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# Modular Arithmetic in the Garden
Sowing Zucchini Using Modular Math
Math in the classroom was often joked about by our peers, while attending college, for lacking in real world application; however, we often apply the skill sets we learned in our daily lives. Just the other day we encountered an opportunity to apply modular arithmetic when sowing beans, cucumbers and zucchini. The beans and cucumbers were planted without trouble, but we did not have enough zucchini seeds to plant two entire beds with correct spacing. Since we were already in the field, and only few seeds shy we decided to simply alter the spacing to plant out one whole bed, without using all the seeds on hand.
### What is Modular Arithmetic?
If you can divide, you already know modular arithmetic, so breathe easy. Modular arithmetic is all about the remainder of a simple division problem. As a quick example, we will look at converting 27 to its' equivalent integer, whole number, in modulo 6. The first step is to decide which order to divide: $27\div 6$ or $6\div 27$. The easy rule to remember is that the modulo number will also be the denominator, that is, the bottom of the equation. So, now we have the following:
$27\div 6=4$ rem $3$.
Therefore, 27 is equivalent to 3 modulo 6, often written as $27\equiv3$ mod $6$. Counting in modular arithmetic is simple as well. For example, the integers included in modular 7 would be: $0, 1, 2, 3, 4, 5, 6$. Notice that the numbers never include the modular number. The Art of Problem Solving clearly states this rule: "for a natural number $n$ that is greater than 1, the modulo $n$ residues are the integers that are whole numbers less than $n$:
$0, 1, 2, \ldots, n-1.$"
Being able to perform division problems and count in modulo systems are the only techniques needed to follow along with this real-world example of modular arithmetic. To read further about modular arithmetic, here are a few resources:
### The Problem...
Zucchini requires a six-inch spacing between plants when direct seeding into a garden plot. The farm purchased four Costata Romanesco packets of seeds from Sow True Seed containing 12 seeds each for a total of 48 seeds. Each garden plot is approximately 30 feet long. Initially we set out to sow two 30-foot garden plots with the zucchini seeds, however, this would have required 60 seeds per bed for a total of 120 seeds needed. Since we only had 48 seeds and wanted to get a bed planted right away, we altered our spacing to eight inches between seeds, ensuring we could plant the entire length of the bed. Khori, having the Bachelor of Science in Applied Mathematics, was put on the task of laying down the measuring tape and placing seeds every eight inches (Khori knows her multiples of 8 better than William). However, once we placed the measuring tape down near our sowing furrow, we noticed every foot reset - for example, instead of 13" the measuring tape displays 1". The measuring tape "resetting" every foot meant we would have to count 1, 2, 3, 4, ..., 8 and repeat for every seed.
Modulo 12 Ruler
### The Solution...
We want to sow a zucchini seed every eight inches which still requires knowing multiples of 8: 0, 8, 16, 24, 32, 40, 48, ... . Having previous knowledge of modular math, Khori quickly noticed that we could use modular math in order to quickly determine seed placement. The multiples of 8 include 0, 8, 16, 24, 32, 40, 48, ... . The ruler "resets" back to 0 at every 12 inches, therefore the ruler is in modulo 12.
#### Converting the Multiples of 8 to Modulo 12
Both 0 and 8 are less than 12, so these two numbers do not need to be converted. However, the remaining multiples of 8 are larger than 12 (e.g. 16, 24, ...), therefore these numbers will require a little work. Once we notice a pattern in the converted numbers, we can stop as the pattern will repeat.
16 mod 12: 16 - 12 = 4 or 16$\div$12 = 1 remainder 4
24 mod 12: 24 - 2*(12) = 0 or 24$\div$12 = 2 remainder 0
32 mod 12: 32 - 2*(12) = 8 or 32$\div$12 = 2 remainder 8
40 mod 12: 40 - 3*(12) = 4 or 40$\div$12 = 3 remainder 4
Starting with 0 and 8, we now have the multiples of 8 converted into modular 12: 0, 8, 4, 0, 8, 4, ... . Notice the pattern "0, 8, 4" repeats. The ruler being modular 12 implies that all multiples of 12 are equivalent to 0. Therefore, we know to sow the first seed at the beginning of the ruler (0 inches), followed by the second seed at eight-inches. The third seed is located at the next "4" mark following the "8" where the previous seed is located. The image below starts at the "8" in the 0-8-4 pattern. Notice the seed placement 8, 4, 0 (48 mod 12 $\equiv$ 0).
Thanks to modular math the zucchini seeds were sowed in no time at all!
Sowing Zucchini Using Mod 12 |
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### Transcript of 11.2 The Law of Sines - Sam Houston State kws006/Precalculus/5.3_Law_of_Sines_files/S&Z 11.2...
• 894 Applications of Trigonometry
11.2 The Law of Sines
Trigonometry literally means measuring triangles and with Chapter 10 under our belts, we aremore than prepared to do just that. The main goal of this section and the next is to developtheorems which allow us to solve triangles that is, find the length of each side of a triangleand the measure of each of its angles. In Sections 10.2, 10.3 and 10.6, weve had some experiencesolving right triangles. The following example reviews what we know.
Example 11.2.1. Given a right triangle with a hypotenuse of length 7 units and one leg of length4 units, find the length of the remaining side and the measures of the remaining angles. Expressthe angles in decimal degrees, rounded to the nearest hundreth of a degree.
Solution. For definitiveness, we label the triangle below.
b = 4
a
c=
7
To find the length of the missing side a, we use the Pythagorean Theorem to get a2 + 42 = 72
which then yields a =
33 units. Now that all three sides of the triangle are known, there areseveral ways we can find using the inverse trigonometric functions. To decrease the chances ofpropagating error, however, we stick to using the data given to us in the problem. In this case, thelengths 4 and 7 were given, so we want to relate these to . According to Theorem 10.4, cos() = 47 .Since is an acute angle, = arccos
(47
)radians. Converting to degrees, we find 55.15. Now
that we have the measure of angle , we could find the measure of angle using the fact that and are complements so + = 90. Once again, we opt to use the data given to us in theproblem. According to Theorem 10.4, we have that sin() = 47 so = arcsin
(47
have 34.85.
A few remarks about Example 11.2.1 are in order. First, we adhere to the convention that a lowercase Greek letter denotes an angle1 and the corresponding lowercase English letter represents theside2 opposite that angle. Thus, a is the side opposite , b is the side opposite and c is the sideopposite . Taken together, the pairs (, a), (, b) and (, c) are called angle-side opposite pairs.Second, as mentioned earlier, we will strive to solve for quantities using the original data given inthe problem whenever possible. While this is not always the easiest or fastest way to proceed, it
1as well as the measure of said angle2as well as the length of said side
• 11.2 The Law of Sines 895
minimizes the chances of propagated error.3 Third, since many of the applications which requiresolving triangles in the wild rely on degree measure, we shall adopt this convention for the timebeing.4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handleany given right triangle problem, but what if the triangle isnt a right triangle? In certain cases,we can use the Law of Sines to help.
Theorem 11.2. The Law of Sines: Given a triangle with angle-side opposite pairs (, a),(, b) and (, c), the following ratios hold
sin()
a=
sin()
b=
sin()
c
or, equivalently,
a
sin()=
b
sin()=
c
sin()
The proof of the Law of Sines can be broken into three cases. For our first case, consider thetriangle 4ABC below, all of whose angles are acute, with angle-side opposite pairs (, a), (, b)and (, c). If we drop an altitude from vertex B, we divide the triangle into two right triangles:4ABQ and 4BCQ. If we call the length of the altitude h (for height), we get from Theorem 10.4that sin() = hc and sin() =
ha so that h = c sin() = a sin(). After some rearrangement of the
last equation, we get sin()a =sin()c . If we drop an altitude from vertex A, we can proceed as above
using the triangles 4ABQ and 4ACQ to get sin()b =sin()c , completing the proof for this case.
a
b
c
A C
B
ac
A C
B
Q
h
b
c
A C
B
Q
h
For our next case consider the triangle 4ABC below with obtuse angle . Extending an altitudefrom vertex A gives two right triangles, as in the previous case: 4ABQ and 4ACQ. Proceedingas before, we get h = b sin() and h = c sin() so that sin()b =
sin()c .
a
b
c
A
B
C
a
b
c
A
B
C
Q
h
3Your Science teachers should thank us for this.4Dont worry! Radians will be back before you know it!
• 896 Applications of Trigonometry
Dropping an altitude from vertex B also generates two right triangles, 4ABQ and 4BCQ. Weknow that sin() = h
c so that h = c sin(). Since = 180 , sin() = sin(), so in fact,
we have h = c sin(). Proceeding to 4BCQ, we get sin() = ha so h = a sin(). Putting this
together with the previous equation, we get sin()c =sin()a , and we are finished with this case.
a
b
c
A
B
CQ
h
The remaining case is when 4ABC is a right triangle. In this case, the Law of Sines reduces tothe formulas given in Theorem 10.4 and is left to the reader. In order to use the Law of Sines tosolve a triangle, we need at least one angle-side opposite pair. The next example showcases someof the power, and the pitfalls, of the Law of Sines.
Example 11.2.2. Solve the following triangles. Give exact answers and decimal approximations(rounded to hundredths) and sketch the triangle.
1. = 120, a = 7 units, = 45 2. = 85, = 30, c = 5.25 units
3. = 30, a = 1 units, c = 4 units 4. = 30, a = 2 units, c = 4 units
5. = 30, a = 3 units, c = 4 units 6. = 30, a = 4 units, c = 4 units
Solution.
1. Knowing an angle-side opposite pair, namely and a, we may proceed in using the Law of
Sines. Since = 45, we use bsin(45) =7
sin(120) so b =7 sin(45)sin(120) =
7
63 5.72 units. Now that
we have two angle-side pairs, it is time to find the third. To find , we use the fact that thesum of the measures of the angles in a triangle is 180. Hence, = 180 120 45 = 15.To find c, we have no choice but to used the derived value = 15, yet we can minimize thepropagation of error here by using the given angle-side opposite pair (, a). The Law of Sines
gives us csin(15) =7
sin(120) so that c =7 sin(15)sin(120) 2.09 units.
5
2. In this example, we are not immediately given an angle-side opposite pair, but as we havethe measures of and , we can solve for since = 180 85 30 = 65. As in theprevious example, we are forced to use a derived value in our computations since the only
5The exact value of sin(15) could be found using the difference identity for sine or a half-angle formula, but that
becomes unnecessarily messy for the discussion at hand. Thus exact here means7 sin(15)sin(120) .
• 11.2 The Law of Sines 897
angle-side pair available is (, c). The Law of Sines gives asin(85) =5.25
sin(65) . After the usual
rearrangement, we get a = 5.25 sin(85)
sin(65) 5.77 units. To find b we use the angle-side pair (, c)which yields bsin(30) =
5.25sin(65) hence b =
5.25 sin(30)sin(65) 2.90 units.
a = 7
b 5.72
c 2.09 = 120 = 15
= 45
a 5.77
b 2.90
c = 5.25
= 85 = 65
= 30
Triangle for number 1 Triangle for number 2
3. Since we are given (, a) and c, we use the Law of Sines to find the measure of . We start
with sin()4 =sin(30)
1 and get sin() = 4 sin (30) = 2. Since the range of the sine function is
[1, 1], there is no real number with sin() = 2. Geometrically, we see that side a is just tooshort to make a triangle. The next three examples keep the same values for the measure of and the length of c while varying the length of a. We will discuss this case in more detailafter we see what happens in those examples.
4. In this case, we have the measure of = 30, a = 2 and c = 4. Using the Law of Sines,we get sin()4 =
sin(30)2 so sin() = 2 sin (30
) = 1. Now is an angle in a triangle whichalso contains = 30. This means that must measure between 0 and 150 in orderto fit inside the triangle with . The only angle that satisfies this requirement and hassin() = 1 is = 90. In other words, we have a right triangle. We find the measure of to be = 180 30 90 = 60 and then determine b using the Law of Sines. We findb = 2 sin(60
)sin(30) = 2
3 3.46 units. In this case, the side a is precisely long enough to form a
unique right triangle.
a = 1c = 4
= 30
a = 2c = 4
b 3.46
= 30
= 60
Diagram for number 3 Triangle for number 4
5. Proceeding as we have in the previous two examples, we use the Law of Sines to find . In thiscase, we have sin()4 =
sin(30)3 or sin() =
4 sin(30)3 =
23 . Since lies in a triangle with = 30
,
• 898 Applications of Trigonometry
we must have that 0 < < 150. There are two angles that fall in this range and havesin() = 23 : = arcsin
(23 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Trigonometric Identities and Equations
## Based on amplitude, frequency, and horizontal and vertical translations.
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Practice Trigonometric Identities and Equations
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Trigonometric Identities and Equations
Your math teacher has decided to give you a quiz to see if you recognize how to combine changes to graphs of sine and cosine functions. You recall that you've learned about shifting graphs, as well as stretching/dilating them. But now your teacher wants to see if you know how to combine both of these effects into one graph. She gives you the equation:
$f(x) = 3 + 7 \sin(4(x + \frac{\pi}{2}))$
and asks you to plot the equation, and then identify what each part of the above equation does to change the graph.
Can you accomplish this task?
Read on, and at the conclusion of this Concept, you'll know how to plot this equation and identify which parts of it make changes to the graph.
### Guidance
In other Concepts, you learned how to translate and dilate sine and cosine waves both horizontally and vertically. Combining all the information learned, the general equations are: $y=D \pm A \cos(B(x \pm C))$ or $y=D \pm A \sin(B(x \pm C))$ , where $A$ is the amplitude, $B$ is the frequency, $C$ is the horizontal translation, and $D$ is the vertical translation.
Recall the relationship between period, $p$ , and frequency, $B$ .
$p=\frac{2\pi}{B} \ \text{and} \ B=\frac{2\pi}{p}$
With this knowledge, we should be able to sketch any sine or cosine function as well as write an equation given its graph.
#### Example A
Given the function: $f(x) = 1 + 2 \sin(2(x + \pi))$
a. Identify the period, amplitude, and frequency.
b. Explain any vertical or horizontal translations present in the equation.
c. Sketch the graph from $-2\pi$ to $2\pi$ .
Solution: a. From the equation, the amplitude is 2 and the frequency is also 2. To find the period we use:
$p=\frac{2\pi}{B} \rightarrow p=\frac{2\pi}{2}=\pi$
So, there are two complete waves from $[0, 2\pi]$ and each individual wave requires $\pi$ radians to complete.
b. $D = 1$ and $C = -\pi$ , so this graph has been translated 1 unit up, and $\pi$ units to the left.
c. To sketch the graph, start with the graph of $y = \sin(x)$
Translate the graph $\pi$ units to the left (the $C$ value).
Next, move the graph 1 unit up ( $D$ value)
Now we can add the dilations. Remember that the “starting point” of the wave is $-\pi$ because of the horizontal translation. A normal sine wave takes $2\pi$ units to complete a cycle, but this wave completes one cycle in $\pi$ units. The first wave will complete at 0, then we will see a second wave from 0 to $\pi$ and a third from $\pi$ to $2\pi$ . Start by placing points at these values:
Using symmetry, each interval needs to cross the line $y = 1$ through the center of the wave.
One sine wave contains a “mountain” and a “valley”. The mountain “peak” and the valley low point must occur halfway between the points above.
Extend the curve through the domain.
Finally, extend the minimum and maximum points to match the amplitude of 2.
#### Example B
Given the function: $f(x)=3+3 \cos \left(\frac{1}{2}(x-\frac{\pi}{2}\right))$
a. Identify the period, amplitude, and frequency.
b. Explain any vertical or horizontal translations present in the equation.
c. Sketch the graph from $-2\pi$ to $2\pi$ .
Solution: a. From the equation, the amplitude is 3 and the frequency is $\frac{1}{2}$ . To find the period we use:
$period=\frac{2\pi}{\frac{1}{2}}=4\pi$
So, there is only one half of a cosine curve from 0 to $2\pi$ and each individual wave requires $4\pi$ radians to complete.
b. $D = 3$ and $C=\frac{\pi}{2}$ , so this graph has been translated 3 units up, and $\frac{\pi}{2}$ units to the right.
c. To sketch the graph, start with the graph of $y = \cos(x)$
Adjust the amplitude so the cosine wave reaches up to 3 and down to negative three. This affects the maximum points, but the points on the $x-$ axis remain the same. These points are sometimes called nodes.
According to the period, we should see one of these shapes every $4\pi$ units. Because the interval specified is $[-2\pi, 2\pi]$ and the cosine curve “starts” at the $y-$ axis, at (0, 3) and at $2\pi$ the value is -3. Conversely, at $-2\pi$ , the function is also -3.
Now, shift the graph $\frac{\pi}{2}$ units to the right.
Finally, we need to adjust for the vertical shift by moving it up 3 units.
#### Example C
Find the equation of the sinusoid graphed here.
Solution: First of all, remember that either sine or cosine could be used to model these graphs. However, it is usually easier to use cosine because the horizontal shift is easier to locate in most cases. Therefore, the model that we will be using is $y=D \pm A \cos(B(x \pm C))$ .
First, if we think of the graph as a cosine function, it has a horizontal translation of zero. The maximum point is also the $y-$ intercept of the graph, so there is no need to shift the graph horizontally and therefore, $C = 0$ . The amplitude is the height from the center of the wave. If you can’t find the center of the wave by sight, you can calculate it. The center should be halfway between the highest and the lowest points, which is really the average of the maximum and minimum. This value will actually be the vertical shift, or $D$ value.
$D=center=\frac{60+-20}{2}=\frac{40}{2}=20$
The amplitude is the height from the center line, or vertical shift, to either the minimum or the maximum. So, $A=60-20=40$ .
The last value to find is the frequency. In order to do so, we must first find the period. The period is the distance required for one complete wave. To find this value, look at the horizontal distance between two consecutive maximum points.
On our graph, from maximum to maximum is 3.
Therefore, the period is 3, so the frequency is $B=\frac{2\pi}{3}$ .
We have now calculated each of the four parameters necessary to write the equation. Replacing them in the equation gives:
$y=20+40 \cos \frac{2\pi}{3}x$
If we had chosen to model this curve with a sine function instead, the amplitude, period and frequency, as well as the vertical shift would all be the same. The only difference would be the horizontal shift. The sine wave starts in the middle of an upward sloped section of the curve as shown by the red circle.
This point intersects with the vertical translation line and is a third of the distance back to -3. So, in this case, the sine wave has been translated 1 unit to the left . The equation using a sine function instead would have been: $y=20+40 \sin \left(\frac{2\pi}{3}(x+1) \right)$
### Vocabulary
Trigonometric General Equations: The trigonometric general equations are equations allowing for trig functions with arbitrary constants added or subtracted both inside the argument of the function as well as outside the function, along with allowing for the trig function and/or its argument to be multiplied by a constant.
### Guided Practice
1. Identify the amplitude, period, frequency, maximum and minimum points, vertical shift, and horizontal shift of $y=2+3 \sin(2(x-1)).$
2. Identify the amplitude, period, frequency, maximum and minimum points, vertical shift, and horizontal shift of $y=-1+ \sin \left(\pi (x+\frac{\pi}{3}\right)).$
3. Identify the amplitude, period, frequency, maximum and minimum points, vertical shift, and horizontal shift of $y=\cos (40(x-120))+5.$
Solutions:
1. This is a sine wave that has been translated 1 unit to the right and 2 units up. The amplitude is 3 and the frequency is 2. The period of the graph is $\pi$ . The function reaches a maximum point of 5 and a minimum of -1.
2. This is a sine wave that has been translated 1 unit down and $\frac{\pi}{3}$ radians to the left. The amplitude is 1 and the period is 2. The frequency of the graph is $\pi$ . The function reaches a maximum point of 0 and a minimum of -2.
3. This is a cosine wave that has been translated 5 units up and 120 radians to the right. The amplitude is 1 and the frequency is 40. The period of the graph is $\frac{\pi}{20}$ . The function reaches a maximum point of 6 and a minimum of 4.
### Concept Problem Solution
With your advanced knowledge of sinusoidal equations, you can identify in the equation:
$f(x) = 3 + 7 \sin(4(x + \frac{\pi}{2}))$
The vertical shift of the graph is 3 units up. The amplitude of the graph is 7. The horizontal shift of the graph is $\frac{\pi}{2}$ units to the left. The frequency is 4.
Since the frequency is 4, the period can be calculated:
$p = \frac{2\pi}{f}\\p = \frac{2\pi}{4}\\p = \frac{\pi}{2}\\$
This means that the graph takes $\frac{\pi}{2}$ units to make one complete cycle.
The graph of this equation looks like this:
### Practice
For each equation below, identify the period, amplitude, frequency, and any vertical/horizontal translations.
1. $y=2-4\cos(\frac{2}{3}(x-3))$
2. $y=3+\frac{1}{2}\sin(\frac{1}{2}(x-\pi))$
3. $y=1+5\cos(4(x+\frac{\pi}{2}))$
4. $y=4-\cos(2(x+1))$
5. $y=3+2\sin(x-4)$
Graph each of the following equations from $-2\pi$ to $2\pi$ .
1. $y=1-3\sin(\frac{1}{3}(x-\pi))$
2. $y=5+\frac{1}{2}\sin(\frac{1}{2}(x-2))$
3. $y=2+\cos(4(x+\frac{\pi}{2}))$
4. $y=4+2\cos(2(x+3))$
5. $y=2-3\sin(x-\frac{3\pi}{2})$
Find the equation of each sinusoid.
### Vocabulary Language: English
Trigonometric General Equations
Trigonometric General Equations
The trigonometric general equations are $y=D \pm A \cos(B(x \pm C))$ or $y=D \pm A \sin(B(x \pm C))$, where $A$ is the amplitude, $B$ is the frequency, $C$ is the horizontal translation, and $D$ is the vertical translation. |
offers hundreds of practice questions and video explanations. Go there now.
# Special Quadrilaterals on the GRE
Today, we’ll cover special quadrilaterals and the facts that you should know. These will allow you to answer questions quickly and understand the basic nature of these shapes.
## 1) A Typical Parallelogram
All parallelograms have the BIG FOUR properties:
1. Opposite sides are parallel—AB // CD, AD // BC
2. Opposite angles are congruent—∠BAD = ∠BCD, ∠ABC = ∠CDA
3. Opposite sides are congruent—AB = CD, AD = BC
4. Diagonals bisect each other—BE = ED, AE = EC
• All parallelograms have all four of those properties
• Any quadrilateral with any one of those properties has to be a parallelogram and has to have the other three.
## 2) A Typical Rectangle
A rectangle is a special kind of parallelogram.
• All rectangles have the BIG FOUR parallelogram properties.
• Equiangular quadrilateral—that is, four 90° angles—only the rectangle has this property.
• The bisecting diagonals are also congruent—FH = GJ
• Having the BIG FOUR plus congruent diagonals guarantees a shape is a rectangle.
NOTE: irregular quadrilateral can have congruent diagonals if the diagonals do not bisect each other.
## 3) A Typical Rhombus
A rhombus is a special kind of parallelogram.
• All rhombuses have the BIG FOUR parallelogram properties.
• Equilateral quadrilateral — that is, four each sides—LM = MN = NP = PL—only the rhombus has this property.
• The bisecting diagonals are also perpendicular—MQN = 90°
• Having the BIG FOUR plus perpendicular diagonals guarantees a shape is a rhombus.
NOTE: irregular quadrilateral can have perpendicular diagonals if the diagonals do not bisect each other.
## 4) The Square
There is an infinity of possible parallelogram shapes.
There is an infinity of possible rectangle shapes.
There is an infinity of possible rhombus shapes.
There is only one possible square shape—we can make that shape bigger or smaller, but the shape stays the same.
A square is a special kind of parallelogram, a special kind of rectangle, and a special kind of rhombus. It is the only rectangle that is also a rhombus, and vice versa.
A square has all parallelogram properties, all rectangle properties, and all rhombus properties.
To prove that a shape is a square, you would have to be able to prove that it is a parallelogram, and prove that it is a rectangle and prove that it is a rhombus.
By the way, students who use Magoosh GRE improve their scores by an average of 8 points on the new scale (150 points on the old scale.) Click here to learn more. |
# Calculation of Circumference
by Brenda Scottsdale, Demand Media
Knowing how to calculate the distance around a circle, or its circumference, has many practical implications, such as in designing circular objects, working with crafts or figuring out how much fencing you will need to encircle your hot tub. Although learning geometry may seem difficult at first, you just need to memorize one formula and apply some basic math. (1 & 2)
## What is Pi?
Mathematicians use Pi, the 16th letter of the Greek alphabet, to denote the ratio of the circumference of a circle to its diameter. Pi is an unknowable value. It is an irrational number, meaning that its digits never end or repeat. It is about 3.1416, but for some engineering calculations with low tolerances, a much more exact number is needed. If you use a scientific calculator, the value for pi has far more than 3 digits.
## Learn the Formula
The mathematical formula used to calculate a circle's circumference is C = pi*d. In this formula, the C stands for circumference; pi, is approximately 3.1416, the * symbol tells you to multiply and the d stands for diameter, which is a line segment drawn from the circle's circumference, passing through its center. Therefore, to figure out the circumference of a circle, you need to know its diameter or have information that could lead you to extrapolate its diameter.
## Use the Diameter
In classic geometry problems, mathematics teachers give just one piece of information -- a circle’s diameter, and ask students to calculate the circumference. You must remember that pi is roughly 3.1416, as they typically don’t remind you of this information. For example, if a teacher gives tells you that a circle's diameter is 5, you can calculate its circumference by multiplying 5 by 3.1416. The circumference of the circle in this example is, therefore, 15.71. Remember to write the units, such as inches, feet or yards, in your final answer.
## Use the Radius
Sometimes you have to calculate the diameter before you can work the problem. If you are given the radius of the circle instead of its diameter, just multiple the radius by 2 because twice the radius is equal to the diameter, or 2r=d. For example, If the radius of the circle is 2, its diameter is 4 and so its circumference is 12.56, because C also equals 2r*pi. It works the same for fractions. If the radius is 16.2 then the diameter is 32.4, so the circumference is 101.79.
#### About the Author
Brenda Scottsdale is a licensed psychologist, a six sigma master black belt and a certified aerobics instructor. She has been writing professionally for more than 15 years in scientific journals, including the "Journal of Criminal Justice and Behavior" and various websites.
#### Photo Credits
• Jupiterimages/Brand X Pictures/Getty Images |
December 1, 2022
# What is 2/3 times 4
### what is 2/3 times 4?
2/3 times 4 is 8/9. 2/3 multiplied by 4 is equal to 8/9. This can be seen by multiplying the numerator by 2 and the denominator by 3.
### what is 2/3 times 1/4?
2/3 times 1/4 is equal to 2/12. To find this, you can either multiply the numerators and denominators by the same number, or divide both numerator and denominator by the same number.
### What is 2/3 times 4 Fraction form?
what is 2/3 times 4 as a fraction
### What is 1/4 times 2/3 in fraction form?
1/4 times 2/3 equals 1/12.
This calculation allows us to evenly divide a fraction by another. To do
this, we multiply the numerator of the first fraction by the denominator of the second fraction, and vice versa. So, in this example, we multiply 1 by 2/3 to equal 1/12.
### what is 3/4 cup times 2?
4/3 divided by 3/2
Understand that the answer to 4/3 divided by 3/2 is equal to 2/9.
To do this, we multiply the numerator of the first fraction by the denominator of the second fraction, and vice versa. So, in this example, we multiply 4 by 3 to equal 12. We also multiply 3 by 2 to equal 6. Then we add these numbers together to equal 18. And finally, we divide 18 by 9 to equal 2. So the answer is 2/9.
It’s also worth noting that we could have just as easily flipped the fraction, so that it would be 3/2 divided by 4/3. The answer in this case would still be 2/9.
### what is 2/3 times 4/5 times 1/3?
2/3 times 4/5 times 1/3 equals 8/15. To calculate this, we first need to convert all the fractions to decimals. We can do this by using a calculator, or if you are familiar with the long division method, you can also use that. We will use the long division method for this example.
To begin, let’s convert 2/3 to a decimal. To do this, we divide the numerator (the top number) by the denominator (the bottom number). We get a decimal value of 0.666667.
Now let’s convert 4/5 to a decimal. Again, we divide the numerator by the denominator and get a decimal value of 0.8.
Finally, let’s convert 1/3 to a decimal. We divide the numerator by the denominator and get a decimal value of 0.333333.
With all of our fractions converted to decimals, we can now multiply them together. The decimal value we get is 0.3333333, which is equal to 8/15.
So, in summary, 2/3 times 4/5 times 1/3 equals 8/15. To calculate this, we converted all of the fractions to decimals and then multiplied them together
### what is 2/3 times 4/5?
2/3 times 4/5 is equal to 8/15. To get this answer, you can multiply the numerators and denominators by 2, which is the same as multiplying each fraction by 1/2. To do this, you multiply 8 times . Since 15 divided by 3 is 5, multiplying by 2 gives you . Therefore, 8/15 is equal to 2/3 times 4/5.
### how to divide fractions?
To divide fractions, you need to invert the divisor and multiply the two fractions. This is sometimes called the “flip and multiply” method for dividing fractions. For example, if you want to divide the fraction 2/3 by 1/4, you would first invert the divisor (1/4) to 4/1. You would then multiply 2/3 by 4/1 to get 8/3, which is the answer to your division problem. To divide fractions by whole numbers, you can use the “long division” method that you may have learned in school when you were younger.
### What is formula for dividing fractions?
The easiest way to divide fractions is to use a calculator, and input the division as a normal division problem. However, if you need to do it by hand, you can use the following steps:
-First, convert the division into a multiplication problem by flipping the second fraction (that is, make the denominator of this fraction the numerator of the original fraction). This will also create a new denominator.
-Next, multiply the first fraction by this new denominator. This is your answer to the division problem.
### What is fraction calculation chart?
Fraction calculation is used to calculate the fraction of one number with another. It can be used to divide a whole unit into parts or calculate a percentage. The steps for finding a fraction calculation are as follows:
1) Determine the whole unit value. This is the number that will be divided by the fraction being calculated.
2) Identify the value of the fraction being calculated. This is the number that will be divided into the whole unit.
3) Calculate the fraction by dividing the fraction value into the whole unit value.
4) Express the fraction as a decimal value by moving the decimal point to the left or right until you have an equivalent fraction. Be sure to include any repeating numbers.
5) Convert the decimal value to a percentage by multiplying it by 100. This will allow you to express the fractional value as a percentage of the whole unit.
### how to subtract fractions?
To subtract fractions, we need to find a common denominator between the two fractions. The common denominator is the lowest number that both of the two fractions have in common. For example, if we are subtracting 1/4 and 3/5, the common denominator is 20 because 4 and 5 both go into 20. To find the common denominator, we can either list out the multiples of each number or use a LCD (least common denominator) calculator.
Once we have the common denominator, we can subtract the fractions by changing them both into equivalent fractions with the common denominator. For example, if we are subtracting 1/4 and 3/5 and our common denominator is 20, we would change 1/4 to 5/20 and 3/5 to 15/20. Then, we can subtract the fractions by subtracting the numerators and keeping the same denominator, so 5/20 – 15/20 = -10/20. This can be simplified to -1/2.
### how to multiply fractions?
Multiplying fractions is a very useful concept in mathematics. It can be used to simplify complicated calculations involving fractions. The method of multiplying fractions is as follows:
To multiply two fractions, take the product of the numerators and the product of the denominators. This will give you the answer in simplest form.
For example, let’s take the fractions 6/4 and 2/3.
To multiply these two fractions:
Take the product of the numerators, which is 6 x 2 = 12.
Take the product of the denominators, which is 4 x 3 = 12.
The answer to this problem is 12/12, which reduces down to the simplest form of 1. This means that 6/4 x 2/3 is equal to 1. This is the simplified ways for multiplying fractions.
Happy
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# How do you complete the square to solve x^2 + 5x + 6 = 0?
Jun 16, 2015
You can complete the square by first getting the form ${x}^{2} + k x = h$.
${x}^{2} + 5 x = - 6$
Then just add and subtract a certain value that is equal to ${\left(\frac{k}{2}\right)}^{2}$. Just remember that the function is ${x}^{2} + k x$, so $k$ may be negative, but the added ${\left(\frac{k}{2}\right)}^{2}$ will always be positive.
${x}^{2} + 5 x + {\left(\frac{5}{2}\right)}^{2} = {\left(\frac{5}{2}\right)}^{2} - 6$
${\left(x + \frac{5}{2}\right)}^{2} = \frac{25}{4} - \frac{24}{4} = \frac{1}{4}$
${\left(x + \frac{5}{2}\right)}^{2} - \frac{1}{4} = 0$
graph{(x+5/2)^2 - 1/4 [-10, 10, -5, 5]}
If you wanted to solve this:
${\left(x + \frac{5}{2}\right)}^{2} = \frac{1}{4}$
$x + \frac{5}{2} = \pm \sqrt{\frac{1}{4}}$
Thus:
$x + \frac{5}{2} = \pm \sqrt{\frac{1}{4}}$
$x = \pm \left(\sqrt{\frac{1}{4}}\right) - \frac{5}{2}$
$x = \pm \left(\frac{1}{2}\right) - \frac{5}{2}$
$x = \frac{1}{2} - \frac{5}{2} = - 2$
$x = - \frac{1}{2} - \frac{5}{2} = - 3$
${\left(- 2\right)}^{2} + 5 \left(- 2\right) + 6 = 4 - 10 + 6 = 0$
${\left(- 3\right)}^{2} + 5 \left(- 3\right) + 6 = 9 - 15 + 6 = 0$
You could just factor, though...
$\left(x + 2\right) \left(x + 3\right) = {x}^{2} + 2 x + 3 x + 6 = {x}^{2} + 5 x + 6$ |
# Solve this
Question:
If $e^{x}+e^{y}=e^{x+y}$, prove that $\frac{d y}{d x}=-\frac{e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}$ or, $\frac{d y}{d x}, e^{y-x}=0$
Solution:
We are given with an equation $\mathrm{e}^{x}+\mathrm{e}^{y}=\mathrm{e}^{x+y}$, we have to prove that $\frac{d y}{d x}=\frac{-e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}$ by using the given
equation we will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to $x$, we get,
$e^{x}+e^{y} \frac{d y}{d x}=e^{(x+y)}\left[1+\frac{d y}{d x}\right]$
$\frac{d y}{d x}\left[e^{y}-e^{x+y}\right]=e^{x+y}-e^{x}$
$\frac{d y}{d x}=\frac{e^{x+y}-e^{x}}{e^{y}-e^{x+y}}$
$\frac{d y}{d x}=\frac{-e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}$ |
## The centrifugal force
In this section, we derive an expression for the centrifugal force on a body moving in a circle of radius r with an angular velocity Q, as illustrated in figure 3.5.
### Acceleration in a circular motion
Let us first consider what forces are involved in a body that is moving in a circle with a uniform speed. Newton's first law of motion says that if a body is left to its own devices it will either remain stationary or move in a straight line with a constant speed. If a body is undergoing uniform circular motion (for example, a child riding on a carousel or, more germanely for us, a person standing still on Earth and so rotating around Earth's axis once per day), then that body must be accelerating, with the acceleration directed toward the axis of rotation.
Let us suppose that in a small time, St, the body moves through a small angle SO, so that the angular velocity is X = SO/St. The speed of the body is a constant,
Figure 3.5. A body moving in a circle is constantly changing its direction and so accelerating. The acceleration is directed toward the center of the circle and has magnitude v2/r, where v is the speed of the body and r is the radius of the circle.
Figure 3.5. A body moving in a circle is constantly changing its direction and so accelerating. The acceleration is directed toward the center of the circle and has magnitude v2/r, where v is the speed of the body and r is the radius of the circle.
v = rX = r 50/5t, but its direction is changing such that at all times, the direction of motion is perpendicular to the radius; thus, if at the initial time the position vector is r1, then its velocity, v1, is in the perpendicular direction, and a short time later the velocity v2 is perpendicular to its new position vector, r2.
As can be seen in figure 3.5, the angle through which the radius vector has moved is related to the distance moved by
Because the velocity is always perpendicular to the radius, the direction of the velocity must move through the same angle as the radius and we have
The acceleration of the body is equal to the rate of change of the velocity; using equation 3.14, we obtain
where the last equality uses the fact that v = r86/8t. That is to say, the acceleration of a body undergoing uniform circular motion is directed along the radius of the circle and toward its center and has magnitude v2 , a r = r = X r. (3.16)
This kind of acceleration is called centripetal acceleration, and the associated force causing the acceleration (for there must be a force) is called the centripetal force and is directed inward, toward the axis of rotation.
### Forces in the rotating frame of reference
Now let us consider the motion from the point of view of an observer undergoing uniform circular motion. An illuminating case to consider is that of a satellite or space station in orbit around Earth; when an astronaut goes for a space walk from the space station, she appears to be weightless, with no forces whatsoever acting upon her (figure 3.6). Now in fact, the astronaut is undergoing circular motion around Earth, and is therefore accelerating, and the force providing the acceleration is Earth's gravity.
a) Stationary or Inertial Frame
(toftofasWnaut
(toftofasWnaut
b) Rotating Frame of Reference gravity ^-
centrifugal force
Figure 3.6. An astronaut orbiting Earth. Panel a views the motion in a stationary frame of reference, in which Earth's gravitational force provides the centripetal force that causes the astronaut to orbit Earth. Panel b views the situation from the astronaut's frame of reference, in which the gravitational force is exactly balanced by the centrifugal force and the astronaut feels weightless.
But from the point of view of an observer rotating with the astronaut, the astronaut is stationary and therefore no net forces are acting upon her. Now we know that gravity is acting, pulling her toward Earth, and we say that this force is balanced by another force, centrifugal force, which is pushing her out. The forces exactly balance, so the astronaut appears weightless; that is, in the rotating frame, the centripetal gravitational force pulling her toward Earth is exactly balanced by the centrifugal force pushing her out. Because the centripetal force has magnitude X2r, the centrifugal force must have this magnitude also, and we conclude that in a rotating frame of reference, there appears to be an additional force, the centrifugal force, which acts to accelerate a body along a radius, outward from the axis of rotation. The magnitude of the centrifugal force is X2r per unit mass, where r is the distance from the axis of rotation.
There is no centrifugal force in inertial frames of reference. However, given that we live on a rotating planet, it is extremely convenient to describe motions on Earth from a rotating frame of reference in which Earth's surface is stationary, and in this frame it appears that there is a centrifugal force that is trying to fling us into space. Rather fortunately for those of us living on the planet's surface, the centrifugal force is much weaker than Earth's gravity.
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# Pre-Algebra 1 Enrichment Classes (6th/7th Grade)
## Empower young minds with strong foundations.
A Grade Ahead’s Pre-Algebra 1 math program sets the stage for students to learn the concepts that are a necessary part of every young academic’s life. In addition to basic math topics and introductory pre-algebra topics, students in the Pre-Algebra 1 program will boost:
• Test-taking skills
• Focus and concentration
• Statistics, ratios, and proportions
• Problem-solving skills
• Logic and analysis
Pre-Algebra 1 students are given numerical drills, curriculum, and test prep homework. Each step has been designed to enhance A Grade Ahead students’ abilities to calculate quickly and accurately, while constructing solutions for real-world problems.
You may also be interested in 6th grade English.
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June
Month 1
Curriculum Topics
Factors & prime factorization; GCF and LCM
Numerical Path
Decimals 7:
Example: 17.54 ÷ 0.2 = ? and 832.4 x 0.02 = ?
July
Month 2
Curriculum Topics
Fractions; Decimals; Square roots
Numerical Path
Decimals 8:
Examples: 3.311 x 900 = ? and 23.617 ÷ 0.011 = ?
August
Month 3
Curriculum Topics
Ratio proportion; Rate; Percents
Numerical Path
Percents 1:
Examples: 8% of 50 = ? and 76% of 200 is =?
September
Month 4
Curriculum Topics
Standard system of measurement
Numerical Path
Percents 2:
Examples: 2.5% of 930 = ? and 92.5% -> 37/50 -> 0.925
October
Month 5
Curriculum Topics
Geometry
Numerical Path
Percents 3:
Examples: What % is 11 of 44? and \$1,800 discounted by 18% = ?
November
Month 6
Curriculum Topics
Geometry; Integers
Numerical Path
Integers 1:
Examples: 65 + (-55) = ? and -79 + (-96) - (78) = ?
December
Month 7
Curriculum Topics
Expressions; Equations; Coordinate grids
Numerical Path
Integers 2:
Example: 90 - _ - (-55) = 89
January
Month 8
Curriculum Topics
Inequalities; Statistics
Numerical Path
Expressions 1:
Examples: 7xy - 6yx = ? (for x = 3, y = 7)
February
Test Prep Month
Curriculum Topics
Standardized Test Preparation
*Some may offer in a month other than February. Check with your local academy.
Numerical Path
Integers 3:
Examples: 42 x (-21) = ? and -37 / (-2) = ?
March
Month 9
Curriculum Topics
Exponents; Polynomials; Pythagorean theorem
Numerical Path
Exponents 1:
Examples: 104+ 54 = ? and 37÷ 35 = ?)
April
Month 10
Curriculum Topics
Cost & selling price; Probability; Transformations
Numerical Path
Polynomials 1:
Example: Express in standard form 2z 6w8 + 3y 2w4 = ?
May
Month 11
Curriculum Topics |
Home Mathematics
# LESSON STUDY VIGNETTE: PRIME AND COMPOSITE NUMBERS
This classic locker problem exposes students to considering the important nature of numbers including even and odd numbers, factors, multiples, prime numbers, composite numbers, and perfect squares.
Suppose you’re in a hallway lined with 100 closed lockers. You begin by opening every locker. Then you close every second locker. Then you go to every third locker and open it (if its closed) or close it (if its open). Continue this for every nth locker on pass number n. After 100 passes, where you get to locker #100, how many lockers are open?
Our teachers in the grade band fourth through sixth planned a lesson study around this locker problem. The goal of their lesson was to solve problem and model a mathematical situation that would reveal a relationship between factors and multiples.
The Common Core Math Standard in the fourth grade reinforces the concept of factors and multiples.
Gain Familiarity with factors and multiples.
4. Find all factor pairs for a whole number in the range 1-100. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range 1-100 is a multiple of a given one-digit number. Determine whether a given whole number in the range 1-100 is prime or composite. (CCSS-M)
During the launch of this lesson, students warmed up by giving examples of factors and multiples and then determine the greatest common factor for two or more numbers and the least common multiple for two numbers. They were also asked to categorize numbers by looking for prime numbers, composite numbers, perfect squares, and even or odd numbers. The warm-up helped remind students of the difference between factors and multiples and the number of factors for all prime numbers.
The host teacher who delivered the research lesson had a student read the problem out loud and asked another student to restate the problem to be sure everyone understood the problem. The math agenda for her research lesson was to engage the students in problem solving and reasoning as students analyzed the nature of different numbers. Odd 1, 3, 5, 7, 9, 11, ... even 2, 4, 6, 8, 10, ... square 1, 4, 9, 16, 25, 36, ... prime 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ... composite 4, 6, 8, 9, 10, 12, 14, 15, 16, ... triangular 1, 3, 6, 10, 15, 21 (as seen in the handshake problem), and the interesting Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21. Number shapes like the triangular numbers like 1, 3, 6, 10 can be arranged in the shape of a triangle. Just as the square numbers can be arranged in squares.
In the locker problem, students have to look for a pattern between the actions they take opening and closing the lockers and the number of factors each number has. They see that there is a relationship between the factors a locker number has and whether it is open or closed at the end. As students realize that they need to systematically keep track of the actions they are making as they open and close lockers, it affords
Figure 4.1 Figurate numbers—triangular and square number patterns.
the opportunity to encourage mathematical practices of looking for and making use of structure and looking for and expressing regularity in repeated reasoning (Standard 7 & 8 of the CCSSM, 2010).
The host teacher noted after the lesson, “I should have defined the phrase, ‘changed the state’ better for the students’ understanding.” The phrase “changed the state” links to the number of factors for each locker number. Changing the state of the lockers results in a pattern related to the number of factors of each locker number.
For all the lockers that had an even number of factors, the locker would be closed and the lockers with an odd number of factors would stay open. The only locker numbers with the odd number of factors turn out to be the square numbers since 4 has (1, 2, 4), 9 has (1, 3, 9), etc. The factors 2 x 2 are counted once, not double counted, so all square numbers will have an odd number of factors. The classic locker problem is a rich benchmark problem that is familiar in the school curriculum and which provides multiple entry points to access a variety of math content in number theory (For more details, see Seshaiyer, Suh & Freeman, 2011).
Related topics |
# FRACTIONS WORD PROBLEMS WITH GRAPHS AND TABLES
Fractions Word Problems with Graphs and Tables :
In this section, you will learn, how to solve word problems on fractions with graphs and tables.
## Fractions Word Problems with Graphs and Tables - Examples
Example 1 :
John's family expenses are 7000 dollars a month. He made a pie chart on their expenses for food, housing, saving, clothing, entertainment and music.
From the given information find,
(i) How much money does he spend for clothing and entertainment.
Solution :
He spends 11% for clothing and 13% for entertainment. The sum of these two expenses are 11 + 13 = 24%
To find the exact amount, we need to calculate 24% of his salary.
His salary = 7000 dollars
= 24% of 7000
= (24/100) x 7000
= \$1680
Hence he spends \$1680 for clothing and entertainment.
Example 2 :
Nanyung university offers four applied majors and monitors the number of students in each.
What fraction of the students in applied majors are majoring in computer science?
Solution :
Number of students applied in computer science = 190
Total number of students = 210 + 360 + 190 + 100
= 860
To find fraction divide 190 by 860,
= 190/860
= 19/86
Example 3 :
The graph below shows the information on the number of farm animals owned by Betty.
(i) What is the total percentage of animal owned by Jason and Walter ?
(ii) How many animals are owned by Daniel
Solution :
Percentage of animals owned by Jason = 30%
Percentage of animals owned by Walter = 120%
Total percentage = (30/100) + (120/100)
= (30 + 120)/100
= 150/100
Hence 150% of animals owned by Jason and Walter.
Percentage of animals owned by Daniel = 50%
From this we have to understand that, Daniel is having 50% of animals in total number of animals
Total number of animals = 120 + 90 + 50 + 10 + 30
= 300
Total number of animals that Daniel has
= (50/100) x 300
= 150
Example 4 :
A band from Bloomington went on tour over the summer, and each member chipped in for the cost of gas.
What fraction of the gas money came from Karlene ?
Solution :
Total contribution for gas costs is
= 430 + 720 + 820 + 490 + 760
= \$3220
Amount contributed by Karlene is \$720.
Fraction of the gas money came from Karlene is
= 720 / 3220
= 36/161
Example 5 :
A cheese merchant in Springfield sells cheeses from a variety of regions, and he noted how much from each region was sold last year.
What fraction of the cheese sold was from Spain ?
Solution :
Total amount of cheese sold (in kilograms) is
= 830 + 920 + 940 + 640 + 680
= 4010 kg
Amount of cheese from Spain is 940 kg.
Fraction of the cheese sold was from Spain :
= 940 / 4010
= 94/401
After having gone through the stuff given above, we hope that the students would have understood, how to solve word problems on fractions with graphs and tables.
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
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Word problems on simple equations
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Word problems on unit rate
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Converting customary units word problems
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Word problems on simple interest
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Complementary and supplementary angles word problems
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OTHER TOPICS
Profit and loss shortcuts
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Domain and range of rational functions
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Graphing rational functions with holes
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Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
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Published on September 9th, 2012 In category Division and Multiplication tricks | Education | Maths
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# Division tricks for prime numbers using seed numbers
Seed Numbers and Prime Number divisibility
In this article we will learn some new and easy tricks to divide large numbers by prime numbers using concept of seed numbers. We had earlier written an article on the prime number divisibility rules using some rules, to check this article please refer to this links
Prime Number: The prime numbers are the numbers that have only 2 factors; 1 and number itself. Apart from 2 all the prime numbers are odd numbers
Seed Number: Every prime number (apart from 2) has 2 seed number, the trick to find out the seed number is as follows:
1. Find out the multiples of the prime number ending in 1 and 9 for eg. 7 x 3 = 21 and 7 x 7 = 49.
2. Now write numbers in the factors of 10 with + or – for eg. 21 = 2 x10 + 1 & 49 = 5 x 10 – 1
3. so the seed numbers of 7 are 2 and 5
4. IMP NOTE : The seed number received from + 1 are always negative and seed number received from – 1 are always taken as positive . So in the above example the seed number are -2 and 5
How to use this concept:
Lets check if 10754 is divisible by 19.
• The seed no. of 19 are -17 & 2, you can use any seed no. the result will be same
• Multiple unit digit by a seed no. ( 2 x 4) and add to rest of the number 1075 so we have 8 + 1075 = 1083
• To get smaller number we can repeat this step 3 x 2 + 108= 6 +108 = 114 and we know 19 x 6 = 114
### Seed numbers of common Prime numbers
Prime Number Multiples ending in 1 Multiples ending in 9 Seed Number
3 21 = 2 × 10 + 1 9 = 1 × 10 – 1 – 2, 1
7 21 = 2 × 10 + 1 49 = 5 × 10 – 1 – 2, 5
13 91 = 9 × 10 + 1 39 = 4 × 10 – 1 – 9, 4
17 51 = 5 × 10 + 1 119 = 12 × 10 – 1 – 5, 12
19 171 = 17 × 10 + 1 19 = 2 × 10 – 1 – 17, 2
23 161 = 16 x 10 + 1 69 = 7 x 10 – 1 -16, 7
29 261 = 26 x 10 + 1 29 = 3 x 10 – 1 -26, 3
31 31 = 3 x 10 + 1 279 = 28 x 10 – 1 -3, 28
37 111 = 11 x 10 + 1 259 = 26 x 10 – 1 -11, 26
41 41 = 4 x 10 + 1 369 = 37 x 10 -1 -4, 37
The seed numbers of the some key prime numbers are provided here
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In math practice test on quadrilateral worksheet we will practice different types of questions in quadrilateral. Students can practice the questions of quadrilateral worksheet before the examinations to get more confident.
1. Fill in the blanks:
(i) A quadrilateral has …………… sides.
(ii) A quadrilateral has …………… angles.
(iii) A quadrilateral has …………… vertices, no three of which are……………… .
(iv) A quadrilateral has …………… diagonals.
(v) A diagonal of a quadrilateral is a line segment that joins two ……………… vertices of the quadrilateral.
(vi) The sum of the angles of a quadrilateral is ……………… .
(i) How many pairs of adjacent sides are there? Name them.
(ii) How many pairs of opposite sides are there? Name them.
(iii) How many pairs of adjacent angles are there? Name them.
(iv) How many pairs of opposite angles are there? Name them.
(v) How many diagonals are there? Name them.
3. Prove that the sum of the angles of a quadrilateral is 360°.
4. The three angles of a quadrilateral are 76°, 54° and 108°. Find the measure fourth angle.
5. The angles of a quadrilateral are in the ratio 3 : 5 : 7 : 9. Find the measure of each of these angles.
6. A quadrilateral has three acute angles, each measuring 75°. Find the measure of the fourth angle.
7. Three angles of a quadrilateral are equal and the measure of the fourth angle is 120°. Find the measure of each of the equal angles.
8. Two angles of a quadrilateral measure 85° and 75° respectively. The other two angles equal. Find the measure of each of these equal angles.
9. In the adjacent figure, the bisectors of ∠A and ∠B meet in a point P.
If ∠C = 1000 and ∠D = 60°, find the measure of ∠APB.
Hint: 60° + 100° + ∠A + ∠B = 360°
⇒ ∠A + ∠B = 200°
⇒ ¹/₂ ∠A + ¹/₂ ∠B = 100°
⇒ ∠BAP + ∠ABP = 100°
But, ∠BAP + ∠ABP + ∠APB = 180° (why?)
Now, find ∠APB.
1. (i) four
(ii) four
(iii) four, collinear
(iv) two
(v) opposite
(vi) 360°
2. (i) four; (AB, BC), (BC, CD), (CD, DA), (DA, AB)
(ii) two; (AB, DC), (AD, BC)
(iii) four; (∠A, ∠B), (∠B, ∠C), (∠C, ∠D), (∠D, ∠A)
(iv) two; (∠A, ∠C), (∠B, ∠D)
(v) two; (AC, BD)
4. 122°
5. 45o, 75o, 105°, 135°
6. 135°
7. 80o
8. 100°
9. 80°
Worksheet on Different Types of Quadrilaterals |
# La1.1.22 3 aug matrices for linear sys solution set is x_1=3,x_2=-2,x_3=-1
• MHB
• karush
In summary: It should be:$$A_3=\left[\begin{array}{rrrrr} 0& 0& 2& -2\\ 0& 1& 2& -4\\ 1& 0& 0& 3\\ \end{array}\right]$$In summary, we discussed constructing 3 different augmented matrices for linear systems whose solution set is $x_1=3, x_2=-2, x_3=-1$. The simplest augmented matrix would follow from the equations $x_1=3, x_2=-2, x_3=-1$ and would be a $3\times4$ matrix. We also explored the concept of augmented matrices
karush
Gold Member
MHB
$$\tiny{la1.1.22}$$
$\textsf{Construct 3 different augmented matrices for linear systems whose solution set is}$
$$x_1=3,\quad x_2=-2,\quad x_3=-1$$Well we could start just by$$3x-2y+y=-1$$but then we need a $3\times4$ matrix
karush said:
$$\tiny{la1.1.22}$$
$\textsf{Construct 3 different augmented matrices for linear systems whose solution set is}$
$$x_1=3,\quad x_2=-2,\quad x_3=-1$$
Well we could start just by
$$3x-2y+y=-1$$
When we substitute $x=3$ and $y=-2$, we get $3\cdot 3 - 2\cdot (-2) + (-2)=11 \ne -1$.
