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SSC BOARD PAPERS IMPORTANT TOPICS COVERED FOR BOARD EXAM 2024
### If [2a+b3a-bc+2d2c-d]=[234-1], find a, b, c and d.
Exercise 2.2 | Q 14 | Page 47
#### QUESTION
If $\left[\begin{array}{cc}2\text{a}+\text{b}& 3\text{a}-\text{b}\\ \text{c}+2\text{d}& 2\text{c}-\text{d}\end{array}\right]=\left[\begin{array}{cc}2& 3\\ 4& -1\end{array}\right]$, find a, b, c and d.
#### SOLUTION
$\left[\begin{array}{cc}2\text{a}+\text{b}& 3\text{a}-\text{b}\\ \text{c}+2\text{d}& 2\text{c}-\text{d}\end{array}\right]=\left[\begin{array}{cc}2& 3\\ 4& -1\end{array}\right]$
∴ By equality of matrices, we get
2a + b = 2 ....(i)
3a – b = 3 ....(ii)
c + 2d = 4 ....(iii)
2c –d = – 1 ....(iv)
Adding (i) and (ii), we get
5a = 5
∴ a = 1
Substituting a = 1 in (i), we get
2(1) + b = 2
∴ b = 0
By (iii) + (iv) x 2, we get
5c = 2
∴ c = $\frac{2}{5}$
Substituting c = $\frac{2}{5}$ i (iii), we get
$\frac{2}{5}+2d$ = 4
∴ 2d = $4-\frac{2}{5}$
∴ 2d = $\frac{18}{5}$
∴ d = $\frac{9}{5}$.
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# How many integer pairs $(x,y)$ satisfy $|x|^m+|y|^m=r$
First let me put the question succinctly:
For whole numbers $m$ and $r$ how many integer pairs $(x,y)$ satisfy the equation $|x|^m+|y|^m=r$?
Now for exposition:
For some motivations to this questions check out: Computing $\beta(\frac{1}{m},\frac{1}{m})$
For a $m=2$ the result can be given in general which I will do below but the sake of example consider the case $m=2$ and $r=45$. Then there are $8$ integer pairs (x,y) such that $x^2+y^2=45$ and this can be seen in the picture below:
Generally speaking for the number of integer pairs that satisfy $x^2+y^2=r$ is equal to $$4\sum_{d|r} \chi_2(r)$$ where $$\chi_2(x) = \begin{cases} -1 & \quad \text{if } x \text{ is congruent to 3 mod 4}\\ 1 & \quad \text{if } x \text{ is congruent to 1 mod 4}\\ 0 & \quad \text{if } x \text{ is congruent to 0 mod 2} \end{cases}$$
Now let's look at a case for $m\neq 2$. Consider the case $m=4$ and $r=97$. There are $8$ integer pairs that satisfy the equation $x^4+y^4=97$ as seen in the diagram below.
Is there a way to know in general how many solutions there are to $|x|^m+|y|^m=r$?
• Why do you think there must be a solution that takes the form of a divisor sum? The case $m=2$ is special because of its relation to complex multiplication (which has $x^2+y^2$ factoring as $(x+iy)(x-iy)$). I see no reason to believe that this generalizes to higher powers, although it's conceivable there is still an answer to the title question by different methods. – Mario Carneiro Apr 23 '18 at 2:27
• I think you are off by a factor of 4 in your definition of $\chi_2$. – Mario Carneiro Apr 23 '18 at 2:39
• It doesn't look like $\chi_3$ is very nice. Here's a table of the first hundred values: {2, -1, -2, -1, -2, 1, -2, 2, 2, 1, -2, 1, -2, 1, 2, -1, -2, -2, -2, 1, 2, 1, -2, -2, 0, 1, 0, 3, -2, -1, -2, -1, 2, 1, 4, 0, -2, 1, 2, -2, -2, -1, -2, 1, -2, 1, -2, 1, 0, 0, 2, 1, -2, 1, 2, -4, 2, 1, -2, -1, -2, 1, -2, 2, 4, -1, -2, 1, 2, -3, -2, 2, -2, 1, 0, 1, 2, -1, -2, 1, -2, 1, -2, -3, 2, 1, 2, -2, -2, 2, 4, 1, 2, 1, 2, 1, -2, 0, -2, 0} – Mario Carneiro Apr 23 '18 at 2:45
• See also: oeis.org/A175362 ($m=3$) – Mario Carneiro Apr 23 '18 at 3:40
• I used mathematica, not wolfram alpha, and I've had little success running arbitrary mathematica code in alpha without it getting confused by the natural language stuff. Here's a straightforward calculation of those numbers (I didn't keep the original code, but this works too): count[n_] := With[{t = Floor[n^(1/3)]}, Sum[Boole[x^3 + y^3 == n], {x, -t, t}, {y, -t, t}]] chi3[n_] := chi3[n] = count[n] - Sum[chi3[k], {k, Complement[Divisors[n], {n}]}] Table[chi3[n], {n, 100}] – Mario Carneiro Jul 3 '18 at 3:47
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## Algebra 1: Common Core (15th Edition)
The x-intercept is $4$. The y-intercept is $\frac{-8}{5}$
To find the x-intercept and y-intercept of the line, we first need to find the equation of the line. We can use the two points given to formulate the point-slope form. Let's first find the slope: $m=\frac{y_2-y_1}{x_2-x_1}$, where $m$ is the slope and $(x_1,y_1)$ and $(x_2,y_2)$ are points on the line. Let's plug in our two points into this formula: $m=\frac{-2-0.4}{-1-5}=\frac{2}{5}$ Since we have the slope and two points, we can use the point-slope form, which is given by the following formula: $y-y_1=m(x-x_1)$ Let's plug in the slope and a point into this formula: $y-(-2)=\frac{2}{5}(x-(-1))$ $y+2=\frac{2}{5}(x+1)$ This is the point-slope formula of the equation. To find the x-intercept, we set y equal to 0: $0+2=\frac{2}{5}(x+1)$ Use the distributive property on the right side of the equation: $\frac{2}{5}x+\frac{2}{5}=2$ $x=4$ To find the y-intercept, we set x equal to 0: $y+2=\frac{2}{5}(0+1)$ $y+2=\frac{2}{5}$ $y=\frac{-8}{5}$ The x-intercept is $4$. The y-intercept is $\frac{-8}{5}$.
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# ordinary differential equation
mathematics
Also known as: ODE
Related Topics:
differential equation
ordinary differential equation (ODE), in mathematics, an equation relating a function f of one variable to its derivatives. (The adjective ordinary here refers to those differential equations involving one variable, as distinguished from such equations involving partial derivatives of several variables, called partial differential equations.)
The derivative, written f′ or df/dx, of a function f expresses its rate of change at each point—that is, how fast the value of the function increases or decreases as the value of the variable increases or decreases. For the function f = ax + b (representing a straight line), the rate of change is simply its slope, expressed as f′ = a. For other functions, the rate of change varies along the curve of the function, and the precise way of defining and calculating it is the subject of differential calculus. In general, the derivative of a function is again a function, and therefore the derivative of the derivative can also be calculated, (f′)′ or simply f″ or d2f/dx2, and is called the second-order derivative of the original function. Higher-order derivatives can be similarly defined.
More From Britannica
analysis: Ordinary differential equations
The order of a differential equation is defined to be that of the highest order derivative it contains. The degree of a differential equation is defined as the power to which the highest order derivative is raised. The equation (f‴)2 + (f″)4 + f = x is an example of a second-degree, third-order differential equation. A first-degree equation is called linear if the function and all its derivatives occur to the first power and if the coefficient of each derivative in the equation involves only the independent variable x.
Some equations, such as f′ = x2, can be solved by merely recalling which function has a derivative that will satisfy the equation, but in most cases the solution is not obvious by inspection, and the subject of differential equations consists partly of classifying the numerous types of equations that can be solved by various techniques.
The Editors of Encyclopaedia BritannicaThis article was most recently revised and updated by Erik Gregersen.
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# RS Aggarwal Chapter 10 Class 9 Maths Exercises 10.2 (ex 10b) Solutions
RD Sharma Chapter 10 Class 9 Maths Exercise 10.2 Solutions: You have studied many properties of a triangle in Chapters 6 and 7 and you know that on joining three non-collinear points in pairs, the figure so obtained is a triangle. Now, let us mark four points and see what we obtain on joining them in pairs in some order. Although most of the objects we see around are of the shape of a special quadrilateral called a rectangle, we shall study more about quadrilaterals and especially parallelograms because a rectangle is also a parallelogram and all properties of a parallelogram are true for a rectangle as well.
EXERCISE 10B
## Important Definition for RS Aggarwal Chapter 10 Class 9 Maths Ex 10b Solutions
A parallelogram is a type of quadrilateral that contains parallel opposite sides.
Area of parallelogram = Base × HeightArea of Triangle = (1/2)× Base × Height
Quadrilaterals- In this chapter, you will learn about a four-point closed figure. The figure formed by joining four points in order is called a quadrilateral. A quadrilateral has the following properties: Four sides, Four angles, and Four vertices.
The chapter begins by giving details about the Angle Sum Property of a Quadrilateral.
The sum of the angles of a quadrilateral is 360°.
Another Condition for a Quadrilateral to be a Parallelogram is explained. A quadrilateral is a parallelogram if a pair of contrary sides are equal and parallel.
After this, the Mid-factor Theorem is discussed.
The line segment becoming a member of the mid-points of two aspects of a triangle is parallel to the third side.
A line via the midpoint of a triangle parallel to another side bisects the third side.
Two theorems are given in this section. A project for students is also given in which they learn the mid-point theorem.
While the definition states “parallelogram”, it is enough to say, “A quadrilateral is a rhombus if and only if it has 4 congruent sides”, considering any quadrilateral with four congruent aspects is a parallelogram.
Know more here.
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# ISEE Upper Level Math : How to find mode
## Example Questions
### Example Question #11 : Mode
What is the mode of the following set of numbers?
Explanation:
To determine the mode, you need to identify which number appears most frequently in the set of numbers.
In this set of numbers, the number 2 shows up three times, and no other number occurs three or more times. Thus, it is the only mode.
### Example Question #12 : Mode
What is the mode value in the set below?
Explanation:
The first step is to convert the expressions in the set to their simplified forms. This gives us:
While the value of 4 appears twice, given that the value 3 appears 3 times, it is the mode in this set.
### Example Question #13 : Mode
Find the mode of the following data set:
Explanation:
Find the mode of the following data set:
Begin by putting your numbers in increasing order:
We can identify the mode simply by finding the most common term.
In this data set, we have two 67's. There are no other duplicate numbers, so we have our answer: 67
### Example Question #14 : Mode
Find the mode of the following data set:
Explanation:
Find the mode of the following data set:
Let's begin by rearranging our terms from least to greatest:
Now, the mode will be our most common term.
In this case, it is 76, because there are two 76's.
### Example Question #15 : Mode
Find the mode of the following data set:
Explanation:
Find the mode of the following data set:
First, let's put our terms in increasing order:
Now, our mode will just be our most common term. In this case it is 33.
### Example Question #16 : Mode
Find the mode of the following data set:
Explanation:
Find the mode of the following data set:
First, let's put our terms in ascending order.
Next, identify the mode by choosing the most common term.
In this case, we have 2 78's. Therefore, our mode is 78.
### Example Question #11 : Mode
Use the following data set to answer the question:
Find the mode.
Explanation:
To find the mode of a data set, we will find the number that occurs most often.
So, given the data set
we can see the number 5 occurs most often (it occurs 4 times.)
Therefore, the mode of the data set is 5.
### Example Question #18 : Mode
Use the following data set to answer the question:
Find the mode.
Explanation:
To find the mode of a data set, we will find the number that occurs most often.
So, given the data set
we can see the number that occurs most often is 8 (it occurs 4 times).
Therefore, the mode of the data set is 8.
### Example Question #11 : How To Find Mode
Find the mode of the following data set:
Explanation:
Find the mode of the following data set:
To find the mode, first put the numbers in increasing order
Next, find the mode by simply choosing the most common number.
In this case, 12 appears three times, so it is our most common term.
### Example Question #111 : Data Analysis
Use the following data set to answer the question:
Find the mode.
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# Variable Acceleration
Variable acceleration is a situation where there is a difference in the average acceleration within different points along the path of an object in motion.
#### Create learning materials about Variable Acceleration with our free learning app!
• Flashcards, notes, mock-exams and more
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Differences could be either in magnitude, direction, or in both magnitude and direction. Variable acceleration occurs when changes in the velocity of an object is not the same at equal time intervals. Thus, it is dependent on both velocity and time.
You can also learn about Constant Acceleration.
## Examples of variable acceleration
Imagine a policeman is chasing a criminal and stumbles on a crowd. The policeman would have to reduce his pace and eventually increase it again once he gets to a less occupied place. The same occurs when a car is accelerating on the highway and meets traffic. The car would accelerate very little in traffic, and when the road becomes free its acceleration increases.
## Tools needed in variable acceleration
In order to solve variable acceleration questions, you must have a background knowledge of calculus. This is essential because in order to derive velocity when displacement has been given, you are expected to differentiate the displacement. Meanwhile, to determine your displacement with a given velocity, you are expected to integrate the velocity. This process is repeated between velocity and acceleration and vice versa.
A circular toolbox for variable acceleration problems, Njoku - StudySmart Originals
## The function of time in variable acceleration
Remember that variable acceleration is dependent on time because a change in acceleration occurs within time intervals. However, we shall begin by solving examples that expresses velocity and displacement as functions of time.
The displacement s of a particle in motion on a straight line from a point O at time t seconds is given as: $$s = 4t^3 - 9t$$ where 0 <t. Find:
a. The displacement when t = 3
b. The time it takes the particle to return to point O.
Solution:
a. In order to find the displacement s, just substitute the value of t = 3 in the equation
$$s = 4t^3 - 9t$$
$$s = 4 (3^3) - 9(3) = 108 - 27 = 81 \space m$$
b. To find the time taken to return to point O, it means the displacement is zero, and would be put in the equation.
Thus:
$$0 = 4t^3 - 9t$$
Factorise
$$0 = t(4t^2 - 9)$$
Let us factories $$4t^2 - 9$$ separately
$$4t^2 - 9 = (2t -3)(2t+3)$$
Thus:
$$0 = t (2t - 3) (2t + 3)$$
$$t = 0, 2t - 3 = 0$$ or $$2t + 3 = 0$$
$$t = 0, \frac{3}{2} \text{ or } \frac{-3}{2}$$
Recall that 0 <t
That means the answer is $$\frac{3}{2}$$ seconds.
The motion of a toy car on a straight track has been modelled to follow the equation $$s = -t^3 + 4t^2$$. If this toy leaves at time t = 0 and returns at the start of the track, prove the restriction 0 ≤ t ≤ 4.
Solution:
From the start of the motion,
s ≥ 0
So,
$$-t^3 + 4t^2 \geq 0, \quad 4t^2 - t^3 \geq 0, \quad t^2(4-t) \geq 0, \qquad 4-t \geq 0$$
$$t^2 \geq 0$$ and $$4 \geq t$$
Since time cannot be in negative. We can consider only t ≥ 0 in this case.
thus; t ≤ 4
t ≥ 0 and t ≤ 4
which proves the restriction 0 ≤ t ≤ 4.
The relationship between the velocity and time of an object moving in a straight line is given by the expression: $$v = 2t^2 - 16t + 24$$ for t ≥ 0
Find:
1. The initial velocity.
2. The time the object is instantaneously at rest.
3. The time the velocity is at 64 m/s.
4. The greatest speed with the interval 0 ≤ t ≤ 5.
Solution
a. At the initial velocity the time is zero because the motion just began.
t = 0
substitute the value of t in the equation to find the velocity
$$v = 2t^2 - 16t + 24$$
v = 24 m / s.
b. At the instantaneous velocity at rest, v = 0
Substitute that in the equation to find the time.
$$0 = 2t^2 - 16t + 24 = 2(t^2-8t+12)$$
Divide both sides by 2
$$0 = t^2 - 8t + 12$$
Factorise
$$0 = (t - 6) (t - 2)$$
t = 6 or 2
Thus, the object comes to an instantaneous rest at 2 seconds and 6 seconds.
c. At the velocity = 64 m / s, the time is:
$$64 = 2t^2 - 16t+24 = 2 (t^2 - 8t + 12)$$
Divide both sides by 2
$$32 = t^2 - 8t + 12$$
$$0 = t^2-8t + 12 -32 = t^2 - 8t - 20$$
Factorise
$$0 = (t - 10) (t + 2)$$
t = 10 or -2
Thus, the object moves at 64 m/s after 10 seconds.
d. The greatest speed between 0 ≤ t ≤ 5 is the highest speed attained by the object within t = 0 and t = 5.
To derive this, a graph of the equation could be plotted with the values between 0 to 5, with v in the y-axis and t in the x-axis.
Initially, the value for t = 0 has been derived in question a) as 24 m / s. Use the same approach and fill in the table for values 1 to 5.
$$x = t$$ 0 1 2 3 4 5 $$y = v = 2t^2 - 16t+24$$ 24 10 0 -6 -8 -6
The graph below reveals the max velocity. This is seen in the table and confirmed by the graph that the highest velocity is 24 m / s. This is obtained at time t = 0.
A velocity-time graph showing the max velocity, Njoku - StudySmarter Originals
## Applications of differentiation in variable acceleration
Velocity is defined as the change in displacement with time. We have explained the function of time in displacement and velocity, thus, $$v = \frac{ds}{dt}$$.
In a similar way, acceleration is defined as the change in velocity with time. If velocity is overexpressed as a function of t to derive the acceleration, then, $$a = \frac{dv}{dt} = \frac{d^2 s}{dt^2}$$.
This knowledge of differential calculus would be needed when determining the velocity if the displacement is expressed as a function of time. In the same way, the acceleration can be derived when the velocity is expressed as a function of time. Let's look at some examples to make this clearer:
A tricycle is moving horizontally. The displacement q from rest O at time t is given as:
$$q = t^4-32t+12$$
a. Calculate the velocity of the tricycle when t is 4 seconds.
b. Find the time it attains an instantaneous velocity.
c. Calculate the acceleration when t is 1.5 seconds.
Solution:
a. q is the displacement given as
$$q = t^4-32t+12$$
Remember that you differentiate the displacement to get the velocity.
$$v = \frac{dq}{dt} = \frac{d(q = t^4 -32t+12)}{dt}$$
$$v = 4t^3 -32$$
Since the t function of the velocity has been derived, substitute the value of t = 4 in the equation.
$$v = 4(4^3)-32 = 256-32 = 224$$
Therefore the velocity of the tricycle is 224 m/s
b. To find the time at which the tricycle experiences instantaneous velocity, always remember that the object is momentarily at rest. It means that velocity is zero at that instant.
Input the value of v = 0 in the equation $$v = 4t^3 - 32$$ to find time t.
$$0 = 4t^3 - 32$$
$$4t^3 = 32$$
Divide both sides by 4
$$t^3= 8$$
Find the cube root of both sides: t = 2
Therefore the tricycle experiences instantaneous velocity after 2 s.
Since $$v = 4t^3 - 32$$
$$a = \frac{dv}{dt} = \frac{d(v=4t^3-32)}{dt}$$
$$a = 12t^2$$
Now you are asked to find the acceleration. Just substitute the value of t = 1.5 in the acceleration equation.
$$a = 12(1.5^2); \quad a = 27 ms^{-2}$$
Therefore the acceleration at time t = 1.5 s is 27 ms-2
## Using variable acceleration in minimum and maximum instances
The idea of variable acceleration also has profitable applications in finding the minimum and maximum values of displacement, velocity and acceleration.
A woman has a spring that leaves her hand at time t = 0 seconds, and moves vertically in a straight line before returning back to her hand. The distance y between the spring and her hand after t seconds is given by:
$$y = -0.2t^3 + 0.4t^2 + 0.6 t$$, 0 ≤ t ≤ 3
a. Prove the restriction 0 ≤ t ≤ 3.
b. Calculate the maximum distance between the woman's hand and the spring.
Solution:
a. $$y = -0.2t^3 + 0.4t^2 + 0.6 t$$
Factorise:
$$y = -0.2t(t^2 - 2t - 3)$$
Factorise further:
$$y = -0.2t (t - 3) (t + 1)$$
when y = 0, then
$$-0.2t = 0, \space t - 3 = 0 \text{ or } t + 1 = 0$$
t = 0, 3 or -1
There are no negative values for time, so, t = 0 or 3. This justifies the restriction of t, 0 ≤ t ≤ 3.
b. Remember that when a particle reaches its maximum displacement, it becomes momentarily at rest. We say its velocity is instantaneously at rest and v = 0.
Thus, to find the max displacement, we need to know the time the object becomes instantaneously at rest. To do this, we need to find the t function of v.
$$y = -0.2t^3+0.4t^2+0.6t$$
$$v = \frac{dy}{dt} = \frac{d(y = -0.2t^3 + 0.4 t^2 + 0.6t)}{dt}$$
$$v = -0.6t^2 + 0.8t + 0.6$$
Now we have the t function of our velocity in the equation, let us find t when v is instantaneously at rest.
v = 0
$$0 = -0.6t^2 + 0.8t + 0.6$$
Factorise
$$0 = -0.2(3t^2-4t-3)$$
Divide both sides by -0.2
$$0 = 3t^2-4t-3$$
Using the formula for quadratic equations $$\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$
Where a = 3, b = -4 and c = -3
After substituting the values and solving the equation,
t = 1.8685 or -0.5351.
Remember no negative value of t is valid, thus, t = 1.8685
This means the distance y is maximum after 1.8685 seconds.
Substitute the value of t to find ymax.
$$y = -0.2t^3 + 0.4t^2 + 0.6t$$
$$y_{max} = -0.2(1.8685)^3 + 0.4(1.8685)^2 + 0.6(1.8685) = 1.2129$$
The maximum distance between the woman's hand and the spring is 1.21 meters.
## Using integration in variable acceleration problems
Initially, we have seen how differentiation is used to find the velocity from a given displacement with respect to time. We have also seen how this process is used to find the acceleration when the velocity is given as a function of time. To carry out the reverse of these situations such as from velocity to displacement as well as acceleration to velocity, integration is used.
You know that:
$$v = \frac{ds}{dt}$$ other $$a = \frac{dv}{dt}$$
Then:
$$s = \int{v \space dt}$$ other $$v = \int{a \space dt}$$
$$(t-5)ms^{-2}$$ is the acceleration of a basketball bounced by a teenager. The ball starts with a velocity of 8 m/s. Determine the time (s) the basketball is instantaneously at rest.
Solution:
An illustration of the basketball instantaneously at rest, Njoku - StudySmarter Originals
$$a = t - 5$$
$$v = \int{a \space dt}$$
$$v = \int{(t-5) dt} = \frac{t^2}{2} - 5t +c$$ do not forget to add the constant c after each integration.
We were told that the ball began with a velocity of 8 m / s, which means that at time t = 0, v = 8
Thus substitute the value of t and v in the equation $$8 = \frac{0^2}{2} - 5(0) + c$$
c = 8
This means the t function of the velocity is:
$$v = \frac{t^2}{2} - 5t + 8$$
Now we have a complete equation for the velocity, we can determine the time (s) the ball is instantaneously at rest.
v = 0 because the object is momentarily at rest.
$$v = \frac{t^2}{2} - 5t + 8$$
$$0 = \frac{t^2}{2} - 5t +8$$
Multiply the equation by 2 to get rid of the fraction
$$0 = t^2 - 10t + 16$$
Factorise
$$(t - 8) (t - 2) = 0$$
t = 8 or 2
Thus, the basketball is instantaneously at rest in 2 and 8 seconds.
## Variable Acceleration - Key takeaways
• Variable acceleration takes place when velocity changes are not equal for the same time intervals. Thus, the acceleration is not constant.
• Differential and Integral Calculus are tools used to solve problems in variable acceleration.
• You differentiate to find velocity when the displacement has been expressed as a function of time. The same applies to acceleration when velocity is given.
• You integrate to find displacement when the velocity has been expressed as a function of time. The same applies to velocity when acceleration is given.
• The velocity of an object instantaneously at rest is 0.
• You can find the maximum displacement by differentiating the displacement to find the t function of the velocity. Then take the time at which the velocity is instantaneously at rest and substitute it in the displacement equation.
#### Flashcards in Variable Acceleration 5
###### Learn with 5 Variable Acceleration flashcards in the free StudySmarter app
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What two variables is acceleration dependent on?
Acceleration is dependent on velocity and time.
How do you calculate variable acceleration?
You calculate variable acceleration by differentiating the velocity which has been expressed as a function of time. Then, you substitute the value of t into the equation.
Is acceleration an independent variable?
No, acceleration is not independent since it depends on the velocity and time intervals.
What is a variable acceleration?
Variable acceleration takes place when velocity changes are not equal for the same time intervals.
How do you find max displacement variable acceleration?
To find the max displacement, you have to determine the time taken for the body to be instantaneously at rest. Afterwards, input your value of t into the displacement equation.
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The teaching of number operation Mixed Operations
MIXED OPERATIONS
In this chapter,
34 8 2 = ?
Introduction
Examples of mixed operation situation
The conventional in mixed operations
- Multiplication and Division
- Subtraction and Multiplication
- Subtraction and Division
- Bracket.
Summary
LEARNING OUTCOMES
At the end of the topic, the students are able to
to plan teaching and learning activities to
- state basic facts for the four numbers (mixed) operations spontaneously
- perform the four numbers (mixed) operations in the standard form
- solve daily problems involving the four number (mixed) operations.
- Perform micro/macro teaching and make reflection.
INTRODUCTION
CONCEPT
In this chapter, we will see questions like these:
3 + 7 x 4 = 12 - 6 2 =
Can you see that these questions involved more than one mathematical operations. Yes,
they are known as mixed operations. Mixed operations are the combination of more than one
basic operations ( +, - , x and ) in one mathematical sentence.
In the example of 3 + 7 x 4 = , there may be two possible ways to solve it.
First, you add 3 to 7. You will get the total of 10. Then 10 is multiplied to 4 giving the answer
40.
Or
Multiply 7 to 4, then add 3 to the product, that is 28, giving you the answer 34
Which one is correct?
This is the focus of our discussion in this chapter which are mostly based on the
convertion that had been agreed by the mathematicians.
1.1 Examples of Mixed Operational Situations;
Situations that involved mixed operations
Can be introduced to students to give
them the ideas of what mixed operations
Here are some situations that involved
Activity 1.1 To introduce an example of
mixed Operation
Teacher prepares a series of pictures as
in Figure 3.2.75
For each picture,
(a) tell story of what has happened
in the picture
(b) write it in a mathematical
sentence
For picture series (Figure 3.2.75)
Students are to refer to the pictures
A,B, and C in the Figure 3.2.75
Students are guided in writing a story
for each picture and then translate it
into a mathematical sentence.
What can you see in picture A?
( 2 group of birds,containing 4 birds
and 3 birds respectively are flying
to rest on the electrical wire).
What can you see in picture B?
(The birds were sitting on the electrical
wire).
What can you see in picture C?
(Two of the birds were flying away
from the electrical wire).
Then the students are asked to write
one mathematical sentence for the
above situation based on the pictures
in diagram 1.1
Figure 3.2.75
4 + 3 2 = ?
Diagram 1.1
For picture series Figure 3.2.76
Students are to refer to pictures
A,B and C in Figure 3.2.76
For each picture,
(c) tell story of what has happened
in the picture
(d) write it in a mathematical sentence
Students are guided in writing a story
for each picture and then translate it
into a mathematical sentence.
What can you see in picture A?
What can you see in picture B?
What can you see in picture C?
Then the students are asked to write
one mathematical sentence for the
above situation based on the pictures
in Figure 3.2.76
6 + 5 - 3 = ?
Figure 3.2.76
For picture series Figure 3.2.77
Students are to refer to pictures
A,B C and D in Figure 3.2.77
As before,students are guided
in writing a story for each picture
and then translate it into
a mathematical sentence.
What can you see in picture A?
What can you see in picture B?
What can you see in picture C and D?
Then the students are asked to write
one mathematical sentence for the
above situation based on the pictures
in Figure 3.2.77
7 - 3 + 5 = ?
Diagram 1.3
Figure 3.4.77
For picture series (Figure 3.4.78)
Students are to refer to pictures
A,B C and D in the (Figure 3.4.78)
As before, students are guided
in writing a story for each picture
and then translate it into
a mathematical sentence.
What can you see in picture A?
What can you see in picture B?
What can you see in picture C and D ?
Then the students are asked to write
one mathematical sentence for the
above situation based on the pictures
in (Figure 3.4.78)
8 5 = _____ + ____ = 9
Diagram 1.4
Figure 3.4.78
For each pictures series, we can write one mathematical equation involving mixed
operation.
EXTRA ACTIVITIES
In reverse, try to make a story if you are given a mathematical equation
Examples:
(a) 7 + 5 6 = ?
(b) 13 4 + 7 = ?
For each mathematical equation that involved mixed operations, we can make a story out of
it.
1.2 Conventional method in Mixed Operations
All mathematician had came to cosensus to use one fixed procedure on
the steps in solving the mixed operations
(a) For mixed operations that involved +,and , do it from left
to right.
Examples:
(i) 8 + 4 - 5 = 12 5
= 7
(ii) 9 5 + 2 = 4 + 2
= 6
(iii) 16 9 2 = 7 2
= 5
( b) For mixed operations that involved x,and , do it from left
to right.
(i) 3 x 8 4 = 24 4
= 6
(ii) 18 3 x 2 = 6 x 2
= 12
( c) For mixed operations that involved +,and x
The following situation can be used
To solve the problem, we have to solve for the x first, then followed by the operation + ,
no matter whether we use equation (i) or (ii).
So for the mixed operation that involved + and x, solve for x (multiplication) first then
Example 2:
(i) 5 x 6 + 4 = 30 + 4
= 34
(ii) 5 + 6 x 4 = 5 + 24
= 29
(d) For the mixed operation that involved and x, first solve for x
(multiplication) then followed by the
(subtraction).
(i) 7 x 8 6 = 56 6
= 50
(ii) 12 6 x 2 = 12 8
= 4
(e) For mixed operations that involved +,and , the following situation can
be used
Alina has 7 marbles. She received another 8 boxes of marble
containing 3 marbles in each box. How many marbles does Alina
has now?
7 + 8 x 3 = _________ ....... (i)
or
8 x 3 + 7 = _________ ........(ii)
To solve the problem, we have to solve for the first, then followed by the operation + ,
no matter whether we use equation (i) or (ii).
So for the mixed operation that involved + and , solve for (division) first then
Examples:
(j) 24 6 + 2 = 4 + 2
= 6
(ii) 24 + 6 2 = 24 + 3
= 27
(g) For the mixed operation that involved and , first solve for
(division) then followed by the (subtraction).
Examples:
(i) 48 8 4 = 6 4
= 2
(ii) 48 8 4 = 48 2
= 46
(h) For the mixed operation that involved +, , x and ,
Solve for x and , from left to right
Solve for + and , from left to right
Examples:
(i) 7 x 9 + 6 4 2 = 63 + 6 2
Azmi and Anwar are the school badminton team. Before the final
game, Azmi has RM4.00. The couple has won the final game and
received the cash prize of RM26.00.The money was divided
equally between them.How much money that Azmi has now?
4 + 26 2 = _________ ....... (i)
or
26 2 + 4 = _________ ........(ii)
= 69 2
= 67
(ii) 27 3 2 x 4 + 3 = 9 8 + 3
= 1 + 3
= 4
(iii) 7 7 + 9 3 x 3 = 1 + 9 9
= 10 9
= 1
(i) For the mixed operation that involved bracket, solve using the following steps:
Bracket ( )
Solve for x and , from left to right
Solve for + and , from left to right
Examples:
(i) 15 5 ( 9 7) = 15 5 2
= 3 2
= 1
(ii) 12 ( 6 + 4) 2 = 12 10 2
= 12 5
= 7
(iii) (9 3 x 2) x 7 = ( 9 6 ) x 7
= 3 x 7
= 21
1.3 Sample exercises on mixed operation
Example 1:
Complete the following mathematical equation by filling in the basic
operation in the spaces provided.
(1) 4 + 8 7 = 15
(2) ( 9 x 6 ) 4 = 58
(3) (10 5) + 14 = 16
(4) 8 4 2 = 10
(5) ( 4 3 ) 9 = 3
(6) ( 8 3 ) 7 = 35
(7) 9 2 4 = 72
(8) 24 4 2 = 4
SUMMARY
1. The correct procedure to solve mixed operations without bracket
Solve for x and , from left to right
Then solve for + and , from left to right
2. The correct procedure to solve mixed operations with bracket
First solve the operation in the bracket
If there are some other brackets in the bracket, start with the innest bracket.
Then solve for x and , from left to right
Lastly,solve for + and , from left to right
3. The following Mnemonik can be used to recall on the operation procedure
BODMAS (Bracket Of Division Multiplication Addition Subtraction)
4. The bracket can be in the form of () or [ ] or { }
Glossory
Subtraction
Minuend
Subtrahend
Remainder
Difference
Take away
Part-whole
Partitioned
Comparative Subtraction
Subtraction fact
Undoing
Subtraction matrix
Subtraction Algorithd
No Regouping
Regrouping
Decomposition
Associative property
Compensation
Counting on
Doubles
Indentity
Make-a-ten
Near doubles
One-more
Sharing numbers
Sum families
Excess of nines
Upper and lower boundaries
Basic facts algorithm
Counting algorithm
No-regrouping
Partial sums
Regrouping
Scratch
Naming a set
Union of disjoint sets
Associative property
Commutative property
Identity property
Whole number multiplication
Whole number division
Whole number subtraction
Basic fact strategies
Division
Multiplication
subtraction
Bibliography
Bahagian Pendidikan Guru Kementerian Pendidikan Malaysia (1998). Konsep dan
Aktiviti Pengajaran Pembelajaran Matematik: Nombor Bulat untuk sekolah rendah .
Kuala Lumpur : Dewan Bahasa Dan Pustaka.
C. Alan. Riedesel. (1990). Teaching elementary School mathematics. PrenticeHall
Randall,J. Souviney (1989) Learning To Teach Mathematics. Merrill Publishing
Company
Schminke, C.W. (1981) Math Activities for Child involvement.London: Allyn and Bacon
Thomas R. Post. (1985). Teaching mathematics in Grades k-8. Allyn and Bacon, Inc
Willians, Elizabeth & Shuard, Hilary ( 1979). Primary Mathematics Today.
ELBS Edition, London: Longman Group.
|
# What is mean data?
## What is the mean of the data?
The mean is the same as the average value of a data set and is found using a calculation. Add up all of the numbers and divide by the number of numbers in the data set. The median is the central number of a data set. … Count how many times each number occurs in the data set.
## What is the mean of data in math?
Mean is just another name for average. To find the mean of a data set, add all the values together and divide by the number of values in the set. The result is your mean!
## What does data example mean?
Data is defined as facts or figures, or information that's stored in or used by a computer. An example of data is information collected for a research paper. An example of data is an email. … Statistics or other information represented in a form suitable for processing by computer.
## What do you mean by data in statistics?
data are individual pieces of factual information recorded and used for the purpose of analysis. It is the raw information from which statistics are created. Statistics are the results of data analysis – its interpretation and presentation. … Often these types of statistics are referred to as 'statistical data'.
## What is data in maths class 7?
Data is a collection of numbers gathered to give some information. It could be the marks of students, shopping done in an year, attendance of students, weight of some individuals and so on. … The collection, recording and presentation of data help us organize our experiences and draw inferences from them.
## What is data handling in maths class 6?
CBSE Class 6 Maths Notes Chapter 9 Data Handling Data is a collection of numbers gathered to give some information. Recording Data. Sometimes some information is required very quickly. It is possible only when we adopt some suitable system of collecting data. Raghav.
## What is mean in data handling?
The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest. The mode is the number that occurs most often in a data set.
## What is data in statistics class 11?
Data is a collection of information gathered by observations, measurements, research or analysis.
|
# Question #db8e2
Jul 10, 2016
For increasing you have to find out where the first derivative is positive, for decreasing where first derivative is negative
#### Explanation:
Let's calculate the first derivative of the function $f \left(x\right)$:
$\left[\left(x + 2\right) \cdot {e}^{-} x\right] '$ = $\left(x + 2\right) ' \cdot {e}^{-} x + \left(x + 2\right) \cdot \left({e}^{-} x\right) ' = 1 \cdot {e}^{-} x + \left(x + 2\right) \left(- {e}^{-} x\right) = {e}^{-} x - x {e}^{-} x - 2 {e}^{-} x = - x {e}^{-} x - {e}^{-} x = {e}^{-} x \left(- x - 1\right)$
Now, regardless of the value of $x$, ${e}^{-} x$ is always positive, so the sign of the first derivative depends on whether $\left(- x - 1\right)$ is positive or negative. That is:
• If $\left(- x - 1\right) > 0$, the first derivative is positive and the function is increasing. Thus, if $- 1 > x$ the function is increasing
• Conversely, if $\left(- x - 1\right) < 0$, the first derivative is negative and the function is decreasing. Thus, if $- 1 < x$ the function is decreasing
|
# Search by Topic
#### Resources tagged with Working systematically similar to Children's Mathematical Writing:
Filter by: Content type:
Stage:
Challenge level:
### There are 321 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Round the Dice Decimals 1
##### Stage: 2 Challenge Level:
Use two dice to generate two numbers with one decimal place. What happens when you round these numbers to the nearest whole number?
### Round the Three Dice
##### Stage: 2 Challenge Level:
What happens when you round these three-digit numbers to the nearest 100?
### Round the Two Dice
##### Stage: 1 Challenge Level:
This activity focuses on rounding to the nearest 10.
### Round the Dice Decimals 2
##### Stage: 2 Challenge Level:
What happens when you round these numbers to the nearest whole number?
### Two Spinners
##### Stage: 1 Challenge Level:
What two-digit numbers can you make with these two dice? What can't you make?
##### Stage: 1 and 2 Challenge Level:
How could you put these three beads into bags? How many different ways can you do it? How could you record what you've done?
### Dice in a Corner
##### Stage: 2 Challenge Level:
How could you arrange at least two dice in a stack so that the total of the visible spots is 18?
### Multiply Multiples 3
##### Stage: 2 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Multiply Multiples 2
##### Stage: 2 Challenge Level:
Can you work out some different ways to balance this equation?
### Multiply Multiples 1
##### Stage: 2 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### What Could it Be?
##### Stage: 1 Challenge Level:
In this calculation, the box represents a missing digit. What could the digit be? What would the solution be in each case?
### Spell by Numbers
##### Stage: 2 Challenge Level:
Can you substitute numbers for the letters in these sums?
### Jigsaw Pieces
##### Stage: 1 Challenge Level:
Imagine that the puzzle pieces of a jigsaw are roughly a rectangular shape and all the same size. How many different puzzle pieces could there be?
### Magic Vs
##### Stage: 2 Challenge Level:
Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total?
### Dodecamagic
##### Stage: 2 Challenge Level:
Here you see the front and back views of a dodecahedron. Each vertex has been numbered so that the numbers around each pentagonal face add up to 65. Can you find all the missing numbers?
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Two Egg Timers
##### Stage: 2 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### Name the Children
##### Stage: 1 Challenge Level:
Can you find out in which order the children are standing in this line?
### A Bag of Marbles
##### Stage: 1 Challenge Level:
Use the information to describe these marbles. What colours must be on marbles that sparkle when rolling but are dark inside?
### Junior Frogs
##### Stage: 1 and 2 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### Trebling
##### Stage: 2 Challenge Level:
Can you replace the letters with numbers? Is there only one solution in each case?
### Ordered Ways of Working Lower Primary
##### Stage: 1 Challenge Level:
These activities lend themselves to systematic working in the sense that it helps to have an ordered approach.
### ABC
##### Stage: 2 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### Broken Toaster
##### Stage: 2 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### Jumping Squares
##### Stage: 1 Challenge Level:
In this problem it is not the squares that jump, you do the jumping! The idea is to go round the track in as few jumps as possible.
### The Add and Take-away Path
##### Stage: 1 Challenge Level:
Two children made up a game as they walked along the garden paths. Can you find out their scores? Can you find some paths of your own?
### Knight's Swap
##### Stage: 2 Challenge Level:
Swap the stars with the moons, using only knights' moves (as on a chess board). What is the smallest number of moves possible?
### Late Again
##### Stage: 1 Challenge Level:
Moira is late for school. What is the shortest route she can take from the school gates to the entrance?
### Button-up
##### Stage: 1 Challenge Level:
My coat has three buttons. How many ways can you find to do up all the buttons?
### Button-up Some More
##### Stage: 2 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### Centred Squares
##### Stage: 2 Challenge Level:
This challenge, written for the Young Mathematicians' Award, invites you to explore 'centred squares'.
### Dart Target
##### Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards.
### Jumping Cricket
##### Stage: 1 Challenge Level:
El Crico the cricket has to cross a square patio to get home. He can jump the length of one tile, two tiles and three tiles. Can you find a path that would get El Crico home in three jumps?
##### Stage: 2 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Mixed-up Socks
##### Stage: 1 Challenge Level:
Start with three pairs of socks. Now mix them up so that no mismatched pair is the same as another mismatched pair. Is there more than one way to do it?
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
##### Stage: 1 Challenge Level:
If you put three beads onto a tens/ones abacus you could make the numbers 3, 30, 12 or 21. What numbers can be made with six beads?
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Whose Sandwich?
##### Stage: 1 Challenge Level:
Chandra, Jane, Terry and Harry ordered their lunches from the sandwich shop. Use the information below to find out who ordered each sandwich.
### Briefcase Lock
##### Stage: 1 Challenge Level:
My briefcase has a three-number combination lock, but I have forgotten the combination. I remember that there's a 3, a 5 and an 8. How many possible combinations are there to try?
### One of Thirty-six
##### Stage: 1 Challenge Level:
Can you find the chosen number from the grid using the clues?
### Six Is the Sum
##### Stage: 2 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### Mrs Beeswax
##### Stage: 1 Challenge Level:
In how many ways could Mrs Beeswax put ten coins into her three puddings so that each pudding ended up with at least two coins?
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Pouring the Punch Drink
##### Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### The School Trip
##### Stage: 1 Challenge Level:
Lorenzie was packing his bag for a school trip. He packed four shirts and three pairs of pants. "I will be able to have a different outfit each day", he said. How many days will Lorenzie be away?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
|
# Executive Assessment: Quant Strategies for Faster Solutions – Part 1
by on October 27th, 2017
The Executive Assessment (EA) shares a lot of roots with the GMAT, GMAC’s flagship graduate business school exam. The Quant section covers almost all of the same material and uses the same question types, and the Integrated Reasoning section is identical.
Luckily, we get to use all of the same test-taking techniques that can make the GMAT easier to take. (Not easy…but easier!) We’ll explore the major techniques in this series.
Give yourself ~2 minutes to try the below problem and then we’ll talk. All problems in this series are from the free problem sets that appear on the official Executive Assessment website.
Note: This is a Data Sufficiency (DS) problem. If you’ve never seen anything like this before—and are wondering, for example, where the answer choices are?—then follow this link first to a GMAT article explaining how DS works. DS is exactly the same on the GMAT and the EA.
“Is the integer n odd?
“(1) n is divisible by 3.
“(2) n is divisible by 5.”
Got an answer? Even if you’re not sure, guess—that’s what you want to do on the real test, too, so practice that now (even if your practice consists of saying “I have no idea, so I’m randomly picking B!”).
Glance: This is a DS problem.
Read: They told us that n is an integer and they’re asking whether it’s odd. This is a Yes/No question (as opposed to a Value question). We don’t need to figure out what the value of n is, just whether it is odd.
Pause for a second: What’s the difference between odd and even, mathematically speaking?
If you divide an integer by 2 and get another integer, then that starting number was even. But if you divide the integer by 2 and get a decimal, then that starting number was odd.
Jot:
Time to Reflect. The statements look about equally hard, so just start with the first one. What does divisible by mean again?
If something is divisible by 3, then you get an integer when you divide by 3. Okay. I’m noticing that this question keeps talking about n but never provides a real value for n anywhere. It just keeps giving characteristics of n (or asking about characteristics of n).
That’s a great clue that I should Test Cases on this question. Testing Cases is a super-useful technique that you will use repeatedly on the EA. When the question is abstract, as in this case, you can just try real numbers to see what happens in the problem.
Let’s dive into this to see how it Works! You’re going to be working under Statement (1) on your scratch paper. Set up a little table that lists out the steps that you need to do.
The first rule of Testing Cases is to choose a value that makes the statement—and any other facts in the problem—true. Also, you want to start from a fact, not from the question itself.
The question stem gives one fact (n is an integer) and statement (1) gives another (that n / 3 is an integer). You can start with whichever one you like. Most people will probably want to start with n—it’s easier.
Set up your table. The first two columns are n and n / 3. I’ll explain the V thing in a minute. Finally, put a column for the question asked.
Okay, let’s pick a value for n. It needs to be an integer, so start with a small positive integer. Try 1. What happens?
If n is 1, then n / 3 is 1/3. In that third column, V stands for Valid. Pause to make sure that everything you’ve done so far fits with all of the facts given in the problem.
n is an integer? Check.
n / 3 is an integer? Uh-oh … it’s not.
What does this mean? You have to toss out this case. You’re only allowed to try cases that fit all of the facts in the problem—so this is one of the skills to practice to get better at Testing Cases.
Start again. What kind of number do you need to make both of the first two columns work?
In order to get the second column to work, n has to be a multiple of 3.
Great! Now, everything is valid and we can get to the real question: Is n odd? In this case, yes; the number 3 is odd.
Will it always be odd, no matter what value you pick? We don’t know yet—we have to test some more cases.
You’ve already found one Yes answer, so the goal now is to see whether you can find a No answer. This is called Proving Insufficiency. If you can find one Yes answer and one No answer, using Valid numbers, then you have proved that the statement is not sufficient to answer the question.
Look back at how the math worked. Can you think of something to try that might give you a No answer?
In this problem, what does a No answer actually mean? The number n has to be an integer no matter what, so there are two possible classifications: odd or even. If integer n is not odd, then it has to be even. Can you get it to be even, given the restrictions that you were given?
Sure! Try this:
Now, let’s take a look at statement (2). It uses a different number, but the set-up is the same, so test cases again. (But be careful: You are only testing the second statement now. Ignore the first statement completely.)
Follow the process yourself before you look at my work below.
Great! We can prove this one not sufficient, too. Cross off answer (B) in your answer grid.
Now, try the two statements together. You can only pick values that work for both statements. That could be tricky, so take a look back at your prior work.
For statement 1, the values for n were multiples of 3. In statement 2, n was multiples of 5.
If you’re going to use the two statements together, then, the values of n are going to have to be multiples of 3 × 5 = 15.
If you’ve gotten this far, you may also feel comfortable enough with the process to just think it through. For example:
n = 15
Valid? It is an integer. It is divisible by 3. It is divisible by 5. Valid!
Is it odd? Yes.
n = 30
Valid? It is an integer. It is divisible by 3. It is divisible by 5. Valid!
Is it odd? No!
Since you once again got Yes and No answers, using the two statements together is still not sufficient to answer the question. Cross off (C); only one answer is left!
|
# All Divisors of 8
The divisors of 8 are those numbers that completely divide 8 with the remainder zero. In this section, we will discuss about divisors of 8.
## Highlights of Divisors of 8
• Divisors of 8: 1, 2, 4 and 8
• Negative divisors of 8: -1, -2, -4 and -8
• Prime divisors of 8: 2
• Number of divisors of 8: 4
• Sum of divisors of 8: 15
• Product of divisors of 8: 82
## What are Divisors of 8
A number n is a divisor of 8 if $\dfrac{8}{n}$ is an integer. Note that if 8/n=m is an integer, then both m and n will be the divisors of 8.
We have:
No numbers other than 1, 2, 4, and 8 can divide 8. So we conclude that
Thus, the total number of divisors of 8 is four.
## Negative Divisors of 8
We know that if m is a divisor of a number, then -m is also a divisor of that number.
As the divisors of 8 are 1, 2, 4, and 8, we can say that the negative divisors of 8 are -1, -2, -4, and –8.
## Prime Divisors of 8
The divisors of 8 are 1, 2, 4, and 8. Among these numbers, only 2 is a prime number. So we obtain that:
The only prime divisor of 8 is 2.
Video solution of Divisors of 8:
## Sum, Product & Number of Divisors of 8
The prime factorization of 8 is given below.
8 = 23
(i) By the number of divisors formula, we have that the number of divisors of 8 is
=(3+1)=4.
(ii) By the sum of divisors formula, we have that the sum of the divisors of 8 is
$=\dfrac{2^4-1}{2-1}$
$=\dfrac{16-1}{1}$
$=15$
(iii) By the product of divisors formula, we have that the product of the divisors of 8 is
=8(Number of divisors of 8)/2
=84/2
=82
## Question Answer on Divisors of 8
Question 1: Is 3 a divisor of 8?
|
Stage 2 whole numbers
Strategies
Students can:
• use place value to read, represent and order numbers up to four digits
• record numbers using expanded notation
Activities to support the strategies
Students in Stage 1 and 2 need to develop an understanding of place value. For example, in the number 3450, the ‘four’ represents four hundred. Students need to understand how a number is constructed and the value each digit holds within the total number. We use expanded notation to show this, for example, 3450 = 3000 + 400 + 50.
However, we also need students to understand that place value is more than just position value. For example, if I ask “How many hundreds are in 3450?” some students may answer, “There are four, as there is a four in the hundreds column”. This is not entirely accurate, because there are 34 hundreds in 3450 as 3000 is made up of 30 hundreds. We need to focus on the whole number not just on column values. This is important for addition and subtraction as there are different ways to break up 3450 depending on what we are adding it to.
If I need to solve:
3450 + 1 400 =
I could use standard decomposition to split the thousands and hundreds,
3000 + 1 000 + 400 + 400 + 50
But say I was asked to solve
3450 + 2 450 =
I could see 3450 as 2450 + 1000 (non-standard decomposition)
Therefore it may be easier to double 2450 then add on the remaining 1000
Students also need to have opportunities to create numbers using concrete materials. Students can build numbers using a variety of resources such as Base 10 Blocks, unifix cubes, ten strips or ten frames and bundles of pop sticks (for one- and two-digit numbers). Unifix cubes are particularly useful as students can build rows of ten and can break them up when necessary- this supports addition and subtraction skills.
Activity 1 – three and four digit numbers
In small groups, students use a pack of playing cards with the tens and picture cards removed. The aces are retained and count as 1.
• student A turns over the first 3 cards and each player makes a different three digit number.
• student A records the three numbers.
• student A then puts the cards at the bottom of the pile.
• students each take a turn in turning over three cards and recording the group's three digit numbers.
• When each student has had a turn they sort and order their numbers and write them in ascending order
• students can then check each other’s working out.
Students extend the game by making four digit numbers.
Possible questions include:
• Can you read each number aloud?
• Can you order the numbers in ascending and descending order?
• Can you state the place value of each numeral?
• What is the largest/smallest number you can make using three cards/four cards?
• What is the next largest/smallest number you can make using three cards/four cards?
• Can you identify the number before/after one of your three digit/four-digit numbers?
• Can you find a pattern? How can you describe your pattern? How can you continue the pattern?
• How many different ways can you represent each number? (expanded notation, in words)
• Can you count forwards/backwards by tens/hundreds from one of your three-digit/four-digit numbers?
• Can you round one of your three-digit or four-digit numbers to the nearest hundred? To the nearest thousand?
Activity 2 – how many ways?
The teacher selects a four digit number and records it on the board. Students express and present the number in as many ways as they can. A time limit may be imposed. Students can show the class what each variation looks like using concrete materials or drawings.
For example:
Activity 3 – highest number
Students play in pairs, sharing one score sheet. Players take turns to roll a die to try to make the highest number they can. Once a number has been placed in a column its position cannot be changed. The student who makes the higher number wins that game.
Students play several games to determine an overall winner.
The teacher ties the lesson together by asking:
• What is the largest possible number you can score? (9999 if you are using 0–9 dice and playing a 4-digit game)
• Who scored closest to this?
• What was your highest number?
• What was your lowest number?
Some of the results may be written on cards and pinned onto a “clothesline” to help students order 3-digit and 4-digit numbers.
Highest number score sheet - view/print (PDF 43.33KB)
Activity 4 – place value Bingo
Materials required: pen, A4 paper and place value cards (PDF 28.1KB)
• On the top of the card write 'ones' then write a '0'.
• On the next card write 'ones' and put a '1' on it. Continue this through to 9.
• Then start a set of cards for the 'tens' from 0 to 9, the 'hundreds' from 0 to 9 and the 'thousands' from 0 to 9. Alternately you can use different coloured paper to represent each value, e.g. red = thousands, green = hundreds, yellow = tens, orange = ones.
On a piece of paper the students draw a place value chart. On the chart they write a three- or four-digit number of their choice.
Sample chart for printing (PDF 26.5KB)
To play the game
The teacher calls out a number, for example, 9 hundreds. If the student has a 9 in the hundreds column they circle the number. The teacher places or writes the number on the board to keep track of past numbers. The teacher continues calling numbers, for example, 3 tens, and so on, until a student has all digits circled in one number. The student then calls out 'bingo' and says their numbers using place value.
The student reads read the whole number, for example, two thousand, nine hundred and fifty-one. If they are correct, they win. A new game starts with the students writing another number on their bingo card.
References
Australian curriculum
ACMNA052: Recognise, model, represent and order numbers to at least 10 000
ACMNA054: Recognise and explain the connection between addition and subtraction
NSW syllabus
MA2-4NA: Applies place value to order, read and represent numbers up to five digits
MA2-5NA: Addition and subtractions. Uses mental and written strategies for addition and subtraction involving 2, 3.4 and 5 digit numbers
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1. ## Two Math Questions.
1. (-2, 2) reflected through the origin is:
A. (2, 2)
B. (2, -2)
C. (-2, 2)
D. (-2, -2)
20. A picture measures 16 inches wide by 20 inches long. If the frame is to be 20 inches wide. What is the scale factor of the frame as a dilation of the picture, and what is the length of the frame?
A. scale factor is 5/4, length of frame is 25 inches
B. scale factor is 4/5, length of frame is 16 inches
C. scale factor is 1/1, length of frame is 16 inches
D. scale factor is 4/1, length of frame is 24 inches
2. Originally Posted by dgenerationx2
1. (-2, 2) reflected through the origin is:
A. (2, 2)
B. (2, -2)
C. (-2, 2)
D. (-2, -2)
Try graphing the point. (-2, 2) is 2 units left of the origin and 2 units up. Its reflection will be 2 units right of the origin and 2 units down.
Originally Posted by dgenerationx2
20. A picture measures 16 inches wide by 20 inches long. If the frame is to be 20 inches wide. What is the scale factor of the frame as a dilation of the picture, and what is the length of the frame?
A. scale factor is 5/4, length of frame is 25 inches
B. scale factor is 4/5, length of frame is 16 inches
C. scale factor is 1/1, length of frame is 16 inches
D. scale factor is 4/1, length of frame is 24 inches
Scale factor (dilation factor) is 16/20 or 4/5. We find this by using the ratio of the two widths.
To find the new length solve this: $\displaystyle \frac{20}{L}=\frac{4}{5}$
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# Solving Equations with Two Variables
Related Topics:
More Lessons for GRE Math
Math Worksheets
This lesson is part of a series of lessons for the quantitative reasoning section of the GRE revised General Test. In this lesson, we will learn:
• Linear Equations in Two Variables
• Solving Simultaneous Equations
• Using the Substitution Method
• Using the Elimination Method
## Linear Equation in Two Variables
A linear equation in two variables, x and y, can be written in the form
ax + by = c
where x and y are real numbers and a and b are not both zero.
For example, 3x + 2y = 8 is a linear equation in two variables.
A solution of such an equation is an ordered pair of numbers (x, y) that makes the equation true when the values of x and y are substituted into the equation.
For example, both (2, 1) and (0, 4) are solutions of the equation but (2, 0) is not a solution. A linear equation in two variables has infinitely many solutions.
The following video shows how to complete ordered pairs to make a solution to linear equations.
## Simultaneous Equations
If another linear equation in the same variables is given, it is usually possible to find a unique solution of both equations. Two equations with the same variables are called a system of equations, and the equations in the system are called simultaneous equations. To solve a system of two equations means to find an ordered pair of numbers that satisfies both equations in the system.
There are two basic methods for solving systems of linear equations, by substitution or by elimination.
### Substitution Method
In the substitution method, one equation is manipulated to express one variable in terms of the other. Then the expression is substituted in the other equation.
For example, to solve the system of equations
3x + 2y = 2
y + 8 = 3x
Isolate the variable y in the equation y + 8 = 3x to get y = 3x – 8.
Then, substitute 3x – 8 for y into the equation 3x + 2y = 2.
3x + 2 (3x – 8) = 2
3x + 6x – 16 = 2
9x – 16 = 2
9x = 18
Substitute x = 2 into y = 3x – 8.to get the value for y
y = 3 (2) – 8
y = 6 – 8 = – 2
Answer: x = 2 and y = –2
This video shows how to solve simultaneous equations using substitution.
### Elimination Method
In the elimination method, the object is to make the coefficients of one variable the same in both equations so that one variable can be eliminated either by adding the equations together or by subtracting one from the other.
Consider the following example:
2x + 3y = –2
4x – 3y = 14
In this example the coefficients of y are already opposites (+3 and –3). Just add the two equations to eliminate y.
6x = 12
To get the value of y, we need to substitute x = 2 into the equation 2x + 3y = –2
2(2) + 3y = –2
4 + 3y = –2
3y = –6
y = –2
Answer: x = 2 and y = –2
The following video shows how to solve simultaneous equations using the substitution method and elimination (or combination) method.
The following video shows an example of the GRE Quantitative Comparison question that involves simultaneous equations.
The following video shows how to solve (linear) linear simultaneous equations by the method of elimination. Four examples are given whereby the last example requires the multiplying of both of the equations before one of the variable can be eliminated.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# (7a+5b)−(5a−7b) = I do not know the method to answering this question.
embizze | Certified Educator
First, you want to look at the instructions for the problem or set of problems. Look for the words evaluate, simplify, or solve. These are the key words that tell you what actions to take and the form of your answer.
If the instructions are to evaluate then you will be given an input number or expression. You will then substitute (plug in) the given value. Then perform the arithmetic operations to arrive at an answer. The answer will be an expression, usually be in the form of a number. For example evaluate the expression x^2 at x=-2; the result is (-2)^2=4. Another example is to evaluate 2xy at x=3; the result is 2(3)y=6y.
If the instruction is to simplify, the answer will be an expression, sometimes a number. Here you use the order of operations to reduce the given expression. For example simplify 3x^2-2(x^2+x); first eliminate the parantheses using the distributive property to get 3x^2-2x^2-2x; then add like terms to get x^2-2x.
Finally, if the instructions are to solve then you will be given an equation or inequality. The answer will be in the form of a set of numbers that satisfy the given conditions. (Note the difference -- evaluating and simplifying yield expressions and solving has solutions.) For example, solve 2x=x+3; subtracting x from both sides gives x=3 which is the only solution.
--------------------------------------------------------------------------
For your problem I suspect the instructions are to simplify the given expression. Use the order of operations:
(7a+5b)-(5a-7b): Use the distributive property
(7a+5b)-5a+7b Note that -(-7b) is 7b (read as the opposite of the opposite of 7b.)
Now the first set of parantheses can be dropped to get:
7a+5b-5a+7b Now add like terms to get:
***********************************************************
7a+5b-5a+7b Use the commutative property of addition
7a-5a+5b+7b Now use the distributive property
(7-5)a+(5+7)b Evaluate the expressions in the parantheses
2a+12b as above.
malkaam | Student
To solve this you can either follow the BODMAS or PEMDAS,
(7a + 5b) - (5a - 7b)
First we need to check that can we solve within the brackets, the answer here is no because only similar terms can be added or subtracted, so we need to open the brackets,
(7a + 5b) - (5a - 7b)
7a + 5b - 5a + 7b
(the sign of 5a became minus and the sign of 7b became plus, because different signs when multiplied become minus and similar signs when multiplied become plus)
Arrange them so that similar terms are written together and it is easier to solve then add or subtract,
7a - 5a + 5b + 7b
PrunTuns | Student
You'll want to first reference PEMDAS (Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction). Notice that the problem has two sets of parenthesis. They have to be addressed somehow. There are no exponents, so therefore, moving onto Multiplication of PEMDAS, we distribute that subtraction sign in the middle of the two sets of terms:
(7a+5b)−(5a−7b)
becomes
7a+5b-5a+7b.
Noting that there is no Division involved, Add the like terms, a and b:
7a-5a=2a
and
5b+7b=12b
Because we've simplified the original problem as much as we could, the final expression is 2a+12b.
nisarg | Student
(7a+5b)-(5a-7b)
This is your problem so the first step is to move the negative sign into the parenthesis
(7a+5b)+(-5a+7b)
Now you want to put like numbers and variables together
7a-5a+5b+7b
now you simply add with order of operations
2a+5b+7b
2a+12b
ccbbrenes | Student
(7a+5b)-(5a-7b)
First you must distribute the negative sign through out the second set of parenthesis:
(7a+5b)+(-5a+7b)
Now you combine the a terms:
7a-5a= 2a
Then the b terms:
5b+7b= 12b
2a+12b
sid-sarfraz | Student
QUESTION:-
(7a+5b)−(5a−7b)
SOLUTION:-
In this problem we have to simplify the equation till it cannot be simplified further. "a" and "b" are separate values that cannot be added, hence;
(7a+5b)−(5a−7b)
Open the brackets, Multiply minus sign to both values, this will change the signs of both digits;
= 7a + 5b - 5a + 7b
= 7a - 5a + 5b + 7b
= 2a +12b
``
atyourservice | Student
The first step of this problem is to distribute the negative sign. Because it is in front of the (5a-7b) it means that, you multiply everything in that parenthesis by the negative sign.
This means that : (7a+5b)−(5a−7b) becomes (7a+5b)−5a+7b
The reason the - became a + is because a negative x a negative = positive
After spreading the - you can get rid of the parenthesis
7a + 5b - 5a + 7b
you can't add the a terms and the b terms because they have different exponents, so you add the terms that are alike, so the As with the As and the Bs with the Bs
7a - 5a + 5b + 7b
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For Educational Use Only © 2010 10.5 Factoring x 2 + bx + c Brian Preston Algebra 1 2009-2010.
Presentation on theme: "For Educational Use Only © 2010 10.5 Factoring x 2 + bx + c Brian Preston Algebra 1 2009-2010."— Presentation transcript:
For Educational Use Only © 2010 10.5 Factoring x 2 + bx + c Brian Preston Algebra 1 2009-2010
For Educational Use Only © 2010 Real World Application How wide is the stone border?
For Educational Use Only © 2010 Lesson Objectives 1) Factor a quadratic expression of the form x 2 + bx + c. 2) Solve quadratic equations by factoring.
