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# 6.1 Graphs of the sine and cosine functions (Page 7/13)
Page 7 / 13
## Determining a rider’s height on a ferris wheel
The London Eye is a huge Ferris wheel with a diameter of 135 meters (443 feet). It completes one rotation every 30 minutes. Riders board from a platform 2 meters above the ground. Express a rider’s height above ground as a function of time in minutes.
With a diameter of 135 m, the wheel has a radius of 67.5 m. The height will oscillate with amplitude 67.5 m above and below the center.
Passengers board 2 m above ground level, so the center of the wheel must be located $\text{\hspace{0.17em}}67.5+2=69.5\text{\hspace{0.17em}}$ m above ground level. The midline of the oscillation will be at 69.5 m.
The wheel takes 30 minutes to complete 1 revolution, so the height will oscillate with a period of 30 minutes.
Lastly, because the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a vertically reflected cosine curve.
• Amplitude: $\text{\hspace{0.17em}}\text{67}\text{.5,}\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}A=67.5$
• Midline: $\text{\hspace{0.17em}}\text{69}\text{.5,}\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}D=69.5$
• Period: $\text{\hspace{0.17em}}\text{30,}\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}B=\frac{2\pi }{30}=\frac{\pi }{15}$
• Shape: $\text{\hspace{0.17em}}\mathrm{-cos}\left(t\right)$
An equation for the rider’s height would be
$y=-67.5\mathrm{cos}\left(\frac{\pi }{15}t\right)+69.5$
where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is in minutes and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is measured in meters.
Access these online resources for additional instruction and practice with graphs of sine and cosine functions.
## Key equations
Sinusoidal functions $\begin{array}{l}f\left(x\right)=A\mathrm{sin}\left(Bx-C\right)+D\\ f\left(x\right)=A\mathrm{cos}\left(Bx-C\right)+D\end{array}$
## Key concepts
• Periodic functions repeat after a given value. The smallest such value is the period. The basic sine and cosine functions have a period of $\text{\hspace{0.17em}}2\pi .$
• The function $\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is odd, so its graph is symmetric about the origin. The function $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is even, so its graph is symmetric about the y -axis.
• The graph of a sinusoidal function has the same general shape as a sine or cosine function.
• In the general formula for a sinusoidal function, the period is $\text{\hspace{0.17em}}P=\frac{2\pi }{|B|}.\text{\hspace{0.17em}}$ See [link] .
• In the general formula for a sinusoidal function, $\text{\hspace{0.17em}}|A|\text{\hspace{0.17em}}$ represents amplitude. If $\text{\hspace{0.17em}}|A|>1,\text{\hspace{0.17em}}$ the function is stretched, whereas if $\text{\hspace{0.17em}}|A|<1,\text{\hspace{0.17em}}$ the function is compressed. See [link] .
• The value $\text{\hspace{0.17em}}\frac{C}{B}\text{\hspace{0.17em}}$ in the general formula for a sinusoidal function indicates the phase shift. See [link] .
• The value $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ in the general formula for a sinusoidal function indicates the vertical shift from the midline. See [link] .
• Combinations of variations of sinusoidal functions can be detected from an equation. See [link] .
• The equation for a sinusoidal function can be determined from a graph. See [link] and [link] .
• A function can be graphed by identifying its amplitude and period. See [link] and [link] .
• A function can also be graphed by identifying its amplitude, period, phase shift, and horizontal shift. See [link] .
• Sinusoidal functions can be used to solve real-world problems. See [link] , [link] , and [link] .
## Verbal
Why are the sine and cosine functions called periodic functions?
The sine and cosine functions have the property that $\text{\hspace{0.17em}}f\left(x+P\right)=f\left(x\right)\text{\hspace{0.17em}}$ for a certain $\text{\hspace{0.17em}}P.\text{\hspace{0.17em}}$ This means that the function values repeat for every $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ units on the x -axis.
How does the graph of $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ compare with the graph of $\text{\hspace{0.17em}}y=\mathrm{cos}\text{\hspace{0.17em}}x?\text{\hspace{0.17em}}$ Explain how you could horizontally translate the graph of $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to obtain $\text{\hspace{0.17em}}y=\mathrm{cos}\text{\hspace{0.17em}}x.$
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
what is a complex number used for?
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
Is there any rule we can use to get the nth term ?
how do you get the (1.4427)^t in the carp problem?
A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola
A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug?
Find the domain of the function in interval or inequality notation f(x)=4-9x+3x^2
hello
Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of ?105°F??105°F? occurs at 5PM and the average temperature for the day is ?85°F.??85°F.? Find the temperature, to the nearest degree, at 9AM.
if you have the amplitude and the period and the phase shift ho would you know where to start and where to end?
rotation by 80 of (x^2/9)-(y^2/16)=1
thanks the domain is good but a i would like to get some other examples of how to find the range of a function
what is the standard form if the focus is at (0,2) ?
a²=4 |
# 2010 AMC 12B Problems/Problem 11
The following problem is from both the 2010 AMC 12B #11 and 2010 AMC 10B #21, so both problems redirect to this page.
## Problem
A palindrome between $1000$ and $10,000$ is chosen at random. What is the probability that it is divisible by $7$?
$\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}$
## Solution
View the palindrome as some number with form (decimal representation): $a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$. But because the number is a palindrome, $a_3 = a_0, a_2 = a_1$. Recombining this yields $1001a_3 + 110a_2$. 1001 is divisible by 7, which means that as long as $a_2 = 0$, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 ($9 \cdot 10$) possibilities for palindromes. However, if $a_2 = 7$, then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to $18/90 = \boxed {\frac{1}{5} } = \boxed {E}$
## Addendum (Alternate)
$7\mid 1001a^3+110b^2$ and $1001 \equiv 0 \pmod 7$. Knowing that $a$ does not factor (pun intended) into the problem, note 110's prime factorization and $7\mid b$. There are only 10 possible digits for $b$, 0 through 9, but $7\mid b$ only holds if $b=0, 7$. This is 2 of the 10 digits, so $\frac{2}{10}=\boxed{\textbf{E)}\frac{1}{5}}$
~BJHHar
~IceMatrix
## See also
2010 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 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 12 Problems and Solutions
2010 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. |
# C3 Chapter 1 Algebraic Fractions Dr J Frost Last modified: 13 th May 2014.
## Presentation on theme: "C3 Chapter 1 Algebraic Fractions Dr J Frost Last modified: 13 th May 2014."— Presentation transcript:
C3 Chapter 1 Algebraic Fractions Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 13 th May 2014
7 3 =2 + 1 3 dividend divisor quotient remainder RECAP: Terminology dividend divisor quotientremainder ? ? ??
? ? ? ? ? 1 2 3 4 5 6 7 Simplify the following fractions. 8 Bro Tip: When you have a fraction within a fraction, multiply the top and bottom of the outer fraction by the denominator of the inner one. ? ? ?
Test Your Understanding ? ? ? ?
Exercises 1g 1i 1j 1l 1n Exercise 1B Exercise 1A 1a 1g 1i 1j 1k 1o Exercise 1C 1b 1c 1g 1k 1l 1n 1o 1q ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
x + 56x 3 + 28x 2 – 7x + 15 6x 2 6x 3 + 30x 2 – 2x 2 – 7x - 2x – 2x 2 – 10x 3x + 15 + 3 3x + 15 0 The Anti-Idiot Test: You can check your solution by expanding (x+5)(6x 2 – 2x + 3) The Anti-Idiot Test: You can check your solution by expanding (x+5)(6x 2 – 2x + 3) RECAP: Algebraic Division
x - 42x 3 – 5x 2 – 16x + 10 2x 2 2x 3 – 8x 2 3x 2 – 16x + 3x 3x 2 – 12x -4x + 10 – 4 -4x + 16 -6 Find the remainder. Q: Is (x-4) a factor of 2x 3 – 5x 2 – 16x + 10? Test Your Understanding
Alternative Method: Remainder Theorem RECAP ? ? ? ? ?
Alternative Method: Remainder Theorem ? Q |
# Algebra Multiplication Rules
0
381
There are some algebra multiplication rules of algebraic expressions that help in simplifying the equations clearly without any errors or wrong results. Following, implementing and practicing them on different types of questions helps in achieving mastery over the topic. We shall learn the most important rules that are must be known for solving problems.
## Basic Algebra Multiplication Rules Involved In Every Algebraic Equations
Here are the basic rules of how the signs are considered in multiplying algebraic expressions. These signs concept comes in every simplification. Remembering these 4 sign rules of multiplication helps in solving every algebraic expression.
### Sign Rule Multiplication In Algebra
– × – = + + × + = + – × + = – + × – = –
### Examples for Multiplication Rules Of Algebraic Equations
Multiply the algebraic expressions -ax and -by
Sol:
(-ax) × (-by)
first, multiply the signs – × – = + and then coefficients and variables
+(abxy)
Multiply the algebraic expressions -ax and +by
Sol:
(-ax) × (+by)
first, multiply the signs – × + = – and then coefficients and variables
-(abxy)
### Rules Of Exponents Or Powers
Exponent or Power is defined as the number of times a number is multiplied by itself.
1. When bases are same in multiplication we add the powers
\begin{align*} a^{m}\times a^{n}=a^{m+n} \end{align*}
2. When bases are same in division we subtract the numerator power with denominator.
\begin{align*} \begin{aligned}\dfrac {a^{m}}{a^{n}}=a^{m-n}\\ where \ a\neq 0,m >n\end{aligned}
\end{align*}
3. For same base if there are one or more exponents then, multiply all the exponents.
\begin{align*} \left( a^{m}\right) ^{n}=a^{mn} \end{align*}
4. If exponents are same and bases are different then multiply the bases for same exponents.
\begin{align*} a^{m}\times b^{m}=\left( ab\right) ^{m} \end{align*}
5. When exponents are same in division with different bases then divide the bases with same exponent.
\begin{align*} \dfrac {a^{m}}{b^{m}}=\left( \dfrac {a}{b}\right) ^{m} \end{align*}
6. For any value of base if power is zero then the total value is always ‘1’.
\begin{align*} a^{0}=1 \end{align*}
7. Negative power can also be represented in denominator with positive power in fraction form.
\begin{align*} \begin{aligned}a^{-m}=\dfrac {1}{a^{m}}\\ a\neq 0\end{aligned} \end{align*} |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Interpretation of Circle Graphs
## Connect circle graph percents to degrees of a circle.
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Interpretation of Circle Graphs
Credit: CK-12 Foundation
Monica is opening a restaurant. She wants to offer at least two kinds of ice cream flavors for dessert. She uses a data collection company to find out what flavors people prefer. The company uses this circle graph to show the results of the survey. Of the four flavors, which two flavors would most people prefer?
In this concept, you will learn how to interpret circle graphs.
### Interpreting Circle Graphs
A circle graph is a visual way of displaying data that is written in percents. The circle represents 100%. The circle is divided into sections. Each section shows what part of 100 that item represents. The circle graph uses different colors for each category being described. The legend on the right shows what each color corresponds to.
This circle graph describes a student’s spending habits.Each color represents what his money is spent on. Blue is savings, red is baseball cards, and green is food. By looking at this graph, you may determine that this student saves half (or 50%) of all his money. He spends a smaller fraction on baseball cards and the rest on food.
Notice that all three percents add up to 100%. A circle graph shows information out of 100.
### Examples
#### Example 1
Earlier, you were given a problem about Monica and her ice cream flavors.
Monica needs to pick two flavors that would satisfy the most people. Monica needs to use the circle graph to make her decision.
First, Monica needs to find the two flavors with biggest section or the largest percents.
She sees that the red and blue sections make up most of the circle graph.
Then, she looks at the legend to see what the colors represent.
Red represents chocolate and blue represents vanilla.
Monica should pick chocolate and vanilla to satisfy the most people.
#### Example 2
Use the circle graph to answer the questions.
Based on this graph, which foreign language is the most popular?
First, look at the circle graph. Find the largest wedge. The percent is also written each wedge.
The pink wedge with 55% is the largest wedge.
Then, look at the legend to find what language pink represents.
Pink represents Spanish.
Spanish has the largest percent, 55%. This means 55% of all the students have studied Spanish.
Spanish is the most popular foreign language studied.
#### Example 3
Looking at the graph above, about what percent is not spent on saving?
First, look at the wedge that represents savings.
The blue wedge represents the amount spent on savings.
Then, determine about how much of the total is spent on savings.
The blue wedge takes up about half or 50% of the circle.
#### Example 4
True or false: the following percents could be part of one circle graph: 25%, 10%, and 65%.
25%+10%+65%=100%\begin{align*}25\% + 10\% + 65\% = 100\%\end{align*}
A circle graph represents 100%.
True, 25%, 10% and 65% can be part of one circle graph.
#### Example 5
True or false: the following percents could be part of the same circle graph: 35%, 70%, and 5%.
35% + 70% + 5% = 110%
A circle graph cannot be more than 100%.
False, 35%, 70% and 5% cannot be part of the same circle graph.
### Review
Use the circle graph to answer the following questions.
1. Which foreign language is the least popular?
2. If 30% of the students chose French, what percent did not choose French?
3. What percent of the students chose Italian?
4. What percent of the students did not chose Italian?
5. What percent of the students chose Latin?
6. What percent of the students did not choose Latin or French?
7. What percent of the students did not choose French or Spanish?
8. What percent of the students did not choose German?
1. What percent of Patrick’s budget is spent of transportation?
2. What percent of the budget is spent on food?
3. What percent of the budget is not spent on food or savings?
4. If 35% of the budget is spent on rent and utilities, what percent is not spent on this?
5. True or false. Patrick does not spend any money on clothing.
6. What percent is spent on savings and food together?
7. If Patrick was going to increase his savings by 15% what would the new percent of savings be?
To see the Review answers, open this PDF file and look for section 8.13.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Decimal
In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
fraction
A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.
Percent
Percent means out of 100. It is a quantity written with a % sign.
Sector
A sector of a circle is a portion of a circle contained between two radii of the circle. Sectors can be measured in degrees.
sector angle formula
The sector angle formula is used to calculate how many degrees of the circle should be allocated to a given value and is calculated by dividing the frequency of the data in the sector by the total frequency of the data all multiplied by 360. |
4 Ijesha Close, Ilupeju, Lagos
+2347 086 296 002
#### Pythagoras and Trigonometry
##### PYTHAGORAS THEOREM
• A right-angle triangle is a triangle with one of its angles as 90°.
• The Pythagoras theorem states the square of the hypotenuse of a right-angle triangle is equal to the sum of the square of its sides.
• The side of the triangle that is opposite the right angle is the hypotenuse.
• The side of the triangle that is opposite the arc is the opposite.
• The last side is called the adjacent.
• The hypotenuse is the longest side of any right angle triangle.
• Mathematically: hypotenuse² = opposite² + adjacent²; AB² = BC² + AC²
##### PYTHAGOREAN TRIPLES
• Pythagorean triples are a set of three integers, a, b, c, that describe the sides of a right angle triangle.
• The integers satisfy the Pythagoras theorem; a² + b² = c²; where c is the hypotenuse.
• Examples of Pythagorean triples include:
• 3, 4, 5
• 15, 8, 17
• If a set of Pythagorean triples is multiplied by a factor, the result is also a set of Pythagorean triples. For example:
• (3, 4, 5) × 2 = (6, 8, 10) ⇒ is a Pythagorean triple
• (5, 12, 13) × 3 = (15, 36, 39) ⇒ is a Pythagorean triple
##### TRIGONOMETRY
• Trigonometry basically studies the relationship between side lengths and angles of triangles.
• Sine (sin), cosine (cos), tangent (tan) are functions used in trigonometry.
• The ratios of a right angle triangle are:
• SOH: sin θ = opposite/hypotenuse
• CAH: cos θ = adjacent/hypotenuse
• TOA: tan θ = opposite/adjacent
• If you cannot use the Pythagoras theorem to solve for the unknown side, as long as you are given an angle, you can use any of the ratios above to solve the problem. |
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# 7.9: Sums of Finite Geometric Series
Difficulty Level: At Grade Created by: CK-12
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Anna is on a progressive workout plan, every day she adds 5% to her exercise time. If she starts by exercising 15mins on the first day, how many minutes will she have exercised all together on day 45?
This is a geometric series, since the difference between the exercise time on any two days is greater than the difference between any prior two days. You could just add: 15+(151.05)+[(151.05)1.05]...\begin{align*}15 + (15 \cdot 1.05) + [(15 \cdot 1.05) \cdot 1.05]...\end{align*} and so on up to 45, but that would be horribly tedious. In this lesson, you will learn how to answer a question like this will little effort.
Embedded Video:
### Guidance
A finite geometric series is simply a geometric series with a specific number of terms. For example, consider the series: 50 + 25 + 12.5 + ....The series is geometric: the first term is 50, and the common ration is (1/2).
The sum of the first two terms is 50 + 25 = 75. We can write this as S2 = 75
The sum of the first three is 50 + 25 + 12.5 = 87.5. We can write this as S3 = 87.5
To find the value of Sn in general, we could simply add together the first n terms in a series. However, this would obviously be tedious for a large value of n. Given the regular pattern in a geometric series - every term is (1/r) of the previous term, and the nth term is an = a1rn - 1 , we can use induction to prove a formula for Sn .
The sum of the first n terms in a geometric series is Sn=a1(1rn)1r\begin{align*}S_n = \frac{a_1(1 - r^n)} {1 - r}\end{align*}
For example, for the series 50 + 25 + 12.5 + ... , the sum of the first 6 terms is:
Sn=a1(1rn)1r=50(1(12)6)112=50(1164)12=50(6364)12=50(6364)(21)=98716\begin{align*}S_n = \frac{a_1(1 - r^n)} {1 - r} = \frac{50 \left(1 - (\frac{1} {2})^6 \right)} {1 - \frac{1} {2}} = \frac{50 \left(1 - \frac{1} {64} \right)} {\frac{1} {2}} = \frac{50 \left(\frac{63} {64} \right)} {\frac{1} {2}} = 50 \left(\frac{63} {64} \right) \left(\frac{2} {1} \right) = 98 \frac{7} {16}\end{align*}
The figure below shows the same calculation on a TI-83/4 calculator:
We can use this formula as long as the series in question is geometric.
#### Example A
Find the sum of the first 10 terms of a geometric series with a1 = 3 and r = 5.
Solution
The sum is 58,593.
Sn=a1(1rn)1r=3(157)15=3(178,125)4=3(78,124)4=58,593\begin{align*}S_n = \frac{a_1(1 - r^n)} {1 - r} = \frac{3(1 - 5^7)} {1 - 5} = \frac{3(1 - 78,125)} {-4} = \frac{3(-78,124)} {-4} = 58,593\end{align*}
Notice that because the common ratio in this series is 5, the terms get larger and larger. This means that for increasing values of n the sums will also get larger and larger. In contrast, in the series with common ratio (1/2), the terms gets smaller and smaller. This situation implies something important about the sum.
#### Example B
Find the sum of each series:
a) The first term of a geometric series is 4, and the common ratio is 3. Find S8.
b) The first term of a geometric series is 80, and the common ratio is (1/4). Find S7.
Solution
a) S8=4(138)13=13,120\begin{align*}S_{8}=\frac {4(1-3^8)}{1-3}=13,120\end{align*}
b) S7=80(1(14)7)114106.66\begin{align*}S_{7}= \frac {80 \left ( {1- \left ( \frac{1}{4} \right )^7} \right )}{1-\frac {1}{4}} \approx 106.66 \end{align*}
#### Example C
Prove the formula Sn=a1(1rn)1r\begin{align*}S_{n}=\frac{a_1(1 - r^n)} {1 - r}\end{align*} by induction
Solution
1. If n = 1, the nth sum is the first sum, or a1 . Using the hypothesized equation, we get S1=a1(1r1)1r=a1(1r)1r=a1\begin{align*}S_1 = \frac{a_1(1 - r^1)} {1 - r} = \frac{a_1(1 - r)} {1 - r} = a_1\end{align*}. This establishes the base case.
2. Assume that the sum of the first k terms in a geometric series is Sk=a1(1rk)1r\begin{align*}S_k = \frac{a_1(1 - r^k)} {1 - r}\end{align*}.
3. Show that the sum of the first k+1 terms in a geometric series is Sk+1=a1(1rk+1)1r\begin{align*}S_{k + 1} = \frac{a_1(1 - r^{k + 1})} {1 - r}\end{align*}.
Sk+1=Sk+ak+1\begin{align*}S_{k + 1} = S_k + a_{k + 1}\end{align*} The k+1\begin{align*}k + 1\end{align*} sum is the kth\begin{align*}k^{th}\end{align*} sum, plus the k+1\begin{align*}k + 1\end{align*} term
=a1(1rk)1r+a1rk+11\begin{align*} = \frac{a_1(1 - r^k)} {1 - r} + a_1r^{k + 1 - 1}\end{align*} Substitute from step 2, and substitute the k+1\begin{align*}k + 1\end{align*} term
=a1(1rk)1r+a1rk(1r)1r\begin{align*}= \frac{a_1(1 - r^k)} {1 - r} + \frac{a_1r^k(1 - r)} {1- r}\end{align*} The common denominator is 1r\begin{align*}1 - r\end{align*}
=a1(1rk)+a1rk(1r)1r\begin{align*}= \frac{a_1(1 - r^k) + a_1r^k(1 - r)} {1 - r}\end{align*} Simplify the fraction
=a1[1rk+rk(1r)]1r\begin{align*}= \frac{a_1 \left[1 - r^k + r^k(1 - r) \right]} {1 - r}\end{align*}
=a1[1rk+rkrk+1]1r\begin{align*}= \frac{a_1 \left[1 - r^k + r^k - r^{k + 1} \right]} {1 - r}\end{align*}
=a1[1rk+1]1r\begin{align*}= \frac{a_1 \left[1 - r^{k + 1} \right]} {1 - r}\end{align*} It is proven.
Therefore we have shown that Sn=a1(1rn)1r\begin{align*}S_{n}=\frac{a_1(1 - r^n)} {1 - r}\end{align*} for a geometric series. Now we can use this equation to find any sum of a geometric series.
Concept question wrap-up "Anna is on a progressive workout plan, every day she adds 5% to her exercise time. If she starts by exercising 15mins on the first day, how many minutes will she have exercised all together on day 45?"
Use the formula: Sn=a1(1rn)1r\begin{align*}S_{n}=\frac{a_1(1 - r^n)} {1 - r}\end{align*}
Sn=15(11.0545)11.05\begin{align*}S_{n} = \frac{15 (1 - 1.05^{45})}{1 - 1.05}\end{align*}
Sn=2395.5\begin{align*}S_{n} = 2395.5\end{align*} minutes.
### Vocabulary
A finite series has a defined ending value.
An infinte series does not have a defined ending value.
A geometric series is a series where the difference between terms increases or decreases between each pair of terms.
### Guided Practice
Questions
1) Find the sum: 5+10+20+...+640\begin{align*}5 + 10 + 20 +...+ 640\end{align*} (Hint: if an = 640 , what is n?)
2) Use a geometric series to answer the question:
In January, a company’s sales totaled 11,000. It is predicted that the company’s sales will increase 5% each month for the next year. At this rate, what will be the total sales for the year? 3) Write the first 5 terms of the sequence: 534n\begin{align*}-5 \cdot \frac{3}{4}^n\end{align*} 4) Write the 3rd, 4th, and 6th terms of: (3)(N2)\begin{align*}(3)^{\left(\frac{N}{2}\right)}\end{align*} 5) Find the sum of the series: n=16(32)n1\begin{align*}\sum_{n = 1}^6 \left(-\frac{3}{2}\right)^{n-1}\end{align*} Solutions 1) S8=5(128)12=1275\begin{align*}S_{8}=\frac {5(1-2^8)}{1-2} = 1275\end{align*} 2) S12=11000(11.0512)11.05=175,088.39\begin{align*}S_{12}=\frac {11000(1-1.05^{12})}{1-1.05} = 175,088.39\end{align*}
3) Just do the multiplication for each term n=0n=4\begin{align*}n = 0 \to n = 4\end{align*}
5340 to515\begin{align*}-5 \cdot \frac{3}{4}^0 \ to -5 \cdot 1 \to -5\end{align*} ..... for n=0\begin{align*}n = 0\end{align*}
5341 to5341543.75\begin{align*}-5 \cdot \frac{3}{4}^1 \ to -5 \cdot \frac{3}{4} \to -\frac{15}{4} \to -3.75\end{align*} ..... for n=1\begin{align*}n = 1\end{align*}
5342 to591645162.8\begin{align*}-5 \cdot \frac{3}{4}^2 \ to -5 \cdot \frac{9}{16} \to -\frac{45}{16} \to -2.8\end{align*} ..... for n=2\begin{align*}n = 2\end{align*}
5343 to52764135642.1\begin{align*}-5 \cdot \frac{3}{4}^3 \ to -5 \cdot \frac{27}{64} \to -\frac{135}{64} \to -2.1\end{align*} ..... for n=3\begin{align*}n = 3\end{align*}
5344 to5812564052561.6\begin{align*}-5 \cdot \frac{3}{4}^4 \ to -5 \cdot \frac{81}{256} \to -\frac{405}{256} \to -1.6\end{align*} ..... for n=4\begin{align*}n = 4\end{align*}
\begin{align*}\therefore\end{align*} the first 5 terms are: 5,3.75,2.8,2.1,1.6\begin{align*}-5, -3.75, -2.8, -2.1, -1.6\end{align*}
4) As with problem 3, just perform the operations on the indicated values of n:
33233275.2\begin{align*}3^{\frac{3}{2}} \to \sqrt{3^3} \to \sqrt{27} \to 5.2\end{align*} ..... for n=3\begin{align*}n = 3\end{align*}
3429\begin{align*}3^{\frac{4}{2}} \to 9\end{align*} ..... for n=4\begin{align*}n = 4\end{align*}
3623327\begin{align*}3^{\frac{6}{2}} \to 3^3 \to 27\end{align*} ..... for \begin{align*}n = 6\end{align*}
\begin{align*}\therefore\end{align*} the 3rd, 4th, and 6th terms are: \begin{align*}5.2, 9, 27\end{align*}
5) To find the sum of the series \begin{align*}\sum_{n = 1}^6 \left(-\frac{3}{2}\right)^{n-1}\end{align*}
We could calculate all of the values for \begin{align*}n = 1 \to 6\end{align*} and add them, getting:
\begin{align*}1+ \frac{-3}{2} + \frac{9}{4} + \frac{-27}{8} + \frac{81}{16} + \frac{-243}{32} = \frac{-133}{32}\end{align*}
Or we can use the formula: \begin{align*}\left( \frac{1 - r^k}{1 - r} \right)\end{align*}
\begin{align*}\left( \frac{1 - \left(\frac{-3}{2}\right)^6}{1 - \left(\frac{-3}{2}\right)} \right) = \frac{-133}{32}\end{align*}
### Practice
Find the sum of the finite series. You may simply calculate the individual terms and add them, or you may use the formula: \begin{align*}S_n = \frac{a_1 (1- r^n)}{1- r}\end{align*}
1. \begin{align*}1 + \left(-\frac{1}{2}\right) + \frac{1}{4} + ... + \frac{1}{64}\end{align*}
2. \begin{align*}-6 + 12 -24 + ... -6144\end{align*}
3. \begin{align*}(-4) + (-12) + (-36) + ... + (-2916)\end{align*}
4. \begin{align*}9 + (-45) + 225 + ... + (-17,578,125)\end{align*}
5. \begin{align*}\sum_{n=1}^5 -(2)^{n-1}\end{align*}
6. \begin{align*}\sum_{n=1}^{10} 6\left(\frac{1}{2}\right)^{n-1}\end{align*}
7. \begin{align*}\sum_{n=1}^6 8 \cdot 3^{n-1}\end{align*}
8. \begin{align*}(-5) + 10 + (-20) + ... + (-1280)\end{align*}
9. \begin{align*}(-9) + \frac{9}{2} + \left(-\frac{9}{4}\right) + ... + \left(-\frac{9}{16}\right)\end{align*}
10. \begin{align*}\sum_{n=1}^9 4 \cdot \left(\frac{1}{2}\right)^{n-1}\end{align*}
11. \begin{align*}\sum_{n=1}^{10} 4 \cdot \left(-\frac{2}{3}\right)^{n-1}\end{align*}
12. \begin{align*}(-3) + \left(-\frac{3}{2}\right) + \left(-\frac{3}{4}\right) + ... + \left(-\frac{3}{1024}\right)\end{align*}
13. \begin{align*}\sum_{n=1}^{11} -9 \cdot (-2)^{(n-1)}\end{align*}
14. \begin{align*}\sum_{n=1}^9 -9 \cdot \left(-\frac{5}{3}\right)^{n-1}\end{align*}
15. \begin{align*}\sum_{n=1}^6 -2 \cdot \left(-\frac{5}{4}\right)^{n-1}\end{align*}
16. \begin{align*}\sum_{n=1}^{11} 8 \cdot \left(-\frac{1}{3}\right)^{n-1}\end{align*}
### Vocabulary Language: English
finite series
finite series
A series is finite if it has a defined ending value.
geometric series
geometric series
A geometric series is a geometric sequence written as an uncalculated sum of terms.
induction
induction
Induction is a method of mathematical proof typically used to establish that a given statement is true for all positive integers.
infinite series
infinite series
An infinite series is the sum of the terms in a sequence that has an infinite number of terms.
series
series
A series is the sum of the terms of a sequence.
## Date Created:
Nov 01, 2012
Jun 08, 2015
Files can only be attached to the latest version of Modality |
# Power basics consumer math
Here, we debate how Power basics consumer math can help students learn Algebra. Our website can help me with math work.
## The Best Power basics consumer math
Keep reading to understand more about Power basics consumer math and how to use it. Solving quadratic equations by factoring is a process that can be used to find the roots of a quadratic equation. In order to solve a quadratic equation by factoring, the first step is to rewrite the equation in standard form. The next step is to factor the equation. Once the equation is factored, the roots of the equation can be found by setting each factor equal to zero and solving for x. Solving quadratic equations by factoring is a useful tool that can be used to find the roots of any quadratic equation.
Two equation solvers are a type of calculator that can be used to solve two equations at once. They are typically used in situations where two equations need to be solved simultaneously, such as when finding the intersection of two lines. Two equation solvers can be either stand-alone devices or software applications. While stand-alone devices are usually more expensive, they often offer more features and flexibility than software applications. Two equation solvers typically have a number of input methods, including keypads, touchscreens, and handwriting recognition. They also have a variety of output methods, including displays, printers, and projection systems. Two equation solvers can be used in a wide range of applications, from simple mathematical problems to complex engineering calculations.
The distance formula is generally represented as follows: d=√((x_2-x_1)^2+(y_2-y_1)^2) In this equation, d represents the distance between the points, x_1 and x_2 are the x-coordinates of the points, and y_1 and y_2 are the y-coordinates of the points. This equation can be used to solve for the distance between any two points in two dimensions. To solve for the distance between two points in three dimensions, a similar equation can be used with an additional term for the z-coordinate: d=√((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2) This equation can be used to solve for the distance between any two points in three dimensions.
The next step is to use matrix operations to simplify the matrix. Finally, the solution to the system of linear equations can be found by solving the simplified matrix. By using a matrix to represent a system of linear equations, it is possible to solve the equation quickly and efficiently. |
# Critical Number Trig Problem
• Oct 17th 2013, 12:13 PM
Jason76
Critical Number Trig Problem
Find critical numbers:
$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$ http://www.freemathhelp.com/forum/im...s/confused.png How do we solve for $\displaystyle \theta$ at this point?
• Oct 17th 2013, 12:23 PM
SlipEternal
Re: Critical Number Trig Problem
How do you normally find critical numbers? Set it equal to zero and solve for $\displaystyle \theta$.
$\displaystyle -18\sin \theta + 18\sin \theta \cos \theta = 0$.
Factor out $\displaystyle 18\sin \theta$:
$\displaystyle 18\sin \theta (\cos \theta-1) = 0$
Hence, either $\displaystyle 18\sin \theta = 0$ or $\displaystyle \cos \theta -1 = 0$.
• Oct 17th 2013, 12:38 PM
Jason76
Re: Critical Number Trig Problem
Find critical numbers:
$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta = 0$ Sorry left the $\displaystyle 0$ out.
$\displaystyle f '(\theta) = -18 \sin \theta + 18 (\sin \theta) \cos \theta = 0$
$\displaystyle f '(\theta) = 18 (\sin \theta) \cos \theta = 18 \sin \theta$
• Oct 17th 2013, 12:51 PM
SlipEternal
Re: Critical Number Trig Problem
I stand by my initial response.
• Oct 17th 2013, 05:58 PM
Jason76
Re: Critical Number Trig Problem
Quote:
I stand by my initial response.
I agree with you my approach to the problem was all wrong. The way to go is finding things to factor out.
Again:
Find critical numbers:
$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta = 0$
$\displaystyle f '(\theta) = 18[(-\sin \theta) + (\sin \theta) \cos \theta] = 0$
$\displaystyle f '(\theta) = -18[(\sin \theta) + (\sin \theta) \cos \theta] = 0$
$\displaystyle f '(\theta) = (\sin \theta) [-18 + \cos \theta] = 0$
$\displaystyle f '(\theta) = (\sin \theta) -18 + \cos \theta = 0$ ?? Next move?
• Oct 17th 2013, 06:17 PM
Prove It
Re: Critical Number Trig Problem
Did you not read SlipEternal's post at all? He left you with ONE step...
• Oct 17th 2013, 06:32 PM
Jason76
Re: Critical Number Trig Problem
Find critical numbers:
$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$
$\displaystyle f '(\theta) = -[18 (\sin \theta) + 18 (\sin \theta) \cos \theta]$
$\displaystyle f '(\theta) = - 18 (\sin \theta)[ 1 + \cos \theta]$ ?? Now?
• Oct 17th 2013, 06:38 PM
SlipEternal
Re: Critical Number Trig Problem
You seem to have tremendous difficulty with algebra. If I multiply out what you wrote:
$\displaystyle -18(\sin \theta)[1+\cos \theta] = -18\sin\theta - 18 \sin\theta \cos\theta$. That is not equal to what you started with. As Prove It said, it appears you either did not read or did not understand my post at all.
• Oct 17th 2013, 06:46 PM
Jason76
Re: Critical Number Trig Problem
Find critical numbers:
$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$
$\displaystyle 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta = 0$
$\displaystyle 18 (\sin \theta)[cos \theta - 1] = 0$ ??
• Oct 17th 2013, 06:49 PM
SlipEternal
Re: Critical Number Trig Problem
Let's multiply out again:
$\displaystyle 18(\sin\theta)[\cos\theta-1] = 18\sin\theta \cos\theta - 18\sin\theta$ Yes, this is correct. Next step, set the derivative equal to zero.
• Oct 17th 2013, 06:55 PM
Jason76
Re: Critical Number Trig Problem
Find critical numbers:
$\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$
$\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$
$\displaystyle 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta = 0$
$\displaystyle 18 (\sin \theta)[cos \theta - 1] = 0$
$\displaystyle 18 (\sin \theta)(cos \theta) - 18 (\sin \theta) = 0$ ??
• Oct 17th 2013, 06:58 PM
SlipEternal
Re: Critical Number Trig Problem
Quote:
Originally Posted by Jason76
Find critical numbers:
1. $\displaystyle f(\theta) = 18 \cos \theta + 9 \sin^{2} \theta$
2. $\displaystyle f '(\theta) = 18 (-\sin \theta) + 9 (u) du$
3. $\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (u) \cos \theta$
4. $\displaystyle f '(\theta) = 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta$
5. $\displaystyle 18 (-\sin \theta) + 18 (\sin \theta) \cos \theta = 0$
6. $\displaystyle 18 (\sin \theta)[\cos \theta - 1] = 0$ ??
7. $\displaystyle 18 (\sin \theta)(\cos \theta) - 18 (\sin \theta) = 0$ ??
From line 6, you can find solutions. You have the product of two expressions is equal to 0. The expressions are $\displaystyle [18\sin\theta]$ times $\displaystyle [\cos\theta -1]$ equals 0. This means $\displaystyle 18\sin\theta=0$ or $\displaystyle \cos\theta-1=0$.
So, when does $\displaystyle 18\sin\theta=0$? You can divide both sides by 18, so really, I am asking when is $\displaystyle \sin\theta=0$?
Next, when is $\displaystyle \cos\theta -1=0$? |
# Tag Archives: binomial
## Binomial Distribution Basics
The Binomial Distribution is a discrete probability distribution of the number of successes in n independent trials with a binary outcome and probability of success p. There are three important conditions that must be met for the binomial distribution to be used correctly:
• Trials are independent
• The probability of success p is constant for all trials
• Each trial has only two possible outcomes (binary)
One way to describe a Binomial random variable is in terms of Bernoulli Trials. A Bernoulli trial is a random experiment in which there are only two possible outcomes – success and failure. If we let success = 1 and failure = 0, then a Bernoulli Random Variable looks like this:
P(X = 1) = p
P(X = 0) = 1 – p
So really a Binomial experiment is just a sequence of n repeated Bernoulli trials. Formally, a random variable X that counts the number of successes, k, in n trials, has a Binomial Distribution with parameters n and p, written X~Bin(n, p). Let p = Probability that success occurs at any given trial. Then the probability of k successes is:
• The first part of the equation is “n choose k” or a combination of k objects from a population of n objects. This expression is the number of ways we can arrange k successes in n trials, where the order of the k successes is unimportant. This is called a Binomial Coefficient.
• The second part of the equation is the expression p raised to the power k, where p is the probability of success in any given trial. This expression takes advantage of the independence assumption – independent probabilities are multiplicative, so the probability of p happening k times is p*p*p*…*p (k times) = p raised to the k power.
• The third part of the equation is the probability of realizing n-k failures where trials are independent and the probability of failure is (1-p). Intuitively, if there are n trials and k of those are successes, the remaining trials, of which there are n-k, must result in failure.
It might help to think backwards and ignore the binomial coefficient for a moment. If k independent events occur with probability p, and n – k independent events occur with probability (1 – p), then the probability that all mentioned events occur is obviously the product (pk)*(1-p)n-k. While what we’ve just stated is accurate for one specific sequence of k successes and n – k failures, we must consider that there are a number of combinations of outcomes that yield k successes and n – k failures. If we flip a coin three times and define success as landing on heads, there is more than one outcome for k = 2. We could have HHT, HTH, or THH. The exact number of outcomes that yield k successes and n – k failures is the binomial coefficient “n choose k”. That is why it is multiplied by probability of success and failure.
Ex) Three coins are tossed, each with probability p that the coin lands on heads (it may not be a fair coin, so p may not be 0.5). Since coin tosses are independent, we can write:
The table below shows the probabilities associated with each sequence:
The chart above gives the probability for each outcome – that is a sequence of heads and tails. It would be incorrect to look at row 6 and then assume that the probability of obtaining 2 heads is the probability in column 4. The probability there is the probability of obtaining exactly the sequence TTH, not the more general case of obtaining two heads.
But if our main interest is the number of heads that occur, the sequence HHT is the same as HTH. Note that there are three outcomes with exactly two heads, for example, each with probability p2(1 – p). This is where the “n choose k” term becomes necessary. Exactly k successes may be achieved in a number of different ways. In the context of this question, the binomial probability is:
P(k heads) = (# ways to arrange k heads and n-k tails)*(prob. of any specific sequence w/ k heads and n-k tails)
P(k heads) = (# ways to arrange k heads and n-k tails)*pk (1-p)n-k
Implicit in the operations just performed is that a random variable was used to map outcomes in a sample space to the real numbers. Basically, we used a random variable to assign a real number (0, 1, 2, or 3 in this case) to the outcomes (sequences of H and T). A random variable is not random nor is it a variable – it is a function that maps sample outcomes to the real numbers. This is a very simplified definition that will not suffice in a mathematical statistics course most of the time, but that’s basically what’s going on here. The outcomes were grouped in such a way to facilitate problem solving – in this case, it was used to give us the number of heads rather than a specific sequence.
Just as the probability function assigns probabilities to sample outcomes in the sample space, a probability density function assigns probabilities to values that a random variable takes on. Let X be the number of heads that occur in a sequence of 3 coin tosses. Then the probability density function is:
For example:
And here is a chart of the actual probabilities:
It should be clear now why independence of trials is such an important criterion in using the Binomial distribution. If trials are not independent, i.e. the outcome of one trial affects the probability of subsequent outcomes, then the binomial model is not applicable. In this situation, the Hypergeometric Distribution should be used instead.
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Mixture or Alligation method of Quantitative AptitudeAlligation means linking. Today I'm going to discuss alligation method with you all. This is one of an important chapters of Quantitative exams. Basically, alligation is a rule that enables us to find the ratio of the quantities of the ingredients at given prices and are mixed to produce a mixture of given price.
## Alligation
Alligation method is basically for two purpose:
• to find the mean value of the mixtures when prices and proportions of both the ingredients are given
• to find the proportion of an ingredients of mixture when prices of both ingredients and mixture are given
### Rule of Alligation
Let two ingredients are mixed, out of that one is dearer and other is cheaper comparatively. Let Cost price (C.P.) of dearer one = d and Cost price of cheaper one = c
Study the following figure, if you remember this, you can easily solve the questions related to this chapter. Always remember that you will always consider cost price, not the selling price
Therefore, So, you can easily find the ratio of both the ingredients if prices of both the ingredients are given.
## Examples with Solution
Example1: In what proportion rice @Rs3/kg be mixed with rice @ Rs 3.50/kg, so that mixture worth Rs. 3.15/kg?
Solution: C.P. of cheaper rice C.P. of dearer rice Mean price i.e. price of mixture
Applying the Alligation method,
Therefore, Proportion in which both will be added is 7:3
Example2: A certain quantity of milk is mixed with 20 litres of water and the worth becomes Rs 15/litre. If price of pure milk is Rs 20/litre, then how much milk is there in the mixture?
Solution: Mixture $=MILK+WATER$
C.P. of pure milk
C.P. of water Mean price i.e. price of mixture
Applying the Alligation method,
Ratio of water to milk $=1:3$ Therefore, quantity of milk in mixture = 3 times quantity of water i.e. $3×20=60$ litres
Example3: In what ratio must water be added to milk to gain 20% by selling the mixture at cost price?
Solution: In this ques, we will assume the cost price of milk so that we can simplify the problem.
Let assumed price (C.P.) of milk
Then, according to ques, S.P. of milk Gain by selling mixture $=20\mathrm{%}$ Therefore selling price of mixture
By alligation method, Ratio of water and milk $=1:3$
So, in this way we can calculate these types of problems. One more thing, if in case percentage is given instead of cost price, then also you can do the problem in the same way.
Hope you understand the topic. Will soon update more examples of related topic.
Do you have a similar question? |
# Algebra 1 : How to find standard deviation
## Example Questions
2 Next →
### Example Question #11 : How To Find Standard Deviation
Find the standard deviation of the data set:
Explanation:
Write the formula of standard deviation.
Determine the mean, .
Subtract each number in the dataset from the mean and square each result.
This is the term.
Find the mean of the squared differences, or the variance.
This is the term.
Square root the variance for the standard deviation.
### Example Question #12 : How To Find Standard Deviation
Find the standard deviation. Round your answer to the nearest tenth.
Explanation:
To find standard deviation, we apply this formula: .
n represents how many data in the set.
represents the average of the data in the set.
is any data in the set.
is summation of the difference between average and any data value squared.
So the mean is . Now, we apply the formula.
### Example Question #13 : How To Find Standard Deviation
Find the standard deviation. Round your answer to the nearest tenth.
Explanation:
To find standard deviation, we apply this formula: .
n represents how many data in the set.
represents the average of the data in the set.
is any data in the set.
is summation of the difference between average and any data value squared.
So the mean is . Now, we apply the formula.
### Example Question #14 : How To Find Standard Deviation
Find the standard deviation of the following set of numbers:
Explanation:
To find standard deviation, we will follow these steps:
1. Find the mean of the original set of numbers.
2. Subtract the mean from each of the original numbers.
3. Square each number.
4. Find the mean of the new set of numbers.
5. Take the square root of the mean.
So, we will take this step by step.
## STEP 2
7, 10, 13
(7-10), (10-10), (13-10)
-3, 0, 3
9, 0, 9
## STEP 5
Thefore, the standard deviation is .
### Example Question #1991 : Algebra 1
Find the standard deviation of the following set of numbers:
Explanation:
To solve, you must use the following equation:
where,
Thus,
### Example Question #11 : How To Find Standard Deviation
Find the standard deviation of the following set of numbers.
Explanation:
To solve, simply use the formula for standard deviation. Thus,
where xm is the mean. Thus,
### Example Question #17 : How To Find Standard Deviation
Find the standard deviation of the following data set:
4, 2, 6, 8, 8, 2, 8, 10, 6.
Explanation:
To find the standard deviation, we will go through a few steps.
## Step 1: Find the mean.
We will find the mean of the data set.
## Step 2: Subtract the mean from each number in the data set.
We will now take each number in the data set and subtract the mean from it.
Giving us a new data set of -2, -4, 0, 2, 2, -4, 2, 4, 0.
## Step 3: Square each number in the data set.
Giving us a new data set of 4, 16, 0, 4, 4, 16, 4, 16, 0.
## Step 4: Add each number in the data set together.
Giving us an answer of 64.
## Step 5: Take the square root of the answer.
That is the standard deviation! So, the standard deviation of the data set
4, 2, 6, 8, 8, 2, 8, 10, 6
is 8.
### Example Question #18 : How To Find Standard Deviation
If the variance of a set of sample data is , what is the standard deviation?
Explanation:
The formula for finding the standard deviation given the variance is:
Substitute the variance.
The standard deviation is .
### Example Question #19 : How To Find Standard Deviation
Approximate the standard deviation of the given set of numbers to three decimal places:
Explanation:
Write the formula for standard deviation.
represents the total numbers in the data set:
is the mean:
Find the mean of the squared differences. Subtract each number in the data set with the mean, square the quantities, and take the average.
This will be the variance:
The standard deviation is the square root of the variance, which is the square root of this fraction.
### Example Question #20 : How To Find Standard Deviation
If the variance of a data set is 60, what must be the standard deviation? Round to three decimal places. |
+0
0
18
2
+1582
Find the sum of the squares of the roots of \$2x^2+4x-1=x^2-8x+3\$.
Apr 2, 2024
#1
+961
+1
Find the sum of the squares of the roots of \$2x^2+4x-1=x^2-8x+3\$.
2x2 + 4x – 1 = x2 – 8x + 3
Combine like terms x2 + 12x – 4 = 0
Complete the square x2 + 12x + 36 = 40
(x + 6)2 = 40
x + 6 = +sqrt(40)
x = +sqrt(40) – 6
Square each root [ +sqrt(40) – 6 ]2 = 40 – 12sqrt(40) + 36 (eq 1)
[ –sqrt(40) – 6 ]2 = 40 + 12sqrt(40) + 36 (eq 2)
Add (eq 1) & (eq 2) 80 + 72
Sum of squares = 152
.
Apr 2, 2024
#2
+6
+1
The given equation is:
2x2+4x−1=x2−8x+3
Rearranging terms, we get:
2x2+4x−1−x2+8x−3=0
x2+12x−4=0
Now, the sum of the squares of the roots of this equation can be found using the formula:
Sum of the squares of the roots=(Sum of the roots)2−2(Product of the roots)
For the equation \$x^2 + bx + c = 0\$, the sum of the roots is \$-b\$ and the product of the roots is \$c\$. Therefore, for our equation \$x^2 + 12x - 4 = 0\$, the sum of the roots is \$-12\$ and the product of the roots is \$-4\$.
Plugging these values into the formula, we get: Sum of the squares of the roots=(−(−12*12))−2(−4)
=(12*12)+8
=144+8
=152
So, the sum of the squares of the roots of the given equation is 152.
Apr 3, 2024 |
Basics of Calculus
In this calculus essentials series, I introduce and explain in detail the basic concepts that you will need to learn calculus.
This is my favorite Calculus book on Amazon, if you are interested in learning Calculus I highly recommend it
Rates of Change
In the examples that are to follow I try to provide any explanations that seem necessary. I want to be clear and comment what I do and why I do it. I am hoping this method will help others understand. If it seems a little long or self-explanatory then I apologize. My goal is to be consistent and thorough. We are dealing with functions in this first chapter and the rate at which they change.
To get the most out of these examples I would encourage you to try the problem first by yourself. If you have issues then refer to how it is worked. Hopefully that, along with my comments, will fill in any gaps that you might have in your knowledge.
Problem 1
Find the average rate of change of the function over the given intervals. $f(x)= 3x+3$ Using the intervals: [2,4] [-2,2]
Use the given function and the first interval.
You are evaluating the function at both points in order to find the difference.
Once you do that you are dividing by the difference of the interval length.
For the interval [2,4]
$\Longrightarrow \frac{f(4)-f(2)}{4-2}$
Substitute in all of your values to find the rate of change.
$\Longrightarrow \frac{((3)(4)^3 +3)-((3)(2)^3 +3)}{4-2} = 84$
For the interval [-2,2]
$\Longrightarrow \frac{f(2)-f(-2)}{2+2}$
Evaluate the equation with those values.
$\Longrightarrow \frac{((3)(2)^3+3)-((3)(-2)^3+3)}{2+2} = 12$
Problem 2
Find the average rate of change of the $R\theta$ function over the given interval: [0,8]
$\Longrightarrow r \theta = \sqrt{3\theta+1}$
Using this formula, we want to use the intervals given to us and evaluate them into the function. How this is done is simple. In the numerator we are using the intervals within the function itself. In the denominator, we just have the interval difference. Then we are just dividing the two.
$\Longrightarrow \frac{r \theta_2 - r \theta_1}{\theta_2 - \theta_1}$
I will now go ahead and set up the equation.
$\Longrightarrow \frac{r (8) - r (0)}{8-0}$
Now just substitute the intervals into the formula and start evaluating.
$\Longrightarrow \frac{\sqrt{3*8+1}-\sqrt{3*0+1}}{8-0}$
This gives us a simple equation to solve.
$\Longrightarrow \frac {5-1}{8-0}$
We are left with :
$\Longrightarrow \frac {1}{2}$
This is your rate of change.
Problem 3
Find the slope of the curve at the given point and an equation of the tangent line at that point. $$(-2,-9)$$
$y=7-4x^2$
Approach this problem as change in y divided by change in x.
$\Longrightarrow \frac{(7-4(-2+h)^2) -( 7-4(-2)^2)}{h}$
Distribute everything out.
$\Longrightarrow \frac{16h-4h^2}{h}$
Now simplify and set h=0.
The slope at (-2,-9) is: 16
Now to find the equation of the tangent line. You use this formula for that:
$\Longrightarrow y-y_1 = m(x-x_1)$
Substitute in your points and you get:
$\Longrightarrow y-(-9) = 16(x-(-2)$
Simplify a bit.
$\Longrightarrow y+9 = 16(x+2)$
Keep simplifying. This is the equation of the tangent line.
$\Longrightarrow y = 16x + 23$
Problem 4
Find the slope of the curve at the given point by finding the limiting value of the slope of the secants through the point. Also find an equation of the tangent line to the curve at the same point.
$y = x^3 - 6x$
Point = (1,-5)
This is just another form of what we have done before. You have an equation and a point to go by. Put the point into the function and evaluate it.
$\Longrightarrow \frac{(1+h)^3 - 6(1+h) - (1)^3 6(1)}{h}$
Now distribute. Pay careful attention to where everything goes as it is a lot to deal with.
$\Longrightarrow \frac{(1+3h+3h^2 +h^3-6-6h) - (-5)}{h}$
Reduce it down some.
$\Longrightarrow \frac{h^3 + 3h^2 -3h}{h}$
Simplify it now.
$\Longrightarrow h^2 + 3h - 3$
Then just assign h=0 and see what you have left. That is your slope of course as h approaches 0.
$\Longrightarrow = -3$
Now to find the equation of the tangent line.
We do the same process as before. We have the point-slope equation.
$\Longrightarrow y-y_1 = m(x-x_1)$
Use your slope that you got above. Then put in your point when it asks for an x or a y.
$\Longrightarrow y-(-5) = -3(x-1)$
Simply the equation.
$\Longrightarrow y + 5 = -3x + 3$
Move everything to one side of the equal sign. This is your equation of the tangent line.
$\Longrightarrow y = -3x - 2$
Limits
Limits of a function have been around in concept for a very long time. Mathematicians first started using the Limit concept somewhere around 400 years ago but our modern form of a Limit has only been used since the early 1800's or so. They are the foundation to learning derivatives and are subsequently used in modern Calculus quite often.
What Is A Limit
Lets say you are starting at 0,0 on this graph. That means that your X value =0 and your Y value =0. If you then move to some value to your right like X=4 then your X value =4. What does your Y value equal? Well that depends on whether you traveled in a straight line or some type of diagonal path up. Why does that matter?
Well, the Limit = the Y value!
So if you drew a crazy line that just went to X=4 but your Y value =7 then your Limit at X=4 =7. Does that make more sense now?
Wherever you end up on the X plane you will have some Y value as well. If you are still on Y =0 then your Limit =0.
So now that you should hopefully understand what the Limit is lets look at some problems.
Problem 1
$\tag{1}\lim_{x \to 3} f(x)$
When you look at this graph think about what you see. You see the X-axis marked off and the Y-axis is too. Now what is the question asking? It is asking what is the limit of f(x) as X approaches 3. What is the limit then? The limit is the value that the Y-axis is when X is at or very near 3. When you look at the graph again you will see that Y = 0 so the limit is 0.
$\tag{2}\bbox[red,2pt]{\lim_{x \to 3} f(x) = 0}$
Problem 2
What is
$$\tag{1}\lim_{x \to 9} f(x)$$ :
This is the same graph as above and looking at it we can see our X and Y values clearly marked. Our question is asking what is the limit of f(x) as X approaches 9. Now we look at X=9 on the graph and then see what Y values are there. Do you see the issue yet? Yes there are two values for Y at X=9. These are one sided limits which we will get into another time but the thing to take away here is those two Y values are not equal. This means the limit in our problem does not exist.
$\tag{2}\bbox[red,2pt]{\lim_{x \to 9} f(x) = }$ Does Not Exist!
Problem 3
What is the
$$\tag{1}\lim_{x \to -2} ( -x^2 + 9x -1 )$$
This is a limit problem that does not have a graph to look at but instead uses an equation. This equation is a simple polynomial. What can you do here? Well in math it is a good idea to try the simplest idea first to rule it out unless you have some reason not to. Lets try substitution here since that is barely algebra and it looks like it could work and give us a value. What we will want to do is substitute X=-2 into our equation. Why X=-2? That is because it is the point at which X is approaching.
$$\tag{2} ( -(-2)^2 + 9(-2) -1 )$$
You should notice now that everywhere there was an (X) value in the equation I put in a (-2).
$$\tag{3}(-4 -18 -1)$$
Multiply the values together and remember to keep the signs right.
$$\tag{3}\bbox[red,2pt] {-23}$$
Problem 4
Find $$\tag{1}\lim_{y \to -11}(16-y)^4/3$$
Here we have another substitution problem where our variable is Y. It is a little different because our equation has an exponent. So lets take it step by step.
$$\tag{2}\lim_{y \to -11}(16 - (-11))^4/3$$
Now we have substituted -11 for our Y variable. Then we simplify this a bit.
$$\tag{3}\lim_{y \to -11}(16 + 11)^4/3$$
$$\tag{4}\lim_{y \to -11}(27)^4/3$$
When dealing with an unusual exponent like this you can remember this little trick. Take your base number to the 4th power and then take the 3rd root of it. The 3rd root is also called the cube root. This is all the same as taking 27 to the $$4/3$$ power.