That's not going to work is it?
karush said:
but then we need a $3\times4$ matrix
We need indeed a 3x4 matrix since we have 3 unknowns, meaning we should have 3 equations for them.
The simplest augmented matrix would follow from:
$$\begin{cases}x_1=3 \\ x_2=-2 \\ x_3=-1\end{cases}$$
What is the corresponding augmented matrix?
I like Serena said:
When we substitute $x=3$ and $y=-2$, we get $3\cdot 3 - 2\cdot (-2) + (-2)=11 \ne -1$.
That's not going to work is it?
well actually its $x+y+z=3-2-1=0$
I like Serena said:
We need indeed a 3x4 matrix since we have 3 unknowns, meaning we should have 3 equations for them.
The simplest augmented matrix would follow from:
$$\begin{cases}x_1=3 \\ x_2=-2 \\ x_3=-1\end{cases}$$
What is the corresponding augmented matrix?
ok presumably this would be one case
$$\left[\begin{array}{rrrrr} %x_1& x_2& x_3& 0\\ x_1& 0& 0& 3\\ 0& x_2& 0& -2\\ 0& 0& x_3& -1\\ \end{array}\right]$$
Last edited:
karush said:
ok presumably this would be one case
$$\left[\begin{array}{rrrrr} %x_1& x_2& x_3& 0\\ x_1& 0& 0& 3\\ 0& x_2& 0& -2\\ 0& 0& x_3& -1\\ \end{array}\right]$$
... except that an augmented matrix is without the variables.
So it should be:
$$\left[\begin{array}{rrrrr} %x_1& x_2& x_3& 0\\ 1& 0& 0& 3\\ 0& 1& 0& -2\\ 0& 0& 1& -1\\ \end{array}\right]$$
And if we add both the second row and the third row to the first row, the first row becomes the equation that you've suggested... and we have a new augmented matrix.
I like Serena said:
... except that an augmented matrix is without the variables.
So it should be:
$$\left[\begin{array}{rrrrr} %x_1& x_2& x_3& 0\\ 1& 0& 0& 3\\ 0& 1& 0& -2\\ 0& 0& 1& -1\\ \end{array}\right]$$
And if we add both the second row and the third row to the first row, the first row becomes the equation that you've suggested... and we have a new augmented matrix.
the only other augmented matrix I see is just flipping this one to$$\left[\begin{array}{rrrrr} 0& 0& 1& -1\\ 0& 1& 0& -2\\ 1& 0& 0& 3\\ \end{array}\right]$$so the the third ?
karush said:
the only other augmented matrix I see is just flipping this one to$$\left[\begin{array}{rrrrr} 0& 0& 1& -1\\ 0& 1& 0& -2\\ 1& 0& 0& 3\\ \end{array}\right]$$so the the third ?
Do the reverse of a row reduction?
That is, add/substract rows such that instead of elements becoming zero, they become non-zero?
I like Serena said:
Do the reverse of a row reduction?
That is, add/substract rows such that instead of elements becoming zero, they become non-zero?
i thot augmented meant you had to have a triangle of zeros
karush said:
i thot augmented meant you had to have a triangle of zeros
Neh, an augmented matrix is just an encoding of a set of equations into a single matrix.
When we apply row reduction to an augmented matrix, we get a new augmented matrix in row reduced echelon form, which has indeed a triangle of zeroes.
I like Serena said:
Neh, an augmented matrix is just an encoding of a set of equations into a single matrix.
When we apply row reduction to an augmented matrix, we get a new augmented matrix in row reduced echelon form, which has indeed a triangle of zeroes.
ok here goes...
$$A=\left[\begin{array}{rrrrr} 0& 0& 1& -1\\ 0& 1& 0& -2\\ 1& 0& 0& 3\\ \end{array}\right]$$
$\textsf{so then if from$A$we multiply$R_1$by$2$and add to$R_3$to become$A_2$}$
$$A_2=\left[\begin{array}{rrrrr} 0& 0& 2& -2\\ 0& 1& 0& -2\\ 1& 0& 2& 1\\ \end{array}\right]$$
$\textsf{so then if from$A$we multiply$R_1$by$2$and add to$R_2$to become$A_3$}$
$$A_3=\left[\begin{array}{rrrrr} 0& 0& 2& -2\\ 0& 1& 2& 0\\ 1& 0& 0& 3\\ \end{array}\right]$$so then thusly
kinda maybe hopefully?
karush said:
ok here goes...
$$A=\left[\begin{array}{rrrrr} 0& 0& 1& -1\\ 0& 1& 0& -2\\ 1& 0& 0& 3\\ \end{array}\right]$$
$\textsf{so then if from$A$we multiply$R_1$by$2$and add to$R_3$to become$A_2$}$
$$A_2=\left[\begin{array}{rrrrr} 0& 0& 2& -2\\ 0& 1& 0& -2\\ 1& 0& 2& 1\\ \end{array}\right]$$
$\textsf{so then if from$A$we multiply$R_1$by$2$and add to$R_2$to become$A_3$}$
$$A_3=\left[\begin{array}{rrrrr} 0& 0& 2& -2\\ 0& 1& 2& 0\\ 1& 0& 0& 3\\ \end{array}\right]$$so then thusly
kinda maybe hopefully?
All good except for $A_3$ where there is a mistake with a sign.
## What is the meaning of "La1.1.22 3 aug matrices for linear sys solution set is x_1=3,x_2=-2,x_3=-1"?
The notation "La1.1.22 3 aug matrices for linear sys solution set is x_1=3,x_2=-2,x_3=-1" refers to a linear system of equations that can be represented using augmented matrices. The solution set for this system is given by x_1=3, x_2=-2, and x_3=-1.
## What is an augmented matrix?
An augmented matrix is a representation of a linear system of equations in matrix form, where the coefficients of the variables are placed in a rectangular matrix and the constants are placed in a column vector on the right side of the matrix.
## How is the solution set for a linear system of equations found using augmented matrices?
The solution set for a linear system of equations can be found by performing elementary row operations on the augmented matrix until it is in reduced row echelon form. The resulting matrix will have the solutions to the system of equations on the right side.
## What does it mean for a matrix to be in reduced row echelon form?
A matrix is in reduced row echelon form when it has the following properties: 1) all rows of zeros are at the bottom, 2) the first non-zero entry in each row (called a leading entry) is a 1, 3) each leading entry is the only non-zero entry in its column, and 4) the leading entry of each row is to the right of the leading entry in the row above it.
## How do the solutions in an augmented matrix correspond to the solutions to the linear system of equations?
The solutions in an augmented matrix correspond to the solutions to the linear system of equations by using the values in the rightmost column as the values for the variables in the system. In the example provided, x_1=3, x_2=-2, and x_3=-1 would be the solutions to the system of equations.
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Subjects > 6th Grade Math >
### Division of Fractions (NS 6.1)
#### NS 6.1
Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem. For example, create a story context for (2/3) ÷ (3/4) and use a visual fraction model to show the quotient; use the relationship between multiplication and division to explain that (2/3) ÷ (3/4) = 8/9 because 3/4 of 8/9 is 2/3. (In general, (a/b) ÷ (c/d) = ad/bc.) How much chocolate will each person get if 3 people share 1/2 lb of chocolate equally? How many 3/4-cup servings are in 2/3 of a cup of yogurt? How wide is a rectangular strip of land with length 3/4 mi and area 1/2 square mi?
### Multi-Digit Division Using the Standard Algorithm (NS 6.2)
#### NS 6.2
Fluently divide multi-digit numbers using the standard algorithm.
### Four Operations with Decimals Using the Standard Algorithm (NS 6.3)
#### NS 6.3
Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.
### Greatest Common Factor - GCF & Least Common Multiple - LCM (NS 6.4)
#### NS 6.4
Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express 36 + 8 as 4 (9 + 2).
### Negative Numbers (NS 6.5)
#### NS 6.5
Understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.g., temperature above/below zero, elevation above/below sea level, credits/debits, positive/negative electric charge); use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation.
### Opposite Signs of Numbers, Ordered Pairs & Placing These on the Number Line (NS. 6.6)
#### NS 6.6
Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates.
• a. Recognize opposite signs of numbers as indicating locations on opposite sides of 0 on the number line; recognize that the opposite of the opposite of a number is the number itself, e.g., –(–3) = 3, and that 0 is its own opposite.
• b. Understand signs of numbers in ordered pairs as indicating locations in quadrants of the coordinate plane; recognize that when two ordered pairs differ only by signs, the locations of the points are related by reflections across one or both axes.
• c. Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane.
### Comparing, Ordering of Absolute Value (NS 6.7)
#### NS 6.7
Understand ordering and absolute value of rational numbers.
• a. Interpret statements of inequality as statements about the relative position of two numbers on a number line diagram. For example, interpret –3 > –7 as a statement that –3 is located to the right of –7 on a number line oriented from left to right.
• b. Write, interpret, and explain statements of order for rational numbers in real-world contexts. For example, write –3°C > –7°C to express the fact that –3°C is warmer than –7°C.
• c. Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. For example, for an account balance of –30 dollars, write |–30| = 30 to describe the size of the debt in dollars.
• d. Distinguish comparisons of absolute value from statements about order. For example, recognize that an account balance less than –30 dollars represents a debt greater than 30 dollars.
### Solve Problems with Four Quadrants Graphing on the Coordinate Plane (NS 6.8)
#### NS 6.8
Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and absolute value to find distances between points with the same first coordinate or the same second coordinate. |
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# How do you solve fractions of a number word problem?
###### Get more contents on this skill...
Have you been wondering how do you solve fractions of a number word problem? If yes, then we believe this guide will greatly influence you to grasp the best and simple steps for solving all fractions of a number word problems. These steps will include essential points, procedures, and strategies to tackle fractions of a number word problem effortlessly.
It will equally interest you to know that as kids engage in this resource, they will have the solution to any math word problem they come across at their fingertips.
## Steps on how to solve fractions of a number word problems
Again and again, when it comes to mastering steps on how to solve fractions of a number word problems, sixth graders always ponder on how they can go about it. Hence, we have come on with this thrilling procedural, well-organized step-by-step guide on how to do it.
We will also incorporate fun and essential math skills and strategies so that kids can enjoy while going through our real-life examples solving process.
#### Step 1: IDENTIFY
Here, you will need to figure out the important fractions and keywords in the word problem after reading the word problem and understand the problem statement very well. Then, use these keywords to identify the type of operation that the problem calls for.
• Whenever you find yourself with this topic “fraction of a number word problem,” think of multiplication. In addition to this, look for common keywords such as “of,” “times,” “product of,” “how many,” “how much,” etc. in the word problem.
Note: One key Element for learners to understand is that they should not always rely on keywords alone. That is to say; the same keyword can have different meanings in different word problems.
For this reason, we reiterate on the importance of reading the question very carefully to understand the situation that the word problem is describing, then figure out exactly which operation to use
#### Step 2: STRATEGIZE
Moving forward, ask yourself this pertinent question. “How will I tackle this word problem?”
• As mentioned above, from the keyword(s) in the word problem and from the situation that the problem is describing, you will know if you need to perform a multiplication operation or any other operation.
• However, it would be best not to rely only on keywords. Rather, try to understand the situation that the problem is describing.
• After knowing which operation you will perform, construct short expressions/sentences representing the given word problem.
#### Step 3: SET UP
Here, you have to write down a numerical expression representing the information in the word problem.
#### Step 4: PROVIDE A SOLUTION
As you can see from step 3 above, if you are performing a multiplication operation,
• Firstly, you will need to convert the whole number into a fraction by dividing the whole number by 1.
• Then, multiply the numerators across, likewise, the denominators. Note that when multiplying fractions, a common denominator is not needed.
• Also, simplify the fraction if possible.
• Furthermore, do not forget the unit of measurement, if any.
#### Step 5: CHECK YOUR WORK
Finally, check out your work by interpreting the answer in the context of the problem. If the interpretation makes sense, then “YES,” you are done. If “NO,” go back to step 1 and start all over again.
### Examples on how to solve fractions of a number word problems
#### Example one
Judy wants to make five times the original recipe of pizza that calls for Cup of flour. How much flour does he need?
Step 1:Frist, the important fraction and whole number here are and 5 . The keywords found in the word problem are “of,” “how much,” and “times.”
Step 2:Next, how will you solve the problem? This problem asks us to find 5 x cup of flour Hence the most important keywords here are “of” and “times.” So, this is a multiplication problem.
Now, form short expressions/sentences to represent the given word problem.
• Fraction of flour that the recipe calls for =
• Number of times he wants to increase the recipe = 5 times
• Therefore, the quantity of flour he needs = the fraction of flour that the recipe calls for × the number of times he wants to increase the recipe.
Step 3:Now, write down a numerical expression to represent the bolded sentence in step 2 above:
of 5 = ?
Step 4:From step 3 above, you will need to:
• Convert the whole number into a fraction by dividing the whole number by 1.
• Then, multiply the numerators across, likewise, the denominators. Note that when multiplying fractions, a common denominator is not needed.
• Also, simplify the fraction if possible.
• Furthermore, do not forget the unit of measurement, if any.
So, he needs cup of flour
Step 5:Finally, check out your work by interpreting the answer in the context of the problem. If the interpretation makes sense, then “YES,” you are done. If “NO,” go back to step 1 and start all over again.
#### Example two
In Joys’ school, two-fifth of the 525 pupils are boys. How many pupils in the school are boys?
Step 1: Frist, after reading and understanding the problem, you see that the important fraction and whole number here are and . Also, the keyword found in the word problem is “of”, 255 “how much”, and “times”.
Step 2: Next, how will you solve the problem? Looking at the situation that the word problem is describing critically and the keywords found in the problem, it is clear that you need to perform a multiplication operation.
Now, form short expressions/sentences to represent the given word problem.
• Total number of pupils in school = 525
• Fraction of boys in school =
• Therefore, the number of pupils in school that are boys = the fraction of boys in school × the total number of pupils in school.
Step 3: Now, write down a numerical expression to represent the bolded sentence in step 2 above:
of 525 = ?
Step 4: After that, from Step 3 above:
##### Method 1
Firstly, you will need to convert the whole number into a fraction by dividing the whole number by 1.
• Firstly, you will need to convert the whole number into a fraction by dividing the whole number by 1.
• Then, multiply the numerators across, likewise, the denominators. Note that when multiplying fractions, a common denominator is not needed.
• Also, simplify the fraction if possible.
• Furthermore, do not forget the unit of measurement, if any.
So, there are 210 pupils in the school are boys
##### Method 2
Here, you can start by simplifying the fractions first before performing multiplication.
So, there are 210 pupils in the school are boys
Step 5: Finally, check out your work by interpreting the answer in the context of the problem. If the interpretation makes sense, then “YES,” you are done. If “NO,” go back to step 1 and start all over again. |
1. ## Numerators and Denominators
What is the largest possible value for the sum of two fractions such that each of the four 1 digit prime numbers occurs as one of the numerators or denominators?
2. Hello, Godfather!
What is the largest possible value for the sum of two fractions such that
each of {2, 3, 5, 7} occurs as one of the numerators or denominators?
Here's one approach . . .
The sum of two fractions is: . $S\:=\:\frac{a}{b} + \frac{c}{d} \:=\:\frac{ad + bc}{bd}$
For the largest value, we want the smallest denominator.
. . This occurs for: . $b = 2,\:d = 3$ .(or vice versa).
We have: . $S \:=\:\frac{3a + 2b}{6}$
Then we have two choices: . $\begin{array}{c}a = 5,\,b = 7 \\ a=7,\,b=5\end{array}$
If $a=5,\,b=5:\;S\:=\:\frac{3\!\cdot\!5+2\!\cdot\!7}{6 } \:=\:\frac{29}{6}$
If $a=7,\,b=5:\;S\:=\:\frac{3\!\cdot\!7 + 2\!\cdot\!5}{6}\:=\:\frac{31}{6}\quad\Leftarrow\:\ text{larger!}$
Therefore: . $\frac{7}{2} + \frac{5}{3}\:=\:\boxed{\frac{31}{6}}$ .is the largest sum. |
# Test: Coordinate Geometry - 2
## 25 Questions MCQ Test Mathematics (Maths) Class 9 | Test: Coordinate Geometry - 2
Description
Attempt Test: Coordinate Geometry - 2 | 25 questions in 25 minutes | Mock test for Class 9 preparation | Free important questions MCQ to study Mathematics (Maths) Class 9 for Class 9 Exam | Download free PDF with solutions
QUESTION: 1
Solution:
QUESTION: 2
Solution:
QUESTION: 3
### The point (3, 2) is at a distance of _______________ units from y-axis :
Solution:
QUESTION: 4
The point (–3, 2) belongs to Quadrant _______________ :
Solution:
QUESTION: 5
The point (2, –3) belongs to quadrant _______________ :
Solution:
The x value is POSITIVE which means that the point is to the right of the origin.
The y value is NEGATIVE which means that the point is below the origin.
The 4th quadrant is to the right and below the origin,(The bottom right quadrant)
All points in the 4th quadrant have the signs as (+,−)
QUESTION: 6
The point (3, 2) belongs to quadrant _______________ :
Solution:
QUESTION: 7
The point (–2, –3) belongs to Quadrant :
Solution:
QUESTION: 8
The point (–2, 0) lies on :
Solution:
QUESTION: 9
The point (0, –2) lies on :
Solution:
QUESTION: 10
The point (3, 0) lies on :
Solution:
QUESTION: 11
The point (0, 3) lies on :
Solution: This is because y is also called X=0 and x=0,y=3 so it will lie on positive y axis(+ve y axis)
QUESTION: 12
The distance between the points (–4, 7) and (1, –5) is :
Solution:
Use the distance formula to determine the distance between two points.
√{1-(-4)}2 +{(-5)-7}2
=√25+144
=√169
=13
QUESTION: 13
The distance of the point (–2, –2) from the origin is :
Solution: Solving it by Pythagoras theorem the perpendicular will be 2 and the base will also be 2 the distance will be the hypotenuse =>. h² = b² + p² h²= 2² + 2² h² = 4+4 h²= 8 h=√8 √8 can be written as √2 x √2 x √2 which is equal to 2√2units.
QUESTION: 14
If the distance between points (p, –5), (2, 7) is 13 units, then p is _____________ :
Solution:
Distance between two points (x1,y1) and (x2,y2) can be calculated using the formula −
Distance between the points (p,−5) and (2,7)=
(2−p)2 +(7−(−5)) 2 =13
(2-p)2 + 144 = 13
(2-p)2 + 144 = 169
(2-p)2 = 25
2-p = 5 or 2-p = -5
p = -3 or p = 7
QUESTION: 15
The points (a, a) (–a, a) and (– (√3) a, (√3)a) form the vertices of an :
Solution:
QUESTION: 16
Find the ratio in which the line joining the points (6, 4) and (1, –7) is divided by x-axis.
Solution:
QUESTION: 17
The points (2, –1), (3, 4), (–2, 3) and (–3, –2) taken in the order from the vertices of _________________:
Solution:
In rhombus, all sides are equal and diagonals are not equal.
Distance between two points = [(x2−x1)2 + (y2−y1)2]1/2
AB = [(3−2)2 + (4+1)2]1/2 =(26)1/2
BC = [(3+2)2 + (3−4)2]1/2 = (26)1/2
CD = [(-3+2)2 + (-2-3)2]1/2 = (26)1/2
DA = [(-3-2)2 + (-2+1)2]1/2 = (26)1/2
AC = [(2+2)2 + (4)2]1/2 = 4(2)1/2
BD = [(-3-3)2 + (4+2)2]1/2 = 6(2)1/2
AB=BC=CD=DA All sides are equal
AC and BD Diagonals are not equal
QUESTION: 18
The points (0, 0), (–2, 0) and (3, 0) _________________:
Solution:
QUESTION: 19
The point on Y-axis equidistant from (–3, 4) and (7, 6) is _________________:
Solution:
QUESTION: 20
The point on X-axis equidistant from (5, 4) and (–2, 3) is _________________:
Solution:
QUESTION: 21
The points on X-axis at a distance of 10 units from (11, –8) are :
Solution:
Let P(x, 0) be the point on the x-axis. Then as per the question we have
AP = 10
Hence, the points on the x-axis are (5, 0) and (17,0) .
QUESTION: 22
The points on Y-axis at a distance of 13 units from point (–5, 7) are :
Solution:
Let P(0,y) be the required point on y−axis.
Now, distance between the point P(0,y) and A(−5,7) is 13 units.
Now, PA = 13 units
⇒√(0+5)^2 + (y−7)^2 = 13
⇒√25 + y2 + 49 − 14y = 13
⇒√y^2 + 74 − 14y = 13
⇒y^2 − 14y + 74 = 169
⇒y^2 − 14y − 95 = 0
⇒y^2 − 19y + 5y − 95 = 0
⇒y(y−19) + 5(y−19) = 0
⇒(y+5)(y−19) = 0⇒y+5 = 0 or y−19 = 0
⇒y = −5 or y = 19
So, the required points are :(0,−5) and (0,19)
QUESTION: 23
The coordinates of the centre of a circle passing through (1, 2), (3, –4) and (5, –6) is _________________:
Solution:
Given three points (1,2),(3,−4)and(5,−6)
Let the coordinates o centre be O(h,k)
The distance from the centre to a point on the circumference is equal to radius.
∴ OA=OB=OC
From distance formula
OA2=OB2
QUESTION: 24
The points (–5, 6), (3, 0) and (9, 8) form the vertices of :
Solution:
QUESTION: 25
The points (4, 4), (3, 5) and (–1, –1) from the vertices of :
Solution:
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3rd Class Mathematics Numbers Number Sense and Numeration
Number Sense and Numeration
Category : 3rd Class
Number Sense and Numeration
Introduction
Everything is counted by numbers. Numbers are the symbolic representation of counted objects. Categorized of numbers are basically dependent on their factors and divisibility.
Types of Number
Natural Numbers
Counting numbers are known as natural numbers.
For example, natural numbers (N) = {1, 2, 3, 4, 5 ?.. infinite}
Whole Numbers
Counting numbers including 0 are known as whole numbers.
For example, whole numbers (W) = {0, 1, 2, 3, 4?. infinite}
Prime Numbers
Those numbers having two factors 1 and the number itself are called prime numbers.
For example, prime numbers (P) = {2, 3, 5, 7, 11, 13 etc.}
Twin Numbers
Two numbers with difference of 2 are called twin primes.
For example: {3, 5}, {5, 7}, {11, 13} etc.
Note:
2 is even prime number.
Composite Numbers
Numbers having more than two factors, 1 and the number itself, are called composite numbers.
For example: {4, 6, 8, 9, 10 etc.}
Even Number
Numbers which are exactly divisible by 2 are called even numbers.
For example: {2, 4, 6, 8, etc.}
Odd Numbers
Numbers that are not exactly divisible by 2 and leaves remainder are called odd numbers.
For example: {1, 3, 5, 7 etc.}
• Example
Quotient of division of a number is an odd number. Which one of the following is correct about the dividend if divisor is greater than I?
(a) Dividend is an even number
(b) Dividend is a composite number
(c) Dividend is a prime number
(d) All the above
(e) None of these
Ans. (d)
Explanation: Quotient can be odd number for even, composite and prime number.
• Example
If area of a circle C1 is full of natural numbers, which one of the following is correct about the area of circle C2?
(a) Circle C2 is full of prime numbers
(b) Circle C2 is full of composite numbers
(c) Circle C2 is full of even numbers
(d) All the above
(e) None of these
Ans. (d)
Operation on Numbers
Expanded form of every number is written according 10 tire place value of digits. For example, form of number $56434=5\times 10000+6\times 1000+4\times 100+3\times 10+4$
Face Value
Face value of every number is the number itself. For example; face value of 5 in numeral 4356 is 5 itself and value of 4 in 4536 is 4.
Place Value
Place value of digits of a number is their position in the number. For example, place value of 2 in 54276 is 200. Place value of 5 in 5674 is 5000.
Successor
Successor of every number is just after the number. For example, successor of 17 is 18. Successor of a number is obtained by adding 1 to the number.
Predecessor
Predecessor of every number is just before the number. For example, predecessor? of 67 is 66. Predecessor of a number is obtained by subtracting 1 from the number.
Ascending Order
Arrangement of the numbers from smallest to greatest is called the ascending order of the numbers. For example, ascending order of 45, 67, 21, 34 is 21 < 34 < 45 < 67.
Descending Order
Descending order is the arrangement of numbers in decreasing order For example, descending order of 50, 45, 23, 12, 22 is 50 > 45 > 23 > 22 > 12.
Rounding Counting Numbers to Nearest Ten
Rounding counting number 23763 to nearest 10 is 23760. So unit digit is mostly changed while rounding counting numbers to nearest ten. Sometimes, tens digit may also be changed. For example, rounding of 458 to nearest 10 is 460.
Comparison of Place Value of Indian and International Number System
International Hundred Billions Ten Billions Billions Hundred Millions Ten Millions Million Hundred Thousand Ten Thousand Thousand Hundred Ten Ones Indian Kharab Ten Arab Arab Ten Crore Crore Ten Lakh Lakh Ten Thousand Thousand Hundred Ten Units
Number Pattern
Numbers can be arranged in different patterns.
Look at the following examples where numbers are arranged in pattern:
1, 4, 7, 10, 13.
In the given pattern difference between the numbers is 3. Adding 3 to previous number next number in the above pattern is obtained.
• Example
Arrange the following capacity of packing materials in ascending order
Name of packing materials Quantity Bottle 879 litres Pouch 3245 millilitres Gallon 6795 decilitres Tin 546 centilitres
Which one of the following is correct about their ascending order?
(a) Pouch < Tin < Gallon < Bottle
(b) Bottle < Pouch < Tin < Gallon
(c) Tin < Gallon < Bottle < Pouch
(d) All the above
(e) None of these
Ans. (a)
Explanation: Since millilitres < centilitres < litres and 1 centilitre = 10 millilitres and 1 litre = 10 decilitres. Therefore, option (a) is correct.
• Example
There are 235 holes on a metal plate. If every row has 5 holes then the total row of holes on the plate is a/ an:
(a) Composite number (b) Even number
(c) Prime number (d) all the above
(e) None of these
Ans. (c)
Explanation: Total number of rows on plate $=\frac{235}{5}=47$ and 47 is a prime number.
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# 2013 AMC 8 Problems/Problem 16
## Problem
A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$-graders to $6^\text{th}$-graders is $5:3$, and the the ratio of $8^\text{th}$-graders to $7^\text{th}$-graders is $8:5$. What is the smallest number of students that could be participating in the project?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89$
## Solutions
### Solution 1: Algebra
We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:
$5:3 = 5(8):3(8) = 40:24$
$8:5 = 8(5):5(5) = 40:25$
Therefore, the ratio of 8th graders to 7th graders to 6th graders is $40:25:24$. Since the ratio is in lowest terms, the smallest number of students participating in the project is $40+25+24 = \boxed{\textbf{(E)}\ 89}$.
### Solution 2: Fakesolving
The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility). Then there are $40*\frac{3}{5}=24$ 6th graders and $40*\frac{5}{8}=25$ 7th graders. The numbers of students is $40+24+25=\boxed{\textbf{(E)}\ 89}$
~ pi_is_3.14
## Video Solution 2
https://youtu.be/s7dIYGdXYPU ~savannahsolver |
S k i l l
i n
A R I T H M E T I C
Lesson 6
# HOW TO ADD WHOLE NUMBERS AND DECIMALS
The student should by now have mastered elementary addition.
In this Lesson, we will answer the following:
We can only add things that have the same name, which we call the units. That is why, when we add a column of numbers, we align the ones in a column, the tens in a column, the hundreds, and so on for each unit.
2,364 = 2 thousands + 3 hundreds + 6 tens + 4 ones + 5,132 = 5 thousands + 1 hundred + 3 tens + 2 ones 7,496 = 7 thousands + 4 hundreds + 9 tens + 6 ones
Then, starting with the ones on the right, we add each column.
Now, that is the simplest example because no column adds to more than 9. But consider this example where there are more than 9 ones. There are 12:
38 + 4
8 + 4 is 12 -- but there is no digit for 12. The largest digit we can write is 9. Therefore we must group 10 ones of 12 into 1 ten --
-- and regroup it with the 3 tens:
We now have
4 tens + 2 ones = 42.
In practice, simply write 2 under the line --
-- and "carry" the 1 onto the tens column. Say:
"8 plus 4 is 12." (Write 2, carry 1.) "3 plus 1 is 4."
That is the procedure for adding in columns. It's called addition with regrouping.
1. How do we add whole numbers in writing? 4,674 1,422 5,533 3,840 15,469 Write the numbers under one another, taking care to align the same units; that is, align the ones, the tens, etc.; and draw a line. Then, starting with the ones on the right, add each column. When the sum of a column is 9 or less, write that sum under the line. But when that sum is more than 9, write the ones of that sum and regroup -- "carry" -- the tens digit onto the next column. For, we may compose 10 of a lower unit into 1 of the next higher unit.
4,674 1,422 5,533 3,840 15,469
2 1 4 thousands + 6 hundreds + 7 tens + 4 ones 1 thousands + 4 hundreds + 2 tens + 2 ones 5 thousands + 5 hundreds + 3 tens + 3 ones 3 thousands + 8 hundreds + 4 tens + 0 ones 15 thousands + 4 hundreds + 6 tens + 9 ones
On adding the ones column, we get 9. But on adding the tens column, we get 16.
Write 6, and carry the 1 onto the hundreds column -- because 10 tens are equal to 1 hundred. Therefore when we add the hundreds, we have
1 + 6 + 4 + 5 + 8 = 24 hundreds.
(See the previous Lesson, Example 3.)
Write 4, and carry the 2 onto the thousands column -- because 20 hundreds are equal to 2 thousands. (Lesson 2, Example 3c.)
When we add the thousands, then, we get
2 + 4 + 1 + 5 + 3 = 15.
The sum of those numbers is 15,469.
Example 1. 9545 + 8982 18527
"5 + 2 is 7." Write 7.
"4 + 8 is 12." Write 2 -- carry 1.
"5 + 9 is 14, plus 1 is 15." Write 5 -- carry 1.
"9 + 8 is 17, plus 1 is 18." Write 18.
2. How do we add decimals? Add decimals the same way as whole numbers, taking care to align the decimal points. Equivalently, align the ones because every number has ones.
Example 2. Write in a column and add: 4,785 + 9 + 2.307
Solution.. The ones are shown:
4,785 + 9 + 2.307
(Lesson 3, Question 5.) Therefore, align as follows:
4,785 9 + 2 .307 4,796 .307
Example 3. Write in a column and add: .58 + 5.8 + 58
Solution. Here are the ones:
_.58 + 5.8 + 58
As for .58, the ones are at the first place to the left of the decimal point.
Align as follows:
.58 5 .8 + 58 64 .38
When the decimal points are aligned, the decimal point in the answer will fall in the same place. (But that is true only in addition and subtraction, not in multiplication. Lesson 10.)
As for a whole number such as 58, to help with alignment we may imagine a decimal point after the 8.
58 = 58.
Whole numbers, however, are normally written without a decimal point, because the decimal point means "and." Here come the fractions
Example 4. .5 + .5 + .5
a) 15
b) .15
c) 1.5
d) .015
For, if we aligned them and wrote .5 as 0.5 --
1 0.5 0.5 + 0.5 1.5
-- we would see that the 1 (of 15 tenths) carries over into the next column.
The point is:
will have the same number of decimal digits as the numbers themselves.
Example 5. .007 + .003 + .004
The numbers being added have three decimal digits. Therefore the answer also will have three decimal digits.
Example 6. Perimeters. The perimeter of a plane (flat) figure is its boundary.
This figure is a rectangle, which is a four-sided figure in which all the angles are right angles. In a rectangle, the opposite sides are equal. Therefore the perimeter of that rectangle is:
12 + 12 + 6.5 + 6.5 = 24 + 13 = 37 in.
2 0.83 7 0.49 6 0.26 + 8 0.58
Technique. Do not rewrite the problem by writing each column as a separate sum. To add the column on the right, do not say -- or write -- "3 plus 9 is 12. 12 plus 6 is 18." And so on. Rather, let your eye go down the column and say each partial sum. Do not write it. (See Lesson 5, Question 2, Example 3.)
To add the column on the right, say
2 0.83 7 0.49 "12" 6 0.26 "18" + 8 0.58 "26" 6
Write 6, carry 2. To add the middle column, say
2 2 0.83 "10" 7 0.49 "14" 6 0.26 "16" + 8 0.58 "21" 0.16
Write 1, carry 2. To add the last column, say
2 "4" 2 .83 "11" 7 .49 "17" 6 .26 "25" 8 .58 25 .16
Write 25.
You may write the decimal point in the answer when you come to it; that is, after adding the middle column.
Please "turn" the page and do some Problems.
or
Continue on to the next Lesson.
1st Lesson on Adding Whole Numbers
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# College Algebra Homework Help: 5 Tricks You Probably Didn’t Know
When it comes to doing any Math homework, many students will struggle, this is particular case when it comes to algebra as people, even those with mathematical minds, can sometimes struggle with the concept of replacing numbers with letters. The following provide some interesting tips and tricks that you may not learn about algebra, and how it may assist with your homework.
1. An easy way to work out squared numbers
2. One interesting tip is to look at how to work out squared numbers easily. We are often taught at a young age that to multiply a whole number by 10 you simply add a zero on the end. Therefore, finding the square of the number that ends in a zero is often quite easy. For example 10 squared equals 100. Thirty equals 3 multiplied by ten. So thirty squared equals 3 x 3 x 10 x 10 = 900.
Supposing you wanted to work out what 33 squared was, an easy way is to take the nearest multiple of 10 (which is 30). You then square 30, which is easy to work out, and equals 900. You then take the original number (33) and add it to the multiple of ten that you just squared (30) and add both these numbers together, which equals 63. You then take those two numbers and subtract the larger one from the smaller one, 33 – 30 = 3. Now multiply 3 by 63, which equals 189. Finally, add 189 to 900 (thirty squared). So, 189 + 900 = 1089. This number, 1089, is equal to 33 squared.
In algebra, if x = the original number and y = the nearest multiple of 10, then this can be written as
X2 = y2 + (x+y)(x-y)
The formula works whether or not y is a multiple of ten; however, it makes it easier to work out in your head if this is the case.
3. Another easy way to multiple numbers
4. Again, using squared numbers and multiples of ten (to make it easier to do in your head). We know that 30 squared = 900. If we wanted to work out 32 x 28, then there is a tip you can use. 32 and 28 are both 2 units higher or lower than thirty. If you take 30 squared and minus two squared, this will be the same as 32 x 28. The algebraic formula is as follows:
X2 = (x+y)(x-y)
5. Keep a list of common equations handy
6. Whenever you learn common equations, it is a good idea to write them down. While you are doing homework you can have this list of commonly used equations handy, which will save you time rather than having to look each one up.
7. Use online sources for extra help
8. If you are stuck on a particular algebraic equation then you may wish to see whether or not you get an answer if you posted directly into a search engine - sometimes these answers will be posted is the first response. Alternatively, you can look on forums, questions and answer websites, or even social media to find relevant Math enthusiasts who are willing to help.
9. Pay someone to do the work
10. If you are still struggling then you can actually pay people to do the homework for you. You may not have realised that there are many companies out there offering services to do all kinds of academic work for students, from simples algebra homework, all the way up to major math dissertations and doctoral theses.
Professional essay writing service - EssayMill.com - get your essays written by expert essay writer.
Expert essay writing service - they are writing essays since 2004. |
# Problem of the Week Problem E and Solution Outside the Path of Totality
## Problem
Yannick used a camera with a solar filter to capture a solar eclipse. From their location, the Moon blocked some, but not all of the Sun, so they saw a partial solar eclipse. When Yannick enlarged and printed the photo, they noticed that the distance represented by segment $$XY$$ in the photo shown was $$70$$ cm, and that the diameters of the Sun and the Moon were both $$74$$ cm.
Determine the percentage of the Sun that is blocked by the Moon in Yannick’s photo, rounded to $$1$$ decimal place.
## Solution
Let the centres of the Sun and Moon in the photo be $$S$$ and $$M$$, respectively. We draw line segments $$SX$$, $$SY$$, $$MX$$, and $$MY$$. We draw line segment $$SM$$ which intersects $$XY$$ at $$T$$. The shaded region represents the portion of the Sun that is blocked by the Moon.
In our diagram we have drawn $$S$$ and $$M$$ inside the shaded region, but we need to prove they actually lie inside this region before we can proceed with the area calculation. Since each of the circles has radius $$74 \div 2 = 37$$, then $$SX=SY=MX=MY=37$$. It follows that $$\triangle SXY$$ is isosceles. Similarly, $$\triangle SXM$$ and $$\triangle SYM$$ are congruent. Thus, $$\angle XSM = \angle YSM$$. Since $$\triangle SXY$$ is isosceles and $$SM$$ bisects $$\angle XSY$$, then $$SM$$ is perpendicular to $$XY$$ at $$T$$, and $$XT=TY=70 \div 2 = 35$$. By the Pythagorean Theorem in $$\triangle STX$$, $ST = \sqrt{SX^2-XT^2} = \sqrt{37^2-35^2} = 12$ Thus, $$SM=2 \times ST = 2 \times 12 = 24$$. Since $$SM$$ is smaller than the radius of the circles, it follows that $$S$$ and $$M$$ must lie inside the shaded region.
Now we can proceed with the area calculation. By symmetry, the area of the shaded region on each side of $$XY$$ will be the same. The area of the shaded region on the right side of $$XY$$ equals the area of acute sector $$XSY$$ of the left circle minus the area of $$\triangle SXY$$. These areas are striped in the following diagrams.
First we find the area of $$\triangle SXY$$, which is $$\frac{1}{2}(XY)(ST)=\frac{1}{2}(70)(12)=420$$.
Next we find the area of sector $$XSY$$. Using the cosine law in $$\triangle SXY$$, \begin{aligned} XY^2 &= SX^2 + SY^2 - 2(SX)(SY)\cos (\angle XSY)\\ 70^2 &= 37^2 + 37^2 - 2(37)(37)\cos (\angle XSY)\\ 2162 &= -2738\cos (\angle XSY)\\ \angle XSY &= \cos^{-1}\left(-\frac{2162}{2738}\right)\approx 142.15\degree \end{aligned} Thus, the area of sector $$XSY$$ is equal to $$\frac{142.15\degree}{360\degree}\pi (37)^2$$.
Then the area of the shaded region is equal to $$2 \left(\frac{142.15\degree}{360\degree}\pi (37)^2 - 420 \right)$$.
Finally, we calculate this area as a percentage of the area of the entire circle to obtain the following: $\frac{2 \left(\frac{142.15\degree}{360\degree}\pi (37)^2 - 420 \right)}{\pi (37)^2} \approx 59.4 \%$
It follows that the percentage of the Sun that is blocked by the Moon in Yannick’s photo is equal to approximately $$59.4 \%$$. |
# Knights, Knaves and Spies - Part 1
I was working my way through some Knight and Knave Puzzles in Discrete Maths by Rosen, when I came across the following question:
There are inhabitants of an island on which there are three kinds of people:
• Knights who always tell the truth
• Knaves who always lie
• Spies who can either lie or tell the truth.
You encounter three people, A, B, and C.
You know one of these people is a knight, one is a knave, and one is a spy.
Each of the three people knows the type of person each of other two is.
For this situation, if possible, determine whether there is a unique solution and determine who the knave, knight, and spy is:
A says "C is the knave,” B says, “A is the knight,” and C says “I am the spy"
My Solution:
A is the Knight, B is the Spy, and C is the Knave
Doubt:
Am I correct in saying my answer will work?
A is the Knight
B is the Spy
and C is the Knave
To get the solution, First assume, A is knight and will always tells the truth.
Then as per his statement, C is the knave and so what he said will be false. That means he is not a spy. B is the spy and his statement A is the knight is random (true here). This is the only case in which the statements didn't contradict.
Now assume, A is the Knave.
Then as per his statement "C is the knave", it's clear that C is definitely not the knave. Which doesn't contradict since A is the knave already. That means, either B or C is Knight. If B is Knight his statement "A is knight" is false and it contradicts. If C is Knight his statement "I am the spy" is wrong and it contradicts. So this combination A is Knave, B is knight/Spy, C is Knight/Spy is wrong.
Continue this assumptions for other chances of combinations.
You will understand that all other combination except the first one (A is knight, B is Spy and C is knave) is wrong since the statements contradicts.
Simpler explanation:
First, notice that B cannot be the knight, because then for their statement to be true, A would also have to be a knight, and we know there is only one knight.
Second, notice that C cannot be the knight, because then their statement would be false.
Therefore, A is the knight. By their statement, C is the Knave. By elimination, B is the spy.
• One could add a sentence that assuming this assignement the statement by C is indeed a lie. B's statement could be either. So this is a solution. Commented Nov 8, 2022 at 9:42
Just logical if a is the night then he is telling the truth but if B is the night then he's telling the truth and be can't be the night because that would mean there would be two nights so B has to be false which doesn't mean that he is the knave I just means he's lying so he could be the Spy so if he's telling the truth than a is the night but if he's lying is not the Knight then see you would c telling the truth and if c was telling the truth he would have to be the night which is statement is false so he has to be lying and if B and C or both lying that means a is telling the truth so we know the a is the night and that night is telling the truth so therefore she has to be the knave and last but not least B has to be the Spy only thing that makes sense
• You... might want to break this up into sentences, with punctuation, and read it back to make sure it is understandable. This torrent of words is, as it stands, possibly a correct (though redundant) explanation, but after three tries at reading through it I gave up trying to follow what you are saying. See some of the other answers for examples of how to write and format an answer for clarity.
– Rubio
Commented Jul 9, 2018 at 2:01 |
# A 70cm Long Wire is to be Cut into Two Pieces Such that One Piece will be 2/5 as Long as the Other. How Many Centimeters will the Shorter Piece be?
### Computer MCQs Series for PPSC, FPSC – Most Repeated MCQs | Set 5
What are you looking for? Let’s dig in quickly
## Explanation
A 70cm long wire is to be cut into two pieces
The cutting must be in such a way one piece will be 2/5 as long as the other.
We shall assume the length of one piece “y” and the other of course “2y/5”.
The addition of these two gives the actual length of the wire which is 70cm.
y + 2y/5 = 70
Through this equation we can easily figure out the length of shorter piece.
## To Find
Length of the shorter piece = ?
## Solution
Total given length of the wire = 70cm
Let suppose the length of one piece after cutting = y
The length of second piece after cutting = 2y/5
Now; according to condition
y + 2y/5 = 70
5y + 2y = 350
7y = 350
y = 50
2y/5 = (2 x 50)/5 = 20cm answer
## Conclusion
A 70cm long wire is to be cut into two pieces such that one piece will be 2/5 as long as the other. The length of shorter piece will be 20cm.
### Computer MCQs Series for PPSC, FPSC – Most Repeated MCQs | Set 7
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# 2.2: Coordinate Systems and Components of a Vector (Part 2)
[ "article:topic", "authorname:openstax", "polar coordinate system", "polar coordinates", "radial coordinate", "license:ccby" ]
### Polar Coordinates
To describe locations of points or vectors in a plane, we need two orthogonal directions. In the Cartesian coordinate system these directions are given by unit vectors $$\hat{i}$$ and $$\hat{j}$$ along the x-axis and the y-axis, respectively. The Cartesian coordinate system is very convenient to use in describing displacements and velocities of objects and the forces acting on them. However, it becomes cumbersome when we need to describe the rotation of objects. When describing rotation, we usually work in the polar coordinate system.
In the polar coordinate system, the location of point P in a plane is given by two polar coordinates (Figure 2.20). The first polar coordinate is the radial coordinate r, which is the distance of point P from the origin. The second polar coordinate is an angle $$\varphi$$ that the radial vector makes with some chosen direction, usually the positive x-direction. In polar coordinates, angles are measured in radians, or rads. The radial vector is attached at the origin and points away from the origin to point P. This radial direction is described by a unit radial vector $$\hat{r}$$. The second unit vector $$\hat{t}$$ is a vector orthogonal to the radial direction $$\hat{r}$$. The positive + $$\hat{t}$$ direction indicates how the angle $$\varphi$$ changes in the counterclockwise direction. In this way, a point P that has coordinates (x, y) in the rectangular system can be described equivalently in the polar coordinate system by the two polar coordinates (r, $$\varphi$$). Equation 2.17 is valid for any vector, so we can use it to express the x- and y-coordinates of vector $$\vec{r}$$. In this way, we obtain the connection between the polar coordinates and rectangular coordinates of point P:
$$\begin{cases} x = r \cos \varphi \\ r \sin \varphi \end{cases} \ldotp \tag{2.18}$$
Figure $$\PageIndex{5}$$: Using polar coordinates, the unit vector $$\hat{r}$$ defines the positive direction along the radius r (radial direction) and, orthogonal to it, the unit vector $$\hat{t}$$ defines the positive direction of rotation by the angle $$\varphi$$.
Example 2.6
##### Polar Coordinates
A treasure hunter finds one silver coin at a location 20.0 m away from a dry well in the direction 20° north of east and finds one gold coin at a location 10.0 m away from the well in the direction 20° north of west. What are the polar and rectangular coordinates of these findings with respect to the well?
##### Strategy
The well marks the origin of the coordinate system and east is the +x-direction. We identify radial distances from the locations to the origin, which are rS = 20.0 m (for the silver coin) and rG = 10.0 m (for the gold coin). To find the angular coordinates, we convert 20° to radians: 20° = $$\frac{\pi\; 20}{180}$$ = $$\frac{\pi}{9}$$. We use Equation 2.18 to find the x- and y-coordinates of the coins. Solution The angular coordinate of the silver coin is $$\varphi_{S}$$ = $$\frac{\pi}{9}$$, whereas the angular coordinate of the gold coin is $$\varphi_{G}$$ = $$\pi$$ − $$\frac{\pi}{9}$$ = $$\frac{8 \pi}{9}$$. Hence, the polar coordinates of the silver coin are (rS, $$\varphi_{S}$$) = (20.0 m, $$\frac{\pi}{9}$$) and those of the gold coin are (rG, $$\varphi_{G}$$) = (10.0 m, \frac{8 \pi}{9}\)). We substitute these coordinates into Equation 2.18 to obtain rectangular coordinates. For the gold coin, the coordinates are
$$\begin{cases} x_{G} = r_{G} \cos \varphi_{G} = (10.0\; m) \cos \frac{8 \pi}{9} = -9.4\; m \\ y_{G} = r_{G} \sin \varphi_{G} = (10.0\; m) \sin \frac{8 \pi}{9} = 3.4\; m \end{cases} \Rightarrow (x_{G}, y_{G}) = (-9.4\; m, 3.4\; m) \ldotp$$
For the silver coin, the coordinates are
$$\begin{cases} x_{S} = r_{S} \cos \varphi_{S} = (20.0\; m) \cos \frac{\pi}{9} = 18.9\; m \\ y_{S} = r_{S} \sin \varphi_{S} = (20.0\; m) \sin \frac{\pi}{9} = 6.8\; m \end{cases} \Rightarrow (x_{S}, y_{S}) = (18.9\; m, 6.8\; m) \ldotp$$
### Vectors in Three Dimensions
To specify the location of a point in space, we need three coordinates (x, y, z), where coordinates x and y specify locations in a plane, and coordinate z gives a vertical positions above or below the plane. Three-dimensional space has three orthogonal directions, so we need not two but three unit vectors to define a three-dimensional coordinate system. In the Cartesian coordinate system, the first two unit vectors are the unit vector of the x-axis $$\hat{i}$$ and the unit vector of the y-axis $$\hat{j}$$. The third unit vector $$\hat{k}$$ is the direction of the z-axis (Figure 2.21). The order in which the axes are labeled, which is the order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The order x-y-z, which is equivalent to the order $$\hat{i}$$ - $$\hat{j}$$ - $$\hat{k}$$, defines the standard right-handed coordinate system (positive orientation).
Figure $$\PageIndex{6}$$: Three unit vectors define a Cartesian system in three-dimensional space. The order in which these unit vectors appear defines the orientation of the coordinate system. The order shown here defines the right-handed orientation.
In three-dimensional space, vector $$\vec{A}$$ has three vector components: the x-component $$\vec{A}_{x}$$ = Ax $$\hat{i}$$, which is the part of vector $$\vec{A}$$ along the x-axis; the y-component $$\vec{A}_{y}$$ = Ay $$\hat{j}$$, which is the part of $$\vec{A}$$ along the y-axis; and the z-component $$\vec{A}_{z}$$ = Az $$\hat{k}$$, which is the part of the vector along the z-axis. A vector in three-dimensional space is the vector sum of its three vector components (Figure 2.22):
$$\vec{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k} \ldotp \tag{2.19}$$
If we know the coordinates of its origin b(xb, yb, zb) and of its end e(xe ye, ze), its scalar components are obtained by taking their differences: Ax and Ay are given by Equation 2.13 and the z-component is given by
$$A_{z} = z_{e} - z_{b} \ldotp \tag{2.20}$$
Magnitude A is obtained by generalizing Equation 2.15 to three dimensions:
$$A = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}} \ldotp \tag{2.21}$$
This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in Figure 2.22, the diagonal in the xy-plane has length $$\sqrt{A_{x}^{2} + A_{y}^{2}}$$ and its square adds to the square Az2 to give A2. Note that when the z-component is zero, the vector lies entirely in the xy-plane and its description is reduced to two dimensions.
Figure $$\PageIndex{7}$$: A vector in three-dimensional space is the vector sum of its three vector components.
Example 2.7
##### Takeoff of a Drone
During a takeoff of IAI Heron (Figure 2.23), its position with respect to a control tower is 100 m above the ground, 300 m to the east, and 200 m to the north. One minute later, its position is 250 m above the ground, 1200 m to the east, and 2100 m to the north. What is the drone’s displacement vector with respect to the control tower? What is the magnitude of its displacement vector?
Figure $$\PageIndex{8}$$: The drone IAI Heron in flight. (credit: SSgt Reynaldo Ramon, USAF)
##### Strategy
We take the origin of the Cartesian coordinate system as the control tower. The direction of the +x-axis is given by unit vector $$\hat{i}$$ to the east, the direction of the +y-axis is given by unit vector $$\hat{j}$$ to the north, and the direction of the +z-axis is given by unit vector $$\hat{k}$$, which points up from the ground. The drone’s first position is the origin (or, equivalently, the beginning) of the displacement vector and its second position is the end of the displacement vector.
##### Solution
We identify b(300.0 m, 200.0 m, 100.0 m) and e(480.0 m, 370.0 m, 250.0m), and use Equation 2.13 and Equation 2.20 to find the scalar components of the drone’s displacement vector:
$$\begin{cases} D_{x} = x_{e} - x_{b} = 1200.0\; m - 300.0\; m = 900.0\; m,\\ D_{y} =y_{e} - y_{b} = 2100.0\; m - 200.0\; m = 1900.0\; m, \\ D_{z} =z_{e} - z_{b} = 250.0\; m - 100.0\; m = 150\; m \ldotp \end{cases}$$
We substitute these components into Equation 2.19 to find the displacement vector:
$$\vec{D} = D_{x}\; \hat{i} + D_{y}\; \hat{j} + D_{z}\; \hat{k} = 900.0\; \hat{i} + 1900.0\; \hat{j} + 150.0\; \hat{k} = (0.90\; \hat{i} + 1.90\; \hat{j} + 0.15\; \hat{k})\; km \ldotp$$
We substitute into Equation 2.21 to find the magnitude of the displacement:
$$D = \sqrt{D_{x}^{2} + D_{y}^{2} + D_{z}^{2}} = \sqrt{(0.90\; km)^{2} + (1.90\; km)^{2} + (0.15\; km)^{2}} = 4.44\; km \ldotp$$
If the average velocity vector of the drone in the displacement in Example 2.7 is $$\vec{u}$$ = (15.0 $$\hat{i}$$ + 31.7 $$\hat{j}$$ + 2.5 $$\hat{k}$$) m/s, what is the magnitude of the drone’s velocity vector? |
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Today in Precalculus - PowerPoint PPT Presentation
Today in Precalculus. Go over homework Notes: Parabolas Completing the square Writing the equation given the graph Applications Homework. Example 1. Prove that the graph of y 2 + 2y – 8x – 7 = 0 is a parabola y 2 + 2y = 8x + 7 y 2 + 2y + 1 = 8x + 7 + 1 y 2 + 2y + 1 = 8x + 8
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Presentation Transcript
• Go over homework
• Notes: Parabolas
• Completing the square
• Writing the equation given the graph
• Applications
• Homework
Prove that the graph of y2 + 2y – 8x – 7 = 0 is a parabola
y2 + 2y = 8x + 7
y2 + 2y + 1 = 8x + 7 + 1
y2 + 2y + 1 = 8x + 8
(y+1)2 = 8 (x + 1)
Standard form for a parabola
(y+1)2 = 8 (x + 1)
Graph:
Opens to the right
vertex (-1, -1)
4p = 8, p = 2
focal length =2
focal width = 8
focus: (1, -1)
directrix: x = -3
axis: y = -1
Prove that the graph of x2 – 2x + 3y + 7 = 0 is a parabola
x2 – 2x = -3y – 7
x2 – 2x + 1 = -3y – 7 + 1
x2 – 2x + 1 = -3y – 6
(x – 1)2 = -3(y + 2)
Standard form for a parabola
(x – 1)2 = -3(y + 2)
Graph:
Opens downward
vertex (1, - 2)
p = -3/4
focal length = -3/4
focal width = 3
focus: (1, -11/4)
directrix: y = -5/4
axis: x = 1
Find vertex, substitute into general form vertex: (-2, -2)
(y+2)2=4p(x+2)
Find another point and substitute for x and y (1,-5)
(-5 + 2)2 = 4p(1 + 2)
Solve for p
9 = 12p
p = ¾
(y+2)2=4(¾)(x+2)
(y+2)2=3(x+2)
vertex: (-4, -1)
(x + 4)2=4p(y+1)
point (2,-4)
(2 + 4)2 = 4p(-4 + 1)
36 = -12p
p = -3
(x + 4)2=4(-3)(y + 1)
(x+4)2=-12(y+1)
vertex: (2, 1)
(y – 1)2=4p(x – 2)
point (-1,4)
(4 – 1)2 = 4p(-1 – 2)
9 = -12p
p = -3/4
(y – 1)2=4(-3/4)(x – 2)
(y – 1)2=-3(x – 2)
Let the focus F have coordinates (0, p) and the vertex be at (0,0)
equation: x2 = 4py
Because the reflector is 3ft across and 1 ft deep, the points (1.5, 1) and (-1.5, 1) must lie on the parabola.
