2)20-log(2)5
thank you
• Jan 15th 2008, 12:09 PM
wingless
A summary of the basic properties:
1- $\displaystyle \text{log}_{c}a + \text{log}_{c}b = \text{log}_{c}ab$
2- $\displaystyle \text{log}_{c}a - \text{log}_{c}b = \text{log}_{c}(\frac{a}{b})$
3- $\displaystyle log_{a}{b} = \frac{\text{log}_{c}{b}}{\text{log}_{c}{a}}$ (c can be anything you want, don't forget that $\displaystyle c > 0$ and $\displaystyle c \neq 1$)
4- $\displaystyle a \text{log}_{b}c = \text{log}_{b}(c^a)$
5- $\displaystyle \frac{1}{a}\text{log}_{b}c = \text{log}_{(b^a)}c$
Last two properties combined:
6- $\displaystyle \frac{c}{d}\text{log}_{a}{b} = \text{log}_{(a^d)}{(b^c)}$
7-Let's say that $\displaystyle x = \text{log}_{a}{b}$
If I multiply it by $\displaystyle \frac{c}{c}$
$\displaystyle x = \text{log}_{a}{b} = \frac{c}{c}\text{log}_{a}{b}$
From property 6,
$\displaystyle x = \text{log}_{(a^c)}{(b^c)}$
Which means,
$\displaystyle \text{log}_{a}{b} = \text{log}_{(a^2)}{(b^2)} = \text{log}_{(a^3)}{(b^3)} = \text{log}_{(a^4)}{(b^4)} = \text{log}_{(a^{1/2})}{(b^{1/2})}....$
---------------------------
---------------------------
A)
$\displaystyle 4\text{log}_{25}{15} - 4\text{log}_{25}{3}$
$\displaystyle 4(\text{log}_{25}{15} - \text{log}_{25}{3})$
$\displaystyle 4\text{log}_{25}{\frac{15}{3}}$ ...(From property 2)
$\displaystyle 4\text{log}_{25}{5}$
$\displaystyle 4\frac{1}{2} = \boxed{2}$
---------------------------
B)
$\displaystyle \frac{1}{3}(\text{log}_{5}{x} + \text{log}_{5}{y}) - 4 \text{log}_{5}{z}$
$\displaystyle \text{log}_{(5^3)}{x} + \text{log}_{(5^3)}{y} - 4 \text{log}_{5}{z}$
$\displaystyle \text{log}_{(125)}{x} + \text{log}_{(125)}{y} - 4 \text{log}_{125}{z^3}$
$\displaystyle \text{log}_{(125)}{xy} - 4 \text{log}_{125}{z^3}$
$\displaystyle \text{log}_{(125)}{xy} - \text{log}_{125}{z^{12}}$
$\displaystyle \text{log}_{(125)}{\frac{xy}{z^{12}}}$
---------------------------
C)
$\displaystyle \text{log}_{2}{20} - \text{log}_{2}{5}$
$\displaystyle \text{log}_{2}{4}$
$\displaystyle \boxed{2}$
• Jan 15th 2008, 12:28 PM
johett
how do you get the $\displaystyle 4\frac{1}{2}$ =4 answer?
• Jan 15th 2008, 12:37 PM
wingless
Quote:
Originally Posted by johett
how do you get the $\displaystyle 4\frac{1}{2}$ =4 answer?
Typo :p
|
Question 1: A person invests Rs. 10000 for 3 years at a certain rate of interest compounded annually. At the end of one year this sum amounts to Rs. 11200. Calculate:
1. the rate of interest
2. the amount at the end of second year
3. the amount at the end of third year [2006]
The rate of interest:
$\displaystyle P=10000 \text{ Rs.; } A= 11200 \text{ Rs.; } r=x\%$
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 11200=10000 \Big(1+ \frac{x}{100} \Big)^1 \Rightarrow x=12\%$
Amount at the end of the second year
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = 10000 \Big(1+ \frac{12}{100} \Big)^2 = 12544 \text{ Rs. }$
Amount at the end of third year
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = 10000 \Big(1+ \frac{12}{100} \Big)^3 = 14049.28 \text{ Rs. }$
$\displaystyle \\$
Question 2: A sum of Rs. 9600 is invested for 3 years at 10% per annum at compound interest.
1. What is the sum due at the end of the first year?
2. What is the sum due at the end of the second year?
3. Find the difference between the answers of the 2) and 1) and find the interest on this sum (difference) for one year
4. Hence, write down the compound interest for the third year [1996]
Sum due at the end of the first year?
$\displaystyle P=9600 \text{ Rs.; } A= ? \text{ Rs.; } r=10\%; n=1$
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = 9600 \Big(1+ \frac{10}{100} \Big)^1 = 10560 \text{ Rs. }$
Sum due at the end of the second year?
$\displaystyle P=9600 \text{ Rs.; } A= ? \text{ Rs.; } r=10\%; n=2$
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = 9600 \Big(1+ \frac{10}{100} \Big)^2 = 11616 \text{ Rs. }$
Difference between the answers of the 2) and 1)
$\displaystyle 11616-10560=1056 \text{ Rs. }$
$\displaystyle Interest = 1056 \times \frac{10}{100} \times 1 = 105.6 \text{ Rs. }$
Compound Interest for third year
$\displaystyle =1056+105.6 =1161.60 \text{ Rs. }$
$\displaystyle \\$
Question 3: What sum of money will amount to Rs. 9261 in 3 years at 5% per annum compound interest? [2009]
Given
$\displaystyle P=x \text{ Rs.; } A= 9261 \text{ Rs.; } r=5\%; n=3$
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 9261 = x \Big(1+ \frac{5}{100} \Big)^3 \Rightarrow x = 8000 \text{ Rs. }$
$\displaystyle \\$
Question 4: In what period of time will Rs. 12000 yield Rs. 3972 as compound interest at 10%, if compounded on a yearly basis. [2011]
Given $\displaystyle P=12000 \text{ Rs.; } A= (12000+3972)= 15972 \text{ Rs.; } r=10\%; n=n$
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 15972 = 12000 \Big(1+ \frac{10}{100} \Big)^n \Rightarrow n = 3 \text{ years }$
$\displaystyle \\$
Question 5: On what sum of money will the difference between the compound interest and the simple interest for two years be equal to Rs. 25, if the rate of interest charged for both is 5% p.a.? [2012]
Let the sum be $\displaystyle x \text{ Rs. }$
Simple Interest for 2 years $\displaystyle = x \times \frac{5}{100} \times 2 = 0.1x$
Amount Compound Interest
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = x \Big(1+ \frac{5}{100} \Big)^2 \Rightarrow A = 1.1025x \text{ Rs. }$
Given difference = 25 Rs.
Therefore $\displaystyle C.I. - S.I. = 25 \Rightarrow (1.1025x-x)-(0.1x) = 25 \Rightarrow x = 10000 \text{ Rs. }$
$\displaystyle \\$
Question 6: The simple interest on a sum of money for 2 years at 4% p.a. is Rs. 340. Find
1. the sum of the money
2. the compound interest on this sum for one year payable half-yearly at the same rate [2008]
Simple Interest
Given: $\displaystyle I = 340 \text{ Rs.; } n=2 \text{ years; } r=4\%; P=x \text{ Rs.; }$
$\displaystyle 340 = x \times \frac{4}{100} \times 2 \Rightarrow x = 4250 \text{ Rs. }$
Compound Interest
$\displaystyle P= 4250 \text{ Rs.; } n=1 \text{ years; } r=4\% \text{ compounded half yearly;}$
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = 4250 \Big(1+ \frac{4}{2 \times 100} \Big)^2 \Rightarrow A = 4421.7 \text{ Rs. }$
$\displaystyle C.I. = 4421.7-4250 = 171.70 \text{ Rs. }$
$\displaystyle \\$
Question 7: Simple interest on a sum of money for 2 years at 4% is Rs. 450. Find compound interest on the same sum and at the same rate for 1 year, if the interest on the same sum and at the same rate for year, if the interest is reckoned half-yearly. [1997]
Simple Interest
$\displaystyle P=x \text{ Rs.; } T=2 \text{years; } r=4\%$
$\displaystyle S.I=x \times \frac{4}{100} \times 2 = \frac{2}{25}x$
$\displaystyle 450 = \frac{2}{25}x \Rightarrow x= 5625 \text{ Rs. }$
Compounded Half Yearly
$\displaystyle P=5625 \text{ Rs.; } r=4\%; \text{ Compounded half yearly } n=1 \text{ year }$
$\displaystyle A=P \Big(1+ \frac{r}{2 \times 100} \Big)^{1 \times 2} = 5625 \Big(1+ \frac{4}{2 \times 100} \Big)^{1 \times 2} = 5852.25 \text{ Rs. }$
$\displaystyle \text{ Compound Interest }=5852.25-5625= 227.25 \text{ Rs. }$
$\displaystyle \\$
Question 8: Rohit borrows Rs. 86000 from Arun for 2 years at 5% per annum simple interest. He immediately lends his money to Akshay at 5% compounded interest annually for the same period. Calculate Rohits profit at the end of two years. [2010]
Simple Interest for 2 years
$\displaystyle S.I. = P \times \frac{r}{100} \times 2 \text{ Rs. }$
$\displaystyle S.I. = 86000 \times \frac{5}{100} \times 2 = 8600 \text{ Rs. }$
Compound Interest for 2 years
$\displaystyle P=8600 \text{ Rs.; } r=5\%; \text{ Compounded yearly } n=2 \text{ year }$
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^{2} \Rightarrow A= 86000 \Big(1+ \frac{5}{100} \Big)^{2} \Rightarrow A = 9481.50 \text{ Rs. }$
Gain $\displaystyle =(94815-86000)-8600 = 215 \text{ Rs. }$
$\displaystyle \\$
Question 9: Nikita invests Rs.6000 for two years at a certain rate of interest compounded annually, At the end of first year it amounts to Rs.6720. Calculate;
1. The rate of interest;
2. The amount at the end of the second year. [2010]
Compound Interest for 1 year
$\displaystyle P=6000 \text{ Rs.; } r=x\%; \text{ Compounded yearly } n=1 \text{ year }$
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^{1} \Rightarrow A= 6000 \Big(1+ \frac{x}{100} \Big)^{1}$
Given $\displaystyle 6000 \Big(1+ \frac{x}{100} \Big)^{1}=6720 \Rightarrow x= 12\%$
Amount at the end of second year
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^{1} \Rightarrow A= 6000 \Big(1+ \frac{12}{100} \Big)^{2} = 7526.40 \text{ Rs. }$
$\displaystyle \\$
Question 10: If the interest is compounded half-yearly, calculate the amount when principal is Rs.7400; the rate of interest is 5% per annum and the duration is one year. [2005]
$\displaystyle P=7400 \text{ Rs.; } r=5\%; \text{ Compounded half yearly } n=1 \text{ year }$
$\displaystyle A=P \Big(1+ \frac{r}{2 \times 100} \Big)^{n \times 2} = 12000 \Big(1+ \frac{5}{2 \times 100} \Big)^{1 \times 2} = 7774.63 \text{ Rs. }$
$\displaystyle \\$
Question 11: At what rate per cent will a sum of Rs.4000 yield Rs.1324 as compound interest in 3 years? [2013]
$\displaystyle P=4000 \text{ Rs.; } r=x\%; n=3 \text{ year; Interest } =1324 \text{ Rs. }$
$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 4000 \Big(1+ \frac{x}{100} \Big)^3.$
Given $\displaystyle \text{ Interest } = 1324 \text{ Rs. }$
$\displaystyle \Rightarrow 4000 \Big(1+ \frac{x}{100} \Big)^3 - 4000 = 1324 \Rightarrow x= 10\%$
$\displaystyle \\$
Question 12: The difference between compound interest for a year payable half-yearly and simple interest on a certain sum of money lent out at $\displaystyle 10\%$ for a year is $\displaystyle Rs.15$ . Find the sum of money lent out. [1998]
Simple Interest
$\displaystyle P=x \text{ Rs.; } T=1 \text{ Years; } r=10\%$
$\displaystyle S.I=x \times \frac{10}{100} \times 1 = 0.1x$
Compound Interest
$\displaystyle P=x; A= A; r=10\%; n=2 \text{ half } \text{ years }$
$\displaystyle A= x \times \Big(1+ \frac{10}{200} \Big)^2 = x \times \Big( \frac{21}{20} \Big)^2$
$\displaystyle C.I. = x \times \Big( \frac{21}{20} \Big)^2 - x$
Given $\displaystyle C.I. - S.I. = x \times \Big( \frac{21}{20} \Big)^2 - x - 0.1x = 15$
$\displaystyle \Rightarrow x= 6000 \text{ Rs. }$
$\displaystyle \\$
Question 13: The compound interest, calculated yearly, on a certain sum of money for the second year is $\displaystyle Rs. 1320$ and for the third year is $\displaystyle Rs. 1452$ . Calculate the rate of interest and the original sum of money. [2014]
Difference between the Compound interest of two successive years.
$\displaystyle = 1452-1320 = Rs. 132$
$\displaystyle \Rightarrow Rs. 132$ is the interest on $\displaystyle Rs. 1320$
$\displaystyle \therefore \text{ Rate of Interest } = \frac{100 \times I}{P \times T} \% = \frac{100 \times 132}{1320 \times 1} \% = 10\%$
Let the sum of money $\displaystyle = Rs. 100$
Therefore Interest on it for 1st Year $\displaystyle = 10\% of Rs. 100 = Rs. 10$
$\displaystyle \Rightarrow \text{ Amount in one year } = 100 + 10 = Rs. 110$
$\displaystyle \therefore \text{ Interest on it for } 2^{nd} \text{ Year } = 10\% of 110 = Rs. 11$
$\displaystyle \Rightarrow \text{ Amount in 2nd year } = 110 + 11 = Rs. 121$
When interest of 2nd year $\displaystyle = Rs. 11, sum is Rs. 100$
$\displaystyle \Rightarrow \text{ When interest of 2nd year } = Rs. 1320$ , then
$\displaystyle \text{ sum } = \frac{100}{11} \times 1320 = Rs. 12000$
$\displaystyle \\$
Question 14: Ramesh invests $\displaystyle Rs. 12800$ for three years at the rate of $\displaystyle 10\%$ per annum compound interest. Find;
1. The sum due to that person at the end of the first year.
2. The interest he earns for the second year.
3. The total amount due to him at the end of the third year. [2007]
For 1st year: $\displaystyle P = Rs. 12800; R=10\% and T=1 year$
$\displaystyle \text{ Therefore Interest }= \frac{12800 \times 10 \times 1}{100} = Rs. 1280$
$\displaystyle \text{ and, Amount }= 12800+1280 = Rs. 14080$
For 2nd year: $\displaystyle P = Rs. 14080; R=10\% and T=1 year$
$\displaystyle \text{ Therefore Interest } = \frac{14080 \times 10 \times 1}{100} = Rs. 1408$
$\displaystyle \text{ and, Amount } = 14080+1408 = Rs. 15488$
For 3rd year: $\displaystyle P = Rs. 15488; R=10\% and T=1 year$
$\displaystyle \text{ Therefore Interest } = \frac{15488 \times 10 \times 1}{100} = Rs. 1548.80$
$\displaystyle \text{ and, Amount } = 15488+1548.80 = Rs. 17036.80$
$\displaystyle \\$
Question 15: The compound interest, calculated yearly, on a certain sum of money for the second year is $\displaystyle Rs. 880$ and for the third year is $\displaystyle Rs. 968$ . Calculate the rate of interest and the sum of money. [1995]
Difference between the Compound interest of two successive years
$\displaystyle = 968-880 = Rs. 88$
$\displaystyle \Rightarrow Rs. 88$ is the interest on $\displaystyle Rs. 880$
$\displaystyle \therefore \text{ Rate of Interest } = \frac{100 \times I}{P \times T} \% = \frac{100 \times 88}{880 \times 1} \% = 10\%$
For 1st year: $\displaystyle P = Rs. x; R=10\% and T=1 year$
$\displaystyle \text{Therefore Interest } = \frac{x \times 10 \times 1}{100} = Rs. 0.1x$
$\displaystyle \text{ and, Amount } = x+ 0.1x = Rs. 1.1x$
For 2nd year: $\displaystyle P = Rs. 1.1x; R=10 \% and T=1 year$
$\displaystyle \text{ Therefore Interest } = \frac{1.1x \times 10 \times 1}{100} = Rs. 0.11x$
Given $\displaystyle 0.11x = 880 \Rightarrow x=Rs. 8000$
$\displaystyle \\$
Question 16: Mr. Kumar borrowed $\displaystyle Rs. 15000$ for two years. The rate of interest for the two successive years are $\displaystyle 8\% \text{ and } 10\%$ respectively. If the repays $\displaystyle Rs. 6200$ at the end of the first year, find the outstanding amount at the end of the second year. [2011]
Principal $\displaystyle = 15000 Rs$
The rate of interest for the two successive years are $\displaystyle 8\% \text{ and } 10\%$ respectively.