Our little problem now becomes much easier and the solution is $$\tag{5}\bbox[red,2pt]{\lim_{y \to -11}(16 - y)^4/3 = 81}$$
Problem 5
Find $$\tag{1}\lim_{h \to 0}(\frac{6}{\sqrt{6h + 4} +4})$$
This is a nice problem because it combines a few different styles and techniques to finish. This one is still quick to finish but it will be the start of harder problems that look like this one.
We have a limit, a fraction, and and a square root to deal with so this will be fun while it lasts.
Now for this and any other problem involving a fraction, we can substitute as long as the bottom part of the fraction does not equal 0. We should be safe here by the looks of it so lets try that.
$$\tag{2}\lim_{h \to 0}(\frac{6}{\sqrt{6(0) + 4} + 4})$$
Simplifying this now makes it easier to see what is happening.
$$\tag{3}\lim_{h \to 0}(\frac{6}{\sqrt{4} + 4})$$
We still are not doing anything with the top but we are almost there. Lets finish with the bottom now.
$$\tag{4}\lim_{h \to 0}(\frac{6}{2 + 4})$$
$$\tag{5}\lim_{h \to 0}(\frac{6}{6})$$
Well this is shaping up nicely.
$$\tag{6}\bbox[red,2pt]{\lim_{h \to 0} = 1}$$
Problem 6
Find $$\tag{1}\lim_{h \to 0}(\frac{\sqrt{19h + 1 } -1 }{h})$$
Ok different looking problem here. You can see we have the [h] on the bottom now. Since this is a fraction and we already know you can never divide by 0, we now know you must do something entirely different to solve this problem. Your main clue is that there is a square root on the top. Remember that square roots multiplied by themselves gives the value inside the root and the root disappears. However, what you do to the top will also have to be done on the bottom. This is called the conjugate method and it is used to solves problems like this.
$$\tag{2}\lim_{h \to 0}(\frac{\sqrt{19h + 1} -1 }{h}) * (\frac{\sqrt{19h + 1} + 1}{\sqrt{19h + 1} +1})$$
Now we are just multiplying the left side by the right side. Watch your signs carefully here. Always good to keep everything spaced apart a lot so that you can see what your doing and avoid making mistakes.
$$\tag{3}\lim_{h \to 0} (\frac{19h + 1 -1}{{h}{\sqrt{19h+1}+1}})$$
Now that we have multiplied both sides together we can subsequently start simplifying the top and bottom of our equation.
$$\tag{4}\lim_{h \to 0}(\frac{19h}{{h}{\sqrt{19h+1}+1}})$$
Since we have an [h] on top and bottom being multiplied by another expression we can cancel those [h] out.
$$\tag{5}\lim_{h \to 0} (\frac{19}{\sqrt{19h+1}+1})$$
Substitute 0 for h on the bottom part of the equation.
$$\tag{5}\lim_{h \to 0}(\frac{19}{0+1+1})$$
We are basically done so lets just make it look nice and final. The answer is:
$$\tag{6}\bbox[red,2pt]{\lim_{h \to 0}=(\frac{19}{2})}$$\
Problem 7
Find $$\tag{1}\lim_{x \to 3}(\frac{x-3}{x^2-9})$$
Here is another type of problem involving a limit however it just uses algebra to solve. You should be able to recognize instantly that the top and bottom parts of the equation are related to each. So break the problem up into parts and simplify from there.
$$\tag{2}\lim_{x \to 3}(\frac{x-3}{(x+3)(x-3)})$$
Since you will have an x-3 on both the top and the bottom you can just cancel them out in this situation. That will leave you:
$$\tag{3}\lim_{x \to 3}(\frac{1}{x+3})$$
Now you just apply the limit and you will have your answer.
$$\tag{4}\lim_{x \to 3}(\frac{1}{3+3})$$
This will give us a nice fraction as the answer.
$$\tag{5}\bbox[red,2pt]{\lim_{x \to 3}=(\frac{1}{6})}$$
Problem 8
Find $$\tag{1}\lim_{x \to 4} (\frac{x^2-2x-8}{x-4})$$
Here is another problem that is a fraction and it looks different than the one before you will notice. Actually, they should all look slightly different than the one before because there can be many variations. This is yet another variation with a polynomial. These problems are arranged this way to teach you the steps in troubleshooting a calculus problem and to help you recognize common forms of problems and what to do with them. If you are new to calculus just keep practicing until you instantly see what to do. Now here we go!
If you look at this problem it might seem weird but it is just a factoring calculus problem. You will obviously need to factor the top and hopefully that is clear to you. However, what beginners might not recognize is that the problem is already half factored for you. At this level of calculus problems are not going to be given to you that do not factor easily.
So, with that in mind, remember how the previous problems were dealt with? Yeah we factored and then cancelled out a top and bottom factor. We will do the same here. The bottom part of the fraction is one of the factors of the top part of the fraction.
$$\tag{2}\lim_{x \to 4}(\frac{(x+2) (x-4)}{x-4})$$
So see my point there how that works? We now have x-4 on both the top and bottom so we can just cancel them out.
$$\tag{3}\lim_{x \to 4}(x+2)$$
Now we just apply the limit.
$$\tag{4}\lim_{x \to 4}(4+2)$$
We now have a solution.
$$\tag{5}\bbox[red,2pt]{\lim_{x \to 4}=(6)}$$
Problem 9
Find the limit of $$\tag{1}\lim_{u \to 1}(\frac{u^3-1}{u^4-1})$$
There is a lot of factoring to do in this problem and you should be familiar with how to deal with numbers to the 3rd and 4th power. It is not hard at all though and furthermore it is good practice for harder problems later on. First, factor top and bottom and then see what we can do after that.
$$\tag{2}\lim_{u \to 1}(\frac{(u^2+ 1u +1)(u-1)}{(u^2+1)(u+1)(u-1)})$$
There we have both the top and the bottom parts of the fraction factored. Lets see what we can do with it now. We can cancel expressions on both top and bottom subsequently really make this simpler.
$$\tag{3}\lim_{u \to 1}(\frac{u^2 + 1u + 1}{(u^2 + 1)(u + 1) })$$
We are almost there now and we just need to substitute to apply the limit.
$$\tag{4}\lim_{u \to 1}(\frac{1^2 + 1(1) +1 }{(1^2 +1)(1 + 1)})$$
Now just do the simple math and we will have our answer.
$$\tag{5}\bbox[red,2pt]{\lim_{u \to 1} = 3/4 }$$
Problem 10
Find $$\tag{1}\lim_{x \to -6} (\frac{5-\sqrt {x^2 - 11}}{x+6})$$
Here is another nice problem but with a twist because now the square root is on the top. From looking at the top we can treat this as another conjugate problem. So lets multiple the top and bottom by the conjugate of the top.
$$\tag{2}\lim_{x \to -6} (\frac{5-\sqrt{x^2 - 11}}{x + 6}) * (\frac{5 + \sqrt{x^2 - 11}} {5 + \sqrt{x^2 - 11}})$$
That is an ugly equation and set up so let us try to make it look a little nicer. Just multiple top and bottom and remember to keep the sings straight.
$$\tag{3}\lim_{x \to -6} (\frac{25 - (x^2 - 11)}{(x + 6)(5 + \sqrt{x^2 - 11})})$$
While this still looks messy and difficult to work with it is getting better. Keep simplifying things out step by step so you do not miss any signs or do something silly.
$$\tag{4}\lim_{x \to -6} (\frac{ -x^2 + 36}{(x + 6)(5 + \sqrt{x^2 - 11})})$$
Rearrange the top so that it makes more sense as to what your supposed to.
$$\tag{5}\lim_{x \to -6} (\frac{36 - x^2}{(x + 6)(5 + \sqrt{x^2 - 11})})$$
Now factor what you can and this problem start to make sense.
$$\tag{6}\lim_{x \to -6} (\frac{(6 + x)(6 - x)}{(x + 6)(5 + \sqrt{x^2 - 11})})$$
Now cancel the similar terms and lets see what is left.
$$\tag{7}\lim_{x \to -6}(\frac{6-x}{5 + \sqrt{x^2 - 11}})$$
Substitute and apply the limit now. We get a good integer answer.
$$\tag{8}\bbox[red,2pt]{\lim_{x \to -6} = (\frac{6}{5})}$$
Problem 11
Find the limit of $$\tag{1}\lim_{x \to 0}(\csc(x))$$
Now things are getting interesting because we now have limits of trig functions. Hopefully you remember your pre-calculus! These are simple though as you just rewrite the harder forms into something that you do recognize.
$$\tag{2}\lim_{x \to 0}(\frac{1}{\sin(x)} )$$
We know that $$(\sin (0) = 0)$$ but if you graph it or do a table of values you will see that values go wildly out of control as they get close to 0. That is our clue there.
$$\tag{3}\bbox[red,2pt]{\lim_{x \to 0} (\csc(x) = \infty)}$$
Problem 12
Find the limit of $$\tag{1}\lim_{h \to 0} (\frac{f(x + h) + f(x)}{h} )$$ when $$f(x) = x^2$$ and $$x = 7$$.
$$\tag{2}\lim_{h \to 0} (\frac {(x + h)^2 -x^2}{h})$$
We just substituted in our functions and the value of x.
$$\tag{3}\lim_{h \to 0} (\frac{(7 + h)^2 - 7^2}{h})$$
We just plugged things in but it does take some attention to detail to keep things straight. We now have:
$$\tag{4}\lim_{h \to 0} (\frac{49 + 14h - 49}{h} )$$
Just factor everything and watch expressions disappear again.
$$\tag{5}\lim_{h \to 0}(\frac{h(14 + h)}{h})$$
Factor out the $$h$$ now so that we no longer have a fraction to deal with.
$$\tag{6}\lim_{h \to 0}(14 + h)$$
Now you just evaluate the expression and you get the answer.
$$\tag{7}\lim_{h \to 0}(14 + 0)$$
$$\tag{8}\bbox[red,2pt]{\lim_{h \to 0} = (14)}$$
Definition of a Limit
The definition of a limit continues on in our learning limits and how they are applied. Specifically we are going to look at some proofs in this section. With a precise definition we can now prove many limit properties.
The $$\delta$$ value will always depend on the $$\epsilon$$ value. That is important to remember in order to understand what is happening in these limits. Once you find a value that works in your problem then smaller values of that $$\delta$$ also work.
Problem 1
For the given function f(x) and values L,c, and $$\epsilon$$ > 0 find the largest open interval about c on which the inequality |f(x)-L| < $$\epsilon$$ holds. Then determine the largest value for $$\delta$$ >0 such that 0<|x-c|<$$\delta$$$$\rightarrow$$ |f(x)-L|<$$\epsilon$$.
f(x) = $$\frac{1}{x}$$, L= 0.125, c = 8, $$\epsilon$$ = 0.015
Solve |f(x) - L| < $$\epsilon$$ to find the largest interval containing c on which the inequality holds.
Write the inequality without absolute value. Substitute $$\frac{1}{x}$$ for f(x), 0.1215 for L, and 0.015 for $$\epsilon$$.
|f(x)-L| < $$\epsilon$$
|$$\frac{1}{x}$$ - 0.125| < 0.015
-0.015 < $$\frac{1}{x}$$ - 0.125 < 0.015
Simplify by adding 0.125 to all expressions.
-0.015 < $$\frac{1}{x}$$ - 0.125 < 0.015
0.11 < $$\frac{1}{x}$$ < 0.14
Simplify further by taking the reciprocals of all terms. Be sure to reverse the inequalities. Make sure you check that the rounded endpoints still allow the inequality |f(x)-L| < $$\epsilon$$ to hold true.
9.0909 > x > 7.1429
Since the interval 7.1429 < x < 9.0909 is not centered on c=8, $$\delta$$ is the distance from 8 to closer endpoint of the interval. The value of $$\delta$$ is then 0.8571.
Problem 2
For the given function $$f(x)$$, the point $$c$$, and a positive number $$\epsilon$$, find $$L=lim f(x)$$. Then find a number $$\delta$$ > 0 such that for all x, 0<|x-c|<$$\delta$$$$\rightarrow$$|$$f(x)-L$$| < $$\epsilon$$.
$$f(x)=8-3x, c=4, \epsilon=0.03$$
Notice that $$f(x)$$ is a linear function. Since a linear function is a simple polynomial function and it is defined for all $$x$$, the limit $$L$$ = lim is the value of $$f(x)$$ at $$c$$.
Evaluate the function at $$c=4$$.
$$L=8-3(4)=-4$$
To find $$\delta$$, begin by solving the inequality $$|f(x)-L|<\epsilon$$ to find an open interval $$(a,b)$$ containing $$c$$ on which the inequality holds for all $$x\neq c$$.
Remove the absolute value sign and rewrite the inequality as a compound inequality.
$$|(8-3x)-(-4)| < 0.03$$
$$-0.03 < (8-3x) - (-4) < 0.03$$
Simplify the center expression and isolate the x-term.
$$-0.03 < -3x+12 < 0.03$$
$$-12.03 < -3x < -11.97$$
Then isolate $$x$$ in the center. Notice that the direction of the inequalities has been changed.
$$4.01 > x > 3.99$$
Therefore, for $$x$$ in the interval $$(3.99,4.01)$$, the inequality $$(8-3x)-(-4)|<0.03$$ holds. Now choose a positive value for $$\delta$$ that places the open interval $$(c-\delta,c+\delta)$$ centered on $$c$$ inside the interval $$(3.99,4.01)$$.
The distance to the first endpoint is $$4-3.99=0.01$$.
The distance to the second endpoint is $$4.01 -4=0.01$$.
The largest possible value for $$\delta$$ is $$0.01$$.
Therefore, for all $$x$$ satisfying $$0<|x-4|<0.01$$, the inequality $$|(8-3x)-(-4)|<0.03$$ holds.
Problem 3
Give an $$\epsilon - \delta$$ proof of the limit fact.
$$lim_{x \to 0}(5x+1)=1$$
We begin by giving the precise meaning of a limit.
To say that $$lim f(x)=L$$ means that for each $$\epsilon>0$$ there is a corresponding $$\delta>0$$ such that $$|f(x)-L|<\epsilon$$, provided that $$0<|x-c|<\delta$$.
To establish the proof, we first perform a preliminary analysis to find the appropriate choice of $$\delta$$.
Let $$\epsilon$$ be any positive number.
We must produce a $$\delta > 0$$ such that $$0<|x-0|<\delta \rightarrow |(5x+1) -1 < \epsilon$$.
By simplifying the inequality on the right, we will determine the value of $$\delta$$ needed for the inequality on the left.
$$|(5x+1)-1| < \epsilon \Leftrightarrow |5x| < \epsilon$$
$$\Leftrightarrow |5| |x| < \epsilon$$
$$\Leftrightarrow 5|x| < \epsilon$$
$$\Leftrightarrow |x| < \frac{\epsilon}{5}$$.
Now we see that we should choose $$\delta=\frac{\epsilon}{5}$$. We can now construct the formal proof of the statement $$\lim_{x \to 0}(5x+1)=1$$.
Let $$\epsilon > 0$$ be given. Choose $$\delta=\frac{\epsilon}{5}$$. Then $$0<|x-0|<\delta$$ implies the following chain of equalities and an inequality.
$$|(5x+1)-1| = |5x|$$ Simplify inside the absolute value bars.
$$= |5| |x|$$ Rewrite as product of absolute values.
$$= 5 |x|$$ Simplify.
$$=5 |x-0|$$ Rewrite expression inside absolute value bars.
$$< 5\delta$$ Apply the condition $$0<|x-0|<\delta$$.
$$= \epsilon$$ Substitute $$\delta=\frac{\epsilon}{5}$$.
The result of the above chain of equalities and an inequality is that $$|(5x+1)-1|< \epsilon$$.
Therefore, we have proven that $$\lim_{x \to 0}(5x+1)=1$$.
Conclusion
The definition of a limit has some nice proofs that can establish. If you go over these problems slowly the ideas behind limits will begin to emerge. While this section is short it is essential to understanding limits and beginning calculus in general.
One-Sided Limits
Our current idea of the limit definition goes back around 200 years ago by Bernard Bolzano. He was instrumental in several mathematical ideas. One of which was the definition of one-sided limits which I will be talking about today. Now on to some problems!
Problem 1
Use the graph to see if the following statements are true or false. Look at the graph carefully because it can be tricky to see what is really going on sometimes.
We can see there is a function $$y=f(x)$$. There is also a small portion of the positive x-axis defined from [3 to 6].
A. True or False: $$lim_{x \to -3+}(f(x)=9$$
F(x) is your [y] value. X is of course your [x] value. Does [y] or [f(x)] get closer to 9 as [x] gets closer to 3 when coming from larger numbers than +3 ?
We can see that it sure does. The answer is [true].
B. True or False: $$lim_{x \to 0-}(f(x)=3$$
F(x) is your [y] value. X is your [x] value. The question is asking if [y=3] as [x] approaches 0 from the negative side of the [x] axis ?
It appears that [y=0] when [x=0] so the answer is false for this question.
C. True or False: $$lim_{x \to 0-}(f(x)$$ = $$lim_{x \to 0+}(f(x)$$ ?
This question is asking if the left limit is the same as the right limit. What does [y] equal when [x] is coming from the negative side? It equals 0. Now what does [y] equal as [x] is coming from the positive side? It equals 0 also. So both one-sides limits equal each other and the answer to the question is [true].
D. True or False: $$lim_{x \to 0}(f(x)$$ exists?
We have basically answered this question already. When the limit is listed without directional signs(- or +) it is asking if the limit exists and is equal from both sides of the graph. It is equivalent to asking if the limit from the left side equal the limit from the right side? As mentioned before we have answered this. We did see that the limit from the left side equaled the limit from the right side. So the answer to this question is [true].
E. True or False: $$lim_{x \to 0}(f(x)=0$$
This is another version of the question above. Since there are no directional signs(- or +) it is implying the full limit. The full limit is each one-sided limit being equal to each other. The left hand limit=0. The right hand limit=0. Therefore the limits equal each other and both equal 0. So the answer is also true for this question.
F. True or False: $$lim_{x \to 3}(f(x)=9$$
F(x) is your [y] value. X is your [x] value. The question is asking if [y=9] whether [x] approaches 3 from the negative or positive side?
When [x] is coming from the positive side of the [x] axis things are pretty clear and defined. However, there is a hole in the graph when [x] approaches 3 from the negative side. Therefore our limits are not equal and the answer is [false].
G. True or False: $$lim_{x \to 6-}(f(x)=6$$
F(x) is your [y] value. X is your [x] value. This is asking if [y=6] when [x] approaches 6 from the negative side? We can plainly see that it does not. When [y=6] it looks like [x] is somewhere between 2 and 3. So the answer to this question is [false].
Problem 2
F(x) = 6-x when x < 2 and (x/2) +1 when x > 2.
Find $$lim_{x \to 2+}(f(x)$$ and $$lim_{x \to 2-}(f(x)$$.
This is what I call a conditional function. Certain sections of the function give different values depending on the condition.
We will start by looking at the one-sided limits. First lets look at the limit as [x] approaches 2 from the positive side of the x-axis. The expresison (x/2) + 1 is used to evaluate when x > 2.
The $$lim_{x \to 2+}(f(x)=((x/2)+1)$$ = (2/2) + 1 = 2.
The expression 6-x is used to evaluate the expression when x < 2.
The $$lim_{x \to 2-}(f(x)=6-x$$ = 6-2 = 4.
Does the limit exist? For a full limit to exist its one-sided limits have to exist and be equal. The two limits above that we evaluated are not equal. Therefore the limit does not exist.
Problem 3
Use the relation $$lim_{\theta \to 0}{\frac{\sin\theta}{\theta}}=1$$ to find the limit of this function.
$$\frac{\sin \sqrt{2} \theta}{\sqrt{2} \theta}$$
As $$\theta$$ gets closer to 0 the square root of $$\theta$$ also gets closer to 0.
We now have : $$lim_{\sqrt{2}\theta \to 0}{\frac{\sin \sqrt{2} \theta}{\sqrt{2} \theta}}$$
So $$x = \sqrt{2} \theta$$. Now substitute in the values.
$$lim_{x \to 0} \frac{\sin x}{x} =1$$. That is our answer using the relation.
Problem 4
Find the limit of this expression. $$lim_{y \to 0}{\frac{\sin 3y}{7y}}$$
Again we will use the theorem $$lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$$
To get the form that we want so we can solve easily we need to multiply the numerator and denominator by $$\frac{3}{7}$$.
Once you see how everything cancels out you will see that your answer is also $$\frac{3}{7}$$.
Problem 5
Use the relation $$lim_{\theta \to 0} \frac{\sin \theta}{\theta} =1$$ for the following one-sided limits problem.
$$f(x) = \frac{2x + 2x\cos(2x)}{\sin(2x)\cos(2x)}$$ as x approaches 0.
We should write this expression as the sum of two terms.
$$\frac{2x}{\sin(2x)\cos(2x)} + {2x}{\sin(2x)}$$
Now rewrite the first fraction as the product of two fractions and then factor the result.
$$\frac{2x}{\sin(2x)} * \frac{1}{\cos(2x)} + \frac{2x}{\sin(2x)}$$
$$= \frac{2x}{\sin(2x)} \left(\frac{1}{\cos(2x)} + 1\right)$$
Multiply this out and then simplify. I did not show all the Algebra steps there because it is a pain but if you are in calculus this should be trivial for you.
You will this expression: $$\frac{1}{1} \left(\frac{1}{1} +1\right)$$
$$=2$$
Problem 6
Use the relation $$lim_{\theta \to 0} \frac{\sin \theta}{\theta} =1$$ to find out the limit of the function below.
$$f(\theta)=\frac{\sin \theta}{\sin (2\theta)}$$
Multiply $$\frac{\sin \theta}{\sin (2\theta)} by \frac{2\theta}{2\theta}$$.
This equals : $$\frac{1}{2} * \frac{\sin \theta}{\theta} * \frac{2\theta}{\sin (2\theta)}$$.
We now have: $$\frac{1}{2} * 1 * \frac{1}{1}$$.
So you should get $$\frac{1}{2}$$ as your answer.
Conclusion
After doing a few problems you should be able to see how useful one-sided limits can be to you. This will be one of your premier tools later on to help you solve problems. Limits often exhibit some wild behavior. That is why we must look at both sides of any particular limit. Sometimes they may be exactly equal on both sides but many times they will not be. Graphing will help when it gets more complicated so I will be using more graphs as we progress.
Solving Rates of Change
One of the easiest techniques to use is called the difference method. When you put a curve on a graph it will have an infinite number of points between its starting and ending points. If you take any two points on this curve you can draw a straight line to connect them. This straight line might be perfectly flat with respect to the x-axis, go up, or go down.
The line that we drew will have a certain length according to the points on the graph. Length will equal the number of points on the graph that it moves through then. The line will also have a certain amount of points it moves up or down. This value could be 0, positive, or negative. If you divide this change in y by the change in x you will get the slope of the curve in between those two points. This will be important to remember as we move towards limits eventually.
Problem 1
Find the slope of the curve at the point (1,5).
$$y= 3x^2 -2x +4$$
The curve looks like this.
We will use the difference formula to solve basic derivatives for now.
$$\frac{3(x + \bigtriangleup x)^2 - 2(x + \bigtriangleup x) + 4 - (3x^2 -2x + 4)}{\bigtriangleup x}$$
Lets multiple out these terms.
$$\frac{3x^2 + 6x\bigtriangleup x + 3(\bigtriangleup x)^2 -2x - 2\bigtriangleup x +4 -3x^2 + 2x - 4} {\bigtriangleup x}$$
We can simplify this a lot now.
$$\frac{6x\bigtriangleup x + 3(\bigtriangleup x)^2 - 2\bigtriangleup x} {\bigtriangleup x}$$
Further reducing this give us:
$$6x + 3\bigtriangleup x - 2$$
This gives us:
$$6x - 2$$
Our point we are evaluating is (1,5). The x-value of this point is 1. Substitute 1 for x in what is left of our equation.
$$6(1) - 2$$
$$4$$ is our slope at this point and the rate of change on the curve we are given.
Problem 2
Find the instantaneous rates of change of :
$$y = \frac{2x}{x + 1}$$
when x = 2.
First we need to find:
$$\frac{\bigtriangleup y}{\bigtriangleup x}$$
This basically says to find the change in y over the change in x. However, there is more to solving this.
What we are really saying is this.
$$\bigtriangleup y = f(x + \bigtriangleup x) - f(x)$$
The original problem tells us that:
$$y = \frac{2x}{x + 1}$$
Remember that y = f(x) and we are trying to find the change in y. So the change in y is:
$$f( x + \bigtriangleup x) = \frac{2(x + \bigtriangleup x)}{x + \bigtriangleup x + 1}$$
Now we substitute what we know into this so far.
$$\bigtriangleup y = \frac{ 2x + 2\bigtriangleup x}{x + \bigtriangleup x + 1} - \frac{2x}{x + 1}$$
Lets start by combining the terms:
$$\bigtriangleup y = \frac{(2x + 2 * \bigtriangleup x)(x + 1) - 2x (x + \bigtriangleup x + 1)}{(x + 1)(x + \bigtriangleup x + 1)}$$
This of course can be simplified. Look for the terms that cancel.
$$\bigtriangleup y = \frac{2 * \bigtriangleup x}{(x \bigtriangleup x + 1)(x + 1)(\bigtriangleup x)}$$
We now get a much better looking version.
$$\frac{2}{(x + \bigtriangleup x + 1)(x + 1)}$$
Since this limit is approaching 0 we will try substituting 0 for change the in x value and see what kind of answer we get.
$$\frac{\bigtriangleup y}{\bigtriangleup x} = \frac{2}{(x + 1)^2}$$
Now substitute 2 for x and see if we get an answer that makes sense. It looks good so I am happy with that. You can see we are edging towars the definition of limits here.
$$= \frac{2}{(2 + 1)^2} = \frac{2}{9}$$
Problem 3
Find the average rates of change of:
$$y = f(x) = x^2 - 2$$ between the points $$x = 3 and x = 4$$
The average rate of change is defined by:
$$\frac{\bigtriangleup y}{\bigtriangleup x} and \bigtriangleup y = f(x + \bigtriangleup x) - f(x)$$
We are given x = 3 and change in x = 4 - 3 = 1. So this gives us:
$$y = f(x) = f(3) = 3^2 - 2 = 7$$
The other point we are given is x = 4 so we evaluate for that as well.
$$= y + \bigtriangleup y = f(x + \bigtriangleup x) = 4^2 - 2 =14$$
This will yield:
$$= \bigtriangleup y = f(x + \bigtriangleup x) - f(x) = f(4) - f(3)$$
Which reduces to:
$$= (4^2 - 2) - (3^2 - 2) = 14 - 7 = 7$$
The average rate of change of this function is 7.
Problem 4
Find the average rates of change for:
$$y = \frac{1}{x}$$
The average rate of change is defined as:
$$\frac{\bigtriangleup y}{\bigtriangleup x} with \bigtriangleup y = f(x + \bigtriangleup x) - f(x)$$
Since:
$$f(x) = \frac{1}{x} and f(x + \bigtriangleup x) = \frac{1}{x + \bigtriangleup x}$$
And:
$$\bigtriangleup y = \frac{1}{x + \bigtriangleup x} - \frac{1}{x} = \frac{x - (x + \bigtriangleup x)}{x(x + \bigtriangleup x)}$$
This will evaluate to:
$$= \frac{- \bigtriangleup x}{x(x + \bigtriangleup x)}$$
This reduces to:
$$\frac{\bigtriangleup y}{\bigtriangleup x} = \frac{- \bigtriangleup x}{x(x + \bigtriangleup x) \bigtriangleup x} = - \frac{1}{x(x + \bigtriangleup x)}$$
So that makes the average rate of change of this particular function as:
$$\frac{-1}{x(x + \bigtriangleup x)}$$
Problem 5
Evaluate the rates of change in y=3x + 7 between x = -1 and x = 1.
$$\bigtriangleup y = f(x + \bigtriangleup x) - f(x)$$
$$\bigtriangleup x = 1 - (-1) = 2.$$
The change in x is equal to 2.
$$\bigtriangleup y = f(x + \bigtriangleup x) - f(x)$$
Since we know the change in x we can now use it in this part of the equation.
$$f(-1 + 2) - f(-1)$$
As you can see I just substituted the values.
$$(3(1) + 7) - (3(-1) + 7) = 10 - 4 = 6$$
So 6 is our answer as to the change in y.
Problem 6
Evaluate the rate of change in Y with respect to X at the point x = 5 when $$2y = x^2 + 3x - 1$$
The rate of change is defined as $$\bigtriangleup y = f(x + \bigtriangleup x) - f(x)$$
So, we do our inputs into the original equation.
$$2\bigtriangleup y = (x + \bigtriangleup x)^2 + 3(x + \bigtriangleup x) - 1 - (x^2 + 3x - 1)$$
As you can see this is a little more difficult version of what we have been working with. Pay attention to the sings and think it through and you will be fine.
We now have:
$$x^2 + 2x * \bigtriangleup x + (\bigtriangleup x)^2 + 3x + 3\bigtriangleup x - 1 - x^2 - 3x + 1$$
Theres a lot going on there but its not too bad.
Simplify things now.
$$2x * \bigtriangleup x + (\bigtriangleup x)^2 + 3\bigtriangleup x$$
Now we need to divide.
$$\frac{2 \bigtriangleup y}{\bigtriangleup x} = \frac{2x * \bigtriangleup x}{\bigtriangleup x} + \frac{(\bigtriangleup x)^2}{\bigtriangleup x} + \frac{3 \bigtriangleup x}{\bigtriangleup x} = 2x + \bigtriangleup x + 3$$
A lot of that will simplify down thank goodness. Here is what we have left after simplifying.
$$\frac{\bigtriangleup y}{\bigtriangleup x} = x + \frac{\bigtriangleup x}{2} + \frac{3}{2}$$
Now we have:
$$\frac{\bigtriangleup y}{\bigtriangleup x} = x + \frac{\bigtriangleup x}{2} + \frac{3}{2} = x + \frac{3}{2}$$
We are almost done now. If x = 5 then:
$$\frac{\bigtriangleup y}{\bigtriangleup x} = 5 + \frac{3}{2} = 6 \frac{1}{2}$$
That is our rate of change of this curve at x = 5.
Problem 7
Find the derivative of the function:
$$y = 2x^2 + 3x$$
The definition of of this type of derivative is:
$$y'(x) = \frac{f(x + \bigtriangleup x) - f(x)}{\bigtriangleup x}$$
Does that definition look familiar? It should. It is what we have been using all along. What we have been doing all along is finding derivatives at points on a graph.
We have been given $$f(x) = 2x^2 + 3x$$
We use our inputs now into the function.
$$f(x + \bigtriangleup x) = 2(x + \bigtriangleup x)^2 + 3(x + \bigtriangleup x)$$
Now we start substituting.
$$y'(x) = \frac{2(x + \bigtriangleup x)^2 + 3(x + \bigtriangleup x) - (2x^2 + 3x)}{\bigtriangleup x}$$
Lets reduce this down some.
$$y'(x) = \frac{4x\bigtriangleup x + 2(\bigtriangleup x)^2 + 3\bigtriangleup x}{\bigtriangleup x}$$
Keep simplifying.
$$y'(x) = 4x + 2\bigtriangleup x + 3$$
The middle term drops because we assume it is approaching 0. We are then left with:
$$y'(x) = 4x + 3$$ as our derivative. |
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# Lesson 1.7 Logical Reasoning and Counterexamples
Lesson 1.7 Logical Reasoning and Counterexamples. Objectives: a) To recognize conditional statements. b) To write converses of conditional statements. I. If – Then Statements. AKA – Conditional Statements Part following “If” – Hypothesis Part following “Then” – Conclusion Example 1:
## Lesson 1.7 Logical Reasoning and Counterexamples
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### Presentation Transcript
1. Lesson 1.7 Logical Reasoning and Counterexamples Objectives: a) To recognize conditional statements. b) To write converses of conditional statements.
2. I. If – Then Statements AKA – Conditional Statements Part following “If” – Hypothesis Part following “Then” – Conclusion Example 1: Conditional: If today is the 1st day of fall, then the month is September. • Hypothesis: • today is the 1st day of fall • Conclusion: • the month is September
3. Example 2: If y = 3 + 5, then y = 8 Hypothesis: y = 3 + 5 Conclusion: y = 8
4. II. Writing Conditionals Example 3: A rectangle has 4 right angles. If a shape is a rectangle then it has 4 right angles. Example 4: An acute angle has a measure less than 90°. If an angle is acute, then it has a measure less than 90°.
5. Not all conditionals are true! If a conditional is true, then every time the hypothesis is true the conclusion will also be true. To prove a conditional is false, you only need to find one counter example. If it is cloudy outside, then it is raining. (T or F) False: Give me a counter example It was cloudy last week but it wasn’t raining.
6. Example 4 and 5 An animal with 4 legs is a dog. Conditional: If an animal has 4 legs then it is a dog. True or False False: Give me a counter example. A
7. III. Converse Flip-flop the hypothesis and the conclusion Example 6: Conditional: If an angle is obtuse, then its measure is more than 90°. (T or F) Converse: If an angle has a measure greater than 90°, then it an obtuse angle. (T or F) If false give a counter example.
8. Example 7: A square has 4 sides. Conditional: If a shape is a square, then it has 4 sides. (Tor F) Converse: If a shape has 4 sides, then it is a square. (T or F) Give me a counter example. A trapezoid
9. Just because the condition is true doesn’t mean the converse is true. Example 8: Condition: If I am asleep, then I am breathing. (T or F) • Converse: • If I am breathing, then I am asleep. (T or F) • Counter Example
10. Notation Conditional: p → q Reads “If p then q” Converse: q → p Reads “If q then p”
11. Conditional Statements Identify the hypothesis and the conclusion: If two lines are parallel, then the lines are coplanar. In a conditional statement, the clause after if is the hypothesis and the clause afterthen is the conclusion. Hypothesis: Two lines are parallel. Conclusion: The lines are coplanar. Quick Check 2-1
12. Write the statement as a conditional: An acute angle measures less than 90º. The subject of the sentence is “An acute angle.” The hypothesis is “An angle is acute.” The first part of the conditional is “If an angle is acute.” The verb and object of the sentence are “measures less than 90°.” The conclusion is “It measures less than 90°.” The second part of the conditional is “then it measures less than 90°.” The conditional statement is “If an angle is acute, then it measures less than 90°.”
13. A counterexample is a case in which the hypothesis is true and the conclusion is false. This counterexample must be an example in which x2 0 (hypothesis true) and x 0 or x < 0 (conclusion false). > > Find a counterexample to show that this conditional is false: If x2 > 0, then x > 0. Because any negative number has a positive square, one possible counterexample is x = –1. Because (–1)2 = 1, which is greater than 0, the hypothesis is true. Because –1 < 0, the conclusion is false. The counterexample shows that the conditional is false.
14. Conditional Statements GEOMETRY LESSON 2-1 Use the Venn diagram below. What does it mean to be inside the large circle but outside the small circle? The large circle contains everyone who lives in Illinois. The small circle contains everyone who lives in Chicago. To be inside the large circle but outside the small circle means that you live in Illinois but outside Chicago.
15. Conditional Converse Hypothesis Conclusion Hypothesis Conclusion x = 9 x + 3 = 12 x + 3 = 12 x = 9 Conditional Statements Write the converse of the conditional: If x = 9, then x + 3 = 12. The converse of a conditional exchanges the hypothesis and the conclusion. So the converse is: If x + 3 = 12, then x = 9.
16. = / The conditional is false. A counterexample is a = –5: (–5)2 = 25, and –5 5. Write the converse of the conditional, and determine the truth value of each: If a2 = 25, a = 5. Conditional: If a2 = 25, then a = 5. The converse exchanges the hypothesis and conclusion. Converse: If a = 5, then a2 = 25. Because 52 = 25, the converse is true.
17. The Mad Hatter states: “You might just as well say that ‘I see what I eat’ is the same thing as ‘I eat what I see’!” Provide a counterexample to show that one of the Mad Hatter’s statements is false. The statement “I eat what I see” written as a conditional statement is “If I see it, then I eat it.” This conditional is false because there are many things you see that you do not eat. One possible counterexample is “I see a car on the road, but I do not eat the car.”
18. By phrasing a conjecture as an if-then statement, you can quickly identify its hypothesis and conclusion.
19. Writing Math “If p, then q” can also be written as “if p, q,” “q, if p,” “p implies q,” and “p only if q.”
20. Example 2B: Writing a Conditional Statement Write a conditional statement from the following. If an animal is a blue jay, then it is a bird. The inner oval represents the hypothesis, and the outer oval represents the conclusion.
21. Use the following conditional for Exercises 1–3. If a circle’s radius is 2 m, then its diameter is 4 m. 1. Identify the hypothesis and conclusion. 2. Write the converse. 3. Determine the truth value of the conditional and its converse. Show that each conditional is false by finding a counterexample. 4. If lines do not intersect, then they are parallel. 5. All numbers containing the digit 0 are divisible by 10. Hypothesis: A circle’s radius is 2 m. Conclusion: Its diameter is 4 m. If a circle’s diameter is 4 m, then its radius is 2 m. Both are true. skew lines Sample: 105
More Related |
NEET > L3 - Introduction to Vectors
# L3 - Introduction to Vectors Video Lecture - Additional Study Material for NEET
## Additional Study Material for NEET
26 videos|287 docs|64 tests
## FAQs on L3 - Introduction to Vectors Video Lecture - Additional Study Material for NEET
1. What is a vector in mathematics?
Ans. A vector is a mathematical object that has both magnitude and direction. It can be represented by an arrow, where the length of the arrow represents the magnitude and the direction of the arrow represents the direction of the vector.
2. How are vectors represented in three-dimensional space?
Ans. In three-dimensional space, vectors can be represented using three coordinates (x, y, z) or by their corresponding components along the x, y, and z axes. For example, a vector v can be represented as v = (x, y, z) or v = xi + yj + zk, where i, j, and k are the unit vectors along the x, y, and z axes respectively.
3. What is the difference between a scalar quantity and a vector quantity?
Ans. A scalar quantity only has magnitude, while a vector quantity has both magnitude and direction. For example, temperature is a scalar quantity as it only has magnitude (e.g., 25 degrees Celsius), whereas displacement is a vector quantity as it has both magnitude (e.g., 10 meters) and direction (e.g., north).
4. How do you add two vectors together?
Ans. To add two vectors together, you can use the parallelogram rule or the head-to-tail method. In the parallelogram rule, you draw the vectors as adjacent sides of a parallelogram and the resultant vector is the diagonal of the parallelogram. In the head-to-tail method, you place the tail of the second vector at the head of the first vector and the resultant vector is the vector from the tail of the first vector to the head of the second vector.
5. Can vectors be multiplied?
Ans. Yes, vectors can be multiplied, but there are different types of vector multiplication. The dot product (or scalar product) of two vectors gives a scalar quantity, while the cross product (or vector product) of two vectors gives a vector quantity. The dot product is defined as the product of the magnitudes of the vectors and the cosine of the angle between them, while the cross product is defined as the product of the magnitudes of the vectors, the sine of the angle between them, and a unit vector perpendicular to the plane of the two vectors.
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# How to Solve a Rubik's Cube!
14,689
When thinking of a Rubik's cube a lot of people believe it is crazy or impossible. In reality a Rubik's cube consists of only several steps with a few algorithms in each step. This is going to be a guide to help solve the impossible Rubik's cube in hopefully the easiest way possible. This guide will help you solve your Rubik's cube for the first time and can be used to help in future solves as well. The guide will consist of several different beginner steps and algorithms to help you understand how the Rubik's cube works.
* If there is an asterisk in front of something or if it is in bold it is important!
### Supplies:
A single Rubik's Cube!
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Learning the Pieces
The Rubik’s cube consists of 3 different pieces, there’s the center pieces, edge pieces, and corner pieces. The center piece is the piece in the center that distinguishes the color of that side. The center piece is only one color and never moves because it is attached to the core on the inside. The edge pieces are the pieces that connect the center pieces. The edge pieces only have 2 colors and can always be moved. The last piece is the corner piece. The corner piece has 3 colors that connect the edge pieces. The corners can always be moved. There are 6 center pieces, 12 edge pieces, and 8 corner pieces. The cube has 3 layers which consist of the first layer, middle layer, and top layer. Please use the picture above to visualize the pieces and layers.
## Step 2: Learning the Notations
All algorithms in a Rubik's cube are made up with certain notations. There are 6 notations and 6 inverses of the notations. All of the regular notations with the side that is specified is turned clockwise, while the inverses are turned counter-clockwise. The inverses will be marked with a lowercase i or an apostrophe ( ' ). If the notation has a "2" next to it then you want to do that notation twice. Example "U2" will be doing the U notation two times. You read these algorithms just like you normally read left to right, top to bottom.
* When doing an algorithm you DO NOT rotate the cube. You will look at the same face throughout the whole algorithm you are doing. Make sure to look at and do the algorithm very carefully!
R = Right, Ri = Right inverse
L = Left, Li = Left inverse
B = Back, Bi = Back inverse
D = Down, Di = Down inverse
F = Front, Fi = Front inverse
U = Up, Ui = Up inverse
Use the image above to visualize how the cubes notations work.
## Step 3: Solving the White Cross
* Please read instructions and look at pictures when prompted to. If you only look at pictures it will be challenging!
The first step to a Rubik's cube is solving the "White cross". This step is the hardest because it is intuitive and requires patience. We are going to be using the "daisy method". The white edge pieces can be anywhere on the cube.
Step 1: The first step in the white cross is to get all of the white edge pieces to match up with the yellow center. The end result of this step is in (Picture 1). It doesn't matter if the white pieces match up with the other centers as long as the 4 white edges are facing straight up. Again this step is intuitive so please be patient when trying complete it.
Step 2: Once it is in this position you want to hold the yellow face up and the blue center to face you (Picture 2).
Step 3: At this point you want to rotate the top layer or "U" notation until the blue/white edge piece lines up with the blue center (Picture 3). If it is already lined up with blue center do the next step.
Step 4: Once lined up you want to do "F2" which will move the blue/white edge piece to the bottom.
Step 5: Next you want to rotate the cube where the yellow face is up and the red center is facing you.
Step 6: You want to rotate the top layer or "U" notation until the red/white edge piece is lined up with the red center. If it is already lined up go to the next step.
Step 7: Once lined up you want to do "F2" which will move the red/white edge piece to the bottom.
Step 8: Rotate the cube where the yellow center is on top and the green center is facing you.
Step 9: You want to rotate the top layer or "U" notation until the green/white edge piece is lined up with the green center. If it is already lined up go to the next step.
Step 10: Once lined up you want to do "F2" which will move the green/white edge piece to the bottom.
Step 11: Rotate the cube where the yellow center is on top and the orange center is facing you.
Step 12: You want to rotate the top layer or "U" notation until the orange/white edge piece is lined up with the orange center. If it is already lined up go to the next step.
Step 13: Once lined up you want to do "F2" which will move the orange/white edge piece to the bottom.
Done!
Your cube should have the white cross where all the white edge pieces line up with their corresponding colors. (Picture 4)
## Step 4: Solving the White Corners
* Please read instructions and look at pictures when prompted to. If you only look at pictures it will be challenging!
Step 1: You want to hold your cube with the white center on the bottom the entire time.
Step 2: Look for any white corner pieces in the top layer. If there are corner pieces in the top layer then move to step 3. If there are no white corner pieces in the top layer and they are all in the bottom layer you want to hold the cube with the yellow center facing up. Then do the following algorithm. R, U, Ri, Ui. Once completing the algorithm the white corner should have been moved up to the top layer. Proceed to step 3.
Step 3: Once you find a white corner piece in the top layer you want to look and see what the other 2 colors on that SAME white corner piece. For example if you have a white corner you find, the other colors on that piece might be green and orange. Then you want to move that corner piece around the top layer using "U" or "Ui" to match the corner in between its centers. The corner can be rotated any way. Example: If you find the white corner piece that has the colors white, green, and orange on it then you want to move it around the top to match it in between the green and orange centers. Again the corner can be rotated any way but as long as you put it in between its correct centers you should be fine. Use the Picture 1 to help understand.
Step 4: Once the piece is in the right place you are going to hold the cube where the yellow center is on top and the corner you just put in the right spot is in the top right position. Use Picture 2 as an example. The corner could be rotated any way but as long as it is between the centers that it matches its good.
Step 5: With the yellow center on top and the corner piece in the top right position you want to do the following algorithm. R, U, Ri, Ui. Upon completing the algorithm the corner should be in the bottom right position. If it is there but not rotated the correct way (where white is match up with the white center on the bottom) then you want to do the algorithm again and again until it is rotated correctly. You want to make sure you hold the cube the same way you did when you did the algorithm the first time. Proceed to step 6 when the corner is rotated correctly in the right position (Picture 3).
Step 6: You want to find another white corner piece in the top layer and do steps 3-5. If there are no white corner pieces in the top layer then do steps 2-5.
Step 7: After completing the previous steps you should have the white corners in their correct places in the correct rotation as seen in Picture 4.
## Step 5: Solving the Middle Layer
* Please read instructions and look at pictures when prompted to. If you only look at pictures it will be challenging!
When solving the middle layer you will be inserting a non-yellow edge piece into the middle layer between the correct centers. Make sure when doing this step you know that a non-yellow edge piece will have 2 other colors besides yellow e.g. blue/red or green/orange.
* Note: Make sure the white face is facing down for the entire step.
Step 1:First you need to locate a non-yellow edge piece in your top layer. For example a edge piece that is non-yellow could be orange/blue or red/green. If there are no non-yellow edge pieces in the top layer then go to step 2. Once you found a non-yellow edge, you want to rotate the top layer or "U" notate until the front matches with its center. This will create a vertical line of that color as shown in case 1 and 2 in picture 1. Example: If you found a blue/orange edge piece in the top layer then you want to rotate the top layer until the blue face is matched with the blue center. Once completed move to step 3.
Step 2: This step is only if you have no non-yellow edge pieces in your top layer! From here you will need to do case 1's algorithm which should pop a non-yellow edge piece to the top layer. You may have a special case where an edge piece is in the proper location but is backwards. Use picture 1 case 3 to see what this looks like. If you have it then all you have to do is rotate your cube where the backwards edge in the right middle position like in picture 1 case 3, and do case 2's algorithm and it should pop it into the top layer. Once you have a non-yellow edge piece then you can move to step 3.
Step 3: Once the edge piece is matched up with its center color you need to determine if the color on the top of that edge piece matches with the center on the left or the center on the right. To understand please use Picture 1 above. In the picture above the blue center is matched with the blue edge in case 1 and 2 but it has 2 different colors on the top. The first case it goes to the left and in the second case it goes to the right. The color on top will determine if it needs to go to the right or the left. Once you find out which direction the edge piece goes proceed to step 4.
Step 4: Now that you know which way your edge piece should go in the middle layer (right or left) please use the picture above and locate which case you have. If you have case 1 where the edge goes to the left then do the algorithm for case 1. If you have case 2 where the edge goes to the right them do the algorithm for case 2.
Step 5: Now your edge piece should be in the correct position in your middle layer! Now you will repeat the process from steps 1 - 4 until you have solved the middle layer.
Once the middle layer is solved please move to the next step!
## Step 6: Solving the Yellow Cross
* Please read instructions and look at pictures when prompted to. If you only look at pictures it will be challenging!
After completing the last step you should have the first and middle layer solved! If so then you're now solving the yellow cross on the top.
* When doing any algorithm for this step you must keep the white face on the bottom.
Step 1: Locate which case you have from the image above. Once located proceed to step 2.
Step 2: If you have case 1 you are good and can proceed to step 3. If you have case 2 then you must complete the algorithm next to it. Once completing the algorithm you should end up with case 3 or 4. If you have case 3 then you must complete the algorithm next to it. Once completed proceed to step 3. If you have case 4 then you must complete the algorithm next to it. Once completed proceed to step 3.
Step 3: You should have a yellow cross on your top face. Your next goal is to finish solving the yellow face. Please proceed to the next section to finish solving the yellow face!
## Step 7: Finishing the Yellow Face
* Please read instructions and look at pictures when prompted to. If you only look at pictures it will be challenging!
Now that you have the yellow cross it is time to finish the yellow face.
Step 1: Your yellow face could have multiple patterns along with the yellow cross. You only want to have a certain pattern though. That pattern is called the "fish". The pattern is in picture 1. If you already have this pattern then go to step 3. If not go to step 2.
* Note: There are two fish patterns one that has the yellow corner facing you when the fish is pointing to the bottom left, and the fish you don't want will not have the yellow corner facing you. The correct one will look exactly like the picture above where the fish is pointing to the bottom left and the yellow corner is facing you.
Step 2: It doesn't matter what your pattern looks like all you have to do is the algorithm in picture 1 over and over until you get the fish pattern. If you do the algorithm and the pattern isn't changing then just do 1 or 2 "U" notations to rotate the pattern around and repeat the algorithm until you get the fish.
Step 3: Once you have the correct fish pattern you want to hold the cube with the white face down. You want to make sure that the fish is pointing to the bottom left of the yellow face and the yellow corner is facing you. Once you are in this position you want to do the same algorithm only one more time and your yellow face should be solved!
Once the yellow face is solved please move to the next section!
## Step 8: Solving the Yellow Corners
* Please read instructions and look at pictures when prompted to. If you only look at pictures it will be challenging!
Now that you have a completed yellow face you need to solve the yellow corners.
Step 1: You will have 2 cases. Case 1 you will have 2 yellow corners that are in their correct spot. This forms something called headlights which has 2 colors that are the same with one random color in the middle and is case 1 in Picture 1 above. If you have headlights then proceed to step 3. Case 2 you will have a solid bar. This means the one side of the top layer will have a bar that is all one color. If you have case 2 then proceed to step 3. If you do not have either case 1 or case 2 then proceed to step 2.
Step 2: This step is only if you don't have case 1 or 2 where no yellow corners are in their correct spot. If you do not have case 1 or 2 then you need to simply do the algorithm in the picture above one time and it will give you case 1 or 2. Once you have case 1 or 2 then you can move on to step 3.
Step 4: At this point you just need to do the algorithm that is in Picture 1 and your corners should be solved.
Once all of your yellow corners are in their correct locations please proceed to the last section!
## Step 9: Solving the Yellow Edges
* Please read instructions and look at pictures when prompted to. If you only look at pictures it will be challenging!
Now you've made it to the last step! You now have to rotate the yellow edges into their correct positions to match their color.
Step 1: You can have 2 cases. Case 1 will be that you have 3 yellow edge pieces in the wrong spot and you will have 1 bar. If you have case 1 proceed to step 3. Case 2 will be that you have all 4 yellow edges in the wrong spot. If you have case 2 proceed to step 2.
Step 2: If you have case 2 and all 4 yellow edges are in the wrong spot and you have no bar then all you have to do is one of the algorithms above. It does not matter which one as long as the white face is down and the yellow face is up. This should give you case 1 and then you can proceed to step 3.
Step 3: You should have case 1 when moving on to this step. You now need to decide if your edge pieces need to move clockwise or counter-clockwise. In other words rotate the entire cube to where the bar is in the back. Now do a "U" notation and see it the yellow edge piece matches to the center of that face. If it does then you are going to rotate them clockwise. First you need to do "Ui" to move the yellow corners to their right locations again. Now holding the cube with the bar in the back and of course white face on bottom and yellow on top, do the clockwise algorithm in Picture 1. If the yellow edge does not match the center of that face then you will be rotating them counter-clockwise. First you need to do "Ui" to move the yellow corners to their right locations again. Now holding the cube with the bar in the back and of course white face on bottom and yellow on top, do the counter-clockwise algorithm in Picture 1.
After completing the correct algorithm your cube should be solved!!
Congratulations!
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Abacus Maths | Vedic Maths | Memory Techniques | Calligraphy
Towards Better & Brighter Future
# Vedic Math Magic
Vedic Math >> Vedic Math Tricks >> Predic Date Of Birth >>
## Predict Date Of Birth
With this technique you can predict the date of birth of any number of people simultaneously..
## Vedic Math Magic
#### How to predict person's date of birth
With this technique you can predict the date of birth of any number of people simultaneously.
## Steps to predit date of birth
1) Ask the people to take the number of the month in which they were born (January is '1', February is '2', and so on..)
2) Next, ask them to double the number
3) Add'5' to it
4) Multiply it by '5'
5) Put a '0' behind the answer
6) Add their date of birth (If they are born on 5th January then add '5')
After the steps are over, ask them to tell you the final answer. And Lo! just by listening to their final answer you can predict their date of birth!
## Secret Magical Calculation
From the answers that you get from each member of the audience.
1) Mentally subtract '50' from the last two digits and you will have the date
2) Subtract '2' from the remaining digits and you will have the month
Thus, you will easily get his date of birth
## Example
Let us suppose, the date of birth of a person is 26th June. Then the steps to be worked out are as follows
1) Take the month number '6'
2) Double the answer '12'
3) Add '5' to it
4) Multiply it by '5'
5) Put a zero behind the answer '850'
6) Add the date of birth = 850 + 26 = 876
Thus our final answer is '876'. Now, let us see how we can deduct date of birth from the final answer. We will subtract '50' from the last two digits to get the date. Next, we will subtract'2' from the remaining digits to get the month.
8 / 76 - 2 / 50 = 6 / 26 ; Month = 6 & Date = 26
With the knowledge of this technique you can predict the birth-date of hundreds of people simultaneously.
Note: There are many such mathematical ways by which you can predict a person's date of birth but the method given above is one of the simplest methods. |
# Integrating Trigonometric Functions by Recognition
## Example 1
Find the derivative of $\sin (2x-5)$ and use this result to deduce $\displaystyle \int 10 \cos (2x-5) dx$.
\begin{align} \displaystyle \frac{d}{dx} \sin (2x-5) &= \cos (2x-5) \times \frac{d}{dx} (2x-5) &\color{green}{\text{Chain Rule}} \\ &= \cos (2x-5) \times 2 \\ &= 2 \cos (2x-5) \\ \sin (2x-5) &= \int 2 \cos (2x-5) dx \\ \int 10 \cos (2x-5) dx &= 5 \int 2 \cos (2x-5) dx \\ \require{AMSsymbols} \therefore \int 10 \cos (2x-5) dx &= 5 \sin (2x-5) + C \end{align}
## Example 2
Find the derivative of $x \cos x$ and use this result to find $\displaystyle \int x \sin x dx$.
\begin{align} \displaystyle \frac{d}{dx} x \cos x &= x \times \frac{d}{dx} \cos x + \frac{d}{dx} x \times \cos x &\color{green}{\text{Product Rule}} \\ &= x \times (-\sin x) + 1 \times \cos x \\ &= -x \sin x + \cos x \\ x \cos x &= \int (-x \sin x + \cos x) dx \\ x \cos x &=-\int x \sin x dx + \int \cos x dx \\ \int x \sin x dx &= \int \cos x dx-x \cos x \\ \require{AMSsymbols} \therefore \int x \sin x dx &= \sin x-x \cos x + C &\color{green}{\int \cos x dx = \sin x + C} \end{align}
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The Ultimate Q51 Guide [Expert Level]
Author Message
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8769
GMAT 1: 760 Q51 V42
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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20 Oct 2019, 18:15
[GMAT math practice question]
(function) f(x) is a function. What is the value of f(2006)?
1) f(11)=11
2) f(x+3)=(f(x)-1)/(f(x)+1)
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have many variables to determine a function and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Since we have f(x+3) = (f(x)-1)/(f(x)+1) and f(11)=11, we have f(14)=(f(11)-1)/(f(11)+1)=10/12=5/6 when we substitute 11 for x.
We have f(17) = (f(14)-1)/(f(14)+1) = ((5/6)-1)/(5/6)+1) = (-(1/6))/(11/6) = -1/11 when we substitute 14 for x.