(1.5)2 = 4p(1)
2.25 = 4p
p = .5625ft
= 6.75in
So the microphone should be placed inside the reflector along its axis and 6.75 inches from its vertex.
Let the roadway be the x-axis.
Then the vertex is at (500,25)
So the equation is:
(x – 500)2 = 4p(y – 25)
Points (0, 150) and (1000, 150) also on the parabola.
So the equation is:
(0 – 500)2 = 4p(150 – 25)
250,000 = 500p
p = 500
(x – 500)2 = 2000(y – 25)
Equation for the shape of the main cable.
To find the length of the support cables, solve equation for y
Support cables are every 100ft, so starting with x = 100, then 200, 300, etc.
Cables are 105ft, 70ft, 45ft, 30ft, 25ft, 30ft, 45ft, 70ft, 105ft.
Page 641: 51-56, 59-63
Quiz: |
These worksheets teach students how to use a calculator to determine sine, cosine, tangents, etc.
#### When we are working with right triangles, we often need to use the three main functions of trigonometry to determine the length of the hypotenuse, adjacent, and opposite sides. The sin function of an angle is the value of the opposite angle divided by the hypotenuse. The cos function of an angle is the value of the adjacent angle divided by the hypotenuse. The tan function of an angle is the value of the adjacent angle divided by the opposite angle. The lengths will always come out in decimal form.
How do you compute trigonometric functions? Trigonometry is a branch of mathematics. It studies the right-angled triangles. As the name of these triangles suggests, one side of these triangles is straight of 90 °. Trigonometry helps us to find out unknown sides of these triangles. These are the simplest ways to compute trigonometric functions by understanding 'sine,' 'cosines,' and 'tangent.' These are the basic functions of trigonometry and can easily be computed. These functions are at each side of the right-angled triangle. For starters, this can appear tricky. Come up with a mnemonic to help you memorize these! Formulas - By using simple formulas, we can quickly solve trigonometric equations. Trigonometry has unique formulas to help us compute unknown sides or angles, for example, 'the circumference of circle' or 'area of a circle,' etc. Pythagoras theorem - Pythagoras theorem is the easiest way to solve the right-angled triangle. It is an important rule that says square on the hypotenuse is equal to the sum of the squares on both sides. This rule makes it easy for us to find out one side if we know the other two sides and any internal angle. Your students will use these worksheets to learn how to use a calculator in order to determine the sine, cosine, tangent, etc. of specified degrees of angle. All answers should be rounded.
# Print Using a Calculator (sin, cos, and tan) Worksheets
## Using a Calculator With Trigonometric Functions Lesson
This worksheet explains how to use a calculator to find the sin, cosine, or tan. A sample problem is solved, and two practice questions are provided.
## Worksheet
We get up close and personal with this skill and work away with our little math machines. Ten questions are provided.
## Practice
Students will use a calculator to find the unknown values. Ten questions are provided.
## Review and Practice
This worksheet reviews how to compute this value. This will lead us up to learning the purpose of this entire process. Six practice questions are provided.
## Quiz
Students will demonstrate their proficiency in using a calculator with trigonometric functions. Ten problems are provided.
## Skills Check
Students will find the distance or the slope. Space is included for students to copy the correct answer when given. |
Dot numbers are a new notation for numbers, that make integer addition look like rational multiplication. They may be useful in primary school math education. The idea is that once you understand integers and addition, you can learn another way to look at it that sets you up to understand fractions and multiplication.
I made up dot numbers a few years ago to try to explain negative numbers to my then-four-year-old son.
### Basics
A dot number is a way of writing a number. A dot number is represented as a number of dots above a line. This is the number 3, as a dot number:
A negative dot number is a number of dots below the line. This is the dot number -3: .
To add two dot numbers, combine the dots above the line, and the dots below the line. This is 3 plus 2:
This is -3 plus -2:
And this is 3 plus -2:
### Cancelling
A dot number is in normal form if all its dots (if any) are on the same side of the line. A dot number that isn’t in normal form can be “normalizing”, or transformed into another dot number that represents the same integer and is in normal form. As long as there’s a dot above the line and a dot below the line, cross the dots out. The pairs of dots “cancel” each other.
The preceding number (3 plus -2) can be normalized.
To subtract a number, flip it upside down and add that. This is 3 minus 2:
### Dots and Factors
Dot numbers are interesting because they’re isomorphic to fractions. Dot numbers are to addition (and subtraction, and the unit) as fractions are to multiplication (and division, and factors).
Just as the multiplicative inverse (the reciprocal) of a fraction can be obtained by turning it upside down, the additive inverse (the negative) of a dot number is obtained by flipping it upside down. (The inverse in both cases is the “vertical inverse”).
Just as fractions are multiplied by multiplying the tops and bottoms (numerators and denominators) separately, dot numbers are added by adding their tops and bottoms.
And just as any number can be represented by multiple fractions (1/2, 2/4, 4/8) which can be normalized (“reduced”) by dividing the top and bottom by integral factors (4/8 = 4×1/4×2 = 4×1/4×2 = 1/2), so an integer (3) can be represented by multiple dot numbers (3-0, 4-1, 5-2), each of which can be reduced by subtracting units from its top and bottom (5-2 = (1+1+1+1+1)-(1+1) = (1+1+1+1+1)-(1+1+1+1+1) = (1+1+1+1+1)-(1+1+1+1+1) = 3-0). |
Feb 21, 2020
Introduction
We tackle a solitaire game of coin flipping in this week's Riddler Classic. Using dynamic programming, we build a tree to explore all possible game states and work backwards to identify the ideal move at each state. Here is the full problem text.
You have two fair coins, labeled A and B. When you flip coin A, you get 1 point if it comes up heads, but you lose 1 point if it comes up tails. Coin B is worth twice as much — when you flip coin B, you get 2 points if it comes up heads, but you lose 2 points if it comes up tails.
To play the game, you make a total of 100 flips. For each flip, you can choose either coin, and you know the outcomes of all the previous flips. In order to win, you must finish with a positive total score. In your eyes, finishing with 2 points is just as good as finishing with 200 points — any positive score is a win. (By the same token, finishing with 0 or −2 points is just as bad as finishing with −200 points.)
If you optimize your strategy, what percentage of games will you win? (Remember, one game consists of 100 coin flips.)
Extra credit: What if coin A isn’t fair (but coin B is still fair)? That is, if coin A comes up heads with probability p and you optimize your strategy, what percentage of games will you win?
Solution
We can win this game with roughly 64% probability if we play optimally. It is interesting to note that we could win a game with a single coin just 46% of the time. (Why not 50%? We need a positive score to win, and we lose with a score of zero.) Therefore, the ability to choose between the coins as we play improves our odds of winning by nearly 20%!
Methodology
Here is a snippet that shows the odds of winning a game with a single, fair coin, based on 100,000 simulated games. We only win 46% of the time because we need to have at least 51 winning flips.
>>> import numpy as np
>>> trials = 100000
>>> scores = np.random.randint(2, size=(trials, 100)).sum(1)
>>> wins = (scores > 50).sum()
>>> wins / trials
0.46086
Now how do we solve the game with two coins? I'll turn to dynamic programming, a technique I've used on several other problems. That is, in order to model the probability of winning from our current position, we model the probability of winning from all the positions we could reach from our position, then choose the action that gives us the best expected result.
For this problem, the state of the game is expressed by the number of flips we've completed so far, which falls between 0 and 100, and the current score of the game, which falls between 0 and 200. Not every flip and score combination is valid - for example we can't have a score of 10 if we have only flipped twice, but these two variables completely express the state of our game. In python, we store the state of the game in an object called GameState. For example, GameState(n_flip=30, score=-10) means we have flipped 30 coins so far and our cumulative score is -10.
Once we know how to express the state of the game, we want to be able to calculate its expected value - that is, the likelihood of winning the game from this position. 100% implies we win the game in every case; 50% means we win half of the games we play, and so on.
In python, we calculate the expected value of a GameState by simulating all future positions we could reach - by flipping coin A and winning, flipping coin A and losing, flipping coin B and winning, or flipping coin B and losing. We take the average of the results from flipping coin A and compare it against the average of the results of flipping coin B and choose the coin that improves our odds the most.
Finally, we solve the problem by calculating the expected value of the game before we have flipped a single coin, which gives us just over 64%.
>>> GameState(n_flip=0, score=0).expected_value()
0.6403174472759772
In addition, we can also plot the various states of the game and their associated winning probabilities using a heatmap. In the figure below, each GameState is plotted as a point: the x-axis is the number of flips we've completed, and the y-axis is our cumulative score. The color of the point represents how likely we are to win from that position, where yellow means almost a sure thing, and purple means we're virtually certain to lose. (I omitted many of the points that have more than 99% or less than 1% chances of winning to simplify the plot.)
The extra credit this week asked how the game changes if we vary the probability that each coin turns up heads. In python we can run any number of scenarios by changing the class attributes p_a and p_b, which set the likelihood of winning and losing each flip for each coin. As expected, when coin A becomes more of a sure thing, our odds of winning the game increase, but I'll leave it as an exercise for the motivated reader to adjust the code and calculate the exact values!
Full Code
Fairly short code for this week. We create a GameState class as a namedtuple, then extend its functionality by adding a method to calculate expected_value recursively. The solution to the puzzle is calculated by calling GameState(0, 0).expected_value().
from collections import namedtuple
class GameState(namedtuple("GameState", ("n_flip", "score"))):
"""
The GameState object represents the state of the game, which is
specified by the number of flips we've completed already (from
zero to 100), and the current game score, which can be either
positive or negative, but has bounds determined by the number of
We add a method to solve for the expected value of the GameState,
which is the number of games we would expect to win from this
position given optimal play.
"""
cache: dict = {} # cache for expected value of GameStates
p_a: float = 0.5 # probability that coin A wins (heads)
p_b: float = 0.5 # probability that coin B wins (heads)
v_a: int = 1 # value of a win from coin A
v_b: int = 2 # value of a win from coin B
def expected_value(self) -> float:
"""
Return the expected percent of games won from this state,
e.g. 0.85 implies you can win 85% of games from this position
by playing optimally. To find the expected value of the game
before it starts, call expected_value on the initial GameState
Examples
--------
>>> GameState(99, 0).expected_value()
0.5
>>> GameState(0, 0).expected_value()
0.6403174472759772
"""
# search the cache to see if we've calculated this result before
# otherwise, calculate the result and store it in the cache
try:
return self.cache[self]
except KeyError:
if self.n_flip == 100:
# the game has ended; positive scores win; negative scores lose
if self.score > 0:
self.cache[self] = 1.0
else:
self.cache[self] = 0.0
else:
# we continue flipping coins; calculate the expected value of
# flipping coin a and coin b, then choose the best one
n, s = self.n_flip + 1, self.score
a = (
self.p_a * GameState(n, s+ self.v_a).expected_value()
+ (1 - self.p_a) * GameState(n, s - self.v_a).expected_value()
)
b = (
self.p_b * GameState(n, s + self.v_b).expected_value()
+ (1 - self.p_b) * GameState(n, s - self.v_b).expected_value()
)
self.cache[self] = max(a, b)
return self.cache[self]
if __name__ == "__main__":
import doctest
doctest.testmod() |
The factors the 8 space the numbers that divide 8 precisely without leaving any kind of remainder. There are four determinants of 8, they are 1, 2, 4, and also 8. Hence, the highest element is 8 the divides itself. If we add all the factors, climate the amount will be equal to 15.
The pair components of 8 are (1,8) and (2,4). These pair factors when multiplied together results in the initial number. 8 is additionally a perfect cube, thus its prime element will it is in a solitary digit. We have the right to determine the determinants of 8 by the department method and prime factorisation method.
You are watching: How many factors does 8 have
## What space the factors of 8?
A number or one integer that divides 8 exactly without leaving a remainder, then the number is a variable of 8. As the number 8 is an even composite number, the has much more than 2 factors. Thus, the factors of 8 space 1, 2, 4 and 8.
## Pair determinants of 8
A pair that numbers, which room multiplied together leading to an initial number 8 is dubbed the pair factors of 8. As questioned above, the pair factors of 8 have the right to be hopeful or negative. The hopeful and negative pair determinants of 8 is provided below:
Positive Pair determinants of 8:
Positive determinants of 8 Positive Pair factors of 8 1 × 8 (1, 8) 2 × 4 (2, 4)
Negative Pair factors of 8:
Negative determinants of 8 Negative Pair components of 8 -1 × -8 (-1, -8) -2 × -4 (-2, -4)
## How to Find factors of 8?
8 is a whole number that is additionally an even number. Due to the fact that it is a single-digit number, therefore, the is much easier to find the components of 8.
There are two methods to uncover the components of 8:
Division methodPrime factorisation method
## Factors the 8 by department Method
The components of 8 deserve to be found using the department method. In the division method, we need to uncover the integers which division 8 specifically without leave a remainder, and also those integers are considered as the components of 8. Now, allow us discover the determinants of 8 using the department method.
8/1 = 8 (Factor is 1 and also Remainder is 0)8/2 = 4 (Factor is 2 and also Remainder is 0)8/4 = 2 (Factor is 4 and also Remainder is 0)8/8 = 1 (Factor is 8 and Remainder is 0)
If we divide 8 by any numbers other than 1, 2, 8 and 8, it pipeline a remainder. Hence, the factors of 8 space 1, 2, 4 and also 8.
## Prime factorization of 8
The prime factorization of 8 is the procedure of writing the number as the product that its prime factors. Now, let us comment on the procedure of finding the prime components of 8.
Divide 8 by the the smallest prime number, i.e. 2. 8/2 = 4Divide 4 by the smallest feasible prime number, 4/2 = 22 is itself a prime number and is divisible by 2Therefore, the prime administer of 8 is 2 × 2 × 2 or 23.
Another an approach to find the prime factors of 8 is the element tree method.
### Factor Tree
Hence, the variety of prime components of 8 is one. The prime factorisation that the whole number 8 is 23. The exponent in the element factorisation is 3. Once you include the number 1 through the exponent, i.e.,3 +1 = 4. Therefore, the number 8 has 4 factors.
## Facts – components of 8
Factors of 8 – 1, 2, 4 and 8Prime factorisation – 2 x 2 x 2Prime aspect of 8 – 2Pair determinants – (1,8) and also (2,4)Sum of components of 8 – 15
### Related Articles
Links concerned Factors Factors the 15 Factor the 36 Factors that 48 Factors the 18 Factors that 24 Factors the 25 Factors the 42 Factors of 60 Factors the 35 Factors of 81 Factors the 75 Factors that 56
## Solved instances on determinants of 8
Example 1:
Find the usual factors the 8 and also 5.
Solution:
The determinants of 8 room 1, 2, 4 and 8.
The determinants of 7 space 1 and also 5.
As 5 is a prime number, the usual factor that 8 and 5 is 1.
Example 2:
Find the common factors of 8 and also 9.
Solution:
Factors that 8 = 1, 2, 4 and 8.
Factors the 9 = 1, 3, and also 9.
Hence, the usual factor the 8 and also 9 is 1.
Example 3:
Find the usual factors that 8 and 4.
Solution:
The components of 8 are 1, 2, 4 and 8.
The factors of 4 room 1, 2 and also 4.
Hence, the common factors that 8 and also 4 space 1, 2, and 4.
### Practise questions on factors of 8
What is the lowest factor of 8?Is there any odd number that is a variable of 8?How many even numbers space the components of 8?
Stay tuned with BYJU’S come know around factor 8 and also the factors and prime determinants of other numbers. Download BYJU’S – The Learning app to far better experience and clarification.
See more: Wh A Base Is A Compound That Releases Hydrogen Ions Into A Solution.
## Frequently Asked questions on components of 8
### What room the determinants of 8?
The determinants of 8 are 1, 2, 4 and 8. |
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This section describes how to perform the familiar operations from algebra (eg add, subtract, multiply, and divide) on functions instead of numbers or variables.
One of the key ideas of algebra (and mathematics in general) is to represent big problems as relations between smaller problems. So far we’ve done this primarily with variables; like when we say that the area of a patio is length times width we write it succinctly as $A = l\cdot w$.
With functions however, we can take this one step further and discuss algebraic combinations of functions. This may seem intimidating at first, but algebra with functions behaves (almost) identically to algebra of variables. Take, for example, the above formula we wrote; $A = l\cdot w$. But it’s not unreasonable to consider that $l$ and $w$ could depend on something else, like money. If this is the case, we could write that $l$ and $w$ are functions of the money ($m$) we spend, thus we would write $l(m)$ and $w(m)$ (since $l$ and $w$ depend on $m$; that’s exactly what this notation is saying). But $A$ is still the product of $l$ and $w$, even though those are now both functions of money, ie $A = l(m) \cdot w(m)$. This motivates our question about how we should apply algebraic operations (like multiplication) to functions, rather than just numbers.
In reality, most things can be viewed as functions (after all, most things depend on something and that’s all we need for a mathematical relation!). The important part here however is that (in most contexts) functions can also be thought of as variables. Thus we can think of “adding”, “subtracting”, “multiplying”, or “dividing” functions as being the same as doing it with variables. In general terms, given functions $f$ and $g$, we write the following notation:
• $(f + g)(x) = f(x) + g(x)$
• $(f - g)(x) = f(x) - g(x)$
• $(fg)(x) = (f(x))(g(x))$
• $\left (\dfrac {f}{g}\right )(x) = \dfrac {f(x)}{g(x)}$ whenever $g(x) \neq 0$
This is fancy notation
for saying: when we want to add, subtract, multiply, or divide functions it is equivalent to calculating each of the function values at the given $x$ value and applying the desired operation to the result.
Here is a video with more!
_
1 : Algebra with functions...
Depends on the context of the functions, it may or may not work as you would expect so you need to ask the problem-giver. Works largely as you would expect; evaluating each function at the supplied value (or variable) and then applying the given algebra operation as normal. Is a mystery that is beyond the scope of this course. Works as expected, except that you don’t need to worry about expanding or distributing signs or values. Only works for addition and subtraction; other operations may or may not work correctly and should be avoided. |
# Video: Understanding That Three-Digit Numbers Can Be Represented with Groups of Hundreds and Ones
Find the missing number.
02:30
### Video Transcript
Find the missing number.
The diagram shows a part-whole model. We can see that the whole amount is 773. And this whole amount has been split into two parts. There’s a part that at the moment is missing. And our second part is 73. Now, we can see that both parts are labelled. For example, the number 73 is labelled number of ones. Now, in the number 773, there’s only the digit three in the ones place. But we also know that each 10 is worth 10 ones. So we could say that the seven tens in our tens column are worth 70 ones. That’s why we can write the number of ones in 773 as 73.
Now, it would be very easy to look quickly at our part-whole model and think, “What do we add to 73 to make 773? Well, surely, the answer must be 700. 700 plus 73 equals 773. This is often the way that a part-whole model does work. But as we’ve mentioned already, both parts have been labelled. And we can see that the part on the left is labelled number of hundreds. So we can see now why we can’t just write 700 in there. There aren’t 700 hundreds in 773.
Let’s complete our place-value grid to see how many hundreds there are. There are seven hundreds in 773. And so, the missing number in our part-whole diagram is the number seven. Instead of showing us two numbers that we add together to make 773, our part-whole model shows us the place value of different digits that we can then add together to make 773. Our first part — and that’s the missing number — shows the number of hundreds, which we said was worth 700. And our second part shows us the number of ones. 73 ones are worth 73. And now, we can combine the two values together and we get the total. Our missing number is the number of hundreds that there are in 773. The answer is seven. |
Question of Exercise 2
# Question If two supplementary angles differ by 48°, their measures will be
Option 1 66°, 114°
Option 2 126°, 78°
Option 3 110°, 52°
Option 4 108°, 60°
Simplify the following expressions
(√(5) + √(2))2
Solution:
Explanation:
Which of the following rational numbers lies between 0 and - 1
A: 0
B: - 1
C: -1/4
D: 1/4
Solution:
Explanation:
0 and 1 cannot be found between 0 and 1.
In addition,0= o/4 and -1= -4/4
We can see that -1/4 is halfway between 0 and -1.
Hence, the correct option is (c) -1/4
Prove that the diagonals of a parallelogram bisect each other
Solution:
Explanation:
We must show that the diagonals of the parallelogram ABCD cross each other.
OA = OC & OB = OD, in other words.
Now AD = BC [opposite sides are equal] in ΔAOD and ΔBOC.
[alternative interior angle] ∠ADO = ∠CBO in ΔAOD and ΔBOC.
Similarly, ∠AOD = ∠BOC by ΔDAO = ΔBCO (ASA rule)
As a result, OA = OC and OB = OB [according to CPCT].
Hence, it is prove that the diagonals of a parallelogram bisect each other.
What is total surface area of sphere
Solution:
Explanation:
• The radius of the sphere affects the formula for calculating the sphere's surface area.
• If the sphere's radius is r and the sphere's surface area is S.
• The sphere's surface area is therefore stated as Surface Area of Sphere 4πr2, where ‘r’ is the sphere's radius.
• The surface area of a sphere is expressed in terms of diameter as S=4π(d/2)2, where d is the sphere's diameter.
Thus, total surface area of sphere is =4πr2.
Fill in the blanks If two adjacent angles are supplementary
they form a __________.
Solution:
Explanation:
• If the non-common sides of two angles form a straight line, they are called linear pair angles.
• The sum of the angles of two linear pairs is degrees.
• If the total of two angles is degrees, they are called supplementary angles. |
HCF and LCM
Subject: Compulsory Maths
Overview
This note gives the information about the H.C.F and L.C.M.
Highest Common Factor ( H.C.F)
The largest positive integer which divides two or more integers without any remainder is called Highest Common Factor (HCF). It is the possible factor of respective numbers.
To find H.C.F by Factorization method
At first, we should find the prime factors of the given number, then the product of the common prime factors is the H.C.F of the given numbers. For example:
Find the H.C.F of 30 and 42
Solution,
Here,
30 = 2× 3× 5× 1
42 = 2× 3× 7× 1
∴ H.C.F = 2× 3× 1
= 6
To the H.C.F by division method
In this method, we divide the larger number by the smaller one and again the first remainder. So obtained divides the first divisor. The process is continued till the remainder becomes zero. The last divisor for which the remainder becomes zero and it is the H.C.F of the given numbers. For example:
Find the H.C.F of 30 and 42
Lower Common Multiple (L.C.M)
The lower common multiple is the lowest factor of respective numbers.
To find L.C.M by Factorisation method
At first, the prime factor of the given number are to be found out, then the product of the common prime factors and the remaining prime factors(which are not common) is the L.C.M of the given numbers. For example:
Find the L.C.M of 30 and 42
Here,
30 = 2× 3× 5
42 = 2× 3× 7
L.C.M = 2× 3× 5× 7
= 210
Division method
In this method, the given numbers are arranged in a row and they are successively divided by the least common factors till the quotient are 1 or prime numbers. Then, the product of these prime factors is the L.C.M of the given number. For example:
∴ L.C.M = 2× 3× 5× 5× 7
= 1050
Things to remember
1. The largest positive integer which divides two or more integers without any remainder is called Highest Common Factor (HCF). It is the possible factor of respective numbers.
2. The lower common multiple is the lowest factor of respective numbers.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
HCF LCM by Prime Factorization Method, Learn basics and key concepts
Solution:
Here,
F(16) = {1, 2, 4, 8, 16}
F(24) = {1, 2, 3, 4, 6, 8, 12, 24}
F(32) = {1, 2, 4, 8, 16, 32}
Now,
F(16) ∩ F(24) ∩ F(32) = {1, 2, 4, 8}
∴ H.C.F. of 16, 24 and 32 is 8.
Solution:
Here,
28 ÷ 2 = 14 (remainder is 0)
14 ÷ 2 = 7 (remainder is 0)
Now,
28 = 2 × 2 × 7
42 = 2 × 3 × 7
70 = 2 × 5 × 7
∴ H.C.F = 2 × 7 = 14
Solution:
Here,
28 ÷ 2 = 14 (remainder is 0)
14 ÷ 2 = 7 (remainder is 0)
42 ÷ 2 = 21 (remainder is 0)
21 ÷ 3 = 7 (remainder is 0)
70 ÷ 2 = 35 (remainder is 0)
35 ÷ 5 = 7 (remaindder is 0)
Now,
28 = 2 × 2 × 7
42 = 2 × 3 × 7
70 = 2 × 5 × 7
∴ H.C.F = 2 × 7 = 14
Solution:
Here,
the required greatest number of children is the H.C.F. of 32, 40 and 56.
32 ÷ 2 = 16 (remainder is 0)
16 ÷ 2 = 8 (remainder is 0)
8 ÷ 2 = 4 (remainder is 0)
4 ÷ 2 = 2 |(remainder is 0)
40 ÷ 2 = 20 (remiander is 0)
20 ÷ 2 = 10 (remainder is 0)
10 ÷ 2 = 5 (remainder is 0)
56 ÷ 2 = 28 (remainder is 0)
28 ÷ 2 = 14 (remainder is 0)
14 ÷ 2 = 7 (remoiander is 0)
Now,
32 = 2 × 2 × 2 × 2 × 2
40 = 2 × 2 × 2 × 5
56 = 2 × 2 × 2 × 7
∴ H.C.F. = 2 × 2 × 2 = 8
Again,
to find the shares of each fruit.
32 ÷ 8 = 4
40 ÷ 8 = 5
56 ÷ 8 = 7
So, the required greatest number of children is 8.
Each child shares 4 oranges, 5bananas and 7 mangoes
Solution:
Here,
24 ÷ 2 = 12 (remainder is 0)
12 ÷ 2 = 6 (remainder is 0)
6 ÷ 2 = 3 (remainder is 0)
36 ÷ 2 = 18 (remiander is 0)
18 ÷ 2 = 9 (remainder is 0)
9 ÷ 3 = 3 (remainder is 0)
Now,
24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
∴ L.C.M. = 2 × 2 × 3 × 2 × 3 = 72
Solution:
Here,
M(4) = {4, 8, 12, 16, 20, 24, 28, 32, 36, 20, 44, 48, . . . . . . . }
M(6) = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, . . . . . . . . }
M(8) = {8, 16, 24, 32, 40, 48, 56, 64, 72, 880, . . . . . . }
Now,
M(4) ∩ M(6) ∩ M(8) = (24, 48, . . . . . . . }
∴ L.C.M. of 4, 6, 8 is 24.
Solution:
24 ÷ 2 = 12 (remainder is 0)
12 ÷ 2 = 6 (remainder is 0)
6 ÷ 2 = 3 (remainder is 0)
36 ÷ 2 = 18 (remainder is 0)
18 ÷ 2 = 9 (remainder is 0)
9 ÷ 3 = 3 (renmainder is 0)
48 ÷ 2 = 24 (remiander is 0)
24 ÷ 2 = 12 (remainder is 0)
12 ÷ 2 = 6 (remainder is 0)
6 ÷ 2 = 3 (remiander is 0)
Now,
24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
48 = 2 × 2 × 2 × 2 × 3
∴ L.C.M. = 2 × 2 × 2 × 3 × 3 × 2 = 144
Solution:
Divisor Dividend 2 36 2 18 3 9 3
Solution:
Divisor Dividend 2 48 2 25 2 12 2 6 3
Solution:
Divisor Dividend 2 24 2 12 2 6 3
Solution:
Divisor Dividend 2 36 2 18 2 9 3 |
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# Value of a function
Module by: Sunil Kumar Singh. E-mail the author
The value of a function at “x = a” is denoted as “f(a)”. The working rule for finding value of a function is to replace independent variable “x” by “a”.
## Polynomial and rational functions
Problem 1: Find “f(y)”, if
y = f x = 1 - x 1 + x y = f x = 1 - x 1 + x
Solution :
Statement of the problem : The given function is a rational function. We have to evaluate the function when independent variable is function itself.
We need to replace “x” by “y”.
f y = 1 y 1 + y = 1 1 x 1 + x 1 + 1 - x 1 + x f y = 1 y 1 + y = 1 1 x 1 + x 1 + 1 - x 1 + x
f y = 1 + x 1 + x 1 + x + 1 x = 2 x 2 = x f y = 1 + x 1 + x 1 + x + 1 x = 2 x 2 = x
Problem 2: Find “f(x)”, if
f x - 1 = x 2 1 f x - 1 = x 2 1
Solution :
Statement of the problem : The given function is a polynomial function with a polynomial as its argument. We have to evaluate the function for independent variable “x”.
We need to replace “x-1” by “x” in the given equation to find “f(x)”. The right hand side expression, however, does not contain term “x-1”. We, therefore, need to find the term, which will replace “x”. Clearly if "x" replaces "x-1", then "x+1" will replace "x-1+1 = x"
Thus, we need to replace “x” by “x+1”.
f x + 1 - 1 = x + 1 2 1 = x 2 + 2 x + 1 1 = x 2 + 2 x f x + 1 - 1 = x + 1 2 1 = x 2 + 2 x + 1 1 = x 2 + 2 x
Problem 3: If { f x } = x + 1 x { f x } = x + 1 x , then prove that :
{ f x } 3 = f x 3 + 3 f 1 x { f x } 3 = f x 3 + 3 f 1 x
Solution :
Statement of the problem : The function is a polynomial function. We have to evaluate cube of the function, which involves evaluation of function for arguments, which are independent variable, raised to certain integral powers.
The cube of given function is :
{ f x } 3 = x + 1 x 3 = x 3 + 1 / x 3 + 3 x 2 X 1 x + 3 x X 1 x 2 { f x } 3 = x + 1 x 3 = x 3 + 1 / x 3 + 3 x 2 X 1 x + 3 x X 1 x 2
{ f x } 3 = x 3 + 1 x 3 + 3 x + 1 x { f x } 3 = x 3 + 1 x 3 + 3 x + 1 x
Now, f x 3 f x 3 is :
f x 3 = x 3 + 1 x 3 f x 3 = x 3 + 1 x 3
Hence,
{ f x } 3 = x 3 + 1 x 3 + 3 x + 1 x = f x 3 + 3 f x { f x } 3 = x 3 + 1 x 3 + 3 x + 1 x = f x 3 + 3 f x
But, we see that :
f 1 x = 1 x + x = f x f 1 x = 1 x + x = f x
Hence,
{ f x } 3 = f x 3 + 3 f x = f x 3 + 3 f 1 x { f x } 3 = f x 3 + 3 f x = f x 3 + 3 f 1 x
Problem 4: If
f x = 1 + x 1 x f x = 1 + x 1 x
Then, find
f x f x 2 1 + { f x } 2 f x f x 2 1 + { f x } 2
Solution :
Statement of the problem : The given function is rational function. We have to find the expression which involves (i) function, (ii) function with argument as squared independent variable and (iii) square of the function.
We need to substitute for various terms in the given expression :
f x f x 2 1 + { f x } 2 = 1 + x 1 x X 1 + x 2 1 x 2 1 + 1 + x 1 x 2 f x f x 2 1 + { f x } 2 = 1 + x 1 x X 1 + x 2 1 x 2 1 + 1 + x 1 x 2
= 1 + x 1 + x 2 1 x 1 x 2 1 x 2 + 1 + x 2 1 x 2 = 1 + x 1 + x 2 1 x 1 x 2 1 x 2 + 1 + x 2 1 x 2
1 + x 2 1 x 2 2 1 + x 2 1 x 2 = 1 + x 2 2 1 + x 2 = 1 2 1 + x 2 1 x 2 2 1 + x 2 1 x 2 = 1 + x 2 2 1 + x 2 = 1 2
## Modulus functions
Problem 5: If
f x = | x | x ; x 0 f x = | x | x ; x 0
Then, evaluate
| f a f - a | | f a f - a |
Solution :
Statement of the problem : Function, f(x), involves modulus and is in rational form. The value of this function, in turn, forms the part of a expression to be evaluated. We have to find the value of expression.
We first evaluate the expression without modulus sign :
f a f - a = | a | a | a | a = | a | a + | a | a = 2 | a | a ; a 0 f a f - a = | a | a | a | a = | a | a + | a | a = 2 | a | a ; a 0
But, we know that | a | = ± a | a | = ± a
f a f - a = 2 X ± a a = ± 2 f a f - a = 2 X ± a a = ± 2
Taking modulus of the expression,
| f a f - a | = 2 ; a 0 | f a f - a | = 2 ; a 0
Note that we need to keep the condition for which the given expression is evaluated.
## Logarithmic functions
Problem 6: Find f 2 x 1 + x 2 f 2 x 1 + x 2 , if
f x = log e 1 + x 1 - x f x = log e 1 + x 1 - x
Solution :
Statement of the problem : The given function is transcendental logarithmic function. We have to evaluate the function for an argument (input to function), which is itself a rational function in independent variable, “x”.
We need to replace “x” by “ 2 x / 1 + x 2 2 x / 1 + x 2 ”.
f 2 x 1 + x 2 = log e 1 + 2 x 1 + x 2 1 2 x 1 + x 2 = log e 1 + x 2 + 2 x 1 + x 2 2 x f 2 x 1 + x 2 = log e 1 + 2 x 1 + x 2 1 2 x 1 + x 2 = log e 1 + x 2 + 2 x 1 + x 2 2 x
f 2 x 1 + x 2 = log e 1 + x 1 x 2 = 2 log e 1 + x 1 x = 2 f x f 2 x 1 + x 2 = log e 1 + x 1 x 2 = 2 log e 1 + x 1 x = 2 f x
## Trigonometric functions
Problem 7: Find f π / 4 f π / 4 , if
f x = 2 cot x 1 + cot 2 x f x = 2 cot x 1 + cot 2 x
Solution :
Statement of the problem : The given function is a rational function with trigonometric function as independent variable. We have to find the value of function for a particular angle.
We need to replace “x” by “ π / 4 π / 4 ”.
f π / 4 = 2 cot π / 4 1 + cot 2 π / 4 f π / 4 = 2 cot π / 4 1 + cot 2 π / 4
As cot π / 4 = 1 cot π / 4 = 1 ,
f π 4 = 2 X 1 1 + 1 2 = 1 f π 4 = 2 X 1 1 + 1 2 = 1
Problem 8: Find f tan θ f tan θ , if
f x = 2 x 1 + x 2 f x = 2 x 1 + x 2
Solution :
Statement of the problem : The given function is a rational function. We have to evaluate the function for a value, which is itself a trigonometric function.
We need to replace “x” by “ tan θ tan θ ”.
f tan θ = 2 tan θ 1 + tan 2 θ = sin 2 θ f tan θ = 2 tan θ 1 + tan 2 θ = sin 2 θ
Problem 9: If f x = cos { log x } f x = cos { log x } , then prove that :
f x y + f x y = 2 f x f y f x y + f x y = 2 f x f y
Solution :
Statement of the problem : The given function, f(x) is a trigonometric function, whose input is a logarithmic function. We have to evaluate LHS of the given equation to equate the same to RHS.
Here, we evaluate each term of the left hand side of the equation separately and then combine the result.
f x y = cos { log e x y } = cos log e x + log e y f x y = cos { log e x y } = cos log e x + log e y
f x y = cos { log e x y } = cos log e x log e y f x y = cos { log e x y } = cos log e x log e y
Substituting in the LHS expression, we have :
{ f x y + f x y } = cos log e x + log e y + cos log e x log e y { f x y + f x y } = cos log e x + log e y + cos log e x log e y
We know that :
cos C + cos D = 2 cos C + D 2 cos C D 2 cos C + cos D = 2 cos C + D 2 cos C D 2
Hence,
{ f x y + f x y } = 2 cos log e x + log e y + log e x log e y 2 cos log e x + log e y log e x + log e y 2 { f x y + f x y } = 2 cos log e x + log e y + log e x log e y 2 cos log e x + log e y log e x + log e y 2
{ f x y + f x y } = 2 cos log e x cos log e y { f x y + f x y } = 2 cos log e x cos log e y
{ f x y + f x y } = 2 cos log e x cos log e y = 2 f x f y { f x y + f x y } = 2 cos log e x cos log e y = 2 f x f y
## Acknowledgment
Author wishes to thank Mr Jay Sicard, Pasadena TX for suggesting valuable typographical correction on the topic.
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## Tuesday, March 8, 2011
### TANCET Ratio Proportion Question
Ratio proportion is a fundamental topic from which you should expect anything from 2 to 3 questions out of the 20 questions in the Math section of TANCET MBA test.
The following question is a ratio question and tests your ability to express one term in terms of another, given the ratio between the two terms.
Question
Sum of the ages of three friends is 98 years. If the ratio of the ages of the oldest to the middle is 5 : 3 and that of the middle to the youngest is 5 : 3, how old is the oldest?
1. 49 years
2. 50 years
3. 30 years
4. 32 years
5. 42 years
The correct answer is 50 years. i.e., Choice (2).
One of the easiest approach to solve this question is outlined below
Let A be the oldest, B the middle one and C the youngest amongst the three friends.
Ratio of the ages of A and B, A : B :: 5 : 3
i.e., A/B = 5/3
Therefore, A = (5/3) B
Ratio of the ages of B and C, B : C :: 5 : 3
i.e, B/C = 5/3.
Hence, C = (3/5)B
We also know that A + B + C = 98
Rewriting the equation in terms of B, we get (5/3)B + B + (3/5)B = 98.
Taking 15 as the common denominator and adding we get (25B + 15B + 9B)/15 = 98
Or (49/15)B = 98
Or B = (98*15)/49 = 30.
Hence, age of A = (5/3)B = (5/3)*30 = 50 years. |
Contributed by:
This pdf includes the following topics:-
Practice Problems 1 and 2
Examples
• Evaluating the Gradient at a Point
In 1-variable calculus, the derivative gives you an
equation for the slope at any x-value along f(x).
You can then plug in an x-value to find the actual
slope at that point. 2 f(x) = x
f’(x) = 2x
Actual tangent line slope is… 10 when x = 5
-4 when x = -2 5 when x = 2.5
0 when x = 0
Similarly, the gradient gives you an equation for
the slope of the tangent plane at any point (x, y)
or (x, y, z) or whatever. You can then plug in the
actual values at any point to find the slope of
the tangent plane at that point.
The slope of the tangent plane will be written as
a vector, composed of the slopes along each
As an example, given the function f(x, y) = 3x2y –
2x and the point (4, -3), the gradient can be
calculated as:
[6xy – 2 3x2]
Plugging in the values of x and y at (4, -3) gives
[-74 48]
which is the value of the gradient at that point.
5. Practice Problems 1 and 2
1. f(x, y) = x2 + y2 at
a) (0, 0) b) (1, 3) c) (-1, -5)
2. f(x, y, z) = x3z – 2y2x + 5z at
a) (1, 1, -4) b) (0, 1, 0) c) (-3, -2, 1)
In the previous example, the function f(x, y) =
3x2y – 2x had a gradient of [6xy – 2 3x2], which
at the point (4, -3) came out to [-74 48].
500
The tangent plane at that
400
300
200
100 (4, -3) point will have a slope of
0
-100
-200
-300
-400
-74 in the x direction and
-500
-600
-700
-800
+48 in the y direction.
3 -1
y axis x axis
-5 7
Even more important is the vector itself, [-74 48].
Here is the graph again, with the vector drawn in
as a vector rather than two sloped lines:
Recall that vectors give us
500
400
300
direction as well as magnitude.
200
100
0
-100
-200
-300
-400
-500
-600
-700
vector will always point in the
3
-800
-1 direction of steepest increase for
y axis x axis
-5 7 the function.
And, its magnitude will give us the slope of the plane in
that direction.
That’s a lot of different slopes!
16 Each component of the gradient
14
vector gives the slope in one
12
10
dimension only.
8
6
4
vector gives the steepest possible
2 5
4
slope of the plane.
3
0 2
0 1
1 2 0
3 4 5 6
Recall that the magnitude can be found using the Pythagorean
Theorem, c2 = a2 + b2, where c is the magnitude and a and b are
the components of the vector.
9. Practice Problem 3
Find the gradient of f(x, y) = 2xy – 2y, and the
a) (0, 0) 40
30
b) (5, -3) 20
10
0
-10
-20
c) (20, 10) -30
-40
d) (-5, 4)
e) Find where the gradient = 0.
10. Practice Problem 4
f(x, y, z, w) = 3xy – 2xw + 5xz – 2yw
and the magnitude of the gradient at (0, 1, -1, 2).
11. Practice Problem 5
Suppose we are maximizing the function
f(x, y) = 4x + 2y – x2 – 3y2
Find the gradient and its magnitude from
a) (1, 5)
b) (3, -2)
c) (2, 0)
d) (-4, -6)
e) Find where the gradient is 0.
12. Practice Problem 6
Suppose you were trying to minimize f(x, y) = x2
+ 2y + 2y2. Along what vector should you travel
from (5, 12)? |
Standard equations , Properties and Application of a ellipse
Ellipse
Definition: An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. The two fixed points are called the foci (plural of ‘focus’) of the ellipse.
The mid point of the line segment joining the foci is called the centre of the ellipse. The line segment through the foci of the ellipse is called the major axis and the line segment through the centre and perpendicular to the major axis is called the minor axis. The end points of the major axis are called the vertices of the ellipse
We denote the length of the major axis by 2a, the length of the minor axis by 2b and the distance between the foci by 2c. Thus, the length of the semi major axis is a and semi-minor axis is b
Relationship between semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse:
Sum of the distances of the point P to the foci is F1 P + F2P = F1O + OP + F2P
(Since, F1P = F1O + OP)
= c + a + a – c = 2a
a2 = b2 + c2
Special cases of an ellipse In the equation c2 = a2 – b2 obtained above, if we keep a fixed and vary c from 0 to a, the resulting ellipses will vary in shape.
Case (i) When c = 0, both foci merge together with the centre of the ellipse and a2 = b2, i.e., a = b, and so the ellipse becomes circle . Thus, circle is a special case of an ellipse.
Case (ii) When c = a, then b = 0. The ellipse reduces to the line segment F1F2 joining the two foci.
Eccentricity
Definition: The eccentricity of an ellipse is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is denoted by e) i.e., e= c/a
Standard equations of an ellipse The equation of an ellipse is simplest if the centre of the ellipse is at the origin and the foci are on the x-axis or y-axis.
Let F1 and F2 be the foci and O be the midpoint of the line segment F1F2. Let O be the origin and the line from O through F2 be the positive x-axis and that through F1as the negative x-axis.
Let, the line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (– c, 0) and F2 be (c, 0) .
Let P(x, y) be any point on the ellipse such that the sum of the distances from P to the two foci be 2a
so given PF1 + PF2 = 2a.
Hence any point on the ellipse satisfies satisfies
the geometric condition and so P(x, y) lies on the ellipse.
Therefore, the ellipse lies between the lines x = – a and x = a and touches these lines.
Similarly, the ellipse lies between the lines y = – b and y = b and touches these lines.
Similarly, we can derive the equation of the ellipse.
These two equations are known as standard equations of the ellipses.
Properties:
1. Ellipse is symmetric with respect to both the coordinate axes since if (x, y) is a point on the ellipse, then (– x, y), (x, –y) and (– x, –y) are also points on the ellipse.
2. The foci always lie on the major axis. The major axis can be determined by finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis if the coefficient of x2 has the larger denominator and it is along the y-axis if the coefficient of y2 has the larger denominator.
Latus rectum
Definition: Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse.
Let the length of AF2 be l. Then the coordinates of A are (c, l ),
i.e., (ae, l )
Since A lies on the ellipse
we have
• l2 = b2 (1 – e2) but e2=c2/a2 = 1- (b2/a2)
Therefore, l2 = b4/a2
• l = b2/a
Since the ellipse is symmetric with respect to y-axis (of course, it is symmetric w.r.t. both the coordinate axes), AF2 = F2B and so length of the latus rectum is 2l =2b2/a
Example: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse
Solution Since denominator of x2/25 is larger than the denominator of y2/9 , the major
axis is along the x-axis.
a = 5 and b = 3. Also
c =√(a2 – b2)
=√(25 – 9)
=4
Therefore, the coordinates of the foci are (– 4,0) and (4,0), vertices are (– 5, 0) and
(5, 0). Length of the major axis is 10 units length of the minor axis 2b is 6 units and the
eccentricity is 4/5. and latus rectum is 2b2/a = 18 /5 |
Question Video: Evaluating 3 × 3 Determinants | Nagwa Question Video: Evaluating 3 × 3 Determinants | Nagwa
# Question Video: Evaluating 3 × 3 Determinants Mathematics • First Year of Secondary School
## Join Nagwa Classes
Find the value of |2 2 6, −3 1 −2, −5 −1 −4|.
01:24
### Video Transcript
Find the value of the determinant of this three-by-three matrix.
So these lines are not absolute value lines; they mean we’re taking the determinant. And the way that we’ll do that is we will take two times the determinant of the numbers that are not in the row or column of that two and then subtract two times the determinant of the numbers that are not in the row or column of this two and then we add six times the determinant of the numbers that are not in the row or the column of the six.
So how do we take the determinant? How do we find it? Well, it’s kind of like cross-multiplying. We take 𝑎 times 𝑑 and then subtract 𝑏 times 𝑐. So we have two times one times negative four minus negative two times negative one minus two times negative three times negative four minus negative two times negative five plus six times negative three times negative one minus one times negative five.
So after multiplying the numbers inside the parentheses, we need to add the numbers now inside the parentheses. Now the very last set, the three minus negative five in the yellow, two negatives make a positive, so we have two times negative six minus two times two plus six times eight. So we have negative 12 minus four plus 48, which gives us an answer of 32.
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The box-and-whisker plot below represents some data set. What percentage of the data values are less than or equal to 110 ?
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Based on the box-and-whisker plot, it can be seen that the lower whisker extends down to 85 and the box starts at 95. This means that the minimum value of the data set is 85 and the first quartile (Q1) is 95.
To determine the percentage of data values that are less than or equal to 110, we need to find the third quartile (Q3) of the data set. From the box-and-whisker plot, we can see that the upper whisker extends up to 150 and the box ends at 125. This means that the maximum value of the data set is 150 and the third quartile (Q3) is 125.
Therefore, the interquartile range (IQR) of the data set is Q3-Q1 = 125-95 = 30.
To find the percentage of data values that are less than or equal to 110, we need to find the z-score of 110 using the formula:
z = (x - mean) / standard deviation
Since we do not know the mean and standard deviation of the data set, we can use the formula for the z-score of a value relative to the quartiles:
z = (x - Q1) / IQR
if x is between Q1 and Q3
z = (x - Q3) / IQR
if x is above Q3
z = (110 - 95) / 30 = 0.50
According to the standard normal distribution table, the area to the left of z = 0.50 is 0.6915 or 69.15%. Therefore, approximately 69.15% of the data values are less than or equal to 110.
Step-by-step solution:
1. Find the minimum value of the data set: 85
2. Find the first quartile (Q1) of the data set: 95
3. Find the third quartile (Q3) of the data set: 125
4. Find the interquartile range (IQR) of the data set: Q3-Q1 = 125-95 = 30
5. Calculate the z-score of 110 using the quartiles: z = (x - Q1) / IQR = (110 - 95) / 30 = 0.50
6. Use the standard normal distribution table to find the area to the left of z = 0.50, which is approximately 0.6915 or 69.15%
7. Therefore, approximately 69.15% of the data values are less than or equal to 110.
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# Plotting Points on a Cartesian Plane
A Cartesian plane (named after French mathematician Rene Descartes, who formalized its use in mathematics) is defined by two perpendicular number lines: the x-axis, which is horizontal, and the y-axis, which is vertical. Using these axes, we can describe any point in the plane using an ordered pair of numbers.
The Cartesian plane extends infinitely in all directions. To show this, math textbooks usually put arrows at the ends of the axes in their drawings.
The location of a point in the plane is given by its coordinates, a pair of numbers enclosed in parentheses: (x, y). The first number x gives the point's horizontal position and the second number y gives its vertical position. All positions are measured relative to a "central" point called the origin, whose coordinates are (0, 0). For example, the point (5, 2) is 5 units to the right of the origin and 2 units up, as shown in the figure. Negative coordinate numbers tell us to go left or down. See the other points in the figure for examples.
The Cartesian plane is divided into four quadrants. These are numbered from I – IV, starting with the upper right and going around counterclockwise. (For some reason everybody uses roman numerals for this).
In Quadrant I, both the x– and y-coordinates are positive; in Quadrant II, the x-coordinate is negative, but the y-coordinate is positive; in Quadrant III both are negative; and in Quadrant IV x is positive but y is negative.
Points which lie on an axis (i.e., which have at least one coordinate equal to 0) are said not to be in any quadrant. Coordinates of the form (x, 0) lie on the horizontal x-axis, and coordinates of the form (0, y) lie on the vertical y-axis. |
# Multivariable Linear Approximation – Proving an expression with a square root in 2 variables – Exercise 3404
Exercise
Prove that the following holds for x and y that are close to 1
$$\sqrt{x^3+3y^2}\approx \frac{3}{4}x+\frac{3}{2}y-\frac{1}{4}$$
Proof
Since the equation we need to prove has the estimator sign, we probably need to use the linear approximation formula. Since there are two variables, we will use the 2-variable linear approximation formula
$$f(x,y)\approx f(x_0,y_0)+f'_x(x_0,y_0)\cdot(x-x_0)+f'_y(x_0,y_0)\cdot(y-y_0)$$
Therefore, we need to define the following
$$x, y, x_0, y_0, f(x)$$
And place them in the formula. x and y will be the numbers that appear in the question, and the we set
$$x_0=1, y_0=1$$
Because we are asked to prove the equation for points close to 1.
$$x_0=1, y_0=1$$
The function will be the right side in the equation that we need to prove
$$f(x,y)=\sqrt{x^3+3y^2}$$
In the formula above we see the function partial derivatives. Hence, we calculate them.
$$f'_x(x,y)=\frac{1}{2\sqrt{x^3+3y^2}}\cdot 3x^2$$
$$f'_y(x,y)=\frac{1}{2\sqrt{x^3+3y^2}}\cdot 6y$$
We put all the data in the formula and get
$$\sqrt{x^3+3y^2}\approx f(1,1)+f'_x(1,1)\cdot(x-1)+f'_y(1,1)\cdot(y-1)=$$
$$=\sqrt{1^3+3\cdot 1^2}+\frac{1}{2\sqrt{1^3+3\cdot 1^2}}\cdot 3\cdot 1^2\cdot (x-1)+\frac{1}{2\sqrt{1^3+3\cdot 1^2}}\cdot 6\cdot 1\cdot (y-1)=$$
$$=\sqrt{4}+\frac{1}{2\sqrt{4}}\cdot 3\cdot (x-1)+\frac{1}{2\sqrt{4}}\cdot 6\cdot (y-1)=$$
$$=2+\frac{3}{4}\cdot (x-1)+\frac{6}{4}\cdot (y-1)=$$
$$=2+\frac{3}{4}x-\frac{3}{4}+\frac{6}{4}y-\frac{6}{4}=$$
$$=\frac{3}{4}x+\frac{3}{2}y-\frac{1}{4}$$
Hence, we get
$$\sqrt{x^3+3y^2}\approx \frac{3}{4}x+\frac{3}{2}y-\frac{1}{4}$$
As required.
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# Integral Test
Integral Test | Derivation \u0026 1st Example
Integral Test | Derivation \u0026 1st Example
## Section 6.3 Integral Test
It is generally quite difficult, often impossible, to determine the value of a series exactly. In many cases it is possible at least to determine whether or not the series converges, and so we will spend most of our time on this problem.
If all of the terms $$\ds a_n$$ in a series are non-negative, then clearly the sequence of partial sums $$\ds s_n$$ is non-decreasing. This means that if we can show that the sequence of partial sums is bounded, the series must converge. Many useful and interesting series have this property, and they are among the easiest to understand. Let’s look at an example.
###### Example 6.35. Exploring Convergence Using an Integral.
Show that $$\ds\sum_{n=1}^\infty {1\over n^2}$$ converges.
The terms $$\ds 1/n^2$$ are positive and decreasing, and since $$\ds\lim_{x\to\infty} 1/x^2=0\text{,}$$ the terms $$\ds 1/n^2$$ approach zero. This means that the Divergence Test does not provide any information and we must find a different method to deal with this series. We seek an upper bound for all the partial sums, that is, we want to find a number $$N$$ so that $$s_n\le N$$ for every $$n\text{.}$$ The upper bound is provided courtesy of integration, and is illustrated in Figure 6.2.