$\displaystyle \text{Formula: } A = P \Big(1+ \frac{r}{100} \Big)^n$
$\displaystyle \text{Therefore Amount after } 1^{st} \text{ year } = 15000 \times \Big(1+ \frac{8}{100} \Big)^1 = 16200 \text{ Rs. }$
$\displaystyle \text{Principal at the start of } 2^{nd}$ year after repayment $\displaystyle = 16200 - 6200 = 10000 \text{ Rs. }$
Amount outstanding at the end of second year
$\displaystyle = 10000 \times \Big(1+ \frac{10}{100} \Big)^1 = 11000 \text{ Rs. }$
$\displaystyle \\$
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# Roots of Complex Numbers
Recall that a consequence of the Fundamental Theorem of Algebra is that a polynomial equation of degree n has n solutions in the complex number system. Each solution is an nth root of the equation. The nth root of a complex number is defined as follows.
Definition of nth Root of a Complex Number
The complex number u = a + bi is an nth root of the complex number z if z = un = (a + bi)n.
STUDY TIP
The nth roots of a complex number are useful for solving some polynomial equations. For instance, you can use DeMoivre’s Theorem to solve the polynomial equation x4 + 16 = 0 by writing -16 as 16(cos π + i sin π).
To find a formula for an nth root of a complex number, let u be an nth root of z, where u = s(cos β + i sin β) and z = r(cos θ + i sin θ).
By DeMoivre’s Theorem and the fact that un = z you have
sn(cos nβ + i sin nβ) = r (cos θ + i sin θ).
Taking the absolute values of both sides of this equation, it follows that sn = r. Substituting back into the previous equation and dividing by r, you get
cos nβ + i sin nβ = cos θ + i sin θ
Thus, it follows that
cos nβ = cos θ and sin nβ = sin θ.
Because both sine and cosine have a period of 2π, these last two equations have solutions if and only if the angles differ by a multiple of 2π. Consequently, there must exist an integer k such that
By substituting this value for into the polar form of u, you get the following result.
NOTE
When k exceeds n - 1 the roots begin to repeat. For instance, if k = n the angle is coterminal with θ/n which is also obtained when k = 0.
Theorem
nth Roots of a Complex Number
For a positive integer n, the complex number z = r(cos θ + i sin θ) has exactly n distinct nth roots given by
where k = 0, 1, 2, ..., n - 1.
This formula for the nth roots of a complex number z has a nice geometrical interpretation, as shown in the figure below.
Note that because the nth roots of z all have the same magnitude they all lie on a circle of radius with center at the origin. Furthermore, because successive nth roots have arguments that differ by 2π/n, the n roots are equally spaced along the circle.
Example 1
Finding the nth Roots of a Complex Number
Find the three cube roots of z = -2 + 2i.
Solution
Because z lies in Quadrant II, the polar form for z is
By the formula for nth roots, the cube roots have the form
Finally, for k = 0, 1, and 2, you obtain the roots
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# Hexacontagon facts for kids
Kids Encyclopedia Facts
Regular hexacontagon
A regular hexacontagon
Type Regular polygon
Edges and vertices 60
Schläfli symbol {60}, t{30}, tt{15}
Coxeter diagram
Symmetry group Dihedral (D60), order 2×60
Internal angle (degrees) 174°
Dual polygon Self
Properties Convex, cyclic, equilateral, isogonal, isotoxal
A hexacontagon or 60-gon is a shape with 60 sides and 60 corners.
## Regular hexacontagon
A regular hexacontagon is represented by Schläfli symbol {60} and also can be constructed as a truncated triacontagon, t{30}, or a twice-truncated pentadecagon, tt{15}. A truncated hexacontagon, t{60}, is a 120-gon, {120}.
One interior angle in a regular hexacontagon is 174°, meaning that one exterior angle would be 6°.
### Area
The area of a regular hexacontagon is (with t = edge length)
$A = 15t^2 \cot \frac{\pi}{60}$
and its inradius is
$r = \frac{1}{2}t \cot \frac{\pi}{60}$
The circumradius of a regular hexacontagon is
$R = \frac{1}{2}t \csc \frac{\pi}{60}$
This means that the trigonometric functions of π/60 can be expressed in radicals.
### Constructible
Since 60 = 22 × 3 × 5, a regular hexacontagon is constructible using a compass and straightedge. As a truncated triacontagon, it can be constructed by an edge-bisection of a regular triacontagon.
## Dissection
60-gon with 1740 rhombs
Coxeter states that every zonogon (a 2m-gon whose opposite sides are parallel and of equal length) can be dissected into m(m-1)/2 parallelograms.
In particular this is true for regular polygons with evenly many sides, in which case the parallelograms are all rhombi. For the regular hexacontagon, m=30, and it can be divided into 435: 15 squares and 14 sets of 30 rhombs. This decomposition is based on a Petrie polygon projection of a 30-cube.
## Images for kids
Hexacontagon Facts for Kids. Kiddle Encyclopedia.
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##### Basic Math & Pre-Algebra All-in-One For Dummies (+ Chapter Quizzes Online)
The percent circle is a simple visual aid that helps you make sense of percent problems so that you can solve them easily. The three main types of percent problems are finding the ending number, finding the percentage, and finding the starting number.
The trick to using a percent circle is to write information into it. For example, the following figure shows how to record the information that 50% of 2 is 1.
Notice that as you fill in the percent circle, you change the percentage, 50%, to its decimal equivalent, 0.5.
Here are the two main features of the percent circle:
• When you multiply the two bottom numbers together, they equal the top number:
0.5 2 = 1
• If you make a fraction out of the top number and either bottom number, that fraction equals the other bottom number:
These features are the heart and soul of the percent circle. They enable you to solve any of the following three types of percent problems quickly and easily.
Most percent problems give you enough information to fill in two of the three sections of the percent circle. But no matter which two sections you fill in, you can find out the number in the third section.
## Problem type 1: Find the ending number from the percent and starting number
Suppose you want to find out the answer to this problem:
What is 75% of 20?
You’re given the percent and the starting number and asked to find the ending number. To use the percent circle on this problem, fill in the information as shown in the following figure.
Because 0.75 and 20 are both bottom numbers in the circle, multiply them to get the answer:
So 75% of 20 is 15.
As you can see, this method involves translating the word of as a multiplication sign. You still use multiplication to get your answer, but with the percent circle, you’re less likely to get confused.
## Problem type 2: Find the percentage from the starting and ending numbers
In the second type of problem, you start with both the starting and ending numbers, and you need to find the percentage. Here’s an example:
What percent of 50 is 35?
In this case, the starting number is 50 and the ending number is 35. Set up the problem on the percent circle as shown in the following figure.
This time, 35 is above 50, so make a fraction out of these two numbers:
This fraction is your answer, and all you have to do is convert the fraction to a percent. First, convert 35/50 to a decimal:
Now convert 0.7 to a percent:
0.7 = 70%
## Problem type 3: Find the starting number from the percentage and ending number
In the third type of problem, you get the percentage and the ending number, and you have to find the starting number. For example,
15% of what number is 18?
This time, the percentage is 15% and the ending number is 18, so fill in the percent circle as shown in the following figure.
Because 18 is above 0.15 in the circle, make a fraction out of these two numbers:
This fraction is your answer; you just need to convert to a decimal:
In this case, the “decimal” you find is the whole number 120, so 15% of 120 is 18.
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A (4, 2), B(6, 5) and C (1, 4) are the vertices of ΔABC.
Question:
A (4, 2), B(6, 5) and C (1, 4) are the vertices of ΔABC.
(i) The median from A meets BC in D. Find the coordinates of the point D.
(ii) Find the coordinates of point P and AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of the points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1
(iv) What do you observe?
Solution:
We have triangle $\triangle \mathrm{ABC}$ in which the co-ordinates of the vertices are $\mathrm{A}(4,2) ; \mathrm{B}(6,5)$ and $\mathrm{C}(1,4)$
(i)It is given that median from vertex A meets BC at D. So, D is the mid-point of side BC.
In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,
$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$
Therefore mid-point D of side BC can be written as,
$\mathrm{D}(x, y)=\left(\frac{6+1}{2}, \frac{5+4}{2}\right)$
Now equate the individual terms to get,
$x=\frac{7}{2}$
$y=\frac{9}{2}$
So co-ordinates of $D$ is $\left(\frac{7}{2}, \frac{9}{2}\right)$
(ii)We have to find the co-ordinates of a point P which divides AD in the ratio 2: 1 internally.
Now according to the section formula if any point $\mathrm{P}$ divides a line segment joining $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ in the ratio $\mathrm{m}$ : $\mathrm{n}$ internally than,
$\mathrm{P}(x, y)=\left(\frac{m x_{1}+m x_{2}}{m+n}, \frac{m y_{1}+m y_{2}}{m+n}\right)$
P divides AD in the ratio 2: 1. So,
$\mathrm{P}(x, y)=\left(\frac{2\left(\frac{7}{2}\right)+4(1)}{1+2}, \frac{2\left(\frac{9}{2}\right)+1(2)}{1+2}\right)$
$=\left(\frac{11}{3}, \frac{11}{3}\right)$
(iii)We need to find the mid-point of sides AB and AC. Let the mid-points be F and E for the sides AB and AC respectively.
Therefore mid-point F of side AB can be written as,
$\mathrm{F}(x, y)=\left(\frac{6+4}{2}, \frac{5+2}{2}\right)$
So co-ordinates of $\mathrm{F}$ is $\left(5, \frac{7}{2}\right)$
Similarly mid-point E of side AC can be written as,
$\mathrm{E}(x, y)=\left(\frac{1+4}{2}, \frac{4+2}{2}\right)$
So co-ordinates of $E$ is $\left(\frac{5}{2}, 3\right)$
Q divides BE in the ratio 2: 1. So,
$\mathrm{Q}(x, y)=\left(\frac{2\left(\frac{5}{2}\right)+6(1)}{1+2}, \frac{2(3)+1(5)}{1+2}\right)$
$=\left(\frac{11}{3}, \frac{11}{3}\right)$
Similarly, R divides CF in the ratio 2: 1. So,
$\mathrm{R}(x, y)=\left(\frac{2(5)+1(1)}{1+2}, \frac{2\left(\frac{7}{2}\right)+1(4)}{1+2}\right)$
$=\left(\frac{11}{3}, \frac{11}{3}\right)$
(iv)We observe that that the point P, Q and R coincides with the centroid. This also shows that centroid divides the median in the ratio 2: 1.
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# untitled
(Barré) #1
Solution : 5. 23 457 = 5. 23457457457 .........
So, 5. 23 457 u 100000 = 523457. 457457
and 5. 23 457 u 100 = 523. 457457
Subtracting, 5. 23 457 u 99900 = 522934
Therefore, 5. 234 57 =
49950
``````261467
99900``````
(^522934)
Required fraction is
49950
261467
Explanation : Here in the decimal part the recurring decimal has been multiplied first
by 100000 (5 zeros at the right side of 1) as there are two digits at the left side of
recurring part in the decimal portion, the recurring decimal has been multiplied by 100
(two zeros at the right side of 1). The seco nd product has been subtracted from the first
product. In one side of the result of subtraction is a whole number and at the other side
of the result is ( 100000 100 ) = 99900 times of the value of the given recurring
decimal. Dividing both the sides by 99900 , the required fraction is obtained.
Activity : Express 0. 4 1 and 3. 046 23 into fractions.
Rules of Transformation of Recurring Decimals into Simple Fractions
Numerator of the required fraction = the result by subtracting the number obtained from
exempting the decimal point of the given decimal point and the non-recurring part.
Denominator of the required fraction = Numbers formed by putting the number of 9
equal to the number of digits in the recurring part of the from the number of zeros equal
to the number of digits in the non-recurring part. Here the above rules are directly
applied to convert some recurring decimals into simple factions.
Example 10. Express 45. 23 46 into simple fraction.
Solution : 45. 23 46 =
4995
1172
45
4995
225947
9990
451894
9990
452346 452
Required fraction is
4995
1172
45
Example 11. Express 32. 5 67 into simple fraction.
Solution :
37
21
32
37
1205
111
3615
999
32535
999
32567 32
32. 567
Required fraction is
37
21
32.
Activity : Express 0. 01 2 and 3. 312 4 into fraction.