For Educational Use Only © 2010 Now we are going to learn how to do this backwards. Review (x + 4)(x + 5)5 x5x 4 4 x x 1) x2x2 + 4x + 5x+ 20 x2x2 + 9x + 20 First Outside Inside Last
For Educational Use Only © 2010 2) x 2 + 3x + 2 1 1 + 2 Factors of 2 1 2 + 3 = Factor the trinomial. Example
For Educational Use Only © 2010 +2 +1 +2 +1 2) x 2 + 3x + 2 1x 1 1 + 2 2 1 ( + 2 + 1 +3 = Factor the trinomial. 1 2 Factors of 2 )( 1x ) 1x 2 1x Example
For Educational Use Only © 2010 3) Mike is building a stone border along two sides of a rectangular Japanese garden that measures 6 yards by 15 yards. His budget limits him to only enough stone to cover 46 square yards. How wide should the border be? border 46 6 15 (15)(6) Are of border 46(x + 6) 46 6 Real World Application = Garden area – (x + 15) Total area 15 border 6 15
For Educational Use Only © 2010 3)(x + 15)(x + 6) – (15)(6) = 46 6156 – 46 Real World Application – 46 = + 21x x2x2 0 – 46 x2x2 + 15x + 6x+ 90 x2x2 + 21x + 0 – 90 = 46 x x xx
For Educational Use Only © 2010 3) x 2 + 21x – 46 = 0 1 Factors of 46 1 46 + 21 = 2 23 Factor the trinomial. Real World Application 1 + 46 47 = 46 – 1 45 =
For Educational Use Only © 2010 3) x 2 + 21x – 46 = 0 1 – 2 + 23 +21 = Factor the trinomial. 1 46 2 23 Factors of 46 Real World Application 2 + 23 25 = 23 – 2 21 =
For Educational Use Only © 2010 3) x 2 + 21x – 46 = 0 +23 -2 +23 -2 1 1x 2 1x – 2 + 23 23 – 2 ( + 23 – 2 +21 = Factor the trinomial. 1 46 2 23 Factors of 46 )( ) Real World Application = 0= 0
For Educational Use Only © 2010 – 23 + 2 (x + 23) (x – 2) (x + 23) (x – 2) x + 23 Example Solve the equation by factoring. 3) = 0 x – 2 x – 23 = ( ) ( ) x + 2 = 2 2 yards wide
For Educational Use Only © 2010 Real World Application How wide is the stone border? 2 yards wide
For Educational Use Only © 2010 What is factoring? Factoring is another way to solve for variables in a quadratic equations.
For Educational Use Only © 2010 4) x 2 – 5x + 6 1 1 + – 6 Factors of 6 1 6 – 5 -5 = 2 3 Factor the trinomial. Example 1 + 6 7 = 6 – 1 5 = (–1) to all 1 + – 6
For Educational Use Only © 2010 Rule x 2 – 2x – 8 One (-) & One (+) Patterns for factoring trinomials. (x + 2) (x - 4) x 2 + 6x + 8 (x + 2) (x + 4) x 2 – 6x + 8 (x – 2) (x – 4) Two (+) Two (-) x 2 + 2x – 8 (x + 4) (x - 2)
For Educational Use Only © 2010 4) x 2 – 5x + 6 1 – 2 + – 3 -5 = Factor the trinomial. 1 6 2 3 Factors of 6 Example 2 + 3 5 = 3 – 2 1 = (–1) to all
For Educational Use Only © 2010 – 3 -2 – 3 -2 4) x 2 – 5x + 6 1x 1 – 2 + – 3 – 3 – 2 ( – 3 – 2 -5 = Factor the trinomial. 1 6 2 3 Factors of 6 )( 1x ) 1x 2 1x Example
For Educational Use Only © 2010 5) x 2 – 2x – 8 1 Factors of 8 1 8 – 2 -2 = 2 4 Factor the trinomial. Example 1 + 8 9 = 8 – 1 7 =
For Educational Use Only © 2010 5) x 2 – 2x – 8 1 2 + – 4 -2 = Factor the trinomial. 1 8 2 4 Factors of 8 Example 2 + 4 6 = 4 – 2 2 = (–1) to all
For Educational Use Only © 2010 5) x 2 – 2x – 8 – 4 1x 1 +2 2 + – 4 – 4 2 ( + 2 -2 = Factor the trinomial. 1 8 2 4 Factors of 8 )( 1x ) 1x 2 1x Example
For Educational Use Only © 2010 1 6) x 2 + 7x – 18 Factors of 18 1 18 + 7 = 2 9 Factor the trinomial. Example 3 6 1 + 18 19 = 18 – 1 17 =
For Educational Use Only © 2010 6) x 2 + 7x – 18 1 – 2 + 9 +7 = Factor the trinomial. 1 18 2 9 Factors of 18 Example 3 6 2 + 9 11 = 9 – 2 7 =
For Educational Use Only © 2010 +9 -2 +9 -2 6) x 2 + 7x – 18 1x 1 – 2 + 9 9 – 2 ( + 9 – 2 +7 = Factor the trinomial. 1 18 2 9 Factors of 18 )( 1x ) 1x 2 1x Example 3 6
For Educational Use Only © 2010 1 7) w 2 + 13w + 36 Factors of 36 + 13 = Factor the trinomial. Example 1 36 2 18 3 12 4 9 6 6 1 + 36 37 = 36 – 1 35 =
For Educational Use Only © 2010 7) w 2 + 13w + 36 1 Factors of 36 +13 = Factor the trinomial. Example 1 36 2 18 3 12 4 9 6 6 2 + 18 20 = 18 – 2 16 =
For Educational Use Only © 2010 7) w 2 + 13w + 36 1 Factors of 36 +13 = Factor the trinomial. Example 1 36 2 18 3 12 4 9 6 6 3 + 12 15 = 12 – 3 9 =
For Educational Use Only © 2010 7) w 2 + 13w + 36 1 4 + 9 +13 = Factor the trinomial. Factors of 36 Example 1 36 2 18 3 12 4 9 6 6 4 + 9 13 = 9 – 4 5 =
For Educational Use Only © 2010 7) w 2 + 13w + 36 +4 1w +9 1 4 + 9 9 4 ( + 9 + 4 +13 = Factor the trinomial. Factors of 36 )( ) 1w 2 Example 1 36 2 18 3 12 4 9 6 6
For Educational Use Only © 2010 8) 32 + 12n + n 2 (4 + n) (3 + n) (n + 1) (n + 2) Example Factor the trinomial.
For Educational Use Only © 2010 8) 32 + 12n + n 2 Example Factors of 32 1 32 + 12 = 2 16 Factor the trinomial. 1 3 12 4 8 6 6 1 + 32 33 = 32 – 1 31 =
For Educational Use Only © 2010 8) 32 + 12n + n 2 +12 = Factor the trinomial. Example 1 36 2 18 3 12 4 8 6 6 Factors of 32 1 2 + 18 20 = 18 – 2 16 =
For Educational Use Only © 2010 8) 32 + 12n + n 2 +12 = Factor the trinomial. Example 1 36 2 18 3 12 4 8 6 6 1 Factors of 32 3 + 12 15 = 12 – 3 9 =
For Educational Use Only © 2010 8) 32 + 12n + n 2 1 4 + 8 +12 = Factor the trinomial. Example 1 36 2 18 3 12 4 8 6 6 Factors of 32 4 + 8 12 = 8 – 4 4 =
For Educational Use Only © 2010 +8 8 + 8) 32 + 12n + n 2 1n +4 1 4 + 8 8 4 ( 4 + +12 = Factor the trinomial. )( ) Example 1 36 2 18 3 12 4 8 6 6 Factors of 32 1n 2
For Educational Use Only © 2010 Example 3 3 x 2 + 3x – 4 – 4 (1) (– 4) a (3) 9) Tell whether the trinomial can be factored. 1 ax 2 + bx + c = 0 a = b = 3 1 1 c = – 4 b 2 – 4ac 2 – 4 b c 1 Standard form
For Educational Use Only © 2010 (1) (– 4) (3) 9) 2 – 4 9 + 16 25 25 is a perfect square, so Yes Example
For Educational Use Only © 2010 – 6 3 3 x 2 + 3x – 6 (1) (– 6) a (3) 10) Tell whether the trinomial can be factored. 1 ax 2 + bx + c = 0 a = b = 3 1 1 c = – 6 b 2 – 4ac 2 – 4 b c 1 Standard form Example
For Educational Use Only © 2010 (1) (– 6) (3) 10) 2 – 4 9 + 24 33 33 is not a perfect square, so No Example
For Educational Use Only © 2010 11) b 2 + 7b + 10 = 0 1 Factors of 10 1 10 + 7 = 2 5 Solve the equation by factoring. Example Standard form 1 + 10 11 = 10 – 1 9 =
For Educational Use Only © 2010 11) b 2 + 7b + 10 = 0 1 2 + 5 +7 = Solve the equation by factoring. 1 10 2 5 Factors of 10 Example 2 + 5 7 = 5 – 2 3 =
For Educational Use Only © 2010 +5 1b+ 5 + 2 1b 11) b 2 + 7b + 10 = 0 1 +2 2 + 5 5 2 ( +7 = Solve the equation by factoring. 1 10 2 5 Factors of 10 )( ) 1b 2 Example = 0
For Educational Use Only © 2010 – 5 – 2– 5 – 2 (b + 5) (b + 2) (b + 5) (b + 2) b + 5 Example Solve the equation by factoring. 11) = 0 b + 2 b – 5 = ( ) ( ) b – 2 =
For Educational Use Only © 2010 12) x 2 + 5x – 14 = 0 1 Factors of 14 1 14 + 5 = 2 7 Solve the equation by factoring. Example Standard form 1 + 14 15 = 14 – 1 13 =
For Educational Use Only © 2010 12) x 2 + 5x – 14 = 0 1 – 2 + 7 +5 = Solve the equation by factoring. 1 14 2 7 Factors of 14 Example 2 + 7 9 = 7 – 2 5 =
For Educational Use Only © 2010 +7 -2 +7 -2 1x 12) x 2 + 5x – 14 = 0 + 7 – 2 1 – 2 + 7 7 – 2 ( +5 = Solve the equation by factoring. 1 14 2 7 Factors of 14 )( ) 1x 2 Example = 0
For Educational Use Only © 2010 – 7 + 2– 7 + 2 (x + 7) (x – 2) (x + 7) (x – 2) x + 7 Example Solve the equation by factoring. 12) = 0 x – 2 x – 7 = ( ) ( ) x + 2 = 2
For Educational Use Only © 2010 + 4 Example 13) x 2 – 2x – 19 = – 4 Solve the equation by factoring. + 4 = – 2x x2x2 0 – 15 Standard form
For Educational Use Only © 2010 13) x 2 – 2x – 15 = 0 1 Factors of 15 1 15 – 2 -2 = 3 5 Solve the equation by factoring. Example 1 + 15 16 = 15 – 1 14 =
For Educational Use Only © 2010 13) x 2 – 2x – 15 = 0 1 3 + – 5 -2 = Solve the equation by factoring. 1 15 3 5 Factors of 15 Example 3 + 5 8 = 5 – 3 2 = (–1) to all
For Educational Use Only © 2010 -5 +3 13) x 2 – 2x – 15 = 0 1x – 5 + 3 1 3 + – 5 – 5 3 ( -2 = Solve the equation by factoring. 1 15 3 5 Factors of 15 )( ) 1x 2 Example = 0
For Educational Use Only © 2010 + 5 – 3+ 5 – 3 (x – 5) (x + 3) x – 5 Example Solve the equation by factoring. 13) = 0 x + 3 x + 5 = 5 ( ) ( ) x – 3 = (x – 5)
For Educational Use Only © 2010 + 15 Example 14) x 2 + 16x = – 15 Solve the equation by factoring. = + 16x x2x2 0 + 15 Standard form
For Educational Use Only © 2010 14) x 2 + 16x + 15 = 0 1 1 + 15 Factors of 15 1 15 + 16 = 3 5 Solve the equation by factoring. Example 1 + 15 16 = 15 – 1 14 =
For Educational Use Only © 2010 +1 +15 ( 1 + 15 14) x 2 + 16x + 15 = 0 1x + 15 + 1 1 15 1 ( +16 = Solve the equation by factoring. 1 15 3 5 Factors of 15 ) ) 1x 2 Example = 0
For Educational Use Only © 2010 – 1 – 15 (x + 15) (x + 1) (x + 15) (x + 1) x + 15 Example Solve the equation by factoring. 14) = 0 x + 1 x – 15 = ( ) ( ) x – 1 =
For Educational Use Only © 2010 1) Don’t forget the negative signs. 2) Make sure the expressions or equations are in standard form before factoring. Key Points & Don’t Forget
For Educational Use Only © 2010 pg. 462-463 #’s 10-37, 40-46 even The Assignment
For Educational Use Only © 2010 Please email brianspowerpoints@gmail.com with errors, confusing slides, improvements, complications, or any other comments or questions.brianspowerpoints@gmail.com The template is from www.spiralgraphics.bizwww.spiralgraphics.biz http://www.worldofteaching.comhttp://www.worldofteaching.com is home to over a thousand powerpoints submitted by teachers. This is a completely free site and requires no registration. Please visit and I hope it will help in your teaching. Bibliography
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# INTRODUCTION TO SINGAPORE MATH
Welcome to Singapore Math! The math curriculum in Singapore has been recognized worldwide for its excellence in producing students highly skilled in mathematics.
Students in Singapore have ranked at the top in the world in mathematics on the Trends in International Mathematics and Science Study (TIMSS) in 1993, 1995, 2003, and
2008. Because of this, Singapore Math has gained in interest and popularity in the United States.
Singapore Math curriculum aims to help students develop the necessary math concepts and process skills for everyday life and to provide students with the ability to
formulate, apply, and solve problems. Mathematics in the Singapore Primary (Elementary) Curriculum cover fewer topics but in greater depth. Key math concepts are
introduced and built-on to reinforce various mathematical ideas and thinking. Students in Singapore are typically one grade level ahead of students in the United States.
The following pages provide examples of the various math problem types and skill sets taught in Singapore.
At an elementary level, some simple mathematical skills can help students 3. Addition Number Bond (double and single digits)
understand mathematical principles. These skills are the counting-on, counting-
back, and crossing-out methods. Note that these methods are most useful when 2 + 15
the numbers are small.
5 10
1. The Counting-On Method
Used for addition of two numbers. Count on in 1s with the help of a picture or = 2 + 5 + 10 Regroup 15 into 5 and 10.
number line. = 7 + 10
7 + 4 = 11 = 17
## +1 +1 +1 +1 4. Subtraction Number Bond (double and single digits)
7 8 9 10 11 12 7
2 10
2. The Counting-Back Method
Used for subtraction of two numbers. Count back in 1s with the help of a 10 7 = 3
picture or number line. 3+2=5
16 3 = 13
5. Subtraction Number Bond (double digits)
1 1 1
13 14 15 16 20 15
10 10 10 5
3. The Crossing-Out Method
Used for subtraction of two numbers. Cross out the number of items to be 10 5 = 5
taken away. Count the remaining ones to find the answer. 10 10 = 0
5+0=5
20 12 = 8
Students should understand that multiplication is repeated addition and that
division is the grouping of all items into equal sets.
Mackenzie eats 2 rolls a day. How many rolls does she eat in 5 days?
2 + 2 + 2 + 2 + 2 = 10
5 2 = 10
She eats 10 rolls in 5 days.
A number bond shows the relationship in a simple addition or subtraction problem. 2. The Grouping Method (Division)
The number bond is based on the concept part-part-whole. This concept is useful Mrs. Lee makes 14 sandwiches. She gives all the sandwiches equally to 7
in teaching simple addition and subtraction to young children. friends. How many sandwiches does each friend receive?
whole
part part
## To find a whole, students must add the two parts.
To find a part, students must subtract the other part from the whole.
14 7 = 2
The different types of number bonds are illustrated below.
1. Number Bond (single digits)
9 One of the basic but essential math skills students should acquire is to perform the
4 operations of whole numbers and fractions. Each of these methods is illustrated
3 6 below.
## 9 (whole) 3 (part) = 6 (part) H T O O: Ones
3 2 1
9 (whole) 6 (part) = 3 (part) T : Tens
+ 5 6 8
8 8 9 H: Hundreds
2. Addition Number Bond (single digits)
Since no regrouping is required, add the digits in each place value accordingly.
9 + 5
1 4 H T O O: Ones
41
9 2
T: Tens
=9+1+4 Make a ten first. + 1 5 3
6 4 5 H: Hundreds
= 10 + 4
= 14 In this example, regroup 14 tens into 1 hundred 4 tens.
## 70 Must-Know Word Problems Level 3 3
_1 2+ _1 3= ___
H T O O: Ones 6 2 4 3 12 12
3= __
2+ ___ 5
12
12 18 6
T : Tens Always remember to make the denominators common before adding the
+ 3 6 5
6 5 1 H: Hundreds fractions.
Regroup twice in this example. 12. The Subtraction-of-Fractions Method
First, regroup 11 ones into 1 ten 1 one.
Second, regroup 15 tens into 1 hundred 5 tens. __
1 5
__ 1 2
__
5 2= 3
= ___ ___
2 5 5 2 10 10 10
4. The Subtracting-Without-Regrouping Method Always remembers to make the denominators common before subtracting the
fractions.
H T O O: Ones
7 3 9 13. The Multiplication-of-Fractions Method
T : Tens
3 2 5
4 1 4 H: Hundreds
1
5 9 __
3 __
__ 1 =
3
151
Since no regrouping is required, subtract the digits in each place value
accordingly. When the numerator and the denominator have a common multiple, reduce
them to their lowest fractions.
5. The Subtracting-by-Regrouping Method
14. The Division-of-Fractions Method
H T O O: Ones
5 78 111 7 __
1 = __ 6 = ___
7 __ 14 = 4 2
9 6 9 1 3
__
2
_
T : Tens 3
2 4 7 3
3 3 4 H: Hundreds
When dividing fractions, first change the division sign () to the multiplication
sign (). Then, switch the numerator and denominator of the fraction on the
In this example, students cannot subtract 7 ones from 1 one. So, regroup the
right hand side. Multiply the fractions in the usual way.
tens and ones. Regroup 8 tens 1 one into 7 tens 11 ones.
6. The Subtracting-by-Regrouping-Twice Method Model drawing is an effective strategy used to solve math word problems. It is
a visual representation of the information in word problems using bar units. By
H T O O: Ones drawing the models, students will know of the variables given in the problem, the
78 90 100 variables to find, and even the methods used to solve the problem.
T : Tens
5 9 3
2 0 7 H: Hundreds Drawing models is also a versatile strategy. It can be applied to simple word
problems involving addition, subtraction, multiplication, and division. It can also
In this example, students cannot subtract 3 ones from 0 ones and 9 tens from be applied to word problems related to fractions, decimals, percentage, and ratio.
0 tens. So, regroup the hundreds, tens, and ones. Regroup 8 hundreds into 7
The use of models also trains students to think in an algebraic manner, which uses
hundreds 9 tens 10 ones.
symbols for representation.
7. The Multiplying-Without-Regrouping Method
The different types of bar models used to solve word problems are illustrated below.
T O
2 4 O: Ones 1. The model that involves addition
2 T : Tens Melissa has 50 blue beads and 20 red beads. How many beads does she
4 8 have altogether?
?
Since no regrouping is required, multiply the digit in each place value by the
multiplier accordingly.
50 20
8. The Multiplying-With-Regrouping Method
50 + 20 = 70
H T O O: Ones
13 24 9 2. The model that involves subtraction
T : Tens
3 Ben and Andy have 90 toy cars. Andy has 60 toy cars. How many toy cars
1, 0 4 7 H: Hundreds does Ben have?
90
In this example, regroup 27 ones into 2 tens 7 ones, and 14 tens into 1 hundred
4 tens.
60 ?
9. The Dividing-Without-Regrouping Method
90 60 = 30
2 4 1
2 4 8 2 3. The model that involves comparison
4 Mr. Simons has 150 magazines and 110 books in his study. How many more
8 magazines than books does he have?
8
2 Magazines 150
2
0
Books 110 ?
Since no regrouping is required, divide the digit in each place value by the
divisor accordingly. 150 110 = 40
10. The Dividing-With-Regrouping Method 4. The model that involves two items with a difference
1 6 6 A pair of shoes costs \$109. A leather bag costs \$241 more than the pair of
5 8 3 0 shoes. How much is the leather bag?
5 ?
3 3
3 0 Bag \$241
3 0
3 0 Shoes \$109
0
\$109 + \$241 = \$350
In this example, regroup 3 hundreds into 30 tens and add 3 tens to make 33
tens. Regroup 3 tens into 30 ones.
## 4 70 Must-Know Word Problems Level 3
5. The model that involves multiples ?
Mrs. Drew buys 12 apples. She buys 3 times as many oranges as apples. She
also buys 3 times as many cherries as oranges. How many pieces of fruit does Tie
Belt
Apples 12
?
Oranges ? \$539 7 = \$77
Cherries Tie (2 units) 2 x \$77 = \$154
Belt (5 units) 5 x \$77 = \$385
13 12 = 156
11. The model that involves comparison of fractions
6. The model that involves multiples and difference Jacks height is __ 3 of Lindsays height. If
2 of Leslies height. Leslies height is __
3 4
There are 15 students in Class A. There are 5 more students in Class B than in Lindsay is 160 cm tall, find Jacks height and Leslies height.
Class A. There are 3 times as many students in Class C than in Class A. How ?
many students are there altogether in the three classes?
Jack
Class A 15 ?
Class B 5 ? Leslie
Class C Lindsay
160 cm
(5 15) + 5 = 80
1 unit 160 4 = 40 cm
7. The model that involves creating a whole Leslies height (3 units) 3 40 = 120 cm
Ellen, Giselle, and Brenda bake 111 muffins. Giselle bakes twice as many
muffins as Brenda. Ellen bakes 9 fewer muffins than Giselle. How many muffins Jacks height (2 units) 2 40 = 80 cm
does Ellen bake?
Thinking skills and strategies are important in mathematical problem solving.
Ellen ? 9 These skills are applied when students think through the math problems to solve
them. Below are some commonly used thinking skills and strategies applied in
Giselle 111 + 9 mathematical problem solving.
1. Comparing
Brenda
Comparing is a form of thinking skill that students can apply to identify
similarities and differences.
(111 + 9) 5 = 24
(2 24) 9 = 39 When comparing numbers, look carefully at each digit before deciding if a
number is greater or less than the other. Students might also use a number line
8. The model that involves sharing for comparison when there are more numbers.
There are 183 tennis balls in Basket A and 97 tennis balls in Basket B. How Example:
many tennis balls must be transferred from Basket A to Basket B so that both
baskets contain the same number of tennis balls?
0 1 2 3 4 5 6 7 8
183
3 is greater than 2 but smaller than 7.
2. Sequencing
Basket B 97 A sequence shows the order of a series of numbers. Sequencing is a form
of thinking skill that requires students to place numbers in a particular order.
There are many terms in a sequence. The terms refer to the numbers in a
183 97 = 86 sequence.
86 2 = 43 To place numbers in a correct order, students must first find a rule that
generates the sequence. In a simple math sequence, students can either add
9. The model that involves fractions
or subtract to find the unknown terms in the sequence.
George had 355 marbles. He lost __ 1of the remaining
1of the marbles and gave __
5 4 Example: Find the 7th term in the sequence below.
marbles to his brother. How many marbles did he have left? 1, 4, 7, 10, 13, 16 ?
355 1st 2nd 3rd 4th 5th 6th 7th
term term term term term term term
L: Lost
L B
R R R B: Brother Step 1: This sequence is in an increasing order.
R: Remaining Step 2: 4 1 = 3 74=3
? The difference between two consecutive terms is 3.
5 parts 355 marbles Step 3: 16 + 3 = 19
The 7th term is 19.
1 part 355 5 = 71 marbles
3 parts 3 71 = 213 marbles 3. Visualization
Visualization is a problem solving strategy that can help students visualize a
10. The model that involves ratio problem through the use of physical objects. Students will play a more active
role in solving the problem by manipulating these objects.
Aaron buys a tie and a belt. The prices of the tie and belt are in the ratio
2 : 5. If both items cost \$539, The main advantage of using this strategy is the mobility of information in the
process of solving the problem. When students make a wrong step in the
(a) what is the price of the tie?
process, they can retrace the step without erasing or canceling it.
(b) what is the price of the belt?
The other advantage is that this strategy helps develop a better understanding
of the problem or solution through visual objects or images. In this way,
students will be better able to remember how to solve these types of problems.
## 70 Must-Know Word Problems Level 3 5
Some of the commonly used objects for this strategy are toothpicks, straws, assumptions will eliminate some possibilities and simplifies the word problems
cards, strings, water, sand, pencils, paper, and dice. by providing a boundary of values to work within.
4. Look for a Pattern Example: Mrs. Jackson bought 100 pieces of candy for all the students in her
This strategy requires the use of observational and analytical skills. Students class. How many pieces of candy would each student receive if
have to observe the given data to find a pattern in order to solve the problem. there were 25 students in her class?
Math word problems that involve the use of this strategy usually have repeated In the above word problem, assume that each student received the same
numbers or patterns. number of pieces. This eliminates the possibilities that some students would
Example: Find the sum of all the numbers from 1 to 100. receive more than others due to good behaviour, better results, or any other
reason.
Step 1: Simplify the problem.
Find the sum of 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. 8. Representation of Problem
Step 2: Look for a pattern. In problem solving, students often use representations in the solutions to
1 + 10 = 11 2 + 9 = 11 3 + 8 = 11 show their understanding of the problems. Using representations also allow
4 + 7 = 11 5 + 6 = 11 students to understand the mathematical concepts and relationships as well
as to manipulate the information presented in the problems. Examples of
Step 3: Describe the pattern.
representations are diagrams and lists or tables.
When finding the sum of 1 to 10, add the first and last numbers to get a
result of 11. Then, add the second and second last numbers to get the Diagrams allow students to consolidate or organize the information given in the
same result. The pattern continues until all the numbers from 1 to 10 problems. By drawing a diagram, students can see the problem clearly and
are added. There will be 5 pairs of such results. Since each addition solve it effectively.
equals 11, the answer is then 5 11 = 55.
A list or table can help students organize information that is useful for analysis.
Step 4: Use the pattern to find the answer. After analyzing, students can then see a pattern, which can be used to solve
Since there are 5 pairs in the sum of 1 to 10, there should be (10 5 = the problem.
50 pairs) in the sum of 1 to 100.
Note that the addition for each pair is not equal to 11 now. The addition 9. Guess and Check
for each pair is now (1 + 100 = 101). One of the most important and effective problem-solving techniques is Guess
50 101 = 5050 and Check. It is also known as Trial and Error. As the name suggests, students
have to guess the answer to a problem and check if that guess is correct. If the
The sum of all the numbers from 1 to 100 is 5,050.
guess is wrong, students will make another guess. This will continue until the
5. Working Backward guess is correct.
The strategy of working backward applies only to a specific type of math word It is beneficial to keep a record of all the guesses and checks in a table. In
problem. These word problems state the end result, and students are required addition, a Comments column can be included. This will enable students to
to find the total number. In order to solve these word problems, students have analyze their guess (if it is too high or too low) and improve on the next guess.
to work backward by thinking through the correct sequence of events. The Be careful; this problem-solving technique can be tiresome without systematic
strategy of working backward allows students to use their logical reasoning or logical guesses.
and sequencing to find the answers.
Example: Jessica had 15 coins. Some of them were 10-cent coins and the rest
Example: Sarah has a piece of ribbon. She cuts the ribbon into 4 equal parts. were 5-cent coins. The total amount added up to \$1.25. How many
Each part is then cut into 3 smaller equal parts. If the length of each coins of each kind were there?
small part is 35 cm, how long is the piece of ribbon?
Use the guess-and-check method.
3 35 = 105 cm
4 105 = 420 cm Number of Number of
Total
Value Value Number of Total Value
The piece of ribbon is 420 cm. 10 Coins 5 Coins
Coins
6. The Before-After Concept 7 7 10 = 70 8 8 5 = 40 7 + 8 = 15
70 + 40 = 110
= \$1.10
The Before-After concept lists all the relevant data before and after an event.
Students can then compare the differences and eventually solve the problems. 80 + 35 = 115
8 8 10 = 80 7 7 5 = 35 8 + 7 = 15
Usually, the Before-After concept and the mathematical model go hand in = \$1.15
hand to solve math word problems. Note that the Before-After concept can be 10 10 = 100 + 25 = 125
applied only to a certain type of math word problem, which trains students to 10 5 5 5 = 25 10 + 5 = 15
100 = \$1.25
think sequentially.
Example: Kelly has 4 times as much money as Joey. After Kelly uses some There were ten 10-cent coins and five 5-cent coins.
money to buy a tennis racquet, and Joey uses \$30 to buy a pair of
pants, Kelly has twice as much money as Joey. If Joey has \$98 in the 10. Restate the Problem
beginning, When solving challenging math problems, conventional methods may not be
(a) how much money does Kelly have in the end? workable. Instead, restating the problem will enable students to see some
(b) how much money does Kelly spend on the tennis racquet? challenging problems in a different light so that they can better understand
them.
Before
The strategy of restating the problem is to say the problem in a different
Kelly
and clearer way. However, students have to ensure that the main idea of the
Joey \$98
problem is not altered.
After How do students restate a math problem?
? ? First, read and understand the problem. Gather the given facts and unknowns.
Kelly Note any condition(s) that have to be satisfied.
Joey \$30 Next, restate the problem. Imagine narrating this problem to a friend. Present
the given facts, unknown(s), and condition(s). Students may want to write the
(a) \$98 - \$30 = \$68 revised problem. Once the revised problem is analyzed, students should
2 \$68 = \$136 be able to think of an appropriate strategy to solve it.
Kelly has \$136 in the end.
(b) 4 \$98 = \$392 11. Simplify the Problem
\$392 \$136 = \$256 One of the commonly used strategies in mathematical problem solving is
Kelly spends \$256 on the tennis racquet. simplification of the problem. When a problem is simplified, it can be broken
down into two or more smaller parts. Students can then solve the parts
7. Making Supposition
systematically to get to the final answer.
Making supposition is commonly known as making an assumption. Students
can use this strategy to solve certain types of math word problems. Making
|
## Precalculus (6th Edition) Blitzer
We have the expression on the left side $\frac{1}{\sin t-1}+\frac{1}{\sin t+1}$, which can be simplified by multiplying and dividing the expression by $\sin t+1$ and $\sin t-1$, and then further simplifying by using the algebraic formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ \begin{align} & \frac{1}{\sin t-1}+\frac{1}{\sin t+1}=\frac{1}{\sin t-1}.\frac{\sin t+1}{\sin t+1}+\frac{1}{\sin t+1}.\frac{\sin t-1}{\sin t-1} \\ & =\frac{\sin t+1}{{{\sin }^{2}}t-1}+\frac{\sin t-1}{{{\sin }^{2}}t-1} \\ & =\frac{\sin t+1+\sin t-1}{{{\sin }^{2}}t-1} \\ & =\frac{2\sin t}{{{\sin }^{2}}t-1} \end{align} We know the Pythagorean identity, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$; therefore, ${{\sin }^{2}}x-1=-{{\cos }^{2}}x$ and using the quotient identity $\tan x=\frac{\sin x}{\cos x}$ and the reciprocal identities $\frac{1}{\cos x}=\sec x$, the expression can be simplified as: \begin{align} & \frac{2\sin t}{{{\sin }^{2}}t-1}=\frac{2\sin t}{-{{\cos }^{2}}t} \\ & =-2.\frac{\sin t}{\cos t}.\frac{1}{\cos t} \\ & =-2\tan t.\sec t \end{align} Hence, the expression on the left side is equal to the expression on the right side.
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# Objectives The student will be able to:
## Presentation on theme: "Objectives The student will be able to:"— Presentation transcript:
Objectives The student will be able to:
1. solve equations with variables on both sides. 2. solve equations containing grouping symbols. Designed by Skip Tyler, Varina High School
1) Solve. 3x + 2 = 4x - 1 You need to get the variables on one side of the equation. It does not matter which variable you move. Try to move the one that will keep your variable positive.
1) Solve 3x + 2 = 4x - 1 - 3x - 3x 2 = x - 1 + 1 + 1 3 = x
3 = x 3(3) + 2 = 4(3) - 1 9 + 2 = Draw “the river” Subtract 3x from both sides Simplify Add 1 to both sides Check your answer
2) Solve 8y - 9 = -3y + 2 + 3y + 3y 11y – 9 = 2 + 9 + 9 11y = 11 11 11
11y = y = 1 8(1) - 9 = -3(1) + 2 Draw “the river” Add 3y to both sides Simplify Add 9 to both sides Divide both sides by 11 Check your answer
What is the value of x if 3 - 4x = 18 + x?
3) Solve 4 = 7x - 3x 4 = 4x 4 4 1 = x 4 = 7(1) - 3(1) Draw “the river”
1 = x 4 = 7(1) - 3(1) Draw “the river” – Notice the variables are on the same side! Combine like terms Divide both sides by 4 Simplify Check your answer
4) Solve -7(x - 3) = -7 -7x + 21 = -7 - 21 - 21 -7x = -28 -7 -7 x = 4
-7x = -28 x = 4 -7(4 - 3) = -7 -7(1) = -7 Draw “the river” Distribute Subtract 21 from both sides Simplify Divide both sides by -7 Check your answer
What is the value of x if 3(x + 4) = 2(x - 1)?
-14 -13 13 14 Answer Now
5) Solve 3 - 2x = 4x – 6 + 2x +2x 3 = 6x – 6 + 6 + 6 9 = 6x 6 6
Draw “the river” Clear the fraction – multiply each term by the LCD Simplify Add 2x to both sides Add 6 to both sides Divide both sides by 6 Check your answer 3 - 2x = 4x – 6 + 2x +2x = 6x – 6 = 6x or 1.5 = x
-2x -2x 5 = -3 This is never true! No solutions
Special Case # ) 2x + 5 = 2x - 3 -2x x 5 = -3 This is never true! No solutions Draw “the river” Subtract 2x from both sides Simplify
Infinite solutions or identity
Special Case # ) 3(x + 1) - 5 = 3x - 2 3x + 3 – 5 = 3x - 2 3x - 2 = 3x – 2 -3x x -2 = -2 This is always true! Infinite solutions or identity Draw “the river” Distribute Combine like terms Subtract 3x from both sides Simplify
What is the value of x if -3 + 12x = 12x - 3?
4 No solutions Infinite solutions Answer Now
Challenge! What is the value of x if -8(x + 1) + 3(x - 2) = -3x + 2?
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# NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions InText Questions
These NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions InText Questions and Answers are prepared by our highly skilled subject experts.
## NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions InText Questions
NCERT In-text Question Page No. 230
Question 1.
Describe how the following expressions are obtained:
(i) 7xy + 5
(ii) x2
(ii) 4x2 – 5x
Answer:
(i) 7xy + 5
To obtain this expression, we first multiply two variables x and y, i.e. x × y = xy. Then we multiply their product by 7 to get 7xy. Next, we add 5 to 7xy to obtain 7xy + 5.
(ii) x:y
First, x is multiplied by itself i.e., x × x = x2 Then multiply x2 by y to get the required expression, i.e. x2 × y = x2y
(iii) 4x2 – 5x
Here, the variable x is multiplied by itself,
i. e. x × x = x2
Then x2 is multiplied by, i.e. x2 × 4 = 4x2
Next, we multiply the variable x by 5, i.e. x × 5 = 5x
Now, we subtract 5x from 4x2 to get 4x2 – 5x
NCERT In-text Question Page No. 231
Question 1.
What are the terms in the followig expressions? Show how the terms are formed. Draw a tree diagram for each expression:
(i) 8y + 3x2
(ii) 7mn – 4
(iii) 2x2y
Answer:
(i) 8y + 3x2
Terms of 8y + 3x2 are: 8y and 3x2 the term 8y is formed by multiplying the variable y by 8.
The term 3x2 is formed by first multiplying the variable x with itself to get x × x = x2 and then multiplying x2 by 3.
Tree diagram:
Terms of 7mn – 4 are 7 mn and -4. The term 7 mn formed by first multiplying variables m and n to get m × n = mn
Then multiplying mn by constant 7 to get 7 × mn = 7 mn. The term -4 is a constant.
Tree diagram:
(iii) 2x2y
This expression has only one term, i.e. 2x2y. To form this term first we multiply the variable x by itself to get x × x = x2. Then the variable y is multiplied by x2 to get
y × x2 = x2y
Next, the product x2y is multiplied by 2 to ‘ get
2 × x2y = 2x2y
Tree diagram:
Question 11.
Write three expressions each having 4
terms.
Answer:
(i) 2x3 – 4x2 + 9xy + 8
(ii) 6x3 + 9y2 – 3xy2 – 12
(iii) 9x2 – 3x + 12xy – 1
NCERT In-text Question Page No. 231
Question 1.
Identify the coefficients of the terms of following expressions:
Answer:
(i) 4x – 3y
(ii) a + b + 5
(iii) 2y + 5
(iv) 2xy
Answer:
(i) 4x – 3y
The coefficient of x in 4x is 4.
The coefficient of y in -3y is (-3)
(ii) a + b + 5
The coefficient of a is 1.
The coefficient of b is 1.
The coefficient of 5 is 1.
(iii) 2y + 5
The coefficient of y in 2y is 2.
The coefficient of 5 is 1.
(iii) 2xy
The coefficient of xy in 2xy is 2.
The coefficient of y in 2xy is 2x.
The coefficient of x in 2xy is 2y.
NCERT In-text Question Page No. 233
Question 1.
Group the like terms together from the following:
12x, 12, -25x, -25, -25y, 1, x, 12y, y
Answer:
We have:
(i) 12x, -25x and x are like terms.
(ii) -25y, 12y and y are like terms.
(iii) 12, -25 and 1 are like terms.
NCERT In-text Question Page No. 233
Question 1.
Classify the following expressions as a monomial, a binomial or a trianomial: a, a + b, ab + a + b, ab + a + b -5, xy, xy + 5, 5x2– x + 2,4pq – 3q + 5p, 7,4m – 7n + 10, 4mn + 7.
Answer:
(i) a is containing 1 term. It is a monomial.
(ii) a + b is containing 2 terms. It is a binomial.
(iii) ab + a + b is containing 3 terms, it is a trinomial.
(iv) ab + a + b – 5 is containing 4 terms. It is a polynomial.
(v) xy is containing only 1 term. It is a monomial.
(vi) xy + 5 is containing 2 terms. It is a binomial.
(vii) 5x2 – x + 2 is containing 3 terms. It is a trinomial.
(viii) 4pq – 3q + 5p is containing 3 terms. It is a trinomial.
(ix) 7 is containing only 1 term. It is a monomial.
(x) 4 m – 7n + 10 is containing 3 terms. It is a trinomial.
(xi) 4mn + 7 is containing 2 terms. It is a binomial.
NCERT In-text Question Page No. 236
Question 1.
Think of at least two situations in which you need to form two algebraic expressions and add or subtract them.
Answer:
Situation-I
Rohan’s per month earning is twice the per month earning of Bibha. Kavita’s per month earnings is RS. 400 more than the sum of Rohan’s and Vibha’s per month earnings. What is the montly earning of Kavita?
Situation-II
Mahesh has twice the number of toys Kanta has: Lata has 5 toys more than twice the number of toys Mahesh and Kanta together have. What is the number of toys that Kanta has?
NCERT In-text Question Page No. 238
Question 1.
Add and subtract:
(i) m – n, m + n
(ii) mn + 5-2, mn + 3
Answer:
(i) Adding (m – n) and (m + n), we have:
(m-n) + (m + n) = m- n + m + n
Collecting like terms, we have:
(m + m) + (-n + n)
= (1 + 1)m + (-1 + 1)n
= 2m + (0)n
= 2m
Now subtracting (m – n) from (m + n), we have:
(m + n)-(m-n) = m + n- m + n
= (m – m) + (n + n)
= (1 – 1)m + (1 + 1)n
= 0m + 2n
= 2n
(ii) Adding mn + 5-2 and mn + 3, we have:
(mn + 5 – 2) + (mn + 3)
= mn + 5 – 2 + mn + 3
= (mn + mn) + (5 – 2 + 3)
= (1 + 1)mn + (6)
= 2 mn + 6
Now, subtracting (mn + 3) from (mn + 5 – 2), we have:
(mn + 5 – 2) – (mn + 3)
= mn + 5 – 2 – mn – 3
= (mn – mn) + (5 – 2 – 3)
= (0)mn + 0 = 0 + 0 = 0
NCERT In-text Question Page No. 245
Question 1.
Make similar pattern with basic figure as shown
(The number of segments required to make the figure is given to the right. Also, the expression for the number of segments required to make n shapes is also given.)
Answer:
Discover more such patterns as follow.
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# System of First Order Differential Equations
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## Transcription
1 CHAPTER System of First Order Differential Equations In this chapter, we will discuss system of first order differential equations. There are many applications that involving find several unknown functions simultaneously. Those unknown functions are related by a set of equations that involving the unknown functions and their first derivatives. For example, in Chapter Two, we studied the epidemic of contagious diseases. Now if S(t) denotes number of people that is susceptible to the disease but not infected yet. I(t) denotes number of people actually infected. R(t) denotes the number of people have recovered. If we assume The fraction of the susceptible who becomes infected per unit time is proportional to the number infected, b is the proportional number. A fixed fraction rs of the infected population recovers per unit time, r. A fixed fraction of the recovers g become susceptible and infected, g. proportional function. The system of differential equations model this phenomena are S = bis + gr I = bis ri R = ri gr The numbers of unknown function in a system of differential equations can be arbitrarily large, but we will concentrate ourselves on to 3 unknown functions.. Principle of superposition Let a ij (t), b j (t) i =,,, n and j =,,, n be known function, and x i t, i =,,, n be unknown functions, the linear first
2 . SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS order system of differential equation for x i (t) is the following, x (t) = a (t)x (t) + a (t)x (t) + + a n (t)x n (t) + b (t) x (t) = a (t)x (t) + a (t)x (t) + + a n (t)x n (t) + b (t) x 3(t) = a 3 (t)x (t) + a 3 (t)x (t) + + a 3n (t)x n (t) + b 3 (t). x n(t) = a n (t)x (t) + a n (t)x (t) + + a nn (t)x n (t) + f (t) Let x(t) be the column vector of unknown functions x i t, i =,,, n, A(t) = (a ij (t), and b(t) be the column vector of known functions b i t, i =,,, n, we can write the first order system of equations as () x (t) = A(t)x(t) + b(t) When n =, the linear first order system of equations for two unknown functions in matrix form is, x (t) a (t) a x = (t) x (t) b (t) + (t) a (t) a (t) x (t) b (t) When n = 3, the linear first order system of equations for three unknown functions in matrix form is, x (t) x (t) = a (t) a (t) a 3 a (t) a (t) a 3 x (t) x (t) + b (t) b (t) x 3(t) a 3 (t) a 3 (t) a 33 x 3 t b 3 (t) A solution of equation () on the open interval I is a column vector function x(t) whose derivative (as a vector-values function) equals A(t)x(t) + b(t). The following theorem gives existence and uniqueness of solutions, Theorem.. If the vector-valued functions A(t) and b(t) are continuous over an open interval I contains t, then the initial value problem { x (t) = A(t)x(t) + b(t) x(t ) = x has an unique vector-values solution x(t) that is defined on entire interval I for any given initial value x. When b(t), the linear first order system of equations becomes x (t) = A(t)x(t), which is called a homogeneous equation. As in the case of one equation, we want to find out the general solutions for the linear first order system of equations. To this end, we first have the following results for the homogeneous equation,
3 . PRINCIPLE OF SUPERPOSITION 3 Theorem.. Principle of Superposition Let x (t), bx (t),, x n (t) be n solutions of the homogeneous linear equation x (t) = A(t)x(t) on the open interval I. If c, c,, c n are n constants, then the linear combination is also a solution on I. c x (t) + c x (t) + c 3 x 3 (t) + + c n x n (t) Example.. Let x (t) = x(t) e t, x (t) = and x (t) = e t are two solutions, as (e bx (t) = t ) e t e t = = and [ bx (t) = [ (e t ) = [ e t = [ e t By the Principle of Superposition, for any two constants c and c e t c e x(t) = c x (t) + c x (t) = c + c e t = t c e t is also solution. We shall see that it is actually the general solution. The next theorem gives the general solution of linear system of equations, Theorem.3. - Let x (t), x (t),, bx n (t) be n linearly independent (as vectors) solution of the homogeneous system x (t) = A(t)x(t), then for any solution x c (t) there exists n constants c, c,, c n such that x c (t) = c x (t) + c x (t) + + c n x n (t). We call x c (t) the general solution of the homogeneous system.
4 4. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS If x p (t) is a particular solution of the nonhomogeneous system, x(t) = B(t)x(t) + b(t), and x c (t) is the general solution to the associate homogeneous system, x(t) = B(t)x(t) then x(t) = x c (t) + x p (t) is the general solution. Example.. Let [ 4 3 x 4t (t) = x(t) + + 5t 6 7 6t x(t) + 7t + 3e t e 5t, x (t) = e t and x (t) = 3e 5t are two linearly independent t solutions. and x p (t) = is a particular solution. By Theorem t.3, 3c e x(t) = c x (t) + c x (t) + x p (t) = t + c e 5t + t () c e t + 3c e 5t + t is the general solution. Now suppose we want to [ find a particular solution that satisfies the initial condition x() =, then let t = in (), we have x() = [ 3c + c = c + 3c which can be written in matrix form, [ 3 c = 3 c c Solve this equation, we get = c [ 3e solution is x(t) = t e 5t + t e t 3e 5t. + t, [,. So the particular From the above example, we can summarize the general steps in find a solution to initial value problem, { x (t) = A(t)x(t) + b(t) x(t ) = x
5 . HOMOGENEOUS SYSTEM 5 Step One: Find the general solution x c = c x (t) + c x (t) + + c n x n (t), where x (t), x (t),, x n (t) are a set of linearly independent solutions, to the associate homogeneous system, x (t) = A(t)x(t). Step Two: Find a particular solution x p (t)to the nonhomogeneous system, x (t) = A(t)x(t) + b(t). Step Three: Set x(t) = x c (t) + x p (t) and use the equation x(t ) = x, to determine c, c,, c n.. Homogeneous System We will use a powerful method called eigenvalue method to solve the homogeneous system x (t) = Ax(t) where A is a matrix with constant entry. We will present this method for A is either a or 3 3 cases. The method can be used for A is an n n matrix. The idea is to find solutions of form (3) x(t) = ve λt, a straight line that passing origin in the direction v. Now taking derivative on x(t), we have (4) x (t) = λve λt put (3) and (.) into the homogeneous equation, we get So x (t) = λve λt = Ave λt Av = λv, which indicates that λ must be an eigenvalue of A and v is an associate eigenvector... A is a matrix. Suppose a a A = a a Then the characteristic polynomial p(λ) of A is p(λ) = A λi = (a λ) (a λ) a a = λ (a +a )+(a a a a. So p(λ) is a quadratic polynomial of λ. From Algebra, we know that p(λ) = has either distinct real solutions, or a double solution, or conjugate complex solutions. The following theorem summarize the solution to the homogeneous system,
6 6. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS Theorem.. Let p(λ) be the characteristic polynomial of A, for x (t) = Ax(t), Case : p(λ) = has two [ distinct real solutions [ λ and λ. v v Suppose v = and v v = are associate eigenvector (i.e, Av = λ v and Av = λ v ) Then the general v solution is And x c (t) = c v e λ t + c v e λ t v e Φ(t) = λ t v e λ t v e λ t v e λ t is called the fundamental matrix(a fundamental matrix is a square matrix whose columns are linearly independent solutions of the homogeneous system). Case : p(λ) = has a double solutions λ. In this case p(λ) = (λ λ ) and λ is a zero of p(λ) with multiplicity. () λ has [ two linearly independent eigenvectors: v v Suppose v = and v v = are associate linearly v independent eigenvectors. Then the general solution is And x c (t) = (c v + c v )e λ t Φ(t) = e λ t v v v v () λ has [ only one associate eigenvector: v Suppose v = is the only associated eigenvector and v v v = is a solution of v (λ I A)v = v. Then the general solution is, And x c (t) = (c v + c (tv + v )e λ t [ Φ(t) = e λ t v (v t + v ) v (v t + v ) is the fundamental solution matrix.
7 . HOMOGENEOUS SYSTEM 7 Case 3: p(λ) = has two [ conjugate complex solutions a+bi and a bi. v + iv Suppose v = is the associate complex eigenvector[ with respect to a [ + bi, then the general solution is, v + iv v v v = and v v = v x c (t) = [c (v cos(bt) v sin(bt))c (v cos(bt) + v sin(bt))e at. And [ Φ(t) = e at v cos(bt) v sin(bt) v cos(bt) + v sin(bt) v cos(bt) v sin(bt) v cos(bt) + v sin(bt) is the fundamental matrix. From Theorem., let Φ(t) be the fundamental [ matrix, the general c solution is given by x c (t) = Φ(t)c, with c = and the solution c that satisfies a given initial condition x(t ) = x is given by x(t) = Φ(t)Φ(t ) -t x Example.. [ Two distinct eigenvalues case Find the general 3 solution to x (t) = x(t) 5 Using Mathcad, functions eigenvals() and eigen- Solution vecs() In Mathcad, eigenvecs(m) Returns a matrix containing the eigenvectors. The nth column of the matrix returned is an eigenvector corresponding to the nth eigenvalue returned by eigenvals. we find,λ = [ and λ = 3 6 with associated eigenvectors v = and v [ = respectively. So the fundamental matrix is [ ( 7 6)e ( 3 Φ(t) = + 6)t ( 7 + 6)e ( 3 6)t e ( 3 + 6)t e ( 3 6)t c and the general solution is, for c =, c x c (t) = Φ(t)c
8 8. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS Example.. One double eigenvalues with two linearly [ in-dependent eigenvectors Find the general solution to x (t) = x(t). Solution [ The eigenvalue is λ = and associated [ eigenvectors c e are and, so the general solution is x c = t c e t Example.3. One double eigenvalues with only one eigenvector Find the solution to x (t) = x(t) and x() = Solution Using Mathcad, functions eigenvals() and [ eigenvecs() we can find a double eigenvalue λ = 5 and eigenvector Notice, the symbolic operator (bring up by either [Shift[Ctrl[. or [Ctrl[.) will not work with eigenvecs() this time, but since multiply an eigenvector by a nonzero constant still get an eigenvector, we can 3 choose v =. 3 To [ find w that [ satisfies (A λ I)w = v λ we will solve (A w λ I) =. That is, w 3 3 w 3 = 3 3 w 3 One solution is w = and w = So the fundamental matrix is 3 3t + Φ(t) = e 5t 3 3t c and the general solution is, c =, c x c (t) = Φ(t)c 3 3() + Now, Φ() = e 5() = 3 3() [ 3 3 and Φ() - = Hence, the particular solution is x(t) = Φ(t)Φ() - x = e 5t [ 3t + 4 3t 3.
9 . HOMOGENEOUS SYSTEM 9 Example.4. Two conjugate [ complex eigenvalues case Find 3 the general solution to x (t) = x(t) Solution Using Mathcad, functions eigenvals() and eigenvecs() we find two conjugate complex eigenvalues, λ = + i 3 and λ = i [ 3 3 with associated eigenvector v = with respect to i λ. Compare this with the Theorem., we have a =, b = 3, v = 3, v =, v =, and v =. So the fundamental matrix is [ 3 cos(bt) sin(bt) Φ(t) = e t 3 sin(bt) cos(bt) and the general solution is, c = [ c c, [ 3 cos( 3t) sin( 3t) x c (t) = Φ(t)c = e t c 3 sin( 3t) cos( 3t) c [ = e t 3c cos( 3t) c sin( 3t) 3c sin( 3t) c cos( 3t) Suppose we want to find a solution such that x() =, then x(t) = Φ(t)Φ() - x() [ [ 3 cos( 3t) sin( 3t) = e t 3 [ 3 sin( 3t) cos( 3t) [ 3 cos( 3t) sin( 3t) 3 = e t 3 sin( 3t) cos( 3t) = e t [ cos( 3t) + + sin( 3t) sin( 3t) + cos( 3t).. A is a 3 3 matrix. Suppose A = a a a 3 a a a 3 a 3 a 3 a 33 Then the characteristic polynomial p(λ) of A given by p(λ) = A λi, - [
10 . SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS is a cubic polynomial of λ. From Algebra, we know that p(λ) = has either 3 distinct real solutions, or distinct solutions and one is a double solution, or one real solution and conjugate complex solutions, or a triple solution. The following theorem summarize the solution to the homogeneous system, Theorem.. Let p(λ) be the characteristic polynomial of A, for x (t) = Ax(t), Case : p(λ) = has three distinct real solutions λ, λ, and λ 3. Suppose v = v v, v = v v, and v 3 = v 3 v 3 v 3 are associate eigenvector (i.e, Av = λ v, Av = λ v, and Av 3 = λ 3 v 3 ) Then the general solution is v 3 x c (t) = c v e λ t + c v e λ t + c 3 v 3 e λ 3t And the fundamental matrix is Φ(t) = v e λ t v e λ t v e λ t v e λ t v 3 e λ 3t v 3 e λ 3t v 3 e λ t v 3 e λ t v 33 e λ 3t Case : p(λ) = has a double solutions λ. So p(λ) = (λ λ ) (λ λ ), and λ has multiplicity. Let v 3 = v v is the eigenvector associated with λ. v 3 [ λ has [ two linearly independent eigenvectors: v v Suppose v = and v v = are associate linearly v independent eigenvectors. Then the general solution is And x c (t) = (c v + c v )e λ t + c 3 v 3 e λ t Φ(t) = v e λ t v e λ t v e λ t v e λ t v 3 e λ t v 3 e λ t v 3 e λ t v 3 e λ t v 33 e λ t. v 33
11 . HOMOGENEOUS SYSTEM [ λ has one eigenvector: Suppose v = v v is the associated eigenvector with respect to λ and v = v v 3 v is a solution of v 3 (λ I A)v = v. Then the general solution is, x c (t) = (c v + c (tv + v ))e λ t + c 3 v 3 e λ And Φ(t) = v e λ t v e λ t (v t + v )e λ t (v t + v )e λ t v 3 e λ v 3 e λ v 3 e λ t (v 3 t + v 3 )e λ t v 33 e λ is the fundamental solution matrix. Case 3: p(λ) = has two conjugate complex solutions a ± bi and a real solution λ. Suppose v = v + iv v + iv is the associate complex eigenvector with respect to a + bi, then the general solution is, let v 3 + iv 3 v 3 = v 3 v 3, are associated eigenvectors with respect to λ, V 33 x c (t) = [c (v cos(bt) v sin(bt))c (v cos(bt)+v sin(bt))e at +c 3 v 3 e λ. Φ(t) = e at And v cos(bt) v sin(bt) v cos(bt) + v sin(bt) v 3 e λ v cos(bt) v sin(bt) v cos(bt) + v sin(bt) v 3 e λ v 3 cos(bt) v 3 sin(bt) v 3 cos(bt) + v 3 sin(bt) v 33 e λ is the fundamental matrix. Case 4: p(λ) = has solution λ with multiplicity 3. In this case, p(λ) = (λ λ ) 3. [ λ has three linearly independent eigenvectors. Let v = v v v 3, v = v v V 3, and v 3 = v 3 v 3 V 33 be the three linearly independent eigenvectors. Then the general
12 . SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS solution is x c (t) = (c v + c v + c 3 v 3 )e λt and fundamental v v v 3 v v V 3 matrix is Φ(t) = e λ t v 3 v 3 V 33 [ λ has two linearly independent eigenvectors. Suppose v = v v, v = v v are the linearly independent eigenvectors. Let v 3 = v v 3 V 3 3 v 3, then only one of the two equations, (A λ I)v 3 = v or (A λ I)v 3 = v can has a solution that is linearly independent with v, v. Suppose (A λ I)v 3 = v generates such a solution. Then the general solution is x c (t) = [c v + c v + c 3 (tv + v 3 )e λ t and fundamental matrix is Φ(t) = e λ t v v tv + v 3 v v tv + V 3 v 3 v 3 tv 3 + V 33 [3 λ has only one eigenvector. Let v = v v be the linearly independent eigenvectors. Let v = satisfies v 3 v v V 3 and v 3 = V 33 v 3 v 3 V 33 (A λ I)v = v and (A λ I)v 3 = v. be two vectors that Then the general solution is x c (t) = [c v + c (tv + v ) + c 3 (t v + tv + v 3 )e λ t and fundamental matrix is Φ(t) = e λ t v tv + v t v + tv + v 3 v tv + v t v + tv + V 3 v 3 tv 3 + v 3 t v 3 + tv 3 + V 33 Remark.. Suppose A is an n n matrix, for the homogeneous system x (t) = Ax(t), three general case would happen Case : A has n distinct eigenvalues λ i, i =,,, n with linearly independent eigenvectors v i, i =,,, n then the general solution will be x c (t) = c v e λ + c v e λ + + c n v n e λn
13 . HOMOGENEOUS SYSTEM 3 Case: A has m < n distinct eigenvalues, in this case some eigenvalues would have multiplicity greater than. Suppose λ r has multiplicity r. Depending on how many linearly independent eigenvectors are associated with λ r the situation could be very complex. Let p be the number of linearly eigenvectors associated with λ r, then d = r p is called the deficit of λ r. The simply cases are either d = or d = r. When < d < r the situation could be very complex. Suppose d = r and v is the only eigenvector associate with λ r, then one will have to solve r equations (A λ r ) i v i+ = v i, i =,,, r. And the general solution would contains terms like [c v + c (v t + v ) + c 3 (v t + v t + v 3 ) + + c r (v r + v t r + + v r )e λr. Case 3: A complex root a+bi with associated eigenvector v a +iv b, then the general solution contains term, [c (v a cos(bt) v b sin(bt))+ c (v a sin(bt) + v b cos(bt))e at. Remark.. Suppose x (t), x (t), x 3 (t),, x n (t) are n linearly independent solution for n n homogeneous system, x (t) = Ax(t), the fundamental matrix Φ(t) is a matrix whose columns are x i (t), i =,,, n. Example.5. (Two distinct eigenvalues) Find the general solution to x = 3x + 4x x 3 x = x + x 4x 3 x 3 = x + x Solution Let x(t) = x (t) x (t) x 3 (t) and The equations can be written in matrix form x (t) = Ax(t). Using Mathcad, functions eigenvals() and eigenvecs() we find,λ = and λ = with associated eigenvectors v = 4 and v = respectively. Since λ has multiplicity as appeared twice in the result of eigenvals() function, we need to solve the equation (A λ I)v 3 = v.
14 4. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS To use Mathcad, () you first compute (A λ I)v 3 using the following sequences of key stroke, [* type ([Ctrl[M, set the rows and columns in the matrix definition popup menu, input the data for A, [* type -[Ctrl[M set the row and column number and input data for λ I, [* type )[Ctrl[M, now set as column number, enter a, b, c in the place holders, [* type [Ctrl[. to compute symbolically and you get. () Using the Given Find block to find a solution. Type Given in a blank space, type a+b-c[ctrl= and a-4c[ctrl= in two rows, then type key word Find following by typing (a,b)[ctrl[. you will get the solution in terms of c. Set c =, we get v 3 = 4. So the fundamental matrix is Φ(t) = and the general solution is, 4et e t (t + 4)e t e t e t e t e t (t + )e t x c (t) = c v e t + c v e t + c 3 (tv + v 3 )e t Example.6. (One eigenvalue with deficit ) Find the solution to x (t) = 3 x(t) and x() = Solution Using Mathcad, functions eigenvals() (Notice the eigenvecs() will not find a good result in this case due to the rounding error.) we find, λ = is the only eigenvalue. To find the associate eigenvectors we compute (Using (A λ I)v = ) (A λ I)v = v v 4 v 3 v + v 3 v + v 3 = 4v v 3
15 . HOMOGENEOUS SYSTEM 5 We have only v + v 3 = for three variables v, v, v 3, this indicates that v can be any value, and set v = find v 3 =, So v = and v = are two eigenvectors. To find the generalize eigenvector associated with λ we will have to solve two equations (A λ I) w w =, w 3 and From (.), (A λ I) 4 w w w 3 = [ w w =, we get two inconsistent equations w + w 3 = and w + w 3 =. So now solution can be found in this case. From (.), 4 [ w w = we get one equation w + w 3 = choose w 3 = we get w =, since w can be anything, we set w =. So v 3 = and we can verify that v, v, and v 3 are linearly independent. So the fundamental matrix is Φ(t) = e t t t + t + and the general solution is,,, x c (t) = [c v + c v + c 3 (tv + v 3 )e t
16 6. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS Now, Φ() = and Φ() = Hence, the particular solution is x(t) = [v + 3v e t 3. Example.7. (One eigenvalue with deficit ) Find the general solution to x (t) = 4 x(t). 3 Solution Using Mathcad, functions eigenvals() (Notice the eigenvecs() will not find a good result in this case due to the rounding error.) we find, λ = 3 is the only eigenvalue. To find the associate eigenvectors we compute (Using (A λ I)v = ) (A 3I)v = v v v 3 v + v + v 3 v + v + v 3 = v v We have only one eigenvector v =. To find the generalize eigenvector associated with λ we will have to solve two equations and From (.), (A 3I)v = v, (A 3I)v 3 = v, a b c =, we have two equations { b a = a + b + c = Choosing a =, we get b =, c =. Hence v =
17 . HOMOGENEOUS SYSTEM 7 From (.), a b c =, we have two equations { b a = a + b + c = Choosing a =, we get b =, c =. So v 3 = that v, v, and v 3 are linearly independent. So the fundamental matrix is Φ(t) = e t + t t + t t t t t + and the general solution is, and we can verify x c (t) = [c v + c (tv + v ) + c 3 (t v + tv + v 3 )e 3t Example.8. (Two conjugate complex eigenvalues case) Find the general solution to x (t) = 3 3 x(t) Solution Using Mathcad, functions eigenvals() and eigenvecs() we find two conjugate complex eigenvalues and one real eigenvalue, λ =, λ = + i 9, and λ3 = i 9 with associated eigenvector v = and v = + i 9 + i 9 = + 9 i 9 with respect to λ 3. Compare this with the Theorem.3,
18 8. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS we have a =, b = 9, v = 9 9 (imaginary part of v). The general solution is, x c (t) = c v e t +c (v cos( (real part of v), and v 3 = 9) v3 sin( 9))e t +c 3 (v sin( 9)+v3 cos( 9))e t 3. Nonhomogeneous System of Equations To find solutions to the initial value problem of nonhomogeneous equations x (t) = Ax(t) + b(t), x(t ) = x we follow the steps below, () Find the general solution x c (t) = Φ(t)c to homogeneous equation x (t) = Ax(t), where Φ(t) is the fundamental matrix. () Find a particular solution x p to x (t) = Ax(t) (3) The general solution to the nonhomogeneous equation x (t) = Ax(t) is x(t) = x c (t) + x p (t). Using x(t ) = x to determine the coefficient vector c. The following theorem gives one way to find a particular solution based on the fundamental matrix, Theorem 3.. Let Φ(t) be a fundamental matrix of x (t) = Ax(t), a particular solution to x (t) = Ax(t) + b(t) is given by x p (t) = Φ(t) Φ(t) - b(t) dt. Example 3.. Find the general solution to Solution Let x(t) = b(t) = t t x = 3x + 4x x 3 + t x = x + x 4x 3 x 3 = x + x t x (t) x (t) x 3 (t), A = 3 4 4, and. The equations can be written in matrix form x (t) =
19 3. NONHOMOGENEOUS SYSTEM OF EQUATIONS 9 Ax(t)+b(t). From Example.5, we know that the fundamental matrix to x(t) = Ax(t) is Φ(t) = 4et e t (t + 4)e t e t e t e t e t (t + )e t To find a particular solution, we first compute Φ - (t)b(t) = then we compute Φ(t) Φ - 5 (t)b(t) dt = Φ(t) t e t dt ( 5 t3 + 5 t )e t dt 5 t e t dt 5 = t + t t 3t t + 4t And so the general solution is, x(t) = c 4 e t + c + c 3 (t + 4 e t ) e t + 5 t + t t 3 5 t t t The following is a screen shot that shows how to carry out the computation in Mathcad, 5 t e t ( 5 t3 + 5 t )e t 5 t e t To use Mathcad, () Define fundamental matrix A(t) and b(t) in the same line (not as shown in graph), and compute in the next line A b(t) () type A(t)*[Ctrl[M choose column as, at each place holder, type [Ctrl[I to get the indefinite integral, (3) and put the corresponding entry of A b(t) in the integrant position. (4) press [Shift[Ctrl[. type key work simplify
20 . SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS Theorem 3.. If Φ(t) is the fundamental matrix for x (t) = Ax(t),, and x p (t) = Φ - (t)b(t) dt, then x(t) = Φ(t)Φ - (t )(x x p (t )) + x p (t) is the solution to the nonhomogeneous initial value problem, x (t) = Ax(t) + b(t), x(t ) = x Example 3.. Find the solution to x (t) = e t e t and x() = [ x(t) + Solution From Example.3 is the fundamental matrix is 3 3t + Φ(t) = e 5t 3 3t and Φ() - = e Now b(t) = t e t, using the formula x p (t) = Φ(t) Φ - (t)b(t) dt and Mathcad, we have x p (t) = 3 et Therefore, Φ() - (x() x p ()) = ( [ and the solution is x(t) = Φ(t)Φ() - (x() x p ()) + x p (t) = e 5t [ [ 3 ) = [ t 6 4t + 3 e 3t 4. Higher order differential equations One can transform equations that involving higher order derivatives of unknown functions to system of first order equations. For example, suppose x(t) is an unknown scalar function that satisfies mx (t) + cx (t) + kx(t) = f(t) an equation can be used to model a spring system with external force f(t) or an RCL electronic circuit with an energy source f(t).