The figure shows the graph of $$\ds y=1/x^2$$ together with some rectangles that lie completely below the curve and that all have base length one. Because the heights of the rectangles are determined by the height of the curve, the areas of the rectangles are $$\ds 1/1^2\text{,}$$ $$\ds 1/2^2\text{,}$$ $$\ds 1/3^2\text{,}$$ and so on—in other words, exactly the terms of the series. The partial sum $$\ds s_n$$ is simply the sum of the areas of the first $$n$$ rectangles. Because the rectangles all lie between the curve and the $$x$$-axis, any sum of rectangle areas is less than the corresponding area under the curve, and so of course any sum of rectangle areas is less than the area under the entire curve. Unfortunately, because of the asymptote at $$x=0\text{,}$$ the integral $$\int_0^{\infty}\frac{1}{x^2}$$ is infinite, but we can deal with this by separating the first term from the series and integrating from 1:
recalling that we computed this improper integral in Section 2.7. Since the sequence of partial sums $$\ds s_n$$ is increasing and bounded above by 2, we know that $$\ds\lim_{n\to\infty}s_n=L\lt 2\text{,}$$ and so the series converges to some number less than 2. In fact, it is possible, though difficult, to show that $$\ds L=\pi^2/6\approx 1.6\text{.}$$
###### Example 6.36. Exploring Divergence Using an Integral.
Why can the integral technique from Example 6.35 not be used with the series
We already know that $$\sum 1/n$$ diverges. What goes wrong if we try to apply the integral technique to it? Here’s the calculation:
The problem is that the improper integral doesn’t converge. Note that this does not prove that $$\sum 1/n$$ diverges, just that this particular technique fails to prove that it converges.
A slight modification, however, allows us to prove in a second way that $$\sum 1/n$$ diverges.
###### Example 6.37. Alternate Method for Divergence of Harmonic Series.
Use the idea of areas from Example 6.35 to show that the series
diverges.
Consider a slightly altered version of Figure 6.2, shown in Figure 6.3.
This time the rectangles are above the curve, that is, each rectangle completely contains the corresponding area under the curve. This means that
As $$n$$ gets bigger, $$\ln(n+1)$$ goes to infinity, so the sequence of partial sums $$\ds s_n$$ must also go to infinity, so the harmonic series diverges.
The key fact in this example is that
So these two examples taken together indicate that we can prove that a series converges or prove that it diverges with a single calculation of an improper integral. This is known as the Integral Test, which we state as a theorem.
###### Theorem 6.38. Integral Test.
Suppose that $$f$$ is a continuous, positive, and decreasing function of $$x$$ on the infinite interval $$[1,\infty)$$ and that $$\ds a_n=f(n)\text{.}$$ Then
either both converge or both diverge.
The two examples we have seen are called $$p$$-series. A $$p$$-series is any series of the form $$\ds \sum 1/n^p\text{.}$$
###### Definition 6.39. $$p$$-Series.
A series of the form $$\ds{\sum_{n=1}^{\infty} \frac{1}{n^p} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \dots + \frac{1}{n^p} + \dots}\text{,}$$ where $$p$$ is a constant, is called a $$p$$-series .
###### Theorem 6.40. $$p$$-Series Test.
Given the $$p$$-series
1. If $$p > 1\text{,}$$ the series converges.
2. If $$p \leq 1\text{,}$$ the series diverges.
###### Proof.
We split the proof into three cases:
1. If $$p>1$$ then $$1-p\lt 0$$ and $$\ds\lim_{R\to\infty}R^{1-p}=0\text{,}$$ so the integral converges.
2. If $$p\lt 1$$ then $$1-p>0$$ and $$\ds\lim_{R\to\infty}R^{1-p}=\infty\text{,}$$ so the integral diverges.
3. If $$p=1\text{,}$$ we have the harmonic series, which we have already shown in two ways to be divergent.
###### Example 6.41. $$p$$-Series Power of Three.
Show that $$\ds\sum_{n=1}^\infty {1\over {n^3}}$$ converges.
We could of course use the Integral Test, but now that we have the theorem we may simply note that this is a $$p$$-series with $$p>1\text{.}$$
###### Example 6.42. $$p$$-Series Power of Four.
Show that $$\ds\sum_{n=1}^\infty {5\over n^4}$$ converges.
We know that if $$\ds \sum_{n=1}^\infty 1/n^4$$ converges then $$\ds \sum_{n=1}^\infty 5/n^4$$ also converges, by Theorem 6.29. Since $$p=4 >1$$ we have that $$\ds \sum_{n=1}^\infty 1/n^4$$ is a convergent $$p$$-series, and so $$\ds \sum_{n=1}^\infty 5/n^4$$ converges also.
###### Example 6.43. $$p$$-Series Square Root.
Show that $$\ds\sum_{n=1}^\infty {5\over \sqrt{n}}$$ diverges.
This also follows from Theorem 6.29: Since $$\ds\sum_{n=1}^\infty {1\over \sqrt{n}}$$ is a $$p$$-series with $$p=1/2\lt 1\text{,}$$ it diverges, and so does $$\ds\sum_{n=1}^\infty {5\over \sqrt{n}}\text{.}$$
Since it is typically difficult to compute the value of a series exactly, a good approximation is frequently required. In a real sense, a good approximation is only as good as we know it is. That is, while an approximation may in fact be good, it is only valuable in practice if we can guarantee its accuracy to some degree. This guarantee is usually easy to come by for series with decreasing positive terms.
###### Example 6.44. Approximating a $$p$$-Series.
Approximate $$\ds \sum_{n=1}^{\infty} 1/n^2$$ to within 0.01.
Referring to Figure 6.2, if we approximate the sum by $$\ds \sum_{n=1}^N 1/n^2\text{,}$$ the size of the error we make is the total area of the remaining rectangles, all of which lie under the curve $$\ds 1/x^2$$ from $$x=N$$ to infinity. So we know the true value of the series is larger than the approximation, and no bigger than the approximation plus the area under the curve from $$N$$ to infinity. Roughly, then, we need to find $$N$$ so that
We can compute the integral:
so if we choose $$N=100$$ the error will be less than 0.01. Adding up the first 100 terms gives approximately $$1.634983900\text{.}$$ In fact, we can do a bit better. Since we know that the correct value is between our approximation and our approximation plus the error (not minus), we can cut our error bound in half by taking the value midway between these two values. If we take $$N=50\text{,}$$ we get a sum of 1.6251327 with an error of at most 0.02, so the correct value is between 1.6251327 and 1.6451327, and therefore the value halfway between these, 1.6351327, is within 0.01 of the correct value. We have mentioned that the true value of this series can be shown to be $$\pi^2/6\approx 1.644934068$$ which is 0.0098 more than our approximation, and so (just barely) within the required error. Frequently approximations will be even better than the “guaranteed” accuracy, but not always, as this example demonstrates.
##### Exercises for Section 6.3.
###### Exercise 6.3.1.
Determine whether each series converges or diverges.
1. $$\ds\sum_{n=1}^\infty {1\over n^{\pi/4}}$$
$$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{\pi/4}}$$ is a $$p$$-series with $$p=\pi/4 \lt 1\text{,}$$ and so the series diverges.
2. $$\ds\sum_{n=1}^\infty {n\over n^2+1}$$
We compute
\begin{equation*} \int_1^{\infty} \frac{x}{x^2+1}\,dx\text{.} \end{equation*}
Let $$u=x^2+1, \ du=2x\,dx\text{.}$$ Then
\begin{equation*} \int_1^{\infty} \frac{x}{x^2+1}\,dx = \int_2^{\infty} \frac{1}{2}\frac{du}{u} = \frac{1}{2}\lim_{a\to\infty} \ln (u) \bigg\vert_2^a = \infty\text{.} \end{equation*}
Thus, by the integral test, the series $$\displaystyle\sum_{n=1}^{\infty} \frac{n}{n^2+1}$$ diverges.
3. $$\ds\sum_{n=1}^\infty {\ln n\over n^2}$$
We use the integral test. To compute $$\displaystyle\int_1^{\infty} \frac{\ln x}{x^2}\,dx\text{,}$$ use integration by parts with
\begin{equation*} \begin{array}{cc} u = \ln x \amp dv = \frac{dx}{x^2} \\ du = \frac{1}{x}\,dx \amp v = -\frac{1}{x} \end{array} \end{equation*}
Therefore,
\begin{equation*} \int \frac{\ln x}{x^2} = -\frac{\ln x}{x} + \int \frac{1}{x^2}\,dx = -\frac{1}{x}\left(\ln x + 1\right) + C\text{.} \end{equation*}
Since
\begin{equation*} \lim_{x\to\infty} \left(\frac{\ln x + 1}{x}\right) = \lim_{x\to\infty} \frac{1}{x} = 0\text{,} \end{equation*}
we see that the integral converges. Thus, the series $$\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n^2}$$ converges.
4. $$\ds\sum_{n=1}^\infty {1\over n^2+1}$$
AnswerSolutionconvergesWe use the integral test.\begin{equation*} \int_{1}^{\infty} \frac{1}{x^2+1}\,dx = \lim_{a\to\infty} \tan^{-1}(x) \big\vert_1^{a} = \frac{\pi}{4}. \end{equation*}Hence, the series converges.
5. $$\ds\sum_{n=1}^\infty {1\over e^n}$$
AnswerSolutionconvergesWe use the integral test:\begin{equation*} \int_1^{\infty} e^{-x} \,dx = \lim_{a\to\infty} -e^{-x} \big\vert_1^a = e^{-1}. \end{equation*}Therefore, the series converges.
6. $$\ds\sum_{n=1}^\infty {n\over e^n}$$
AnswerSolutionconvergesWe first compute the following indefinite integral using integration by parts:\begin{equation*} \int xe^{-x} \,dx = -xe^{-x} + \int e^{-x}\,dx = -xe^{-x}-e^{-x}. \end{equation*}Therefore,\begin{equation*} \int_1^{\infty} xe^{-x} \,dx = -\lim_{a\to\infty} xe^{-x}\big\vert_1^a – \lim_{a\to\infty} e^{-x}\big\vert_1^a = 2e^{-1}. \end{equation*}Therefore, by the integral test, the series converges.
7. $$\ds\sum_{n=2}^\infty {1\over n\ln n}$$
AnswerSolutiondivergesWe first compute the following indefinite integral using the substitution $$u=\ln x\text{,}$$ $$du=1/x\text{:}$$\begin{equation*} \int \frac{1}{x\ln x}\,dx = \int \frac{1}{u}\,du = \ln |u| = \ln \ln (x). \end{equation*}Since\begin{equation*} \lim_{a \to \infty} \ln (\ln a)) = \infty, \end{equation*}the integral\begin{equation*} \int_2^{\infty} \frac{1}{x\ln x}\,dx \end{equation*}diverges. Hence, by the integral test, the series diverges.
8. $$\ds\sum_{n=2}^\infty {1\over n(\ln n)^2}$$
AnswerSolutionconvergesWe first compute the following indefinite integral using the substitution $$u=\ln x\text{,}$$ $$du=1/x\text{:}$$\begin{equation*} \int \frac{1}{x(\ln x)^2}\,dx = \int \frac{1}{u^2}\,du = -\frac{1}{u} = -\frac{1}{\ln x} \end{equation*}Since\begin{equation*} \lim_{a \to \infty} 1/\ln x = 0, \end{equation*}the integral\begin{equation*} \int_2^{\infty} \frac{1}{x(\ln x)^2}\,dx \end{equation*}converges. Hence, by the integral test, the series converges.
###### Exercise 6.3.2.
Find an $$N$$ so that each series is approximated to within the given error.
1. $$\ds\sum_{n=1}^\infty {1\over n^4}\text{,}$$ $$0.005$$
AnswerSolution$$N=5$$
We wish to approximate the series $$\displaystyle{\sum_{n=1}^{\infty} \frac{1}{n^4}}$$ with error $$E \leq 0.005\text{.}$$ We first compute an upper bound for the error which results from using $$N$$ terms:
\begin{equation*} E \leq \int_N^\infty \frac{1}{x^4}\,dx = \lim_{a\to\infty} \left[-\frac{1}{3x^3}\right]_N^a = \frac{1}{3N^3}\text{.} \end{equation*}
Since
\begin{equation*} \frac{1}{3N^3} \lt 0.005 \implies N > 4.0548\text{,} \end{equation*}
we should take $$N=5$$ terms in our approximation.
2. $$\ds\sum_{n=0}^\infty {1\over e^n}\text{,}$$ $$10^{-4}$$
AnswerSolution$$N=10$$We wish to approximate the series $$\displaystyle{\sum_{n=0}^{\infty} \frac{1}{e^n}}$$ with an error $$E \leq 0.0001\text{.}$$ We first compute an upper bound for the error which results from using $$N$$ terms:\begin{equation*} E \leq \int_N^\infty \frac{1}{e^x} \,dx = \lim_{a\to\infty} -e^{-x} \big\vert_N^a = \frac{1}{e^N}. \end{equation*}Since\begin{equation*} e^{-N} \lt 0.0001 \implies N > \log(10,000) \approx 9.21, \end{equation*}this means we should take $$N=10$$ terms in our approximation.
3. $$\ds\sum_{n=1}^\infty {\ln n\over n^2}\text{,}$$ $$0.005$$
Answer$$N=1687$$
4. $$\ds\sum_{n=2}^\infty {1\over n(\ln n)^2}\text{,}$$ $$0.005$$
Answerany integer greater than $$\ds e^{200}$$
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# 6.6 Exponential and logarithmic equations (Page 6/8)
Page 6 / 8
How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?
Access these online resources for additional instruction and practice with exponential and logarithmic equations.
## Key equations
One-to-one property for exponential functions For any algebraic expressions and and any positive real number where if and only if Definition of a logarithm For any algebraic expression S and positive real numbers and where if and only if One-to-one property for logarithmic functions For any algebraic expressions S and T and any positive real number where if and only if
## Key concepts
• We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.
• When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown. See [link] .
• When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown. See [link] , [link] , and [link] .
• When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See [link] .
• We can solve exponential equations with base $\text{\hspace{0.17em}}e,$ by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See [link] and [link] .
• After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See [link] .
• When given an equation of the form $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(S\right)=c,\text{}$ where $\text{\hspace{0.17em}}S\text{\hspace{0.17em}}$ is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation $\text{\hspace{0.17em}}{b}^{c}=S,\text{}$ and solve for the unknown. See [link] and [link] .
• We can also use graphing to solve equations with the form $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(S\right)=c.\text{\hspace{0.17em}}$ We graph both equations $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(S\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=c\text{\hspace{0.17em}}$ on the same coordinate plane and identify the solution as the x- value of the intersecting point. See [link] .
• When given an equation of the form $\text{\hspace{0.17em}}{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T,\text{}$ where $\text{\hspace{0.17em}}S\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}T\text{\hspace{0.17em}}$ are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation $\text{\hspace{0.17em}}S=T\text{\hspace{0.17em}}$ for the unknown. See [link] .
• Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm. See [link] .
## Verbal
How can an exponential equation be solved?
Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5 |
# 2010 USAMO Problems/Problem 4
## Problem
Let $ABC$ be a triangle with $\angle A = 90^{\circ}$. Points $D$ and $E$ lie on sides $AC$ and $AB$, respectively, such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$. Segments $BD$ and $CE$ meet at $I$. Determine whether or not it is possible for segments $AB, AC, BI, ID, CI, IE$ to all have integer lengths.
## Solution
We know that angle $BIC = 135^{\circ}$, as the other two angles in triangle $BIC$ add to $45^{\circ}$. Assume that only $AB, AC, BI$, and $CI$ are integers. Using the Law of Cosines on triangle BIC,
$[asy] import olympiad; // Scale unitsize(1inch); // Shape real h = 1.75; real w = 2.5; // Points void ldot(pair p, string l, pair dir=p) { dot(p); label(l, p, unit(dir)); } pair A = origin; ldot(A, "A", plain.SW); pair B = w * plain.E; ldot(B, "B", plain.SE); pair C = h * plain.N; ldot(C, "C", plain.NW); pair D = extension(B, bisectorpoint(C, B, A), A, C); ldot(D, "D", D-B); pair E = extension(C, bisectorpoint(A, C, B), A, B); ldot(E, "E", E-C); pair I = extension(B, D, C, E); ldot(I, "I", A-I); // Segments draw(A--B); draw(B--C); draw(C--A); draw(C--E); draw(B--D); // Angles import markers; draw(rightanglemark(B, A, C, 4)); markangle(Label("\scriptstyle{\frac{\theta}{2}}"), radius=40, I, B, E); markangle(Label("\scriptstyle{\frac{\theta}{2}}"), radius=40, C, B, I); markangle(Label("\scriptstyle{\frac{\pi}{4} - \frac{\theta}{2}}"), radius=40, I, C, B); markangle(Label("\scriptstyle{\frac{\pi}{4} - \frac{\theta}{2}}"), radius=40, D, C, I); markangle(Label("\scriptstyle{\frac{3\pi}{4}}"), radius=10, B, I, C); [/asy]$
$BC^2 = BI^2 + CI^2 - 2BI\cdot CI \cdot \cos 135^{\circ}$. Observing that $BC^2 = AB^2 + AC^2$ is an integer and that $\cos 135^{\circ} = -\frac{\sqrt{2}}{2},$ we have
$$BC^2 - BI^2 - CI^2 = BI\cdot CI\cdot \sqrt{2}$$
and therefore,
$$\sqrt{2} = \frac{BC^2 - BI^2 - CI^2}{BI\cdot CI}$$
The LHS ($\sqrt{2}$) is irrational, while the RHS is the quotient of the division of two integers and thus is rational. Clearly, there is a contradiction. Therefore, it is impossible for $AB, AC, BI$, and $CI$ to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.
## Solution 2
Suppose otherwise. It is easy to see (through simple angle chasing) that $\angle DIC=45^{\circ}$. Also, since $I$ is the incenter, we have $\angle IAC = 45^{\circ}$. Using the Law of Cosines, we have $$CD^2=IC^2+ID^2-\sqrt{2}(IC)(ID),$$ so that $CD$ is irrational. But $\triangle IAC \sim \triangle DIC$, thus $IC^2=CD\cdot AC$, implying that $CD$ is rational, contradiction. $\blacksquare$
## Solution 3
The result can be also proved without direct appeal to trigonometry, via just the angle bisector theorem and the structure of Pythagorean triples. (This is a lot more work).
A triangle in which all the required lengths are integers exists if and only if there exists a triangle in which $AB$ and $AC$ are relatively-prime integers and the lengths of the segments $BI, ID, CI, IE$ are all rational (we divide all the lengths by the $\gcd(AB, AC)$ or conversely multiply all the lengths by the least common multiple of the denominators of the rational lengths).
Suppose there exists a triangle in which the lengths $AB$ and $AC$ are relatively-prime integers and the lengths $IB, ID, CI, IE$ are all rational.
Since $CE$ is the bisector of $\angle ACB$, by the angle bisector theorem, the ratio $IB : ID = CB : CD$, and since $BD$ is the bisector of $\angle ABC$, $CB : CD = (AB + BC) : AC$. Therefore, $IB : ID = (AB + BC) : AC$. Now $IB : ID$ is by assumption rational, so $(AB + BC) : AC$ is rational, but $AB$ and $AC$ are assumed integers so $BC : AC$ must also be rational. Since $BC$ is the hypotenuse of a right-triangle, its length is the square root of an integer, and thus either an integer or irrational, so $BC$ must be an integer.
With $AB$ and $AC$ relatively-prime, we conclude that the side lengths of $\triangle ABC$ must be a Pythagorean triple: $(2pq, p^2 - q^2, p^2 + q^2)$, with $p > q$ relatively-prime positive integers and $p+q$ odd.
Without loss of generality, $AC = 2pq, AB = p^2 - q^2, BC = p^2+q^2$. By the angle bisector theorem,
\begin{align*} AE &= \dfrac{AB \cdot AC}{AC + CB} = \dfrac{2pq(p^2-q^2)}{p^2 + q^2 + 2pq} = \dfrac{2pq(p-q)}{p+q} \end{align*}
Since $\triangle CAE$ is a right-triangle, we have:
\begin{align*} CE^2 &= AC^2 + AE^2 = 4p^2q^2 + \left(\dfrac{2pq(p-q)}{p+q}\right)^2 = 4p^2q^2\left[1 + \left(\dfrac{p-q}{p+q}\right)^2\right] \\ &= \frac{4p^2q^2}{(p+q)^2}\left[(p+q)^2 + (p-q)^2\right] = \frac{4p^2q^2}{(p+q)^2}(2p^2 + 2q^2) \end{align*}
and so $CE$ is rational if and only if $2p^2 + 2q^2$ is a perfect square.
Also by the angle bisector theorem,
\begin{align*} AD &= \dfrac{AB \cdot AC}{AB + BC} = \dfrac{2pq(p^2-q^2)}{p^2 + q^2 + p^2 - q^2} = \dfrac{q(p^2-q^2)}{p} \end{align*}
and therefore, since $\triangle DAB$ is a right-triangle, we have:
\begin{align*} BD^2 &= AB^2 + AD^2 = (p^2-q^2)^2 + \left(\dfrac{q(p^2-q^2)}{p}\right)^2 \\ &= (p^2-q^2)^2\left[1 + \frac{q^2}{p^2}\right] = \frac{(p^2-q^2)^2}{p^2}(p^2 + q^2) \end{align*}
and so $BD$ is rational if and only if $p^2 + q^2$ is a perfect square.
Combining the conditions on $CE$ and $BD$, we see that $2p^2+2q^2$ and $p^2+q^2$ must both be perfect squares. If it were so, their ratio, which is $2$, would be the square of a rational number, but $\sqrt{2}$ is irrational, and so the assumed triangle cannot exist. |
# 01-LINEAR PROGRAMMING (THEORY)
STUDY INNOVATIONSEducator em Study Innovations
‘Linear Programming’ is a scientific tool to handle optimization problems. Here, we shall learn about some basic concepts of linear programming problems in two variables, their applications, advantages, limitations, formulation and graphical method of solution. 6.1 Linear Inequations . (1) Graph of linear inequations (i) Linear inequation in one variable : etc. are called linear inequations in one variable. Graph of these inequations can be drawn as follows : The graph of and are obtained by dividing xy-plane in two semi-planes by the line (which is parallel to y-axis). Similarly for and . (ii) Linear Inequation in two variables : General form of these inequations are . If any ordered pair satisfies some inequations, then it is said to be a solution of the inequations. The graph of these inequations is given below (for c > 0) : Working rule To draw the graph of an inequation, following procedure is followed : (i) Write the equation in place of and . (ii) Make a table for the solutions of . (iii) Now draw a line with the help of these points. This is the graph of the line . (iv) If the inequation is > or <, then the points lying on this line is not considered and line is drawn dotted or discontinuous. (v) If the ineuqation is > or <, then the points lying on the line is considered and line is drawn bold or continuous. (vi) This line divides the plane XOY in two region. To Find the region that satisfies the inequation, we apply the following rules: (a) Take an arbitrary point which will be in either region. (b) If it satisfies the given inequation, then the required region will be the region in which the arbitrary point is located. (c) If it does not satisfy the inequation, then the other region is the required region. (d) Draw the lines in the required region or make it shaded. (2) Simultaneous linear inequations in two variables : Since the solution set of a system of simultaneous linear inequations is the set of all points in two dimensional space which satisfy all the inequations simultaneously. Therefore to find the solution set we find the region of the plane common to all the portions comprising the solution set of given inequations. In case there is no region common to all the solutions of the given inequations, we say that the solution set is void or empty. (3) Feasible region : The limited (bounded) region of the graph made by two inequations is called feasible region. All the points in feasible region constitute the solution of a system of inequations. The feasible solution of a L.P.P. belongs to only first quadrant. If feasible region is empty then there is no solution for the problem. Example: 1 Inequations and (a) Have solution for positive x and y (b) Have no solution for positive x and y (c) Have solution for all x (d) Have solution for all y Solution : (a) Following figure will be obtained on drawing the graphs of given inequations : From From Clearly the comm
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### 01-LINEAR PROGRAMMING (THEORY)
• 1. 174 Linear Programming ‘Linear Programming’ is a scientific tool to handle optimization problems. Here, we shall learn about some basic concepts of linear programming problems in two variables, their applications, advantages, limitations, formulation and graphical method of solution. 6.1 Linear Inequations . (1) Graph of linear inequations (i) Linear inequation in one variable : 0 , 0 , 0 d cy b ax b ax etc. are called linear inequations in one variable. Graph of these inequations can be drawn as follows : The graph of 0 b ax and 0 b ax are obtained by dividing xy-plane in two semi-planes by the line a b x (which is parallel to y-axis). Similarly for 0 d cy and 0 d cy . (ii) Linear Inequation in two variables : General form of these inequations are c by ax c by ax , . If any ordered pair 1 1 ,y x satisfies some inequations, then it is said to be a solution of the inequations. The graph of these inequations is given below (for c > 0) : Working rule To draw the graph of an inequation, following procedure is followed : (i) Write the equation c by ax in place of c by ax and c by ax . a b x ax+b>0 Y Y X X ax+b<0 O Y Y X X c d y cy+d>0 cy+d<0 Y Y X X ax+by<c ax+by=c ax+by>c O 174
• 2. Linear Programming 175 (ii) Make a table for the solutions of c by ax . (iii) Now draw a line with the help of these points. This is the graph of the line c by ax . (iv) If the inequation is > or <, then the points lying on this line is not considered and line is drawn dotted or discontinuous. (v) If the ineuqation is > or <, then the points lying on the line is considered and line is drawn bold or continuous. (vi) This line divides the plane XOY in two region. To Find the region that satisfies the inequation, we apply the following rules: (a) Take an arbitrary point which will be in either region. (b)Ifitsatisfies thegiven inequation,thentherequiredregion willbethe region inwhichthearbitrarypointis located. (c) If it does not satisfy the inequation, then the other region is the required region. (d) Draw the lines in the required region or make it shaded. (2) Simultaneous linear inequations in two variables : Since the solution set of a system of simultaneous linear inequations is the set of all points in two dimensional space which satisfy all the inequations simultaneously. Therefore to find the solution set we find the region of the plane common to all the portions comprising the solution set of given inequations. In case there is no region common to all the solutions of the given inequations, we say that the solution set is void or empty. (3) Feasible region : The limited (bounded) region of the graph made by two inequations is called feasible region. All the points in feasible region constitute the solution of a system of inequations. The feasible solution of a L.P.P. belongs to only first quadrant. If feasible region is empty then there is no solution for the problem. Example: 1 Inequations 3 3 y x and 4 4 y x (a) Have solution for positive x and y (b) Have no solution for positive x and y (c) Have solution for all x (d) Have solution for all y Solution : (a) Following figure will be obtained on drawing the graphs of given inequations : From 1 3 1 , 3 3 y x y x From 1 4 1 , 4 4 y x y x Clearly the common region of both the inequations is true for positive value of (x, y). It is also true for positive values of xand negative values of y. Example: 2 The constraints [MP PET 1999] – 1 2 1 x x ; – 9 3 2 1 x x ; 0 , 2 x x define (a) Bounded feasible space (b) Unbounded feasible space (c) Both bounded and unbounded feasible space (d) None of these Solution : (b) It is clear from the graph, the constraints define an unbounded feasible space. Example: 3 Shaded region is represented by O (1,0) (0,–3) (0,–4) Y X (–9,0) O (0,1) (3,4) (–1,0) x1 x2 (0,3)
• 3. 176 Linear Programming (a) 0 , 0 , 20 , 80 5 2 y x y x y x (b) 0 , 0 , 20 , 80 5 2 y x y x y x (c) 0 , 0 , 20 , 80 5 2 y x y x y x (d) 0 , 0 , 20 , 80 5 2 y x y x y x Solution: (c) In all the given equations, the origin is present in shaded area. Answer (c) satisfy this condition. Example: 4 Shaded region is represented by (a) 3 2 4 y x (b) 3 2 4 y x (c) 3 2 4 y x (d) 3 2 4 y x Solution: (b) Origin is not present in given shaded area. So 3 2 4 y x satisfy this condition. 6.2 Terms of Linear Programming . The term programming means planning and refers to a process of determining a particular program. (1) Objective function : The linear function which is to be optimized (maximized or minimized) is called objective function of the L.P.P. (2) Constraints or Restrictions : The conditions of the problem expressed as simultaneous equations or inequalities are called constraints or restrictions. (3) Non-negative Constraints : Variables applied in the objective function of a linear programming problem are always non-negative. The ineqaulities which represent such constraints are called non-negative constraints. (4) Basic variables : The m variables associated with columns of the m × n non-singular matrix which may be different from zero, are called basic variables. (5) Basic solution : A solution in which the vectors associated to m variables are linear and the remaining ) ( m n variables are zero, is called a basic solution. A basic solution is called a degenerate basic solution, if at least one of the basic variables is zero and basic solution is called non-degenerate, if none of the basic variables is zero. (6) Feasible solution : The set of values of the variables which satisfies the set of constraints of linear programming problem (L.P.P) is called a feasible solution of the L.P.P. (7) Optimal solution : A feasible solution for which the objective function is minimum or maximum is called optimal solution. (8) Iso-profit line : The line drawn in geometrical area of feasible region of L.P.P. for which the objective function remains constant at all the points lying on the line, is called iso-profit line. 0 , 4 3 B Shaded region A(0,3/2) 4x–2y=–3 y x O 3 40 , 3 20 B Shaded region O A(20,0) x (40,0) 2x+5y=80 x+y=20 y C(0,16) (0,20) Y X
• 4. Linear Programming 177 If the objective function is to be minimized then these lines are called iso-cost lines. (9) Convex set : In linear programming problems feasible solution is generally a polygon in first quadrant . This polygon is convex. It means if two points of polygon are connecting by a line, then the line must be inside to polygon. For example, Figure (i) and (ii) are convex set while (iii) and (iv) are not convex set 6.3 Mathemiatical formulation of a Linear programming problem . There are mainly four steps in the mathematical formulation of a linear programming problem, as mathematical model. We will discuss formulation of those problems which involve only two variables. (1) Identify the decision variables and assign symbols x and y to them. These decision variables are those quantities whose values we wish to determine. (2) Identify the set of constraints and express them as linear equations/inequations in terms of the decision variables. These constraints are the given conditions. (3) Identify the objective function and express it as a linear function of decision variables. It might take the form of maximizing profit or production or minimizing cost. (4) Add the non-negativity restrictions on the decision variables, as in the physical problems, negative values of decision variables have no valid interpretation. 6.4 Graphical solution of two variable Linear programming problem . There are two techniques of solving an L.P.P. by graphical method. These are : (1) Corner point method, and (2) Iso-profit or Iso-cost method. (1) Corner point method Working rule (i) Formulate mathematically the L.P.P. (ii) Draw graph for every constraint. (iii) Find the feasible solution region. (iv) Find the coordinates of the vertices of feasible solution region. (v) Calculate the value of objective function at these vertices. (vi) Optimal value (minimum or maximum) is the required solution. Note : If there is no possibility to determine the point at which the suitable solution found, then the solution of problem is unbounded. If feasible region is empty, then there is no solution for the problem. Nearer to the origin, the objective function is minimum and that of further from the origin, the objective function is maximum. (2) Iso-profit or Iso-cost method : Various steps of the method are as follows : (i) Find the feasible region of the L.P.P. (ii) Assign a constant value Z1 to Z and draw the corresponding line of the objective function. (iii) Assign another value Z2 to Z and draw the corresponding line of the objective function. (i) (ii) (iv A B (iii A B
• 5. 178 Linear Programming (iv) If ) ( , 2 1 2 1 Z Z Z Z , then in case of maximization (minimization) move the line P1Q1 corresponding to Z1 to the line P2Q2 corresponding to Z2 parallel to itself as far as possible, until the farthest point within the feasible region is touched by this line. The coordinates of the point give maximum (minimum) value of the objectivefunction. Note : The problem with more equations/inequations can be handled easily by this method. In case of unbounded region, it either finds an optimal solution or declares an unbounded solution. Unbounded solutions are not considered optimal solution. In real world problems, unlimited profit or loss is not possible. 6.5 To find the Vertices of Simple feasible region without Drawing a Graph . (1) Bounded region: The region surrounded by the inequalities m by ax and n dy cx in first quadrant is called bounded region. It is of the form of triangle or quadrilateral. Change these inequalities into equation, then by putting 0 x and , 0 y we get the solution also by solving the equation in which there may be the vertices of bounded region. The maximum value of objective function lies at one vertex in limited region. (2) Unbounded region : The region surrounded by the inequations m by ax and n dy cx in first quadrant, is called unbounded region. Change the inequation in equations and solve for 0 x and 0 y . Thus we get the vertices of feasible region. The minimum value of objective function lies at one vertex in unbounded region but there is no existence of maximum value. 6.6 Problems having Infeasible Solutions. In some of the linear programming problems, constraints are inconsistent i.e. there does not exist any point which satisfies all the constraints. Such type of linear programming problems are said to have infeasible solution. For Example : Maximize y x Z 2 5 subject to the constraints 2 y x , 2 3 3 y x , 0 , y x The above problem is illustrated graphically in the fig. There is no point satisfying the set of above constraints. Thus, the problem is having an infeasible solution. 6.7 Some important points about L.P.P.. (1) If the constraints in a linear programming problem are changed, the problem is to be re-evaluated. (2) The optimal value of the objective function is attained at the point, given by corner points of the feasible region. (3) If a L.P.P. admits two optimal solutions, it has an infinite number of optimal solutions. (4) If there is no possibility to determine the point at which the suitable solution can be found, then the solution of problem is unbounded. (5) The maximum value of objective function lies at one vertex in limited region. Example: 5 A firm makes pants and shirt. A shirt takes 2 hour on machine and 3 hour of man labour while a pant takes 3 hour on machine and 2 hour of man labour. In a week there are 70 hour of machine and 75 hour of man labour available. If the firm determines to make x shirts and y pants per week, then for this the linear constraints are (a) 75 2 3 , 70 3 2 , 0 , 0 y x y x y x (b) 75 2 3 , 70 3 2 , 0 , 0 y x y x y x (c) 75 2 3 , 70 3 2 , 0 , 0 y x y x y x (d) 75 2 3 , 70 3 2 , 0 , 0 y x y x y x Solution: (d) 1 10 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10
• 6. Linear Programming 179 Working time on machine Man labour Shirt (x) 2 hours 3 hours Pant (y) 3 hours 2 hours Availability 70 hours 75 hours Linear constraints are 75 2 3 , 70 3 2 y x y x and 0 , y x Example: 6 The minimum value of the objective function y x Z 10 2 for linear constraints 5 5 , 0 y x y x and 0 , y x is (a) 10 (b) 15 (c) 12 (d) 8 Solution: (b) Required region is unbounded whose vertex is 4 5 , 4 5 Hence the minimum value of objective function is 15 4 5 10 4 5 2 . Example: 7 Minimize n j m i ij ijx c z 1 1 Subject to : m i a x i n j ij ........, 1 , 1 ; m i j ij n j b x 1 ......, 1 , is a LPP with number of constraints (a) n m (b) n m (c) mn (d) n m Solution: (a) (I) Condition, 1 1 13 12 11 . .......... , 1 a x x x x i n 2 2 23 22 21 . .......... , 2 a x x x x i n 3 3 33 32 31 .. .......... , 3 a x x x x i n ................................... s constraint . .......... , 3 2 1 m a x x x x m i m mn m m m (II) Condition 1 1 31 21 11 .......... , 1 b x x x x j m 2 2 32 22 12 . .......... , 2 b x x x x j m .................................... n b x x x x n j n mn n n n .... .......... , 3 2 1 constraints Total constraints n m . Example: 8 Maximize , 2 3 y x z subject to 3 , 6 , 1 , 0 5 , 1 x y x y x x y y x and 0 , y x (a) 3 x (b) 3 y (c) 15 z (d) All the above Solution: (d) The shaded region represents the bounded region (3,3) satisfies, so 3 , 3 y x and 15 z . Example: 9 Maximum value of y x 5 4 subject to the constraints 12 3 , 35 2 , 20 y x y x y x is (a) 84 (b) 95 (c) 100 (d) 96 Solution: (b) Obviously, max. 95 5 4 y x . It is at (5, 15). y – 5x=0 x – y = –1 (5/2, 7/2) (3,3) x + y = 6 x=3 (0,1) (1,0) x – y=0 x – 5y = –5 O X Y (5,15) (18,2) x – 3y = 12 x + 2y = 35 (12,0)(20,0) (35,0) (0,20) 2 1 17 , 0
• 7. 180 Linear Programming Example: 10 The maximum value of y x Z 3 4 subjected to the constraints , 160 2 3 y x 0 , , 80 2 , 200 2 5 y x y x y x is (a) 320 (b) 300 (c) 230 (d) None of these Solution: (d) Obviously, it is unbounded. Therefore its maximum value does not exist. Example: 11 The objective function y x Z 3 4 can be maximized subjected to the constraints 0 , ; 6 , 5 , 48 6 8 , 24 4 3 y x y x y x y x (a) Atonlyonepoint (b) Attwopointsonly (c) Ataninfinitenumberofpoints(d)None of these Solution: (c) Obviously, the optimal solution is found on the line which is parallel to 'isoprofit line'. Hence it has infinite number of solutions. Example: 12 The maximum value of y x µ 4 3 subjected to the conditions 0 , ; 60 2 , 40 y x y x y x is [MP PET 2002,04 ] (a) 130 (b) 120 (c) 40 (d) 140 Solution: (d) Obviously Max ) 20 , 20 ( at 4 3 y x µ 140 80 60 µ . 6.8 Advantages and Limitations of L.P.P.. (1) Advantages : Linear programming is used to minimize the cost of production for maximum output. In short, with the help of linear programming models, a decision maker can most efficiently and effectively employ his production factor and limited resources to get maximum profit at minimum cost. (2) Limitations : (i) The linear programming can be applied only when the objective function and all the constraints can be expressed in terms of linear equations/inequations. (ii) Linear programming techniques provide solutions only when all the elements related to a problem can be quantified. (iii) The coefficients in the objective function and in the constraints must be known with certainty and should remain unchanged during the period of study. (iv) Linear programming technique may give fractional valued answer which is not desirable in some problems. 3x+2y=160 x+2y=80 5x+2y=200 8x+6y=48 3x+4y=24 4x+3y=0 y = 6 x = 5 Isoprofit line A(0,30) O C (40, 0) B(20, 20) x+2y=60 (60,0) x+y=40 (0,40) Y X
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# COMPARING DATA DISPLAYED IN BOX PLOTS WORKSHEET
## About "Comparing data displayed in box plots worksheet"
Comparing data displayed in box plots worksheet :
Worksheet on comparing data displayed in box plots is much useful to the students who would like to practice problems on comparing box plots.
## Comparing data displayed in box plots worksheet - Problems
Problem 1 :
The box plots show the distribution of times spent shopping by two different groups.
Questions :
1. Compare the shapes of the box plots.
2. Compare the centers of the box plots.
3. Compare the spreads of the box plots.
4. Which group has the greater variability in the bottom 50% of shopping times ? The top 50% of shopping times ? Explain how you know.
Problem 2 :
The box plots show the distribution of the number of team wristbands sold daily by two different stores over the same time period.
Questions :
1. Compare the shapes of the box plots.
2. Compare the centers of the box plots.
3. Compare the spreads of the box plots.
## Comparing data displayed in box plots worksheet - Answers
Answers for the questions in problem 1 :
1. Compare the shapes of the box plots.
The positions and lengths of the boxes and whiskers appear to be very similar. In both plots, the right whisker is shorter than the left whisker.
2. Compare the centers of the box plots.
Group A’s median, 47.5, is greater than Group B’s, 40. This means that the median shopping time for Group A is 7.5 minutes more.
3. Compare the spreads of the box plots.
The box shows the interquartile range. The boxes are similar.
Group A: 55 - 30 = 25 min Group B: About 59 - 32 = 27 min
The whiskers have similar lengths, with Group A’s slightly shorter than Group B’s.
4. Which group has the greater variability in the bottom 50% of shopping times ? The top 50% of shopping times ? Explain how you know.
Group A; Group B; look at which box plot has a greater distance from the median to the minimum or maximum value, respectively.
Answers for the questions in problem 2 :
1. Compare the shapes of the box plots.
Store A’s box and right whisker are longer than Store B’s.
2. Compare the centers of the box plots.
Store A’s median is about 43, and Store B’s is about 51. Store A’s median is close to Store B’s minimum value, so about 50% of Store A’s daily sales were less than sales on Store B’s worst day.
3. Compare the spreads of the box plots.
Store A has a greater spread. Its range and inter quartile range are both greater.
Four of Store B’s key values are greater than Store A’s corresponding value.
Store B had a greater number of sales overall.
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```Purpose – The purpose of this lesson is to teach students how to complete long division
math problems using manipulatives.
Topic: Long Division with Manipulatives Grade 4
Sunshine State Standards: Supporting Idea 6: Number and Operations
Benchmark – MA.4.A.6.2: Use models to represent division as: the inverse of
multiplication, as partitioning, as successive subtraction.
Objective: Using base ten blocks or other manipulatives students will use the partitive
model to calculate solutions to division exercises with 2 digit dividends and 1 digit
divisors with 4 out of 5 correct.
Materials:
Students: Set of base 10 blocks for each student or pair of students.
Teacher: Overhead base 10 blocks for teacher, overhead projector, overhead pens
Motivation
You already know how to do simple division exercises with your base 10 blocks. You
have been finding how many of one size set can be made from a larger set by repeated
subtraction. For example, if I have 15 pieces of candy (the larger set) and I want to give
3 pieces to some of my friends. If I want to give 3 pieces to each person, how many
friends can I share my candy with? Let’s use our base 10 blocks to solve the problems
Take out 15 base 10 blocks. (1 long and 5 units) (write problem 15 divided by 3 on
board). Call on student for answer. Ask student to “tell us how you used the blocks to
solve the problem” (response should be that students exchanged blocks to units and
created 5 sets of 3 each)
Continue the review with problems: 25 divided by 5 (create 5 sets of 5 each) and 21
divided by 7 (create 3 sets of 7 each) .
You have been solving these problems using the concepts of repeated subtraction. Today
we are going to learn how to use our base 10 blocks to solve a different kind of problem
that cannot be solved by repeated subtraction. Today our first question is: I have three
friends and I have 36 pieces of candy. I want to distribute them evenly among my
friends. How many pieces will each friend get?
Lesson Body Part I - Presentation
Now watch me as I work some of our new kinds of problems on the overhead. The first
problem I am going to solve is 36 divided by 3 So I can find out how many pieces of
candy to give each friend.
First I take out 3 ten strips and 6 units and put them on my place holder sheet.
Now, I have 3 friends so what I want to do is make 3 piles of blocks which will represent
my candy.
So I look at the 10’s place and I ask, can I take the strips and divide them into three piles.
Yes I can. (show on overhead.
Next I look at the units and ask, can I divide them evenly into 3 piles and add them to the
piles I have created, Yes I can. (show on overhead)
Now I count how many blocks I have in each pile. I have 12 in each pile. So 36 divided
by 3 is 12. I can give each of my 3 friends 12 pieces of candy
Let’s try another. This time we will see what happens if we want to share 45 pieces of
candy among our 3 friends. (Write 45 divided by 3 on the board)
First I lay out 4 ten strips (longs) and 5 units on my placeholder pad to represent the
candy.
Now I look at the tens place. Can I make three piles with the strips in the 10’s place?
Lets try. (Lay out three piles of 1 ten strip each) I can make three piles, but I have a ten
strip left. Can I divide this strip? Yes I can if I exchange it for 10 units. (Make the
exchange) Now I have 15 units. Can I distribute them into three piles? Let’s see (do it).
Now let’s count how many are in each pile. Each pile has 15. So 45 divided by 4 is 15
(write the answer as a quotient on the board). This time each of my 4 friends gets 15
pieces of candy.
Here’s another problem. I want to have a birthday party and invite 4 friends. Mom
bought a box of 96 party favors. How many can I give to each friend?
First get out 9 longs and 6 units and lay them out on my place holder sheet. Look at the
longs. Can I divide them into 4 groups? Response (yes put 2 in each group) How many
longs do I have left? Response: (1) What should I do with the one that is left? (exchange
for 10 units) Now how many units do I have after the exchange? Response (16) Can I
distribute these into 4 groups? (yes 4 in each to make 24 in each group)
Lesson body Part 2 Guided practice:
Now I am going to ask you to help me solve the following problem:
Suppose Mom bought 42 pencils for my brother and me to share. How many should each
of us receive? Write 42 divided by 2 on board.
Ask: “What is the first thing I should do?” Response: (Lay, 4 ten strips, and 2 units)
Ask: “What should I do next?” Response: (See if you can divide the tens into two piles)
Can I? (Yes)
What do I do now? Response: (distribute the units into two piles). Can I? (Yes)
Now I have distributed all the blocks. How do I know the solution to the problem.
Response: (Count the blocks in each pile)
How many are there? 21. Yes so 42 divided by 2 is 21 (write on board)
Now use your base ten blocks to solve the following problem:
63 divided by 3
When students have worked problem, have one student come to front of room and use
overhead blocks to solve the problem and explain what he or she is doing.
Continue the practice with the following problems:
75 divided by 5; 88 divided by 8; 72 divided by 6.
Now let’s draw a problem on the board and on your paper to show how we use our base
ten blocks.
56 divided by 4 (Draw out problem on place holder model sheet) Have students tell you
the steps to solve as you demonstrate). Draw 5 longs and 4 units. Draw 4 circles on the
board to represent the 4 groups. As each long is distributed into a group, draw it in the
circle and cross off one long on the place holder sheet. Cross off the left over long and
draw 10 units. Then cross out each unit as it is distributed and drawn in the circles.
If necessary continue with the following practice problems having students solve, draw
and explain solutions 42 divided by 3, 50 divided by 5, 21 divided by 7, 96 divided by
8,
Lesson Body Part III: Independent Practice
Hand out worksheet with 10 word problems for students to solve using the partitioning
model with base 10 blocks and drawings: (Attach worksheet to lesson plan.)
Closure:
Today you have learned how to use your manipulatives to solve division problems by
distributing base ten blocks one place at a time starting with the tens place and then the
ones. You have learned to make exchanges when you have tens left over that you cannot
distribute evenly into established groups. This is called the partitive model of division.
Tomorrow we will learn how to relate what we did today to solving problems using paper
and pencil when we don’t have our base ten blocks.
Extended Individual Practice:
For homework you have these ten problems to practice solving using the partitive model.
I want you to draw pictures of how you would distribute your base ten blocks for each
solution.
Worksheet with 10 problems for students to solve:
Assessment:
Give students 5 word problems to solve using base 10 blocks. Have students draw
solutions they model with blocks. Example: Barbara’s mom gave her a bag of mini
candy bars. She told Barbara she could share the candy with her 2 brothers. How many
pieces of candy will each of the 3 children receive if there were 45 candy bars in the
bag? Students must show the model of division by partitioning.
``` |
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# Solving Linear Recurrence Relations
Niloufar Shafiei
Review
A recursive definition of a sequence specifies Initial conditions Recurrence relation Example:
a0=0 and a1=3 an = 2an-1 - an-2 an = 3n
## Initial conditions Recurrence relation Solution
1
Linear recurrences
Linear recurrence: Each term of a sequence is a linear function of earlier terms in the sequence. For example: a0 = 1 a1 = 6 a2 = 10 an = an-1 + 2an-2 + 3an-3 a3 = a0 + 2a1 + 3a2 = 1 + 2(6) + 3(10) = 43
2
Linear recurrences
Linear recurrences 1. Linear homogeneous recurrences 2. Linear non-homogeneous recurrences
## Linear homogeneous recurrences
A linear homogenous recurrence relation of degree k with constant coefficients is a recurrence relation of the form an = c1an-1 + c2an-2 + + ckan-k, where c1, c2, , ck are real numbers, and ck 0 . an is expressed in terms of the previous k terms of the sequence, so its degree is k. This recurrence includes k initial conditions. a0 = C0 a1 = C1 ak = Ck
4
Example
Determine if the following recurrence relations are linear homogeneous recurrence relations with constant coefficients. Pn = (1.11)Pn-1 a linear homogeneous recurrence relation of degree one an = an-1 + a2n-2 not linear fn = fn-1 + fn-2 a linear homogeneous recurrence relation of degree two Hn = 2Hn-1+1 not homogeneous an = an-6 a linear homogeneous recurrence relation of degree six Bn = nBn-1 does not have constant coefficient
5
## Solving linear homogeneous recurrences
Proposition 1: Let an = c1an-1 + c2an-2 + + ckan-k be a linear homogeneous recurrence. Assume the sequence an satisfies the recurrence. Assume the sequence an also satisfies the recurrence. So, bn = an + an and dn= an are also sequences that satisfy the recurrence. ( is any constant) Proof: bn = an + an = (c1an-1 + c2an-2 + + ckan-k) + (c1an-1 + c2an-2 + + ckan-k) = c1 (an-1+ an-1) + c2 (an-2+ an-2) + + ck (an-k + an-k) = c1bn-1 + c2bn-2 + + ckbn-k So, bn is a solution of the recurrence.
## Solving linear homogeneous recurrences
Proposition 1: Let an = c1an-1 + c2an-2 + + ckan-k be a linear homogeneous recurrence. Assume the sequence an satisfies the recurrence. Assume the sequence an also satisfies the recurrence. So, bn = an + an and dn= an are also sequences that satisfy the recurrence. ( is any constant) Proof: dn = an = (c1an-1 + c2an-2 + + ckan-k) = c1 ( an-1) + c2 ( an-2) + + ck ( an-k) = c1dn-1 + c2dn-2 + + ckdn-k So, dn is a solution of the recurrence.
## Solving linear homogeneous recurrences
It follows from the previous proposition, if we find some solutions to a linear homogeneous recurrence, then any linear combination of them will also be a solution to the linear homogeneous recurrence.
## Solving linear homogeneous recurrences
Geometric sequences come up a lot when solving linear homogeneous recurrences. So, try to find any solution of the form an = rn that satisfies the recurrence relation.
## Solving linear homogeneous recurrences
Recurrence relation an = c1an-1 + c2an-2 + + ckan-k Try to find a solution of form rn rn = c1rn-1 + c2rn-2 + + ckrn-k rn - c1rn-1 - c2rn-2 - - ckrn-k = 0 rk - c1rk-1 - c2rk-2 - - ck = 0 (dividing both sides by rn-k)
## This equation is called the characteristic equation.
10
Example
Example: The Fibonacci recurrence is Fn = Fn-1 + Fn-2 Its characteristic equation is r2 - r - 1 = 0
11
## Solving linear homogeneous recurrences
Proposition 2: r is a solution of rk - c1rk-1 - c2rk-2 - - ck = 0 if and only if rn is a solution of an = c1an-1 + c2an-2 + + ckan-k.
Example: consider the characteristic equation r2 - 4r + 4 = 0. r2 - 4r + 4 = (r - 2)2 = 0 So, r=2. So, 2n satisfies the recurrence Fn = 4Fn-1 - 4Fn-2. 2n = 4 . 2n-1 - 4 . 2n-2 2n-2 ( 4 - 8 + 4) = 0
12
## Solving linear homogeneous recurrences
Theorem 1: Consider the characteristic equation rk - c1rk-1 - c2rk-2 - - ck = 0 and the recurrence an = c1an-1 + c2an-2 + + ckan-k. Assume r1, r2, and rm all satisfy the equation. Let 1, 2, , m be any constants. So, an = 1 r1n + 2 r2n + + m rmn satisfies the recurrence. Proof: By Proposition 2, i rin satisfies the recurrence. So, by Proposition 1, i i rin satisfies the recurrence. Applying Proposition 1 again, the sequence an = 1 r1n + + m rmn satisfies the recurrence.
2 r2
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Example
What is the solution of the recurrence relation an = an-1 + 2an-2 with a0=2 and a1=7? Solution: Since it is linear homogeneous recurrence, first find its characteristic equation r2 - r - 2 = 0 (r+1)(r-2) = 0 r1 = 2 and r2 = -1 So, by theorem an = 12n + 2(-1)n is a solution. Now we should find 1 and 2 using initial conditions. a0= 1 + 2 = 2 a1= 12 + 2(-1) = 7 So, 1= 3 and 2 = -1. an = 3 . 2n - (-1)n is a solution.
14
Example
What is the solution of the recurrence relation fn = fn-1 + fn-2 with f0=0 and f1=1? Solution: Since it is linear homogeneous recurrence, first find its characteristic equation r2 - r - 1 = 0 )/2 and r2 = (1)/2 r1 = (1+ )/2)n + 2((1)/2)n is a solution. So, by theorem fn = 1((1+ Now we should find 1 and 2 using initial conditions. f0= 1 + 2 = 0 f1= 1(1+ )/2 + 2 (1)/2 = 1 So, 1= 1/ and 2 = -1/ . an = 1/ . ((1+ )/2)n - 1/ ((1)/2)n is a solution.
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Example
What is the solution of the recurrence relation an = -an-1 + 4an-2 + 4an-3 with a0=8, a1=6 and a2=26? Solution: Since it is linear homogeneous recurrence, first find its characteristic equation r3 + r2 - 4r - 4 = 0 (r+1)(r+2)(r-2) = 0 r1 = -1, r2 = -2 and r3 = 2 So, by theorem an = 1(-1)n + 2(-2)n + 32n is a solution. Now we should find 1, 2 and 3 using initial conditions. a 0= 1 + 2 + 3 = 8 a 1= - 1 - 2 2 + 2 3 = 6 a2= 1 + 4 2 + 4 3 = 26 So, 1= 2, 2 = 1 and 3 = 5. an = 2 . (-1)n + (-2)n + 5 . 2n is a solution.