We have f(20) = (f(17)-1)/(f(17)+1) = (-(1/11)-1)/(-(1/11)+1) = (-(12/11))/(10/11) = -12/10 = -6/5, when we substitute 17 for x.
We have f(23) = (f(20)-1)/(f(20)+1) = (-(6/5)-1)/(-(6/5)+1) = (-(11/5))/(-(1/5))=11, when we substitute 20 for x.
Then we have the following patterns.
f(11) = f(23) = f(35) = … = f(12k-1) = 11
f(14) = f(26) = f(38) = … = f(12k+2) = 5/6
f(17) = f(29) = f(41) = … = f(12k+5) = -1/11
f(20) = f(32) = f(44) = … = f(12k+8) = -6/5
So, f(2006) = f(12*167+2) = 5/6.
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions when the answer is A, B, C, or D.
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8769 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The Ultimate Q51 Guide [Expert Level] [#permalink] Show Tags 21 Oct 2019, 19:02 [GMAT math practice question] (Number) What is a positive integer p? 1) p is a prime number 2) p^2+2 is a prime number => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. Since we have 1 variable (p) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first. Condition 1) Since we have an infinite number of prime numbers, we don’t have a unique value of p, and condition 1) is not sufficient. Condition 2) If p has a remainder 1 when it is divided by 3 or p=3k+1 for some integer k, then p^2+2 = (3k+1)^2+2 = 9k^2+6k+1+2 = 3(3k^2+2k+1) is a multiple and it is a prime number. We have 3k^2+2k+1=1, 3k^2+2k=0, k(3k+2)=0 and k=0 or k=-2/3. However, k is an integer so only k=0 works. Then p=3(0)+1 = 1. However, p = 1 is not a solution since 1 is not a prime number. If p has a remainder 2 when it is divided by 3 or p=3k+2 for some integer k, then p^2+2 = (3k+2)^2+2 = 9k^2+12k+4+2 = 3(3k^2+4k+2) is a multiple and it is a prime number. Since we have 3k^2+4k+2=1, 3k^2+4k+1=0 or (3k+1)(k+1)=0 and we have k =-1 and k=-1/3. However, k must be an integer so then p=3(-1)+2 = -1. However, p = -1 is not a solution since -1 is negative. Assume p has a remainder 0 when it is divided by 3. If p=3, then p^2+2=11 is a prime number. If p=9, then p^2+2=83 is a prime number. Since condition 2) does not yield a unique solution, it is not sufficient. Conditions 1) & 2) p is a multiple of 3 from condition 2), and p is a prime number from condition 1). Then p = 3. Since both conditions together yield a unique solution, it is sufficient. Therefore, C is the answer. Answer: C If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations,” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8769
GMAT 1: 760 Q51 V42
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
Show Tags
22 Oct 2019, 17:39
[GMAT math practice question]
(number properties) m and n are integers. What is the value (-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}?
1) m = n + 1
2) m = 3
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have 2 variables (m and n) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Since we have m = n + 1 and m = 3, we can substitute m = 3 into m = n + 1 to get 3 = n + 1 and n = 2.
(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}
=(-1)^{3-2} +(-1)^{3+2} +(-1)^{3*2} +(-1)^{2*2}
=(-1)^1 +(-1)^5 +(-1)^6 +(-1)^4
=(-1) + (-1) + 1 + 1 = 0
Since both conditions together yield a unique solution, they are sufficient.
Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
Since m = n + 1, m and n are consecutive integers, they have different parities, which means that if m is an odd integer, then n is an even integer, and if m is an even integer, then n is an odd integer.
Case 1: m is an odd integer and n is an even integer.
Then, m+n is an odd integer, m – n is an odd integer, mn is an even integer and 2n is an even integer.
(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}
=(-1)^{odd} +(-1)^{odd} +(-1)^{even} +(-1)^{even}
=(-1) + (-1) + 1 + 1 = 0
Case 2: m is an even integer and n is an odd integer.
Then, m+n is an odd integer, m – n is an odd integer, mn is an even integer and 2n is an even integer.
(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}
=(-1)^{odd} +(-1)^{odd} +(-1)^{even} +(-1)^{even}
=(-1) + (-1) + 1 + 1 = 0
Since condition 1) yields a unique solution, it is sufficient.
Condition 2)
Case 1: n is an even integer.
Then, m+n is an odd integer, m – n is an odd integer, mn is an even integer and 2n is an even integer, since m = 3.
(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}
=(-1)^{odd} +(-1)^{odd} +(-1)^{even} +(-1)^{even}
=(-1) + (-1) + 1 + 1 = 0
Case 2: n is an odd integer.
Then, m+n is an even integer, m – n is an even integer, mn is an odd integer and 2n is an even integer.
(-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}
=(-1)^{even} +(-1)^{even} +(-1)^{odd} +(-1)^{even}
=1 + 1 + (-1) + 1 = 2
Since condition 2) does not yield a unique solution, it is not sufficient.
If the question has both C and A as its answer, then A is an answer rather than C by the definition of DS questions. Also, this question is a 50/51 level question and can be solved by using the Variable Approach and the relationship between Common Mistake Type 3 and 4 (A or B).
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8769 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The Ultimate Q51 Guide [Expert Level] [#permalink] Show Tags 24 Oct 2019, 18:08 [GMAT math practice question] (Geometry) The figure shows that line m is parallel to the line n, and l is parallel to k. Moreover, ∠CBD=45°, ∠FAE=80°. What is ∠BDC? Attachment: 10.14ps.png [ 19.74 KiB | Viewed 349 times ] A. 40° B. 45° C. 50° D. 55° E. 60° => Attachment: 10.21PS(A).png [ 29.3 KiB | Viewed 349 times ] Since lines m and n are parallel, we have <FBG=<FAE=80° and 80°+<ABD+45°=180°. Then we have 125°+<ABD=180 and <ABD=55°. Since <ABD and <BDC are alternate interior angles, they are congruent and <BDC=55°. Therefore, D is the answer. Answer: D _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8769
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
Show Tags
27 Oct 2019, 18:54
[GMAT math practice question]
(number properties) p and q are integers. Is (p-1)(q-1) an even number?
1) p+q is an odd number
2) pq is an even number
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
The following reasoning shows that in the question, either p or q is an odd integer.
(p-1)(q-1) is an even integer
=> p – 1 or q – 1 is an even integer
=> p or q is an odd integer
Therefore, either p and q is an odd number, and the other one is an even number, according to condition 1. So, condition 1) is sufficient.
Condition 2)
If p is an odd number and q is an even number, then p-1 is an even number, q-1 is an odd number, and (p-1)(q-1) is an even number, which means the answer is ‘yes’.
If both p and q are even numbers, then (p-1)(q-1) is an odd number, and the answer is ‘no’ since both p-1 and q-1 are odd numbers.
Since condition 2) does not yield a unique solution, it is not sufficient.
_________________
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8769 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The Ultimate Q51 Guide [Expert Level] [#permalink] Show Tags 28 Oct 2019, 22:34 [GMAT math practice question] (geometry) The figure shows that OA = 20, OB = 30 and OC = x and □OCDE is a rectangle. What is the area of rectangle OCDE? 1) x = 10 2) OE = 15 Attachment: 10.23DS.png [ 10.14 KiB | Viewed 309 times ] => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. Since the triangle OAB and the triangle CAD are similar, we have OA:OB = 2:3 and CA:CD = 2:3. Then we have 3CA = 2CD or CD = (3/2)(20-x). So the area of the rectangle OCDE is x*(3/2)(20-x). Therefore, we have one variable in this question. Since we have 1 variable (x) and 0 equations, D is the most likely answer. So, we should consider each condition separately first. Condition 1) is sufficient, since it yields a unique solution. Condition 2) Since CD = OE = 15 from condition 2), and from the original condition we know CD = (3/2)(20-x). =>15 = (3/2)(20-x) =>10 = 20-x =>x = 10 =>3CA = 2CD =>3CA = 2(15) =>3CA = 30 =>CA = 10 We have CA = 10 and x = 10. So, condition 2) is also sufficient, because it is equivalent to condition 1). Therefore, D is the answer. Answer: D If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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29 Oct 2019, 17:33
[GMAT math practice question]
(propotional) What is z^2/xy + x^2/yz + y^2/zx ?
1) x:y = 2:3
2) x:z = 1:2
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have 3 variables (x, y, and z) and 0 equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Since x:y = 2:3 and x:z = 1:2, we have x:y:z = 2:3:4.
Then we have x = 2k, y = 3k, and z = 4k for some number k.
z^2/xy + x^2/yz + y^2/zx
= (4k)^2/(2k)(3k) + (2k)^2/(3k)(4k) + (3k)^2/(4k)(2k)
= 16k^2/6k^2 + 4k^2/12k^2 + 9k^2/8k^2
= 16/6 + 4/12 + 9/8
= 64/24 + 8/24 + 27/24 = 99/24 = 33/8.
Since both conditions together yield a unique solution, they are sufficient.
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8769 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The Ultimate Q51 Guide [Expert Level] [#permalink] Show Tags 30 Oct 2019, 17:24 [GMAT math practice question] (equation) What are the values of x+y and xy? 1) x + y + xy = -2 2) (1/x) + (1/y) = 1 => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. Since we have 2 variables (x and y) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2) Since 1/x + 1/y = 1 from condition 2), we have y + x = xy by multiplying both sides of the equation by xy, which rearranges to get xy – (x+y) = 0. Since xy + (x+y) = -2 from condition 1), we have xy - (x+y) + xy + (x+y) = 0 + -2 by adding the two equations. Then 2xy = -2 or xy = -1. Then we have x+y=-1. Since both conditions together yield a unique solution, they are sufficient. Therefore, C is the answer. Answer: C Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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31 Oct 2019, 17:38
[GMAT math practice question]
(geometry) The figure shows the rectangle ABCD. What is ∠x - ∠y?
Attachment:
10.23PS.png [ 20.7 KiB | Viewed 236 times ]
A. 20°
B. 17°
C. 15°
D. 13°
E. 12°
=>
Attachment:
10.23ps(a).png [ 21.61 KiB | Viewed 236 times ]
Since AP, RP and CD are parallel, we have <ABP = <BPR and <DQP = <RPQ. Since <BPQ = <BPR + <RPQ, we have <x = 15° + <y.
So, we have <x - <y = 15°.
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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8769 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The Ultimate Q51 Guide [Expert Level] [#permalink] Show Tags 03 Nov 2019, 18:39 [GMAT math practice question] (algebra) What is the value of x/(x+y) + y/(x-y)? 1) (x+y):y = 3:1 2) x + y = 8 => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. The question x/(x+y) + y/(x-y) is equivalent to (x^2+y^2)/(x^2-y^2) for the following reason x/(x+y) + y/(x-y) => x(x-y)/(x+y)(x-y) + y(x+y)/(x+y)(x-y) => (x^2-xy+xy+y^2)/(x^2-y^2) => (x^2+y^2)/(x^2-y^2) => (x^2/y^2+1)/(x^2/y^2-1) by dividing the top and bottom by y^2 => [(x/y)^2+1)/[(x/y)^2-1] When a question asks for a ratio, if one condition includes a ratio and the other condition just gives a number, the condition including the ratio is most likely to be sufficient. This tells us that A is most likely to be the answer to this question. Condition 1) The condition (x+y):y = 3:1 is equivalent to x = 2y since x + y = 3y from (x+y):y = 3:1. Then (x^2+y^2)/(x^2-y^2) = ((2y)^2+y^2)/((2y)^2-y^2) = (4y^2+y^2)/(4y^2-y^2) = 5y^2/3y^2 = 5/3. Since condition 1) yields a unique solution, it is sufficient. Condition 2) If x = 5 and y = 3, then we have x/(x+y) + y/(x-y) = 5/8 + 3/2 = 5/8 + 12/8 = 17/8. If x = 6 and y = 2, then we have x/(x+y) + y/(x-y) = 6/8 + 2/4 = 3/4 + 2/4 = 5/4. Since condition 2) does not yield a unique solution, it is not sufficient. Therefore, A is the answer. Answer: A _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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04 Nov 2019, 17:42
[GMAT math practice question]
(algebra) max{x, y} denotes the maximum of x and y, and min{x, y} denotes the minimum of x and y. What is the value of x + y?
1) max{x, y} = x + y
2) min{x, y} = 2x + y - 2
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have 2 variables (x and y) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Case 1: x ≥ y
Since max(x,y) = x and max(x,y) = x + y, we have x = x + y or y = 0.
Since min(x,y) = y and min(x,y) = 2x + y - 2, we have y = 2x + y - 2, 0 = 2x - 2, 2x = 2, or x = 1.
Then we have x + y = 0 + 1 = 1.
Case 2: x < y
Since max(x,y) = y and max(x,y) = x + y, we have y = x + y or x = 0.
Since min(x,y) = x, min(x,y) = 2x + y - 2 and x = 0, we have x = 2x + y - 2 or y = 2.
Then we have x + y = 0 + 2 = 2.
Since both conditions together do not yield a unique solution, they are not sufficient.
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8769 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The Ultimate Q51 Guide [Expert Level] [#permalink] Show Tags 05 Nov 2019, 18:11 [GMAT math practice question] (equation) What is the value of a + b? 1) ax + by = 2(ax - by) - 3 = x + y + 7 2) x = 3, y = 1 => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. We have 4 variables (a, b, x and y). However, since both conditions have 4 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2) Since x = 3 and y = 1, we have 3a + b = 2(3a - b) - 3 = 3+1+7 = 11. Then we have 3a + b = 11 and 6a - 2b = 14 or 3a – b = 7. When we add those equations we have 3a + b + 3a - b = 11 + 7, 6a = 18 or a = 3. Then we have 3(3) + b = 11, 9 + b = 11 or b = 2. Since both conditions together yield a unique solution, they are sufficient. Therefore, C is the answer. Answer: C Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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06 Nov 2019, 20:35
[GMAT math practice question]
(geometry) The figure shows that triangle ABC is inscribed in circle O. If ∠AOB = x + 60°, ∠BOC = 2x + 20°, ∠AOC = 3x - 20°, what is ∠ACB?
Attachment:
10.28ps.png [ 16.76 KiB | Viewed 535 times ]
A. 45°
B. 50°
C. 55°
D. 60°
E. 65°
=>
Since the sum of central angles of those three sectors equals to 360°, we have x + 60°+ 2x + 20°+ 3x - 20°= 360°, 6x + 60°= 360°, 6x = 300°, or x = 50°
Then we have ∠AOB = x + 60° = 50° + 60° = 110°, ∠BOC = 2x + 20° = 2(50°) + 20° = 120°, ∠AOC = 3x - 20° = 3(50°) - 20° = 130°.
Attachment:
10.28ps(a).png [ 16.45 KiB | Viewed 528 times ]
∠ACO = (180°- 130°)/2 = 25° and ∠BCO = (180°- 120°)/2 = 30°.
Then ∠ACB = ∠AOC + ∠BOC = 25°+ 30°= 55°.
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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8769 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The Ultimate Q51 Guide [Expert Level] [#permalink] Show Tags 07 Nov 2019, 18:36 [GMAT math practice question] (probability) The figure shows that l is parallel to m and points A, B, C, D, E are on line l, and points F, G, H, I are on the line m. How many quadrilaterals are possible with 4 points out of the above 8 points? A.48 B. 56 C. 60 D. 72 E. 108 Attachment: 10.30ps.png [ 15.36 KiB | Viewed 508 times ] => To create a quadrilateral, we should choose 2 points out of 5 points on line l, and 2 points out of 4 points on the line m and connect them. Then we have 5C2*4C2 = [(5*4)/(1*2)][(4*3)/(1*2)] = 10*6 = 60. Therefore, C is the answer. Answer: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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10 Nov 2019, 20:32
[GMAT math practice question]
b = (-1)+(-1)^2+(-1)^3+….+(-1)^a. What is the value of b?
1) a = 2019
2) a is an odd number.
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have $$2$$ variables ($$a$$ and $$b$$) and $$1$$ equation, D is most likely to be the answer. So, we should consider each condition on its own first.
Condition 1)
We have $$(-1) = (-1)^3 = (-1)^5 = … =(-1)^{2019} = -1 and (-1)^2 = (-1)^4 = (-1)^6 = … = (-1)^{2018} = 1. (-1) + (-1)^2 + (-1)^3+….+(-1)^a. = ((-1)+(-1)^2) + ((-1)^3+(-1)^4) + … + ((-1)^{2017} +(-1)^{2018}) + (-1)^{2019} = 0 + 0 + … + 0 + (-1) = -1$$
Since condition 1) yields a unique solution, it is sufficient.
Condition 2)
We have $$(-1) = (-1)^3 = (-1)^5 = … =(-1)^a = -1 and (-1)^2 = (-1)^4 = (-1)^6 = … = (-1)^{a-1} = 1. (-1) + (-1)^2 + (-1)^3+….+(-1)^a. = ((-1)+(-1)^2) + ((-1)^3+(-1)^4) + … + ((-1)^{a-2} +(-1)^{a-1}) + (-1)^a = 0 + 0 + … + 0 + (-1) = -1$$
Since condition 2) yields a unique solution, it is sufficient.
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A, B, C or E.
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8769 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The Ultimate Q51 Guide [Expert Level] [#permalink] Show Tags 11 Nov 2019, 21:20 [GMAT math practice question] (algebra) What is the value of (4x-3xy+4y)/(3x+3y)? 1) x = y 2) (1/x) + (1/y) = 3 => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. The question asks the value of (4/3) – xy/(x+y) for the following reason (4x-3xy+4y)/(3x+3y) = [4(x+y)-3xy]/[3(x+y)] = [4(x+y)]/[3(x+y)] - [3xy]/[3(x+y)] = (4/3) – xy/(x+y) Since we have (x+y)/xy = 3 from condition 2),) for the following reason (1/x)+(1/y) = 3 (y/xy) + (x/xy) = 3 (x+y)/xy = 3 Then we have xy/(x+y) = 1/3. Then (4/3) – xy/(x+y) = (4/3) – (1/3) = 1. Since condition 2) yields a unique solution, it is sufficient. Condition 1) Since we have x=y, (4/3) – xy/(x+y) = (4/3) – x^2/2x = (4/3)-x/2. If x = y = 1, then (4/3) - x/2 = 5/6. If x = y = 2, then (4/3) - x/2 = 1/3. Since condition 1) does not yield a unique solution, it is not sufficient. Therefore, B is the answer. Answer: B _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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12 Nov 2019, 18:37
[GMAT math practice question]
(Equation) 2ax - 3b = a - bx is an equation in terms of x. What is its solution?
1) -3/2 is a solution of (b-a)x - (2a-3b) = 0
2) a = 3b
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
The question asks the value of (a+3b)/(2a+b) for the following reason.
2ax - 3b = a - bx
=> 2ax + bx = a + 3b
=> x(2a+b) = a+3b
=> x = (a+3b)/(2a+b)
Since we have a = 3b from condition 2), we have x = (a+3b)/(2a+b) = (3b+3b)/(6b+b) = (6b)/(7b) = 6/7.
Thus, condition 2) is sufficient.
Condition 1)
When we substitute -3/2 for x, we have (b-a)(-3/2) - (2a-3b) = 0 or (-3)(b-a) = 2(2a-3b). We have -3b+3a = 4a-6b or a = 3b.
Condition 1) is equivalent to condition 2), and it is also sufficient.
When a question asks for a ratio, if one condition includes a ratio and the other condition just gives a number, the condition including the ratio is most likely to be sufficient. This tells us that D is most likely to be the answer to this question, since each condition includes a ratio.
Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).
This question is a CMT4(B) question: condition 2) is easy to work with, and condition 1) is difficult to work with. For CMT4(B) questions, D is most likely to be the answer.
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8769 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The Ultimate Q51 Guide [Expert Level] [#permalink] Show Tags 13 Nov 2019, 17:41 [GMAT math practice question] The figure shows that l is parallel to m, ABCD is a square, and the point D is on l and B is on m. What is ∠y - ∠x? A. 15° B. 25° C. 30° D. 35° E. 40° Attachment: 11.4.png [ 20.8 KiB | Viewed 455 times ] => When we draw an additional line k on the figure as follows, Attachment: 11.4ps(a).png [ 37.31 KiB | Viewed 455 times ] Since ∠BCD = 90°, we have ∠x + 8∠x = 90° and ∠x = 10°. Since ∠x + ∠y = 45°, we have ∠y = 35°. Thus ∠y - ∠x = 25°. Therefore, B is the answer. Answer: B _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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14 Nov 2019, 17:41
[GMAT math practice question]
(geometry) The figure shows that the lines AB and AC are the tangential lines to the circle O. What is ∠ABC?
Attachment:
11.5ps.png [ 11.72 KiB | Viewed 443 times ]
A. 55
B. 60
C. 65
D. 70
E. 75
=>
Since AB and AC are tangent lines to the circle, we have AB = AC.
Since the triangle ABC is isosceles with AB = AC, we have ∠ABC = (1/2)(180-∠BAC) = 70.
_________________
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8769 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The Ultimate Q51 Guide [Expert Level] [#permalink] Show Tags 17 Nov 2019, 17:43 [GMAT math practice question] (number properties) p and q are positive integers and relative primes. Is p divisible by 1979? 1) p is a multiple of 1979. 2) p/q = 1 - (1/2) + (1/3) - (1/4) +…- (1/1318) + (1/1319). => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. The question “is p divisible by 1979” is equivalent to condition 1) “p is a multiple of 1979”. Condition 2) Remember that -1/2k = 1/2k – 1/k for k = 1, 2, 3, …, 659. -1/2 = 1/2 – 1/1 -1/4 = 1/4 – 1/2 -1/6 = 1/6 – 1/3 -1/1318 = 1/1318 – 1/659 p/q = 1 + 1/2 + 1/3 + 1/4 + … + 1/1317 + 1/1318 + 1/1319 – 2(1/2 + 1/4 + … + 1/1318) = 1 + 1/2 + 1/3 + 1/4 + … + 1/1317 + 1/1318 + 1/1319 – (1/1 + 1/2 + … + 1/659) = 1/660 + 1/661 + … + 1/1318 + 1/1319 = (1/660 + 1/1319) + (1/661 + 1/1318) + … + (1/989 + 1/990) = 1979/(660*1319) + 1979/(661*1318) + … + 1979/(989*990) = (1979*k)/(660*661*…*1318*1319) Then, we have p(660*661*…*1318*1319) = q(1979*k). Since 1979 is a prime number, p is a multiple of 1979. Therefore, D is the answer. Answer: D This question is a CMT4 (B) question: condition 1) is easy to work with, and condition 2) is difficult to work with. For CMT4 (B) questions, D is most likely to be the answer. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink] 17 Nov 2019, 17:43
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The Ultimate Q51 Guide [Expert Level]
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How do I find the value of x in geometry?
Orval Ambler asked, updated on July 2nd, 2021; Topic: how to find the value of x
👁 556 👍 33 ★★★★☆4
Since the value of all angles within a triangle must equal 180 degrees, if you know at least two angles, you can subtract them from 180 to find the missing third angle. If you are working with equilateral triangles, divide 180 by three to find the value of X.
Beside that, how do you find the value of X outside a circle?
Even in the case, how do you find the value of the variable in a circle?
Well, how do you find the value of x in chords and arcs?
How do u solve for x?
Isolate x by dividing each term by the x coefficient. Just divide 3x and 9 by 3, the x term coefficient, to solve for x. 3x/3 = x and 3/3 = 1, so you're left with x = 1.
How do you solve an inscribed angle?
By the inscribed angle theorem, the measure of an inscribed angle is half the measure of the intercepted arc. The measure of the central angle ∠POR of the intercepted arc ⌢PR is 90°. Therefore, m∠PQR=12m∠POR =12(90°) =45°.
circumference
How do you do the circle theorem?
Circle theorems: where do they come from?
• The angle at the centre is twice the angle at the circumference.
• The angle in a semicircle is a right angle.
• Angles in the same segment are equal.
• Opposite angles in a cyclic quadrilateral sum to 180°
• The angle between the chord and the tangent is equal to the angle in the alternate segment.
How do you find the measure of an arc?
A circle is 360° all the way around; therefore, if you divide an arc's degree measure by 360°, you find the fraction of the circle's circumference that the arc makes up. Then, if you multiply the length all the way around the circle (the circle's circumference) by that fraction, you get the length along the arc.
Are all chords in a circle congruent?
If two chords of a circle are congruent, then they determine central angles which are equal in measure. If two chords of a circle are congruent, then their intercepted arcs are congruent. Two congruent chords in a circle are equal in distance from the center.
How do you find the congruence of a circle?
Two circles are congruent if they have the same size. The size can be measured as the radius, diameter or circumference. They can overlap.
|
# Key Concepts
#### 1.1 Use the Language of Algebra
• Divisibility Tests
A number is divisible by:
2 if the last digit is 0, 2, 4, 6, or 8.
3 if the sum of the digits is divisible by 3.
5 if the last digit is 5 or 0.
6 if it is divisible by both 2 and 3.
10 if it ends with 0.
• How to find the prime factorization of a composite number.
1. Step 1. Find two factors whose product is the given number, and use these numbers to create two branches.
2. Step 2. If a factor is prime, that branch is complete. Circle the prime, like a bud on the tree.
3. Step 3. If a factor is not prime, write it as the product of two factors and continue the process.
4. Step 4. Write the composite number as the product of all the circled primes.
• How To Find the least common multiple using the prime factors method.
1. Step 1. Write each number as a product of primes.
2. Step 2. List the primes of each number. Match primes vertically when possible.
3. Step 3. Bring down the columns.
4. Step 4. Multiply the factors.
• Equality Symbol
a=ba=b is read “a is equal to b.”
The symbol “=” is called the equal sign.
• Inequality
• Inequality Symbols
• G
• anan means multiply a by itself, n times.
The expression anan is read a to the nthnth power.
• Simplify an Expression
To simplify an expression, do all operations in the expression.
• How to use the order of operations.
1. Step 1. Parentheses and Other Grouping Symbols
• Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first.
2. Step 2. Exponents
• Simplify all expressions with exponents.
3. Step 3. Multiplication and Division
• Perform all multiplication and division in order from left to right. These operations have equal priority.
4. Step 4. Addition and Subtraction
• Perform all addition and subtraction in order from left to right. These operations have equal priority.
• How to combine like terms.
1. Step 1. Identify like terms.
2. Step 2. Rearrange the expression so like terms are together.
3. Step 3. Add or subtract the coefficients and keep the same variable for each group of like terms.
• Figure 1.9
1.5 Properties of Real Numbers |
# Pre-calculus
prove sin(x+y)+sin(x-y)/cos(x+y)+cos(x-y)=tanx
1. 👍 0
2. 👎 0
3. 👁 774
1. Use trigonometric identities:
sin ( x + y ) = sin x cos y + cos x sin y
sin ( x - y ) = sin x cos y - cos x sin y
cos ( x + y ) = cos x cos y - sin x sin y
cos ( x - y ) = sin x sin y + cos x cos y
sin ( x + y ) + sin ( x - y ) =
sin x cos y + cos x sin y + sin x cos y - cos x sin y =
sin x cos y + sin x cos y + cos x sin y - cos x sin y =
2 sin x cos y
cos ( x + y ) + cos ( x - y ) =
cos x cos y - sin x sin y + sin x sin y + cos x cos y =
cos x cos y + cos x cos y - sin x sin y + sin x sin y =
2 cos x cos y
[ sin ( x + y ) + sin ( x - y ) ] / [ cos ( x + y ) + cos ( x - y ) ] =
2 sin x cos y / 2 cos x cos y =
sin x / cos x = tan x
1. 👍 0
2. 👎 0
2. Watch those brackets, you must have meant:
(sin(x+y)+sin(x-y))/(cos(x+y)+cos(x-y))=tanx
LS =
(sinxcosy + cosxsiny + sinxcosy - sinxcosy)/(cosxcosy - sinxsiny + cosxcosy + sinxsiny)
= 2sinxcosy/(2cosxcosy)
= (sinx/cosx)(cosy/cosy)
= tanx
= RS
1. 👍 1
2. 👎 1
## Similar Questions
1. ### Trig
Find sin(s+t) and (s-t) if cos(s)= 1/5 and sin(t) = 3/5 and s and t are in quadrant 1. =Sin(s)cos(t) + Cos(s)Sin(t) =Sin(1/5)Cos(3/5) + Cos(-1/5)Sin(3/5) = 0.389418 Sin(s-t) =sin(s)cos(t) - cos(s)sin(t) =sin(-3/5)cos(1/5) -
2. ### calculus
Find the points on the curve y= (cos x)/(2 + sin x) at which the tangent is horizontal. I am not sure, but would I find the derivative first: y'= [(2 + sin x)(-sin x) - (cos x)(cos x)]/(2 + sin x)^2 But then I don't know what to
3. ### Pre-calculus
Prove the following identities. 1. 1+cosx/1-cosx = secx + 1/secx -1 2. (tanx + cotx)^2=sec^2x csc^2x 3. cos(x+y) cos(x-y)= cos^2x - sin^2y
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1. ### solving trig. equations
tan(3x) + 1 = sec(3x) Thanks, pretend 3x equals x so tanx + 1 = secx we know the law that 1 + tanx = secx so tanx + 1 becomes secx and... secx = secx sec(3x) = sec(3x) [just put 3x back in for x- you don't really have to change 3x
1.)Find dy/dx when y= Ln (sinh 2x) my answer >> 2coth 2x. 2.)Find dy/dx when sinh 3y=cos 2x A.-2 sin 2x B.-2 sin 2x / sinh 3y C.-2/3tan (2x/3y) D.-2sin2x / 3 cosh 3yz...>> my answer. 2).Find the derivative of y=cos(x^2) with
3. ### Precalculus
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4. ### Trigonometry
Solve the equation for solutions in the interval 0
1. ### trig
The expression 4 sin x cos x is equivalent to which of the following? (Note: sin (x+y) = sin x cos y + cos x sin y) F. 2 sin 2x G. 2 cos 2x H. 2 sin 4x J. 8 sin 2x K. 8 cos 2x Can someone please explain how to do this problem to
2. ### Trigonometry
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3. ### calc
Where do I start to prove this identity: sinx/cosx= 1-cos2x/sin2x please help!! Hint: Fractions are evil. Get rid of them. Well, cos2x = cos2x - sin2x, so 1-coscx = 1 - cos2x - sin2x = 1 - cos2x + sin2x You should be able to
4. ### math
Can you please check my work. A particle is moving with the given data. Find the position of the particle. a(t) = cos(t) + sin(t) s(0) = 2 v(0) = 6 a(t) = cos(t) + sin(t) v(t) = sin(t) - cos(t) + C s(t) = -cos(t) - sin(t) + Cx + D |
# What Is 7/26 as a Decimal + Solution With Free Steps
The fraction 7/26 as a decimal is equal to 0.269.
Division is one of the four fundamental operations of arithmetic and is the opposite of multiplication. Therefore, if p and q are two numbers, then rather than finding p groups of q, it finds p parts of q. Division can result in either an integer or decimal value.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 7/26.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 7
Divisor = 26
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 7 $\div$ 26
This is when we go through the Long Division solution to our problem.
Figure 1
## 7/26 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 7 and 26, we can see how 7 is Smaller than 26, and to solve this division, we require that 7 be Bigger than 26.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 7, which after getting multiplied by 10 becomes 70.
We take this 70 and divide it by 26; this can be done as follows:
 70 $\div$ 26 $\approx$ 2
Where:
26 x 2 = 52
This will lead to the generation of a Remainder equal to 70 – 52 = 18. Now this means we have to repeat the process by Converting the 18 into 180 and solving for that:
180 $\div$ 26 $\approx$ 6Â
Where:
26 x 6 = 156
This, therefore, produces another Remainder which is equal to 180 – 156 = 24. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 240.
240 $\div$ 26 $\approx$ 9Â
Where:
26 x 9 = 234
Finally, we have a Quotient generated after combining the three pieces of it as 0.269, with a Remainder equal to 6.
Images/mathematical drawings are created with GeoGebra. |
# What is the rule used in this given pattern 3 6 9 12 15?
What is the pattern rule for 3 6 9 12 15
This is an arithmetic sequence since there is a common difference between each term. In this case, adding 3 to the previous term in the sequence gives the next term.
What is the pattern of 3 3 6 9 15
The pattern tells us that each number is the previous number plus the one before it. Or a cleaner way to say it is, each number is the sum of the previous two numbers. Therefore the next number is 15 + 24 = 39.
What is the answer of 3 6 9 12 15
3,6,9,12,15,18,21,24,27,30,33…
What is the general rule for the nth term of 3 6 9 12
a = 3 and d = 3 where a is first term of an AP and d is common difference of an AP. ⇒ an = 3n. Hence, nth term of the sequence, 3,6,9,12… is an = 3n.
What is the sixth term in the pattern 3 6 9 15
So at each iteration we alternately add or subtract a number that increases by 3. So the next delta should be +12, and the sixth term is 21.
What is the rule for 3 5 9 15 pattern
In the first two stages, The series follows the rule of adding the first two numbers and adding 1 in the result. After then The series follows the rule of adding the first two numbers and subtracting 1 in the result. So, complete the series with this rule. And you will see 3, 5, 9, 15, 23, 37.
What is the pattern rule for 3 6 10 15
triangular numbers: 1, 3, 6, 10, 15, … (these numbers can be represented as a triangle of dots). The term to term rule for the triangle numbers is to add one more each time: 1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10 etc. Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, …
What is the pattern of 3 6 10 15
The triangular number sequence is the representation of the numbers in the form of equilateral triangle arranged in a series or sequence. These numbers are in a sequence of 1, 3, 6, 10, 15, 21, 28, 36, 45, and so on. The numbers in the triangular pattern are represented by dots.
What is the sequence of 3 6 9 15 24 39
Detailed Solution
The pattern followed here is; The given series is an example of the Fibonacci series. Hence, "102" is the correct answer.
What are the missing terms in multiple of 3 3 6 9 _ 15
⇒Multiples of three include: 3, 6, 9, 12, 15, 18, 21, 24, 27, and thirty. Therefore, 12 is missing numbers in the given sequence.
What is the nth term of 6 9 12 15
Let's find the nth term of the sequence, 15, 12, 9, 6, … Thus, the expression for the nth term of the sequence, 15, 12, 9, 6, … is an = 18 – 3n. We can use Cuemath's Online Arithmetic sequence calculator to find the arithmetic sequence using the first term and the common difference between the terms.
What is the 7th term of the sequence 3 6 9 12
Hence, the 7th term is 21.
What is the next number in the sequence 3 6 9 12
3,6,9,12,15,18.
What is the rule of 3 6 10 15 21 pattern
– the nth term is. triangular numbers: 1, 3, 6, 10, 15, … (these numbers can be represented as a triangle of dots). The term to term rule for the triangle numbers is to add one more each time: 1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10 etc.
What is the sequence of 3 9 5 15 11
The next number in the series 3, 9, 5, 15, 11, 33, 29, is 81.
What is the pattern rule for 1 3 6 10 15 21
These numbers are in a sequence of 1, 3, 6, 10, 15, 21, 28, 36, 45, and so on. The numbers in the triangular pattern are represented by dots. The sum of the previous number and the order of succeeding number results in the sequence of triangular numbers.
What is the pattern rule of this sequence 1 3 5 7 9 11
The sequence that is given to us is 1, 3, 5, 7, 9, … a5 – a4 = 9 – 7 = 2. Hence, from the above simplification we can see that the common difference is 2. Therefore, the general term for the sequence 1, 3, 5, 7, 9, . . . is 2n – 1.
What kind of a number pattern is this 9 12 15 18
Hence, the given sequence of numbers is a composite number pattern.
Is the sequence 3 6 9 12 15 an example of arithmetic sequence
Arithmetic Sequence Example
Consider the sequence 3, 6, 9, 12, 15, …. is an arithmetic sequence because every term is obtained by adding a constant number (3) to its previous term. Here, The first term, a = 3. The common difference, d = 6 – 3 = 9 – 6 = 12 – 9 = 15 – 12 = …
How many terms are there in the sequence 3 6 9 12 and so on 111
37 terms
Thus, the given sequence contains 37 terms.
What is the pattern 6 9 12 15
(b) 3, 6, 9, 12, 15, . . . (a) This sequence is a list of even numbers, so the next three numbers will be 12, 14, 16. (b) This sequence is made up of the multiples of 3, so the next three numbers will be 18, 21, 24.
What is the missing term in the sequence 1 3 3 6 7 9 _ 12 21
13
Given sequence: 1, 3, 3, 6, 7, 9, __, 12, 21. Hence, the correct answer is "13".
What is the pattern rule for 3 4 6 9 13
Sequence is defined as a list of numbers (or items) that exhibits a particular pattern. Given: 3, 4, 6, 9, 13, 18, 24, Thus, we see that every resulting term in the given sequence is getting added to consecutive integers. Hence, the next number is 31.
What is the rule for 1 3 6 10 15
These numbers are in a sequence of 1, 3, 6, 10, 15, 21, 28, 36, 45, and so on. The numbers in the triangular pattern are represented by dots. The sum of the previous number and the order of succeeding number results in the sequence of triangular numbers.
What is the answer to 1 3 5 7 9 11 13 15
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64. Q. Q. Find the sum of the following numbers without actually adding the numbers. |
# Position and Displacement
## Factors related to physical quantities of motion.
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Position and Displacement
Students will learn the meaning of an object's position, the difference between distance and displacement and some basic graphing of position vs. time.
### Key Equations
Symbols{Δ(anything)anything0Final value - initial valueValue at time 0\begin{align*}\text{Symbols}\begin{cases} \Delta \text{(anything)} & \text{Final value - initial value}\\ \text{anything}_0 & \text{Value at time 0} \end{cases}\end{align*}
Scalorstd=|Δx1|+|Δx2|v=|v|Time in seconds, sDistance (in meters, m)Speed (in meters per second, m/s)\begin{align*}\text{Scalors}\begin{cases} t & \text{Time in seconds, s}\\ d = |{\Delta x_1}| + |{\Delta x_2}| & \text{Distance (in meters, m}\text{)}\\ v = |{v}| & \text{Speed (in meters per second,}\ \text{m}/\text{s}\text{)} \end{cases}\end{align*}
Vectors{x=x(t)Δx=xfxiPositionDisplacement\begin{align*}\text{Vectors}\begin{cases} x = x(t) & \text{Position} \\ \Delta x = x_f-x_i & \text{Displacement}\\ \end{cases}\end{align*}
When beginning a one dimensional problem, define a positive direction. The other direction is then taken to be negative. Traditionally, "positive" is taken to mean "to the right"; however, any definition of direction used consistently throughout the problem will yield the right answer.
Guidance
Position is the location of the object (whether it's a person, a ball or a particle) at a given moment in time. Displacement is the difference in the object's position from one time to another. Distance is the total amount the object has traveled in a certain period of time. Displacement is a vector quantity (direction matters), where as distance is a scalor (only the amount matters). Distance and displacement are the same in the case where the object travels in a straight line and always moving in the same direction.
#### Example 1
Problem: An indecisive car goes 120 m North, then 30 m south then 60m North. What is the car's distance and displacement?
Solution:
Distance is the total amount traveled. Thus distance = 120 + 30 + 60 m = 210 m
Displacement is the amount displaced from the starting position. Thus displacement = 120 - 30 + 60 m = 150 m.
#### Example 2
Problem: An 8th\begin{align*}8^{\text{th}}\end{align*} grader is timed to run 24 feet in 12 seconds, what is her speed in meters per second?
Solution:
D24 ftvv=vt=v(12 s)=24 ft/12 s=2 ft/s=2 ft/s(1m/3.28 ft)=0.61 m/s\begin{align*}D & = vt\\ 24 \ ft & = v(12 \ s)\\ v & = 24 \ ft/12 \ s = 2 \ ft/s\\ v & = 2 \ ft/s \ast (1 m/3.28 \ ft) = 0.61 \ m/s\end{align*}
### Time for Practice
1. Does the odometer reading in a car measure distance or displacement?
2. Imagine a fox darting around in the woods for several hours. Can the displacement x\begin{align*}\triangle x\end{align*} of the fox from his initial position ever be larger than the total distance d\begin{align*}d\end{align*} he traveled? Explain.
3. Your brother borrowed the scissors from your room and now you want to use them. Do you care about the distance the scissors have traveled or their displacement? Explain your answer.
4. You’re trying to predict how long it’s going to take to get to Los Angeles for the long weekend. Do you care about the distance you’ll travel or your displacement? Explain your answer.
1. distance
2. No, displacement is 'as the crow flies' so to speak, while distance takes into account the curves and turns that the fox takes.
3. displacement
4. distance
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# Find the sum of the first 123 even natural numbers.
Last updated date: 11th Aug 2024
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Hint: Here, we are required to find the sum of the first 123 even natural numbers. Thus, we will find the sequence representing the first 123 even natural numbers. We will observe that this is an arithmetic progression and hence, we will use the formula of sum of first $n$ terms to find the required answer.
Formula Used:
Sum of $n$ terms of an AP, ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Where, $a$ is the first term, $d$ is the common difference and $n$represents the total number of terms in the AP
Complete step by step solution:
As we know, natural numbers are positive integers or those numbers which are greater than equal to 1 or in other words, all positive integers except 0 are natural numbers.
Now, even natural numbers are those numbers which are divisible by 2.
Hence, the sequence of even natural numbers can be written as:
$2,4,6,8,10,....$
Here, clearly this sequence is an Arithmetic Progression because each term is greater than the preceding term by 2.
Thus, the first term, $a = 2$
Common difference, $d = 2$
And, according to the question, the total number of even natural numbers to be considered in this sequence, $n = 123$
Now, in an AP, the sum of first $n$ terms is given as:
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Hence, the sum of first 123 even natural numbers is:
${S_{123}} = \dfrac{{123}}{2}\left[ {2\left( 2 \right) + \left( {123 - 1} \right)\left( 2 \right)} \right]$
$\Rightarrow {S_{123}} = \dfrac{{123}}{2}\left[ {4 + \left( {122} \right)\left( 2 \right)} \right] = \dfrac{{123}}{2}\left[ {4 + 244} \right] = \dfrac{{123}}{2} \times 248$
Hence, solving further, we get,
$\Rightarrow {S_{123}} = 123 \times 124$
$\Rightarrow {S_{123}} = 15252$
Therefore, the required sum of first 123 even natural numbers is 15252
Hence, this is the required answer.
Note:
An Arithmetic Progression or A.P. is a sequence in which the difference between two consecutive terms is the same. Arithmetic progressions are also used in real life such as adding the same amount as our pocket money in our money bank. Since, we add the same amount each time, that amount or that pocket money will become our common difference in this case. Similarly, we hire a taxi, we are charged an initial rate and then rate per kilometer. That rate per kilometer becomes our common difference and each addition gives us an A.P. Hence, this is used in our day to day life. |
# Right Prism MCQ Quiz - Objective Question with Answer for Right Prism - Download Free PDF
Last updated on Apr 12, 2024
Testbook provides Right Prism MCQ Quiz with logical and easy explanations to all the questions. Detailed solutions for all the Right Prism Objective questions have been provided so that the candidates can get the strategies and shortcuts to approach a question and solve it in less time. The Right Prism Question Answers will help the candidates understand the concept better and grasp faster making it easier for them to ace exams.
## Latest Right Prism MCQ Objective Questions
#### Right Prism Question 1:
The base of a right prism is a square. The height of the prism is 20 cm and its total surface area is 768 cm2. Find the volume of the prism?
1. 1250 cm3
2. 1180 cm3
3. 1500 cm3
4. More than one of the above
5. None of the above
#### Answer (Detailed Solution Below)
Option 5 : None of the above
#### Right Prism Question 1 Detailed Solution
Given:
Height of prism = 20 cm
The surface area of prism = 768 cm2
Concept:
Using the total surface area, first calculate the perimeter of the base and then the side of the square base. After that, calculate the volume of the prism.
Formula used:
Surface area of prism = [(Perimeter of base) × height] + 2(Area of Base)
Volume of prism = (Area of base) × Height
Area of square = side × side
Perimeter of square = 4 × side
Calculation:
Let, the side of square base = a cm
Surface area of prism = [(Perimeter of base) × height] + 2(Area of base)
⇒ 768 = (4a × 20) + 2a2
⇒ 768 = 80a + 2a2
⇒ 384 = 40a + a2
⇒ a2 + 40a – 384 = 0
⇒ a2 + 48a – 8a – 384 = 0
⇒ a(a + 48) – 8(a + 48) = 0
⇒ (a – 8)(a + 48) = 0
⇒ a – 8 = 0 or a + 48 = 0
⇒ a = 8 or a = (-48)
Side of the square can’t be negative
⇒ a = 8 cm
Volume of prism = (Area of base) × Height
= (8 × 8) × 20
= 1280 cm3
∴ Required volume of the prism is 1280 cm3.
#### Right Prism Question 2:
The volume of the right prism with an area of base 121 m2 and height 23 m is
1. 5.26 m3
2. 2783 m3
3. 529 m3
4. More than one of the above
5. None of the above
#### Answer (Detailed Solution Below)
Option 2 : 2783 m3
#### Right Prism Question 2 Detailed Solution
Given:
Area of base = 121 m3
Height = 23 m
Formula used:
Volume of Prism = area of base × height
Calculation:
⇒ Volume = area × height
⇒ Volume = 121 × 23
⇒ Volume = 2783
∴ The volume of the right prism is 2783 m3.
#### Right Prism Question 3:
The base of a prism is a right angled isoceles triangle whose hypotenuse is 3√2 cm. If the height of the prism is 12 cm, find the volume of the prism?
1. 54 cm3
2. 44 cm3
3. 64 cm3
4. 52 cm3
5. 50 cm3
#### Answer (Detailed Solution Below)
Option 1 : 54 cm3
#### Right Prism Question 3 Detailed Solution
Given:
Hypotenuse of base triangle = 3√2 cm
Height of prism = 12 cm
Formula used:
In a right angled triangle;
(Hypotenuse)2 = (Base)2 + (Height)2
Area of triangle = (1/2) × Base × Height
Volume of Prism = Area of base × Height
Calculation:
In Isoceles triangle; two sides are equal.
Let the equal sides = a cm
(3√2)2 = a2 + a2
⇒ 18 = 2a2
⇒ a2 = 9 cm2
⇒ a = 3 cm
Area of triangular base = (1/2) × 3 × 3
⇒ 9/2 cm2
Volume of Prism = 12 × (9/2) cm3
⇒ 54 cm3
∴ The volume of prism is 54 cm3.
#### Right Prism Question 4:
What is the number of faces in a triangular prism?
1. 3
2. 5
3. 6
4. 7
Option 2 : 5
#### Right Prism Question 4 Detailed Solution
Formula used:
Number of faces = Number of triangular bases + Number of lateral faces
Calculation:
According to question,
Number of faces = 2 (triangular bases) + 3 (lateral faces) = 5
∴ The correct answer is 5.
#### Right Prism Question 5:
How many edges does a triangular pyramid have ?
1. 2
2. 3
3. 6
4. 8
Option 3 : 6
#### Right Prism Question 5 Detailed Solution
Calculation:
A triangular pyramid has 6 edges
∴ The correct answer is 6
## Top Right Prism MCQ Objective Questions
#### Right Prism Question 6
The base of a right prism is an equilateral triangle whose side is 10 cm. If height of this prism is 10$$\sqrt{3}$$ cm, then what is the total surface area of prism ?
1. 125$$\sqrt{3}$$ cm2
2. 325$$\sqrt{3}$$ cm2
3. 150$$\sqrt{3}$$ cm2
4. 350$$\sqrt{3}$$ cm2
#### Answer (Detailed Solution Below)
Option 4 : 350$$\sqrt{3}$$ cm2
#### Right Prism Question 6 Detailed Solution
Given:
The base of a right prism is an equilateral triangle whose side is 10 cm.
The height of this prism is 10$$√{3}$$ cm
Concept used:
TSA of Prism = [2(area of triangular base)] + [(Base Perimeter × height)]
Area of Equilateral Triangle = (√3/4)a2
The perimeter of the Equilateral triangle = 3 × side.
Calculation:
According to the concept,
⇒ Area of Equilateral Triangle = (√3/4)(10)2 = (100/4)√3 = 253
⇒ Base Perimeter = 10 × 3 = 30
Then,
TSA = [2(253)] + [30 × 10√3]
⇒ TSA = 50√3 + 3003
⇒ TSA = 350√3
∴ The total surface area of the prism is 3503.
#### Right Prism Question 7
The base of a prism is a right angled isoceles triangle whose hypotenuse is 3√2 cm. If the height of the prism is 12 cm, find the volume of the prism?
1. 54 cm3
2. 44 cm3
3. 64 cm3
4. 52 cm3
#### Answer (Detailed Solution Below)
Option 1 : 54 cm3
#### Right Prism Question 7 Detailed Solution
Given:
Hypotenuse of base triangle = 3√2 cm
Height of prism = 12 cm
Formula used:
In a right angled triangle;
(Hypotenuse)2 = (Base)2 + (Height)2
Area of triangle = (1/2) × Base × Height
Volume of Prism = Area of base × Height
Calculation:
In Isoceles triangle; two sides are equal.
Let the equal sides = a cm
(3√2)2 = a2 + a2
⇒ 18 = 2a2
⇒ a2 = 9 cm2
⇒ a = 3 cm
Area of triangular base = (1/2) × 3 × 3
⇒ 9/2 cm2
Volume of Prism = 12 × (9/2) cm3
⇒ 54 cm3
∴ The volume of prism is 54 cm3.
#### Right Prism Question 8
A prism has a regular hexagonal base with side 8 cm and the total surface area of the prism is 912√3 cm2, then what is the height of the prism?
1. 13√6 cm
2. 15√6 cm
3. 13√3 cm
4. 15√3 cm
#### Answer (Detailed Solution Below)
Option 4 : 15√3 cm
#### Right Prism Question 8 Detailed Solution
Given:
The side of the prism = 8cm
The total surface area of the prism = 912√3 cm2
Formula used:
Area of the regular hexagon = 3√3/2 × side2
The total surface area of the prism = 2 × Area of the base + 6 × (Side of the base) × (Height of prism)
Calculation:
Let the height of the prism be h cm
Area of the regular hexagon = 3√3/2 × side2
⇒ 3√3/2 × 82
⇒ 96√3 cm2
The total surface area of the prism = 2 × Area of the base + 6 × (Side of the base) × (Height of prism)
⇒ 2 × 96√3 + 6 × 8 × h = 912√3
⇒ 6 × 8 × h = 720√3
⇒ h = 15√3
∴ The height of the prism is 15√3 cm
#### Right Prism Question 9
The base of a right prism is a regular hexagon of a side 5 cm. If its height is 12√3 cm, then its volume (in cm3) is:
1. 1800
2. 900
3. 1350
4. 670
Option 3 : 1350
#### Right Prism Question 9 Detailed Solution
Given :-
The base of a right prism is a regular hexagon of side 5 cm
height is 12 √3 cm
Concept :-
Prism is a part of cylinder so,
Volume of prism = Base × Height
As base of prism is shan so
Volume of prism with base hexagonal = base area × height
Base area = Area of hexagonal is equal to area of 6 equilateral triangle = 6 × (√3/4) × side2
Calculation :-
⇒ Base area = 6 × (√3/4) × 52
⇒ Base area = 150 × (√3/4)
⇒ Volume = 150 × (√3/4) × 12√3
⇒ Volume = (1800 × 3)/4
⇒ Volume = 1350 cm3
∴ Volume = 1350 cm3
#### Right Prism Question 10
The lateral surface area of a right triangular prism is 288 cm2. If the lengths of the smaller bases are 6 cm and 8 cm respectively, find the height of the prism.
1. 14 cm
2. 12 cm
3. 24 cm
4. 22 cm
Option 2 : 12 cm
#### Right Prism Question 10 Detailed Solution
Length of the larger base = √(62 + 82) = 10 cm (Using Pythagoras theorem)
Let the length of the height of the prism be X cm.
Perimeter of the given prism = 6 cm + 8 cm + 10 cm = 24 cm
Area of the right triangular prism = Perimeter × Height
⇒ 24 × X = 288
⇒ X = 12 cm
#### Right Prism Question 11
The height of a right triangular prism is 21 cm and the ratio of the sides of its base is 8 : 15 : 17. If the total area of the three lateral surfaces is 840 cm2, then what is the volume of the prism in cubic centimetre?
1. 1300
2. 1200
3. 1269
4. 1260
Option 4 : 1260
#### Right Prism Question 11 Detailed Solution
Given,
Height of the triangular prism is h = 21 cm
The ratio of sides of triangular prism is = 8x : 15x : 17x
The total area of three lateral surfaces is = 840 cm2
Perimeter of the triangle × height = 840
⇒ (8x + 15x + 17x) × 21 = 840
⇒ 40x = 840/21
⇒ x = 40/40
⇒ x = 1
Sides of the triangular prism are 8, 15 and 17.
As we know,
⇒ s = (a + b + c)/2
⇒ s = (8 + 15 + 17)/2
⇒ s = 40/2 = 20
Area of the triangle = √[s (s – a) (s – b) (s – c)]
Area of the triangle = √[20 × (20 – 8) (20 – 15) (20 – 17)]
Area of the triangle = √[20 × 12 × 5 × 3]
Area of the triangle = 60
As we know,
Volume of the prime = Area of Base × height
Volume of triangular prism = 60 × 21 = 1260 cm3
#### Right Prism Question 12
The base of a right prism is a triangle whose sides are 8 cm, 15 cm and 17 cm, and its lateral surface area is 480 cm2. What is the volume (in cm3) of the prism?
1. 540
2. 600
3. 720
4. 640
Option 3 : 720
#### Right Prism Question 12 Detailed Solution
Given:
The base of a right prism is a triangle whose sides are 8 cm, 15 cm and 17 cm
The lateral surface area of right prism 480 cm2
Formula Used:
Lateral surface area of prism = perimeter of base × height
Volume of prism = Area of base × height
Calculation:
Since base side length is 8, 15, 17 [which is a triplet]
It means the base is a right angled triangle
Perimeter of triangle = 8 + 15 + 17 = 40 cm
Let the height of the prime is h cm
So, 40 × h = 480
⇒ h = 12 cm
Now, Area of base = (1/2) × 8 × 15 = 60 cm2
Volume of prism = area × height
∴ Volume of prism = 60 × 12 = 720 cm3.
#### Right Prism Question 13
Find the altitude of a right prism for which the area of the lateral surface is 143 sq.cm and the perimeter of the base is 13 cm.
1. 22 cm
2. 11 cm
3. 16 cm
4. 15 cm
Option 2 : 11 cm
#### Right Prism Question 13 Detailed Solution
Given:
Area of the lateral surface of Prism(L.S.A) = 143 sq.cm
Perimeter of the base, P = 13 cm
Concept:
Lateral surface area of a prism is the product of its base perimeter and altitude
Formula used:
Lateral surface area of Prism = Perimeter of base × Altitude (L.S.A = P × h)
Calculation:
L.S.A = P × h
⇒ h = L.S.A/P = 143/13 = 11 cm
∴ Altitude of the Right Prism = 11 cm
#### Right Prism Question 14
The base of a right prism is a square having side of 15 cm. If its height is 8 cm, then the total surface area.
1. 920 cm2
2. 900 cm2
3. 940 cm2
4. 930 cm2
#### Answer (Detailed Solution Below)
Option 4 : 930 cm2
#### Right Prism Question 14 Detailed Solution
Shortcut TrickIn general a prism having a square or rectangular base is called a cuboid.
Now, we have a cuboid of l = 15, b = 15, h = 8
Total surface area = 2 (lb + bh + hl)
⇒ Total surface area = 2 (15 × 15 + 15 × 8 + 8 × 15)
⇒ Total surface area = 2 (465) = 930
∴ The total surface area is 930 cm2
Alternate Method
Given:
The base of a right prism is a square having side of 15 cm.