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Home probability
Probability We deal with probability every day, from the roll of a dice or a flip of a coin, probability allows us to determine the likely outcomes of an event. The probability that an event will occur is a number between 0 and 1, with 0 being impossible, and 1 being definite. If a coin is flipped it can either land on heads or tails. There are only two outcomes and assuming the coin is fair (i.e. its not weighted so that it lands on heads more often than tails) then we would expect, over a period of time, for the number of flips resulting in heads to be the same as the number of flips resulting in tails (you can try this - you may discover that the more number of trials you run, the nearer the results converge to 0.5 - this is due to the law of large numbers). We can therefore find the probability than a random flip of a coin would produce a heads as oppose to tails. There are 2 outcomes, therefore our denominator is 2. In one trial a heads can only be flipped once, ergo our numerator is one. Thus, written in notation, the probability that a head is flipped is a half: P(H) = 1/2 This can also be written as 0.5 or as 50%. The opposite outcome, is that a tail is flipped. The probability of this is also a half. Remember the total probability of an event occurring will total one. Henc, we can says that P(H') = 1/2. ' means the probability of an event not occurring. Example 1 A fair die is rolled, find the probability that it lands on 2. [P(X=2)] There are 6 possible outcomes, so our denominator is 6, we are asked to find the probability that when rolled the die will land of the face of a 2. On a die there is only 1 face with a 2 on it meaning our numerator is 1. Hence P(X=2) = 1/6 Example 2 A fair die is rolled, find the probability that it lands on a number greater than 3. [P(X>3)] The numbers on a die which are greater than 3, are 4, 5 and 6. Again our denominator is 6, but this time our numerator will be 3, because there are possible chances that our die will land on a number greater than 3: P(X>3) = 3/6 = 1/2 Try some more questions at the bottom of the page. If we are asked to find out multiple probabilities then we can use tree diagrams. Tree diagrams are used to show multiple events which are independent, this means that the outcome of one event has no bearing on the outcome of another event. For example, tossing a coin is an independent event, the first toss has no bearing on what I toss the second time. After drawing a tree diagram, we may be asked to find out the probability of 2 events occurring. For example, what is the probability that I roll a 2 and then a 3 on a die? If we draw the tree diagram for this we come up with: As you can see we have simplified the tree, by saying that on the first roll, we can either roll a 2, or we don't. We could draw the same tree diagram with 6 branches for the first roll, followed by 6 branches on the second roll for every branch of the first. This would give us a complete picture of what we could roll with a 6-sided die. Now we were asked for the probability that a 2 is rolled, followed by a 3, this can be written as: P(2AND3) or as P(2∩3) [more on this notation later]. The probability that we roll a 2 [P(2)] is 1/6 and separately the probability that we roll a 3 [P(3)] is also 1/6. To find the probability that we roll a 2 AND a 3, we multiply both the individual probabilities. So we do 1/6*1/6 to get 1/36. P(A and B) = P(A)*P(B) when the events are independent Example 1 I flip a coin twice, what is the probability that I land on a head followed by a tail: Step 1 - Draw the tree diagram: Step 2 - Write the question: P(H AND T) Step 3 - Solve: P(H) = 1/2; P(T) = 1/2; P(H AND T) = 1/2*1/2 => 1/4. Therefore, the answer is one quarter. If I had to find the probability of Event 1 occurring OR Event 2 occurring then I would add the probabilities. For example if P(Event 1)=4/10 and P(Even 2)=2/10; then the probability that E1 AND E2 would occur is 4/10*2/10 = 8/100 => 2/25. If asked to find the probability that E1 occurs OR E2 occurs we would do 4/10+/2/10 = 6/10 =>3/5. P(A or B) = P(A)+P(B) Example 2 I have a bag with 5 red counters, 4 blue counters and 1 green counter. I take out a counter and then place it back in the bag. a) What is the probability that I pick a red counter followed by a blue counter? b) What is the probability that I pick 2 red counters? c)What is the probability than I don't pick any red counters? d) What is the probability that I pick a red and blue counter? My tree diagram is going to have 3 branches for the first pick, and then 3 additional branches (for every first branch) for the second pick as shown below. The total number of counters in the bag which I can pick from is 10 (5+4+1) so this is our denominator. I can pick 5 red, so P(R) = 5/10, I can pick 4 blues so P(B)=4/10 and I can pick 1 green so P(G)=1/10. a) The probability that I first of all pick a red counter is 5/10, if we follow the R branch we can see that the probability of then picking a blue counter is 4/10. Hence the probability of picking a red followed by a blue is 5/10*4/10 = 20/100 => 1/5. b) The probability of picking 2 reds will simply be 5/10*5/10 = 25/100 => 1/4. c) If I don't pick any red counters then I could pick: (a Green AND a Green) OR (a Green AND a Blue) OR (a Blue AND a Blue) OR (a Blue AND a Green): (1/10*1/10)+(1/10*4/10)+(4/10*4/10) = 21/100. d) This question looks similar to part a, but it is different. In part a, we were constrained by the order, it hand to be Red followed by Blue. Now we are asked to find the probability that a red and blue counter are chosen in any order. Therefore we would do Red and Blue OR Blue and Red: (5/10*4/10)+(4/10*5/10) = 2/5. Venn Diagrams Venn diagrams can be created to show the possibilities of different events occurring, and their overlap. A basic 2 circle Venn diagram would look like: We would write the sample space (the total number of possibilities of an event occurring) on the outside of the Venn diagram. On the inside (outside the circles) we would write the probabilities that are not included in Events A and B. Therefore if all probabilities are calculated from inside the square box it should equal one. The total number of possible events accumulated from inside the square box (called the sample space) would equal the value for the sample space. (see Example 1). When using Venn Diagrams we use certain notation (which is also used for Rules): P(A ∩ B) means the probability that A and B both occur, it is called the intersection of A and B, and can be seen as the overlap between the 2 events IMAGE 1 P(A ∪ B) means the probability that either A or B, or both occur and is called A union B, it is shown as both circles of A and B IMAGE 2 P(A') means the probability of not A (hence A doesn't occur) and can be seen as everything but A (note: this includes the overlap of A and B, as it is still A) IMAGE 3 Using this information it is also possible to come up with other notation, for example P(A' ∩ B) means the probability of not A intersect B, and is hence just B with no overlap IMAGE 4 Example 1 There is a pack of playing cards containing 52 cards. A is the even that the card is the number 3, and B is the event that the card is a heart. Fill in the Venn diagram to show the events A and B and the sample space: IMAGE 5 Venn diagrams can be composed to show the possibility of up to 3 events occurring. When filling in a Venn diagram with 3 circles it is best to start with P(A ∩ B ∩ C) and then discount from there, as shown below: Example 2 There are 125 people in a bar, the findings below show who had crisps, who had beer and how had soft drinks: 15 ordered all 3 items 43 had crisps 40 had beer 44 had soft drinks 20 had beer and a soft drink 26 ordered crisps and soft drinks 25 had beer and crisps a.) Compose a Venn diagram for this, b.) find the probability that a randomly selected person had i.) all three items ii.) beer but not soft drink or crisps iii.) none of these items a.) IMAGE 6 Start by filling in 15, for all 3 items. Then we look at who had beer and a soft drink (20), because these people are included in the 15 people who ordered all 3 items we have to discount them. Therefore only 5 people had JUST beer and a soft drink, hence we can fill in this. We continue doing this (don't forget to discount) for JUST beer and crisps and JUST crisps and soft drink. Then we move out to those who JUST ordered beer (discounting 10, 15 and 5 from the 40 beer drinkers, because they are included in the other measures), JUST ordered crisps and those who JUST ordered soft drinks, until we have a complete picture. We know the total number of pub-goers was 125 people, but only 71 ordered beer, crisps and or soft drinks. Therefore 54 people didn't order anything. bi.) P(all 3 items) = 15/125 [the number who had all 3 divided by the total number]= 0.12 bii.) P(beer but not soft drink or crisps = JUST beer) = 10/125 = 0.08 biii.) P(none of the above) = 54/125 = 0.432 Rules We have already seen some of the rules below, but here they are formalised into formulae and expanded upon. P(A) + P(A') = 1 The probability that A occurs and the probability that A doesn't occur add up to one. P(A') = 1 - P(A) The probability that A doesn't occur is 1 minus the probability that A does occur. Addition Rule: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) This means that the probability of A union B (A OR B) is the probability of A add the probability of B take away the probability of A intersection B (which is A AND B). This rule applies to Venn diagrams, we have already seen a slight variation of this rule for Tree Diagrams: Addition Rule for Independent Events: P(A ∪ B) = P(A) + P(B) This formula can be rearranged to find the intersection if necessary: P(A ∩ B) = P(A) + P(B) - P(A ∪ B) Multiplication Rule: P(A ∩ B) = P(A|B) * P(B) Alternatively - P(B ∩ A) = P(B|A) * P(B) This rule states that the probability of A intersection B (A AND B) is equal to the probability of A given B multiplied by the probability of B. Conditional Probability: P(B|A) = [P(B ∩ A)]/P(A) Alternatively - P(A|B) = [P(A ∩ B)]/P(B) Find out more about conditional probabilities below. Now try some examples: Example 1 P(A) = 0.4, P(B) = 0.3, P(A ∪ B) = 0.6. Find a) P(A ∩ B) b) P(A') c) P(A' ∩ B') d) Draw these events out on a Venn Diagram a) P(A ∩ B) = P(A) + P(B) - P(A ∪ B) P(A ∩ B) = 0.4 + 0.3 - 0.6 P(A ∩ B) = 0.1 b) P(A') = 1 - P(A) P(A') = 1 - 0.4 P(A') = 0.6 c) P(A' ∩ B') - this means that event A doesn't occur AND event B doesn't occur. We know that P(A ∪ B) = 0.6; this means the probability of event A OR B occurring is 0.6. Therefore the probability that event A AND B DON'T occur, must be 1 - P(A ∪ B). This equals 0.4. d) Example 2 The probability that a child in a school has green eyes is 0.37, and the probability they have black hair is 0.45. The probability that the child has either green eyes or black hair or both is 0.40. A child is randomly selected from the school, what is the probability that the child has a) black hair and green eyes b) black hair but not green eyes c) neither black hair or green eyes? P(BH) = 0.45; P(GE) = 0.37; P(BH ∪ GE) = 0.8 a) P(BH AND GE) = P(BH ∩ GE) P(BH ∩ GE) = P(BH) + P(GE) - P(BH ∪ GE) P(BH ∩ GE) = 0.45 + 0.37 - 0.8 P(BH ∩ GE) = 0.02 b) P(BH AND NOT GE) = P(BH ∩ GE') If we visualise a Venn diagram, we want the BH circle without the intersection with GE circle. Therefore we want BH minus P(BH ∩ GE): P(BH ∩ GE') = P(BH) - P(BH ∩ GE) P(BH ∩ GE') = 0.45 - 0.02 P(BH ∩ GE') = 0.43 We can look at this another way, by rearranging formulae: P(BH ∩ GE') = P(BH) + P(GE') - P(BH ∪ GE') P(GE') = 1 - P(GE) P(GE') = 1 - 0.37 => 0.63 P(BH ∪ GE') - is the probability of black hair and not green eyes, which means we can add P(BH) with the sample space. The sample space would be 1 - P(BH ∪ GE) => 1 - 0.8 => 0.2. P(BH ∪ GE') = 0.65 P(BH ∩ GE') = 0.45 + 0.63 - 0.65 P(BH ∩ GE') = 0.43 This method is more convoluted. It is probably best to either picture, or draw the Venn diagram and solve from there. Mutually Exclusive and Independent When 2 events have no common outcomes they are said to be mutually exclusive. This means they can never happen at the same time. For example it is impossible to be dead and alive at the same time, hence they are mutually exclusive events. This can be shown on a Venn diagram as two separate circles: IMAGE 7 As we can see there is no intersection: P(A ∩ B) = 0. Hence the additional rule becomes P(A ∪ B) = P(A) + P(B) When an event has no bearing on the outcome of another event they are said to be independent. Therefore the probability of A occurring is the same, regardless of whether B has occurred. Thus the conditional rule (shown below) changes to P(A|B) = P(A). The multiplication rules becomes P(A ∩ B) = P(A)*P(B). Which is what we would expect from a tree diagram with independent events. Conditional Probability We have already seen that the formula for conditional probability is: Conditional Probability: P(B|A) = [P(B ∩ A)]/P(A) Alternatively - P(A|B) = [P(A ∩ B)]/P(B) We can use a Venn Diagram to explain why this is the case: We are finding P(B|A). The event A has already happened therefore we can shade in the A circle. The total probability is a, therefore our denominator will be the value of a. Our numerator is going to be the bit where B and A both occur, because we are looking for B given A. Hence this will be P(A ∩ B) which is i. So P(B|A) can be seen on our Venn diagram as i/a. If we formalise this we get P(B|A) = [P(B ∩ A)]/P(A). Example 1 P(A) = 0.3; P(B) = 0.5; P(A|B) = 0.4 Find a) P(A ∩ B) b) P(A ∪ B) c) P(B|A) a) P(A|B) = [P(A ∩ B)]/P(B) 0.4 = [P(A ∩ B)]/0.5 0.4*0.5 = P(A ∩ B) P(A ∩ B) = 0.2 b) P(A ∪ B) = P(A) + P(B) - P(A ∩ B) P(A ∪ B) = 0.3 + 0.5 - 0.2 P(A ∪ B) = 0.6 c) P(B|A) = [P(B ∩ A)]/P(A) P(B ∩ A) = P(A ∩ B) => 0.2 P(B|A) = 0.2/0.3 P(B|A) = 0.666... Example 2 A card is selected randomly from a deck of 52 cards. Find the probability that the card is an ace, given that the card is a diamond. We need to find P(ace|diamond) P(ace|diamond) = [P(diamond AND ace)]/P(diamond) P(diamond) = 13/52 - in the deck of 52 cards, we would expect a quarter of them to be diamond, therefore 13 out of 52. P(diamond AND ace) = 1/52 P(ace|diamond) = (1/52)/(13/52) P(ace|diamond) = 1/13 Questions 1. A coin is flipped 10 times. How many times should it land on heads? 2. The probability that an event occurs is 0.4, the event occurs 20 times, how many times should it occur? 3. A card is taken from a standard deck of playing cards (there are 52 cards in a pack), what is the probability that: a) a heart is picked b) the number 4 is selected c) the 4 of hearts is picked d)a diamond is not selected e) a king is not selected. 4. Two dice are thrown and the product of the numbers (both numbers multiplied together) is written down. Find the probability that the product of the dice is a.) exactly 6 and b.) more than 4; given that one die lands on 2. 5. 1. 5 (half) 2. 8 3. 4. Page last updated on 03/11/13
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## Definition
function is a relation for which each value from the set the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair.
### References
“Algebra – The Definition Of A Function “. 2022. tutorial.math.lamar.edu. https://tutorial.math.lamar.edu/classes/alg/functiondefn.aspx.
“What Is A Function? Definition, Types And Notation”. 2022. BYJUS. https://byjus.com/maths/what-is-a-function/.
## Vertical Line Test, Horizontal Line Test, One-to-One Function
### Vertical Line Test
If no two different points in a graph have the same first coordinate, this means that vertical lines cross the graph at most once. This is known as the vertical line test. Graphs that pass the vertical line test are graphs of functions.
### Horizontal Line Test
If no two different points in a graph have the same second coordinate, this means that horizontal lines cross the graph at most once. This is known as the horizontal line test. Functions whose graphs pass the horizontal line test are called one-to-one.
### One-to-One Function
Graphs that pass both the vertical line and horizontal line tests are one-to-one functions. These are exactly those functions whose inverse relation is also a function. One-to-one functions have an inverse.
### References
“Vertical and Horizontal Line Tests”. 2022. hardycalculus.com. http://hardycalculus.com/calcindex/IE_verticalline.htm.
## Inverse of Functions
A function and its inverse function can be described as the “DO” and the “UNDO” functions. A function takes a starting value, performs some operation on this value, and creates an output answer. The inverse function takes the output answer, performs some operation on it, and arrives back at the original function’s starting value. This “DO” and “UNDO” process can be stated as a composition of functions.
A function composed with its inverse function yields the original starting value. Think of them as “undoing” one another and leaving you right where you started.
If functions f and g are inverse functions, f(g(x)) = g(f(x)) = x.
Basically speaking, the process of finding an inverse is simply the swapping of the x and y coordinates. This newly formed inverse will be a relation, but may not necessarily be a function. The original function must be a one-to-one function to guarantee that its inverse will also be a function.
DEFINITION: An inverse relation is the set of ordered pairs obtained by interchanging the first and second elements of each pair in the original function. If the graph of a function contains a point (a, b), then the graph of the inverse relation of this function contains the point (b, a). Should the inverse relation of a function f(x) also be a function, this inverse function is denoted by f -1(x).
If we reflect this graph over the line y = x, the point (1, 0) reflects to (0, 1) and the point (4, 2) reflects to (2, 4). Sketching the inverse on the same axes as the original graph gives us the result in Figure 10.
Note: If the original function is a one-to-one function, the inverse will be a function.
### References
⭐ Roberts, Donna. 2022. “Inverse Of Functions – MathBitsNotebook (A2 – CCSS Math)”. mathbitsnotebook.com. https://mathbitsnotebook.com/Algebra2/Functions/FNInverseFunctions.html.
“Use the graph of a function to graph its inverse”. 2022. Course Hero. https://www.coursehero.com/study-guides/collegealgebra1/use-the-graph-of-a-function-to-graph-its-inverse/.
## Transformations of Function f (x)
Transformation of functions means that the curve representing the graph either “moves to left/right/up/down” or “it expands or compresses” or “it reflects”. Function transformations are very helpful in graphing the functions just by moving/expanding/compressing/reflecting the curve without actually needing to graph it from scratch.
### References
“Function Transformations”. 2022. mathsisfun.com. https://www.mathsisfun.com/sets/function-transformations.html.
“Functions Transformations – Graphing, Rules, Tricks”. 2022. CUE MATH. https://www.cuemath.com/calculus/transformation-of-functions/.
“Transformations Of Functions | College Algebra”. 2022. courses.lumenlearning.com. https://courses.lumenlearning.com/wmopen-collegealgebra/chapter/introduction-transformations-of-functions/.
## The Average Rate of Change of a Function
Given a function that models a certain phenomenon, it’s natural to ask such questions as “how is the function changing on a given interval” or “on which interval is the function changing more rapidly?” The concept of average rate of change enables us to make these questions more mathematically precise. Initially, we will focus on the average rate of change of an object moving along a straight-line path.
For a function s that tells the location of a moving object along a straight path at time t, we define the average rate of change of s on the interval [a, b] to be the quantity
Note particularly that the average rate of change of s on [a, b] is measuring the change in position divided by the change in time.
The value of the average rate of change tells us how much the function rises or falls, on average, for each additional unit we move to the right on the graph. For instance, if AV[3,7] = 0.75, this means that for an additional 1-unit increase in the value of x on the interval [3,7], the function increases, on average, by 0.75 units. In applied settings, the units of AV[3,7] are “units of output per unit of input”.
The value of the average rate of change is also the slope of the line that passes through the points (a, f(a)) and (b, f(b)) on the graph of f, as shown in the figure below.