21 4. HIGHER ORDER DIFFERENTIAL EQUATIONS Now if we set x (t) = x(t) and x (t) = x (t) we then get an system of first order equations (5) x (t) = x (t) (6) x (t) = c m x (t) k m x (t) + f(t) m In general, if we have an differential equation that involving nth order derivative x (n) (t) of unknown function x(t), x (n) = a x(t) + a x (t) + + a n x (n ) + f(t), we can transform it into an system of first order equations of n unknown functions x (t) = x(t), x (t) = x (t), x 3 (t) = x () (t),, x n (t) = x (n ) (t), and using the eigenvalue method for system of differential equation to solve the higher order equation. Example 4.. Transform the differential equation x (3) + 3x () 7x (t) 9x = sin(t) into system of first order equations. Solution Here the highest order of derivative is third derivative x (3) of x(t). So we transfer it into system of 3 equations. Let x (t) = x(t), x (t) = x (t), x 3 (t) = x (t), we have (7) (8) x (t) = x (t) x (t) = x 3 (t) (9) x 3(t) = 3x 3 (t) + 7x (t) + 9x (t) sin(t) Let x(t) = x (t) x (t), A =, and b(t) = x 3 (t) f(t) we can write the system of equation in matrix form x (t) = Ax(t) + b(t). Example 4.. Find the general solution for the 3rd order differential equation x (3) + 3x () 7x (t) 9x = sin(t). Solution From previous example, Example 4., Let x(t) = x (t) x (t), A =, and b(t) = we can write x 3 (t) f(t) the system of equation in matrix form x (t) = Ax(t) + b(t). Using Mathcad we find the eigenvalues are λ =, λ = +, λ 3 = with associate eigenvectors, v =, v = +,
22 . SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS and v 3 = + respectively (after multiply the results of Mathcad by some constants). So the fundamental matrix is Φ(t) = e t e ( + )t e (+ )t e t ( + )e ( + )t ( + )e (+ )t e t ( )e ( + )t ( + )e (+ )t From Φ(t) we find a particular solution 3 x p (t) = Φ(t) Φ - 5 (t)b(t) dt = cos(t) 5 sin(t) cos(t) sin(t) cos(t) 8 sin(t) 5 39 Hence the general solution to the system is x (t) x (t) = x 3 (t) c e t + c e ( + )t + c 3 e (+ )t cos(t) 5 39 sin(t) c e t + c ( + e ( + )t c 3 ( + )e (+ )t cos(t) 56 sin(t) c e t + c ( )e ( + )t + c 3 ( + )e (+ )t 3 5 cos(t) 8 39 sin(t) and x (t) = c e t + c e ( + )t + c 3 e (+ )t + 3 cos(t) 5 sin(t) is 5 39 the general solution to the third order ordinary differential equation x (3) + 3x () 7x (t) 9x = sin(t). Example 4.3. Find the solution to the initial value problem x x + 9x = te t, x() =, x () =. Solution Since the given equation is of second order, we will have two unknowns x (t) = x(t), x (t) = x (t) to transform the equation into a system of first order equations, x (t) = x x (t) = x 9x + te t, and the initial conditions [ are x () = [ x() = x () = x () = [. x (t) Now let x(t) =, A =, and b(t) = x (t) 9 te t. We have the matrix version of this equation, x (t) = Ax(t) + b(t)
23 4. HIGHER ORDER DIFFERENTIAL EQUATIONS 3 Using Mathcad, we find[ the eigenvalues [ λ =, λ = 9, and associate eigenvectors v =, v =. And fundamental 9 e t e matrix Φ(t) = 9t e t 9e 9t. From Φ(t) we find a particular solution x p (t) = Φ(t) Φ - (t)b(t) dt = 5 (3t + 8t + )e t 5 (3t + 7t + 9)e t The solution with initial values x = is given by x(t) = Φ(t)Φ ( - ()(x b()) + ) b(t) = 6 t ( 6 t e t 7 5 e9t ) e t 43 5 e9t Hence the solution to the initial value ( problem of the) second order differential equation is x(t) = x (t) = 6 t 639 t+ e t e9t. Project At beginning you should enter: Project title, your name, ss#, and due date in the following format Project One: Define and Graph Functions. John Doe SS# -- Due: Mon. Nov. 3rd, 3 You should format the text region so that the color of text is different than math expression. You can choose color for text from Format >Style select normal and click modify, then change the settings for font. You can do this for headings etc. () Solutions To System of Equations Finding solution to linear system using Mathcad and study the long time dynamic behavior of the solutions. Find general solution to { x = y y = x
24 4. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS Plot several solutions with different initial values in [- xt-plane, yt-plane xy-plane. Here you will need to define range variable t =,. 7 and set X := x(t), Y := y(t). The graph in xy-plane is called the trajectory. If this models the movement of a satellite, what is its trajectory. Find general solution to { x = 8y y = 8x Plot several solutions with different initial values in [- xt-plane, yt-plane xy-plane. Here you will need to define range variable t =,. 7 and set X := x(t), Y := y(t). The graph in xy-plane is called the trajectory. If this models the movement of a satellite, what is its trajectory. Find general solution to { x = x y y = y 3x Plot several solutions with different initial values in [- xt-plane, yt-plane xy-plane. Here you will need to define range variable t =,. 7 and set X := x(t), Y := y(t). The graph in xy-plane is called the trajectory. If this models the movement of a system of two species, what is you conclusion about interdependency of these species? Can you find initial value such that x(t) = (distinct) for some t? what about y(t). () Solution of Higher order equation In general mx + cx + kx = f(t) models a object with mass m attached to a spring with constant k and damping force that is proportional to the velocity x, c, k >. Suppose m = and f(t) = Ae at sin(bt), that is the external force is oscillatory (b > ) and diminishing (a > ) Find solutions and graph the solutions. - c = b = Find general solution and graph some particular solutions. - c =, k =, a =, b =, A = 4 - c =, k = 3, A =, a =, b =
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Rd Sharma 2019 Solutions for Class 8 Math Chapter 14 Compound Interest are provided here with simple step-by-step explanations. These solutions for Compound Interest are extremely popular among Class 8 students for Math Compound Interest Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 Book of Class 8 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2019 Solutions. All Rd Sharma 2019 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.
#### Question 1:
Compute the amount and the compound interest in each of the following by using the formulae when:
(i) Principal = Rs 3000, Rate = 5%, Time = 2 years
(ii) Principal = Rs 3000, Rate = 18%, Time = 2 years
(iii) Principal = Rs 5000, Rate = 10 paise per rupee per annum, Time = 2 years
(iv) Principal = Rs 2000, Rate = 4 paise per rupee per annum, Time = 3 years
(v) Principal = Rs 12800, Rate = $7\frac{1}{2}%$, Time = 3 years
(vi) Principal = Rs 10000, Rate 20% per annum compounded half-yearly, Time = 2 years
(vii) Principal = Rs 160000, Rate = 10 paise per rupee per annum compounded half-yearly, Time = 2 years.
#### Question 2:
Find the amount of Rs 2400 after 3 years, when the interest is compounded annually at the rate of 20% per annum.
#### Question 3:
Rahman lent Rs 16000 to Rasheed at the rate of $12\frac{1}{2}%$ per annum compound interest. Find the amount payable by Rasheed to Rahman after 3 years.
#### Question 4:
Meera borrowed a sum of Rs 1000 from Sita for two years. If the rate of interest is 10% compounded annually, find the amount that Meera has to pay back.
#### Question 5:
Find the difference between the compound interest and simple interest. On a sum of Rs 50,000 at 10% per annum for 2 years.
#### Question 6:
Amit borrowed Rs 16000 at $17\frac{1}{2}%$ per annum simple interest. On the same day, he lent it to Ashu at the same rate but compounded annually. What does he gain at the end of 2 years?
#### Question 7:
Find the amount of Rs 4096 for 18 months at $12\frac{1}{2}%$ per annum, the interest being compounded semi-annually.
#### Question 8:
Find the amount and the compound interest on Rs 8000 for $1\frac{1}{2}$ years at 10% per annum, compounded half-yearly.
#### Question 9:
Kamal borrowed Rs 57600 from LIC against her policy at $12\frac{1}{2}%$ per annum to build a house. Find the amount that she pays to the LIC after $1\frac{1}{2}$ years if the interest is calculated half-yearly.
#### Question 10:
Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs 64000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half.
#### Question 11:
Rakesh lent out Rs 10000 for 2 years at 20% per annum, compounded annually. How much more he could earn if the interest be compounded half-yearly?
#### Question 12:
Romesh borrowed a sum of Rs 245760 at 12.5% per annum, compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest, but compounded semi-annually. Find his gain after 2 years.
#### Question 13:
Find the amount that David would receive if he invests Rs 8192 for 18 months at $12\frac{1}{2}%$ per annum, the interest being compounded half-yearly.
#### Question 14:
Find the compound interest on Rs 15625 for 9 months, at 16% per annum, compounded quarterly.
#### Question 15:
Rekha deposited Rs 16000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year.
#### Question 16:
Find the amount of Rs 12500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year.
#### Question 17:
Ramu borrowed Rs 15625 from a finance company to buy a scooter. If the rate of interest be 16% per annum compounded annually, what payment will he have to make after $2\frac{1}{4}$ years?
#### Question 18:
What will Rs 125000 amount to at the rate of 6%, if the interest is calculated after every 3 months?
#### Question 19:
Find the compound interest at the rate of 5% for three years on that principal which in three years at the rate of 5% per annum gives Rs 12000 as simple interest.
#### Question 20:
A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs 482 more. Find the sum.
#### Question 21:
Simple interest on a sum of money for 2 years at $6\frac{1}{2}%$ per annum is Rs 5200. What will be the compound interest on the sum at the same rate for the same period?
#### Question 22:
Find the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 1200 as simple interest.
#### Question 1:
On what sum will the compound interest at 5% per annum for 2 years compounded annually be Rs 164?
#### Question 2:
Find the principal if the interest compounded annually at the rate of 10% for two years is Rs 210.
#### Question 3:
A sum amounts to Rs 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.
#### Question 4:
What sum will amount to Rs 4913 in 18 months, if the rate of interest is $12\frac{1}{2}%$ per annum, compounded half-yearly?
#### Question 5:
The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50. Find the sum.
#### Question 6:
Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs 1290 as interest compounded annually, find the sum she borrowed.
#### Question 7:
The interest on a sum of Rs 2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20.
#### Question 8:
In how much time would Rs 5000 amount to Rs 6655 at 10% per annum compound interest?
#### Question 9:
In what time will Rs 4400 become Rs 4576 at 8% per annum interest compounded half-yearly?
#### Question 10:
The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20. Find the sum.
#### Question 11:
In what time will Rs 1000 amount to Rs 1331 at 10% per annum, compound interest?
#### Question 12:
At what rate percent compound interest per annum will Rs 640 amount to Rs 774.40 in 2 years?
#### Question 13:
Find the rate percent per annum if Rs 2000 amount to Rs 2662 in $1\frac{1}{2}$ years, interest being compounded half-yearly?
#### Question 14:
Kamala borrowed from Ratan a certain sum at a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years she received Rs 210 as compound interest, but paid Rs 200 only as simple interest. Find the sum and the rate of interest.
#### Question 15:
Find the rate percent per annum, if Rs 2000 amount to Rs 2315.25 in an year and a half, interest being compounded six monthly.
#### Question 16:
Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.
#### Question 17:
Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half-yearly.
#### Question 18:
A certain sum amounts to Rs 5832 in 2 years at 8% compounded interest. Find the sum.
#### Question 19:
The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360. Find the sum.
#### Question 20:
The difference in simple interest and compound interest on a certain sum of money at $6\frac{2}{3}%$ per annum for 3 years is Rs 46. Determine the sum.
#### Question 21:
Ishita invested a sum of Rs 12000 at 5% per annum compound interest. She received an amount of Rs 13230 after n years. Find the value of n.
#### Question 22:
At what rate percent per annum will a sum of Rs 4000 yield compound interest of Rs 410 in 2 years?
#### Question 23:
A sum of money deposited at 2% per annum compounded annually becomes Rs 10404 at the end of 2 years. Find the sum deposited.
#### Question 24:
In how much time will a sum of Rs 1600 amount to Rs 1852.20 at 5% per annum compound interest?
#### Question 25:
At what rate percent will a sum of Rs 1000 amount to Rs 1102.50 in 2 years at compound interest?
#### Question 26:
The compound interest on Rs 1800 at 10% per annum for a certain period of time is Rs 378. Find the time in years.
#### Question 27:
What sum of money will amount to Rs 45582.25 at $6\frac{3}{4}%$ per annum in two years, interest being compounded annually?
#### Question 28:
Sum of money amounts to Rs 453690 in 2 years at 6.5% per annum compounded annually. Find the sum.
#### Question 1:
The present population of a town is 28000. If it increases at the rate of 5% per annum, what will be its population after 2 years?
#### Question 2:
The population of a city is 125000. If the annual birth rate and death rate are 5.5% and 3.5% respectively, calculate the population of city after 3 years.
#### Question 3:
The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.
#### Question 4:
Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 4%, 5% and 3% respectively, find the present population.
#### Question 5:
There is a continuous growth in population of a village at the rate of 5% per annum. If its present population is 9261, what it was 3 years ago?
#### Question 6:
In a factory the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters.
#### Question 7:
The annual rate of growth in population of a certain city is 8%. If its present population is 196830, what it was 3 years ago?
#### Question 8:
The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population.
#### Question 9:
The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours?
#### Question 10:
The population of a certain city was 72000 on the last day of the year 1998. During next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000?
#### Question 11:
6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year?
#### Question 12:
Aman started a factory with an initial investment of Rs 100000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.
#### Question 13:
The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago.
#### Question 14:
The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years?
#### Question 15:
The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (i) 2001 (ii) 1997.
#### Question 16:
Jitendra set up a factory by investing Rs 2500000. During the first two successive years his profits were 5% and 10% respectively. If each year the profit was on previous year's capital, compute his total profit.
#### Question 1:
Ms. Cherian purchased a boat for Rs 16000. If the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years.
#### Question 2:
The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 100000? Also, find the total depreciation during this period.
#### Question 3:
Pritam bought a plot of land for Rs 640000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years?
#### Question 4:
Mohan purchased a house for Rs 30000 and its value is depreciating at the rate of 25% per year. Find the value of the house after 3 years.
#### Question 5:
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43740, find its purchase price.
#### Question 6:
The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9680, for how much was it purchased?
#### Question 7:
The cost of a T.V. set was quoted Rs 17000 at the beginning of 1999. In the beginning of 2000 the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning of 2001. What was the cost of the T.V. set in 2001?
#### Question 8:
Ashish started the business with an initial investment of Rs 500000. In the first year he incurred a loss of 4%. However during the second year he earned a profit of 5% which in third year rose to 10%. Calculate the net profit for the entire period of 3 years.
#### Question 1:
Find the compound interest when principal = Rs 3000, rate = 5% per annum and time = 2 years.
#### Question 2:
What will be the compound interest on Rs 4000 in two years when rate of interest is 5% per annum?
#### Question 3:
Rohit deposited Rs 8000 with a finance company for 3 years at an interest of 15% per annum. What is the compound interest that Rohit gets after 3 years?
#### Question 4:
Find the compound interest on Rs 1000 at the rate of 8% per annum for $1\frac{1}{2}$ years when interest is compounded half-yearly.
#### Question 5:
Find the compound interest on Rs 160000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.
#### Question 6:
Swati took a loan of Rs 16000 against her insurance policy at the rate of $12\frac{1}{2}%$ per annum. Calculate the total compound interest payable by Swati after 3 years.
#### Question 7:
Roma borrowed Rs 64000 from a bank for $1\frac{1}{2}$ years at the rate of 10% per annum. Compute the total compound interest payable by Roma after $1\frac{1}{2}$ years, if the interest is compounded half-yearly.
#### Question 8:
Mewa Lal borrowed Rs 20000 from his friend Rooplal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.
#### Question 9:
Find the compound interest on Rs 8000 for 9 months at 20% per annum compounded quarterly.
#### Question 10:
Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum gives Rs 200 as simple interest.
#### Question 11:
Find the compound interest on Rs 64000 for 1 year at the rate of 10% per annum compounded quarterly.
#### Question 12:
Ramesh deposited Rs 7500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months.
#### Question 13:
Anil borrowed a sum of Rs 9600 to install a handpump in his dairy. If the rate of interest is $5\frac{1}{2}%$ per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years.
#### Question 14:
Surabhi borrowed a sum of Rs 12000 from a finance company to purchase a refrigerator. If the rate of interest is 5% per annum compounded annually, calculate the compound interest that Surabhi has to pay to the company after 3 years.
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$2(5c^3-3d)(5c^3+3d)$
$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $50c^6-18d^2 .$ Then use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The $GCF$ of the terms is $2$ since it is the highest number that can evenly divide (no remainder) all the given terms. Factoring the $GCF,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 50c^6-18d^2 \\\\= 2(25c^6-9d^2) .\end{array} The expressions $25c^6$ and $9d^2$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $25c^6-9d^2$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2(25c^6-9d^2) \\\\= 2(5c^3-3d)(5c^3+3d) .\end{array}
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## Do the angles in a triangle always add to 180?
A triangle’s angles add up to 180 degrees because one exterior angle is equal to the sum of the other two angles in the triangle.
In other words, the other two angles in the triangle (the ones that add up to form the exterior angle) must combine with the third angle to make a 180 angle..
## What is the sum of all three angles in a triangle?
The sum of the three angles of any triangle is equal to 180 degrees.
## Which set of angles could be the interior angles of a triangle?
Answer: 19° ,70° and 91° only one represent the interior angle of triangle.
## What is the sum of a triangle sides?
The sum of all internal angles of a triangle is always equal to 1800. This is called the angle sum property of a triangle. The sum of the length of any two sides of a triangle is greater than the length of the third side. The side opposite to the largest angle of a triangle is the largest side.
## What are the angles in a isosceles triangle?
In an isosceles triangle, there are two base angles and one other angle. The two base angles are equal to each other. So say you have an isosceles triangle, where only two sides of that triangle are equal to each other. And then you have 36 degrees as one of your base angles.
## What are alternate interior angles equal to?
The angles are positioned at the inner corners of the intersections and lie on opposite sides of the transversal. Alternate interior angles are equal if the lines intersected by the transversal are parallel. Alternate interior angles formed when a transversal crosses two non-parallel lines have no geometrical relation.
## What should the angles of a triangle add up to?
The angles in the triangle you drew all add up to 180°. Remember, all angles on a straight line add up to 180°.
## What is the sum of interior angles in a triangle?
180°Triangle/Sum of interior angles
## How do you find three angles of a triangle?
“SSS” is when we know three sides of the triangle, and want to find the missing angles. To solve an SSS triangle: use The Law of Cosines first to calculate one of the angles. then use The Law of Cosines again to find another angle.
## What is the formula for interior angles?
An interior angle is located within the boundary of a polygon. The sum of all of the interior angles can be found using the formula S = (n – 2)*180. It is also possible to calculate the measure of each angle if the polygon is regular by dividing the sum by the number of sides.
## Can a triangle have two right angles?
Answer and Explanation: Because of the fact that the sum of the three interior angles of a triangle must be 180 degrees, a triangle could not have two right angles.
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# Question 4:
The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
## Solution:
Let the required three digits in A.P. of three digit number = (a – d), a, (a + d)
Now, a – d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
According to the question number is :
100(a – d)+10a + a + d = 111 a – 99 d
Number on reversing the digits is:
100(a + d) 10 a + (a-d) = 111a +99 d
Now, as per given condition in question,
(111 a -99d) -(111a + 99d) = 594
⇒ – 198 d = 594
⇒ d = -3
So, digits of the number are [5-(-3), 5, (5+(-3)]
= 8, 5 , 2
Hence, the required number = 852.
# Question 1:
Ramkali would require Rs 5000 for getting her daughter admitted in a school after a year. She saved Rs 150 in the first month and increased her monthly by Rs 50 every month. Find, if she will be able to arrange the required money after 12 months. Which value is reflected in her efforts ?
# Question 2:
The first term of an A.P. is -5 and the last term is 45. If the sum of the terms of the A.P. is 120, then find the number of terms and the common difference.
# Question 3:
If the sum of the first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of the first 10 terms.
[CBSE 2016, 13, 12]
# Question 5:
If the sum of first m terms of an A.P. is some as the sum of its first n terms, show that the sum of its first (m + n) terms is zero. [CBSE 2019]
# Question 6:
If the sum of first ‘p’ terms of an A.P. is ‘q’ and the sum of first ‘q’ terms is ‘p’ ; then show that the sum of the first (p + q) terms is {-(p + q)}.
# Question 7:
Show that the sum of all terms of an A.P. whose first term is a, the second term is b and the last term is c is equal to $\dfrac{(a + c)(b + c -2a)}{2(b-a)}$. [CBSE 2020]
# Question 8:
Yasmeen saves ₹ 32 during the first month ₹ 36 in the second month and ₹ 40 in the third month. If she continues to save in this manner, in how many months will she save ₹ 2000?
# Question 9:
A child puts one five-rupees coin of her savings in the piggy bank on the first day. She increases her saving by one five-rupee coin daily. If the piggy bank can hold 190 coins of five rupees in all, find the number of days she can continue to put the five -rupee coins into it and find the total money she saved. [CBSE 2017]
# Question 10:
A sum of ₹ 4,250 is to be used to give 10 cash prizes to students of a school for their overall academic performance. If each prize is ₹ 50 less than its preceding prize, find the value of each of the prizes.
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# How you can Practice Chebyshev’s Theorem in Excel
Chebyshev’s Theorem states that for any quantity ok more than 1, a minimum of 1 – 1/ok2 of the information values in any formed distribution lie inside of ok usual deviations of the cruel.
As an example, for any formed distribution a minimum of 1 – 1/32 = 88.89% of the values within the distribution will lie inside of 3 usual deviations of the cruel.
This instructional illustrates a number of examples of easy methods to practice Chebyshev’s Theorem in Excel.
Instance 1: Utility Chebyshev’s Theorem to seek out what proportion of values will fall between 30 and 70 for a dataset with a cruel of fifty and usual redirection of 10.
First, decide the price for ok. We will be able to do that through learning what number of usual deviations away 30 and 70 are from the cruel:
(30 – cruel) / usual redirection = (30 – 50) / 10 = -20 / 10 = -2
(70 – cruel) / usual redirection = (70 – 50) / 10 = 20 / 10 = 2
The values 30 and 70 are 2 usual deviations underneath and above the cruel, respectively. Thus, ok = 2.
We will be able to later worth please see components in Excel to seek out the minimal proportion of values that fall inside of 2 usual deviations of the cruel for this dataset:
The share of values that fall inside of 30 and 70 for this dataset might be a minimum of 75%.
Instance 2: Utility Chebyshev’s Theorem to seek out what proportion of values will fall between 20 and 50 for a dataset with a cruel of 35 and usual redirection of five.
First, decide the price for ok. We will be able to do that through learning what number of usual deviations away 20 and 50 are from the cruel:
(20 – cruel) / usual redirection = (20 – 35) / 5 = -15 / 5 = -3
(50 – cruel) / usual redirection = (50 – 35) / 5 = 15 / 5 = 3
The values 20 and 50 are 3 usual deviations underneath and above the cruel, respectively. Thus, ok = 3.
We will be able to later worth please see components in Excel to seek out the minimal proportion of values that fall inside of 3 usual deviations of the cruel for this dataset:
The share of values that fall inside of 20 and 50 for this dataset might be a minimum of 88.89%.
Instance 3: Utility Chebyshev’s Theorem to seek out what proportion of values will fall between 80 and 120 for a dataset with a cruel of 100 and usual redirection of five.
First, decide the price for ok. We will be able to do that through learning what number of usual deviations away 80 and 120 are from the cruel:
(80 – cruel) / usual redirection = (80 – 100) / 5 = -20 / 5 = -4
(120 – cruel) / usual redirection = (120 – 100) / 5 = 20 / 5 = 4
The values 80 and 120 are 4 usual deviations underneath and above the cruel, respectively. Thus, ok = 4.
We will be able to later worth please see components in Excel to seek out the minimal proportion of values that fall inside of 4 usual deviations of the cruel for this dataset:
The share of values that fall inside of 80 and 120 for this dataset might be a minimum of 93.75%.
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# Proving Average Rate of Change
## Presentation on theme: "Proving Average Rate of Change"— Presentation transcript:
Proving Average Rate of Change
Key Concepts: The rate of change is a ratio describing how one quantity changes as another quantity changes. Slope can be used to describe the rate of change. The slope of a line is the ratio of the change in y-values to the change in x-values. A positive rate of change expresses an increase over time. A negative rate of change expresses a decrease over time.
Key Concepts, continued.
Linear functions have a constant rate of change, meaning values increase or decrease at the same rate over a period of time. Not all functions change at a constant rate. The rate of change of an interval, or a continuous portion of a function, can be calculated. The rate of change of an interval is the average rate of change for that period.
Key Concepts, continued.
Intervals can be noted using the format [a, b], where a represents the initial x value of the interval and b represents the final x value of the interval. Another way to state the interval is a ≤ x ≤ b. A function or interval with a rate of change of 0 indicates that the line is horizontal. Vertical lines have an undefined slope. An undefined slope is not the same as a slope of 0. This occurs when the denominator of the ratio is 0.
Calculating Rate of Change from a Table
Choose two points from the table. Assign one point to be (x1, y1) and the other point to be (x2, y2). Substitute the values into the slope formula. The result is the rate of change for the interval between the two points chosen.
Calculating Rate of Change from an Equation of a Linear Function
Transform the given linear function into slope- intercept form, f(x) = mx + b. Identify the slope of the line as m from the equation. The slope of the linear function is the rate of change for that function.
Calculating Rate of Change of an Interval from an Equation of an Exponential Function
Determine the interval to be observed. Determine (x1, y1) by identifying the starting x-value of the interval and substituting it into the function. Solve for y. Determine (x2, y2) by identifying the ending x-value of the interval and substituting it into the function. Substitute (x1, y1) and (x2, y2) into the slope formula to calculate the rate of change. The result is the rate of change for the interval between the two points identified.
Remember… The rate of change between any two points of a linear function will be equal The rate of change between any two points of any other function will not be equal, but will be an average for that interval.
Practice In 2008, about 66 million U.S. households had both landline phones and cell phones. This number decreased by an average of 5 million households per year. Use the table to the right to calculate the rate of change for the interval [2008, 2011].
The Solution Determine (x1, y1) and (x2, y2).
(x1, y1) is (2008, 66) (x2, y2) is (2011, 51) Using the slope formula = –5 The rate of change for the interval [2008, 2011] is 5 million households per year.
Thanks for Watching! ~Ms. Dambreville
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We show another application of Menalaus’s Theorem, that, together with some infinitesimal calculus, will yield an unexpectedly simple result.
Consider a square of side $$1$$ and divide the left and bottom sides into $$n$$ segments of equal length. Connect then their end points as shown in the Figures below, for $$n=3, 5, 8$$. We want to know what is the shaded area, as a function of $$n$$, and in particular when $$n\to\infty$$ (red line in the bottom Figure). As a final step, we also want to determine the function whose graph is represented by that red line.
We could approach the problem using coordinate geometry, but it will be instructive to use again basic theorems of Euclidean geometry.
To this purpose, let us label the main points of the square as represented below, for a generic value of $$n$$. We aim at determining the required area by adding the area of triangles $$\triangle A_iA_{i-1}C_i$$. Two of them are dashed below as an example.
• Note that the above mentioned triangles have all base equal to $$\frac1{n}$$. We therefore only need to compute the altitude $$\overline{C_iH_i}$$.
• At this aim, for $$i=1,2,\dots,n-1$$, consider the triangle $$\triangle A_nA_{i-1}B_i$$ cut by the line $$\overline{A_iB_{i+1}}$$. Apply Menelaus’s Theorem to find $\frac{\overline{A_{i-1}C_i}}{\overline{C_iB_i}}=\frac{i+1}{n-i}.$
• Use the similarity $$\triangle A_nA_{i-1}B_i \sim \triangle H_iA_{i-1}C_i$$ to determine $\overline{C_iH_i} = \frac{i(i+1)}{n(n+1)}.\tag{1}\label{eq2739:1}$
• Observe that equation \eqref{eq2739:1} is valid also for the leftmost triangle, whose altitude is equal to $$1$$.
• Conclude that the shaded area is equal to $\mathcal A(n)=\frac1{2n^2(n+1)}\sum_{i=1}^n i(i+1).$
• Use the known results $$\sum_{i=1}^n i = \frac{n(n+1)}2$$ and $$\sum_{i=1}^n i^2 = \frac{(2n+1)n(n+1)}6$$ to determine the following expression, $\mathcal A(n) = \frac{n+2}{6n},$ so that the shaded area tends to $$\frac16$$ when $$n\to\infty$$.
We now introduce cartesian coordinates with the origin set in $$O\equiv A_n$$.
Consider the piecewise linear function whose graph is obtained by connecting the points $$B_n\equiv C_n, C_{n-1},\dots,C_2,C_1,C_0\equiv A_0$$ in the Figure above. We will call $$f_n(x)$$ this function.
Our purpose is that of finding the analytical expression of $$f(x)$$, defined as
$f(x) = \lim_{n\to\infty} f_n(x).\tag{2}\label{eq2739:2}$
• For $$n=1,2,\dots,n$$, call $$x_i$$ and $$y_i$$ abscissae and ordinates, respectively, of the point $$M_i$$, midpoint of the segment $$C_{i-1}C_i$$.
• Use equation \eqref{eq2739:1} and the symmetry of the graph to calculate $x_i = \frac{i^2}{n(n+1)}$ and $y_i = \frac{(n-i)^2}{n(n+1)}.$
• Write $$y_i$$ in terms of $$x_i$$. You will get \begin{eqnarray}y_i &=& \left(\frac{n}{\sqrt{n(n+1)}}-\sqrt{x_i}\right)^2.\end{eqnarray} When $$n\to\infty$$ the above expression tends to $f(x) = \left(1-\sqrt x\right)^2.$
• Verify that $\mathcal A_{\infty} = \int_0^1 f(x) dx = \frac16.$
We want to show more formally the convergence \eqref{eq2739:2}, and that the convergence is also uniform. Fix $$\varepsilon>0$$.
• Using previous results, show that \begin{eqnarray}|f_n(x_i) -f(x_i)|&\leq& \left| \frac{n^2}{n(n+1)}-1+\right.\\&+&\left. 2\left(1-\frac{n}{\sqrt{n(n+1)}}\right) \right|\underset{n\to\infty}{\to} 0.\end{eqnarray}Choose $$N_1$$ so that the above quantity is less then $$\frac{\varepsilon}3$$, for all $$n> N_1$$.
• Given a point with coordinates $$(x,f_n(x))$$, lying between, say, $$C_{i-1}$$ and $$C_i$$, show that \begin{eqnarray}|f_n(x_i)-f_n(x)| &\leq& \overline{C_nC_{n-1}} =\\&=& \frac{2n}{n(n+1)}\underset{n\to\infty}{\to} 0.\end{eqnarray}Choose $$N_2$$ so that the above quantity is less then $$\frac{\varepsilon}3$$, for all $$n> N_2$$.
• $$f(x)$$ is uniformly continuous in $$[0,1]$$ (why?), so that we can guarantee $|f(x_i) – f(x)| < \frac{\varepsilon}3,$ provided that $|x_i-x| < \delta,\tag{3}\label{eq2739:3}$ for some $$\delta$$ independently of $$x$$. Note that $|x_i-x| \leq \overline{C_1A_n} = \frac{2n}{n(n+1)}\underset{n\to\infty}{\to} 0,$so that \eqref{eq2739:3} is satisfied for sufficiently high $$n$$, say $$n> N_3$$.
• Using triangular inequality, show that \begin{eqnarray}|f_n(x) -f(x)| &\leq& |f_n(x)-f_n(x_i)| + \\&+&|f_n(x_i) – f(x_i)| +\\&+& |f(x_i) – f(x)|< \varepsilon\end{eqnarray} for all $$n>\max(N_1,N_2,N_3)$$, for all $$x\in [0,1]$$, and conclude that the sequence $$(f_n(x))$$ converges to $$f(x)$$ uniformly.
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# What is the sum and product of roots of cubic equation?
## What is the sum and product of roots of cubic equation?
Find the sum of the squares of the roots of the cubic equation x 3 + 3 x 2 + 3 x = 3 x^3 + 3x^2 + 3x = 3 x3+3×2+3x=3….Relation between coefficients and roots:
Root expression Equals to
p q r pqr pqr − d a -\frac{d}{a} −ad
## What is the sum of roots in a cubic equation?
The sum of roots is -b/a and the product of roots is -d/a.
What is the formula of sum of roots and product of roots?
For any quadratic equation ax2 + bx + c = 0, the sum of the roots = -b/a. the product of the roots = c/a.
### How do you find the sum and product of a cubic polynomial?
Hint: A cubic polynomial is the polynomial whose degree is 3 and it has 3 roots. We will use the sum, sum of the products and products given in the question to find the cubic polynomial. sum of products = α+β+γ=−ba, where b is the coefficient of x2 and a is the coefficient of x3.
### How do you find the product of a cubic polynomial?
α,β & γ are the zeroes of cubic polynomial P(x)=ax3+bx2+cx+d,(a=0) then product of their zeroes [α. β.
What is the formula of product of cubic polynomial?
We know that the general form of a cubic polynomial is ax3 + bx2 + cx + d and the zeroes are α, β, and γ. Let’s look at the relation between sum, and product of its zeroes and coefficients of the polynomial. α + β + γ = – b / a. αβ + βγ + γα = c / a. α x β x γ = – d / a.
#### What is the sum to product formula?
The sum-to-product formulas allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine. Let u + v 2 = α u + v 2 = α and.
#### How do you find the roots of a cubic equation?
By the Fundamental Theorem of Algebra, we have ax^3 + bx^2 + cx + d, which can be expressed as a(x-r)(x-s)(x-t). WLOG let the equation give r. Then, simply divide the cubic by (x-r) and we get a quadratic whose roots are the remaining two roots.
What is the sum of cubic polynomial?
## What is the product of cubic polynomial?
Let p(x)=ax3+bx2+cx+d be the cubic polynomial. Then, the product of zeroes of p(x) is given by a−d. Thus, zeros of a cubic polynomial is given by. Coefficient of x3−(The constant Term) Hence, option D is correct.
## How do you find the roots of a cubic equation in Class 10?
Complete step-by-step answer: We start the solution by getting one root of the equation ${{x}^{3}}-23{{x}^{2}}+142x-120=0$ by hit and trial method. Next step is to solve the equation obtained in (i) which is a quadratic equation, to get the remaining two roots.
How do you write the roots of a cubic equation?
Approach: Let the root of the cubic equation (ax3 + bx2 + cx + d = 0) be A, B and C. Then the given cubic equation can be represents as: ax3 + bx2 + cx + d = x3 – (A + B + C)x2 + (AB + BC +CA)x + A*B*C = 0. Therefore using the above relation find the value of X, Y, and Z and form the required cubic equation.
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35 percent of 4050
Here we will show you how to calculate thirty-five percent of four thousand fifty. Before we continue, note that 35 percent of 4050 is the same as 35% of 4050. We will write it both ways throughout this tutorial to remind you that it is the same.
35 percent means that for each 100, there are 35 of something. This page will teach you three different methods you can use to calculate 35 percent of 4050.
We think that illustrating multiple ways of calculating 35 percent of 4050 will give you a comprehensive understanding of what 35% of 4050 means, and provide you with percent knowledge that you can use to calculate any percentage in the future.
To solidify your understanding of 35 percent of 4050 even further, we have also created a pie chart showing 35% of 4050. On top of that, we will explain and calculate "What is not 35 percent of 4050?"
Calculate 35 percent of 4050 using a formula
This is the most common method to calculate 35% of 4050. 4050 is the Whole, 35 is the Percent, and the Part is what we are calculating. Below is the math and answer to "What is 35% of 4050?" using the percent formula.
(Whole × Percent)/100 = Part
(4050 × 35)/100 = 1417.5
35% of 4050 = 1417.5
Get 35 percent of 4050 with a percent decimal number
You can convert any percent, such as 35.00%, to 35 percent as a decimal by dividing the percent by one hundred. Therefore, 35% as a decimal is 0.35. Here is how to calculate 35 percent of 4050 with percent as a decimal.
Whole × Percent as a Decimal = Part
4050 × 0.35 = 1417.5
35% of 4050 = 1417.5
Get 35 percent of 4050 with a fraction function
This is our favorite method of calculating 35% of 4050 because it best illustrates what 35 percent of 4050 really means. The facts are that it is 35 per 100 and we want to find parts per 4050. Here is how to illustrate and show you the answer using a function with fractions.
Part 4050
=
35 100
Part = 1417.5
35% of 4050 = 1417.5
Note: To solve the equation above, we first multiplied both sides by 4050 and then divided the left side to get the answer.
35 percent of 4050 illustrated
Below is a pie chart illustrating 35 percent of 4050. The pie contains 4050 parts, and the blue part of the pie is 1417.5 parts or 35 percent of 4050.
Note that it does not matter what the parts are. It could be 35 percent of 4050 dollars, 35 percent of 4050 people, and so on. The pie chart of 35% of 4050 will look the same regardless what it is.
What is not 35 percent of 4050?
What is not 35 percent of 4050? In other words, what is the red part of our pie above? We know that the total is 100 percent, so to calculate "What is not 35%?" you deduct 35% from 100% and then take that percent from 4050:
100% - 35% = 65%
(4050 × 65)/100 = 2632.5
Another way of calculating the red part is to subtract 1417.5 from 4050.
4050 - 1417.5 = 2632.5
That is the end of our tutorial folks. We hope we accomplished our goal of making you a percent expert - at least when it comes to calculating 35 percent of 4050.
Percent of a Number
Go here if you need to calculate the percent of a different number.
35 percent of 4060
Here is the next percent tutorial on our list that may be of interest.
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Perimeter
In Geometry, the word Perimeter is used either to indicate path or length. The word Perimeter comes from two Greek words "PERI" (which means around) and "METER"(which means measure). A Perimeter of any figure can be defined in various ways.
• A Perimeter of any polygon is the total distance along the outside of the polygon.
• A Perimeter of a figure is total length of all its sides.
• A Perimeter is the length of the boundary of a figure.
• A Perimeter is the length of the outline of a shape.
Computing Perimeter has many practical applications. For example, it could be used to find out the revolutions of a wheel, length of a fence surrounding a yard or garden, the amount of wool wound around a spool is related to perimeter of the spool, the distance ran by a jogger around a path, etc.
Perimeter formulas (small description of triangle and rectangle and their formulas and diagram)
Perimeter of Regular Shapes:
The perimeter of regular polygons (a figure which has all of its sides equal and all angles equal) is given by the product of number of sides with length of its side.
Perimeter of few regular polygons is given below with diagrams and formulas.
Let "s" be the length of each side of these figures.
Equilateral triangle:
Perimeter = s + s + s
Perimeter = 3 x s
Rhombus:
Perimeter = s + s + s + s
Perimeter = 4 x s
Square:
Perimeter = s + s + s + s
Perimeter = 4 x s
Regular pentagon:
Perimeter = s + s + s + s + s
Perimeter = 5 x s
Regular hexagon:
Perimeter = s + s + s + s + s + s
Perimeter = 6 x s
Perimeter of other Shapes:
Other shapes like Rectangle, Triangle, Parallelogram, Trapezium, Kite also have Perimeter. The following are the diagrams and formulas for the same.
Rectangle:
Perimeter = 2(l+b)
Where,
l = length of the rectangle
b = width of the rectangle
Triangle:
Perimeter = a + b + c
Where, a, b, c are lengths of each side.
Parallelogram:
Perimeter = 2(a+b)
Where,
a = length of the parallelogram
b = width of the parallelogram
Trapezium:
Perimeter = a + b + c+ d
Where,
a, b, c are lengths of each side.
Kite:
Perimeter = 2(a+b)
Where,
a is the length of each side in one pair
b is the length of each side in the other pair
Circle:
The perimeter of a circle is generally known as the circumference.
Circumference of a Circle = 2pi r, where: r is the radius of circle
Circumference of a Circle = pi d, where: d is the diameter of circle.
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Elementary Algebra
$9$
We start with the given expression: $3xy-8y+5x$ We plug in the given values for $x$ and $y$: $3(7)(-2)-8(-2)+5(7)$ Order of operations states that first, we perform operations inside grouping symbols, such as parentheses, brackets, and fraction bars. Then, we simplify powers. Then, we multiply and divide from left to right. Then, we add and subtract from left to right. We follow order of operations to simplify: First, we multiply from left to right: $21(-2)-8(-2)+5(7)=-42-8(-2)+5(7)=-42-(-16)+5(7)=-42-(-16)+35$ Next, we add and subtract from left to right: $-26+35=9$
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Categories
## Ratio of the area of Square and Pentagon | AMC 8, 2013
Try this beautiful problem from Geometry: Ratio of the area between Square and Pentagon.
## Ratio of the area between Square and Pentagon – AMC-8, 2013 – Problem 24
Squares ABCD ,EFGH and GHIJ are equal in area .Points C and D are the mid points of the side IH and HE ,respectively.what is the ratio of the area of the shaded pentagon AJICB to the sum of the areas of the three squares?
• $\frac{1}{4}$
• $\frac{1}{3}$
• $\frac{3}{8}$
### Key Concepts
Geometry
Area of square
Area of Triangle
Answer:$\frac{1}{3}$
AMC-8(2013) Problem 24
Pre College Mathematics
## Try with Hints
extend IJ until it hits the extension of AB .
Can you now finish the problem ……….
find the area of the pentagon
can you finish the problem……..
First let L=2 (where L is the side length of the squares) for simplicity. We can extend IJ until it hits the extension of AB . Call this point X.
Then clearly length of AX=3 unit & length of XJ = 4 unit .
Therefore area of $\triangle AXJ= (\frac{1}{2} \times AX \times XJ)=(\frac{1}{2} \times 4 \times 3)=6$ sq.unit
And area of Rectangle BXIC= $( 1 \times 2)$=2 sq.unit
Therefore the of the pentagon ABCIJ=6-2=4 sq.unit
The combined area of three given squares be $(3 \times 2^2)$=12 sq.unit
Now, the ratio of the shaded area(pentagon) to the combined area of the three squares is $\frac{4}{12}=\frac{1}{3}$
.
Categories
## Area of Triangle and Square | AMC 8, 2012 | Problem 25
Try this beautiful problem from AMC-8-2012 (Geometry) based on area of a Triangle and square and congruency.
## Area of a Triangle- AMC 8, 2012 – Problem 25
A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length a, and the other of length b. What is the value of ab?
• $\frac{1}{5}$
• $\frac{2}{5}$
• $\frac{1}{2}$
### Key Concepts
Geometry
Square
Triangle
Answer:$\frac{1}{2}$
AMC-8, 2014 problem 25
Challenges and Thrills of Pre College Mathematics
## Try with Hints
Find the area of four triangles
Can you now finish the problem ……….
Four triangles are congruent
can you finish the problem……..
Total area of the big square i.e ABCD is 5 sq.unit
and total area of the small square i.e EFGH is 4 sq.unit
So Toal area of the $(\triangle AEH + \triangle BEF + \triangle GCF + \triangle DGH)=(5-4)=1$ sq.unit
Now clearly Four triangles$( i.e \triangle AEH , \triangle BEF , \triangle GCF , \triangle DGH )$ are congruent.
Total area of the four congruent triangles formed by the squares is 5-4=1 sq.unit
So area of the one triangle is $\frac{1}{4}$ sq.unit
Now “a” be the height and “b” be the base of one triangle
The area of one triangle be $(\frac{1}{2} \times base \times height )$=$\frac{1}{4}$
i.e $(\frac{1}{2} \times b \times a)$= $\frac{1}{4}$
i.e $ab$= $\frac{1}{2}$
Categories
## Area of star and circle | AMC-8, 2012|problem 24
Try this beautiful problem from Geometry: Ratio of the area of the star figure to the area of the original circle
## Area of the star and circle – AMC-8, 2012 – Problem 24
A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?
• $\frac{1}{\pi}$
• $\frac{4-\pi}{\pi}$
• $\frac{\pi – 1}{\pi}$
### Key Concepts
Geometry
Circle
Arc
Answer:$\frac{4-\pi}{\pi}$
AMC-8 (2012) Problem 24
Pre College Mathematics
## Try with Hints
Clearly the square forms 4-quarter circles around the star figure which is equivalent to one large circle with radius 2.
Can you now finish the problem ……….
find the area of the star figure
can you finish the problem……..
Draw a square around the star figure. Then the length of one side of the square be 4(as the diameter of the circle is 4)
Clearly the square forms 4-quarter circles around the star figure which is equivalent to one large circle with radius 2.
The area of the above circle is $\pi (2)^2 =4\pi$
and the area of the outer square is $(4)^2=16$
Thus, the area of the star figure is $16-4\pi$
Therefore $\frac{(the \quad area \quad of \quad the \quad star \quad figure)}{(the \quad area \quad of \quad the \quad original \quad circle )}=\frac{16-4\pi}{4\pi}$
= $\frac{4-\pi}{\pi}$
Categories
## Hexagon and Triangle |AMC 8- 2015 -|Problem 21
Try this beautiful problem from Geometry based on hexagon and Triangle.
## Area of Triangle | AMC-8, 2015 |Problem 21
In The given figure hexagon ABCDEF is equiangular ,ABJI and FEHG are squares with areas 18 and 32 respectively.$\triangle JBK$ is equilateral and FE=BC. What is the area of $\triangle KBC$?
• 9
• 12
• 32
### Key Concepts
Geometry
Triangle
hexagon
Answer:$12$
AMC-8, 2015 problem 21
Pre College Mathematics
## Try with Hints
Clearly FE=BC
Can you now finish the problem ……….
$\triangle KBC$ is a Right Triangle
can you finish the problem……..
Clearly ,since FE is a side of square with area 32
Therefore FE=$\sqrt 32$=$4\sqrt2$
Now since FE=BC,We have BC=$4\sqrt2$
Now JB is a side of a square with area 18
so JB=$\sqrt18$=$3\sqrt2$. since $\triangle JBK$ is equilateral BK=$3\sqrt2$
Lastly $\triangle KBC$ is a right triangle ,we see that
$\angle JBA + \angle ABC +\angle CBK +\angle KBJ$ =$360^\circ$
i.e$90^\circ + 120^\circ +\angle CBK + 60^\circ=360^\circ$
i.e $\angle CBK=90^\circ$
So $\triangle KBC$ is a right triangle with legs $3\sqrt 2$ and $4\sqrt2$
Now its area is $3\sqrt2 \times 4\sqrt 2 \times \frac {1}{2}$=$\frac{24}{2}$=12
Categories
## Area of a Triangle -AMC 8, 2018 – Problem 20
Try this beautiful problem from Geometry based on Area of a Triangle Using similarity
## Area of Triangle – AMC-8, 2018 – Problem 20
In $\triangle ABC$ , a point E is on AB with AE = 1 and EB=2.Point D is on AC so that DE $\parallel$ BC and point F is on BC so that EF $\parallel$ AC.
What is the ratio of the area of quad. CDEF to the area of $\triangle ABC$?
• $\frac{2}{3}$
• $\frac{4}{9}$
• $\frac{3}{5}$
### Key Concepts
Geometry
Area
similarity
Answer:$\frac{4}{9}$
AMC-8, 2018 problem 20
Pre College Mathematics
## Try with Hints
$\triangle ADE$ $\sim$ $\triangle ABC$
Can you now finish the problem ……….
$\triangle BEF$ $\sim$ $\triangle ABC$
can you finish the problem……..
Since $\triangle ADE$$\sim$ $\triangle ABC$
$\frac{ \text {area of} \triangle ADE}{ \text {area of} \triangle ABC}$=$\frac{AE^2}{AB^2}$
i.e $\frac{\text{area of} \triangle ADE}{\text{area of} \triangle ABC}$ =$\frac{(1)^2}{(3)^2}$=$\frac{1}{9}$
Again $\triangle BEF$ $\sim$ $\triangle ABC$
Therefore $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$=$\frac{BE^2}{AB^2}$
i.e $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$ =$\frac{(2)^2}{(3)^2}$=$\frac{4}{9}$
Therefore Area of quad. CDEF =$\frac {4}{9}$ of area $\triangle ABC$
i.e The ratio of the area of quad.CDEF to the area of $\triangle ABC$ is $\frac{4}{9}$
Categories
## Area of cube’s cross section |Ratio | AMC 8, 2018 – Problem 24
Try this beautiful problem from Geometry: Ratio of the area of cube’s cross section . You may use sequential hints to solve the problem.
## Area of cube’s cross section – AMC-8, 2018 – Problem 24
In the cube ABCDEFGH with opposite vertices C and E ,J and I are the mid points of segments FB and HD respectively .Let R be the ratio of the area of the cross section EJCI to the area of one of the faces of the cube .what is $R^2$ ?
• $\frac{5}{4}$
• $\frac{3}{2}$
• $\frac{4}{3}$
### Key Concepts
Geometry
Area
Pythagorean theorem
Answer:$\frac{3}{2}$
AMC-8(2018) Problem 24
Pre College Mathematics
## Try with Hints
EJCI is a rhombus by symmetry
Can you now finish the problem ……….
Area of rhombus is half product of its diagonals….
can you finish the problem……..
Let Side length of a cube be x.
then by the pythagorean theorem$EC=X \sqrt {3}$
$JI =X \sqrt {2}$
Now the area of the rhombus is half product of its diagonals
therefore the area of the cross section is $\frac {1}{2} \times (EC \times JI)=\frac{1}{2}(x\sqrt3 \times x\sqrt2)=\frac {x^2\sqrt6}{2}$
This shows that $R= \frac{\sqrt6}{2}$
i.e$R^2=\frac{3}{2}$
Categories
## Radius of a Semi Circle -AMC 8, 2017 – Problem 22
Try this beautiful problem from Geometry based on the radius of a semi circle and tangent of a circle.
## AMC-8(2017) – Geometry (Problem 22)
In the right triangle ABC,AC=12,BC=5 and angle C is a right angle . A semicircle is inscribed in the triangle as shown.what is the radius of the semi circle?
• $\frac{7}{6}$
• $\frac{10}{3}$
• $\frac{9}{8}$
### Key Concepts
Geometry
congruency
similarity
Answer:$\frac{10}{3}$
AMC-8(2017)
Pre College Mathematics
## Try with Hints
Here O is the center of the semi circle. Join o and D(where D is the point where the circle is tangent to the triangle ) and Join OB.
Can you now finish the problem ……….
Now the $\triangle ODB$and $\triangle OCB$ are congruent
can you finish the problem……..
Let x be the radius of the semi circle
Now the $\triangle ODB$ and $\triangle OCB$ we have
OD=OC
OB=OB
$\angle ODB$=$\angle OCB$= 90 degree`
so $\triangle ODB$ and $\triangle OCB$ are congruent (by RHS)
BD=BC=5
And also $\triangle ODA$ and $\triangle BCA$ are similar….
$\frac{8}{12}$=$\frac{x}{5}$
i.e x =$\frac{10}{3}$
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Q:
A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If the coin turns up heads, what is the probability that it is the two-headed coin? Is the answer 1/3? Was the answer intuitive?
Accepted Solution
A:
Answer: Our required probability is $$\dfrac{1}{2}$$Step-by-step explanation:Since we have given that Number of coins = 3Number of coin has 2 heads = 1Number of fair coins = 2Probability of getting one of the coin among 3 = $$\dfrac{1}{3}$$So, Probability of getting head from fair coin = $$\dfrac{1}{2}$$Probability of getting head from baised coin = 1Using "Bayes theorem" we will find the probability that it is the two headed coin is given by$$\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}$$Hence, our required probability is $$\dfrac{1}{2}$$No, the answer is not $$\dfrac{1}{3}$$
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# 0.10 Ch 10: hypothesis testing of two means and two proportions
Page 1 / 1
This module is the complementary teacher's guide for the "Hypothesis Testing: Two Population Means and Two Population Proportions" chapter of the Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean.
The comparison of two groups is done constantly in business, medicine, and education, to name just a few areas. You can start this chapter by asking students if they have read anything on the Internet or seen on television any studies that involve two groups. Examples include diet versus hypnotism, Bufferin® with aspirin versus Tylenol®, Pepsi Cola® versus Coca Cola®, and Kellogg's Raisin Bran® versus Post Raisin Bran®. There are hundreds of examples on the Internet, in newspapers, and in magazines.
This chapter covers independent groups for two population means and two population proportions and matched or paired samples. The module relies heavily on technology. Instructions for the TI-83/84 series of calculators are included for each example. If you and your class are interested, the formulas for the test statistics are included in the text.
Doing problems 1 - 10 in the Homework helps the students to determine what kind of hypothesis test they should perform.
## Matched or paired samples
A course is designed to increase mathematical comprehension. In order to evaluate the effectiveness of the course, students are given a test before and after the course. The sample data is:
Before Course 90 100 160 112 95 190 125 After Course 120 95 150 150 100 200 120
## Two proportions, independent groups
Suppose in the last local election, among 240 30-45 year olds, 45% voted and among 260 46-60 year olds, 50% voted. Does the data indicate that the proportion of 30-45 year olds who voted is less than the proportion of 46-60 year olds? Test at a 1% level of significance.
## Firm a:
• ${N}_{A}=\text{20}$
• ${S}_{A}=\text{100}$ $\overline{{X}_{A}}=\text{1500}$
## Firm b:
• ${N}_{B}=\text{22}$
• ${S}_{B}=\text{200}$ $\overline{{X}_{B}}=\text{1900}$
Test the claim that the average price of Firm A's laptop is no different from the average price of Firm B's laptop.
## Calculator instructions
If you use the TI83/84 series, the functions are located in STATS TESTS. The function for two proportions is 2-PropZTest, the function for two means is 2-SampTTest if the population standard deviations are not known and 2-SampZTest if the population standard deviations are known (highly unlikely). The function for matched pairs is T-test (the same test used for test of a single mean) because you combine two measurements for each object into a single set of "difference" data. For the function 2-SampTTest, answer "NO" to "Pooled."
## Assign practice
Have students do the Practice 1 and Practice 2 collaboratively in class. These practices are for two proportions and two means. For matched pairs, you could have them do Example 10-7 in the text.
## Assign homework
Assign Homework . Suggested homework problems: 1 - 10, 11, 13, 15, 17, 19, 23, 25, 31, 39 - 52.
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Tallying it Up
15 teachers like this lesson
Print Lesson
## Objective
SWBAT generate data flipping coins, record the data using tally marks, and count the tally marks by skip counting by fives.
#### Big Idea
In this collaborative lesson students will use "Head, or Tails" to learn how to tally count by fives and add on left over numbers.
## Warm Up
20 minutes
I start this lesson by saying, “How many of you would like a penny?”
After that, I say, “Today you will learn how to tally and how to count by fives using pennies!” Then, I invite students to the carpet. I give them all a penny. I ask them to look at their pennies carefully. I ask them, “What do you notice about your penny?” If students notice that each penny has two sides, I ask, “Does anyone know what they are called?” (Heads and tails).
I ask, “Why do you think they are called that?
Some students note the side with the face on it is the head, and the side without the face is the tail. I tell students we will be flipping their pennies to see how many times it falls on heads or tails. They all seem pretty excited about flipping coins!
I want students to connect making bunches of five while tallying to the idea of bunching five pennies together to make a nickel - all within the context of exploring a problem.
MP1- Make Sense of Problems & persevering in solving them.
I collect the pennies from students and ask them to move into smaller groups no larger than six.
## Collaborate
20 minutes
After the students move into their small groups, I tell them we are going to try a fun experiment.
"Each group will be given five pennies, one per student, to shake in your hand. Then each of you will open your hands and gently put the pennies on your desk. After that each group will count how many times the pennies land heads up and how many times the pennies land tails up."
I demonstrate how to shake the pennies and how to mark their tally marks on their tally sheetWhile I am demonstrating I ask students howto determine how many tally marks to mark. I also ask them to tell me how to correctly count their tally marks (Counting by fives).
When a level of understanding is reached, I let students go to work.
While students are working, I keep track of the number of heads and the number of tails on the dry erase board that I placed at each group. Since this is their first time taking tally marks, I want them to check their marks with the marks I placed on the board. I hope that they will use my marks as a guide to assist them in the learning process. To reinforce learning I continue asking probing questions to check for understanding:
• Can some explain what you are going to be doing?
• How many heads/tails are there?
• Do you agree? Why or why not?
• Does anyone have the same answer but a different way to explain it?
• Have you compared your work with anyone else’s?
• What did other members of your group try?
• How would you describe the problem in your own words?
## Making the Connection
10 minutes
After students have reached a level of comfort in this exercise, I call them back to the carpet. I ask several questions to check for understanding, and to help them make the connection between tallying and counting by fives:
• Do you think the pennies landed heads up more often or tails up more often?
• How do you know?
Some students are able to make the connection by counting by fives using the tally marks. However, some students continue to add by ones. I make a note of this to address throughout the lesson.
Then, I ask students to count to check how many tally marks are in each group? We started counting the tally marks in the heads column first from each group (I used a large tally board to write the tally marks on). Then we count the tally marks for the tails column. After we finished writing all of the tally marks, I ask students to think of a way to use skip counting to help us count faster. I tell them the more they practice counting the better they will become. They all seem to agree that practice makes perfect.
I ask student volunteers to write next to each group of five tally marks as you count by fives (see: example chart). I repeat the same for the tails column. I ask students, “What do you notice about our chart?” I allowed them time to observe the chart, and to make connections (each complete tally marks represent five, when we to count tally marks we can do it faster by counting by fives.)
I ask students to count by five with me as I write the numbers. I begin at the bottom by writing 0 and then 5 above it (see: counting strip). I tell students we will use this strip to help us count forward and backwards by fives each morning. I post this strip on the board, and ask students to return to their seats. MP6-Attending to precision.
## Closing
10 minutes
For the closing students are back at their seats and eager to talk about what they have learned. I have students model and explain what they learned about tallying.
I ask students who would like to show us how to write 15 using tally marks? I ask them to demonstrate how to use tally marks to show 15 on the board. After that, I ask students to count by fives to check their answer. Then, I ask students to tell how many groups of five tally marks do we have? (3) How many extra tally marks do we have? (0) I repeat with 7, 25, 16, 38, and 41.
I pay particular attention to some of my students struggling to explain how to add on left over numbers by counting by ones. This tells me I need to adjust future lessons on skip counting with marks left over ( MP6).
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# How to Understand the Properties of Isosceles and Equilateral Triangles
Triangles, while seemingly simple, exhibit an array of fascinating characteristics. Among these are the symmetrical wonders - the isosceles and equilateral triangles. Steeped in congruence and perfect proportionality, these triangles stand out with their unique properties and applications. Join us as we delve deeper into the captivating world of these special triangles and reveal their mathematical secrets.
## Step-by-step Guide: Isosceles and Equilateral Triangles
1. Isosceles Triangle: Definition and Properties:
An isosceles triangle has two sides of equal length, known as the legs. The third side is called the base.
• The angles opposite the equal sides are congruent.
• The altitude drawn to the base bisects both the base and the vertex angle.
2. Equilateral Triangle: Definition and Properties:
An equilateral triangle has all three sides of equal length.
• All interior angles are congruent and each measure $$60^\circ$$.
• All altitudes (or heights) are congruent.
• Every equilateral triangle is also equiangular.
3. Area and Perimeter Formulas:
For an isosceles triangle with base $$b$$ and height $$h$$:
$$\text{Area} = \frac{1}{2} b \times h$$
$$\text{Perimeter} = b + 2 \times \text{length of one leg}$$
For an equilateral triangle with side $$a$$:
$$\text{Area} = \frac{\sqrt{3}}{4} \times a^2$$
$$\text{Perimeter} = 3a$$
### Examples
Example 1:
Given an isosceles triangle with a base of $$10 \text{ cm}$$ and legs of $$12 \text{ cm}$$ each, find its height.