Height of the prism = 8 cm
Formula used:
Total surface area of a prism = (2 × Area of a base) + (Perimeter of a base × Height of a prism)
Calculations:
Area of a base of the prism = (side)2
⇒ (15)2
⇒ 225 cm2
Perimeter of a base of the prism = 4 × (side)
⇒ 4 × (15)
⇒ 60 cm
Now,
Total surface area of a prism = (2 × Area of a base) + (Perimeter of a base × Height of a prism)
⇒ ( 2 × 225) + (60 × 8)
⇒ 450 + 480
⇒ 930 cm2
∴ The total surface area of the prism is 930 cm2.
#### Right Prism Question 15
The base of a right prism is a triangle with sides 16 cm, 30 cm and 34 cm. Its height is 32 cm. The lateral surface area (in cm2) and the volume (in cm3) are, respectively:
1. 2624 and 7040
2. 2560 and 7680
3. 2688 and 7680
4. 2560 and 6400
#### Answer (Detailed Solution Below)
Option 2 : 2560 and 7680
#### Right Prism Question 15 Detailed Solution
Given:
The base of a right prism is a triangle with sides = 16 cm, 30 cm and 34 cm.
Height = 32 cm
Formula used:
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
Volume of Prism = Area of base × height = ($$\frac{1}{{2}}$$ × b × h) × l
The lateral surface area of prism = Perimeter of base × height
Calculation:
According to the question
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
⇒ (34)2 = (16)2 + (30)2
⇒ 1156 = (256 + 900)
⇒ 1156 = 1156
Now, we can say triangle is a right angle triangle
As we know,
Volume of Prism = $$\frac{1}{{2}}$$ × b × h × l
⇒ $$\frac{1}{{2}}$$ × 16 × 30 × 32 cm2
⇒ (8 × 30 × 32) cm3
⇒ 7680 cm3
Now,
The lateral surface area of prism = Perimeter of base × height
⇒ (16 + 30 + 34) × 32 cm2
⇒ (80 × 32) cm2
⇒ 2560 cm2
∴ The required value is 2560 and 7680. |
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1 Graphing Module II UCCS Physics Labs Table of Contents Error bars 2 Equation of Straight line 3 Best-fit line with error bars 7 Plotting an equation 7 FYI FYI On average, 100 people choke to death on ball-point pens every year. Graphing Module II - 1
2 Error in Graphs - Error bars We have already talked about how to record an error inherent with any measurement. Now the question is how do we graphically represent the error? The point on a graph is the data itself. We also need a way to graphically show the uncertainty in that data point. The solution is error bars. Error bars are the lines that extend from the data point a distance equal to the value of the uncertainty. A data point is made up of two parts, the x coordinate and y coordinates. Both coordinates can contain uncertainty. Therefore, error bars can extend away from the data point in both directions. Any point within the error bars is a valid value for the data point. Below is a sample set of data with error and the resulting graph. Close-up of error bars extending from a data point. Only an example. Each situation will be different. Graphing Module II - 2
3 The Equation of a Straight Line At this point all you know is how to plot data points. For a graph to become truly useful we need to find the function that interrelates the data points. The simplest functional relationship is the linear relationship. A linear relationship will produce data that lie along a straight line. The data points will most likely not lie perfectly along a straight line because of error in the measurements. We will need to make a line that best-fits the data, which is called a best-fit line, surprise! Let s try an example now. Examine the following graph: y These two lines are other possible best-fit lines for this data. They don t appear to do as good of a job representing all the data as the solid black line. This line divides the data points evenly. The line also gives the general trend the data is following. data points 0 x Drawing a best-fit line is an art form that you will quickly learn. Just remember that the best-fit line should be as close as possible to all the data points. Graphing Module II - 3
4 Equation of the best-fit line Once we have a best-fit line, we need to know how to represent it mathematically. A line is a functional relationship between the two variables plotted on the vertical and horizontal axis. The equation of a line is as follows: y = m x + b y the variable located on the vertical axis. m the slope of the line. (Explained below, wait for it!) x the variable located on the horizontal axis. b y-intercept. (Also explained below, try to be more patient!) To find the equation of any line is to find the value of the slope and the y-intercept. The slope (m) represents how rapidly the line will rise or fall, the larger the value for the slope the steeper the line. If the value of the slope is positive, then the line will be increasing in y as x is increasing. If the slope is negative, the line will be decreasing in y as x is increasing. If the slope is zero the line will not increase or decrease but will stay at a constant value of y no matter the value of x. The y-intercept (b) is the value of the vertical axis where the line intersects that axis. If the line does not reach the vertical axis, then simply extend the line until it does and record the value. The slope is defined by the ratio rise over the run, which is how many vertical units the line will rise or fall divided by the number of units the line will run in the horizontal direction. A more mathematical way of looking at the slope is: The change in the y coordinate ( y) divided by the change in the x coordinate ( x) as you travel along the line. Graphing Module II - 4
5 Let s try calculating the slope of a straight line: To find the rise or the y, just take the difference between the two y coordinates: rise = 3.8 m m = 2.3 m To find the run or the x, just do the same difference calculation for the x coordinates: run = 4.3 s - 1 s = 3.3 s Now to find the slope divide these two numbers: "rise" 2.3 m slope = = = 0.7 m "run" 3.3s s Graphing Module II - 5
6 Some tips and tricks for calculating the slope Try to spread out the two points you are using to calculate the slope over the length of the line. This will give a more accurate slope calculation. Remember you are measuring the slope of the line, so use the x and y coordinates of the line. If it is possible use the coordinates of data points that lie on the line. This will reduce any extra error gained in reading coordinates off the line because you already have the exact coordinates from the data. We now have a value for the slope of our line. All we need to complete the equation of this line is to find the value of the y-intercept (b). This should be a simple task. The line crosses the y-axis at a value of about 0.8 m, so b = 0.8 m. The equation of a line is: y = m x + b Filling in the values for the slope and the y-intercept from our example gives: y = (0.7 m ) x m That s it! s Graphing Module II - 6
7 Find the best-fit line with error bars If your data contains a known error, finding the best-fit line isn t any different. You still have to draw a line so that it is as close as possible to all the data points, or at least the error bars. Error bars represent all the possible values for that data point, so you have a little bigger target for the best-fit line. Example: The line comes as close as possible to all the data points and/or error bars. Plotting an equation In many cases, once you have the equation of the best-fit line, you will need something for comparison. What we would like to compare our best-fit line to is the expected result (theoretical function). For an easier comparison we need to plot this function on the same graph as our data. What we have at this point is the theoretical function, but no data, and therefore no way to plot a line. To generate data, assign a value to one of the variables and then get out your calculator and algebra book and solve for the remaining variable. Repeat this process until you have enough data points to draw a line connecting the data points. Graphing Module II - 7
8 Let s try graphing this theoretical equation: y = 2 x + 4 Let s assign values to x and solve for y. This will limit the amount of math steps needed to get the data. You can assign y values and solve for x. This will give the same results, but it will just be more work. assign x calculate y 0 y = 2(0) + 4 = 4 1 y = 2(1) + 4 = 6 2 y = 2(2) + 4 = 8 4 y = 2(4) + 4 = y = 2(10.2) + 4 = 24.4 You can assign any value you want to x, but integers seem to make the math easier! y x Your best-fit line should lie exactly over the data points. Remember, it was an equation of a straight line you were graphing in the first place! Graphing Module II - 8
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# Q. As a project,the fifth grade students at Scott Elementary School collected newspapers to recycle. During the first week,they collected 212 pounds of newspapers. For each of the next four weeks,they doubled the previous week's amount of newspaper collected. What 6th was the total number of pounds of newspapers collected during the five weeks?
A. 212 X 4 = 848 X 5 = 4,240
That was the total number of pounds collected,am I right?
## Online, “*” is used to indicate multiplication to avoid confusion with “x” as an unknown.
212 + 212*2 + 212*4 + 212*8 + 212*16 = ?
212 + 424 + 848 + 1696 + 3392 = ?
## To find the total number of pounds of newspapers collected during the five weeks, you need to add up the pounds collected each week.
First, they collected 212 pounds in the first week.
In the second week, they doubled the previous week's amount, so you need to multiply 212 by 2:
212 x 2 = 424 pounds
In the third week, they doubled the previous week's amount again, so you need to multiply 424 by 2:
424 x 2 = 848 pounds
In the fourth week, they doubled the previous week's amount again, so you need to multiply 848 by 2:
848 x 2 = 1696 pounds
In the fifth week, they doubled the previous week's amount one more time, so you need to multiply 1696 by 2:
1696 x 2 = 3392 pounds
To find the total number of pounds collected during the five weeks, you need to add up the weights of each week:
212 + 424 + 848 + 1696 + 3392 = 6572 pounds
Therefore, the total number of pounds of newspapers collected during the five weeks is 6572 pounds. |
# Group Homomorphisms Map the Identity to the Identity (Proof) | Summary and Q&A
1.1K views
February 20, 2023
by
The Math Sorcerer
Group Homomorphisms Map the Identity to the Identity (Proof)
## TL;DR
This video provides a proof that in a group homomorphism, the identity element is mapped to the identity element.
## Key Insights
• 🛀 The proof shows that a group homomorphism maps the identity element in G to the identity element in H.
• 👥 Basic group theory concepts are utilized in the proof.
• 👥 The proof highlights the importance of preserving the group structure in group homomorphisms.
• 👥 Inverse elements and the closure property of groups are crucial in the proof.
• ❓ Demonstrating all the steps in the proof provides a thorough understanding of each calculation.
• 💦 The video emphasizes the significance of writing down the reasoning and justifications in words during working on mathematical proofs.
• 🤩 Group homomorphisms exhibit a key property of mapping the identity element to the identity element.
## Transcript
hello in this video we're going to do a proof we have a function f from g into H and we're told it is a group homomorphism and we're going to prove this equation here F of e sub G is equal to e sub H let me briefly explain this so by a group homomorphism we mean a map that has the property that's f of x y is equal to f of x times F of Y and this ha... Read More
### Q: What is a group homomorphism?
A group homomorphism is a function that preserves the group structure, meaning that the result of applying the function to the product of two elements is equal to the product of their images under the function.
### Q: Why do we write F of EG as e sub G times e sub G?
To utilize the property of the identity element in G, which states that any element multiplied by the identity element results in the same element. Therefore, writing F of EG as e sub G times e sub G allows us to use this property in the proof.
### Q: What is the significance of multiplying both sides by the inverse of F of EG?
The inverse exists because H is a group, and multiplying both sides by the inverse is a way to isolate F of EG on one side of the equation, allowing us to manipulate the elements and simplify the expression.
### Q: Why is it important to show all the steps in the proof?
Showing all the steps helps understand the logic and reasoning behind each calculation. It ensures clarity and eliminates any confusion that may arise from skipping steps.
## Summary & Key Takeaways
• The video presents a proof for the equation: F of e sub G is equal to e sub H, where F is a group homomorphism, e sub G is the identity element in group G, and e sub H is the identity element in group H.
• The proof utilizes basic group theory concepts and shows step-by-step calculations.
• It concludes that group homomorphisms map the identity element to the identity element, establishing a key property of these mappings. |
# Position in One Dimension
In order to begin our study of motion, we must first consider the tools that we will be using to describe the objects that we are studying.
### Describing Objects on an Axes
To more easily describe the motion of objects that we are studying, we simplfiy the object by imagining that the object - regardless of its size or shape - is a tiny particle that we treat as a single point.
For describing the position of a particle in one dimension, we imagine that the particle is sitting in a coordinate system, where the position of the particle can be anywhere along a line that we call an axis, with respect to an arbitrary point that we call the origin.
The origin is considered the zero point, and the number of units from the origin to the object on the axis is used to represent the position of that point on the coordinate system.
### Coordinate Systems
In physics, the Cartesian coordinate system is used to specify the position of objects in space at a given time. The x-axis typically represents time, while the y-axis represents position with respect to the origin. At any point in time along the x-axis, there is a corresponding point on the y-axis that represents the distance of that object from the origin.
The following graph represents an object moving away from the origin as time increases. At time 5, we see that position from the origin is 5, and at time 10, the position from the origin is 10.
### Graphing the Position of an Object
The following chart on the left describes the motion of the object on the right. Between 0 - 4 time units, the object is moving away from the origin until it reaches a posiiton of 4 units to the right. Between 4 - 8 time units, the object moves towards the origin, and we see the graph slopes downwards until it reaches position 0.
### The Position Function
Tying this all together, we can describe the position of a particle in one dimension at a given time using a single function - called the position function - that we represent with the symbol $$p(t)$$, where $$t$$ refers to the time, and $$p$$ refers to the position of the particle relative to the origin at the given time.
$$p(t) = position$$
Looking at the above chart, we can come up with an equation that can be used to describe the motion of the object.
Between time 0 - 4, we can use the following equation:
$$p(t) = t$$
Between time 4 - 8, we can use the following equation:
$$p(t) = 8 - t$$
We can validate this ourselves by substituting $$t$$ into any of the above equations, using the correct function for a given time, and see that we get the correct position on the graph.
If we know what $$p(t)$$ is as a function, then we know everything we need to know about the motion of the particle, as we will see in later sections. |
In this page area of quadrilateral we are going to see the formula to be used to find the area of the quadrilateral.To find the area we are given four vertices.We need to consider the given vertices as
(x1,y1), (x2,y2), (x3,y3),(x4,y4)
The formula to be used to find the area is
Required area = 12 {x1y2+x2y3+x3y4+x4y1)-(x2y1+x3y2+x4y3+x1y4)}
We need to apply the the appropriate values in the formula to get the area.Now let us see the example problems to understand this topic more clear
Example 1:
Find the area formed by the points (-3,-9) (-1,6) (3,9) and (5,-8).
Now we have
x1 = -3 , y1 = -9 , x2 = -1 , y2 = 6 , x3 = 3 , y3 = 9 , x4 = 5 , y4 = -8
= 1/2 {(-3(6)+(-1)9+3(-8)+5(-9))-((-1)(-9)+3(6)+5(9)+(-3)(-8))}
= 1/2 {(-18-9-24-45) - (9+18+45+24)}
= 1/2 {-96 -96}
= 1/2 {-192}
= -192/2
= -96
Area cannot be negative. So the answer is 96 Square Units.
Example 2:
Find the area formed by the points (5,8) (6,3) (3,1) and (2,6)
Now we have
x1 = 5 , y1 = 8 , x2 = 6 , y2 = 3 , x3 = 3 , y3 = 1 , x4 = 2 , y4 = 6
= 1/2 {(5(3)+6(1)+3(6)+2(8))-(6(8)+3(3)+2(1)+5(6)}
= 1/2 {(15+6+18+16) - (48+9+2+30)}
= 1/2 {55 -89}
= 1/2 {-34}
= -34/2
= -17
Area cannot be negative. So the answer is 17 Square units.
Related Topics |
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3
# NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3
## NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 9 Areas of Parallelograms and Triangles Ex 9.3.
Ex 9.3 Class 9 Maths Question 1.
In the given figure, E is any point on median AD of a ΔABC. Show that
ar (ΔABE) = ar (ΔACE).
Solution.
Given: AD is a median of ΔABC and E is any point on AD.
To prove: ar (ΔABE) = ar (ΔACE)
Proof: Since AD is the median of ΔABC.
ar (ΔABD) = ar (ΔACD) …..…(i)
[Since a median of a triangle divides it into two triangles of equal areas]
Also, ED is the median of ΔEBC.
ar (ΔBED) = ar (ΔCED) …..… (ii)
[Since a median of a triangle divides it into two triangles of equal areas]
On subtracting eq. (ii) from eq. (i), we get
ar (ΔABD) – ar (ΔBED) = ar (ΔACD) – ar (ΔCED)
ar (ΔABE) = ar (ΔACE)
Hence proved.
Ex 9.3 Class 9 Maths Question 2.
In a ΔABC, E is the mid-point of median AD. Show that
ar (ΔBED) = ¼ ar (ΔABC).
Solution.
Given: ABC is a triangle and E is the mid-point of the median AD.
To prove: ar (ΔBED) = ¼ ar (ΔABC)
Proof: We know that the median a triangle divides it into two triangles of equal areas.
ar (ΔABD) = ½ ar (ΔABC) …..…(i)
In ΔABD, BE is the median because E is the mid-point of AD.
ar (ΔBED) = ar (ΔBAE)
ar (ΔBED) = ½ ar (ΔABD)
ar (ΔBED) = ½ . ½ ar (ΔABC) [From eq. (i)]
ar (ΔBED) = ¼ ar (ΔABC)
Hence proved.
Ex 9.3 Class 9 Maths Question 3.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution.
Given: ABCD is a parallelogram and its diagonals AC and BD intersect each other at O.
To prove: Diagonals AC and BD divide parallelogram ABCD into four triangles of equal areas.
i.e., ar (ΔOAB) = ar (ΔOBC) = ar (ΔOCD) = ar (ΔOAD)
Proof: We know that the diagonals of a parallelogram bisect each other, so we have
OA = OC and OB = OD.
Also, we know that a median of a triangle divides it into two triangles of equal areas.
Now, as in ΔABC, BO is a median.
ar (ΔOAB) = ar (ΔOBC) ….… (i)
In ΔABD, AO is a median.
ar (ΔOAB) = ar (ΔOAD) ….… (ii)
Similarly, in ΔACD, DO is the median.
ar (ΔOAD) = ar (ΔOCD) …..…(iii)
From eqs. (i), (ii) and (iii), we get
ar (ΔOAB) = ar (ΔOBC) = ar (ΔOCD) = ar (ΔOAD)
Thus, the diagonals of a parallelogram divide it into four triangles of equal area.
Hence proved.
Ex 9.3 Class 9 Maths Question 4.
In the given figure, ABC and ABD are two triangles on the same base AB. If the line-segment CD is bisected by AB at O, show that ar (ΔABC) = ar( ΔABD).
Solution.
Given: ABC and ABD are two triangles on the same base AB.
To prove: ar (ΔABC) = ar (ΔABD)
Proof: Since the line segment CD is bisected by AB at O.
OC = OD
In ΔACD, we have OC = OD
So, AO is the median of ΔACD.
Since we know that the median divides a triangle into two triangles of equal areas.
ar (ΔAOC) = ar (ΔAOD) …..… (i)
Similarly, in ΔBCD, BO is the median.
ar (ΔBOC) = ar (ΔBOD) …..… (ii)
On adding eqs. (i) and (ii), we get
ar (ΔAOC) + ar (ΔBOC) = ar (ΔAOD) + ar (ΔBOD)
ar (ΔABC) = ar (ΔABD)
Hence proved.
9.3 Class 9 Maths Question 5.
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (ΔDEF) = ¼ ar (ΔABC).
(iii) ar (||gm BDEF) = ½ ar (ΔABC).
Solution.
Given: ABC is a triangle in which the mid-points of sides BC, CA and AB
are respectively D, E and F.
To prove:
(i) BDEF is a parallelogram.
(ii) ar (ΔDEF) = ¼ ar (ΔABC)
(iii) ar (||gm BDEF) = ½ ar (ΔABC)
Proof:
(i) Since, E and F are the mid-points of AC and AB.
BC || FE and FE = ½ BC = BD [By mid-point theorem]
Thus, BD || FE and BD = FE
Similarly, BF || DE and BF = DE
Hence, BDEF is a parallelogram.
[
a pair of opposite sides are equal and parallel]
(ii) Similarly, we can prove that both FDCE and AFDE are also parallelograms.
Now, BDEF is a parallelogram, so its diagonal FD divides it into two triangles of equal areas.
Therefore, ar (ΔBDF) = ar (ΔDEF) ……….(i)
Since FDCE is also parallelogram.
Therefore, ar (ΔDEF) = ar (ΔDEC) ……….(ii)
Again, AFDE is also parallelogram.
ar (ΔAFE) = ar (ΔDEF) ……….(iii)
From eqs. (i), (ii) and (iii), we get
ar (ΔDEF) = ar (ΔBDF) = ar (ΔDEC) = ar (ΔAFE) ……….(iv)
Now, ar (ΔABC) = ar (ΔDEF) + ar (ΔBDF) + ar (ΔDEC) + ar (ΔAFE) ……….(v)
ar (ΔABC) = ar (ΔDEF) + ar (ΔDEF) + ar (ΔDEF) + ar (ΔDEF) [Using (iv) and (v)]
ar (ΔABC) = 4 × ar (ΔDEF)
ar (ΔDEF) = ¼ ar (ΔABC)
(iii) ar (||gm BDEF) = ar (ΔBDF) + ar (ΔDEF)
= ar (ΔDEF) + ar (ΔDEF) [Using (iv)]
ar (||gm BDEF) = 2 ar (ΔDEF)
ar (||gm BDEF) = 2 × ¼ ar (ΔABC)
ar (||gm BDEF) = ½ ar (ΔABC)
Ex 9.3 Class 9 Maths Question 6.
In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that
(i) ar (ΔDOC) = ar (ΔAOB)
(ii) ar (ΔDCB) = ar (ΔACB)
(iii) DA || CB or ABCD is a parallelogram.
Solution:
We have a quadrilateral ABCD whose diagonals AC and BD intersect at O.
We have also given that OB = OD and AB = CD.
Let us draw DE AC and BF AC.
(i) In ∆DEO and ∆BFO, we have
DO = BO [Given]
DOE = BOF [Vertically opposite angles]
DEO = BFO [Each 90°]
∆DEO ∆BFO [By AAS congruency rule]
DE = BF [By C.P.C.T.]
and ar (∆DEO) = ar (∆BFO) ….…(1)
Now, in ∆DEC and ∆BFA, we have
DEC = BFA [Each 90°]
DE = BF [Proved above]
DC = BA [Given]
∆DEC ∆BFA [By RHS congruency rule]
ar (∆DEC) = ar (∆BFA) ….…(2)
and
1 = 2 [By C.P.C.T.] ….…(3)
Adding eq. (1) and (2), we have
ar (∆DEO) + ar (∆DEC) = ar (∆BFO) + ar (∆BFA)
ar (∆DOC) = ar (∆AOB)
(ii) Since, ar (∆DOC) = ar (∆AOB) [Proved above]
Adding ar (∆BOC) on both sides, we have
ar (∆DOC) + ar (∆BOC) = ar (∆AOB) + ar (∆BOC)
ar (∆DCB) = ar (∆ACB)
(iii) Since, ∆DCB and ∆ACB are both on the same base CB and having equal areas.
They lie between the same parallels CB and DA.
CB || DA
Also
1 = 2, [By eq. (3)]
which are alternate interior angles.
So, AB || CD
Hence, ABCD is a parallelogram.
Ex 9.3 Class 9 Maths Question 7.
D and E are points on sides AB and AC respectively of ∆ABC such that ar (DBC) = ar (EBC). Prove that DE || BC.
Solution:
We have ∆ABC and points D and E are such that ar (DBC) = ar (EBC).
Since ∆DBC and ∆EBC are on the same base BC and having the same area.
They must lie between the same parallels DE and BC.
Hence, DE || BC.
Ex 9.3 Class 9 Maths Question 8.
XY is a line parallel to side BC of a ∆ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF)
Solution:
We have a ∆ABC such that XY || BC, BE || AC and CF || AB.
Since, XY || BC and BE || CY
BCYE is a parallelogram.
Now, the parallelogram BCYE and ∆ABE are on the same base BE and between the same parallels BE and AC.
ar (∆ABE) = ½ ar (gm BCYE) ……..(1)
Again, CF || AB [Given]
XY || BC [Given]
CF || BX and XF || BC
BCFX is a parallelogram.
Now, ∆ACF and parallelogram BCFX are on the same base CF and between the same parallels AB and CF.
ar (∆ACF) = ½ ar (gm BCFX) ………(2)
Also, parallelogram BCFX and parallelogram BCYE are on the same base BC and between the same parallels BC and EF.
ar (||gm BCFX) = ar (||gm BCYE) ………(3)
From eqs. (1), (2) and (3), we get
ar (∆ABE) = ar (∆ACF)
Hence proved.
Ex 9.3 Class 9 Maths Question 9.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then A parallelogram PBQR is completed (see figure). Show that ar (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]
Solution:
Join AC and PQ.
ABCD is a parallelogram. [Given]
AC is its diagonal.
We know that diagonal of a parallelogram divides it into two triangles of equal areas.
ar (∆ABC) = ½ ar (gm ABCD) …..…(1)
Also, PBQR is a parallelogram. [Given]
QP is its diagonal.
ar (∆BPQ) = ½ ar (gm PBQR) …..…(2)
Since, ∆ACQ and ∆APQ are on the same base AQ and between the same parallels AQ and CP.
ar (∆ACQ) = ar (∆APQ)
ar (∆ACQ) – ar (∆ABQ) = ar (∆APQ) – ar (∆ABQ)
[Subtracting ar (∆ABQ) from both sides]
ar (∆ABC) = ar (∆BPQ) …..…(3)
From eqs. (1), (2) and (3), we get
½ ar (
gm ABCD) = ½ ar (gm PBQR)
ar (||gm ABCD) = ar (||gm PBQR)
Hence proved.
Ex 9.3 Class 9 Maths Question 10.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC)
Solution:
We have a trapezium ABCD having AB || DC and its diagonals AC and BD intersect each other at O.
Since, triangles on the same base and between the same parallels have equal areas.
∆ABD and ∆ABC are on the same base AB and between the same parallels AB and DC.
ar (∆ABD) = ar (∆ABC)
Subtracting ar (∆AOB) from both sides, we get
ar (∆ABD) – ar (∆AOB) = ar (∆ABC) – ar (∆AOB)
ar (∆AOD) = ar (∆BOC)
Hence proved.
Ex 9.3 Class 9 Maths Question 11.
In figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
Solution:
We have a pentagon ABCDE in which BF || AC and DC is produced to F.
(i) Since, the triangles on the same base and between the same parallels are equal in areas.
∆ACB and ∆ACF are on the same base AC and between the same parallels AC and BF.
ar (∆ACB) = ar (∆ACF)
(ii) Since, ar (∆ACB) = ar (∆ACF) [Proved above]
ar (∆ACB) + ar (quad. AEDC) = ar (∆ACF) + ar (quad. AEDC)
ar (ABCDE) = ar (AEDF)
ar (AEDF) = ar (ABCDE) Hence proved.
Ex 9.3 Class 9 Maths Question 12.
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Solution:
We have a plot in the form of a quadrilateral ABCD.
Let us draw DF || AC and join AF and CF.
Now, ∆DAF and ∆DCF are on the same base DF and between the same parallels AC and DF.
ar (∆DAF) = ar (∆DCF)
Subtracting ar (∆DEF) from both sides, we get
ar (∆DAF) – ar (∆DEF) = ar (∆DCF) – ar (∆DEF)
The portion of ∆ADE can be taken over by the Gram Panchayat by adding the land (∆CEF) to his (Itwaari) land so as to form a triangular plot, i.e. ∆ABF.
Let us prove that ar (∆ABF) = ar (quad. ABCD).
We have, ar (CEF) = ar (ADE) [Proved above]
ar (∆ABF) = ar (quad. ABCD)
Ex 9.3 Class 9 Maths Question 13.
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY). [Hint: Join CX]
Solution:
We have a trapezium ABCD such that AB || DC.
XY || AC meets AB at X and BC at Y. Let us join CX.
∆ADX and ∆ACX are on the same base AX and between the same parallels AX and DC.
ar (∆ADX) = ar (∆ACX) …..…(1)
∆ACX and ∆ACY are on the same base AC and between the same parallels AC and XY.
ar (∆ACX) = ar (∆ACY) …..…(2)
From eq. (1) and (2), we have
Ex 9.3 Class 9 Maths Question 14.
In figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
Solution:
We have, AP || BQ || CR
∆BCQ and ∆BQR are on the same base BQ and between the same parallels BQ and CR.
ar (∆BCQ) = ar (∆BQR) …..…(1)
∆ABQ and ∆PBQ are on the same base BQ and between the same parallels AP and BQ.
ar (∆ABQ) = ar (∆PBQ) …..…(2)
Adding eqs. (1) and (2), we get
Ar (∆BCQ) + ar (∆ABQ) = ar (∆BQR) + ar (∆PBQ)
ar (∆AQC) = ar (∆PBR) Hence proved.
Ex 9.3 Class 9 Maths Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
Solution:
We have a quadrilateral ABCD and its diagonals AC and BD intersect at O such that
ar (∆AOD) = ar (∆BOC) [Given]
Adding ar (∆AOB) to both sides, we have
ar (∆AOD) + ar (∆AOB) = ar (∆BOC) + ar (∆AOB)
ar (∆ABD) = ar (∆ABC)
Also, they are on the same base AB.
Since, the triangles are on the same base and having equal area.
They must lie between the same parallels.
AB || DC
Now, ABCD is a quadrilateral having a pair of opposite sides parallel.
So, ABCD is a trapezium.
Ex 9.3 Class 9 Maths Question 16.
In figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Solution:
We have, ar (∆DRC) = ar (∆DPC) [Given]
And they are on the same base DC.
∆DRC and ∆DPC must lie between the same parallels.
So, DC || RP, i.e., a pair of opposite sides of quadrilateral DCPR is parallel.
Again, we have ar (∆BDP) = ar (∆ARC) [Given] …..…(1)
Also, ar (∆DPC) = ar (∆DRC) [Given] …..…(2)
Subtracting eq. (2) from (1), we get
ar (∆BDP) – ar (∆DPC) = ar (∆ARC) – ar (∆DRC) |
How to Find the Maximum or Minimum Value of a Quadratic Function Easily
For a variety of reasons, you may need to be able to define the maximum or minimum value of a selected quadratic function. You can find the maximum or minimum if your original function is written in general form, ${\displaystyle f(x)=ax^{2}+bx+c}$, or in standard form, ${\displaystyle f(x)=a(x-h)^{2}+k}$. Finally, you may also wish to use some basic calculus to define the maximum or minimum of any quadratic function.
Method 1
Method 1 of 3:
Beginning with the General Form of the Function
1. 1
Set up the function in general form. A quadratic function is one that has an ${\displaystyle x^{2}}$ term. It may or may not contain an ${\displaystyle x}$ term without an exponent. There will be no exponents larger than 2. The general form is ${\displaystyle f(x)=ax^{2}+bx+c}$. If necessary, combine similar terms and rearrange to set the function in this general form.[1]
• For example, suppose you start with ${\displaystyle f(x)=3x+2x-x^{2}+3x^{2}+4}$. Combine the ${\displaystyle x^{2}}$ terms and the ${\displaystyle x}$ terms to get the following in general form:
• ${\displaystyle f(x)=2x^{2}+5x+4}$
2. 2
Determine the direction of the graph. A quadratic function results in the graph of a parabola. The parabola either opens upward or downward. If ${\displaystyle a}$, the coefficient of the ${\displaystyle x^{2}}$ term, is positive, then the parabola opens upward. If ${\displaystyle a}$ is negative, then the parabola opens downward.[2] Look at the following examples:[3]
• For ${\displaystyle f(x)=2x^{2}+4x-6}$, ${\displaystyle a=2}$ so the parabola opens upward.
• For ${\displaystyle f(x)=-3x^{2}+2x+8}$, ${\displaystyle a=-3}$ so the parabola opens downward.
• For ${\displaystyle f(x)=x^{2}+6}$, ${\displaystyle a=1}$ so the parabola opens upward.
• If the parabola opens upward, you will be finding its minimum value. If the parabola opens downward, you will find its maximum value.
3. 3
Calculate -b/2a. The value of ${\displaystyle -{\frac {b}{2a}}}$ tells you the ${\displaystyle x}$ value of the vertex of the parabola. When the quadratic function is written in its general form of ${\displaystyle ax^{2}+bx+c}$, use the coefficients of the ${\displaystyle x}$ and ${\displaystyle x^{2}}$ terms as follows:
• For a function ${\displaystyle f(x)=x^{2}+10x-1}$, ${\displaystyle a=1}$ and ${\displaystyle b=10}$. Therefore, find the x-value of the vertex as:
• ${\displaystyle x=-{\frac {b}{2a}}}$
• ${\displaystyle x=-{\frac {10}{(2)(1)}}}$
• ${\displaystyle x=-{\frac {10}{2}}}$
• ${\displaystyle x=-5}$
• As a second example, consider the function ${\displaystyle f(x)=-3x^{2}+6x-4}$. In this example, ${\displaystyle a=-3}$ and ${\displaystyle b=6}$. Therefore, find the x-value of the vertex as:
• ${\displaystyle x=-{\frac {b}{2a}}}$
• ${\displaystyle x=-{\frac {6}{(2)(-3)}}}$
• ${\displaystyle x=-{\frac {6}{-6}}}$
• ${\displaystyle x=-(-1)}$
• ${\displaystyle x=1}$
4. 4
Find the corresponding f(x) value. Insert the value of x that you just calculated into the function to find the corresponding value of f(x). This will be the minimum or maximum of the function.
• For the first example above, ${\displaystyle f(x)=x^{2}+10x-1}$, you calculated the x-value for the vertex to be ${\displaystyle x=-5}$. Enter ${\displaystyle -5}$ in place of ${\displaystyle x}$ in the function to find the maximum value:
• ${\displaystyle f(x)=x^{2}+10x-1}$
• ${\displaystyle f(-5)=(-5)^{2}+10(-5)-1}$
• ${\displaystyle f(-5)=25-50-1}$
• ${\displaystyle f(-5)=-26}$
• For the second example above, ${\displaystyle f(x)=-3x^{2}+6x-4}$, you found the vertex to be at ${\displaystyle x=1}$. Insert ${\displaystyle 1}$ in place of ${\displaystyle x}$ in the function to find the maximum value:
• ${\displaystyle f(x)=-3x^{2}+6x-4}$
• ${\displaystyle f(1)=-3(1)^{2}+6(1)-4}$
• ${\displaystyle f(1)=-3+6-4}$
• ${\displaystyle f(1)=-1}$
5. 5
Report your results. Review the question you have been asked. If you are asked for the coordinates of the vertex, you need to report both the ${\displaystyle x}$ and ${\displaystyle y}$ (or ${\displaystyle f(x)}$) values. If you are only asked for the maximum or minimum, you only need to report the ${\displaystyle y}$ (or ${\displaystyle f(x)}$) value. Refer back to the value of the ${\displaystyle a}$ coefficient to be sure if you have a maximum or a minimum.
• For the first example, ${\displaystyle f(x)=x^{2}+10x-1}$, the value of ${\displaystyle a}$ is positive, so you will be reporting the minimum value. The vertex is at ${\displaystyle (-5,-26)}$, and the minimum value is ${\displaystyle -26}$.
• For the second example, ${\displaystyle f(x)=-3x^{2}+6x-4}$, the value of ${\displaystyle a}$ is negative, so you will be reporting the maximum value. The vertex is at ${\displaystyle (1,-1)}$, and the maximum value is ${\displaystyle -1}$.
Method 2
Method 2 of 3:
Using the Standard or Vertex Form
1. 1
Write your quadratic function in standard or vertex form. The standard form of a general quadratic function, which can also be called the vertex form, looks like this:[4]
• ${\displaystyle f(x)=a(x-h)^{2}+k}$
• If your function is already given to you in this form, you just need to recognize the variables ${\displaystyle a}$, ${\displaystyle h}$ and ${\displaystyle k}$. If your function begins in the general form ${\displaystyle f(x)=ax^{2}+bx+c}$, you will need to complete the square to rewrite it in vertex form.
• To review how to complete the square, see Complete the Square.
2. 2
Determine the direction of the graph. Just as with a quadratic function written in its general form, you can tell the direction of the parabola by looking at the coefficient ${\displaystyle a}$. If ${\displaystyle a}$ in this standard form is positive, then the parabola opens upward. If ${\displaystyle a}$ is negative, then the parabola opens downward.[5] Look at the following examples:[6]
• For ${\displaystyle f(x)=2(x+1)^{2}-4}$, ${\displaystyle a=2}$, which is positive, so the parabola opens upward.
• For ${\displaystyle f(x)=-3(x-2)^{2}+2}$, ${\displaystyle a=-3}$, which is negative, so the parabola opens downward.
• If the parabola opens upward, you will be finding its minimum value. If the parabola opens downward, you will find its maximum value.
3. 3
Identify the minimum or maximum value. When the function is written in standard form, finding the minimum or maximum value is as simple as stating the value of the variable ${\displaystyle k}$. For the two example functions given above, these values are:
• For ${\displaystyle f(x)=2(x+1)^{2}-4}$, ${\displaystyle k=-4}$. This is the minimum value of the function because this parabola opens upward.
• For ${\displaystyle f(x)=-3(x-2)^{2}+2}$, ${\displaystyle k=2}$. This is the maximum value of the function, because this parabola opens downward.
4. 4
Find the vertex. If you are asked for the coordinates of the minimum or maximum value, the point will be ${\displaystyle (h,k)}$. Note, however, that in the standard form of the equation, the term inside the parentheses is ${\displaystyle (x-h)}$, so you need the opposite sign of the number that follows the ${\displaystyle x}$.
• For ${\displaystyle f(x)=2(x+1)^{2}-4}$, the term inside the parentheses is (x+1), which can be rewritten as (x-(-1)). Thus, ${\displaystyle h=-1}$. Therefore, the coordinates of the vertex for this function are ${\displaystyle (-1,-4)}$.
• For ${\displaystyle f(x)=-3(x-2)^{2}+2}$, the term inside the parentheses is (x-2). Therefore, ${\displaystyle h=2}$. The coordinates of the vertex are (2, 2).
Method 3
Method 3 of 3:
Using Calculus to Derive the Minimum or Maximum
1. 1
Start with the general form. Write your quadratic function in general form, ${\displaystyle f(x)=ax^{2}+bx+c}$. If necessary, you may need to combine like terms and rearrange to get the proper form.[7]
• Begin with the sample function ${\displaystyle f(x)=2x^{2}-4x+1}$.
2. 2
Use the power rule to find the first derivative. Using basic first-year calculus, you can find the first derivative of the general quadratic function to be ${\displaystyle f^{\prime }(x)=2ax+b}$.[8]
• For the sample function ${\displaystyle f(x)=2x^{2}-4x+1}$, find the derivative as:
• ${\displaystyle f^{\prime }(x)=4x-4}$
3. 3
Set the derivative equal to zero. Recall that derivative of a function tells you the slope of the function at that selected point. The minimum or maximum of a function occurs when the slope is zero. Therefore, to find where the minimum or maximum occurs, set the derivative equal to zero. Continue with the sample problem from above:[9]
• ${\displaystyle f^{\prime }(x)=4x-4}$
• ${\displaystyle 0=4x-4}$
4. 4
Solve for x. Use basic rules of algebra to rearrange the function and solve the value for x, when the derivative equals zero. This solution will tell you the x-coordinate of the vertex of the function, which is where the maximum or minimum will occur.[10]
• ${\displaystyle 0=4x-4}$
• ${\displaystyle 4=4x}$
• ${\displaystyle 1=x}$
5. 5
Insert the solved value of x into the original function. The minimum or maximum value of the function will be the value for ${\displaystyle f(x)}$ at the selected ${\displaystyle x}$ position. Insert your value of ${\displaystyle x}$ into the original function and solve to find the minimum or maximum.[11]
• For the function ${\displaystyle f(x)=2x^{2}-4x+1}$ at ${\displaystyle x=1}$,
• ${\displaystyle f(1)=2(1)^{2}-4(1)+1}$
• ${\displaystyle f(1)=2-4+1}$
• ${\displaystyle f(1)=-1}$
6. 6
Report your solution. The solution gives you the vertex of the maximum or minimum point. For this sample function, ${\displaystyle f(x)=2x^{2}-4x+1}$, the vertex occurs at ${\displaystyle (1,-1)}$. The coefficient ${\displaystyle a}$ is positive, so the function opens upward. Therefore, the minimum value of the function is the y-coordinate of the vertex, which is ${\displaystyle -1}$.[12]
Search
• Question
How do you tell if a parabola is maximum or minimum?
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Jake Adams is an academic tutor and the owner of Simplifi EDU, a Santa Monica, California based online tutoring business offering learning resources and online tutors for academic subjects K-College, SAT & ACT prep, and college admissions applications. With over 14 years of professional tutoring experience, Jake is dedicated to providing his clients the very best online tutoring experience and access to a network of excellent undergraduate and graduate-level tutors from top colleges all over the nation. Jake holds a BS in International Business and Marketing from Pepperdine University.
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First solve for a. If the value of a is a positive number, you'll have an upward-facing parabola and you'll need to find its minimum value. If a is a negative number, you'll have a downward-facing parabola and you'll need to find its maximum value.
• Question
How do you tell if a parabola is up or down?
Academic Tutor & Test Prep Specialist
Jake Adams is an academic tutor and the owner of Simplifi EDU, a Santa Monica, California based online tutoring business offering learning resources and online tutors for academic subjects K-College, SAT & ACT prep, and college admissions applications. With over 14 years of professional tutoring experience, Jake is dedicated to providing his clients the very best online tutoring experience and access to a network of excellent undergraduate and graduate-level tutors from top colleges all over the nation. Jake holds a BS in International Business and Marketing from Pepperdine University.
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You can remember this concept by thinking about smiles and frowns. If someone is positive they smile, and if someone is negative, they frown. Similarly, a positive number will have an upward-facing parabola, and a negative number will have a downward-facing parabola.
• Question
How do I graph a quadratic function?
First, create a data table with multiple experimental values for x. Sub in those x coordinates and get y coordinates. Plot these along the x and y axis and join the dots with a smooth curve.
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• The parabola's axis of symmetry is x = h.
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This article was co-authored by Jake Adams. Jake Adams is an academic tutor and the owner of Simplifi EDU, a Santa Monica, California based online tutoring business offering learning resources and online tutors for academic subjects K-College, SAT & ACT prep, and college admissions applications. With over 14 years of professional tutoring experience, Jake is dedicated to providing his clients the very best online tutoring experience and access to a network of excellent undergraduate and graduate-level tutors from top colleges all over the nation. Jake holds a BS in International Business and Marketing from Pepperdine University. This article has been viewed 2,396,695 times.
Co-authors: 33
Updated: February 3, 2023
Views: 2,396,695
Categories: Algebra
Article SummaryX
To find the maximum or minimum value of a quadratic function, start with the general form of the function and combine any similar terms. For example, if you’re starting with the function f(x) = 3x + 2x - x^2 + 3x^2 + 4, you would combine the x^2 and x terms to simplify and end up with f(x) = 2x^2 + 5x + 4. Now figure out which direction the parabola opens by checking if a, or the coefficient of x^2, is positive or negative. If it’s positive, the parabola opens upward. If it’s negative, the parabola opens downward. In the function f(x) = 2x^2 + 5x + 4, the coefficient of x^2 is positive, so the parabola opens upward. Next, find the x value of the vertex by solving -b/2a, where b is the coefficient in front of x and a is the coefficient in front of x^2. In the function f(x) = 2x^2 + 5x + 4, b = 5 and a = 2. Therefore, you would divide -5 by 2 times 2, or 4, and get -1.25. Finally, plug the x value into the function to find the value of f(x), which is the minimum or maximum value of the function. The function f(x) = 2x^2 + 5x + 4 would become f(-1.25) = 2(-1.25)^2 + 5(-1.25) + 4, or f(-1.25) = 0.875. If the parabola opens upward, your answer will be the minimum value. If the parabola opens downward, your answer is the maximum value. In this example, since the parabola opens upward, f(-1.25) = 0.875 is the minimum value of the function. If you want to learn how to use standard or vertex form for your formula, keep reading the article!
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MATH 221 Lecture 2
## MATH 221 Lecture 2
Last updated: 10 July 2012
## Angles
Measure angles according to the distance traveled on a circle of radius 1.
Sketch both $x$ and $y$ to get a circle of radius $r$.
The distance $2\pi$ around a circle of radius 1 stretches to $2\pi r$ around a circle of radius $r$. So the circumference of a circle is $2\pi r$ if the circle is radius $r$.
To find the area of a circle first approximate with a polygon inscribed in the circle. The eight triangles form an octagon ${P}_{8}$ in the circle. The area of the octagon ${P}_{8}$ is almost the same as the area of the circle. Unwrap the octagon.
The area of the octagon is the area of the 8 triangles. The area of each triangle is $\frac{1}{2}bh$. So the area of the octagon is $\frac{1}{2}Bh$.
Take the limit as the number of triangles in the interior polygon gets larger and larger (the polygon gets closer and closer to being the circle). Then
Where $B$ is the total base, $h$ is the height of the triangle, $2\pi$ is the length of an unwrapped circle and $r$ is the radius of the circle.
So the area of a circle is $\pi {r}^{2}$ if the circle is radius $r$.
## Trigonometric functions
${$ $}$ $}$ $θ$ $cosθ$ $sinθ$ $←$ $\mathrm{sin}\theta$ is the $y$-coordinate of a point at distance $\theta$ on a circle of radius 1 $\mathrm{cos}\theta$ is the $x$-coordinate of a point at distance $\theta$ on a circle of radius 1
$\begin{array}{cccc}\mathrm{tan}\theta =\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }\text{,}& \mathrm{cot}\theta =\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }\text{,}& \mathrm{sec}\theta =\frac{1}{\mathrm{cos}\theta }\text{,}& \mathrm{csc}\theta =\frac{1}{\mathrm{sin}\theta }\end{array}$
Since the equation of a circle of radius 1 is ${x}^{2}+{y}^{2}=1$ this forces ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$.
The pictures
show that
$sin(-θ)=-sinθandcos(-θ)=cosθ$
Also
show that
$sin0=0andsinπ2=1 cos0=1cosπ2=0$
Draw the graphs
$y=\mathrm{sin}\theta$
$y=\mathrm{cos}\theta$
by seeing how the $x$ and $y$ coordinates change as you walk around the circle.
Example: Verify $\frac{\mathrm{sec}B}{\mathrm{cos}B}-\frac{\mathrm{tan}B}{\mathrm{cot}B}=1$
$secBcosB-tanBcotB=1cosBcosB-sinBcosBcosBsinB=1cos2B-sin2Bcos2B =1-sin2Bcos2B=cos2Bcos2B =1$
Example: Verify $\mathrm{cot}\alpha -\mathrm{cot}\beta =\frac{\mathrm{sin}\left(\beta -\alpha \right)}{\mathrm{sin}\alpha \mathrm{sin}\beta }$
$Left hand side=cotα-cotβ=cosαsinα-cosβsinβ =cosαsinβ-cosβsinαsinαsinβ Right hand side=sin(β-α)sinαsinβ=sinβcos(-α)+cosβsin(-α)sinαsinβ =sinβcosα+cosβ(-sinα)sinαsinβ=sinβcosα-cosβsinαsinαsinβ$
So
$\text{Left hand side}=\text{Right hand side}$
Example: Verify $\frac{\mathrm{tan}A-\mathrm{sin}A}{\mathrm{sec}A}=\frac{{\mathrm{sin}}^{3}A}{1+\mathrm{cos}A}$
## References
[Ram] A. Ram, MATH 221 Lecture 2, September 8, 2000, University of Wisconsin.
page history |
New Zealand
Level 8 - NCEA Level 3
# Manipulate Ellipse Equations
Lesson
Typically we are given the equation of an ellipse in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 or $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(xh)2b2+(yk)2a2=1 where $a>b$a>b. Ideally we would like to express any equation of an ellipse in such a way, since we can immediately identify its centre and the length of the semi-major and semi-minor axes:
• $\left(h,k\right)$(h,k) is the centre of the ellipse.
• $a$a is the length of semi-major axis.
• $b$b is the length of semi-minor axis.
But generally speaking, the equation of an ellipse can be expressed in many ways. For instance, the following pair of equations both represent the same ellipse.
$9\left(x-2\right)^2+4\left(y-3\right)^2=36$9(x2)2+4(y3)2=36 or $9x^2-36x+4y^2-24x=-36$9x236x+4y224x=36
In order to express the above equations in standard form, we can manipulate both sides by multiplying or by completing the square.
#### Worked examples
##### Example 1
Consider the equation of the ellipse $x^2-12x+36y^2=0$x212x+36y2=0
Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 where $a>b$a>b.
Think: Completing the square allows us to rewrite the quadratic expressions in $x$x as a perfect square of the form $\left(x-h\right)^2$(xh)2. Notice that the quadratic expression in $y$y is already a perfect square.
To complete the square in $x$x we look at the expression $x^2-12x$x212x, halve and square the coefficient of $x$x, and then add the result to both sides of the equation.
Do: Halving and squaring $-12$12 gives a result of $36$36, so we add this result to both sides of the equation:
$x^2-12x+36+36y^2=36$x212x+36+36y2=36
So altogether we get the following steps:
$x^2-12x+36y^2$x2−12x+36y2 $=$= $0$0 (Writing down the equation) $x^2-12x+36+36y^2$x2−12x+36+36y2 $=$= $36$36 (Completing the square for $x$x) $\left(x-6\right)^2+36y^2$(x−6)2+36y2 $=$= $36$36 (Factorising the perfect square) $\frac{\left(x-6\right)^2}{36}+y^2$(x−6)236+y2 $=$= $1$1 (Dividing both sides by $36$36)
The equation of the ellipse is then $\frac{\left(x-6\right)^2}{36}+y^2=36$(x6)236+y2=36. Optionally, the standard form can be made more obvious by writing it as $\frac{\left(x-6\right)^2}{6^2}+\frac{y^2}{1^2}=36$(x6)262+y212=36.
Reflect: The equation of the ellipse centred at $\left(6,0\right)$(6,0) with a vertical semi-major axis of length $6$6 units and a horizontal semi-minor axis of length $1$1 unit can be expressed by the equation $x^2-12x+36y^2=0$x212x+36y2=0 or the equation $\frac{\left(x-6\right)^2}{36}+y^2=1$(x6)236+y2=1.
##### Example 2
Consider the equation of the ellipse $9x^2-36x+4y^2-24x=-36$9x236x+4y224x=36.
Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(xh)2b2+(yk)2a2=1 where $a>b$a>b.
Think: Completing the square allows us to rewrite the terms containing $x$x and $y$y in the form $\left(x-h\right)^2$(xh)2 and $\left(y-k\right)^2$(yk)2. Before we do so, it will be more convenient if we factorise so that the coefficients of $x^2$x2 and $y^2$y2 are $1$1.
Do: In this case, the coefficient of $x^2$x2 is $9$9 the coefficient of $y^2$y2 is $4$4:
$9\left(x^2-4x\right)+4\left(y^2-6x\right)=-36$9(x24x)+4(y26x)=36
Now we can complete the square in $x$x and $y$y:
$9\left(x^2-4x+4\right)+4\left(y^2-6x+9\right)=-36+9\times4+4\times9$9(x24x+4)+4(y26x+9)=36+9×4+4×9
Putting this together, we get the following steps:
$9x^2-36x+4y^2-24x$9x2−36x+4y2−24x $=$= $-36$−36 (Writing down the equation) $9\left(x^2-4x\right)+4\left(y^2-6x\right)$9(x2−4x)+4(y2−6x) $=$= $-36$−36 (Factorising the leading coefficients of each variable) $9\left(x^2-4x+4\right)+4\left(y^2-6x+9\right)$9(x2−4x+4)+4(y2−6x+9) $=$= $-36+9\times4+4\times9$−36+9×4+4×9 (Completing the square for $x$x and $y$y) $9\left(x^2-4x+4\right)+4\left(y^2-6x+9\right)$9(x2−4x+4)+4(y2−6x+9) $=$= $36$36 (Simplifying the constant terms) $9\left(x-2\right)^2+4\left(y-3\right)^2$9(x−2)2+4(y−3)2 $=$= $36$36 (Rewriting the perfect squares in factorised form) $\frac{\left(x-2\right)^2}{4}+\frac{\left(y-3\right)^2}{9}$(x−2)24+(y−3)29 $=$= $1$1 (Dividing both sides by $36$36)
So the equation of the ellipse can be expressed in the form $\frac{\left(x-2\right)^2}{4}+\frac{\left(y-3\right)^2}{9}=1$(x2)24+(y3)29=1.
Reflect: The centre of the ellipse is $\left(h,k\right)=\left(2,3\right)$(h,k)=(2,3), the length of the vertical semi-major axis is $3$3 units and the length of the horizontal semi-minor axis is $2$2 units. From this, we can draw the equation of the ellipse as shown below.
Graph of the ellipse $9x^2-36x+4y^2-24x=-36$9x2−36x+4y2−24x=−36.
#### Practice questions
##### question 1
Consider the ellipse with equation $\left(x+3\right)^2+4\left(y-1\right)^2=16$(x+3)2+4(y1)2=16.
1. Write the equation in standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 where $a>b$a>b.
2. Draw the graph of the ellipse.
##### question 2
The graph of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$x29+y24=1 is given below.
1. Write $4x^2+9\left(y-2\right)^2=36$4x2+9(y2)2=36 in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 where $a>b$a>b.
2. What transformation of the ellipse drawn gives the ellipse $4x^2+9\left(y-2\right)^2=36$4x2+9(y2)2=36?
Translated to the right by $2$2 units
A
Translated to the left by $2$2 units
B
Translated up by $2$2 units
C
Translated down by $2$2 units
D
Translated to the right by $2$2 units
A
Translated to the left by $2$2 units
B
Translated up by $2$2 units
C
Translated down by $2$2 units
D
##### question 3
Consider the ellipse with equation $4x^2+24x+5y^2+20y+36=0$4x2+24x+5y2+20y+36=0.
1. Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 where $a>b$a>b.
2. What are the coordinates of the centre of the ellipse?
3. What are the coordinates of the vertices?
### Outcomes
#### M8-1
Apply the geometry of conic sections
#### 91573
Apply the geometry of conic sections in solving problems |
# The scope of integration and derivation in mathematics
0
0
Integration is the process of giving numbers to the various functions it is commonly used in calculus. The assigning numbers will define displacement, volume, and area. Integral is a given function in derivatives which is further produced a function that is used for differentiation and calculating the area lies under the curve of a graph in a function. Integration is an important step that is specific in finding central points, areas, volumes, etc.
How to calculate integral?
Integral is calculated using a formula. It involved multiple steps. For example, we can find the function by dividing the area into small pieces and then add the width of these slices by Δx. But the addition of all dividing slices will mess the graph and won’t define function properly.
Despite if we have decreased the width of differentiated x i.e. (Δx) and then add up all the divided areas then the accuracy will become better. We can achieve the maximum accuracy by decreasing the width of all slices up to zero. It will give us the definite result.
How does an integral represent?
Integral is represented by symbols “S” and “s” in the sum of the parts to find the area under the curve or you can use online integration by parts calculator.
∫2xdx
The above formula is used to find integral. Where ∫is the symbol of integral 2x is the function which we want to integrate and dx indicates the direction along the x-axis. It is dependent on the plane.
If there is a y plane then dx will be replaced by dy. dx is used to indicate that area slices are in the direction of the x-axis, and the width is approaching zero. We can write a complete equation like this
∫2x+1dx
The two important entities of integral “definite integral” and “double integral”
Integral is a function which is a summation. There is a need to find the net area between function and x-axis. The theme of a definite integral is closely related to the notation used for indefinite integral. There are two limits of integral. I.e Upper limit and lower limit.
The lower limit is indicated by the value of “a” which is present at the bottom of the integral sign. And the “upper limit” is assigned to the value of b which is present at top. Multiple integrals show more than one real variable.
Double integral is a little bit more complex than others. For understanding the property of double integral, let’s consider an example.
Considering a function f (a, b) present in a 3D space in the XY plane. And R presents in any region in AB or XY plane. If we divide the R into smaller parts than the double integral in r region will be:
ʃʃR ƒ (a, b) dadb=limn ͢ 0 Σ (n, i=1) ƒ (ai, bi) δai δbi
Integration is a basic but complex function. Its complexity increases as we move on different variable integrals. The complexity of integration can be simplified by using an integration calculator. It will not only save you from the hustle of remembering long equations but derive accurate results too. And you can save time and solve such tricky questions very easily.