### Example
1. Graph the equation -4x2 +3x – 4.
2. Using the given x values [-1, 3], calculate the corresponding y values, and then plot the lines.
(-1, -11), (3, -31)
3. Calculate the slope of the line using the calculated (x, y) pairs.
m = -5
4. Calculate the value of b using the slope-intercept form and one of the calculated (x, y) pairs.
y = mx + b
-11 = -5(-1) + b
b = -11 – (-5)(-1) = -11 – 5 = -16
Therefore, the equation of the line is y = –5x – 16
5. Now plot the line.
### References
⭐ “APC The Average Rate Of Change Of A Function”. 2022. activecalculus.org. https://activecalculus.org/prelude/sec-changing-aroc.html.
## Increasing, Decreasing and Constant Functions
In mathematics, as we know that a function is a relation between input and output. A function can be increasing, decreasing, or constant for the given intervals throughout their entire domain, and they are continuous and differentiable in the given interval. Now, what is an interval: so, an interval is known as a continuous or connected part or portion on the real line. This increasing or decreasing of function is generally used in the application of derivatives. So, if you want to find that the given function is increasing or decreasing in the given interval then you can easily find it with the help of derivatives.
### Increasing
When a function is increasing in the given interval, then such type of function is known as increasing function. Or in other words, when a function, f(x), is increasing, the values of f(x) are increasing as x increases. Or, let us considered I be an interval which presents in the domain of a real valued function f. Then the function f is increasing on I, if x1 < x2 in I ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ I. Or, in terms of derivative, a function is increasing when the derivative at that point is positive. The graphical representation of an increasing function is:
### Decreasing
When a function is decreasing in the given interval, then such type of function is known as decreasing function. Or in other words, when a function, f(x), is decreasing, the values of f(x) are decreasing as x increases. Or, let us considered I be an interval which presents in the domain of a real valued function f. Then
• The function f is decreasing on I, if x1, x2 in I ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ I.
• The function f is decreasing on I, if x1 < x2 in I ⇒ f (x1) ≥ f(x2)∀ x1, x2 ∈ I.
• The function f is strictly decreasing on I, if x1 < x2 in I ⇒ f(x1) > f(x2)∀ x1, x2 ∈ I.
Or, in terms of derivative, a function is decreasing when the derivative at that point is negative. The graphical representation of a decreasing function is:
### Constant
When a function is neither increasing nor decreasing in the given interval, then such type of function is known as constant function. Or in other words, when a function, f(x), is constant, the value of f(x) does not change as x increases. Or, let us considered I be an interval which presents in the domain of a real valued function f. Then the function f is constant on I, if f(x) = c ∀ x ∈ I. Here, c is a constant. Or, in terms of derivative, a function is constant (i.e. neither increasing nor decreasing) when the derivative is zero. The graphical representation of constant function is:
### References
“Increasing And Decreasing Functions – GeeksForGeeks”. 2021. GeeksForGeeks. https://www.geeksforgeeks.org/increasing-and-decreasing-functions/.
“3.3: Increasing And Decreasing Functions”. 2017. Mathematics Libretexts. https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(Apex)/03%3A_The_Graphical_Behavior_of_Functions/3.03%3A_Increasing_and_Decreasing_Functions.
“How to Find the Increasing or Decreasing Functions?” 2022. EffortlessMath. https://www.effortlessmath.com/math-topics/how-to-find-the-increasing-or-decreasing-functions/.
“Increasing And Decreasing Functions – Definition, Rules, Examples”. 2022. CUEMATH. https://www.cuemath.com/calculus/increasing-and-decreasing-functions/.
“Increasing And Decreasing Functions”. 2022. mathsisfun.com. https://www.mathsisfun.com/sets/functions-increasing.html.
“Increasing/Decreasing Functions”. 2022. CliffsNotes. https://www.cliffsnotes.com/study-guides/calculus/calculus/applications-of-the-derivative/increasing-decreasing-functions.
## Function: Continuous or Discontinuous (Infinite, Jump or Removeable)
In calculus, a function is continuous at x = a if and only if all three of the following conditions are met:
1. The function is defined at x = a; that is, f(a) equals a real number
2. The limit of the function as x approaches a exists
3. The limit of the function as x approaches a is equal to the function value at x = a
There are three basic types of discontinuities:
1. Removable (point) discontinuity – the graph has a hole at a single x-value. Imagine you’re walking down the road, and someone has removed a manhole cover (Careful! Don’t fall in!). This function will satisfy condition #2 (limit exists) but fail condition #3 (limit does not equal function value).
2. Infinite discontinuity – the function goes toward positive or negative infinity. Imagine a road getting closer and closer to a river with no bridge to the other side
3. Jump discontinuity – the graph jumps from one place to another. Imagine a superhero going for a walk: he reaches a dead end and, because he can, flies to another road.
Both infinite and jump discontinuities fail condition #2 (limit does not exist), but how they fail is different. Recall for a limit to exist, the left and right limits must exist (be finite) and be equal. Infinite discontinuities have infinite left and right limits. Jump discontinuities have finite left and right limits that are not equal.
### Examples
#### Example 1
Is f(x) continuous at x = 0?
To check for continuity at x = 0, we check the three conditions:
1. Is the function defined at x = 0? Yes, f(0) = 2
2. Does the limit of the function as x approaches 0 exist? Yes
3. Does the limit of the function as x approaches 0 equal the function value at x = 0? Yes
Since all three conditions are met, f(x) is continuous at x = 0.
#### Example 2
Is f(x) continuous at x = -4?
To check for continuity at x = -4, we check the same three conditions:
1. The function is defined; f(-4) = 2
2. The limit exists
3. The function value does not equal the limit; point discontinuity at x = 4
Since all three conditions are NOT met, f(x) is not continuous at x = 4.
### References
“Continuity in Calculus: Definition, Examples & Problems”. 2022. study.com. https://study.com/academy/lesson/continuity-in-calculus-definition-examples-problems.html.
“Calculus I – Continuity”. 2022. tutorial.math.lamar.edu. https://tutorial.math.lamar.edu/Classes/CalcI/Continuity.aspx.
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# 5 Math Tricks To Speed Up Calculations
Here we are going to discuss some of the Math tricks which will help you do calculations at the faster speed. These tricks are a kind of helping hand in your exam time, saving a lot of time and scoring more. Efficiency come with practice but these math tricks will increase your speed and help you do math calculations faster.
### Square and square root of a number ending with 6
Example: 762
Step 1. Put down 6
Step 2. Multiply 2 with (7 + 1) = 16 and add 16+1= 17. Put down 7 and carry 1
Step 3. Multiply 7 with (7 + 1) = 56 + carry 1 = 57 put down 57
### Multiplication of 11 with any number of 3 digits.
Let me explain this rule by taking examples
1. 352*11 = 3—(3+5)—(5+2)—2 = 3872
Means insert the sum of first and second digits, then sum of second and third digits between the two terminal digits of the number
213*11 = 2—(2+1)—(1+3)—3 = 2343
### PERCENTAGE COMPUTATION
Calculate 65% of 460:
Step 1: 50% of 460 = 230
Step 2: 10% of 460 = 46
Step 3: 5% of 460 = 23 (half 46)
Result: 65% of 460 = 299 (sum)
### Addition of common difference in a series of numbers
We need to add the larger number with smaller number, After that multiply this add result by count of numbers in the series of numbers, at last we have to do that we divide the result by 2.
39, 49, 59, 69 and 79
Step 1: Add the larger number with small number that is 79 + 39 = 118
Step 2: Then multiply the add result 118 × 5 = 590
Step 3: At last divide the number by 2, 590/2 = 295
### Five Digit cube and cube root trick
Need to remember 1 to 10 cube.
Example: 3√13824 =?
Step 1: Last digit of cube number from right side is 4 that we consider 64 = 43 we put down 4.
Then
Step 2: Take the number whose cube is nearest to 13.
That is 13 is nearest to 23 and 33, we take small one cube digit that is 2.
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# 1. Integers
Before we talk about integers, let’s think about the numbers we first learned as children.
## Natural Numbers
The natural numbers arise from counting. We can have
no objects: 0
one object: 1
two objects: 2
... etc.
We can represent the natural numbers on a one-dimensional number line. Here is a graph of the first 4 natural numbers 0, 1, 2, and 3:
We put a dot on those numbers that are included. In this case, we have graphed 0, 1, 2, and 3, but we have not included 4 to illustrate the point.
We can also write the natural numbers as a set:
{0, 1, 2, 3, 4, ...}
The natural numbers are sometimes called whole numbers or counting numbers. There is some disagreement among mathematicians about whether or not zero should be included. [See Is 0 a Natural Number?]
### Less Than and Greater Than
On the number line, LEFT is LESS. We use the "<" (less than) sign to indicate one number is smaller than another number.
So for example:
1 < 3 (We say this as "one is less than three".)
We use the ">" (greater than) sign to indicate one number is bigger than another.
4 > 0 (We say "four is greater than zero".)
## Negative Numbers
One of the biggest problems people have in mathematics is with negative numbers. Take great care with negatives!
A negative number is any number whose value is less than zero. We write a minus in front of negative numbers.
### Examples of Negative Numbers
(a) -3.97
(b) -pi
(c) -2
(d) -3/5
The negative numbers are represented on the left half of the number line. The four negative numbers above are indicated on the following number line like this:
Once again, we have put dots on the number line representing the numbers in the set.
### Uses of Negative Numbers
• Temperatures (below zero)
• Bank balances which are overdrawn
• Golf ("under par" means the number of shots less than par for the golf course)
• Engineering (forces acting in different directions, for example gravity is usually written as a negative force, since upwards forces are usually regarded as positive)
• Science
Now we're ready to dive in.
## Integers
Defined as:
• The negative natural numbers (...,−4, −3, −2, −1) and
• 0; and
• The positive natural numbers (1, 2, 3, 4, ...).
Notice that the set does not include decimals or fractions − just whole numbers.
They are represented on the number line like this:
Recall from above that on the number line, LEFT is LESS.
### Examples of Comparing Integers
(a) −4 < 2 (we say "negative 4 is less than 2")
(b) −103 < −45
Similarly, if a number is to the RIGHT of second number, it is greater than the second number.
(c) 2 > −2 (we say "2 is greater than negative 2")
(d) 3 > −150
## Absolute Value
The distance from 0 to a number is its absolute value (written using vertical line brackets around the number).
### Examples of Absolute Value
(a) |-4| = 4
The distance from -4 to 0 is 4 units:
Similarly:
(b) |3|=3
(c) |-12.85| = 12.85
In each example, the number on the right is the distance from 0.
## Opposite of an Integer
The opposite of an integer is obtained by changing its sign. (That is, change − to + or change + to −).
### Examples of Opposite of an Integer
(a) The opposite of -3 is 3 and
(b) The opposite of 4 is -4.
Notice that opposite is not the same as absolute value.
Use the number line and think of "journeys".
(a) -2 + 5 means "start at -2 and go 5 in the positive direction"
So we have:
-2 + 5 = 3
Similarly
(b) 3 + -7 means "start at 3 and go 7 in the negative direction".
Our answer is: 3 + -7 = -4
(c) -5 + 12 means "start at -5 and go 12 in the positive direction"
The answer is -5 + 12 = 7
### Practical Example of Integer Addition
It is -4° and snowing. The forecast for tomorrow is for a rise in temperature of 6°. What will it be tomorrow?
Answer: -4 + 6 = 2. It will be 2° tomorrow.
## Integer Subtraction
We can change the subtraction into a more familiar addition by realising that subtracting an integer is the same as adding its opposite.
### Examples of Integer Subtraction
(a) -4 - (-3) = -4 + (+3) = -1
(We added +3 because the opposite of -3 is +3.)
(b) 5 - (+7) = 5 + (-7) = -2.
(We added -7 because the opposite of +7 is -7. Eventually you’ll see the question is the same as "5 - 7" and we can do this as a journey: Start at 5 and move 7 units to the left. Answer: -2.)
## Integer Multiplication
When multiplying integers, we can think of multiplying "blocks" of negative numbers.
### Examples of Integer Multiplication
(a) Here is an integer times an integer:
5 xx -3 = -15
Why is this so? We are simply taking "5 lots of −3", like this:
Notice that we were multiplying a positive number by a negative number and our result was negative.
Similarly, we can show:
(b) -6 xx 2 = -12 (negative times positive gives negative)
and
(c) -3 xx -7 = 21 (multiplying 2 negatives gives a positive)
### Practical example of Integer Multiplication
The acceleration due to gravity (1 g) on the earth is about −10 ms−2. (It’s negative because it acts downwards. This is a convention.)
On the planet Jupiter, the acceleration due to gravity is 2.65 times the acceleration due to gravity on the Earth.
On Jupiter, the acceleration is around:
2.65 xx -10 "ms"^-2 ~~ -26.5\ "ms"^-2
## Summary
We can summarise the results for multiplying:
positive × positive = positive
+ × + = +
Example: 2 × 3 = 6
negative × positive = negative
− × + = −
Example: −8 × 2 = −16
positive × negative = negative
+ × − = −
Example: 5 × −2 = −10
negative × negative = positive
− × − = +
Example: −5 × −3 = 15
### Challenge: Integer Multiplication
What is a practical application for −3 × −7 = 21?
If it is 0° now and the temperature is dropping 3° per hour, what was the temperature 7 hours ago?
## Integer Division
When we divide with negative numbers, we have similar results to those for multiplication:
positive ÷ positive = positive
+ ÷ + = +
Example: 15 ÷ 3 = 5
negative ÷ positive = negative
− ÷ + = −
Example: −8 ÷ 2 = −4
positive ÷ negative = negative
+ ÷ − = −
Example: 21 ÷ −7 = −3
negative ÷ negative = positive
− ÷ − = +
Example: −50 ÷ −5 = 10
We can rewrite division problems as multiplication problems, as in the following example.
### Example of Integer Division
−32 ÷ 4
is the same question as
4 × what? = −32
### Challenge: Integer Division
What is a practical application for −10 ÷ 5 = −2?
1. In financial accounting, a loss of $10 million could be written as -$10 million.
If we make a loss of $10 million in 5 years, then the average rate of loss is −10 ÷ 5 = −2, or −$2 million per year.
2. (Similar, but for the sporty among you:) In golf, "1 under par" means that you get the ball in the hole in one stroke less than expected for that hole.
So your score may be 10 under par after 5 holes of golf. Your average per hole is
−10 ÷ 5 = −2,
or 2 under par per hole.
## Integer Properties
The set of integers is closed, commutative, associative and has an identity under both addition and multiplication.
The following table gives examples and explains what this means in plain English.
Closed
3 + −7 = −4
When we add 2 integers, we get an integer.
−5 × −3 = 15
When we multiply 2 integers, we get an integer.
Commutative
4 + −5 = −5 + 4
It doesn’t matter what order we add integers, we get the same answer.
2 × −5 = −5 × 2
It doesn’t matter what order we multiply integers, we get the same answer.
Associative
(4 + −2) + −5 = 4 + (−2 + −5)
When adding 3 integers, it doesn’t matter if we start by adding the first pair or the last pair; the answer is the same.
(4 × −2) × −5 = 4 × (−2 × −5)
When multiplying 3 integers, it doesn’t matter if we start by multiplying the first pair or the last pair; the answer is the same.
Identity
−5 + 0 = 0 + −5 = −5
Zero is the identity element for addition. By adding zero on either side, we don’t change the number.
−3 × 1 = 1 × −3 = −3
One is the identity element for multiplication. By multiplying by 1 on either side, we don’t change the number.
The Distributive Law over addition and subtraction holds for integers:
Distributive
3(2 + −4)
= 3 × 2 + 3 × (−4)
We multiply each number inside the brackets by the number outside, retaining the plus in the middle.
−2(5 − 7)
= (−2 × 5) − (−2 × 7)
We multiply each number inside the brackets by the number outside, retaining the minus in the middle.
## Game
### Magic Square
In a magic square, all the rows, all the columns and the 2 diagonals must add to the same number.
1. Complete the magic square, using only the positive integers 1 to 9:
1 4 9 2
8 1 6 3 5 7 4 9 2
2. Complete the magic square, using only the integers:
−10, −8, −6, −4, 0, 2, 4, 6
−2
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# SAT Math : How to find the volume of a sphere
## Example Questions
← Previous 1
### Example Question #1 : How To Find The Volume Of A Sphere
A cube with sides of 4” each contains a floating sphere with a radius of 1”. What is the volume of the space outside of the sphere, within the cube?