Solution:
Using Pythagoras’ theorem for one half of the triangle (a right triangle):
$$h = \sqrt{\text{leg}^2 – \left(\frac{\text{base}}{2}\right)^2}$$
$$h = \sqrt{12^2 – 5^2}$$
$$h = \sqrt{119}$$ or approximately $$10.9 \text{ cm}$$
Example 2:
Find the area of an equilateral triangle with a side length of $$8 \text{ cm}$$.
Solution:
Using the area formula for an equilateral triangle:
$$\text{Area} = \frac{\sqrt{3}}{4} \times 8^2 = 55.4 \text{ cm}^2$$
### Practice Questions:
1. Calculate the area and perimeter of an isosceles triangle with a base of $$14 \text{ cm}$$ and legs of $$15 \text{ cm}$$ each.
2. Determine the height of an equilateral triangle with a side length of $$9 \text{ cm}$$.
1. Using the Pythagoras’ theorem, height $$h = \sqrt{15^2 – 7^2} = \sqrt{164} \approx 12.8 \text{ cm}$$. Area $$= \frac{1}{2} \times 14 \times 12.8 = 89.6 \text{ cm}^2$$. Perimeter $$= 14 + 2 \times 15 = 44 \text{ cm}$$.
2. Using the area formula relationship, height $$h = \sqrt{9^2 – 4.5^2} = \sqrt{56.25} = 7.5 \text{ cm}$$.
### What people say about "How to Understand the Properties of Isosceles and Equilateral Triangles - Effortless Math: We Help Students Learn to LOVE Mathematics"?
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# How do you find the equation of the line that passes through (2,-3) and (-7, -3)?
Jun 11, 2018
color(blue)(y = -3
color(blue)( " is the equation of horizontal line with y-intercept = -3"
#### Explanation:
color(crimson)("Equation of a line that passes through two points is given by "
color(crimson)((y-y_1) / (y_2 - y_1) = (x - x_1) / (x_2 - x_1)
$\left({x}_{1} , {y}_{1}\right) = \left(2 , - 3\right) , \left({x}_{2} , {y}_{2}\right) = \left(- 7 , - 3\right)$
$\frac{y + 3}{- 3 + 3} = \frac{x - 2}{- 7 - 2}$
$- 9 \cdot \left(y + 3\right) = 0 \cdot \left(x - 2\right)$
$- 9 \cdot \left(y + 3\right) = 0$
color(blue)(y = -3 " is the equation of the line"
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# Let ${a_1},{a_2}.........$ be positive real numbers in geometric progression. For each $n$ , let ${A_n},{G_n},{H_n}$ be respectively, the arithmetic mean, geometric mean & harmonic mean of ${a_1},{a_2}.........{a_n}$ . Find the expression for the geometric mean of ${G_1},{G_2},{G_3}........{G_n}$ in terms of ${A_1},{A_2},{A_3}.........{A_n},{H_1},{H_2},{H_3}..........{H_n}$
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Hint: From the question student should understand that this sum is an application of formulae related to Arithmetic Mean , Geometric Mean , Harmonic Mean. First step towards solving this sum is noting down the formulae for sum upto $n$ terms . Bring it in the simplest possible form in the next step. After this the student should remove the common terms and bring the relation between these means.
In order to solve the numerical first step is to list down the formulae for Arithmetic Mean , Geometric Mean & Harmonic Mean.
${G_k} = {({a_1} \times {a_2} \times {a_3}.........{a_k})^{1/k}}..............(1)$
Where $k$ is the last term of the expression.
We can simplify equation $1$ as below
${G_k} = {({a_1}r)^{\dfrac{{k - 1}}{2}}}..............(2)$
Following is the formula for Arithmetic progression upto $k$ terms.
${A_k} = \dfrac{{{a_1} + {a_2} + ......{a_k}}}{k}..........(3)$
${A_k} = \dfrac{{{a_1}(1 + r + .......{r^{k - 1}})}}{k}..........(4)$
${A_k} = \dfrac{{{a_1}({r^k} - 1)}}{{(r - 1)k}}..........(5)$
Noting down the formula for Harmonic Progression upto $k$ terms.
${H_k} = \dfrac{k}{{\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} + \dfrac{1}{{{a_3}}} + .....\dfrac{1}{{{a_k}}}}}..........(6)$
${H_k} = \dfrac{{{a_1}k}}{{1 + \dfrac{1}{r} + ....... + \dfrac{1}{{{r^{k - 1}}}}}}..........(7)$
${H_k} = \dfrac{{{a_1}k(r - 1) \times {r^{k - 1}}}}{{{r^{k - 1}}}}..........(8)$
From Equations $2,5,8$ ,we get the following relation between ${G_k},{H_k},{A_k}$
${G_k} = {({A_k}{H_k})^{\dfrac{1}{2}}}$
Considering there are infinite number of terms , equation will transform as follows
${\prod\limits_{k = 1}^n G _k} = \prod\limits_{k = 1}^n {{{({A_k}{H_k})}^{\dfrac{1}{2}}}} ................(9)$
Thus expanding RHS of equation $9$ we get following relation
${\prod\limits_{k = 1}^n G _k} = {({A_1}{A_2}.......{A_n} \times {H_1}{H_2}........{H_n})^{\dfrac{1}{{2n}}}}$
Thus the relation of geometric mean in terms of arithmetic mean and Harmonic mean is
${\prod\limits_{k = 1}^n G _k} = {({A_1}{A_2}.......{A_n} \times {H_1}{H_2}........{H_n})^{\dfrac{1}{{2n}}}}$
So, the correct answer is “${\prod\limits_{k = 1}^n G _k} = {({A_1}{A_2}.......{A_n} \times {H_1}{H_2}........{H_n})^{\dfrac{1}{{2n}}}}$
”.
Note: Though this sum looks extremely complicated and difficult to solve, it is easy if the approach is correct. Students are advised to memorize the formula for Arithmetic Mean , Geometric Mean , Harmonic mean for sum upto $n$ terms. The sum from this chapter should be picked up last if it is of similar type. This is because if the approach is wrong for the sum , it will lead to complete waste of time. This sum is important for Students who are good with application and like to take up challenging numericals.
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# Mental Math to Add and Subtract Decimals
## Group decimal and whole number parts to make mental addition and subtraction easier.
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Practice Mental Math to Add and Subtract Decimals
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Mental Math to Add and Subtract Decimals
Have you ever had to add decimals without a paper and a pencil? Well, there is an easy way to do it. Take a look.
Justin and Kiley went to the mall. At lunch, they stopped at the food court to get something to eat. Justin ordered a hamburger for $2.99 and a drink for$1.09. Kiley ordered the same thing.
How much did they spend in all?
This Concept will show you how to add decimals using mental math. At the end of the Concept, you will know how to figure out the total cost for lunch.
### Guidance
Sometimes, you don’t need to go through all of the work of lining up decimal points and filling in the zeros. Sometimes you can use mental math to figure out a sum.
When is mental math most helpful with decimal sums and differences?
When you have a decimal where the decimal parts can easily add up to be one whole, you can use mental math to figure out the sum. Think about this. If you had .30 + .70, you know that 3 + 7 is 10, therefore you know that .30 + .70 is 1.00.
Let’s apply this information.
5.30 + 6.70 = _____
Here we can start by looking at the decimals, since .30 + .70 is 1. Then we combine the whole numbers and add the total of the decimals to get an answer:
5 + 6 = 11 + 1 = 12
We can use mental math to complete subtraction problems too.
We just look for which decimals add up to be wholes and go from there.
25.00 - 22.50 = _____
We are subtracting 25.00 - 22.50, we can think about this problem in reverse to make the mental math simpler. “What plus 22.50 will give us 25.00?" Think: 2.50 plus what equals 5.00?
25.00 - 22.50 = 2.50
Not all problems will be able to be solved mentally, but when we can mental math makes things a whole lot simpler!!
Here are few for you to work on. Add or subtract using mental math.
#### Example A
33.50 + 5.50 = _____
The total cost of lunch is $8.16. ### Vocabulary Sum the answer in an addition problem. Difference the answer in a subtraction problem. ### Guided Practice Here is one for you to try on your own. Kiley went shopping and spent$5.60. She gave the clerk a ten dollar bill. What was her change?
To figure this out, we can write a subtraction problem.
\begin{align*}10 - 5.60\end{align*}
Then we can use mental math to solve it.
### Practice
Directions: Use mental math to compute each sum or difference.
1. .50 + 6.25 = _____
2. 1.75 + 2.25 = _____
3. 3.50 + 4.50 = _____
4. 7.25 + 1.25 = _____
5. 8.75 + 3.25 = _____
6. 8.50 + 2.50 = _____
7. 10 + 4.50 = _____
8. 12 + 3.75 = _____
9. 15.50 - 5.25 = _____
10. 20 - 15.50 = _____
11. 10 - 4.50 = _____
12. 30 - 15.50 = _____
13. 40 - 16.40 = _____
14. 75 - 50.50 = _____
15. 80 - 40.25 = _____
### Vocabulary Language: English
Decimal
Decimal
In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
Difference
Difference
The result of a subtraction operation is called a difference.
Sum
Sum
The sum is the result after two or more amounts have been added together.
### Explore More
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# 8.3: Series of Real Numbers
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Learning Objectives
In this section, we strive to understand the ideas generated by the following important questions:
• What is an infinite series?
• What is the nth partial sum of an infinite series?
• How do we add up an infinite number of numbers? In other words, what does it mean for an infinite series of real numbers to converge?
• What does is mean for an infinite series of real numbers to diverge?
In Section 8.2, we encountered several situations where we naturally considered an infinite sum of numbers called a geometric series. For example, by writing
$N = 0.1212121212 · · · = \dfrac{12}{100} + \dfrac{12}{100} · \dfrac{1}{100} + \dfrac{12}{100} · \dfrac{1}{1002} + · · · \nonumber$
as a geometric series, we found a way to write the repeating decimal expansion of \9N\) as a single fraction: $$N = \dfrac{4}{33}$$. There are many other situations in mathematics where infinite sums of numbers arise, but often these are not geometric. In this section, we begin exploring these other types of infinite sums. Preview Activity $$\PageIndex{1}$$ provides a context in which we see how one such sum is related to the famous number e.
Preview Activity $$\PageIndex{1}$$
Have you ever wondered how your calculator can produce a numeric approximation for complicated numbers like $$e$$, $$π$$ or $$\ln(2)$$? After all, the only operations a calculator can really perform are addition, subtraction, multiplication, and division, the operations that make up polynomials. This activity provides the first steps in understanding how this process works. Throughout the activity, let $$f (x) = e^x$$.
1. Find the tangent line to $$f$$ at $$x = 0$$ and use this linearization to approximate $$e$$. That is, find a formula $$L(x)$$ for the tangent line, and compute $$L(1)$$, since $$L(1) \approx f (1) = e$$.
2. The linearization of $$e^x$$ does not provide a good approximation to $$e$$ since 1 is not very close to 0. To obtain a better approximation, we alter our approach a bit. Instead of using a straight line to approximate $$e$$, we put an appropriate bend in our estimating function to make it better fit the graph of $$e^x$$ for $$x$$ close to 0. With the linearization, we had both $$f (x)$$ and $$f' (x)$$ share the same value as the linearization at $$x = 0$$. We will now use a quadratic approximation $$P_2(x)$$ to $$f (x) = e^x$$ centered at $$x = 0$$ which has the property that $$P_2(0) = f (0)$$, $$P'_2 (0) = f 0 (0)$$, and $$P''_2 (0) = f''(0)$$.
1. Let $$P_2(x) = 1 + x + \dfrac{x^2}{2}$$. Show that $$P_2(0) = f (0)$$, $$P'_2 (0) = f'(0)$$, and $$P''_2 (0) = f''(0)$$. Then, use $$P_2(x)$$ to approximate $$e$$ by observing that $$P_2(1) \approx f (1)$$.
2. We can continue approximating e with polynomials of larger degree whose higher derivatives agree with those of $$f$$ at 0. This turns out to make the polynomials fit the graph of $$f$$ better for more values of $$x$$ around 0. For example, let $$P_3(x) = 1 + x + \dfrac{x^2}{2} + \frac{x^3}{6}$$. Show that $$P_3(0) = f (0), P'_3(0) = f'(0), P''_3 (0) = f''(0)$$, and $$P'''_3 (0) = f'''(0)$$. Use $$P_3(x)$$ to approximate e in a way similar to how you did so with $$P_2(x)$$ above.
Preview Activity $$\PageIndex{1}$$ shows that an approximation to e using a linear polynomial is 2, an approximation to e using a quadratic polynomial is 2.5, and an approximation using a cubic polynomial is 2.6667. As we will see later, if we continue this process we can obtain approximations from quartic (degree 4), quintic (degree 5), and higher degree polynomials giving us the following approximations to $$e$$:
• linear: $$e \approx 1 + 1 = 2$$
• quadratic: $$e \approx 1 + 1 + \dfrac{1}{2} =2.5$$
• cubic: $$e \approx 1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} = 2.6$$
• quartic: $$e \approx 1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} = 2.7083$$
• quintic: $$e \approx 1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} + \dfrac{1}{120} = 2.716$$
We see an interesting pattern here. The number $$e$$ is being approximated by the sum
$1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} + \dfrac{1}{120} + · · · + \dfrac{1}{n!} \tag{8.11}\label{8.11}$
for increasing values of $$n$$. And just as we did with Riemann sums, we can use the summation notation as a shorthand4 for writing the sum in Equation $$\ref{8.11}$$ so that
$e \approx 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + · · · + \dfrac{1}{ n!} \sum_n^{k=0} \dfrac{1}{k!}. \label{8.12}$
We can calculate this sum using as large an $$n$$ as we want, and the larger $$n$$ is the more accurate the approximation (Equation \ref{8.12}) is. Ultimately, this argument shows that we can write the number e as the infinite sum:
$e = \sum_{k=0}^\infty \dfrac{1}{k!}.\tag{8.13} \label{8.13}$
Note that 0! appears in Equation $$\ref{8.12}$$ and by definition, 0! = 1. This sum is an example of a series (or an infinite series). Note that the series in Equation \ref{8.13} is the sum of the terms of the (infinite) sequence {$$\dfrac{1}{n!}$$}. In general, we use the following notation and terminology. Definition 8.3. An infinite series of real numbers is the sum of the entries in an infinite sequence of real numbers. In other words, an infinite series is sum of the form
$a_1 + a_2 + · · · + a_n + · · · = \sum_{k=1}^\infty ak, \nonumber$
where $$a_1, a_2, . . .,$$ are real numbers.
We will normally use summation notation to identify a series. If the series adds the entries of a sequence {$$a_n$$}$$n\geq1$$, then we will write the series as
$\sum_{k \geq1} a_k \nonumber$
or
$\sum_{k =1} a_k \nonumber$
Note well: each of these notations is simply shorthand for the infinite sum $$a_1 + a_2 + · · · + a_n + · · ·$$.
Is it even possible to sum an infinite list of numbers? This question is one whose answer shouldn’t come as a surprise. After all, we have used the definite integral to add up continuous (infinite) collections of numbers, so summing the entries of a sequence might be even easier. Moreover, we have already examined the special case of geometric series in the previous section. The next activity provides some more insight into how we make sense of the process of summing an infinite list of numbers.
Activity $$\PageIndex{2}$$:
Consider the series
$\sum_{k = 1} \dfrac{1}{k^2}. \nonumber$
While it is physically impossible to add an infinite collection of numbers, we can, of course, add any finite collection of them. In what follows, we investigate how understanding how to find the nth partial sum (that is, the sum of the first n terms) enables us to make sense of the infinite sum.
a. Sum the first two numbers in this series. That is, find a numeric value for
$\sum_{k=1}^a \dfrac{1}{k^2} \nonumber$
b. (b) Next, add the first three numbers in the series.
c. Continue adding terms in this series to complete Table 8.4. Carry each sum to at least 8 decimal places.
$$\sum_{k=1}^1 \dfrac{1}{k^2} = 1$$ $$\sum_{k=1}^6 \dfrac{1}{k^2} =$$ $$\sum_{k=1}^2 \dfrac{1}{k^2} =$$ $$\sum_{k=1}^7 \dfrac{1}{k^2} =$$ $$\sum_{k=1}^3 \dfrac{1}{k^2} =$$ $$\sum_{k=1}^8 \dfrac{1}{k^2} =$$ $$\sum_{k=1}^4 \dfrac{1}{k^2} =$$ $$\sum_{k=1}^9 \dfrac{1}{k^2} =$$ $$\sum_{k=1}^5 \dfrac{1}{k^2} =$$ $$\sum_{k=1}^10 \dfrac{1}{k^2} =$$
Table 8.4: Sums of some of the first terms of the series $$\sum_{k=1}^\infty \dfrac{1}{k^2}$$
(d) The sums in the table in (c) form a sequence whose $$n$$th term is $$S_n = \sum_{k=1}^\infty \dfrac{1}{k^2}$$. Based on your calculations in the table, do you think the sequence {$$S_n$$} converges or diverges? Explain. How do you think this sequence {$$S_n$$} is related to the series $$\sum_{k=1}^\infty \dfrac{1}{k^2}$$?
The example in Activity $$\PageIndex{2}$$ illustrates how we define the sum of an infinite series. We can add up the first n terms of the series to obtain a new sequence of numbers (called the sequence of partial sums). Provided that sequence converges, the corresponding infinite series is said to converge, and we say that we can find the sum of the series.
Definition
The $$n$$th partial sum of the series $$\sum_{k=1}^\infty a_k$$ is the finite sum $$S_n = \sum_{k=1}^\infty a_k$$. In other words, the nth partial sum Sn of a series is the sum of the first n terms in the series, or
$S_n = a_1 + a_2 + · · · + a_n. \nonumber$
We then investigate the behavior of a given series by examining the sequence
$S_1, S_2, . . ., S_n, . . . \nonumber$
of its partial sums. If the sequence of partial sums converges to some finite number, then we say that the corresponding series converges. Otherwise, we say the series diverges. From our work in Activity $$\PageIndex{2}$$, the series
$\sum_{k=1}^\infty \dfrac{1}{k^2} \nonumber$
appears to converge to some number near 1.54977. We formalize the concept of convergence and divergence of an infinite series in the following definition.
Definition
The infinite series
$\sum_{k=1}^\infty a_k \nonumber$
converges (or is convergent) if the sequence {$$S_n$$} of partial sums converges, where
$S_n = \sum_{k=1}^\infty a_k. \nonumber$
If $$\lim_{n \rightarrow \infty} S_n = S$$, then we call $$S$$ the sum of the series $$\sum_{k=1}^\infty a_k$$. That is,
$\sum_{k=1}^\infty a_k = \lim_{n \rightarrow \infty} S_n = S. \nonumber$
If the sequence of partial sums does not converge, then the series
$\sum_{k=1}^\infty a_k \nonumber$
diverges (or is divergent).
The early terms in a series do not contribute to whether or not the series converges or diverges. Rather, the convergence or divergence of a series
$\sum_{k=1}^\infty a_k \nonumber$
is determined by what happens to the terms $$a_k$$ for very large values of $$k$$. To see why, suppose that $$m$$ is some constant larger than 1. Then
$\sum_{k=1}^\infty a_k = (a_1+a_2+...a_m) + \sum_{k=m+1}^\infty a_k. \nonumber$
Since $$a_1 + a_2 + · · · + a_m$$ is a finite number, the series $$\sum_{k=1}^\infty a_k$$ will converge if and only if the series $$\sum_{k=m+1}^\infty a_k$$ converges. Because the starting index of the series doesn’t affect whether the series converges or diverges, we will often just write
$\sum a_k \nonumber$
when we are interested in questions of convergence/divergence and not necessarily the exact sum of a series.
In Section 8.2 we encountered the special family of infinite geometric series whose convergence or divergence we completely determined. Recall that a geometric series is a special series of the form $$\sum_{k=1}^\infty a_k$$ where $$a$$ and $$r$$ are real numbers (and $$r \neq 1$$ ). We found that the $$n$$th partial sum $$S_n$$ of a geometric series is given by the convenient formula
$S_n = \dfrac{1 − r^n}{1 − r}, \nonumber$
and thus a geometric series converges if |$$r$$| < 1. Geometric series diverge for all other values of $$r$$. While we have completely determined the convergence or divergence of geometric series, it is generally a difficult question to determine if a given nongeometric series converges or diverges. There are several tests we can use that we will consider in the following sections.
### The Divergence Test
The first question we ask about any infinite series is usually “Does the series converge or diverge?” There is a straightforward way to check that certain series diverge; we explore this test in the next activity.
Activity $$\PageIndex{3}$$:
If the series $$\sum a_k$$ converges, then an important result necessarily follows regarding the sequence {$$a_n$$}. This activity explores this result. Assume that the series $$\sum_{k=1}^\infty a_k$$ converges and has sum equal to $$L$$.
1. What is the $$n$$th partial sum $$S_n$$ of the series $$\sum_{k=1}^\infty a_k$$ ?
2. What is the $$(n − 1)$$st partial sum $$S_n−1$$ of the series $$\sum_{k=1}^\infty a_k$$ ?
3. What is the difference between the nth partial sum and the $$(n − 1)$$st partial sum of the series $$\sum_{k=1}^\infty a_k$$ ?
4. Since we are assuming that $$\sum_{k=1}^\infty a_k = L$$, what does that tell us about $$\lim_{n \rightarrow \infty} S_n$$? Why? What does that tell us about $$\lim_{n \rightarrow \infty} S_{n-1}$$? Why?
5. Combine the results of the previous two parts of this activity to determine $$\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} (S_n - S_{n-1}$$
The result of Activity $$\PageIndex{3}$$ is the following important conditional statement:
If the series $$\sum_{k=1}^\infty a_k$$ converges, then the sequence {$$a_k$$} of $$k$$th terms converges to 0. It is logically equivalent to say that if the sequence {ak } of n terms does not converge to 0, then the series $$\sum_{k=1}^\infty a_k$$ cannot converge. This statement is called the Divergence Test.
THe divergence Test
If $$\lim_{k \rightarrow \infty}a_k \neq 0,$$, then the series $$\sum ak$$ diverges.
Activity $$\PageIndex{4}$$:
Determine if the Divergence Test applies to the following series. If the test does not apply, explain why. If the test does apply, what does it tell us about the series?
a. $$\sum \dfrac{k}{k+1}$$
b. $$\sum (-1)^k$$
c. $$\sum \frac{1}{k}$$
Note well: be very careful with the Divergence Test. This test only tells us what happens to a series if the terms of the corresponding sequence do not converge to 0. If the sequence of the terms of the series does converge to 0, the Divergence Test does not apply: indeed, as we will soon see, a series whose terms go to zero may either converge or diverge.
## The Integral Test
The Divergence Test settles the questions of divergence or convergence of series $$\sum a_k$$ in which $$\lim_{k \rightarrow \infty}a_k \neq 0$$. Determining the convergence or divergence of series $$\sum a_k$$ in which $$\lim_{k \rightarrow \infty}a_k = 0$$ turns out to be more complicated. Often, we have to investigate the sequence of partial sums or apply some other technique.
As an example, consider the harmonic series
$\sum_{k=1}^\infty \dfrac{1}{k}. \nonumber$ 5
5 This series is called harmonic because each term in the series after the first is the harmonic mean of the term before it and the term after it. The harmonic mean of two numbers $$a$$ and $$b$$ is $$\dfrac{2ab}{a+b}$$. See “What’s Harmonic about the Harmonic Series", by David E. Kullman (in the College Mathematics Journal, Vol. 32, No. 3 (May, 2001), 201-203) for an interesting discussion of the harmonic mean.
Table 8.3 shows some partial sums of this series.
$$\sum_{k=1}^1 \dfrac{1}{k}$$. 1 $$\sum_{k=1}^6 \dfrac{1}{k}$$. 2.45 $$\sum_{k=1}^2 \dfrac{1}{k}$$. 1.5 $$\sum_{k=1}^7 \dfrac{1}{k}$$. 2.59286 $$\sum_{k=1}^3 \dfrac{1}{k}$$. 1.83333 $$\sum_{k=1}^8 \dfrac{1}{k}$$. 2.71786 $$\sum_{k=1}^4 \dfrac{1}{k}$$. 2.08333 $$\sum_{k=1}^9 \dfrac{1}{k}$$. 2.82897 $$\sum_{k=1}^5 \dfrac{1}{k}$$. 2.28333 $$\sum_{k=1}^10 \dfrac{1}{k}$$. 2.92897
This information doesn’t seem to be enough to tell us if the series$$\sum_{k=1}^\infty \dfrac{1}{k}$$ converges or diverges. The partial sums could eventually level off to some fixed number or continue to grow without bound. Even if we look at larger partial sums, such as $$\sum_{n=1}^{1000} \approx 7.485470861$$, the result doesn’t particularly sway us one way or another. The Integral Test is one way to determine whether or not the harmonic series converges, and we explore this further in the next activity.
Activity $$\PageIndex{5}$$:
Consider the harmonic series $$\sum_{k=1}^\infty \dfrac{1}{k}$$. Recall that the harmonic series will converge provided that its sequence of partial sums converges. The $$n$$th partial sum $$S_n$$ of the series $$\sum_{k=1}^\infty \dfrac{1}{k}$$ is
$S_n = \sum_{k=1}^n \dfrac{1}{k} \nonumber$
$= 1 + \dfrac{1}{2} + \dfrac{1}{3}+...\dfrac{1}{n} \nonumber$
$1(1) + (1)(\dfrac{1}{2})+(1)(\dfrac{1}{3})+...+(1)\dfrac{1}{n}. \nonumber$
Through this last expression for $$S_n$$, we can visualize this partial sum as a sum of areas of rectangles with heights $$\dfrac{1}{m}$$ and bases of length 1, as shown in Figure 8.3, which uses the 9th partial sum. The graph of the continuous function $$f$$ defined by $$f (x) = \frac{1}{x}$$ is overlaid on this plot.
1. Explain how this picture represents a particular Riemann sum.
2. What is the definite integral that corresponds to the Riemann sum you considered in (a)?
Figure 8.3: A picture of the 9th partial sum of the harmonic series as a sum of areas of rectangles.
1. Which is larger, the definite integral in (b), or the corresponding partial sum $$S_9$$ of the series? Why?
2. If instead of considering the 9th partial sum, we consider the nth partial sum, and we let $$n$$ go to infinity, we can then compare the series $$\sum_{k=1}^\infty \dfrac{1}{k}$$ to the improper integral $$\int_{1}^{\infty} \frac{1}{x} dx$$. Which of these quantities is larger? Why?
3. Does the improper integral $$\int_{1}^{\infty} \frac{1}{x} dx$$ converge or diverge? What does that result, together with your work in (d), tell us about the series $$\sum_{k=1}^\infty \dfrac{1}{k}$$ ?
The ideas from Activity $$\PageIndex{5}$$ hold more generally. Suppose that $$f$$ is a continuous decreasing function and that $$a_k = f (k)$$ for each value of $$k$$. Consider the corresponding series $$\sum_{k=1}^\infty a_k$$. The partial sum
$S_n = \sum_{k=1}^n a_k \nonumber$
can always be viewed as a left hand Riemann sum of $$f (x)$$ using rectangles with heights given by the values $$a_k$$ and bases of length 1. A representative picture is shown at left in Figure 8.4. Since $$f$$ is a decreasing function, we have that
$S_n > \int_{1}^{n} (x) dx. \nonumber$
Taking limits as $$n$$ goes to infinity shows that
$\sum_\infty^{k=1} a_k >\int_{1}^{\infty} f(x) dx. \nonumber$
Therefore, if the improper integral $$\int_{1}^{\infty} f(x) dx.$$ diverges, so does the series $$\sum_{k=1}^\infty a_k$$.
Figure 8.4: Comparing an improper integral to a series
What’s more, if we look at the right hand Riemann sums of f on [1, $$n$$] as shown at right in Figure 8.4, we see that
$\int_{1}^{\infty} f(x) dx > \sum_{k=2}^\infty a_k. \nonumber$
So if $$\int_{1}^{\infty} f(x) dx$$ converges, then so does $$\sum_{k=2}^\infty a_k$$, which also means that the series $$\sum_{k=1}^\infty a_k$$ converges. Our preceding discussion has demonstrated the truth of the Integral Test.
### The Integral Test
Let $$f$$ be a real valued function and assume $$f$$ is decreasing and positive for all $$x$$ larger than some number $$c$$. Let $$a_k = f (k)$$ for each positive integer $$k$$.
1. If the improper integral $$\int_{c}^{\infty} f(x) dx$$ converges, then the series $$\sum_{k=1}^\infty a_k$$ converges.
2. If the improper integral $$\int_{c}^{\infty} f(x) dx$$ diverges, then the series $$\sum_{k=1}^\infty a_k$$ diverges.
Note that the Integral Test compares a given infinite series to a natural, corresponding improper integral and basically says that the infinite series and corresponding improper integral both have the same convergence status. In the next activity, we apply the Integral Test to determine the convergence or divergence of a class of important series.
Activity $$\PageIndex{6}$$:
The series $$\sum \frac{1}{k^p}$$ are special series called $$p$$-series. We have already seen that the $$p$$-series with $$p = 1$$ (the harmonic series) diverges. We investigate the behavior of other p-series in this activity.
1. Evaluate the improper integral $$\int_{1}^{\infty} \frac{1}{x^2} dx$$. Does the series $$\sum_{k=1}^\infty \frac{1}{k^P}$$ converge or diverge? Explain.
2. Evaluate the improper integral $$\int_{1}^{\infty} \frac{1}{x^P} dx$$ where $$p > 1$$. For which values of $$p$$ can we conclude that the series $$\sum_{k=1}^\infty \frac{1}{k^P}$$ converges?
3. Evaluate the improper integral \int_{1}^{\infty} \frac{1}{x^P} dx where $$p < 1$$. What does this tell us about the corresponding p-series $$\sum_{k=1}^\infty \frac{1}{k^P}$$ ?
4. Summarize your work in this activity by completing the following statement. The $$p$$-series $$\sum_{k=1}^\infty \frac{1}{k^P}$$ converges if and only if ___________________________.
## The Limit Comparison Test
The Integral Test allows us to determine the convergence of an entire family of series: the p-series. However, we have seen that it is, in general, difficult to integrate functions, so the Integral Test is not one that we can use all of the time. In fact, even for a relatively simple series like $$\sum \frac{k^2+1}{k^4_2k+2'}$$, the Integral Test is not an option. In this section we will develop a test that we can use to apply to series of rational functions like this by comparing their behavior to the behavior of $$p$$-series.
Activity $$\PageIndex{7}$$:
Consider the series $$\sum \frac{k+1}{k^3+ 2}$$. Since the convergence or divergence of a series only depends on the behavior of the series for large values of $$k$$, we might examine the terms of this series more closely a $$k$$ gets large.
a. By computing the value of $$\frac{k+1}{k^3+ 2}$$for $$k = 100$$ and $$k = 1000$$, explain why the terms $$\frac{k+1}{k^3+ 2}$$ are essentially $$\dfrac{k}{k^3}$$ when $$k$$ is large.
b. Let’s formalize our observations in (a) a bit more. Let $$a_k = \frac{k+1}{k^3+2}$$ and $$b_k = \frac{k}{ k^3}$$.
Calculate
$\lim_{k \rightarrow \infty} \frac{a_k}{b_k}. \nonumber$
What does the value of the limit tell you about $$a_k$$ and $$b_k$$ for large values of $$k$$? Compare your response from part (a).
c. Does the series $$\sum \dfrac {k}{k^3}$$ converge or diverge? Why? What do you think that tells us about the convergence or divergence of the series $$\sum \dfrac{k+1}{k^3+2}$$? Explain.
Activity $$\PageIndex{7}$$ illustrates how we can compare one series with positive terms to another whose behavior (that is, whether the series converges or diverges) we know. More generally, suppose we have two series $$\sum a_k$$ and $$\sum b_k$$ with positive terms and we know the behavior of the series $$\sum a_k$$. Recall that the convergence or divergence of a series depends only on what happens to the terms of the series for large values of k, so if we know that ak and bk are essentially proportional to each other for large $$k$$, then the two series $$\sum a_k$$ and $$\sum b_k$$ should behave the same way. In other words, if there is a positive finite constant c such that
$\lim_{k \rightarrow \infty} \frac{b_k}{a_k} = c , \nonumber$
then $$b_k \approx ca_k$$ for large values of $$k$$. So
$\sum b_k \approx \sum ca_k = c \sum a_k. \nonumber$
Since multiplying by a nonzero constant does not affect the convergence or divergence of a series, it follows that the series P ak and P bk either both converge or both diverge. The formal statement of this fact is called the Limit Comparison Test.
### The Limit Comparison Test.
Let $$\sum a_k$$ and $$\sum b_k$$ be series with positive terms. If
$\lim_{k \rightarrow \infty} \frac{b_k}{a_k} = c \nonumber$
for some positive (finite) constant $$c$$, then $$\sum a_k$$ and $$\sum b_k$$either both converge or both diverge.
In essence, the Limit Comparison Test shows that if we have a series $$\sum \dfrac{p(k)}{q(k)}$$ of rational functions where $$p(k)$$ is a polynomial of degree $$m$$ and $$q(k)$$ a polynomial of degree $$l$$, then the series $$\sum \dfrac{p(k)}{q(k)}$$ will behave like the series $$\sum \frac{k^m}{k^l}$$. So this test allows us to quickly and easily determine the convergence or divergence of series whose summands are rational functions.
Activity $$\PageIndex{8}$$:
Use the Limit Comparison Test to determine the convergence or divergence of the series
$\sum \dfrac{3k^2 + 1}{5k^4 + 2k + 2} . \nonumber$
by comparing it to the series $$\sum \dfrac{1}{k^2}$$.
## The Ratio Test
The Limit Comparison Test works well if we can find a series with known behavior to compare. But such series are not always easy to find. In this section we will examine a test that allows us to examine the behavior of a series by comparing it to a geometric series, without knowing in advance which geometric series we need.
Activity $$\PageIndex{9}$$:
Consider the series defined by
$\sum_{k=1}^{\infty} \dfrac{2^k}{3^k - k}.\tag{4.18} \label{8.14}$
This series is not a geometric series, but this activity will illustrate how we might compare this series to a geometric one. Recall that a series $$\sum a_k$$ is geometric if the ratio $$\dfrac{a_{k+1}}{ak}$$ is always the same. For the series in Equation $$\ref{8.14}$$, note that $$a_k = \dfrac{2^k}{3^k−k}$$.
a. To see if $$\sum \dfrac{2^k}{3^k - k }$$ is comparable to a geometric series, we analyze the ratios of successive terms in the series. Complete Table 8.6, listing your calculations to at least 8 decimal places.
k $$\dfrac{a_{k+1}}{a_k}$$ 5 10 20 21 22 23 24 25
Table 8.6: Ratios of successive terms in the series $$\sum \dfrac{2^k}{3^k - k }$$
b. Based on your calculations in Table 8.6, what can we say about the ratio $$\dfrac{a_{k+1}}{a_k}$$ if $$k$$ is large?
c. Do you agree or disagree with the statement: “the series $$\sum \dfrac{2^k}{3^k - k}$$ is approximately geometric when $$k$$ is large”? If not, why not? If so, do you think the series $$\sum \dfrac{2^k}{3^k - k}$$ converges or diverges? Explain.
We can generalize the argument in Activity 8.14 in the following way. Consider the series $$\sum a_k$$. If
$$\dfrac{a_{k+1}}{a_k} \approx r$$
for large values of $$k$$, then $$a_{k+1} \approx ra_k$$ for large $$k$$ and the series $$\sum ak$$ is approximately the geometric series $$\sum ar^k$$for large $$k$$. Since the geometric series with ratio r converges only for $$−1 < r < 1$$, we see that the series $$\sum a_k$$ will converge if
$l\lim_{k \rightarrow \infty}\dfrac{a_{k+1}}{a_k} = r \nonumber$
for a value of $$r$$ such that $$|r| < 1$$. This result is known as the Ratio Test.
### The Ratio Test
Let $$\sum a_k$$ be an infinite series. Suppose
$l\lim_{k \rightarrow \infty}\dfrac{a_{k+1}}{a_k} = r \nonumber$
1. If $$0 ≤ r < 1$$, then the series $$\sum a_k$$ converges.
2. If $$1 < r$$, then the series $$\sum a_k$$ diverges.
3. If $$r = 1$$, then the test is inconclusive.
Note well: The Ratio Test takes a given series and looks at the limit of the ratio of consecutive terms; in so doing, the test is essentially asking, “is this series approximately geometric?” If the series can be thought of as essentially geometric, the test use the limiting common ratio to determine if the given series converges.
We have now encountered several tests for determining convergence or divergence of series. The Divergence Test can be used to show that a series diverges, but never to prove that a series converges. We used the Integral Test to determine the convergence status of an entire class of series, the $$p$$-series. The Limit Comparison Test works well for series that involve rational functions and which can therefore by compared to $$p$$-series. Finally, the Ratio Test allows us to compare our series to a geometric series; it is particularly useful for series that involve nth powers and factorials. Two other tests, the Direct Comparison Test and the Root Test, are discussed in the exercises. Now it is time for some practice.
Activity $$\PageIndex{10}$$:
Determine whether each of the following series converges or diverges. Explicitly state which test you use.
a. $$\sum \dfrac{k}{2^k}$$
b. $$\sum \dfrac{k^3+2}{k^2+1}$$
c. $$\sum \dfrac{10^k}{k!}$$
d. $$\sum \dfrac{k^3-2k^2+1}{k^6+4}$$
## Summary
In this section, we encountered the following important ideas:
• An infinite series is a sum of the elements in an infinite sequence. In other words, an infinite series is a sum of the form
$a_1 + a_2 + · · · + a_n + · · · = \sum_{k=1}^\infty a_k \nonumber$
where $$a_k$$ is a real number for each positive integer $$k$$.
• The $$n$$t partial sum $$S_n$$ of the series $$\sum_{k=1}^\infty a_k$$ is the sum of the first $$n$$ terms of the series. That is,
$S_n = a_1 + a_2 + · · · + a_n = \sum_{k=1}^n a_k \nonumber$
• The sequence {$$S_n$$} of partial sums of a series $$\sum_{k=1}^\infty a_k$$ tells us about the convergence or divergence of the series. In particular
• The series $$\sum_{k=1}^\infty a_k$$ converges if the sequence {$$S_n$$} of partial sums converges. In this case we say that the series is the limit of the sequence of partial sums and write
$\sum_{k=1}^\infty a_k = \lim_{n \rightarrow \infty} Sn. \nonumber$
• The series $$\sum_{k=1}^\infty a_k$$diverges if the sequence {$$S_n$$} of partial sums diverges.
This page titled 8.3: Series of Real Numbers is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Matthew Boelkins, David Austin & Steven Schlicker (ScholarWorks @Grand Valley State University) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# 2.2.E: Problems on Natural Numbers and Induction (Exercises)
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Exercise $$\PageIndex{1}$$
Complete the missing details in Examples $$(\mathrm{a}),(\mathrm{b}),$$ and $$(\mathrm{d})$$.
Exercise $$\PageIndex{2}$$
Prove Theorem 2 in detail.
Exercise $$\PageIndex{3}$$
Suppose $$x_{k}<y_{k}, k=1,2, \ldots,$$ in an ordered field. Prove by induction on $$n$$ that
(a) $$\sum_{k=1}^{n} x_{k}<\sum_{k=1}^{n} y_{k}$$
(b) if all $$x_{k}, y_{k}$$ are greater than zero, then
$\prod_{k=1}^{n} x_{k}<\prod_{k=1}^{n} y_{k}$
Exercise $$\PageIndex{4}$$
Prove by induction that
(i) $$1^{n}=1$$;
(ii) $$a<b \Rightarrow a^{n}<b^{n}$$ if $$a>0$$.
Hence deduce that
(iii) $$0 \leq a^{n}<1$$ if $$0 \leq a<1$$;
(iv) $$a^{n}<b^{n} \Rightarrow a<b$$ if $$b>0 ;$$ proof by contradiction.
Exercise $$\PageIndex{5}$$
Prove the Bernoulli inequalities: For any element $$\varepsilon$$ of an ordered field,
(i) $$(1+\varepsilon)^{n} \geq 1+n \varepsilon$$ if $$\varepsilon>-1$$;
(ii) $$(1-\varepsilon)^{n} \geq 1-n \varepsilon$$ if $$\varepsilon<1 ; n=1,2,3, \ldots$$
Exercise $$\PageIndex{6}$$
For any field elements $$a, b$$ and natural numbers $$m, n,$$ prove that
$\begin{array}{ll}{\text { (i) } a^{m} a^{n}=a^{m+n} ;} & {\text { (ii) }\left(a^{m}\right)^{n}=a^{m n}} \\ {\text { (iii) }(a b)^{n}=a^{n} b^{n} ;} & {\text { (iv) }(m+n) a=m a+n a} \\ {\text { (v) } n(m a)=(n m) \cdot a ;} & {\text { (vi) } n(a+b)=n a+n b}\end{array}$
[Hint: For problems involving two natural numbers, fix $$m$$ and use induction on $$n ]$$.
Exercise $$\PageIndex{7}$$
Prove that in any field,
$a^{n+1}-b^{n+1}=(a-b) \sum_{k=0}^{n} a^{k} b^{n-k}, \quad n=1,2,3, \ldots$
Hence for $$r \neq 1$$
$\sum_{k=0}^{n} a r^{k}=a \frac{1-r^{n+1}}{1-r}$
(sum of $$n$$ terms of a geometric series).
Exercise $$\PageIndex{8}$$
For $$n>0$$ define
$\left(\begin{array}{l}{n} \\ {k}\end{array}\right)=\left\{\begin{array}{ll}{\frac{n !}{k !(n-k) !},} & {0 \leq k \leq n} \\ {0,} & {\text { otherwise }}\end{array}\right.$
Verify Pascal's law,
$\left(\begin{array}{l}{n+1} \\ {k+1}\end{array}\right)=\left(\begin{array}{l}{n} \\ {k}\end{array}\right)+\left(\begin{array}{c}{n} \\ {k+1}\end{array}\right).$
Then prove by induction on $$n$$ that
(i) $$(\forall k | 0 \leq k \leq n)\left(\begin{array}{l}{n} \\ {k}\end{array}\right) \in N ;$$ and
(ii) for any field elements $$a$$ and $$b$$,
$(a+b)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l}{n} \\ {k}\end{array}\right) a^{k} b^{n-k}, \quad n \in N \text { (the binomial theorem). }$
What value must $$0^{0}$$ take for (ii) to hold for all $$a$$ and $$b ?$$
Exercise $$\PageIndex{9}$$
Show by induction that in an ordered field $$F$$ any finite sequence $$x_{1}, \ldots, x_{n}$$ has a largest and a least term (which need not be $$x_{1}$$ or $$x_{n} ) .$$ Deduce that all of $$N$$ is an infinite set, in any ordered field.
Exercise $$\PageIndex{10}$$
Prove in $$E^{1}$$ that
(i) $$\sum_{k=1}^{n} k=\frac{1}{2} n(n+1)$$;
(ii) $$\sum_{k=1}^{n} k^{2}=\frac{1}{6} n(n+1)(2 n+1)$$;
(iii) $$\sum_{k=1}^{n} k^{3}=\frac{1}{4} n^{2}(n+1)^{2}$$;
(iv) $$\sum_{k=1}^{n} k^{4}=\frac{1}{30} n(n+1)(2 n+1)\left(3 n^{2}+3 n-1\right)$$.
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Introduction to Polynomial Expressions - Polynomial Expressions - High School Algebra I Unlocked (2016)
## High School Algebra I Unlocked (2016)
### Chapter 3. Polynomial Expressions
GOALS
By the end of this chapter, you will be able to:
•Define and explain polynomial expressions
•Combine polynomial expressions through addition and subtraction
•Multiply polynomial expressions using the distributive property and FOIL
•Factor polynomial expressions by reversing the FOIL process or by completing the square
### Lesson 3.1. Introduction to Polynomial Expressions
REVIEW
BEFORE BEGINNING THIS CHAPTER, YOU SHOULD BE FAMILIAR WITH:
•general number properties
•the parts of an equation or expression
•basic mathematical operations
•solving numeric and variable expressions
•the rules of exponents and manipulating exponential terms
In the first two chapters, we discussed the roles of both variables and exponents. Now we’ll discuss polynomial expressions, which result when the world of variables collides with the world of exponents.
A single-variable polynomial is an expression of the sum of two or more terms that contain different powers of the same variable. All of the following are single-variable polynomial expressions:
y2 − y + 6 x6 + 2x4 + x2 18z10 + 2z 4 + z − 14
In Latin, the term poly
means “many” and
nomial means “term.”
Therefore, a polynomial
is “many terms.”
A polynomial can have 2 terms, 20 terms, or 200 terms. Polynomials having one, two, or three terms are referred to as monomials, binomials, and trinomials, respectively.
Monomial Binomial Trinomial Number of Terms 1 2 3 Example 2x 2x + 6 12x2 + 2x + 6
The degree of a single-variable polynomial is equal to the largest exponent associated with that variable. Take a look at the following table to see how the degree of a polynomial relates to the polynomial itself:
Polynomial Expression Variable Term with Largest Exponent Degree y2 − y + 6 y2 2 x6 + 2x4 + x2 x6 6 18z10 + 2z4 + z − 14 18z10 10
Polynomials can also be named for their degree: A quadratic is a second-degree polynomial, a cubic is a third-degree polynomial, and a quartic is a fourth-degree polynomial. In algebra, you will focus quite a bit on quadratics, particularly quadratic equations, which are often used to represent rates of change. But you don’t have to worry about this yet—we will tackle quadratic equations in more detail in Chapter 6.
Quadratic Cubic Quartic Form ax2 + bx + c ax3 + bx2 + cx + d ax4 + bx3 + cx2 + dx + e Example 12x2 + 2x + 6 12x3 + 2x + 6 12x4 + 2x + 6 Degree 2 3 4
A polynomial with two variables is an expression of the sum of more than two terms that contain different powers of the same variables. The following are polynomial expressions with two variables:
y2 − x + 6 2x4y4 + x6 + y2 2z14y + 18z10 + y − 14
The degree of a term in a multi-variable polynomial is equal to the sum of the exponents in that term, and the degree of a multi-variable polynomial is the greatest sum. The degrees of the two-variable polynomial expressions are shown in the following table.
When working with
polynomials, you will
encounter variables
without coefficients, such
as x, y2, and z3. If a variable
does not have a coefficient
in front of it, the coefficient
is assumed to be 1.
Polynomial Expression Variable Term with Largest Exponent Degree y2 − x + 6 y2 2 2x4y4 + x6 + y2 2x4y4 8 2z14y + 18z10 + y − 14 2z14y 15
To express polynomials in standard form, you put the terms in order of degree and combine like terms. (See Lesson 3.2 for more on like terms.) You will notice that all of the previous polynomial expressions place the terms in descending order according to the term’s degree. For example, in y2y + 6, the term y2 has a degree of 2 and is placed first; the term y has a degree of 1 and is placed second; and the final term, 6, is a constant with no degree, and is placed last.
While you may not always be required to express polynomials in standard form, doing so helps organize the terms and, in turn, reduces errors stemming from adding, subtracting, multiplying, or dividing terms.
Finally, there are some characteristics of polynomials that you should be familiar with.
A Polynomial Can… Examples have constants, variables, and exponents 12x2 + 2x + 6 have only positive or non-negative exponents x2 or y3 or z0 have only a finite number of terms x2 + x + 14
A Polynomial Cannot… Examples have variables in the denominator of a fraction or have fractional exponents x or y have negative exponents x−2 or y−3 have roots of variables or have an infinite number of terms x + x2 + …
Now that we’ve had a crash course in polynomials, let’s see if we can answer a few questions about them.
EXAMPLE
Which of the following is a polynomial expression?
A) 2x4 + x−6 + 4
B) + 2x2 + 3
C) x8 + 10x2 + 1
D) 4x4 +
To answer this question, you need to understand the properties of polynomials, so take a look at each polynomial expression to see if it fits the criteria. Choice (A) gives you the expression 2x4 + x−6 + 4. While a polynomial can have terms with coefficients, variables, exponents, and constants, a polynomial cannot have a term with a negative exponent. Therefore, the term x−6 does not allow the expression to be classified as a polynomial; eliminate (A). The expression in (B), + 2x2 + 3, has a term with variables in the denominator, ; since polynomials cannot have variables in the denominator, eliminate (B). The expression in (C), x8 + 10x2 + 1, has terms that adhere to the rules of polynomials; keep (C) as an option. Finally, the expression in (D), 4x4 + , contains a term with a variable under a radical, , which prohibits the expression from being classified as a polynomial. Therefore, the only expression that is a polynomial expression is (C), x8 + 10x2 + 1.
EXAMPLE
What is the degree of the polynomial 2x4 + x6 + 4 ?
In order to answer this question, you must know that the degree of a single-variable polynomial is equal to the largest exponent associated with that variable. In the expression 2x4 + x6 + 4, the term x6 is the term with the largest exponent. Therefore, the degree of the expression 2x4 + x6 + 4 is 6.
Now let’s look at a question that deals with expressing polynomials in standard form.
EXAMPLE
Express the polynomial 4 + 2x4 + x6 in standard form.
To express polynomials in standard form, you need to put the terms in order of degree from highest to lowest. In the expression 4 + 2x4 + x6, the term x6 has a degree of 6, the term 2x4 has a degree of 4, and the term 4 is a constant and has a degree of 0. Therefore, the polynomial 4 + 2x4 + x6 would be expressed in standard form as x6 + 2x4 + 4.
Let’s do one more question before moving on.
EXAMPLE
Evaluate 4 + 2x4 + x6 when x = 2.
Are you thinking to yourself, “Wait a minute! You never mentioned anything about evaluating polynomials!” Don’t panic when you see a question like this. You evaluate polynomials in the same way that you evaluate other expressions. Here, they ask you to evaluate 4 + 2x4 + x6 when x = 2, so replace x with 2 in the expression and solve:
4 + 2x4 + x6 = 4 + 2(2)4 + (2)6
= 4 + 2(16) + 64
= 4 + 32 + 64
= 100
So, when x = 2, the value of 4 + 2x4 + x6 is 100.
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# Superposition Theorem Statement and Theory
Superposition Theorem is one of the simplified circuit analysis techniques.
Suppose there is an active network. Since it is an active network, there may be numbers of active sources acting simultaneously on the network.
## Statement of Superposition Theorem
Superposition theorem states if each of the sources acts on the network independently then current through any branch of the network is the sum of the currents due to each source.
### Explanation of Superposition Theorem
Suppose there is a branch in the network through which current ‘I’ is flowing due to the sources connected to the network. Now if we make dead all the sources of the network except one source, there will still be a certain current through the same branch. This current is due to the live source alone.
Then we go to another source and make it live. At the same time, we keep other sources of the network including the previously lived source as dead. At that time also there is one current in the same branch due to the second source. After that, we will go to the next source and do the same thing. In this condition, there will be still a certain current through the same branch. In this way, we get as many as currents through the said branch as many the network has the sources (current sources and voltage sources).
The actual current through the said branch is the current through it when all the sources are acting on the network. According to superposition theorem, the actual current through the said branch is the sum of all those individual currents.
### Equation of Superposition Theorem
The equation of theorem is
Where ‘I’ be the current through the branch when all sources are acting on the network. On the other hand, I1, I2, I3,…. In are the currents through the same branch when each of the n numbers of sources is acting on the network individually.
### Example of Superposition Theorem
In figure 1, ‘I’ represents the current through the resistance R when both voltage sources of emf E1 and E2 act on the circuit simultaneously.
Besides the above figure, figure 2 shows the current through the resistance R when only the voltage source of emf E1 acts on the network. This current is I’. Here we have replaced the second source with its internal resistance, r2.
Similarly, the current through the resistance R when only the voltage source of emf E2 acting on the network is I”. Here also we have replaced the source E1 with its internal resistance, r1.
Now as per superposition theorem
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# Adding Fractions in 5 Simple Steps
Adding fractions (if they have different denominators) is not something you can easily work out how to do. You have to know a method. The method is not difficult and becomes second nature with practice.
Fractions with different denominators are incompatible, you cannot add them before you have made them compatible by converting them so that have the same (common) denominator. A foolproof way to find a common denominator is to multiply them. You could say that, to make fractions compatible, go forth and multiply the denominators!
For example:-
5/4 + 4/6 — a common denominator is 4 x 6 = 24
You will have possibly noticed that 4 & 6 have a lower common denominator, namely 12. If you notice a lower common denominator, go ahead and use it; it will save you having to simplify your answer at the end. BUT don’t get hung up about finding the lowest common denominator. Just be assured that by multiplying the denominators you have a secure way to find a common denominator. You can always simplify your answer at the end.
Mixed number - a whole number and a fraction. For example 1¼.
Improper Fraction– a fraction where the top number (the numerator) is greater than the bottom number (the denominator). For example 5/4.
Numerator- The top number of a fraction. For example, the numerator of 5/4 is 5.
Denominator- The bottom number of a fraction. For example, the denominator of 5/4 is 4.
Here is a video that walks you through the 5 simple steps to add any two fractions. I suggest you read the rest of this article first and then use the video to make sure you understand and, more importantly, remember the 5 steps to add fractions:
## Adding Fractions in 5 Simple Steps
We know now the underlying principles of adding fractions and have defined some key terms. Here’s an addition of fractions question to show the 5 simple steps that can be used to add any two fractions.
Question: 1¼ + 4/6
### Step 1 Convert Any Mixed Numbers to Improper Fractions
In this question 1¼ is a mixed number. To convert a mixed number to improper fractions, multiply the whole units (in this case, 1) by the denominator of the fraction (in this case, 4). Add the answer to the numerator of the fraction (in this case, 1) and place this answer over the denominator:
((1 x4 ) + 1 ) / 4 = 5/4
So we have revised the fractions to add to:-
5/4 + 4/6
### Step 2 Find a Common Denominator
5/4 + 4/6
To find a common denominator simply multiply the 2 denominators:
4 x 6 =24
### Step 3 Convert the Numerators
To do this multiply the numerators by the same amount as you multiplied the denominators.
Taking the first fraction 5/4
to get the common denominator we multiplied by 6:- 4 x 6 =24
So we need to multiply the numerator by the same amount:-
5 x 6 = 30
So our first fraction becomes (5 x 6)/(4 x 6) = 30/24
Similarly for the second fraction 4/6
to get the common denominator we multiplied by 4:- 6 x 4 =24
So we need to multiply the numerator by the same amount:-
4 x 4 = 16
So our second fraction becomes (4 x 4)/(6 x 4) = 16/24
And we now have compatible fractions to add:-
30/24 + 16/24
### Step 4 Add the Numerators
We can now simply add the numerators:
30/24 + 16/24 = 46/24
### Step 5– If Possible Simplify the Fraction and If Answer is an Improper Fraction Convert to a Mixed Number
So from step 4 we have 30/24 + 16/24 = 46/24
First simplify 46/24 = 23/12
Now we have an improper fraction (the numerator is greater than the denominator) so we need to convert to a mixed number. To do this divide the numerator by the denominator (23 ÷ 12) and show the answer as a whole number with the remainder as a fraction:-
23/12 = 1 11/12
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# What Times What Equals 44? Solve the Mystery Here!
Do you find it difficult to solve mathematical equations? While most of us probably won’t be able to solve complicated mathematical problems, simple ones like “what times what equals 44” are quite easy to solve. In this article, we’ll solve the mystery behind the equation and explain how to find the answer using different methods.
## What’s the Basic Approach to Solving Equations?
If you’re not used to solving mathematical equations, you might find them intimidating at first. But equations follow a few basic rules, which makes solving them easier. Mathematically, solving an equation means finding its roots, which is the value or values of the unknown that makes the equation true. The basic approach to solving equations is to keep both sides of the equals sign balanced.
### What is an Equation?
An equation is a mathematical statement that two expressions are equal, with at least one unknown value. We use equations to solve problems that involve unknown quantities. Equations can be written in various forms, but the most common form is the linear equation. Linear equations are equations whose graph forms a straight line. Understanding linear equations is essential in solving simple mathematical problems like “what times what equals 44”.
### What is a Multiplication Equation?
A multiplication equation is an equation that involves multiplying two or more numbers or variables. Multiplication equations come in different forms, but the most basic form is when two numbers are being multiplied to give a product. The product is the answer to the equation. Multiplication equations are used when we want to find the value of the unknown that produces the given product.
### What is the General Technique to Solve a Multiplication Equation?
The general technique to solve a multiplication equation involves finding the value of the unknown that satisfies the equation. To solve a multiplication equation, we need to divide both sides of the equation by one of the factors. For example, if the equation is 5 x N = 20, we divide both sides by 5 to get N = 4. This means that 4 times 5 equals 20, and the value of the unknown (N) is 4.
## How to Solve “What Times What Equals 44”?
Let’s look at the basic approach to solve the equation “what times what equals 44”. We can write the equation in the general multiplication form as:
X * Y = 44
Here, X and Y are the factors that must be multiplied to give the product 44. To solve for X and Y, we need to find the pair of numbers whose product is 44. One way to do this is to list all the factors of 44 and look for the pair of factors.
### What are Factors?
In mathematics, factors are the numbers that can be multiplied together to get another number. For instance, the factors of 12 are 1, 2, 3, 4, 6, and 12. Factors can be determined by long division or through prime factorization. Factoring is important in solving mathematical problems that involve multiplication, division, addition or subtraction. In this case, we’ll use factoring to solve the equation “what times what equals 44”.
### Listing Factors
We can list all the factors of 44 in ascending order:
1 2 4 11 22 44
Now we need to determine which pair of these factors gives us the answer 44. We can start by trying pairs of factors that are close to each other.
• 1 x 44 = 44
• 2 x 22 = 44
• 4 x 11 = 44
Out of these, the pair of factors that gives us 44 is 4 x 11. Therefore, the answer is:
4 * 11 = 44
We have found the values of X and Y. So, the equation “what times what equals 44” is solved.
## What if There are Negative Numbers?
The above method works well if we’re only dealing with positive numbers. But what if we include negative numbers? For instance, we could ask the question “what times what equals -44?”.
To solve an equation where one or both of the factors are negative, we need to remember a basic rule: the product of two negative numbers is a positive number. For example, -3 x -4 is equal to 12.
### Solving “What Times What is Equal to -44?”
If we want to solve “what times what is equal to -44”, we need to consider the factors of both 44 and -44. The factors of 44 are 1, 2, 4, 11, 22, and 44. The factors of -44 are -1, -2, -4, -11, -22, and -44.
So, we have to find the pair of numbers among these factors that gives us the product -44. Upon doing so, we obtain the values of X and Y as:
X Y -4 11 4 -11 -1 44 1 -44
Hence, we have two sets of solutions i.e (-4, 11) and (4, -11).
## What are the Factors of 44?
As we have already discussed, factors are the numbers that can be multiplied together to get another number. The factors of a number are essential in solving multiplication equations. The factors of 44 are 1, 2, 4, 11, 22, and 44.
### How to Find the Factors of a Number?
The easiest way to find the factors of a number is by listing all the possible combinations of two numbers whose product equals the given number. For example, if we want to find the factors of 12, we list all possible pairs such as 1 x 12, 2 x 6, and 3 x 4. The factors of 12 are thus 1, 2, 3, 4, 6, and 12.
## What is Prime Factorization?
Prime factorization is the process of finding a set of prime numbers that multiply to a given number. Prime numbers are those numbers that are only divisible by 1 and themselves. For instance, 2, 3, 5, 7, 11, 13, 17, 19… are prime numbers. In contrast, a composite number is a number that is not a prime number i.e. it has at least one factor other than 1 and itself.
### Prime Factorization of 44
To find the prime factorization of a number, we write the number as a product of prime numbers. We can find the prime factorization of 44 by dividing it by its smallest prime factor.
• 44 ÷ 2 = 22
• 22 ÷ 2 = 11
Therefore, the prime factorization of 44 is:
44 = 2 x 2 x 11
The prime factorization tells us that 44 can be expressed as a product of prime numbers i.e. 2 and 11. Thus, any number that is a factor of 44 must be a combination of these prime numbers.
## Conclusion
“What times what equals 44” is a simple multiplication equation that can be solved using different methods such as listing all factors, prime factorization or general multiplication technique. Additionally, we learned that we can solve problems involving negative numbers by remembering the rule that the product of two negative numbers is positive.
## Most Common Questions
• What is an equation? — An equation is a mathematical statement that two expressions are equal, with at least one unknown value.
• What is a multiplication equation? — A multiplication equation is an equation that involves multiplying two or more numbers or variables.
• What is prime factorization? — Prime factorization is the process of finding a set of prime numbers that multiply to a given number.
• What are the factors of 44? — The factors of 44 are 1, 2, 4, 11, 22, and 44.
• How to solve “What times what is equal to -44”? — We can solve the equation by considering the factors of both 44 and -44.
## References
1. “Mathematical equation – Wikipedia.” Wikipedia, Wikimedia Foundation, 2021, https://en.wikipedia.org/wiki/Mathematical_equation.
2. “What is equation? | List of Mathematical formulae | Vedantu.” Vedantu, Vedantu Innovations Pvt. Ltd, 2021, https://www.vedantu.com/definition/what-is-equation.
3. “Factor (mathematics) – Wikipedia.” Wikipedia, Wikimedia Foundation, 2021, https://en.wikipedia.org/wiki/Factor_(mathematics).
4. “Prime factorization – Wikipedia.” Wikipedia, Wikimedia Foundation, 2021, https://en.wikipedia.org/wiki/Prime_factorization.
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# Percent Error – Explanation & Examples
Percent Error is used to calculate the relative or percent error between the experimental and the actual value. For example, we are trying to measure air pressure, and we know the actual value is 760mm Hg, but our experimental or measured value is 758 mm Hg. The relative difference between 760 mm Hg and 758mm Hg is calculated using the percent error formula.
The answer in percent error is represented in percentage, so we first need to understand a percentage concept. When we express a number as a fraction of 100 is said to be a percentage. For example, 10 percent (i.e., 10%) is equal to $\dfrac{10}{100}$; similarly, 2 percent is $\dfrac{2}{100}$. The percentage sign is denoted by “%,” and it is equal to 1/100.
Percent Error is the ratio of the absolute error and actual value multiplied by 100.
You should refresh the following concepts to understand the material discussed here.
1. Percentage.
2. Basic Arithmetic.
## What is Percent Error
Percent error is calculated when there is a reference or actual value against which we compare our measured values. The difference between these two values is treated as the error.
These errors arise due to certain limitations in technology or human mistakes/misjudgments, and calculation of these errors during experiments is necessary. Percent error is used to calculate the error and present the error in percentage. As we stated above percent error is the ratio of the absolute error and actual value. Absolute error is the absolute value of the difference of the measured and the actual value, So percent error can be represented as.
Absolute error = |Actual value – Experimental value|
Percent error = [Absolute error/Actual Value] * 100.
We have discussed percent error so far, but there are other closely related terms and the difference between them is very subtle. You should know the difference between the following terms.
1. Absolute Error
2. Relative Error
3. Percent Error
Absolute Error: It is the difference between the actual value and the observed or measured value. The difference is given as an absolute value which means that we are interested in the magnitude of the error and ignore the sign.