What is a derivative?
The derivative of a function has a specific entity in mathematics. The process by which a derivative is solved is called differentiation.
When a function changes at a specific rate at some point it is called derivation or derivation of a function. The rate of change can be predicted by calculating the ratio of change of Y against the independent variable X.
The ratio of change in a derivative is considered the limit as X approaches zero. i.e. Δx0
Leibniz’s notation
It is a specific type of representing the derivative of a function. According to this the derivative of “a” is represented as:
Function y = f(x) as af / dx or dy / dx.
The derivative of a function can be found by following these steps.
• Delta is the differentiation quotient hence Δy/Δx = f(x0+Δx) −f(x0) / Δx
• The quotient can be simplified by canceling Δx. (cancel if possible).
• After finding the differentiation off (x0), apply the limit to the quotient. The function would be differentiated at xo only if the limit exists.
These steps are used to find the derivative. But if you are confused between these terms you can use a leibniz notation calculator and solve the differentiation very easily. |
One of the benefits of graphing calculators and graphing software is that it allows you to learn things about functions and equations that would have been tedious to explore via pencil and graph paper. Families of graphs is a concept that starts with a “parent” function or equation and asks you to compare “child” functions or equations to it to determine how they have translated or transformed from the original.
Consider, for example one of the easiest equations to graph:
y = x or f(x) = x in functional notation.
When you plug values into the independent variable, x, the dependent variable, y, has the same value. When x is 1, y is 1. When x is 2, y is 2. When x is -4, y is -4. Plotting the points on a coordinate plane result in a line that goes through the origin, rising to the right as much as it runs.
So if y = x is the parent equation, compare the following to it.
y = x +5
y = x – 2
y = x + 3
y = x – 1
Next graph y = |x|, the parent and compare it to the following children:
y = |x| + 1
y = |x| + 2
y = |x| – 3
y = |x| – 4
What do you notice?
Compare y = |x| to the following:
y = |x – 2|
y = |x – 1|
y = |x + 4|
y = |x + 2|
Did they behave as you expected?
Compare the graph of the parent y = x² with the graphs of the following equations. How are they changed from the original?
y = x² – 3
y = x² + 1
y = (x – 4)²
y = (x + 2)²
y = (x + 2)² + 3
Returning to our original line y = x, what happens when we multiply x by a number?
y = ½ x
y = 3x
y = -¾ x
y = -2 x
Do your explanations hold for the following?
y = |x| and y = -|x|
y = x² and y = ¼ x²
y = x² and y = -3x²
y = x³ and y = -x³
Use the Families of Graphs Exploration to sketch the graphs and note your observations.
A Parent Functions Foldable is available for note taking.
A Functions Transformations Foldable is available for note taking. Inspiration by Sarah Hagan. |
Math Geek
Math Geek
Home Topics
Often in probability we wish to count the number of ways something can be done such as chosing a number of items from a set. This is frequently difficult to do manually as the numbers of ways can be extremely large. Because there can be various conditions on how the items are chosen, there are various techniques available to count the many ways. One such method is called permutations.
Permutations are used when we are picking items from a set and the items are not put back so they can be picked again. This is know as choosing without replacement. So the pool of items is being reduced each time a choice is made. But another critical feature that must be present when we use permuations is the order in which the items are picked makes a difference. There is some special significance for the first one picked, another significance for the second one picked, and so forth. For example, suppose we have a club and we want to pick a president and vice president. Suppose also that the first person picked, say Jane, becomes president while the second one picked, say Joe, becomes vice president. Jane is president because she was picked first. Now suppose Joe was picked first and Jane was picked second. Joe now is president and Jane is vice president. The order in which the two were picked makes a difference in the outcome.
The formula we are going to use to calculate permutations, denoted by nPr, is
where n is the number of items availble for selection and r is the number chosen. Incidentally, n! means to multiply together every integer from 1 to n. The same applies to r!. So the first step is to calculate the numerator n!. The second step is to calculate the denominator (n-r)!. Finally, we divide the two. Now we will look at an example.
Find 12P9
1. Find n! 12! = 479001600 2. Find (n-r)! ( 12 - 9 )! = 3! = 6 3. Divide the numerator by the denominator 479001600 / 6 = 79833600
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# Ex. 6.6 Q8 Triangles Solution - NCERT Maths Class 10
Go back to 'Ex.6.6'
## Question
In Fig. below, two chords $$AB$$ and $$CD$$ of a circle intersect each other at the point $$P$$ (when produced) outside the circle.
Prove that:
(i) $$\Delta {\text{ }}PAC{\text{ }}\text{~}{\text{ }}\Delta {\text{ }}PDB$$
(ii) $$PA. PB = PC. PD$$
Video Solution
Triangles
Ex 6.6 | Question 8
## Text Solution
Reasoning:
(i) Exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.
Steps:
Draw $$AC$$
(i) In, $$\Delta PAC,\,\,\Delta PDB$$
\begin{align} & \angle APC=\angle BPD\,\,\, \\ & (\text{Common}\,\text{angle}) \\ & \\ & \angle PAC=\angle PDB\,\, \\ & \left( \begin{array} & \text{Exterior angle of a cyclic} \\ \,\text{quadrilateral is equal to the} \\ \text{interior opposite angle} \\ \end{array} \right)\, \\ & \\ & \Rightarrow \Delta PAC~\sim \Delta \text{PDB}\, \\ \end{align}
(ii) In, $$\Delta PAC,\,\,\Delta PDB$$
\begin{align} &\frac{{PA}}{{PD}} = \frac{{PC}}{{PB}} = \frac{{AC}}{{BD}}\\ &\frac{{PA}}{{PD}} = \frac{{PC}}{{PB}}\\ &PA \cdot PB = PC \cdot PD \end{align}
Learn from the best math teachers and top your exams
• Live one on one classroom and doubt clearing
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# 2020 AIME I Problems/Problem 8
Note: Please do not post problems here until after the AIME.
## Solution 1 (Coordinates)
We plot this on the coordinate grid with point $O$ as the origin. We will keep a tally of the x-coordinate and y-coordinate separately.
First move: The ant moves right $5$. Second move: We use properties of a $30-60-90$ triangle to get $\frac{5}{4}$ right, $\frac{5\sqrt{3}}{4}$ up. Third move: $\frac{5}{8}$ left, $\frac{5\sqrt{3}}{8}$ up. Fourth move: $\frac{5}{8}$ left. Fifth move: $\frac{5}{32}$ left, $\frac{5\sqrt{3}}{32}$ down. Sixth move: $\frac{5}{64}$ right, $\frac{5\sqrt{3}}{64}$ down.
Total of x-coordinate: $5 + \frac{5}{4} - \frac{5}{8} - \frac{5}{8} - \frac{5}{32} + \frac{5}{64} = \frac{315}{64}$. Total of y-coordinate: $0 + \frac{5\sqrt{3}}{4} + \frac{5\sqrt{3}}{8} + 0 - \frac{5\sqrt{3}}{32} - \frac{5\sqrt{3}}{64} = \frac{105\sqrt{3}}{64}$.
After this cycle of six moves, all moves repeat with a factor of $(\frac{1}{2})^6 = \frac{1}{64}$. Using the formula for a geometric series, multiplying each sequence by $\frac{1}{1-\frac{1}{64}} = \frac{64}{63}$ will give us the point $P$.
$\frac{315}{64} \cdot \frac{64}{63} = 5$, $\frac{105\sqrt{3}}{64} \cdot \frac{64}{63} = \frac{5\sqrt{3}}{3}$. Therefore, the coordinates of point $P$ are $(5,\frac{5\sqrt{3}}{3})$, so using the Pythagorean Theorem, $OP^2 = \frac{100}{3}$, for an answer of $\boxed{103}$.
-molocyxu
## Solution 2 (Complex)
We put the ant in the complex plane, with its first move going in the positive real direction. Take $$|\sum_{k=0}^{\infty} (5\frac{e^{k\pi i / 3}}{2^i})|^2$$ and this is an infinite geometric series. Summing using $\frac{a}{1-r}$ gives $\boxed{103}.$ ~awang11 |
# Modelling by numbers: Part Two AModelling by numbers: Part Two A
An introduction to procedural geometry. Part 2A: The cylinder. Welcome to part two. This is where we move on to our second basic shape: the cylinder.
Jayelinda Suridge, Blogger
September 5, 2013
# Modelling by numbers
### An introduction to procedural geometry
(This is the third part of a four part tutorial. If you haven't already, you should check out Modelling by numbers: Part One A)
# Part Two A: The cylinder
Unity assets - Unity package file containing scripts and scenes for this tutorial (all 4 parts).
Source files - Just the source files, for those that don’t have Unity.
Welcome to part two. This is where we move on to our second basic shape: the cylinder.
## A note about maths
Curvy things are a little more maths-heavy than the shapes we’ve covered so far. Some basic maths (particularly trigonometry) knowledge will be helpful here. Nothing fancy, and you can always just copy the code if you want. But if you can remember the basic trig ratios and know the difference between degrees and radians, then you’ll be better equipped to actually understand what’s going on.
## The basic cylinder
This is the second of our basic building blocks, and a very handy one, too.
Building a cylinder requires a similar approach to building the ground plane in part one of this tutorial. Except that instead of rows and columns we’re now dealing with rings and columns.
The first step, therefore, is learning how to make a ring. Going back to high school maths, the X and Y values of a unit vector will always be the cosine of the angle between the vector and X axis, and the sine of that angle, respectively. As the angle moves from 0 to 360 degrees, the points will form a circle.
Alternatively, I’ll explain in code:
float radiansInACircle = Mathf.PI * 2;
for (int i = 0; i < segmentCount; i++)
{
float angle = radiansInACircle * i / segmentCount;
Vector2 position = new Vector2(Mathf.Cos(angle), Mathf.Sin(angle));
}
The result of this will be that all of those positions will form a circle (with a radius of one).
Something worth pointing out at this point: you’ll notice that this code gives the angle in radians. This is required for the Mathf trigonometric functions, unlike the entire rest of Unity, which expects angles to be in degrees. In our scripts, we’re going to follow the same pattern and work with degrees, converting when entering trigonometric code.
Right, let’s build a ring:
void BuildRing(MeshBuilder meshBuilder, int segmentCount, Vector3 centre, float radius,
float v, bool buildTriangles)
{
float angleInc = (Mathf.PI * 2.0f) / segmentCount;
for (int i = 0; i <= segmentCount; i++)
{
float angle = angleInc * i;
Vector3 unitPosition = Vector3.zero;
unitPosition.x = Mathf.Cos(angle);
unitPosition.z = Mathf.Sin(angle);
meshBuilder.UVs.Add(new Vector2((float)i / segmentCount, v));
if (i > 0 && buildTriangles)
{
int baseIndex = meshBuilder.Vertices.Count - 1;
int vertsPerRow = segmentCount + 1;
int index0 = baseIndex;
int index1 = baseIndex - 1;
int index2 = baseIndex - vertsPerRow;
int index3 = baseIndex - vertsPerRow - 1;
}
}
}
This is a modified version of the grid-building code from our ground mesh in part one. Instead of incrementing a position offset, we’re incrementing an angle. From that angle we can get a unit position (a position one unit away from the centre) in the XZ plane. Multiply by the radius and add the centre offset and we have the position of our vertex.
The normals of this mesh all face directly outward from the centre axis. Happily for us, this means that the normal and the unitPosition are the same.
If we call this function once for the bottom of the cylinder, and once for the top, we’ll have a basic cylinder:
BuildRing(meshBuilder, m_RadialSegmentCount, Vector3.up * m_Height, m_Radius, 1.0f, true);
This is perfectly adequate if we never want to deform the cylinder, although the lighting may have a little bit of trouble with the very long triangles. Let’s modify our code to divide the height into segments.
float heightInc = m_Height / m_HeightSegmentCount;
for (int i = 0; i <= m_HeightSegmentCount; i++)
{
Vector3 centrePos = Vector3.up * heightInc * i;
float v = (float)i / m_HeightSegmentCount;
BuildRing(meshBuilder, m_RadialSegmentCount, centrePos, m_Radius, v, i > 0);
}
Again, just like our ground mesh code, except we’re only incrementing the height, and we’re building rings rather than rows.
### Hang on, that’s not a cylinder! What about the caps?
You may want to cap the ends of this mesh and make it a proper cylinder. This is likely to be necessary if your cylinder is going to stay as it is and not become something else. Usually, the cylinder will be the first stage of a more complicated mesh, and much of the time caps will not be required. In fact, you know what? I’ve never had to cap one before.
But, if you insist, let’s do this. A simple way would be create a vertex at the center of the cap, and build a triangle between that and each side.
void BuildCap(MeshBuilder meshBuilder, Vector3 centre, bool reverseDirection)
{
Vector3 normal = reverseDirection ? Vector3.down : Vector3.up;
//one vertex in the center:
int centreVertexIndex = meshBuilder.Vertices.Count - 1;
//vertices around the edge:
float angleInc = (Mathf.PI * 2.0f) / m_RadialSegmentCount;
for (int i = 0; i <= m_RadialSegmentCount; i++)
{
float angle = angleInc * i;
Vector3 unitPosition = Vector3.zero;
unitPosition.x = Mathf.Cos(angle);
unitPosition.z = Mathf.Sin(angle);
Vector2 uv = new Vector2(unitPosition.x + 1.0f, unitPosition.z + 1.0f) * 0.5f;
//build a triangle:
if (i > 0)
{
int baseIndex = meshBuilder.Vertices.Count - 1;
if (reverseDirection)
meshBuilder.AddTriangle(centreVertexIndex, baseIndex - 1,
baseIndex);
else
baseIndex - 1);
}
}
}
This is a modification of the BuildRing() code. Instead of building a quad between the current vertex, the previous vertex, and the previous ring, we build a triangle between the current vertex, the previous vertex, and the centre vertex.
Our UVs use the unitPosition, halved and offset by 0.5, meaning that the cap will be mapped in a circle across the UV space, centred on [0.5, 0.5].
The reverseDirection argument does two things: reverse the normal, and switch the winding order of the triangles. The end result being that the cap faces in the opposite direction. Now we can use this to cap both the top and bottom of our cylinder:
BuildCap(meshBuilder, Vector3.zero, true);
BuildCap(meshBuilder, Vector3.up * m_Height, false);
## Cylinder deformation - bending
Slightly more complicated is bending the cylinder. This requires a second, vertical circle to be calculated. Our cylinder, instead of pointing straight up, will follow the edge of this circle.
We need to calculate some things first:
Our bend angle is how much we want the cylinder to bend. A bend of 90 degrees will give us a right angle bend, and 360 degrees will give us a doughnut shape. m_BendAngle is, following the convention we described before, defined in degrees. As we’re about to enter trigonometry land, we need it converted.
This is the radius of our second circle. This has to be such that an arc defined by our bend angle will equal the height of the cylinder. This way the cylinder stays the same length as it bends. As radians, by definition, are the arc length of a unit circle, we just need to divide the height by this length.
float angleInc = bendAngleRadians / m_HeightSegmentCount;
This is how much we want increment the angle each time the height loop runs. It works much the same way as angleInc inside the BuildRing() function. Because we’re dividing the angle by the segment count, rather than dividing 2PI, we’ll get the arc rather than a full circle.
Vector3 startOffset = new Vector3(bendRadius, 0.0f, 0.0f);
This is the position on our vertical circle where the angle is equal to zero. X is equal to the cosine of zero times the radius (ie. bendRadius * 1) and Y is the sine of zero times the radius (bendRadius * 0). This is where the base of our cylinder would go, if we let it. The problem is that we want the origin of our mesh at zero, not off to the side of the vertical circle, so all the rings need to be pulled over to the center. We will use startOffset to do this.
Now on to our height loop:
for (int i = 0; i <= m_HeightSegmentCount; i++)
{
Vector3 centrePos = Vector3.zero;
centrePos.x = Mathf.Cos(angleInc * i);
centrePos.y = Mathf.Sin(angleInc * i);
centrePos -= startOffset;
float v = (float)i / m_HeightSegmentCount;
BuildRing(meshBuilder, m_RadialSegmentCount, centrePos, m_Radius, v, i > 0);
}
Here we have the familiar circle code, applied in the XY plane. The position of the point on the circle becomes the center of that ring.
We subtract the start offset from the centre position to force the base of the cylinder (where the angle is zero) back onto the origin. If we wanted, we could just subtract bendRadius from the X position, rather than using a Vector3. However the Vector3 code leaves us ready for starting angles other than zero (which we will need later for other shapes).
Let’s have a look:
Well, the cylinder as a whole is following the arc, but each ring is flat, as if the cylinder was still vertical. Also the lighting looks strange. This is because all the normals are still acting as though there was no bend.
What we need to do is rotate each ring:
for (int i = 0; i <= m_HeightSegmentCount; i++)
{
Vector3 centrePos = Vector3.zero;
centrePos.x = Mathf.Cos(angleInc * i);
centrePos.y = Mathf.Sin(angleInc * i);
float zAngleDegrees = angleInc * i * Mathf.Rad2Deg;
Quaternion rotation = Quaternion.Euler(0.0f, 0.0f, zAngleDegrees);
centrePos -= startOffset;
float v = (float)i / m_HeightSegmentCount;
BuildRing(meshBuilder, m_RadialSegmentCount, centrePos, m_Radius, v, i > 0, rotation);
}
We calculate a rotation using the current angle around the Z axis. Note that we need to convert the angle back to degrees for this.
So, how does the BuildRing() function then use the rotation?
void BuildRing(MeshBuilder meshBuilder, int segmentCount, Vector3 centre, float radius,
float v, bool buildTriangles, Quaternion rotation)
{
float angleInc = (Mathf.PI * 2.0f) / segmentCount;
for (int i = 0; i <= segmentCount; i++)
{
float angle = angleInc * i;
Vector3 unitPosition = Vector3.zero;
unitPosition.x = Mathf.Cos(angle);
unitPosition.z = Mathf.Sin(angle);
unitPosition = rotation * unitPosition;
meshBuilder.UVs.Add(new Vector2((float)i / segmentCount, v));
if (i > 0 && buildTriangles)
{
int baseIndex = meshBuilder.Vertices.Count - 1;
int vertsPerRow = segmentCount + 1;
int index0 = baseIndex;
int index1 = baseIndex - 1;
int index2 = baseIndex - vertsPerRow;
int index3 = baseIndex - vertsPerRow - 1;
}
}
}
A very simple change. All we are doing is rotating the unit position. Both the vertex positions and the normals are now properly offset:
## Important: Why, hello, Divide-by-Zero.
Something important to remember: This code breaks if the bend angle is zero. The vertical circle would have an infinite radius, as the cylinder can never bend around to meet itself. Because of this, our script needs to check the bend angle and just build a normal cylinder if it’s zero:
if (m_BendAngle == 0.0f)
{
//Build a normal cylinder
}
else
{
//Build a cylinder with a bend deformation
}
## A common cylinder deformation, the sphere
Spheres can also be made by deforming cylinders. The method is not completely dissimilar to the way we made our bent cylinder. We use a second, vertical circle, but instead of using it to offset the centre of each ring, we use it to offset the radius.
int heightSegmentCount = m_RadialSegmentCount / 2;
float angleInc = Mathf.PI / heightSegmentCount;
for (int i = 0; i <= heightSegmentCount; i++)
{
Vector3 centrePos = Vector3.zero;
Vector3 centrePos.y = -Mathf.Cos(angleInc * i) * m_Radius;
float radius = Mathf.Sin(angleInc * i) * m_Radius;
float v = (float)i / heightSegmentCount;
BuildRing(meshBuilder, m_RadialSegmentCount, centrePos, radius, v, i > 0);
}
Here, we are applying the coordinates of our second circle to the height and radius of the ring. Essentially, the second circle defines a vertical cross-section.
A minor point: to keep the number of segments in the vertical plane the same as in the horizontal, we use half the number of height segments as radial segments. Our vertical curve defines a semi-circle rather than a full one.
Looking at the result, we have a similar story to the cylinder bend. This time the shape is correct, but the normals are wrong. They are all pointing straight outward in the XZ plane, when we actually need a vertical tilt as they get toward the poles. We need to calculate them differently.
Fortunately, this is a sphere with its origin in the centre. That means that the normal at any point on the surface is equal to the normalised position of that point:
void BuildRingForSphere(MeshBuilder meshBuilder, int segmentCount, Vector3 centre, float radius,
float v, bool buildTriangles)
{
float angleInc = (Mathf.PI * 2.0f) / segmentCount;
for (int i = 0; i <= segmentCount; i++)
{
float angle = angleInc * i;
Vector3 unitPosition = Vector3.zero;
unitPosition.x = Mathf.Cos(angle);
unitPosition.z = Mathf.Sin(angle);
Vector3 vertexPosition = centre + unitPosition * radius;
meshBuilder.UVs.Add(new Vector2((float)i / segmentCount, v));
if (i > 0 && buildTriangles)
{
int baseIndex = meshBuilder.Vertices.Count - 1;
int vertsPerRow = segmentCount + 1;
int index0 = baseIndex;
int index1 = baseIndex - 1;
int index2 = baseIndex - vertsPerRow;
int index3 = baseIndex - vertsPerRow - 1;
}
}
}
If we call this instead of BuildRing() in our sphere code, we’ll have a sphere with correct normals:
## Another deformation - taper
Another possible deformation is a taper. This requires two radii, and our code interpolates from one to the other.
float heightInc = m_Height / m_HeightSegmentCount;
for (int i = 0; i <= m_HeightSegmentCount; i++)
{
Vector3 centrePos = Vector3.up * heightInc * i;
float v = (float)i / heightSegmentCount;
i > 0, Quaternion.identity);
}
We’re now calculating the radius inside the height loop, and interpolating from the start radius to the end radius. Setting either radius to zero will give us a cone.
But we’re not quite done yet. Our normals are still facing straight outward, instead of being tilted by the slope of the cylinder. We can’t fix this by rotating each ring, as that would rotate every vertex in the ring uniformly. What we need is an offset, so each normal can be adjusted relative to its base, horizontal value.
float heightInc = m_Height / m_HeightSegmentCount;
Vector2 slope = new Vector2(m_RadiusEnd - m_RadiusStart, m_Height);
slope.Normalize();
for (int i = 0; i <= m_HeightSegmentCount; i++)
{
Vector3 centrePos = Vector3.up * heightInc * i;
float v = (float)i / heightSegmentCount;
i > 0, Quaternion.identity, slope);
}
We are using the difference between the start and end radii and the height of our cylinder to calculate the slope (or rise and run, or tangent) of the cylinder.
Now to tell BuildRing() what to do with this value:
void BuildRing(MeshBuilder meshBuilder, int segmentCount, Vector3 centre, float radius,
float v, bool buildTriangles, Quaternion rotation, Vector2 slope)
{
float angleInc = (Mathf.PI * 2.0f) / segmentCount;
for (int i = 0; i <= segmentCount; i++)
{
float angle = angleInc * i;
Vector3 unitPosition = Vector3.zero;
unitPosition.x = Mathf.Cos(angle);
unitPosition.z = Mathf.Sin(angle);
float normalVertical = -slope.x;
float normalHorizontal = slope.y;
Vector3 normal = unitPosition * normalHorizontal;
normal.y = normalVertical;
normal = rotation * normal;
unitPosition = rotation * unitPosition;
meshBuilder.UVs.Add(new Vector2((float)i / segmentCount, v));
if (i > 0 && buildTriangles)
{
int baseIndex = meshBuilder.Vertices.Count - 1;
int vertsPerRow = segmentCount + 1;
int index0 = baseIndex;
int index1 = baseIndex - 1;
int index2 = baseIndex - vertsPerRow;
int index3 = baseIndex - vertsPerRow - 1;
}
}
}
First, we calculate vertical and horizontal lengths for our normal value. This is done using the slope vector, rotated 90 degrees. Our unit position is horizontal, so we multiply this by the horizontal scale, and plug the vertical scale directly into the normal’s Y value.
That done, we apply the rotation value, so that the normal is rotated along with the position.
And that’s our taper.
## Combining bend and taper
To get some more interesting results, we can combine two of the deformations we’ve learned so far, bend and taper:
float angleInc = bendAngleRadians / m_HeightSegmentCount;
Vector3 startOffset = new Vector3(bendRadius, 0.0f, 0.0f);
Vector2 slope = new Vector2(m_RadiusEnd - m_RadiusStart, m_Height);
slope.Normalize();
for (int i = 0; i <= m_HeightSegmentCount; i++)
{
Vector3 centrePos = Vector3.zero;
centrePos.x = Mathf.Cos(angleInc * i);
centrePos.y = Mathf.Sin(angleInc * i);
float zAngleDegrees = angleInc * i * Mathf.Rad2Deg;
Quaternion rotation = Quaternion.Euler(0.0f, 0.0f, zAngleDegrees);
centrePos -= startOffset;
float v = (float)i / m_HeightSegmentCount;
BuildRing(meshBuilder, m_RadialSegmentCount, centrePos, radius, v, i > 0, rotation, slope);
}
This really is a straight mash-up of the two deformations. We’re using our bend code, plus an interpolated radius and a slope offset, taken straight from our taper code.
That’s it for this tutorial. See you in the next, and final, part, where we make some much more interesting shapes: mushrooms and flowers.
On to part 2B: Making cylinders interesting |
# Calculate lim_(x ->x_o) (x^2- x_o x)/(x^2-x_o^2) for every value of x_o in RR ?
Feb 18, 2018
lim_(x→x_0) (x^2-x_0x)/(x^2-x_0^2) = 1/2
#### Explanation:
lim_(x→x_0) (x^2-x_0x)/(x^2-x_0^2) = lim_(x→x_0) (x(x-x_0))/((x-x_0)(x+x_0))=lim_(x→x_0) x/(x+x_0)
We can now let $x = {x}_{0}$ to get the limit as
${x}_{0} / \left({x}_{0} + {x}_{0}\right) = {x}_{0} / \left(2 {x}_{0}\right) = \frac{1}{2}$
Feb 18, 2018
${\lim}_{x \to {x}_{0}} \frac{{x}^{2} - {x}_{0} x}{{x}^{2} - {x}_{0}^{2}} = \left\{\begin{matrix}\frac{1}{2} \text{ for " x_0 !=0 \\ 1 " for } {x}_{0} = 0\end{matrix}\right.$
#### Explanation:
Simplify the function:
$\frac{{x}^{2} - {x}_{0} x}{{x}^{2} - {x}_{0}^{2}} = \frac{x \left(x - {x}_{0}\right)}{\left(x + {x}_{0}\right) \left(x - {x}_{0}\right)} = \frac{x}{x + {x}_{0}}$
so for ${x}_{0} \ne 0$:
${\lim}_{x \to {x}_{0}} \frac{{x}^{2} - {x}_{0} x}{{x}^{2} - {x}_{0}^{2}} = {\lim}_{x \to {x}_{0}} \frac{x}{x + {x}_{0}} = {x}_{0} / \left({x}_{0} + {x}_{0}\right) = \frac{1}{2}$
while for ${x}_{0} = 0$:
${\lim}_{x \to {x}_{0}} \frac{{x}^{2} - {x}_{0} x}{{x}^{2} - {x}_{0}^{2}} = {\lim}_{x \to 0} {x}^{2} / {x}^{2} = 1$ |
# Class 9 NCERT Mathematics Solutions - Class IX CBSE Board Maths Questions | Lines and Angles - Exercise 6.3 Answers
CBSE Class IX NCERT Mathematics Solutions
Chapter 6 - Lines and Angles
IXth NCERT Mathematics Textbook Exercise 6.3 Solved
(Page 107, 108)
Q1: In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If SPR = 135º and PQT = 110º, find PRQ.
Ans: In ∆ PQR side QP and RQ are produced to point S and T respectively such that / SPR = 135O, / PQT = 110O. We have to find / PRQ.
As / SPR + / RPQ = 180O (linear pair of angles).
Or, 135O + / RPQ = 180O
Or, / RPQ = 180O – 135O = 45O
Now, / TPQ = / QPR + / PRQ
Or, 110O = 45O + / PRQ
Or, / PRQ = 65O
Q2: In the given figure, X = 62º, XYZ = 54º. If YO and ZO are the bisectors of XYZ and XZY respectively of ΔXYZ, find OZY and YOZ.
Ans:
As the sum of all interior angles of a triangle is 180º, therefore, for ΔXYZ,
X + Y + Z = 180º
Or, 62º + 54º + Z = 180º
Or, ∠Z = 180º − 116º = 64O
Now, ∠OZY = 64O/2 = 32º (as OZ is the angle bisector of Z)
Similarly, OYZ = 54O/2 = 27O
Consider ΔOYZ,
OYZ + YOZ + OZY = 180º
Or, 27º + YOZ + 32º = 180º
Or, ∠YOZ = 180º − 59º = 121º
Q3: In the given figure, if AB || DE, BAC = 35º and CDE = 53º, find DCE.
Ans:
/ DEC = / BAC = 35O ………. (i) [Alternate interior angles]
/ CDE = 53O ………. (ii) [given]
In ∆CDE using angle sum property we have,
/ CDE + / DEC + / DCE = 180O
Or, 53O + 35O + / DCE = 180O
Or, / DCE = 92O
Q4: In the given figure, if lines PQ and RS intersect at point T, such that PRT = 40º, RPT = 95º and TSQ = 75º, find SQT.
Ans: Applying angle sum property in ∆PRT we have,
/ PTR + / PRT + / RPT = 180O
Or, / PTR + 40O + 95O = 180O
Or, / PTR = 45O
Or, / QTS = / PTR = 45O [vertically opposite angles]
Applying angle sum property in ∆TSQ we have,
/ QTS + / TSQ + / SQT = 180O
Or, / SQT + 45O + 75O = 180O
Or, / SQT = 60O
Q5: In the given figure, if PQ PS, PQ || SR, SQR = 28º and QRT = 65º, then find the values of x and y.
Ans:
/ QRT = / RQS + / QSR [as the exterior angle is equal to the sum of the two interior opposite angles].
Or, 65O = 28O + / QSR
Or, / QSR = 37O
Given that PQ PS i.e. / QPS = 90O
Or, PQ || SR.
So, / QPS + / PSR = 180o [sum of the consecutive interior angles on the same side of the traversal is 180O].
Therefore, 90O + / PSR = 180O
Or, / PSR = 90O
Or, / PSR + / QSR = 90O
Or, y + / QSR = 90O
Or, y + 37O = 90O
Or, y = 53O
Now consider ∆PQS,
/ PQS + / QSP + / QPS = 180O
Or, x + 53O + 90O = 180O
Or, x = 37O
Q6: In the given figure, the side QR of ΔPQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR= ½ QPR.
Ans:
/ QTR = ½ / QPR
As RT and QT are the bisectors of / PRS and / PQR respectively,
Therefore, / PRT = / TRS and / PQT = / TQR.
=> / TRS = / T + / TQR
=> 2/ TRS = 2/ T + 2/ TQR
=> / PRS = 2/ T + / PQR
=> / P + / PQR = 2/ T + / PQR
Or, / P = 2/ T
Or, / T = / P/2
[Exterior angle = Sum of opposite interior angle]. |
Probability Class 12th Notes - Free NCERT Class 12 Maths Chapter 13 Notes - Download PDF
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# Probability Class 12th Notes - Free NCERT Class 12 Maths Chapter 13 Notes - Download PDF
Edited By Ramraj Saini | Updated on Apr 23, 2022 01:32 PM IST
Probability Class 12 notes belong to the 13 chapter of NCERT. The NCERT Class 12 Maths chapter 13 notes are entirely based on the important topics required for the exam. Class 12 Maths chapter 13 notes nicely define the important formulas and their required derivations. The notes for Class 12 Maths chapter 13 helps a student to get a last-minute revision before the exam. Notes for Class 12 Maths chapter 13 is made in such a way that no students face any difficulty during their preparation. NCERT Notes for Class 12 Maths chapter 13 not only cover the NCERT notes but also covers the CBSE Class 12 Maths chapter 13 notes.
After going through Class 12 Probability notes
students can also refer to,
## NCERT Class 12 Maths Chapter 13 Notes
In general terms, probability is defined as a measurement of the uncertainty of events in random experiments. Mathematically it is the ratio of the number of outcomes to the total number of possible outcomes.
Conditional Probability
It gives us the way to find a reason for the experiment based on partial information. Let us think of such a situation that like
1. In a consecutive roll of dice, the sum is 9. What is the probability that the first roll is 6 and many more?
Thus, we can define the conditional probability that if A and B are two events of sample space S and P(A)≠0, then the probability of B after the event A has occurred is called conditional Probability.
Properties of Conditional probability
Property 1: P(F|F) = P(S|F) = 1
Property 2: If A and B are two events in the sample space S and F being an event of S such that P(F) ≠ 0, then
P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)
Property 3: P(E′|F) = 1 − P(E|F)
## Multiplication Theorem on Probability
For Given two dependent events A and B
P(A ∩ B) = P(A) P(B | A)
The multiplication rules of probability for more than two given events are as follows
Let E, F and G are three events given in a sample space, then we have
P(E ∩ F ∩ G) = P(E) P(F|E) P(G|(E ∩ F)) = P(E) P(F|E) P(G|EF)
Similarly, this rule can be extended for four or more events.
Independent Events
We have introduced conditional probability where partial information that event B provides about event A. Now we found an interesting case where the occurrence of B does not provide information on A.
i.e. P(B|A)=P(B)
Now we can deduce that
Example:
In a dice. If M is the event in which number appears as a multiple of 3 and N be the event in which even numbers occur. So, find that event M and N are independent or not?
Solution:
We know that the sample space is S = {1, 2, 3, 4, 5, 6}
There are even numbers in event M also.
Therefore,
So,
From the above relation, we can say M and N are independent events.
Bayes’ Theorem
Bayes theorem is a theorem in the probability that is used to determine the probability of the event that is related to any event that has already occurred.
The formula is
Total Probability Theorem
## Probability Distribution of The Random Variable
Let S be a sample space associated with a random experiment. A function R: S→R is termed a random variable.
Example:
From 52 cards well-shuffled deck two cards are drawn. Find the probability for a number of aces.
Solution:
Let the number of aces be A.
Therefore, 0,1 or 2 will be the values of A.
When ace doesn’t occur
When ace occurs once
When ace occurs twice
X 0 1 2 P(X) 144/169 24/169 1/169
Mean of a random variable
The mean of a random variable is used to locate the middle or the average value of the random variable.
The variance of a random variable
It is the expectation of the squared deviation of a random variable from its sample mean.
## Bernoulli Trails and Binomial Distribution
Bernoulli Trails
Trials that contain only two outcomes usually referred to as ‘success’ or ‘failure’ are known as Bernoulli trials.
Example:
If there are 7 white and 9 red balls in a container 6 balls are drawn successively. Find out the trails of 6 balls from the container will be Bernoulli’s trials after each draws of the ball as replaced or not replaced in the urn.
Solution:
1. If the trial numbers are finite then the drawing of the ball with replacement will be the success for drawing a white ball is p = 7/16. It will be the same for all six trials. So, the balls drawing with replacements are Bernoulli trials.
2. If the drawing of the ball is done without replacement then the probability of success of drawing white ball will be 7/16 for the first trial, 6/15 for the second trial or if a red ball is drawn for 1st time will be 7/15 and so on. From the above statement, we can say that the success will not be the same for all the trials, so the trials are not Bernoulli trials.
Binomial Distribution
It is a type of distribution that entire up the likely values that will take one or two independent values given under a set of assumptions.
Example:
Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.
Solution:
Let X denote the number of defective eggs in the 10 eggs drawn. Since the drawing is done with replacement, the trials are Bernoulli trials. Clearly, X has the binomial distribution with
This brings us to the end of the chapter
## Significance of NCERT Class 12 Math Chapter 13 Notes
Class 12 Probability notes will be really helpful for the revision of the entire chapter and getting a list of the important topics covered in the notes. Also, Class 12 Math chapter 13 Notes is useful for getting a glance of Class 12 CBSE Syllabuses and also for national competitive exams like BITSAT, and JEE MAINS. Class 12 Maths chapter 13 notes pdf download can be used for getting a hard copy and to prepare for the exam.
### NCERT Class 12 Notes Chapter Wise.
NCERT Class 12 Maths Chapter 1 Notes NCERT Class 12 Maths Chapter 2 Notes NCERT Class 12 MathsChapter 3 Notes NCERT Class 12 Maths Chapter 4 Notes NCERT Class 12 Maths Chapter 5 Notes NCERT Class 12 Maths Chapter 6 Notes NCERT Class 12 Maths Chapter 7 Notes NCERT Class 12 Maths Chapter 8 Notes NCERT Class 12 Maths Chapter 9 Notes NCERT Class 12 Maths Chapter 10 Notes NCERT Class 12 Maths Chapter 11 Notes NCERT Class 12 Maths Chapter 12 Notes NCERT Class 12 Maths Chapter 13 Notes
## NCERT Books and Syllabus
NCERT Book for Class 12 NCERT Syllabus for Class 12
1. Define Probability according to Probability Class 12 notes.
It is a department of maths that deals with the occurrence of random event
2. What is probability’s formula
It is defined as the possibility of an event to happen is equal to the ratio of the number of outcomes and the total number of outcomes.
3. Name the textbook that to be followed
Class 12 Math chapter 13 textbook should be followed.
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Get answers from students and experts
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
Option 1) Option 2) Option 3) Option 4)
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms−2 :
Option 1) 2.45×10−3 kg Option 2) 6.45×10−3 kg Option 3) 9.89×10−3 kg Option 4) 12.89×10−3 kg
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
Option 1) Option 2) Option 3) Option 4)
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
Option 1) Option 2) Option 3) Option 4)
In the reaction,
Option 1) at STP is produced for every mole consumed Option 2) is consumed for ever produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2
If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will
Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.
With increase of temperature, which of these changes?
Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.
Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is
Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023
A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is
Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9
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Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.
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GIS officer work on various GIS software to conduct a study and gather spatial and non-spatial information. GIS experts update the GIS data and maintain it. The databases include aerial or satellite imagery, latitudinal and longitudinal coordinates, and manually digitized images of maps. In a career as GIS expert, one is responsible for creating online and mobile maps.
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A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.
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If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi
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The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.
Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.
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A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts.
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Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.
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The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.
Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.
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A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.
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An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.
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Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.
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Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.
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A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.
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An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.
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Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.
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Carpenters are typically construction workers. They stay involved in performing many types of construction activities. It includes cutting, fitting and assembling wood. Carpenters may help in building constructions, bridges, big ships and boats. Here, in the article, we will discuss carpenter career path, carpenter salary, how to become a carpenter, carpenter job outlook.
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An individual who opts for a career as a welder is a professional tradesman who is skilled in creating a fusion between two metal pieces to join it together with the use of a manual or fully automatic welding machine in their welder career path. It is joined by intense heat and gas released between the metal pieces through the welding machine to permanently fix it.
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Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems.
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Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.
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A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.
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A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.
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When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.
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An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.
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Those who wish to make a dentist career in India must know that dental training opens up a universe of expert chances. Notwithstanding private practice, the present dental school graduates can pick other dental profession alternatives, remembering working in medical clinic crisis rooms, leading propelled lab examinations, teaching future dental specialists, or in any event, venturing to the far corners of the planet with International health and relief organizations.
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Individuals following a career as health inspectors have to face resistance and lack of cooperation while working on the sites. The health inspector's job description includes taking precautionary measures while inspecting to save themself from any external injury and the need to cover their mouth to avoid toxic substances. A health inspector does the desk job as well as the fieldwork. Health inspector jobs require one to travel long hours to inspect a particular place.
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For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.
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Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.
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Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.
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The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.
If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.
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Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.
A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.
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An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story.
They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.
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Fashion bloggers use multiple social media platforms to recommend or share ideas related to fashion. A fashion blogger is a person who writes about fashion, publishes pictures of outfits, jewellery, accessories. Fashion blogger works as a model, journalist, and a stylist in the fashion industry. In current fashion times, these bloggers have crossed into becoming a star in fashion magazines, commercials, or campaigns.
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Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.
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In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.
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Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.
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For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.
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In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.
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##### Editor
Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.
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Fashion journalism involves performing research and writing about the most recent fashion trends. Journalists obtain this knowledge by collaborating with stylists, conducting interviews with fashion designers, and attending fashion shows, photoshoots, and conferences. A fashion Journalist job is to write copy for trade and advertisement journals, fashion magazines, newspapers, and online fashion forums about style and fashion.
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A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.
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Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.
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A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.
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A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.
A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.
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Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.
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Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.
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A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.
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Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.
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A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.
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Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.
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ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.
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Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack
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Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.
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##### Product Manager
A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.
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An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.
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The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.
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##### UX Architect
A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.
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These worksheets will introduce your beginning geometry students to the concept of proportions of lengths.
#### How to Use Proportions to Determine the Length - Ratios and proportions are widely used in math and everyday lives. A ratio is a comparison of two numbers by division. It is the quotient that you get when you divide the first number by the second, non-zero number. Ratios represent proportions, and they are the quotients of numbers divided in a definite order. For this reason, you should be careful to write each ratio in the intended order. For instance, if you want to write the ratio of 2 to 4, you will write it as: 2/4 (as a fraction) or 2:4 (using a colon). To find the ratio of two quantities, such as length, you should express both quantities in the same unit of measure before you determine their quotient. For instance, to compare the value of lengths expressed in meters and centimeters, you should either convert the length in meters to centimeters or the length in centimeters to meters so that you can form a ratio with comparable quantities. If two articles have a similar shape, they are classified as "comparable." When two figures are comparable, the proportions of the lengths of their relating sides are equivalent. If you realize that two articles are comparable, you can utilize extents and cross items to discover the length of an obscure side. The scale factor is utilized to discover relative estimations to make sensible models. Relative methods have a similar proportion. A scale factor is the proportion of the model estimation to the genuine estimation in the easiest structure. In contrast to a scale proportion, the scale factor doesn't think about various units of estimation. A model vehicle with the scale factor of 1:20 implies that the vehicle is multiple times the size of the genuine vehicle. It likewise implies that the vehicle is multiple times the size of the model.
This collection of lessons and worksheets works off of the proportionality of similar figures. You will be given measures of the sides of lengths of a fixed similar polygon and be asked to find the length of corresponding sides. If two figures are similar corresponding sides should have proportional measures of length. This proportion is called the scale factor. You are trying to determine that scale factor. Scale factors are paramount to the process of modeling and taking a design from a screen to a live object. Your beginning geometry students will use these activity sheets to learn how to calculate the length of a given segment by applying the same proportion that is evidenced in other given line pairs.
# Print Proportions to Determine Length Worksheets
## Proportions to Determine Length Lesson
This worksheet explains how to use proportions to determine an unknown length. A sample problem is solved, and two practice problems are provided.
## Worksheet
Students will use proportions that are presented to find all unknown lengths. Ten problems are provided.
## Practice Worksheets
This worksheet explains how to use proportions to determine these measures. A sample problem is solved, and two practice problems are provided.
## Review and Practice
Students review how to do use measures that are unknown to find known measures. Six practice problems are provided.
## Skill Quiz
Students will demonstrate their proficiency with this skill. Ten problems are provided.
## Determine Length Check
You will show this skill in a number of ways. Three problems are provided, and space is included for students to copy the correct answer when given. |
# RS Aggarwal Solutions for Class 9 Maths Chapter 5 - Congruence of Triangles Ex 5A (5.1)
## RS Aggarwal Class 9 RS Aggarwal Solutions for Class 9 MathsChapter 5 - Congruence of Triangles Ex 5A (5.1) Solutions Free PDF
Question 1:
In the adjoining figure, PA $\perp$ AB, QB $\perp$ AB and PA = QB. If PQ intersects AB at O, show that O id the mid-point of AB as well as that of PQ.
Solution:
Given: PA $\perp$ AB, QB $\perp$ AB, and PA = QB
To prove: AO = OB and PO = OQ
Proof: In $\Delta$APO and $\Delta$BPO,
$\angle$PAO = $\angle$QBO = 90o [Given]
PA = QB [Given]
$\angle$PAO = $\angle$QBO [Since PA $\perp$ AB, and QB $\perp$ AB, PA || QB, and thus PQ is a transversal, therefore, alternate angles are equal]
So, by Angle-Side-Angle criterion of congruence, we have
$\Delta$APO $\cong$ $\Delta$BPO
$\Rightarrow$ AO = OB and PO = OQ [Since corresponding parts of congruent triangles are equal]
Thus, we have O is the midpoint of AB and PQ.
Question 2:
Let the line segments AB and CD intersect at O in such a way that OA = OD and OB = OC. Prove that AC = BD but AC may not be parallel to BD.
Solution:
Given: Line segments AB and CD intersect at O such that OA = OD and OB = OC.
To prove: AC = BD
Proof: In $\Delta$AOC and $\Delta$BOD, we have
AO = OD [Given]
$\angle$AOC Â = $\angle$BOD [Vertically opposite angles are equal]
OC = OB [Given]
So, by Side-Angle-Side criterion of congruence, we have,
$\Rightarrow$ AOC $\cong$ $\Delta$BOD
$\Rightarrow$ AC = BD [Since the corresponding parts of the congruent parts of the congruent triangles are equal]
$\Rightarrow$ $\angle$CAO = $\angle$BDO [by c.p.c.t]
Thus, we have, AC = BD
In case   $\angle$ODB = $\angle$OBD, then $\angle$CAO = $\angle$OBD which means alternate angles made by lines AB and BD with transversal AB are equal and then lines AC and BD will be parallel.
Question 3:
In the given figure, l||m and M is the mid-point of AB. Prove that M is also the mid-point of any line segment CD having its end points at l and m respectively.
Solution:
Given: Two lines l and m are parallel to each other. M is the midpoint of segment AB. The line segment CD meets AB at M.
To prove: M is the midpoint of CD, that is, CM = MD
Proof:
In $\Delta$AMC and $\Delta$BMD, we have
$\angle$MAC = $\angle$MBD [Since l and m are parallel, AB is the transversal, and thus, alternate angles are equal]
AM = MB [Given]
$\angle$AMC = $\angle$BMD [vertically opposite angles are equal]
So, by Angle-Side-Angle criterion of congruence, we have
$\Delta$AMC $\cong$ $\Delta$BMD
Therfore, by corresponding parts of the congruent triangles are equal, we have, CM = MD
Question 4:
In the given figure, AB = AC and OB = OC. Prove that $\angle$ABO = $\angle$ACO
Solution:
Given: AB = AC and O is an interior point of the triangle such that OB = OC
To prove: $\angle$ABO = $\angle$ACO
Construction: Join AO
Proof:
In $\Delta$AOB and $\Delta$AOC, we have
AB = AC [Given]
AO = AO [Common]
OB = OC [Given]
So, by Side-Side-Side criterion of congruence, we have,
$\Delta$ABO $\cong$ $\Delta$ACO
$\Rightarrow$ $\angle$ABO = $\angle$ACO [by corresponding parts of congruent triangles are equal]
Question 5:
In the given figure, ABC is a triangle in which AB = AC and D is a point AB. Through D, a line DE is drawn parallel to BC and meeting AC at E. Prove that AD = AE.
Solution:
Given: A $\Delta$ABC in which;
AB = AC
and, DE|| BC
Proof:
Since DE || BC and AB is atransversal.
So, $\angle$ADE = $\angle$ABC —— (i) [These are corresponding angles]
Also DE || BC and AC is a transversal
So, $\angle$AED = $\angle$ACB —— (ii) [These are corresponding angles]
But, AB = AC [Given]
So, $\angle$ABC = ACB ——- (iii)
as opposite angles are also equal in case sides are equal
So from (i), (ii) and (iii) we have
$\angle$ADE = $\angle$AED
and in $\Delta$ADE, this implies that AD = AE.
Question 6:
In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of $\Delta$ABC such that AX = AY. Prove that CX = BY.
Solution:
Given: AX = AY
To prove: CX = BY
Proof: In $\Delta$AXC and $\Delta$AYB, we have
AX = AY [Given]
$\angle$A = $\angle$A [Common angle]
AC = AB [Two sides are equal]
So, by Side-Angle-Side criterion of congruence, we have
$\Delta$AXC $\cong$ $\Delta$AYB
$\Rightarrow$ XC = YB [Since corresponding parts of congruent triangles are equal]
Question 7:
In the given figure, C is the mid-point of AB. If $\angle$DCA = $\angle$ECB and $\angle$DBC = $\angle$EAC, prove that DC = EC.
Solution:
Given: C is the mid-point of a line segment AB, and D is point such that,
$\angle$DCA = $\angle$ECB
and $\angle$DBC = $\angle$EAC
To prove: DC = EC
Proof:
In $\Delta$ACE and $\Delta$DCB we have;
AC = BC [Given]
$\angle$EAC = $\angle$DBC [Given]
Also, $\angle$DCA = $\angle$CDB + $\angle$DBA because exterior $\angle$DCA in $\Delta$DCB is equal to sum of interior opposite angles.
Again in $\angle$ACE, we have
ext. $\angle$BCE = $\angle$CAE + $\angle$AEC
But, $\angle$DCA = $\angle$BCE [Given]
$\Rightarrow$ $\angle$CDB + $\angle$DBA = $\angle$CAE + $\angle$AEC
$\Rightarrow$ $\angle$CDB = $\angle$AEC [$\angle$DBA = $\angle$CAE (Given)]
Thus in $\Delta$ACE and $\Delta$DCB,
$\angle$EAC = $\angle$DBC
AC = BC
and, $\angle$AEC = $\angle$CDB
Thus by Angle-Side-Angle criterion of congruence, we have
$\Delta$ACE $\cong$ $\Delta$DCB
The corresponding parts of the congruent triangles are equal.
So, DC = CE [by c.c.p.c.t]
Question 8:
In the given figure, BA $\perp$ AC and DE $\perp$ EF such that BA = DE and BF = DC. Prove that AC = EF.
Solution:
Given: AB $\perp$ AC and DE $\perp$ FE such that,
AB = DE and BF = CD
To prove: AC = EF
Proof:
In $\Delta$ABC, we have,
BC = BF + FC
and, in $\Delta$DEF
FD = FC + CD
But, BF = CD [Given]
So, BC = BF + FC
and, FD = FC + BF
$\Rightarrow$ BC = FD
So, in $\Delta$ABC and $\Delta$DEF, we have,
$\angle$BAC = $\angle$DEF = 90o [Given]
BC = FD [Proved above]
AB = DE [Given]
Thus, by Right angle-Hypotenuse-Side criterion of congruence, we have
$\Delta$ABC $\cong$ $\Delta$DEF
The corresponding parts of the congruent triangle are equal.
So, AC = EF [c.p.c.t]
AE = $\angle$BCD [Proved above]
Thus by Angle-Side-Angle criterion of congruence, we have
$\Delta$BCD $\cong$ $\Delta$BBAE
The corresponding parts of the congruent triangles are equal.
So, CD = AE [Proved]
#### Practise This Question
Preeti told Pavani that if the alternate interior angles are equal then lines need not be parallel. Is Preeti's statement true or false? |
# 5.1 Number Theory. The study of numbers and their properties. The numbers we use to count are called the Natural Numbers or Counting Numbers.
## Presentation on theme: "5.1 Number Theory. The study of numbers and their properties. The numbers we use to count are called the Natural Numbers or Counting Numbers."— Presentation transcript:
5.1 Number Theory
The study of numbers and their properties. The numbers we use to count are called the Natural Numbers or Counting Numbers.
Factors The natural numbers that are multiplied together to equal another natural number are called factors of the product. Example: The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Divisors If a and b are natural numbers and the quotient of b divided by a has a remainder of 0, then we say that a is a divisor of b or a divides b.
Prime and Composite Numbers A prime number is a natural number greater than 1 that has exactly two factors (or divisors), itself and 1. A composite number is a natural number that is divisible by a number other than itself and 1. The number 1 is neither prime nor composite, it is called a unit.
Rules of Divisibility OMIT THIS PART
The Fundamental Theorem of Arithmetic Every composite number can be written as a unique product of prime numbers. This unique product is referred to as the prime factorization of the number.
Finding Prime Factorizations Branching Method: Select any two numbers whose product is the number to be factored. If the factors are not prime numbers, then continue factoring each number until all numbers are prime.
Example of branching method Therefore, the prime factorization of 3190 = 2 5 11 29
1. Divide the given number by the smallest prime number by which it is divisible. 2.Place the quotient under the given number. 3.Divide the quotient by the smallest prime number by which it is divisible and again record the quotient. 4.Repeat this process until the quotient is a prime number. Division Method
Write the prime factorization of 663. The final quotient 17, is a prime number, so we stop. The prime factorization of 663 is 3 13 17 Example of division method 13 3 17 221 663
Greatest Common Divisor The greatest common divisor (GCD) of a set of natural numbers is the largest natural number that divides (without remainder) every number in that set.
Finding the GCD Determine the prime factorization of each number. Find each prime factor with smallest exponent that appears in each of the prime factorizations. Determine the product of the factors found in step 2.
Example (GCD) Find the GCD of 63 and 105. 63 = 3 2 7 105 = 3 5 7 Smallest exponent of each factor: 3 and 7 So, the GCD is 3 7 = 21
Least Common Multiple The least common multiple (LCM) of a set of natural numbers is the smallest natural number that is divisible (without remainder) by each element of the set.
Finding the LCM Determine the prime factorization of each number. List each prime factor with the greatest exponent that appears in any of the prime factorizations. Determine the product of the factors found in step 2.
Example (LCM) Find the LCM of 63 and 105. 63 = 3 2 7 105 = 3 5 7 Greatest exponent of each factor: 3 2, 5 and 7 So, the GCD is 3 2 5 7 = 315
Example of GCD and LCM Find the GCD and LCM of 48 and 54. Prime factorizations of each: 48 = 2 2 2 2 3 = 2 4 3 54 = 2 3 3 3 = 2 3 3 GCD = 2 3 = 6 LCM = 2 4 3 3 = 432
Next Steps Read Examples 2-7 Work Problems in text on p. 216 15-20, all; 35-55, odds; 63-67, all Do Online homework corresponding to this section
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# EX.2.2 Q2 Linear Equations in One Variable Solutions - NCERT Maths Class 8
Go back to 'Ex.2.2'
## Question
The perimeter of a rectangular swimming pool is $$154 \,\rm{m}$$. Its length is $$2 \,\rm{m}$$ more than twice its breadth. What are the length and the breadth of the pool?
## Text Solution
What is known?
(i) Perimeter of a rectangular swimming pool is $$154\,\rm{m}$$.
(ii) Its length is $$2\,\rm{m}$$ more than twice its breadth.
What is unknown?
Length and breadth of the pool.
Reasoning:
Form a linear equation by using the formula for the perimeter of a rectangle. Assume either the breadth or the length to be a variable.
Steps:
Let the breadth of swimming pool be $$x \,\rm{m}$$.
Therefore, the length of the swimming pool will be $$(2x + 2)\,\rm{m}$$
Perimeter of rectangular swimming pool: $$2(\text{Length + Breadth})$$
Therefore, $$2(x + 2x + 2) = 154$$
\begin{align}3x + 2 &= \frac{{154}}{2}\\3x + 2 &= 77\\3x &= 77 - 2\\3x &= 75\\x &= \frac{{75}}{3}\\x &= 25\\{\rm{breadth}} &= 25\,{\rm{m}}\\{\rm{length}} &= 2x + 2\\ &= 2(25) + 2\\ &= 52\,{\rm{m}}\end{align}
Length of the pool $$= 52 \,\rm{m}$$
Breadth of the pool $$= 25 \;\rm{m}$$
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### Essential Calculus: Early Transcendentals
Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280
# To d$${\bf{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle$$escribe all set of points for condition $$\left| {{\bf{r}} - {{\bf{r}}_0}} \right| = 1$$.