11.813 in3
4.187 in3
11.813 in3
59.813 in3
64 in3
59.813 in3
Explanation:
Volume of Cube = side3 = (4”)3 = 64 in3
Volume of Sphere = (4/3) * π * r3 = (4/3) * π * 13 = (4/3) * π * 13 = (4/3) * π = 4.187 in3
Difference = Volume of Cube – Volume of Sphere = 64 – 4.187 = 59.813 in3
### Example Question #1 : How To Find The Volume Of A Sphere
If a sphere's diameter is doubled, by what factor is its volume increased?
2
16
3
4
8
8
Explanation:
The formula for the volume of a sphere is 4πr3/3. Because this formula is in terms of the radius, it would be easier for us to determine how the change in the radius affects the volume. Since we are told the diameter is doubled, we need to first determine how the change in the diameter affects the change in radius.
Let us call the sphere's original diameter d and its original radius r. We know that d = 2r.
Let's call the final diameter of the sphere D. Because the diameter is doubled, we know that D = 2d. We can substitute the value of d and obtain D = 2(2r) = 4r.
Let's call the final radius of the sphere R. We know that D = 2R, so we can now substituate this into the previous equation and write 2R = 4r. If we simplify this, we see that R = 2r. This means that the final radius is twice as large as the initial radius.
The initial volume of the sphere is 4πr3/3. The final volume of the sphere is 4πR3/3. Because R = 2r, we can substitute the value of R to obtain 4π(2r)3/3. When we simplify this, we get 4π(8r3)/3 = 32πr3/3.
In order to determine the factor by which the volume has increased, we need to find the ratio of the final volume to the initial volume.
32πr3/3 divided by 4πr3/3 = 8
The volume has increased by a factor of 8.
### Example Question #101 : Solid Geometry
A sphere increases in volume by a factor of 8. By what factor does the radius change?
4
6
2
10
8
2
Explanation:
Volume of a sphere is 4Πr3/3. Setting that equation equal to the original volume, the new volume is given as 8*4Πr3/3, which can be rewritten as 4Π8r3/3, and can be added to the radius value by 4Π(2r)3/3 since 8 si the cube of 2. This means the radius goes up by a factor of 2
### Example Question #1 : How To Find The Volume Of A Sphere
A foam ball has a volume of 2 units and has a diameter of x. If a second foam ball has a radius of 2x, what is its volume?
2 units
4 units
8 units
16 units
128 units
128 units
Explanation:
Careful not to mix up radius and diameter. First, we need to identify that the second ball has a radius that is 4 times as large as the first ball. The radius of the first ball is (1/2)x and the radius of the second ball is 2x. The volume of the second ball will be 43, or 64 times bigger than the first ball. So the second ball has a volume of 2 * 64 = 128.
### Example Question #1 : Solid Geometry
A cross-section is made at the center of a sphere. The area of this cross-section is 225π square units. How many cubic units is the total volume of the sphere?
13500π
4500π
1687.5π
3375π
4500π
Explanation:
The solution to this is simple, though just take it step-by-step. First find the radius of the circular cross-section. This will give us the radius of the sphere (since this cross-section is at the center of the sphere). If the cross-section has an area of 225π, we know its area is defined by:
A = 225π = πr2
Solving for r, we get r = 15.
From here, we merely need to use our formula for the volume of a sphere:
V = (4 / 3)πr3
For our data this is: (4 / 3)π * 153 = 4π * 152 * 5 = 4500π
### Example Question #2 : How To Find The Volume Of A Sphere
At x = 3, the line y = 4x + 12 intersects the surface of a sphere that passes through the xy-plane. The sphere is centered at the point at which the line passes through the x-axis. What is the volume, in cubic units, of the sphere?
4896π
816π√(11)
4896π√(17)
2040π√(7)
4896π√(17)
Explanation:
We need to ascertain two values: The center point and the point of intersection with the surface. Let's do that first:
The center is defined by the x-intercept. To find that, set the line equation equal to 0 (= 0 at the x-intercept):
0 = 4x + 12; 4x = –12; x = –3; Therefore, the center is at (–3,0)
Next, we need to find the point at which the line intersects with the sphere's surface. To do this, solve for the point with x-coordinate at 3:
y = 4 * 3 + 12; y = 12 + 12; y = 24; therefore, the point of intersection is at (3,24)
Reviewing our data so far, this means that the radius of the sphere runs from the center, (–3,0), to the edge, (3,24). If we find the distance between these two points, we can ascertain the length of the radius. From that, we will be able to calculate the volume of the sphere.
The distance between these two points is defined by the distance formula:
d = √( (x1 – x0)2 + (y1 – y0))
For our data, that is:
√( (3 + 3)2 + (24 – 0)) = √( 62 + 24) = √(36 + 576) = √612 = √(2 * 2 * 3 * 3 * 17) = 6√(17)
Now, the volume of a sphere is defined by: V = (4/3)πr3
For our data, that would be: (4/3)π * (6√(17))3 = (4/3) * 63 * 17√(17) * π = 4 * 2 * 62 * 17√(17) * π = 4896π√(17)
### Example Question #1 : How To Find The Volume Of A Sphere
The surface area of a sphere is 676π in2. How many cubic inches is the volume of the same sphere?
8788π
2028π
(2197π)/3
(2028π)/3
(8788π)/3
(8788π)/3
Explanation:
To begin, we must solve for the radius of our sphere. To do this, recall the equation for the surface area of a sphere: A = 4πr2
For our data, that is: 676π = 4πr2; 169 = r2; r = 13
From this, it is easy to solve for the volume of the sphere. Recall the equation:
V = (4/3)πr3
For our data, this is: V = (4/3)π * 133 = (4π * 2197)/3 = (8788π)/3
### Example Question #1 : How To Find The Volume Of A Sphere
What is the difference between the volume and surface area of a sphere with a radius of 6?
216π
720π
133π
288π
144π
144π
Explanation:
Surface Area = 4πr2 = 4 * π * 62 = 144π
Volume = 4πr3/3 = 4 * π * 63 / 3 = 288π
Volume – Surface Area = 288π – 144π = 144π
### Example Question #2 : How To Find The Volume Of A Sphere
A sphere is perfectly contained within a cube that has a volume of 216 units. What is the volume of the sphere?
Explanation:
To begin, we must determine the dimensions of the cube. This is done by solving the simple equation:
We know the volume is 216, allowing us to solve for the length of a side of the cube.
Taking the cube root of both sides, we get s = 6.
The diameter of the sphere will be equal to side of the cube, since the question states that the sphere is perfectly contained. The diameter of the sphere will be 6, and the radius will be 3.
We can plug this into the equation for volume of a sphere:
We can cancel out the 3 in the denominator.
Simplify.
### Example Question #1 : How To Find The Volume Of A Sphere
The surface area of a sphere is . Find the volume of the sphere in cubic millimeters.
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# 1TTO §8.5 multiplying fractions
Dagopening:
Grip
Clip Lauren Daigle Hold on to me
Gebed
Mededeling
1 / 28
Slide 1: Slide
WiskundeMiddelbare schoolhavo, vwoLeerjaar 1
This lesson contains 28 slides, with interactive quizzes and text slides.
Lesson duration is: 30 min
## Items in this lesson
Dagopening:
Grip
Clip Lauren Daigle Hold on to me
Gebed
Mededeling
#### Slide 1 -Slide
§8.5 Multiplying fractions
After + and - we now look at
the operation: x !
By the way, this is a pretty long LessonUp.
#### Slide 2 -Slide
First we multiply a fraction by a whole number:
#### Slide 4 -Slide
D
Do: 2 X 33 : 3 = .......
32
x 33 = ...
solution:
2
#### Slide 7 -Slide
another one for you:
3/13 x 52 = ....
A
10
B
9
C
14
D
12
Solution:
#### Slide 10 -Slide
Finally it's
Fraction times Fraction ...
#### Slide 11 -Slide
2nd important point:
2. (in fact good news to some of you, who didn't know it yet!)
When multiplying fractions with unequal denominators,
you don't have to make them equal!!
Watch the next slides that show how simple it really is!
3/8 x 3/8 = ...
A
9/8
B
9/64
C
6/16
D
3/64
Solution:
#### Slide 17 -Slide
Fotovraag: work out on paper, take a pic and send in in next slide!
2nd This a task for ALL!
#### Slide 18 -Slide
2/7 + 1/7 x 1/2 = ......
Solution:
#### Slide 20 -Slide
Now it's time for a complicated one:
Think about it for a while. After that we look at the solution.
#### Slide 21 -Slide
This is the Solution:
#### Slide 22 -Slide
For example this one. You might think:
2 x 3 = 6 and x =
So the final answer is: 6 + = 6 However this is WRONG!
21
41
81
81
81
#### Slide 23 -Slide
Let's look back at the previous slide for a sec to see if both methods are good!
#### Slide 24 -Slide
again work out on paper and send it in.
This time for every-every-every one!
#### Slide 25 -Slide
1 1/8 x 1 1/3 = ......
Solution:
Homework:
§8.5 Make 8.5.
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# Plus One Math's Solution Ex 3.3 Chapter 3 Trigonometric Functions
The solutions for the
Trigonometric Functions are not readily available. While many schools may have a ready-made solution for this set in their school textbook, some might not. This is where our solution will be useful. In this article, you will find detailed solutions provided by us for the above set.Trigonometric Functions (Key Concept Reference) describes some basic and advanced uses of trigonometric functions, including identities, graph transformations, inverse functions, solutions of triangles, and polar coordinates.
Ncert Plus one Maths chapter-wise textbook solution for chapter 3 Trigonometric Functions Exercise 3.3. It contains detailed solutions for each question which have prepared by expert teachers to make each answer easily understand the students. they are well arranged solutions so that students would be able to understand easily.
Board SCERT, Kerala Text Book NCERT Based Class Plus One Subject Math's Textbook Solution Chapter Chapter 3 Exercise Ex 3.3 Chapter Name Trigonometric Functions Category Plus One Kerala
## Kerala Syllabus Plus One Math's Textbook Solution Chapter 3 Trigonometric Functions Exercises 3.3
### Chapter 3 Trigonometric Functions Textbook Solution
Kerala plus One maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.
### Chapter 3 Trigonometric Functions Exercise 3.3
Question 1:
Prove that
$LHS=\frac{cos&space;(\pi&space;+&space;x)cos(-x)}{sin&space;(\pi&space;-&space;x)&space;cos&space;(\frac{\pi&space;}{2}&space;+&space;x)}$
L.H.S. =
Prove that
L.H.S. =
Prove that
L.H.S. =
Prove that
L.H.S =
Find the value of:
(i) sin 75°
(ii) tan 15°
(i) sin 75° = sin (45° + 30°)
= sin 45° cos 30° + cos 45° sin 30°
[sin (x + y) = sin x cos y + cos x sin y]
(ii) tan 15° = tan (45° – 30°)
Prove that:
Prove that:
It is known that
∴L.H.S. =
L.H.S. =
Prove that
It is known that.
∴L.H.S. =
Prove that sin2 6x – sin2 4x = sin 2x sin 10x
It is known that
∴L.H.S. = sin26x – sin24x
= (sin 6x + sin 4x) (sin 6x – sin 4x
= (2 sin 5x cos x) (2 cos 5x sin x)
= (2 sin 5x cos 5x) (2 sin x cos x)
= sin 10x sin 2x
= R.H.S.
Prove that cos2 2x – cos2 6x = sin 4x sin 8x
It is known that
∴L.H.S. = cos2 2x – cos2 6x
= (cos 2x + cos 6x) (cos 2– 6x)
= [2 cos 4x cos 2x] [–2 sin 4(–sin 2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S.
Prove that sin 2x + 2sin 4x + sin 6x = 4cos2x sin 4x
L.H.S. = sin 2x + 2 sin 4x + sin 6x
= [sin 2x + sin 6x] + 2 sin 4x
= 2 sin 4x cos (– 2x) + 2 sin 4x
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
= 2 sin 4x (2 cos2x – 1 + 1)
= 2 sin 4x (2 cos2x)
= 4cos2x sin 4x
= R.H.S.
Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
L.H.S = cot 4x (sin 5x + sin 3x)
= 2 cos 4x cos x
R.H.S. = cot x (sin 5x – sin 3x)
= 2 cos 4x. cos x
L.H.S. = R.H.S.
Prove that
It is known that
∴L.H.S. =
Prove that
It is known that
∴L.H.S. =
Prove that
It is known that
∴L.H.S. =
Prove that
L.H.S. =
Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x
= cot x cot 2x – cot 3x (cot 2x + cot x)
= cot x cot 2x – cot (2x) (cot 2x + cot x)
= cot x cot 2– (cot 2cot x – 1)
= 1 = R.H.S.
Prove that
It is known that.
∴L.H.S. = tan 4x = tan 2(2x)
Prove that cos 4x = 1 – 8sinx cosx
L.H.S. = cos 4x
= cos 2(2x)
= 1 – 2 sin2 2x [cos 2A = 1 – 2 sin2A]
= 1 – 2(2 sin x cos x)2 [sin2A = 2sin A cosA]
= 1 – 8 sin2x cos2x
= R.H.S.
Prove that: cos 6x = 32 cos6x – 48 cos4x + 18 cos2x – 1
L.H.S. = cos 6x
= cos 3(2x)
= 4 cos3 2x – 3 cos2x [cos 3A = 4 cos3A – 3 cosA]
= 4 [(2 cos2– 1)3 – 3 (2 cos2x – 1) [cos 2x = 2 cos2– 1]
= 4 [(2 cos2x)3 – (1)3 – 3 (2 cos2x)2 + 3 (2 cos2x)] – 6cos2x + 3
= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3
= 32 cos6x – 4 – 48 cos4x + 24 cos2x – 6 cos2x + 3
= 32 cos6– 48 cos4x + 18 cos2x – 1
= R.H.S.
#### Chapter 3 Trigonometric Functions EX 3.3 Solution- Preview
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### Plus Two Math's Related Links
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Contemporary Mathematics
Key Concepts
4.1Hindu-Arabic Positional System
• Exponents are used to represent repeated multiplication of a base.
• In arithmetic, exponents are computed before multiplication, division, addition, and subtraction. Computing an exponent is done by multiplying the base by itself the number of times equal to the exponent.
• The system of numbers currently used is the Hindu-Arabic system. Digits in this system take on values based on their place in the number. The place values are determined by multiplying the digit by 10 raised to the appropriate power.
• The expanded form of a Hindu-Arabic number is the sum of each digit times 10 raised to the exponent for that place value.
4.2Early Numeration Systems
• Historically, there have been many systems for numbering. One system is an additive system, in which symbols are repeated to express larger numbers. Another system is a positional system, in which the digits and their positions determine the quantity being represented.
• The Babylonian system was a combination of a positional and additive system. It used 60 as its base. Using that in the positional system makes it possible to convert between Babylonian and Hindu-Arabic numbers.
• The Mayan system was a combination of a positional and additive system. It used 20 as its base. Using that in the positional system makes it possible to convert between Mayan and Hindu-Arabic numbers.
• The Roman system was an additive system. Knowing what each symbol represents makes it possible to convert between Roman and Hindu-Arabic numbers.
4.3Converting with Base Systems
• The system we use is the base 10 system. Base 10 is not the only base that can be used. To use another base, one could start with a list of numbers in that base.
• To indicate that a number is written in a base other than 10, a subscript is appended to the end of the number. That subscript indicates the base for the number.
• Numbers written in a base smaller than 10 use the same symbols as base 10. However, when using bases larger than 10, the symbols A, B, C, … are used to represent digits larger than 9.
• To convert from a number written in a base other than 10 into a base 10 number, the number is written in expanded form and then that expression is computed.
• To convert a number from base 10 into another base, the base 10 number is repeatedly divided by the new base. The remainders when performing these divisions become the digits for the number in the new base.
• Common errors can be detected when performing base conversions.
4.4Addition and Subtraction in Base Systems
• Addition tables for bases other than 10 can be built using the same processes that are used in base 10, including using a number line.
• Addition in bases other than base 10 use the same processes as addition in base 10, but use the addition table for that base.
• Subtraction in bases other than base 10 use the same processes as subtraction in base 10, but use the addition table for that base.