$\color{blue}\mathbf{Absolute\hspace{2mm} Error = \left | Actual\hspace{2mm} value – Estimated\hspace{2mm} Value \right | }$
Relative Error: When we divide the absolute value by the actual value, it is called relative error. Here actual value is also taken as the absolute value. Hence the relative error cannot be negative.
$\color{blue}\mathbf{Relative\hspace{2mm} Error = \left | \dfrac{Absolute\hspace{2mm} Error}{Actual\hspace{2mm} value} \right | }$
Percent Error: When a relative error is multiplied by 100, it is known as percent error.
$\color{blue}\mathbf{Percent\hspace{2mm} Error = Relative\hspace{2mm} Error \times 100\%}$
## How to Calculate Percent Error
Calculation of the percent difference is pretty simple and easy. But, first, you need to follow the steps given below.
1. Identify the real or actual value of the quantity you are going to measure or observe.
2. Take the experimental value of the quantity.
3. Calculate the absolute error by subtracting the experimental value from the actual value
4. Now divide the absolute error by the actual value, and the resulting value is also an absolute value, i.e., it cannot be negative.
5. Express the final answer in percentage by multiplying the result in step 4 by $100$.
### Percent Error Formula:
We can calculate percent error by using the formula given below.
$\mathbf{Percentage Difference = [\dfrac{\left | A.V\hspace{1mm} -\hspace{1mm} M.V \right |}{A.V}]\times 100}$
Here,
A.V = Actual value
M.V = Measured value or Estimated value.
### Percent Error Mean Formula:
The percent error mean is the average of all the means calculated for a given problem or data. Its formula is given as.
$\mathbf{\sum_{i=1}^{n}[\dfrac{\left|A.V\hspace{1mm} -\hspace{1mm}M.V \right|}{\left|A.V \right|}]\times \frac{100}{n}\%}$
### Difference between Percent Error, Standard Error, and Margin of Error:
Some terms are closely related, and students can confuse one term with the other. This section will explain the difference between percent, standard, and margin of error.
Percent Error: Percent error is used to measure error or discrepancy between the actual and the measured value.
Standard Error: This term is used in statistics to calculate the error between a sample and a population. When a sample is taken from a population, the standard error is used to measure the accuracy of that sample with a given population.
Margin of Error: The margin of error is also related to the population’s standard deviation and sample size. It is calculated by multiplying the standard error by the standard score.
Example 1: Allan bought a new football. The radius of the football is 8 inches. The actual radius of a football used internationally is 8.66 inches. You are required to calculate the percent error between these two values.
Solution:
$Actual \hspace{1mm}Value = 8.66 \hspace{1mm}and\hspace{1mm} Measured\hspace{1mm} or\hspace{1mm} observed\hspace{1mm} value = 8$
$Percentage\hspace{1mm} Error = \left |\dfrac{ Actual\hspace{1mm} Value \hspace{1mm}-\hspace{1mm} Observed\hspace{1mm} Value }{Actual\hspace{1mm} Value}\right|\times 100$
$A.V\hspace{1mm}- \hspace{1mm}O.V = 8.66\hspace{1mm} – \hspace{1mm}8 = 0.66$
$Percentage\hspace{1mm} error = \left|\dfrac{ 0.66 }{8.66}\right|\times 100$
$Percent\hspace{1mm} error = 0.0762\times 100 = 7.62\%$
Example 2: Calculate the percent error between the actual and experimental values in the table given below.
Actual Value Experimental Value Percent Error $10$ $7$ $11$ $13$ $15$ $18$ $6$ $4$
Solution:
1).$Actual\hspace{1mm} Value = 10\hspace{1mm} and\hspace{1mm} Measured\hspace{1mm} or\hspace{1mm} observed\hspace{1mm} value = 7$
$Percentage\hspace{1mm} error = \left|\dfrac{ Actual\hspace{1mm} Value\hspace{1mm}-\hspace{1mm} Observed\hspace{1mm} Value }{Actual \hspace{1mm}Value}\right|\times 100$
$A.V\hspace{1mm}-\hspace{1mm} M.V = 10 \hspace{1mm}-\hspace{1mm}7 = 3$
$Percentage\hspace{1mm} error = \left |\dfrac{ 3 }{10}\right|\times 100$
$Percent\hspace{1mm} error = 0.3\times 100 = 30\%$
2). $Actual\hspace{1mm} Value = 11\hspace{1mm} and\hspace{1mm} Measured\hspace{1mm} or\hspace{1mm} observed\hspace{1mm} value = 13$
$Percentage\hspace{1mm} error = \left|\dfrac{ Actual\hspace{1mm} Value\hspace{1mm}-\hspace{1mm} Observed \hspace{1mm}Value }{Actual \hspace{1mm}Value}\right|\times 100$
$A.V\hspace{1mm}-\hspace{1mm} M.V = 11 \hspace{1mm}-\hspace{1mm} 13 = -2$
$Percentage\hspace{1mm} error = \left |\dfrac{ -2 }{11}\right|\times 100$
$Percent\hspace{1mm} error = 0.1818\times 100 = 18.18\%$
3). $Actual\hspace{1mm} Value = 15\hspace{1mm} and\hspace{1mm} Measured\hspace{1mm} or\hspace{1mm} observed\hspace{1mm} value = 18$
$Percentage\hspace{1mm} error = \left|\dfrac{ Actual\hspace{1mm} Value\hspace{1mm}-\hspace{1mm} Observed \hspace{1mm}Value }{Actual \hspace{1mm}Value}\right|\times 100$
$A.V\hspace{1mm}-\hspace{1mm} M.V = 15 \hspace{1mm}-\hspace{1mm} 18 = -3$
$Percentage\hspace{1mm} error = \left|\dfrac{ -3 }{15}\right|\times 100$
$Percent\hspace{1mm} error = 0.2\times 100 = 20\%$
4).$Actual \hspace{1mm}Value = 6\hspace{1mm} and\hspace{1mm} Measured\hspace{1mm} or\hspace{1mm} observed\hspace{1mm} value = 4$
$Percent\hspace{1mm} Error = \left|\dfrac{ Actual\hspace{1mm} Value\hspace{1mm}-\hspace{1mm} Observed \hspace{1mm}Value }{Actual \hspace{1mm}Value}\right|\times 100$
$A.V\hspace{1mm}-\hspace{1mm} M.V = 16 \hspace{1mm}-\hspace{1mm} 20 = -4$
$Percentage\hspace{1mm} Error = \left|\dfrac{ -4 }{16}\right|\times 100$
$Percent\hspace{1mm} difference = 0.25\times 100 = 25\%$
Actual Value Experimental Value Percent Error $10$ $7$ $30\%$ $11$ $13$ $18.18\%$ $15$ $18$ $20\%$ $16$ $20$ $25\%$
Example 3: William wants to buy a new car for his son. Due to the pandemic, the estimated increased price at which the car is available is 130,000 dollars while the actual value of the car is 100,000 dollars. You are required to help William in the calculation of the percent error between these two prices.
Solution:
$Actual \hspace{1mm}Value = 15\hspace{1mm} and\hspace{1mm} Measured \hspace{1mm} or\hspace{1mm} observed \hspace{1mm} value = 18$
$Percentage\hspace{1mm} error = \left|\dfrac{ Actual\hspace{1mm} Value\hspace{1mm}-\hspace{1mm} Observed\hspace{1mm} Value }{Actual\hspace{1mm} Value}\right|\times 100$
$A.V\hspace{1mm}-\hspace{1mm} M.V = 15\hspace{1mm} -\hspace{1mm} 18 = -3$
$Percentage\hspace{1mm} error = \left|\dfrac{ -3 }{15}\right|\times 100$
$Percent\hspace{1mm} error = 0.2\times 100 = 20\%$
Example 4: Mayer held a birthday party. Mayer estimated that 200 people will attend his birthday party but the actual amount of people who attended the function were 180. You are required to calculate the absolute error, relative error, and percent error.
Solution:
$Actual\hspace{1mm} Value = 180 \hspace{1mm}and\hspace{1mm} Estimated\hspace{1mm} value = 200$
$Absolute\hspace{1mm} error = |Actual \hspace{1mm}value\hspace{1mm} – \hspace{1mm}Measured\hspace{1mm} value| = |180\hspace{1mm} -\hspace{1mm} 200| = |-20| = 20$
$Relative\hspace{1mm} error = \left|\dfrac{Absolute\hspace{1mm} error }{Actual\hspace{1mm} Value}\right|$
$Relative\hspace{1mm} error = \left|\frac{20 }{180}\right|= 0.1111$
$Percent\hspace{1mm} error = Realtive error\times 100 = 20\%$
$Percent\hspace{1mm} error = 0.1111\times 100 = 11.11\%$
Example 5: Mason started a restaurant in August 2021 and invested a lot of money as he expected to generate good revenue through this restaurant. The expected and actual income of the first four months is given below. You are required to calculate the percent error mean.
Month Expected Income (Dollars) Actual Income (Dollars) Percent Error August $2500$ $1700$ September $3500$ $2500$ October $4000$ $2800$ November $5000$ $3900$
Solution:
We can give a percent error calculation for the first four months as.
Month Absolute Difference Relative Error Percent Error August $800$ $0.47$ $47\%$ September $1000$ $0.4$ $40\%$ October $1200$ $0.42$ $42\%$ November $1100$ $0.282$ $28.2\%$
P.E.M = $\dfrac{$47\%\hspace{1mm}+\hspace{1mm}40\%\hspace{1mm}+\hspace{1mm}42\%\hspace{1mm}+\hspace{1mm}28.2\%$}{$4$} = 39.3\ %$
we can also calculate percent error mean by using relative error values.
P.E.M = $[\dfrac{$0.47\hspace{1mm}+\hspace{1mm}0.40\hspace{1mm}+\hspace{1mm}0.42\hspace{1mm}+\hspace{1mm}0.282$}{$4$}] \times 100 = 39.3\ %$
### Practice Questions
1. The estimated height of a shopping mall is $290$ ft, while its actual height is $320$ ft. Which of the following shows the percent error between these two values?
2. Alice is $25$ years old according to her identity card, while her actual age is $27$ years. Which of the following shows the percent error between the given values?
3. Fabian does morning exercise daily to keep himself healthy and fit. The estimated time duration for morning exercise is $30$ minutes, while the actual time duration of morning exercise is $29$ minutes. Which of the following shows the percent error between these two values?
4. M&N’s is a multi-national company. A newspaper published an article regarding the company and mentioned that the number of people working in the company is estimated to be $6000$ while the actual strength of employees is $7000$. Which of the following shows the percent error between these two values?
5. Nina held a birthday party. Nina estimated that $300$ people would attend his birthday party, but the actual number of people attending the function was $250$. What is the absolute error?
6. Nina held a birthday party. Nina estimated that $300$ people would attend his birthday party, but the actual number of people attending the function was $250$. What is the relative error?
7. Nina held a birthday party. Nina estimated that $300$ people would attend his birthday party, but the actual number of people attending the function was $250$. What is the percent error?
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# Decimals and Fractions: Multiplying Decimals Cont’d
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When we multiply decimals it’s usually easier to turn them into ordinary numbers first. We do this by simply multiplying them by numbers of 10,100,1000 etc. This allows us to do normal multiplication. When we find out the answer, we reverse the process to find out the final answer.
### Example 1
To make it an ordinary number we need to multiply 0.7 by 10 which equals 7.
We then do 5 × 7 = 35
Now reverse the process: Divide 35 by 10, making the final answer 3.5
### Example 2
To make it an ordinary number we need to multiply 0.07 by 100 which equals 7.
We then do 6 × 7 = 42
Now reverse the process: Divide 42 by 100, making the final answer 0.42
### Example 3
To make it an ordinary number we need to multiply 2.7 by 10 which equals 27.
We then do 5 × 27 = 135
Now reverse the process: Divide 135 by 10, making the final answer 13.5
### Example 4
When multiplying two decimals the process is slightly different. Both numbers need to be divided by 10 × 10 = 100
To make it an ordinary number we need to multiply both numbers by 10 and 5 × 3 = 15
Now reverse the process: Divide 15 by 100, making the final answer 0.15
### Example 5
When multiplying two decimals the process is slightly different. Both numbers need to be divided by 10 × 100 = 1000
To make it an ordinary number we need to multiply both numbers by 10 and 100 and 6 × 8 = 48
Now reverse the process: Divide 48 by 1000, making the final answer 0.048
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# Graphs of Exponential Functions
## Growth and decay functions with varying compounding intervals
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Exponential Functions
In a laboratory, one strain of bacteria can double in number every 15 minutes. Suppose a culture starts with 60 cells, use your graphing calculator or a table of values to show the sample’s growth after 2 hours. Use the function b=60×2q\begin{align*}b=60 \times 2^q\end{align*}where b\begin{align*}b\end{align*} is the number of cells after q quarter hours.
In this concept, you will learn to recognize, evaluate and graph exponential functions.
### Exponential Functions
An exponential function is any function that can be written in the form y=abx\begin{align*}y=ab^x\end{align*} , where a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} are constantsa0,b>0\begin{align*}a \neq 0,b > 0\end{align*}, and b1\begin{align*}b\neq 1\end{align*}.
Two girls in a small town once shared a secret, just between the two of them. They couldn’t stand it though, and each of them told three friends. Of course, their friends couldn’t keep secrets, either, and each of them told three of their friends. Those friends told three friends, and those friends told three friends, and so on... and pretty soon the whole town knew the secret. There was nobody else to tell!
These girls experienced the startling effects of an exponential function. If you start with the two girls who each told three friends, you can see that they told six people or 2×3\begin{align*}2 \times 3\end{align*}. Those six people each told three others, so that 6×3\begin{align*}6 \times 3\end{align*} or 2×3×3\begin{align*}2 \times 3 \times 3\end{align*}. They told 18 people. Those 18 people each told 3, so that now is 18×3\begin{align*}18 \times 3\end{align*} or 2×3×3×3\begin{align*}2 \times 3 \times 3 \times 3\end{align*} or 54 people.
You can see how this is growing and you could show the number of people told in each round of gossip with a function: y=abx\begin{align*} y=ab^x\end{align*} where y\begin{align*}y \end{align*} is the number of people told, a\begin{align*}a\end{align*} is the two girls who started the gossip, b\begin{align*}b\end{align*} is the number of friends that they each told, and x\begin{align*}x\end{align*} is the number of rounds of gossip that occurred.
You could make a table of values and calculate the number of people told after each round of gossip. Use the function y=2×3x\begin{align*}y=2 \times 3^x\end{align*} where y\begin{align*}y\end{align*} is the number of people told and x\begin{align*}x\end{align*} is the number of rounds of gossip that occurred.
x\begin{align*}x\end{align*} rounds of gossip 0 1 2 3 4 5 y\begin{align*}y\end{align*} people told 2 6 18 54 162 486
Next, graph the relationship between the rounds of gossip and the number of people told.
How can you tell if a function is an exponential function?
If your function can be written in the form y=abx\begin{align*}y=ab^x\end{align*}, where a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} are constants, a0,b>0\begin{align*} a \neq 0,b > 0\end{align*}, and b1\begin{align*}b \neq 1\end{align*}, then it must be exponential. In quadratic equations, your functions were always to the 2nd power. In exponential functions, the exponent is a variable. Their graphs will have a characteristic curve either upward or downward.
Let’s look at some examples of exponential functions.
1. y=2x\begin{align*}y=2^x\end{align*}
2. c=4×10a\begin{align*}c=4 \times 10^a\end{align*}
3. y=2×(23)x\begin{align*}y=2 \times \left( \frac{2}{3} \right)^x\end{align*}
4. t=4×10y\begin{align*}t=4 \times 10^y\end{align*}
Now, here are some examples that are not exponential functions
1. y=3×1x\begin{align*}y=3 \times 1^x\end{align*} because b=1\begin{align*}b = 1\end{align*}.
2. n=0×3p\begin{align*}n= 0 \times 3^p\end{align*} because a=0\begin{align*}a = 0\end{align*}.
3. y=(4)x\begin{align*}y=(-4)^x\end{align*} because b<0\begin{align*}b<0\end{align*}.
4. y=6×0x\begin{align*}y=-6 \times 0^x\end{align*} because b1\begin{align*} b \le 1\end{align*}.
Exponential functions can be graphed by using a table of values like you did for quadratic functions. Substitute values for x\begin{align*}x\end{align*} and calculate the corresponding values for y\begin{align*}y\end{align*}.
Let’s look at an example.
Graph y=2x\begin{align*}y=2^x\end{align*}.
First, fill in the table of values.
x\begin{align*}x \end{align*} y=2x\begin{align*}y=2^x\end{align*} y\begin{align*}y\end{align*} −3\begin{align*}-3\end{align*} y=2−3\begin{align*}y=2^{-3} \end{align*} 18\begin{align*}\frac{1}{8}\end{align*} −2\begin{align*}-2\end{align*} y=2−2\begin{align*}y=2^{-2}\end{align*} \begin{align*}\frac{1}{4}\end{align*} \begin{align*}-1\end{align*} \begin{align*}y=2^{-1} \end{align*} \begin{align*}\frac{1}{2}\end{align*} \begin{align*}y=2^{0} \end{align*} \begin{align*}1\end{align*} \begin{align*}1\end{align*} \begin{align*}y=2^{1}\end{align*} \begin{align*}2\end{align*} \begin{align*}2\end{align*} \begin{align*}y=2^{2} \end{align*} \begin{align*}4\end{align*} \begin{align*}3\end{align*} \begin{align*}y=2^{3} \end{align*} \begin{align*}8\end{align*}
Next, graph the function.
Notice that the shapes of the graphs are not parabolic like the graphs of quadratic functions. Also, as the \begin{align*}x\end{align*} value gets lower and lower, the \begin{align*}y\end{align*} value approaches zero but never reaches it. As the \begin{align*}x\end{align*} value gets even smaller, the \begin{align*}y\end{align*} value may get infinitely close to zero but will never cross the \begin{align*}x\end{align*}-axis.
### Examples
#### Example 1
Earlier, you were given a problem about the scientists and the bacteria. The scientists are studying a strain of bacteria that doubles in number every 15 minutes. The function \begin{align*}b=60 \times 2^q\end{align*} represents the growth rate of the bacteria where \begin{align*}b\end{align*} is the number of cells there are after \begin{align*}q\end{align*} quarter hours.
First, create a t-table to go with the equation of the function.
\begin{align*}q\end{align*} \begin{align*}b=60 \times 2^q \end{align*} \begin{align*}b\end{align*} 0 \begin{align*}b=60 \times 2^0 \end{align*} 60 1 \begin{align*}b=60 \times 2^1\end{align*} 120 2 \begin{align*}b=60 \times 2^2 \end{align*} 240 3 \begin{align*}b=60 \times 2^3 \end{align*} 480 4 \begin{align*}b=60 \times 2^4 \end{align*} 960 5 \begin{align*}b=60 \times 2^5 \end{align*} 1920 6 \begin{align*}b=60 \times 2^6 \end{align*} 3840 7 \begin{align*}b=60 \times 2^7 \end{align*} 7680 8 \begin{align*}b=60 \times 2^8 \end{align*} 15360
Next, graph the function.
#### Example 2
Graph the following.
\begin{align*}y= 2 \times \left( \frac{2}{3} \right)^x\end{align*}
First, fill in the table of values.
\begin{align*}x\end{align*} \begin{align*}-3\end{align*} \begin{align*}-2\end{align*} \begin{align*}-1\end{align*} \begin{align*}1\end{align*} \begin{align*}2\end{align*} \begin{align*}3\end{align*} \begin{align*}y=2 \times \left( \frac{2}{3} \right)^x\end{align*} \begin{align*}y=2 \left( \frac{2}{3} \right)^{-3}\end{align*} \begin{align*}y=2 \left( \frac{2}{3} \right)^{-2}\end{align*} \begin{align*}y=2 \left( \frac{2}{3} \right)^{-1}\end{align*} \begin{align*}y=2 \left( \frac{2}{3} \right)^{0}\end{align*} \begin{align*}y=2 \left( \frac{2}{3} \right)^{1}\end{align*} \begin{align*}y=2 \left( \frac{2}{3} \right)^{2}\end{align*} \begin{align*}y=2 \left( \frac{2}{3} \right)^{3}\end{align*} \begin{align*}y\end{align*} \begin{align*}\frac{27}{4}\end{align*} \begin{align*}\frac{9}{2}\end{align*} \begin{align*}3\end{align*} \begin{align*}2\end{align*} \begin{align*}\frac{4}{3}\end{align*} \begin{align*}\frac{8}{9}\end{align*} \begin{align*}\frac{16}{27}\end{align*}
Next, graph the function.
#### Example 3
Identify the function \begin{align*}y=4^x\end{align*}.
\begin{align*}y=4^x\end{align*} is an exponential function.
#### Example 4
Identify the function \begin{align*}y=3x-1\end{align*}.
\begin{align*}y=3x-1\end{align*} is a linear function.
#### Example 5
Identify the function \begin{align*}y=ax^2-bx+c\end{align*}.
\begin{align*}y=ax^2-bx+c\end{align*} is a quadratic function.
### Review
Classify the following functions as exponential or not exponential. If it is not exponential, state the reason why.
1. \begin{align*}y=7^x\end{align*}
2. \begin{align*}c=-2 \times 10^d\end{align*}
3. \begin{align*}y=1^x\end{align*}
4. \begin{align*} y=4^x\end{align*}
5. \begin{align*}n=0 \times \left( \frac{1}{2} \right)^x\end{align*}
6. \begin{align*}y=5 \times \left( \frac{4}{3} \right)^x\end{align*}
7. \begin{align*}y=(-7)^x\end{align*}
8. Use a table of values to graph the function \begin{align*}y=3^x\end{align*}.
9. Use a table of values to graph the function \begin{align*}y=\left( \frac{1}{3} \right)^x\end{align*}.
10. What type of graph did you make in number 7?
11. What type of graph did you make in number 8?
12. Use a table of values to graph the function \begin{align*}y=-2^x\end{align*}.
13. Use a table of values to graph the function \begin{align*}y=5^x\end{align*}.
14. Use a table of values to graph the function \begin{align*}y=-5^x\end{align*}.
15. Use a table of values to graph the function \begin{align*}y=6^x\end{align*} .
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Asymptotic A function is asymptotic to a given line if the given line is an asymptote of the function.
Exponential Function An exponential function is a function whose variable is in the exponent. The general form is $y=a \cdot b^{x-h}+k$.
grows without bound If a function grows without bound, it has no limit (it stretches to $\infty$).
Horizontal Asymptote A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote.
Transformations Transformations are used to change the graph of a parent function into the graph of a more complex function.
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Home » Pythagoras theorem class 10 » Pythagoras theorem class 10
# Pythagoras Theorem for Class 10- Formula and Proof
The Pythagoras theorem is one of the most crucial and fundamental concepts in mathematics. Pythagoras theorem is named after the Greek philosopher Pythagoras.
## Pythagoras Theorem Class 10
Pythagoras theorem basically illustrates how a right-angled triangle’s sides are related to one another. The Pythagoras theorem states that the hypotenuse square of a triangle is equal to the sum of the squares of the other two sides. The Pythagoras theorem helps us to find out a triangle’s angle along with the length of the hypotenuse, perpendicular, and base of the right angle triangle. Here we learn more about the Pythagoras theorem, its formula, its equations, and derivations with solved examples, Let’s start learning…
## Pythagoras Theorem Formula Class 10
Pythagoras theorem formula: Before learning the formula of Pythagoras theorem, let’s get a quick review of a right-angle triangle because the Pythagorean Theorem mainly shows how the sides of a right-angled triangle relate to one another. A right-angle triangle. Right angles always have a 90° angle. Hypotenuse, Base, and Perpendicular are the names of the three sides that make up a right angle. The longest side is always the hypotenuse. The hypotenuse is the side with the 90° angle opposite The other two inner angles add up to 90 degrees.
According to the Pythagoras theorem, the square of the hypotenuse of a right-angle triangle equals the sum of the squares of the other two sides. This theorem can be expressed as the Pythagorean equation, which is a relationship between the lengths of the three sides of a right angle triangle a, b, and the hypotenuse c.
Perpendicular2 + Base2 = Hypotenuse2
a² + b² = c²
let’s assume a triangle ABC, where BC² = AB²+ AC² is present. The base is represented by AB, the altitude by AC, and the hypotenuse by BC in this equation.
## State Pythagoras Theorem
The Pythagoras theorem was first introduced by the Greek mathematician Pythagoras of Samos. He was a philosopher of the Classic Greek tradition. Born on the Greek island of Samos in 569 BC, Pythagoras of Samos travelled extensively in Egypt where he studied mathematics and other subjects. He established a community of mathematicians who practised strict discipline. Finally, the Greek mathematician proved the Pythagoras theorem; as a result, it is known as the “Pythagoras theorem” in his honour.
## Pythagoras Theorem Proof for Class 10
There are numerous ways to demonstrate Pythagoras Theorem. The algebraic method and the method utilising similar triangles are two of the most popular methods to prove the Pythagoras theorem. To comprehend the proof of this theorem, let’s take a deeper look at each of these approaches separately.
## Pythagoras Theorem Proof using Algebraic Method
The Pythagorean Theorem was proved using algebra using the diagram below. A big square and an inner square with sides equal to the hypotenuses of the four right triangles are constructed.
The triangles’ legs are of lengths a and b, and their hypotenuse is of length c.The sides of the big square that is produced measure (a + b). Its area is therefore equivalent to (a+b)²
Given that the length of the inner square sides is c, its area is. c².We can also observe that the area of the big square is the same as the sum of the areas of the four triangles and the inner square. As a result, we have:
(a+b)² =4(½×a× b) + c²
a² + b² +2ab = 2ab+ c²
a² + b² =c² (Proved)
## Pythagoras Theorem Proof using similar triangles
When the corresponding sides and angles of two triangles are the same size and ratio, then we can say that both two triangles are similar. Let’s learn how we can prove the Pythagoras theorem using similar triangles.
1. In a right triangle, the hypotenuse is divided into two pieces by an altitude taken from the right-angled vertex
2. The two smaller right triangles that resulted are similar to one another.
3. Measure the three triangles’ three sides collectively. The corresponding sides of the three triangles are proportional to one another because they are comparable triangles.
4. Consider the ratios of the corresponding sides. From the equations, get the squares of the original triangle’s sides.
5. Use the squares of the sides to prove the Pythagorean Theorem. Then, determine whether the sum is equal to the square of the hypotenuse.
## Pythagoras Theorem- Pythagorean triples Explained
Pythagorean triples consist of the three positive numbers a, b, and c, where a2+b2 = c2. The symbols for these triples are (a,b,c). Here, a represents the right-angled triangle’s hypotenuse, b its base, and c its perpendicular. The smallest and most well-known triplets are (3,4,5).
Pythagorean Triples Table (3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25) (20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53) (11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73) (13, 84, 85) (36, 77, 85) (39, 80, 89) (65, 72, 97)
## Pythagoras Theorem Derivation of Hypotenuse Formula
Imagine if the triangle ABC, which is right-angled at B. Create a perpendicular BD meeting AC at D. Use related triangles to prove the Pythagoras theorem.
In triangles △ABD and △ACB,
∠A = ∠A (common angle of both triangles)
∠ABC = ∠ADB (both angles are right angles)
Therefore, △ABD ∼△ACB (by AA similarity criterion)
We can also demonstrate similarly that△ BCD =△ ACB.
As a result,△ABD ∼ △ACB,
Henec, AD/AB = AB/AC. We can state that AD × AC = AB².
Similarly, we can prove that △BCD ∼ △ACB.
Hence, CD/BC = BC/AC Additionally, We can say, CD × AC = BC².
By combining these two equations, we obtain
AB² + BC² = (AD × AC) + (CD × AC)
AB² + BC² =AC² [ Proved ]
## Pythagoras Theorem Model
In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides, according to Pythagoras’s Theorem. These triangle’s three sides are known as the Perpendicular, Base, and Hypotenuse.
This theorem can be expressed as the Pythagorean equation, which is a relationship between the lengths of the three sides of a right angle triangle a, b, and the hypotenuse c.
Perpendicular2 + Base2 = Hypotenuse2
a² + b² = c²
## Pythagoras Theorem Calculator
The main application of Pythagoras theorem is that it helps us to determine the lengths and angles of a right triangle. Along with mathematics, there are various uses of Pythagoras theorm in our real life as mentioned below. We express the pythagorus theorem in a simple Pythagoras equation, which is
Perpendicular2 + Base2 = Hypotenuse2
Or, a² + b² = c², where the lengths of the three sides of a right angle triangle a, b, and the hypotenuse c.
## Pythagoras Theorem Applications
Several examples of how the Pythagoras Theorem is applied include
1. The Pythagorean Theorem is frequently used to determine the lengths of a right-angled triangle’s sides. We can get the formula for the base, perpendicular, and hypotenuse using the Pythagoras Theorem.
2. To determine the diagonal length of a rectangle, square, etc., use the theorem.
3. A triangle’s correctness is determined using the converse of Pythagoras Theorem.
4. In trigonometry, Pythagoras Theorem is used to determine trigonometric ratios such as sin, cos, tan, cosec, sec, and cot.
5. In the disciplines of engineering and building, architects apply the Pythagoras Theorem approach.
6. Applying the Pythagoras Theorem to mountain surveying
7. Finding the shortest path is another application for it in navigation.
8. Calculating how steep a mountain or hill’s slope requires the theorem.
## Pythagoras Theorem Based Questions
Q.A triangle with two sides that are 4 cm and 3 cm long. what is the length of the hypotenous of the triangle?
Solution.According to formula of Pythagoras theorem ,Perpendicular2 + Base2 = Hypotenuse2
In the given problem, Base = 4 cm , Perpendicular = 3 cm.
As per the Pythagoras theorem,
Hypotenuse2 = 4 2 + 32
Hypotenuse2 = 16+ 9 = 25
Hypotenuse =√25 = 5 cm
Q.Given is a triangle with sides measuring 5 cm, 13 cm, and 12cm. Determine if these would make up the triangle’s right-angled sides.
Solution: A right-angled triangle with these sides must satisfy the Pythagoras theorem.
We must determine whether 52 + 122 = 132 [ Perpendicular2 + Base2 = Hypotenuse2 ]
Now, L.H.S= (52 + 122 )=(25 +144)= 169
R.H.S= (13)= 169
L.H.S = R.H.S
So, this triangle fully satisfies the conditions of the Pythagoras theorem. So that it is a right-angle triangle.
Q. A triangle that has a hypotenosis of 17 cm and the base of 8 cm . Find the length of the perpendicular of the triangle.
Solution:
Length of the Base (a)= 8 cm
Length of the Hypotenous (c) = 17 cm
Let’s assume the length of the perpendicular is b cm.
According to Pythagoras theorem, a² + b² = c²
Or, b² = c² – a²
Or, b² = √( 17²- 8² )
Or, b = √ ( 289- 64) =√ 225
Or, b = 15
So,the length of the perpendicular is 15 cm.
## Pythagoras Theorem Class 10 worksheet
Q.1.A ladder is placed 12 cm away from the wall such that the top of the ladder is 5 cm above the floor. Find the wall’s length.
Q.2.Given is a triangle with sides measuring 9 cm, 40 cm, and 41cm. Determine if these would make up the triangle’s right-angled sides.
Q.3.A triangle with two sides that are 16cm and 63 cm long. what is the length of the hypotenous of the triangle?
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## FAQs
### What is the Pythagoras theorem?
According to the Pythagoras theorem, the square of the hypotenuse of a right-angle triangle equals the sum of the squares of the other two sides. This theorem can be expressed as the Pythagorean equation, which is a relationship between the lengths of the three sides of a right angle triangle a, b, and the hypotenuse c.
Perpendicular² + Base² = Hypotenuse²
a² + b² = c²
### Q.How do you prove the Pythagoras theorem?
There are numerous ways to prove Pythagoras Theorem. The algebraic method and the method utilizing similar triangles are two of the most popular methods to prove the Pythagoras theorem. Both of the methods are clearly explained in the article
### Is it possible to apply the Pythagorean theorem to any triangle?
Not at all. Any triangle has nothing to do with the Pythagorean theorem. Only a right-angled triangle can be used to use the Pythagoras theorem to get the relationship where the sum of two squared sides equals the square of the third side.
### Are there any hard Pythagoras questions?
There are some hard Pythagoras questions. but if you have clearly understood the basic concept of the Pythagoras theorem. Then you easily solve most of the questions.
### What is the Pythagorean equation for a triangle?
Any triangle that has a 90° angle on one side is subject to the Pythagoras equation. The square of the hypotenuse (AC²) of an ABC is equal to the sum of the squares of the sides (AB² + BC²), according to the Pythagorean theorem formula.
### Which is the longest side of the Pythagoras theorem?
The Pythagoras Theorem states that the side opposed to the right angle (90°) is the longest side because the longest side is that which is opposite the greatest angle.
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Find for which value of the parameter $k$ a function is bijective
I have to draw (by hand obviously) the plot of the following function:
$$f(x)= 13\ln(\frac{x}{|x+1|})-12\ln (x+x^2) +kx,$$
for $k \in \mathbb{R}$. To do so, I have to study the first and second derivative, limits at infinity, and so on. Normally, I do these exercises with functions without parameters. Could you show me how to proceed in this case?
Also, I have to determine for which values of $k$ this function is bijective. However, I don’t have a clue about how to proceed, as I’ve never been shown exercises of this kind.
Solutions Collecting From Web of "Find for which value of the parameter $k$ a function is bijective"
Due to the first term, the function is only defined for $x>0$, because $\frac{x}{|x+1|}>0$ if and only if $x>0$. So we’ll assume $f\colon(0,\infty)\to\mathbb{R}$.
So first of all rewrite it as
$$f(x)=13\ln x-13\ln(x+1)-13\ln x-12\ln(1+x)+kx=\ln x-25\ln(x+1)+kx$$
We easily have
$$\lim_{x\to0}f(x)=-\infty,$$
In order that the function is onto $\mathbb{R}$, we need that $\lim_{x\to\infty}f(x)=\infty$. Now
$$f(x)=\ln\frac{x}{(x+1)^{25}}+kx$$
and the limit at $\infty$ of the first term is $-\infty$. So this forces $k>0$, because for $k\le0$ we have $\lim_{x\to-\infty}f(x)=-\infty$. When $k>0$ we indeed have $\lim_{x\to\infty}f(x)=\infty$ (verify it).
Now we need that the function is monotonic, so we look at the derivative
$$f'(x)=\frac{1}{x}+\frac{25}{1+x}+k$$
Under the hypothesis that $k>0$ this is everywhere positive, so the function is indeed increasing.
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CLAT > Squares, Square Roots, Cubes, Cube Roots
# Squares, Square Roots, Cubes, Cube Roots - Quantitative Techniques for CLAT
Table of contents SQUARES, SQUARE ROOTS, CUBES, CUBE ROOTS To determine whether a given number is a perfect square. Properties of Squares Some interesting facts: Square Roots Finding the square root of a perfect square Finding the number of digits in the square root of a perfect square. Finding square root of a perfect square decimal number by long division. Cubes How to determine whether a given number is a perfect cube Properties of cubes: Cube roots:
## SQUARES, SQUARE ROOTS, CUBES, CUBE ROOTS
Squares: When the exponent of a base is 2, the numbers obtained are called squares or squared numbers.
Perfect Square: A natural number is called a perfect square if it is the square of a number.
1, 4, 9, 16, 25, 36 --------- are the perfect square of the numbers
1, 2, 3, 4, 5, 6 ------------ respectively.
5, 6, 7, 8, 11, 50, and 125 are not perfect squares of any natural number.
Examples of Perfect Square
## To determine whether a given number is a perfect square.
Step I: Express the given number as the product of its prime numbers.
Step II: Express the prime factors as pairs of same number
Step III: Check if there are any unpaired factors
Step IV: If there is no unpaired factor, the given number is a perfect square.
## Properties of Squares
1) Squares of even numbers are even.
2) Squares of odd numbers are odd.
3) A number ending with 2, 3, 7, or 8 is never a perfect square. e.g. 23, 347, 2818 cannot be perfect squares.
4) A number ending with an odd number of zeros is never a perfect square. Thus 150, 4400000, and 48000 cannot be perfect squares.
5) The square of a natural number (other than 1) is either a multiple of 3 or exceeds a multiple of 3 by 1.
6) The square of a natural number (other than 1) is either a multiple of 4 or exceeds a multiple of 4 by 1
7) The difference between the squares of two consecutive natural numbers is equal to their sum.
For example: 22 - 12 = 4 – 1 = 3 which is equal to 2 + 1 = 3
52 - 42 = 25 – 16 = 9 which is equal to 5 + 4 = 9
112 - 102 = 121 – 100 = 21 which is equal to 11 + 10 = 21
Question for Squares, Square Roots, Cubes, Cube Roots
Try yourself:The least perfect square, which is divisible by each of 21, 36 and 66 is:
## Some interesting facts:
1) Pythagorean Triplets: For every natural number m (other than 1) there is a triplet such as (2m), (m2-1), and (m2+1)
These triplets are called Pythagorean Triplets. They obey the relation (2m)2+(m2-1)2=(m2+1)2
For m = 2, Pythagorean triplets are 2 x 2, 22-1, 22+1 or 4, 3 and 5 and (4)2 + (3)2 = 52
For m = 4, triplets are 4 x 2, 4- 1, 4+ 1
i.e. 8, 15, 17 and 82 + 152 = 172
and so on it is true for every natural number.
2) The square of any natural number will have either 0, 1, 4, 5, 6, or 9 at the unit place.
3) The square of a natural number (n) is always equal to the sum of the first (n) odd natural numbers.
If x = 4 Then 42 = 1 + 3 + 5 + 7
If x = 6 Then 62 = 1 + 3 + 5 + 7 + 9 + 11
If x = 9 Then 92 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17
4) If n is a perfect square then 2n can never be a perfect square.
5) A perfect square number is never negative.
## Square Roots
The square root of a number n is that number which when multiplied by itself gives n as the product. For example 7 x 7 = 49, So 7 is the square root of 49. It is written as
## Finding the square root of a perfect square
(a) By finding prime factors.
Example: If the square root of 576 is to be found out we find the prime factors of 576
576 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
Make pairs of factors and pick up one factor from each pair.
= 2 x 2 x 2 x 3 = 24
Example: Simplify
Solution: 900 = 30 x 30
0.09 = 0.3 x 0.3 = 0.3
0.003
= 30.303
(b) Finding the square root by long division method
Example1: Find the square root of 7396
Solution:
Step I: Beginning from the right, reduce the digits in pairs from right to left.
Step II: Take the first pair of digits and find the nearest sq. root. The nearest square root of 73 is 8. Write 8 as quotient and also as divisor. Write the product 8 x 8 = 6 below 73 and Subtract.
The remainder is 9
Step III: Bring down the next pair 96. Double 8, the number in the quotient and place the double i.e. 16 as the next divisor. Now 99 divided by 16 is approximately 6. So put 6 with 8 in the quotient also with 16 as the divisor and multiply 166 by 6 and get 996. Now the remainder is zero. So 86 is the exact square root.
Therefore
Example 2: Find the sq. root of
Solution:
## Finding the number of digits in the square root of a perfect square.
Rule: If a perfect square contains n digits then, its square root will contain digits if n is even and digits if n is odd.
Question for Squares, Square Roots, Cubes, Cube Roots
Try yourself:is equal to
## Finding square root of a perfect square decimal number by long division.
While finding the square root of a natural number we make pairs from right to left and if in the last, one digit is left, we leave it by itself. In case of decimal number we make pairs from right to left for the integral part and from left to right for the decimal portion.
If the last period of the decimal number contains only one digit, we add zero to it. The square root of a decimal number will contain as many decimal places as there are pairs in the decimal part of the given number. The method to be followed is the same as for the whole number.
Example 3: If = 56. Find the value of +
Solution: Since = 56
= 5.6
and = 0.56
+ .= 5.6 + 0.56 = 6.16
Example 4: Find the value of
If
Solution:
Example 5: Evaluate
Solution: Given expression =
Example 6: Evaluate
Solution: The sum of decimal places both in the numerator as well as denominator is 6. So we can remove the decimals
Example 7: Find the least number which must be added to 4562 to make it a perfect square.
Solution:
8 x 128 = 1024
If in place of 962 it is 1024 we will get a perfect square so add
1024 – 962 = 62
## Cubes
A cube of a number, is that number raised to the power 3. The cube of a number is obtained by multiplying it with itself 3 times.
Cube of 5 is 5 x 5 x 5 = 125
Cube of 6 is 6 x 6 x 6 = 216
## How to determine whether a given number is a perfect cube
Step 1 Make prime factors of the given number
Step 2 Make groups of the prime factors – a group containing 3 same numbers.
Step 3 Check the ungrouped factor
Step 4 If there is no ungrouped factor, the number is a perfect cube.
Example: Is 1728 a perfect cube
Solution: Prime factors of 1728 are = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3
The prime factors can clearly be grouped into groups of 3 each and no factor is left ungrouped
So 1728 is a perfect cube.
## Properties of cubes:
(i) Cube of a natural number is a natural number
(ii) Cube of all even natural numbers are even.
(iii) Cube of all odd natural numbers are odd.
(iv) Cubes of all negative numbers are negative.
(v) For any rational number
(vi) Cube of a natural number, which is a multiple of 3, is a multiple of 27
(vii) Cube of a natural number of the form 3n+1 is also a natural number of the form 3n+1
(viii) Cube of a natural number of the form 3n+2 is also a natural number of the form 3n+2
(ix) Cube of a number, having a zero at the unit place, has three zeros at the end.
For example (10)3 = 1000
Example: What is the smallest number by which 2400 be multiplied so that the product is a perfect cube?
a) 70 b) 80 c) 90 d) 60
Solution: 2400 = 2 x 2 x 2 x 2 x 2 x 3 x 5 x 5
After making the groups of three for each prime number we need one 2’s, two 3’s and one 5.
So we need a number 2 x 3 x 3 x 5 i.e. 90
So option (c) is correct.
## Cube roots:
A natural number n is a cube root of the number m if m = n3
The symbol for cube root is
To find the cube root of a perfect cube number
Step 1: Make prime factors of the given number
Step 2: Make groups of three each of the prime factors
Step 3: Choose one out of the group of three factors and find their product
Step 4: This product is the required cube root of the given number.
Example: Find the cube root of -4096
Solution: Prime factors of 4096 are 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Question for Squares, Square Roots, Cubes, Cube Roots
Try yourself:The cube root of .000216 is:
The document Squares, Square Roots, Cubes, Cube Roots | Quantitative Techniques for CLAT is a part of the CLAT Course Quantitative Techniques for CLAT.
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## Quantitative Techniques for CLAT
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# How do you find the amplitude, period, vertical and phase shift and graph y=sintheta+0.25?
Jul 26, 2018
Below
#### Explanation:
$y = \sin \theta + 0.25$ can also be written as $y = 0.25 + \sin \theta$ which is in the form $y = b + a \sin \left(n \theta\right)$
where $a$ is the amplitude and $b$ is the shift upwards or downwards and $n$ is the phase shift
Now when we compared $y = 0.25 + \sin \theta$ and $y = b + a \sin \left(n \theta\right)$, we can say that:
$b = 0.25$
$a = 1$
$p e r i o d \left(T\right) = \frac{2 \pi}{n} = \frac{2 \pi}{1} = 2 \pi$
Therefore, we know that $y = \sin \theta + 0.25$ is the graph $y = \sin \theta$ shifted UPWARDS by $0.25$ units and they also have the same phase shift $\left(2 \pi\right)$
Below is graph $y = \sin \theta$
graph{sinx [-10, 10, -5, 5]}
Below is graph $y = \sin \theta + 0.25$
graph{0.25+sinx [-10, 10, -5, 5]}
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Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 10 Subquestion (vii)
Discontinuous
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left\{\begin{array}{c} \frac{2|x|+x^{2}}{x}, \text { if } x \neq 0 \\\\ 0, \quad \text { if } x=0 \end{array}\right.$
$f(x)=\left\{\begin{array}{c} \frac{2 x+x^{2}}{x}, \text { if } x>0 \\\\ \frac{-2 x+x^{2}}{x}, \text { if } x<0 \\\\ 0, \quad \text { if } x=0 \end{array}\right.$
$f(x)=\left\{\begin{array}{c} (x+2), \text { if } x>0 \\\\ (x-2), \text { if } x<0 \\\\ 0, \quad \text { if } x=0 \end{array}\right.$
We observe
[LHL at $x=0$ ]
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}(-h-2)=-2$
[RHL at $x=0$ ]
\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}(h+2)=2 \\\\ &\lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \end{aligned}
Thus, $f\left ( x \right )$ is discontinuous at $x=0$ .
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Mathematics Class VII
Integers
Fractions and Decimals
Exponents and Powers
Algebraic Equations
Simple Linear Equations
Lines and Angles
Comparing Quantities
Congruence of Triangles
Rational Numbers
Perimeter and Area
Data Handling
Practical Geometry
Symmetry
Visualising Solid Shapes
Applications Of Simple Equations To Practical Situations
If we have statements related to a practical situation then first we have to convert it in the form of the equation then solve it to find the solution.
Example: 1
Radha’s Mother’s age is 5 years more than three times Shikha’s age. Find Shikha’s age, if her mother is 44 years old.
Solution:
Let Shikha’s age = y years
Her mother’s age is 3y + 5 which is 44.
Hence, the equation for Shikha’s age is 3y + 5 =44
3y + 5 = 44
3y = 44 – 5 (by transposing 5)
3y = 39
y = 13 (by dividing both sides by 3)
Hence, Shikha’s age = 13 years.
Example: 2
A number consists of two digits. The digit in the tens place is twice the digit in the units place. If 18 be subtracted from the number, the digits are reversed. Find the number.
Solution:
Let the digit at units place = y
So, the digit in the tens place = 2y
So, the number is (2y) y.
As it is given that if 18 is subtracted from the number, the digits are reversed.
So, we have
(2y) y – 18 = y(2y)
Solve it
20y + y – 18 = 10y + 2y
21y – 18 = 12y
21y – 18 = 12y
9y = 18 (By dividing both sides by 9)
y = 2
The digit at the units place is y = 2.
And the digit at the tens place is 2y
= 4
Hence, the required number is 42.
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Proof of $(a-b)^2$ formula in Geometrical Method
The expansion of a minus b whole squared algebraic identity can be derived in algebraic form by the geometrical approach. The concept of areas of geometrical shapes such as squares and rectangles are used for proving the a minus b whole square formula in algebraic form.
Find the Area of a square
Take a square and assume the length of each side of this square is represented by $a$.
We have to calculate the area of this geometric shape mathematically. According to the mathematics, the area of a square is equal to the square of the length of the side.
In this case, it is taken that the length of the each side is equal to $a$. Therefore, the area of this square is equal to $a^2$, which is very important to remember for deriving the $a-b$ whole square algebraic identity.
Basic steps to split the square
For proving the $a$ minus $b$ whole squared formula geometrically, we have to split the square as four different geometrical figures.
1. Draw a straight line vertically across the pair of opposite sides for dividing the square as two rectangles. If the length of one rectangle is denoted by $b$, then the length of remaining rectangle will be $a-b$.
2. Divide the both rectangles horizontally by drawing a straight line such that the width of two shapes are equal to $b$, then the widths of remaining two shapes are equal to $a-b$. It is useful for creating a square, whose area is equal to $a-b$ whole square.
Geometrically, we have successfully divided a square as two small squares and two rectangles.
Evaluate the Areas of Geometric shapes
Now, calculate the area of each geometrical shape mathematically.
1. The length of each side of one square is $a-b$, then its area is equal to $(a-b)^2$.
2. The length and width of one rectangle are $b$ and $a-b$ respectively, then its area is equal to $b \times (a-b)$, which can be written as $b(a-b)$ simply.
3. The length and width of second rectangle are $a-b$ and $b$ respectively, then its area is equal to $(a-b) \times b$, which is simply written as $(a-b)b$ in mathematics.
4. The length of the remaining square is $b$. Therefore, its area is equal to $b^2$.
Thus, the areas of all geometrical shapes are calculated and expressed in algebraic form.
Equality of the Areas of the Geometric shapes
It is time to prove the expansion of the $a$ minus $b$ whole squared formula geometrically.
Geometrically, a square is divided as four different geometrical shapes. So, the area of the square is equal to the sum of the areas of the four geometrical shapes. It can be written in mathematical form as an equation.
$a^2$ $\,=\,$ $(a-b)^2$ $+$ $b(a-b)$ $+$ $(a-b)b$ $+$ $b^2$
In this derivation, we have to find the expansion of $(a-b)^2$ identity. So, shift all the terms to other side of the equation for finding the equivalent value of the $a-b$ whole squared.
$\implies$ $a^2$ $-$ $b(a-b)$ $-$ $(a-b)b$ $-$ $b^2$ $\,=\,$ $(a-b)^2$
$\implies$ $(a-b)^2$ $\,=\,$ $a^2$ $-$ $b(a-b)$ $-$ $(a-b)b$ $-$ $b^2$
In the right hand side of the equation, the second and third terms $b(a-b)$ and $(a-b)b$ are equal mathematically as per commutative property of multiplication.
$\implies$ $(a-b)^2$ $\,=\,$ $a^2$ $-$ $b(a-b)$ $-$ $b(a-b)$ $-$ $b^2$
$\implies$ $(a-b)^2$ $\,=\,$ $a^2-2b(a-b)-b^2$
$\implies$ $(a-b)^2$ $\,=\,$ $a^2-2ba+2b^2-b^2$
$\,\,\, \therefore \,\,\,\,\,\,$ $(a-b)^2$ $\,=\,$ $a^2-2ab+b^2$
It is also written as $(a-b)^2$ $\,=\,$ $a^2+b^2-2ab$
It is read as the $a$ minus $b$ whole squared is equal to $a$ squared plus $b$ squared minus $2$ times product of $a$ and $b$. Thus, the $a-b$ whole square algebraic identity is proved in algebraic form geometrically.
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The Difference Quotient: The Bridge between Algebra (Slope) and Calculus (the Derivative) - dummies
# The Difference Quotient: The Bridge between Algebra (Slope) and Calculus (the Derivative)
One of the cornerstones of calculus is the difference quotient. The difference quotient — along with limits — allows you to take the regular old slope formula that you used to compute the slope of lines in algebra class and use it for the calculus task of calculating the slope (or derivative) of a curve. Here’s how it works.
In the following example, you want to find the slope at a point on the parabola.
To compute the slope, you need two points to plug into this formula. For a line, this is easy. You just pick any two points on the line and plug them in.
You can see the line drawn tangent to the curve at (2, 4), and because the slope of the tangent line is the same as the slope of the parabola at (2, 4), all you need is the slope of the tangent line. But you don’t know the equation of the tangent line, so you can’t get the second point — in addition to (2, 4) — that you need for the slope formula.
Here’s how the inventors of calculus got around this roadblock.
The above figure is the graph of y = x2 with a tangent line and a secant line. It shows the tangent line again and a secant line intersecting the parabola at (2, 4) and at (10, 100).
A secant line is a line that intersects a curve at two points. This is a bit oversimplified, but it’ll do.
The slope of this secant line is given by the slope formula:
You can see that this secant line is quite a bit steeper than the tangent line, and thus the slope of the secant, 12, is higher than the slope you’re looking for.
Now add one more point at (6, 36) and draw another secant using that point and (2, 4) again. See the above figure.
Calculate the slope of this second secant:
You can see that the slope of this secant line is a better approximation of the slope of the tangent line than the slope of the first secant was.
Now, imagine what would happen if you grabbed the point at (6, 36) and slid it down the parabola toward (2, 4), dragging the secant line along with it. Can you see that as the point gets closer and closer to (2, 4), the secant line gets closer and closer to the tangent line, and that the slope of this secant thus gets closer and closer to the slope of the tangent?
So, you can get the slope of the tangent if you take the limit of the slope of this moving secant.
So here’s the limit you need:
Watch what happens to this limit when you plug in three more points on the parabola that are closer and closer to (2, 4):
When the point slides to (2.01, 4.0401), the slope is 4.01
When the point slides to (2.001, 4.004001), the slope is 4.001
Sure looks like the slope is headed toward 4.
As with all limit problems, the variable in this problem, the run, approaches but never actually gets to zero. If it got to zero — which would happen if you slid the point you grabbed along the parabola until it was actually on top of (2, 4) — you’d have a slope of 0/0, which is undefined. But, of course, that’s precisely the slope you want — the slope of the line when the point does land on top of (2, 4). Herein lies the beauty of the limit process.
And the slope of the tangent line is — you guessed it — the derivative.
The derivative of a function, f(x), at some number x = c, written as f’(c), is the slope of the tangent line to f drawn at c.
Okay, here’s the most common way of writing the difference quotient (you may run across other, equivalent ways).
Take a look at the following figure, which shows how a limit produces the slope of the tangent line at (2, 4).
Doing the math gives you, at last, the slope of the tangent line at (2, 4):
So the slope is 4. (By the way, it’s a meaningless coincidence that the slope at (2, 4) happens to be the same as the y-coordinate of the point.)
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# How do you find the first derivative of y=(sinx/(1+cosx))^2?
May 22, 2015
Let's use the chain rule, by naming $u = \frac{\sin x}{1 + \cos x}$.
The chain rule states that
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$
Thus,
$\frac{\mathrm{dy}}{\mathrm{du}} = 2 u$
$\frac{\mathrm{du}}{\mathrm{dx}}$ - here, we have to apply quocient rule, which states that
Be $y = \frac{f \left(x\right)}{g \left(x\right)}$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{f \left(x\right)} ^ 2$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\left(\left(\cos x\right) \left(1 + \cos x\right)\right) - \left(\sin x\right) \left(- \sin x\right)}{1 + \cos x} ^ 2$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\left({\cos}^{2} x + \cos x\right) + {\sin}^{2} x}{1 + \cos x} ^ 2$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\textcolor{g r e e n}{{\cos}^{2} x + {\sin}^{2} x} + \cos x}{1 + \cos x} ^ 2$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\cancel{1 + \cos x}}{1 + \cos x} ^ \cancel{2}$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{1 + \cos x}$
Now, combining $\frac{\mathrm{du}}{\mathrm{dx}}$ and $\frac{\mathrm{dy}}{\mathrm{du}}$:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 u\right) \left(\frac{1}{1 + \cos x}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(\sin \frac{x}{1 + \cos x}\right) \left(\frac{1}{1 + \cos x}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{g r e e n}{\frac{2 \sin x}{1 + \cos x} ^ 2}$
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# Dissections and Proof Homework
## Session 5, Homework
### Problem H1
You can make your own tangram set from construction paper. Start with a large square of construction paper and follow the directions below:
Step 1: Fold the square in half along the diagonal; unfold and cut along the crease. What observations can you make about the two pieces you have? How could you prove that your observations are correct?
Step 2: Take one of the triangles you have, and find the midpoint of the longest side. Connect this to the opposite vertex and cut along this segment.
Step 3: Take the remaining half and lightly crease to find the midpoint of the longest side. Fold so that the vertex of the right angle touches that midpoint, and cut along the crease. Continue to make observations.
Step 4: Take the trapezoid and find the midpoints of each of the parallel sides. Fold to connect these midpoints, and cut along the fold.
Step 5: Fold the acute base angle of one of the trapezoids to the adjacent right base angle and cut on the crease. What shapes are formed? How do these pieces relate to the other pieces?
Step 6: Fold the right base of the other trapezoid to the opposite obtuse angle. Cut on the crease. You should now have seven tangram pieces. Are there any other observations you can make?
### Problem H2
Try this with a very long and skinny parallelogram.
When people work on the dissection problems, they often create a figure that looks like a rectangle, but they can’t explain why the process works. Or sometimes they perform a cutting process that they think should work, but the result doesn’t look quite right. Reasoning about the geometry of the process allows you to be sure. A cutting process is outlined below. Your job is to analyze if the cuts really work. Does this algorithm turn any parallelogram into a rectangle? If so, provide the justifications. If not, explain what goes wrong.
First, cut out the parallelogram. Then fold along both diagonals. Cut along the folds, creating four triangles as shown.
Slide the bottom triangle (number 4) straight up, aligning its bottom edge with the top edge of triangle 2. Slide the left triangle (number 1) to the right, aligning its left edge with the right edge of triangle 3.
Here are some new cutting problems. When you find a process that you think works, justify the steps to be sure.
Form two angles that are smaller than the smallest one in the original triangle, and a third angle that is larger than the largest angle in the original.
### Problem H3
Start with a scalene triangle. Find a way to dissect it into pieces that you can rearrange to form a new triangle, but with three different angles. That is, no angles of the new triangle have the same measure as any angle of the original triangle. You should be able to demonstrate the following:
a. The final figure is a triangle. (It has exactly three sides.) b. The two joined edges match. c. All three angles are different from those of the original triangle. d. As a challenge, can you solve this with just one cut?
### Problem H4
Start with a scalene triangle. Find a way to dissect it into pieces that you can rearrange to form a new triangle, but with three different sides. That is, no sides of the new triangle have the same length as any side of the original triangle. You should be able to demonstrate the following:
a. The final figure is a triangle. (It has exactly three sides.) b. The two joined edges match. c. All three sides are different from those of the original triangle.
#### Take it Further
Remember from Session 3 that you can divide any quadrilateral into two triangles.
### Problem H5
Start with an arbitrary quadrilateral. Find a way to dissect it into pieces that you can rearrange to form a rectangle. Test your method on quadrilaterals like these. Try to justify why it will always work.
Problems H3-H5 adapted from Connected Geometry, developed by Educational Development Center, Inc. pp.169, 171. © 2000 Glencoe/McGraw-Hill. Used with permission. www.glencoe.com/sec/math
Senechal, Marjorie (1990). Shape. In On the Shoulders of Giants: New Approaches to Numeracy. Edited by Lynn Arthur Steen (pp. 148-160). Washington, D.C.: National Academy Press.
Shape
Continued…
Continued…
Continued…
### Solutions
Problem H1
Step 1: The two shapes are congruent isosceles right triangles since all three pairs of corresponding sides are congruent, and two of the square’s right angles are preserved, one in each triangle.
Step 2: Again, we get two congruent isosceles right triangles since the two legs of each are half the length of the original diagonal of the square, and since the angle between the two legs is the right angle (SAS congruence).
Step 3: We get an isosceles right triangle and a trapezoid. The base of the triangle is half the length of the longer base of the trapezoid (midline theorem), and is of same length as the shorter base of the trapezoid.
Step 4: The two shapes are congruent right trapezoids, each one of which has two right angles, one angle of 45° and one of 135°.
Step 5: We get a square and an isosceles right triangle. These shapes are similar to those encountered in previous steps; i.e., their corresponding sides are proportional.
Step 6: The final two shapes are a parallelogram and an isosceles right triangle. The triangle is congruent to the one created in Step 5. The parallelogram has one pair of sides congruent to the leg of the triangle created, and one pair of sides congruent to the hypotenuse of the triangle just created.
### Problem H2
The construction does not work in general. To see this, consider triangle number 4: In the final arrangement, one of its angles is also an angle of the final shape, which is supposed to be a rectangle. The angle in question is formed by the intersecting diagonals, so the construction only works if the diagonals of the original parallelogram intersect at an angle of 90°. This happens when the original parallelogram is a rhombus.
### Problem H3
Start with a scalene triangle ABC. Find the midpoint D of the side opposite the smallest angle.
Cut along the segment DB, then rotate the triangle DCB about the vertex D by 180° (see picture). The final figure has three sides. Sides AD and DC match since they are of equal length, and points B, D, and B’ are collinear since B’D is obtained by rotating DB by 180°. The angle at A is larger than any of the angles in the original triangle (since it is the sum of the two largest angles of the original triangle). The angles at B and B’ are smaller than (hence different from) any of the angles in the original triangle since they are both smaller than the smallest angle in the original triangle. We only used one cut.
### Problem H4
Let’s say AB is the longest side of the original triangle.
Draw the altitude CD (from vertex C to the side AB). Notice that it is shorter than any of the three sides of the original triangle.
Make the triangle into a rectangle using one of the methods we found in the session. Use side AB for the base. The other side will have length 1/2 CD (half the altitude).
Now, cut along a diagonal of the rectangle. (Note that the diagonal is longer than side AB, and so it is longer than all three sides of the original triangle, since AB was longest.) This creates two small right triangles; call them FAB and BEF (they share two vertices; the right angles are at A and E).
Form an isosceles triangle by flipping and translating triangle BEF so that angles A and E (the right angles) are adjacent and sides FA and EB align, thus forming a new shortest side of the new triangle.
The shortest side of this triangle is twice the smaller side of our rectangle. That means it’s the same as the altitude of the original triangle, and hence smaller than any of the original sides. The other two sides are the same, and since they are diagonals of the rectangle, they are longer than any of the original three sides.
### Problem H5
Split the quadrilateral in half along an interior diagonal. You have two triangles that have a side in common. For each triangle, use the common side as the base, and turn it into a rectangle using your algorithm from Problem B4. Now stack the two rectangles on top of each other. They match up because they have a common base (the diagonal of the original quadrilateral).
|
Equable Cylindrical Cone
EQUABLE CYLINDRICAL CONE
Balmoral Software
Solutions: 1
Consider the "cylindrical cone" consisting of a cone of radius R and height h surmounting a cylinder of the same radius and height. There is one such polytope that is equable. The total volume of the cone + cylinder is V, where
V/π = (1/3)R2h + R2h = (4/3)R2h
The lateral surface area of the cone is Scone, where
The lateral surface area of the cylinder is Scyl, where
Scyl/π = 2Rh [1]
The surface area of the cylinder base is Sbase, where
Sbase/π = R2 [2]
The above four quantities are positive integers by convention. We have
so R is rational. By [2], R2 is an integer, so it follows that R is an integer.
The equability assumption is
Any equable solution will have a positive left side. Squaring both sides, we have
(16R2 - 48R + 27)h = 24R2 - 36R [3]
If R = 1, then h = 12/5 and Scyl/π = 24/5 is not an integer. If R = 2, then h < 0. Therefore, R ≥ 3 and
so both the right side and the coefficient of h in [3] are strictly positive, and a rational solution for h is
[4]
From [1],
so
for some integer k. We previously established that the denominator is positive, and the numerator is also positive since R ≥ 3, so k is positive. The preceding equation can be written as a quadratic in R:
(16k - 72)R2 - (48k - 81)R + 27k = 0 [5]
Since R is an integer, the discriminant of this quadratic must be a square:
576k2 + 6561 = m2 for some integer m
(m - 24k)(m + 24k) = 6561
The possibilities are summarized below:
m - 24k m + 24k Sum = 2m m < k = [(m + 24k) - m]/24 1 < 6561 6562 3281 (6561 - 3281)/24 is not an integer 3 < 2187 2190 1095 (2187 - 1095)/24 is not an integer 9 < 729 738 369 (729 - 369)/24 = 15 27 < 243 270 135 (243 - 135)/24 is not an integer
The only solution is k = 15, and thus [5] can be written
168R2 - 639R + 405 = 0,
which has the sole integer solution R = 3. From [4], we have h = 4 and
V = 48 π
Scone = 15 π
Scyl = 24 π
Sbase = 9 π
The equable cylindrical cone is a solid of revolution of a non-equable quadrilateral consisting of a 4 x 3 right triangle with its shortest side appended to a 4 x 3 rectangle, having perimeter 20 and area 18 and rotated along its longest side.