All set of points for condition $$\left| {{\bf{r}} - {{\bf{r}}_0}} \right| = 1$$ is described and it forms a sphere with radius $$1$$.
See the step by step solution
## Step 1: Concept of Subtraction of Vector
Subtraction of vectors:
Consider the two three-dimensional vectors such as $$a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle$$ and .
The vector subtraction of two vectors $$(a - b)$$ is,
\begin{aligned}{l}(a - b) &= \left\langle {{a_1},{a_2},{a_3}} \right\rangle - \left\langle {{b_1},{b_2},{b_3}} \right\rangle \\ &= \left\langle {{a_1} - {b_1},{a_2} - {b_2},{a_3} - {b_3}} \right\rangle \end{aligned}
Given:
Two three-dimensional vectors $${\bf{r}} = \langle x,y,z\rangle$$ and $${{\bf{r}}_0} = \left\langle {{x_0},{y_0},{z_0}} \right\rangle$$.
## Step 2: Calculate Subtraction of Vector
Consider the expression for magnitude of vector $$a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle (|{\rm{a}}|)$$.
$$|{\rm{a}}| = \sqrt {a_1^2 + a_2^2 + a_3^2} .......(1)$$
Write the expression for sphere with center $$C(h,k,l)$$ and radius $$r$$.
$${(x - h)^2} + {(y - k)^2} + {(z - l)^2} = {r^2}.......(2)$$
Write the expression for relation between vectors $$r$$ and $${{\bf{r}}_0}$$.
$$\left| {{\bf{r}} - {{\bf{r}}_0}} \right| = 1$$
From definition, substitute $$\langle x,y,z\rangle$$ for $$r$$ and $$\left\langle {{x_0},{y_0},{z_0}} \right\rangle$$ for $${{\bf{r}}_0}$$,\begin{aligned}{l}\left| {\langle x,y,z\rangle - \left\langle {{x_0},{y_0},{z_0}} \right\rangle } \right| = 1\\\left| {\left\langle {x - {x_0},y - {y_0},z - {z_0}} \right\rangle } \right| = 1.......(3)\end{aligned}
Find the value of $$\left| {\left\langle {x - {x_0},y - {y_0},z - {z_0}} \right\rangle } \right|$$ by using equation (1).
$$\left| {\left\langle {x - {x_0},y - {y_0},z - {z_0}} \right\rangle } \right| = \sqrt {{{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2} + {{\left( {z - {z_0}} \right)}^2}}$$
Substitute $$\sqrt {{{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2} + {{\left( {z - {z_0}} \right)}^2}}$$ for $$\left| {\left\langle {x - {x_0},y - {y_0},z - {z_0}} \right\rangle } \right|$$ in equation (3), $$\sqrt {{{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2} + {{\left( {z - {z_0}} \right)}^2}} = 1$$
Take square on both sides of equation.
\begin{aligned}{l}{\left( {\sqrt {{{\left( {x - {x_0}} \right)}^2} + {{\left( {y - {y_0}} \right)}^2} + {{\left( {z - {z_0}} \right)}^2}} } \right)^2} &= {1^2}\\{\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} + {\left( {z - {z_0}} \right)^2} &= {1^2}.......(4)\end{aligned}
Compare the equations (2) and (4), the equation (4) represents a sphere with radius $$1$$.
Thus, all set of points for condition $$\left| {{\bf{r}} - {{\bf{r}}_0}} \right| = 1$$ is described and it forms a sphere with radius $$1$$. |
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• RD Sharma Class 11 Solutions for Maths
# Class 11 RD Sharma Solutions – Chapter 19 Arithmetic Progressions- Exercise 19.7 | Set 2
### Question 11. A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount he will pay in the 30th installment?
Solution:
In the first installment, the man pays Rs. 100, so a1 = 100 and in next installment the man pays Rs. 105, so a2 = 105.
So we can conclude, common difference, d = 105 – 100 = 5
Thus, the money he will pay in 30th installment, a30 = a1 + 29d = 100 + 29 x 5 = 100 + 145 = 245
Hence, he will pay Rs. 245 in the 30th installment.
### Question 12. A carpenter was hired to build 192 window frames. The first day he made five frames and each day thereafter he made two more frames than he made the day before. How many did it take him to finish the job?
Solution:
On the first day, the carpenter makes 5 frames, so a = 5 and thereafter each day he makes 2 extra frames, so d = 2.
Total frames to be made, Sn = 192
We know the formula,
We will neglect n= -16 and consider n = 12
Therefore, the carpenter took 12 days to make 192 frames.
### Question 13. We know that the sum of interior angled of a triangle is 180°. Show that the sums of the interior angles of polygons with 3,4,5,6…. form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.
Solution:
The sum of interior angles of a polygon having n-sides is = (n-2) x 180°
Thus, for a polygon with 3 sides, we have a3 = (3-2) x 180° = 180°
similarly, for a polygon with 4 sides, we have a4 = (4-2) x 180° = 360°
for a triangle with 5 sides, we have a5 = (5-2) x 180° = 540°
Now, a4 – a3 = 360° – 180° = 180°
also, a5 – a4 = 540° – 360° = 180°, we are getting the same common difference for every succeeding and preceding term, hence we can say that it forms an AP.
Now, the sum of the interior angles for a 21 sided polygon = (21-2) x 180° = 3420°
### Question 14. In a potato race 20 potatoes are placed in a line at intervals of 4 m with the first potato 24 m from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
Solution:
Since, there are in total 20 potatoes and they are placed in a line at intervals of 4 m, therefore. n = 20 and d=4
First potato is placed at a distance of 24 m, so a1 = 24, now a2 = 28, similarly a20 = 24+ 19 x 4 = 100.
We know the formula,
therefore,
The contestant has to bring back the potatoes hence, he will run back and forth, so total distance covered = 1240 x 2 =2480 m.
### Question 15. A man accepts a position with an initial salary of Rs. 5200 per month. It is understood that he will receive an automatic increase of Rs. 320 in the very next month and each month thereafter.
(i) Find his salary for the tenth month.
(ii) What is total earnings during the first year?
Solution:
Given: initial salary, a = Rs. 5200 and d = Rs. 320.
(i) his salary in 10th month = a + 9d = 5200 + 9 x 320 = 8080
(ii) his total earnings during the first year
### Question 16. A man saved Rs. 66000 in 20 years. In each succeeding year after the first year he saved Rs. 200 more than what he saved in the previous year. How much did he save in the first year?
Solution:
Given: Sn = 66000, n=20, d=200.
We know the formula,
The man saved Rs. 1400 in the first year.
### Question 17. In a cricket team tournament 16 teams participated. A sum of Rs 8000 is to be awarded among themselves as prize money. If the last place team is awarded Rs 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?
Solution:
Total number of teams participating in the tournament, n = 16
Total prize money, Sn = 8000
The last placed team received prize money, a = 275
Suppose, for every successive team the prize money increases by d, then by the sum formula
therefore, the amount received by the first placed team = a+15d = 275 + 15 x 30 = 725
Hence, the first placed team received Rs 725 as the prize money.
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# The driver of a car traveling with a speed of 30 m/s towards a hill sounds a horn of frequency 600 Hz. If the velocity of sound in air is 330 m/s, the frequency of reflected sound as heard by the driver is then:A). 555.5HzB). 720HzC). 500HzD). 550Hz
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Hint: In this question, we will use the relation between frequency and speed; this will give us the required result of the frequency of reflected sound as heard by the driver. Further, we will discuss the basic concept of sound and frequency, for our better understanding.
Formula used:
$n' = \dfrac{v}{{v - {v_s}}} \times n$
$n'' = \dfrac{{v + {v_1}}}{v} \times n'$
Complete step-by-step solution
Here, as we know that the source is moving towards the hill, so apparent frequency of horn striking the wall is given by:
$n' = \dfrac{v}{{v - {v_s}}} \times n$
Substituting the given values in the above equation, we get:
\eqalign{ & n' = \dfrac{{330}}{{330 - 30}} \times 600 \cr & \Rightarrow n' = 330 \times 2 \cr & \Rightarrow n' = 660Hz \cr}
Now, as we know that for the reflected sound, driver acts as listener moving towards source, frequency is given as:
$n'' = \dfrac{{v + {v_1}}}{v} \times n'$
$n'' = \dfrac{{v + {v_1}}}{v} \times n'$
Substituting the given values in the above equation, we get:
\eqalign{& n’’ = \dfrac{{330 + 30}}{{330}} \times 600 \cr & \Rightarrow n’’ = 360 \times 2 \cr & \therefore n’’ = 720Hz \cr}
Therefore, we get the required result which gives the frequency of reflected sound as heard by the driver is given by the above result.
Additional information: We know the basic difference between speed and velocity i.e., speed is the measure of how fast an object can travel, whereas velocity tells us the direction of this speed. Speed is a scalar quantity that means it has only magnitude, whereas velocity is a vector quantity that means it has both magnitude and direction. The S.I unit of velocity is meter per second (m/sec).
Further, the frequency is defined as the number of waves that pass a fixed point in unit time. It can also be defined as the number of cycles or vibrations undergone during one unit of time.
The S.I unit of frequency is Hertz or Hz and the unit of wavelength is meter or m. further, we also know the S.I unit of time which is given by second or s.
Note: Here we should notice that the speed of sound is different in different mediums. Further, the speed that is more than the speed of sound is termed differently like- supersonic, hypersonic, etc. we should also remember that the speed of light cannot be achieved by any other transport or object. |
GED Mathematical Reasoning: Inequalities I | Open Window Learning
Algebraic Equations and Inequalities
# GED Mathematical Reasoning: Inequalities I
• An inequality looks like an equation, but instead of an equal sign it contains an inequality sign. The four basic inequality signs are: the “greater than” sign, the “less than” sign, the “greater than or equal to” sign and the “less than or equal to” sign.
“greater than”
“less than”
“greater than or equal to”
“less than or equal to”
• You may be happy to know that solving an inequality involves pretty much the same process as solving an equation, except for one small twist: If you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign.
• Why do we flip the inequality sign when we multiply or divide by a negative number?It all goes back to the nature of negative numbers and the fact that the larger the number part of a negative number is, the smaller the number actually is. Take for example the TRUE statement: 20 greater than 10
What happens when we divide both sides of this inequality by -2?
20 divided by -2 is equal to -10
10 divided by -2 is equal to -5
The resulting statement is: -10 greater than -5
However notice that this statement isn’t TRUE unless we flip the inequality sign. -10 is NOT greater than -5, it’s less than -5.
Flip the inequality sign
is TRUE
And the same scenario would be true if we had multiplied both sides of the inequality by a negative number. The resulting statement wouldn’t be true unless we flipped the inequality sign.
• In general, an open dot is used on the graph when the inequality sign is “greater than” or “less than” and a closed dot is used when the inequality sign is “greater than or equal to” or “less than or equal to.”
or open dot
or closed dot
Example 1
Solve:
Notice that the statement given in example one looks a lot like an equation, but instead of an equal sign we see a “greater than” sign. We will solve this inequality like an equation and all these things still hold true:
First, solving an inequality means to find the values that make the inequality true. These values make up the solution “set.”
Also, just like with an equation, to solve an inequality we’ll use the concept of opposite operations to get the variable by itself on one side of the inequality sign.
It is still a good idea to check the solution to make sure the result is a true statement.
To isolate the variable “x” on the left side, we will need two steps involving opposite operations. We first need to subtract 1 from both sides. And then as a second step, we will divide both sides by -2.
Now we are left with the one-step inequality: -2x greater than 10. To solve for x, we will divide both sides by -2.
On the left side, -2 divided by -2 equals one, leaving just “x” on the left side.
On the right side, we need to divide 10 by -2, which is equal to -5.
And since we divided both sides BY a negative number, we need to flip the inequality sign to “less than.”
The resulting solution is: “x” less than -5.
The solution: “x” less than -5 means that any number less than -5 will make the original inequality true. Let’s check this to be sure. Pick any number less than -5 and substitute it back into the original inequality for “x.” I’ll choose “x” equal to -7. When we let “x” equal -7, we get the statement: -2 times -7 plus 1 is greater than 11. Is this true?
Check: let
is TRUE
Since 15 is, in fact, greater than 11, yes! The statement is true and we can be confident in our solution.
Another thing you should know is that we can represent the solution set of an inequality visually using a number line. We can draw a picture of the solution set “x” less than -5 by shading all the numbers less than -5 on a number line.
Notice that I used an open dot ON the number -5. This is to denote that the number -5 is NOT included in the solution set, since it does not make the inequality true.
Let’s try it. If we let “x” equal -5, is -2 times -5 plus 1 greater than 11?
No, the left side simplifies to 11 and 11 is not greater than 11. So the statement is false, meaning that the number -5 should not be included in the solution set.
Let:
is FALSE
Example 2
Solve:
As we consider solving the inequality in example 2, let’s review the steps for solving multi-step equations. Remember – these are the same steps we need to use for solving inequalities, except for that additional step of flipping the inequality sign if we multiply or divide by a negative number.
• First, simplify each side of the equal sign separately, if possible, by distributing to get rid of any parenthesis and combining like terms.
• Second, add and subtract to get the variable terms on one side of the equation and the constant terms on the other side.
• Third, multiply or divide to get the variable by itself on one side of the equal sign.
The inequality in example 2 contains parenthesis so let’s first distribute the -4 on left side.
Next, let’s combine the “like terms”
Now that we’ve completed step one of the process by distributing and simplifying what we can, let’s move on to step 2 and solve for “n” by first adding 20 to both sides. This will put the variable term on the left side of the inequality and the constant terms on the right side.
Now we are left with a one-step inequality: 3n less than or equal to 30.
Following step 3 of the process, we will solve for “n” by dividing both sides by 3.
We did not divide by a negative number, so we will not flip the inequality sign. The resulting solution is: “n” less than or equal to 10
Visually, this solution set looks like this, where the numbers less than 10 are shaded and there’s a closed dot on the number 10 to denote that the number 10 IS included in the solution set.
We can check our solution by choosing a number in the shaded region and substituting it into our original inequality. Let’s choose “n” equal to zero. When we simplify the left side according to the order of operations, is it less than or equal to 10?
Check:
is TRUE
Yes! The resulting statement is: -20 less than or equal to 10, which is a true statement.
As a final note, I would like to show you why we include the endpoint of 10 in the solution set. Let’s look at what happens when we substitute “n” equals 10 back into the original inequality.
Check:
is TRUE
When we simplify the left side according to the order of operations, we get: 10 less than or equal to 10. While 10 is not less than 10, it IS equal to 10. Therefore, the statement is true and the number 10 is included in the solution set.
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Description
Rational functions are often difficult for students because they require understanding of zeroes, asymptotes, point discontinuities, and end behaviors. We will investigate behaviors of rational functions symbolically and graphically. Examples will include tasks that indicate what students need to know for work beyond their high school classes.
Description
Rational functions are often difficult for students because they require understanding of zeroes, asymptotes, point discontinuities, and end behaviors. We will investigate behaviors of rational functions symbolically and graphically. Examples will include tasks that indicate what students need to know for work beyond their high school classes.
What do students need beyond high school?
Definitions
Rational function: A quotient of two polynomials
Polynomial: A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients.
What is simpliest rational function?
$$\dfrac{a}{b} \hspace{2em} \dfrac{x+a}{b} \hspace{2em} \dfrac{a}{x+b} \hspace{2em} \dfrac{x+a}{x+b} \hspace{2em} \ldots$$
Summary: Zeros
Conjecture: Sometimes a rational function is zero when numerator is zero
Reason: $$\dfrac{P(x)}{Q(x)} = 0$$, multiply both sides by $$Q(x)$$, so $$P(x) = 0$$
Exception: Values must be in domain. Consider $$\dfrac{x^3 + x^2 + x + 1}{x + 1}$$. Notice $$x^3 + x^2 + x + 1 = 0$$ when $$x = -1$$, which is not in the domain. (see point discontinuity)
Summary: Vertical asympototes
Conjecture: Sometimes a rational function has a vertial asymptote when the denominator is zero
Reason: $$\dfrac{P(x)}{Q(x)} = DNE$$, when $$Q(x) = 0$$
Exception: "Zeros" may factor out. Consider $$\dfrac{(x - 1) (x - 2)}{x - 1} = x - 2$$ when $$x \neq 1$$. (see point discontinuity)
Removable ("point") discontinuity
Conjecture: Rational function has a point that is undefined when both the numerator and denomonator are zero.
Reason: Consider $$\dfrac{(x - 1) (x - 2)}{x - 1}$$. This simplifies to $$x - 2$$ when $$x \neq 1$$.
Exception: multiplicities in denominator
End behaviors
Conjecture: Divide the polynomials to identify how the function behaves as $$x \to \infty$$
Reason: After dividing the remainder goes to zero for large x values, e.g., $$\dfrac{1+x+x^2}{1+x} = x + \dfrac{1}{1+x}$$
Conjecture: There are other ways to write rational functions.
Desmos example
What I want as a professor:
exact answers, bad example: $$x^2 = 2$$ when $$x = 1.414$$
algebra skills, $$x = 0$$ vs. $$f(x) = 0$$
math questions and healthy skepticism
conjectures, then justification
(DO NOT WANT) memorized rules like horizontal asymptotes
What do students need beyond highschool?
- ability to find patterns
- ability to disect complex ideas
Last thing: Desmos review
Plesantly surprised
Removeable discontinuities could be better
Works off line |
# Question #04717
Mar 23, 2017
Given: $\ln \left(\frac{x - 5}{x - 1}\right) = \ln \left(6\right)$
Subtract $\ln \left(6\right)$ from both sides:
$\ln \left(\frac{x - 5}{x - 1}\right) - \ln \left(6\right) = 0$
Use the property of logarithms where subtracting is equivalent to division within the argument:
$\ln \left(\frac{x - 5}{6 \left(x - 1\right)}\right) = 0$
Make both sides the exponent of e:
${e}^{\ln \left(\frac{x - 5}{6 \left(x - 1\right)}\right)} = {e}^{0}$
The logarithm and the exponential on the left side cancel and the right side becomes 1:
$\frac{x - 5}{6 \left(x - 1\right)} = 1$
Now, it is easy to solve for x:
$x - 5 = 6 x - 6$
$5 x = 1$
$x = \frac{1}{5}$
Check:
$\ln \left(\frac{\frac{1}{5} - 5}{\frac{1}{5} - 1}\right) = \ln \left(6\right)$
$\ln \left(\frac{1 - 25}{1 - 5}\right) = \ln \left(6\right)$
$\ln \left(\frac{- 24}{- 4}\right) = \ln \left(6\right)$
$\ln \left(6\right) = \ln \left(6\right)$
This checks.
$x = \frac{1}{5}$
Mar 24, 2017
$x = \frac{1}{5}$
#### Explanation:
The contents of the two logarithms, which are the only things present on either side of the equation, are equal:
$\frac{x - 5}{x - 1} = 6$
$\implies x - 5 = 6 \left(x - 1\right)$
$\implies x - 5 = 6 x - 6$
$\implies 1 = 5 x$
$\implies x = \frac{1}{5}$ |
Circles, Cylinders and Circular Shapes
Circles, cylinders and circular shapes follows on from area of 2D shapes and surface area of 3D cuboids and prisms which students study in Term 2 of Year 8.
In this unit students learn how to calculate the circumference and area of circles both as decimals and in terms of π. Learning progresses from 2D circles to finding the total surface area and volume of cylinders.
Prerequisite Knowledge
• Derive and apply formulae to calculate and solve problems involving: perimeter and area of triangles, parallelograms, trapezia, volume of cuboids (including cubes) and other prisms
• Calculate and solve problems involving: perimeters of 2-D shapes and composite shapes.
Key Concepts
• The radius is the distance from the centre to any point on the circumference. The plural of radius is radii.
• The diameter is the distance across the circle through the centre.
• π is a Greek letter used to represent the value of the circumference of a circle divided by its diameter.
• The circumference is the distance about the edge of a circle. The circumference of a circle can be calculated as:
• C = πD where D is the diameter, or,
• C = 2πr where r is the radius.
• The area of a circle can be calculated using the formula
• A = πr2 where r is the radius.
• A cylinder is a circular prism.
Working mathematically
Develop fluency
• Use language and properties precisely to analyse numbers, algebraic expressions, 2-D
and 3-D shapes, probability and statistics.
• Use algebra to generalise the structure of arithmetic, including to formulate
mathematical relationships
• Substitute values in expressions, rearrange and simplify expressions, and solve
equations
Reason mathematically
• Make and test conjectures about patterns and relationships; look for proofs or counterexamples
• Begin to reason deductively in geometry, number and algebra, including using geometrical constructions
Solve problems
• Begin to model situations mathematically and express the results using a range of
formal mathematical representations
• Select appropriate concepts, methods and techniques to apply to unfamiliar and nonroutine
problems.
Circles, Cylinders and Circular Shapes Subject Content
Shape
• Derive and illustrate properties of triangles, quadrilaterals, circles, and other plane figures [for example, equal lengths and angles] using appropriate language and technologies
• Calculate and solve problems involving: perimeters of 2-D shapes (including circles), areas of circles and composite shapes
• Derive and apply formulae to calculate and solve problems involving: perimeter and area of circles and cylinders.
Mr Mathematics Blog
Teaching Mathematics for a Growth Mindset
Inspiring students to enjoy maths and feel the success that comes with attempting a difficult challenge is why I teach. The feeling of success is addictive. The more students experience it the more they want it and the further out of their comfort zone they are willing to go to get more of it. Teaching […]
Equation of Straight Line Graphs
To find the equation of straight line graphs students need to calculate the gradient using two pairs of coordinates and the intercept which is the y value of where it crosses the vertical axis. Examiner’s reports of past exam questions show students are more able to find the intercept of a straight line than they […]
Arithmetic and Geometric Sequences
Students learn how to generate and describe arithmetic and geometric sequences on a position-to-term basis. Learning progresses from plotting and reading coordinates in the first quadrant to describing geometric sequences using the nth term. This unit takes place in Term 4 of Year 10 and is followed by the equations of straight line graphs. Arithmetic […] |
# Curriculum: Autumn Term 2018
#### Maths
In this term, we will learn about:
• Finding pairs with a total of 100; adding to the next multiple of 100 and subtracting to the previous multiple of 100; subtract by counting up to find a difference; adding several numbers
• Read, write 4-digit numbers and know what each digit represents; compare 4-digit numbers using < and > and place on a number line; add 2-digit numbers mentally; subtract 2-digit and 3-digit numbers
• Learn × and ÷ facts for the 6 and 9 times-table and identify patterns; multiply multiples of 10 by single-digit numbers; multiply 2-digit numbers by single-digit numbers (the grid method); find fractions of amount
• Tell and write the time to the minute on analogue and digital clocks; calculate time intervals; measure in metres, centimetres and millimetres; convert lengths between units; record using decimal notation
• Add two 3-digit numbers using column addition; subtract a 3-digit number from a 3-digit number using an expanded column method (decomposing only in one column)
• Double 3-digit numbers and halve even 3-digit numbers; revise unit fractions; identify equivalent fractions; reduce a fraction to its simplest form; count in fractions (each fraction in its simplest form)
• Look at place value in decimals and the relationship between tenths and decimals; add two 4-digit numbers; practise written and mental addition methods; use vertical addition to investigate patterns
• Convert multiples of 100 g into kilograms; convert multiples of 100 ml into litres; read scales to the nearest 100 ml; estimate capacities; draw bar charts, record and interpret information
• Round 4-digit numbers to the nearest: 10, 100 and 1000; subtract 3-digit numbers using the expanded written version and the counting up mental strategy and decide which to use
• Use the grid method to multiply 3-digit by single-digit numbers and introduce the vertical algorithm; begin to estimate products; divide numbers (up to 2 digits) by single-digit numbers with no remainder, then with a remainder
#### English
In this unit, the children explore the Big Question: Who killed Tutankhamen? They read the interactive eBook, retrieving and collating information and identifying evidence in the text to support their theories. They investigate main and subordinate clauses and are introduced to the perfect tense. In their writing tasks, they write letters, paragraphs, and finally compose an explanation text in response to the Big Question.
The children listen to the story and predict what may happen at various points. They focus on character, setting and mood, asking questions as the story progresses and locating evidence in the text to answer specific questions. They understand why descriptive sentences are important and how setting affects mood. They revise and develop punctuating direct speech and then work on nouns, adjectives and expanded noun phrases. The writing tasks involve planning and writing a description of setting, and planning and writing a conversation using direct speech and correct punctuation.
#### Science
States Of Matter
By the end of the unit the children will be able to…
• Sort materials into solids, liquids and gases.
• Explain that heating causes melting, and cooling causes freezing.
• Identify the melting and freezing point of water.
• Describe evaporation and condensation using practical examples.
• Describe the effect of temperature on evaporation referring to their investigation.
• Identify the stages of the water cycle. • Predict what will happen in an investigation.
• Make observations.
• Explain thebehaviourof the particles in solids, liquids and gases.
• Explain how heating and cooling causes materials to melt and freeze.
• Explain why a material’s melting and freezing point is the same temperature.
• Explain how heating and cooling can cause materials to evaporate and condense.
• Explain why a higher temperature will speed up evaporation.
• Use the water cycle to explain why the water we have on Earth today is the same water that has been here for millions of years.
• Set up reliable and accurate investigations.
• Make and explain predictions.
• Make and record accurate observations.
• Use scientific language to explain their findings.
• Be able to ask and answer questions based on their learning using scientific language.
• Describe the properties of solids, liquids and gases.
• Explain that melting and freezing are opposite processes that change the state of a material.
• Identify the melting and freezing point of several different materials.
• Explain that heating causes evaporation and cooling causes condensation.
• Explain that evaporation and condensation are opposite processes that change the state of a material.
• Explain that the higher the temperature, the quicker water evaporates.
• Explain what happens to water at the different stages of the water cycle.
• Make observations and conclusions.
• Be able to answer questions based on their learning.
#### History
Crime and Punishment
By the end of the unit the children will be able to…
• Talk about some of the key facts about punishments that were used during the Roman, Anglo-Saxon, Tudor and Victorian times.
• Recall key facts about the life of Dick Turpin and talk about differences in how he is portrayed in various historical sources.
• Talk about and compare the punishments that were used during the Roman, AngloSaxon, Tudor and Victorian times and give some reasons for them.
• Explain some key terms in the history of crime and punishment in Britain, such as wergild, trial by ordeal, tithings, hue and cry, treason, transportation and hard labour.
• Use primary sources to decide what are facts, what opinions can be formed from the evidence, and identify the questions they have about the life of the highway man Dick Turpin.
• Compare modern day crime and punishment with those from the past, and talk about the legacy of past methods of crime prevention and detection with those of the present day.
• Explain their understanding of the different experiences of people who may have committed crimes according to their status in society e.g. a slave compared with a noble during the Roman period.
• Compare and contrast a variety of historical sources to form their own conclusions and questions regarding the life of the highway man Dick Turpin.
• Imagine and write about the experiences of people living during the historical periods studied based on factual evidence.
#### Islamic Studies
In this term, we will learn about
• Names of Allah
• Angels and their duties
• Characteristics of the Prophet (SAW)
• Miracles
• Signs of the last day
• Day of Judgement
Surahs
• Surah Naas
• Surah Fajar
Duas
• Grade 4
#### P.E.
This term we are concentrating on games. We will be looking at the following objectives:
• Can choose the most appropriate throw to use within a game
• Can call out for a catch in a game showing they know they are in the best place
• Can hit a ball into a space to help increase the score within a game; can dribble a ball in different directions and avoid obstacles
• Can help other members of the team to find space within the team game
• Can use a range of attacking and defending skills when playing a team game.
# Class Teachers
Apa Parveen Akhtar
Class Teacher
Apa Maria Khan
Quran & Islamic Studies |
# ECON 360 Probability Theory and Distributions Assignment
### ECON 360 Probability Theory and Distributions Assignment
In ECON 360 Probability Theory and Distributions Assignment students can clear the probability question because this assignment has different type of probability question.
## Question
Below are a number of problems. You need to solve these individually, and write the solution in the report that you submit. The problems are:
A lottery game contains 49 balls, each with a number, ranging from 1 through 49. A player has to choose six numbers. When the lottery is being held, six balls from the set of balls are chosen at random.
• Calculate the probability that the player has picked all six balls correctly.
• Calculate the probability that the player has picked?ve balls correctly.
• Calculate the probability that the player has picked four balls correctly.
• Calculate the probability that the player has picked three balls correctly.
• One lottery ticket costs \$1. The payout is as follows:
• Six correct numbers: 2.5 million dollars.
• Five correct numbers: Ten thousand dollars.
• Four correct numbers: Two hundred dollars.
• Three correct numbers: Ten dollars.
Assuming a player plays the lottery 10 million times, how much can he expect to win or lose?
So, if he were to play inde?nitely, how much can he expect to win or lose for every dollar he wagers?
A blank page consists of a number of horizontal lines, located with a vertical distanceL from each other. You drop a needle of lengthLon the page. What is the probability that the needle will cross one of the lines, if the needle has fallen fully on the page? Note that the needle has to cross a line, lying on top of a line does not count as crossing it.
For Probability Help: Statistical Methods
A lunch place next to an of?ce building is open every day from 11 AM to 3 PM (4 hours). The cost to run the place is \$200 per hour. On average the shop receives 11 customers per hour, who spend on average \$20 each. On a given day, what is the probability that the shop will work with a loss?
A roulette wheel has 21 red numbers, 21 black numbers, and four zeros. A player places \$20 bets on the red numbers. If the roulette spins a red number, the player get his \$20 back, and wins another \$20. If the roulette does not spin a red number, the player loses his bet. The player sits down with \$40, and keeps on playing until he has no more money. How many spins can he expect to be able to play until he has no more money?
### Some tips:
Write down the?rst couple of combinations with the lowest number of spins that will bust him. For example, the following combination of spins will bust him after two spins:
loss, loss. Because after the?rst spin he has \$20 left, after the second he is broke. The following combinations will bust him after four spins, loss, win, loss, loss, win, loss, loss, loss. In the?rst case, after the?rst spin, he has \$20. After the second spin he has \$40 again, which he loses after two more spins. In the second case, after the?rst spin, he has \$60, which he loses after three spins.
Calculate the probability that the player is bust after 2 spins, after 4 spins, 6 spins, and so on, and add these probabilities up. |
# document.write (document.title)
Math Help > Number Theory > Squares > Quadratic Residue
. . . . . . reference other pages in the "number theory" section with many of these subjects
## Modular Arithmetic and Quadratic Equations
How does one determine the solvability and the solutions of an equation such as
ax2 + bx + c = 0 (mod R)
The problem boils down to the question: Which nonzero residues (mod R) have square roots (mod R)? That is, for which a in NR are there x in NR such that
x2 = a (mod R)?
The following two definitions apply only to nonzero elements of NR: If a is a nonzero residue (mod R) which has a square root (mod R), then a is said to be a quadratic residue (mod R); otherwise, a is said to be a non-quadratic residue (mod R) (textbooks often say quadratic non-residue instead of non-quadratic residue).
It is always the case that 1 is a quadratic residue mod R; the main interest is in determining if there are any others.
Examples:
R Quadratic Residues Non-quadratic Residues 4 1 2, 3 5 1, 4 2, 3 9 1, 4, 7 2, 3, 5, 6, 8 15 1, 4, 6, 9, 10, 2, 3, 5, 7, 8, 11, 12, 13, 14
The following theorems provide us with computational tools to determine if for odd prime number P, a given number in NP has a square root mod P.
Theorem (Euler) Let P be an odd prime number and let a be a nonzero number in NP. Then
a(p-1)/2 ≡�1 (mod P). (Recall that in NP, -1 is the same thing as P-1.)
Theorem (Euler's Criterion): Let P be an odd prime number and let a be a nonzero number in NP. Then the following are equivalent:
1. a is a quadratic residue modulo P.
2. a(p-1)/2 mod P = 1.
Hence we can also conclude that a is a nonquadratic residue if, and only if, a(p-1)/2 mod P = P-1.
### Legendre Symbol
The quantity a(p-1)/2 mod P is denoted by a special symbol, called the Legendre symbol, which is
(a/P) = a(P-1)/2 mod P.
(a/P) is 1 iff a is a quadratic residue (mod P). The Legendre symbol has the following properties which are quite useful in computing whether one number is a quadratic residue modulo another, which is an odd prime:
Theorem: Let P be an odd prime number and let a and b be nonzero elements of NP. Then:
1. (ab/P) = (a/P) (b/P).
2. (1/P)=1.
3. (-1/P) = (-1)(p-1)/2.
From item 3 in the last theorem one gleans:
Corollary: Let P be an odd prime number. Then
1. (-1/P) = 1 if p mod 4 = 1, i.e. if p is a "Pythagorean Prime"
2. (-1/P) = -1 if p mod 4 = 3.
Moreover, there are for a given odd prime number P as many quadratic residues mod P as there are non-quadratic residues mod P:
Theorem: For an odd prime number P, there are (P-1)/2 quadratic residues mod P, and there are (P-1)/2 non-quadratic residues mod P.
The following important theorem is due to Gauss:
Theorem (Gauss): If P is an odd prime number, then (2/P) is equal to (-1) raised to the power (P2-1)/8.
The deepest of the results regarding quadratic residues was first proven rigorously by Gauss, and is known as the Quadratic Reciprocity Law. It states:
Theorem (Quadratic Reciprocity Law): Let P and Q be odd prime numbers. Then
(P/Q) (Q/P) = (-1)[(P-1)/2][(Q-1)/2] .
Another way to express the quadratic reciprocity law is:
If p and q are primes, at least one of which can be expressed as 4k+1, then p is a quadratic residue (mod q) if and only if q is a quadratic residue (mod p).
If p and q are primes, both of which can be expressed as 4k+3, then p is a quadratic residue (mod q) if and only if q is not a quadratic residue (mod p).
Alan Baker's book "A concise introduction to the theory of numbers" uses this as an example of how to use the law of quadratic reciprocity to show that (-3/p) = (p/3). First, by quadratic reciprocity,
(p/3) (3/p) = (-1)[(p-1)/2][(3-1)/2] = (-1)(p-1)/2, so
(-1)(p-1)/2(3/p)=(p/3).
Then, by the identities (ab/P) = (a/P) (b/P) and (-1/P) = (-1)(p-1)/2,
(-3/p) = (-1/p)(3/p)=(-1)(p-1)/2(3/p)=(p/3).
Now, to find the primes, p, for which -3 is a quadratic residue (mod p) -- which are also the primes, p, that are a quadratic residue (mod 3),
(p/3) = p(3-1)/2 (mod 3) = p (mod 3)
if p = 6k+1, then p (mod 3)=1, and -3 is a quadratic residue (mod p).
but if p = 6k-1 (the only other possibility) then p (mod 3)=-1, and -3 is a quadratic nonresidue (mod p).
Primes which have a given number, d, as a quadratic residue:
Mathworld's article, Quadratic Residue includes a table giving the primes which have a given number, d, as a quadratic residue (left). One of the many delightful fractal images (right) shows (y/x), i.e. a dot is shown if y is a quadratic residue (mod x). The dark horizontal lines are the numbers that are square in any modulus -- 1, 4, 9, 16, ... The diagonal lines are the squares that are between 1 and 2 times the modulus. That is, the points for which (y/x)=1 because y+x is a square. If you look very carefully, you can also see some steeper diagonal lines (with a slope of -2) that represent the points (y/x) for which y+2x is a square. And, of course, the snowstorm of black dots in between these lines represent points (y/x) for which y+mx is a square, for some value of m, so in a sense, all the dots are part of some diagonal with slope -m.
d Primes -6 24k+1, 5, 7, 11 -5 20k+1, 3, 7, 9 -3 6k+1 -2 8k+1, 3 -1 4k+1 2 8k±1 3 12k±1 5 10k±1 6 24k±1, 5
How do you compute a square root mod P?
We now have a computational method for determining for an odd prime number P whether a given a in NP is a quadratic residue mod P, or not. Once it is known that such a square root exists, one can proceed to try and find such a square root. How is this done?
A second, equally important question, is: suppose we know that a is a quadratic residue modulo the prime number P: Can we find a quadratic residue b modulo P such that b2 = a mod P? And what about the case when P is a composite number?
In the case when P = 3 mod 4, or where the modulus is a product of two distinct prime numbers, each equal to 3 mod 4, there is a simple algorithm for this:
First, one solves the quadratic equation x2 = a mod P for the case when P is an odd prime equal to 3 mod 4 and a is a quadratic residue mod P. The method is given in the following theorem.
Theorem: Let P be a prime number such that P mod 4 = 3. Let a be a quadratic residue mod P. Define k by (P-1)/2 = 2*k + 1. Then
x = ak+1 mod P
is a solution for x2 = a mod P. Moreover, x is a quadratic residue mod P, and is the only such solution to this equation.
Then, using the method for doing this, one solves the equation x2 mod R = a where R = P*Q and P and Q are two prime numbers, each equal to 3 mod 4. This is done by first solving each of the equations x2 = a mod P and x2 mod Q = a, and then using the Chinese remainder theorem to find the final result.
Thus, say we have found p in NP and q in NQ such that
p2 mod P = a and
q2 mod Q = a.
Then we use the technique of the Chinese Remainder Theorem to compute from p and q an element x of NR such that
x2 mod R = a.
(Tonelli-Shanks algorithm)
### Internet references
Mathpages: The Jewel of Arithmetic: Quadratic Reciprocity -- Euler's Criterion for quadratic residue is that
a is a quadratic residue (mod p) iff a(p-1)/2 = 1 (mod p).
Mathworld: Quadratic Residue includes a table giving the primes which have a given number d as a quadratic residue:
### Related Pages in this website
Irrationality Proofs -- Proofs that π and e are irrational.
Pythagorean Theorem
Arithmetic Sequence of Perfect Squares, page 3 -- If a2, b2, c2 are in arithmetic sequence, why is their constant difference a multiple of 24? Look at the second answer to this question for the relationship between Pythagorean Triples and this arithmetic sequence of squares.
Legendre Symbol - (a/p) = 1 iff a is a quadratic residue, i.e. nonzero square, (mod p), where p is an odd prime
The webmaster and author of this Math Help site is Graeme McRae. |
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# Find the number of ways in which 5 beads, chosen from 8 different beads can be threaded on a ring.
Last updated date: 13th Jul 2024
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Hint: At first count the number of ways we can choose 5 beads from 8 different beads. Then count in how many ways we can arrange them.
It is given in the question that we have 8 different beads.
We know that the combination is a way of selecting items from a collection, such that the order of selection does not matter.
At first when we are selecting 5 beads, from 8 different beads then the order of selection does not matter.
So, we can choose 5 beads from 8 different beads in ${}^{8}{{C}_{5}}$ ways.
${}^{8}{{C}_{5}}=\dfrac{8!}{\left( 8-5 \right)!\times 5!}=\dfrac{8!}{3!\times 5!}=\dfrac{5!\times 6\times 7\times 8}{1\times 2\times 3\times 5!}=\dfrac{6\times 7\times 8}{2\times 3}=56$
Therefore, we can select 5 beads from 8 different beads in 56 ways.
Now we have to arrange those 5 beads on a circular ring.
When we will arrange these 5 beads then the order of arranging them plays a very significant role. If the order changes, that will become a different arrangement. Therefore, we have to use permutation.
Permutation is arranging all the members of a set into some sequence or order, where the order of selection matters.
Now, we can arrange $n$ elements in a circular permutation in $\left( n-1 \right)!$ ways.
So we can arrange 5 beads in a circular ring in $\left( 5-1 \right)=4!=1\times 2\times 3\times 4=24$ ways.
But here we can arrange 5 beads in two directions. One is clockwise, another is anticlockwise. Here in both directions we will get the same arrangement. So, we have to divide 24 by 2.
Therefore the total number of different ways of arranging 5 beads is $\dfrac{24}{2}=12$ .
Now, we can select 5 beads from 8 different beads in 56 ways.
In each way we can arrange them in 12 ways.
So, the total number of ways of selecting 5 beads from 8 different beads and arranging them on a circular ring is $56\times 12=672$
Therefore the total number of ways in which 5 beads, chosen from 8 different beads be threaded on a circular ring is 672.
Note: For circular permutation we have to be careful about the clockwise and anticlockwise directions.
Generally we make a mistake here. If the direction does not matter then we have to divide $\left( n-1 \right)!$ by 2. |
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# Remainder Theorem
Remainder theorem is a very important and interesting topic in number system and can be learnt easily. In this article we will try to learn some concepts regarding remainders with examples. Here we go! Definition of remainder If a and d are natural numbers, with d # 0, it can be proved that there exist unique integers q and r, such that a = qd + r and 0 r < d. The number q is called the quotient, while r is called the remainder. By definition remainder cannot be negative. Dividend = Divisor Quotient + Remainder. Now just to give an example, 17 = 3 * 5 + 2, which means 17 when divided by 5 will give 2 as remainder. Well that was simple! What is remainder of (12 * 13 * 14) / 5 = 2184 / 5, 2184 = 436 * 5 + 4, hence remainder 4. But this method is not the right one for us :) In order to find the remainder of an expression find the individual remainder and replace each term with the respective remainders.In the above case 12, 13 and 14 will give remainders 2, 3 and 4 respectively when divided by 5. So replace them with the remainders in the expression and find the remainder again. (2 * 3 * 4) / 5 = 24 / 5 gives remainder 4. What is the remainder of 1421 * 1423 * 1425 when divided by 12? 1421, 1423 and 1425 gives 5, 7 and 9 as remainders respectively when divided by 12. The given expression can be written as (5 * 7 * 9) / 12, Gives a remainder of 3. Find the reminder when 1! + 2! + 3! + . . . . .. . .. 99! + 100! is divided by the product of first 7 natural numbers From 7! the remainder will be zero. Why ? because 7! is nothing but product of first 7 natural numbers and all factorial after that will have 7! as one of the factor. so we are concerned only factorials till 7!, i.e, 1! + 2! + 3! + 4! + 5! + 6! 1! + 2! + 3! + 4! + 5! + 6! = 873 and as 7! > 873 our remainder will be 873 What is the remainder when 64999 is divided by 7? Many of us get intimidated with such numbers, but as we told earlier all it takes to crack quant is a strong hold of basic concepts and keeping them closer.
64999 is nothing but 64 * 64 * . 64 (999 times) We need to find the remainder when (64 * 64 * 999 times) is divided by 7 64 gives a remainder 1 when divided by 7, hence it is 1 * 1999 times / 7 which gives a remainder 1. Some useful concepts while dealing with remainder are given below. (ax + 1)n / a gives a remainder 1 for all values of n. Find the remainder when 6499 is divided by 7 6499 / 7 = (7 * 9 + 1)99 / 7, remainder =1. (ax - 1)n / a gives a remainder 1 when n is even and (a-1) when n is odd. Find the remainder when 21875 is divided by 17. 21 / 17 gives a remainder of 4, so we need to find 4875/ 17 42 = 16 = (17 1), we can write the expression as (42)437 * 4 / 17 = (17 1)437* 4 / 17 = (17-1) * 4 / 17 = 64 / 17, gives remainder 13. (an + bn) is divisible by (a + b) when n is odd. (2101 + 3101) is completely divisible by 5 (= 2 + 3). (an - bn) is divisible by (a + b) when n is even. (5100 2100) is completely divisible by 7 (= 5 + 2) (an - bn) is always divisible by (a - b). Now we can say (10175 7675) / 25 gives a remainder zero within no time..! The remainder when f(x) = a + bx + cx2+ dx3+ is divided by (x-a) is f(a).. Remainder [(3x2 + 4x + 1) / (x-2)] = 3 * 22 + 4 * 2 + 1 = 21 Eulers Remainder Theorem In number theory, Euler's totient or phi function, (n) is an arithmetic function that counts the number of positive integers less than or equal to n that are relatively prime to n. If n is a positive integer, then (n) is the number of integers k in the range 1 k n which are co primes to n ( GCD(n, k) = 1). Take n = 9, then 1, 2, 4, 5, 7 and 8, are relatively prime to 9. Therefore, (9) = 6.
How to find Eulers totient? Say n = P1a x P2b x P3c (n) = n x (1 - 1/P1) x (1 - 1/P2) x (1 - 1/P3) (100) = 100 x (1-1 / 2) x (1- 1 / 5) = 100 x 1 / 2 x 4 / 5 = 40. (9) = 9 x (1 1/3) = 6. Eulers Remainder theorem states that for co prime numbers M and N, Remainder [M(N) / N] = 1 What is the remainder when 21875 is divided by 17 21 and 17 are co prime numbers. (Always check whether the numbers given in the problem are co primes are not. Eulers theorem is applicable only for co prime numbers) (17) = 17 x ( 1 1 / 17) = 16. So Eulers theorem says Remainder [ 2116/ 17 ] = 1 Now 21875 can be written as 2116 * 54 x 2111 First part gives remainder 1 on division by 17 we need to find only the remainder of 2111 / 17 21 on division by 17 gives a remainder of 4, so it comes to 4 11 / 17. 42 = 17 1 so it is still reduced to (17-1)5 x 4 / 17 (ax - 1)n / a gives a remainder(a-1) when n is odd Above expression can be written as 16 x 4 / 17 which gives a remainder 13 Fermats little theorem Eulers theorem says that if p is a prime numberand a and p are co -primes then a(p) / p gives a remainder of 1. Now we know for any prime number p, (p) = p - 1 . ( why so ? ) So, Remainder of a( p 1 )/ p is 1, which is Fermats little theorem. We can derive other useful results like Remainder of ap / p is a.
(ap a) is always divisible by a. Wilsons theorem Remainder of (p-1)! / p is (p-1), if p is a prime number. Here also we can derive some useful results like Remainder of [(p-1)! + 1] / p is zero. Remainder of (p-2)! / p is 1. Remainder of [50!/51] = 50 Remainder of [49!/51] = 1 Concept of negative remainder We saw earlier that by definition remainder cannot be negative. But considering negative remainder is a very useful exam trick. For example, What is the remainder when 211 divided by 3? We can use some of the above mentioned methods here, but the easiest one will be using the concept of negative remainders. Here 2 when divided by 3 gives a remainder of -1. (Say) 2 = 3 * 1 + (-1), remainder is -1, which is not theoretically correct but lets cheat! So we are asked to find (-1) * (-1) * 11 times divided by 3. Which is -1/3, remainder -1. Whenever you are getting negative number as a remainder, make it positive by adding the divisor to the negative remainder. Here required answer is 3 + (-1) = 2. Remainder when (41 * 42) is divided by 43 Use negative remainder concept, We can write the above expression as (-2) * (-1) = -2 ( as 41 = 43 * 1 2 and 42 = 43 * 1 1) Answer = 2 (here we got a positive remainder itself, so no need of correction)
Now for your brain cells, what is the remainder when 1499 is divided by 15? Cyclic property of remainders Sometimes it is easy to find the remainder by using the cyclic property of remainders, remainders forming a pattern. As a thumb rule if we divide pn with q, the remainder will follow a pattern. For example, Remainder when 21 is divided by 3 = 2 Remainder when 22 is divided by 3 = 1 Remainder when 23 is divided by 3 = 2 Remainder when 24 is divided by 3 = 1 and so on. Pattern repeats in cycles of 2. Remainder is 2 when n is odd and 1 when n is even. Now if we are asked to find remainder when 23276 divided by 3 we can answer it very quickly because we know the pattern :) One more, Remainder when 91 is divided by 11 = 9 Remainder when 92 is divided by 11 = 4 Remainder when 93 is divided by 11 = 3 Remainder when 94 is divided by 11 = 5 Remainder when 95 is divided by 11 = 1 Remainder when 96 is divided by 11 = 9 Remainder when 97 is divided by 11 = 4 Remainder when 98 is divided by 11 = 3 Pattern repeats in cycles of 5. So if we are asked to find the remainder when 9100 by 11 we know it is 1. (100 is in the form 5n and we know remainder for 5 is 1.. cool na :) ) Note: Remainder [93/11] = Remainder [Remainder [92/11] * Remainder [92/11] ) / 11] = Remainder [ 4 * 9 / 11] = 3. This funda comes very handy in scenarios like this. Like we dont have to solve Rem [98 /11] because we already know Rem[94 /11] as 5.. Rem [98 /11] = Rem [ (5 * 5) / 11 ] = 3 Also, Rem [97 /11] = Rem [ ( 3 * 5 ) / 11 ] = 4 ( as Rem [93 /11] = 3 and Rem [94 /11] = 5 )
What is the remainder when 7100 is divided by 4? 71 divided by 4 gives a remainder of 3 72 divided by 4 gives a remainder of 1 73 divided by 4 gives a remainder of 3 74 divided by 4 gives a remainder of 1 and so on Pattern repeats in cycles of 2. 7n when divided by 4 gives a remainder of 3 when n is odd and a remainder 1 when n is even. 7100 when divided by 4 gives a remainder of 1. (Same can be solved using other methods also) Find the remainder when 399^99 is divided by7 Find the pattern of remainder when 3n is divided by 7. 31 / 7 gives remainder 3 32 / 7 gives remainder 2 33 / 7 gives remainder 6 (Dont calculate Rem[33/7] we already have Rem[31/7] & Rem[32/7]) 34 / 7 gives remainder 4 (using Rem[32/7]) 35 / 7 gives remainder 5 (using Rem[33/7] and Rem[32/7] ) 36 / 7 gives remainder 1 (using Rem[33/7]) 37 / 7 gives remainder 3 (using Rem[33/7] and Rem[34/7] ) 38 / 7 gives remainder 2 (using Rem[34/7]) Pattern repeats in cycles of 6. (We can do this easily from Eulers theorem, as (7) = 6, hence Remainder [36/7] = 1. Explained just to get the idea of patterns in remainders) Now our task is to find the remainder of 9999/6 Remainder [9999/6] = Remainder [399/6] 31/6 gives remainder 3 32/6 gives remainder 3 33/6 gives remainder 3 and so on..!
Hence 9999 can be written as 6n + 3. So 399^99 / 7 is in the form 3(6n + 3)/ 7 we have found out the pattern repeats in cycles of 6. So the remainder is same as 33/7 = 6 :) Find the remainder when 9797^97 is divided by 11 Remainder [9797^97/11] = Remainder [997^97/11] (as Remainder [97/11] = 9) From Eulers theorem, Remainder [910/11] = 1 97 and 10 are again co primes. So (10) = 10 (1-1/2) (1-1/5) = 4 Remainder [974/11] = 1 97 = 4n + 1, So Remainder [9797/11] = Remainder [97/10] = 7 Means 9797 can be written as 10n + 7 Now our original question, Remainder [997^97/11] = Remainder [910n+7/11] = Remainder [97/11] = 4 :) Note: One common mistake while dealing with remainders is when we have common factors in both dividend and divisor. Example, what is the remainder when 15 is divided by 9 15 / 9 is same as 5 / 3, remainder 2. Correct? No 15/9 will give a remainder of 6. Where we slipped? Always remember that if we find remainder after cancelling common terms make sure we multiply the remainder obtained with the common factors we removed. In previous case we will get correct answer (6) when we multiply the remainder obtained (2) with the common factor we removed (3). I hope the explanations are clear are correct. Please let me know if any concepts regarding remainders are missed out or incase of any errors.. Happy learning :) |
## find the largest number which divides 615 and 963 leaving remainder 6 in each case (euclids division lemma)
Question
find the largest number which divides 615 and 963 leaving remainder 6 in each case (euclids division lemma)
in progress 0
4 weeks 2021-11-02T05:13:26+00:00 2 Answers 0 views 0
1. Step-by-step explanation:
To find the largest number which divides 615 and 963 leaving remainder 6 in each case.
We have to find HCF.
615 = 3*3*29
963 = 3*11*29
HCF = 3*29 = 87
87 is the largest number which divides 615 and 963 leaving remainder 6 in each case.
2. Firstly, the required numbers which on dividing doesn’t leave any remainder are to be found.
This is done by subtracting 6 from both the given numbers.
So, the numbers are 615 – 6 = 609 and 963 – 6 = 957.
Now, if the HCF of 609 and 957 is found, that will be the required number.
957 = 609 x 1+ 348
609 = 348 x 1 + 261
348 = 261 x 1 + 87
261 = 87 x 3 + 0.
⇒ H.C.F. = 87. |
# Tools in Geogebra
My blog has gotten a little lofty lately, and it’s been a while since I just posted some plain old good ideas you could use tomorrow. Here’s one if you have access to a class set of laptops or a computer lab: have your students make tools in geogebra. I’m not going to try to frame this in a lesson plan – it’s just a tutorial for you. Open geogebra to follow along.
We’ll make a midpoint tool1 in two different ways today. The first way will be geometrically, via construction. The second way will use an algebraic formula. This might be a fun way to connect geometry and algebra! If you already know both of these methods, skip down to the “Toolify” section. If you already know how to make tools, skip down to “The Point” section at the bottom!
## Midpoint via Construction
To make tools in geogebra, you first do what you want the tool to do, and then tell geogebra about it. So to make our midpoint tool via construction, we first have to actually do the construction.
2. Use the circle tool to draw a circle with center A and perimeter point B. Draw a second one with center B and perimeter point A.
3. The circles intersect at two points. Use the line tool to draw the line between them. Also draw the line from A to B.
4. Of course, these two lines intersect at the midpoint between A and B. Use the point tool to give it its own name.
5. A crucial step: test your construction by moving the points A and B. The entire construction should move, but E should still be the midpoint. Do not move on if your construction does not pass this “wiggle test.”When your construction passes the wiggle test, go to the “Toolifying” section below.
## Midpoint via Algebra
We’ll do this construction entirely from the input bar. Text in bold is text you can type directly into the input bar.
1. A = (2, 4)
B = (5, 6)
Typing these commands creates two points, A and B, at the specified coordinates.
2. x_A = x(A)
y_A = y(A)
x_B = x(B)
y_B = y(B)
These commands create variables with which you can access the coordinates of points A and B. The thing on the left of the equal sign is the NAME of the variable. The thing on the right of the equal sign is the VALUE of the variable.
3. Do the wiggle test on your variables. When you wiggle points A and B, all four of the variables from step 2 should change. You can move point A by dragging it with the mouse, or by redefining it with something like A = (1, 4). Do not move on until your variables have passed the wiggle test.
4. E = ( (x_A + x_B) / 2 , (y_A + y_B) / 2 )
This command creates the point $( \frac{x_A + x_B} {2}, \frac{y_A + y_B}{2} )$. If all has gone well, you should see the midpoint appear.
When this point
E passes the wiggle test, move on to “Toolifying” below.
## Toolifying
Regardless of HOW your construction was made – via algebra, geometry, or even calculus2 – if it passes the wiggle test, you can make it into a tool.
1. From the “Tools” menu, choose “Create New Tool.” You’ll be presented with a dialog like the one below.
2. The most crucial part here is to identify to geogebra your output object. What is the RESULT of your tool supposed to be? In our case, we were trying to make a tool that finds the midpoint of two points. The output is that midpoint. We called it point E. Select that object from the list. You could also click on that object from the graph view.
3. Go to the “Input Objects” tab. On this tab you will select the objects that your tool needs to work. Our tool is supposed to create the midpoint from two starting points, so those two starting points must be listed as input objects.
So far, in my experience, Geogebra has always guessed the necessary input objects for me. Point A and B are already listed because geogebra knows that they are at the root of your construction. This will save a lot of confusion with your students.
4. Head to the “Name & Icon” tab to personalize your tool. The “Tool Name” is what will appear on the tool bar. The “Command name” is what you would type on the input bar to use your command. The “Tool help” will appear in the toolbar when your tool is selected.