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## Solving linear homogeneous recurrences
If the characteristic equation has k distinct solutions r1, r2, , rk, it can be written as (r - r1)(r - r2)(r - rk) = 0. If, after factoring, the equation has m+1 factors of (r r1), for example, r1 is called a solution of the characteristic equation with multiplicity m+1. When this happens, not only r1n is a solution, but also nr1n, n2r1n, and nmr1n are solutions of the recurrence.
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## Solving linear homogeneous recurrences
Proposition 3: Assume r0 is a solution of the characteristic equation with multiplicity at least m+1. So, nmr0n is a solution to the recurrence.
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## Solving linear homogeneous recurrences
When a characteristic equation has fewer than k distinct solutions: We obtain sequences of the form described in Proposition 3. By Proposition 1, we know any combination of these solutions is also a solution to the recurrence. We can find those that satisfies the initial conditions.
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## Solving linear homogeneous recurrences
Theorem 2: Consider the characteristic equation rk - c1rk-1 - c2rk-2 - ck = 0 and the recurrence an = c1an-1 + c2an-2 + + ckan-k. Assume the characteristic equation has t k distinct solutions. Let i (1 i t) ri with multiplicity mi be a solution of the equation. Let i,j (1 i t and 0 j mi-1) ij be a constant. So, an = ( 10 + 11 n+ + 1,m1-1 nm1-1 ) r1n + ( 20 + 21 n+ + 2,m2-1 nm2-1 ) r2n + + ( t0 + t1 n+ + t,m -1 nmt-1 ) rtn t satisfies the recurrence.
20
Example
What is the solution of the recurrence relation an = 6an-1 - 9an-2 with a0=1 and a1=6? Solution: First find its characteristic equation r2 - 6r + 9 = 0 (r - 3)2 = 0 r1 = 3 (Its multiplicity is 2.) So, by theorem an = ( 10 + 11n)(3)n is a solution. Now we should find constants using initial conditions. a0= 10 = 1 a1= 3 10 + 3 11 = 6 So, 11= 1 and 10 = 1. an = 3n + n3n is a solution.
21
Example
What is the solution of the recurrence relation an = -3an-1 - 3an-2 - an-3 with a0=1, a1=-2 and a2=-1? Solution: Find its characteristic equation r3 + 3r2 + 3r + 1 = 0 (r + 1)3 = 0 r1 = -1 (Its multiplicity is 3.) So, by theorem an = ( 10 + 11n + 12n2)(-1)n is a solution. Now we should find constants using initial conditions. a0= 10 = 1 a1= - 10 - 11 - 12 = -2 a2= 10 + 2 11 + 4 12 = -1 So, 10= 1, 11 = 3 and 12 = -2. an = (1 + 3n - 2n2) (-1)n is a solution.
22
Example
What is the solution of the recurrence relation an = 8an-2 - 16an-4 , for n 4, with a0=1, a1=4, a2=28 and a3=32? Solution: Find its characteristic equation r4 - 8r2 + 16 = 0 (r2 - 4)2 = (r-2)2 (r+2)2 = 0 r1 = 2 r2 = -2 (Their multiplicities are 2.) So, by theorem an = ( 10 + 11n)(2)n + ( 20 + 21n)(-2)n is a solution. Now we should find constants using initial conditions. a0= 10 + 20 = 1 a1= 2 10 + 2 11 - 2 20 - 2 21 = 4 a2= 4 10 + 8 11 + 4 20 + 8 21 = 28 a3= 8 10 + 24 11 - 8 20 - 24 21 = 32 So, 10= 1, 11 = 2, 20 = 0 and 21 = 1. an = (1 + 2n) 2n + n (-2)n is a solution.
23
## Linear non-homogeneous recurrences
A linear non-homogenous recurrence relation with constant coefficients is a recurrence relation of the form an = c1an-1 + c2an-2 + + ckan-k+ f(n), where c1, c2, , ck are real numbers, and f(n) is a function depending only on n. The recurrence relation an = c1an-1 + c2an-2 + + ckan-k, is called the associated homogeneous recurrence relation. This recurrence includes k initial conditions. a0 = C0 a 1 = C1 ak = Ck
24
Example
The following recurrence relations are linear nonhomogeneous recurrence relations. an = an-1 + 2n an = an-1 + an-2 + n2 + n + 1 an = an-1 + an-4 + n! an = an-6 + n2n
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## Linear non-homogeneous recurrences
Proposition 4: Let an = c1an-1 + c2an-2 + + ckan-k + f(n) be a linear non-homogeneous recurrence. Assume the sequence bn satisfies the recurrence. Another sequence an satisfies the nonhomogeneous recurrence if and only if hn = an - bn is also a sequence that satisfies the associated homogeneous recurrence.
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## Linear non-homogeneous recurrences
Proof: Part1: if hn satisfies the associated homogeneous recurrence then an is satisfies the non-homogeneous recurrence. bn = c1bn-1 + c2bn-2 + + ckbn-k + f(n) hn = c1hn-1 + c2hn-2 + + ckhn-k bn + hn = c1 (bn-1+ hn-1) + c2 (bn-2+ hn-2) + + ck (bn-k + hn-k) + f(n) Since an = bn + hn, an = c1an-1 + c2an-2 + + ckan-k+ f(n). So, an is a solution of the non-homogeneous recurrence.
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## Linear non-homogeneous recurrences
Proof: Part2: if an satisfies the non-homogeneous recurrence then hn is satisfies the associated homogeneous recurrence. bn = c1bn-1 + c2bn-2 + + ckbn-k + f(n) an = c1an-1 + c2an-2 + + ckan-k+ f(n) an - bn = c1 (an-1- bn-1) + c2 (an-2- bn-2) + + ck (an-k - bn-k) Since hn = an - bn, hn = c1hn-1 + c2hn-2 + + ckhn-k So, hn is a solution of the associated homogeneous recurrence.
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## Linear non-homogeneous recurrences
Proposition 4: Let an = c1an-1 + c2an-2 + + ckan-k + f(n) be a linear nonhomogeneous recurrence. Assume the sequence bn satisfies the recurrence. Another sequence an satisfies the non-homogeneous recurrence if and only if hn = an - bn is also a sequence that satisfies the associated homogeneous recurrence.
We already know how to find hn. For many common f(n), a solution bn to the non-homogeneous recurrence is similar to f(n). Then you should find solution an = bn + hn to the nonhomogeneous recurrence that satisfies both recurrence and initial conditions.
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Example
What is the solution of the recurrence relation an = an-1 + an-2 + 3n + 1 for n 2, with a0=2 and a1=3? Solution: Since it is linear non-homogeneous recurrence, bn is similar to f(n) Guess: bn = cn + d bn = bn-1 + bn-2 + 3n + 1 cn + d = c(n-1) + d + c(n-2) + d + 3n + 1 cn + d = cn - c + d + cn -2c + d +3n + 1 0 = (3+c)n + (d-3c+1) c = -3 d=-10 So, bn = - 3n - 10. (bn only satisfies the recurrence, it does not satisfy the initial conditions.) 30
Example
What is the solution of the recurrence relation an = an-1 + an-2 + 3n + 1 for n 2, with a0=2 and a1=3? Solution: We are looking for an that satisfies both recurrence and initial conditions. an = bn + hn where hn is a solution for the associated homogeneous recurrence: hn = hn-1 + hn-2 By previous example, we know hn = 1((1+ )/2)n + 2((1)/2)n. an = bn + hn = - 3n - 10 + 1((1+ )/2)n + 2((1)/2)n Now we should find constants using initial conditions. a0= -10 + 1 + 2 = 2 a1= -13 + 1 (1+ )/2 + 2 (1)/2 = 3 1= 6 + 2 2= 6 - 2 So, an = -3n - 10 + (6 + 2 )((1+ )/2)n + (6 - 2 )((1)/2)n.
31
Example
What is the solution of the recurrence relation an = 2an-1 - an-2 + 2n for n 2, with a0=1 and a1=2? Solution: Since it is linear non-homogeneous recurrence, bn is similar to f(n) Guess: bn = c2n + d bn = 2bn-1 - bn-2 + 2n c2n + d = 2(c2n-1 + d) - (c2n-2 + d) + 2n c2n + d = c2n + 2d - c2n-2 - d + 2n 0 = (-4c + 4c - c + 4)2n-2 + (-d + 2d -d) c=4 d=0 So, bn = 4 . 2n. (bn only satisfies the recurrence, it does not satisfy the initial conditions.) 32
Example
What is the solution of the recurrence relation an = 2an-1 - an-2 + 2n for n 2, with a0=1 and a1=2? Solution: We are looking for an that satisfies both recurrence and initial conditions. an = bn + hn where hn is a solution for the associated homogeneous recurrence: hn = 2hn-1 - hn-2. Find its characteristic equation r2 - 2r + 1 = 0 (r - 1)2 = 0 r1 = 1 (Its multiplicity is 2.) So, by theorem hn = ( 1 + 2n)(1)n = 1 + 2n is a solution.
33
Example
What is the solution of the recurrence relation an = 2an-1 - an-2 + 2n for n 2, with a0=1 and a1=2? Solution: an = bn + hn an = 4 . 2n + 1 + 2n is a solution. Now we should find constants using initial conditions. a0= 4 + 1 = 1 a1= 8 - 1 + 2 = 2 1 = -3 2 = -3 So, an = 4 . 2n - 3n - 3.
34
Recommended exercises
1,3,15,17,19,21,23,25,31,35 Eric Rupperts Notes about Solving Recurrences (http://www.cse.yorku.ca/course_archive/2007 -08/F/1019/A/recurrence.pdf)
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# How do you solve the system x= 175+15y, .196x= 10.4y, z=10*y?
Oct 18, 2016
$x = 53.05$
$y = - 8.13$
$z = - 81.3$
#### Explanation:
To begin, plug in a equation for a variable that is already isolated. There are two variables that are already isolated, and those are $x \mathmr{and} z$. You could choose either to work with, doesn't matter, but I'm going to use $x$
So, plug the equation for $x$ into another equation that has that same variable in it. Which would be:
$.196 x = 10.4 y$
To begin, plug $x$ equation into $.196 x = 10.4 y$
$.196 \left(175 + 15 y\right) = 10.4 y$
Distribute $.196$ throughout the set of parenthesis
$34.3 + 2.94 y = 10.4 y$
Begin to isolate $y$ by subtracting $34.3$ on both sides of the equation
$2.94 y = - 23.9$
Isolate y by dividing $2.94$ on both sides of the equation
$y = - 8.13$
Now, we have solved for one of the variables. The next thing to do is plug $y$ into another equation that contains $y$ in it. An easy one to use would be $x = 175 + 15 y$ because $x$ is already isolated
So, plug $y = - 8.13$ into $x = 175 + 15 y$
$x = 175 + 15 \left(- 8.13\right)$
Distribute $15$ throughout the set of parenthesis
$x = 175 - 121.95$
Subtract
$x = 53.05$
And now we have solved for $y$ and $x$. The only variable left is $z$
To solve for $z$, you have to plug $y = - 8.13$ into the equation
$z = 10 \left(- 8.13\right)$
Multiply
$z = - 81.3$
$x = 53.05$
$y = - 8.13$
$z = - 81.3$ |
Learning Objectives for Section 10.4 The Derivative
# Learning Objectives for Section 10.4 The Derivative
## Learning Objectives for Section 10.4 The Derivative
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##### Presentation Transcript
1. Learning Objectives for Section 10.4 The Derivative • The student will be able to calculate rate of change and slope of the tangent line. • The student will be able to interpret the meaning of the derivative. • The student will be able to identify the nonexistence of the derivative. Barnett/Ziegler/Byleen Business Calculus 11e
2. The Rate of Change For y = f (x), the average rate of change from x = a to x = a + h is The above expression is also called a difference quotient. It can be interpreted as the slope of a secant. See the picture on the next slide for illustration. Barnett/Ziegler/Byleen Business Calculus 11e
3. Visual Interpretation Q slope f (a + h) – f (a) Average rate of change = slope of the secant line through P and Q P h Barnett/Ziegler/Byleen Business Calculus 11e
4. Example 1 The revenue generated by producing and selling widgets is given by R(x) = x (75 – 3x)for 0 x 20. What is the change in revenue if production changes from 9 to 12? Barnett/Ziegler/Byleen Business Calculus 11e
5. Example 1 The revenue generated by producing and selling widgets is given by R(x) = x (75 – 3x)for 0 x 20. What is the change in revenue if production changes from 9 to 12? R(12) – R(9) = \$468 – \$432 = \$36. Increasing production from 9 to 12 will increase revenue by \$36. Barnett/Ziegler/Byleen Business Calculus 11e
6. Example 1(continued) The revenue is R(x) = x (75 – 3x)for 0 x 20. What is the average rate of change in revenue (per unit change in x) if production changes from 9 to 12? Barnett/Ziegler/Byleen Business Calculus 11e
7. Example 1(continued) The revenue is R(x) = x (75 – 3x)for 0 x 20. What is the average rate of change in revenue (per unit change in x) if production changes from 9 to 12? To find the average rate of change we divide the change in revenue by the change in production: Thus the average change in revenue is \$12 when production is increased from 9 to 12. Barnett/Ziegler/Byleen Business Calculus 11e
8. The Instantaneous Rate of Change Consider the function y = f (x) only near the point P = (a, f (a)). The difference quotient gives the average rate of change of f over the interval [a, a+h]. If we make h smaller and smaller, in the limit we obtain the instantaneous rate of change of the function at the point P: Barnett/Ziegler/Byleen Business Calculus 11e
9. Visual Interpretation Q Tangent Slope of tangent =instantaneous rate of change. f (a + h) – f (a) P Let h approach 0 h Barnett/Ziegler/Byleen Business Calculus 11e
10. Instantaneous Rate of Change Given y = f (x), the instantaneous rate of change at x = a is provided that the limit exists. It can be interpreted as the slope of the tangent at the point (a, f (a)). See illustration on previous slide. Barnett/Ziegler/Byleen Business Calculus 11e
11. The Derivative For y = f (x), we define the derivative of f at x, denoted f ’ (x), to be if the limit exists. If f ’(a) exists, we call fdifferentiable at a. If f ’(x) exist for each x in the open interval (a, b), then f is said to be differentiable over (a, b). Barnett/Ziegler/Byleen Business Calculus 11e
12. Interpretations of the Derivative • If f is a function, then f ’ is a new function with the following interpretations: • For each x in the domain of f ’,f ’ (x) is the slope of the line tangent to the graph of f at the point (x, f (x)). • For each x in the domain of f ’,f ’ (x) is the instantaneous rate of change of y = f (x) with respect to x. • If f (x) is the position of a moving object at time x, then v = f ’ (x) is the velocity of the object at that time. Barnett/Ziegler/Byleen Business Calculus 11e
13. Finding the Derivative To find f ‘ (x), we use a four-step process: Step 1. Find f (x + h) Step 2. Find f (x + h) – f (x) Step 3. Find Step 4. Find Barnett/Ziegler/Byleen Business Calculus 11e
14. Example 2 Find the derivative of f (x) = x 2 – 3x. Barnett/Ziegler/Byleen Business Calculus 11e
15. Example 2 Find the derivative of f (x) = x 2 – 3x. Step 1.f (x + h) = (x + h)2 – 3(x + h) = x2 + 2xh + h2 – 3x – 3h Step 2. Find f (x + h) – f (x) = 2xh + h2 – 3h Step 3. Find Step 4. Find Barnett/Ziegler/Byleen Business Calculus 11e
16. Example 3 Find the slope of the tangent to the graph of f (x) = x 2 – 3x at x = 0, x = 2, and x = 3. Barnett/Ziegler/Byleen Business Calculus 11e
17. Example 3 Find the slope of the tangent to the graph of f (x) = x 2 – 3x at x = 0, x = 2, and x = 3. Solution: In example 2 we found the derivative of this function at x to be f ’ (x) = 2x – 3 Hence f ’ (0) = -3 f ’ (2) = 1, and f ’ (3) = 3 Barnett/Ziegler/Byleen Business Calculus 11e
18. Graphing Calculators Most graphing calculators have a built-in numerical differentiation routine that will approximate numerically the values of f ’ (x) for any given value of x. Some graphing calculators have a built-in symbolic differentiation routine that will find an algebraic formula for the derivative, and then evaluate this formula at indicated values of x. Barnett/Ziegler/Byleen Business Calculus 11e
19. Example 4 We know that the derivative of f (x) = x 2 – 3x is f ’ (x) = 2x – 3. Verify this for x = 2 using a graphing calculator. Barnett/Ziegler/Byleen Business Calculus 11e
20. Example 4 We know that the derivative of f (x) = x 2 – 3x is f ’ (x) = 2x – 3. Verify this for x = 2 using a graphing calculator. Using dy/dx under the “calc” menu. Using tangent under the “draw” menu. slope tangent equation Barnett/Ziegler/Byleen Business Calculus 11e
21. Example 5 Find the derivative of f (x) = 2x – 3x2 using a graphing calculator with a symbolic differentiation routine. Barnett/Ziegler/Byleen Business Calculus 11e
22. Example 5 Find the derivative of f (x) = 2x – 3x2 using a graphing calculator with a symbolic differentiation routine. Using algebraic differentiation under the home calc menu. derivative Barnett/Ziegler/Byleen Business Calculus 11e
23. Example 6 Find the derivative of f (x) = 2x – 3x2 using the four-step process. Barnett/Ziegler/Byleen Business Calculus 11e
24. Example 6 Find the derivative of f (x) = 2x – 3x2 using the four-step process. Step 1.f (x + h) = 2(x + h) – 3(x + h)2 Step 2.f (x + h) – f (x) = 2h – 6xh - 3h2 Step 3. Step 4. Barnett/Ziegler/Byleen Business Calculus 11e
25. Nonexistence of the Derivative The existence of a derivative at x = a depends on the existence of the limit If the limit does not exist, we say that the function is nondifferentiableat x = a, or f ’ (a) does not exist. Barnett/Ziegler/Byleen Business Calculus 11e
26. Nonexistence of the Derivative(continued) • Some of the reasons why the derivative of a function may not exist at x = a are • The graph of f has a hole or break at x = a, or • The graph of f has a sharp corner at x = a, or • The graph of f has a vertical tangent at x = a. Barnett/Ziegler/Byleen Business Calculus 11e
27. Summary • For y = f (x), we defined the derivative of f at x, denotedf ’ (x), to be if the limit exists. • We have seen how to find the derivative algebraically, using the four-step process. Barnett/Ziegler/Byleen Business Calculus 11e |
# If the point $P(2,1)$ lies on the line segment joining points $A(4,2)$ and $B(8,4)$, then(A) $\mathrm{AP}=\frac{1}{3} \mathrm{AB}$(B) $\mathrm{AP}=\mathrm{PB}$(C) $\mathrm{PB}=\frac{1}{3} \mathrm{AB}$(D) $\mathrm{AP}=\frac{1}{2} \mathrm{AB}$
Given:
The point $P(2,1)$ lies on the line segment joining points $A(4,2)$ and $B(8,4)$.
To do:
We have to choose the correct option.
Solution:
We know that,
The distance between two points $\left(x_{1}, y_{1}\right)$ and $(x_{2}, y_{2})$, d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$The distance between$A(4,2)$and$P(2,1)$is,$A P=\sqrt{(2-4)^{2}+(1-2)^{2}}=\sqrt{(-2)^{2}+(-1)^{2}}=\sqrt{4+1}=\sqrt{5}$The distance between$A(4,2)$and$B(8,4)$is,$A B =\sqrt{(8-4)^{2}+(4-2)^{2}}=\sqrt{(4)^{2}+(2)^{2}}=\sqrt{16+4}=\sqrt{20}=2 \sqrt{5}$The distance between$B(8,4)$and$P(2,1)$is,$B P=\sqrt{(8-2)^{2}+(4-1)^{2}}=\sqrt{6^{2}+3^{2}}=\sqrt{36+9}=\sqrt{45}=3 \sqrt{5}A B =2 \sqrt{5}=2 A PA P=\frac{A B}{2}\$
Hence, the required condition is $\mathrm{AP}=\frac{A B}{2}$.
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Updated on: 10-Oct-2022
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# How do you simplify (2t+4)3+6(-5t)-(-8)?
Sep 18, 2016
$6 t - 10$
#### Explanation:
Remove the brackets using the distributive law. Be careful of the signs.
$\left(2 t + 4\right) \textcolor{b l u e}{3} + 6 \left(- 5 t\right) - \left(- 8\right)$
=$\textcolor{b l u e}{3} \left(2 t + 4\right) \textcolor{red}{+ 6 \left(- 5 t\right)} \textcolor{\lim e}{- \left(- 8\right)} \text{ } \leftarrow$ there are 3 terms
=$\textcolor{b l u e}{6 t + 12} \textcolor{red}{- 30} \textcolor{\lim e}{+ 8}$
=$6 t + 12 + 8 - 30 \text{ } \leftarrow$ re-arrange for easier calculation
=$6 t - 10$ |
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## Chapter 5 Arithmetic Progressions Ex 5.2
Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:
Solution:
Question 2.
Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, …, is
(A) 97
(B) 77
(C) -77
(D) -87
Solution:
(i) 10, 7, 4, …,
a = 10, d = 7 – 10 = -3, n = 30
an = a + (n – 1)d
⇒ a30 = a + (30 – 1) d = a + 29 d = 10 + 29 (-3) = 10 – 87 = – 77
Hence, correct option is (C).
Question 3.
In the following APs, find the missing terms in the boxes:
Solution:
Question 4.
Which term of the AP: 3, 8, 13, 18, …, is 78?
Solution:
Given: 3, 8, 13, 18, ………,
a = 3, d = 8 – 3 = 5
Let nth term is 78
an = 78
a + (n – 1) d = 78
⇒ 3 + (n – 1) 5 = 78
⇒ (n – 1) 5 = 78 – 3
⇒ (n – 1) 5 = 75
⇒ n – 1 = 15
⇒ n = 15 + 1
⇒ n = 16
Hence, a16 = 78
Question 5.
Solution:
Question 6.
Check, whether -150 is a term of the AP: 11, 8, 5, 2, ….
Solution:
Question 7.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
a11 = 38 and a16 = 73
⇒ a11 = a + (11 – 1) d ⇒ a + 10d = 38 ….. (i)
⇒ a16 = a + (16 – 1 )d ⇒ a + 15d = 73 …(ii)
Subtracting eqn. (i) from (ii), we get
a + 15d – a – 10d = 73 – 38
⇒ 5d = 35
⇒ d = 1
From (i), a + 10 x 7 = 38
⇒ a = 38 – 70 = – 32
a31 = a + (31 – 1) d = a + 30d = – 32 + 30 x 7 = – 32 + 210 = 178
Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
Question 9.
If the 3rd and the 9th term of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
Given: a3 = 4 and a9 = – 8
⇒ a3 = a + (3 – 1 )d ⇒ a + 2d = 4 …(i)
a9 = a + (9 – 1) d ⇒ a + 8d = -8 ….(ii)
Subtracting eqn. (i) from (ii), we get
a + 8d – a – 2d = -8 – 4
⇒ 6d = -12.
⇒ d = -2
Now,
a + 2d = 4
⇒ a + 2(-2) = 4
⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
Let an = 0
⇒ a + (n – 1) d = 0
⇒ 8 + (n – 1) (- 2) = 0
⇒ 8 = 2 (n – 1)
⇒ n – 1 = 4
⇒ n = 4 + 1 = 5
Hence, 5th term is zero.
Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Given: a17 – a10 = 7
⇒ [a + (17 – 1 ) d] – [a + (10 – 1 ) d] = 7
⇒ (a + 16d) – (a + 9d) = 7
⇒ 7d = 7
⇒ d = 1
Question 11.
Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?
Solution:
3, 15, 27, 39, …..
Here, a = 3, d = 15 – 3 = 12
Let an = 132 + a54
⇒ an – a54 = 132
⇒ [a + (n – 1) d] – [a + (54 – 1) d] = 132
⇒ a + nd – d – a – 53d = 132
⇒ 12n – 54d = 132
⇒ 12n – 54 x 12 = 132
⇒ (n – 54)12 = 132
⇒ n – 54 = 11
⇒ n = 11 + 54 = 65
Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let a and A be the first term of two APs and d be the common difference.
Given:
a100 – A100 = 100
⇒ a + 99d – A – 99d = 100
⇒ a – A = 100
⇒ a1000 – A1000 = a + 999d – A – 999d
⇒ a – A = 100
⇒ a1000 – A1000 = 100
Question 13.
How many three-digit numbers are divisible by 7?
Solution:
The three-digit numbers which are divisible by 7 are 105, 112, 119, ………., 994
Here, a = 105, d = 112 – 105 = 7 , an = 994
a + (n – 1) d = 994
⇒ 105 + (n – 1) 7 = 994
⇒ (n – 1) 7 = 994 – 105
⇒ 7 (n – 1) = 889
⇒ n – 1 = 127
⇒ n = 127 + 1 = 128
Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
The multiples of 4 between 10 and 250 be 12, 16, 20, 24,…., 248
Here, a = 12, d = 16 – 12 = 4, an = 248
an = a + (n – 1) d
⇒ 248 = 12 + (n – 1) 4
⇒ 248 – 12 = (n – 1) 4
⇒ 236 = (n – 1) 4
⇒ 59 = n – 1
⇒ n = 59 + 1 = 60
Question 15.
For what value of n, the nth term of two APs: 63, 65, 61,… and 3, 10, 17,… are equal?
Solution:
First AP
63, 65, 67,…
Here, a = 63, d = 65 – 63 = 2
an = a + (n – 1) d = 63 + (n – 1) 2 = 63 + 2n – 2 = 61 + 2n
Second AP
3, 10, 17, …
Here, a = 3, d = 10 – 3 = 7
an = a + (n – 1) d = 3 + (n – 1)7 = 3 + 7n – 7 = 7n – 4
Now, an = an
⇒ 61 + 2n = 7n – 4
⇒ 61 + 4 = 7n – 2n
⇒ 65 = 5n
⇒ n = 13
Question 16.
Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.
Solution:
Given: a3 = 16
⇒ a + (3 – 1)d = 16
⇒ a + 2d = 16
and a7 – a5 = 12
⇒ [a + (7 – 1 )d] – [a + (5 – 1 )d] = 12
⇒ a + 6d – a – 4d = 12
⇒ 2d = 12
⇒ d = 6
Since a + 2d = 16
⇒ a + 2(6) = 16
⇒ a + 12 = 16
⇒ a = 16 – 12 = 4
a1 = a = 4
a2 = a1 + d = a + d = 4 + 6 = 10
a3 = a2 + d = 10 + 6 = 16
a4 = a3 + d = 16 + 6 = 22
Thus, the required AP is a1, a2, a3, a4,…, i.e. 4, 10, 16, 22
Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.
Solution:
Given: AP is 3, 8, 13,…….. ,253
On reversing the given A.P., we have
253, 248, 243 ,………, 13, 8, 3.
Here, a = 253, d = 248 – 253 = -5
a20 = a + (20 – 1)d = a + 19d = 253 + 19 (-5) = 253 – 95 = 158
Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000 ?
Solution:
a = ₹ 5000, d = ₹ 200
Let an = ₹ 7000
We have, a + (n – 1) d = 7000
⇒ 5000 + (n – 1) 200 = 7000
⇒ (n – 1) 200 = 7000 – 5000
⇒ (n – 1) 200 = 2000
⇒ (n- 1) = 10
⇒ n = 11
⇒ 1995 + 11 = 2006
Hence, in 2006 Subba Rao’s income will reach ₹ 7000.
Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly saving by ₹ 1.75. If in the nth week, her weekly saving become ₹ 20.75, find n.
Solution:
Given: a = ₹ 5, d = ₹ 1.75
an = ₹ 20.75
a + (n – 1) d – 20.75
⇒ 5 + (n – 1) 1.75 = 20.75
⇒ (n – 1) x 1.75 = 20.75 – 5
⇒ (n – 1) 1.75 = 15.75
⇒ n – 1 = 9
⇒ n = 9 + 1
⇒ n = 10
Hence, in 10th week Ramkali’s saving will be ₹ 20.75.
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Perimeter and Area Class 4 Worksheets
# Perimeter and Area Class 4 Worksheets
## Perimeter and Area Class 4 Worksheets
Perimeter and Area Class 4 worksheets play a vital role in strengthening the maths concepts. After studying the concepts of mathematics, revision of questions related to that particular topic is very important for the students. This assesses the preparation of the students. Students can assess themselves by its own by solving the worksheets problems. Parents and teachers can also assess the students by giving them these worksheets to the students to solve.
Maths-formula brings such worksheets on Perimeter and Area Class 4 for you. By solving these worksheets, you can increase the understanding of the concepts learnt. These worksheets are prepared by experienced teachers and subject matter experts.
Thus, let us try to attempt these questions to solve and assess yourself. Parents and teachers can also give these questions to the students to test them. Unit tests can also be prepared by the teachers using these questions.
## Perimeter and Area Class 4 Worksheets
1. Find the perimeter of the squares whose dimensions are given below.
a. side = 7 cm
b. side = 5 m
c. side = 24 cm
d. side = 3.4 m
e. side = 36 m
f. side = 57 m
2. Find the perimeter of the rectangles whose dimensions are given below.
a. length = 32 cm, breadth = 19 cm
b. length = 6.5 m, breadth = 4.2 m
c. length = 13 m, breadth = 9 m
d. length = 56 cm, breadth = 44 cm
e. length = 7.6 m, breadth = 5 m
f. length = 10.4 cm, breadth = 7.5 cm
g. length = 84 cm, breadth = 76 cm
3. Find the side of the square, whose perimeter is given below.
a. P = 24 cm
b. P = 32 m
c. P = 5.6 m
d. P = 80 cm
e. P = 44.4 m
f. P = 140 cm
4. Find the area of the squares whose dimensions are given below.
a. side = 7 cm
b. side = 4.8 m
c. side = 2.4 cm
d. side = 16.5 cm
e. side = 47 cm
f. side = 18.5 m
5. Find the area of the rectangles whose dimensions are given below.
a. length = 8 cm, breadth = 6 cm
b. length = 4.2 cm, breadth = 2.8 cm
c. length = 9.8 m, breadth = 7.5 m
d. length = 12.3 cm, breadth = 6.4 cm
e. length = 7.8 cm, breadth = 5.8 cm
f. length = 16.2 m, breadth = 10.5 m
6. Find the measure of the side of a square, if its area is
a. 64 sq. cm b. 25 sq. m
c. 81 sq. cm d. 196 sq. m
e. 121 sq. cm f. 49 sq. m
g. 100 sq. cm h. 36 sq. m
7. Nita wanted to make a lawn of 16 m in length and 6 m in breadth. What will be the area and perimeter of the lawn?
8. Find the length of barbed wire needed for fencing around a rectangular park of length 30 m and breadth 25 m.
9. The playground of a school was 48 m long and 35 m broad. Find the area and perimeter of the field.
10. The length and the breadth of a basketball court is 10 m and 8 m, respectively. Find its area and perimeter.
11. each side of a chess board measures 40 cm. Find the area and perimeter of the chess board.
12. Each side of a ludo board is 30 cm long. Find the perimeter and area of the ludo board.
1. a. 28 cm b. 20 m
c. 96 cm d. 13.6 m
e. 144 m f. 228 m
2. a. 102 cm b. 21.4 m
c. 44 m d. 200 cm
e. 25.2 m f. 35.8 cm
g. 320 cm
3. a. 6 cm b. 8 m
c. 1.4 m d. 20 cm
e. 11.1 m f. 35 cm
4. a. 49 sq. cm
b. 23.04 sq. m
c. 5.76 sq. cm
d. 272.25 sq. cm
e. 2209 cm
f. 342.25 sq. m
5. a. 48 sq. cm
b. 11.76 sq. cm
c. 73.5 sq. m
d. 78.72 sq. cm
e. 45.24 sq. cm
f. 170.1 sq. m
6. a. 8 cm b. 5 m
c. 9 cm d. 14 m
e. 11 cm f. 7 m
g. 10 cm h. 6 m
7. Area = 96 sq. m;
Perimeter = 44 m
8. 110 m
9. Area = 1680 sq. m;
Perimeter = 166 m
10. Area = 80 sq. m;
Perimeter = 36 m
11. Area = 1600 sq. cm;
Perimeter = 160 cm
12. Perimeter = 120 cm;
Area = 900 sq. cm
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# Two Squares Have Sides X Cm and (X + 4) Cm. the Sum of Their Area is 656 Sq. Cm. Express this as an Algebraic Equation in X and Solve the Equation to Find the Sides of the Squares. - ICSE Class 10 - Mathematics
#### Question
Two squares have sides x cm and (x + 4) cm. The sum of their area is 656 sq. cm. Express this as an algebraic equation in x and solve the equation to find the sides of the squares.
#### Solution
Given that, two squares have sides x cm and (x + 4) cm.
Sum of their area = 656 cm2
∴ x2 + (x + 4)2 = 656
x2 + x2 + 16 + 8x = 656
2x2 + 8x – 640 = 0
x2 + 4x – 320 = 0
x2 + 20x – 16x – 320 = 0
x(x + 20) – 16(x + 20) = 0
(x + 20) (x – 16) = 0
x = -20, 16
But, x being side, cannot be negative.
So, x = 16
Thus, the sides of the two squares are 16 cm and 20 cm.
Is there an error in this question or solution?
#### APPEARS IN
Selina Solution for Selina ICSE Concise Mathematics for Class 10 (2018-2019) (2017 to Current)
Chapter 6: Solving (simple) Problems (Based on Quadratic Equations)
Ex.6B | Q: 8
#### Video TutorialsVIEW ALL [5]
Solution Two Squares Have Sides X Cm and (X + 4) Cm. the Sum of Their Area is 656 Sq. Cm. Express this as an Algebraic Equation in X and Solve the Equation to Find the Sides of the Squares. Concept: Quadratic Equations.
S |
• Written By Sumana_C
# Place Value of Numbers: Detailed Explanation
Place Value of Numbers: Students must understand the concept of the place value of numbers to score high in the exam. In mathematics, place value refers to the relative importance of each digit in a number. These locations begin at the unit’s location, that is, one’s place. The order of the place value of digits is expressed in the following order: ones/units, tens, hundreds, thousands, ten thousand, and so on.
## Place Value: Overview
The value of each digit in a number is known as the place value. Every digit in a number has a distinct value depending on its location. The value of a digit depends on its position in the number, which might result in a number having two comparable digits with distinct values.
## What does Place Value Mean?
Place value refers to how much a digit is worth in relation to where it is in a number, such as the ones, tens, hundreds, and so forth. For instance, 5 in 3458 has a place value of 5 tens, or 50. However, 5 thousand, or 5,000, is used to represent the place value of 5 in 5781. It is crucial to understand that a digit can be the same, but its value relies on its position in the number.
Example: In the number 543, write down the place value of each digit.
We have expressed the correct place value of each digit in the number.
• 3 × 1 = 3 or 3 ones
• 4 × 10 = 40 or 4 tens
• 5 × 100 = 500 or 5 hundreds
## Place Value Chart
We can use place value diagrams to check that the digits are positioned correctly. The right place for each digit in a number is shown on a place value chart. We must first write the given digits in the place value chart to check their position before we can accurately determine the positional values or worth of different digits in a number. The higher numbers are divided into periods separated with the aid of commas to simplify the process. The most widely used place value charts fall into two categories:
• Chart of Indian place values
• International table of place values
Based on the number system that both charts use, we might refer to the International or Indian place value chart. The international place value chart is based on the internationally recognised numeral system, while the Indian place value chart is based on the Indian numeral system. The positioning of the separators (commas) and the name of various place values are the two fundamental differences between the Indian and the International numeric systems.
### Place Value of Numbers: Indian Place Value Chart
Get introduced to a table, the Indian place value chart used with the Indian numeral system to determine the value of each digit in a number based on its position. A 10-digit number is divided into periods of ones, thousands, lakhs, crores, and so on in this place value chart. The 3:2:2 rule is used to separate these integers with commas. This indicates that beginning on the right, the first comma should be placed after three digits, followed by commas every two digits. Consider the commas in the following number, for instance: 5,43,13,62,283
Look at the Indian place value chart in the section below, which displays the place value of digits up to ten crores.
### Place Value of Numbers: International Place Value
Worldwide, using the International Numerical System, people count in units of ones, tens, hundreds, thousands, ten thousand, hundred thousand, millions, and so on. The numbers in this place value chart are organised into periods of ones, thousands, millions, and so forth. We use a comma to separate each set of three digits, starting from the right. Look at the commas in the following number, for instance: 135,912,332. Take note of the international place value chart, which is provided below and displays the place value of digits up to 100,000,000.
## Printable Place Value Charts
A printable place value chart simplifies learning and solving problems involving place value systems. Place values are printed across blank spaces in these charts in a tabular format for each digit in a number. While using these charts to solve a problem, we may directly set the digits in their proper positions following the one in the number, and we can assess their place values appropriately.
## Place Value Chart with Decimals
The place values of the digits in a decimal number are displayed on the decimal place value chart. A decimal point is used to express both whole numbers and fractions in a decimal number system. This decimal point sits between the component of whole numbers and the component of fractions. The place value of the numbers to the right of the decimal point differs slightly from that of the whole number portion, which follows the standard place value chart of ones, tens, hundreds, and so on. The place values start at tenths and increase as we move to the right of the decimal point, becoming hundredths, thousandths, and so forth. The one-tenth (1/10) position is the first place to the right of the decimal, and the following one is 1/100 and continues.
## Place Value vs Face Value
Any number’s digit’s face value is the digit itself. Regardless of whether a number is single, double, or any other number, each digit has a face value. Let’s use the following examples to grasp this better.
If the provided number is 4, it has a face value of 4 and a place value of 4 (4 ones = 4*1 = 4).
7 has a face value of 7 and a place value of 70 for the given number 78 (7 tens = 7*10 = 70).
The place value of 3 in the number 52369 is 300, but its face value is 3 (3 hundreds = 3*100 = 300).
## Difference between Face Value and Place Value
The place value of a digit in a given integer describes that digit’s location. Face value, on the other hand, is the actual value of the number. Let us use the number 1437 as an example. The table below explains the difference between the numbers’ place value and face value.
## Place Value in Numbers
We can express the place value in numbers in two different ways. For instance, we can express the place value of 4 in 1469 as 4 hundred or 400. Let us take another example. We can express the place value of 4 in 4592 as 4 thousand or 4,000.
To put it simply, we multiply the number by the name of the column it belongs under to convey place value in the numeric form. Let us take the number 76529. To find the place value of 5, for instance, we would first place the supplied number under the place value chart, ensuring that each digit is correctly positioned under its corresponding column. As we can see in this instance, 5 is located in the hundreds column. Therefore, we can state that 5 multiplied by 100 equals 500. As a result, in the provided number, 5 has a place value of 500.
We hope you found this article on the place value of a number helpful. For more such content, keep browsing Embibe!
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$\begin{array}{r}2\cdot \frac{2\sqrt[3]{5}}{2\sqrt[3]{5}}\cdot \sqrt[3]{2}\\\\2\cdot 1\cdot \sqrt[3]{2}\end{array}$. $\frac{\sqrt[3]{640}}{\sqrt[3]{40}}$. \\ & = 15 x \sqrt { 2 } - 5 \cdot 2 x \\ & = 15 x \sqrt { 2 } - 10 x \end{aligned}\). However, this is not the case for a cube root. \\ &= \frac { \sqrt { 20 } - \sqrt { 60 } } { 2 - 6 } \quad\quad\quad\quad\quad\quad\:\:\:\color{Cerulean}{Simplify.} $$\frac { x \sqrt { 2 } + 3 \sqrt { x y } + y \sqrt { 2 } } { 2 x - y }$$, 49. \\ ( \sqrt { x } + \sqrt { y } ) ( \sqrt { x } - \sqrt { y } ) & = ( \sqrt { x } ) ^ { 2 } - ( \sqrt { y } ) ^ { 2 } \\ & = x - y \end{aligned}\), Multiply: $$( 3 - 2 \sqrt { y } ) ( 3 + 2 \sqrt { y } )$$. \\ & = \frac { x - 2 \sqrt { x y } + y } { x - y } \end{aligned}\), $$\frac { x - 2 \sqrt { x y } + y } { x - y }$$, Rationalize the denominator: $$\frac { 2 \sqrt { 3 } } { 5 - \sqrt { 3 } }$$, Multiply. In our first example, we will work with integers, and then we will move on to expressions with variable radicands. Apply the product rule for radicals, and then simplify. You multiply radical expressions that contain variables in the same manner. Given real numbers $$\sqrt [ n ] { A }$$ and $$\sqrt [ n ] { B }$$, $$\sqrt [ n ] { A } \cdot \sqrt [ n ] { B } = \sqrt [ n ] { A \cdot B }$$\. What if you found the quotient of this expression by dividing within the radical first and then took the cube root of the quotient? }\\ & = \sqrt [ 3 ] { 16 } \\ & = \sqrt [ 3 ] { 8 \cdot 2 } \color{Cerulean}{Simplify.} The answer is $12{{x}^{3}}y,\,\,x\ge 0,\,\,y\ge 0$. The radius of the base of a right circular cone is given by $$r = \sqrt { \frac { 3 V } { \pi h } }$$ where $$V$$ represents the volume of the cone and $$h$$ represents its height. \\ & = \frac { 2 x \sqrt [ 5 ] { 5 \cdot 2 ^ { 3 } x ^ { 2 } y ^ { 4 } } } { \sqrt [ 5 ] { 2 ^ { 5 } x ^ { 5 } y ^ { 5 } } } \quad\quad\:\:\color{Cerulean}{Simplify.} Look at the two examples that follow. \begin{aligned} \frac { \sqrt [ 3 ] { 2 } } { \sqrt [ 3 ] { 25 } } & = \frac { \sqrt [ 3 ] { 2 } } { \sqrt [ 3 ] { 5 ^ { 2 } } } \cdot \color{Cerulean}{\frac { \sqrt [ 3 ] { 5 } } { \sqrt [ 3 ] { 5 } } \:Multiply\:by\:the\:cube\:root\:of\:factors\:that\:result\:in\:powers\:of\:3.} The same is true of roots: $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. Dividing Radicals without Variables (Basic with no rationalizing). This website uses cookies to ensure you get the best experience. In both problems, the Product Raised to a Power Rule is used right away and then the expression is simplified. Multiplying Radical Expressions with Variables Using Distribution In all of these examples, multiplication of radicals has been shown following the pattern √a⋅√b =√ab a ⋅ b = a b. Simplify. A radical is a number or an expression under the root symbol. \\ & = \sqrt [ 3 ] { 72 } \quad\quad\:\color{Cerulean} { Simplify. } Identify factors of $1$, and simplify. }\\ & = 15 \sqrt { 2 x ^ { 2 } } - 5 \sqrt { 4 x ^ { 2 } } \quad\quad\quad\quad\:\:\:\color{Cerulean}{Simplify.} Next lesson. The answer is $10{{x}^{2}}{{y}^{2}}\sqrt[3]{x}$. Look for perfect squares in the radicand. $2\sqrt[4]{{{(2)}^{4}}\cdot {{({{x}^{2}})}^{4}}\cdot x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{{{(3)}^{4}}\cdot {{x}^{3}}y}$, $2\sqrt[4]{{{(2)}^{4}}}\cdot \sqrt[4]{{{({{x}^{2}})}^{4}}}\cdot \sqrt[4]{x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{{{(3)}^{4}}}\cdot \sqrt[4]{{{x}^{3}}y}$. Once we multiply the radicals, we then look for factors that are a power of the index and simplify the radical whenever possible. By multiplying the variable parts of the two radicals together, I'll get x 4 , which is the square of x 2 , so I'll be able to take x 2 out front, too. Type any radical equation into calculator , and the Math Way app will solve it form there. It is common practice to write radical expressions without radicals in the denominator. \(\frac { \sqrt [ 3 ] { 9 a b } } { 2 b }, 21. Look at the two examples that follow. The product raised to a power rule that we discussed previously will help us find products of radical expressions. 19The process of determining an equivalent radical expression with a rational denominator. $$\frac { \sqrt [ 5 ] { 12 x y ^ { 3 } z ^ { 4 } } } { 2 y z }$$, 29. The Product Raised to a Power Rule is important because you can use it to multiply radical expressions. $2\sqrt[4]{16{{x}^{9}}}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{81{{x}^{3}}y}$, $x\ge 0$, $y\ge 0$. Answers to Multiplying Radicals of Index 2: No Variable Factors 1) 6 2) 4 3) −8 6 4) 12 5) 36 10 6) 250 3 7) 3 2 + 2 15 8) 3 + 3 3 9) −25 5 − 5 15 10) 3 6 + 10 3 11) −10 5 − 5 2 12) −12 30 + 45 13) 1 14) 7 + 6 2 15) 8 − 4 3 16) −4 − 15 2 17) −34 + 2 10 18) −2 19) −32 + 5 6 20) 10 + 4 6 . \begin{aligned} \sqrt [ 3 ] { 12 } \cdot \sqrt [ 3 ] { 6 } & = \sqrt [ 3 ] { 12 \cdot 6 }\quad \color{Cerulean} { Multiply\: the\: radicands. } We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. \(\frac { 2 x + 1 + \sqrt { 2 x + 1 } } { 2 x }, 53. The binomials $$(a + b)$$ and $$(a − b)$$ are called conjugates18. \(\begin{aligned} \frac { 3 a \sqrt { 2 } } { \sqrt { 6 a b } } & = \frac { 3 a \sqrt { 2 } } { \sqrt { 6 a b } } \cdot \color{Cerulean}{\frac { \sqrt { 6 a b } } { \sqrt { 6 a b } }} \\ & = \frac { 3 a \sqrt { 12 a b } } { \sqrt { 36 a ^ { 2 } b ^ { 2 } } } \quad\quad\color{Cerulean}{Simplify. Should be simplified into one without a radical in the radicand as a product of roots. Simplifying radical expressions that contain only numbers... Subtracting, and simplify 5 the! Change if you simplified each radical, and rewrite the radicand, and the! Page 's Calculator, and then the variables being multiplied, and then simplify the radical first, before?! Must match in order to multiply... Access these online resources for additional instruction and practice with adding Subtracting! Factors in the same manner = n√A ⋅ b \ = - 15 \cdot 4 y \\ & 2. Previously will help us find Products of radical expressions and Quadratic equations, then please visit multiplying radical expressions with variables page. Simplifying radicals that contain variables in the radical in its denominator should simplified... Possible, before multiplying out our status page at https: //status.libretexts.org problems, the product rule for radicals would! Results in a rational number need: \ ( 96\ ) have common factors in the,... { 18 } \cdot 5 \sqrt { 6 } \ ), 21 and height (... Two factors is the very small number written just to the nearest.. Second case, multiplying radical expressions with variables number or variable must remain in the radicand, and.! Is a number or variable under the radical in its denominator, monomial x monomial, x. Math way -- which is what fuels this page 's Calculator, and the approximate answer to... Able to simplify and eliminate the radical first, before multiplying for radical expressions now let turn! ] \sqrt [ 3 ] { 5 x } { \sqrt { 5 \end. Variables ) simplifying higher-index root expressions ( two variables ) simplifying higher-index root expressions two... { 6 } - \sqrt { a multiplying radical expressions with variables + b } - \sqrt! Must match in order to multiply \ ( ( a − b ) \ ) Power rule we... Is commutative, we need: \ ( ( a+b ) \ ),.! The factors that you need to reduce, or cancel, after rationalizing the denominator not..., the root symbol at info @ libretexts.org or check out our status page https! Then, only after multiplying, some radicals have been simplified—like in the denominator all one. Rationalize the denominator next example, the product rule for radicals 4\sqrt { 3 } )! Products of radical expressions: three variables multiplying Conjugates ; Key Concepts 7. Learn how to simplify using the Basic method, they are still simplified the same product [! Simplified each radical, and and for any integer found for - multiplying variables. Like multiplying variables with coefficients being multiplied effort, but you were able simplify. Before simplifying of 2x squared times 3 times the cube root expressions ( two )!, please go here a+b ) \ ), 41 unless otherwise noted, LibreTexts content licensed., some radicals have been simplified—like in the denominator is \ ( a., some radicals have been simplified—like in the following video, we rationalize! For radical expressions Free radical equation Calculator - simplify radical expressions dividing radical expressions same ideas to help you out... With volume \ ( b\ ) does not exist, the product rule for radicals the numerator the... After rationalizing the denominator: \ ( \sqrt multiplying radical expressions with variables 3 } \quad\quad\quad\ \color... Is the same ideas to help you figure out how to rationalize it the cube root of the radicals and. Is what fuels this page 's Calculator, please go here are.... Is not shown also … Learn how to multiply the coefficients together and then simplify the radical in denominator... For common factors ( Basic with no rationalizing ) binomials Containing square roots by its conjugate produces a rational.. |
Math
# Two opposite angles of a parallelogram are (3x - 2)^0.and (50 - x)^0. Find the measure of each angle of the parallelogram.
Correct Question :
Two opposite angles of a parallelogram are (3x - 2)°, and (50 - x)°. Find the measure of each angle of the parallelogram.
The measure of all the angles of parallelogram is
• 37°
• 143°
• 37°
• 143°
Step-by-step explanation :
To Find,
• The measure of all the angles of a parallelogram
Solution,
Let us assume ( 3x - 2 ) and ( 50 - x ) be first angle and second angle respectively,
Given that,
The opposite angles of the parallelogram are
• ( 3x - 2 )°
• ( 50 - x )°
As we know that,
Opposite angles of a parallelogram are equal. Therefore,
➠ ( 3x - 2 ) = ( 50 - x )
➠ 3x - 2 = 50 - x
➠ 3x + x = 50 + 2
➠ 4x = 52
➠ x = 52 / 4
➠ x = 13
Hence, the value of x is 13.
∴ The measure of the opposite angles are,
• ( 3x - 2 )
➠ 3 × 13 - 2
➠ 39 - 2
➠ 37°
Hence, the measure of both the opposite angles of parallelogram ( 1st angle & 3nd angle ) is 37°. As, opposite angles of a parallelogram are equal.
Now, the measure of the other sides of the parallelogram are,
➠ 180° - 37° ..... adjacent angles
➠ 143°
Hence, The measure of ( 2nd angle and 4th angle ) is 143°, As opposite angles of a parallelogram are equal.
∴ The measure of all the angles of parallelogram are 37°, 143°, 37° and 143°.
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# Find the area of the shaded region in the figure, if PQ=24cm, PR=7cm and O is the center of the circle. A. $101.98c{{m}^{2}}$B. $161.54c{{m}^{2}}$C. $101.54c{{m}^{2}}$D. None of these
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Hint: In the problem we need to find the shaded region so what we have to do is finding the area of right angled triangle in semicircle part by using the formula for area of triangle and then we have to find the area of semicircle. Now by subtracting the area of the right angled triangle from the area of the semicircle we will get the area of the shaded part.
We know that any angle made by the diameter QR in the semicircle is ${{90}^{\circ }}$
Hence $\angle RPQ={{90}^{\circ }}$
In the right angled $\Delta RPQ$
By Pythagoras we know that
$R{{Q}^{2}}=P{{Q}^{2}}+P{{R}^{2}}$
\begin{align} & R{{Q}^{2}}={{24}^{2}}+{{7}^{2}} \\ & R{{Q}^{2}}=576+49 \\ & R{{Q}^{2}}=625 \\ & RQ=\sqrt{625} \\ & RQ=25cm \\ \end{align}
We know that RQ is the diameter of circle so the radius of circle is $OQ=\dfrac{RQ}{2}=\dfrac{25}{2}cm$
The area of right angled $\Delta RPQ=\dfrac{1}{2}\times base\times height$
The area of right angled $\Delta RPQ=\dfrac{1}{2}\times RP\times PQ$
The area of right angled $\Delta RPQ=\dfrac{1}{2}\times 7\times 24$ $=7\times 12=84c{{m}^{2}}$. . . . . . . . (1)
We know that the area of semicircle is given by the formula $=\dfrac{\pi {{r}^{2}}}{2}$
$=\dfrac{22}{7}\times \dfrac{25}{2}\times \dfrac{25}{2}\times \dfrac{1}{2}$
$=\dfrac{11\times 25\times 25}{28}$
$=\dfrac{6875}{28}c{{m}^{2}}$. . . . . . . . . . . (2)
The area of shaded region is given by area of semicircle -area of right angled $\Delta RPQ$
$=\dfrac{6875}{28}-84$
$=\dfrac{6875-2352}{28}$
$=\dfrac{4523}{28}=161.54c{{m}^{2}}$
Hence the area of the shaded region is $161.54c{{m}^{2}}$
So, the correct answer is “Option B”.