4.5Multiplication and Division in Base Systems
• Multiplication tables for bases other than 10 can be built using the same processes that are used in base 10, including using repeated addition and the addition table for the base.
• Multiplication in bases other than base 10 use the same processes as multiplication in base 10, but use the multiplication table for that base.
• Basic division in bases other than base 10 use the same processes as basic division in base 10, where the missing factor process is used.
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# NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2 – Quadratic Equations
Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2 – Quadratic Equations. This Exercise contains 6 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.
### NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2 – Quadratic Equations
NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2 – Quadratic Equations
1. Find the roots of the following quadratic equations by factorization:
(i) x2-3x-10 = 0
Solution:
x2-3x-10 = 0
x2-5x+2x-10 = 0
x(x-5)+2(x-5) = 0
(x-5)(x+2) = 0
⇒ x-5 = 0 or, x+2 = 0
x = 5 or x =-2
Hence the required roots are 5, -2.
(ii) 2x2+x-6=0
Solution:
2x2+x-6 = 0
2x2+4x-3x-6 = 0
2x(x+2)-3(x+2) = 0
(2x-3)(x+2) = 0
⇒ 2x-3 = 0 or, x+2 = 0
x = 3/2 or x = -2
Hence the required roots are 3/2, -2.
(iii) √2x2+7x+5√2 = 0
Solution:
√2x2+7x+5√2 = 0
√2x2+2x+5x+ 5√2 = 0
√2x(x+√2)+5(x+√2) = 0
(x+√2)(√2x +5) = 0
⇒ x+√2 = 0 or, √2x +5 = 0
x = -√2 or x = -(5/√2)
Hence the required roots are -√2, -(5/√2).
(iv) 2x2-x+(1/8) =0
Solution:
2x2-x+ (1/8) =0
16x2-8x+1 = 0
(4x)2-2.4x.1+12 = 0
(4x-1)2= 0
⇒ 4x-1 = 0 or 4x-1 = 0
x = (1/4) or x = (1/4)
Hence the required roots are 1/4, 1/4
(v) 100x2-20x+1 = 0
Solution:
100x2-20x+1 = 0
(10x)2-2×10x.1+12= 0
(10x-1)2 = 0
10x-1 = 0 or 10x-1 = 0
x = 1/10 or x = 1/10
Hence the required roots are 1/10 or 1/10
2. Solve the problem given in example 1.
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each others, and product of the marbles they now have is 124. We would like to find how many marbles they had to start with?
Solution: Let the number of marbles John had be x.
Then the number of marbles Jivanti had be 45-x
They lost 5 marbles to each others.
Hence now the number of marbles of John is x-5 and that of Jivanti is 45-x-5 = 40-x.
Then by given condition we have
(x-5)(40-x) = 124
40x-200-x2+5x = 124
x2-45x+324 = 0
x2-36x-9x+324 = 0
x(x-36)-9(x-36) = 0
(x-36)(x-9) = 0
x-36 = 0
x = 36
or x-9 = 0
x = 9
Therefore, they had 36 and 9 marvels respectively.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. We would like to find out the number of toys produced on that day.
Solution: Let the number of toys produced on that day is x.
Therefore, the cost of production of each toy on that day is (55-x)
So, the total cost of production on that day is = x(55-x)
Then using the given condition, we have,
x(55-x) = 750
55x-x2 =750
x2-55x+750 = 0
x2-25x-30x+750 = 0
x(x-25)-30(x-25) = 0
(x-25)(x-30) = 0
x-25 = 0
x =25
or x-30 =0
x = 30
Therefore, the number of toys produced on that day is 25 or 30.
3. Find the two numbers whose sum is 27 and product is 182.
Solution: Let one number is x. Then another number is 27-x.
Then by the given condition,
x(27-x) = 182
27x- x2-182 = 0
x2-27x+182 = 0
x2-14x-13x+182 = 0
x(x-14)-13(x-14) = 0
(x-14)(x-13) = 0
x-14 = 0
x =14
or, x-13 = 0
x =13
Hence the required two numbers are 13 and 14.
4. Find the two consecutive positive integers, sum of whose squares is 365.
Solution: Let the two consecutive positive integers be x and x+1
Then by the given condition,
x2+(x+1)2 = 365
x2+ x2+2x+1 = 365
2x2+2x-364 = 0
x2+x-182 = 0
x2+14x-13x-182 = 0
x(x+14)-13(x+14) = 0
(x+14)(x-13) = 0
x = -14 or 13
Since the numbers are positive so x = -14 not possible.
So, the numbers are 13 and 13+1 = 14.
5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution: Let base of the right triangle is x cm. Then the altitude is x-7 cm.
Then by the property of right triangle and using the given property we have
x2+(x-7)2 = (13)2
x2+ x2-14x+49 = 169
2x2-14x-120 = 0
x2-7x-60 = 0
x2-12x+5x-60 = 0
x(x-12)+5(x-12) = 0
(x-12)(x+5) = 0
x-12 = 0
x =12
or, x+5 = 0
x = -5
Since length can’t be negative, so x = -5 is not possible.
Hence the base is 12 cm and the altitude is (12-7)cm = 5 cm
6. A cottage industry produces a certain number of pottery articles in a day. It was observed that on a particular day that the cost of production of each articles (in rupees) was 3 more than twice the number of articles produced on that day. If the total number of production on that day was ₹90, find the number of articles produced and the cost of each article.
Solution: Let the number of articles produced on that day is x. then the price on that day is 2x+3
Then using the given condition we have,
x(2x+3) = 90
2x2+3x-90 = 0
2x2+15x-12x-90 = 0
x(2x+15)-6(2x+15)=0
(2x+15)(x-6) = 0
2x+15 = 0
x = – (15/2)
or, x-6 = 0
x = 6
Since Articles can’t be negative then x = -(15/2) is not possible.
Therefore, the number of articles produced on that day = 6
Cost of each article = 2×6+3 = ₹15
NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2 – Quadratic Equations, has been designed by the NCERT to test the knowledge of the student on the topic – Solution of a Quadratic Equation by Factorisation
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The Number System - Posters for All 8.NS Standards
Subject
Common Core Standards
Product Rating
File Type
PDF (Acrobat) Document File
2 MB|10 pages
Product Description
Number System– Math 8 – Posters for All Standards
................................................................................................................................
Are you looking for a way to display the 8th grade number system standards? Try these FREE posters.
................................................................................................................................
CONTENTS:
• 8.NS.A.1 Poster
• 8.NS.A.2 Poster
................................................................................................................................
HOW CAN YOU USE THIS RESOURCE?
• Option 1: Hang in classroom.
• Option 2: Put in a binder for quick reference for you and/or your students.
................................................................................................................................
Check Out Some Related Activities:
Approximating Square Roots to the Nearest Whole Number – Two Matching Puzzles – 8.NS.A.2
CCSS 8.NS.A.1 and 8.NS.A.2 Irrational and Rational Number Stations
Squares and square Roots – Two Triangle Matching Puzzles
................................................................................................................................
COMMON CORE STANDARDS: 8th Grade Number System
• Know that there are numbers that are not rational, and approximate them by rational numbers.
CCSS.MATH.CONTENT.8.NS.A.1
• Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number.
CCSS.MATH.CONTENT.8.NS.A.2
• Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., π2). For example, by truncating the decimal expansion of √2, show that √2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations.
................................................................................................................................
Check out some station activities:
Area of Polygons (CCSS 6.G.1)
Integer Operations
Slope of a Line Middle School Math Stations
................................................................................................................................
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Domain of a function explained
In mathematics, the domain of definition (or simply the domain) of a function is the set of "input" or argument values for which the function is defined. That is, the function provides an "output" or value for each member of the domain.[1] Conversely, the set of values the function takes on as output is termed the image of the function, which is sometimes also referred to as the range of the function.
For instance, the domain of cosine is the set of all real numbers, while the domain of the square root consists only of numbers greater than or equal to 0 (ignoring complex numbers in both cases).
If the domain of a function is a subset of the real numbers and the function is represented in a Cartesian coordinate system, then the domain is represented on the x-axis.
Formal definition
f\colonX\toY
, the set
X
is the domain of
f
; the set
Y
is the codomain of
f
. In the expression
f(x)
,
x
is the argument and
f(x)
is the value. One can think of an argument as a member of the domain that is chosen as an "input" to the function, and the value as the "output" when the function is applied to that member of the domain.
The image (sometimes called the range) of
f
is the set of all values assumed by
f
for all possible
x
; this is the set
\left\{f(x)|x\inX\right\}
. The image of
f
can be the same set as the codomain or it can be a proper subset of it; it is the whole codomain if and only if
f
is a surjective function, and otherwise it is smaller.
A well-defined function must map every element of its domain to an element of its codomain. For example, the function
f
defined by
f(x)= 1 x
has no value for
f(0)
. Thus, the set of all real numbers,
R
, cannot be its domain. In cases like this, the function is either defined on
R\setminus\{0\}
or the "gap is plugged" by explicitly defining
f(0)
.If one extends the definition of
f
to the piecewise function
f(x)=\begin{cases} 1/x&x\not=0\\ 0&x=0 \end{cases}
then f is defined for all real numbers, and its domain is
R
.
Any function can be restricted to a subset of its domain. The restriction of
g\colonA\toB
to
S
, where
S\subseteqA
, is written
\left.g\right|S\colonS\toB
.
Natural domain
The natural domain of a function is the maximum set of values for which the function is defined, typically within the reals but sometimes among the integers or complex numbers. For instance the natural domain of square root is the non-negative reals when considered as a real number function. When considering a natural domain, the set of possible values of the function is typically called its range.[2]
Domain of a partial function
See main article: Partial function. There are two distinct meanings in current mathematical usage for the notion of the domain of a partial function from X to Y, i.e. a function from a subset X' of X to Y. Most mathematicians, including recursion theorists, use the term "domain of f" for the set X' of all values x such that f(x) is defined. But some, particularly category theorists, consider the domain to be X, irrespective of whether f(x) exists for every x in X.
Category theory
Category theory deals with morphisms instead of functions. Morphisms are arrows from one object to another. The domain of any morphism is the object from which an arrow starts. In this context, many set theoretic ideas about domains must be abandoned or at least formulated more abstractly. For example, the notion of restricting a morphism to a subset of its domain must be modified. See subobject for more.
Other uses
Rn
where a problem is posed, that is, where the unknown function(s) are defined.
More common examples
As a partial function from the real numbers to the real numbers, the function
x\mapsto\sqrt{x}
has domain
x\geq0
. However, if one defines the square root of a negative number x as the complex number z with positive imaginary part such that z2 = x, the function
x\mapsto\sqrt{x}
has as its domain the entire real line (but now with a larger codomain).The domain of the trigonometric function
\tanx=
\sinx \cosx
is the set of all (real or complex) numbers not of the form
\pi 2
+k\pi,k=0,\pm1,\pm2,\ldots
.
|
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#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation8.3 Percentages
To find 50% of a number, find half of that number. To find 25% of a number, find one quarter of that number. To find 10% of a number, divide the number by ten. Examples: 50% of 50 is 25 50% of 60 is 30 50% of 200 is 100 50% of 12 is 6 25% of 40 is 10 25% of 16 is 4 25% of 100 is 25 25% of 240 is 60 10% of 10 is 1 10% of 20 is 2 10% of 30 is 3 10% of 100 is 10 Note the following relationships between percentages and fractions: 100% = 1 whole 50% = 1/2 = one half 25% = 1/4 = one quarter 10% = 1/10 You can use the above facts to solve problems involving other percentages. Examples: Since 10% of 120 is 12, 30% of 120 is 3 x 12 (3 times 12) = 36 Since 25% of 100 is 25, then 75% of 100 is 3 x 25 (3 times 25) = 75 Directions: Answer the following questions. Also write at least ten examples of your own.
Q 1: What number is 10 percent greater than 20?24202210 Q 2: Bob scored 40 points out of 60 in his math test. Ron scored 50% in the same test. Who did better?NeitherRonBob Q 3: If 10% of 130 is 13, then 30% of 130 is ______.36432439 Q 4: 70% of the children in a school buy lunch at the cafeteria and the rest bring their lunch from home. What percentage of the children bring lunch from home?30%unknown40%32% Q 5: A dress is selling for \$100after 20% discount. What is the original selling price?\$80\$200\$120\$125 Q 6: If the sales tax on a \$30.00 item is \$1.80, what is the sales tax rate?5%6%10%8% Q 7: Eric invited 50 friends to his birthday party. 50% of them could not attend the party. How many friends attended the party?50302540 Q 8: A television is on sale for \$900. If the sale price is 10 % less than the regular price. What was the regular price? \$1000\$910\$1100\$800 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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Sometimes you need to find the roots of a function, also known as the zeroes. Sometimes finding the zeroes is pretty easy. Other times, that isn't the case. Take for example the 6th degree polynomial shown below. How do you find the zeroes to that equation? You could make a rough estimate with the graph, or you could use Newton's method.
## Step 1: How It Works
Newton's method uses tangent lines to approximate the zeroes of a function. It's difficult to explain only in word how it works, take a look at the picture below. The graph is e-x. We start at [5,148] and follow the line tangent to the curve. When the tangent line hits the x-axis (y=0) we do it again, at the x-value we found by following the tangent line. Eventually we would expect to reach a 0. In this case, we never would since e-x never touches the x-axis. It illustrates the point though.
Convince yourself that f(a)=f'(a)(a-b). f'(a) is the derivative of the function at point a. It's the slope of the function at a. If you multiply this by a certain number (a-b), b is undefined, it will equal the function at that point. If we know the value of the function at point a, and the derivative of the function at point a, then we can rewrite the equation in terms of b (the undefined point on the x-axis.) Rewritten as such:
b=a-f(a)/f'(a). Using this equation, we can use any point on the x-axis, find the next point, then use that one, and repeat until we hit one of the zeroes.
## Step 2: Let's Find the Roots!
Polynomials work really well for this. Finding their derivatives is easy, and if they're large polynomials, there's no other way to get an accurate estimate of the zeroes.
First things first, graph your function. You can use a spreadsheet program, your calculator, whatever works for you. Make a rough estimate as to where the zeroes are. Looking at the graph below, I'm going to estimate that there are zeroes at -2, -1.2, 1.2, and 3. Let's see how accurate my guess is.
## Step 3: Let Excel Do the Work
We're going to let excel take care of all the guesswork for us. In your spreadsheet you're going to need 3 columns. One for your x-values, one for your function, and one for its derivative.
In cell A1 put in your guess for one of the zeroes. In cell B2 put in your equation as a function of cell A1 (shown in the picture.) In cell C1 put in your derivative as a function of A1.
In cell A2 type in "=A1-B1/C1" this is the equation I showed you a couple steps earlier. Now drag all three cells down 10 or so spaces. The last cell in column A should be your root. To check your answer, make sure that the last cell in column A matches the next to last cell. If any of the numbers are different, drag all three cells down even further. Congrats, there's the first root.
## Step 4: Find the Other Roots
Now that you've found your first root, it's time to find the other ones. Put your next guess in cell A1, make sure your spreadsheet program has calculated your next root. Sometimes you need to hit a CALCULATE button or something. I attached a sample spreadsheet below.
That's how Newton's method works.
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# GRE Math : Decimal Operations
## Example Questions
← Previous 1
### Example Question #1 : How To Add Decimals
Solve for :
Explanation:
To add decimals, simply treat them like you would any other number. Any time two of the digits in a particular place (i.e. tenths, hundredths, thousandths) add up to more than ten, you have to carry the one to the next greatest column. Therefore:
So .
### Example Question #1 : Decimals
Solve for :
Explanation:
To solve this problem, subtract from both sides of the eqution,
Therefore, .
If you're having trouble subtracting the decimal, mutliply both numbers by followed by a number of zeroes equal to the number of decimal places. Then subtract, then divide both numbers by the number you multiplied them by.
### Example Question #3 : Decimals
Solve for :
Explanation:
To solve, first add to both sides of the equation:
### Example Question #4 : Decimals
Solve for :
Explanation:
To solve, you need to do some algebra:
Isolate x by adding the 4.150 to both sides of the equation.
Then add the decimals. If you have trouble adding decimals, an effective method is to place one decimal over the other, and add the digits one at a time. Remember to carry every time the digits in a given place add up to more than .