5. After you click Finish, your tool is created and ready to use. Let’s test it! First, make two new points.
6. Then, choose your tool from the toolbar and click those two points, one after the other. The order you click is important in some tools: the first object you click becomes the first input object, the second you click becomes the second input object, etc. If everything is working, you should see the midpoint appear between your two new input points!
Remember to try the wiggle test by pressing escape and dragging F and G around! You will not be able to drag H around – geogebra cannot (yet?) run tools backwards like that. Note that geogebra does not create all the intermediate objects needed for the construction. If you want those objects to appear, include them in the list of output objects when you create the tool.
7. Your students will enjoy this and feel a sense of ownership of the math. It’s fun to use the tool on its own output, for multiple nested midpoints and things like that.
You can even create other objects with the output of your tools for extra fun. Below is an actual geogebra applet – drag the blue points around for fun!
{{Sorry, the GeoGebra Applet could not be started. If you're seeing this in a reader application click here to see the post live.. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)}} Riley Lark, Created with GeoGebra
## The point (heh)
When your students can program geogebra to perform a mathematical feat:
1. They will feel (and be) smart and empowered.
2. They will learn about programming computers – an invaluable tool not just for “the future” but for exploring mathematical concepts later in your class.
3. They will necessarily have mastered the concept at least one time with enough specificity that a computer can understand what they mean
4. They will have fun!
1. Yes, geogebra does have a built-in midpoint tool.
2. define an f(x) in geogebra, and then type f'(x). Geogebra automatically calculates the derivative! |
# How can three resistors of resistances 2Ω, 3Ω and 6Ω be connected to give a total resistance of 1Ω?
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The three resistors of resistances 2Ω, 3Ω and 6Ω have to be connected in parallel to give a total resistance of 1Ω. Steps to calculate:
Step 1: Given Data:
The given resistances are:
• R1 = 2Ω
• R2 = 3Ω
• R3 = 6Ω
Due to the equivalent resistance’s lower value than each resistor’s resistance, the resistors R1, R2 and R3 must be connected in parallel.
Step 2: Formula used:
Therefore, their equivalent resistance:
1/Req = 1/R1 + 1/R2 + 1/R3
Step 3: Calculate the equivalent resistance:
1/Req = ½ + ⅓ + ⅙
= (3 + 2 + 1)/6
= 6/6
= 1
Req = 1Ω
### Factors affecting resistance
1. The conductor’s cross-sectional area
2. the conductor’s length
3. the conductor’s composition
4. The conducting material’s temperature
### Applications of resistor
1. When accurate measurement, high sensitivity, and balanced current regulation are needed, wire wound resistors are used in applications like shunts with ampere meters.
2. Flame detectors, burglar alarms, photographic equipment, etc. all use photoresistors.
3. Temperature and voltmeter control is accomplished by resistors.
4. Amplifiers, telephony, oscillators, and digital multimeters all employ resistors.
5. Additionally, they are utilized in transmitters, demodulators, and modulators.
Therefore, three resistors of resistances 2Ω, 3Ω and 6Ω have to be connected in parallel to give a total resistance of 1Ω
Summary:
## How can three resistors of resistances 2Ω, 3Ω and 6Ω be connected to give a total resistance of 1Ω?
To get a total resistance of 1Ω, the three resistors with resistances of 2Ω, 3Ω and 6Ω must be linked in parallel. When current flows through a light bulb or conductor, the conductor presents some obstacle to the current. This impairment is called electrical resistance and is represented by R.
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Contenuto principale
# Writing linear equations in all forms
## Trascrizione del video
A line passes through the points negative 3, 6 and 6, 0. Find the equation of this line in point slope form, slope intercept form, standard form. And the way to think about these, these are just three different ways of writing the same equation. So if you give me one of them, we can manipulate it to get any of the other ones. But just so you know what these are, point slope form, let's say the point x1, y1 are, let's say that that is a point on the line. And when someone puts this little subscript here, so if they just write an x, that means we're talking about a variable that can take on any value. If someone writes x with a subscript 1 and a y with a subscript 1, that's like saying a particular value x and a particular value of y, or a particular coordinate. And you'll see that when we do the example. But point slope form says that, look, if I know a particular point, and if I know the slope of the line, then putting that line in point slope form would be y minus y1 is equal to m times x minus x1. So, for example, and we'll do that in this video, if the point negative 3 comma 6 is on the line, then we'd say y minus 6 is equal to m times x minus negative 3, so it'll end up becoming x plus 3. So this is a particular x, and a particular y. It could be a negative 3 and 6. So that's point slope form. Slope intercept form is y is equal to mx plus b, where once again m is the slope, b is the y-intercept-- where does the line intersect the y-axis-- what value does y take on when x is 0? And then standard form is the form ax plus by is equal to c, where these are just two numbers, essentially. They really don't have any interpretation directly on the graph. So let's do this, let's figure out all of these forms. So the first thing we want to do is figure out the slope. Once we figure out the slope, then point slope form is actually very, very, very straightforward to calculate. So, just to remind ourselves, slope, which is equal to m, which is going to be equal to the change in y over the change in x. Now what is the change in y? If we view this as our end point, if we imagine that we are going from here to that point, what is the change in y? Well, we have our end point, which is 0, y ends up at the 0, and y was at 6. So, our finishing y point is 0, our starting y point is 6. What was our finishing x point, or x-coordinate? Our finishing x-coordinate was 6. Let me make this very clear, I don't want to confuse you. So this 0, we have that 0, that is that 0 right there. And then we have this 6, which was our starting y point, that is that 6 right there. And then we want our finishing x value-- that is that 6 right there, or that 6 right there-- and we want to subtract from that our starting x value. Well, our starting x value is that right over there, that's that negative 3. And just to make sure we know what we're doing, this negative 3 is that negative 3, right there. I'm just saying, if we go from that point to that point, our y went down by 6, right? We went from 6 to 0. Our y went down by 6. So we get 0 minus 6 is negative 6. That makes sense. Y went down by 6. And, if we went from that point to that point, what happened to x? We went from negative 3 to 6, it should go up by 9. And if you calculate this, take your 6 minus negative 3, that's the same thing as 6 plus 3, that is 9. And what is negative 6/9? Well, if you simplify it, it is negative 2/3. You divide the numerator and the denominator by 3. So that is our slope, negative 2/3. So we're pretty much ready to use point slope form. We have a point, we could pick one of these points, I'll just go with the negative 3, 6. And we have our slope. So let's put it in point slope form. All we have to do is we say y minus-- now we could have taken either of these points, I'll take this one-- so y minus the y value over here, so y minus 6 is equal to our slope, which is negative 2/3 times x minus our x-coordinate. Well, our x-coordinate, so x minus our x-coordinate is negative 3, x minus negative 3, and we're done. We can simplify it a little bit. This becomes y minus 6 is equal to negative 2/3 times x. x minus negative 3 is the same thing as x plus 3. This is our point slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope intercept form. Let's do that. So let's do slope intercept in orange. So we have slope intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2/3, so you get y minus 6 is equal to-- I'm just distributing the negative 2/3-- so negative 2/3 times x is negative 2/3 x. And then negative 2/3 times 3 is negative 2. And now to get it in slope intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side, so let's add 6 to both sides of this equation. Left-hand side of the equation, we're just left with a y, these guys cancel out. You get a y is equal to negative 2/3 x. Negative 2 plus 6 is plus 4. So there you have it, that is our slope intercept form, mx plus b, that's our y-intercept. Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2/3 x to both sides of this equation. So I'll start it here. So we have y is equal to negative 2/3 x plus 4, that's slope intercept form. Let's added 2/3 x, so plus 2/3 x to both sides of this equation. I'm doing that so it I don't have this 2/3 x on the right-hand side, this negative 2/3 x. So the left-hand side of the equation-- I scrunched it up a little bit, maybe more than I should have-- the left-hand side of this equation is what? It is 2/3 x, because 2 over 3x, plus this y, that's my left-hand side, is equal to-- these guys cancel out-- is equal to 4. So this, by itself, we are in standard form, this is the standard form of the equation. If we want it to look, make it look extra clean and have no fractions here, we could multiply both sides of this equation by 3. If we do that, what do we get? 2/3 x times 3 is just 2x. y times 3 is 3y. And then 4 times 3 is 12. These are the same equations, I just multiplied every term by 3. If you do it to the left-hand side, you can do to the right-hand side-- or you have to do to the right-hand side-- and we are in standard form. |
# How do you divide (-4x^3-15x^2-4x-12)/(x-4) ?
Jul 20, 2018
The remainder is $= - 524$ and the quotient is $= - 4 {x}^{2} - 31 x - 128$
#### Explanation:
Let's perform the synthetic division
$\textcolor{w h i t e}{a a a a}$$4$$|$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a}$$- 15$$\textcolor{w h i t e}{a a a a a a}$$- 4$$\textcolor{w h i t e}{a a a a a}$$- 12$
$\textcolor{w h i t e}{a a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a a a a}$$- 16$$\textcolor{w h i t e}{a a a a}$$- 124$$\textcolor{w h i t e}{a a a a}$$- 512$
$\textcolor{w h i t e}{a a a a a a a a a}$_________
$\textcolor{w h i t e}{a a a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a}$$- 31$$\textcolor{w h i t e}{a a a a}$$- 128$$\textcolor{w h i t e}{a a a a}$$\textcolor{red}{- 524}$
The remainder is $= - 524$ and the quotient is $= - 4 {x}^{2} - 31 x - 128$
ALSO,
Apply the remainder theorem
When a polynomial $f \left(x\right)$ is divided by $\left(x - c\right)$, we get
$f \left(x\right) = \left(x - c\right) q \left(x\right) + r$
Let $x = c$
Then,
$f \left(c\right) = 0 + r$
Here,
$f \left(x\right) = - 4 {x}^{3} - 15 {x}^{2} - 4 x - 12$
Therefore,
$f \left(4\right) = - 4 \cdot {4}^{3} - 15 \cdot {4}^{2} - 4 \cdot 4 - 12$
$= - 256 - 240 - 16 - 12$
$= - 524$
The remainder is $= - 524$ |
# NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2
Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 – Real Numbers. This Exercise contains 7 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.
### NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 – Real Numbers
NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 – Real Numbers
1. Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:
(i) 140 = 2 × 2 × 5 × 7
= 22 × 5 × 7
Prime factors of 140 are 2, 5, 7
(ii) 156 = 2 × 2 × 3 × 13
= 22 × 3 × 13
Prime factors of 156 are 2, 3, 13.
(iii) 3825 = 5 × 5 × 3 × 3 ×17
= 52 × 32 ×17
Prime factors of 3825 are 5, 3, 17.
(iv) 5005 = 5 × 7 × 11 × 13
Prime factors of 5005 are 5, 7, 11, 13.
(v) 7429 = 17 × 19 × 23
Prime factors of 7429 are 17, 19, 23.
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
We know that HCF of two numbers equal to the product of the smallest power of each common prime factor in the numbers.
And LCM of two numbers equal to product of the greatest power of each prime factor, involved in the numbers.
(i) 26 = 2 × 13
91 = 7 × 13
Here 13 is common in both the prime factors of 26 and 91.
HCF of 26 and 91 is 13
LCM of these two numbers = 2 × 7 × 13 =182.
HCF × LCM = 13 × 182 = 2366
Product of 26 and 91 = 26 × 91 = 2366.
Hence, LCM × HCF = Product of the given two numbers
(ii) 510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
Here, 2 is common in both the prime factors of 510 and 92
HCF of 510 and 92 is 2
LCM = 2 × 2 × 3 × 5 17 × 23 = 23460
HCF × LCM = 2 × 23460 = 46920
Product of 510 and 92 is = 510 × 92 = 46920.
Hence, LCM × HCF = Product of the given two numbers
(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
Here, 2 and 3 are common prime factors of 336 and 54.
HCF = 2×3 = 6
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
HCF × LCM = 6 × 3024 = 18144
Product of336 and 54 = 336 × 54 = 18144
Hence, LCM × HCF = product of the given two numbers
3. Find the LCM and HCF of the following integers by prime factorization method:
(i) 12, 15 and 21
(ii) 17, 23, 29
(iii) 8, 9, 25
Solution: We know that HCF of two numbers equal to the product of the smallest power of each common prime factor in the numbers.
And LCM of two numbers equal to product of the greatest power of each prime factor, involved in the numbers.
(i) 12 = 2 × 2 × 3, 15 = 3 × 5 and 21 = 3 × 7
Therefore HCF (12, 15, 21) = 3, since 3 is the only common factor.
And LCM (12, 15, 21) = 2 × 2 × 3 × 5 × 7 = 420
(ii) Here the numbers 17, 23, 29 are all primes
So no common prime factors are there.
Hence HCF (17, 23, 29) = 1
And LCM (17, 23, 29) = 17 × 23 × 29 = 11339.
(iii) 8 = 23 , 9 = 32 and 25 = 52
Since no common prime factors are there so HCF (8, 9, 25) = 1
And LCM (8, 9, 25) = 8 × 9 × 25 = 1800.
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
We know that,
HCF × LCM = Product of the given two numbers.
Therefore 9 × LCM (306, 657) = 306 × 657
i.e., LCM (306, 657) = (306 x 657)/9
LCM (306, 657) = 22338
5. Check whether 6n can end with the digit 0 for any natural number n.
Solution:
Let 6n can end with the digit 0. Then 5 must be a prime factor of it.
Now 6n = (2 × 3)n, i.e., 2 and 3 are the only prime factors, which is a contradiction to the fact that 5 is a prime factor of 6n. Hence our assumption is wrong.
Therefore, 6n can not end with the digit 0 for any natural number n.
6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1+5 are composite numbers.
Solution:
There are two types of numbers – Prime numbers and Composite Numbers.
Prime number is a natural number greater than 1 that has no positive divisors other than 1 and the number itself.
Composite numbers are numbers divisible by one or more numbers other than 1 and itself.
7 × 11 × 13 + 13 = 13 ( 7 × 11 + 1 ) = 13 × ( 77 + 1 ) = 13 × 78 = 13 × 13 × 6
Hence, given expression is divisible by 13 and 6
Therefore, 7 × 11 × 13 + 13 is a composite numbers.
Similarly, 7 × 6 × 5 × 4 × 3 × 2 × 1+5 = 5 ( 7 × 6 × 4 × 3 × 2 + 1) = 5 × (1008 + 1 ) = 5 × 1009
Hence, given expression is divisible by 5 and 1009.
Therefore, 7 × 6 × 5 × 4 × 3 × 2 × 1+5 is a composite numbers.
7. There is circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
To find the required time we have to find the LCM of these two time which they takes to drive one round.
18 = 2 × 3 × 3
12 = 2 × 2 × 3.
LCM (18, 12) = 2 × 2 × 3 × 3 = 36
Hence, after 36 minutes they will meet again.
NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.2 – Real Numbers, has been designed by the NCERT to test the knowledge of the student on the topic – The Fundamental Theorem of Arithmetic
### 1 thought on “NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2”
1. Hi, very nice website |
# What is the range of the function f(x)=2/(x+3) -4?
Apr 30, 2017
$y \in \mathbb{R} , y \ne - 4$
#### Explanation:
$\text{Rearrange f(x) to make x the subject}$
$y = f \left(x\right) = \frac{2}{x + 3} - \frac{4 \left(x + 3\right)}{x + 3}$
$\Rightarrow y = \frac{2 - 4 x - 12}{x + 3} = \frac{- 4 x - 10}{x + 3}$
$\textcolor{b l u e}{\text{cross-multiply}}$
$\Rightarrow y x + 3 y = - 4 x - 10$
$\Rightarrow y x + 4 x = - 10 - 3 y$
$\Rightarrow x \left(y + 4\right) = - 10 - 3 y$
$\Rightarrow x = \frac{- 10 - 3 y}{y + 4}$
The denominator cannot be zero as this would make the function $\textcolor{b l u e}{\text{undefined}} .$Equating the denominator to zero and solving gives the value that y cannot be.
$\text{solve "y+4=0rArry=-4larrcolor(red)" excluded value}$
$\text{range } y \in \mathbb{R} , y \ne - 4$ |
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
Center of the circle is $\left( -4,3 \right)$ and radius is $r=\sqrt{40}$
Standard equation of the circle is: ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ (equation - 1) And equation of the circle is ${{x}^{2}}+{{y}^{2}}+8x-6y=15$ (equation - 2) Now add $16$ and $9$ on both the sides of equation $\left( 2 \right)$ to complete the square twice. \begin{align} & {{x}^{2}}+{{y}^{2}}+8x-6y+16+9=15+16+9 \\ & \left( {{x}^{2}}+8x+16 \right)+\left( {{y}^{2}}-6y+9 \right)=40 \\ & {{\left( x+4 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{\left( \sqrt{40} \right)}^{2}} \end{align} Compare the standard equation with the equation${{\left( x+4 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{\left( \sqrt{40} \right)}^{2}}$. Center coordinate of circle is $\left( h=-4,k=3 \right)$. And radius of circle is $r=\sqrt{40}$. To graph, we plot the points $\left( -4,9.325 \right)$, $\left( -4,-3.325 \right)$, $\left( -10.325,3 \right)$, and $\left( 2.325,3 \right)$ which are, respectively, $\sqrt{40}$units above, below, left and right of $\left( -4,3 \right)$. |
# Engaging students: Slope-intercept form of a line
In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.
I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).
This student submission comes from my former student Jessica Williams. Her topic, from Algebra I: the point-slope intercept form of a line.
A.2 How could you as a teacher create an activity or project that involves your topic?
In order to teach a lesson regarding slope intercept form of a line, I believe it is crucial to use visual learning to really open the student’s minds to the concept. Prior to this lesson, students should know how to find the slope of a line. I would provide each student with a piece of graph paper and small square deli sheet paper. I would have them fold their deli sheet paper into half corner to corner/triangle way). I would ask each student to put the triangle anywhere on the graph so that it passes through the x and the y-axis. Then I will ask the students to trace the side of the triangle and to find two points that are on that line. For the next step, each student will find the slope of the line they created. Once the students have discovered their slope, I will ask each of them to continue their line further using the slope they found. I will ask a few students to show theirs as an example (picking the one who went through the origin and one who did not). I will scaffold the students into asking what the difference would look like in a formula if you go through the origin or if you go through (0,4) or (0,-3) and so on. Eventually the students will come to the conclusion how the place where their line crosses the y-axis is their y intercept. Lastly, each student will be able to write their equation of the line they specifically created. I will then introduce the y=mx+b formula to them and show how the discovery they found is that exact formula. This is a great way to allow the students to work hands on with the material and have their own individual accountability for the concept. They will have the pride of knowing that they learned the slope intercept formula of a line on their own.
E.1 How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic?
Graphing calculators are a very important aspect of teaching slope-intercept form of a line. It allows the students to visually see where the y-intercept is and what the slope is. Also, another good program to use is desmos. It allows the students to see the graph on the big screen and you can put multiple graphs on the screen at one time to see the affects that the different slopes and y intercept have on the graph. This leads students into learning about transformations of linear functions. Also, the teacher can provide the students with a graph, with no points labeled, and ask them to find the equation of the line on the screen. This could lead into a fun group activity/relay race of who can write the formula of the graph in the quickest time. Also, khan academy has a graphing program where the students are asked to create the graph for a specific equation. This allows the students to practice their graphing abilities and truly master the concept at home. To engage the students, you could also use Kahoot to practice vocabulary. For Kahoot quizzes, you can set the time for any amount up to 2 minutes, so you could throw a few formula questions in their as well. It is an engaging way to have each student actively involved and practicing his or her vocabulary.
B1. How can this topic be used in your students’ future courses in mathematics or science?
Learning slope intercept form is very important for the success of their future courses and real world problems. Linear equations are found all over the world in different jobs, art, etc. By mastering this concept, it is easier for students to visualize what the graph of a specific equation will look like, without actually having to graph it. The students will understand that the b in y=mx+b is the y-intercept and they will know how steep the graph will be depending on the value of m. Mastering this concept will better prepare them to lead into quadratic equations and eventually cubic. Slope intercept form is the beginning of what is to come in the graphing world. Once you grasp the concept of how to identify what the graph will look like, it is easier to introduce the students to a graph with a higher degree. It will be easier to explain how y=mx+b is for linear graphs because it is increases or decreases at a constant rate. You could start by asking,
1.What about if we raise the degree of the graph to x^2?
2.What will happen to the graph?
3.Why do you think this will happen, can you explain?
4.What does squaring the x value mean?
It really just prepares the students for real world applications as well. When they are presented a problem in real life, for example, the student is throwing a bday party and has \$100 dollars to go to the skating rink. If they have to spend \$20 on pizza and each friend costs \$10 to take, how many friends can you take? Linear equations are used every day, and it truly helps each one of the students.
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# Triangle
A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted ${\displaystyle \triangle ABC}$.
## Parts of a Triangle
• A triangle has 3 sides. In triangle ABC, the sides are AB, BC, and CA.
• The angle formed by any two sides of a triangle is the angle of the triangle, denoted by the symbol ∠. A triangle has three angles. The three angles of the triangle ABC are ∠ABC, ∠BCA, and ∠CAB. These angles are also called ∠B, ∠C, and ∠A, respectively.
• The point of intersection of any two sides of a triangle is known as a vertex. A triangle has three vertices. In triangle ABC, the vertices are A, B, and C.
## Properties of a Triangle
• The sum of all three interior angles of a triangle is always equal to 180⁰.
• The sum of the length of any two sides of a triangle is always greater than the length of the third side.
• The area of a triangle is equal to half of the product of its base and height.
## Types of Triangles
Triangles can be classified based on the length of the sides or their angle measurements.
To classify triangles according to their angles, we measure each of their interior angles. Triangles can be classified by angles, as:
• Acute Triangle or Acute-angled Triangle
• Right Triangle or Right-angled Triangle
• Obtuse Triangle or Obtuse-angled Triangle
Right Obtuse Acute ${\displaystyle \quad \underbrace {\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad } _{}}$ Oblique
The types of triangles based on the length of the sides are –
• Scalene triangle
• Isosceles triangle
• Equilateral triangle
Equilateral Triangle
Isosceles triangle
Scalene triangle
Area of a Triangle
The area of a triangle is the region that the triangle occupies in 2d space. The area of different triangles differs based on their size. If we know the base length and height of a triangle, we can determine its area. It is expressed in square units.
So, the Area of a triangle = ½ (Product of base and height of a triangle)
In the triangle PQR, PQ, QR, and RP are the sides. QR is the triangle’s base, and PS is the triangle’s height. PS is perpendicular from vertex P to the side QR. So, to find the area of △PQR, we use the following formula:
Area △PQR = ½ (Product of base and height of a triangle)
Or, Area △PQR = ½ (QR X PS)
The perimeter of a Triangle
The perimeter of a triangle is the sum of the length of all sides of the triangle.
So, the perimeter of the triangle = Sum of all three sides.
In triangle PQR, the perimeter will be the sum of the three sides, i.e., PQ, QR, and RP.
So, Perimeter of △PQR = PQ + QR + RP. |
The math portion of the SAT is designed to make you apply math concepts you’ve used before in new ways. To get a high score on the test, you’ll need to know some basic math formulas.
You will be given some formulas, but to save time during SAT testing, it can be helpful to get to know the formulas below:
## Circles
For circles, it will be helpful for you to know how to find the length of an arc. The formula is:
Larc=(2πr)(degree measure center of arc/360)
Next, familiarize yourself with the formula for finding the area of an arc sector:
Aarcsector=(πr2)(degree measure center of arc/360)
Formulas for the area of a circle and circumference are generally given on the test, as well.
## Slopes and Graphs
It’s vital to know the slope formula that is used to find the slope of a line. Remember “rise over run” – that is, the vertical change will be divided by the horizontal change of the line.
You’ll also want to know how to write the equation of a line:
y=mx+b
If there is an equation that is not in this form, remember to re-write it so that it is! On the SAT, it’s common for an equation to be given in a specific form, then to ask whether the intercept and slope are negative or positive so that students are required to flip around the terms of the original formula.
If you don’t re-write the equation into the form y=mx+b, you will get those questions wrong.
If you are given two points — A(x1,y1), B(x2,y2) — you must find the midpoint of the line that connects them. This is known as the midpoint formula, which is written as:
Given two points, A(x1,y1), B(x2,y2), you may be asked to find the distance between them. The distance formula is written as:
## Algebra Formulas
There are a few crucial algebra formulas you will want to know for the SAT, as well.
The quadratic equation asks you to solve for x when given ax2+bx+c. It is written as:
Some of the polynomials you will see on the SAT are easier to factor than others. For the ones that are difficult to factor, the quadratic equation will prove to be invaluable.
## Percentages
For the SAT, you’ll need to know how to find x percent of a given number (n). Use the following formula:
To find what percent a certain number, n, is of another number, m, the formula is:
Finally, make sure you know how to find out what number (n) is a certain percent of (x) with this formula:
## Averages
For the SAT, you will need to remember that the mean is the same thing as the average. With this in mind, if you are asked to find the mean or average of a set of numbers, use this formula:
If you are asked to find the average speed of something, use the following formula:
## Probabilities
When you come across problems in which you need to work with probabilities, keep in mind that probabilities represent the odds of something specific happening. The formula is:
Keep in mind that a probability of 0 will never happen, while a probability of 1 will be guaranteed to happen.
## SAT Prep with Jantzi
There is so much more to doing well on the SAT than just knowing the right math formulas. Earning a high SAT score is vital to getting into the college of your choice. Contact the team at Jantzi today for more information on our test prep services so that you can be fully prepared. |
# What does 1 red 1 unit mean
## What is 1 / i? - The mathematical expression simply explained
"1 / i" is a strange expression and you can hardly believe that it should have anything to do with mathematics. "I" is the so-called imaginary unit that was "invented" by mathematicians in order to be able to extract the root from negative numbers.
### What you need:
• Basic knowledge of "roots"
### Root of -1 - mathematicians define the "i"
• Mathematics has made expansions in the entire range of numbers if a type of calculation required it. For example, negative numbers were "invented" in order to post debit amounts or to always be able to carry out subtractions. Fractions also owe their existence to the desire to be able to divide without remainder.
• However, it is very unsatisfactory not to be able to take roots from negative numbers. So you simply defined a new type of number, namely the complex numbers with which this succeeds.
• The complex numbers are based on the imaginary unit "i", which was defined as follows: i = root (-1), consequently i² = -1.
• Roots from negative numbers can thus be solved, because it is, for example, the root (-4) = 2i.
### But what does 1 / i mean?
• Of course, you can calculate with imaginary and complex numbers, that is, those with real and imaginary parts such as 2-3i, almost as well as with real (the "correct") numbers.
• So you can add and subtract, but also multiply and even divide.
• 1 / i is initially nothing other than that the number "1" is divided by "i" or the reciprocal of "i".
• With a little skill, this division or this reciprocal value can be converted into an expression that is easier to understand and with which one can calculate better.
• The trick is to expand the fraction with "i", that is, to multiply it by i / i (so that the value doesn't change). The following applies: 1 / i = 1 / i * i / i = i / -1 = -i (because i² = -1, see above).
Bottom line: The complicated looking expression 1 / i is nothing more than -i. |
# How do you simplify cotB + (sinB / (1+cosB))?
##### 2 Answers
Nov 5, 2015
$\csc \beta$
#### Explanation:
$\cot \beta$+$\sin \frac{\beta}{1 + \cos \beta}$
1) Turn everything into sin and cos.
$\cos \frac{\beta}{\sin} \beta + \sin \frac{\beta}{1 + \cos \beta}$
2) Use LCD and create a common numerator.
$\frac{1 + \cos \beta}{1 + \cos \beta} \cdot \cos \frac{\beta}{\sin} \beta + \sin \frac{\beta}{1 + \cos \beta} \cdot \sin \frac{\beta}{\sin} \beta$
$\frac{\left(1 + \cos \beta\right) \cos \beta + {\sin}^{2} \beta}{\sin \beta \left(1 + \cos \beta\right)}$
3) Distibute $\cos \beta$ in the denominator.
$\frac{\cos \beta + \left({\cos}^{2} \beta + {\sin}^{2} \beta\right)}{\sin \beta \left(1 + \cos \beta\right)}$
4) Pythagorean Identity in the denominator. Then cancel out matching denominator and numerator.
$\frac{\cancel{\cos \beta + \left(1\right)}}{\sin \beta \left(\cancel{1 + \cos \beta}\right)}$
5) Left with $\frac{1}{\sin} \beta$ = $\csc \beta$
Nov 5, 2015
$\csc \left(B\right)$
#### Explanation:
$\cot \left(B\right) + \frac{\sin \left(B\right)}{1 + \cos \left(B\right)} = \frac{\cos \left(B\right)}{\sin \left(B\right)} + \frac{\sin \left(B\right)}{1 + \cos \left(B\right)} \frac{1 - \cos \left(B\right)}{1 - \cos \left(B\right)}$
$= \frac{\cos \left(B\right)}{\sin \left(B\right)} + \frac{\sin \left(B\right) \left[1 - \cos \left(B\right)\right]}{1 - {\cos}^{2} \left(B\right)}$
$= \frac{\cos \left(B\right)}{\sin \left(B\right)} + \frac{\sin \left(B\right) \left[1 - \cos \left(B\right)\right]}{{\sin}^{2} \left(B\right)}$
$= \frac{\cos \left(B\right)}{\sin \left(B\right)} + \frac{1 - \cos \left(B\right)}{\sin \left(B\right)}$
$= \frac{1}{\sin \left(B\right)}$
$= \csc \left(B\right)$ |
# How do you use implicit differentiation to find dy/dx given x^3+3x^2y+y^3=8?
Nov 5, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{x}^{2} + 2 x y}{{x}^{2} + {y}^{2}}$
#### Explanation:
differentiate $\textcolor{b l u e}{\text{implicitly with respect to x}}$
Use the $\textcolor{b l u e}{\text{product rule"" on the term }} 3 {x}^{2} y$
$\Rightarrow 3 {x}^{2} + 3 {x}^{2.} \frac{\mathrm{dy}}{\mathrm{dx}} + 6 x y + 3 {y}^{2.} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {x}^{2} + 3 {y}^{2}\right) = - 3 {x}^{2} - 6 x y$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 {x}^{2} + 6 x y}{3 {x}^{2} + 3 {y}^{2}}$
$= \frac{- \cancel{3} \left({x}^{2} + 2 x y\right)}{\cancel{3} \left({x}^{2} + {y}^{2}\right)}$
$= - \frac{{x}^{2} + 2 x y}{{x}^{2} + {y}^{2}}$ |
# Development of a Hyperbola from the Definition
A hyperbola is the set of all points in a plane such that the difference of the distances from two fixed points (foci) is constant.
When the major axis is horizontal, the foci are at (-c,0) and at (0,c).
Let d1 be the distance from the focus at (-c,0) to the point at (x,y). Since this is the distance between two points, we'll need to use the distance formula.
Similarly, d2 will involve the distance formula and will be the distance from the focus at the (c,0) to the point at (x,y).
We can use the fact that the vertices are on the hyperbola to find out what the difference of the distances is.
If we take the vertex on the right, then d1 = c + a and d2 = c - a.
d1 - d2 = ( c + a ) - ( c - a ) = c + a - c + a = 2a
Therefore, the constant is 2a and d1 - d2 = 2a for every point on a hyperbola
### Finding the equation of the hyperbola
Now that we know what the difference of the distances is, we can set about finding the equation of the hyperbola.
We start with d1 - d2 = 2a and substitute the formulas for d1 and d2. Because it is difficult to work with absolute values, we'll assume that d1 is greater than d2, although the process doesn't change much if it's the other way around.
We want to get rid of the radicals, so we'll move one term to the other side and then square both sides of the equation.
Let's multiply out the squared terms
If we move everything except for the square root term to the left side, a lot of this will cancel out and we'll get
Let's divide everything by 4 to get
Square both sides to get rid of the radical
Expand the (x-c)2 on the right hand side
Distribute the a2 on the right side
Move all the variable terms to one side and the constants to the other.
Factor and x2 out of the first two terms on the left and an a2 out of the right side.
Notice how there is an c2 - a2 on both sides. Let's define b2 = c2 - a2 and make the substitution into the equation.
Finally, divide everything by a2b2 so the right side is 1.
That gives us the standard form for a hyperbola with a transverse axis and it also gives us the Pythagorean relationship c2 = a2+ b2. |
# Math Tricks | Mensuration Formulas | Qualitative | Aptitude
## Prism Mensuration Formulas concept-Aptitude Tricks
Prism mensuration formulas concept and tricks are very important, and we should know these clearly because several types of prism problems are asked in the competitive math section. Here, we are going to discuss the prism properties, mensuration formulas and some solved problems related to this chapter. Look at below for properties of prisms.
We hope, you got the concept by looking at the above photo. If not, let me clear the property of prism. Then we will discuss mensuration formulas related to the prism.
## Mensuration formulas of a prism
What is Prism?
Prism a type of solid shape which base and top properties are the same and all others properties are flat. If you looked at the above photo, you can see that every shape have the same base and top properties which are yellow coloured. The top and base property may be formed by a triangle, quadrilateral, pentagonal, hexagonal or octagonal or so on geometrical shapes. And the formula for finding volume or surface area(lateral and total surface area) of a prism depending on base, top and lateral faces properties. This is the core concept for mensuration problems of the prism. Now, we are showing you how different Prism mensuration formulas made and also how to remember these mensuration formulas? To understand the concept of the prism, we used too many words. But in reality, it helps you to clearly understand the prism mensuration formulas and tricks for this chapters problems.
Prism volume formulas(General): The general volume formula of the prism is (Area of base X Height). Prism has various shapes, according to shapes, formulas are below with the images. We take only those shapes which are important for competitive maths.
Prism area formulas(General): The general area formula of prism(total surface) is (2 X area of the Base + perimeter of base X height)
The volume of Triangular Prism: Look at below for volume formula of a triangular prism.
We recommend you to read the tutorial on Triangle mensuration formulas. When you have to find volume or area of a triangular prism then you need the formula of triangle area. As you know triangle are various types according to its arms and angles, we briefly discussed that on triangle tutorial.
Volume formula of Rectangular Prism: Here, we are not discussing briefly. Just use the formula of shortcut cuboid volume formulas. Because Cuboid is a prism. But note, when you got non-right angular prism then you have to find the height of the prism.
Soon, we will come back on this topic with other important formulas and concepts.
Problems on trains
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2014 AMC 8 Problems/Problem 25
Problem
A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5 miles per hour, how many hours will it take to cover the one-mile stretch? $[asy]size(10cm); pathpen=black; pointpen=black; D(arc((-2,0),1,300,360)); D(arc((0,0),1,0,180)); D(arc((2,0),1,180,360)); D(arc((4,0),1,0,180)); D(arc((6,0),1,180,240)); D((-1.5,-1)--(5.5,-1));[/asy]$ Note: 1 mile = 5280 feet
$\textbf{(A) }\frac{\pi}{11}\qquad\textbf{(B) }\frac{\pi}{10}\qquad\textbf{(C) }\frac{\pi}{5}\qquad\textbf{(D) }\frac{2\pi}{5}\qquad\textbf{(E) }\frac{2\pi}{3}$
Video Solution
https://youtu.be/zi01koyAVhM ~savannahsolver
Solution
Solution 1
There are two possible interpretations of the problem: that the road as a whole is $40$ feet wide, or that each lane is $40$ feet wide. Both interpretations will arrive at the same result. However, let us stick with the first interpretation for simplicity. Each lane must then be $20$ feet wide, so Robert must be riding his bike in semicircles with radius $20$ feet and diameter $40$ feet. Since the road is $5280$ feet long, over the whole mile, Robert rides $\frac{5280}{40} =132$ semicircles in total. Were the semicircles full circles, their circumference would be $2\pi\cdot 20=40\pi$ feet; as it is, the circumference of each is half that, or $20\pi$ feet. Therefore, over the stretch of highway, Robert rides a total of $132\cdot 20\pi =2640\pi$ feet, equivalent to $\frac{\pi}{2}$ miles. Robert rides at 5 miles per hour, so divide the $\frac{\pi}{2}$ miles by $5$ mph (because $t = \frac{d}{r}$ and time = distance/rate) to arrive at $\boxed{\textbf{(B) }\frac{\pi}{10}}$ hours.
Solution 2
If Robert rides in a straight line, it will take him $\frac{1}{5}$ hours. When riding in semicircles, let the radius of the semicircle $r$, the circumference of a semicircle is $\pi r$. The ratio of the circumference of the semicircle to its diameter is $\frac{\pi}{2}$, so the time Robert takes is $\frac{1}{5} \cdot \frac{\pi}{2}$, which equals to $\boxed{\textbf{(B) }\frac{\pi}{10}}$ hours.
See Also
2014 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. |
Indefinite Integration
1.1. Simple Indefinite Integrals
Indefinite integration, also known as antidifferentiation, is the reversing of the process of differentiation. Given a function f, one finds a function F such that F' = f.
Finding an antiderivative is an important process in calculus. It is used as a method to obtain the area under a curve and to obtain many physical and electrical equations that scientists and engineers use everyday. For example, the equation for the current through a capacitor is , where I is current in Amperes, C is capacitance in Farads, V is voltage in Volts and t is time in seconds. To obtain an unknown (like V), one would have to use integration to obtain a voltage at a certain time interval.
While a true integral exists between a given boundary, taking the indefinite integral is simply reversing differentiation in much the same way division reverses multiplication. Instead of having a set of boundary values, one only finds an equation that would produce the integral due to differentiation without having to use the values to get a definite answer.
Suppose we have the equation f(x) = 3x2. We wish to find an equation F(x) so that F'(x) = 3x2. One method that could be used is the power rule from differentiation in reverse to obtain F(x) = x3. However, this is not the only answer. Remember, when one differentiates a constant, the result is zero (0). Therefore, the function could be any of the following:
F(x) = x3 - 16
F(x) = x3 + 1234567
F(x) = x3 + p
As seen, if one differentiates each one of the equations, the result becomes the same: F(x) = 3x2. Clearly, there are an infinite amount of results that one could obtain, and they all differ by the constant--the constant of integration (C). If F is the antiderivative of f, then (F + C) is the antiderivative of f. This is summarized into the following equation:
[F(x) + C]' = F'(x) + 0 = f(x).
THEOREM 1: Antiderivatives differ by a constant
If F is an antiderivative of the continuous function f, then any other antiderivative must have the form
G(x) = F(x) + C.
This says that two derivatives of the same function differ by the value of the constant of integration, which could be zero. Proof: If F and G are both antiderivatives of f, then F' = f and G' and f and the Constant Difference Theorem states that G(x) - F(x) = C, so G(x) = F(x) + C.
Before attempting some examples, it is necessary to define some basic integration rules that will allow one to take the antiderivative of a differential function. The following chart lists the basic rules. More complex methods will follow.
THEOREM 2: Basic Integration Rules
Procedural Rules Differentiation Formulas Integration Formulas Constant Multiple ò cf(u) du = cò f(u) du Sum Rule ò [f(u) + g(u)]du = ò f(u)du + ò g(u)du Difference Rule ò [f(u) - g(u)]du = ò f(u)du - ò g(u)du Constant Rule ò 0du = c Power Rule n ¹ -1 Trigonometric Rules ò sin(u) du = -cos(u) + C ò cos(u) du = sin(u) + C ò sec2 u du = tan u + C eu Rule n is a constant ò neu du = enu + C ; n is a cosntant ln(u) Rule
Proof: Each of these formulas can be derived by reversing the corresponding derivative formula. For example, to obtain the Power Rule, then
so that is an antiderivative of un and
for n ¹ -1.
The next few examples demonstrate how to apply the rules given in Theorem 2.
EXAMPLE 1: Find ò x dx.
example11.html
EXAMPLE 2: Find ò cos(x) dx
example12.html
EXAMPLE 3: Find ò (4x2 + 1)dx
example13.html
EXAMPLE 4: Find ò 3e3x dx.
example14.html
1.2. Indefinite Integrals Using the Substitution Method
Often, integrals are too complex to simply use a rule. One method for solving complex integrals is the method of substitution, where one substitutes a variable for part of the integral, integrates the function with the new variable and then plugs the original value in place of the variable. This is the integration version of the chain rule. Recall that according to the chain rule, the derivative of (x2 + 3x + 5)3 is
Thus, 3(x2 + 3x + 5)2(2x + 3)dx = (x2 + 3x + 5)3 + C
Note that the product is of the form , where g(u) = 3u2 and u = x2 + 3x + 5. The first example will go over this problem in detail.
Theorem 3: Integration by substitution
Let f, g and u be differentiable functions of x so that
Then
where G is the antiderivative of g.
Proof: If G is an antiderivative of g, then G'(u) = g(u) and, through use of the chain rule
Integrating both sides of the equation yields
Example 1: Find ò 3(x2 + 3x + 5)2(2x + 3)dx.
example21.html
Example 2: Find
Example 3: Find ò (sin3(x) cos2(x))dx.
example23.html
1.3. Integration by Parts
With some complex integrals, like xex dx both methods of inspection and the method of simple substitution can not be used. Simple substitution involves substituting for only one value and its derivative. When one uses the product rule of differentiation, the result is a function that is too complex to solve by the aforementioned means. A new method must be derived to handle these types of problems.
The product rule of differentiation is as follows:
If we substitute u for f(x) and v for g(x), we have
We can reverse the product rule to be a statement about antiderivatives
uv = ò v du + ò u dv.
This equation can then be manipulated to produce the formula for integration by parts
ò u dv = uv - ò v du.
There are several steps one must go through in order to properly use the formula:
Step 1: Let u = f(x) and dv = g(x) dx, where f(x) g(x) dx is the original integrand. Good choices to make are integrals dv = g(x) dx, which are easy to integrate.
Step 2: Compute du = f'(x) dx and v = g(x) dx.
Step 3: Substitute u, v, du and dv into the formula
ò u dv = uv - ò v du.
Step 4: Calculate uv - ò v du. If v du becomes too difficult to integrate, try choosing different values for u and dv.
Step 5: Check your solution by differentiating it.
Example 1: Find ò xex dx.
example31.html
Example 2: Find ò ln(x) dx.
example32.html
Example 3: Find ò arctan(x) dx.
example33.html
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Quick Math Homework Help
Master the 7 pillars of school success that I have learned from 25 years of teaching.
Common Core Standard G.CO.1
Collinear is when you have three or more points on a straight line. Points A,B, and C are collinear.
C
B
A
Technically a line is a group of points on a straight line. A line segment is different than a line because it has endpoints compared to a line that does not have endpoints, and extends in both directions infinitely. Check out the video to the left for a visual picture of a line.
point marks an exact location. In Geometry class we usually mark a point by making a dot with our pencil, but in reality a point has no size. The point does not have any size or dimensions because it marks a place, and is not an object.
Points are named using an upper case letter.
# Names of Lines and Rays
In Geometry a line has these properties:
• One-dimensional
• Has length
• No width
• No height,
• No curves
• Extends in both directions infinitely
In order to name a line or a line segment you have to first name a point. In this example you would have points A, B, and C.
A capital letter is used when naming a point.
Step 1. Pick two points
Step 2. Use Capital letters
Step 3. At this point you can label a line by drawing an arrow over the capital letters, or draw a straight line for a line segment
Line 2
Line 1
Line2
Line 1
Line 1 Line q
Line 2 Line m
A
C
m
q
B
q
C
B
A
m
Line 1
Line2
## What is a ray in Geometry
A ray is a line with a starting point called an endpoint. A ray travels in one direction for infinity and can pass through other points along the way to infinity.
Properties of a Ray in Geometry
• Has a starting point called the endpoint
• Has length
• No height
• No width
• Extends in one direction infinitly
• Sometimes called a "half line"
Naming a Ray in Geometry
• Start at the endpoint
• Use an arrow for the direction of the ray
• If it passes through an endpoint it is included in it's name
A
B
D
Endpoint
Name of Ray =
Name of Ray =
Math has it's own unique language and method for communicating, and here are some tips on how lines are expressed.
Line Segment 1
Line Segment 2 |
NCERT Class 6 Mathematics Chapter 11 Algebra MCQ with Solutions
NCERT Class 6 Mathematics Chapter 11 Algebra MCQ | Multiple Choice Questions and Answers with Solutions.
Students can practice the below provided MCQ’s of Class 6 Mathematics Chapter 11 Algebra. So that you can score good marks in the examination.
Page 1 of 3
1. 2x-3 may be expressed as
2. Pick out the solution from the values given in the bracket next to each equation. p – 5 = 5 (0, 10, 5 – 5)
3. The rule, which gives the number of matchsticks required to make the matchstick pattern L, is
4. Age of Avneet is y years. Avishi is four years younger than Avneet. Therefore age of Avishi is
5. A basket has x mangoes, how many mangoes are there in 5 baskets?
6. The rule, which gives the number of matchsticks required to make the matchstick pattern C, is
7. The rule, which gives the number of matchsticks required to make the matchstick pattern E. is
8. Solution of equation 3q/2=5
9. The expression for the statement: “5 times the sum of x and y” is
10. The rule, which gives the number of matchsticks required to make the matchstick pattern F, is
11. The side of a square is l. Its perimeter is
12. Equation for the statement -2 multiplies by p and then subtracted from 5 is 10 - is
13. 5 ÷ x has the operation
14. The rule, which gives the number of matchsticks required to make the matchstick pattern U, is
15. The side of an equilateral triangle is l. Its perimeter is
16. Number of variables used in the expression x2 +1 is
17. Perimeter of the square, whose each side is ‘n’ cm is
18. The side of a regular pentagon is l. Its perimeter is
19. x = 5 satisfies the equation
20. The coefficient of y³ in the expression y – y³ + y² is
21. The value of p- q +pq for p= -1, q= -2 is
22. The expression for sum of numbers a and b subtracted from their product is
23. The length of an edge of a cube is l. The total length of its edges is
24. Number of matchsticks required to make a pattern of z
25. The sum of mn + 5 – 2 and mn + 3 is
We hope the above provided NCERT Class 6 Mathematics Chapter 11 Algebra MCQ with Solutions are useful for your preparation. Keep visiting our site schools.freshersnow.com for more study material and useful information. |
# If the volumes of two cones are in the ratio of 1:4
Question:
If the volumes of two cones are in the ratio of 1:4 and their diameters are in the ratio of 4:5, then find the ratio of their heights.
Solution:
Let $r$ and $R$ be the base radii, $h$ and $H$ be the heights, $v$ and $V$ be the volumes of the two given cones.
We have,
$\frac{2 r}{2 R}=\frac{4}{5}$ or $\frac{r}{R}=\frac{4}{5} \quad \cdots$ (i)
and
$\frac{v}{V}=\frac{1}{4}$
$\Rightarrow \frac{\left(\frac{1}{3} \pi r^{2} h\right)}{\left(\frac{1}{3} \pi R^{2} H\right)}=\frac{1}{4}$
$\Rightarrow \frac{r^{2} h}{R^{2} H}=\frac{1}{4}$
$\Rightarrow\left(\frac{r}{R}\right)^{2} \times \frac{h}{H}=\frac{1}{4}$
$\Rightarrow\left(\frac{4}{5}\right)^{2} \times \frac{h}{H}=\frac{1}{4} \quad[$ Using $(\mathrm{i})]$
$\Rightarrow\left(\frac{16}{25}\right) \times \frac{h}{H}=\frac{1}{4}$
$\Rightarrow \frac{h}{H}=\frac{1 \times 25}{4 \times 16}$
$\Rightarrow \frac{h}{H}=\frac{25}{64}$
$\therefore h: H=25: 64$
So, the ratio of their heights is 25:64. |
# Chapter 21: Savings Models Lesson Plan Arithmetic Growth and Simple Interest Geometric Growth and Compound Interest A Limit to Compounding A Model for.
## Presentation on theme: "Chapter 21: Savings Models Lesson Plan Arithmetic Growth and Simple Interest Geometric Growth and Compound Interest A Limit to Compounding A Model for."— Presentation transcript:
Chapter 21: Savings Models Lesson Plan Arithmetic Growth and Simple Interest Geometric Growth and Compound Interest A Limit to Compounding A Model for Saving Present Value and Inflation 1 Mathematical Literacy in Today’s World, 8th ed. For All Practical Purposes © 2009, W.H. Freeman and Company
Chapter 21: Savings Models Arithmetic Growth and Simple Interest 2 Principal – The initial balance of the savings account. Interest – Money earned on a savings account or a loan. Example: The amount of interest on 10% of the principal of \$1000 is 10% × \$1000 = 0.10 × \$1000 = \$100 Simple Interest The method of paying interest only on the initial balance in an account, not on any accrued interest. Example: The following shows the simple interest of a savings account with a principal of \$1000 and a 10% interest rate: End of first year, you receive \$100 interest. The account total at the start of the second year is \$1100. End of second year, you receive again only \$100, which is the interest from the original balance of \$1000. Account total at the beginning of the third year is \$1200. At the end of each year you receive just \$100 in interest.
Chapter 21: Savings Models Arithmetic Growth and Simple Interest 3 Bonds An obligation to repay a specified amount of money at the end of a fixed term, with simple interest usually paid annually. Interest Rate Formula: I = Prt The total amount accumulated: A = P(1 + rt ) Interest Rate Formula – I = Prt Where: I = Simple interest earned P = Principal amount r = Annual rate of interest t = Time in years Total Amount Accumulated A = P (1 + rt ) Arithmetic Growth – (linear growth) A = P (1 + rt ) Example: Say you bought a 10-year T-note (U.S. Treasure note) today. What would be the total amount accumulated in 10 years at 4.0% simple interest? Answer: P = \$10,000, r = 4.0% = 0.40, and t = 10 yrs. Interest, I = Prt = (10,000)(0.04)(10) = \$4000 Total A= P(1 + rt ) = 10,000(1 + (0.04)(10)) = 10,000(1.40) = \$14,000
Chapter 21: Savings Models Geometric Growth and Compound Interest 4 Geometric Growth Geometric growth is the growth proportional to the amount present (also called exponential growth). Compound Interest Interest paid on both the principal and on the accumulated interest. Compound Interest Formula – A = P (1 + i ) nt Where: A = Amount earned after interest is made P = Principal amount i = Interest rate per compounding period, which is computed as i = r /n n = Number of compounding periods t = Time of the loan in years Rate per Compounding Period – For a nominal annual rate of interest r compounded n times per year, the rate per compounding period is: i = r / n
Chapter 21: Savings Models Geometric Growth and Compound Interest 5 Compounding Period The amount of time elapsing before interest is paid. For the examples below (annual, quarterly, and monthly compounding), the amount earned increases when interest is paid more frequently. Example: Suppose the initial balance is \$1000 (P = \$1000) and the interest rate is 10% (r = 0.10). What is the amount earned in 10 years (t = 10) for the following compounding periods, n? To answer this problem you need to use the following equations: Rate per compounding period, i = r / n Compound Interest Formula, A = P (1 + i ) nt Annual compounding: i = 0.10, and nt = (1)10 years A = \$1000(1 + 0.10) 10 = \$1000(1.10) 10 = \$2593.74 Quarterly compounding: i = 0.10/4 = 0.025, and nt = (4)(10) = 40 quarters A = \$1000(1 + 0.025) 40 = \$1000(1.025) 40 = \$2685.06 Monthly compounding: i = 0.10/12 = 0.008333, and nt = (12)(10) = 120mo. A = \$1000(1 + 0.10/12) 120 = \$2707.04
Chapter 21: Savings Models Geometric Growth and Compound Interest 6 The equation is obtained by manipulating the formula for finding the amount to be paid in the future after compound interest is earned. With the compound interest formula A = P (1 + i ) nt, we can solve for the present value, P. Present Value Formula – P = Where: P = Present value (principal amount) A = Amount to be paid in the future after compounding interest is earned i = Interest rate per compounding period which is computed as i = r /n n = Number of compounding periods in a year t = The years of the loan A (1 + i ) nt Present Value The present value P of an amount A to be paid in the future, after earning compound interest for n compounding periods at rate i per period is the future amount paid, A divided by (1 + i) nt.
Chapter 21: Savings Models Geometric Growth and Compound Interest Effective Rate The effective rate of interest is: effective rate where i=r/m and n=mt. Annual Percentage Yield (APY) The amount of interest earned in 1 year with a principal of \$1. The annual effective rate of interest. Example: With a nominal rate of 6% compounded monthly, what is the APY? Solution: 7
Chapter 21: Savings Models Geometric Growth and Compound Interest 8 Compound Interest Compared to Simple Interest The graph compares the growth of \$1000 with compound interest and with simple interest. The straight line explains why growth simple interest is also known as linear growth. Example of geometric and arithmetic growth: Thomas Robert Malthus (1766– 1843), an English demographer and economist, claimed that human population grows geometrically but food supplies grow arithmetically—which he attributed to future problems.
Chapter 21: Savings Models A Limit to Compounding 9 A Limit to Compound Interest The following table shows a trend: More frequent compounding yields more interest. As the frequency of compounding increases, the interest tends to reach a limiting amount (shown in the right columns). Comparing Compound Interest The Value of \$1000 at 10% Annual Interest, for Different Compounding Periods Compounded YearsYearlyQuarterlyMonthlyDailyContinuously 11100.001103.811104.711105.161105.17 51610.511638.621645.311648.611648.72 102593.742685.062707.042717.912718.28
Chapter 21: Savings Models A Limit to Compounding 10 Continuous Compounding As n gets very large, (1+ 1 / n ) n approaches the constant e ≈ 2.71828. For a principal P deposited in an account at a nominal annual rate r, compounded continuously, the balance after t years is: A = P e r t Example: For \$1000 at an annual rate of 10%, compounded n times in the course of a single year, what is the balance at the end of the year? As the quantity gets closer and closer to \$1000( e 0.1 ) = \$1105.17. No matter how frequently interest is compounded, the original \$1000 at the end of one year cannot grow beyond \$1105.17. Yield of \$1 at 100% Interest ( i = 1) Compounded n Times per Year n(1+ 1 / n ) n 12.0000000 52.4883200 102.5937424 502.6915880 1002.7048138 1,0002.7169239 10,0002.7181459 100,0002.7182682 1,000,0002.7182818 10,000,0002.7182818 It approaches e ≈ 2.71828 (which is the base of the natural logarithms).
Chapter 21: Savings Models A Model for Saving 11 A Savings Plan To have a specified amount of money in an account at a particular time in the future, you need to determine what size deposit you need to make regularly into an account with a fixed rate of interest. Savings Formula Where: A = Amount accumulated in the future after compounding interest is earned d = Uniform deposits (or payments made) i = Interest rate per compounding period which is computed as i = r /m n = Number of compounding, n=mt t = The years of the savings plan (or loan) Savings Formula The amount A accumulated after a certain period of time can be calculated by stating a uniform deposit of d per compounding period (deposited at the end of the period) and using a certain interest rate i per compounding period.
Chapter 21: Savings Models A Model for Saving Payment Formula: Solving for d in the Savings Formula so we can calculate how much our periodic payment should be in order to have \$A in the future yields: Note: The periodic payment for a sinking fund may be calculated using the Payment Formula. An annuity is a specified number of (usually equal) periodic payments. A sinking fund is a savings plan to accumulate a fixed sum by a particular date, usually through equal periodic payments. 12
Chapter 21: Savings Models Present Value and Inflation 13 Exponential Decay Exponential decay is geometric growth with a negative rate of growth. Present Value of a Dollar a Year from Now with Inflation Rate a. Example: Suppose a 25% annual inflation rate from mid-2009 through mid-2013. What will be the value of a dollar in mid-2013 in constant mid-2009 dollars? Answer: a=0.25, so i=-a/(1+a)=-0.25/(1+.25)=-0.25/1.25=-0.20 and The quantity i=–a/(1+a) behaves like a negative interest rate so we can use the compound interest formula to find the present value of P dollars t years from now.