### Non-equable case
It's interesting that without the equability restriction, the integer surface area sections have the property that
Scone2 = Sbase2 + (Scyl/2)2
Therefore, any Pythagorean triple a2 + b2 = c2, for which
[6]
generates the non-equable solution
For any Pythagorean triple, an alternate solution with a and b exchanged is possible as long as [6] is satisfied. Applying the limits on a and b, we have
so
h2 > 9/16 + 81/624 ≈ 0.6923
Any element of a Pythagorean triple is at least 3, so R2 = a ≥ 3 and
which reduces to
V > Scone
Therefore, all solutions for which Scone ≤ M for some maximum M will include as a subset all solutions with V ≤ M. The 23 solutions having volume less than 2000 π are listed below:
RhSconeSbaseScylV/π
23/25468
341592448 *
4320162464
65/2393630120
340/341980160
415/2341660160
69/2453654216
1217/1214514434272
463/46516126336
68603696384
5126525120400
86806496512
677/68536154616
127/215014484672
635/211136210840
125156144120960
1015/21251001501000
680/3164363201280
815136642401280
912135812161296
1021/21451002101400
724175493361568
1291801442161728
*: Equable
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How do you write the partial fraction decomposition of the rational expression (x^4+1)/(x^5+6 x^3)?
Oct 11, 2016
(x^4+1)/(x^5+6 x^3)=1/(6 x^3) - 1/(36 x) + 37/(72 ( x-i sqrt[6] )) + 37/( 72 (x+i sqrt[6]))
Explanation:
The proposition
$\frac{{x}^{4} + 1}{{x}^{5} + 6 {x}^{3}} = \frac{{x}^{4} + 1}{{x}^{3} \left({x}^{2} + 6\right)} = {c}_{1} / x + {c}_{2} / {x}^{2} + {c}_{3} / {x}^{3} + {c}_{4} / \left(x + i \sqrt{6}\right) + {c}_{5} / \left(x - i \sqrt{6}\right)$
The ${c}_{k}$ can be determined according to some techniques. The most elementar is
$\frac{{x}^{4} + 1}{{x}^{5} + 6 {x}^{3}} = \frac{1 - 6 {c}_{3} - 6 {c}_{2} x - \left(6 {c}_{1} + {c}_{3}\right) {x}^{2} + \left(i \sqrt{6} {c}_{4} - i \sqrt{6} {c}_{5} - {c}_{2}\right) {x}^{3} + \left(1 - {c}_{1} - {c}_{4} - {c}_{5}\right) {x}^{4}}{{x}^{3} \left(6 + {x}^{2}\right)}$
2) Choosing ${c}_{k}$ such that
${x}^{4} + 1 = \left(1 - 6 {c}_{3} - 6 {c}_{2} x - \left(6 {c}_{1} + {c}_{3}\right) {x}^{2} + \left(i \sqrt{6} {c}_{4} - i \sqrt{6} {c}_{5} - {c}_{2}\right) {x}^{3} + \left(1 - {c}_{1} - {c}_{4} - {c}_{5}\right) {x}^{4}\right) , \forall x \in \mathbb{R}$
3) Solving the conditions
{(1 - 6 c_3=0), (-6 c_2=0), (6 c_1 + c_3=0), (-c_2 + i sqrt[6] c_4 - i sqrt[6] c_5=0), (1 - c_1 - c_4 - c_5=0):}
obtaining
${c}_{1} = - \frac{1}{36} , {c}_{2} = 0 , {c}_{3} = \frac{1}{6} , {c}_{4} = \frac{37}{72} , {c}_{5} = \frac{37}{72}$
(x^4+1)/(x^5+6 x^3)=1/(6 x^3) - 1/(36 x) + 37/(72 ( x-i sqrt[6] )) + 37/( 72 (x+i sqrt[6]))
Note. We can avoid the complex expansion doing
$\frac{c {'}_{3} x + c {'}_{4}}{{x}^{2} + 6}$ instead of ${c}_{4} / \left(x + i \sqrt{6}\right) + {c}_{5} / \left(x - i \sqrt{6}\right)$
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# Linear Functions,Slope and Applications
•A function f is a linear function if it can be written as
f(x) = mx + b
where m and b are constants.
If m = 0, the function is the constant function f(x) = b. If m = 1 and b = 0, the
function is the identity function f(x) = x.
•Vertical and Horizontal Lines:
1. Horizontal lines are given by equations of the type y = b or f(x) = b. Are
horizontal lines functions ? What is the slope of a horizontal line?
2. Vertical lines are given by equations of the type x = a. Are vertical lines
functions? What is the slope of a vertical line?
Linear Functions and Slope: Mathematically we define a line’s steepness or slope,
as the ratio of its vertical change (rise) to the corresponding horizontal change (run).
The slope m of a line containing the points (x1, y1) and (x2, y2) is given by:
Example: Find the slope of the line containing the points (-3, 7) and (5,-1)
•Slope- Intercept Equation : The linear function f given by f(x) = mx + b has a
graph with slope m and y-intercept (0, b). May also be written as y = mx + b.
Example: Write the slope- intercept equation for a line with m = -2 and passing
through (-5, 1).
•The point-slope equation of the line with slope m passing through the point (x1, y1)
is
y - y1 = m(x - x1)
Example: Write the point-slope equation of the line that passes through (-3, 7) and
(-1,-5).
•Vertical lines are parallel. non-vertical lines are parallel if and only if they have the
same slope and different y -intercepts.
Two lines with slopes m1 and m2 are perpendicular if and only if the product of
their slopes is -1.
m1 * m2 = -1
In other words, if m1 is the opposite reciprocal of m2 then the two lines are perpendicular.
Lines are also perpendicular if one is vertical (x = a) and the other horizontal
(y = b).
Example: Write an equation for a line passing through the point (-1, 6) and parallel
to the line Then write a second equation for the line perpendicular to
and passing through (-1, 6).
Try these: pp93-94: 12, 28, 38, 48
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# CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Ex 6(a)
Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Ex 6(a) Textbook Exercise Questions and Answers.
## CHSE Odisha Class 12 Math Solutions Chapter 6 Probability Exercise 6(a)
Question 1.
Two balls are drawn from a bag containing 5 white and 7 black balls. Find the probability of selecting 2 white balls if
Solution:
Two balls are drawn from a bag containing 5 white and 7 black balls.
∴ |S| = 12.
(i) the first ball is replaced before drawing the second.
Solution:
The 1st ball is replaced before the 2nd ball is drawn. We are to select 2 white balls. So in both the draws we will get white balls. Drawing a white ball in 1st draw and in 2nd draw are independent events.
Probability of getting 2 white balls = $$\frac{5}{12}$$ × $$\frac{5}{12}$$ = $$\frac{25}{144}$$
(ii) the first ball is not replaced before drawing the second.
Solution:
Here the 1st ball is not replaced before the 2nd ball is drawn. Since we are to get 2 white balls in each draw, we must get a white ball.
Now probability of getting a white ball in 1st draw = $$\frac{5}{12}$$.
Probability of getting a white ball in 2nd = $$\frac{4}{11}$$.
Since the two draws are independent, we have the probability of getting 2 white balls
= $$\frac{5}{12}$$ × $$\frac{4}{11}$$ = $$\frac{20}{132}$$.
Question 2.
Two cards are drawn from a pack of 52 cards; find the probability that
(i) they are of different suits.
(ii) they are of different denominations.
Solution:
Two cards are drawn from a pack of 52 cards. The cards are drawn one after another. Each suit has 13 cards.
|S| = 52C2
(i) As the two cards are of different suits, their probability
= $$\frac{52}{52}$$ × $$\frac{39}{51}$$
(ii) Each denomination contains 4 cards. As the two cards drawn are of different denominations, their probability
= $$\frac{52}{52}$$ × $$\frac{48}{51}$$
Question 3.
Do both parts of problem 2 if 3 cards drawn at random.
Solution:
(i) 3 cards are drawn one after another. As they are of different suits, we have their probability
= $$\frac{52}{52}$$ × $$\frac{39}{51}$$ × $$\frac{26}{50}$$.
(ii) As the 3 cards are of different denominations, we have their probability
= $$\frac{52}{52}$$ × $$\frac{48}{51}$$ × $$\frac{44}{50}$$.
Question 4.
Do both parts of problem 2 if 4 cards are drawn at random.
Solution:
(i) 4 cards are drawn one after another. As they are of different suits, we have their probability
= $$\frac{52}{52}$$ × $$\frac{39}{51}$$ × $$\frac{26}{50}$$ × $$\frac{13}{49}$$.
(ii) As the cards are of different denominations, we have their probability
= $$\frac{52}{52}$$ × $$\frac{48}{51}$$ × $$\frac{44}{50}$$ × $$\frac{40}{49}$$.
Question 5.
A lot contains 15 items of which 5 are defective. If three items are drawn at random, find the probability that
(i) all three are defective
(ii) none of the three is defective.
Do this problem directly.
Solution:
(i) A lot contains 15 items of which 5 are defective. Three items are drawn at random. As the items are drawn one after another.
Their probability = $$\frac{5}{15}$$ × $$\frac{4}{14}$$ × $$\frac{3}{13}$$
(ii) As none of the 3 items are defective, we have to draw 3 non-defective items one after another.
Their probability = $$\frac{10}{15}$$ × $$\frac{9}{14}$$ × $$\frac{8}{13}$$
Question 6.
A pair of dice is thrown. Find the probability of getting a sum of at least 9 if 5 appears on at least one of the dice.
Solution:
A pair of dice is thrown. Let A be the event of getting at least 9 points and B, the event that 5 appears on at least one of the dice.
∴ B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
A = {(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}
∴ A ∩ B = {(4, 5), (5, 4), (5, 5), (5, 6), (6, 5)}
∴ P (A | B) = $$\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}$$ = $$\frac{\frac{5}{36}}{\frac{31}{36}}$$ = $$\frac{5}{11}$$
Question 7.
A pair of dice is thrown. If the two numbers appearing are different, find the probability that
(i) the sum of points is 8.
(ii) the sum of points exceeds 8.
(iii) 6 appears on one die.
Solution:
A pair of dice is thrown as two numbers are different
We have |S| = 30
(i) Let A be the. event that the sum of points on the dice is 8, where the numbers on the dice are different.
A = {(2, 6), (3, 5), (5, 3), (6, 2)}
P(A) = $$\frac{|\mathrm{A}|}{|\mathrm{S}|}$$ = $$\frac{4}{30}$$
(ii) Let B be the event that sum of the points exceeds 8.
B = {(3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5), (4, 6), (6, 4)}
P(B) = $$\frac{|\mathrm{B}|}{|\mathrm{S}|}$$ = $$\frac{8}{30}$$
(iii) Let C be the event that 6 appears on one die.
C = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
P(C) = $$\frac{|\mathrm{C}|}{|\mathrm{S}|}$$ = $$\frac{10}{30}$$ = $$\frac{1}{3}$$
Question 8.
In a class 30% of the students fail in Mathematics, 20% of the students fail in English and 10% fail in both. A student is selected at random.
Solution:
In a class 30% of the students fail in Mathematics, 20% of the students fail in English and 10% fail in both. Let A be the event that a student fails in Mathematics and B be the events that he fails in English.
P(A) = $$\frac{30}{100}$$, P(B) = $$\frac{20}{100}$$
Where |S| = 100, P (A ∩ B) = $$\frac{10}{100}$$
(i) If he has failed in English, what is the probability that he has failed in Mathematics?
Solution:
If he has failed in English, then the probability that he has failed in Mathematics.
i.e., P$$\left(\frac{A}{B}\right)$$ = $$\frac{P(A \cap B)}{P(B)}$$ = $$\frac{\frac{10}{100}}{\frac{20}{100}}$$ = $$\frac{1}{2}$$
(ii) If he has failed in Mathematics, what is the probability that he has failed in English?
Solution:
If he has failed in Mathematics, then the probability that he has failed in English
i.e., P$$\left(\frac{B}{A}\right)$$ = $$\frac{P(A \cap B)}{P(A)}$$ = $$\frac{\frac{10}{100}}{\frac{30}{100}}$$ = $$\frac{1}{3}$$
(iii) What is the probability that he has failed in both?
Solution:
Probability that he has failed in both
i.e., P (A ∩ B) = $$\frac{10}{100}$$ = $$\frac{1}{10}$$
Question 9.
IfA, B are two events such that
P(A) = 0.3, P(B) = 0.4, P (A ∪ B) = 0.6
Find
(i) P (A | B)
(ii) P (B | A)
(iii) P (A | Bc)
(iv) P (B | Ac)
Solution:
A and B are two set events such that
P(A) = 0.3, P(B) = 0.4, P (A ∪ B) = 0.6
We have
P (A ∪ B) = P(A) + P(B) – P (A ∩ B)
or, 0.6 = 0.3 + 0.4 – P (A ∩ B)
or, P (A ∩ B) = 0.7 – 0.6 = 0.1
Question 10.
If A, B are events such that P(A) = 0.6, P(B) = 0.4 and P (A ∩ B) = 0.2, then find
(i) P (A | B)
(ii) P (B | A)
(iii) P (A | Bc)
(iv) P (B | Ac)
Solution:
A and B are events such that
P(A) = 0.6, P(B) = 0.4, P (A ∩ B) = 0.2
Question 11.
If A and B are independent events, show that
(i) Ac and Bc are independent,
(ii) A and Bc are independent,
(iii) Ac and B are independent.
Solution:
A and B are independent events.
(i) We have P (A ∩ B) = P(A). P(B)
P (A’ ∩ B’) = P (A ∪ B)’ = 1 – P (A ∪ B)
= 1 – [P(A) + P(B) – P (A ∩ B)]
= 1 – P(A) – P(B) + P(A) P(B)
= 1 [1 – P(A)] – P(B) [1 – P(A)]
= [1 – P(A)] [1 – P(B)] = P(A’) P(B’)
∴ A’ and B’ are independent events.
(ii) P (A ∩ Bc) = P (A – B)
= P(A) – P (A ∩ B)
= P(A) – P(A) P(B)
= P(A) [1 – P(B)]
= P(A). P(Bc).
∴ A and Bc are independent events.
(iii) P (Ac ∪ B) = P (B – A)
= P(B) – P (A ∩ B)
= P(B) – P(A) P(B)
= P(B) [1 – P(A)] = P(B) P(Ac)
∴ Ac and B are independent events.
Question 12.
Two different digits are selected at random from the digits 1 through 9.
(i) If the sum is even, what is the probability that 3 is one of the digits selected?
(ii) If the sum is odd, what is the probability that 3 is one of the digits selected?
(iii) If 3 is one of the digits selected, what is the probability that the sum is odd?
(iv) If 3 is one of the digits selected, what is the probability that the sum is even?
Solution:
Two different digits are selected at random from the digits 1 through 9.
(i) Let A be the event that the sum is even and B be the event that 3 is one of the number selected.
We have to find P (B | A).
There are 4 even digits and 5 odd digits.
∴ The sum is even if both the numbers are odd or both are even.
∴ |A| = 4C2 + 5C2 = 6 + 10 = 16
∴ P(A) = $$\frac{16}{{ }^9 \mathrm{C}_2}$$ = $$\frac{16}{36}$$
Also B = {(1, 3), (5, 3), (7, 3), (9, 3), (3, 2), (3, 4), (3, 8), (3, 6)}
∴ A ∩ B = {(1, 3), (5, 3), (7, 3), (9, 3)}
P(B) = $$\frac{8}{36}$$ P (A ∩ B) = $$\frac{4}{36}$$
P($$\frac{B}{A}$$) = $$\frac{P(A \cap B)}{P(A)}$$ = $$\frac{\frac{4}{36}}{\frac{16}{36}}$$ = $$\frac{1}{4}$$
(ii) Let A be the event that the sum is odd. The sum is odd if one of the numbers selected is odd and other is even.
∴ P(A) = $$\frac{5}{9}$$ × $$\frac{4}{8}$$ + $$\frac{4}{9}$$ × $$\frac{5}{8}$$ = $$\frac{20}{36}$$
Let B be the event that one of the numbers selected is 3.
∴ B = {(1, 3), (2, 3), (4, 3), (5, 3), (6, 3), (7, 3), (8, 3), (9, 3)}
∴ A ∩ B = {(2, 3), (4, 3), (6, 3), (8, 3)}
Question 13.
If P(A) = 0.4, P (B | A) = 0.3 and P (Bc | Ac) = 0.2. find
(i) P (A | B)
(ii) P (B | Ac)
(iii) P(B)
(iv) P(Ac)
(v) P (A ∪ B)
Solution:
P(A) = 0.4, P (B | A) = 0.3 and P (Bc | Ac) = 0.2.
(iii) P(B) = 0.6
= $$\frac{6}{10}$$
= $$\frac{3}{5}$$
(iv) P(Ac) = 1 – P(A)
= 1 – 0.4 = 0.6
= $$\frac{6}{10}$$ = $$\frac{3}{5}$$
(v) P (A ∪ B) = P(A) + P(B) – P (A ∩ B)
= 0.4 + 0.6 – 0.12
= 1.0 – 0.12 = 0.88
Question 14.
If P(A) = 0.6, P (B | A) = 0.5, find P (A ∪ B) if A, B are independent.
Solution:
P(A) = 0.6, P (B | A) = 0.5
We have P (B | A) = $$\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}$$ = 0.5
or, P (B ∩ A) = 0.5 × P(A)
= 0.5 × 0.6 = 0.3
As A and B are independent events, we have
P (B ∩ A) = P(B) P(A) = 0.3
or, P(B) = $$\frac{0.3}{\mathrm{P}(\mathrm{A})}$$
= $$\frac{0.3}{0.6}$$
= $$\frac{1}{2}$$ = 0.5
P (A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.6 + 0.5 – 0.3 = 0.8
Question 15.
Two cards are drawn in succession from a deck of 52 cards. What is the probability that both cards are of denomination greater than 2 and less than 9?
Solution:
Two cards are drawn in succession from a deck of 52 cards.
There are 6 denominations which are greater than 2 and less than 9. So there are 24 cards whose denominations are greater than 2 and less than 9.
∴ Their probability = $$\frac{24}{52}$$ × $$\frac{23}{51}$$.
Question 16.
From a bag containing 5 black and 7 white balls, 3 balls are drawn in succession. Find the probability that
(i) all three are of the same colour.
(ii) each colour is represented.
Solution:
From a bag containing 5 black and 7 white balls, 3 balls are drawn in succession.
(i) The 3 balls drawn are of same colour.
∴ Probability of drawing 3 balls of black colour
= $$\frac{5}{12}$$ × $$\frac{4}{11}$$ × $$\frac{3}{10}$$ = $$\frac{1}{22}$$
Probability of drawing 3 white balls
= $$\frac{7}{12}$$ × $$\frac{6}{11}$$ × $$\frac{5}{10}$$ = $$\frac{7}{44}$$
∴ Probability of drawing 3 balls of same colour
= $$\frac{5}{12}$$ × $$\frac{4}{11}$$ × $$\frac{3}{10}$$ + $$\frac{7}{12}$$ × $$\frac{6}{11}$$ × $$\frac{5}{10}$$ = $$\frac{9}{44}$$
(ii) Balls of both colour will be drawn. If B represents black ball and W represents the white ball.
∴ The possible draws are WWB, WBW, BWW.
Question 17.
A die is rolled until a 6 is obtained. What is the probability that
(i) you end up in the second roll
(ii) you end up in the third roll.
Solution:
A die is rolled until a 6 is obtained
(i) We are to end up in the 2nd roll i.e., we get 6 in the 2nd roll. Let A be the event of getting a 6 in one roll of a die.
∴ P(A) = $$\frac{1}{6}$$ ⇒ P(A’) = 1 – $$\frac{1}{6}$$ = $$\frac{5}{6}$$
∴ Probability of getting a 6 in the 2nd roll
= $$\frac{5}{6}$$ × $$\frac{1}{6}$$ = $$\frac{5}{36}$$
(ii) Probability of getting a 6 in the 3rd roll
= $$\frac{5}{6}$$ × $$\frac{5}{6}$$ × $$\frac{1}{6}$$ = $$\frac{25}{216}$$
Question 18.
A person takes 3 tests in succession. The probability of his (her) passing the first test is 0.8. The probability of passing each successive test is 0.8 or 0.5 according as he passes or fails the preceding test. Find the probability of his (her) passing at least 2 tests.
Solution:
A person takes 3 tests in succession. The probability of his passing the 1st test is 0.8. The probability of passing each successive test is 0.8 or 0.5 according as he passes or fails the preceding test.
Let S denotes the success (passing) in a test and F denotes the failure in a test.
∴ P(S) = 0.8
∴ P(F) = 1 – P(S) = 1 – 0.8 = 0.2
We have the following mutually exclusive cases:
Event Probability S S S 0.8 × 0.8 × 0.8 = 0.512 S S F 0.8 × 0.8 × 0.2 = 0.128 S F S 0.8 × 0.2 × 0.5 = 0.080 F S S 0.2 × 0.5 x 0.8 = 0.080
∴ Probability of atleast 2 successes
= 0.512 + 0.128 + 0.080 + 0.080
= 0.8 = $$\frac{4}{5}$$
Question 19.
A person takes 4 tests in succession. The probability of his passing the first test is p, that of his passing each succeeding test is p or y depending on his passing or failing the preceding test. Find the probability of his passing
(i) at least three test
(ii) just three tests.
Solution:
A person takes 4 tests in succession. The probability of his passing the 1st test is P, that of his passing each succeeding test is P or P/2 depending bn his passing or failing the preceding test. Let S and F denotes the success and failure in the test.
∴ P(S) = P, P(F) = 1 – P
We have the following mutually exclusive tests:
Question 20.
Given that all three faces are different in a throw of three dice, find the probability that
(i) at least one is a six
(ii) the sum is 9.
Solution:
Three dice are thrown once showing different faces in a throw.
|S| = 63 = 216
Let A be the event that atleast one is a six.
Let B be the event that all three faces are different.
|B| = 663
(i) Now Ac is the event that there is no six. Ac ∩ B is the event that all 3 faces are different and 6 does not occur.
|Ac ∩ B| = 5C3
P (Ac | B) = $$\frac{P\left(A^C \cap B\right)}{P(B)}$$
= $$\frac{{ }^5 \mathrm{C}_3 / 216}{{ }^6 \mathrm{C}_3 / 216}$$ = $$\frac{1}{2}$$
(ii) Let A be the event that the sum is 9.
A ∩ B = {(1,3, 5), (1,5, 3), (3, 5,1), (3, 1, 5), (5, 1, 3), (5, 3, 1), (1, 2, 6), (1, 6, 2), (2, 1, 6), (2, 6, 1), (6, 2, 1), (6, 1, 2), (2, 3, 4), (2, 4, 3), (3, 2, 4), (2, 3, 4), (3, 4, 2), (4, 3, 2)}
|A ∩ B| = 18
P (A | B) = $$\frac{P(A \cap B)}{P(B)}$$
= $$\frac{18 / 216}{20 / 216}$$ = $$\frac{9}{10}$$
Question 21.
From the set of all families having three children, a family is picked at random.
(i) If the eldest child happens to be a girl, find the probability that she has two brothers.
(ii) If one child of the family is a son, find the probability that he has two sisters.
Solution:
A family is picked up at random from a set of families having 3 children.
(i) The eldest child happens to be a girl. We have to find the probability that she has two brothers. Let G denotes a girl and B denotes a boy.
∴ P(B) = $$\frac{1}{2}$$, P(G) = $$\frac{1}{2}$$
P(BB | G) = P(B) × P(G) = $$\frac{1}{2}$$ × $$\frac{1}{2}$$ = $$\frac{1}{4}$$
(ii) The one child of the family is a son. We have to find the probability that he has two sisters. We have the following mutually exclusive events:
BGG, GBG, GGB.
∴ The required probability
= P(B) × P(G) × P(G) + P(G) × P(B) + P(G) + P(G) × P(G) × P(B)
= $$\frac{1}{2}$$ × $$\frac{1}{2}$$ × $$\frac{1}{2}$$ + $$\frac{1}{2}$$ × $$\frac{1}{2}$$ × $$\frac{1}{2}$$ + $$\frac{1}{2}$$ × $$\frac{1}{2}$$ × $$\frac{1}{2}$$ = $$\frac{3}{8}$$
Question 22.
Three persons hit a target with probability $$\frac{1}{2}$$, $$\frac{1}{3}$$ and $$\frac{1}{4}$$ respectively. If each one shoots at the target once,
(i) find the probability that exactly one of them hits the target
(ii) if only one of them hits the target what is the probability that it was the first person?
Solution:
Three persons hit a target with probability
$$\frac{1}{2}$$, $$\frac{1}{3}$$ and $$\frac{1}{4}$$
Let A, B, C be the events that the 1st person, 2nd person, 3rd person hit the target respectively.
(i) As the events are independent, the probability that exactly one of them hit the target
= P(AB’C’) + P(A’BC’) + P(A’B’C)
= P(A) P(B’) P(C’) + P(A’) P(B) P(C’) + P(A’) P(B’) P(C)
(ii) Let E1 be the event that exactly one person hits the target.
∴ P(E1) = $$\frac{11}{24}$$
Let E2 be the event that 1st person hits the target
∴ P(E2) = P(A) = $$\frac{1}{2}$$
∴ E1 ∩ E2 = AB’C’
⇒ P(E1 ∩ E2)
= P(A) × P(B’) P(C’) = $$\frac{6}{24}$$
∴ P(E2 | E1) = $$\frac{6 / 24}{11 / 24}$$ = $$\frac{6}{11}$$
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How do you solve and write the following in interval notation: -2 ≤ x + 4 OR -1 + 3x > -8?
May 26, 2018
$\left[- 6 , \setminus \infty\right)$
Explanation:
You can rewrite $- 2 \setminus \le x + 4$ as
$- 2 - 4 \setminus \le x + 4 - 4 \setminus \implies - 6 \setminus \le x \setminus \implies x \setminus \ge - 6$
Similarly, rewrite $- 1 + 3 x \succ 8$ as
$- 1 + 3 x + 1 \succ 8 + 1 \setminus \implies 3 x \succ 7 \setminus \implies x > - \frac{7}{3}$
Since $- 6 < - \frac{7}{3}$, any number which is greater than $- \frac{7}{3}$ will automatically be greater than $- 6$. And since the OR condition requires at least one of the conditions to be true, it is sufficient to ask $x \setminus \ge - 6$
To write it in interval notation, we must find the boundaries of the desired region. We want $x$ to be greater than $- 6$, but we have no upper bound, which means it's $\setminus \infty$.
So, the interval is $\left[- 6 , \setminus \infty\right)$
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# Multi-Step Inequalities
## Solve inequalities with fractions and distribution
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Progress
Practice Multi-Step Inequalities
Progress
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Algebraic Solutions to One Variable Inequalities
The Morgan Silver Dollar is a very valuable American dollar minted between 1978 and 1921. When placed on a scale with twenty 1-gram masses, the scale tips toward the Morgan Dollar. Draw a picture to represent this scenario and then write the inequality.
### Guidance
Linear equations are of the form \begin{align*}ax + b = 0\end{align*}, where \begin{align*}a \ne 0\end{align*}. With linear equations there is always an equals sign. Linear inequalities are mathematical statements relating expressions by using one or more inequality symbols \begin{align*} <, >, \le\end{align*}, or \begin{align*}\ge\end{align*}. In other words, the left side no longer equals the right side, it is less than, greater than, less than or equal to, or greater than or equal to.
Recall that one method for solving an equation is to use the balance method. To solve for \begin{align*}a + 2 = 5\end{align*}, you would draw the following balance:
If you were to use the balance method to solve the linear inequality version, it would look more like this: \begin{align*}a + 2 >5\end{align*}
Think about it this way. It is like having someone heavier on the \begin{align*}(a + 2)\end{align*} side of the balance and someone light on the (5) side of the balance. The \begin{align*}(a + 2)\end{align*} person has a weight greater than (>) the (5) person and therefore the balance moves down to the ground.
The rules for solving inequalities are basically the same as you used for solving linear equations. If you have parentheses, remove these by using the distributive property. Then you must isolate the variable by moving constants to one side and variables to the other side of the inequality sign. You also have to remember that whatever you do to one side of the inequality, you must do to the other. The same was true when you were working with linear equations. One additional rule is to reverse the sign of the inequality if you are multiplying or dividing both sides by a negative number.
It is important for you to remember what the symbols mean. Always remember that the mouth of the sign opens toward the larger number. So \begin{align*}8 >5\end{align*}, the mouth of the > sign opens toward the 8 so 8 is larger than 5. You know that’s true. \begin{align*}6b-5 < 300\end{align*}, the mouth opens toward the 300, so 300 is larger than \begin{align*}6b - 5\end{align*}.
#### Example A
Solve: \begin{align*}15 < 4 + 3x\end{align*}
Solution: Remember that whatever you do to one side of the inequality sign, you do to the other.
Do a quick check to see if this is true. \begin{align*}\frac{11}{3}\end{align*} is approximately 3.67. Try substituting 0, 3, and 4 into the equation.
The only value out of 0, 3, and 4 where \begin{align*}x > \frac{11}{3}\end{align*} is 4. If you look at the statements above, it was the only inequality that gave a true statement.
#### Example B
Solve: \begin{align*}2y + 3 > 7\end{align*}
Solution: Use the same rules as if you were solving any algebraic expression. Remember that whatever you do to one side of the inequality sign, you do to the other.
Do a quick check to see if this is true. Try substituting 0, 4, and 8 into the equation.
The values out of 0, 4, and 8 where \begin{align*}y>2\end{align*} are 4 and 8. If you look at the statements above, these values when substituted into the inequality gave true statements.
#### Example C
Solve: \begin{align*}-2c-5 < 8\end{align*}
Solution: Again, to solve this inequality, use the same rules as if you were solving any algebraic expression. Remember that whatever you do to one side of the inequality sign, you do to the other.
Note: When you divide by a negative number, the inequality sign reverses.
Do a quick check to see if this is true. \begin{align*}\frac{-13}{2}\end{align*} is equal to –6.5. Try substituting –8, 0, and 2 into the equation.
The values out of –8, 0, and 2 where \begin{align*}c>\frac{-13}{2}\end{align*} are 0 and 2. If you look at the statements above, these values when substituted into the inequality gave true statements.
#### Concept Problem Revisited
The Morgan Silver Dollar is a very valuable American dollar minted between 1978 and 1921. When placed on a scale with twenty 1-gram masses, the scale tips toward the Morgan Dollar. Draw a picture to represent this scenario and then write the inequality.
Let \begin{align*}m =\end{align*} Morgan dollar
Since the weight of the Morgan dollar is greater than 20 g, the mouth of the inequality sign would open towards the variable, \begin{align*}m\end{align*}. Therefore the inequality equation would be:
\begin{align*}m>20\end{align*}
### Vocabulary
Linear Inequality
Linear inequalities are mathematical statements relating expressions by using one or more inequality symbols \begin{align*} <, >, \le\end{align*}, or \begin{align*}\ge\end{align*}.
### Guided Practice
Solve each inequality.
1. \begin{align*}4t + 3 > 11\end{align*}
2. \begin{align*}2z + 7 \le 5z + 28\end{align*}
3. \begin{align*}9(j - 2) \ge 6(j + 3) - 9\end{align*}
1. \begin{align*}t>2\end{align*}. Here are the steps:
2. \begin{align*}z\ge -7\end{align*}. Here are the steps:
3. \begin{align*}j \ge 9\end{align*}. Here are the steps:
### Practice
Solve for the variable in the following inequalities.
1. \begin{align*}a+8>4\end{align*}
2. \begin{align*}4c-1>7\end{align*}
3. \begin{align*}5-3k<6\end{align*}
4. \begin{align*}3-4t \le -11\end{align*}
5. \begin{align*}6 \ge 11-2b\end{align*}
6. \begin{align*}\frac{e}{5}-3 >-1\end{align*}
7. \begin{align*}\frac{1}{5}(r-3) <-1\end{align*}
8. \begin{align*}\frac{1}{3}(f+2) <4\end{align*}
9. \begin{align*}\frac{p+3}{4} \ge -2\end{align*}
10. \begin{align*}\frac{1}{2}(5-w) \le -3\end{align*}
11. \begin{align*}3(2x-5)<2(x-1)+3\end{align*}
12. \begin{align*}2(y+8)+5(y-1)>6\end{align*}
13. \begin{align*}2(d-3)<-3(d+3)\end{align*}
14. \begin{align*}3(g+3) \ge 2(g+1)-2\end{align*}
15. \begin{align*}2(3s-4)+1 \le 3(4s+1)\end{align*}
### Vocabulary Language: English
linear equation
linear equation
A linear equation is an equation between two variables that produces a straight line when graphed.
Linear Inequality
Linear Inequality
Linear inequalities are inequalities that can be written in one of the following four forms: $ax + b > c, ax + b < c, ax + b \ge c$, or $ax + b \le c$.
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Square Root
What Is the Square Root of 3? How to Find the Square Root of 3?
Written by Prerit Jain
Updated on: 05 Aug 2024
What Is the Square Root of 3? How to Find the Square Root of 3?
The square root of 3 is 1.732. The square root of a value is given when a particular number is multiplied by itself and gives the number or a number when they have squared equals to a specified number. The number 3 is a positive real number that when multiplied by itself produces the number 3. The square root of 3 is mentioned as √3.
To find the square root of any number without using a calculator we have to check whether the number is first a perfect square or not. For a number that is perfectly square, we can find the square root using the prime factorization method and for those that end with a first odd number, the square root can be found using repeated subtraction. Since 3 is not a perfect square number we follow long division to find the square root of 3.
Steps to find the square root of 3
• Step 1: Find the smallest integer that can divide the number. The smallest integer is closest and can be divided using the number 1.
• Step 2: Keep following the long division using divisor and dividend.
• Step 3: When the particular number of satisfaction is reached the quotient is the square root of the number. Hence, the square root of 3 is 1.732
Solved examples
Q 1: Find the square root of 3.
A 1: Hence, the √3 = 1.732
Q 2: What is the square root of 5?
A 2: √5 = 2.23
Q 3: Find the square root of 2.
A 3: √2 = 1.4142
Q 4: Find the square root of 36.
A4: The prime factors of 36 = 2² × 3²
√36 = 2× 3
Hence, √36 = 6
Q 5: What is the square root of 64?
The prime factors of 64 = 2×2×2×2×2×2
The common square factors = 2×2×2 = 8.
Hence, 64 = 8.
What is the square root of 3?
√3 = 1.732
Which method is used to find the square root of 3?
The common method that is used to find the square root of any number is the long division method.
How to write the square root of 3 in exponential form?
312 or 30.5 is the exponential way to write √3.
How to write the square root of 3 in radical form?
The radical form of writing square root is √3.
Can I find the square root of 3 using other methods?
Finding the square root for 3 is not possible using the prime factorization method or repeated subtraction because √3 is not a perfect square number.
Is 3 an irrational number?
Yes, 3 is an irrational number since the number is not equal to zero.
What are the methods to find the square root of a number?
There are three methods to find the square root of a number
Prime factorization method
Long division method
Repeated subtraction
Written by by
Prerit Jain
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Triangle Degrees Lesson
Objective
Learn about 3 different types of triangles
Primary Market
Education, Primary Ed
In this triangles lesson, we’ll discuss the three different types of triangles. There are equilateral triangles, isosceles triangles, and right triangles.
Do you know what it is to be an isosceles, right, and equilateral triangle?
Your challenge is to build one isosceles triangle, one right triangle, and one equilateral triangle.
Your child should find that it is not possible to build an equilateral triangle. At this point is a good time to figure out why. See if they can figure out how many degrees should be in each equilateral triangle. Ask them to find two different ways to prove that it is impossible to build equilateral triangle other than just having attempted to do so.
Proof one: there are 180° in a triangle. 180° divided by three angles of a triangle equals 60° per angle. There are 360° in a circle. The circular ball has eight different holes. 360÷8 = 45. Because the angles and available are 45 and greater than or equal to 90 it is impossible to build a 60° angle and therefore you cannot build an equilateral triangle.
Proof two: we know that two of the points on the ball are 90° apart 90÷2 = 45. And equilateral triangle is made up of 60° in each angle totaling 180 degrees. 60° is not equal to 0°, 45°, or 90° and therefore you cannot build an equilateral triangle.
– Brandi
Joe Donahue is the manager of day-to-day operations and the person who originally conceived of the TOOBEEZ product.
Joe had a vision to create a life-sized Tinker Toy(TM) for families to play with to help reconnect with each other. It took four years of business planning, market research, manufacturing trips, endless conference calls with engineers, designers and patent attorneys but finally on September 21, 2003, the Toobeez product was born!
Often referred to as Giant Tinkertoys(TM), Toobeez is a life-sized construction building system comprised of interlocking tubes and spheres that can be linked together to create anything imaginable, from a simple cube shape to a complex structure like a lemonade stand, house, airplane or submarine. Just connect, twist and create!
Initially, Toobeez was sold exclusively to the specialty toy market, but due to its open-ended nature, it didn’t take long to realize the versatility of the product and the many different ways it could be used.
Today, the Toobeez system is currently being used by educators, team building/training professionals, occupational and recreational therapists, camp counselors, Boy and Girl Scout troops, church youth groups and the list grows every day. There is even a group of forward thinking dog trainers who use Toobeez to create challenging courses for agility training (we're not making this up)!
Armed with the knowledge that there was more to Toobeez than just a toy, Joe reached out to industry experts and educators with the goal of developing a collection of materials that provide activities, educational tools and lesson plans for use with Toobeez. Among the topics covered: team building, math, language arts, occupational therapy and special needs and senior needs. Currently under development is a corporate team building workbook and a physical education activity book.
Whether in a playroom, classroom, corporation, camp or healthcare setting, Toobeez aims to provide every participant with a positive, healthy and fun learning experience!
Since the inception of Toobeez in 2001, it has been the company’s mission to connect people, inspire creativity and foster imagination. As a testament to their roots, Joe and the entire Toobeez team continuously strive to deliver a quality product that promotes connection and teamwork to ultimately bring people together.
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# SAT II Math I : Surface Area
## Example Questions
← Previous 1 3
### Example Question #1 : Surface Area
A circular swimming pool has diameter 32 meters and depth meters throughout. Which of the following expressions gives the total area of the inside of the pool, in square meters?
None of the other responses is correct.
Explanation:
The bottom of the pool is a circle with diameter 32, and, subsequently, radius half this, or 16; its area is
The side of the pool is the lateral surface of a cylinder with radius 16 and height ; the area of this is
The area of the inside of the pool is the sum of these two, or
### Example Question #2 : Surface Area
A circular swimming pool at an apartment complex has diameter 50 feet and depth six feet throughout.
The apartment manager needs to get the interior of the swimming pool painted. The paint she wants to use covers 350 square feet per gallon. How many one-gallon cans of paint will she need to purchase?
The correct answer is not given among the other responses.
Explanation:
The pool can be seen as a cylinder with depth (or height) six feet and a base with diameter 50 feet - and radius half this, or 25 feet.
The bottom of the pool - the base of the cylinder - is a circle with radius 25 feet, so its area is
square feet.
Its side - the lateral face of the cylinder - has area
square feet.
Their sum - the total area to be painted - is square feet. Since one gallon of paint covers 350 square feet, divide:
Eight cans of paint and part of a ninth will be required, so the correct response is nine.
### Example Question #1 : Surface Area
Figure not drawn to scale.
Find the surface area of the cylinder above.
94.25 in2
73.44 in2
87.25 in2
122.14 in2
56.55 in2
94.25 in2
Explanation:
In order to find the surface area of a cylinder, you need to find the surface areas of the circles that are the top and bottom of the cylinder (2 x pi x radius2) and add it to the surface are of the rectangle that is the side of the cylinder (diameter x height).
### Example Question #4 : Surface Area
Figure not drawn to scale
# What is the surface area of the sphere above?
615.75 yd2
712.12 yd2
512.63 yd2
615.75 yd3
815.44 yd2
615.75 yd2
Explanation:
In order to find the surface area of a sphere, you must use the equation below:
### Example Question #101 : Geometry
What is the surface area of the following cone?
Explanation:
The formula for the surface area of a cone is:
,
where represents the radius of the cone base and represents the slant height of the cone.
Plugging in our values, we get:
### Example Question #1 : How To Find The Surface Area Of A Cone
The surface area of cone is . If the radius of the base of the cone is , what is the height of the cone?
Explanation:
To figure out , we must use the equation for the surface area of a cone, , where is the radius of the base of the cone and is the length of the diagonal from the tip of the cone to any point on the base's circumference. We therefore first need to solve for by plugging what we know into the equation:
This equation can be reduced to:
For a normal right angle cone, represents the line from the tip of the cone running along the outside of the cone to a point on the base's circumference. This line represents the hypotenuse of the right triangle formed by the radius and height of the cone. We can therefore solve for using the Pythagorean theorem:
so
Our is therefore:
The height of cone is therefore
### Example Question #1 : Surface Area
A circle of radius five is cut into two pieces, and . The larger section is thrown away. The smaller section is curled until the two straight edges meet, and a bottom is made for the cone.
What is the area of the bottom?
Explanation:
When the smaller portion of the circle is curled in, it will make the top of a cone. The circumfrence of the circle on the bottom is (where r is the radius of the circle on the bottom). The circumference of the bottom is also of the circumfrence of the original larger circle, which is (where R is the radius of the original, larger circle)
Therefore we use the circumference formula to solve for our new r:
Substituting this value into the area formula, the area of the small circle becomes:
### Example Question #1 : Solid Geometry
A cone has a bottom area of and a height of , what is the surface area of the cone?
Explanation:
The area of the bottom of the cone yields the radius,
The height of the cone is , so the Pythagorean Theorem will give the slant height,
The area of the side of the cone is and adding that to the given as the area of the circle, the surface area comes to
### Example Question #1 : Advanced Geometry
If the surface area of a right angle cone is , and the distance from the tip of the cone to a point on the edge of the cone's base is , what is the cone's radius?
Explanation:
Solving this problem is going to take knowledge of Algebra, Geometry, and the equation for the surface area of a cone: , where is the radius of the cone's base and is the distance from the tip of the cone to a point along the edge of the cone's base. First, let's substitute what we know in this equation:
We can divide out from every term in the equation to obtain:
We see this equation has taken the form of a quadratic expression, so to solve for we need to find the zeroes by factoring. We therefore need to find factors of that when added equal . In this case, and :
This gives us solutions of and . Since represents the radius of the cone and the radius must be positive, we know that is our only possible answer, and therefore the radius of the cone is .
### Example Question #1 : Surface Area
For a right circular cone , the radius is and the height of the cone is . What is the surface area of the cone in terms of ?
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# Using Mental Math to Solve One-Step Problems: Lesson for Kids
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
After reading this lesson, solving problems in your head will be easier, and you'll be able to answer your teacher or friends quickly and easily. Learn the technique that will help you quickly solve problems in your head.
## One-Step Problems
As you start getting better at your math and you learn more and more math skills, you'll start doing more and more one-step problems. One-step problems are math problems that use only one operation to solve them. On paper, one-step problems will look something like this:
Sometimes, your teacher won't give you a piece of paper with problems on it. Instead, your teacher may simply use words to ask you the same problems. She might say something like this:
'I am thinking of a number that when you subtract 8 from it, you'll get 15. What is this number?'
## Using Mental Math
If your teacher happens to ask you the problem verbally (with words), then you'll need to solve the problem using mental math. You perform mental math when you use your brain to solve problems. This means you use your brain to do all the work instead of writing your work down. The only thing that you'll write down, if anything, is the answer.
When using mental math to solve your one-step problems, you can use a technique that involves using the opposite operation. If you look at the one-step example problems above, you'll see that you have the number you are looking for written down as a letter, and then you see a number being added to or subtracted from this number. To find your answer using mental math and this technique, you'll use the opposite operation to combine the two numbers that you see. Addition is the opposite operation to subtraction. If you see subtraction, you'll use addition to solve the problem. If you see addition, then you'll use subtraction to solve the problem.
For example, you can solve the problem that your teacher just gave ('I am thinking of a number that when you subtract 8 from it, you'll get 15. What is this number?') by adding the 8 to the 15. Because 8 + 15 is equal to 23, your answer is 23.
Whenever you do mental math, though, you'll want to make sure that you always check your work. You can do this in your head, too. So, for the problem we were working on, you can check by subtracting the 8 from the 23 and seeing if it actually equals 15. Let's see: 23 - 8 = 15. Yes, that's right. So, 23 is a correct answer.
## Examples
Let's try a couple more.
'I am thinking of a number that equals 7 when you add 3 to it. What is this number?'
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# 2.1: Sampling Distribution of the Sample Mean
Inferential testing uses the sample mean ($$\bar{x}$$) to estimate the population mean ($$μ$$). Typically, we use the data from a single sample, but there are many possible samples of the same size that could be drawn from that population. As we saw in the previous chapter, the sample mean ($$\bar{x}$$) is a random variable with its own distribution.
• The distribution of the sample mean will have a mean equal to µ.
• It will have a standard deviation (standard error) equal to $$\frac{\sigma}{\sqrt {n}}$$
Because our inferences about the population mean rely on the sample mean, we focus on the distribution of the sample mean. Is it normal? What if our population is not normally distributed or we don’t know anything about the distribution of our population?
The Central Limit Theorem (CLT)
The Central Limit Theorem states that the sampling distribution of the sample means will approach a normal distribution as the sample size increases.
So if we do not have a normal distribution, or know nothing about our distribution, the CLT tells us that the distribution of the sample means () will become normal distributed as n (sample size) increases. How large does n have to be? A general rule of thumb tells us that n ≥ 30.
The Central Limit Theorem tells us that regardless of the shape of our population, the sampling distribution of the sample mean will be normal as the sample size increases.
## Sampling Distribution of the Sample Proportion
The population proportion ($$p$$) is a parameter that is as commonly estimated as the mean. It is just as important to understand the distribution of the sample proportion, as the mean. With proportions, the element either has the characteristic you are interested in or the element does not have the characteristic. The sample proportion ($$\hat {p}$$) is calculated by
$$\hat {p} = \frac{x}{n} \label{sampleproption}$$
where $$x$$ is the number of elements in your population with the characteristic and n is the sample size.
Example $$\PageIndex{1}$$: sample proportion
You are studying the number of cavity trees in the Monongahela National Forest for wildlife habitat. You have a sample size of n = 950 trees and, of those trees, x = 238 trees with cavities. Calculate the sample proportion.
A naturally formed tree hollow at the base of the tree. Image used with permission (CC BY 2.0; Lauren "Lolly" Weinhold).
Solution
This is a simple application of Equation \ref{sampleproption}:
$$\hat {p} = \frac {238}{950} =0.25 \nonumber$$
The distribution of the sample proportion has a mean of $$\mu_{\hat{p}} = p$$
and has a standard deviation of $$\sigma_{\hat {p}} = \sqrt {\frac {p(1-p)}{n}}.$$
The sample proportion is normally distributed if $$n$$ is very large and $$\hat{p}$$ is not close to 0 or 1. We can also use the following relationship to assess normality when the parameter being estimated is p, the population proportion:
$$n\hat {p} (1- \hat {p}) \ge 10$$
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Math Worksheets Ordering Rational Numbers
A Reasonable Figures Worksheet can help your son or daughter be more familiar with the concepts behind this ratio of integers. With this worksheet, pupils are able to remedy 12 diverse problems associated with rational expressions. They may figure out how to grow 2 or more phone numbers, class them in sets, and find out their items. They are going to also practice simplifying realistic expressions. As soon as they have perfected these ideas, this worksheet will be a important instrument for furthering their research. Math Worksheets Ordering Rational Numbers.
Rational Numbers can be a proportion of integers
There are 2 varieties of numbers: irrational and rational. Realistic phone numbers are considered entire amounts, in contrast to irrational numbers will not recurring, and possess an limitless number of numbers. Irrational phone numbers are non-absolutely no, non-terminating decimals, and sq roots which are not excellent squares. These types of numbers are not used often in everyday life, but they are often used in math applications.
To determine a logical number, you must know such a realistic variety is. An integer is actually a whole quantity, along with a rational variety is actually a proportion of two integers. The percentage of two integers is definitely the amount on top split with the amount at the base. If two integers are two and five, this would be an integer, for example. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
They are often manufactured in to a small percentage
A rational quantity has a denominator and numerator that are not no. Which means that they may be conveyed like a small percentage. In addition to their integer numerators and denominators, realistic phone numbers can in addition have a negative benefit. The adverse importance ought to be positioned on the left of as well as its complete worth is its range from zero. To make simpler this case in point, we will claim that .0333333 can be a small fraction which can be composed being a 1/3.
As well as negative integers, a logical number can be made in to a small percentage. For instance, /18,572 can be a reasonable quantity, when -1/ is just not. Any small fraction consisting of integers is realistic, so long as the denominator does not have a and might be published as an integer. Also, a decimal that ends in a position is yet another realistic number.
They make sensation
Regardless of their title, reasonable amounts don’t make much sensation. In mathematics, they are solitary organizations with a special span in the amount range. Consequently if we matter one thing, we are able to buy the size by its percentage to its initial amount. This contains correct even if you will find endless reasonable phone numbers among two particular phone numbers. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
If we want to know the length of a string of pearls, we can use a rational number, in real life. To find the time period of a pearl, by way of example, we could add up its size. One particular pearl weighs about 15 kilograms, which is a reasonable variety. In addition, a pound’s weight equals twenty kilos. Therefore, we should be able to split a lb by twenty, without be worried about the duration of just one pearl.
They are often indicated as a decimal
If you’ve ever tried to convert a number to its decimal form, you’ve most likely seen a problem that involves a repeated fraction. A decimal quantity might be composed being a numerous of two integers, so 4x 5 is equivalent to 8-10. The same problem necessitates the recurring small percentage 2/1, and both sides needs to be divided up by 99 to obtain the proper answer. But how do you have the conversion? Here are some illustrations.
A logical amount may also be developed in great shape, such as fractions plus a decimal. One method to represent a logical amount inside a decimal is always to divide it into its fractional counterpart. You can find three ways to separate a logical amount, and all these ways results in its decimal equal. One of those ways is always to split it into its fractional equal, and that’s what’s known as the terminating decimal.
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Factoring Trinomials
🏆Practice factoring trinomials
I present to you the following trinomial
$ax^2+bx+c$
Regardless of whether the coefficients of the terms are positive or negative, as long as they appear in the style of a trinomial, the exercise will be called "trinomial".
The factorization will look like this:
$(x+solution \space one)(x+solution\space two)$
or with subtractions, depending on the solutions.
Test yourself on factoring trinomials!
$$x^2+6x+9=0$$
What is the value of X?
The first way to factor a trinomial
We will look for two numbers whose product is $a\times c$ and whose sum is $b$
We will ask ourselves: which number multiplied by which other will give us $a\times c$ or $c$ (if $a$ equals $1$).
and what plus what would add up to $b$.
In fact, we need to find a pair of numbers that meet these two conditions at the same time.
We can plot it as follows:
The second way to factor a trinomial - quadratic formula
$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$
$a$ The coefficient of the first term
$b$ The coefficient of the second term
$c$ The constant term
In the first step, we will use only addition to find the first solution, and then, we will use only subtraction to find the second.
Again, the factorization will look as follows:
$(x+solution \space one)(x+solution\space two)$
or with subtractions, depending on the solutions.
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What is a trinomial?
$ax^2+bx+c$
The trinomial represents an expression in which $x$ is squared, preceded by a coefficient (which can be positive or negative), but it must not be $0$ (sometimes the coefficient is equal to $1$ and therefore we will not see the $a$), to this term may be added or subtracted some other $bx$ when $b$ represents the coefficient (under the same conditions as $a$) and the independent variable (number $c$) is added or subtracted.
Regardless of whether the coefficients of the terms are positive or negative, as long as they appear in the form of a trinomial, the exercise will be called "trinomial".
The first way to factor a trinomial
We will look for two numbers whose product is $a*c$ and whose sum is $b$
We will ask ourselves: which number multiplied by which other number will give us $a\times c$ or $c$ (if $a$ equals $1$).
and what plus what would add up to $b$.
In fact, we have to find a pair of numbers that meet these two conditions at the same time.
We can plot it as follows:
AC Method:
We will find all the numbers whose products are $a\times c$ and write them down.
Then, we will see which pair of numbers among those we found will result in $B$.
The two numbers that meet both conditions are the solutions to the trinomial.
Important
• If A were different from $1$, it would appear before the parentheses and then there would be a multiplication.
• If any of the solutions or both were negative, we would not add them to the $X$ but subtract them instead.
Do you know what the answer is?
Let's look at an example of the use of factoring trinomials in the first way.
$x^2+8x+12$
Let's find all the numbers whose products are $12$ (and remember them in negative as well)
we will obtain:
$12,1$
$2,6$
$3,4$
Now let's see which pair of numbers among those we already found will give us a total of $8$
The pair that meets both conditions is $2,6$.
Let's write the factorization:
$(x+2)(x+6)$
The second way to factor a trinomial
Let's look at an example of the use of factoring trinomials in the second way:
$x^2+4x+4=$
Let's find our parameters:
$a$ The coefficient of the first term $1$
$b$ The coefficient of the second term $4$
$c$ The constant term $4$
First, we will place them in the formula with the plus sign and it will give us:
$\frac{-4+\sqrt{4^2-4\times 1\times 4}}{2\times 1}=$
$\frac{-4+\sqrt{16-16}}{2}=$
$\frac{-4+\sqrt{0}}{2}=$
$\frac{-4}{2}=-2$
We will place them in the formula with the minus sign and we will get:
$\frac{-4-\sqrt{0}}{2}=$
$-\frac{4}{2}=-2$
The factorization is:
$(x-2)(x-2)$
If you are interested in this article, you might also be interested in the following articles:
In the Tutorela blog, you will find a variety of articles about mathematics.
Examples and exercises with solutions for factoring trinomials
Exercise #1
$x^2+6x+9=0$
What is the value of X?
Step-by-Step Solution
The equation in the problem is:
$x^2+6x+9=0$We want to solve this equation using factoring,
First, we'll check if we can factor out a common factor, but this is not possible, since there is no common factor for all three terms on the left side of the equation, we can identify that we can factor the expression on the left side using the quadratic formula for a trinomial squared, however, we prefer to factor it using the factoring method according to trinomials, let's refer to the search for Factoring by trinomials:
Let's note that the coefficient of the squared term (the term with the second power) is 1, so we can try to perform factoring according to the quick trinomial method: (This factoring is also called "automatic trinomial"),
But before we do this in the problem - let's recall the general rule for factoring by quick trinomial method:
The rule states that for the algebraic quadratic expression of the general form:
$x^2+bx+c$We can find a factorization in the form of a product if we can find two numbers $m,\hspace{4pt}n$such that the following conditions are met (conditions of the quick trinomial method):
$\begin{cases} m\cdot n=c\\ m+n=b \end{cases}$If we can find two such numbers $m,\hspace{4pt}n$then we can factor the general expression mentioned above into the form of a product and present it as:
$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$which is its factored form (product factors) of the expression,
Let's return now to the equation in the problem that we received in the last stage after arranging it:
$x^2+6x+9=0$Let's note that the coefficients from the general form we mentioned in the rule above:
$x^2+bx+c$are:$\begin{cases} c=9 \\ b=6 \end{cases}$where we didn't forget to consider the coefficient together with its sign,
Let's continue, we want to factor the expression on the left side into factors according to the quick trinomial method, above, so we'll look for a pair of numbers $m,\hspace{4pt}n$ that satisfy:
$\begin{cases} m\cdot n=9\\ m+n=6 \end{cases}$We'll try to identify this pair of numbers through logical thinking and using our knowledge of the multiplication table, we'll start from the multiplication between the two required numbers $m,\hspace{4pt}n$ that is - from the first row of the pair of requirements we mentioned in the last stage:
$m\cdot n=9$We identify that their product needs to give a positive result, and therefore we can conclude that their signs are identical,
Next, we'll consider the factors (integers) of the number 9, and from our knowledge of the multiplication table we can know that there are only two possibilities for such factors: 3 and 3, or 9 and 1, as we previously concluded that their signs must be identical, a quick check of the two possibilities regarding the fulfillment of the second condition:
$m+n=6$ will lead to a quick conclusion that the only possibility for fulfilling both of the above conditions together is:
$3,\hspace{4pt}3$That is - for:
$m=3,\hspace{4pt}n=3$(It doesn't matter which one we call m and which one we call n)
It is satisfied that:
$\begin{cases} \underline{3}\cdot \underline{3}=9\\ \underline{3}+\underline{3}=6 \end{cases}$ From here - we understood what the numbers we are looking for are and therefore we can factor the expression on the left side of the equation in question and present it as a product:
$x^2+6x+9 \\ \downarrow\\ (x+3)(x+3)$
In other words, we performed:
$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$
If so, we have factored the quadratic expression on the left side of the equation into factors using the quick trinomial method, and the equation is:
$x^2+6x+9=0 \\ \downarrow\\ (x+3)(x+3)=0\\ (x+3)^2=0\\$Where in the last stage we noticed that in the expression on the left side the term:
$(x+3)$
multiplies itself and therefore the expression can be written as a squared term:
$(x+3)^2$
Now that the expression on the left side has been factored into a product form (in this case not just a product but actually a power form) we will continue to quickly solve the equation we received:
$(x+3)^2=0$
Let's pay attention to a simple fact, on the left side there is a term raised to the second power, and on the right side the number 0,
and only 0 squared (to the second power) will give the result 0, so we get that the equation equivalent to this equation is the equation:
$x+3=0$(In the same way we could have operated algebraically in a pure form and taken the square root of both sides of the equation, we'll discuss this in a note at the end)
We'll solve this equation by moving the free number to the other side and we'll get that the only solution is:
$x=-3$Let's summarize then the stages of solving the quadratic equation using the quick trinomial factoring method, we got that:
$x^2+6x+9=0 \\ \downarrow\\ (x+3)(x+3)=0\\ (x+3)^2=0\\ \downarrow\\ x+3=0\\ x=-3$Therefore, the correct answer is answer C.
Note:
We could have reached the final equation by taking the square root of both sides of the equation, however - taking a square root involves considering two possibilities: positive and negative (it's enough to consider this only on one side, as described in the calculation below), that is, we could have performed:
$(x+3)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{8pt}}\\ \downarrow\\ \sqrt{(x+3)^2}=\pm\sqrt{0} \\ x+3=\pm0\\ x+3=0$Where on the left side the root (which is a half power) and the second power canceled each other out (this follows from the law of powers for power over power), and on the right side the root of 0 is 0, and we considered two possibilities positive and negative (this is the plus-minus sign indicated) except that the sign (which is actually multiplication by one or minus one) does not affect 0 which remains 0 in both cases, and therefore we reached the same equation we reached with logical and unambiguous thinking earlier - in the solution above,
In any other case where on the right side was a number different from 0, we could have solved only by taking the root etc. and considering the two positive and negative possibilities which would then give two different possibilities for the solution.
3-
Exercise #2
Complete the equation:
$(x+3)(x+\textcolor{red}{☐})=x^2+5x+6$
Step-by-Step Solution
Let's simplify the expression given in the left side:
$(x+3)(x+\textcolor{purple}{\boxed{?}})$ For ease of calculation we will replace the square with the question mark (indicating the missing part that needs to be completed) with the letter $\textcolor{purple}{k}$, meaning we will perform the substitution:
$(x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \\ \downarrow\\ (x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\$Next, we will expand the parentheses using the expanded distribution law:
$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$Let's note that in the formula template for the distribution law mentioned we assume by default that the operation between the terms inside the parentheses is addition, so we won't forget of course that the sign preceding the term is an integral part of it, and we will also apply the laws of sign multiplication and thus we can represent any expression in parentheses, which we expand using the aforementioned formula, first, as an expression where addition is performed between all terms (if necessary),
Therefore, we will first represent each of the expressions in parentheses in the multiplication on the left side as an expression where addition exists:
$(x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ \big(x+(+3)\big)\big(x+(\textcolor{purple}{+k})\big)=x^2+5x+6 \\$Now for convenience, let's write down again the expanded distribution law mentioned earlier:
$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$And we'll apply it to our problem:
$\big (\textcolor{red}{x}+\textcolor{blue}{(+3)}\big)\big(x+(+\textcolor{purple}{k})\big)=x^2+5x+6 \\ \downarrow\\ \textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\$We'll continue and apply the laws of multiplication signs, remembering that multiplying expressions with identical signs will yield a positive result, and multiplying expressions with different signs will yield a negative result:
$\textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ x^2+\textcolor{purple}{k}x+3x+3\textcolor{purple}{k}=x^2+5x+6 \\$Now, we want to present the expression on the left side in a form identical to the expression on the right side, that is - as a sum of three terms with different exponents: second power (squared), first power, and zero power (i.e., the free number - not dependent on x). To do this - we will factor out the part of the expression on the left side where the terms are in the first power:
$x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\ \downarrow\\ x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6$
Now in order for equality to hold - we require that the coefficient of the first-power term on both sides of the equation be identical and at the same time - we require that the free term on both sides of the equation be identical as well:
$x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}}$In other words, we require that:
$\begin{cases} \textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\ 3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2} \end{cases}$
Let's summarize the solution steps:
$(x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \leftrightarrow\textcolor{red}{\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{k}}} \\ \downarrow\\ (x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\ \downarrow\\ x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6\\ \downarrow\\ x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}} \\ \begin{cases} \textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\ 3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2} \end{cases} \\ \textcolor{red}{\bm{\rightarrow}\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{2}}}$Therefore, the missing expression is the number $2$meaning - the correct answer is a'.
2
Exercise #3
How many solutions does the equation have?
$x^4+12x^3+36x^2=0$
Step-by-Step Solution
Let's solve the given equation:
$x^4+12x^3+36x^2=0$We note that it is possible to factor the expression which is in the left side of the given equation, this is done by taking out the common factor $x^2$ which is the greatest common factor of the numbers and letters in the expression:
$x^4+12x^3+36x^2=0 \\ \downarrow\\ x^2(x^2+12x+36) =0$We will focus on the left side of the equation and then on the right side (the number 0).
Since the only way to get the result 0 from a product is to multiply by 0, at least one of the expressions in the product on the left side, must be equal to zero,
Meaning:
$x^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \boxed{x=0}$
Or:
$x^2+12x+36=0$In order to find the additional solutions to the equation we must solve the equation:
Note that the first coefficient is 1, so we can try to solve it using the trinomial formula.