### Example Question #5 : Decimals
Solve for :
Explanation:
To solve for , first add to both sides of the equation, so that you isolate the variable:
### Example Question #6 : Decimals
Solve for :
Explanation:
To solve, first add to both sides of your equation, so you isolate the variable:
### Example Question #1 : Decimals
Quantity A:
Quantity B:
The two quantities are equal.
Quantity B is greater.
Quantity A is greater.
The relationship cannot be determined.
The relationship cannot be determined.
Explanation:
Use the values of .5 and 2 as possible x values. If x = 2, Quantity A is positive and greater, but if x = .5, Quantity A is negative and therefore smaller. As such, the relationship cannot be determined.
### Example Question #1 : Decimal Operations
Solve for :
Explanation:
To solve for , subtract from both sides of the equation. , therefore, .
If you have trouble subtracting the decimals, you can multiply both of them by to get whole numbers, then subtract as normal, then divide your result by
### Example Question #2 : Decimal Operations
Explanation:
Make sure to follow the order of operations. Begin by combining all terms inside the parentheses:
Convert the fraction to a decimal and once again complete the operations inside the parentheses:
### Example Question #3 : Decimal Operations
find 0.72
0.0049
49
4.9
0.49
0.049
0.49
Explanation:
0.7 * 0.7 = 0.49
Trick: do the numbers without the decimals (7*7)
49; move the decimal of the answer the total number of spaces per each number (one for each 0.7)
0.49
← Previous 1
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# The Decimal Dungeon - Part 4
8-10 yrs old
Math & Economics
Explore the Decimal Dungeon in a five-part unit on Numbers & Operations in Base Ten where students observe and build math models to solve problems.
Submitted By: Minecraft Education
February 12, 2020
#### Skills
• Creativity
• Critical Thinking
### External References
Associated Engage NY module and lesson.
### Learning Objectives
• CCSS.MATH.CONTENT.5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths, 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 and explain the reasoning used.
### Guiding Ideas
Watch this short video and have the students follow along by writing the problems in their notebooks.
1) How are adding and subtracting decimals different than adding and subtracting whole numbers?
2) How are adding and subtracting decimals the same as adding and subtracting whole numbers?
3) What are real world situations where we would need to add and subtract fractions?
### Student Activities
Warm up:
1) 1,458.35 + 385.81 =
2) 81.23 - 7.42 =
Students can take the real world situations they just discussed earlier and use them to write a solve a word problem that involves adding or subtracting decimals.
Play the Game
Students will load the attached single player Minecraft world Decimal Dungeon. They will watch videos, observe and build math models, and complete challenges to advance. The world is sequential, but you can warp to any area from the starting Hub. To begin this lesson, speak to the NPC named Level 5.
Challenge 1) Observe Math Model
Observe and Solve Math Models that Represent Addition and Subtraction Problems with Decimals.
Challenge 2) Build a Math Model of Addition with Decimals
Given two numbers with decimals, add, build the problem and take a picture and document in your portfolio. Then regroup the problem, take a picture and document in your portfolio. Answer the NPCs question to finish the level.
Debrief
1) How is adding and subtracting decimals different than adding and subtracting whole numbers?
2) How is adding and subtracting decimals the same as adding and subtracting whole numbers?
3) What strategies can you use to make adding and subtracting with fractions easy.
### Performance Expectations
1) The student was able to observe math models and identify what numbers are being added or subtracted.
2) The student was able to build math models and identify what numbers are being added.
3) The student was able to regroup between place values.
#### Skills
• Creativity
• Critical Thinking
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Into Math Grade 1 Module 17 Lesson 3 Answer Key Use Nonstandard Units to Measure Length
We included HMH Into Math Grade 1 Answer Key PDF Module 17 Lesson 3 Use Nonstandard Units to Measure Length to make students experts in learning maths.
HMH Into Math Grade 1 Module 17 Lesson 3 Answer Key Use Nonstandard Units to Measure Length
I Can measure the length of objects using units that are the same size.
How do you use objects to measure length?
Supply a variety of nonstandard units to be used for measuring such as paper clips, ones blocks, and color tiles. Have children choose an object they would like to measure. Children will draw to show how they measured the length of the object using nonstandard units.
Build Understanding
Theo measures a marker. Find and measure the length of a marker using .
A. Draw to show how to measure the marker.
About how long is the marker? __________
Explanation:
Theo measures a marker
He places under the marker to measure it
So, The marker is 7 long.
B. Choose a different unit to measure the length of a marker. Draw to show how to measure the marker.
About how long is the marker? __________ units
Explanation:
I used to measure the marker
I placed under the marker to measure it
So, the marker is 2 long.
Turn and Talk Why are the measurements different? How are the units different?
The measurements are different because the units of measurement are different one is a color tile and the other is a paperclip.
Step It Out
Question 1.
Nima measures the length of a crayon. How long will the crayon be using ?
A. Match the end of a with the end of the crayon.
B. Place end to end without gaps or overlaps to measure the crayon.
C. Draw to show how to measure the crayon.
About how long is the crayon? __________
Explanation:
I drew to show how to measure a crayon using
A crayon is 5 long.
Turn and Talk Will the measurement be correct if there are gaps between the units? Explain.
No, the measurement will not be correct if there are gaps between the units. If there are gaps the number of units will decrease.
Check Understanding
Question 1.
Use to measure the length of the ribbon.
The ribbon is about __________ long.
Explanation:
I used to measure the length of the ribbon
I placed the color tiles under the ribbon
The ribbon is about 7 long.
Question 2.
Construct Arguments Franco measures the length of the pencil below using paper clips.
Franco says the pencil is 6 long. Did Franco measure the pencil correctly using paper clips? Explain.
Explanation:
Franco measures the length of the pencil below using paper clips
Franco says the pencil is 6 long
He placed the paperclips overlapping on one another
He is not correct
The pencil is about 5 long.
Question 3.
Measure the length of the rectangle below two different ways.
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# AP C UNIT 1 1-Dimension A quick review of basic kinematic variables
## Presentation on theme: "AP C UNIT 1 1-Dimension A quick review of basic kinematic variables"— Presentation transcript:
AP C UNIT 1 1-Dimension A quick review of basic kinematic variables
Difference in math application towards physics Differentiation (Calculus)
Definitions Distance (x) Displacement ( ) Speed (v) Velocity ( )
Avg = 2 pts in time, Instantaneous = 1pt in time
Definitions Acceleration ( ) If v = constant, then a = ?
Velocity is maximum when acceleration is Position is maximum when velocity is Be careful of signs zero zero
Calculus vs Algebra In HP, all problems were solved using a constant variable in the equations. Graphical analysis used lines of constant slope. In APC, it requires utilizing the tool of differential & integral calculus to further probe physics principles where variables might be changing. The “down & dirty” calculus skills presented here are not meant to replace those you learned/about to learn in your calculus class. Note a mixed ability group of Calc AB and BC.
Calculus – Differentiation & Slope
Last year we formulated linear regression lines of best fit, chose 2 points on the line, and determined the slope. No matter what 2 points were chosen, obviously, the slope was always the same since it was constant.
But what happens when we have a non-linear function
But what happens when we have a non-linear function? In the graph below, the slope varies from point to point. Trying to calculate Δy/Δx from A to B we would yield a negative slope, from B to C it would yield a positive slope, and from A to C, a zero slope. To find the slope at ‘A’, we can examine a point just above and just below ‘A’. The closer those points are to ‘A’ (and to each other), the more accurately they would describe the slope at that point.
This idea is to compute the rate of change as the limiting value of the ratio of the differences Δy/Δx as Δx becomes infinitely small. It follows that this limit is the exact slope of the tangent. Such a tiny or infinitesimal change in x and y is denoted by dx and dy, where we replace Δy/Δx with dy/dx.
Differentiating or taking the derivative, f’(x), of a function, f (x), equates to the slope of the tangent line at a point when the run is reduced to such a small value…an instantaneous point To find instantaneous velocity at a point in time, we use the following notation (recall Δ was for 2pts in time) lim Δt dx is a differential distance and dt is a differential time which is a fancy way of saying very, very small. In technical terms, dx is what happens to Δx in the limit when Δx approaches zero.
1st derivative of velocity or 2nd derivative of displacement
To find instantaneous acceleration at a point in time, we use the following notation 1st derivative of velocity or 2nd derivative of displacement
Calculus Rules - Differentiation (down & dirty method…your calc teacher will fill in the details)
Rule #1: Derivative of a Constant *Derivative of a constant is zero If f(x) = 5, then f’(x) = 0. Rule #2: Power Rule If f(x) = xn , then f’(x) = nxn-1. If f(x) = 5x2 + 3, then f’(x) = 2(5x1) + 0 = 10x…using derivative of constant & power rule. NOTE that we are just finding slopes of a tangent using these rules
Maximum and Minimum points (ie; when does max velocity occur?)
When a function reaches its maximum or minimum value, it must turn around where its slope = 0 Relative maxima y relative minimum x x x x Thus, the derivative when evaluated at x1 , x2 ,x3 will equal zero and allows us to determine where max or min occurs.
Example: f(x) = x3 – 6x2 + 9x + 1 To find max or min values, set f’(x) = 0 and solve. f’ (x) = 3x2 – 12x + 9 = 0 = 3(x2 – 4x + 3) = 0 x = 1 and x = 3 (extreme points) This is where the extreme points are….are they a max or min? 2nd Derivative Test: To find if it’s max or min (concavity), take 2nd derivative of function f’’(x) = 6x-12 Plug in extrema value. If f’’(x) < 0, then it’s a Max If f’’(x) > 0, then it’s a Min x = 1 is max since you get -6 x = 3 is min since you get +6 You could also plug in values above and below the extrema values (1 & 3) into the 1st derivative to determine if slope is incr or decreasing to determine concavity. Essentially, the 2nd derivative is looking at how the 1st derivative is changing.
Example The position of a particle is given by:
x(t) = t – 2.1t3 a) Find the velocity of particle at t = 3.0s. b) Find the maximum position of particle. Practice worksheet
Calculus - Integration
Integration is the reverse of differentiation where an integral is called the anti-derivative. Much like finding the slope of the tangent line at a point for differentiation, integration is finding the area under the curve or function. Like finding area for a v-t graph which equals displacement. Last year we only looked area for linear functions such as:
However, if the function is not linear, we must use calculus
However, if the function is not linear, we must use calculus. We could estimate the area by breaking the curve up into many rectangles and summing them up which can approximate the area. The more rectangles we use (smaller width), the better the estimation.
Consider the function at the left where we wish to find area from A to C. The problem is that the height keeps changing as we move from A to C along the function. In order to minimize this problem (height changing), we focus on a very small region of the graph, where the height is relatively stable. Start by picking a point x along x-axis and another point extremely close but beyond it, (x+dx). Drawing vertical lines at these two points where they meet the function yields a shaded region.
If we treat this region as a rectangle, its area is equal to the height f(x) times the tiny width dx. Obviously the shaded region isn't a rectangle, but as dx becomes smaller—as we bring the right side toward the left—the height change becomes less significant, and the region more closely resembles a rectangle. As dx approaches zero, this approximation becomes perfect: the height will equal f(x) on both left and right ends of rectangle…the area of the shaded region = f(x)*dx. We now just use an infinite amount of tiny rectangles and sum them up.
‘S’ looking symbol means to sum up
Calculus Rules - Integration
Rule #1: Power Rule where C is a constant and n ≠ -1 If f(x) = 6x2, then the integral of that function is:
(these integrals are evaluated between 2 values)
DEFINITE INTEGRALS (these integrals are evaluated between 2 values) Consider f(x) = 4x2 + 5x Evaluate the integral between the values of 1 and 2. For each of the following expressions determine the indefinite integral with respect to time: (a) v(t) = 6t3 − 5t (b) a(t) = 3t2 − 4t + 7 (c) a(t) = 10 + t−2 (d) v(t) = 2t5 − t4/3
Using the Constant, C. Acceleration of a bus is given by: a(t) = 1.2t
a) If vbus = 5.0m/s at t = 1.0s, what is vbus at t = 2.0s? b) If position of bus is 6.0m at t = 1.0s, find position at t = 2.0s?
Constant Acceleration Eqns
From above Use substitution to get other two equations
Example: A car sits at a light
Example: A car sits at a light. When light turns green, it accelerates at a constant rate of 2.5m/s2. At the moment the acceleration begins, a truck moves past the car with a uniform speed of 15m/s. At what position beyond the starting point does the car overtake the truck? 23
QUESTION A pair of identical balls are simultaneously released on a pair of equal length tracks, A and B, as shown. Friction is minimal. Both balls reach the ends of their tracks at the same time speed Both of these None of these 24
Graphical Analysis Position vs Time (x vs t):
Describes position of particle with respect to time Slope (Δx/Δt) indicates the velocity of the particle (mag & dirn) Average Velocity (displacement/time) is slope of the secant line Instantaneous Velocity is slope of the tangent line 25
26
Velocity vs Time (v vs t):
Describes velocity of particle with respect to time Slope (Δv/Δt) indicates the acceleration of the particle (mag & dirn) Instantaneous Acceleration is slope of the tangent line Average Acceleration (velocity/time) is slope of the secant line 27
Note that particle can have negative velocity and positive acceleration
28
Area indicates the displacement of particle
29
Acceleration vs Time (a vs t):
i) Describes acceleration of particle with respect to time ii) Area equates to change in velocity 30
Sketch the x vs t, v vs t, and a vs t graphs for the diagram assuming the ball stays in contact with the surface of ramp at all times. a v x *Graph worksheet 31
Freefall Assume no air friction where acceleration due to gravity on Earth is given by a = ‘g’ = -9.8m/s2. Objects in freefall are always accelerating downwards towards center of Earth. Can change based on location. In 1D, what is the value of velocity at apex of flight? What is the value of acceleration at apex?
A balloonist, riding in the basket of a hot air balloon that is rising at 10.0m/s, releases a sandbag when the balloon is 40.8m above the ground. Determine the velocity of the bag when it hits the ground. Sketch a graph of x, v, and a vs t for motion of sandbag
A model rocket is fired from rest vertically and ascends with a constant vertical acceleration of 4.0m/s2 for 6.0s. Its fuel is then exhausted and it continues as a free-fall particle. What is the total time elapsed from takeoff to striking the Earth?
VECTORS A. TERMS DEFINED *SCALAR *VECTOR
B. ADDITION / SUBTRACTION METHODS *2 OR MORE VECTORS COMBINED YIELDS RESULTANT *GRAPHICAL HEAD 2 TAIL, PARALLELOGRAM *ANALYTICAL 35
*SUBTRACTION C. COMPONENTS aX= acosθ ay= asinθ 36
Vector with magnitude = 1 w/ direction Lacks dimension & unit
E. UNIT VECTOR Vector with magnitude = 1 w/ direction Lacks dimension & unit Purpose is to ‘point’ LABELED ˆ = hat, replaces these are vector components of and ax and ay are scalar components of 37
2-D Motion The position of a particle that moves in both the x & y plane at the same time can be described by: < position vector > Find displacement of particle as it moves from to .
Similarly, it’s velocity is described as:
has a direction that is always tangent to path of particle It’s acceleration is described as: Example: find and for arbitrary times.
Projectiles
Neglecting air resistance,
Horizontal velocity = constant Vertical velocity changes Horizontal acceleration = zero Vertical acceleration = ‘g’
a) Rank the paths (use only the symbols > or = , for example, a>b=c) according to time of flight, greatest first. b) Rank the paths (use only the symbols > or = , for example, a>b=c) according to initial speed, greatest first.
Derive an expression to solve for minimum initial v to just clear gap (ignoring bike width)
y x θ
Relative Velocity vAC = vAB + vBC
Used whenever you see “velocity relative to” or “velocity with respect to”, addition and subtraction of velocities is done so by utilizing a subscript method. vAC = vAB + vBC Label each object with its F.O.R. You want the inner subscripts to match up Switching the order of subscripts is like multiplying by negative 1 where vAB = -vBA vAC refers to velocity of A relative to C 45
Example: A car is moving north at 88km/h when a truck approaches it from the other direction moving at 104km/h. a) What is the truck’s velocity relative to the car? b) What is the car’s velocity relative to truck? c) How do relative velocities change after the pass each other? 46
2D example: A supersonic aircraft is moving in still air with a constant velocity of Suddenly, at t = 0 a wind gusts with a velocity of Assuming the pilot makes no attempt to correct for wind, find plane’s displacement after 1hr relative to ground.