Chapter 21: Savings Models Present Value and Inflation Depreciation Example Suppose you bought a car at the beginning of 2009 for \$12,000 and its value in current dollars depreciates steadily at a rate of 15% per year. What will be its value at the beginning of 2012 in current dollars? Answer: Using the compound interest formula, P = \$12,000, i = −0.15, and n = 3. The projected price is A = P (1 + r) n = \$12,000 (1 - 0.15) 3 = \$7369.50 14
Chapter 21: Savings Models Present Value and Inflation 15 Consumer Price Index The official measure of inflation is the Consumer Price Index (CPI), prepared by the Bureau of Labor Statistics. This index represents all urban consumers (CPI-U) and covers about 80% of the U.S. population. This is the index of inflation that is referred to on television news broadcasts, in newspapers, and magazine articles. Each month, the Bureau of Labor Statistics determines the average cost of a “market basket” of goods, including food, housing, transportation, clothing, and other items. The base period used to construct the CPI-U is from 1982–1984 and is set to 100. CPI for other year cost of market basket in other year 100 cost of market basket in base year From this proportion calculation, you can also compute the cost of an item in dollars for one year to what it would cost in dollars in a different year. cost in year A CPI for year A cost in year B CPI for year B = =
Chapter 21: Savings Models Present Value and Inflation 16 Real Growth If your investments are growing at say 6% a year and inflation is growing at 3% a year, you cannot simply subtract the two to find the purchasing power of your investments. It’s not that simple. Where P is the initial investment principle and m is the price of goods. The investment rate is r and the inflation is a. q new = Here, you see the two influences on the investment (investment growth rate r and inflation a) have directly opposite effects. Fisher’s effect – Named after the economist Irving Fisher (1867–1947). To understand why you cannot simply find the difference between interest and inflation, you must realize that the gain itself is not in original dollars but in deflated dollars. At the beginning of the year, your investment would buy the quantity: q old = P/m At the end of the year, the investment would be: P(1 + r ) m(1 + a)
Chapter 21: Savings Models Present Value and Inflation Real Rate of Growth The real (effective) annual rate of growth of an investment at annual interest rate r with annual inflation rate a is: Example: Suppose you have an investment earning 8% per year and the annual inflation rate is 3% per year. What is your real rate of growth? Answer: r=.008 and a=0.03, so g=(0.08-0.03)/(1+0.03)=0.05/1.03=0.0485=4.85% 17
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# SOLVING SURFACE AREA PROBLEMS WORKSHEET
Solving Surface Area Problems Worksheet :
Worksheet given in this section will be much useful for the students who would like to practice solving real-world problems on surface area.
## Solving Surface Area Problems Worksheet - Problems
Problem 1 :
Erin is making a jewelry box of wood in the shape of a rectangular prism. The jewelry box will have the dimensions shown below. The cost of painting the exterior of the box is \$0.50 per square in. How much does Erin have to spend to paint the jewelry box ?
Problem 2 :
A metal box that is in the shape of rectangular prism has the following dimensions. The length is 9 inches, width is 2 inches, and height is 1 1/2 inches. Find the total cost of silver coating for the entire box.
## Solving Surface Area Problems Worksheet - Solutions
Problem 1 :
Erin is making a jewelry box of wood in the shape of a rectangular prism. The jewelry box will have the dimensions shown below. The cost of painting the exterior of the box is \$0.50 per square in. How much does Erin have to spend to paint the jewelry box ?
Solution :
To know that total cost of painting, first we have to know the Surface area of the jewelry box.
Find surface area of the box.
Step 1 :
Identify a base, and find its area and perimeter.
Any pair of opposite faces can be the bases. For example, we can choose the bottom and top of the box as the bases.
Find base area.
B = l x w
B = 12 x 15
B = 180 square in.
Find perimeter of the base.
P = 2(12) + 2(15)
P = 24 + 30
P = 54 in.
Step 2 :
Identify the height, and find the surface area.
The height h of the prism is 6 inches. Use the formula to find the surface area.
S = Ph + 2B
S = 54(6) + 2(180)
S = 684 square inches
Step 3 :
Total cost = Area x Cost per square in.
Total cost = 684 x \$0.50
Total cost = \$342
Hence, Erin has to spend \$342 to paint the jewelry box.
Problem 2 :
A metal box that is in the shape of rectangular prism has the following dimensions. The length is 9 inches, width is 2 inches, and height is 1 1/2 inches. Find the total cost of silver coating for the entire box.
Solution :
To know that total cost of silver coating, first we have to know the Surface area of the metal box.
Find surface area of the box.
Step 1 :
Identify a base, and find its area and perimeter.
Any pair of opposite faces can be the bases. For example, we can choose the bottom and top of the box as the bases.
Find base area.
B = l x w
B = 9 x 2
B = 18 square in.
Find perimeter of the base.
P = 2(9) + 2(2)
P = 18 + 4
P = 22 in.
Step 2 :
Identify the height, and find the surface area.
The height h of the prism is 1 1/2 inches. Use the formula to find the surface area.
S = Ph + 2B
S = 22(1 1/2) + 2(18)
S = 22(3/2) + 36
S = 33 + 36
S = 69 square inches
Step 3 :
Total cost = Area x Cost per square in.
Total cost = 69 x \$1.50
Total cost = \$103.50
Hence, the total cost of silver coating for the entire box is \$103.50.
After having gone through the stuff given above, we hope that the students would have understood, how to solve surface area problems.
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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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 |
# How do you solve 13p^2-3=4209?
Mar 11, 2017
See the entire solution process below:
#### Explanation:
First, add $\textcolor{red}{3}$ to each side of the equation to isolate the $p$ term while keeping the equation balanced:
$13 {p}^{2} - 3 + \textcolor{red}{3} = 4209 + \textcolor{red}{3}$
$13 {p}^{2} - 0 = 4212$
$13 {p}^{2} = 4212$
Now, divide each side of the equation by $\textcolor{red}{13}$ to isolate ${p}^{2}$ while keeping the equation balanced:
$\frac{13 {p}^{2}}{\textcolor{red}{13}} = \frac{4212}{\textcolor{red}{13}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{13}}} {p}^{2}}{\cancel{\textcolor{red}{13}}} = 324$
${p}^{2} = 324$
Now, take the square root of each side of the equation to solve for $p$. Remember, the square root of a number produces a negative and positive result:
$\sqrt{{p}^{2}} = \pm \sqrt{324}$
$p = \pm 18$ |
# Mathematical Reasoning and Statements
## What is Mathematical Reasoning?
Mathematical reasoning or the principle of mathematical reasoning is a part of mathematics where we determine the truth values of the given statements. These reasoning statements are common in most of the competitive exams like JEE and the questions are extremely easy and fun to solve. Let us understand what reasoning in maths is in this article and know how to solve questions easily.
## Mathematically Acceptable Statements
Consider the following Statement:
“The sum of two prime numbers is always even.”
The given statement can either be true or false since the sum of two prime numbers can be either be an even number or an odd number. Such statements are mathematically not acceptable for reasoning as this sentence is ambiguous. Thus, a sentence is only acceptable mathematically when it is “Either true or false, but not both at the same time.” Therefore, the basic entity required for mathematical reasoning is a statement. This is the mathematical statement definition.
## Types of Reasoning in Maths
In terms of mathematics, reasoning can be of two major types which are:
1. Inductive Reasoning
2. Deductive Reasoning
The other types of reasoning are intuition, counterfactual thinking, critical thinking, backwards induction and abductive induction. These are the 7 types of reasoning which are used to make a decision. But, in mathematics, the inductive and deductive reasoning are mostly used which are discussed below.
Note: Inductive reasoning is non-rigorous logical reasoning and statements are generalized. On the other hand, deductive reasoning is rigorous logical reasoning, and the statements are considered true if the assumptions entering the deduction are true. So, in maths, deductive reasoning is considered to be more important than inductive.
### Inductive Reasoning
In the Inductive method of mathematical reasoning, the validity of the statement is checked by a certain set of rules and then it is generalized. The principle of mathematical induction uses the concept of inductive reasoning.
As inductive reasoning is generalized, it is not considered in geometrical proofs. Here, is an example which will help to understand the inductive reasoning in maths better.
• Example of Inductive Reasoning:
Statement: The cost of goods is Rs 10 and the cost of labour to manufacture the item is Rs. 5. The sales price of the item is Rs. 50.
Reasoning: From the above statement, it can be said that the item will provide a good profit for the stores selling it.
### Deductive Reasoning
The principal of deductive reasoning is the opposite of the principle of induction. On the contrary to inductive reasoning, in deductive reasoning, we apply the rules of a general case to a given statement and make it true for particular statements. The principle of mathematical induction uses the concept of deductive reasoning (contrary to its name). The below-given example will help to understand the concept of deductive reasoning in maths better.
• Example of Deductive Reasoning:
Statement: Pythagorean Theorem holds true for any right-angled triangle.
Reasoning: If triangle XYZ is a right triangle, it will follow Pythagorean Theorem.
## Types of Reasoning Statements
There are three main types of reasoning statements:
• Simple Statements
• Compound Statements
• If-Then Statements
### Simple Statements
Simple statements are those which are direct and do not include any modifier. These statements are more comfortable to solve and does not require much reasoning. An example of a simple statement is:
a: The Sun rises in the east
In this statement, there is no modifier and thus it can be simply concluded as true.
### Compound Statement
With the help of certain connectives, we can club different statements. Such statements made up of two or more statements are known as compound statements. These connectives can be “and”, “or”, etc.
With the help of such statements, the concept of mathematical deduction can be implemented very easily. For a better understanding, consider the following example:
Statement 1: Even numbers are divisible by 2
Statement 2: 2 is also an even number
These two statements can be clubbed together as:
Compound Statement: Even numbers are divisible by 2 and 2 is also an even number
Let us now find the statements out of the given compound statement:
Compound Statement: A triangle has three sides and the sum of interior angles of a triangle is 180°
The Statements for this statement is:
Statement 1: A triangle has three sides.
Statement 2: The sum of the interior angles of a triangle is 180°.
These both statements related to triangles are mathematically true. These two statements are connected using “and.”
### If-Then Statements
According to mathematical reasoning, if we encounter an if-then statement i.e. ‘if a then b’, then by proving that a is true, b can be proved to be true or if we prove that b is false, then a is also false.
If we encounter a statement which says ‘a if and only if b’, then we can give reason for such a statement by showing that if a is true, then b is also true and if b is true, then a is also true.
Example:
a: 8 is multiple of 64
b: 8 is a factor of 64
Since one of the given statements i.e. a is true, therefore, a or b is true.
## How to Deduce Mathematical Statements?
For deducing new statements or for making important deductions from the given statements three techniques are generally used:
1. Negation of the given statement
3. Counter Statements
Let’s take a look at both the methods one by one.
### Negation of the Given Statement
In this method, we generate new statements from the old ones by the rejection of the given statement. In other words, we deny the given statement and express it as a new one. Consider the following example to understand it better.
Statement 1: “Sum of squares of two natural numbers is positive.”
Now if we negate this statement then we have,
Statement 2: Sum of squares of two natural numbers is not positive.
Here, by using “not”, we denied the given statement and now the following can be inferred from the negation of the statement:
There exist two numbers, whose squares do not add up to give a positive number.
This is a “false” statement as squares of two natural numbers will be positive.
From the above discussion, we conclude that if (1) is a mathematically acceptable statement then the negation of statement 1 (denoted by statement 2) is also a statement.
In this method, we assume that the given statement is false and then try to prove the assumption wrong.
Example:
a: The derivative of y = 9x+ sin x w.r.t x is 18x + cos x.
For proving the validity of this statement, let us say that dy/dx ≠ 18x + cos x. We know that the derivative of xn is given by n • xn−1. Therefore, the derivative of 9x2 is 18x and the derivative of sin x is given by cos x.
Also,
d/dx(f(x)+g(x))=df(x)/dx+dg(x)/dx
Therefore, d/dx (9x+ sin x) = 18x + cos x
Hence, our assumption is wrong and the statement “a” is a valid statement.
### Counter Statements
Another method for proving validity is to use a counter statement i.e. giving a statement or an example where the given statement is not valid.
Example:
a: If x is a prime number then x is always odd.
To show that the given statement is false we will try to find a counter statement for this. We know that 2 is a prime number i.e. it is divisible by only itself and 1. Also, 2 is the smallest even number. Therefore, we can say that 2 is a prime number which is even. Hence, we can say that the statement “a” is not true for all prime numbers, therefore, the given statement is not valid.
## Video Lessons
### Example Questions Using the Principle of Mathematical Reasoning
Question 1:
Consider the following set of statements and mention which of these are mathematically accepted statements:
i) The Sun rises in the east.
ii) New Delhi is a country.
iii) Red rose is more beautiful than a yellow rose.
Solution:
When we read the first statement we can straightaway say that the first statement is definitely true and the second one is definitely false. As far as the third statement is considered it may depend upon perceptions of different people. Hence, it can be true for some people and at the same time false for others. But such ambiguous statements are not acceptable for reasoning in mathematics.
Thus a sentence is only acceptable mathematically when it is either true or false but not both at the same time. So, statement 1 and 2 are mathematically accepted statements while statement 3 is not accepted mathematically.
Question 2: The sum of three natural numbers x,y and z is always negative.
Solution:
This statement is acceptable. It can never be true because all natural numbers are greater than zero and therefore the sum of positive numbers can never be negative.
Question 3: The product of three real numbers x,y and z is always zero.
Solution:
In this given statement we cannot figure out if the statement is true or false. Such a sentence is not mathematically acceptable for reasoning.
Question 4: Check whether the given two statements are true with respect to each other.
a: A circle with infinite radius is a line
b: A circle with zero radii is a point
Solution:
Since “a” is true and “b” is also true then both statements a and b are also true.
For two given statements a or b to be true, show that either a is true or prove that b is true i.e. if any one of the statements is true then a or b is also true.
Now it would be clear to you how to use a compound form of statements and negative of a statement to deduce results.
Q1
### What is Mathematical Reasoning?
Mathematical reasoning is one of the topics in mathematics where the validity of mathematically accepted statements is determined using logical and Maths skills.
Q2
### Why is Mathematical Reasoning Important?
Mathematical reasoning is important as it helps to develop critical thinking and understand Maths in a more meaningful way. The concepts of reasoning not only helps the students to have a deeper understanding of the subject but also helps in having a wider perspective to logical statements.
Q3
### What are the Types of Reasoning?
There are two main types of reasoning in Maths:
• Inductive reasoning
• Deductive reasoning |
Length of a Vector – Definition, Formulas, and Examples
The length of a vector allows us to understand how large the vector is in terms of dimensions. This also helps us understand vector quantities such as displacement, velocity, force, and more. Understanding the formula for calculating the length of a vector will help us in establishing the formula for the arc length of a vector function.
The length of a vector (commonly known as the magnitude) allows us to quantify the property of a given vector. To find the length of a vector, simply add the square of its components then take the square root of the result.
In this article, we’ll extend our understanding of magnitude to vectors in three dimensions. We’ll also cover the formula for the arc length of the vector function. By the end of our discussion, our goal is for you to confidently work on different problems involving vectors and vector functions’ lengths.
What Is the Length of a Vector?
The length of the vector represents the distance of the vector in the standard position from the origin. In our previous discussion on vector properties, we’ve learned that the length of a vector is also known as the magnitude of the vector.
Suppose that $\textbf{u} = x \textbf{i}+y \textbf{j}$, we can calculate the length of the vector using the formula for magnitudes as shown below: \begin{aligned}|\textbf{u}| = \sqrt{x^2 +y^2}\end{aligned} We can extend this formula for vectors with three components -$\textbf{u} = x \textbf{i}+ y \textbf{j} + z\textbf{k}$ : \begin{aligned}|\textbf{v}| = \sqrt{x^2 +y^2 + z^2}\end{aligned}
In fact, we can extend our understanding of three-coordinate systems and vectors to prove the formula for the vector length in space.
Proof of Vector Length Formula in 3D
Suppose that we have a vector, $\textbf{u} = x_o \textbf{i} + y_o \textbf{j} +z_o \textbf{k}$, we can rewrite the vector as the sum of two vectors. Hence, we have the following:
\begin{aligned}\textbf{v}_1 &= \\ \textbf{v}_2 &= <0 , 0, z_o>\\\textbf{u} &= <x_o,y_o ,z_o>\\&= +<0 ,0, z_o>\\&=\textbf{v}_1+ \textbf{v}_2\end{aligned}
We can calculate the lengths of the two vectors, $\textbf{v}_1$ and $\textbf{v}_2$, by applying what we know of magnitudes.
\begin{aligned}|\textbf{v}_1| &= \sqrt{x_o^2 +y_o^2}\\ |\textbf{v}_2| &= \sqrt{z_o^2}\end{aligned}
These vectors will form a right triangle with $\textbf{u}$ as the hypotenuse, so we can use Pythagorean theorem to calculate the length of the vector, $\textbf{u}$.
\begin{aligned}|\textbf{u}| &= \sqrt{|\textbf{v}_1|^2 +|\textbf{v}_2|^2}\\&= \sqrt{(x_o^2 + y_o^2) + z_o^2}\\&= \sqrt{x_o^2 +y_o^2 +z_o^2}\end{aligned}
This means that for us to calculate the length of the vector in three dimensions, all we have to do is add the squares of its components then take the square root of the result.
Arc Length of a Vector Function
We can extend this notion of length to vector functions – this time, we’re approximating the distance of vector function over an interval of $t$. The length of the vector function, $\textbf{r}(t)$, within the interval of $[a, b]$ can be calculated using the formula shown below.
\begin{aligned}\textbf{r}(t) &= \left\\\text{Arc Length} &= \int_{a}^{b}\sqrt{[x\prime(t)]^2 + [y\prime(t)]^2]}\phantom{x}dt\\\\\textbf{r}(t) &= \left\\\text{Arc Length} &= \int_{a}^{b}\sqrt{[x\prime(t)]^2 + [y\prime(t)]^2] + [z\prime(t)]^2]}\phantom{x}dt\end{aligned}
From this, we can see that the arc length of the vector function is simply equal to the magnitude of the vector tangent to $\textbf{r}(t)$. This means that we can simplify our arc length’s formula to the equation shown below:
\begin{aligned}L &= \int_{a}^{b} |\textbf{r}\prime(t)| \phantom{x} dt\end{aligned}
We’ve now covered all the fundamental definition of vector lengths and vector function lengths, it’s time for us to apply them to calculate their values.
How To Calculate the Length of a Vector and a Vector Function?
We can calculate the length of a vector by applying the formula for the magnitude. Here’s a breakdown of the steps to calculate the vector’s length:
• List down the components of the vector then take their squares.
• Add the squares of these components.
• Take the square root of the sum to return the length of the vector.
This means that we can calculate the length of the vector, $\textbf{u} = \left<2, 4, -1\right>$, by applying the formula, $|\textbf{u}| = \sqrt{x^2 + y^2 + z^2}$, where $\{x, y, z\}$ represents the components of the vector.
\begin{aligned}|\textbf{u}| &= \sqrt{x^2 + y^2 + z^2}\\ &= \sqrt{(2)^2 + (4)^2 + (-1)^2}\\&=\sqrt{4 + 16 + 1}\\&= \sqrt{21}\end{aligned}
Hence, the length of the vector, $\textbf{u}$, is equal to $\sqrt{21}$ units or approximately equal to $4.58$ units.
As we have shown in our earlier discussion, the arc length of the vector function depends on the tangent vector. Here’s a guideline to help you in calculating the arc length of the vector function:
• List down the components of the vector then take their squares.
• Square each of the derivatives then add the expressions.
• Write the square root of the resulting expression.
• Evaluate the integral of the expression from $t = a$ to $t = b$.
Let’s say we have the vector function, $\textbf{r}(t) = \left<(4t – 1), (2t +4)\right>$. We can calculate its arc length by from $t = 0$ to $t = 4$ using the formula, $L = \int_{a}^{b} |\textbf{r}\prime(t)| \phantom{x} dt$, where $\textbf{r}\prime(t)$ represents the tangent vector.
This means that we’ll need to find $\textbf{r}\prime(t)$ by differentiating each of the vector function’s component.
\begin{aligned}x \prime(t)\end{aligned} \begin{aligned}x\prime(t) &= \dfrac{d}{dt} (4t –1)\\&= 4(1) – 0\\&= 4\end{aligned} \begin{aligned}y \prime(t)\end{aligned} \begin{aligned}y\prime(t) &= \dfrac{d}{dt} (2t +4)\\&= 2(1) – 0\\&= 2\end{aligned} \begin{aligned}\textbf{r}\prime(t) &= \left\\&= \left<4, 2\right>\end{aligned}
Take the magnitude of the tangent vector by squaring the components of the tangent vector then writing down the square root of the sum.
\begin{aligned}|\textbf{r}\prime(t)| &= \sqrt{[x\prime(t)]^2 + [y\prime(t)]^2] }\\&= \sqrt{4^2 + 2^2} \\&= \sqrt{20}\end{aligned}
Now, evaluate the integral of the resulting expression from $t = 0$ to $t = 4$.
\begin{aligned}\int_{0}^{4} \sqrt{20} \phantom{x}dt &=\int_{0}^{4} 2\sqrt{5} \phantom{x}dt\\&= 2\sqrt{5}\int_{0}^{4} \phantom{x}dt\\&= 2\sqrt{5} [t]_0^4\\&= 2\sqrt{5}( 4 -0)\\&= 8\sqrt{5}\end{aligned}
This means that the arc length of $\textbf{r}(t)$ from $t=0$ to $t=4$ is equal to $8\sqrt{5}$ units or approximately $17.89$ units.
These are two great examples of how we can apply the formulas for vector and vector function lengths. We’ve prepared some more problems for you to try, so head over to the next section when you’re ready!
Example 1
The vector $\textbf{u}$ has an initial point at $P(-2, 0, 1 )$ and an endpoint at $Q(4, -2, 3)$. What is the vector’s length?
Solution
We can find the position vector by subtracting the components of $P$ from the components of $Q$ as shown below.
\begin{aligned}\textbf{u} &= \overrightarrow{PQ}\\&= \left<(4 – -2), (-2 – 0), (3 -1)\right>\\&= \left<6, -2, 2\right>\end{aligned}
Use the formula for the vector’s magnitude to calculate the length of $\textbf{u}$.
\begin{aligned}|\textbf{u}| &= \sqrt{(6)^2 + (-2)^2 + (2)^2}\\&= \sqrt{36+ 4+ 4}\\&= \sqrt{44}\\&= 2\sqrt{11}\\&\approx 6.63 \end{aligned}
This means that the vector, $\textbf{u}$, has a length of $2\sqrt{11}$ units or approximately $6.33$ units.
Example 2
Calculate the arc length of the vector-valued function, $\textbf{r}(t) = \left<2\cos t, 2\sin t, 4t\right>$, if $t$ is within the interval, $t \in [0, 2\pi]$.
Solution
We’re now looking for the arc length of the vector function, so we’ll use the formula shown below.
\begin{aligned} \text{Arc Length} &= \int_{a}^{b}\sqrt{[x\prime(t)]^2 + [y\prime(t)]^2] + [z\prime(t)]^2]}\phantom{x}dt\\&= \int_{a}^{b} |\textbf{r}\prime(t)| \phantom{x}dt\end{aligned}
First, let’s take the derivative of each components to find $\textbf{r}\prime(t)$.
\begin{aligned}x\prime(t)\end{aligned} \begin{aligned}x\prime(t) &= \dfrac{d}{dt}(2 \cos t)\\&= 2(-\sin t)\\&= -2\sin t \end{aligned} \begin{aligned}y \prime(t)\end{aligned} \begin{aligned}y\prime(t) &= \dfrac{d}{dt}(2 \sin t)\\&= 2(\cos t)\\&= 2\cos t\end{aligned} \begin{aligned}z\prime(t)\end{aligned} \begin{aligned}y\prime(t) &= \dfrac{d}{dt}(2 4t)\\&= 4(1)\\&= 4\end{aligned} \begin{aligned}\textbf{r}\prime(t) &= \left\\&= \left<-2\sin t, 2\cos t, 4\right>\end{aligned}
Now, take the magnitude of $\textbf{r}\prime(t)$ by adding the squares of the tangent vector’s components. Write the square root of the sum to express the magnitude in terms of $t$.
\begin{aligned}|\textbf{r}\prime(t)| &= \sqrt{(-2 \cos t)^2 + (4\sin t)^2 + 4^2}\\&= \sqrt{4 \cos^2 t + 4\sin^2 t + 16}\\&= \sqrt{4(\cos^2 t + \sin^2 t) + 16}\\&= \sqrt{4(1) + 16}\\&= \sqrt{20}\\&= 2\sqrt{5}\end{aligned}
Integrate $|\textbf{r}\prime(t)|$ from $t = 0$ to $t = 2\pi$ to find the arc length of the vector.
\begin{aligned} \text{Arc Length} &= \int_{a}^{b}|\textbf{r}\prime(t)| \phantom{x}dt\\&= \int_{0}^{2\pi} 2\sqrt{5} \phantom{x}dt\\&= 2\sqrt{5}\int_{0}^{2\pi} \phantom{x}dt\\&= 2\sqrt{5}(2\pi – 0)\\&= 4\sqrt{5}\pi\\&\approx 28.10\end{aligned}
This means that the arc length of the vector function is $4\sqrt{5}\pi$ or approximately $28.10$ units.
Practice Questions
1. The vector $\textbf{u}$ has an initial point at $P(-4, 2, -2 )$ and an endpoint at $Q(-1, 3, 1)$. What is the vector’s length?
2. Calculate the arc length of the vector-valued function, $\textbf{r}(t) = \left<t\cos t, t\sin t, 2t\right>$, if $t$ is within the interval, $t \in [0, 2\pi]$.
1. The vector has a length of $\sqrt{19}$ units or approximately $4.36$ units.
2. The arc length is approximately equal to $25.343$ units. |
# Equation of a Circle
## Equation of a Circle Lesson
### Common Forms of a Circle's Equation
Let's take a look together through three common forms of a circle's equation (standard form, general form, and polar coordinates), so we can become more comfortable working with circles in algebra.
#### Standard Form for the Equation of a Circle
The standard form equation of a circle is given as:
(x - a)2 + (y - b)2 = r2
Where a is x-coordinate of the circle's center, b is the y-coordinate of the circle's center, and r is the radius of the circle.
Let's look at a circle defined by the equation (x - 1)2 + (y - 2)2 = 25. The center of this circle is located at the point (1, 2) since a = 1 and b = 2. The circle's radius is 5 since r2 = 25 and r = 5.
INTRODUCING
#### General Form for the Equation of a Circle
The general form equation of a circle is given as:
x2 + y2 + Ax + By + C = 0
Where A, B, and C are constants.
The general form equation is not formatted to give us a quick glimpse of circle location and radius like the standard form equation is. However, general form can be converted to standard form by using the algebraic process known as completing the square.
#### Polar Coordinates Equation of a Circle
The standard form and general form of a circle's equation are expressed in the Cartesian coordinate system. When dealing with circles, using the polar coordinate system is simpler than the Cartesian coordinate system.
In polar coordinates, the equation of a circle centered over the origin is:
r = radius of the circle
If the circle is not centered over the origin but lies on the x-axis, the equation is:
r = 2acosθ
Where a is the x-coordinate of the circle's center.
If the circle is not centered over the origin but lies on the y-axis, the equation is:
r = 2bsinθ
Where b is the y-coordinate of the circle's center.
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# 1997 AJHSME Problems/Problem 15
## Problem
Each side of the large square in the figure is trisected (divided into three equal parts). The corners of an inscribed square are at these trisection points, as shown. The ratio of the area of the inscribed square to the area of the large square is
$[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((1,0)--(1,0.2)); draw((2,0)--(2,0.2)); draw((3,1)--(2.8,1)); draw((3,2)--(2.8,2)); draw((1,3)--(1,2.8)); draw((2,3)--(2,2.8)); draw((0,1)--(0.2,1)); draw((0,2)--(0.2,2)); draw((2,0)--(3,2)--(1,3)--(0,1)--cycle); [/asy]$
$\text{(A)}\ \dfrac{\sqrt{3}}{3} \qquad \text{(B)}\ \dfrac{5}{9} \qquad \text{(C)}\ \dfrac{2}{3} \qquad \text{(D)}\ \dfrac{\sqrt{5}}{3} \qquad \text{(E)}\ \dfrac{7}{9}$
## Solution 1
Since we are dealing with ratios, let the big square have sides of $3$ and thus an area of $3^2 = 9$. Chosing a multiple of $3$ will avoid fractions in the rest of the answer.
To find the area of the inscribed square, subtract off the areas of the four triangles. Each triangle has an area of $\frac{1}{2}\cdot 2\cdot 1 = 1$.
Thus, the area of the inscribed square is $9 - 4\cdot 1 = 5$, and the ratio of areas is $\frac{5}{9}$, giving option $\boxed{B}$.
## Solution 2
Let the side of the big square be $3x$. The area of this square is $3x\cdot 3x = 9x^2$
You can use the Pythagorean Theorem on the any of the right triangles to find the length of the side of the inscribed square. One leg is $x$, while the other leg is $2x$. Thus, the length of the hypotenuse $c$, which is also the side of the inscrubed square, is:
$a^2 + b^2 = c^2$
$x^2 + (2x)^2 = c^2$
$x^2 + 4x^2 = c^2$
$5x^2 = c^2$
Once you get $c^2$, you can stop, because $c^2$ is the area of the inscribed square. The ratio of the areas of the small square to the big square is $\frac{5x^2}{9x^2} = \frac{5}{9}$, giving option $\boxed{B}$.
## Solution 3
Fill in the gridlines of the diagram.
$[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((1,0)--(1,3)); draw((2,0)--(2,3)); draw((3,1)--(0,1)); draw((3,2)--(0,2)); draw((1,3)--(1,0)); draw((2,3)--(2,0)); draw((0,1)--(3,1)); draw((0,2)--(3,2)); draw((2,0)--(3,2)--(1,3)--(0,1)--cycle); [/asy]$
Let each grid space have length $1$. The inscribed square consists of four congruent triangles and one square.
The triangles each have area $\frac{1}{2} \cdot 2\cdot 1=1$, and in total they cover $4 \cdot 1 = 4$ square units of area. The central square's area is $1 \cdot 1=1$. Therefore, the area of the inscribed square is $4+1=5$.
The big square contains $3 \cdot 3=9$ square units of area. Thus, the ratio is equal to $\frac{5}{9}$, or $\boxed{B}$. |
### Prove that the lengths of tangents drawn from an external point $A$ to the points $P$ and $Q$ on the circle are equal.
Step by Step Explanation:
1. It is given that two tangents are drawn from an external point $A$ to the points $P$ and $Q$ on the circle.
The given situation is represented by the below image.
We have to prove that the length $AP$ is equal to length $AQ$.
2. Let us join the point $O$ to points $P, Q,$ and $A.$
We get
$AP$ is a tangent at $P$ and $OP$ is the radius through $P$.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
$\implies OP \perp AP$
Also, $AQ$ is a tangent at $Q$ and $OQ$ is the radius through $Q$.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
$\implies OQ \perp AQ$
3. In right- angled triangle $OPA$ and $OQA$, we have \begin{aligned} & OP = OQ && \text{[Radius of the same circle]} \\ & OA = OA && \text{[Common]} \\ \implies & \triangle OPA \cong \triangle OQA && \text{[By RHS-congruence]} \end{aligned}
4. As the corresponding parts of congruent triangle are equal, we have $AP = AQ$.
5. Thus, the lengths of tangents drawn from an external point $A$ to the points $P$ and $Q$ on the circle are equal.
You can reuse this answer |
# Irregular Hexagon
A hexagon is regular and irregular based on the measure of its sides and angles.
Let us see how an irregular hexagon is different from a regular one.
A regular hexagon has 6 equal sides, 6 equal interior angles each of 120°. The total sum of its interior angles is 720°, 6 exterior angles, each of 60°, 9 diagonals, and 6 lines of symmetry.
In contrast, an irregular hexagon does not satisfy the above features. As the term ‘irregular‘ suggests, an irregular hexagon is a 6-sided polygon with unequal measures of sides and angles.
The properties of an irregular hexagon are:
1. Has 6 unequal sides
2. The interior angles are not equal but their sum is 720°
3. The exterior angles are not equal but their sum will be 360°
4. Cannot be inscribed in a circle
5. A circle cannot be inscribed in it
6. Has 9 unequal diagonals
7. Can be convex or a concave
8. Diagonals do not intersect at a common point
9. Does not have any line of symmetry
We can find the area and perimeter of a regular hexagon very easily. Let us see how to calculate the area and perimeter of an irregular hexagon.
## Area
There is no definite formula to find the area of an irregular hexagon.
Let us solve an example to understand the concept better.
Find the area of an irregular hexagon shaped land shown below. 6 of its sides and 3 of its diagonals are known.
++++++
Solution:
Area of the hexagon ABCDEF = Area of ΔABC + Area of ΔACF+ Area of ΔFCE + Area of ΔECD
Here we will use Heron’s formula to find the area of each triangle.
Area (A) = √s(s – a)( s– b)(s – c), here s = ½ (a + b + c); a, b, and c are the 3 sides of any triangle
In ΔABC, s = ½ (a + b + c) here a = 4.5 m, b = 4.2 m, c = 6.2 m
= ½ (4.5 + 4.2 + 6.2)
= 7.45 m
Area (A)ΔABC = √s(s – a)( s– b)(s – c), here s = 7.45 m
= √7.45(7.45 – 4.5)(7.45 – 4.2)(7.45 – 6.2)
≈ 9.45 sq. m
In ΔACF, s = ½ (a + b + c), a = 6.2 m, b = 6.3 m, c = 2.1 m
= ½ (6.2 + 6.3 + 2.1)
= 7.3 m
Area (A)ΔACF = √s(s – a)( s– b)(s – c), here s = 7.3 m
= √7.3(7.3 – 6.2)(7.3 – 6.3)(7.3 – 2.1)
= 6.46 sq. m
In ΔFCE, s = ½ (a + b + c), a = 6.3 m, b = 6 m, c = 7 m
= ½ (6.3 + 6 + 7)
= 9.65 m
Area (A)ΔFCE = √s(s – a)( s– b)(s – c), here s = 9.65 m
= √9.65(9.65 – 6.3)( 9.65 -6)( 9.65 – 7)
= 17.68 sq. m
In ΔECD, s = ½ (a + b + c), a = 6 m, b = 5.3 m, c = 3.2 m
= ½ (6 + 5.3 + 3.2)
= 7.25 m
Area (A)ΔECD = √s(s – a)( s– b)(s – c), here s = 7.25 m
= √7.25(7.25 – 6)(7.25 – 5.3)(7.25 – 3.2)
≈ 8.46 sq. m
Area of the hexagon ABCDEF = AΔABC + AΔACF + AΔFCE + AΔECD
= 9.45 + 6.46 + 17.68 + 8.46
= 42.05 sq. m
## Perimeter
There is no definite formula to find the perimeter of an irregular hexagon. We can find it when all the sides are known and add them together.
Let us solve an example to understand the concept better.
Find the perimeter of the hexagon with side lengths 4 cm, 12 cm, 11 cm, 7 cm, 6 cm, and 5 cm.
Solution:
As we know,
P = 4 + 12+ 11 + 7 + 6 + 5
= 45 cm |
## Area Formulas
Geometry and measurement ideas are usually taught together in elementary school. The connection between geometry and measurement is very evident in the development of formulas. Formulas are equations which use measures that are easy to determine such as length or height, to find measures that are more difficult to determine such as area and volume. For example, it is much easier to measure the length and width of a rectangle than it is to find the number of square units that cover the rectangle.
It is important that students participate in the development of formulas rather than just being given the formulas and asked to apply them. Participating in the development will help students make connections between the formulas for different figures. For example, your students already know and understand the formula for finding the area of a rectangle, A = l x w, where A represents the area, l the length, and w the width of the rectangle. They will see the relationship between this formula and the formula for finding the area of a triangle from examples like the following.
1. Start with a triangle. Draw a line parallel to one side of the triangle through the vertex that is opposite that side. 2. Draw line segments that form right angles from the other two vertices of the triangle up to the parallel line. Now your triangle is in a rectangle. 3. Draw a perpendicular line from the vertex of the triangle that is opposite the side that is shared by both the triangle and rectangle. Notice that two sets of congruent triangles are formed.
Look at the relationship between the different parts of the triangle and rectangle above. The height of the triangle, line segment CD, is congruent to the width of the rectangle, sides AX and BY. The base of the triangle, line segment AB, is the same as the length of the rectangle, line segment AB. Triangles ACX and ACD are congruent, and triangles BYC and CDB are also congruent. If you took away one triangle from each pair of congruent triangles, you would remove half the area of the rectangle. So, the area of triangle ABC is half the area of rectangle ABYX. The formula for finding the area of any triangle is A = 1/2 (b x h), where b is the base of the triangle and h is the height. To reinforce students' understanding of the formula for finding the area of a triangle, repeat steps 1-3 above with other triangles.
A similar kind of experience can also come from taking any rectangle and drawing a triangle in it. Look at the examples below. Use one side of the rectangle as the base of the triangle and locate the other vertex of the triangle anywhere on the opposite side of the rectangle. Draw a perpendicular line from the base of the triangle to the vertex to show the height of the triangle. Any triangle drawn this way in the same rectangle will have the same area, since the formula for finding the area of a triangle is A = 1/2(b x h), and the base and height will always be equal to the length and width of the rectangle.
A similar kind of hands-on experience can be provided in the development of the formula for the area of a parallelogram. A line segment joining two vertices of a polygon that is not a side of the polygon is called a diagonal. Two diagonals can be drawn in a parallelogram. Either diagonal divides the parallelogram into two congruent triangles. Line segment NP divides the parallelogram below into two congruent triangles, triangle MNP and triangle PON. The formula for finding the area of each triangle is A = 1/2 (b x h), so the formula for finding the area of the parallelogram is A = 1/2 (b x h) + A = 1/2 (b x h) or A = (b x h).
Another way to see this formula is to cut off the right triangle from one of the corners as shown below. By placing triangle MQP over to the right, shown as triangle NRO, we can see that parallelogram MNOP has the same area as rectangle QROP. So we can use the formula A = (b x h) to find the area of the parallelogram.
It is important to note that by the height we are talking about the perpendicular distance between two parallel sides in a parallelogram. In a triangle, when we refer to the height, we are talking about the perpendicular distance from a vertex to its opposite side. |
# An AP 5, 12, 19, ... has 50 terms. Find its last term. Hence, find the sum of its last 15 terms.
Question:
(i) An AP 5, 12, 19, ... has 50 terms. Find its last term. Hence, find the sum of its last 15 terms.
(ii) An AP 8, 10, 12, ... has 60 terms. Find its last term. Hence, find the sum of its last 10 terms.
Solution:
(i) The given AP is 5, 12, 19, ... .
Here, a = 5, d = 12 − 5 = 7 and n = 50.
Since there are 50 terms in the AP, so the last term of the AP is a50.
$l=a_{50}=5+(50-1) \times 7 \quad\left[a_{n}=a+(n-1) d\right]$
$=5+343$
$=348$
Thus, the last term of the AP is 348.
Now,
Sum of the last 15 terms of the AP
$=S_{50}-S_{35}$
$=\frac{50}{2}[2 \times 5+(50-1) \times 7]-\frac{35}{2}[2 \times 5+(35-1) \times 7] \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$
$=\frac{50}{2} \times(10+343)-\frac{35}{2} \times(10+238)$
$=\frac{50}{2} \times 353-\frac{35}{2} \times 248$
$=\frac{17650-8680}{2}$
$=\frac{8970}{2}$
$=4485$
Hence, the required sum is 4485.
(ii) The given AP is 8, 10, 12, ... .
Here, a = 8, d = 10 − 8 = 2 and n = 60
Since there are 60 terms in the AP, so the last term of the AP is a60.
$l=a_{60}=8+(60-1) \times 2 \quad\left[a_{n}=a+(n-1) d\right]$
$=8+118$
$=126$
Thus, the last term of the AP is 126.
Now,
Sum of the last 10 terms of the AP
$=S_{60}-S_{50}$
$=\frac{60}{2}[2 \times 8+(60-1) \times 2]-\frac{50}{2}[2 \times 8+(50-1) \times 2] \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$
$=30 \times(16+118)-25 \times(16+98)$
$=30 \times 134-25 \times 114$
$=4020-2850$
$=1170$
Hence, the required sum is 1170. |
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Angle between Two Vectors Formula
If the two vectors are given as $\vec{a}$ and $\vec{b}$ then its dot product is expressed as a.b. Suppose these two vectors are separated by angle $\theta$. To know what's the angle measurement we solve with the below formula
The angle between two vectors formula is given by
where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
Related Calculators Angle between Two Vectors Calculator Cross Product of Two Vectors Calculator Calculate Vector Determinant Calculator 2x2
Angle between two vectors Examples
Lets see some examples on angle between two vectors:
Solved Examples
Question 1: Find the angle between two vectors 3i + 4j - k and 2i - j + k.
Solution:
Given: $\vec{a}$ = 3i + 4j - k and $\vec{b}$ = 2i - j + k
The dot product is given by
a.b = (3i + 4j - k)(2i - j + k)
= (3)(2) + (4)(-1) + (-1)(1)
= 6-4-1
= 1
The magnitude of vectors is given by
|a| = $\sqrt{3^2 + 4^2 + (-1)^2}$ = $\sqrt{26}$ = 5.09
|b| = $\sqrt{2^2 + (-1)^2 + (1)^2}$ = $\sqrt{6}$ = 2.449
The angle between two vectors is
$\theta$ = cos-1 $\frac{a.b}{|a||b|}$
= cos-1 $\frac{1}{5.09 \times 2.449}$
= cos-1 $\frac{1}{12.465}$
= cos-1 0.0802
= 85.37o.
Question 2: Find the angle between two vectors 5i - j + k and i + j - k.
Solution:
Given: $\vec{a}$ = 5i - j + k and $\vec{b}$ = i + j - k
The dot product is given by
a.b = (5i - j + k)(i + j - k)
= (5)(1) + (-1)(1) + (1)(-1)
= 5-1-1
= 3
The magnitude of vectors is given by
|a| = $\sqrt{5^2 + (-1)^2 + (1)^2}$ = $\sqrt{26}$ = 5.09
|b| = $\sqrt{1^2 + (1)^2 + (-1)^2}$ = $\sqrt{3}$ = 1.73
The angle between two vectors is
$\theta$ = cos-1 $\frac{a.b}{|a||b|}$
= cos-1 $\frac{3}{5.09 \times 1.73}$
= cos-1 $\frac{3}{8.8057}$
= cos-1 0.372
= 68.16o.
*AP and SAT are registered trademarks of the College Board. |
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# Solution for Find the Interval in Which the Following Function Are Increasing Or Decreasing F(X) = X4 − 4x ? - CBSE (Commerce) Class 12 - Mathematics
ConceptIncreasing and Decreasing Functions
#### Question
Find the interval in which the following function are increasing or decreasing f(x) = x4 − 4x ?
#### Solution
$\text { When } \left( x - a \right)\left( x - b \right)>0 \text { with }a < b, x < a \text { or }x>b.$
$\text { When } \left( x - a \right)\left( x - b \right)<0 \text { with } a < b, a < x < b .$
$f\left( x \right) = x^4 - 4x$
$f'\left( x \right) = 4 x^3 - 4$
$= 4\left( x^3 - 1 \right)$
$\text { For}f(x) \text { to be increasing, we must have }$
$f'\left( x \right) > 0$
$\Rightarrow 4\left( x^3 - 1 \right) > 0$
$\Rightarrow x^3 - 1 > 0$
$\Rightarrow x^3 > 1$
$\Rightarrow x > 1$
$\Rightarrow x \in \left( 1, \infty \right)$
$\text { So,}f(x)\text { is increasing on }\left( 1, \infty \right) .$
$\text { For }f(x) \text { to be decreasing, we must have }$
$f'\left( x \right) < 0$
$\Rightarrow 4\left( x^3 - 1 \right) < 0$
$\Rightarrow x^3 - 1 < 0$
$\Rightarrow x^3 < 1$
$\Rightarrow x < 1$
$\Rightarrow x \in \left( - \infty , 1 \right)$
$\text { So,}f(x)\text { is decreasing on }\left( - \infty , 1 \right).$
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [3]
Solution Find the Interval in Which the Following Function Are Increasing Or Decreasing F(X) = X4 − 4x ? Concept: Increasing and Decreasing Functions.
S |
# Copying an angle
Copying an angle or constructing an angle congruent to a given angle using only a straightedge and a compass is what I will show you here. I will show you the steps to copy an acute angle. The steps are still the same when the angle is right or obtuse.
Therefore, study the steps outlined here carefully and you will be able to copy any angle.
Copy Angle A shown below:
Step 1:
Draw a ray and call the endpoint B.
Step 2: Now go back to the angle you are copying. Then, place the needle of the compass at the vertex of angle A. While keeping the needle of the compass at vertex A, open the compass to a certain distance and draw an arc of any size. We show the arc drawn in blue.
Notice that you can open the compass to any distance that you like. Just make sure that the distance you open the compass is not so big that when you draw the arc, it will not intersect the rays.
After that, still keeping the opening of the compass the same distance, go to the segment that you drew earlier. Then, place the needle of the compass at vertex B and draw an arc that will touch the ray at point C for instance.
While doing step 2, you got to make sure the opening of the compass does not become smaller or bigger while drawing the arc. If you do not do this, you will end up copying the angle incorrectly.
Step 3:
Now, notice the blue segment in the following figure. Use your compass to measure the length of this segment shown in blue.
Then, still keeping the distance the same, place the needle of the compass at point C and draw an arc that will intersect the other arc that passes through C. Call the point the two arc intersects D.
Step 4
Draw a line between points B and D and you are done!
## Copying an angle that is bigger than 90 degrees on your own
Activity
1. Get a piece of paper, a ruler, and a compass.
2. Draw an obtuse angle or an angle that is bigger than 90 degrees. Do you not remember what an obtuse angle look like? Check the lesson about types of angles.
3. Using the steps outlined above, copy the obtuse angle you drew in step 2.
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# In a potentiometer a standard cell of EMF 5V and of negligible resistance maintains a steady current through the potentiometer wire of length 5 meter. Two primary cells of emf ${E_1}$ and ${E_2}$, are joined in series. (i) same polarity, (ii) opposite polarity. The configuration is connected to a galvanometer and a jockey to the potentiometer. The balancing lengths in the two cases are found to be 350 cm and 5 cm respectively.A) Draw the necessary circuit diagram.B) Find the value of the emf’s of the two cells,
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Hint: In a potentiometer the jockey is used to slide over the wire to find the balancing length. The polarity is different in the two cases.
Complete step by step solution:
Here the first step is to understand the questions and the conditions given.
There is wire AB of length 5 meter. It is connected to a 5 volt cell. So let us start drawing the circuit.
The first part of the circuit:
The above diagram is the first part of the circuit.
Now for the second part the two cells are connected in series. So for the next part we are going to add them to the first part directly without drawing it separately.
The above diagram shows the final circuit for the given condition.
Now for the second part of the question;
The potential gradient is to be calculated.
Hence if emf is 5 volts and length is 5 meters the potential gradient of AB is given by;
$k = \dfrac{V}{{{l_{cm}}}}$ here V is potential difference across the length and ${l_{cm}}$ is the length of AB in centimeter
So $k = \dfrac{{5V}}{{500cm}} = \dfrac{{1}}{{100}}$
Now case 1: the cells have same polarity-
Thus ${E_1} + {E_2} = kl$, here, l is the balancing length for the same polarity, so, l= 350cm.
Hence, ${E_1} + {E_2} = \dfrac{1}{{100}} \times 350 = 3.5$ (equation: 1)
Now for Case 2: the polarities of the cell are opposite-
Hence, ${E_1} - {E_2} = kl'$ here $l'$ is the balancing length for opposite polarity, so $l' = 5cm$.]
Thus, ${E_1} - {E_2} = \dfrac{1}{{100}} \times 5 = 0.05$ (equation: 2)
Now we solve the equations 1 and 2;
First adding equation 2 to equation 1, we get;
$2{E_1} = 3.55$
Thus, ${E_1} = \dfrac{{3.55}}{2} = 1.78V$
Now substituting the value of ${E_1}$ in equation 1 we get;
$1.78 + {E_2} = 3.5$
Hence, ${E_2} = 3.5 - 1.78 = 1.72V$
Therefore, the emf’s are 1.78 and 1.72 volts respectively.
Note: The polarity is a very important character. It should be carefully considered.
Whenever a question of a potentiometer like this comes, the first step should be to draw the diagram for better understanding of the conditions and the second step should be finding the potential gradient. |
# Is a composite or prime?
## Is a composite or prime?
If a number less than 121 isn’t divisible by 2, 3, 5, or 7, it’s prime; otherwise, it’s composite. If a number less than 289 isn’t divisible by 2, 3, 5, 7, 11, or 13, it’s prime; otherwise, it’s composite.
## What is the successor of 1432?
Dan II of Wallachia
Dan II
Successor Alexander I Aldea
Born unknown
Died 1 June 1432
Is 14 composite or prime?
For example, the integer 14 is a composite number because it is the product of the two smaller integers 2 × 7. Likewise, the integers 2 and 3 are not composite numbers because each of them can only be divided by one and itself.
### Is 202 composite or prime?
The number 202 is composite and therefore it will have prime factors.
### Is 25 composite or prime?
Yes, since 25 has more than two factors i.e. 1, 5, 25. In other words, 25 is a composite number because 25 has more than 2 factors. Problem Statements: Is 25 a Prime Number?
Why is 2 not a prime number?
Proof: The definition of a prime number is a positive integer that has exactly two distinct divisors. Since the divisors of 2 are 1 and 2, there are exactly two distinct divisors, so 2 is prime. Rebuttal: Because even numbers are composite, 2 is not a prime.
## What is the forming number?
with the Given Digits. In formation of numbers with the given digits we may say that a number is an arranged group of digits. Numbers may be formed with or without the repetition of digits.
## How is a number even?
Even numbers are those numbers that can be divided into two equal groups or pairs and are exactly divisible by 2. For example, 2, 4, 6, 8, 10 and so on. Hence, 10 is an even number.
Is 20 a prime number?
No, 20 is not a prime number. The number 20 is divisible by 1, 2, 4, 5, 10, 20. For a number to be classified as a prime number, it should have exactly two factors. Since 20 has more than two factors, i.e. 1, 2, 4, 5, 10, 20, it is not a prime number.
### Is 203 a prime or composite?
Is 203 a Composite Number? Yes, since 203 has more than two factors i.e. 1, 7, 29, 203. In other words, 203 is a composite number because 203 has more than 2 factors.
What is composite natural number?
Composite numbers can be defined as natural numbers that have more than two factors. In other words, a number that is divisible by a number other than 1 and the number itself, is called a composite number. |
Physics Solution Manual for 1100 and 2101
6 as tan 1 0281 nm l since l is the length of the
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Unformatted text preview: m, its value can be determined from the Pythagorean theorem: L= ( 0.281 nm )2 + ( 0.281 nm )2 = 0.397 nm Chapter 1 Problems 11 Thus, the angle is 0.281 nm −1 0.281 nm = tan = 35.3° L 0.397 nm ______________________________________________________________________________ θ = tan −1 20. REASONING There are two right triangles in the drawing. Each H2 contains the common side that is 21° 52° shown as a dashed line and is D labeled D, which is the distance between the buildings. The H1 hypotenuse of each triangle is one of the lines of sight to the top and base of the taller building. The remaining (vertical) sides of the triangles are labeled H1 and H2. Since the height of the taller building is H1 + H2 and the height of the shorter building is H1, the ratio that we seek is (H1 + H2)/H1. We will use the tangent function to express H1 in terms of the 52° angle and to express H2 in terms of the 21° angle. The unknown distance D will be eliminated algebraically when the ratio (H1 + H2)/H1 is calculated. SOLUTION The ratio of the building heights is H + H2 Height of taller building =1 Height of shorter building H1 Using the tangent function, we have that tan 52° = H1 D tan 21° = H2 D or H1 = D tan 52° or H 2 = D tan 21° Substituting these results into the expression for the ratio of the heights gives H + H 2 D tan 52° + D tan 21° Height of taller building =1 = Height of shorter building H1 D tan 52° = 1+ tan 21° = 1.30 tan 52° 12 INTRODUCTION AND MATHEMATICAL CONCEPTS If the taller building were half again as tall as the shorter building, this ratio would have been 1.50. Therefore, your friend is wrong . 21. SSM REASONING The drawing at the right shows the location of each deer A, B, and C. From the problem statement it follows that C c B b = 62 m β α a γ b 77° 51° c = 95 m A γ = 180° − 51° − 77° = 52° Applying the law of cosines (given in Appendix E) to the geometry in the figure, we have 2 2 2 a − 2 ab cos γ + (b − c ) = 0 which is an expression that is quadratic in a. It can be simplified to Aa 2 + Ba + C = 0 , with A=1 B = –2 b cos γ = – 2(62 m) cos 52 ° = –76 m 2 2 2 2 2 C = (b − c ) = (62 m) − (95 m) = –5181 m This quadratic equation can be solved for the desired quantity a. SOLUTION Suppressing units, we obtain from the quadratic formula a= − ( −76 ) ± ( −76 ) 2 − 4 (1)(–5181) = 1.2 × 10 2 m and – 43 m 2 (1) Discarding the negative root, which has no physical significance, we conclude that the 2 distance between deer A and C is 1.2 × 10 m . Chapter 1 Problems 22. REASONING The trapeze cord is L = 8.0 m long, so the trapeze is initially h1 = L cos 41° meters below the support. At the instant he releases the trapeze, it is h2 = L cos θ meters below the support. The difference in heights is d = h2 – h1 = 13 θ 41° h1 L h2 L 41° θ 0.75 m. Given that the trapeze is released at a lower elevation than the platform, we expect to find θ < 41°. 0.75 m SOLUTION Putting the above relationships together, we have d = h2 − h1 = L cos θ − L cos 41o cos θ = or d + L cos 41o = L cos θ d + cos 41o L d 0.75 m + cos 41o = cos−1 + cos 41o = 32o L 8.0 m θ = cos−1 23. SSM REASONING AND SOLUTION A single rope must supply the resultant of the two forces. Since the forces are perpendicular, the magnitude of the resultant can be found from the Pythagorean theorem. a. Applying the Pythagorean theorem, 475 N F = ( 475 N ) + (315 N) = 5.70 × 10 N 2 2 2 b. The angle θ that the resultant makes with the westward direction is θ = tan −1 θ W 315 F N I = 33.6° G NJ HK 475 315 N F S Thus, the rope must make an angle of 33.6° s outh of west . _____________________________________________________________________________________________ 14 INTRODUCTION AND MATHEMATICAL CONCEPTS 24. REASONING The Pythagorean theorem (Equation 1.7) can be used to find the magnitude of the resultant vector, and trigonometry can be employed to determine its direction. a. Arranging the vectors in tail-to-head fashion, we can see that the vector A gives the resultant a westerly direction and vector B gives the resultant a southerly direction. Therefore, the resultant A + B points south of west. b. Arranging the vectors in tail-to-head fashion, we can see that the vector A gives the resultant a westerly direction and vector –B gives the resultant a northerly direction. Therefore, the resultant A + (–B) points north of west. SOLUTION Using the Pythagorean theorem and trigonometry, we obtain the following results: a. Magnitude of A + B = ( 63 units )2 + ( 63 units )2 = 89 units 63 units = 45° south of west 63 units θ = tan −1 b. Magnitude of A − B = ( 63 units )2 + ( 63 units )2 = 89 units 63 units = 45° north of west 63 units ______________________________________________________________________________ θ = tan −1 25. SSM WWW REASONING a. Since the two force vectors A and B have directions due west and due north, they are perpendicular. Therefore, the resultant vector F = A + B has a magnitude give...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
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