However, we can factor, in this case, also using the short multiplication formula for a binomial:
$(\textcolor{red}{a}+\textcolor{blue}{b})^2 = \textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$The reason for trying factoring in this approach is that we can identify in the left side of the equation we got in the last step, that the two terms which are in the far sides (meaning the term in the first position - it is the squared term and the term in the zero position - it is the free number in the expression) can be presented (simply) as a squared term:
$x^2+12x+36=0 \\ \downarrow\\ x^2+12x+6^2=0$
Equating the expression on the left side in the equation:
$\downarrow\\ x^2+12x+6^2$
To the expression on the right side in the short formula above:
$\textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$
The conclusion from this is that what remains to check is whether the middle term in the equation matches the middle term in the short multiplication formula above, meaning - after identifying $a$$b$which are both in the first position in the short multiplication formula above in which $a$and $b$we check if the middle term in the expression in the left side of the equation can be presented as $2\cdot a \cdot b$So, we start by presenting the equation of the short formula to the given expression:
$\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 \Leftrightarrow \textcolor{red}{x}^2+\underline{12x}+\textcolor{blue}{6}^2$And indeed it holds that:
$2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}=12x$Meaning the middle term in the expression in the equation indeed matches the form of the middle term in the short multiplication formula (highlighted with a line below), mathematically:
$x^2+\underline{12x}+6^2=0 \\ \textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{6}^2=0\\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{6})^2=0$We can now remember that a real root can be calculated only for a positive number or for the number zero (since it is not possible to get a negative number from squaring a real number itself), and therefore for an equation there are two real solutions (or one solution) only if:
Next we note that if: $(x+6)^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ x+6=0\\ \boxed{x=-6}$then the only solution to the equation is:
$x^4+12x^3+36x^2=0 \\ \downarrow\\ x^2(x^2+12x+36) =0 \\ \downarrow\\ x^2=0\rightarrow\boxed{x=0} \\ x^2+12x+36=0\\ x^2+2\cdot2\cdot6+6^2=0\\ \rightarrow(x+6)^2=0\rightarrow\boxed{x=-6}\\ \downarrow\\ \boxed{x=0,-6}$Therefore, we can summarize what was explained using the following:
$ax^2+bx+c =0$in which the coefficients are substituted and the discriminant is calculated:
$a,b$If it holds:
a.$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:
There is no (real) solution to the equation.
b.$\Delta$:
There exists a single (real) solution to the equation.
c.$x_{1,2}=\frac{-b\pm\sqrt{\textcolor{orange}{\Delta}}}{2a} \longleftrightarrow\textcolor{orange}{\Delta}=b^2-4ac$:
There exist two (real) solutions to the equation.
Now let's return to the given equation and extract from it the coefficients:
$\Delta\geq0$We continue and calculate $\Delta=0$:
$x=-\frac{b}{2a}$Therefore for the quadratic equation that we solved, one (real) solution,
and in combination with the solution $ax^2+bx+c =0$(the additional solution we found for the given equation which is indicated in the first step after factoring using the common factor),
Therefore we get that for the given equation:
$\Delta=b^2-4ac$
two real solutions.
Two solutions
Exercise #4
Solve the following equation:
$4x^2-14x-8=0$
Step-by-Step Solution
Let's solve the given equation:
$4x^2-14x-8=0$
Instead of dividing both sides of the equation by the common factor of all terms in the equation (which is the number 2), we will choose to factor it out of the parentheses:
$4x^2-14x-8=0 \\ 2(2x^2-7x-4)=0$
From here we remember that the result of multiplication between expressions will yield 0 only if at least one of the multiplying expressions equals zero,
However, the first factor in the expression we got is the number 2, which is obviously different from zero, therefore:
$2x^2-7x-4 =0$Now let's note that the coefficient of the quadratic term (squared) is different from 1, we can solve the resulting equation of course using the quadratic formula, but we prefer, for the sake of skill improvement, to continue and factor the expression on the left side, this we will do using general trinomial factoring (actually - grouping method),
Similar to the quick factoring method by trinomial (which is actually a special case of the general trinomial factoring) we will look for a pair of numbers $m,\hspace{2pt}n$whose product should give the product of the coefficient of the quadratic term and the free term in the general expression:
$ax^2+bx+c$and their sum, the coefficient of the term in the first power, that is in the general expression mentioned, if so - we will look for a pair of numbers $m,\hspace{2pt}n$ that satisfy:
$m\cdot n=a\cdot c\\ m+n=b$ Once we find the pair of numbers that satisfy both conditions mentioned (if indeed such can be found) we will separate the coefficient of the term in the first power accordingly and factor by grouping,
Let's return then to the problem and demonstrate:
In the equation:
$2x^2-7x-4 =0$ We will look for a pair of numbers $m,\hspace{2pt}n$ that satisfy:
$m\cdot n=2\cdot (-4)\\ m+n=-7 \\ \downarrow\\ m\cdot n=-8\\ m+n=-7 \\$We'll continue, identical to what's done in the quick factoring method by trinomial,
From the first requirement mentioned, that is - from the multiplication, let's note that the product of the numbers we're looking for should yield a negative result and therefore we can conclude that the two numbers have different signs, this is according to multiplication laws, and now we'll remember that the possible factors of the number 8 are 2 and 4 or 8 and 1, fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are different from each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:
$\begin{cases} m=-8\\ n=1 \end{cases}$From here, unlike the quick factoring method by trinomial (where this step is actually skipped and factored directly, but it definitely exists), we will separate the coefficient of the term in the first power according to the pair of numbers $m,\hspace{2pt}n$we found:
$2x^2\underline{\textcolor{blue}{-7x}}-4 =0\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{(-8+1)x}}-4 =0\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{-8x+x}}-4 =0\\$We'll continue - in the next step we will factor by grouping:
We will refer to two groups of terms, so that in each group there is one term in the first power (the choice of groups doesn't matter - as long as this condition is maintained), in each group - we will factor out a common factor so that inside the parentheses, in both groups, we get the same expression:
$\textcolor{green}{2x^2-8x}\textcolor{red}{+x-4} =0\\ \downarrow\\ \textcolor{green}{2x(x-4)}\textcolor{red}{+1(x-4)} =0\\$(In this case, in the second group - which is marked in red, it was not possible to factor further, so we settled for factoring out the number 1 as a common factor for emphasis),
We'll continue, now let's note that the expression in parentheses in both groups is identical and therefore we can refer to it as a common factor for both groups (which is a binomial) and factor it out of the parentheses:
$\textcolor{green}{2x}\underline{(x-4)}\textcolor{red}{+1}\underline{(x-4)}=0\\ \downarrow\\ \underline{(x-4)}(\textcolor{green}{2x}\textcolor{red}{+1})=0$We have thus obtained a factored expression on the left side,
Let's summarize this factoring technique:
$2x^2-7x-4 =0 \\ \begin{cases} m\cdot n=2\cdot (-4)=-8\\ m+n=-7 \\\end{cases}\leftrightarrow m,\hspace{2pt}n=?\\ \downarrow\\ m\stackrel{!}{= }-8\\ n\stackrel{!}{= }1\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{-7x}}-4 =0\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{-8x+x}}-4 =0\\ \downarrow\\ \textcolor{green}{2x^2-8x}\textcolor{red}{+x-4} =0\\ \downarrow\\ \textcolor{green}{2x}\underline{(x-4)}\textcolor{red}{+1}\underline{(x-4)}=0\\ \downarrow\\ \underline{(x-4)}(\textcolor{green}{2x}\textcolor{red}{+1})=0$This technique is very important - it is recommended to review and sharpen the understanding of it,
Let's continue solving the equation, we got:
$(x-4)(2x+1)=0$
From here we remember that the result of multiplication between expressions will yield 0 only if at least one of the multiplying expressions equals zero,
Therefore we will get two simple equations and solve them, in the usual way, by isolating the unknown in the side (by transferring sides and division/multiplication of both sides if necessary):
$x-4=0\\ \boxed{x=4}$Or :
$2x+1=0\\ 2x=-1\text{/}:2\\ \boxed{x=-\frac{1}{2}}$
Let's summarize then the solution of the equation:
$4x^2-14x-8=0 \\ 2(2x^2-7x-4)=0 \\ \downarrow\\ 2x^2-7x-4 =0 \\ \begin{cases} m\cdot n=2\cdot (-4)=-8\\ m+n=-7 \\\end{cases}\leftrightarrow m,\hspace{2pt}n=?\\ \downarrow\\ m\stackrel{!}{= }-8\\ n\stackrel{!}{= }1\\ \downarrow\\ 2x^2\textcolor{blue}{-8x+x}-4 =0 \\ \downarrow\\ 2x\underline{(x-4)}+1\cdot\underline{(x-4)}=0\\ \underline{(x-4)}(2x+1)=0\\ \downarrow\\ x-4=0\rightarrow\boxed{x=4}\\ 2x+1=0\rightarrow\boxed{x=-\frac{1}{2}}\\ \downarrow\\ \boxed{x=4,-\frac{1}{2}}$Therefore the correct answer is answer a.
$4,-\frac{1}{2}$
Exercise #5
A rectangle has an area of
$m^2+4m-12$ cm²
and a length of $m+6$ cm.
Is the rectangle's perimeter 16 cm?
Step-by-Step Solution
First, let's consider the rectangle $ABCD$:
(Sketch - indicating the given information about AB on it)
Let's continue and write down the information about the area of the rectangle and the length of the given side in mathematical form:
$\begin{cases} \textcolor{red}{S_{ABCD}}= m^2+4m-12 \\ \textcolor{blue}{AB}=m+6\\ \end{cases}$(We'll use colors here for clarity of the solution later)
Now let's remember the fact that the area of a rectangle whose side lengths (adjacent) are:
$a,\hspace{2pt}b$is:
$S_{\boxed{\hspace{6pt}}}=a\cdot b$and therefore the area of the rectangle in the problem (according to the sketch we established at the beginning of the solution) is:
$S_{ABCD}=AB\cdot AD$We can now substitute the previously mentioned data into this expression for area to get the equation (for understanding - refer to the marked colors and the data mentioned earlier accordingly):
$\textcolor{red}{ S_{ABCD}}=\textcolor{blue}{AB}\cdot AD \\ \downarrow\\ \boxed{ \textcolor{red}{ m^2+4m-12}=\textcolor{blue}{(m+6)}\cdot AD}$
Now, let's pause for a moment and ask what our goal is?
Our goal is, of course, to obtain the algebraic expression for the perimeter of the rectangle (in terms of m), for this we remember that the perimeter of a rectangle is the sum of the lengths of all its sides (let's denote by $P_{ABCD}$ the perimeter of the rectangle),
In addition, we remember that the opposite sides in a rectangle are equal to each other and therefore the perimeter of the given rectangle is:
$P_{ABCD}=AB+AD+BC+DC\\ \begin{cases} AB=DC\\ AD=BC \end{cases}\\ \downarrow\\ \boxed{\textcolor{purple}{P_{ABCD}=2\cdot(AB+AD)} }$However, we already have the algebraic expression for the rectangle's side $AB$ (from the given information in the problem, previously marked in blue), and therefore all we need in order to calculate the perimeter of the rectangle is the algebraic expression (in terms of m) for the length of side $AD$,
Let's return then to the equation we reached in the step before considering the perimeter (highlighted with a square around the equation) and isolate $AD$ from it, this we'll do by dividing both sides of the equation by the algebraic expression that is the coefficient of $AD$, that is by:$(m+6)$:
$\boxed{ \textcolor{red}{\textcolor{blue}{(m+6)}\cdot AD= m^2+4m-12}} \hspace{4pt}\text{/:}(m+6)\\ \downarrow\\ AD=\frac{m^2+4m-12}{m+6}$Let's continue and simplify the algebraic fraction we got, we'll do this easily by factoring the numerator of the fraction:
$m^2+4m-12$We'll use quick trinomial factoring for this (to review the rules of quick trinomial factoring) and we get:
$m^2+4m-12\leftrightarrow\begin{cases} \boxed{?}\cdot\boxed{?}=-12\\ \boxed{?}+\boxed{?}=4\ \end{cases}\\ \downarrow\\ (m+6)(m-2)$and therefore (we'll return to the expression for $AD$):
$AD=\frac{m^2+4m-12}{m+6} \\ \downarrow\\ AD=\frac{(m+6)(m-2)}{m+6}\\ \downarrow\\ \boxed{AD=m-2}$In the final stage, after we factored the numerator of the fraction, we simplified the fraction,
(Sketch - with the found length of AD)
Let's return then to the expression for the perimeter of the rectangle, which we got earlier and substitute into it the algebraic expressions for the lengths of the rectangle's sides that we got, then we'll simplify the resulting expression:
$\boxed{\textcolor{purple}{P_{ABCD}=2(AB+AD)} } \\ AB=m+6\\ AD=m-2\\ \downarrow\\ P_{ABCD}=2(m+6+m-2) \\ P_{ABCD}=2(2m+4) \\ \boxed{P_{ABCD}=4m+8}$(length units)
We have thus found the expression for the perimeter of the rectangle in terms of m,
Is it possible that the perimeter of the rectangle is 16 length units?
In other words, mathematically- does there exist an m for which:
$P_{ABCD}=16$?
To answer this question, we'll need to perform two steps:
a. Find the value of m for which the stated condition is met.
b. Check if the value of m that meets the requirement is logical in terms of all the problem data (which may be limited in terms of m).
Let's find the value of m that meets the stated requirement,
For this, we'll recall the expression for the perimeter that we found in the previous step (enclosed in a frame), and the stated requirement, then we'll demand that the requirement is met and solve the resulting equation:
$\begin{cases} \boxed{P_{ABCD}=4m+8}\\ P_{ABCD}=16 \end{cases}\\ \downarrow\\ 4m+8=16\\ 4m=8\hspace{6pt}\text{/:}4\\ \boxed{m=2}$We have thus found the value of m that meets the requirement regarding the perimeter of the rectangle.
b. Now let's check if for the value of m we found (which meets the requirement regarding the perimeter) we get logical values,
We'll consider that the realistic values in the problem that depend on m are the area and the lengths of the sides, and therefore can only receive positive values (i.e., greater than 0),
For this, we'll recall the various expressions for the area and the lengths of the sides and we'll also state the question asked here mathematically:
$m\stackrel{?}{= }2\leftrightarrow \begin{cases} \textcolor{red}{S_{ABCD}}=(m+6)(m-2) \\ \textcolor{blue}{AB}=m+6\\ AD=m-2 \end{cases}$Where we wrote the expression for the area in its expanded form,
We'll use substitution and note that for: $m=2$, the length of side $AD$becomes zero, and additionally, as expected - the area of the rectangle also becomes zero, that is mathematically:
$m=2\leftrightarrow \begin{cases} \textcolor{red}{S_{ABCD}}=(\underline{2}+6)(\underline{2}-2)=0 \\ \textcolor{blue}{AB}=\underline{2}+6=8\\ AD=\underline{2}-2=0 \end{cases}$Therefore, we can conclude that the solution m=2, which was obtained from the requirement that the perimeter of the rectangle be equal to: 16 length units,
is not possible in the problem, since for this value of m the length of the rectangle's side (and therefore also its area) becomes zero,
But this value is the only value for which the requirement regarding the perimeter of the rectangle is met,
That is - there does not exist an m for which the perimeter of the rectangle is 16 length units,
Therefore, it is not possible that the perimeter of the given rectangle is: 16 length units.
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# How do you find eigenvalues and eigenvectors examples?
## How do you find eigenvalues and eigenvectors examples?
For example, suppose the characteristic polynomial of A is given by (λ−2)2. Solving for the roots of this polynomial, we set (λ−2)2=0 and solve for λ. We find that λ=2 is a root that occurs twice. Hence, in this case, λ=2 is an eigenvalue of A of multiplicity equal to 2.
## How do you find eigenvalues and eigenvectors of a matrix?
1:Finding Eigenvalues and Eigenvectors. Let A be an n×n matrix. First, find the eigenvalues λ of A by solving the equation det(λI−A)=0. For each λ, find the basic eigenvectors X≠0 by finding the basic solutions to (λI−A)X=0.
What is the formula of eigenvector?
Eigenvector Equation The equation corresponding to each eigenvalue of a matrix is given by: AX = λ X.
### How to determine the eigenvectors of a matrix?
The eigenvectors of a matrix A are those vectors X for which multiplication by A results in a vector in the same direction or opposite direction to X. Since the zero vector 0 has no direction this would make no sense for the zero vector.
How to compute eigenvectors from eigenvalues?
Calculate the eigen vector of the following matrix if its eigenvalues are 5 and -1. Lets begin by subtracting the first eigenvalue 5 from the leading diagonal. Then multiply the resultant matrix by the 1 x 2 matrix of x, equate it to zero and solve it. Then find the eigen vector of the eigen value -1. Then equate it to a 1 x 2 matrix and equate
#### How to find eigenvectors?
How to Find Eigenvector. The following are the steps to find eigenvectors of a matrix: Step 1: Determine the eigenvalues of the given matrix A using the equation det (A – λI) = 0, where I is equivalent order identity matrix as A. Denote each eigenvalue of λ1 , λ2 , λ3 ,…
#### How to find eigenvalue 3×3?
3X3 Eigenvalue Calculator. Calculate eigenvalues. First eigenvalue: Second eigenvalue: Third eigenvalue: Discover the beauty of matrices! Matrices are the foundation of Linear Algebra; which has gained more and more importance in science, physics and eningineering.
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# FINDING THE VOLUME OF A CYLINDER IN A REAL WORLD CONTEXT
## About "Finding the volume of a cylinder in a real world context"
Finding the volume of a cylinder in a real world context :
Finding volumes of cylinders is similar to finding volumes of prisms. We can find the volume V of both a prism and a cylinder by multiplying the height by the area of the base.
Let the area of the base of a cylinder be B and the height of the cylinder be h. Write a formula for the cylinder’s volume V.
V = Bh
The base of a cylinder is a circle, so for a cylinder,
B = r²
Then, we have
V = r²h cubic units
## Finding the volume of a cylinder in a real world context - Examples
Example 1 :
The cylindrical Giant Ocean Tank at the New England Aquarium in Boston is 24 feet deep and has a radius of 18.8 feet. Find the volume of the tank. Use the approximate of value of , that is 3.14 and round your answer to the nearest tenth if necessary.
Solution :
Step 1 :
Because the tank is in the shape of cylinder, we can use the formula of volume of a cylinder to find volume of the tank.
V = r²h cubic units
Step 2 :
Substitute the given measures.
≈ 3.14 · 18.8² · 24
(Here deep 24 feet is considered as height)
Simplify.
≈ 3.14 · 353.44 · 24
≈ 26635.2
Hence, the volume of the tank is about 26635.2 cubic feet.
Let us look at the next example on "Finding the volume of a cylinder in a real world context"
Example 2 :
A standard-size bass drum has a diameter of 22 inches and is 18 inches deep. Find the volume of this drum. Use the approximate of value of , that is 3.14 and round your answer to the nearest tenth if necessary.
Solution :
Step 1 :
Usually the bass drum would be in the shape of cylinder. So, we can use the formula of volume of a cylinder, to find volume of the bass drum.
V = r²h cubic units ----- (1)
Step 2 :
To find the volume, we need the radius of the cylinder. But, the diameter is given, that is 22 in. So, find the radius.
r = diameter / 2
r = 22/2
r = 11
Step 3 :
Plug ∏ ≈ 3.14, r = 11 and h = 18 in (1).
≈ 3.14 · 11² · 18
(Here deep 18 inches is considered as height)
Simplify.
≈ 3.14 · 121 · 18
≈ 6838.9
Hence, the volume of the bass drum is about 6838.9 cubic inches.
Let us look at the next example on "Finding the volume of a cylinder in a real world context"
Example 3 :
A barrel of crude oil contains about 5.61 cubic feet of oil. How many barrels of oil are contained in 1 mile of a pipeline that has an inside diameter of 6 inches and is completely filled with oil ? How much is “1 mile” of oil in this pipeline worth at a price of \$100 per barrel ?
Solution :
Step 1 :
Usually the pipe line would be in the shape of cylinder. So, we can use the formula of volume of a cylinder to find volume of the crude oil in the pipe line.
V = r²h cubic units ----- (1)
Step 2 :
To find the volume, we need the radius of the cylinder. But, the diameter is given, that is 6 in. So, find the radius.
r = diameter / 2
r = 6/2
r = 3 inches
Step 3 :
Convert the inches into feet by multiplying 1/12.
Because,
1 inch = 1/12 feet
So, we have
r = 3 x 1/12 feet
r = 1/4 feet
Step 4 :
Convert the length of the pipeline from miles to feet.
1 mile = 5280 feet
So, we have
length = 1 mile
length = 1 x 5280 feet
length = 5280 feet
Step 5 :
Plug ∏ ≈ 3.14, r = 1/4 and h = 5280 in (1).
≈ 3.14 · (1/4)² · 5280
(Here , the length 5280 feet is considered as height)
Simplify.
≈ 3.14 · (1/16) · 5280
≈ 1036.2 cubic feet
Step 6 :
To find how many barrels of oil are contained in 1 mile of a pipeline, divide the volume of crude oil in the pipeline (1036.2 cu.ft) by 5.61.
Because a barrel of crude oil contains about 5.61 cubic feet of oil.
So, number of barrels of oil are contained in 1 mile of a pipeline is
= 1036.2 / 5.61
= 184.7
There are about 184.7 barrels of oil are contained in 1 mile of a pipeline.
Step 7 :
Find the worth of “1 mile” of oil in the pipeline at a price of \$100 per barrel.
No. of barrels of oil in 1 mile of a pipeline = 184.7
So, the worth of “1 mile” of oil in the pipeline is
= \$100 x 184.7
= \$18,470
The worth of “1 mile” of oil in the pipeline at a price of \$100 per barrel is about \$18,470.
Let us look at the next example on "Finding the volume of a cylinder in a real world context"
Example 4 :
A pan for baking French bread is shaped like half a cylinder as shown in the figure. Find the volume of uncooked dough that would fill this pan. Use the approximate of value of , that is 3.14 and round your answer to the nearest tenth if necessary.
Solution :
Step 1 :
Because the pan is shaped like half a cylinder, we can use the formula of volume of a cylinder to find volume of uncooked dough that would fill this pan
V = 1/2 · r²h cubic units ----- (1)
(Because the pan is shaped like half a cylinder, 1/2 is multiplied by the formula of volume of a cylinder)
Step 2 :
To find the volume, we need the radius of the cylinder. But, the diameter is given, that is 3.5 in. So, find the radius.
r = diameter / 2
r = 3.5/2
r = 1.75
Step 3 :
Plug ∏ ≈ 3.14, r = 1.75 and h = 12 in (1).
≈ 1/2 · 3.14 · 1.75² · 12
Simplify.
≈ 1/2 · 3.14 · 3.0625 · 12
≈ 57.7
Hence, the volume of uncooked dough that would fill the pan is about 57.7 cu.inches.
After having gone through the stuff given above, we hope that the students would have understood "Finding the volume of a cylinder in a real world context".
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# Calculus II Practice Problems 10: Answers Answer
```Calculus II
Practice Problems 10: Answers
In problems 1-4 put the conic in standard form, and find the center, vertices, foci.
1. y
8x2
32x
29
0
Answer. Complete the square:
4
32 0
y 3 8 x 2
This is a parabola with vertex at (2,-3) and axis the line x 2. Since 4p at 2 1 .
y
8 x2
4x
29
2
2. 9x2
4y2
36x
8y
4
8, we have p
2, so the focus is
0
Answer. Complete the squares:
9 x2
4x
4 4 y 2
2y
1 4
36
4
0
x 2 y 1 1 4
9
This is an ellipse centered at 2 1 , with major axis the line x 2. Since b 3, and a 2, the vertices are
at 2 1 3 or (2,2) and (2,-4). We have c b a 5, so the foci are at 2 1 5 .
2
2
3. 4x2
y2
2y
2
2
2
2
2y
5
Answer. Complete the squares:
y
4x2
1
5
y 1 1 4
1
2
x2
This is a hyperbola centered at 0 1 , with major axis the line y 1. Since a 1 and b
at 1 1 . We have c2 a2 b2 1 1 4
5 4, so the foci are at
51.
4. x2
5y2
4x
10y
1
Answer. Complete the squares:
x
2
4x
4
5 y2
2y
1 1 leading to
4
x 2 5 y 1
The graph is the pair of lines intersecting at (2,1): x 2 5 y 1 .
2
2
5
0
2, the vertices are
In problems 5-7, find the equation of the tangent line to the curve at the point x 0 y0 on the curve.
5. x2
5y
0
10 20
Answer. We recall from example 11, Chapter I.5, how to find tangent lines by implicit differentiation. Taking
differentials we have
2xdx 5dy 0
Now replace x y by the coordinates of the point: 10, 20, and dx and dy by the increments along the tangent
line. This gives us
2 10 x 10 5 y 20
0 or 20x 5y 100
6. x2
4y2
16
2xdx
and evaluate at the given point:
22 3 x
y2
2 3 1
7. 4x2
8ydy
8 1 y 1 2 3
0
0 or
4 3 x 8y
32
1
2 2 1
8xdx
2ydy
0 or dy
4xdx
and evaluate at the given point:
y
1
4 2 2 x 2 2
or y
2 2 x
1
In each of problems 8 and 9, the curve described depends upon a parameter. Identify the parameter, and find
the equation of the curve in terms of the parameter.
8. A parabola with axis the x-axis and focus at the origin.
Answer. The equation of a parabola with axis the x-axis and vertex at x 0 0 is
y2
4p x
x0
where p is the separation between the focus and the vertex. Since the focus is at the origin, the vertex is at
p 0 , thus the desired equation is
y2 4p x p
9. A hyperbola with foci at (-1,0), (1,0).
Answer. Since the foci are on the x-axis and the origin is midway between the foci, this hyperbola has as its
axis the x-axis, and its center is the origin. Place the vertices at the points a 0 , with a 1. Then, since
b2 a2 c2 a2 1, the desired equation is
x2
a2
10. Find the point x y on the parabola y 2
an angle of 45 .
y2
a2
1
1
12x for which the line from the focus meets the tangent line at
Answer. By the optical property of the parabola, the tangent line at x y makes an angle of 45 with the
horizontal, so the slope of the tangent line is m tan 45
1. Differentiating the equation of the curve, we
have
dy 6
dy
12 so that m
2y
dx
dx
y
Thus 6 y
1, so y
6 and x
3.
```
|
# Greatest Common Divisor of Three Numbers
## Theorem
In the words of Euclid:
Given three numbers not prime to one another, to find their greatest common measure.
## Proof
Let $A, B, C$ be the three given (natural) numbers not prime to one another.
Let the GCD $D$ of $A, B$ be found, by the Euclidean Algorithm.
Then $D$ either divides or does not divide $C$.
First, suppose $D$ divides $C$.
Then as it also divides $A$ and $B$, it is a common divisor of $A, B$ and $C$.
Suppose $D$ is not the GCD of $A, B, C$.
Then some number $E$ is a common divisor of $A, B$ and $C$ such that $E > D$.
But then $E$ is a common divisor of $A$ and $B$ and so from the porism to Euclid's Algorithm $E$ also divides $D$.
That means $E \le D$ which contradicts $E > D$.
So $D$ is the GCD of $A, B, C$ after all.
Next, suppose $D$ is not a divisor of $C$.
Since $A, B, C$ are not prime to one another, some number will divide them.
That number which divides $A, B, C$ will also divide $A, B$ and so from the porism to Euclid's Algorithm will also divide $D$.
But it also divides $C$.
So some number divides both $D$ and $C$, and so $C$ and $D$ are not prime to one another.
Let the GCD of $C$ and $D$ be $E$, by the Euclidean Algorithm.
Since $E$ divides $D$ and $D$ divides $A, B$, then $E$ divides $A, B$.
But $E$ divides $C$, and so $E$ divides $A, B, C$.
So $E$ is a common divisor of $A, B$ and $C$.
Now suppose $E$ is not the GCD of $A, B, C$.
Then some number $F$ such that $F > E$ divides $A, B, C$.
It follows that $F$ divides $A$ and $B$, and so from the porism to Euclid's Algorithm will also divide $D$, the GCD of $A, B$.
So $F$ divides $D$ and $F$ also divides $C$.
So $F$ is a common divisor of $D$ and $C$.
So by the porism to Euclid's Algorithm, $F$ divides the GCD of $C, D$, which is $E$.
But this contradicts the supposition that $F > E$.
So no such $F$ can exist, and so $E$ is the GCD of $A, B, C$.
$\blacksquare$
## Historical Note
This proof is Proposition $3$ of Book $\text{VII}$ of Euclid's The Elements.
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### Unit 2 post assessment
1. 1. Unit 2 Post Assessment Missed Problems
2. 2. Question 1Solve for x: 3x = 30 Multiplication is the operation being performed on the3x = 30 variable (x) so we need to undo multiplication by3x/3 = 30/3 dividing each side of the equation by 3.x = 10 Check your equation! 3(10) 30
3. 3. Question 2Solve for a: -2a = 16 Multiplication is the operation being performed on the-2a = 16 variable (a) so we need to undo multiplication by-2a/-2 = 16/-2 dividing each side of the equation by -2.a = -8 Check your equation! -2(-8) 16
4. 4. Question 5Solve for d: d – 12 = 10 Subtraction is the operation being performed on thed – 12 = 10 variable (d) so we need to undo subtraction by addingd – 12 + 12 = 10 + 12 each side of the equation by 12d = 22 Check your equation! 22 – 12 10
5. 5. Question 6Solve for f: f – 1.5 = 3.5 Subtraction is the operation being performed on thef – 1.5 = 3.5 variable (f) so we need to undo subtraction by addingf – 1.5 + 1.5 = 3.5 + 1.5 each side of the equation by 1.5f=5 Check your equation! 5 – 1.5 3.5
6. 6. Question 7Solve for t: 3 + t = 17 Addition is the operation being performed on the3 + t = 17 variable (t) so we need to undo addition by subtracting3 + t – 3 = 17 – 3 each side of the equation by 3.t = 14 Check your equation! 3 + 14 17
7. 7. Question 9Solve for l: s = l + w Addition is the operation being performed between ls=l+w and w so we need to subtract w to the other side to get ls–w=l+w–w alone.s–w=l
8. 8. Question 11Tenisha is twice as tall as her little brother. If Tenisha is 60 inches tall, how old is her little brother? Brother’s height is ½ of Tenisha’s 60/2 is 30 so Tenisha’s little brother is 30 inches tall.
9. 9. Question 12You and Michael have a total of \$19.75. IfMichael has \$8.25, how much money do youhave.You + Michael = \$19.75You + \$8.25 = \$19.75Subtract \$8.25 on each side.You = \$11.50
10. 10. Question 13Solve for x: 7x – 10 = 10 This is a two-step equation. We are multiplying 7 to the7x – 10 = 10 variable and subtraction 10 so we need to undo these7x – 10 + 10 = 10 + 10 two operations.7x = 20 Step 1: Undo subtraction by adding7x/7 = 20/7 Add 10 to each sidex = 20/7 Step 2: Undo multiplication by dividing Divide by 7 on each side.
11. 11. Question 14At the grocery store you This is a two-step equation.purchase 2 gallons of milk We are multiplying 2 to theand loaf of bread. You spent variable and adding 2 so wea total of \$8. If you know a need to undo these twoloaf of bread costs \$2, how operations.much does a gallon of milkcost? Step 1: Undo addition by subtracting 2 on each side2x + 2 = 8 Step 2: Undo multiplication by2x = 6 dividing by 2 on each side.x=3
12. 12. Question 16Solve for a: 4a + a – 11 = 24 Multiple Step Equation4a + a – 11 = 24 1) Combine like terms5a – 11 = 24 4a + a = 5a5a = 35 2) Add 11 to each side.a=7 3) Divide by 5 on each side.
13. 13. Question 17Solve for y: 4x – 8y = 16 Multiple Step Equation4x – 8y = 16 Subtract 4x on each side.-8y = -4x + 16 Divide by -8 on each side.y=½x–2
14. 14. Question 18Solve for p: 6p + 2(p – 1) = 22 Multiple Step Equation6p + 2(p – 1) = 22 1) Distribute 26p + 2p – 2 = 22 2) Combine like terms 6p + 2p8p – 2 = 22 3) Add 2 to each side8p = 24 4) Divide by 8 on each side.p=3
15. 15. Question 19Solve for r: 5 + 7r + 3r = 8(r – 4) Multiple Step Equation5 + 7r + 3r = 8(r – 4) 1) Distribute 85 + 7r + 3r = 8r – 32 2) Combine like terms 7r + 3r5 + 10r = 8r – 32 3) Subtract 8r to each side5 + 2r = -32 4) Subtract 5 on each side.2r = -37r = -37/2
16. 16. Question 20Solve for n: 3(n -2) + 1 = n + 4 + 2n Multiple Step Equation3(n – 2) + 1 = n + 4 + 2n 1) Distribute 33n – 6 + 1 = n + 4 + 2n 2) Combine like terms3n – 5 = 3n + 4 3) Subtract 3n on each side-5 = 4 -5 does not equal 4 so this equation does not have aNo Solution solution!
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# 32 and 42 LCM
LCM of 32 and 42 is equal to 672. The comprehensive work provides more insight of how to find what is the lcm of 32 and 42 by using prime factors and special division methods, and the example use case of mathematics and real world problems.
what is the lcm of 32 and 42?
lcm (32 42) = (?)
32 => 2 x 2 x 2 x 2 x 2
42 => 2 x 3 x 7
= 2 x 2 x 2 x 2 x 2 x 3 x 7
= 672
lcm (32 and 42) = 672
672 is the lcm of 32 and 42.
where,
32 is a positive integer,
42 is a positive integer,
672 is the lcm of 32 and 42,
{2} in {2 x 2 x 2 x 2 x 2, 2 x 3 x 7} is the common factors of 32 and 42,
{2 x 2 x 2 x 2 x 3 x 7} in {2 x 2 x 2 x 2 x 2, 2 x 3 x 7} are the uncommon factors of 32 and 42.
Use in Mathematics: LCM of 32 and 42
The below are some of the mathematical applications where lcm of 32 and 42 can be used:
1. to find the least number which is exactly divisible by 32 and 42.
2. to find the common denominator for two fractions having 32 and 42 as denominators in the unlike fractions addition or subtraction.
Use in Real-world Problems: 32 and 42 lcm
In the context of lcm real world problems, the lcm of 32 and 42 helps to find the exact time when two similar and recurring events with different time schedule happens together at the same time. For example, the real world problems involve lcm in situations to find at what time the bells A and B all toll together, if bell A tolls at 32 seconds and bell B tolls at 42 seconds repeatedly. The answer is that all bells A and B toll together at 672 seconds for the first time, at 1344 seconds for the second time, at 2016 seconds for the third time and so on.
Important Notes: 32 and 42 lcm
The below are the important notes to be remembered while solving the lcm of 32 and 42:
1. The common prime factors and the remaining prime factors of 32 and 42 should be multiplied to find the least common multiple of 32 and 42, when solving lcm by using prime factors method.
2. The results of lcm of 32 and 42, and the lcm of 42 and 32 are identical, it means the order of given numbers in the lcm calculation doesn't affect the results.
For values other than 32 and 42, use this below tool:
## How-to: What is the LCM of 32 and 42?
The below solved example with step by step work shows how to find what is the lcm of 32 and 42 by using prime factors method and division method.
Solved example using prime factors method:
What is the LCM of 32 and 42?
step 1 Address the input parameters, values and observe what to be found:
Input parameters and values:
A = 32
B = 42
What to be found:
find the lcm of 32 and 42
step 2 Find the prime factors of 32 and 42:
Prime factors of 32 = 2 x 2 x 2 x 2 x 2
Prime factors of 42 = 2 x 3 x 7
step 3 Identify the repeated and non-repeated prime factors of 32 and 42:
{2} is the most repeated factor and {2 x 2 x 2 x 2 x 3 x 7} are the non-repeated factors of 32 and 42.
step 4 Find the product of repeated and non-repeated prime factors of 32 and 42:
= 2 x 2 x 2 x 2 x 2 x 3 x 7
= 672
lcm(32 and 42) = 672
Hence,
lcm of 32 and 42 is 672
Solved example using special division method:
This special division method is the easiest way to understand the entire calculation of what is the lcm of 32 and 42.
step 1 Address the input parameters, values and observe what to be found:
Input parameters and values:
Integers: 32 and 42
What to be found:
lcm (32, 42) = ?
step 2 Arrange the given integers in the horizontal form with space or comma separated format:
32 and 42
step 3 Choose the divisor which divides each or most of the given integers (32 and 42), divide each integers separately and write down the quotient in the next line right under the respective integers. Bring down the integer to the next line if any integer in 32 and 42 is not divisible by the selected divisor; repeat the same process until all the integers are brought to 1 as like below:
2 32 42 2 16 21 2 8 21 2 4 21 2 2 21 3 1 21 7 1 7 1 1
step 4 Multiply the divisors to find the lcm of 32 and 42:
= 2 x 2 x 2 x 2 x 2 x 3 x 7
= 672
LCM(32, 42) = 672
The least common multiple for two numbers 32 and 42 is 672
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# The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?
Option :
Explanation:
Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{original}\phantom{\rule{thinmathspace}{0ex}}\text{length}=x\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\\ & \text{Original}\phantom{\rule{thinmathspace}{0ex}}\text{breadth}=y\\ & \text{Original}\phantom{\rule{thinmathspace}{0ex}}\text{area}=xy\\ & \text{New}\phantom{\rule{thinmathspace}{0ex}}\text{length}=\frac{x}{2}\\ & \text{New}\phantom{\rule{thinmathspace}{0ex}}\text{breadth}=3y\\ & \text{New}\phantom{\rule{thinmathspace}{0ex}}\text{area}=\left(\frac{x}{2}×3y\right)=\frac{3}{2}xy\\ & \therefore \text{Increase}\phantom{\rule{thinmathspace}{0ex}}\mathrm{%}=\left(\frac{1}{2}xy×\frac{1}{xy}×100\right)\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=50\mathrm{%}\end{array}$
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# 8 boys and 12 girls can finish a piece of work in an annual day celebration in 10 days while 6 boys and 8 girls can finish it in 14 days. Find the time taken by one boy and one girl to finish the work.
Last updated date: 10th Sep 2024
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Hint:
If a person does any work (x) in (y) days, then he will do complete work in $\left( {\dfrac{y}{x}} \right)$days.
To solve two linear equations ${a_1}x + {b_1}y + {c_1} = 0;{a_2}x + {b_2}y + {c_2} = 0$ using cross multiplication method $= \dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$
Complete step by step solution:
Let one boy alone can finish the work in x days and one girl alone can finish the work in y days.
In one day a boy can do $\dfrac{1}{x}th$ of the work, then 8 boys can do $\dfrac{8}{x}th$ of the work.
In one day a girl can do $\dfrac{1}{y}th$ of the work, then 12 girls can do $\dfrac{{12}}{y}th$ of the work.
Given that together they can finish it in 10 days;
$\Rightarrow \dfrac{8}{x} + \dfrac{{12}}{y} = \dfrac{1}{{10}}$
Putting $\dfrac{1}{x} = a$ and $\dfrac{1}{y} = b$ in above equation we get;
$\Rightarrow 8a + 12b = \dfrac{1}{{10}} \\ \Rightarrow 80a + 120b = 1.......Eq.01 \\$
Also, In one day a boy can do $\dfrac{1}{x}th$ of the work, then 6 boys can do $\dfrac{6}{x}th$ of the work.
In one day a girl can do $\dfrac{1}{y}th$ of the work, then 8 girls can do $\dfrac{8}{y}th$ of the work.
Given that together they can finish it in 14 days;
$\Rightarrow \dfrac{6}{x} + \dfrac{8}{y} = \dfrac{1}{{14}}$
Putting $\dfrac{1}{x} = a$ and $\dfrac{1}{y} = b$ in above equation we get;
$\Rightarrow 6a + 8b = \dfrac{1}{{14}} \\ \Rightarrow 84a + 112b = 1.......Eq.02 \\$
Solving Eq.01 and Eq.02 using cross-multiplication method we get;
$80a + 120b = 1 \\ 84a + 112b = 1 \\ \Rightarrow \dfrac{a}{{120(1) - 112(1)}} = \dfrac{b}{{84(1) - 80(1)}} = \dfrac{{ - 1}}{{80(112) - 84(120)}} \\ \Rightarrow \dfrac{a}{{120 - 112}} = \dfrac{b}{{84 - 80}} = \dfrac{{ - 1}}{{8960 - 10080}} \\ \Rightarrow \dfrac{a}{8} = \dfrac{b}{4} = \dfrac{{ - 1}}{{ - 1120}} \\ \Rightarrow \dfrac{a}{8} = \dfrac{b}{4} = \dfrac{1}{{1120}} \\ \Rightarrow a = \dfrac{8}{{1120}};b = \dfrac{4}{{1120}} \\ \Rightarrow a = \dfrac{1}{{140}};b = \dfrac{1}{{280}} \\$
Putting back the values of a and b in $\dfrac{1}{x} = a$ and $\dfrac{1}{y} = b$ respectively we get;
$\Rightarrow \dfrac{1}{x} = \dfrac{1}{{140}};\dfrac{1}{y} = \dfrac{1}{{280}} \\ \Rightarrow x = 140;y = 280 \\$
Note:
This is a question from time and work. Below are some important points to remember while solving such questions.
1) More men can do more work
2) More work means more time required to do work
3) More men can do a piece of work in less time.
|
## Engage NY Eureka Math Grade 6 Module 5 Lesson 17 Answer Key
### Eureka Math Grade 6 Module 5 Lesson 17 Opening Exercise Answer Key
Opening Exercise:
a. Write a numerical equation for the area of the figure below. Explain and identify different parts of the figure.
i.
A = $$\frac{1}{2}$$ (14 cm)(12 cm) = 84cm2
14 cm represents the bose of the figure because 5 cm + 9 cm = 14 cm, and 12 cm represents the altitude of the figure because it forms a right angle with the base.
ii. How would you write an equation that shows the area of a triangle with base b and height h?
A = $$\frac{1}{2}$$ bh
b. Write a numerical equation for the area of the figure below. Explain and identify different parts of the figure.
A = (28 ft.)(18 ft) = 504 ft2
28 ft. represents the base of the rectangle, 18 ft. and 18 ft. represents the height of the rectangle.
ii. How would you write an equation that shows the area of a rectangle with base b and height h?
A = bh.
### Eureka Math Grade 6 Module 5 Lesson 17 Example Answer Key
Example 1:
Use the net to calculate the surface area of the figure. (Note: all measurements are in centimeters.)
→ When you are calculating the area of a figure, what are you finding?
The area of a figure is the amount of space inside a two-dimensional figure.
→ The surface area is similar to the area, but the surface area is used to describe three-dimensional figures. What do you think is meant by the surface area of a solid?
The surface area of a three-dimensional figure is the area of each face added together.
→ What type of figure does the net create? How do you know?
It creates a rectangular prism because there are six rectangular faces.
If the boxes on the grid paper represent a 1 cm × 1 cm box, label the dimensions of the net.
→ The surface area of a figure is the sum of the areas of all faces. Calculate the area of each face, and record this value inside the corresponding rectangle.
→ In order to calculate the surface area, we have to find the sum of the areas we calculated since they represent the area of each face. There are two faces that have an area of 4 cm2 and four faces that have an area of 2 cm2. How can we use these areas to write a numerical expression to show how to calculate the surface area of the net?
The numerical expression to calculate the surface area of the net would be
(1 cm × 2 cm) + (1 cm × 2 cm) + (1 cm × 2 cm) + (1 cm × 2 cm) + (2 cm × 2 cm)+ (2 cm × 2 cm).
→ Write the expression more compactly, and explain what each part represents on the net.
4(1 cm × 2 cm) + 2(2 cm × 2 cm)
The expression means there are 4 rectangles that have dimensions 1 cm × 2 cm on the net and 2 rectangles that have dimensions 2 cm × 2 cm on the net.
→ What is the surface area of the net?
The surface area of the net is 16 cm2.
Example 2:
Use the net to write an expression for surface area. (Note: all measurements are in square feet.)
→ What type of figure does the net create? How do you know?
It creates a square pyramid because one face is a square and the other four faces are triangles.
→ If the boxes on the grid paper represent a 1 ft. × 1 ft. square, label the dimensions of the net.
→ How many faces does the rectangular pyramid have?
5
→ Knowing the figure has 5 faces, use the knowledge you gained in Example ito calculate the surface area of the rectangular pyramid.
Area of Base: 3 ft. × 3 ft. = 9 ft2
Area of Triangles: $$\frac{1}{2}$$ × 3 ft. × 2 ft. = 3 ft2
Surface Area: 9 ft2 + 3 ft2 + 3 ft2 + 3 ft2 + 3 ft2 = 21 ft2
### Eureka Math Grade 6 Module 5 Lesson 17 Exercise Answer Key
Exercises:
Name the solid the net would create, and then write an expression for the surface area. Use the expression to determine the surface area. Assume that each box on the grid paper represents a 1 cm × 1 cm square. Explain how the expression represents the figure.
Exercise 1.
Name of Shape: Rectangular Pyramid, but more specifically a Square Pyramid
Surface Area: 4 cm × 4 cm + 4($$\frac{1}{2}$$ × 4 cm × 3 cm)
Work: 16 cm2 + 4(6 cm2) = 40 cm2
The surface area is 40 cm2. The figure is made up of a square base that measures 4 cm × 4 cm and four triangles, each with a base of 4 cm and a height of 3 cm.
Exercise 2.
Name of Shape: Rectangular Prism
Surface Area: 2(5 cm × 5 cm) + 4(5 cm × 2 cm)
Work: 2(25 cm2 ) + 4(10 cm2) = 90 cm2
The surface area is 90 cm2. The figure has 2 square faces, each of which measures 5 cm × 5 cm, and 4 rectangular faces, each of which measures 5 cm × 2 cm.
Exercise 3.
Name of Shape: Rectangular Pyramid
SurfaceArea: 3 cm × 4 cm + 2($$\frac{1}{2}$$ × 4 cm × 4 cm)+ 2($$\frac{1}{2}$$ × 4 cm × 3 cm)
Work: 12 cm2 + 2(8 cm2) + 2(6 cm2) = 40 cm2
The surface area is 40 cm2. The figure has 1 rectangular base that measures 3 cm × 4 cm, 2 triangular faces, each with a bose of 4 cm and a height of 4 cm, and 2 other triangular faces, each with a base of 3 cm and a height of 4 cm.
Exercise 4.
Name of Shape: Rectangular Prism
Surface Area: 2(6 cm × 5 cm) + 2(5 cm × 1 cm) + 2(6 cm × 1 cm)
Work: 2(30 cm2) + 2(5 cm2) + 2(6 cm2) = 82 cm2
The surface area is 82 cm2. The figure has two 6 cm × 5 cm rectangular faces, two 5 cm × 1 cm rectangular faces, and two 6 cm × 1 cm rectangular faces.
### Eureka Math Grade 6 Module 5 Lesson 17 Problem Set Answer Key
Name the shape, and write an expression for surface area. Calculate the surface area of the figure. Assume each box on the grid paper represents a 1 ft. × 1 ft. square.
Question 1.
Name of Shape: Rectangular Prism
SurfaceArea: (2 ft. × 4 ft.) + (2 ft. × 4 ft.)+ (4 ft. × 7 ft.) + (4 ft. × 7 ft.) + (7 ft. × 2 ft.)+ (7 ft. × 2 ft.)
Work: 2(2 ft. × 4 ft.) + 2(4 ft. × 7 ft.) + 2(7 ft. × 2 ft.)
= 16 ft2 + 56 ft2 + 28 ft2
= 100 ft2
Question 2.
Name of Shape: Rectangular Pyramid
SurfaceArea: (2 ft. × 5 ft.) + ($$\frac{1}{2}$$ × 2ft. × 4ft.)+ ($$\frac{1}{2}$$ × 2 ft. × 4ft.) + ($$\frac{1}{2}$$ × 5 ft. × 4ft.) + $$\frac{1}{2}$$ × 5ft. × 4 ft.)
Work: 2 ft. × 5 ft. + 2($$\frac{1}{2}$$ × 2 ft. × 4 ft.) + 2($$\frac{1}{2}$$ × 5 ft. × 4ft.)
= 10 ft2 + 8 ft2 + 20 ft2 = 38 ft2
Explain the error in each problem below. Assume each box on the grid paper represents a 1 m × 1 m square.
Question 3.
Name of Shape: Rectangular Pyramid, but more specifically a Square Pyramid
Area of Base: 3m × 3m = 9m2
Area of Triangles: 3 m × 4m = 12 m2
SurfaceArea: 9m2 + 12m2 + 12m2 + 12m2 + 12m2 = 57m2
The error in the solution is the area of the triangles. In order to calculate the correct area of the triangles, you must use the correct formula A = $$\frac{1}{2}$$bh. Therefore, the area of each triangle would be 6 m2 and not 12 m2.
Question 4.
Name of Shape: Rectangular Prism or, more specifically. a Cube
Area of Faces: 3m × 3m = 9m2
Surface Area: 9 m2 + 9 m2 + 9 m2 + 9 m2 + 9 m2 = 45m2
The surface area is incorrect because the student did not find the sum of all 6 faces. The solution is shown above only calculates the sum of 5 faces. Therefore, the correct surface area should be 9 m2 + 9 m2 + 9 m2 + 9 m2 + 9 m2 + 9m2 = 54m2 and not 45m2.
Question 5.
Sofia and Ella are both writing expressions to calculate the surface area of a rectangular prism. However, they wrote different expressions.
a. Examine the expressions below, and determine if they represent the same value. Explain why or why not. Sofia’s Expression:
(3 cm × 4 cm) + (3 cm × 4 cm) + (3 cm × 5 cm) + (3 cm × 5 cm) + (4 cm × 5 cm) + (4 cm × 5 cm)
Ella’s Expression:
2(3 cm × 4 cm) + 2(3 cm × 5 cm) + 2(4 cm × 5 cm)
Sofia’s and Ella’s expressions are the same, but Ella used the distributive property to make her expression more compact than Sofia’s.
b. What fact about the surface area of a rectangular prism does Ella’s expression show more clearly than Sofia’s?
A rectangular prism is composed of three pairs of sides with identical areas.
### Eureka Math Grade 6 Module 5 Lesson 17 Exit Ticket Answer Key
Question 1.
Name the shape, and then calculate the surface area of the figure. Assume each box on the grid paper represents a 1 in. × 1 in. square.
Name of Shape: Rectangular Pyramid
Area of Base: 5 in. × 4 in. = 20 in2
Area of Triangles: $$\frac{1}{2}$$ × 4 in. × 4 in. = 8 in2, $$\frac{1}{2}$$ × 5 in. × 4 in. = 10 in2
SurfaceArea: 20 in2 +8 in2 + 8in2 + 10 in2 + 10 in2 = 56 in2
Addition and Subtraction Equations – Round 1:
Directions: Find the value of m in each equation.
Question 1.
m + 4 = 11
m = 7
Question 2.
m + 2 = 5
m = 3
Question 3.
m + 5 = 8
m = 3
Question 4.
m – 7 = 10
m = 17
Question 5.
m – 8 = 1
m = 9
Question 6.
m – 4 = 2
m = 6
Question 7.
m + 12 = 34
m = 22
Question 8.
m + 25 = 45
m = 20
Question 9.
m + 43 = 89
m = 46
Question 10.
m – 20 = 31
m = 51
Question 11.
m – 13 = 34
m = 47
Question 12.
m – 45 = 68
m = 113
Question 13.
m + 34 = 41
m = 7
Question 14.
m + 29 = 52
m = 23
Question 15.
m + 37 = 61
m = 24
Question 16.
m – 43 = 63
m = 106
Question 17.
m – 21 = 40
valuem = 61
Question 18.
m – 54 = 37
m = 91
Question 19.
4 + m = 9
m = 5
Question 20.
6 + m = 13
m = 7
Question 21.
2 + m = 31
m = 29
Question 22.
15 = m + 11
m = 4
Question 23.
24 = m + 13
m = 11
Question 24.
32 = m + 28
m = 4
Question 25.
4 = m – 7
m = 11
Question 26.
3 = m – 5
m = 8
Question 27.
12 = m – 14
m = 26
Question 28.
23.6 = m – 7.1
m = 30.7
Question 29.
14.2 = m – 33.8
m = 48
Question 30.
2.5 = m -41.8
m = 44.3
Question 31.
64.9 = m + 23.4
m = 41.5
Question 32.
72.2 = m + 38.7
m = 33.5
Question 33.
1.81 = m – 15.13
m = 16.94
Question 34.
24.68 = m – 56.82
m = 81.5
Addition and Subtraction Equations – Round 2:
Directions: Find the value of m in each equation.
Question 1.
m + 2 = 7
m = 5
Question 2.
m + 4 = 10
m = 6
Question 3.
m + 8 = 15
m = 7
Question 4.
m + 7 = 23
m = 16
Question 5.
m + 12 = 16
m = 4
Question 6.
m – 5 = 2
m = 7
Question 7.
m – 3 = 8
m = 11
Question 8.
m – 4 = 12
m = 16
Question 9.
m – 14 = 45
m = 59
Question 10.
m + 23 = 40
m = 17
Question 11.
m + 13 = 31
m = 18
Question 12.
m – 23 = 48
m = 25
Question 13.
m + 38 = 52
m = 14
Question 14.
m – 14 = 27
m = 41
Question 15.
m – 23 = 35
m = 58
Question 16.
m – 17 = 18
m = 35
Question 17.
m – 64 = 1
m = 65
Question 18.
6 = m + 3
m = 3
Question 19.
12 = m + 7
m = 5
Question 20.
24 = m + 16
m = 8
Question 21.
13 = m + 9
m = 4
Question 22.
32 = m – 3
m = 35
Question 23.
22 = m – 12
m = 34
Question 24.
34 = m – 10
m = 44
Question 25.
48 = m + 29
m = 19
Question 26.
21 = m + 17
m = 4
Question 27.
52 = m + 37
m = 15
Question 28.
$$\frac{6}{7}$$ + m = $$\frac{4}{7}$$
m = $$\frac{2}{7}$$
Question 29.
$$\frac{2}{3}$$ = m – $$\frac{5}{3}$$
m = $$\frac{7}{3}$$
Question 30.
$$\frac{1}{4}$$ – m = $$\frac{8}{3}$$
m = $$\frac{35}{12}$$
Question 31.
$$\frac{5}{6}$$ = m – $$\frac{7}{12}$$
m = $$\frac{17}{12}$$
Question 32.
$$\frac{7}{8}$$ = m – $$\frac{5}{12}$$
m = $$\frac{31}{24}$$
Question 33.
$$\frac{7}{6}$$ + m = $$\frac{16}{3}$$
m = $$\frac{25}{6}$$
$$\frac{1}{3}$$ + m = $$\frac{13}{15}$$
m = $$\frac{8}{15}$$
|
# At a Glance - The Formal Version
When we graph continuous functions, three things happen:
• We are given a continuous function f and a value c.
• We decide how far we wanted to let f(x) move away from f(c).
• We restrict the values of x until we get what we want, making sure that
• x is the same distance away from c on either side, and that
• we didn't restrict x to just equal c, since then we would find a dot.
Enter the Greek letters—frat bros rejoice.
In symbols,
• We're given a continuous function f and a value c.
• We pick a real number ε > 0 (epsilon) that specifies how far we want to let f(x) move away from f(c).
• We restrict the values of x until we get what we want, ending with c δ
The continuous function guarantee says that no matter what ε > 0 we pick, we'll be able to find a δ > 0.
We use the Greek letters to specify how close various things are to one another. We use the letter ε to specify how close we want f(x) and f(c) to be. The definition of continuity says for any ε we can find an appropriate δ such that if x is within δ of c, we can find our desired value.
We have a great recipe for cooking up δ with the ingredients f, c, and ε. First we combine flour, baking soda, and salt in a bowl, then we...no wait. That's the recipe for Nestle Tollhouse Cookies.
• Write down the inequality f(c) - εf(x) < f(c) + ε and fill in whatever we are given for c, f, and ε.
• Solve the inequality for x.
• Subtract c from all parts of the inequality and find δ.
The super-formal definition of continuity says:
The function f is continuous at c if for any real ε > 0 there exists a real δ > 0 such that if |c| < δ, then |f(x) - f(c)| < ε.
To translate, if f is continuous at c we can pick any real ε > 0 and say we want to have f(x) and f(c) within ε of each other. In symbols, we write this |f(x) - f(c)| < ε.
This is the same thing as saying -εf(x) – f(c) < ε which is the same thing as saying f(c) – εf(x) < f(c) + ε.
In pictures,
Since f is continuous, we have a guarantee that we can find some real δ > 0 such that if x is within δ of c, f(x) is within ε of f(c). In symbols, we say |c| < δ.
But as always, a picture is worth a thousand Greek symbols.
#### Example 1
Let f(x) = 2x, c = 4, and ε = 0.5. Find an appropriate δ > 0, so that if x is within δ of 4, f(x) will be within ε of f(c).
#### Example 2
Let f(x) = x2, c = 2, and ε = 0.1. What is an appropriate value of δ > 0, that will guarantee that if |x – c| < δ, |f(x) – f(c)| < ε?
#### Example 3
Let f(x) = 4x + 1. Find a value of δ such that if |x – 5| < δ, then |f(x) – f(5)| < 0.1.
#### Example 4
Let f(x) = sin(x) Find a value of δ such that if , then .
#### Example 5
Let . Find a value of δ such that if |x – 2| < δ, then |f(x) – 0.5| < 0.005.
#### Example 6
Let . What is a value of δ that will guarantee that if |x – 7| < δ, then ?
#### Exercise 1
For the given continuous function, value of c, and ε, what is f(c) and an appropriate δ that will guarantee the continuity of f?
5x – 2 , c = 3 , and ε = 1.
#### Exercise 2
For the given continuous function, value of c, and ε, find f(c) and a value of δ such that if |xc| < δ, |f(x) – f(c)| < ε.
ex , c = 0 , and ε = 1.
#### Exercise 3
For the given continuous function, value of c, and ε, what is f(c) and a value of δ guaranteed by the continuity of f?
x2 + 2 , c = -2 , and ε = 0.2.
|
# How to Solve Exponential Equations?
An exponential equation is an equation with exponents in which exponents (or) is part of the exponents is variable. Here, you learn more about solving exponential equations problems.
When the power is variable and if it is part of an equation, it is called an exponential equation. It may be necessary to use the relationship between power and logarithm to solve the exponential equations.
## Step-by-step guide to exponential equations
There are three types of exponential equations. They are as follows:
• Equations with the same bases on both sides
• Equations with different bases can be made the same
• Equations with different bases that cannot be made the same
### Exponential equations formulas
When solving an exponential equation, the bases of both sides may be the same or may not be the same. Here are the formulas used in each of these cases.
### Solving exponential equations with the same bases
Sometimes, even though the exponents of both sides are not the same, they can be made the same. To solve exponential equations in each of these cases, we use only the property of equality of exponential equations, by which we equalize the exponentials and solve for the variable.
### Solving exponential equations with different bases
Sometimes the bases on either side of an exponential equation may not be the same (or) cannot be made the same. We solve exponential equations using logarithms when the bases on both sides of the equation are not the same. In such cases, we can do one of the following:
• Convert the exponential equation into the logarithmic form using the formula $$b^x=a⇔log _b\left(a\right)=x$$
• Apply $$log$$ on both sides of the equation and solve for the variable. In this case, we have to use the logarithm property.
### Exponential Equations – Example 1:
solve the equation $$7^x=3$$.
The bases on both sides of the exponential equation are not the same, so must apply $$log$$ on both sides of the exponential equation:
$$log 7^x=log 3$$
Then, use the property of $$log$$: $$log a^m=m \:log a$$
$$x log 7=log 3$$
Now, dividing both sides by $$log 7$$:
$$x=\frac{log 3}{log 7}$$
### Exponential Equations – Example 2:
Solve the equation $$4^{2x-1}=64$$.
The bases are not the same, but we can rewrite $$64$$ as a base of $$4$$ → $$4^3=64$$
Then, rewrite the equation as:
$$4^{2x-1}=4^3$$
With the property of exponential functions, if the bases are the same, the exponents must be equal:
$$2x-1=3 → 2x=3+1 → 2x=4$$
Now, divide each side by $$2$$:
$$x=2$$
## Exercises for Exponential Equations
### Solve exponential equations.
1. $$\color{blue}{\frac{81}{3^{-x}}=3^6}$$
2. $$\color{blue}{5^{3x-2}=125^{2x}}$$
3. $$\color{blue}{5^{2x}=21}$$
4. $$\color{blue}{4^{x-2}=0.125}$$
5. $$\color{blue}{3^{^{2x+1}}=15}$$
1. $$\color{blue}{x=2}$$
2. $$\color{blue}{x=-\frac{2}{3}}$$
3. $$\color{blue}{x=\frac{log\:21}{2\:log\:5}}$$
4. $$\color{blue}{x=\frac{1}{2}}$$
5. $$\color{blue}{x=\frac{log\:5}{2\:log\:3}}$$
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Exam 1 Review Guide
# Exam 1 Review Guide - M10360 Exam 1: Review Guide Chapter 5...
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Unformatted text preview: M10360 Exam 1: Review Guide Chapter 5 5.1: The Natural Logarithm; Differentiation Know the definition: ln( x ) = R x 1 1 t dt ; know what y = ln( x ) looks like and its properties. Know the log rules and how to use them ln( ab ) = ln( a ) + ln( b ) ln( a b ) = ln( a )- ln( b ) ln( a n ) = n ln( a ) Know the derivative d dx [ln( u )] = 1 u u where u is a function of x . Logarithmic Differentiation (take ln of both sides and differentiate implicitly after using log rules to make one side easier). Remember to plug y back in when finished. 5.2: Natural Logarithm: Integration Know how to integrate: R 1 u du = ln( | u | ) + C . Be comfortable with u-substitution to get it into this form. Remember, naming u is the same as naming x if you solve for x . Use long division if the numerator and denominator have same power. Also, you can split up the numerator to make things easier by creating two separate integrals. Know how to integrate the trigonometric functions (see page 337). 5.3: Inverse Functions If f ( a ) = b and f has an inverse, then we know that f- 1 ( b ) = a . Inverse exists if function is 1-1 (for one output, there is only one input). Note x 2 does not have an inverse unless we restrict the domain of x 2 to the the numbers x 0. We can create inverses by restricting domains. If a function is strictly monotonic on its entire domain, it will have an inverse. We can check if a function is monotonic by using the first derivative. If f ( x ) > 0 for all x then f is always increasing (likewise for f ( x ) < 0 and f decreasing). The formula for finding derivatives of inverses is: ( f- 1 ) ( a ) = 1 f ( f- 1 ( a )) so long as the denominator is not 0....
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## Exam 1 Review Guide - M10360 Exam 1: Review Guide Chapter 5...
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## The Number of Points on Two Line Segments
We say that a set is countably infinite if we can pair the elements with set of counting numbers 1, 2, 3, and so on. Believe it or not, the number of positive integers and the number of integers (both negative and positive including 0) have the same number of elements. It is because we can pair them in a one-to-one correspondence such as shown in the below.
As shown on the table, if we continue indefinitely, we know that we can pair each counting number with an integer in a one-to-one correspondence without missing any element.