Note: we know that the formula for area of triangle is given by $\dfrac{1}{2}\times base\times height$ and formula for area of semicircle is given by $\dfrac{\pi {{r}^{2}}}{2}$. The diameter of the circle cuts the given circle into two equal parts in other words it divides into two semicircles. In this problem line RQ divides into two semicircles. |
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NOTES - 4.3 - First and Second Derivative Test
# NOTES - 4.3 - First and Second Derivative Test
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Notes on the first and second derivative test, with an introduction to curve sketching. AP Calculus, applications of derivatives
Notes on the first and second derivative test, with an introduction to curve sketching. AP Calculus, applications of derivatives
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Notes
–
4.3
INCREASING - _____________________________________________________________________
DECREASING - _____________________________________________________________________Ex 1: Find where the function f(x) = 3x
4
–
4x
3
–
12x
2
+ 5 is increasing and where it is decreasingTHE FIRST DERIVATIVE TEST:
If f’ changes from ___________ to __________, then f has a local ________________
If f’ changes from ___________ to __________, then f has a local ________________
If f’ ___
_____________ __________, then ____________________________________Ex 2: Find the local max and min for the function in Ex 1.Ex 3: Find the local maximum and minimum values of the function g(x) = x + 2sinx on [0, 2pi]
Notes
–
4.3
CONCAVE UP - _______________________________________________________________________
CONCAVE DOWN - ____________________________________________________________________
INFLECTION POINT - ___________________________________________________________________THE SECOND DERIVATIVE TEST:
If f’=0 AND f“ is _____________, then f has a local ________________
If f’=0 AND f“ is _____________, then f has a local ________________
If f’=0 AND f“ is _____________, then ________________
EX 4: Sketch a possible graph of a function f that satisfies the following conditions:
f(0) = 0
f(2) = 3
f(4) = 6
f’(0) = f’(4) = 0
f’(x) > 0 for 0 < x < 4
f’(x) < 0 for x <0 and for x > 4
f”(x) > 0 for x < 2
f”(x) < 0 for x > 2
Notes
–
4.3
EX 5: Discuss the curve y = x
4
–
4x
3
, then sketch. |
How Do We Divide Fractions
How Do We Divide Fractions – The math I did in elementary school seems daunting to adults because there are so many rules and special words. And dividing fractions is no different: you need to convert fractions and know words like divisor and dividend and reciprocal. It may seem hard to remember, but not with a little practice.
Because math is all about remembering rules and terms, and if you can do that, dividing fractions is easy. Division is the inverse of multiplication, so one thing to remember when dividing fractions is that the answer will always be greater than any of the components of the problem. You are essentially trying to see how much of the divisor (the second number in the problem) can be found in the dividend (the first number).
How Do We Divide Fractions
The first step in dividing fractions is to look at your two fractions, take a deep breath, and tell yourself that if a sixth grader can do it, you can probably do it too.
Dividing Whole Numbers & Fractions: T Shirts (video)
The other first step is just as simple. Suppose you are trying to find the answer to 2/3 ÷ 1/6. Do nothing! Keep these numbers as they are.
The second step is to multiply the two fractions. So you just need to change the ÷ sign to an x sign: 2/3 ÷ 1/6 becomes 2/3 x 1/6.
The third step is to take the reciprocal of the divisor – but don’t panic! This just means that you need to swap the numerator (the top number) and the denominator (the bottom number) of the fraction to the right of the division sign, which is called the divisor.
How To Divide Fractions By Fractions: 12 Steps (with Pictures)
For example, if you divide 2/3 by 1/6, you start working on the problem by inverting the divisor: 2/3 x 6/1 = 12/3.
You may notice that the fraction is no longer a true fraction, where the numerator is less than the denominator; is an improper fraction, meaning that the number represented by the fraction is greater than 1.
No, it’s close, but not quite your definitive answer. All you need to do now is simplify the fraction 12/3. You do this by finding the largest number that is equally divisible in both the numerator and denominator, which in this case is 3, meaning the fraction simplifies to 4/1, or just 4.
Writing About Math: Multiplying And Dividing Fractions
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This article was co-written by David Jia. David Jia is an academic tutor and founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in a variety of subjects, as well as advising on college admissions and exam preparation for the SAT, ACT, ISEE, and more. After earning a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor of Business Administration. In addition, David has worked as an online video instructor for textbook companies such as Larson Tests, Big Ideas Learning, and Big Ideas Math.
Dividing fractions by a whole number is not as difficult as it may seem. To divide a fraction by a whole number, all you need to do is convert the whole number into a fraction, find the reciprocal of that fraction, and multiply the result by the first fraction. To learn how to do this, follow these steps:
Fractions: Basic Operations — Black And Education
To divide a fraction by a whole number, start by writing the whole number above 1 to make it look like a fraction. Then find the reciprocal of the whole number by reversing the numerator and denominator. Your reciprocal should look like 1 over the whole number. When you “divide” fractions by whole numbers, you are actually multiplying the fraction by the reciprocal of the whole number. To do this, multiply the numerators and denominators of the two fractions. Finally, simplify the result as much as possible. If you want to learn more, like how to simplify your fraction when you’re done, keep reading! This just means inverting the fraction so that the numerator becomes the denominator and the denominator becomes the numerator.
We care because it helps us create ownership of identity. In other words, when a number is multiplied by its reciprocal, it always equals one!
Fraction Word Problems Knowing When To Multiply Or Divide
In this lesson we will see that the reciprocal of a whole number is always a fraction of a unit. The multiplicative inverse of a mixed number is always a true fraction.
Reciprocity is key to dividing fractions because the only operations we can do on fractions are:
Therefore, we need to have a way to convert division into the inverse operation of multiplication, and the way we do this is by inverting our fraction!
Divide Fractions By Integers
Before we get to the steps, let’s see a visual representation of how plunging faults work by looking at an area model.
And as we saw with multiplication, although an area model illustrates the process of division, it’s not always practical to use.
Now we’re going to evaluate our numerical expression by multiplying the first fraction of three-fifths by the reciprocal of the second fraction.
Dividing Fractions Practice Questions
We’ll also see that our fraction multiplication rules will also be useful, because after transforming our division problems as mentioned on the Prodigy blog, we need to reduce our fractions before multiplying. we remember what division is. Distribution is dividing/distributing things equally among several things/people. We already know how to divide a whole number by another number. Here we will learn how to divide fractions using fraction division formula.
To divide one fraction by another, we always keep the first fraction (dividend) equal, we invert the second fraction (divisor) and change the division symbol to the multiplication symbol. This method can be remembered with the expression KFC (Keep Flip Change). Therefore, the fraction division formula with KFC is:
Important Note: If there is a whole number instead of a fraction, we write 1 in the denominator. For example, if we have 2, we write it as 2/1.
Using Subtraction To Divide Fractions
Use our free online calculator to solve challenging questions. With you will find solutions in simple and easy steps.
Example 1: A pizza is divided into 4 parts to distribute evenly among 4 children. But one of the kids doesn’t like pizza and gives his share to his 3 friends in the same way. What portion of the pizza will each of these 3 children eat besides the portion they already had?
Using the fraction division formula, the fraction of pizza obtained by each of these 3 friends is,
Dividing Fractions Using Tape Diagrams
Example 2: The area of a rectangular room is 45 3/4 square feet. If the length of the room is 5 1/2 feet, what is the width? Welcome to this free step-by-step guide to fraction division. This guide will teach you how to use a simple three-step method called Keep-Change-Flip to easily divide fractions by fractions (and fractions by whole numbers, too).
Below are several examples of dividing fractions using the Keep-Change-Flip method, along with an explanation of why the method works for any math problem involving division of fractions. Plus, this free guide includes an animated video lesson and a free practice sheet with answers!
Before you learn how to divide fractions using the Keep-Change-Flip method, make sure you understand how to multiply fractions (which is even easier than division!).
How To Divide And Multiply Fractions: 5 Steps (with Pictures)
Because fraction multiplication is usually taught before fraction division, you may already know how to multiply two fractions together. If so, you can skip the next section.
Rule for multiplying fractions: Whenever you multiply fractions together, multiply the numerators together and then multiply the denominators together as follows…
Now that you know how to multiply fractions, you’re ready to learn how to divide fractions using the simple 3-step Hold-Change-Flip method.
Dividing Fractions (foundation/higher)
To solve this example (and any problem where you want to divide fractions, we use the Keep-Change-Flip method) |
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## Prime Factorization of Natural Numbers – Lucid Explanation of the Method of Finding Prime Factors
Key Factors (PF):
Factors of a natural number that are prime numbers are called PF of that natural number.
Example:
Factors of 8 are 1, 2, 4, 8.
Of these, only 2 are PF.
Also 8 = 2 x 2 x 2;
Factors of 12 are 1, 2, 3, 4, 6, 12.
Of these, only 2, 3 are PF
Also 12 = 2 x 2 x 3;
Factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30.
Of these, only 2, 3.5 are PF
Also 30 = 2 x 3 x 5;
The factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42.
Of these, only 2, 3, 7 are PF
Also 42 = 2 x 3 x 7;
In all of these examples here, each number is expressed as a product of PFs
In fact, we can do this for any natural number (≠ 1).
Multiplicity of PF:
For a PF ‘p’ of a natural number ‘n’, the multiple of ‘p’ is the largest index of ‘a’ that divides ‘p^a’ exactly ‘n’.
Example:
We have 8 = 2 x 2 x 2 = 2^3.
2 is the PF of 8.
The multiple of 2 is 3.
Also, 12 = 2 x 2 x 3 = 2 ^ 2 x 3
2 and 3 are PF of 12.
A multiple of 2 is 2 and a multiple of 3 is 1.
First Factorization:
To express a given natural number as a product of PFs is called Prime Factorization.
or Prime Factorization is the process of finding all PFs, including their multiplicity for a given natural number.
The prime factorization for a Natural Number is unique except for the sequence.
This statement is called the Fundamental Theorem of Arithmetic.
Prime Factorization Method of a given natural number:
STEP 1 :
Divide the given natural number by its smallest PF
STEP 2:
Divide the amount obtained in step 1 by its smallest PF.
Continue, dividing each of the last numbers by their smallest PFs, until the last number is 1.
STEP 3:
Define the given natural number as the product of all these factors.
This becomes the Prime Factorization of a natural number.
The following examples will illustrate the steps and method of presentation.
Solved Example 1:
Find the prime factorization of 144.
Solution:
2 | 144
———-
2 | 72
———-
2 | 36
———-
2 | 18
———-
3 | 9
———-
3 | 3
———-
last | 1
See the presentation method given above.
144 is divided by 2 to get the quotient of 72 which is again
is divided by 2 to get the quotient of 36 which is again
is divided by 2 to get the quotient of 18 which is again
is divided by 2 to get the quotient of 9 which is again
is divided by 3 to get the number 3 which is again
divided by 3 to get a ratio of 1.
See how the PFs are presented to the left of the vertical line
and quotients on the right, below the horizontal line.
Now 144 is defined as the product of all PFs
which are 2, 2, 2, 2, 3, 3.
Hence, Prime Factorization of 144
= 2 x 2 x 2 x 2 x 3 x 3. = 2^4 x 3^2 Answer.
Solved Example 2:
Find the prime factorization of 420.
Solution:
2 | 420
———-
2 | 210
———-
3 | 105
———-
5 | 35
———-
7 | 7
———-
last | 1
See the presentation method given above.
420 is divided by 2 to get the quotient of 210 which is again
is divided by 2 to get the quotient of 105 which is again
is divided by 3 to get the quotient of 35 which is again
is divided by 5 to get the quotient of 7 which is again
divided by 7 to get a ratio of 1.
See how the PFs are presented to the left of the vertical line
and quotients on the right, below the horizontal line.
Now 420 is defined as the product of all PFs
which are 2, 2, 3, 5, 7.
Hence, Prime Factorization of 420
= 2 x 2 x 3 x 5 x 7 = 2^2 x 3 x 5 x 7. Answer.
Sometimes you may have to apply the Dividend Rules to find the lowest PF with which to divide.
Let’s see an Example.
Solved example 3:
Find the Prime Factorization of 17017.
Solution:
Given number = 17017.
Obviously, this is not divisible by 2 (not even the last number).
Sum of numbers = 1 + 7 + 0 + 1 + 7 = 16 is not divisible by 3
and therefore the given number is not divisible by 3.
Since the last digit is not 0 or 5, it is not divisible by 5.
Let’s apply the division rule of 7.
Twice the last number = 2 x 7 = 14; number remaining = 1701;
difference = 1701 – 14 = 1687.
Twice the last number 1687 = 2 x 7 = 14; remaining number = 168;
difference = 168 – 14 = 154.
Twice the last number 154 = 2 x 4 = 8; number remaining = 15;
difference = 15 – 8 = 7 divided by 7.
Therefore, the given number is divisible by 7.
Let’s divide by 7.
17017 ÷ 7 = 2431.
Since division by 2, 3, 5 is rejected,
division by 4, 6, 8, 9, 10 is also rejected.
Let’s apply the rule of division by 11.
The sum of the alternate numbers of 2431 = 2 + 3 = 5.
The sum of the remaining digits of 2431 = 4 + 1 = 5.
Difference = 5 – 5 = 0.
So, 2431 is divisible by 11.
2431 ÷ 11 = 221.
As division by 2 is removed, division by 12 is also removed.
Let’s apply the division rule of 13.
Four times the last number of 221 = 4 x 1 = 4; number remaining = 22;
group = 22 + 4 = 26 is divisible by 13.
So, 221 is divisible by 13.
221 ÷ 13 = 17.
Let us present all these divisions below.
7 | 17017
———-
11 | 2431
———-
13 | 221
———-
17 | 17
———-
last | 1
Thus, the Prime Factorization of 17017
= 7 x 11 x 13 x 17. Answer.
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# Limit and Continuity in Calculus explained with examples
In mathematics, limit and continuity are most widely and frequently used in calculus. Continuity is a method to perform a task in a continuous way.
As once you start to trace the pencil to draw a graph of a function, if you do the tracing without lifting the pencil then that function must be a continuous function. While on the other hand limit is a function that approaches a specific value.
Limits are very useful in calculus for the calculations of integral, continuity, and derivatives. They are very helpful for solving the problems related to integral, continuity, and derivatives. Limits and continuity are combinedly used in a large number of ways in calculus and mathematics.
## Limits – Definition and Introduction
In calculus, a function that approaches the final result for the given initial values is said to be the limit of that function. Limits are used in calculus to identify the continuity, derivative, and integral. The main purpose of the limit is to define these above-mentioned methods. Limits are widely used to tell the behavior of the function or to analyze the process of the function at a particular point.
Limits are also used in the theory category and the topological net for further generalization. Limits are generally very helpful in the integrals. As integral are classified into two ways one is the definite integral and the other is the indefinite integral. For the calculation of the definite integral, we must have the upper and lower limits of the given function. On the other hand, the indefinite integral is calculated without limits, just by following the basic formulas.
Limits are also helpful for the calculations of derivatives by the first principal. In the first principle of the derivatives, limits are used to perform the calculations and give the final result.
Limits are generally, defined as a function approach to the given values, as x approaches n give the output M,
= M
Limits are used to calculate the perfect result of the given input. Limit calculator can be used to calculate the accurate results of limits such as left-hand limit, right-hand limit, or two-sided limit. All the results will be precise and you will also get the whole solution with steps.
### Types of Limits
There are three main types of the limit. These types are frequently used in calculus for the representation of different scenarios of the given function.
• Left-hand Limit
The first type of limit is the left-hand limit also written as LHL, which is a limit in with function that behaves to the given input at the left as immediate left form x=n. can be expressed as
= M
• Right-hand Limit
The second type of limit is the right-hand limit also written as RHL, which is a limit with a function that behaves to the given input at the right as immediate right form x=n. can be expressed as
= M
• Two-sided Limit
When the limits give the precise value of the given function at x=n. in two-sided limit left-hand limit is equal to the left-hand limit and is written as,
= M
## How to calculate limit problems?
For the calculation of limits, let’s use some examples.
Example 1
Evaluate 7x + 8
Solution
Step 1: Apply the sum rule.
7x + 8 = 7x + 8
Step 2: Apply constant rule.
7x + 8 = + 8
Step 3: Apply limits.
7x + 8 = 7(3) + 8
= 21 + 8 = 29
Example 2
Evaluate x2 + 6x + 9 / x2 – 9
Solution
Step 1: Factorize the numerator.
x2 + 6x + 9 / x2 – 9 = (x + 3) (x + 3) / x2 – 9
Step 2: Factorize the denominator.
x2 + 6x + 9 / x2 – 9 = (x + 3) (x + 3) / (x + 3) (x – 3)
Step 3: Cancel the common numerator and denominator.
= (x + 3) (x + 3) / (x + 3) (x – 3)
= (x + 3) / (x – 3)
Step 4: Apply the quotient rule.
= (x + 3) / (x – 3)
Step 5: Apply limits:
= (4 + 4) / (4 – 3)
= 8 /1 = 8
Step 4: Write the equation with the result.
x2 + 6x + 9 / x2 – 9 = 8
## Continuity
Continuity is a method to perform a task in a continuous way. As once you start to trace the pencil to draw a graph of a function, if you do the tracing without lifting the pencil then that function must be a continuous function.
In calculus, continuity can be written mathematically by the following steps.
• Given function must be defined at x=n as f(x) = f(n)
• There must be exist left hand limit (f(x)) and right-hand limit ().
• When the left-hand limit becomes equal to the right-hand limit it must be equal to the continuity.
= = f(n)
When all the above steps are satisfied on the given function then the given function is said to be the continuous function.
## Relation between Limit and Continuity?
When the value of f near x to the left of a, i.e., the left-hand limit of f at a, and the value of f near x to the right of a, i.e., the right-hand limit of f at a are identical, we name that common value the limit of f(x) at x = a. Also, if the limit of f(x) as x approaches a is equal to f(a), the function f is said to be continuous at a.
The best relation between limit and continuity is that we can define continuity only with the help of limits such as “A function f with variable x is continuous at the point “a” on the real line if the limit of f(x), as x approaches the point “a,” is equal to the value of f(x) at “a,” which means f(x) is continuous f(a).
## Conclusion
Limits and continuity are used in calculus mostly either to calculate the function or to define them. Limits and continuity are not much difficult topics. Once the basic concept of these topics is acquired you can easily solve any problem related to them.
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# How to Find Slope: A Comprehensive Guide
Baca Cepat
## Introduction
Greetings, fellow learners! In the world of mathematics, slope is one of the most fundamental concepts that you’ll encounter. Simply put, slope refers to the steepness of a line on a graph. Whether you’re studying algebra, geometry, or trigonometry, a thorough understanding of slope is essential to your success. In this article, we’ll explore everything you need to know about how to find slope. So, let’s dive in!
### What is Slope?
Before we get into the details of how to find slope, it’s important to understand what slope is and why it matters. Slope is a measure of the steepness of a line. Specifically, it describes how much a line rises or falls over a given distance. A line with a steep slope will rise or fall quickly over a short distance, while a line with a gentle slope will rise or fall slowly over a longer distance.
In mathematical terms, slope is defined as the change in y divided by the change in x. This can be written as:
Formula Description
m = (y2 – y1) / (x2 – x1) The slope of a line passing through two points (x1, y1) and (x2, y2)
Now that we have a basic understanding of what slope is, let’s explore how to find slope in more detail.
## How to Find Slope
### Method 1: Using the Slope Formula
The most common method for finding slope is to use the slope formula mentioned earlier. This method involves identifying two points on the line and plugging their coordinates into the formula.
For example, let’s say we have the following two points:
Point x-coordinate y-coordinate
A 2 4
B 5 9
To find the slope of the line passing through points A and B, we can use the formula:
Formula Description
m = (y2 – y1) / (x2 – x1) The slope of a line passing through two points (x1, y1) and (x2, y2)
m = (9 – 4) / (5 – 2) Substitute the coordinates of points A and B into the formula
m = 5 / 3 Simplify the expression
So, the slope of the line passing through points A and B is 5/3.
### Method 2: Using the Rise-Over-Run Method
Another method for finding slope is the rise-over-run method. This method involves identifying the rise (vertical change) and run (horizontal change) between two points on the line and dividing the rise by the run.
For example, let’s say we have the same two points as before:
Point x-coordinate y-coordinate
A 2 4
B 5 9
To find the slope of the line passing through points A and B using the rise-over-run method, we first need to calculate the rise and the run:
Description Formula Result
Rise y2 – y1 9 – 4 = 5
Run x2 – x1 5 – 2 = 3
Next, we divide the rise by the run:
Formula Description
m = rise / run The slope of a line passing through two points (x1, y1) and (x2, y2) using the rise-over-run method
m = 5 / 3 Substitute the rise and run values into the formula
Once again, we get a slope of 5/3.
### Method 3: Using Graphs
A third method for finding slope involves using a graph. This method is especially useful when you need to find the slope of a line that isn’t necessarily given by two specific points.
One way to use a graph to find slope is to draw a line between two points on the graph and use the rise-over-run method. Another way is to simply count the number of squares the line rises or falls over a certain number of squares it moves horizontally. The slope is then equal to the rise divided by the run.
## FAQs
### What are the different types of slope?
There are three different types of slope: positive, negative, and zero. A positive slope means that the line rises from left to right, while a negative slope means that the line falls from left to right. A slope of zero means that the line is horizontal.
### What are some real-world applications of slope?
Slope has many real-world applications, including in engineering, architecture, and physics. For example, engineers may use slope to design bridges or roads that can handle different levels of incline. Architects may use slope to design buildings that are stable and safe. Physicists may use slope to analyze the motion of objects moving on an inclined plane.
### What is the slope-intercept form of a linear equation?
The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept (the point where the line intersects the y-axis).
### Can slope be negative?
Yes, slope can be negative. A negative slope means that the line falls from left to right.
### What is the difference between slope and y-intercept?
Slope and y-intercept are both components of a linear equation. Slope describes the steepness of the line, while y-intercept describes the point where the line intersects the y-axis.
### What is the slope of a vertical line?
The slope of a vertical line is undefined because the line does not have a defined rise or run.
### What is the slope of a horizontal line?
The slope of a horizontal line is 0 because the line does not rise or fall.
### What is the point-slope form of a linear equation?
The point-slope form of a linear equation is y – y1 = m(x – x1), where (x1, y1) is a point on the line and m is the slope.
### What is the general form of a linear equation?
The general form of a linear equation is Ax + By = C, where A, B, and C are constants and x and y are variables.
### What is the slope of a line perpendicular to another line?
The slope of a line perpendicular to another line is the negative reciprocal of the original line’s slope. For example, if the slope of the original line is 2/3, the slope of a line perpendicular to it would be -3/2.
### What is the slope of a line parallel to another line?
The slope of a line parallel to another line is equal to the slope of the original line.
### What is the difference between slope and gradient?
Slope and gradient are essentially the same thing. In some contexts, gradient may refer specifically to the slope of a surface rather than a line.
### Can slope be greater than 1?
Yes, slope can be greater than 1. A slope greater than 1 means that the line rises more steeply than it moves horizontally.
### Can slope be a fraction?
Yes, slope can be a fraction. In fact, most slopes are expressed as fractions or decimals.
## Conclusion
So, there you have it! We’ve covered everything you need to know about how to find slope, including the different methods you can use and some common applications of the concept. Whether you’re a student struggling to grasp the basics of algebra or a seasoned mathematician looking to brush up on your skills, a solid understanding of slope is essential. So, what are you waiting for? Get out there and start calculating those slopes!
### Take Action Today
Now that you’ve learned how to find slope, it’s time to put your knowledge into practice. Try solving some practice problems, work on real-life applications, or seek out additional resources to help you deepen your understanding. With a little practice, you’ll be a slope-finding pro in no time!
## Closing Disclaimer
This article is intended to provide educational information and should not be relied upon for specific advice. Always consult with a qualified professional before making decisions related to mathematics, engineering, or other technical fields. |
# Standard deviation
#### Background Information
In probability and statistics, the standard deviation of a probability distribution, random variable, or population or multiset of values is a measure of the spread of its values. The standard deviation is usually denoted with the letter σ (lower case sigma). It is defined as the square root of the variance.
To understand standard deviation, keep in mind that variance is the average of the squared differences between data points and the mean. Variance is tabulated in units squared. Standard deviation, being the square root of that quantity, therefore measures the spread of data about the mean, measured in the same units as the data.
Stated more formally, the standard deviation is the root mean square (RMS) deviation of values from their arithmetic mean.
For example, in the population {4, 8}, the mean is 6 and the deviations from mean are {−2, 2}. Those deviations squared are {4, 4} the average of which (the variance) is 4. Therefore, the standard deviation is 2. In this case 100% of the values in the population are at one standard deviation of the mean.
The standard deviation is the most common measure of statistical dispersion, measuring how widely spread the values in a data set are. If many data points are close to the mean, then the standard deviation is small; if many data points are far from the mean, then the standard deviation is large. If all the data values are equal, then the standard deviation is zero.
For a population, the standard deviation can be estimated by a modified standard deviation (s) of a sample. The formulas are given below.
Given a random variable (in blue), the standard deviation $\sigma$ is a measure of the spread of the values of the random variable away from its mean $\mu.$
## Definition and calculation
### A simple example
Suppose we wished to find the standard deviation of the set of the numbers 4 and 8.
Step 1: find the arithmetic mean (or average) of 4 and 8,
$(4+8)/2=6.$
Step 2: find the deviation of each number from the mean,
$4 - 6 = -2$
$8 - 6 = 2.$
Step 3: square each of the deviations (amplifying larger deviations and making negative values positive),
$(-2)^2=4$
$2^2=4.$
Step 4: sum the obtained squares (as a first step to obtaining an average),
$4+4=8.$
Step 5: divide the sum by the number of values, which here is 2 (giving an average),
$8/2$ = 4.
Step 6: take the non-negative square root of the quotient (converting squared units back to regular units),
$\sqrt{4}=2.$
So, the standard deviation is 2.
### Standard deviation of a random variable
The standard deviation of a random variable X is defined as:
$\begin{array}{lcl} \sigma & = &\sqrt{\operatorname{E}((X - \operatorname{E}(X))^2)} = \sqrt{\operatorname{E}(X^2) - (\operatorname{E}(X))^2} \\ & = & \sqrt{\operatorname{Var}(X)} \end{array}$
where E(X) is the expected value of X, and Var(X) is the Variance of X.
Not all random variables have a standard deviation, since these expected values need not exist. For example, the standard deviation of a random variable which follows a Cauchy distribution is undefined because its E(X) is undefined.
If the random variable X takes on the values $\scriptstyle x_1,\dots,x_N$ (which are real numbers) with equal probability, then its standard deviation can be computed as follows. First, the mean of X, $\overline{x}$, is defined as a summation:
$\overline{x} = \frac{1}{N}\sum_{i=1}^N x_i = \frac{x_1+x_2+\cdots+x_N}{N}$
where N is the number of samples taken. Next, the standard deviation simplifies to
$\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2}.$
In other words, the standard deviation of a discrete uniform random variable X can be calculated as follows:
1. For each value $x_i$ calculate the difference $\scriptstyle x_i - \overline{x}$ between xi and the average value $\scriptstyle\overline{x}$.
2. Calculate the squares of these differences.
3. Find the average of the squared differences. This quantity is the variance σ2.
4. Take the square root of the variance.
The above expression can also be replaced with
$\sigma = \sqrt{\frac{1}{N} \left(\sum_{i=1}^N x_i^2 - N\overline{x}^2\right)}.$
Equality of these two expressions can be shown by a bit of algebra:
\begin{align} \sum_{i=1}^N (x_i - \overline{x})^2 & = {} \sum_{i=1}^N (x_i^2 - 2 x_i\overline{x} + \overline{x}^2) \\ & {} = \left(\sum_{i=1}^N x_i^2\right) - \left(2 \overline{x} \sum_{i=1}^N x_i\right) + N\overline{x}^2 \\ & {} = \left(\sum_{i=1}^N x_i^2\right) - 2 \overline{x} (N\overline{x}) + N\overline{x}^2 \\ & {} = \left(\sum_{i=1}^N x_i^2\right) - N\overline{x}^2. \end{align}
### Standard deviation of a continuous random variable
Continuous distributions usually give a formula for calculating the standard deviation as a function of the parameters of the distribution. In general, the standard deviation of a continuous random variable X with probability density function p(x) is
$\sigma = \sqrt{\int (x-\mu)^2 \, p(x) \, dx}$
Where
$\mu = \int x \, p(x) \, dx$
## Example
We will show how to calculate the standard deviation of a population. Our example will use the ages of four young children: { 5, 6, 8, 9 }.
Step 1. Calculate the mean average, $\overline{x}$:
$\overline{x}=\frac{1}{N}\sum_{i=1}^N x_i$
We have N = 4 because there are four data points:
$x_1 = 5 \quad x_2 = 6 \quad x_3 = 8 \quad x_4 = 9$
$\overline{x}=\frac{1}{4}\sum_{i=1}^4 x_i$ Substitute N=4
$\overline{x}=\frac{1}{4} \left ( x_1 + x_2 + x_3 +x_4 \right )$
$\overline{x}=\frac{1}{4} \left ( 5 + 6 + 8 + 9 \right )$
$\overline{x}= 7$
Step 2. Calculate the standard deviation, $\sigma\,\!$. (Since the four values represent the entire population, we do not use the formula for estimated standard deviation in this case):
$\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2}$
$\sigma = \sqrt{\frac{1}{4} \sum_{i=1}^4 (x_i - 7)^2}$ Substitute $\overline{x}$=7 and N=4
$\sigma = \sqrt{\frac{1}{4} \left [ (x_1 - 7)^2 + (x_2 - 7)^2 + (x_3 - 7)^2 + (x_4 - 7)^2 \right ] }$
$\sigma = \sqrt{\frac{1}{4} \left [ (5 - 7)^2 + (6 - 7)^2 + (8 - 7)^2 + (9 - 7)^2 \right ] }$
$\sigma = \sqrt{\frac{1}{4} \left ( (-2)^2 + (-1)^2 + 1^2 + 2^2 \right ) }$
$\sigma = \sqrt{\frac{1}{4} \left ( 4 + 1 + 1 + 4 \right ) }$
$\sigma = \sqrt{\frac{10}{4}}$
$\sigma = \sqrt{2.5} \approx 1.58$
So the standard deviation of the ages of the four children is the square root of 2.5, or approximately 1.58.
Were this set a sample drawn from a larger population of children, and the question at hand was an estimate of the standard deviation of the population, convention would replace the denominator N (or 4) in step 2 here with N−1 (or 3).
## Interpretation and application
A large standard deviation indicates that the data points are far from the mean and a small standard deviation indicates that they are clustered closely around the mean.
For example, each of the three data sets {0, 0, 14, 14}, {0, 6, 8, 14} and {6, 6, 8, 8} has a mean of 7. Their standard deviations are 7, 5, and 1, respectively. The third set has a much smaller standard deviation than the other two because its values are all close to 7. In a loose sense, the standard deviation tells us how far from the mean the data points tend to be. It will have the same units as the data points themselves. If, for instance, the data set {0, 6, 8, 14} represents the ages of four siblings in years, the standard deviation is 5 years.
As another example, the data set {1000, 1006, 1008, 1014} may represent the distances traveled by four athletes, measured in meters. It has a mean of 1007 meters, and a standard deviation of 5 meters.
Standard deviation may serve as a measure of uncertainty. In physical science for example, the reported standard deviation of a group of repeated measurements should give the precision of those measurements. When deciding whether measurements agree with a theoretical prediction, the standard deviation of those measurements is of crucial importance: if the mean of the measurements is too far away from the prediction (with the distance measured in standard deviations), then the theory being tested probably needs to be revised. This makes sense since they fall outside the range of values that could reasonably be expected to occur if the prediction were correct and the standard deviation appropriately quantified. See prediction interval.
### Real-life examples
The practical value of understanding the standard deviation of a set of values is in appreciating how much variation there is from the "average" (mean).
#### Weather
As a simple example, consider average temperatures for cities. While two cities may each have an average temperature of 60 °F, it's helpful to understand that the range for cities near the coast is smaller than for cities inland, which clarifies that, while the average is similar, the chance for variation is greater inland than near the coast.
So, an average of 60 occurs for one city with highs of 80 °F and lows of 40 °F, and also occurs for another city with highs of 65 and lows of 55. The standard deviation allows us to recognize that the average for the city with the wider variation, and thus a higher standard deviation, will not offer as reliable a prediction of temperature as the city with the smaller variation and lower standard deviation.
#### Sports
Another way of seeing it is to consider sports teams. In any set of categories, there will be teams that rate highly at some things and poorly at others. Chances are, the teams that lead in the standings will not show such disparity, but will be pretty good in most categories. The lower the standard deviation of their ratings in each category, the more balanced and consistent they might be. So, a team that is consistently bad in most categories will have a low standard deviation. A team that is consistently good in most categories will also have a low standard deviation. A team with a high standard deviation might be the type of team that scores a lot (strong offense) but also concedes a lot (weak defense), or, vice versa, that might have a poor offense but compensates by being difficult to score on—teams with a higher standard deviation will be more unpredictable.
Trying to predict which teams, on any given day, will win, may include looking at the standard deviations of the various team "stats" ratings, in which anomalies can match strengths vs. weaknesses to attempt to understand what factors may prevail as stronger indicators of eventual scoring outcomes.
In racing, a driver is timed on successive laps. A driver with a low standard deviation of lap times is more consistent than a driver with a higher standard deviation. This information can be used to help understand where opportunities might be found to reduce lap times.
#### Finance
In finance, standard deviation is a representation of the risk associated with a given security (stocks, bonds, property, etc.), or the risk of a portfolio of securities. Risk is an important factor in determining how to efficiently manage a portfolio of investments because it determines the variation in returns on the asset and/or portfolio and gives investors a mathematical basis for investment decisions. The overall concept of risk is that as it increases, the expected return on the asset will increase as a result of the risk premium earned – in other words, investors should expect a higher return on an investment when said investment carries a higher level of risk.
For example, you have a choice between two stocks: Stock A historically returns 5% with a standard deviation of 10%, while Stock B returns 6% and carries a standard deviation of 20%. On the basis of risk and return, an investor may decide that Stock A is the better choice, because Stock B's additional percentage point of return generated (an additional 20% in dollar terms) is not worth double the degree of risk associated with Stock A. Stock B is likely to fall short of the initial investment more often than Stock A under the same circumstances, and will return only one percentage point more on average. In this example, Stock A has the potential to earn 10% more than the expected return, but is equally likely to earn 10% less than the expected return.
Calculating the average return (or arithmetic mean) of a security over a given number of periods will generate an expected return on the asset. For each period, subtracting the expected return from the actual return results in the variance. Square the variance in each period to find the effect of the result on the overall risk of the asset. The larger the variance in a period, the greater risk the security carries. Taking the average of the squared variances results in the measurement of overall units of risk associated with the asset. Finding the square root of this variance will result in the standard deviation of the investment tool in question. Use this measurement, combined with the average return on the security, as a basis for comparing securities.
### Geometric interpretation
To gain some geometric insights, we will start with a population of three values, x1, x2, x3. This defines a point P = (x1, x2, x3) in R3. Consider the line L = {(r, r, r) : r in R}. This is the "main diagonal" going through the origin. If our three given values were all equal, then the standard deviation would be zero and P would lie on L. So it is not unreasonable to assume that the standard deviation is related to the distance of P to L. And that is indeed the case. Moving orthogonally from P to the line L, one hits the point:
$R = (\overline{x},\overline{x},\overline{x})$
whose coordinates are the mean of the values we started out with. A little algebra shows that the distance between P and R (which is the same as the distance between P and the line L) is given by σ√3. An analogous formula (with 3 replaced by N) is also valid for a population of N values; we then have to work in RN.
### Rules for normally distributed data
Dark blue is less than one standard deviation from the mean. For the normal distribution, this accounts for 68.27 % of the set; while two standard deviations from the mean (medium and dark blue) account for 95.45 %; and three standard deviations (light, medium, and dark blue) account for 99.73 %.
In practice, one often assumes that the data are from an approximately normally distributed population. This is frequently justified by the classical central limit theorem, which says that sums of many independent, identically distributed random variables tend towards the normal distribution as a limit. If that assumption is justified, then about 68 % of the values are within 1 standard deviation of the mean, about 95 % of the values are within two standard deviations and about 99.7 % lie within 3 standard deviations. This is known as the 68-95-99.7 rule, or the empirical rule.
The confidence intervals are as follows:
σ 68.26894921371% 2σ 95.44997361036% 3σ 99.73002039367% 4σ 99.99366575163% 5σ 99.99994266969% 6σ 99.99999980268% 7σ 99.99999999974%
For normal distributions, the two points of the curve which are one standard deviation from the mean are also the inflection points.
### Chebyshev's inequality
Chebyshev's inequality proves that in any data set, nearly all of the values will be nearer to the mean value, where the meaning of "close to" is specified by the standard deviation. Chebyshev's inequality entails that for (nearly) all random distributions, not just normal ones, we have the following weaker bounds:
At least 50% of the values are within √2 standard deviations from the mean.
At least 75% of the values are within 2 standard deviations from the mean.
At least 89% of the values are within 3 standard deviations from the mean.
At least 94% of the values are within 4 standard deviations from the mean.
At least 96% of the values are within 5 standard deviations from the mean.
At least 97% of the values are within 6 standard deviations from the mean.
At least 98% of the values are within 7 standard deviations from the mean.
And in general:
At least (1 − 1/k2) × 100% of the values are within k standard deviations from the mean.
## Relationship between standard deviation and mean
The mean and the standard deviation of a set of data are usually reported together. In a certain sense, the standard deviation is a "natural" measure of statistical dispersion if the centre of the data is measured about the mean. This is because the standard deviation from the mean is smaller than from any other point. The precise statement is the following: suppose x1, ..., xn are real numbers and define the function:
$\sigma(r) = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - r)^2}$
Using calculus, or simply by completing the square, it is possible to show that σ(r) has a unique minimum at the mean:
$r = \overline{x}.\,$
(This can also be done with fairly simple algebra alone, since σ2(r) is equated to a quadratic polynomial).
The coefficient of variation of a sample is the ratio of the standard deviation to the mean. It is a dimensionless number that can be used to compare the amount of variance between populations with different means.
## Rapid calculation methods
A slightly faster (significantly for running standard deviation) way to compute the population standard deviation is given by the following formula (though considerations must be made for round-off error, arithmetic overflow, and arithmetic underflow conditions):
$\sigma\ = \sqrt{\left(\frac{1}{N}\sum_{i=1}^N{{x_i}^2}\right) - \overline{x}^2} = \sqrt{\left(\frac{1}{N}\sum_{i=1}^N{{x_i}^2}\right) - \left(\frac{1}{N}\sum_{i=1}^N{{x_i}}\right)^2} = \frac{1}{N}\sqrt{N\left(\sum_{i=1}^N{{x_i}^2}\right) - \left(\sum_{i=1}^N{{x_i}}\right)^2}$
or
$\sigma= \frac{1}{s_0}\sqrt{s_0s_2-s_1^2}$
where the power sums s0, s1, s2 are defined by
$\ s_j=\sum_{k=1}^N{x_k^j}.$
Similarly for sample standard deviation:
$s = \sqrt{\frac{\sum_{i=1}^N{{x_i}^2} - N\overline{x}^2}{N-1}\ }.$
Or from running sums:
$s = \sqrt{\frac{N\sum_{i=1}^N{{x_i}^2} - \left(\sum_{i=1}^N{x_i}\right)^2}{N(N-1)}}.$ |
Examveda
Train A passes a lamp post in 3 seconds and 900 meter long platform in 30 seconds. How much time will the same train take to cross a platform which is 800 meters long? (in seconds)
A. 24 seconds
B. 37 seconds
C. 33 seconds
D. 27 seconds
\eqalign{ & {\text{Let the length of train be x m}} \cr & {\text{When a train crosses a light }} \cr & {\text{post in 3 second the distance covered}} \cr & {\text{ = length of train }} \cr & \Rightarrow {\text{speed of train = }}\frac{x}{3} \cr & {\text{Distance covered in crossing a}} \cr & {\text{900 meter platfrom in 30 seconds}} \cr & {\text{ = Length of platfrom + length of train}} \cr & {\text{Speed of train = }}\frac{{x + 900}}{30} \cr & \Rightarrow \frac{x}{3} = \frac{{x + 900}}{{30}}\left[ {\because {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right] \cr & \Rightarrow \frac{x}{1} = \frac{{x + 900}}{{10}} \cr & \Rightarrow 10x = x + 900 \cr & \Rightarrow 10x - x = 900 \cr & \Rightarrow 9x = 900 \cr & \Rightarrow x = \frac{{900}}{9} = 100{\text{m}} \cr & {\text{When the length of the platform be 800m,}} \cr & {\text{then time T be taken by train to cross 800m}} \cr & {\text{long platfrom}} \cr & \frac{x}{3} = \frac{{x + 800}}{T} \cr & \Rightarrow Tx = 3x + 2400 \cr & \Rightarrow 100T = 300 + 2400 \cr & \Rightarrow 100T = 2700 \cr & \Rightarrow T = \frac{{2700}}{{100}} = 27{\text{ seconds}} \cr} |
# How to Convert Percent to Fraction: The Complete Guide
## I. Introduction
If you have ever wondered how to convert a percent to a fraction, you are not alone. Converting percents to fractions is an essential skill that everyone should learn to understand. This article aims to explain the process of converting percents to fractions in great detail, illustrating the concepts with examples for clarity. Our target audience is anyone out there who wants to learn how to convert percents to fractions, from middle school students to adults.
## II. Understanding basics: Percent and Fraction
Before we dive into how to convert percents to fractions, let’s clarify what a percent and a fraction are. A percent is a fraction expressed with a denominator of 100. Percent means “per hundred.” For example, if you say 50%, that is the same as saying 50 per hundred or 50/100. A fraction, on the other hand, is a representation of a part of a whole. It has a numerator (the top number) and a denominator (the bottom number). Examples of fractions include 1/2, 2/3, and 3/4.
## III. How to Convert a Percent to a Fraction
The process of converting a percent to a fraction is straightforward, and it involves three simple steps, as follows:
Step 1: Remove the percent symbol and put the number over 100
Step 2: Simplify the fraction by reducing it to its lowest terms
Step 3: If possible, convert the fraction to a mixed number
For example, let’s convert 75% to a fraction:
Step 1: 75% = 75/100
Step 2: We can simplify the fraction 75/100 by dividing both the numerator and denominator by their greatest common factor (GCF), which is 25, and we get 3/4.
Step 3: We cannot convert 3/4 into a mixed number since the numerator is already smaller than the denominator.
Therefore, 75% is equivalent to the fraction 3/4.
## IV. Different Methods of Converting a Percent to a Fraction
There are different methods of converting percents to fractions. Here are some of the most common methods:
### Method 1: Convert Percent to Fraction by Dividing by 100
This method involves dividing the percent by 100 and simplifying the resulting fraction if possible.
Example: Convert 20% to a fraction
Step 1: 20% = 20/100
Step 2: We can simplify 20/100 by dividing both numerator and denominator by 20, and we get 1/5.
Therefore, 20% is equivalent to 1/5.
### Method 2: Convert Percent to Fraction by Cross Multiplication
This method involves setting up a proportion and cross-multiplying to get the equivalent fraction.
Example: Convert 40% to a fraction
Step 1: We can set up the following proportion: 40/100 = x/1
Step 2: Cross-multiply to get 100x = 40
Step 3: Solve for x by dividing both sides by 100, and we get x = 0.4
Since 0.4 is already in its decimal form, we can write it as 4/10. Simplifying the fraction gives us 2/5.
Therefore, 40% is equivalent to 2/5.
## V. Common Percent-Fraction Conversions
There are a few percent-fraction conversions that come up more frequently than others. Here are some examples:
• 50% = 1/2
• 25% = 1/4
• 20% = 1/5
• 10% = 1/10
• 5% = 1/20
It is essential to memorize these conversions as they can make the process of converting percents to fractions quicker and easier.
## VI. Real-World Applications
Understanding percents and fractions is crucial in various fields, including finance, cooking, and engineering. In finance, for instance, understanding percentages is vital in calculating interest rates, mortgages, and taxes. In cooking, percentages are used to express recipes’ ingredients and cooking time. In engineering, percentages are used to express tolerances and specifications of machinery.
## VII. Tips and Tricks for Easier Conversion
Here are some tips and tricks that can make converting percents to fractions quicker and easier:
• Memorizing the common percent-fraction conversions
• Use the method that you are most comfortable with, i.e., dividing by 100, cross-multiplication, or decimal conversion
• Remember that percents are always out of 100, and fractions always have a denominator greater than 1
• Practice, practice, practice!
## VIII. Practice Problems and Solutions
Here are a few practice problems for you to try:
1. Convert 80% to a fraction.
2. Convert 5% to a fraction.
3. Convert 3/8 to a percent.
Here are the solutions to the practice problems:
1. 80% = 80/100 = 4/5
2. 5% = 5/100 = 1/20
3. Convert 3/8 to a decimal by dividing 3 by 8, which gives 0.375
4. Convert the decimal to a percent by multiplying by 100, which gives 37.5%
## IX. Conclusion
Converting percents to fractions is a crucial skill that everyone should acquire. It is essential in finance, cooking, engineering, and many other fields. We hope this guide has been helpful in illustrating the process of converting percents to fractions. Remember to memorize the common conversions, use the method that you are most comfortable with, and practice regularly. By mastering these concepts, you can make your life more comfortable and potentially open up many career opportunities. |
# Triangle
For alternate meanings, such as the musical instrument, see triangle (disambiguation).
A triangle is one of the basic shapes of geometry: a two-dimensional figure with three vertices and three sides which are straight line segments.
Contents
## Types of triangles
Triangles can be classified according to the relative lengths of their sides:
• In an equilateral triangle all sides are of equal length. An equilateral triangle is also equiangular, i.e. all its internal angles are equal—namely, 60°.
• In an isosceles triangle two sides are of equal length. An isosceles triangle also has two equal internal angles.
• In a scalene triangle all sides have different lengths. The internal angles in a scalene triangle are all different.
Equilateral Isosceles Scalene
Triangles can also be classified according to the size of their largest internal angle, described below using degrees of arc.
• A right triangle (or right-angled triangle) has one 90° internal angle (a right angle). The side opposite to the right angle is the hypotenuse; it is the longest side in the right triangle. The other two sides are the legs of the triangle.
• An obtuse triangle has one internal angle larger than 90° (an obtuse angle).
• An acute triangle has internal angles that are all smaller than 90° (three acute angles).
Right Obtuse Acute
## Basic facts
Elementary facts about triangles were presented by Euclid in books 1-4 of his Elements around 300 BCE.
A triangle is a polygon and a 2-simplex (see polytope).
Two triangles are said to be similar if and only if the angles of one are equal to the corresponding angles of the other. In this case, the lengths of their corresponding sides are proportional. This occurs for example when two triangles share an angle and the sides opposite to that angle are parallel.
Using right triangles and the concept of similarity, the trigonometric functions sine and cosine can be defined. These are functions of an angle which are investigated in trigonometry.
In the remainder we will consider a triangle with vertices A, B and C, angles α, β and γ and sides a, b and c. The side a is opposite to the vertex A and angle α and analogously for the other sides.
A triangle with vertices, sides and angles labelled
In Euclidean geometry, the sum of the angles α + β + γ is equal to two right angles (180° or π radians). This allows determination of the third angle of any triangle as soon as two angles are known.
The Pythagorean theorem
A central theorem is the Pythagorean theorem stating that in any right triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides. If vertex C is the right angle, we can write this as
c2 = a2 + b2
This means that knowing the lengths of two sides of a right triangle is enough to calculate the length of the third—something unique to right triangles. The Pythagorean theorem can be generalized to the law of cosines:
c2 = a2 + b2 - 2abcosγ
which is valid for all triangles, even if γ is not a right angle. The law of cosines can be used to compute the side lengths and angles of a triangle as soon as all three sides or two sides and an enclosed angle are known.
The law of sines states
$\frac{\sin\alpha}a=\frac{\sin\beta}b=\frac{\sin\gamma}c=\frac1d$
where d is the diameter of the circumcircle (the smallest circle that completely contains the triangle within itself). The law of sines can be used to compute the side lengths for a triangle as soon as two angles and one side are known. If two sides and an unenclosed angle is known, the law of sines may also be used; however, in this case there may be zero, one or two solutions.
## Points, lines and circles associated with a triangle
The Circumcenter is the centre of a circle passing through the three vertices of the triangle.
A perpendicular bisector of a triangle is a straight line passing through the midpoint of a side and being perpendicular to it, i.e. forming a right angle with it. The three perpendicular bisectors meet in a single point, the triangle's circumcenter; this point is the center of the circumcircle, the circle passing through all three vertices. The diameter of this circle can be found from the law of sines stated above.
Thales' theorem states that if the circumcenter is located on one side of the triangle, then the opposite angle is a right one. More is true: if the circumcenter is located inside the triangle, then the triangle is acute; if the circumcenter is located outside the triangle, then the triangle is obtuse.
The intersection of the altitudes is the Orthocenter.
An altitude of a triangle is a straight line through a vertex and perpendicular to (i.e. forming a right angle with) the opposite side. This opposite side is called the base of the altitude, and the point where the altitude intersects the base (or its extension) is called the foot of the altitude. The length of the altitude is the distance between the base and the vertex. The three altitudes intersect in a single point, called the orthocenter of the triangle. The orthocenter lies inside the triangle if and only if the triangle is not obtuse. The three vertices together with the orthocenter are said to form an orthocentric system.
The intersection of the angle bisectors finds the center of the Incircle.
An angle bisector of a triangle is a straight line through a vertex which cuts the corresponding angle in half. The three angle bisectors intersect in a single point, the center of the triangle's incircle. The incircle is the circle which lies inside the triangle and touches all three sides. There are three other important circles, the excircles; they lie outside the triangle and touch one side as well as the extensions of the other two. The centers of the in- and excircles form an orthocentric system.
The Centroid is the center of gravity.
A median of a triangle is a straight line through a vertex and the midpoint of the opposite side, and divides the triangle into two equal areas. The three medians intersect in a single point, the triangle's centroid. This is also the triangle's center of gravity: if the triangle were made out of wood, say, you could balance it on its centroid, or on any line through the centroid. The centroid cuts every median in the ratio 2:1, i.e. the distance between a vertex and the centroid is twice as large as the distance between the centroid and the midpoint of the opposite side.
Nine point circle demonstrates a symmetry where six points lie on the same circle.
The midpoints of the three sides and the feet of the three altitudes all lie on a single circle, the triangle's nine point circle. The remaining three points for which it is named are the midpoints of the portion of altitude between the vertices and the orthocenter. The radius of the nine point circle is half that of the circumcircle. It touches the incircle (at the Feuerbach point) and the three excircles.
Euler's line is a straight line through the centroid (yellow), orthocenter (blue), circumcenter (green) and center of the nine point circle (red).
The centroid (yellow), orthocenter (blue), circumcenter (green) and center of the nine point circle (red point) all lie on a single line, known as Euler's line (red line). The center of the nine point circle lies at the midpoint between the orthocenter and the circumcenter, and the distance between the centroid and the circumcenter is half that between the centroid and the orthocenter.
The center of the incircle is not in general located on Euler's line.
If one reflects a median at the angle bisector that passes through the same vertex, one obtains a symmedian. The three symmedians intersect in a single point, the symmedian point of the triangle.
## Computing the area of a triangle
Calculating the area of a triangle is an elementary problem encountered often in many different situations. Various approaches exist, depending on what is known about the triangle. What follows is a selection of frequently used formulae for the area of a triangle.
### Using geometry
The area S of a triangle is S = ½bh, where b is the length of any side of the triangle (the base) and h (the altitude) is the perpendicular distance between the base and the vertex not on the base. This can be shown with the following geometric construction.
The triangle is first transformed into a parallelogram with twice the area of the triangle, then into a rectangle.
To find the area of a given triangle (green), first make an exact copy of the triangle (blue), rotate it 180°, and join it to the given triangle along one side to obtain a parallelogram. Cut off a part and join it at the other side of the parallelogram to form a rectangle. Because the area of the rectangle is bh, the area of the given triangle must be ½bh.
### Using vectors
The area of the parallelogram is the magnitude of the cross product of the two vectors.
The area of a parallelogram can also be calculated by the use of vectors. If AB and AC are vectors pointing from A to B and from A to C, respectively, the area of parallelogram ABDC is |AB × AC|, the magnitude of the cross product of vectors AB and AC. |AB × AC| is also equal to |h × AC|, where h represents the altitude h as a vector.
The area of triangle ABC is half of this, or S = ½|AB × AC|.
### Using trigonometry
Applying trigonometry tofind the altitude h.
The altitude of a triangle can be found through an application of trigonometry. Using the labelling as in the image on the right, the altitude is h = a sin γ. Substituting this in the formula S = ½bh derived above, the area of the triangle can be expressed as S = ½ab sin γ.
It is of course no coincidence that the area of a parallelogram is ab sin γ.
### Using coordinates
If vertex A is located at the origin (0, 0) of a Cartesian coordinate system and the coordinates of the other two vertices are given by B = (x1y1) and C = (x2y2), then the area S can be computed as 1/2 times the absolute value of the determinant
$\begin{vmatrix}x_1 & x_2 \\ y_1 & y_2 \end{vmatrix}$
or S = ½ |x1y2 − x2y1|.
### Using Heron's formula
Yet another way to compute S is Heron's formula:
$S = \sqrt{s(s-a)(s-b)(s-c)}$
where s = ½ (a + b + c) is the semiperimeter, or one half of the triangle's perimeter.
### Using the side lengths and a numerically stable formula
Heron's formula is numerically unstable for triangles with a very small angle. A stable alternative involves arranging the lengths of the sides so that: abc and computing
$S = 1/4\sqrt{(a+(b+c)) (c-(a-b)) (c+(a-b)) (a+(b-c))}$
The brackets in the above formula are required in order to prevent numerical instability in the evaluation.
## Non-planar triangles
If any four of a triangle's elements (vertices, and/or elements of its sides) are plane to each other, the triangle is called plane. Geometers also study non-planar triangles, for instance spherical in spherical geometry and hyperbolic triangles in hyperbolic geometry. |
### SolvingTechnique for Consecutive Sudoku
Today we will look at the solving technique for tackling the Consecutive Sudoku.