A canoe has a velocity of 0. 40m/s SE relative to the Earth
A canoe has a velocity of 0.40m/s SE relative to the Earth. The canoe is on a river flowing 0.50m/s East relative to the Earth. Find velocity of canoe relative to the river.
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# Find an orthonormal basis in 3-dimensional euclidean space by applying the gram-schmidt process on the basis 139
find an orthonormal basis in 3-dimensional euclidean space by applying the gram-schmidt process on the basis 139
Find an orthonormal basis in 3-dimensional Euclidean space by applying the Gram-Schmidt process on the basis {1, 3, 9}.
To find an orthonormal basis in 3-dimensional Euclidean space using the Gram-Schmidt process, we start with a given basis and apply a sequence of orthogonalizations. In this case, our given basis is {1, 3, 9}.
1. Step 1: Normalize the first vector in the basis.
• Normalize it by dividing it by its norm: v1 = 1/√(1^2) = 1/1 = 1.
2. Step 2: Calculate the second vector in the orthonormal basis.
• The second vector v2 is obtained by subtracting the projection of v2 onto v1 from v2.
• v2 = 3 - proj(v2, v1)
• To find the projection of v2 onto v1, we multiply v1 by the dot product of v2 and v1, divided by the squared norm of v1: proj(v2, v1) = (v2 · v1) / ||v1||^2 * v1
• Calculate: proj(v2, v1) = (3 · 1) / (1^2) * 1 = 3 / 1 = 3
• Subtract the projection from v2: v2 = 3 - 3 = 0.
3. Step 3: Normalize the second vector in the basis.
• Since v2 is now zero, we cannot normalize it further. We move on to the next step.
4. Step 4: Calculate the third vector in the orthonormal basis.
• The third vector v3 is obtained by subtracting the projections of v3 onto v1 and v2 from v3.
• v3 = 9 - proj(v3, v1) - proj(v3, v2)
• Calculate the projection of v3 onto v1: proj(v3, v1) = (v3 · v1) / ||v1||^2 * v1
• proj(v3, v1) = (9 · 1) / (1^2) * 1 = 9 / 1 = 9
• Calculate the projection of v3 onto v2: proj(v3, v2) = (v3 · v2) / ||v2||^2 * v2
• proj(v3, v2) = (9 · 0) / (0^2) * 0 = 0
• Subtract the projections from v3: v3 = 9 - 9 - 0 = 0.
5. Step 5: Normalize the third vector in the basis.
• Since v3 is now zero, we cannot normalize it further. We have completed the Gram-Schmidt process.
The resulting orthonormal basis is {1/1, 0, 0}, which can be simplified to {1, 0, 0}.
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# How do you solve and graph 3[4x-(2x-7)]<2(3x-5)?
Jun 28, 2016
$x \in \emptyset$
That is there is no value of $x$ for which the given inequality is true.
#### Explanation:
Given
$\textcolor{w h i t e}{\text{XXX}} 3 \left[4 x - \left(2 x - 7\right)\right] < 2 \left(3 x - 5\right)$
First, simplify both the left and right sides:
$\textcolor{w h i t e}{\text{XXX}} 3 \left[2 x + 7\right] < 6 x - 10$
$\textcolor{w h i t e}{\text{XXX}} 6 x + 21 < 6 x - 10$
Since we can subtract the same amount from both sides without effecting the validity or orientation of the inequality we have
$\textcolor{w h i t e}{\text{XXX}} 21 < - 10$
which is clearly not true for any value of $x$
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# Proof via Induction for A Summation
I'm starting to understand how induction works (with the whole $k \to k+1$ thing), but I'm not exactly sure how summations play a role. I'm a bit confused by this question specifically:
$$\sum_{i=1}^n 3i-2 = \frac{n(3n-1)}{2}$$
Any hints would be greatly appreciate
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Base case:
• Let $n=1$ and test: $$\sum_{i=1}^1(3i-2)=3-2=1=\frac{1(3\cdot 1-1)}{2}$$
• True for $n = 1$
Induction Hypothesis:
• Assume that it is true for $k$: assume that $$\sum_{i=1}^k(3i-2)=\frac{k(3k-1)}{2}.$$
Inductive Step:
• Prove, using the Inductive Hypothesis as a premise, that $$\sum_{i=1}^{k+1}(3i-2)=\left(\sum_{i = 1}^k(3i - 2)\right) + (3(k+1)-2) = \frac{(k+1)(3(k+1)-1)}{2}.$$
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So do I just manipulate the right side of the equation to try to get it to look like the original k(3k-1)/2 ? – K. Barresi Jan 25 '13 at 1:58
No, manipulate the inner third (in the equality chain of last line) to get the right hand side. You know, from the inductive hypothesis, what that the sum $$\left(\sum_{i = 1}^k(3i - 2)\right) = \frac{k(3k - 1)}{2}.$$ Add that to the term $3(k+1)- 2$ to obtain the right hand side. The right hand side is precisely the form you want to exhibit to show that the given sum holds for $k+1$ if it holds for $k$, hence, you will have proven, by induction, that for all $n$, $$\sum_{i=1}^n 3i-2 = \frac{n(3n-1)}{2}$$ – amWhy Jan 25 '13 at 2:01
Ahh ok I think I understand now. Thank you for your help. – K. Barresi Jan 25 '13 at 2:19
Sorry to bug you again, but I appear to be stuck. I've reached: $$(3k-2)+(3[k+1]-2)=\frac{k+1(3[k+1]-1)}{2}$$ which I simplified to $$12k-2=3k^2+5k+2$$ – K. Barresi Jan 25 '13 at 4:05
That's exactly what you want: you want (k+1) in every spot where k occurs in the inductive hypothesis: i.e. you want the right hand side of my last line. That shows that what holds for k, holds when every k is replaced by (k+1): you want the same "form" as on the right hand side. – amWhy Jan 25 '13 at 4:12
For $n=1$ you have $\sum_{i=1}^1(3i-2)=3-2=1=\frac{1(3\cdot 1-1)}{2}$.
Suppose that $$\sum_{i=1}^n(3i-2)=\frac{n(3n-1)}{2}$$ and prove that $$\sum_{i=1}^{n+1}(3i-2)=\frac{(n+1)(3(n+1)-1)}{2}.$$
-
Let's try amWhy's proof for a general summation:
$$\sum_{i=1}^n f(i) = g(n)$$
Base case:
• Let $n=1$ and test: $$\sum_{i=1}^1 f(i) = f(1) =?\ g(1)$$
• True for $\,n = 1\iff \color{#C00}{f(1) = g(1)}$
Induction Hypothesis:
• Assume that it is true for $\, n = k$: assume that $$\sum_{i=1}^k f(i) = g(k).$$
Inductive Step:
• Prove, using the Induction Hypothesis as a premise, that $$\sum_{i=1}^{k+1}f(i)=\left(\sum_{i=1}^k f(i)\right) + f(k\!+\!1) = g(k) + f(k\!+\!1) =?\ g(k\!+\!1).$$
• Inductive step from $k$ to $k+1$ is true $\iff \color{#C00}{ g(k\!+\!1) - g(k) = f(k\!+\!1)}$
Therefore we have proved by induction the following generic summation criterion
Theorem $\displaystyle\,\ \sum_{i=1}^n f(i) = g(n)\iff g(1) = f(1)\ {\rm and}\ g(n\!+\!1)-g(n) = f(n\!+\!1)\$ for $\,n \ge 1.$
Indeed, the induction proves the direction $(\Leftarrow),$ and $(\Rightarrow)$ is clear.
This theorem reduces the inductive proof to simply verifying the RHS equalities, which is trivial polynomial arithmetic when $f(n),g(n)$ are polynomials in $n$. The above theorem is an example of telescopy, also known as the Fundamental Theorem of Difference Calculus, depending on context. You can find many more examples of telescopy and related results in other answers here.
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hint:we know that $$\sum_{"i=1}^ni=\frac{n(n+1)}{2}$$ then write $\sum_{i=1}^n(3i-2)$=$3\sum_{i=1}^n(i) -2n$ substitute above assumption $\sum_{i=1}^n (3i-2 )= \frac{n(3n-1)}{2}$
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I will prove it in two way for you:
1- Mathematical Induction:
If $n=1$ then the left side is $1$ and also the right side is $1$ too.
Now think that we have $\sum_{i=1}^{n}(3i-2)=\frac{n(3n-1)}{2}$, we should show $\sum_{i=1}^{n+1}(3i-2)=\frac{(n+1)(3(n+1)-1)}{2}$ that is $\sum_{i=1}^{n+1}(3i-2)=\frac{(n+1)(3n+2)}{2}$. But we can write; $$\sum_{i=1}^{n+1}(3i-2)=(\sum_{i=1}^{n}(3i-2))+(3(n+1)-2)=\frac{n(3n-1)}{2}+(3n+1)=\frac{3n^{2}-n+6n+2}{2}=\frac{3n^{2}+5n+2}{2}=\frac{3n^{2}+3n+2n+2}{2}=\frac{3n(n+1)+2(n+1)}{2}=\frac{(n+1)(3n+2)}{2}$$. And it finished our work.
2- Without Mathematical Induction:
We know $\sum_{i=1}^{k}i=\frac{k(k+1)}{2}$ and $\sum_{i=1}^{k}c=kc$ for a constant number "c".
Now $$\sum_{i=1}^{n}(3i-2)=3\sum_{i=1}^{n}i-\sum_{i=1}^{n}2=3\frac{n(n+1)}{2}-2n=\frac{n(3n+3-4)}{2}=\frac{n(3n-1)}{2}$$.
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# What is the equation of the line that passes through (1,5) and (-2,14) in slope intercept form?
Mar 25, 2018
$y = - 3 x + 8$
#### Explanation:
$\text{the equation of a line in "color(blue)"slope-intercept form}$ is
•color(white)(x)y=mx+b
$\text{where m is the slope and b the y-intercept}$
$\text{to calculate the slope m use the "color(blue)"gradient formula}$
•color(white)(x)m=(y_2-y_1)/(x_2-x_1)
$\text{let "(x_1,y_1)=(1,5)" and } \left({x}_{2} , {y}_{2}\right) = \left(- 2 , 14\right)$
$\Rightarrow m = \frac{14 - 5}{- 2 - 1} = \frac{9}{- 3} = - 3$
$\Rightarrow y = - 3 x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$
$\text{to find b substitute either of the 2 given points}$
$\text{into the partial equation}$
$\text{using "(1,5)" then}$
$5 = - 3 + b \Rightarrow b = 5 + 3 = 8$
$\Rightarrow y = - 3 x + 8 \leftarrow \textcolor{red}{\text{in slope-intercept form}}$
Mar 25, 2018
The reqd. equn. of the line is
$3 x + y = 8$ or $y = - 3 x + 8$
#### Explanation:
If $A \left({x}_{1} , {y}_{1}\right) \mathmr{and} B \left({x}_{2} , {y}_{2}\right)$,then equation of the line:
color(red)((x-x_1)/(x_2-x_1)=(y-y_1)/(y_2-y_1).
We have,
$A \left(1 , 5\right) \mathmr{and} B \left(- 2 , 14\right)$
So,
$\frac{x - 1}{- 2 - 1} = \frac{y - 5}{14 - 5}$.
$\implies \frac{x - 1}{-} 3 = \frac{y - 5}{9}$
$\implies 9 x - 9 = - 3 y + 15$
$\implies 9 x + 3 y = 15 + 9$
$\implies 9 x + 3 y = 24$
$\implies 3 x + y = 8$ or $y = - 3 x + 8$
graph{3x+y=8 [-20, 20, -10, 10]}
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## Good Luck on Friday the 13th
A friend is taking a business trip today, and I asked if he was worried about flying on Friday the 13th. “I think it’s unlucky to have superstitions,” he replied.
If you’re a baseball player and you see a cross-eyed woman today, you might want to spit in your hat to avoid bad luck. (Or so they say.)
As it turns out, the first Friday the 13th of each year is “Blame Someone Else Day,” so if you don’t like this post, you should defintely let my wife know about it.
To ensure that you point a finger at others on the correct day, wouldn’t it be helpful to know when the first Friday the 13th of the year will occur? Consider yourself lucky, because you’ve stumbled across this post. I have two methods you can use for determining which months contain a Friday the 13th.
Method 1: Look-Up Table
Take the last two digits of the year, yy.
1. Calculate the sum yy + ⌊yy/4⌋. (The notation ⌊x⌋ indicates the floor function, which is the greatest integer less than x, so ⌊π⌋ = 3 and ⌊7.28⌋ = 7, for example.)
2. Determine the remainder when that sum is divided by 7.
3. Look up the remainder in the table below.
For example, in 2013, the calculations would give 13 + ⌊13/4⌋ = 13 + 3 = 16, which has a remainder of 2 when divided by 7. Since 2013 is a non-leap year, the table tells us that Friday the 13th will occur in September and December this year.
Note that the table only works for dates in the 2000’s. To modify the process for dates in the 1900’s, you need to add 1 in the first step; that is, find the sum yy + ⌊yy/4⌋ + 1. Then proceed as described above.
Also note that four lines in the table are highlighted in pink and yellow. Leap years always cause problems. The yellow lines in the table indicate years in which an extra Friday the 13th occurs in January or Feburary because of leap year, and the pink lines indicate years in which a Friday the 13th in January or February does not occur because of leap year.
Method 2: Internet
Go to http://www.timeanddate.com/calendar, enter the year in question, and examine all 12 months to see which contain a Friday the 13th.
[Ed. Note: Though perhaps more efficient, the use of Method 2 is highly discouraged and less fun than Method 1. Using Method 2 instead of Method 1 will result in the automatic revocation of your Geek Card. Plus, there’s a practical issue: What will you do when you have a time-sensitive need to know the month of the first Friday the 13th in the year 2044, say, and you find yourself in a location without Internet access? Shudder. Consequently, learning Method 1 is just as critical to you as learning to calculate change mentally is to a grocery store cashier, who lives in perpetual fear of power outages.]
Incidentally, there is at least one Friday the 13th every year. In addition, every month has four Fridays the 13th in a 28-year period, which means there are an average of 1.71 Fridays the 13th each year. (There is a slight snafu regarding this last fact, because years ending in 00 aren’t leap years if the year is not a multiple of 400, but whatever. It’s true most of the time.)
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The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.
## MJ4MF (offline version)
Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.
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## Geometry: Common Core (15th Edition)
area of deck = $80$ $m^2$
If we want to find the area of the deck in the real world and not according to the scale drawing, let us convert the scale drawing to real world measurements first. We know that the scale is $1$ $in.$ = $2$ $m$, so let's set up the proportion to find what the dimensions are in real life. Let's convert the width of the pool first: $\frac{1}{2} = \frac{2}{x}$ Use the cross products property to eliminate the fractions: $x = 4$ Now, we convert the length of the pool: $\frac{1}{2} = \frac{6}{x}$ Use the cross products property to eliminate the fractions: $x = 12$ m Next, we convert the width of the deck and pool combined: $\frac{1}{2} = \frac{4}{x}$ Use the cross products property to eliminate the fractions: $x = 8$ m Finally, we convert the length of the deck and pool combined: $\frac{1}{2} = \frac{8}{x}$ Use the cross products property to eliminate the fractions: $x = 16$ m We want to find the area of just the deck, which we can do if we subtract the area of the pool from the area of both the pool and the deck together. Let's find the area of the pool by using the following formula for the area of a rectangle: $A = lw$, where $A$ is the area, $l$ is the length, and $w$ is the width of the rectangle. Let's substitute what we know into this formula to find the area of the pool itself: $A = (4 m)(12 m)$ Multiply to solve: $A = 48$ $m^2$ Now, let's find the area of the combined pool and deck: $A = (8 m)(16 m)$ Multiply to solve: $A = 128$ $m^2$ To find the area of the deck as depicted in the scale drawing, we subtract the area of the pool from the combined area of the pool and the deck together: area of deck = $128$ $m^2 - 48$ $m^2$ Subtract to solve: area of deck = $80$ $m^2$
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