Using this concept, we show intuitively that the number of points on two line segments is equal even if they have different lengths. We can do this by showing that for each point on segment $\overline{AB}$, there is a corresponding point on segment $\overline{CD}$. » Read more
## Real Numbers: A Summary
For the past two years, we have talked a lot about real numbers. We have talked about integers and its operations (addition, subtraction, multiplication, and division), we have discussed about rational and irrational numbers, and we have talked about their properties, structure, and wonders. In this post, we are going to summarize what we have learned about them.
Figure 1 - The Number Line
The set of real numbers is the collection of all rational and irrational numbers. By convention, real numbers are represented by a line infinitely long where the positive real numbers are situated at the right hand side of 0, while the negative are at the left hand side. It is also important to note that for each point on the number line, there exists a corresponding real number equivalent to it, and for each real number, there is a corresponding point on the line that represents it. » Read more
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# The final exam of a particular class makes up 40% of the
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Re: Tough airthmetic [#permalink]
### Show Tags
11 Dec 2012, 00:40
VeritasPrepKarishma wrote:
gettinit wrote:
Karishma can you please explain your method a bit more? I don't understand how you solved the problem using it? thanks
What is weighted average?
It is average when each value has a different weight. e.g. a group of friends has 10 boys and 20 girls. Average age of boys is 20 years and average age of girls is 17 years. What is the average age of the group?
Here, the average is weighted since we have different number of boys and girls.
We calculate it as follows:
$$W Avg = \frac{20*10 + 17* 20}{10 + 20}$$
What we are doing instinctively here is using weighted average formula which as given below:
$$C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}$$
You need to find the average of C and W is the weight. In the example above, C is age and W is number of boys and girls.
The alligation method, or the scale method as we call it, is based on the weighted averages formula itself:
$$C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}$$
If I re-arrange the formula, I get
$$\frac{W_1}{W_2} = \frac{C_2 - C_{avg}}{C_{avg} - C_1}$$
So I get that weights will be in the same ratio as difference between higher value of C and average value of C and difference between average value of C and lower value of C.
How does this help? Knowing this, we can directly make a diagram and get the answer.
e.g. A group of friends has 10 boys and some girls. Average age of boys is 20 years and average age of girls is 17 years. The average age of the group is 18 years. How many girls are there?
Draw:
Attachment:
Ques1.jpg
On a scale (number line), mark 17 years as age of girls, 18 years as average and 20 years as age of boys. Now, distance between 17 and 18 is 1 and distance between 18 and 20 is 2, The ratio of W1/W2 will be 2:1 (Note, the numbers 1 and 2, give a ratio of 2:1 for girls:boys as seen by the formula)
Since there are 10 boys, there will be 20 girls.
This method is especially useful when you have the average and need to find the ratio of weights.
[/quote]
Karishma, this was very helpful and your way of doing the scale method makes sense because MGMAT doesn't explain it clearly in the foundation book. My only question is your equation
$$C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}$$
If I re-arrange the formula, I get
$$\frac{W_1}{W_2} = \frac{C_2 - C_{avg}}{C_{avg} - C_1}$$
How did you re-arrange the formula to get that? I don't see it. Thanks in advance!
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dcastan2 wrote:
Karishma, this was very helpful and your way of doing the scale method makes sense because MGMAT doesn't explain it clearly in the foundation book. My only question is your equation
$$C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}$$
If I re-arrange the formula, I get
$$\frac{W_1}{W_2} = \frac{C_2 - C_{avg}}{C_{avg} - C_1}$$
How did you re-arrange the formula to get that? I don't see it. Thanks in advance!
$$C_{avg} = \frac{C_1*W_1 + C_2 * W_2}{W_1 + W_2}$$
$$C_{avg}*(W_1 + W_2) = C_1*W_1 + C_2 * W_2$$ (Cross multiplying)
$$C_{avg}*W_1 + C_{avg}*W_2 = C_1*W_1 + C_2 * W_2$$
$$C_{avg}*W_1 - C_1*W_1 = C_2 * W_2 - C_{avg}*W_2$$
$$W_1(C_{avg} - C_1) = W_2(C_2 - C_{avg})$$
$$\frac{W_1}{W_2} = \frac{C_2 - C_{avg}}{C_{avg} - C_1}$$
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 38846 Followers: 7719 Kudos [?]: 105945 [0], given: 11602 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 13 Jun 2013, 03:07 Expert's post 1 This post was BOOKMARKED Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE All DS Mixture Problems to practice: search.php?search_id=tag&tag_id=43 All PS Mixture Problems to practice: search.php?search_id=tag&tag_id=114 _________________ Intern Joined: 29 Apr 2013 Posts: 8 Followers: 0 Kudos [?]: 0 [0], given: 11 Re: Tough airthmetic [#permalink] ### Show Tags 27 Aug 2013, 15:52 VeritasPrepKarishma wrote: anish319 wrote: karishma how did you derive this: Since the weights are in the ratio 3:2, the distance on the scale will be in the ratio 2:3. So x = 82.5% I followed the whole weighted average discussion. what is C2 for the above question Weights are the number of boys and number of girls. Here we need to find the number of girls so we do not have the ratio of weights. We need to find it. What we do have is C1, CAvg and C2. As shown in the diagram above, ratio of distance between C2 and CAvg and distance between CAvg and C1 gives the ratio of weights. C1 = 17 yrs, CAvg = 18 yrs and C2 = 20 yrs Hence distance on the scale between 17 and 18 is 1 and distance on the scale between 18 and 20 is 2. This gives us a ratio of 2:1 for the weights i.e. for the number of girls:number of boys. Remember C is what you want to find the average of. Here average age is 18 yrs. So age is C. Weights is the number of boys/girls. Is there a document which can explain the Mixture concept in detail. I seem to be having issues with it. Intern Joined: 01 Aug 2013 Posts: 10 Location: United States Concentration: Operations, International Business GMAT 1: 710 Q47 V41 GPA: 3.65 Followers: 0 Kudos [?]: 7 [0], given: 0 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 07 Sep 2013, 12:28 I attacked this in a completely different way: We know that the exam makes up 40%, so it's 2/5 of the grade. Moe has 45% for 60% of the grade. We are trying to find out what he needs for those last 2/5ths in order to get 60% as a final average. So, [(45*3)+(2X)]/5 = 60 gives us the number needed for the final 2 'components' (that 40% or 2/5) to make a final average of 60 Solve for X and get 82.5 Senior Manager Joined: 15 Aug 2013 Posts: 314 Followers: 0 Kudos [?]: 63 [0], given: 23 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 24 Aug 2014, 14:17 VeritasPrepKarishma wrote: anish319 wrote: VeritasPrepKarishma wrote: Since the weights are in the ratio 3:2, the distance on the scale will be in the ratio 2:3. So x = 82.5% If you have time, it would be a good idea to be comfortable with the scale method. It will save you a lot of time and energy. I was actually about the above solution....sorry for the confusion. how did you get "Since the weights are in the ratio 3:2, the distance on the scale will be in the ratio 2:3. So x = 82.5%" Oh ok... I thought I was missing something! In the question above, you have the average of marks. So your C is marks. Here, the difference is that weights are given to you and one of the C (i.e. C2) is missing. We know that ratio of weights will be the distance on the number line. So if you look at the diagram above, the ratio of weights is 3:2 (because weightage of mid terms is 60% and weightage of finals is 40%, so 60:40 = 3:2). This means that the distance on the number line should be in the ratio 2:3 (The ratio on the number line flips). So distance between 45 and 60 is 2 units. This means 1 unit is 15/2 = 7.5 on the number line. Now we need to find what 3 units distance is because C2 (i.e. x in the diagram) will be 3 units away from 60. Since 1 unit is 7.5, 3 units will be 22.5. Adding 22.5 to 60, we get 82.5. So x, the missing extreme right value must be 82.5 Hi karishma, I thought that I understood your line method until I read this statement. Whenever we look at the line method (You usually have an average weight and the two ends(weights) of the data given). So in the example with ages, there are 17yr old girls, 20yr old boys and the group is 18, you have a clear number that's above the average and below the average -- which makes complete sense. Two questions: 1) Line method: If I were to use the line method in this example -- we are already TOLD that the middle weight is 40. In addition to that, i'm a little confused because the "middle weight" is actually not in the middle? I'm not sure if I'm explaining this properly but if we are looking to score a 60% (that should be the average of the two ends, correct?) -- we know that we have scored 40% but we don't have the other end (a number higher than 60). How do we find a ratio without knowing that number, which incidentally, is the same number that the question is asking for. 2) If I use the equation: Weight 1 / Weight 2 = Average 2 - Average. Avg / Average. Avg - A1 We know that the weight of the final is 40% which means that the weight of everything but the exam is 100-40 = 60%. So we have w1=40 and w2 = 60 Additionally, we know that we are looking to score a 60% in the class, which is different than the 60% I outlined above. So we have that the Average. Avg = 60. We also know that currently, he has scored 45%, which is on the non final part, therefore, we are looking to find what he needs to score on the Final. Correct? Meaning, we need to find A1 in the equation above. Correct? This seems to get me to 82.5 but it was extremely difficult for me to draw the gap between what weights correspond to what grades. I've read your "quarter wit" documents but it still eludes me a little. Any help would be appreciated. Thanks. Manager Joined: 26 Feb 2015 Posts: 127 Followers: 0 Kudos [?]: 15 [0], given: 43 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 27 May 2015, 13:24 VeritasPrepKarishma wrote: russ9 wrote: I thought that I understood your line method until I read this statement. Whenever we look at the line method (You usually have an average weight and the two ends(weights) of the data given). So in the example with ages, there are 17yr old girls, 20yr old boys and the group is 18, you have a clear number that's above the average and below the average -- which makes complete sense. Two questions: 1) Line method: If I were to use the line method in this example -- we are already TOLD that the middle weight is 40. In addition to that, i'm a little confused because the "middle weight" is actually not in the middle? I'm not sure if I'm explaining this properly but if we are looking to score a 60% (that should be the average of the two ends, correct?) -- we know that we have scored 40% but we don't have the other end (a number higher than 60). How do we find a ratio without knowing that number, which incidentally, is the same number that the question is asking for. 2) If I use the equation: Weight 1 / Weight 2 = Average 2 - Average. Avg / Average. Avg - A1 We know that the weight of the final is 40% which means that the weight of everything but the exam is 100-40 = 60%. So we have w1=40 and w2 = 60 Additionally, we know that we are looking to score a 60% in the class, which is different than the 60% I outlined above. So we have that the Average. Avg = 60. We also know that currently, he has scored 45%, which is on the non final part, therefore, we are looking to find what he needs to score on the Final. Correct? Meaning, we need to find A1 in the equation above. Correct? This seems to get me to 82.5 but it was extremely difficult for me to draw the gap between what weights correspond to what grades. I've read your "quarter wit" documents but it still eludes me a little. Any help would be appreciated. Thanks. Scale method: w1/w2 = (C2 - Cavg)/(Cavg - C1) Cavg is the thing we need to average - marks here w1 and w2 are the weights allotted to marks - 40% to finals and 60% to rest of the tests Question: The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam. What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class? "The final exam of a particular class makes up 40% of the final grade" - this tells us that weight allotted to final exam is 40% and hence 60% is allotted to exams before finals. So we have w1 and w2. This is the tricky part. "Moe is failing the class with an average (arithmetic mean) of 45%" - This means in exams before finals, Moe has 45% marks. "What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class" - To pass, Moe needs average 60% marks i.e. Cavg should be 60%. So what we need is his marks in finals i.e. C2 w1/w2 = (C2 - Cavg)/(Cavg - C1) 60/40 = (C2 - 60)/(60 - 45) This will give you C2. This is still quite unclear to me. I too understand the boys and girls problem just fine, but like the previous poster metioned, I don't understand how we get that upper limit. Any chance you can show it on a number line too? e-GMAT Representative Joined: 04 Jan 2015 Posts: 697 Followers: 180 Kudos [?]: 1551 [0], given: 114 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 27 May 2015, 23:11 The question can be interpreted in an another way Total marks needed to pass We are given that out of total marks(let's assume it to be $$y$$) Moe should receive 60% to pass i.e. 60% of $$y$$ = $$0.6y$$ Final exam marks Since the final exam makes up for 40% of the total marks, it constitutes 40% of $$y$$ $$= 0.4y$$ marks. Let's assume he should get $$x$$% of marks in his final exam to pass the exam i.e. he should get $$x$$% of $$0.4y$$ Rest of the marks The rest 60% of total marks constitute of 60% of $$y$$ $$= 0.6y$$ marks Out of these $$0.6y$$ marks, Moe has got only 45% i.e. 45% of $$0.6y = 0.45 * 0.6y$$ Writing the equation We can write the equation for marks of Moe as Total marks needed by Moe to pass = Rest of the marks + Final exam marks $$0.6y = 0.45 * 0.6y +$$ $$x$$% $$* 0.4y$$ Solving this would give us $$x = 82.5$$% of marks he needs in his final exam to pass the class Hope this helps Regards Harsh _________________ | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com Manager Joined: 21 Jul 2014 Posts: 71 Location: United States WE: Project Management (Non-Profit and Government) Followers: 0 Kudos [?]: 11 [0], given: 58 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 28 May 2015, 08:46 EgmatQuantExpert wrote: The question can be interpreted in an another way Total marks needed to pass We are given that out of total marks(let's assume it to be $$y$$) Moe should receive 60% to pass i.e. 60% of $$y$$ = $$0.6y$$ Final exam marks Since the final exam makes up for 40% of the total marks, it constitutes 40% of $$y$$ $$= 0.4y$$ marks. Let's assume he should get $$x$$% of marks in his final exam to pass the exam i.e. he should get $$x$$% of $$0.4y$$ Rest of the marks The rest 60% of total marks constitute of 60% of $$y$$ $$= 0.6y$$ marks Out of these $$0.6y$$ marks, Moe has got only 45% i.e. 45% of $$0.6y = 0.45 * 0.6y$$ Writing the equation We can write the equation for marks of Moe as Total marks needed by Moe to pass = Rest of the marks + Final exam marks $$0.6y = 0.45 * 0.6y +$$ $$x$$% $$* 0.4y$$ Solving this would give us $$x = 82.5$$% of marks he needs in his final exam to pass the class Hope this helps Regards, Harsh Quite simple way to interpret & arrive at the final solution. Senior Manager Joined: 15 Oct 2015 Posts: 376 Concentration: Finance, Strategy GPA: 3.93 WE: Account Management (Education) Followers: 5 Kudos [?]: 105 [0], given: 211 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 11 Mar 2016, 09:43 rxs0005 wrote: The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam. What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class? A. 75% B. 82.5% C. 85% D. 90% E. 92.5% $$\frac{40}{100} + \frac{60}{100} = \frac{100}{100}$$ when you weight each of these scores with what More actually scored in each, x and 45 respectively u get 60. $$\frac{40x}{100} + \frac{(60*45)}{100} = 60$$ $$\frac{2x}{5} + 18 = 60$$ Manager Joined: 05 Jul 2016 Posts: 50 Location: China Concentration: Finance, Nonprofit GMAT 1: 680 Q49 V33 GMAT 2: 690 Q51 V31 GMAT 3: 710 Q50 V36 GPA: 3.4 Followers: 0 Kudos [?]: 9 [0], given: 129 The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 24 Jul 2016, 20:10 VeritasPrepKarishma wrote: rxs0005 wrote: The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam. What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class? (A) 75% (B) 82.5% (C) 85% (D) 90% (E) 92.5% Using the scale method (called alligation by some), you can very neatly and quickly solve this question of weighted averages. Weightage of mid terms (or whatever) - 60% Weightage of final exams - 40% Marks obtained in mid term - 45% Average required - 60% So marks obtained in finals - x% Now make a diagram like this: Attachment: Ques2.jpg Since the weights are in the ratio 3:2, the distance on the scale will be in the ratio 2:3. So x = 82.5% If you have time, it would be a good idea to be comfortable with the scale method. It will save you a lot of time and energy. Hi Karishma, thank you so much for your hint that reminded me to recall the weighted average method. Here's the general equation: 2*(45%-60%)+3*(X-60%)=0 X=82.5. Rgds, _________________ It's better to burn out than to fade away. Director Joined: 07 Dec 2014 Posts: 666 Followers: 3 Kudos [?]: 136 [0], given: 3 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 25 Jul 2016, 14:00 let x=final exam grade needed .4*x+.6*.45=.6 x=.825=82.5% Senior Manager Status: DONE! Joined: 05 Sep 2016 Posts: 409 Followers: 3 Kudos [?]: 16 [0], given: 283 Re: The final exam of a particular class makes up 40% of the [#permalink] ### Show Tags 01 Dec 2016, 18:47 .60(45)+.40x = 60 27+0.40x=60 0.40x=33 x=82.5 B. Re: The final exam of a particular class makes up 40% of the [#permalink] 01 Dec 2016, 18:47 Go to page Previous 1 2 [ 33 posts ] Similar topics Replies Last post Similar Topics: 4 A shop owner needs to make a 20% profit on a particular item that cost 4 03 May 2017, 05:09 4 In an engineering class that contains 50 students, the final exam 4 25 Apr 2017, 02:11 9 On a particular exam, the boys in a history class averaged 86 points a 10 10 Dec 2016, 10:26 9 A student is to take her final exams in two subjects. The probability 19 22 May 2016, 12:42 6 A dress is marked up 16 2/3% to a final price of$140. What 6 01 Oct 2016, 13:32
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# The final exam of a particular class makes up 40% of the
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# If the solve the problem
Question:
If $y=a \log x+b x^{2}+x$ has its extreme values at $x=1$ and $x=2$, then $(a, b)=$ ____________
Solution:
It is given that, $y=a \log x+b x^{2}+x$ has its extreme values at $x=1$ and $x=2$.
$\therefore \frac{d y}{d x}=0$ at $x=1$ and $x=2$
$y=a \log x+b x^{2}+x$
Differentiating both sides with respect to x, we get
$\frac{d y}{d x}=\frac{a}{x}+2 b x+1$
Now,
$\left(\frac{d y}{d x}\right)_{x=1}=0$
$\Rightarrow a+2 b+1=0$
$\Rightarrow a+2 b=-1$ .....(1)
Also,
$\left(\frac{d y}{d x}\right)_{x=2}=0$
$\left(\frac{d y}{d x}\right)_{x=2}=0$
$\Rightarrow \frac{a}{2}+4 b+1=0$
$\Rightarrow a+8 b=-2 \quad \ldots(2)$
Subtracting (1) from (2), we get
$6 b=-1$
$\Rightarrow b=-\frac{1}{6}$
Putting $b=-\frac{1}{6}$ in $(1)$, we get
$a+2 \times\left(-\frac{1}{6}\right)=-1$
$\Rightarrow a=-1+\frac{1}{3}=-\frac{2}{3}$
Thus, the values of $a$ and $b$ are $-\frac{2}{3}$ and $-\frac{1}{6}$, respectively.
Hence, the ordered pair $(a, b)$ is $\left(-\frac{2}{3},-\frac{1}{6}\right)$
If $y=\operatorname{alog} x+b x^{2}+x$ has its extreme values at $x=1$ and $x=2$, then $(a, b)=$ $\left(-\frac{2}{3},-\frac{1}{6}\right)$
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# How To Find The Hypotenuse, Knowing The Legs
## Video: How To Find The Hypotenuse, Knowing The Legs
A right-angled triangle is a flat figure in which one of the angles is right, that is, it is ninety degrees. The sides of such a triangle are named: hypotenuse and two legs. The hypotenuse is the side of the triangle opposite the right angle, and the legs, respectively, are adjacent to it. The main mathematical game of the parties is played through the Pythagorean theorem, which states that the sum of the squares of the legs is equal to the square of the hypotenuse. It sounds confusing, but it's actually much simpler.
## Instructions
### Step 1
Let the legs have the designation a and b, and the hypotenuse - c. Then, the Pythagorean theorem can be written in the form: (c) in the second degree = (a) in the second degree + (b) in the second degree. Before you can find the value of the hypotenuse, you need to find the squares of the other two sides. Raise the first leg to the second power, then the second. Example: the legs of a right-angled triangle are 3 and 4 centimeters long. Then (4) squared = 16 and (3) squared = 9
### Step 2
After finding the value of the squares of the legs, find their sum. You should not first summarize expressions that are under the sign of the second degree, this will complicate the task and confuse with the answer. Example: 16 + 9 = 25.
### Step 3
Then extract the total from the square root. Since after addition in the above example, the equation is obtained: (c) squared = 25, therefore, the final answer has not yet been received.
Example: If you take the square root of twenty-five, you get five. This is the numerical value of the hypotenuse.
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### Topic 3 – Circular functions and Trigonometry Mathematics
# TOPIC TITLE
1 Study Plan Study plan – Topic 3 – Circular functions & Trigonometry
Objective: On completion of the course formative assessment a tailored study plan is created identifying the lessons requiring revision.
2 Geometry-circles The equation of a circle: to find radii of circles
Objective: On completion of the lesson the student will be able to describe a circle mathematically given its equation or its graph. Additionally, the student will be able to work out the equation of a circle given its centre and radius.
3 Geometry-circles The semicircle: to select the equation given the semi circle and vice versa
Objective: On completion of the lesson the student will be able to sketch a semicircle given its equation and derive the equation of a given semicircle.
4 Geometry-parabola The parabola: to describe properties of a parabola from its equation
Objective: On completion of the lesson the student will be able to predict the general shape and important features of a parabola and then graph the parabola to check the predictions.
5 Circle Geometry Theorem – Equal arcs on circles of equal radii subtend equal angles at the centre. Theorem – Equal angles at the centre of a circle on equal arcs.
Objective: On completion of the lesson the student will be able to prove that ‘Equal arcs on circles of equal radii, subtend equal angles at the centre’, and that ‘Equal angles at the centre of a circle stand on equal arcs.’ They should then be able to use these pro
6 Circle Geometry Theorem – The perpendicular from the centre of a circle to a chord bisects the chord. Theorem – The line from the centre of a circle to the mid-point of the chord is perpendicular to the chord.
Objective: On completion of the lesson the student will be able to prove that ‘The perpendicular from the centre of a circle to a chord bisects the chord.’ and its converse theorem ‘The line from the centre of a circle to the mid-point of the chord is perpendicular’
7 Circle Geometry Theorem – Equal chords in equal circles are equidistant from the centres. Theorem – Chords in a circle which are equidistant from the centre are equal.
Objective: On completion of the lesson the student will be able to prove that equal chords in equal circles are equidistant from the centre.
8 Circle Geometry Theorem – The angle at the centre of a circle is double the angle at the circumference standing on the same arc.
Objective: On completion of the lesson the student will be able to prove that the angle at the centre of a circle is double the angle at the circumference standing on the same arc.
9 Circle Geometry Theorem – Angles in the same segment of a circle are equal.
Objective: On completion of the lesson the student will be able to prove that the angles in the same segment are equal.
10 Circle Geometry Theorem – The angle of a semi-circle is a right angle.
Objective: On completion of the lesson the student will be able to prove that ‘The angle of a semi-circle is a right-angle.’
11 Circle Geometry Theorem – The opposite angles of a cyclic quadrilateral are supplementary.
Objective: On completion of the lesson the student will be able to prove that the opposite angles of a cyclic quadrilateral are supplementary.
12 Circle Geometry Theorem – The exterior angle at a vertex of a cyclic quadrilateral equals the interior opposite angle.
Objective: On completion of the lesson the student will be able to prove that the exterior angle at a vertex of a cyclic quadrilateral equals the interior opposite.
13 Circle Geometry Theorem – The tangent to a circle is perpendicular to the radius drawn to it at the point of contact.
Objective: On completion of the lesson the student will be able to prove that the tangent and the radius of a circle are perpendicular at the point of contact.
14 Circle Geometry Theorem – Tangents to a circle from an external point are equal.
Objective: On completion of the lesson the student will be able to prove that tangents to a circle from an external point are equal.
15 Circle Geometry Theorem – The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.
Objective: On completion of the lesson the student will be able to prove that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.
16 Circle Geometry-chords Theorem – The products of the intercepts of two intersecting chords are equal.
Objective: On completion of the lesson the student will be able to prove that ‘The product of the intercepts of two intersecting chords are equal.’, and use this result to complete questions that require this knowledge.
17 Circle Geometry-tangents Theorem – The square of the length of the tangent from an external point is equal to the product of the intercepts of the secant passing through this point. [Including Alternate Proof]
Objective: On completion of the lesson the student will be able to prove and apply ‘The square of the length of the tangent from an external point is equal to the product of the intercepts of the secant passing through this point ‘, and use this result to complete q
18 Circle Geometry-cyclic quads Theorem – If the opposite angles in a quadrilateral are supplementary then the quadrilateral is cyclic.
Objective: On completion of the lesson the student will be able to prove that a quadrilateral is cyclic using the supplementary angles theorem.
19 Circle Geometry-subtending Theorem – If an interval subtends equal angles at two points on the same side of it, then the end points of the interval and the two points are concyclic.
Objective: On completion of the lesson the student will be able to prove that ‘ If an interval subtends equal angles at two points on the same side of it, then the end points of the interval and the two points are concyclic’, and use this result to complete the ques
20 Circle Geometry Theorem – When circles touch, the line of the centres passes through the point of contact.
Objective: On completion of the lesson the student will be able to prove that ‘ When two circles touch, the line of the centres passes through the point of contact’, and use this result to complete questions that require it.
21 Circle Geometry-non-collinear Theorem – Any three non-collinear points lie on a unique circle whose centre is the point of concurrency of the perpendicular bisectors of the intervals joining these points.
Objective: On completion of the lesson the student will be able to prove that ‘ Any three non-collinear points lie on a unique circle whose centre is the point of concurrency of the perpendicular bisectors of the intervals joining these points’, and use this knowled
22 Trig-reciprocal ratios Reciprocal ratios.
Objective: On completion of the lesson the student will be able to identify and use the reciprocal trigonometric ratios of sine, cosine and tan, that is, the cosecant, secant and cotangent ratios.
23 Trig complementary angles Complementary angle results.
Objective: On completion of the lesson the student will understand how to establish the complementary angle results for the sine and cosine ratios and then how to use these results to solve trig equations.
24 Trig identities Trigonometric identities
Objective: On completion of the lesson the student will be able to simplify trigonometrical expressions and solve trigonometry equations using the knowledge of trig identities.
25 Trig larger angles Angles of any magnitude
Objective: On completion of the lesson the student will be able to find the trigonometric values of angles of any magnitude by assigning angles to the four quadrants of the circle.
26 Trig larger angles Trigonometric ratios of 0°, 90°, 180°, 270° and 360°
Objective: On completion of the lesson the student will learn how to find the Trigonometric Ratios of 0, 90, 180, 270 and 360 degrees.
27 Graph sine Graphing the trigonometric ratios – I Sine curve.
Objective: On completion of the lesson the student will recognise and draw the sine curve exploring changes in amplitude and period.
28 Graph cosine Graphing the trigonometric ratios – II Cosine curve.
Objective: On completion of the lesson the student will know how to recognise and draw the cosine curve exploring changes in amplitude and period.
29 Graphs tan curve Graphing the trigonometric ratios – III Tangent curve.
Objective: On completion of the lesson the student will know how to recognise and draw the tan curve.
30 Graph reciprocals Graphing the trigonometric ratios – IV Reciprocal ratios.
Objective: On completion of the lesson the student will know how to recognise and draw the curves of the reciprocal ratios: cosec, sec and cot.
31 Trig larger angles Using one ratio to find another.
Objective: On completion of the lesson the student will find other trig ratios given one trig ratio and to work with angles of any magnitude.
32 Trig equations Solving trigonometric equations – Type I.
Objective: On completion of the lesson the student will solve simple trig equations with restricted domains.
33 Trig equations Solving trigonometric equations – Type II.
Objective: On completion of the lesson the student will solve trig equations with multiples of theta and restricted domains.
34 Trig equations Solving trigonometric equations – Type III.
Objective: On completion of the lesson the student will solve trig equations with two trig ratios and restricted domains.
35 Polar coordinates Plotting polar coordinates and converting polar to rectangular
Objective: On completion of the lesson the student will understand the polar coordinate system and relate this to the rectangular coordinate system.
36 Polar coordinates Converting rectangular coordinates to polar form
Objective: On completion of the lesson the student will understand the polar coordinate system and report these from rectangular coordinates.
37 Polar coordinates Write and graph points in polar form with negative vectors (Stage 2)
Objective: On completion of the lesson the student will be using negative angles and negative vector lengths.
38 Trigonometry Sin(A+B) etc sum and difference identities (Stage 2)
Objective: On completion of the lesson the student will be using the reference triangles for 30, 45 and 60 degrees with the sum and difference of angles to find additional exact values of trigonometric ratios.
39 Trigonometry Double angle formulas (Stage 2)
Objective: On completion of the lesson the student will derive and use the double angle trig identities.
40 Trigonometry Half angle identities (Stage 2)
Objective: On completion of the lesson the student will derive and use the power reducing formulas and the half angle trig identities.
41 Trigonometry t Formulas (Stage 2)
Objective: On completion of the lesson the student will solve trig equations using the t substitution.
42 Trigonometry-ratios Trigonometric ratios.
Objective: On completion of the lesson the student will be able to identify the hypotenuse, adjacent and opposite sides for a given angle in a right angle triangle. The student will be able to label the side lengths in relation to a given angle e.g. the side c is op
43 Trigonometry-ratios Using the calculator.
Objective: On completion of the lesson the student will be able to use the calculator to find values for the sine, cosine and tangent ratios of acute angles.
44 Trigonometry-ratios Using the trigonometric ratios to find unknown length. [Case 1 Sine].
Objective: On completion of the lesson the student will be able to use the sine ratio to calculate lengths and distances.
45 Trigonometry-ratios Using the trigonometric ratios to find unknown length. [Case 2 Cosine].
Objective: On completion of the lesson the student will be able to use the cosine ratio to find the length of the adjacent side of a right angle triangle.
46 Trigonometry-ratios Using the trigonometric ratios to find unknown length. [Case 3 Tangent Ratio].
Objective: On completion of the lesson the student will be able to use the tangent ratio to calculate the length of the opposite side in a right angle triangle.
47 Trigonometry-ratios Unknown in the denominator. [Case 4].
Objective: On completion of the lesson the student will understand how to use the trig ratios to calculate lengths and distances when the denominator is unknown.
48 Trigonometry-compass Bearings – the compass.
Objective: On completion of the lesson the student will be able to identify compass bearings, compass bearings with acute angles and 3 figure bearings from true north.
49 Trigonometry-elevation Angles of elevation and depression.
Objective: On completion of the lesson the student will be able to identify angles of depression and angles of elevation, and the relationship between them.
50 Trigonometry-practical Trigonometric ratios in practical situations.
Objective: On completion of the lesson the student will be able to use trigonometric ratios to solve problems involving compass bearings and angles of depression and elevation.
51 Trigonometry-ratios Using the calculator to find an angle given a trigonometric ratio.
Objective: On completion of the lesson the student will be capable of using a calculator to find the value of an unknown angle when given a trigonometric ratio.
52 Trigonometry- ratios Using the trigonometric ratios to find an angle in a right-angled triangle.
Objective: On completion of the lesson the student will be able to find the value of an unknown angle in a right angle triangle given the lengths of 2 of the sides.
53 Trigonometry-exact ratios Trigonometric ratios of 30., 45. and 60. – exact ratios.
Objective: On completion of the lesson the student will be able to find the exact sine, cosine and tangent ratios for the angles 30., 45.and 60.
54 Trigonometry-cosine rule The cosine rule to find an unknown side. [Case 1 SAS].
Objective: On completion of the lesson the student will be able to use the cosine rule to find the length of an unknown side of a triangle knowing 2 sides and the included angle.
55 Trigonometry-cosine rule The cosine rule to find an unknown angle. [Case 2 SSS].
Objective: On completion of the lesson the student will be able to find the size of an unknown angle of a triangle using the cosine rule given the lengths of the 3 sides.
56 Trigonometry-sine rule The sine rule to find an unknown side. Case 1.
Objective: On completion of the lesson the student will be able to use the Sine rule to find the length of a particular side when the student is given the sizes of 2 of the angles and one of the sides.
57 Trigonometry-sine rule The sine rule to find an unknown angle. Case 2.
Objective: On completion of the lesson the student will be able to use the sine rule to find an unknown angle when given 2 sides and a non-included angle.
58 Trigonometry-areas The area formula
Objective: On completion of the lesson the student will be able to use the sine formula for finding the area of a triangle given 2 sides and the included angle.
59 Exam Exam – Topic 3 – Circular functions & Trigonometry
Objective: Exam
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# Inverse Variation
Definition:
we can say two quantities are said to be in inverse variation if one quantity increases,then the other quantity decreases or when one quantdity decreases,the other quantity increases.
Where we use this method:
For example 12 men can do a work in 10 days.8 men can do the same work in 15 days. 16 men can do the same work in 7 days.This example shows that more men take less days and less men and number of days are in this variation.
Example 1:
10 workers can build a wall in 12 days.How many days will 15 workers take to build the same wall?
Solution:
Number of workers Number of days
10 12
15 n
Working rule: Here the number of workers and days is i inverse variation. Because if the number of workers got increased then the number of days taken by the workers to finish the work will become reduced.So it comes under this topic.To proceed this problem we have to multiply straight. But in direct variation we have to do cross multiply .
10 x 12 = 15 x n
n = (10 x 12)/15
n = 8
There fore the number of days required to finish the work is 8 days.
Example 2:
In an army camp provisions were there for 500 men for 28 days. If 400 men attended the camp,then how long did the provisions last?
Solution:
Number of Men Number of days
500 28
400 n
Working rule: Here we have two parts number of men and number of days.If the number of men will be increased in this the provisions will be enough for only less number of days.So it comes under this topic.So we have to multiply straight.
500 x 28 = 400 x n
n = (500 x 28) / 400
n = 35 days
There fore the number of days the provisions lasted for 400 men = 35 days.
Inverse variation to Basic math
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Published on September 24th, 2011 In category Division and Multiplication tricks | Education | Maths
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# Finding factors easily using number 9
While appearing for the competitive exams there is always a need to find the factors of the number. Here we are going to learn how to use the number 9 for the finding the factors of a number. I will explain with example.
a. First the simple one: Suppose we need to find the factor of number 44.
Step 1. – Multiply the digits before unit place by 9 ( In our example we will multiply by 4 x 9 = 36
Step 2Then add the unit digit with the digits in front of unit digits ( In our example = 4 + 4 )
Step 3 – So you will note 44 = ( 4 x 9 ) + (4 + 4)
This will always be true for all the numbers
Lets take the 2nd Example for the 107
Step 1. – Multiply the digits before unit place by 9 ( In our example we will multiply by 10 x 9 = 90 )
Step 2 – Then add the unit digit with the digits in front of unit digits ( In our example = 10 + 7 = 17)
Step 3 – So you will note 107 = ( 10 x 9 ) + (10 + 7)
Now you practice this concept with few example given over here
Question- 1
Find factors of 1236
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# Antisymmetric Relation: Definition, Properties, Conditions, Rules, and Examples
An antisymmetric relation is an important concept in mathematics, particularly in set theory and discrete mathematics. This type of relation is defined on a set, where for any two elements a and b, if both a is related to b and b is related to a, then a must be equal to b. Antisymmetric relations are significant in various mathematical disciplines and are essential in understanding order theory and equivalence relations. Important properties of antisymmetric relations include reflexivity, transitivity, and the uniqueness of elements in the relation. Understanding antisymmetric relations is crucial for solving problems in competitive exams like GATE, IIT-JEE, GRE, and other engineering and mathematical entrance tests. This Antisymmetric Relation blog also involves exploring conditions for a relation to be antisymmetric, graph representations of these relations, specific rules to identify them, and practical examples of antisymmetric relations to solidify the concept.
## What is Antisymmetric Relation?
An antisymmetric relation is a type of binary relation defined on a set that exhibits a specific characteristic involving pairs of elements. In simple terms, a relation R on a set A is called antisymmetric if for all pairs of elements a and b in A, whenever a is related to b (i.e., aRb) and b is related to a (i.e., bRa), it must be the case that a = b.
Important Terms:
• Binary Relation: A binary relation R on a set A is a collection of ordered pairs (a,b)(a, b)(a,b) where a and b are elements of A.
• Set A: A collection of distinct objects, considered as an object in its own right. For instance, A={1,2,3}.
• Ordered Pair: A pair of elements (a,b) where the order of elements matters, meaning (a,b) is different from (b,a) unless a = b.
• Relation R: A subset of the Cartesian product A × A, meaning R ⊆ A×A.
Formal Definition:
A relation R on a set A is antisymmetric if:
Example:
Consider the set A = {1,2,3} and the relation R = {(1,1),(2,2),(3,3),(1,2)}. In this case, the relation is antisymmetric because there are no elements a and b such that a ≠ b and both (a,b) and (b,a) belong to R.
## Properties of Antisymmetric Relation
Antisymmetric relations have several important properties and characteristics that make them significant in various areas of mathematics. Here are the important properties of Antisymmetric Relation:
1. Definition and Basic Property
• Property: A relation R on a set A is antisymmetric if ∀a, b∈A, whenever aRb and bRa, then a = b.
• Example: Let A = {1,2,3} and R = {(1,1),(2,2),(3,3),(1,2)}. Here, the relation R is antisymmetric because there are no pairs (a,b) and (b,a) in R where a ≠ b.
2. Reflexivity
• Property: An antisymmetric relation does not have to be reflexive, but if it is reflexive, it means every element is related to itself.
• Example: Consider the relation R on A = {1,2} where R = {(1,1),(2,2)}. This relation is both reflexive (every element is related to itself) and antisymmetric (there are no pairs (a,b) and (b,a) with a≠b).
3. Transitivity
• Property: Antisymmetric relations can be transitive. Transitivity means if aRb and bRc, then aRc.
• Example: Let A = {1,2,3} and R = {(1,2),(2,3),(1,3)}. This relation is antisymmetric (no pairs (a,b) and (b,a) with a≠b) and transitive (if 1R2 and 2R3, then 1R3).
4. Partial Order
• Property: A relation that is antisymmetric, reflexive, and transitive is called a partial order.
• Example: The relation ≤ (less than or equal to) on the set of integers Z is a partial order. For any integers a, b, and c:
• Antisymmetric: If a ≤ b and b ≤ a, then a = b.
• Reflexive: a ≤ a for any a.
• Transitive: If a ≤ b and b ≤ c, then a ≤ c.
5. Graph Representation
• Property: In graph theory, an antisymmetric relation can be represented as a directed graph where there are no two distinct nodes aaa and bbb with both a→b and b→a.
• Example: In the directed graph of the relation R = {(1,2),(2,3),(1,3)} on A = {1,2,3}, there are no edges going both ways between any pair of nodes, which illustrates antisymmetry.
6. Examples and Non-Examples
• Example: Consider the set A = {a,b} and the relation R = {(a,a),(b,b)}. This relation is antisymmetric because it does not contain any pairs (a,b) and (b,a) with a ≠ b.
• Non-Example: Let A = {1,2} and R = {(1,2),(2,1)}. This relation is not antisymmetric because 1R2 and 2R1 but 1≠2.
7. Connection with Other Relations
• Property: Antisymmetric relations are related to partial orders but are not necessarily equivalence relations.
• Example: The relation “is a subset of” (⊆) on the set of all subsets of a given set is antisymmetric. If A⊆B and B⊆A, then A=B. It is also a partial order because it is reflexive and transitive.
Also Read: All Perfect Cube Numbers
## Conditions of Antisymmetric Relation
For a relation R on a set A to be considered antisymmetric, it must satisfy the following condition:
Antisymmetric Condition
A relation R on a set A is antisymmetric if for all elements a and b in A: If aRb and bRa, then a = b.
Conditions for Antisymmetric Relation
1. Pairwise Condition:
• Condition: For any pairs (a,b) and (b,a) in the relation R, if both pairs exist, then aaa must be equal to b.
• Example: In the relation R = {(1,2),(2,1)} on the set A = {1,2}, the pairs (1,2) and (2,1) violate antisymmetry because 1≠2. Therefore, this relation is not antisymmetric.
2. No Bidirectional Pairs:
• Condition: There should be no two distinct elements a and b in A such that both aRb and bRa hold true, unless a=b.
• Example: In the relation R = {(1,1),(2,2),(1,2)} on the set A = {1,2}, there are no instances where both (1,2) and (2,1) are present. Hence, the relation is antisymmetric.
3. Irreflexivity and Reflexivity:
• Condition: While antisymmetry does not require reflexivity, if a relation is both reflexive and antisymmetric, then it is a partial order.
• Example: The “less than or equal to” relation (≤) on integers is both reflexive (every number is less than or equal to itself) and antisymmetric (if a ≤ b and b ≤ a).
4. Graph Representation:
• Condition: In a directed graph representing the relation, there should be no edges going both ways between any two distinct nodes.
• Example: For the relation R = {(1,2),(2,3)} on A = {1,2,3}, the directed graph shows no bidirectional edges between different nodes, which supports antisymmetry.
5. Relation with Other Properties:
• Condition: Antisymmetry alone does not imply reflexivity or transitivity. However, when combined with reflexivity and transitivity, the relation becomes a partial order.
• Example: The subset relation (⊆) on a set of subsets is antisymmetric, reflexive, and transitive, making it a partial order.
## Rules of Antisymmetric Relation
Understanding the rules governing antisymmetric relations helps in identifying and working with such relations in different mathematical contexts. Here are the important rules of antisymmetric relations:
Rules of Antisymmetric Relations
1. Rule of Non-Existence of Bidirectional Relations:
In an antisymmetric relation R on a set A, if (a,b) ∈ R and (b,a) ∈ R, then a must be equal to b.
• Example: If R = {(1,2),(2,1)} on the set {1,2}, this relation is not antisymmetric because 1 ≠ 2. For a relation to be antisymmetric, it cannot contain such pairs where the elements are distinct.
1. Rule of Reflexivity:
Antisymmetry does not require reflexivity, but if a relation is reflexive and antisymmetric, then it must be a partial order.
• Example: The “less than or equal to” (≤) relation on integers is both reflexive and antisymmetric, thus making it a partial order.
1. Rule of Transitivity:
Antisymmetry does not imply transitivity. However, if a relation is antisymmetric and transitive, then it could be a partial order if it is also reflexive.
• Example: In the set {1,2,3}, consider the relation R = {(1,2),(2,3),(1,3)}. This relation is antisymmetric (no two distinct elements have bidirectional pairs) and transitive. If it were reflexive, it would be a partial order.
1. Rule of Directed Graph Representation:
In the directed graph of an antisymmetric relation, there should be no two distinct nodes with both edges a→b and b→a.
• Example: For the relation R={(1,2),(2,3)} on {1,2,3}, the directed graph shows no bidirectional edges between distinct nodes, which satisfies antisymmetry.
1. Rule of Checking Non-Existence of Pairs:
To check if a relation is antisymmetric, examine whether any pairs (a,b) and (b,a) exist in the relation where a ≠ b. If such pairs exist, the relation is not antisymmetric.
• Example: For the relation R = {(2,1),(1,2),(3,3)} on {1,2,3}, the presence of both (2,1) and (1,2) with 2 ≠ 1 indicates that the relation is not antisymmetric.
1. Rule of Combining with Other Properties:
While antisymmetry alone is a distinct property, combining it with reflexivity and transitivity results in a partial order.
• Example: The subset relation (⊆) on a set of subsets is antisymmetric, reflexive (every set is a subset of itself), and transitive (if A⊆B and B⊆C, then A⊆C), making it a partial order.
## Examples of Antisymmetric Relation
Here are several examples of antisymmetric relations, covering a range of contexts to illustrate the concept:
1. Less Than or Equal To Relation (≤)
• Set: Z (set of all integers)
• Relation: ≤ (less than or equal to)
• Explanation: The relation a ≤ b is antisymmetric because if a ≤ b, then a must be equal to b. For example, if 3 ≤ 5 and 5 ≤ 3, then 3 must equal 5, which is not possible.
2. Subset Relation (⊆)
• Set: The power set of {1,2 (i.e., {∅,{1},{2},{1,2}}
• Relation: ⊆ (subset of)
• Explanation: The relation A⊆B is antisymmetric because if A⊆B and B⊆A, then A must be equal to B. For example, if {1}⊆{1,2} and {1,2}⊆{1}, then {1} must equal {1,2}, which is not possible.
3. Equal to Relation (=)
• Set: Any set, e.g., {a,b,c}
• Relation: = (equality)
• Explanation: The relation a = b is antisymmetric because if a = b and b = a, then a must be equal to b, which is always true.
4. Divisibility Relation
• Set: N (set of natural numbers)
• Relation: / (divides)
• Explanation: The relation a/b (a divides b) is antisymmetric because if a/b and b/a, then aaa must be equal to b. For example, if 2/6 and 6/2, then 2 must equal 6, which is not true.
5. Matrix Representation
• Set: The set of all 2×2 matrices
• Relation: A ≤ B if A is less than or equal to B in terms of matrix entry-wise comparison.
• Explanation: If A ≤ B and B ≤ A, then A must be equal to B, satisfying antisymmetry in matrix comparisons.
6. Order on a Set
• Set: {1,2,3}
• Relation: R = {(1,2),(2,2),(3,3),(1,3)}
• Explanation: This relation is antisymmetric because there are no pairs (a,b) and (b,a) where a ≠ b.
7. Parent-Child Relationship (in a Family Tree)
• Set: The set of people in a family
• Relation: Parent of
• Explanation: If A is a parent of B and B is a parent of A, then A and B must be the same person, which is impossible in a family tree, hence the relation is antisymmetric.
## FAQs
What is an antisymmetric relation?
An antisymmetric relation is a binary relation where if (a, b) and (b, a) are both in the relation, then a must equal b. In other words, if two different elements are related in both directions, then they must be the same element.
What is an example of an asymmetric relation?
Example of an asymmetric relation: “Is less than” between numbers. If 3 is less than 5, then 5 cannot be less than 3.
What is an example of an antisymmetric function?
An antisymmetric function is a function where if f(x) = f(y), then x = y. For example, the function f(x) = x3 is antisymmetric because if x3 = y3, then x = y.
What is the difference between symmetric and antisymmetric?
Symmetric relations mean that if A is related to B, then B is also related to A. Antisymmetric relations mean that if A is related to B and B is related to A, then A must equal B.
This was all about the “Antisymmetric Relation”. For more such informative blogs, check out our Study Material Section, or you can learn more about us by visiting our Indian exams page.
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Maths Without Limits
Opening Young Minds to Endless Possibilities
# 5.1.3 How many ways can I sort, match and order things?
The next most important mathematical skills after counting are sorting, matching and ordering – grouping things together that are similar and arranging things in order of size. The activities in this section explore these ideas. You will need various collections of objects such as cutlery, socks, shells & pebbles, coloured counters etc. Again, the emphasis should be as much on speaking the correct language as on doing the task.
Skill: Choose how to sort a range of objects and explain how you have sorted them.
Suggested language: What do you see? Can you sort them? How did you do it?
Skill: Sort a range of objects by given criteria (size, shape, colour etc).
Suggested language: Can we sort these by colour? Can we sort them by size? Can we do it a different way? Can we sort them by shape? Is that possible?
Skill: Match similar objects (eg) from the beach/garden/woods.
Suggested language: What sets of things can we find? How can we sort them? Can we match things that are similar?
Skill: Match sets of objects (eg socks) and make decisions about what belongs or does not belong to a set.
Suggested language: Can we find all the matching pairs? Are there any odd ones?
Skills: Arrange objects in order according to length, thickness or height. Use a variety of objects to help compare length and height. Use the language of measure to describe what we have found out – long, short, tall, wide, narrow, thick and thin.
Suggested language: Which is the longest? Which is the shortest? Can we arrange them in order of size from longest to shortest? Can we arrange them in a different order? What would happen if we put the thickest ones first and the thinnest ones last? Etc
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1. Consider the function
$$f(x)=\left{\begin{array}{l} \frac{1}{x-1} \text { if } x \neq 1 \ 0 \text { if } x=1 \end{array}\right.$$(a) Does $\lim {x \rightarrow 5} f(x)$ exist? If $s 0$, what is it? Try and establish the validity of your answer formally using an epsilon-delta argunent. If it exists, does it equal $f(5)$ ? Is this function continuous at the point $x=5$ ? (b) Does $\lim {x \rightarrow 1} f(x)$ exist? If so, what is it? Try and establish the validity of your answer formally using an epsilon-delta arguneut. If it exists, does it equal $f(1)$ ? Is this function continuous at the point $x=1$ ?
(c) Is this function contimuous?
Solution:
Consider the function
$$f(x)= \begin{cases}\frac{1}{x-1} & \text { if } x \neq 1 \ 0 & \text { if } x=1\end{cases}$$
Part (a)
First of all, note that
$$\lim {x \rightarrow 5} f(x)=\lim {x \rightarrow 5}\left(\frac{1}{x-1}\right)=\frac{\lim {x \rightarrow \pi} 1}{\lim {x \rightarrow 5}(x-1)}=\frac{1}{4}=f(5) .$$
Thus we know that (i) $\lim {x \rightarrow 5} f(x)=\frac{1}{4}$, and (ii), the function is continuous at the point $x=5$. However, we also want to show that $\lim {x \rightarrow 5} f(x)=\frac{1}{4}$ formally, using an epsilon-delta proof. Note that
$$d(5, x)=\sqrt{(x-5)^{2}}=|x-5|,$$
and
$$d(f(5), f(x))=\sqrt{(f(x)-f(5))^{2}}=|f(x)-f(5)|=\left|f(x)-\frac{1}{4}\right| .$$
Suppose that we want $d(f(5), f(x))<\varepsilon$, where $\varepsilon>0$. Any problems will be caused by small values of $\varepsilon>0$. (Try and explain why this is the case.) Thus we can focus on the cases where
$$d(f(5), f(x))=\left|f(x)-\frac{1}{4}\right|=\left|\frac{1}{(x-1)}-\frac{1}{4}\right|$$
Note that
$$d(f(5), f(x))<\varepsilon \Longleftrightarrow\left|\frac{1}{(x-1)}-\frac{1}{4}\right|<\varepsilon_{.}$$ This requires that both $\frac{1}{(x-1)}-\frac{1}{4}<\varepsilon$ and $-\left(\frac{1}{(x-1)}-\frac{1}{4}\right)\frac{4}{1+4 \epsilon} \Longleftrightarrow x>\frac{4}{(1+4 c)}+1 \Longleftrightarrow x>1+\frac{4}{(1+4 \epsilon)} $$Note that$$ \left(\frac{1}{(x-1)}-\frac{1}{4}\right)<\varepsilon \Longleftrightarrow \frac{1}{(x-1)}-\frac{1}{4}>-\varepsilon $$We have$$ \frac{1}{(x-1)}-\frac{1}{4}>-\varepsilon \Longleftrightarrow \frac{1}{(x-1)}>\frac{1}{4}-\varepsilon \Longleftrightarrow \frac{1}{(x-1)}>\frac{1-4 \varepsilon}{4} . $$This requires that$$ x-1<\frac{4}{1-4 \epsilon} \Longleftrightarrow x<\frac{4}{(1-4 \epsilon)}+1 \Longleftrightarrow x<1+\frac{4}{(1-4 \epsilon)} $$Thus we know that$$ d(f(5), f(x))<\varepsilon \Longleftrightarrow 1+\frac{4}{(1+4 c)}<x<1+\frac{4}{(1-4 \epsilon)} $$Suppose that we waut d(5, x)<\delta. This reyuires that$$ d(5, x)<\delta \Longleftrightarrow|x-5|<\delta \Longleftrightarrow-\delta<x-5<\delta \Longleftrightarrow 5-\delta<x<5+\delta$\$
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# Reflection of a Point in the y-axis
We will discuss here about reflection of a point in the y-axis.
Reflection in the line x = 0 i.e., in the y-axis.
The line x = 0 means the y-axis.
Let P be a point whose coordinates are (x, y).
Let the image of P be P’ in the y-axis.
Clearly, P’ will be similarly situated on that side of OY which is opposite to P. So, the x-coordinates of P’ will be – x while its y-coordinates will remain same as that of P.
The image of the point (x, y) in the y-axis is the point (-x, y).
Symbolically, My (x, y) = (-x, y)
Rules to find the reflection of a point in y-axis:
(i) Change the sign of abscissa i.e. x-coordinate.
(ii) Retain the ordinate i.e., y-coordinate.
Therefore, when a point is reflected in the y-axis, the sign of its abscissa changes.
Examples:
(i) The image of the point (3, 4) in the y-axis is the point (-3, 4).
(ii) The image of the point (-3, -4) in the y-axis is the point (-(-3), -4) i.e., (3, -4).
(iii) The image of the point (0, 7) in the y-axis is the point (0, 7).
(iv) The image of the point (-6, 5) in the y-axis is the point (-(-6), 5) i.e., (6, 5).
(v) The reflection of the point (5, 0) in the y-axis = (-5, 0) i.e., My (5, 0) = (-5, 0)
Solved example to find the reflection of a point in the y-axis:
Find the points onto which the points (11, -8), (-6, -2) and (0, 4) are mapped when reflected in the y-axis.
Solution:
We know that a point (x, y) maps onto (-x, y) when reflected in the y-axis. So, (11, -8) maps onto (-11, -8); (-6, -2) maps onto (6, -2) and (0, 4) maps onto (0, 4).
Reflection
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# 2015 AIME I Problems/Problem 6
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$. Find the degree measure of $\angle BAG$.
$[asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label("A",A,dir(0)); label("B",B,dir(75)); label("C",C,dir(90)); label("D",D,dir(105)); label("E",E,dir(180)); label("F",F,dir(225)); label("G",G,dir(260)); label("H",H,dir(280)); label("I",I,dir(315)); [/asy]$
## Solution 1
Let $O$ be the center of the circle with $ABCDE$ on it.
Let $x$ be the degree measurement of $\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ in circle $O$
and $y$ be the degree measurement of $\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}$ in circle $C$.
$\angle ECA$ is, therefore, $5y$ by way of circle $C$ and $$\frac{360-4x}{2}=180-2x$$ by way of circle $O$. $\angle ABD$ is $180 - \frac{3x}{2}$ by way of circle $O$, and $$\angle AHG = 180 - \frac{3y}{2}$$ by way of circle $C$.
This means that:
$$180-\frac{3x}{2}=180-\frac{3y}{2}+12$$
which when simplified yields $$\frac{3x}{2}+12=\frac{3y}{2}$$ or $$x+8=y$$ Since: $$5y=180-2x$$ and $$5x+40=180-2x$$ So: $$7x=140\Longleftrightarrow x=20$$ $$y=28$$ $\angle BAG$ is equal to $\angle BAE$ + $\angle EAG$, which equates to $\frac{3x}{2} + y$. Plugging in yields $30+28$, or $\boxed{058}$.
## Solution 2
Let $m$ be the degree measurement of $\angle GCH$. Since $G,H$ lie on a circle with center $C$, $\angle GHC=\frac{180-m}{2}=90-\frac{m}{2}$.
Since $\angle ACH=2 \angle GCH=2m$, $\angle AHC=\frac{180-2m}{2}=90-m$. Adding $\angle GHC$ and $\angle AHC$ gives $\angle AHG=180-\frac{3m}{2}$, and $\angle ABD=\angle AHG+12=192-\frac{3m}{2}$. Since $AE$ is parallel to $BD$, $\angle DBA=180-\angle ABD=\frac{3m}{2}-12=$$\overarc{BE}$.
We are given that $A,B,C,D,E$ are evenly distributed on a circle. Hence,
$\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$$=\frac{\angle DBA}{3}=\frac{m}{2}-4$
Here comes the key: Draw a line through $C$ parallel to $AE$, and select a point $X$ to the right of point $C$.
$\angle ACX$ = $\overarc{AB}$ + $\overarc{BC}$ = $m-8$.
Let the midpoint of $\overline{HG}$ be $Y$, then $\angle YCX=\angle ACX+\angle ACY=(m-8)+\frac{5m}{2}=90$. Solving gives $m=28$
The rest of the solution proceeds as in solution 1, which gives $\boxed{058}$
## Solution 3
Let $\angle GAH = \varphi \implies \overset{\Large\frown} {GH} = 2\varphi \implies$ $$\overset{\Large\frown} {EF} = \overset{\Large\frown} {FG} = \overset{\Large\frown} {HI} = \overset{\Large\frown} {IA} = 2\varphi \implies$$ $$\angle AGH = 2\varphi, \angle ACE = 10 \varphi.$$
$$BD||GH \implies \angle AJB = \angle AGH = 2 \varphi.$$ $$\triangle AHG: \hspace{10mm} \angle AHG = \beta = 180^\circ – 3 \varphi.$$ $\hspace{10mm} \triangle ABJ: \hspace{10mm} \angle BAG + \angle ABD = \alpha + \gamma = 180^\circ + 2 \varphi.$
Let arc $\overset{\Large\frown} {AB} = 2\psi \implies$
$\angle ACE = \frac {360^\circ – 8 \psi}{2}= 180^\circ – 4 \psi, \angle ABD = \gamma =\frac {360^\circ – 6 \psi}{2} =180^\circ – 3 \psi.$ $\gamma – \beta = 3(\varphi – \psi) = 12^\circ \implies \psi = \varphi – 4^\circ \implies 10 \varphi = 180^\circ – 4(\varphi – 4^\circ) \implies 14 \varphi = 196^\circ \implies \varphi = 14^\circ.$
Therefore $\gamma = 180^\circ – 3 \cdot (14^\circ – 4^\circ) = 150^\circ \implies \alpha = 180^\circ + 2 \cdot 14^\circ – 150^\circ = \boxed{\textbf{058}}.$
2015 AIME I (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
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# 3 Lines On Top Of Each Other In Math Means What Is Angle of Elevation?
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## What Is Angle of Elevation?
Definition of elevation angle: First we define the elevation angle. Let O and P be two points such that point P is at a higher level. Let OA and PB be horizontal lines through O and P respectively. If an observer is at O and the point P is the object under consideration, then the line OP is called the line of sight of the point P and the angle AOP, between the line of sight and the horizontal line OA, is known as to elevation angle of point P as seen from O. If an observer is at P and the object under consideration is at O, then the angle BPO is known as the angle of depression of O as seen from P .
Elevation angle formula: The formula we use for angle elevation is also known as angle of elevation. We can measure the angle of the sun relative to a right angle by the elevation of the angle. The horizon line drawn from the measurement angle to the sun at right angles is the elevation. Using opposite, hypotenuse and adjacent in a right triangle we can find the angle. elevation Of the right triangle the sin is opposite divided by the hypotenuse; cosine is adjacent divided by the hypotenuse; the tangent is opposite divided by adjacent. To understand the angle of elevation we will do a little
Elevation angle problems. Suppose the height of a tower is 100 sqrt(3) meters given. And we must find the elevation of the angle if its top from a point 100 meters away from its foot. So first we collect information, we know that the height of the given tower is 100 sqrt3 and the distance from the foot of the tower is 100m. Let’s take (theta) the angular elevation of the top of the tower…we’ll use the trigonometric relationship that contains base and perpendicular. This relationship is tangent. Using the tangent to the right triangle we have,
tan (theta) = perpendicular / adjacent
tan (theta) = 100 squared (3)/100 = squared (3).
tan (theta) = tan 60
theta = 60 degrees.
So the elevation angle will be 60 degrees
Example: The angle of elevation of the top of the tower from a point on the ground, which is 30 meters from the foot of the tower, is 30 degrees. Find the height of the tower.
Solution: Let AB be the top A of the tower h meters high and let C be a point on the ground such that the elevation of the angle from the top A of the tower AB is 30 degrees.
In the triangle ABC we have an angle C = 30 degrees and base BC = 30 and we have to find the perpendicular AB. Therefore, we use those trigonometric ratios that contain a base and a perpendicular. It is clear that this relationship is tangential. So we take the tangent of angle C.
In triangle ABC, taking the tangent of angle C, we have
so C = AB/AC
so 30 = AB/AC
1/sqrt(3) = h/30
h = 30/sqrt(3) meters = 10 sqrt(3) meters.
Therefore, the height of the tower is 10 square meters (3) meters.
## Question about 3 Lines On Top Of Each Other In Math Means
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# Solving Multi-Step Inequalities
## Presentation on theme: "Solving Multi-Step Inequalities"— Presentation transcript:
Solving Multi-Step Inequalities
Algebra 1 ~ Chapter 6-3 Solving Multi-Step Inequalities
Inequalities that contain more than one operation require more than one step to solve.
Use inverse operations to undo the operations in the inequality one at a time. ALWAYS check your solution!
Example 1 - Solve the inequality and graph the solutions.
Check, b = 10 45 + 2b > 61 45 + 2(10) > 61 > 61 65 > 61 45 + 2b > 61 – –45 2b > 16 b > 8 Check, b = 6 45 + 2b > 61 45 + 2(6) > 61 > 61 57 > 61 2 4 6 8 10 12 14 16 18 20
Ex. 2 - Solve the inequality and graph the solutions.
– –8 Check, y = -8 8 – 3y ≥ 29 8 – 3(-8) ≥ 29 ≥ 29 32 ≥ 29 –3y ≥ 21 by a negative # y ≤ –7 –10 –8 –6 –4 –2 2 4 6 8 10 –7
Check, x = -7 -12 ≥ 3x + 6 -12 ≥ 3(-7) + 6 -12 ≥ -21 + 6 -12 ≥ -15
Ex. 3 - Solve the inequality and graph the solutions. Check your answer. –12 ≥ 3x + 6 Check, x = -7 -12 ≥ 3x + 6 -12 ≥ 3(-7) + 6 -12 ≥ -12 ≥ -15 –12 ≥ 3x + 6 – – 6 –18 ≥ 3x –6 ≥ x or x ≤ -6 –10 –8 –6 –4 –2 2 4 6 8 10
Ex. 4 - Solve the inequality and graph the solutions. Check your answer.
Check, x = -13 –5 –5 x + 5 < –6 > 3 x < –11 > 3 –20 –12 –8 –4 –16 –11 4 > 3
Example 5 - Solve the inequality and graph the solutions.
Check, x = 2 –4(2 – x) ≤ 8 -4(2 – 2) ≤ 8 -4(0) ≤ 8 0 ≤ 8 −4(2 – x) ≤ 8 –8 + 4x ≤ 8 x ≤ 4 –10 –8 –6 –4 –2 2 4 6 8 10
Example 6 - Solve the inequality and graph the solutions
Example 6 - Solve the inequality and graph the solutions. Check your answer. 3 + 2(x + 4) > 3 3 + 2(x + 4) > 3 Check, x = 0 3 + 2(x + 4) > 3 3 + 2(0 + 4) > 3 3 + 2(4) > 3 3 + 8 > 3 11 > 3 3 + 2x + 8 > 3 2x + 11 > 3 – 11 – 11 2x > –8 x > –4 –10 –8 –6 –4 –2 2 4 6 8 10
Lesson Review Solve each inequality and graph the solutions. x ≤ –4
Assignment Worksheet 6-3 Pages 335-336 #’s 16-28 (evens),
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