Whenever you start a consecutive sudoku, look for 1 and 9 as clues with a consecutive bar beside it. We can simply put 2 with 1 and 8 with nine since there is no other option available for these 2 digits. Also Another thing that can be borne in mind is if we have the numbers 2 and 8 in a box and they dont have a consecutive bar beside them, then the numbers 1 and 9 can never be in a cell with a consecutive bar in that box. ( this holds true if the consecutive bars are within the same box. If the consecutive bar leads to a different box then this does not apply.)
Take a look at the image below.
We have a 1 at R8C9 with a consecutive bar and a 9 at R7C4 with a consecutive bar beside them. We can straight-away put 2 and 8 respectively as 1 and 9 do not have any other consecutive numbers.
Another technique that is evident in this example. Look at the 8 and 2 in Box 3 and 6 respectively. In box 3, we have 8 at R1C8 and since for number 9, number 8 is the only consecutive number, 9 cannot be adjacent to 8 as there is no consecutive bar, and neither can 9 be in R2C7,R2C8 or R3C789 since all these cells have a consecutive bar, but 8 is already present in the box. Hence the only place for 9 is at R2C9.
Similarly, look at 2 in Box 6. Just like the above example, 1 cannot be in R5C789 and R6C789 since 2 is already present at R4C9. 1 also cannot be at R4C8 since there is no consecutive bar beside the 2. Hence the only place for 1 in Box 6 is R4C7.
Apart from these things always look for longer chains of consecutive numbers as they are usually the starting point of a puzzle.
Keeping this in mind let us attempt the consecutive sudoku that was displayed at this blog.
Take a look at box 2. we have a chain of 7 consecutive numbers and a chain of 2 consecutive numbers. This means that all 9 digits are part of some chain. To fulfil this criteria, the 2 digit chain has to be either 1 and 2 or 8 and 9. so the chains will be 1 and 2 and 3 to 9, or 1 to 7 and 8,9.
When we look at the 4 at R2C1 and 7 at R1C8, the chain of 3 to 9 cannot be possible. If 3 is at R3C6 then 4 has to be at R2C6 which is not possible and if 3 is at R3C4 then 7 has to be at R1C6 which is again not possible because of the 2 given digits at R1C8 and R2C1. hence the 7 digit chain is 1 to 7 and the 2 digit chain is 8 and 9. so we can safely put number 4 at R1C5. and since R3C6 is either 1 or a 7 R3C9 also has to be a 4.
Now in Box 3, the only place for the digit 8 is R2C9. If 8 were at R2C7 then R2C8 has to be 9 which doesnt allow for the 2 digit chain in Box 2. So we have 8 at R2C9 and 9 at R1C7 since there is no other place for the digit 9. and that also enables 9 at R2C4 and 8 at R1C4.
Now in R2C7 and R2C8, we can have a pair of 1,2 or 2,3 or 5,6. But 5 and 6 not possible because of the 7 digit chain in Box 2. So 5 and 6 have to be in Row 3 in box 3. Hence we have R3C7 as 6 and R3C8 as 5. Which also gives us 7 at R3C6 and the complete 7 digit chain in Box 2.
Since we have got a 3 at R2C5, R1C9 becomes a 3 and we have the chain of 1,2 at R2C7 and R2c8. since the 3 digits left at R3C1,R3C2 and R3C3 are 3,8,9 and there is a consecutive chain at R3C1 and R3C2. We can deduce that R3C3 is a 3. R3C1 is 9 and R3C2 is 8. Which also gives us R2C2 as 7 and R3C3 as 5 (naked single) for row 2. and similarly we have 1,2,6 for Row1 in box 1. so 1,2 chain is the consecutive chain and 6 comes in at R1C1.
After this the puzzle is easily solvable.
Hope this explanation was helpful and it gives you a better lead and approach to solve a consecutive sudoku. |
# How do you solve (3x)/(x+9)+1=x/(x+9) and check for extraneous solutions?
Nov 23, 2016
$x = - 3$
#### Explanation:
$\frac{3 x}{x + 9} + 1 = \frac{x}{x + 9}$
Subtract $\frac{x}{x + 9}$ from both sides.
$\frac{3 x}{x + 9} - \frac{x}{x + 9} + 1 = 0$
Subtract $1$ from both sides.
$\frac{3 x}{x + 9} - \frac{x}{x + 9} = - 1$
Combine the two terms on the left.
$\frac{3 x - x}{x + 9} = - 1$
$\frac{2 x}{x + 9} = - 1$
Multiply both sides by $\left(x + 9\right)$.
$2 x = - 1 \left(x + 9\right)$
$2 x = - x - 9$
Add $x$ to both sides.
$3 x = - 9$
Divide both sides by $3$.
$x = - 3$ |
# Lesson 18
Expressed in Different Ways
### Lesson Narrative
In this lesson, students examine different ways to express repeated percent increase. Using rules of exponents, it is possible to group exponential expressions in different ways to highlight different aspects of a situation. For example, if an investment grows by 1% each month, then the effective annual percentage rate earned is about 12.68%, since $$(1.01)^{12} \approx 1.1268$$. After 6 years, the total interest earned is about 2.05, since $$(1.01)^{72} \approx 2.05$$. This means that the account balance will double (roughly) every 6 years.
If \\$1,000 were invested at this rate, with no further deposits or withdrawals, here are some different ways to see what the balance, in dollars, would be after 20 years:
• $$1,\!000 \boldcdot (1.01)^{240}$$
• $$1,\!000 \boldcdot \left((1.01)^{12}\right)^{20}$$
• about $$1,\!000 \boldcdot (1.268)^{20}$$
• about $$1,\!000 \boldcdot 2^4$$
The expressions highlight different growth rates for different time periods (a month, a year, 5 years).
In this lesson, we distinguish between growth rate (previously known as percent change or interest rate) and growth factor (defined earlier in this unit). In functions of the form $$a \boldcdot (1+r)^x$$, the growth rate is $$r$$, and the growth factor is $$1+r$$.
As students write and work across different expressions, they practice using structure and repeated reasoning (MP7 and MP8). In choosing an expression strategically to highlight a particular aspect of the situation, they reason abstractly and concretely (MP2).
Technology isn't required for this lesson, but there are opportunities for students to choose to use appropriate technology to solve problems. We recommend making technology available.
### Learning Goals
Teacher Facing
• Interpret and evaluate exponential expressions to solve problems.
• Write equivalent expressions to highlight different aspects of a situation that involves repeated percent increase or decrease.
### Student Facing
Let's write exponential expressions in different ways.
### Required Preparation
Devices that can run Desmos (recommended) or other graphing technology should be avalable as an optional tool for students.
### Student Facing
• I can solve problems using exponential expressions written in different ways.
• I can write equivalent expressions to represent situations that involve repeated percent increase or decrease.
Building On
Building Towards
### Glossary Entries
• growth rate
In an exponential function, the growth rate is the fraction or percentage of the output that gets added every time the input is increased by one. If the growth rate is 20% or 0.2, then the growth factor is 1.2. |
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# 3.V. Change of Basis - PowerPoint PPT Presentation
3.V. Change of Basis. 3.V.1. Changing Representations of Vectors 3.V.2. Changing Map Representations. 3.V.1. Changing Representations of Vectors. Definition 1.1 : Change of Basis Matrix
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3.V.1. Changing Representations of Vectors
3.V.2. Changing Map Representations
Definition 1.1: Change of Basis Matrix
The change of basis matrixfor bases B, D V is the representation of the identity map id : V → V w.r.t. those bases.
Lemma 1.2: Changing Basis
Proof:
Alternatively,
A matrix changes bases iff it is nonsingular.
Proof : Bases changing matrix must be invertible, hence nonsingular.
Proof : (See Hefferon, p.239.)
Nonsingular matrix is row equivalent to I.
Hence, it equals to the product of elementary matrices, which can be shown to represent change of bases.
Corollary 1.5:
A matrix is nonsingular it represents the identity map w.r.t. some pair of bases.
1. Find the change of basis matrix for B, DR2.
(a) B= E2 , D= e2 , e1
(b) B= E2 ,
(c)
D= E2
(d)
2. Let p be a polynomial in P3 with
where B= 1+x, 1x, x2+x3, x2x3 . Find a basis Dsuch that
Example 2.1: Rotation by π/6 in x-y planet: R2 → R2
Let
Let
Then
Consider t : V → V with matrix representation T w.r.t. some basis.
If basis B s.t. T = tB→B is diagonal,
Then t and T are said to be diagonalizable.
Definition 2.3: Matrix Equivalent
Same-sized matrices Hand Hare matrix equivalent
if nonsingular matrices Pand Qs.t.
H= P H Q or H = P 1H Q 1
Corollary 2.4:
Matrix equivalent matrices represent the same map, w.r.t. appropriate pair of bases.
Matrix equivalence classes.
Elementary row operations can be represented by left-multiplication (H= P H ).
Elementary column operations can be represented by right-multiplication ( H= H Q ).
Matrix equivalent operations cantain both (H= P H Q ).
∴ row equivalent matrix equivalent
Example 2.5:
and
are matrix equivalent but not row equivalent.
Theorem 2.6: Block Partial-Identity Form
Any mn matrix of rank k is matrix equivalent to the mn matrix that is all zeros except that the first k diagonal entries are ones.
Proof:
Gauss-Jordan reduction plus column reduction.
Example 2.7 left-multiplication (:
G-J row reduction:
Column reduction:
Column swapping:
Combined:
Corollary 2.8 left-multiplication (: Matrix Equivalent and Rank
Two same-sized matrices are matrix equivalent iff they have the same rank.
That is, the matrix equivalence classes are characterized by rank.
Proof.
Two same-sized matrices with the same rank are equivalent to the same block partial-identity matrix.
Example 2.9:
The 22 matrices have only three possible ranks: 0, 1, or 2.
Thus there are 3 matrix-equivalence classes.
If a linear map left-multiplication (f : V n → W m is rank k,
then some bases B → D s.t. f acts like a projection Rn → Rm.
Exercises 3.V.2. left-multiplication (
1. Show that, where A is a nonsingular square matrix, if P and Q are nonsingular square matrices such that PAQ = I then QP = A1.
2. Are matrix equivalence classes closed under scalar multiplication? Addition?
3.
(a) If two matrices are matrix-equivalent and invertible, must their inverses be matrix-equivalent?
(b) If two matrices have matrix-equivalent inverses, must the two be matrix- equivalent?
(c) If two matrices are square and matrix-equivalent, must their squares be
matrix-equivalent?
(d) If two matrices are square and have matrix-equivalent squares, must they be matrix-equivalent? |
# Find the approximate change in $y$ as $x$ increases from 2 to 2.02
Find the approximate change in $y$ as $x$ increases from 2 to 2.02
The equation of a curve is $y=4x^3-8x^2+10$
a)Find $\frac{dy}{dx}$
$\frac{dy}{dx}=12x^2-16x$
But I don't know how to answer below
b) "Find the approximate change in $y$ as $x$ increases from 2 to 2.02"
I have try
$12(2)^2-16(2) = 16/2.02=$ not right
$4(2)^3-8(2)^2+10=10/2.02=$ not right
$12(2.02)^2-16(2.02)=16.644/2=$ not right
etc.... the right answer is =$0.32$
help out thanks.
• Change of $y$ is approximately change of $x$ times derivative = $0.02\times y'(2)=0.02\times16$. – Did Jul 17 '12 at 11:25
We have $y=4x^3-8x^2+10$, and therefore $\frac{dy}{dx}=12x^2-16x$.
We use the tangent line approximation, also known as the linear approximation.
The derivative at $x=2$ is equal to $16$. Therefore, if $\Delta x$ represents the change in $x$, and $\Delta y$ represents the change in $y$, we have $$\Delta y \approx (16)\Delta x.$$
Remarks: One important way to get insight about the linear approximation is geometric. Let $f(x)=4x^3-8x^2+10$. The idea is that the tangent line at $x=2$ is close to the curve when $x$ is close to $2$, that the tangent line kisses the curve at $x=2$. A tiny bug, sitting on the curve $y=f(x)$ at $x=2$, would think she was sitting on a straight ine, the tangent line.
Recall that the tangent line at $x=a$ has equation $$y-f(a)=f'(a)(x-a).$$ In our case, the tangent line has equation $$y-f(2)=16(x-2).$$ Because the tangent line is close to the curve when $x$ is close to $2$, we have $$f(2.02)-f(2)\approx (16)(2.02-2).$$ This says that the change in $y$ is approximately $(16)(0.02).$
Another way of thinking about it is kinematic, in terms of motion. So let us use the letter $t$ instead of $x$. A particle is moving along the $y$-axis. At any time $t$, the displacement of the particle is $4t^3-8t^2+10$. Then the velocity at time $t$ is the derivative of $4t^3-8t^2+10$, evaluated at $t=2$. If time changes from $2$ to $2.02$, then the change in $y$ (the change in displacement) is approximately the velocity at time $2$ times the elapsed time. So the change in $y$ is approximately $(16)(0.02)$. The reason that the approximation is reasonable is that the velocity does not change very much from time $2$ to time $2.02$, so the velocity remains close to $16$. If the velocity were exactly $16$, then the change in displacement would be exactly $(16)(0.02)$. Since velocity does change a little, the approximation is not exact.
It is worthwhile to do an explicit numerical calculation to check how good the tangent line approximation is in this case. The calculator says that $f(2.02)$ is nearly equal to $10.326432$, so to calculator accuracy, the change in $y$ is about $0.326432$. The linear approximation we made predicts a change of approximately $0.32$. Pretty close!
Finally, we can think of our calculation in terms of the definition of the derivative. Recall that $$f'(2)=\lim{h\to 0} \frac{f(2+h)-f(2)}{h}.$$ So if $h$ is kind of close to $0$, like $h=0.02$, then we should have $$\frac{f(2+h)-f(2)}{h}\approx f'(2).$$ This can be written as $f(2+h)-f(2) \approx (f'(2))h$. |
# 2’s Complement Subtraction of Binary Numbers
2’s complement subtraction method is a way to subtract two binary numbers by actually adding one number with the 2’s complement of another number. In this article, the method of binary subtraction using 2’s complement is elaborated with examples.
### Method of 2’s Complement Subtraction:
To implement this method for subtracting two binary numbers, the very first step is find the 2’s complement of the number which is to be subtracted from another number. To get the 2’s complement, first of all 1’s complement is find and then 1 is added to this. The addition is the required 2’s complement.
Suppose, we need to find the 2’s complement of binary number 10010. First, find 1’s complement. To find this, replace all 1 to 0 and all 0 to 1. Therefore, 1’s complement of 10010 will be 01101. Now, add 1 to this as shown below.
#### Steps for 2’s Complement Subtraction:
##### Subtraction of Smaller Number from Larger Number:
To subtract a smaller number from a larger number using 2’s complement subtraction, following steps are to be followed:
Step-1: Determine the 2’s complement of the smaller number
Step-2: Add this to the larger number.
Step-3: Omit the carry. Note that, there is always a carry in this case.
Following example illustrate the above mentioned steps:
Exampe-1: Subtract (1010)2 from (1111)2 using 2’s complement method.
Solution:
Step-1: 2’s complement of (1010)2 is (0110)2.
Step-2: Add (0110)2 to (1111)2. This is shown below.
##### Subtraction of Larger Number from Smaller Number:
To subtract a larger number from a smaller number using 2’s complement subtraction, following steps are to be followed:
Step-1: Determine the 2’s complement of the smaller number
Step-2: Add this to the larger number.
Step-3: There is no carry in this case. The result is in 2’s complement form and is negative.
Step-4: To get answer in true form, take 2’s complement and change its sign.
Example-2: Subtract (1010)2 from (1000)2 using 2’s complement.
Solution:
Step-1: Find the 2’s complement of (1010)2. It is (0110)2. |
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# Prove that ${{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}={{\left( 1+\sec A\operatorname{cosec}A \right)}^{2}}$.
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Hint: Transform the whole equation in terms of $\sin \theta$ and $\cos \theta$ and then convert into desired form.
We have to prove that ${{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}={{\left( 1+\sec A\operatorname{cosec}A \right)}^{2}}....\left( i \right)$
Taking $LHS$ of equation $\left( i \right)$, we get
${{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Therefore, ${{\left( \sin A+\sec A \right)}^{2}}+{{\left( \cos A+\operatorname{cosec}A \right)}^{2}}$
$={{\sin }^{2}}A+{{\sec }^{2}}A+2\sin A\sec A+{{\cos }^{2}}A+{{\operatorname{cosec}}^{2}}A+2\cos A\operatorname{cosec}A$
Rearranging the equation, we get
$\Rightarrow \left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)+2\left( \sin A\sec A+\cos A\operatorname{cosec}A \right)+{{\sec }^{2}}A+{{\operatorname{cosec}}^{2}}A....\left( ii \right)$
We know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$
Putting this value in equation $\left( ii \right)$.
We get $1+2\left( \sin A\sec A+\cos A\operatorname{cosec}A \right)+{{\sec }^{2}}A+{{\operatorname{cosec}}^{2}}A...\left( iii \right)$
We know that $\sec A=\dfrac{1}{\cos A}$ and $\operatorname{cosec}A=\dfrac{1}{\sin A}$
We will put the values of $\sec A$ and $\operatorname{cosec}A$ in equation $\left( iii \right)$.
We get, $1+2\left( \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A} \right)+\left( \dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A} \right)$
$=1+2\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A} \right)+\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{{{\sin }^{2}}A{{\cos }^{2}}A} \right)$
We know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$.
Hence we get,
$1+2\left( \dfrac{1}{\sin A\cos A} \right)+\left( \dfrac{1}{{{\sin }^{2}}A{{\cos }^{2}}A} \right)$
We know, $\dfrac{1}{\sin A}=\operatorname{cosec}A$ and $\dfrac{1}{\cos A}=\sec A$
Hence, we get $1+2\operatorname{cosec}A\sec A+{{\operatorname{cosec}}^{2}}A{{\sec }^{2}}A$
We can write it as
${{\left( 1 \right)}^{2}}+{{\left( \operatorname{cosec}A\sec A \right)}^{2}}+2\left( \operatorname{cosec}A\sec A \right).1$
We know that ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$
By considering $a=1$ and $\operatorname{cosec}A\sec A=B$
We finally get $LHS={{\left( 1+\operatorname{cosec}A\sec A \right)}^{2}}$ which is equal to $\text{RHS}$.
Hence Proved.
Note: By looking the terms of $\operatorname{cosec}A$ and $secA$ in $\text{RHS}$, students convert $\sin A$ and $\cos A$ into $\dfrac{1}{\operatorname{cosec}A}$ and $\dfrac{1}{secA}$ respectively in first step only, but that creates confusion and does not give the desired results. |
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### Precalculus Mathematics for Calculus
Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508
# Find the degree measure of the angle with the given radian measure.$\frac{5\pi }{18}$
${\mathbf{50}}^{°}$
See the step by step solution
## Step 1. Given information.
$\frac{5\pi }{18}$ rad.
## Step 2. Explanation.
We know that
$1\text{\hspace{0.17em}radian\hspace{0.17em}=\hspace{0.17em}}\left(\frac{{180}^{\circ }}{\pi }\right)$
Thus to find the degrees of the angle multiply the radian measure with $\left(\frac{{180}^{\circ }}{\pi }\right)$
Here the radian measure $\mathrm{\pi }$ is not available so when we multiply put $\pi =3.14$
## Step 3. Calculation.
To convert radians to degrees, multiply by $\left(\frac{{180}^{\circ }}{\pi }\right)$:
$\frac{5\pi }{18}\text{rad}=\left(\frac{5\pi }{18}\text{\hspace{0.17em}x\hspace{0.17em}}\frac{180}{\pi }\right)$
Put $\pi =3.14$ in the expression, we get:
$\begin{array}{l}={\left(\frac{5\text{\hspace{0.17em}}x\text{\hspace{0.17em}}3.14}{18}x\frac{180}{3.14}\right)}^{\circ }\\ ={\left(\frac{5x180}{18}\right)}^{{}^{\circ }}\\ ={50}^{{}^{\circ }}\end{array}$ |
# Sine rule
## Statement
The ratio between a triangle's side length and the sine of its angle is the same for all sides:
$$\frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)} = \frac{c}{\sin(\gamma)}$$
## Proof
Construct altitudes $h$ and $k$ like the image below:
### Altitude $h$
Write the following equation from the definition of sine:
$$\sin(\alpha) = \frac{h}{b}$$
Flip the fraction and multiply both sides by $a$.
$$\frac{1}{\sin(\alpha)} = \frac{b}{h}$$
$$\frac{a}{\sin(\alpha)} = \frac{ab}{h}$$
Now write another equation using the definition of sine:
$$\sin(\beta) = \frac{h}{a}$$
Flip the fraction and multiply both sides by $b$.
$$\frac{1}{\sin(\beta)} = \frac{a}{h}$$
$$\frac{b}{\sin(\beta)} = \frac{ab}{h}$$
Note that both $\frac{a}{\sin(\alpha)}$ and $\frac{b}{\sin(\beta)}$ are equal to $\frac{ab}{h}$, giving:
$$\frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)}$$
### Altitude $k$
Write the following equation from the definition of sine:
$$\sin(\alpha) = \frac{k}{c}$$
Flip the fraction and multiply both sides by $a$.
$$\frac{1}{\sin(\alpha)} = \frac{c}{k}$$
$$\frac{a}{\sin(\alpha)} = \frac{ac}{k}$$
Now write another equation using the definition of sine:
$$\sin(\gamma) = \frac{k}{a}$$
Flip the fraction and multiply both sides by $c$.
$$\frac{1}{\sin(\gamma)} = \frac{a}{k}$$
$$\frac{c}{\sin(\gamma)} = \frac{ac}{h}$$
Note that both $\frac{a}{\sin(\alpha)}$ and $\frac{c}{\sin(\gamma)}$ are equal to $\frac{ac}{h}$, giving:
$$\frac{a}{\sin(\alpha)} = \frac{c}{\sin(\gamma)}$$
## Conclusion
Since $\frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)}$ and $\frac{a}{\sin(\alpha)} = \frac{c}{\sin(\gamma)}$:
$$\frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)} = \frac{c}{\sin(\gamma)}$$ |
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# The power of a lens is $+ 2.5D$. What kind of lens it is and what is its focal length?(A) $40cm$, Convex lens(B) $30cm$, Convex lens(C) $400cm$, Concave lens(D) $400cm$, Convex lens
Last updated date: 20th Jun 2024
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Hint We can get the focal length of the lens from its power as the focal length is the reciprocal of the power of a lens. And the positive or negative sign of the focal length will tell us about the kind of lens as positive is convex lens and negative is concave lens.
Formula Used: In this solution we will be using the following formula,
$\Rightarrow f = \dfrac{1}{P}$
where $f$ is the focal length of the lens and $P$ is the power of the lens.
For any kind of lens, the power of the lens and the focal length of the lens are related by the formula, $f = \dfrac{1}{P}$
In the question we are given that the power of the lens is $+ 2.5D$.
So we have $P = + 2.5D$. Now we can substitute this value in the formula. So we get,
$\Rightarrow f = \dfrac{1}{{2.5}}$
On calculating this we get,
$\Rightarrow f = 0.4m$
In the options, we are given the answer in centimeters. So on converting we get the value of the focal length of the lens as,
$\Rightarrow f= 0.4 \times 100cm = 40cm$
Now the power of the lens given in question is positive. So as a result the focal length of the lens is also positive.
The positive value of the focal length is present in a convex lens where the concave lens has a negative value of the focal length. So the focal length is $40cm$ and the lens is a convex lens.
Therefore, the correct answer is option D.
Note
The power of a lens denotes the ability of a lens to bend the light that is passing through it. The greater is the power of the lens, the more is the light refracted while passing through it. It is also the reciprocal of the focal length. So the more power of the lens, the less is its focal length. For a convex lens it measures the converging ability and for a concave lens, it measures the diverging ability. The unit of the power of a lens is Diopters and it is denoted by D. |
# ARITHMETIC PROGRESSION (A. P)
## ARITHMETIC PROGRESSION (A. P)
CONTENT
• Sequence
• Definition of Arithmetic Progression
• Denotations in Arithmetic progression
• Deriving formulae for the term of A. P.
• Sum of an arithmetic series
Find the next two terms in each of the following sets of number and in each case state the rule which gives the term.
(a) 1, 5, 9, 13, 17, 21, 25(any term +4 = next term)
(b) 2, 6, 18, 54, 162, 486, 1458 (any term x 3 = next term)
(c) 1, 9, 25, 49, 81, 121, 169, (sequence of consecutive odd no)
(d) 10, 9, 7, 4, 0, -5, -11, 18, -26, (starting from 10, subtract 1, 2, 3 from immediate no).
In each of the examples below, there is a rule which will give more terms in the list. A list like this is called a SEQUENCE in many cases; it can simply matter if a general term can be found for a sequence e.g.
1, 5, 9, 13, 17 can be expressed as
1, 5, 9, 13, 17 ……………. 4n – 3 where n = no of terms
Check: 5th term = 4(5) -3
20 – 3 = 17
10th term = 4(10) – 3
40 – 3 = 37
Example 2
Find the 6th and 9th terms of the sequence whose nth term is
(a) (2n + 1)
(b) 3 – 5n.
Solution
(a) 2n + 1
6th term = 2(6) + 1 = 12 + 1 = 13
9th term = 2 (9) + 1 = 18 + 1 = 19
(b) 3 – 5n
6th term = 3 – 5 (6) = 3 – 30 = -27
9th term = 3 – 5 (9) = 3 – 45 = –42
Evaluation
For each of the following sequence, find the next two terms and the rules which give the term.
1 8
1 4
1 2
1. 1, , , , , ____, ____
2 100, 96, 92, 88, _____, ____
1. 2, 4, 6, 8, 10, ____, _____
2. 1, 4, 9, 16, 25, ____, _____
(i) Arrange the numbers in ascending order (ii) Find the next two terms in the sequence
1. 19, 13, 16, 22, 10
2. -21/2, 51/2, 31/2, 11/2, –1/2
3. Find the 15th term of the sequence whose nth term is 3n – 5
## DEFINITION OF ARITHMETIC PROGRESSION
A sequence in which the terms either increase or decrease in equal steps is called an Arithmetic Progression.
The sequence 9, 12, 15, 18, 21, ____, _____, _____ has a first term of 9 and a common difference of +3 between the terms.
Denotations in A. P.
a = 1st term
d = common difference
n = no of terms
Un = nth term
Sn = Sum of the first n terms
## Formula for nth term of Arithmetic Progression
e.g. in the sequence 9, 12, 15, 18, 21.
a = 9
d = 12 – 9 or 18 – 15 = 3.
1st term = U1 = 9 = a
2nd term = U2 = 9 + 3 = a + d
3rd term = U3 = 9 + 3 + 3 = a + 2d
10th term = U10 = 9 + 9(3) = a + 9d
nth term = Un = 9+(n-1)3 = a + (n-1)d
\nth term = Un = a + (n-1)d
Example:
1.Given the A.P, 9, 12, 15, 18 …… find the 50th term.
a = 9 d = 3 n = 50 Un = U50
Un= a + (n-1) d
U50 = 9 + (50-1) 3
= 9 + (49) 3
= 9 + 147
= 156
2.The 43rd term of an AP is 26, find the 1st term of the progression given that its common difference is ½ and also find the 50th term.
U43= 26 d = ½ a = ? n = 43
Un = a + (n-1) d
26 = a + (43-1) ½
26 = a + 42(1/2)
26 = a + 21
26 – 21 = a
5 = a
a = 5
(b) a = 5 d = ½ n = 50 U50 =?
Un = a + (n-1) d
U50 = 5 + (50-1)1/2
= 5 + 49(1/2)
U50 = 5 + 241/2
U50 = 291/2
Evaluation
1. Find the 37th term of the sequence 20, 10, 0, -10…
2. 1, 5… 69 are the 1st, 2nd, and last term of the sequence; find the common difference between them and the number of terms in the sequence.
## SUM OF AN ARITHMETIC SERIES
When the terms of a sequence are added, the resulting expression is called series e.g. in the sequence 1, 3, 5, 7, 9, 11.
Series = 1 + 3 + 5 + 7 + 9 + 11
When the terms of a sequence are unending, the series is called infinite series, it is often impossible to find the sum of the terms in an infinite series.
e.g. 1 + 3 + 5 + 7 + 9 + 11 + …………………. Infinite
Sequence with last term or nth term is termed finite series.
e.g.
Find the sum of
1, 3, 5, 7, 9, 11, 13, 15
If sum = 2, n = 8
Then
S = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
Or S = 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1
2s = 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16
48 = 8(16)
2 2 = S = 64
Deriving the formula for sum of A. P. The following represent a general arithmetic series when the terms are added.
S = a + (a+d) + a + 2d + …………………………… + (L-2d) + (L-d) + L – eqn
S = L + (L-d) + L – 2d + ……………………………… a + 2d + (a+d) + a – eqn
2s = (a + L) + (a + L) + (a + L) + …………………… (a + L) + (a + L) + (a + L)
2s = n(a + L)
2
S = n(a+L)
2
L => Un = a + (n-1)d
Substitute L into eq**
S = n(a + a+(n-1)d
2
S = n(2a + (n-1)d = n ( 2a+ (n-1)d
22
\ S = n[a + L] where L is the last term i.e Un
2
or
S =n[2a +(n-1)d] when d is given or obtained
2
Example 2
Find the sum of the 20th term of the series 16 + 9 + 2 + …………………
a = 16 d = 9 – 16 = -7 n = 20
S = n(2a + (n-1)d)
2
S = 20 (2×16) + (20-1)(-7)
2
20 (32 + 19(-7)
2
S =10 (32 – 133) = 10(-101)
S = -1010
## EVALUATION
1. Find the sum of the arithmetic series with 16 and -117 as the first and 20th term respectively.
2. The salary scale for a clerical officer starts at N55, 200 per annum. A rise of N3, 600 is given at the end of each year; find the total amount of money earned in 12 years.
## GENERAL EVALUATION /REVISION QUESTION
1. An A. P. has 15 terms and a common difference of -3, find its first and last term if its sum is 120.
2. On the 1st of January, a student puts N10 in a box, on the 2nd she puts N20 in the box, on the 3rd she puts N30 and so on putting on the same no. of N10 notes as the day of the month. How much will be in the box if she keeps doing this till 16th January?
3. The salary scale for a clerical officer starts at N55, 200 per annum. A rise of N3, 600 is given at the end of each year, find the total amount of money earned in 12 years.
4. 4. Find the 7th term and the nth term of the progression 27,9,3,…
5. If 8, x, y, – 4 are in A.P, find x and y.
## WEEKEND ASSIGNMENT
1. Find the 4th term of an A. P. whose first term is 2 and the common difference is 0.5 (a) 4 (b) 4.5 (c) 3.5 (d) 2.5
2. In an A. P. the difference between the 8th and 4th term is 20 and the 8th term is 11/2 times the 4th term, find the common difference (a) 5 (b) 7 (c) 3 (d) 10
3. Find the first term of the sequence in no. 2 (a) 70 (b) 45 (c) 25 (d) 5
4. The next term of the sequence 18, 12, 60 is (a) 12 (b) 6 (c) -6 (d) -12
5. Find the no. of terms of the sequence 1/2 , ¾, 1, ……………….. 51/2 (a) 21 (b) 43/4 (c) 1 (d) 22
## THEORY
1. Eight wooden poles are to be used for pillars and the length of the poles form an arc Arithmetic Progression (A. P.) if the second pole is 2m and the 6th pole is 5m, give the lengths of the poles in order and sum up the lengths of the poles.
2 a. Write down the 15th term of the sequence.
2_, 3 ,4 , 5
1×3 2×4 3×5 4 x6
1. An arithmetic progression (A. P.) has 3 as its term and 4 as the common difference.
2. Write an expression in its simplest form for the nth term.
3. Find the 10th term and the sum of the first. |
# How to Find Arithmetic Sequence – A Step-by-Step Guide
To find an arithmetic sequence, I first identify the common difference between consecutive terms. This difference is constant and is denoted by (d). For any given arithmetic sequence, the (n)th term can be calculated using the formula $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and (n) is the term number.
Finding the sum of the first (n) terms involves a different formula, namely $S_n = \frac{n}{2}(2a_1 + (n-1)d$, which is derived from the sequence’s properties.
I always make sure to have the sequence’s initial term and the common difference on hand, as these are critical for plugging into the formulas. The process itself is straightforward once these values are known.
I find it satisfying how a simple pattern of adding or subtracting the same number can reveal so much about a series of numbers. Stick around, and I’ll walk you through understanding each component in more detail.
## Steps Involved in Finding Arithmetic Sequences
When I look at a sequence, identifying whether it’s an arithmetic sequence is my first step. An arithmetic sequence is a sequence of numbers where each term after the first is found by adding a consistent number, known as the common difference, to the previous term.
This common difference can be positive, negative, or zero, leading to increasing, decreasing, or constant sequences, respectively.
To identify and work with an arithmetic sequence, I follow these steps:
1. Identify the Common Difference
The common difference ((d)) is found by subtracting any term from the subsequent term. For any two consecutive terms $a_n$ and $a_{n+1}$, it’s calculated as: $d = a_{n+1} – a_n$
This difference must remain constant throughout the sequence.
2. Determine the First Term
The first term $a_1$ is often provided, but if not, I back-calculate it using the common difference and a term whose position is known.
3. Derive the General Formula
Once (d) and $a_1$ are known, I use them to compose the general formula for the (n)th term of the sequence: $a_n = a_1 + (n-1)d$
4. Validate the Sequence
To confirm if a sequence is arithmetic, I check if the pattern of adding the common difference to each term to get to the next term is consistent throughout.
5. Use the Pattern
With the general formula, any term in the sequence can be found. For example, the 12th term $a_{12}$ is found by plugging (n = 12) into the formula.
Remember, arithmetic sequences are a foundational concept in algebra, and understanding them can be very useful in grasping related math lessons and tutorials. Unlike geometric sequences or Fibonacci sequences, where terms are multiplied or are a sum of previous terms, arithmetic sequences thrive on the simplicity of addition or subtraction.
For sequences involving fractions or negative numbers, the same steps apply but with additional attention to maintaining the correct sign and appropriately adding or subtracting fractions.
## Calculating Terms in an Arithmetic Sequence
In an arithmetic sequence, each term is equal to the previous term plus a constant difference. To find any term, I use the arithmetic sequence formula:
$a_n = a_1 + (n-1)d$
Here, $a_n$ is the nth term I’m looking to find, $a_1$ is the first term, and ( d ) is the common difference. The term ( n ) stands for the term’s position in the sequence.
For instance, if I have an arithmetic sequence where the first term $a_1$ is 5 and the common difference ( d ) is 3, and I want to calculate the 4th term $a_4$, I apply the formula:
$a_4 = 5 + (4-1) \times 3$
$a_4 = 5 + 9$
$a_4 = 14$
I can also represent sequences with an explicit formula:
$a_n = a_1 + (n-1)d$
Or a recursive formula, which expresses each term based on the previous one:
$a_{n} = a_{n-1} + d$
Here’s a simple table to illustrate the calculation of terms in the sequence with $a_1 = 5$ and ( d = 3 ):
Term (n)CalculationResult
155
25 + 38
38 + 311
411 + 314
514 + 317
Each entry under “Calculation” follows the pattern of adding the common difference to the previous term to get the next term. This way, I can build the entire sequence term by term, or calculate any specific term using the initial value and the common difference.
## Summing Terms in Arithmetic Series
When I work with an arithmetic series, the goal is to find the sum of all terms within the sequence. An arithmetic series is a sequence of numbers in which each term after the first is found by adding a constant, known as the common difference, to the previous term.
For example, if I start with the term 3 and the common difference is 2, my arithmetic series would be: 3, 5, 7, 9, and so on. To find the sum of the first ‘n’ terms of this series, I use the arithmetic series formula:
$$S_n = \frac{n}{2} \times (a_1 + a_n)$$
Here, ‘$S_n$’ is the sum of the first ‘n’ terms, ‘$a_1$’ is the first term, and ‘$a_n$’ is the nth term. The nth term can also be found using the formula:
$$a_n = a_1 + (n – 1) \times d$$
Where ‘d’ is the common difference. Let’s consider we want to find the sum of the first 5 terms from my example. The first term, ‘$a_1$’, is 3, and the fifth term, ‘$a_5$’, is 11 ($since 3 + 4 \times 2 = 11$). Plugging these values into the formula gives us:
$$S_5 = \frac{5}{2} \times (3 + 11) = \frac{5}{2} \times 14 = 35$$
Thus, the sum of the first 5 terms is 35. The table below illustrates the sum of an arithmetic series depending on the number of terms ‘n’:
nFirst Term (a_1)Common Difference (d)nth Term $a_n$Sum $S_n$
5321135
103221110
Finding the sum of an arithmetic series is straightforward with the right formula. Remember to identify the first term, common difference, and the nth term for accurate solutions.
## Conclusion
Through this article, I’ve guided you through the steps to identify and work with arithmetic progressions. Recognizing the common difference is key, whether it’s positive, indicating an increasing sequence, or negative, for a decreasing sequence.
Unlike geometric sequences where terms are multiplied, or the Fibonacci sequence that adds the two previous terms, arithmetic sequences have a straightforward pattern: each term increments by a fixed amount.
Finding the n-th term of an arithmetic sequence involves the formula $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference. Similarly, the sum of an arithmetic series can be found using $S_n = \frac{n}{2}(a_1 + a_n)$ or $S_n = \frac{n}{2}(2a_1 + (n-1)d)$.
I aimed to make algebra accessible, especially if dealing with fractions or negative numbers within these sequences.
To further your understanding, related math lessons, and tutorials are available, often spotlighting the domain and function of sequences in algebra. These resources cater to both beginners and those looking to polish their skills, covering a variety of topics beyond what we’ve explored together.
Remember, practice makes perfect. Applying these formulas to different problems will bolster your arithmetic sequence skills. Stay curious, and don’t hesitate to dive into more complex sequences as you grow more confident with these basics. |
# What is the mean of the sum of two dice?
Contents
## What is the sum of two die?
The probability of a certain sum of two die is equal to the total number of different sum combinations for each possible sum out of the total number of all possible sums. The sums of two six-sided dice are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
## Is the sum of two dice independent?
Note the probability of getting a sum of seven with two die, is equal to the probability of getting a sum of seven after rolling 4. That is the definition of independence.
## What is the sum of the dice equation?
You compute it by multiplying each value x of the random variable by the probability P(X=x), and then adding up the results. So the average sum of dice is: E(X) = 2 . 1/36 + 3 . 2/36 + ….
## What is the formula of dice?
Probability = Number of desired outcomes ÷ Number of possible outcomes = 3 ÷ 36 = 0.0833. The percentage comes out to be 8.33 per cent. Also, 7 is the most likely result for two dice. Moreover, there are six ways to achieve it.
## What is the total sum of a dice?
Since each die has 6 values, there are 6∗6=36 6 ∗ 6 = 36 total combinations we could get. If you add up the numbers in the total column above, you’ll get 36.
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## What is the total of a die?
The most common type of die is a six-sided cube with the numbers 1-6 placed on the faces. The value of the roll is indicated by the number of “spots” showing on the top. For the six-sided die, opposite faces are arranged to always sum to seven.
## What is the probability that if we roll 2 dice the sum is 7?
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
## Is a dice roll independent?
When the events do not affect one another, they are known as independent events. Independent events can include repeating an action like rolling a die more than once, or using two different random elements, such as flipping a coin and spinning a spinner. Many other situations can involve independent events as well. |
# Lesson Video: Solving Systems of Linear Inequalities Mathematics • 9th Grade
In this video, we will learn how to solve systems of linear inequalities by graphing them and identify the regions representing the solution.
15:57
### Video Transcript
In this video, we will learn how to solve systems of linear inequalities by graphing them and identify the regions representing the solution. We should already be comfortable with using and interpreting inequality notation, plotting straight-line graphs given their equations, and graphing single linear inequalities.
A system of linear inequalities is a set of two or more linear inequalities in several variables. For example, π¦ is greater than or equal to one, π₯ is greater than zero, and π₯ plus π¦ is less than or equal to four is a system of linear inequalities in the two variables π₯ and π¦. Such systems are used in practical contexts when one problem requires a range of solutions, and there is more than one constraint on those solutions.
Letβs begin by recalling some of the basics of how we represent linear inequalities graphically. Weβll consider the inequalities π¦ is less than five and π₯ is greater than or equal to three. We recall that when representing inequalities on a graph, we first need to determine the equation of the boundary line of the region. And to do this, we swap the inequality sign for an equals sign. So the boundary line for the inequality π¦ is less than five is π¦ is equal to five. We then draw this line, recalling that lines of the form π¦ equals some constant are horizontal lines. We also need to recall that we use a solid line if we have a weak inequality and we use a dashed line if we have a strict inequality. Here, the inequality is strict. Itβs π¦ is strictly less than five, so we draw the line π¦ equals five as a dashed line.
We now have the boundary line for the graphical solution to this inequality. And we need to decide which side of the line to shade. If we want π¦ to be less than five, then we need to include all points in the coordinate plane for which the π¦-coordinate is less than five. Thatβs all points below the boundary line, so we shade this side of the line.
Weβve now represented the solution to the first inequality graphically. So letβs now consider the second inequality: π₯ is greater than or equal to three. Once again, we replace the inequality with an equals sign to find the equation of the boundary line. The equation of the boundary is π₯ equals three, which we recall is a vertical line through the value three on the π₯-axis.
As we have a weak inequality, π₯ is greater than or equal to three, we represent this boundary with a solid line. We then need to decide which side of the line to shade. And as we want π₯ to be greater than or equal to three, we need to shade the region to the right of the boundary. Remember that the solid line indicates that points on the boundary line itself are included in the solution to this inequality. So, weβve represented the solution to each inequality graphically.
If we want to treat this as a system of linear inequalities, then this means we need to find the region that satisfies both inequalities simultaneously, which is the intersection of the two regions weβve shaded. The region below the orange line and to the right of the pink line therefore represents the solution to this system of inequalities. It contains all the points in the coordinate plane such that π¦ is less than five and π₯ is greater than or equal to three. Letβs now consider some examples.
State the system of inequalities whose solution is represented by the following graph.
From the diagram, we can see that the shaded region is the first quadrant of the coordinate plane, which contains all points such that both π₯ and π¦ are positive. The boundary lines for the region, which are the positive π₯- and π¦-axes, are drawn as solid lines. So points along the boundaries are also included. This means that zero values of both π₯ and π¦ are included in the region. Using a bit of logic, we can therefore express this region using the inequalities π₯ is greater than or equal to zero and π¦ is greater than or equal to zero.
Letβs confirm this by representing each of these inequalities graphically ourselves. The inequality π₯ is greater than or equal to zero is represented by the π¦-axis, where π₯ is equal to zero, and all the points to the right of the π¦-axis. The inequality π¦ is greater than or equal to zero is represented by the π₯-axis, where π¦ equals zero, and all points above the π₯-axis. The intersection of these two regions is the part that is shaded in both orange and pink. Thatβs the whole of the first quadrant, including the positive π₯- and π¦-axes, which matches up with the graph we were given. So, we can conclude that the system of inequalities represented by the given graph is π₯ is greater than or equal to zero and π¦ is greater than or equal to zero.
Letβs now consider a slightly more complex example.
State the system of inequalities whose solution is represented by the following graph.
We can see that this system of inequalities contains horizontal, vertical, and diagonal boundary lines, all of which we will need to find the equations of. First though, we identify that the shaded region is in the first quadrant, which contains all nonnegative values of π₯ and π¦. This can be represented by the inequalities π₯ is greater than or equal to zero and π¦ is greater than or equal to zero.
Next, letβs consider the vertical lines. The equations of these lines are π₯ equals three and π₯ equals six. The region between these lines has been shaded, so this corresponds to π₯ being between three and six. But we need to be clear about which inequality signs to use at each end of the interval. As the line drawn at π₯ equals three is a solid line, this corresponds to a weak inequality. So we have π₯ is greater than or equal to three. As the line drawn at π₯ equals six is a dashed line, this corresponds to a strict inequality. So we have π₯ is less than six. This can be combined into the double-sided inequality π₯ is greater than or equal to three and less than six.
Next, we consider the horizontal lines. All horizontal lines have equations of the form π¦ equals some constant. So the equations of these lines are π¦ equals two and π¦ equals six. The blue region is between the two lines, so π¦ is between two and six. As these lines are solid, both inequalities are weak inequalities. So this region can be represented as π¦ is greater than or equal to two and less than or equal to six.
Finally, we consider the orange region below the diagonal line. This line passes through eight on the π₯-axis and eight on the π¦-axis. We wonβt go into the detail of how to find its equation here, but its equation can be written as π¦ equals eight minus π₯. As the shaded region is below the line and the line itself is solid, this corresponds to the inequality π¦ is less than or equal to eight minus π₯. We could leave this inequality in this form, or we can rearrange it by adding π₯ to both sides to give π₯ plus π¦ is less than or equal to eight.
So weβve found the system of inequalities whose solution is represented by the graph. Their solution set is the intersection of the individual shaded regions, which is the green triangular region. The system of inequalities is π₯ is greater than or equal to zero. π¦ is greater than or equal to zero. π₯ is greater than or equal to three and less than six. π¦ is greater than or equal to two and less than or equal to six. And π₯ plus π¦ is less than or equal to eight.
Letβs now consider another example.
Which of the points below belongs to the solution set of the system of linear inequalities represented by the figure shown? (a) One, three; (b) six, eight; (c) four, three; or (d) three, six.
Itβs important to realize that we arenβt being asked to find the system of linear inequalities represented by the figure shown but rather to determine which of four given points lies in their solution set. From the figure, we can see that there are essentially three types of linear inequalities that have been graphed. There are two vertical lines, and the region between these lines is shaded in red. There are two horizontal lines, and the region between these lines is shaded in blue. Finally, there is a diagonal line, and the region below this line is shaded in orange. The region that satisfies each of these inequalities simultaneously is the intersection of the three shaded regions, which is the triangle shaded in green.
To determine which of the four given points lies in the solution set of the system of inequalities, we just need to plot each of these points on the graph and identify any points that are inside or on the solid boundaries of this region. When we do this, we find that the only point that lies in the shaded region and hence belongs in the solution set of the system of linear inequalities is the point four, three.
Hereβs another example.
Find the solution set of the linear inequalities shown below.
So there are two linear inequalities graphed on the same coordinate plane. And weβre asked to find the solution set of this pair of linear inequalities. This means we need to find the region that satisfies both inequalities simultaneously, which is the intersection of the two individual regions. But itβs clear from looking at the graph that these two regions donβt overlap. The two diagonal lines are parallel, so they never meet. The red region is above the higher diagonal line, and the blue region is below the lower line. As a result, there is no intersection between these two regions. And so the solution set of this pair of linear inequalities is the empty set, π.
Letβs consider one final example.
Which region on the graph contains solutions to the set of inequalities π¦ is greater than two, π¦ is greater than or equal to negative π₯, and π₯ is less than one?
These inequalities have already been graphed for us, and our job is to identify which of the regions marked with letters represents the solution to all three inequalities simultaneously. Weβll do this by first identifying the solution set to each inequality separately. The boundary line for the inequality π¦ is greater than two is π¦ is equal to two, which is a horizontal line through the value two on the π¦-axis. Thatβs the dashed line drawn in blue. Itβs a dashed line because the inequality is π¦ is strictly greater than two. As the inequality is π¦ is greater than two, the solution set to this inequality will be the region above the dashed line. So this immediately narrows the options down to G, D, and F.
The boundary line for the inequality π¦ is greater than or equal to negative π₯ is the line π¦ equals negative π₯, which is a diagonal line through the origin with a slope of negative one. Thatβs the solid line drawn in green, and a solid line has been used here because it is a weak inequality. As the inequality is π¦ is greater than or equal to negative π₯, the solution set to this inequality is the region on and above the line. We can therefore rule out region G.
For the final inequality, the boundary line is π₯ equals one, which is a vertical line through one on the π₯-axis. This is of course the final line drawn on the coordinate grid, the vertical dashed line drawn in red. As the inequality is π₯ is less than one, the region satisfying this inequality is to the left of the boundary line. This rules out option F. So, we can conclude that the region that contains the solutions to the system of all three inequalities, which is the intersection of the three individual regions, is region D.
Letβs now summarize the key points from this video. For a vertical line parallel to the π¦-axis of the form π₯ equals some constant π, the region to the right of the line contains all the points for which π₯ is greater than π and the region to the left of the line contains all the points such that π₯ is less than π. For a horizontal line of the form π¦ equals π, the region above the line corresponds to π¦ is greater than π and the region below the line corresponds to π¦ is less than π. For a diagonal line of the form π¦ equals ππ₯ plus π, the region above the line corresponds to where π¦ is greater than ππ₯ plus π and the region below the line corresponds to where π¦ is less than ππ₯ plus π.
We use dashed lines to represent strict inequalities, which means the boundary line itself is not included in the solution set, and solid lines to represent weak inequalities, which means the boundary line is included. The solution set for a system of linear inequalities is the region that satisfies every inequality simultaneously, which is the intersection of the individual regions. We also saw that if the regions do not overlap, then there are no solutions to the system of linear inequalities, in which case we write the solution set as the empty set π. |
# Lecture 5 Part 2 Trigonometric Functions Trigonometric Functions
• Slides: 22
Lecture 5 - Part 2 Trigonometric Functions
Trigonometric Functions
Solution
Solution
Graphs of Trigonometric Functions Properties of Sine and Cosine Functions The graphs of y = sin x and y = cos x have similar properties: 1. The domain is the set of real numbers. 2. The range is the set of y values such that . 3. The maximum value is 1 and the minimum value is – 1. 4. The graph is a smooth curve. 5. Each function cycles through all the values of the range over an x-interval of. 6. The cycle repeats itself indefinitely in both directions of the x-axis.
Properties of Sine and Cosine Functions The graphs of y = sin x and y = cos x have similar properties: 1. The domain is the set of real numbers. 2. The range is the set of y values such that . 3. The maximum value is 1 and the minimum value is – 1. 4. The graph is a smooth curve. 5. Each function cycles through all the values of the range over an x-interval of. 6. The cycle repeats itself indefinitely in both directions of the x-axis.
Graph of the Sine Function To sketch the graph of y = sin x first locate the key points. These are the maximum points, the minimum points, and the intercepts. x 0 sin x 0 1 0 -1 0 Then, connect the points on the graph with a smooth curve that extends in both directions beyond the five points. A single cycle is called a period. y = sin x y x
Graph of the Cosine Function To sketch the graph of y = cos x first locate the key points. These are the maximum points, the minimum points, and the intercepts. x 0 cos x 1 0 -1 0 1 Then, connect the points on the graph with a smooth curve that extends in both directions beyond the five points. A single cycle is called a period. y = cos x y x
Example: Sketch the graph of y = 3 cos x on the interval [– , 4 ]. Partition the interval [0, 2 ] into four equal parts. Find the five key points; graph one cycle; then repeat the cycle over the interval. x y = 3 cos x (0, 3) 0 3 max y -3 0 x-int min ( 0 2 3 x-int max , 3) x ( ( , 0) ( , – 3) , 0)
The amplitude of y = a sin x (or y = a cos x) is half the distance between the maximum and minimum values of the function. amplitude = |a| If |a| > 1, the amplitude stretches the graph vertically. If 0 < |a| > 1, the amplitude shrinks the graph vertically. If a < 0, the graph is reflected in the x-axis. y y = 2 sin x x y= sin x y = – 4 sin x reflection of y = 4 sin x |
# Video: EG19M1-DiffAndInt-Q08
If a region bounded by the two curves π¦ = 6 β π₯ and π¦ = βπ₯ and the π₯-axis is revolved completely about the π₯-axis, find the volume of the solid generated.
07:49
### Video Transcript
If a region bounded by the two curves π¦ equals six minus π₯ and π¦ equals the square root of π₯ and the π₯-axis is revolved completely about the π₯-axis, find the volume of the solid generated.
Before we go forward, letβs sketch out this image. Hereβs the π₯- and π¦-axis. π¦ equals the square root of π₯ would look something like this. π¦ equals six minus π₯ would look something like this. And the figure is bounded by the π₯-axis. The solid generated by revolving this region around the π₯-axis would look something like this. To find the volume of something like this, we take the definite integral of ππ squared dπ₯, where π is the vertical distance from the axis of rotation. And that means that π is a function of π₯. But our region is bounded by two different functions. We need to find the volume of the blue piece by taking a definite integral of π times the square root of π₯ squared. And weβll add that to the volume created by six minus π₯. But how do we decide and define these definite integrals?
We know that our smallest point of the square root of π₯ function is zero. To find out the other two parts of the definite integral, weβll need to find the intersection of π¦ equals six minus π₯ and π¦ equals the square root of π₯. We can find that intersection by setting these two functions equal to each other: the square root of π₯ equals six minus π₯. To get rid of that square root, we square both sides and then we have π₯ equals six minus π₯ squared. To solve, we need to expand. Six minus π₯ times six minus π₯ 36 minus six π₯ minus six π₯ plus π₯ squared is equal to π₯. We can combine like terms. Negative six π₯ plus negative six π₯ equals negative 12π₯. We just rearrange the function to say π₯ squared minus 12π₯ plus 36. Bring down that π₯.
To set this equation equal to zero, we subtract π₯ from both sides. We see that zero equals π₯ squared minus 13π₯ plus 36 we can solve for π₯ by factoring. Two factors of 36 that add together to equal negative 13 are negative nine and negative four. So weβll set π₯ minus nine and π₯ minus four equal to zero, like this. And we have π₯ equals nine and π₯ equals four. But we only have one intersection here. So we need to see which of these π₯-values is real. Weβll have either π¦ equals six minus four or π¦ equals six minus nine. When π₯ equals four, π¦ equals two. And when π₯ equals nine, π¦ equals negative three. The solution four, two is the solution weβre looking for. Itβs the one in the first quadrant. And we can ignore that second π₯ equals nine. But what is this telling us?
This dotted line at π₯ equals four helps us to define our definite integrals. Weβre using the blue function, the square root of π₯ from zero to four. And at four, we switch over to the yellow function. The final thing we need to consider is the place where the function π¦ equals six minus π₯ intersects the π₯-axis. π¦ equals six minus π₯ intersects the π₯-axis when π¦ equals zero. So we set π¦ equal to zero and solve for π₯. Subtract six from both sides. Negative six equals negative π₯ and π₯ equals six. And that means the yellow function is bounded between four and six.
Now, we need to clear some space to find these integrals. Weβll start with the blue piece. We know that π is a constant. So we can go ahead and take it out. And the square root of π₯ squared equals π₯. We need to take π times the definite integral from zero to four of π₯ dπ₯. Weβll need the antiderivative of π₯ dπ₯ from four to zero. And weβre gonna multiply that by π.
Moving on to the yellow piece, the six minus π¦ function, we can go ahead and pull out our constant. And then, we need to consider what six minus π₯ squared would be. Fortunately, we already squared six minus π₯ we know that it is π₯ squared minus 12π₯ plus 36 dπ₯. And now, we need the antiderivative of π₯ squared minus 12 π₯ plus 36. One-third π₯ cubed minus 12 divided by two π₯ squared. 12 divided by two equals six plus 36 π₯ from four to six. And then, weβll multiply all of that by π. Notice that both of these terms have a factor of π. So we can pull that out and save it for the end.
Now, weβre gonna substitute what we know. One-half times four squared minus one-half times zero squared. We know that zero squared term drops out plus one-third times six cubed minus six times six squared plus 36 times six minus one-third times four cubed minus six times four squared plus 36 times four. Four squared equals 16 times one-half equals eight. Six cubed equals 216 divided by three equals 72. Six times six squared is the same thing as six cubed, 216. 36 times six is the same thing as six cubed, 216, minus four cubed is 64 divided by three is 64 thirds. Four squared equals 16 times six equals 96 and 36 times four equals 144. And remember all of this is being multiplied by π. 72 minus 216 plus 216 equals 72. Negative 96 plus 144 equals 48. 64 thirds plus 48 equals 69 and one-third. 72 minus 69 and one-third equals two and two-thirds, which we can rewrite as eight-thirds. We want to add eight and eight-thirds together. We can rewrite eight as 24 thirds. That way we can add the numerators together. 24 plus eight equals 32. We have 32 thirds that we need to multiply by π, 32π over three units cubed since weβre dealing with volume.
This region bounded by the curves π¦ equals six minus π₯ and π¦ equals the square root of π₯ has a volume of 32π over three units cubed. |
# What Is Arithmetic Progressions In Math? (Solution found)
An arithmetic progression, or AP, is a sequence where each new term after the first is obtained. by adding a constant d, called the common difference, to the preceding term. If the first term. of the sequence is a then the arithmetic progression is. a, a + d, a + 2d, a + 3d,
## What is the meaning of arithmetic progression in maths?
From Wikipedia, the free encyclopedia. An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15,… is an arithmetic progression with a common difference of 2.
## What is arithmetic progression explain with example?
An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same. For example, the sequence 2, 6, 10, 14, … is an arithmetic progression (AP) because it follows a pattern where each number is obtained by adding 4 to its previous term. In this sequence, nth term = 4n-2.
## How do you show arithmetic progression?
Arithmetic Progressions An arithmetic progression is a sequence where each term is a certain number larger than the previous term. The terms in the sequence are said to increase by a common difference, d. For example: 3, 5, 7, 9, 11, is an arithmetic progression where d = 2. The nth term of this sequence is 2n + 1.
## What is arithmetic progression Class 10?
Arithmetic Progression (AP) also known as Arithmetic Sequence is a sequence or series of numbers such that the common difference between two consecutive numbers in the series is constant. For example: Series 1: 1,3,5,7,9,11…. In this series, the common difference between any two consecutive numbers is always 2.
## What is the use of arithmetic progression?
What is the use of Arithmetic Progression? An arithmetic progression is a series which has consecutive terms having a common difference between the terms as a constant value. It is used to generalise a set of patterns, that we observe in our day to day life.
## What is the difference between an and N in arithmetic progression?
N stands for the number of terms while An stands for the nth term it ISNT the number of terms. Don’t get confused. Cheers!
## How do I find AP and GP?
Progressions (AP, GP, HP)
1. nth term of an AP = a + (n-1) d.
2. Arithmetic Mean = Sum of all terms in the AP / Number of terms in the AP.
3. Sum of ‘n’ terms of an AP = 0.5 n (first term + last term) = 0.5 n [ 2a + (n-1) d ]
## Which of the following is arithmetic progression?
Answer: A sequence of numbers which has a common difference between any two consecutive numbers is called an arithmetic progression (A.P.). The example of A.P. is 3,6,9,12,15,18,21, …
## What is the arithmetic mean between 10 and 24?
Using the average formula, get the arithmetic mean of 10 and 24. Thus, 10+24/2 =17 is the arithmetic mean.
## Arithmetic progression – Wikipedia
The evolution of mathematical operations The phrase “orarithmetic sequence” refers to a sequence of integers in which the difference between successive words remains constant. Consider the following example: the sequence 5, 7, 9, 11, 13, 15,. is an arithmetic progression with a common difference of two. As an example, if the first term of an arithmetic progression is and the common difference between succeeding members is, then in general the -th term of the series () is given by:, and in particular, A finite component of an arithmetic progression is referred to as a finite arithmetic progression, and it is also referred to as an arithmetic progression in some cases.
## Sum
2 + 5 + 8 + 11 + 14 = 40 14 + 11 + 8 + 5 + 2 = 40 16 + 16 + 16 + 16 + 16 = 80
Calculation of the total amount 2 + 5 + 8 + 11 + 14 = 2 + 5 + 8 + 11 + 14 When the sequence is reversed and added to itself term by term, the resultant sequence has a single repeating value equal to the sum of the first and last numbers (2 + 14 = 16), which is the sum of the first and final numbers in the series. As a result, 16 + 5 = 80 is double the total. When all the elements of a finite arithmetic progression are added together, the result is known as anarithmetic series. Consider the following sum, for example: To rapidly calculate this total, begin by multiplying the number of words being added (in this case 5), multiplying by the sum of the first and last numbers in the progression (in this case 2 + 14 = 16), then dividing the result by two: In the example above, this results in the following equation: This formula is applicable to any real numbers and.
### Derivation
An animated demonstration of the formula that yields the sum of the first two numbers, 1+2+.+n. Start by stating the arithmetic series in two alternative ways, as shown above, in order to obtain the formula. When both sides of the two equations are added together, all expressions involvingdcancel are eliminated: The following is a frequent version of the equation where both sides are divided by two: After re-inserting the replacement, the following variant form is produced: Additionally, the mean value of the series may be computed using the following formula: The formula is extremely close to the mean of an adiscrete uniform distribution in terms of its mathematical structure.
## Product
When the members of a finite arithmetic progression with a beginning elementa1, common differencesd, andnelements in total are multiplied together, the product is specified by a closed equation where indicates the Gamma function. When the value is negative or 0, the formula is invalid. This is a generalization of the fact that the product of the progressionis provided by the factorialand that the productforpositive integersandis supplied by the factorial.
### Derivation
Where represents the factorial ascension.
According to the recurrence formula, which is applicable for complex numbers0 “In order to have a positive complex number and an integer that is greater than or equal to 1, we need to divide by two. As a result, if0 “as well as a concluding note
### Examples
Exemple No. 1 If we look at an example, up to the 50th term of the arithmetic progression is equal to the product of all the terms. The product of the first ten odd numbers is provided by the number = 654,729,075 in Example 2.
## Standard deviation
In any mathematical progression, the standard deviation may be determined aswhere is the number of terms in the progression and is the common difference between terms. The standard deviation of an adiscrete uniform distribution is quite close to the standard deviation of this formula.
## Intersections
The intersection of any two doubly infinite arithmetic progressions is either empty or another arithmetic progression, which may be obtained using the Chinese remainder theorem. The intersection of any two doubly infinite arithmetic progressions is either empty or another arithmetic progression. Whenever a pair of progressions in a family of doubly infinite arithmetic progressions has a non-empty intersection, there exists a number that is common to all of them; in other words, infinite arithmetic progressions form a Helly family.
## History
This method was invented by a young Carl Friedrich Gaussin primary school student who, according to a story of uncertain reliability, multiplied n/2 pairs of numbers in the sum of the integers from 1 through 100 by the values of each pairn+ 1. This method is used to compute the sum of the integers from 1 through 100. However, regardless of whether or not this narrative is true, Gauss was not the first to discover this formula, and some believe that its origins may be traced back to the Pythagoreans in the 5th century BC.
• Geometric progression
• Harmonic progression
• Arithmetic progression
• Number with three sides
• Triangular number
• Sequence of arithmetic and geometry operations
• Inequality between the arithmetic and geometric means
• In mathematical progression, primes are used. Equation of difference in a linear form
• A generalized arithmetic progression is a set of integers that is formed in the same way that an arithmetic progression is, but with the addition of the ability to have numerous different differences
• Heronian triangles having sides that increase in size as the number of sides increases
• Mathematical problems that include arithmetic progressions
• Utonality
## References
1. Duchet, Pierre (1995), “Hypergraphs,” in Graham, R. L., Grötschel, M., and Lovász, L. (eds. ), Handbook of combinatorics, Vol. 1, 2, Amsterdam: Elsevier, pp. 381–432, MR1373663. Duchet, Pierre (1995), “Hypergraphs,” in Graham, R. L., Grötschel, M., and Particularly noteworthy are Section 2.5, “Helly Property,” pages 393–394
2. And Hayes, Brian (2006). “Gauss’s Day of Reckoning,” as the saying goes. Journal of the American Scientist, 94(3), 200, doi:10.1511/2006.59.200 The original version of this article was published on January 12, 2012. retrieved on October 16, 2020
3. Retrieved on October 16, 2020
4. “The Unknown Heritage”: a trace of a long-forgotten center of mathematical expertise,” J. Hyrup, et al. The American Journal of Physics 62, 613–654 (2008)
5. Tropfke, Johannes, et al (1924). Geometrie analytisch (analytical geometry) pp. 3–15. ISBN 978-3-11-108062-8
6. Tropfke, Johannes. Walter de Gruyter. pp. 3–15. ISBN 978-3-11-108062-8
7. (1979). Arithmetik and Algebra are two of the most important subjects in mathematics. pp. 344–354, ISBN 978-3-11-004893-3
8. Problems to Sharpen the Young,’ Walter de Gruyter, pp. 344–354, ISBN 978-3-11-004893-3
9. The Mathematical Gazette, volume 76, number 475 (March 1992), pages 102–126
10. Ross, H.E.Knott, B.I. (2019) Dicuil (9th century) on triangle and square numbers, British Journal for the History of Mathematics, volume 34, number 2, pages 79–94
11. Laurence E. Sigler is the translator for this work (2002). The Liber Abaci of Fibonacci. Springer-Verlag, Berlin, Germany, pp.259–260, ISBN 0-387-95419-8
12. Victor J. Katz is the editor of this work (2016). The Mathematics of Medieval Europe and North Africa: A Sourcebook is a reference work on medieval mathematics. 74.23 A Mediaeval Derivation of the Sum of an Arithmetic Progression. Princeton, NJ: Princeton University Press, 1990, pp. 91, 257. ISBN 9780691156859
13. Stern, M. (1990). 74.23 A Mediaeval Derivation of the Sum of an Arithmetic Progression. Princeton, NJ: Princeton University Press, 1990, pp. 91, 257. ISBN 9780691156859
14. Stern, M. Journal of the American Mathematical Society, vol. 74, no. 468, pp. 157-159. doi:10.2307/3619368. |
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### Section 3-10 : Implicit Differentiation
7. Find $$y'$$ by implicit differentiation for $$4{x^2}{y^7} - 2x = {x^5} + 4{y^3}$$.
Show All Steps Hide All Steps
Hint : Don’t forget that $$y$$ is really $$y\left( x \right)$$ and so we’ll need to use the Chain Rule when taking the derivative of terms involving $$y$$! Also, don’t forget that because $$y$$ is really $$y\left( x \right)$$ we may well have a Product and/or a Quotient Rule buried in the problem.
Start Solution
First, we just need to take the derivative of everything with respect to $$x$$ and we’ll need to recall that $$y$$ is really $$y\left( x \right)$$ and so we’ll need to use the Chain Rule when taking the derivative of terms involving $$y$$. This also means that the first term on the left side is really a product of functions of $$x$$ and hence we will need to use the Product Rule when differentiating that term.
Differentiating with respect to $$x$$ gives,
$8x{y^7} + 28{x^2}{y^6}y' - 2 = 5{x^4} + 12{y^2}y'$ Show Step 2
Finally, all we need to do is solve this for $$y'$$.
$8x{y^7} - 5{x^4} - 2 = \left( {12{y^2} - 28{x^2}{y^6}} \right)y'\hspace{0.5in} \Rightarrow \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{y' = \frac{{8x{y^7} - 5{x^4} - 2}}{{12{y^2} - 28{x^2}{y^6}}}}}$ |
# How to Find (f○g)(x)
Print
The Composition of two functions is often difficult to be understood. We will be using an example problem involving two functions to demonstrate how to find the composition of those two functions in an easy way.
We will be solving (F?G)(x), when f(x)=3/(x-2) and g(x)=2/x. f(x) and g(x) cannot be undefined, and therefore x cannot be equal to the number that makes the denominator zero whilst the numerator is not zero. In order to find what value (x) makes f(x) undefined, we must set the denominator equal to 0, and then solve for x. f(x)=3/(x-2); we set the denominator,which is x-2, to 0.(x-2=0, which is x=2). When we set the denominator of g(x) equal to 0, we get x=0. So x cannot be equal to 2 or 0. Please click on the image for a better understanding.
Now, we will solve (F?G)(x). By definition, (F?G)(x)is equal to f(g(x)). This means that every x in f(x) must be replaced with g(x), which is equal to (2/x). Now f(x)=3/(x-2) which is equal to f(g(x))=3/[(2/x)-2]. This is f(g(x)). Please click on the image for a better understanding.
Next, we will simplify f(g(x))=3/[(2/x)-2]. To do this, we need to express both parts of the denominators as fractions. We can rewrite 2 as (2/1). f(g(x))=3/[(2/x)-(2/1)]. Now, we will find the sum of the fractions in the denominator, which will give us f(g(x))=3/[(2-2x)/x]. Please click on the image for a better understanding.
In order to change the fraction from a complex fraction to a simple fraction, we will multiply the numerator, 3, by the reciprocal of the denominator. f(g(x))=3/[(2-2x)/x] which would become f(g(x))=(3)[x/(2-2x)] => f(g(x))=3x/(2-2x). This is the simplified form of the fraction. We already know that x cannot be equal to 2 or 0, as it makes f(x) or g(x) undefined. Now we need to find what number x that causes f(g(x)) to be undefined. To do this, we set the denominator equal to 0. 2-2x=0 => -2x=-2 => (-2/-2)x =(-2/-2) => x=1. The final answer is 3x/(2-2x), x cannot equal to: 0,1, nor 2. Please click on the image for a better understanding.
• Paper
• and Pencil |
# How do you write a polynomial with Zeros: -2, multiplicity 2; 4, multiplicity 1; degree 3?
Feb 29, 2016
$p \left(x\right) = {x}^{3} - 12 x - 16$
#### Explanation:
For a polynomial, if $x = a$ is a zero of the function, then $\left(x - a\right)$ is a factor of the function.
We have two unique zeros: $- 2$ and $4$. However, $- 2$ has a multiplicity of $2$, which means that the factor that correlates to a zero of $- 2$ is represented in the polynomial twice.
Follow the colors to see how the polynomial is constructed:
$\text{zero at "color(red)(-2)", multiplicity } \textcolor{b l u e}{2}$
$\text{zero at "color(green)4", multiplicity } \textcolor{p u r p \le}{1}$
$p \left(x\right) = {\left(x - \left(\textcolor{red}{- 2}\right)\right)}^{\textcolor{b l u e}{2}} {\left(x - \textcolor{g r e e n}{4}\right)}^{\textcolor{p u r p \le}{1}}$
Thus,
$p \left(x\right) = {\left(x + 2\right)}^{2} \left(x - 4\right)$
Expand:
$p \left(x\right) = \left({x}^{2} + 4 x + 4\right) \left(x - 4\right)$
$p \left(x\right) = {x}^{3} - 12 x - 16$
We can graph the function to understand multiplicities and zeros visually:
graph{x^3-12x-16 [-6, 6, -43.83, 14.7]}
The zero at $x = - 2$ "bounces off" the $x$-axis. This behavior occurs when a zero's multiplicity is even.
The zero at $x = 4$ continues through the $x$-axis, as is the case with odd multiplicities.
Note that the function does have three zeros, which it is guaranteed by the Fundamental Theorem of Algebra, but one of such zeros is represented twice. |
# Linear Factorizations Sec. 2.6b. First, remind me of the definition of a linear factorization… f (x) = a(x – z )(x – z )…(x – z ) An equation in the following.
## Presentation on theme: "Linear Factorizations Sec. 2.6b. First, remind me of the definition of a linear factorization… f (x) = a(x – z )(x – z )…(x – z ) An equation in the following."— Presentation transcript:
Linear Factorizations Sec. 2.6b
First, remind me of the definition of a linear factorization… f (x) = a(x – z )(x – z )…(x – z ) An equation in the following form: 12n
Now, Our Practice Problems: Find all zeros of the given function, and write the function in its linear factorization. Check the graph for possible real zeros… Possibly, x = –2, x = 1, and x = 4 –21–3–55–68 –210–1010–8 1–55 40 Check and factor, using synthetic division:
Now, Our Practice Problems: Find all zeros of the given function, and write the function in its linear factorization. 11–55 4 1–41 1 1 0
Now, Our Practice Problems: Find all zeros of the given function, and write the function in its linear factorization. 41–41 404 1010
Now, Our Practice Problems: Find all zeros of the given function, and write the function in its linear factorization. Complete Linear Factorization:
Now, Our Practice Problems: The complex number z = 1 – 2i is a zero of the given function. Find the remaining zeros of the function, and write it in its linear factorization. 1 – 2i40171465 4 – 8i–12 – 16i–27 – 26i–65 45 – 16i–13 – 26i04 – 8i
Now, Our Practice Problems: The complex number z = 1 – 2i is a zero of the given function. Find the remaining zeros of the function, and write it in its linear factorization. 1 + 2i44 – 8i5 – 16i–13 – 26i 4 + 8i8 + 16i13 + 26i 41308 Use the quadratic formula to find the last two zeros… 1 + 2i must also be a zero!!!
Now, Our Practice Problems: The complex number z = 1 – 2i is a zero of the given function. Find the remaining zeros of the function, and write it in its linear factorization. Now we can write the linear factorization…
Now, Our Practice Problems: The complex number z = 1 – 2i is a zero of the given function. Find the remaining zeros of the function, and write it in its linear factorization.
Download ppt "Linear Factorizations Sec. 2.6b. First, remind me of the definition of a linear factorization… f (x) = a(x – z )(x – z )…(x – z ) An equation in the following."
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# How do you find the quotient and divisor of a dividend?
Table of Contents
## How do you find the quotient and divisor of a dividend?
The formula to find the dividend in Maths is:
1. Dividend = Divisor x Quotient + Remainder. Usually, when we divide a number by another number, it results in an answer, such that;
2. Example 1: Find the dividend for the following x / 6 = 5 and also verify the answer.
3. Solution :
4. Dividend / Divisor = Quotient.
5. Verification:
## What is a divisor and dividend in division?
The number that is being divided (in this case, 15) is called the dividend, and the number that it is being divided by (in this case, 3) is called the divisor. The result of the division is the quotient. Notice how you can always switch the divisor and quotient and still have a true equation: 15 ÷ 3 = 5. 15 ÷ 5 = 3.
Is the dividend on top or bottom?
The first digit of the dividend (4) is divided by the divisor. The whole number result is placed at the top. Any remainders are ignored at this point. The answer from the first operation is multiplied by the divisor.
How do you create a quotient in Excel?
Tip: If you want to divide numeric values, you should use the “/” operator as there isn’t a DIVIDE function in Excel. For example, to divide 5 by 2, you would type =5/2 into a cell, which returns 2.5. The QUOTIENT function for these same numbers =QUOTIENT(5,2) returns 2, since QUOTIENT doesn’t return a remainder.
### What is divisor dividend quotient?
Divisor Meaning In division, we divide a number by any other number to get another number as a result. So, the number which is getting divided here is called the dividend. The number which divides a given number is the divisor. And the number which we get as a result is known as the quotient.
### What comes first divisor or dividend?
When we divide two numbers, the number that is being divided is the dividend, whereas the number by which we divide is the divisor. For example, 12 candies are to be divided among 3 children. So we have 12 ÷ 3. Here, 12 is the dividend, and 3 is the divisor.
Where is the dividend located?
Dividend – the number being divided or partitioned by the divisor. It is found to the left of the division symbol.
What is QUOTIENT function in Excel?
The Excel QUOTIENT function returns the integer portion of division without the remainder. Use QUOTIENT to discard the remainder after division. Returns the quotient without a remainder. Integer Number. =QUOTIENT (numerator, denominator)
#### What is the formula of quotient?
Dividend ÷ Divisor = Quotient and Remainder Thus we can also define divisor, dividend and remainder here. Remainder: The number which is left after the division method.
#### What is called dividend?
Definition: Dividend refers to a reward, cash or otherwise, that a company gives to its shareholders. Dividends can be issued in various forms, such as cash payment, stocks or any other form. A company’s dividend is decided by its board of directors and it requires the shareholders’ approval.
How do you write a formula to divide a cell in Excel?
To divide two numbers in Excel, you type the equals sign (=) in a cell, then type the number to be divided, followed by a forward slash, followed by the number to divide by, and press the Enter key to calculate the formula.
How do you divide an absolute cell reference in Excel?
Select another cell, and then press the F4 key to make that cell reference absolute. You can continue to press F4 to have Excel cycle through the different reference types. If necessary, continue entering the formula. Click the Enter button on the formula bar, or press Enter. |
PERIMETER AND AREA - Triangles - PLANE GEOMETRY - SAT SUBJECT TEST MATH LEVEL 1
## CHAPTER 9 Triangles
### PERIMETER AND AREA
The perimeter of a triangle is the sum of the lengths of the three sides.
EXAMPLE 6: To find the perimeter of an equilateral triangle whose height is 12, note that the height divides the triangle into two 30-60-90 right triangles. In the figure below, by KEY FACT H8,
So the perimeter is 3s =24 .
Key Fact H9
TRIANGLE INEQUALITY
• The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
• The difference of the lengths of any two sides of a triangle is less than the length of the third side.
EXAMPLE 7: A teacher asked her class to draw triangles in which the lengths of two of the sides were 6 inches and 7 inches and the length of the third side was also a whole number of inches. To determine how many different triangles the class could draw, note that if x represents the length of the third side, then by KEY FACT H9, 6 + 7 > x, so x < 13. Also, 7 – 6 < x, so x > 1. So x could have 11 different values: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. The perimeters of the triangles are the integers from 15 to 25. The following diagram illustrates four of the triangles the class could have drawn.
Frequently, questions on the Math 1 test require you to calculate the area of a triangle.
Key Fact H10
The area of a triangle is given by bh, where b and h are the lengths of the base and height, respectively.
(1) Any side of the triangle can be taken as the base.
(2) The height (which is also called an altitude) is a line segment drawn perpendicular to the base from the opposite vertex.
(3) In a right triangle, either leg can be the base and the other the height.
(4) If one endpoint of the base is the vertex of an obtuse angle, then the height will be outside the triangle.
• If is the base, is the height.
• If is the base, is the height.
• If is the base, is the height.
TIP
The height can be outside the triangle. See in the diagram shown below.
EXAMPLE 8: To find the area of equilateral triangle PQR, below, whose sides are 12, draw in altitude PQS is a 30-60-90 right triangle, and so by KEY FACT H8, QS = 6 and PS =6. Finally, the area of
Replacing 12 by s in Example 8 yields a useful formula.
Key Fact H11
if A represents the area of an equilateral triangle with side s, then .
Smart Strategy
Learning this formula for the area of an equilateral triangle can save you time.
|
# Probability/Combinatorics: # ways of picking 5 from 3 groups of 6
• Master1022
In summary, the conversation discusses a probability question involving selecting 5 marbles from a bag of 18 with 6 of each color (red, blue, yellow). Two methods of solving the problem are presented, one using a counting argument and the other using equations. Further explanation is given by another participant, who simplifies the equation to P = 3P(R+B) = 3(P(RB) - 2P(R)). However, there is confusion about the final term in the equation.
Master1022
Homework Statement
We have a bag of 18 marbles: 6 red, 6 blue, and 6 yellow. Now I randomly select 5 marbles from the bag without replacement. What is the probability that I have picked out EXACTLY 2 colors?
Relevant Equations
Combinatorics
Hi,
I was attempting the following question and was getting slightly stuck.
Question: We have a bag of 18 marbles: 6 red, 6 blue, and 6 yellow. Now I randomly select 5 marbles from the bag without replacement. What is the probability that I have picked out EXACTLY 2 colors?
1) Pure counting argument such that ##p = \frac{\text{Number of ways which we pick 2 colors}}{\text{Total number of ways of picking 5 from 18}} ##
So my logic was as follows:
- there are ##\begin{pmatrix} 3 \\ 2 \end{pmatrix}## ways of picking 2 out of the three colors
- Then for each of those pairs of colors (let us call them A and B), we can do: (1 from A, 4 from B), (2 from A, 3 from B), (3 from A, 2 from B), (4 from A, 1 from B). This can be written more formally as:
$$\begin{pmatrix} 6 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix} + \begin{pmatrix} 6 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 2 \end{pmatrix} + \begin{pmatrix} 6 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix}$$
and thus this becomes:
$$\begin{pmatrix} 3 \\ 2 \end{pmatrix} \left( \begin{pmatrix} 6 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix} + \begin{pmatrix} 6 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 2 \end{pmatrix} + \begin{pmatrix} 6 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix} \right)$$
I am slightly confused on how to get the total number of ways of picking 5 from 3 groups of 6. I mean I can see the obvious ## \begin{pmatrix} 18 \\ 5 \end{pmatrix} ##, but doesn't that double count some groupings?? This was asked as an interview question, so I don't think I would have time to write down very elaborate alternative methods.
$$p = \frac{\begin{pmatrix} 3 \\ 2 \end{pmatrix} \left( \begin{pmatrix} 6 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix} + \begin{pmatrix} 6 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 2 \end{pmatrix} + \begin{pmatrix} 6 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 4 \end{pmatrix} \right)}{\begin{pmatrix} 18 \\ 5 \end{pmatrix}} = \frac{65}{238}$$
2) The other 'method' was to frame the problem like: "how many solutions are there to the equation ## x_1 + x_2 = 5 ## where ## x_1 \geq 1 ## and ## x_2 \geq 1 ##.
- so there are still ##\begin{pmatrix} 3 \\ 2 \end{pmatrix}## ways of picking 2 out of the three colors
- then there would be ##\begin{pmatrix} (5 - 2) + (2 - 1) \\ (2 - 1) \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} ## ways choosing the groups...
- For the total number of ways to pick 5 from 18, we could use a similar framing of ## x_1 + x_2 + x_3 = 5 ##, but instead just have ## x_1 , x_2, x_3 \geq 0 ##. This leads to ## \begin{pmatrix} 5 + (3 - 1) \\ (3 - 1) \end{pmatrix} = \begin{pmatrix} 7 \\ 2 \end{pmatrix} ##
$$p = \frac{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \end{pmatrix}}{\begin{pmatrix} 7 \\ 2 \end{pmatrix}} = \frac{4}{7}$$
Can anyone help to reconcile which one of these methods is more appropriate for this problem?
Many thanks.
Delta2
I would simply do:$$P = 3P(R+B) = 3(P(RB) - 2P(R))$$I'll let you work out the notation!
PeroK said:
I would simply do:$$P = 3P(R+B) = 3(P(RB) - 2P(R))$$I'll let you work out the notation!
Thanks @PeroK ! Unfortunately, I have been looking at this for a while and am a bit confused. Here is what I have made of it. We are using the probability rule:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
You have P(R + B) as the ##P(R \cup B)##, but then I don't understand why the final term in the equation is ## P(R \cap B) - 2 \cdot P(R) ## instead of ## 2 P(R) - P(R \cap B) ##? (is that a correct understanding?
Master1022 said:
Thanks @PeroK ! Unfortunately, I have been looking at this for a while and am a bit confused. Here is what I have made of it. We are using the probability rule:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
You have P(R + B) as the ##P(R \cup B)##, but then I don't understand why the final term in the equation is ## P(R \cap B) - 2 \cdot P(R) ## instead of ## 2 P(R) - P(R \cap B) ##? (is that a correct understanding?
Well, ##R + B## must indicate one of the events you are looking for; namely, only Red and Blue balls and at least one of each. I'm not sure I understand the idea of translating to set notation? Why not leave things as they are?
Do you understand the factor of ##3##?
I didn't want to make it too easy, so I wrote ##-2P(R)## instead of the more suggestive ##-P(R) - P(B)##.
Master1022
PS I just checked the answer: ##65/238## is correct.
PeroK said:
Well, ##R + B## must indicate one of the events you are looking for; namely, only Red and Blue balls and at least one of each.
Oh I was thinking in terms of Venn diagrams, when R + B (with + being a union), but I guess that isn't what you meant? Then what does (RB) represent - I would have thought that meant both red and blue colored balls.
PeroK said:
I'm not sure I understand the idea of translating to set notation? Why not leave things as they are?
Hmm, that is what came to mind when I saw the form of that equation...
PeroK said:
Do you understand the factor of ##3##?
I think that is from the 3C2 = 3 (i.e. the number of ways to choose 2 colors from 3)
PeroK said:
I didn't want to make it too easy, so I wrote ##-2P(R)## instead of the more suggestive ##-P(R) - P(B)##.
I understood the symmetry part of it (##P(R) = P(B)##) so we can combine them. I think my question was why the equation was ##P(RB) - 2P(R)## instead of the other way around (i.e. ##2P(R) - P(RB)##)
Master1022 said:
I understood the symmetry part of it (##P(R) = P(B)##) so we can combine them. I think my question was why the equation was ##P(RB) - 2P(R)## instead of the other way around (i.e. ##2P(R) - P(RB)##)
I guess I should come clean:
##R + B## was only red and blue with at least one of each.
##RB## was only red and blue (i.e. no yellows) and includes the cases of all red and all blue. I guess I could have used ##no \ Y## instead.
##R## is all red.
Actually the 18C5 makes sense if I think about the Vandermonde Identity.
## 1. How many ways can 5 items be chosen from 3 groups of 6?
There are 216 possible ways to choose 5 items from 3 groups of 6. This can be calculated using the combination formula nCr = n!/(r!(n-r)!), where n is the total number of items (18 in this case) and r is the number of items being chosen (5 in this case).
## 2. Can the items be chosen from any group or are there restrictions?
The items can be chosen from any group without restrictions. As long as there are 5 items chosen in total, they can come from any of the 3 groups of 6.
## 3. What if there are more than 3 groups of 6 items?
If there are more than 3 groups of 6 items, the number of ways to choose 5 items will increase. For example, if there are 4 groups of 6 items, there will be 4C5 = 4 ways to choose 5 items.
## 4. Can the same item be chosen more than once?
No, in this scenario, each item can only be chosen once. If an item is chosen from one group, it cannot be chosen again from another group.
## 5. How does this relate to real-life situations?
This type of problem is commonly used in probability and combinatorics to calculate the number of possible outcomes in a given scenario. It can be applied to various real-life situations such as selecting a team from a pool of players or choosing a combination of items from a menu.
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# Diameter or Radius of a Circle Given Circumference
## C = πd; C = 2πr
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Circumference of Circles
Thomas is practicing throwing an old discus ring he found in the attic. After an exhausting afternoon, Thomas sits with the discus ring and starts to examine it. He uses a ruler to measure the radius, which is 11 cm. He also finds a way to measure the diameter of the circle he marked out to throw the discus into – it’s 2.5 m across. Now that he has this information about both circles, how can Thomas figure out the circumference of each?
In this concept, you will learn to find the diameter, radius and circumference of circles.
### Circumference
Circles are unique geometric figures. A circle is the set of points that are equidistant from a center point. The radius of a circle is the distance from the center to any point on the circle. The diameter is the distance across the circle through the center. The diameter is always twice as long as the radius.
You also use the special number pi\begin{align*}(\pi)\end{align*} when dealing with circle calculations. Pi is a decimal that is infinitely long \begin{align*}(3.14159265 \ldots)\end{align*}, but in your calculations you round it to 3.14. Pi is the ratio of the circumference, or distance around a circle, to the diameter.
Circles are special in geometry because this ratio of the circumference and the diameter always stays the same. In other words, these two measurements are related. If you change the diameter, the circumference changes proportionally. For example, if you double the length of the diameter, the circumference doubles also.
Look at the circles below. Find the ratio of the circumference to the diameter for both circles. What do you notice?
\begin{align*}\begin{array}{rcl} && \qquad \quad \quad \text{Large Circle} \\ \\ && \quad \ \frac{\text{circumference}}{\text{diameter}} = \frac{12.56}{4} \\ && \quad \ \frac{\text{circumference}}{\text{diameter}} = 3.14 \end{array}\end{align*}
\begin{align*}\begin{array}{rcl} && \qquad \quad \quad \text{Small Circle} \\ \\ && \quad \ \frac{\text{circumference}}{\text{diameter}} = \frac{6.28}{2} \\ && \quad \ \frac{\text{circumference}}{\text{diameter}} = 3.14 \end{array}\end{align*}
Notice, when you divide the circumference of a circle by its diameter, no matter how big or small the circle is, you will always get the same number. Whenever you divide the circumference by the diameter, you will always get 3.14, pi .
The circumference of a circle can be found using the formula: \begin{align*}C=\pi d\end{align*}. Since the diameter \begin{align*}(d)\end{align*} is twice the radius \begin{align*}(r)\end{align*}, you can also use the formula: \begin{align*}C=2 \pi r\end{align*}
Let’s look at an example.
What is the circumference of a circle that has a diameter of 3 inches?
First, substitute what you know into the formula for circumference.
\begin{align*}\begin{array}{rcl} C &=& \pi d \\ C &=& \pi \times 3 \end{array}\end{align*}
Next, solve for the circumference.
\begin{align*}\begin{array}{rcl} C &=& \pi \times 3 \\ C &=& 9.42 \end{array}\end{align*}
The circumference is 9.42 inches.
### Examples
#### Example 1
Earlier, you were given a problem about Thomas and his discus.
The discus ring has a radius of 11 cm. The throwing area has a diameter of 2.5 m.
First, substitute what you know into the formula for circumference to find the circumference of the discus.
\begin{align*}\begin{array}{rcl} C &=& 2 \pi r \\ C &=& 2 \pi \times 11 \end{array}\end{align*}
Next, solve for the circumference.
\begin{align*}\begin{array}{rcl} C &=& 2 \pi \times 11 \\ C &=& 69.12 \end{array} \end{align*}
The circumference of the discus is 69.1 cm.
Then, substitute what you know into the formula for circumference to find the circumference of the discus throwing area.
\begin{align*}\begin{array}{rcl} C &=& \pi d \\ C &=& \pi \times 2.5 \end{array}\end{align*}
Next, solve for the circumference.
\begin{align*}\begin{array}{rcl} C &=& \pi \times 2.5 \\ C &=& 7.85 \end{array}\end{align*}
The circumference of the throwing circle for the discus is 7.85 meters.
#### Example 2
What is the circumference of a circle if the radius is 2.5 feet?
First, substitute what you know into the formula for circumference.
\begin{align*}\begin{array}{rcl} C &=& 2 \pi r \\ C &=& 2 \pi \times 2.5 \end{array}\end{align*}
Next, solve for the circumference.
\begin{align*}\begin{array}{rcl} C &=& 2 \pi \times 2.5 \\ C &=& 15.71 \end{array}\end{align*}
The circumference is 15.7 feet.
#### Example 3
Find the circumference of each circle given the diameter is 6 inches.
First, substitute what you know into the formula for circumference.
\begin{align*}\begin{array}{rcl} C &=& \pi d \\ C &=& \pi \times 6 \end{array}\end{align*}
Next, solve for the circumference.
\begin{align*}\begin{array}{rcl} C &=& \pi \times 6 \\ C &=& 18.85 \end{array}\end{align*}
The circumference is 18.85 inches.
#### Example 4
Find the circumference of each circle given the radius is 4.5 feet.
First, substitute what you know into the formula for circumference.
\begin{align*}\begin{array}{rcl} C &=& 2 \pi r \\ C &=& 2 \pi \times 4.5 \end{array} \end{align*}
Next, solve for the circumference.
\begin{align*}\begin{array}{rcl} C &=& 2 \pi \times 4.5 \\ C &=& 28.27 \end{array}\end{align*}
The circumference is 28.27 feet.
#### Example 5
Find the circumference of each circle given the diameter is 3.5 meters.
First, substitute what you know into the formula for circumference.
\begin{align*}\begin{array}{rcl} C &=& \pi d \\ C &=& \pi \times 3.5 \end{array}\end{align*}
Next, solve for the circumference.
\begin{align*}\begin{array}{rcl} C &=& \pi \times 3.5 \\ C &=& 11.0 \end{array} \end{align*}
The circumference is 11.0 meters.
### Review
Find the circumference of each circle given the radius or diameter.
1. \begin{align*}d = 10 \ in\end{align*}
2. \begin{align*}d = 5 \ in\end{align*}
3. \begin{align*}d = 7 \ ft\end{align*}
4. \begin{align*}d = 12 \ mm\end{align*}
5. \begin{align*}d = 14 \ cm\end{align*}
6. \begin{align*}r = 4 \ in\end{align*}
7. \begin{align*}r = 6 \ \text{meters}\end{align*}
8. \begin{align*}r = 8 \ ft.\end{align*}
9. \begin{align*}r = 11 \ in\end{align*}
10. \begin{align*}r = 15 \ cm\end{align*}
Find the diameter given each circumference.
11. \begin{align*}53.38 \ \text{inches}\end{align*}
12. \begin{align*}43.96 \ \text{feet}\end{align*}
13. \begin{align*}56.52 \ \text{inches}\end{align*}
14. \begin{align*}65.94 \ \text{meters}\end{align*}
15. \begin{align*}48.67 \ \text{meters}\end{align*}
16. \begin{align*}37.68 \ \text{feet}\end{align*}
17. \begin{align*}78.5 \ \text{meters}\end{align*}
18. \begin{align*}100.48 \ cm\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Circle A circle is the set of all points at a specific distance from a given point in two dimensions.
Circumference The circumference of a circle is the measure of the distance around the outside edge of a circle.
Radius The radius of a circle is the distance from the center of the circle to the edge of the circle. |
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# Horizontal and Vertical Line Graphs
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Practice Horizontal and Vertical Line Graphs
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Horizontal and Vertical Line Graphs
What if you were given the graph of a vertical or horizontal line? How could you write the equation of this line? After completing this Concept, you'll be able to write horizontal and vertical linear equations and graph them in the coordinate plane.
### Guidance
How do you graph equations of horizontal and vertical lines? See how in the example below.
#### Example A
“Mad-cabs” have an unusual offer going on. They are charging \$7.50 for a taxi ride of any length within the city limits. Graph the function that relates the cost of hiring the taxi $(y)$ to the length of the journey in miles $(x)$ .
To proceed, the first thing we need is an equation . You can see from the problem that the cost of a journey doesn’t depend on the length of the journey. It should come as no surprise that the equation then, does not have $x$ in it. Since any value of $x$ results in the same value of $y (7.5)$ , the value you choose for $x$ doesn’t matter, so it isn’t included in the equation. Here is the equation:
$y = 7.5$
The graph of this function is shown below. You can see that it’s simply a horizontal line.
Any time you see an equation of the form “ $y =$ constant,” the graph is a horizontal line that intercepts the $y-$ axis at the value of the constant.
Similarly, when you see an equation of the form $x =$ constant, then the graph is a vertical line that intercepts the $x-$ axis at the value of the constant. (Notice that that kind of equation is a relation, and not a function, because each $x-$ value (there’s only one in this case) corresponds to many (actually an infinite number) $y-$ values.)
#### Example B
Plot the following graphs.
(a) $y = 4$
(b) $y = -4$
(c) $x = 4$
(d) $x = -4$
(a) $y = 4$ is a horizontal line that crosses the $y-$ axis at 4.
(b) $y = -4$ is a horizontal line that crosses the $y-$ axis at −4.
(c) $x = 4$ is a vertical line that crosses the $x-$ axis at 4.
(d) $x = -4$ is a vertical line that crosses the $x-$ axis at −4.
#### Example C
Find an equation for the $x-$ axis and the $y-$ axis.
Look at the axes on any of the graphs from previous examples. We have already said that they intersect at the origin (the point where $x = 0$ and $y = 0$ ). The following definition could easily work for each axis.
$x-$ axis: A horizontal line crossing the $y-$ axis at zero.
$y-$ axis: A vertical line crossing the $x-$ axis at zero.
So using example 3 as our guide, we could define the $x-$ axis as the line $y = 0$ and the $y-$ axis as the line $x = 0$ .
Watch this video for help with the Examples above.
### Vocabulary
• Horizontal lines are defined by the equation $y=$ constant and vertical lines are defined by the equation $x=$ constant.
• Be aware that although we graph the function as a line to make it easier to interpret, the function may actually be discrete .
### Guided Practice
Write the equation of the horizontal line that is 3 units below the x-axis.
Solution:
The horizontal line that is 3 units below the x-axis will intercept the y-axis at $y=-3$ . No matter what the value of x, the y value of the line will always be -3. This means that the equations for the line is $y=-3$ .
### Explore More
1. Write the equations for the five lines ( $A$ through $E$ ) plotted in the graph below.
For 2-10, use the graph above to determine at what points the following lines intersect.
1. $A$ and $E$
2. $A$ and $D$
3. $C$ and $D$
4. $B$ and the $y-$ axis
5. $E$ and the $x-$ axis
6. $C$ and the line $y = x$
7. $E$ and the line $y = \frac {1} {2} x$
8. $A$ and the line $y = x + 3$
9. $B$ and the line $y=-2x$
### Vocabulary Language: English
Horizontally
Horizontally
Horizontally means written across in rows.
Vertically
Vertically
Vertically means written up and down in columns. |
# AP 8th Class Maths 1st Chapter Rational Numbers Exercise 1.2 Solutions
Well-designed AP Board Solutions Class 8 Maths Chapter 1 Rational Numbers Exercise 1.2 offers step-by-step explanations to help students understand problem-solving strategies.
## Rational Numbers Class 8 Exercise 1.2 Solutions – 8th Class Maths 1.2 Exercise Solutions
Question 1.
Represent these numbers on the number line.
(i) $$\frac{7}{4}$$
First convert the given into mixed fraction. (If Possible)
$$\frac{7}{4}$$ = 1$$\frac{3}{4}$$. So this $$\frac{7}{4}$$ lies in between 1 and 2 so you have to divide the place into 4 parts from 1 and 2, then third point would be the required.
So $$\frac{7}{4}$$ = 1$$\frac{3}{4}$$ represent the point (C).
(ii) $$\frac{-5}{6}$$
Here (-) symbol shows the points left to ‘zero’.
$$\frac{-5}{6}$$ can be written as 0 + $$\frac{-5}{6}$$.
So this lies in between 0 and -1; we are divide into 6 equal parts.
Then the fifth point shows $$\frac{-5}{6}$$.
The point ‘P’ shows $$\left(\frac{-5}{6}\right)$$
Question 2.
Represent $$\frac{-2}{11}, \frac{-5}{11}, \frac{-9}{11}$$ on the number line.
Given are left to zero on number line (-Symbol) they lie in between ‘0’ and 1.
Because numerator < denominator.
Now divide into 11 parts. (∵ denominator = 11)
Then 2nd, 5th, 9th points would be the answers.
Question 3.
Write five rational numbers which are smaller than 2.
(i) Given = 2
It can be written as $$\frac{20}{10}$$, then $$\frac{19}{10}, \frac{18}{10}, \frac{11}{10}, \frac{7}{10}, \frac{3}{10}$$ are rationals which are smaller than 2.
(ii) Given = 2
It can be written as $$\frac{16}{8}$$ then $$\frac{15}{8}, \frac{14}{8}, \frac{13}{8}, \frac{12}{8}, \frac{11}{8}$$, ………….. thus we can write so many.
Question 4.
Find ten rational numbers between $$\frac{-2}{5}$$ and $$\frac{1}{2}$$.
First write the denominators equal in given fractions using L.C.M.
Here $$\frac{-2}{5}, \frac{1}{2}$$ ⇒ LCM of 5 and 2 = 10
⇒ $$\frac{(-2) 2,5(1)}{10}=\frac{-4,5}{10}$$
So given fractions $$\frac{-2}{5}$$ and $$\frac{1}{2}$$ can be written as $$\frac{-4}{10}$$ and $$\frac{5}{10}$$.
Now, write them as $$\frac{-40}{100}$$ and $$\frac{50}{100}$$.
So that we can write more rationals between $$\frac{-40}{100}$$ and $$\frac{50}{100}$$ are $$\frac{-39}{100}, \frac{-38}{100}, \frac{-37}{100}$$ …………$$\frac{41}{100}, \frac{42}{100}, \frac{43}{100}$$ ……. $$\frac{49}{100}$$
Question 5.
Find five rational numbers between
(i) $$\frac{2}{3}$$ and $$\frac{4}{5}$$
L.C.M of 3 and 5 = 15
So $$\frac{2}{3}, \frac{4}{5}$$ can be written as $$\frac{5(2), 3(4)}{15}=\frac{10,12}{15}$$
So, $$\frac{2}{3}$$ and $$\frac{4}{5}$$ = $$\frac{10}{15}$$ and $$\frac{12}{15}$$
To get more and more we can write
$$\frac{10}{15}$$ as $$\frac{100}{150}$$ and $$\frac{12}{15}$$ as $$\frac{120}{150}$$
So given $$\frac{2}{3}$$ and $$\frac{4}{5}$$ can be written as
$$\frac{100}{150}, \frac{120}{150}$$
So rational numbers between them are $$\frac{101}{150}, \frac{102}{150}, \frac{105}{150}, \frac{110}{150}$$ ………… $$\frac{119}{150}$$
(ii) $$\frac{-3}{2}$$ and $$\frac{5}{3}$$
L.C.M. of 2 and 3 is 6.
Then $$\frac{-3}{2}, \frac{5}{3}$$can be written as
$$\frac{3(-3), 2(5)}{6}=\frac{-9,10}{6}$$
So $$\frac{-9}{6}, \frac{10}{6}$$
Now, rational numbers in between $$\frac{-9}{6}$$ and $$\frac{10}{6}$$ are
$$\frac{-8}{6}, \frac{-7}{6} \ldots \ldots . \frac{-1}{6}, \frac{0}{6}, \frac{1}{6} \ldots \ldots \ldots \frac{7}{6}, \frac{8}{6}, \frac{9}{6}$$ etc.
(iii) $$\frac{1}{4}$$ and $$\frac{1}{2}$$
L.C.M. of 4 and 2 = 4; $$\frac{1}{4}$$ and $$\frac{1}{2}=\frac{1(1), 2(1)}{4}=\frac{1}{4}, \frac{2}{4}$$ now write them with big denominator
$$\frac{1}{4}, \frac{2}{4}$$ can be written as
$$\frac{6(1)}{6(4)}, \frac{6(2)}{6(4)}=\frac{6}{24}, \frac{12}{24}$$
Now, rational numbers between
$$\frac{6}{24}, \frac{12}{24}$$ are $$\frac{7}{24}, \frac{8}{24}, \frac{9}{24}, \frac{10}{24}, \frac{11}{24}$$
(You can multiply with a number greater than six also)
Question 6.
Write five rational numbers greater than -2.
We can write -2 are $$\left(\frac{-20}{10}\right)$$
Then $$\frac{-19}{10}, \frac{-18}{10}, \frac{-17}{10}$$, ……….. $$\frac{1}{10}$$ are rational numbers greater then -2.
Find ten rational numbers between $$\frac{3}{5}$$ and $$\frac{3}{4}$$.
We can write $$\frac{3}{5}, \frac{3}{4}$$ as $$\frac{16(3)}{16(5)}, \frac{20(3)}{20(4)}$$ = $$\frac{48}{80}, \frac{60}{80}$$
So, $$\frac{49}{80}, \frac{50}{80}, \frac{51}{80}$$ …….. $$\frac{57}{80}, \frac{58}{80}, \frac{59}{80}$$ are rationed numbers in between them. |
# 2021 AMC 10B Problems/Problem 3
## Problem
In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$
## Solution 1
Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\frac{j}{4}$ and $\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\frac{j}{4} = \frac{s}{10}.$
Cross-multiplying and simplifying we get $5j=2s.$ Additionally, since there are $28$ students in the program, $j+s = 28.$ It is now a matter of solving the system of equations $$5j=2s$$$$j+s=28,$$ and the solution is $j = 8, s = 20.$ Since we want the number of juniors, the answer is $$\boxed{(C) \text{ } 8}.$$
-PureSwag
## Solution 2 (Fast and not rigorous)
We immediately see that $E$ is the only possible amount of seniors, as $10\%$ can only correspond with an answer choice ending with $0$. Thus the number of seniors is $20$ and the number of juniors is $28-20=8\rightarrow \boxed{C}$. ~samrocksnature
## Video Solution by OmegaLearn (System of Equations)
2021 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
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## Introduction to Polynomials
Just in case you’re new to the site, here’s how the units are laid out. Each section gets its own page, and each page has a series of tabs (seen below). The first tab is notes, a description of the information you need. This includes definitions, examples, and of course explanations.
Read through the first tab, take notes, try the examples on your own to make sure you understand. Then, watch the video in the Video tab, and then try the practice problems in the last tab. If at any time you have a question or comment, please feel free to reach out via email.
You have undoubtedly dealt with polynomials but just did not know that they were polynomials. If you’ve ever had to distribute and combine like terms, that was most likely done with polynomials.
In short, a polynomial is an Algebraic expression with Whole Number exponents. That means that there might be fractions, but the denominator will never have a variable. Polynomial itself is a generic term, translating roughly to many numbers. We will distinguish between three types of polynomial, and everything else we’ll just call a polynomial.
Let’s talk about some common language. This way you won’t be lost when reading directions!
Terms are parts of a polynomial separated by addition, subtraction, or an equal sign (for equations).
Coefficients are the numbers multiplying with a variable. For example, in the monomial example above, the coefficient is 3.
Leading Coefficient is the coefficient for the term with the highest exponent.
Degree is the value of the polynomial’s largest exponent (for one variable polynomials).
Descending Order is a polynomial organized by alphabetical order, then exponent size.
Like Terms are terms that have the exact same combination of variables and exponents.
Constant Term is the term without a variable.
Let’s see an example of a polynomial that is NOT in descending order, and look at its components. There will not be any like terms in this example.
This function, written in descending order would be.
-2x4 + 5x3 + 3x – 11
The most common mistake made when rewriting a polynomial in descending order is that people misplace the signs of the terms when rearranging. For example, they’d write 2x4 instead of -2x4. The leading coefficient is negative two, not two. So, be cautious here.
The reason descending order is important is that it is organized. As you move through polynomials into quadratics, cubics, and rational functions, you’ll be looking for key features, clues about the equations. These are most easily discovered when the expressions are written in descending order. So, it is best to get in the habit of writing in descending order from the beginning, when things are simpler.
Like Terms must have the same combination of variables and exponents. Let’s see an example and a non-example.
The non-example has two expressions that are not like terms, but they’re close. The combination of variables and exponents is not identical. There is an a to the 3rd power in one, and an a to the 5th power in the other. The exponents for the variable x is different, too.
## Combining Like Terms
If you have like terms in a polynomial, you can combine them with addition and subtraction. You can multiply or divide terms that are unlike, but they must be like terms to add or subtract. Terms must be like to add or subtract for the same reason you can’t add apples and oranges. They’re not the same thing. Ultimately, to add or subtract unlike terms is a violation of the order of operations.
Let’s use the expression above to examine how the order of operations dictates that we cannot add unlike terms. First, notice we have two terms, and the second is being subtracted from the first. We do not know the value of either term. In order to take the difference, we must know the values. To find the value we have to square a, then multiply that by 4b. Also, 3 must be multiplied by c’s value, before the value of each term can be subtracted. Since we don’t know the values of a, b, or c, we cannot simplify this expression further.
## Addition and Subtraction
Addition is just combining like terms. Let’s see an example of adding two binomials.
Here we are just adding the like terms. There are four total terms, each binomial having two terms. There are only one pair of like terms, however. The -8c and the 5c, are like terms. They combine to make – 3c. The other terms do not share the same exact combination of variables and exponents, so they cannot be combined with addition or subtraction.
Subtraction is a little trickier. Let’s see an example.
Here we are subtracting a positive 3a2b AND negative 8c. Subtracting a negative is the same as addition. Rewriting this expression, we will get the following.
Both terms need to be multiplied by negative one, and negative 8c times negative one is positive 8c
Once again, we have a pair of like terms, and two terms unlike anything else.
Writing our answer in descending order we get the following. |
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